Solutions Manual - Microelectronic Circuit Design - 4th Ed
By Richard C. Jaeger, Travis N. Blalock - McGraw-Hill (2010)
NOTE: these answers are for the International Edition (?)
But they’re still very similar to the original (sometimes a, b, c, d answers will be switched around, and
some numbers may be a little off. As a general rule, subtract 3 from the answer you are looking for
and that should be the real one)
Special thanks to Moser from NIU for the main files
February 1, 2012
Go to this website for book updates / corrections by the publisher:
http://www.jaegerblalock.com/
1.1
Answering machine
Alarm clock
Automatic door
Automatic lights
ATM
Automobile:
Engine controller
Temperature control
ABS
Electronic dash
Navigation system
Automotive tune-up equipment
Baggage scanner
Bar code scanner
Battery charger
Cable/DSL Modems and routers
Calculator
Camcorder
Carbon monoxide detector
Cash register
CD and DVD players
Ceiling fan (remote)
Cellular phones
Coffee maker
Compass
Copy machine
Cordless phone
Depth finder
Digital Camera
Digital watch
Digital voice recorder
Digital scale
Digital thermometer
Electronic dart board
Electric guitar
Electronic door bell
Electronic gas pump
Elevator
Exercise machine
Fax machine
Fish finder
Garage door opener
GPS
Hearing aid
Invisible dog fences
Laser pointer
LCD projector
Light dimmer
Keyboard synthesizer
Keyless entry system
Laboratory instruments
Metal detector
Microwave oven
Model airplanes
MP3 player
Musical greeting cards
Musical tuner
Pagers
Personal computer
Personal planner/organizer (PDA)
Radar detector
Broadcast Radio (AM/FM/Shortwave)
Razor
Satellite radio receiver
Security systems
Sewing machine
Smoke detector
Sprinkler system
Stereo system
Amplifier
CD/DVD player
Receiver
Tape player
Stud sensor
Talking toys
Telephone
Telescope controller
Thermostats
Toy robots
Traffic light controller
TV receiver & remote control
Variable speed appliances
Blender
Drill
Mixer
Food processor
Fan
Vending machines
Video game controllers
Wireless headphones & speakers
Wireless thermometer
Workstations
Electromechanical Appliances*
Air conditioning and heating systems
Clothes washer and dryer
Dish washer
Electrical timer
Iron, vacuum cleaner, toaster
Oven, refrigerator, stove, etc.
*These appliances are historically based only upon
on-off (bang-bang) control. However, many of the
high end versions of these appliances have now
added sophisticated electronic control.
1-1
©R. C. Jaeger & T. N. Blalock
6/9/06
1.2
B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip
1.3
(a)
0.1977(Y2 −1960)
B2 19.97x10
0.1977(Y2 −Y1 )
0.1977(Y2 −Y1 )
=
= 10
so 2 = 10
0.1977(Y1 −1960)
B1 19.97x10
Y2 − Y1 =
(b)
log2
= 1.52 years
0.1977
Y2 − Y1 =
log10
= 5.06 years
0.1977
1.4
0.1548(2020−1970)
N = 1610x10
1.5
= 8.85 x 1010 transistors/μP
(2
)
N 2 1610x10
0.1548(Y2 −Y1 )
=
= 10
0.1548(Y1 −1970)
N1 1610x10
log2
(a) Y2 − Y1 =
= 1.95 years
0.1548
log10
(b) Y2 − Y1 =
= 6.46 years
0.1548
0.1548 Y −1970
1.6
−0.05806(2020−1970)
F = 8.00x10
μm = 10 nm .
No, this distance corresponds to the diameter of only a few atoms. Also, the wavelength of the
radiation needed to expose such patterns during fabrication is represents a serious problem.
1.7
From Fig. 1.4, there are approximately 600 million transistors on a complex Pentium IV
microprocessor in 2004. From Prob. 1.4, the number of transistors/μP will be 8.85 x 1010. in
2020. Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors.
1-2
©R. C. Jaeger & T. N. Blalock
6/9/06
1.8
(
)
P = 75x106 tubes (1.5W tube)= 113 MW!
I=
1.13 x 108W
= 511 kA!
220V
D, D, A, A, D, A, A, D, A, D, A
1.9
1.10
10.24V
10.24V
10.24V
=
= 2.500 mV
VMSB =
= 5.120V
12
2
2 bits 4096bits
1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500mV )= 5.855 V
VLSB =
1.11
VLSB =
5V
5V
mV
=
= 19.53
bit
2 bits 256bits
8
2.77V
= 142 LSB
mV
19.53
bit
and
14210 = (128 + 8 + 4 + 2) = 100011102
10
1.12
VLSB =
2.5V
2.5V
mV
=
= 2.44
bit
2 bits 1024 bits
10
(
01011011012 = 28 + 26 + 25 + 23 + 22 + 20
)
10
⎛ 2.5V ⎞
VO = 365 ⎜
⎟ = 0.891 V
⎝ 1024 ⎠
= 36510
1.13
(
)
10V
mV
6.83V 14
= 0.6104
and
2 bits = 11191 bits
14
bit
10V
2 bits
1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1)
10
VLSB =
1119110 = 101011101101112
1.14
A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000. The
number of bits must satisfy 2B ≥ 10,000 where B is the number of bits. Here B = 14 bits.
1.15
5.12V
5.12V
mV
V
=
= 1.25
and VO = (1011101110112 )VLSB ± LSB
12
bit
2
2 bits 4096 bits
11
9
8
7
5
4
3
VO = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV ± 0.0625V
VLSB =
(
)
10
VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V
1-3
6/9/06
1.16
IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A
1.17
VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000 t Volts
1.18
vCE = [5 + 2 cos (5000t)] V
1.19
vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V
1.20
V = 10 V, R1 = 22 kΩ, R2= 47 kΩ and R3 = 180 kΩ.
+
V
1
R
I2
1
R2
V
I
+
V
3
R
2
3
-
V1 = 10V
22kΩ
(
)
22kΩ + 47kΩ 180kΩ
= 10V
22kΩ
= 3.71 V
22kΩ + 37.3kΩ
37.3kΩ
= 6.29 V Checking : 6.29 + 3.71 = 10.0 V
22kΩ + 37.3kΩ
⎛
⎞
180kΩ
180kΩ
10V
I2 = I1
=⎜
= 134 μA
⎟
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
⎛
⎞
47kΩ
47kΩ
10V
I3 = I1
=⎜
= 34.9 μA
⎟
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
V2 = 10V
Checking : I1 =
10V
= 169μA and I1 = I2 + I3
22kΩ + 37.3kΩ
1-4
©R. C. Jaeger & T. N. Blalock
6/9/06
1.21
V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ.
V
+
1
R
I
R
V
I3
+
2
1
V
2
R
2
3
-
V1 = 18V
56kΩ
(
)
56kΩ + 33kΩ 11kΩ
= 15.7 V
V2 = 18V
33kΩ 11kΩ
(
)
56kΩ + 33kΩ 11kΩ
= 2.31 V
Checking :V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct.
I1 =
18V
(
)
56kΩ + 33kΩ 11kΩ
I3 = I1
= 280 μA
I2 = I1
11kΩ
11kΩ
= (280 μA)
= 70.0 μA
33kΩ + 11kΩ
33kΩ + 11kΩ
33kΩ
33kΩ
= (280 μA)
= 210 μA
33kΩ + 11kΩ
33kΩ + 11kΩ
1.22
I1 = 5mA
(5.6kΩ + 3.6kΩ) = 3.97 mA
(5.6kΩ + 3.6kΩ)+ 2.4kΩ
(
Checking : I2 + I3 = 280 μA
I2 = 5mA
2.4kΩ
= 1.03 mA
9.2kΩ + 2.4kΩ
= 3.72V
)5.6kΩ3.6kΩ
+ 3.6kΩ
V3 = 5mA 2.4kΩ 9.2kΩ
Checking : I1 + I2 = 5.00 mA
and
I2 R2 = 1.03mA(3.6kΩ)= 3.71 V
1.23
150kΩ
150kΩ
= 125 μA
I3 = 250μA
= 125 μA
150kΩ + 150kΩ
150kΩ + 150kΩ
82kΩ
V3 = 250μA 150kΩ 150kΩ
= 10.3V
68kΩ + 82kΩ
Checking : I1 + I2 = 250 μA and I2 R2 = 125μA(82kΩ)= 10.3 V
I2 = 250μA
(
)
1-5
6/9/06
1.24
+
R
v
v
1
s
+
g v
v
m
th
-
Summing currents at the output node yields:
v
+ .002v = 0 so v = 0 and v th = vs − v = vs
5x10 4
+
v
-
R
1
ix
g v
m
vx
Summing currents at the output node :
v
− 0.002v = 0 but v = −vx
5x10 4
vx
v
1
ix =
+ 0.002vx = 0 Rth = x =
= 495 Ω
4
1
ix
5x10
+ gm
R1
ix = −
Thévenin equivalent circuit:
495 Ω
v
s
1-6
©R. C. Jaeger & T. N. Blalock
6/9/06
1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24,
and
⎛ 1
⎞−1
Rth = ⎜
+ .025⎟ = 39.6 Ω
⎝ 4kΩ
⎠
+
v
R
1
-
vs
g v
m
in
The short circuit current is :
v
in =
+ 0.025v and v = vs
4kΩ
v
i n = s + 0.025vs = 0.0253vs
4kΩ
Norton equivalent circuit:
0.0253v
s
39.6 Ω
1-7
6/9/06
1.26 (a)
+
βi
R
vs
R2
1
v th
i
Vth = Voc = −β i R2
-
i =−
but
vs
R1
and
Vth = β vs
R2
39kΩ
= 120 vs
= 46.8 vs
R1
100kΩ
ix
βi
R
R2
1
v
Rth
x
i
Rth =
vx
;
ix
ix =
vx
+ βi
R2
but
i = 0 since VR 1 = 0.
Rth = R2 = 39 kΩ.
Thévenin equivalent circuit:
39 k Ω
58.5v
s
(b)
+
βi
i
s
R
R2
1
i
v th
-
⎛ i ⎞
Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β ⎜ − s ⎟ R2 = 38700 is
⎝ β + 1⎠
1-8
©R. C. Jaeger & T. N. Blalock
6/9/06
βi
R
R2
1
Rth
v
x
i
Rth =
vx
;
ix
ix =
vx
+ βi but
R2
i + βi = 0 so i = 0
and Rth = R2 = 39 kΩ
Thévenin equivalent circuit:
39 k Ω
38700i s
1.27
βi
R
vs
R2
1
in
i
in = −β i but i = −
vs
R1
and in =
β
R1
vs =
From problem 1.26(a), Rth = R2 = 56 kΩ.
0.00133v
s
100
vs = 1.33 x 10−3 vs
75kΩ
Norton equivalent circuit:
56 k Ω
1-9
6/9/06
1.28
is
v
βi
R
s
R2
1
i
is =
vs
v
v
β +1
− β i = s + β s = vs
R1
R1
R1
R1
R=
vs
R
100kΩ
= 1 =
= 1.24 Ω
is β + 1
81
1.29
The open circuit voltage is vth = −g mv R2 and v = +is R1.
( )( )
vth = −g m R1 R2is = −(0.0025) 105 106 i s= 2.5 x 108 is
For is = 0, v = 0 and Rth = R2 = 1 MΩ
1.30
5V
3V
f (Hz)
0
0
500
1000
1.31
2V
0
f (kHz)
9
10
11
[
]
4
cos(20000πt + 2000πt )+ cos(20000πt − 2000πt )
2
v = 2cos(22000πt )+ 2cos(18000πt )
v = 4sin (20000πt )sin (2000πt )=
1.32
2∠36 o
A = −5 0 = 2x105 ∠36 o
10 ∠0
A = 2x105
∠A = 36o
1-10
©R. C. Jaeger & T. N. Blalock
6/9/06
1.33
(a) A =
10−1∠ −12 o
10−2 ∠ − 45o
o
A
=
=
5∠
−
45
= 100∠ −12 o
(b)
2x10−3 ∠0 o
10−3 ∠0 o
1.34
(a) Av = −
R2
620kΩ
180kΩ
=−
= −44.3 (b) Av = −
= −10.0
14kΩ
R1
18kΩ
(c) Av = −
62kΩ
= −38.8
1.6kΩ
1.35
vo (t ) = −
IS =
R2
v s (t )= (−90.1 sin 750πt ) mV
R1
VS 0.01V
=
= 11.0μA and
R1 910Ω
is = (11.0 sin 750πt ) μA
1.36 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vs.
Therefore Av = 1.
1.37 Since the voltage across the op amp input terminals must be zero, v- = v+ = vs. Also, i- = 0.
v
v− − vo
+ i− + − = 0
R2
R1
or
vs − v o vs
+ =0
R2
R1
and A v =
vo
R
= 1+ 2
vs
R1
1.38 Writing a nodal equation at the inverting input terminal of the op amp gives
v −v
v1 − v− v2 − v−
+
= i− + − o
R1
R2
R3
vo = −
but v- = v+ = 0
and
i- = 0
R3
R
v1 − − 3 v2 = −0.255sin 3770t − 0.255sin10000t volts
R1
R2
1-11
6/9/06
1.39
⎛b b b ⎞
⎛ 0 1 1⎞
⎛ 1 0 0⎞
vO = −VREF ⎜ 1 + 2 + 3 ⎟ (a) vO = −5⎜ + + ⎟ = −1.875V (b) vO = −5⎜ + + ⎟ = −2.500V
⎝2 4 8⎠
⎝ 2 4 8⎠
⎝ 2 4 8⎠
b1b2b
vO (V)
3
000
0
001
-0.625
010
-1.250
011
-1.875
100
-2.500
101
-3.125
110
-3.750
111
-4.375
1.40 Low-pass amplifier
Amplitude
10
f
6 kHz
1-12
©R. C. Jaeger & T. N. Blalock
6/9/06
1.41 Band-pass amplifier
Amplitude
20
f
1 kHz
1.42
5 kHz
High-pass amplifier
Amplitude
16
f
10 kHz
1.43
vO (t ) = 10x5sin (2000πt )+ 10x3cos(8000πt )+ 0x3cos(15000πt )
[
]
vO (t ) = 50sin(2000πt )+ 30cos(8000πt ) volts
1.44
vO (t ) = 20x0.5sin (2500πt )+ 20x0.75cos(8000πt )+ 0x0.6cos(12000πt )
[
]
vO (t ) = 10.0sin (2500πt )+ 15.0cos(8000πt ) volts
1.45 The gain is zero at each frequency:
vo(t) = 0.
1-13
6/9/06
1.46
t=linspace(0,.005,1000);
w=2*pi*1000;
v=(4/pi)*(sin(w*t)+sin(3*w*t)/3+sin(5*w*t)/5);
v1=5*v;
v2=5*(4/pi)*sin(w*t);
v3=(4/pi)*(5*sin(w*t)+3*sin(3*w*t)/3+sin(5*w*t)/5);
plot(t,v)
plot(t,v1)
plot(t,v2)
plot(t,v3)
2
1
0
-1
-2
0
1
2
3
4
5
x10-3
1
2
3
4
5
x10-3
(a)
10
5
0
-5
-10
0
(b)
1-14
©R. C. Jaeger & T. N. Blalock
6/9/06
10
5
0
-5
-10
0
1
2
3
4
(c)
5
x10-3
10
5
0
-5
-10
0
1
2
3
4
5
x10-3
(d)
1.47
(a) 3000(1− .01)≤ R ≤ 3000(1+ .01) or 2970Ω ≤ R ≤ 3030Ω
(b) 3000(1− .05)≤ R ≤ 3000(1+ .05) or 2850Ω ≤ R ≤ 3150Ω
(c) 3000(1− .10) ≤ R ≤ 3000(1+ .10) or 2700Ω ≤ R ≤ 3300Ω
ΔV ≤ 0.05V
0.05
= 0.0200 or 2.00%
2.50
1.48
Vnom = 2.5V
1.49
20000μF (1− .5)≤ C ≤ 20000μF (1+ .2) or 10000μF ≤ R ≤ 24000μF
1.50
8200(1− 0.1)≤ R ≤ 8200(1+ 0.1) or 7380Ω ≤ R ≤ 9020Ω
The resistor is within the allowable range of values.
T=
1-15
6/9/06
1.51
(a) 5V (1− .05)≤ V ≤ 5V (1+ .05) or 5.75V ≤ V ≤ 5.25V
V = 5.30 V exceeds the maximum range, so it is out of the specification limits.
(b) If the meter is reading 1.5% high, then the actual voltage would be
5.30
Vmeter = 1.015Vact or Vact =
= 5.22V which is within specifications limits.
1.015
1.52
Ω
ΔR 6562 − 6066
=
= 4.96 o
100 − 0
ΔT
C
= R o + TCR (ΔT)= 6066 + 4.96(27)= 6200Ω
TCR =
R nom
0 C
1-16
©R. C. Jaeger & T. N. Blalock
6/9/06
1.53
V1
+
I2
R
1
R2
V
I3
+
R
V2
3
-
Let RX = R2 R3
R min
X =
V1max =
I1 =
I2min =
I3 = I1
I3max =
I3min =
R1
=
R1 + RX
47kΩ(0.9)(180kΩ)(0.9)
47kΩ(0.9)+ 180kΩ(0.9)
10(1.05)
33.5kΩ
1+
22kΩ(1.1)
V
R1 + RX
I2max =
then V1 = V
= 4.40V
and I2 = I1
V1
R
1+ X
R1
= 33.5kΩ R max
=
X
V1min =
R3
=
R2 + R3
47kΩ(1.1)(180kΩ)(1.1)
47kΩ(1.1)+ 180kΩ(1.1)
10(0.95)
41.0kΩ
1+
22kΩ(0.9)
= 3.09V
V
R1 + R2 +
R1 R2
R3
10(1.05)
22000(0.9)(47000)(0.9)
22000(0.9)+ 47000(0.9)+
10(0.95)
22000(1.1)+ 47000(1.1)+
R2
=
R2 + R3
= 41.0kΩ
= 158 μA
180000(1.1)
22000(1.1)(47000)(1.1)
= 114 μA
180000(0.9)
V
R1 + R3 +
R1 R3
R2
10(1.05)
22000(0.9)+ 180000(0.9)+
22000(0.9)(180000)(0.9)
10(0.95)
22000(1.1)+ 180000(1.1)+
= 43.1 μA
47000(1.1)
22000(1.1)(180000)(1.1)
= 28.3 μA
47000(0.9)
1-17
6/9/06
1.54
I1 = I
R2 + R3
=I
R1 + R2 + R3
1+
and similarly I2 = I
R1
R2 + R3
mA = 4.12 mA
2400(0.95)
1+
1
R + R3
1+ 2
R1
I1min =
1+
5600(1.05)+ 3600(1.05)
5(1.02)
I2max =
5600(0.95)+ 3600(0.95)
V3 = I2 R3 =
V3min =
1+
5(1.02)
I1max =
V3max =
1
mA = 1.14 mA
2400(1.05)
1+
mA = 3.80 mA
5600(0.95)+ 3600(0.95)
5(0.98)
I2min =
2400(1.05)
5(0.98)
5600(1.05)+ 3600(1.05)
mA = 0.936 mA
2400(0.95)
I
1
1
R
+
+ 2
R1 R3 R1 R3
5(1.02)
5600(0.95)
1
1
+
+
2400(1.05) 3600(1.05) 2400(1.05)(3600)(1.05)
5(0.98)
= 4.18 V
5600(1.05)
1
1
+
+
2400(0.95) 3600(0.95) 2400(0.95)(3600)(0.95)
= 3.30 V
1.55
Rth =
From Prob. 1.24 :
Rthmax =
1
gm +
1
1
0.002(0.8)+
5x10 4 (1.2)
1
R1
= 619 Ω
Rthmin =
1-18
1
1
0.002(1.2)+
5x10 4 (0.8)
= 412 Ω
©R. C. Jaeger & T. N. Blalock
6/9/06
1.56
For one set of 200 cases using the equations in Prob. 1.53.
V = 10 * (0.95 + 0.1* RAND())
R1 = 22000 * (0.9 + 0.2 * RAND())
R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND())
V1
I2
I3
Min
3.23 V
116 μA
29.9 μA
Max
3.71 V
151 μA
40.9 μA
Average
3.71 V
133 μA
35.1 μA
1.57
For one set of 200 cases using the Equations in Prob. 1.54:
I = 0.005* (0.98 + 0.04 * RAND())
R1 = 2400 * (0.95 + 0.1* RAND())
R1 = 5600 * (0.95 + 0.1* RAND()) R3 = 3600 * (0.95 + 0.1* RAND())
I1
I2
V3
Min
3.82 mA
0.96 mA
3.46 V
Max
4.09 mA
1.12 mA
4.08 V
Average
3.97 mA
1.04 mA
3.73 V
1.58
3.29, 0.995, -6.16; 3.295, 0.9952, -6.155
1.59
(a) (1.763 mA)(20.70 kΩ) = 36.5 V (b) 36 V
(c) (0.1021 A)(97.80 kΩ) = 9.99 V; 10 V
1-19
6/9/06
CHAPTER 2
2.1
Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor.
2.2
Based upon Table 2.1, a resistivity of 1015 Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator.
2.3
I max
⎛ 10−8 cm2 ⎞
⎛ 7 A ⎞
= ⎜10
⎟ = 500 mA
⎟(5μm)(1μm)⎜
2
cm 2 ⎠
⎝
⎝ μm ⎠
2.4
EG
⎛
⎞
ni = BT 3 exp⎜ −
⎟
−5
⎝ 8.62 x10 T ⎠
31
For silicon, B = 1.08 x 10 and EG = 1.12 eV:
-10
3
ni = 2.01 x10 /cm
9
3
13
6.73 x10 /cm
30
For germanium, B = 2.31 x 10 and EG = 0.66 eV:
3
13
ni = 35.9/cm
3
2.27 x10 /cm
15
function f=temp(T)
ni=1E14;
f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
14
3
16
ni = 10 /cm3 for T = 739 K
for T = 506 K
2.6
⎛
⎞
EG
ni = BT 3 exp⎜ −
−5 ⎟
⎝ 8.62x10 T ⎠
with
B = 1.27x1029 K −3cm−6
6
3
T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm
3
T = 100 K: ni = 6.03 x 10-19/cm
20
3
8.04 x 10 /cm .
2.5
Define an M-File:
ni = 10 /cm
3
8.36 x 10 /cm .
11
3
T = 500 K: ni = 2.79 x10 /cm
2.7
⎛
cm2 ⎞⎛
V ⎞
6 cm
vn = −μn E = ⎜ −700
⎟⎜ 2500 ⎟ = −1.75x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
V ⎞
cm2 ⎞⎛
5 cm
v p = +μ p E = ⎜ +250
⎟⎜ 2500 ⎟ = +6.25x10
cm ⎠
V − s ⎠⎝
s
⎝
⎛
1 ⎞⎛
cm ⎞
4 A
jn = −qnvn = −1.60x10−19 C ⎜1017 3 ⎟⎜ −1.75x106
⎟ = 2.80x10
s ⎠
cm ⎠⎝
cm2
⎝
⎛
1 ⎞⎛
cm ⎞
−10 A
j p = qnv p = 1.60x10−19 C ⎜103 3 ⎟⎜ 6.25x105
⎟ = 1.00x10
s ⎠
cm 2
⎝ cm ⎠⎝
(
)
(
)
2.8
⎛ E ⎞
ni2 = BT 3 exp⎜ − G ⎟
⎝ kT ⎠
B = 1.08x1031
⎛
⎞
1.12
T 3 exp⎜ −
⎟
⎝ 8.62x10−5 T ⎠
Using a spreadsheet, solver, or MATLAB yields T = 305.22K
(10 ) = 1.08x10
10
2
31
Define an M-File:
function f=temp(T)
f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
Then: fzero('temp',300) | ans = 305.226 K
2.9
v=
j − 1000 A / cm 2
cm
=
= − 105
2
Q
s
0.01C / cm
2.10
C ⎞⎛
cm ⎞
MA
⎛
6 A
j = Qv = ⎜ 0.4 3 ⎟⎜10 7
=4 2
⎟ = 4 x10
2
cm ⎠⎝
sec ⎠
cm
cm
⎝
21
2.11
⎛
V ⎞
cm2 ⎞⎛
6 cm
vn = −μn E = ⎜−1000
⎟⎜ −2000 ⎟ = +2.00x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
V ⎞
cm 2 ⎞⎛
5 cm
v p = +μ p E = ⎜ +400
⎟⎜ −2000 ⎟ = −8.00x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
1 ⎞⎛
cm ⎞
−10 A
jn = −qnvn = −1.60x10−19 C ⎜103 3 ⎟⎜ +2.00x106
⎟ = −3.20x10
s ⎠
cm2
⎝ cm ⎠⎝
⎛
1 ⎞⎛
cm ⎞
4 A
j p = qnv p = 1.60x10−19 C ⎜1017 3 ⎟⎜ −8.00x105
⎟ = −1.28x10
s ⎠
cm ⎠⎝
cm2
⎝
(
)
(
)
2.12
(a )
E=
V
5V
= 5000
−4
cm
10 x10 cm
(b )
(
)
V ⎞
⎛
−4
V = ⎜105
⎟ 10 x10 cm = 100 V
cm
⎝
⎠
2.13
⎛ 1019 ⎞⎛
cm ⎞
7 A
j p = qpv p = 1.60x10−19 C ⎜ 3 ⎟⎜10 7
⎟ = 1.60x10
s ⎠
cm2
⎝ cm ⎠⎝
⎛
A ⎞
i p = j p A = ⎜1.60x10 7 2 ⎟ 1x10−4 cm 25x10−4 cm = 4.00 A
cm ⎠
⎝
(
)
(
)(
)
2.14
For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p )
σ ≥ 1000(Ω − cm) for a conductor
−1
ni ≥
σ
q (μn + μ p )
1000(Ω − cm)
−1
=
cm 2
1.602x10−19 C (100 + 50)
v − sec
39
⎛
⎞
1.73x10
E
n 2i =
= BT 3 exp⎜ − G ⎟ with
6
cm
⎝ kT ⎠
=
4.16x1019
cm3
B = 1.08x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV
This is a transcendental equation and must be solved numerically by iteration. Using the HP
solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above the
melting temperature of silicon.
22
2.15
For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p )
σ ≤ 10−5 (Ω − cm) for an insulator
−1
ni ≥
σ
q (μn + μ p )
10−5 (Ω − cm)
−1
=
⎛ cm 2 ⎞
1.602x10 C (2000 + 750)⎜
⎟
⎝ v − sec ⎠
⎛ E ⎞
5.152x1020
n 2i =
= BT 3 exp⎜ − G ⎟ with
6
cm
⎝ kT ⎠
(
−19
)
=
2.270x1010
cm 3
B = 1.08x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV
Using MATLAB as in Problem 2.5 yields T = 316.6 K.
2.16
Si
Si
Si
P
B
Si
Si
Si
Si
Donor electron
fills acceptor
vacancy
No free electrons or holes (except those corresponding to ni).
2.17
(a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron
and will act as a donor impurity.
(b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in one
electron and will act as an acceptor impurity.
2.18
Since Ge is from column IV, acceptors come from column III and donors come from column
V. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi
23
2.19
(a) Germanium is from column IV and indium is from column III. Thus germanium has one
extra electron and will act as a donor impurity.
(b) Germanium is from column IV and phosphorus is from column V. Thus germanium has
one less electron and will act as an acceptor impurity.
2.20
A ⎞
V
⎛
= jρ = ⎜10000 2 ⎟(0.02Ω − cm ) = 200
, a small electric field.
cm ⎠
σ
cm
⎝
j
E=
2.21
⎛ C ⎞⎛
cm ⎞
A
jndrift = qnμn E = qnv n = 1.602x10−19 1016 ⎜ 3 ⎟⎜10 7
⎟ = 16000 2
s ⎠
cm
⎝ cm ⎠⎝
(
)( )
2.22
⎛ 1015 atoms ⎞
⎛ 10−4 cm ⎞3
N =⎜
⎟(1μm)(10μm)(0.5μm )⎜
⎟ = 5,000 atoms
3
⎝ cm
⎠
⎝ μm ⎠
2.23
N A > N D : N A − N D = 1015 −1014 = 9x1014 /cm3
If we assume N A − N D >> 2ni = 1014 / cm 3 :
p = N A − N D = 9x1014 /cm3 | n =
If we use Eq. 2.12 : p =
9x1014 ±
ni2 251026
=
= 2.78x1012 /cm3
p 9x1014
(9x10 ) + 4(5x10 ) = 9.03x10
14
2
13
2
14
2
and n = 2.77x10 /cm . The answers are essentially the same.
12
3
2.24
N A > N D: N A − N D = 5 x1016 − 1016 = 4 x1016 /cm 3 >> 2ni = 2 x1011 /cm 3
p = N A − N D = 4 x1014 /cm 3 | n =
ni2
10 22
=
= 2.50 x10 5 /cm 3
p 4 x1016
2.25
N D > N A: N D − N A = 3x1017 − 2x1017 = 1x1017 /cm3
2ni = 2x1017 /cm3 ; Need to use Eq. (2.11)
n=
p=
24
1017 ±
( ) ( ) = 1.62x10
2
1017 + 4 1017
2
2
2
i
34
n
10
=
= 6.18x1016 /cm3
n 1.62x1017
17
/cm3
2.26
N D − N A = −2.5x1018 / cm 3
Using Eq. 2.11: n =
−2.5x1018 ±
(−2.5x10 ) + 4(10 )
18
2
10
2
2
ni2
= ∞.
p
No, the result is incorrect because of loss of significant digits
Evaluating this with a calculator yields n = 0, and n =
within the calculator. It does not have enough digits.
2.27
(a) Since boron is an acceptor, NA = 6 x 1018/cm3. Assume ND = 0, since it is not specified.
The material is p-type.
At room temperature, ni = 1010 /cm3 and N A − N D = 6 x1018 / cm3 >> 2n i
ni2
10 20 /cm 6
So p = 6 x10 /cm and n =
=
= 16.7 /cm3
18
3
p 6 x10 /cm
(b)
⎛
⎞
3
1.12
⎟
At 200K, ni2 = 1.08x1031 (200) exp⎜⎜ −
= 5.28x109 /cm6
−5
⎟
⎝ 8.62x10 (200)⎠
18
3
ni = 7.27x10 4 /cm 3
N A − N D >> 2ni , so p = 6x1018 /cm3 and n =
5.28x109
= 8.80x10−10 /cm 3
18
6x10
2.28
(a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The
material is n-type.
At room temperature, n i = 1010 / cm 3 and N D − N A = 3 x1017 / cm 3 >> 2n i
10 20 /cm 6
ni2
=
= 333 /cm 3
17
3
n 3x10 /cm
⎛
⎞
3
1.12
⎟
= 4.53x1015 /cm6
(b) At 250K, ni2 = 1.08x1031 (250) exp⎜⎜−
−5
⎟
⎝ 8.62x10 (250)⎠
So n = 3 x1017 /cm 3 and p =
ni = 6.73x10 7 /cm 3
N D − N A >> 2ni , so n = 3x1017 / cm 3 and n =
4.53x1015
= 0.0151/ cm 3
17
3x10
2.29
(a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3.
Since NA > ND, the material is p-type.
25
(b) At room temperature, n i = 1010 / cm3 and N A − N D = 6 x1018 / cm3 >> 2n i
ni2
10 20 /cm 6
So p = 6 x10 /cm and n =
=
= 16.7 /cm3
18
3
p 6 x10 /cm
18
3
2.30
(a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3.
Since NA > ND, the material is p-type.
(b) At room temperature, ni = 1010 /cm3 and N A − N D = 3x1017 / cm3 >> 2n i
ni2
10 20 /cm 6
So p = 3x10 /cm and n =
=
= 333 /cm3
17
3
p 3x10 /cm
17
3
2.31
ND = 4 x 1016/cm3. Assume NA = 0, since it is not specified.
N D > N A : material is n - type | N D − N A = 4x1016 / cm3 >> 2ni = 2x1010 / cm 3
n = 4x1016 / cm3 | p =
n 2i
1020
=
= 2.5x103 / cm 3
16
n 4x10
N D + N A = 4x1016 / cm3 | Using Fig. 2.13, μn = 1030
ρ=
26
1
qμn n
=
1
⎛
cm 2 ⎞⎛ 4x1016 ⎞
1.602x10 C ⎜1030
⎟⎜
⎟
V − s ⎠⎝ cm 3 ⎠
⎝
(
−19
)
cm2
cm 2
and μp = 310
V −s
V −s
= 0.152 Ω − cm
2.32
NA = 1018/cm3. Assume ND = 0, since it is not specified.
N A > N D : material is p - type | N A − N D = 1018 / cm3 >> 2ni = 2x1010 / cm 3
p = 1018 / cm3
|
n=
n 2i 1020
= 18 = 100 / cm 3
p 10
cm2
cm 2
N D + N A = 10 / cm | Using Fig. 2.13, μn = 375
and μp = 100
V −s
V −s
1
1
ρ=
=
= 0.0624 Ω − cm
⎛
qμ p p
cm 2 ⎞⎛ 1018 ⎞
−19
1.602x10 C⎜100
⎟⎜
⎟
V − s ⎠⎝ cm 3 ⎠
⎝
18
3
2.33
Indium is from column 3 and is an acceptor. NA = 7 x 1019/cm3. Assume ND = 0, since it is not
specified.
N A > N D : material is p - type | N A − N D = 7x1019 /cm3 >> 2ni = 2x1010 /cm3
p = 7x1019 /cm3
|
n=
ni2
1020
=
= 1.43/cm3
19
p 7x10
cm2
cm 2
N D + N A = 7x10 / cm | Using Fig. 2.13, μn = 120
and μp = 60
V −s
V −s
1
1
ρ=
=
= 1.49 mΩ − cm
⎛
qμ p p
cm 2 ⎞⎛ 7x1019 ⎞
−19
1.602x10 C⎜ 60
⎟⎜
3 ⎟
⎝ V − s ⎠⎝ cm ⎠
19
3
2.34
Phosphorus is a donor : N D = 5.5x1016 / cm 3 | Boron is an acceptor : N A = 4.5x1016 / cm 3
N D > N A : material is n - type
|
N D − N A = 1016 / cm3 >> 2ni = 2x1010 / cm 3
ni2 1020
n = 10 /cm | p =
= 16 = 10 4 /cm3
p 10
16
3
cm2
cm 2
N D + N A = 10 / cm | Using Fig. 2.13, μn = 800
and μp = 230
V −s
V −s
1
1
ρ=
=
= 0.781 Ω − cm
⎛
qμ n n
cm 2 ⎞⎛ 1016 ⎞
−19
1.602x10 C⎜ 800
⎟⎜
⎟
V − s ⎠⎝ cm 3 ⎠
⎝
17
27
3
2.35
1
ρ=
qμ p p
| μp p =
1
(1.602x10 C)(0.054Ω − cm)
−19
=
1.16x1020
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
NA
μp
μp p
1018
96.7
9.67 x 1020
1.1 x1018
93.7
1.03 x 1020
1.2 x 1017
91.0
1.09 x 1020
1.3 x 1019
88.7
1.15 x 1020
2.36
8.32x1018
ρ=
| μp p =
=
qμ p p
1.602x10−19 C (0.75Ω − cm) V − cm − s
1
1
(
)
An iterative solution is required. Using the equations in Fig. 2.8:
μp
NA
1016
μp p
406
4.06 x
1018
2 x 1016
363
7.26 x
1018
3 x 1016
333
1.00 x
1019
2.4 x 1016
350
8.40 x 1018
2.37
Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic
because it contains impurities. Addition of the impurities has increased the resistivity.
2.38
ρ=
1
qμn n
| μ n n ≈ μn N D =
1
(1.602x10 C)(2Ω − cm)
−19
=
3.12x1018
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
ND
μn
μnn
1015
1350
1.35 x 1018
2 x 1015
1330
2.67 x 1018
2.5 x 1015
1330
3.32 x 1018
28
2.3 x 1015
29
1330
3.06 x 1018
2.39 (a)
1
1
6.24x1021
ρ=
| μn n ≈ μn N D =
=
qμn n
1.602x10−19 C (0.001Ω − cm) V − cm − s
(
)
An iterative solution is required. Using the equations in Fig. 2.8:
ND
μn
μnn
1019
116
1.16 x 1021
7 x 1019
96.1
6.73 x 1021
6.5 x 1019
96.4
6.3 x 1021
(b)
ρ=
1
qμ p p
| μp p ≈ μp N A =
1
6.24 x10 21
=
(1.602x10−19 C)(0.001Ω − cm) V − cm − s
An iterative solution is required using the equations in Fig. 2.8:
NA
1.3 x 1020
μp
μp p
49.3
6.4 x 1021
2.40
Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but
the hole and electron concentrations remain unchanged. See Problem 2.37 for example.
However, it is physically impossible to add exactly equal amounts of the two impurities.
2.41
(a) For the 1 ohm-cm starting material:
1
1
6.25x1018
ρ=
| μp p ≈ μpN A =
=
qμ p p
1.602x10−19 C (1Ω − cm) V − cm − s
(
)
An iterative solution is required. Using the equations in Fig. 2.8:
NA
μp
μp p
1016
406
4.1 x 1018
1.5 x 1016
383
5.7 x 1018
1.7 x 1016
374
6.4 x 1019
30
To change the resistivity to 0.25 ohm-cm:
1
1
2.5x1019
ρ=
| μp p ≈ μpN A =
=
qμ p p
1.602x10−19 C (0.25Ω − cm) V − cm − s
(
)
NA
μp
μp p
6 x 1016
276
1.7 x 1019
8 x 1016
233
2.3 x 1019
1.1 x 1017
225
2.5 x 1019
17
16
16
3
Additional acceptor concentration = 1.1x10 - 1.7x10 = 9.3 x 10 /cm
(b) If donors are added:
ND
ND + NA
μn
ND - NA
μnn
2 x 1016
3.7 x 1016
1060
3 x 1015
3.2 x 1018
1 x 1017
1.2 x 1017
757
8.3 x 1016
6.3 x 1019
8 x 1016
9.7 x 1016
811
6.3 x 1016
5.1 x 1019
4.1 x 1016
5.8 x 1016
950
2.4 x 1016
2.3 x 1019
16
3
So ND = 4.1 x 10 /cm must be added to change achieve a resistivity of 0.25 ohm-cm. The
silicon is converted to n-type material.
2.42
Phosphorus is a donor: ND = 1016/cm3 and μn = 1250 cm2/V-s from Fig. 2.8.
2.00
σ = qμn n ≈ qμn N D = 1.602x10−19 C (1250) 1016 =
Ω − cm
-1
Now we add acceptors until σ = 5.0 (Ω-cm) :
(
)
( )
5(Ω − cm)
−1
σ = qμ p p
|
3.12x1019
μ p p ≈ μ p (N A − N D )=
=
1.602x10−19 C V − cm − s
NA
ND + NA
μp
NA - ND
μp p
1 x 1017
1.1 x 1017
250
9 x 1016
2.3 x 1019
2 x 1017
2.1 x 1017
176
1.9 x 1017
3.3 x 1019
1.8 x 1017
1.9 x 1017
183
1.7 x 1016
3.1 x 1019
31
2.43
Boron is an acceptor: NA = 1016/cm3 and μp = 405 cm2/V-s from Fig. 2.8.
0.649
σ = qμ p p ≈ qμ p N A = 1.602x10−19 C (405) 1016 =
Ω − cm
-1
Now we add donors until σ = 5.5 (Ω-cm) :
(
)
( )
5.5(Ω − cm)
−1
σ = qμ n n
|
μn n ≈ μn (N D − N A )=
1.602x10−19 C
=
3.43x1019
V − cm − s
ND
ND + NA
μn
ND - NA
μp p
8 x 1016
9 x 1016
832
7 x 1016
5.8 x 1019
7 x 1016
901
5 x 1016
4.5 x 1019
5.5 x 1016
964
3.5 x 1016
3.4 x 1019
6 x 1016
4.5 x 1016
2.44
VT =
kT 1.38x10−23 T
=
= 8.62x10−5 T
q 1.602x10−19
T (K)
50
75
100
150
200
250
300
350
400
VT (mV)
4.31
6.46
8.61
12.9
17.2
21.5
25.8
30.1
34.5
2.45
⎛ dn ⎞
dn
j = −qDn ⎜− ⎟ = qVT μn
dx
⎝ dx ⎠
⎛
cm2 ⎞⎛ 1018 − 0 ⎞ 1
kA
j = 1.602x10−19 C (0.025V )⎜ 350
= −14.0 2
⎟⎜
−4 ⎟
4
V − s ⎠⎝ 0 −10 ⎠ cm
cm
⎝
(
)
2.46
⎛ cm2 ⎞⎛ 1019 / cm 3 ⎞ ⎛
⎞
dp
x
= −1.602x10−19 C ⎜15
⎟⎜ −
⎟ exp⎜−
⎟
−4
−4
s ⎠⎝ 2x10 cm ⎠ ⎝ 2x10 cm ⎠
dx
⎝
⎛
x ⎞ A
j = 1.20x105 exp⎜−5000 ⎟ 2
cm ⎠ cm
⎝
⎛ 10−8 cm2 ⎞
⎛
A ⎞
I (0)= j (0)A = ⎜1.20x105 2 ⎟ 10μm2 ⎜
⎟ = 12.0 mA
2
cm ⎠
⎝
⎝ μm ⎠
j = −qD p
(
)
(
32
)
2.47
j p = qμ p pE − qD p
⎛
dp
1 dp ⎞
1 dp
= qμ p p⎜ E − VT
⎟ = 0 → E = VT
dx
p dx ⎠
p dx
⎝
(
)
−1022 exp −10 4 x
1 dN A
E ≈ VT
= 0.025 14
N A dx
10 + 1018 exp −10 4 x
E (0)= −0.025
(
)
22
10
V
= −250
18
cm
10 + 10
22
10 exp(−5)
V
E 5x10−4 cm = −0.025 14
= −246
18
cm
10 + 10 exp(−5)
(
14
)
2.48
⎛
V ⎞
cm2 ⎞⎛ 1016 ⎞⎛
A
jndrift = qμn nE = 1.60x10−19 C ⎜350
⎟⎜ 3 ⎟⎜ −20 ⎟ = −11.2 2
cm ⎠
V − s ⎠⎝ cm ⎠⎝
cm
⎝
⎛
V ⎞
cm2 ⎞⎛1.01x1018 ⎞⎛
A
drift
−19
−20
j p = qμ p pE = 1.60x10 C ⎜150
⎟⎜
⎟
⎜
⎟ = −484 2
3
cm ⎠
V − s ⎠⎝ cm
cm
⎝
⎠⎝
⎛
A
cm2 ⎞⎛ 10 4 −1016 ⎞
dn
= −70.0 2
jndiff = qDn
= 1.60x10−19 C ⎜ 350 ⋅ 0.025
⎟⎜
−4
4⎟
s ⎠⎝ 2x10 cm ⎠
dx
cm
⎝
⎛
A
cm2 ⎞⎛ 1018 −1.01x1018 ⎞
dp
j pdiff = −qD p
= −1.60x10−19 C ⎜150 ⋅ 0.025
⎟⎜
⎟ = 30.0 2
−4
4
s ⎠⎝ 2x10 cm ⎠
dx
cm
⎝
A
jT = −11.2 − 484 − 70.0 + 30.0 = −535 2
cm
(
)
(
)
(
)
(
2.49
)
NA = 2ND
EC
ED
N
D
ND
ND
EA
NA
NA
NA
E
V
Holes
2.50
(
)(
)
)
−34
8
hc 6.626x10 J − s 3x10 m / s
λ= =
= 1.108 μm
E
(1.12eV ) 1.602x10−19 J / eV
33
(
NA
2.51
Al - Anode
Al - Cathode
Si02
n-type silicon
p-type silicon
2.52
An n-type ion implantation step could be used to form the n+ region following step (f) in Fig.
2.17. A mask would be used to cover up the opening over the p-type region and leave the
opening over the n-type silicon. The masking layer for the implantation could just be
photoresist.
Mask
Ion implantation
Photoresist
Si02
n+
p-type silicon
n-type silicon
p-type silicon
n-type silicon
Structure following ion
implantation of n-type impurity
Structure after exposure and
development of photoresist layer
Mask for ion implantation
Side view
Top View
2.53
⎛ 1⎞ ⎛ 1⎞
1 = 8 atoms
(a) N = 8⎜ ⎟ + 6⎜ ⎟ + 4()
⎝ 8⎠ ⎝ 2⎠
(
) (
3
)
3
(b) V = l 3 = 0.543x10−9 m = 0.543x10−7 cm = 1.60x10−22 cm3
8 atoms
atoms
= 5.00x1022
1.60x1022 cm3
cm3
⎛
g ⎞
(d ) m = ⎜ 2.33 3 ⎟1.60x1022 cm3 = 3.73x10−22 g
cm ⎠
⎝
(c) D =
(e) From Table 2.2, silicon has a mass of 28.086 protons.
mp =
g
3.73x10−22 g
= 1.66x10−24
proton
28.082(8)protons
Yes, near the actual proton rest mass.
34
CHAPTER 3
3.1
1019 ⋅ cm−3 )(1018 ⋅ cm −3 )
(
NA ND
φ j = VT ln 2 = (0.025V )ln
= 0.979V
ni
10 20 ⋅ cm −6
2(11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
w do =
+
⎜ 19 −3 + 18 −3 ⎟ (0.979V)
⎜
⎟ φj =
−19
⎝ 10 cm
q ⎝ NA ND ⎠
1.602x10 C
10 cm ⎠
w do = 3.73 x 10−6 cm = 0.0373μm
w do
0.0373μm
w do
0.0373μm
-3
xn =
=
=
μm
18
−3 = 0.0339 μm | x p =
19
−3 = 3.39 x 10
ND
N
10
10
cm
cm
A
1+
1+
1+ 19 −3
1+ 18 −3
NA
ND
10 cm
10 cm
E MAX =
qN A x p
εs
(1.60x10
=
−19
C )(1019 cm −3 )(3.39x10−7 cm)
11.7 ⋅ 8.854 x10
−14
F /cm
= 5.24 x 10 5
V
cm
3.2
p po = N A =
1018
cm 3
1015
n no = N D =
cm 3
| n po =
n i2 10 20 10 2
=
=
p po 1018 cm 3
n i2 10 20 10 5
| p no =
=
=
n no 1015 cm 3
1018 cm−3 )(1015 cm −3 )
(
NA ND
φ j = VT ln 2 = (0.025V )ln
= 0.748 V
ni
10 20 cm−6
w do =
2(11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
+
φ
=
⎜ 18 −3 + 15 −3 ⎟ (0.748V)
⎜
⎟ j
−19
⎝ 10 cm
10 cm ⎠
q ⎝ NA ND ⎠
1.602x10 C
w do = 98.4 x 10−6 cm = 0.984 μm
3.3
p po = N A =
1018
ni2 1020 102
|
n
=
=
=
po
p po 1018 cm3
cm3
1018
ni2 1020 102
nno = N D = 3 | p no =
=
=
nno 1018 cm3
cm
1018 ⋅ cm−3 )(1018 ⋅ cm −3 )
(
NA ND
φ j = VT ln 2 = (0.025V )ln
= 0.921V
ni
10 20 ⋅ cm −6
w do =
2(11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
+
φ
=
⎜ 18 −3 + 18 −3 ⎟ (0.921V)
⎜
⎟ j
−19
⎝ 10 cm
10 cm ⎠
q ⎝ NA ND ⎠
1.602x10 C
w do = 4.881x10−6 cm = 0.0488 μm
34
3.4
p po = N A =
1018
ni2 1020 102
|
n
=
=
=
po
p po 1018 cm3
cm3
1018
ni2 1020 102
|
p
=
=
=
no
nno 1018 cm3
cm3
(1018 ⋅ cm−3 )(1020 ⋅ cm−3 ) = 1.04V
N N
φ j = VT ln A 2 D = (0.025V )ln
ni
10 20 ⋅ cm −6
nno = N D =
2(11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
w do =
+
⎜ 18 −3 + 20 −3 ⎟ (1.04V)
⎜
⎟ φj =
−19
⎝ 10 cm
10 cm ⎠
q ⎝ NA ND ⎠
1.602x10 C
w do = 0.0369 μm
3.5
p po = N A =
1016
ni2 1020 10 4
|
n
=
=
=
po
p po 1016 cm3
cm3
1019
ni2 1020 10
|
p
=
=
=
no
nno 1019 cm3
cm3
1019 ⋅ cm−3 1016 ⋅ cm −3
N AND
φ j = VT ln
= (0.025V )ln
= 0.864V
ni2
1020 ⋅ cm −6
nno = N D =
(
(
)(
)
)
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1 ⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
wdo =
+
⎜
⎟ φj =
⎜ 19 −3 + 16 −3 ⎟ (0.864V)
−19
q ⎝ N A ND ⎠
10 cm ⎠
1.602x10 C
⎝ 10 cm
wdo = 0.334 μm
3.6
wd = wdo 1+
VR
φj
| (a) wd = 2wdo requires VR = 3φ j = 2.55 V | wd = 0.4μm 1+
5
= 1.05 μm
0.85
3.7
wd = wdo 1+
VR
φj
| (a) wd = 3wdo requires VR = 8φ j = 4.80 V | wd = 1μm 1+
10
= 4.20 μm
0.6
3.8
jn = σE , σ =
1
ρ
=
1
2
j 1000 A⋅ cm −2
V
=
| E= n =
= 500
−1
0.5 Ω ⋅ cm Ω ⋅ cm
cm
σ
2(Ω ⋅ cm)
35
3.9
j p = σE
|
E=
jn
σ
(
)
= jn ρ = 5000 A⋅ cm −2 (2Ω ⋅ cm)= 10.0
kV
cm
3.10
⎛ 4x1015 ⎞⎛ 10 7 cm ⎞
A
j ≅ jn = qnv = 1.60x10−19 C ⎜
⎟ = 6400 2
3 ⎟⎜
cm
⎝ cm ⎠⎝ s ⎠
(
)
3.11
⎛ 5x1017 ⎞⎛ 10 7 cm ⎞
kA
j ≅ j p = qpv = 1.60x10−19 C ⎜
⎟ = 800 2
3 ⎟⎜
cm
⎝ cm ⎠⎝ s ⎠
(
3.12
j p = qμ p pE − qD p
)
⎛ D ⎞ 1 dp
⎛ kT ⎞ 1 dp
dp
= 0 → E = −⎜⎜ p ⎟⎟
= −⎜ ⎟
dx
⎝ q ⎠ p dx
⎝ μ p ⎠ p dx
⎛ x⎞
p( x) = N o exp⎜ − ⎟ |
⎝ L⎠
1 dp 1
V
0.025V
V
=
| E = − T = − −4
= −250
L
p dx L
cm
10 cm
The exponential doping results in a constant electric field.
3.13
j p = qDn
dn
dn
dn
2000 A/ cm 2
1.00 x 1021
= qμnVT
|
=
=
dx
dx
dx 1.60x10−19 C 500cm2 /V − s (0.025V )
cm 4
(
3.14
10 = 10 4 ⋅10−16 exp(40VD )−1 + VD
[
]
3.15
f = 10 −10 4 I D − 0.025ln
ID + IS
IS
)(
)
and the solver yields VD = 0.7464 V
| f ' = −10 4 −
0.025
ID + IS
| I'D = I D −
-13
Starting the iteration process with ID = 100 μA and IS = 10 A:
ID
1.000E-04
9.275E-04
9.426E-04
9.426E-04
36
f
f'
8.482E+0
0
1.512E01
3.268E06
9.992E16
-1.025E+04
-1.003E+04
-1.003E+04
-1.003E+04
f
f'
3.16
(a)
Create the following m-file:
function fd=current(id)
fd=10-1e4*id-0.025*log(1+id/1e-13);
Then: fzero('current',1) yields ans = 9.4258e-04
-15
(b) Changing IS to 10 A:
function fd=current(id)
fd=10-1e4*id-0.025*log(1+id/1e-15);
Then: fzero('current',1) yields ans = 9.3110e-04
3.17
−19
qVT 1.60x10 C (0.025V )
T=
=
= 290 K
k
1.38x10−23 J / K
3.18
(
)
−23
kT 1.38x10 J / K T
VT =
=
= 8.63x10−5 T
−19
q
1.60x10 C
For T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV
3.19
⎡ ⎛ 40V ⎞ ⎤
D
Graphing I D = I S ⎢exp⎜
⎟ −1⎥ yields :
n
⎝
⎠ ⎦
⎣
6
5
(b)
4
(a)
3
2
1
(c)
0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
37
3.20
(
)
1.38x10−23 J / K (300)
kT
= 1.04
= 26.88 mV
nVT = n
q
1.60x10−19 C
T = 26.88mV
1.602x10-19
= 312 K
1.38x10-23
3.21
⎡ ⎛v ⎞ ⎤
⎛ i ⎞
vD
= ln⎜1+ D ⎟
iD = I S ⎢exp⎜ D ⎟ −1⎥ or
nVT
⎝ IS ⎠
⎣ ⎝ nVT ⎠ ⎦
⎛i ⎞
⎛ 1 ⎞
v
For i D >> I S , D ≅ ln⎜ D ⎟ or ln (I D )= ⎜
⎟v D + ln (IS )
nVT
⎝ IS ⎠
⎝ nVT ⎠
which is the equation of a straight line with slope 1/nVT and x-axis intercept at -ln (IS). The
-4
values of n and IS can be found from any two points on the line in the figure: e. g. iD = 10 A
-9
for vD = 0.60 V and iD = 10 A for vD = 0.20 V. Then there are two equations in two
unknowns:
⎛ 40 ⎞
⎛8⎞
ln 10-9 = ⎜ ⎟.20 + ln (IS ) or 9.21 = ⎜ ⎟ + ln (IS )
⎝n⎠
⎝n⎠
⎛ 40 ⎞
⎛ 24 ⎞
ln 10-4 = ⎜ ⎟.60 + ln (IS ) or 20.72 = ⎜ ⎟ + ln (IS )
⎝n⎠
⎝n⎠
-12
Solving for n and IS yields n = 1.39 and IS = 3.17 x 10 A = 3.17 pA.
( )
( )
3.22
⎡ ⎛V ⎞ ⎤
⎛ I ⎞
VD = nVT ln⎜1+ D ⎟ | I D = I S ⎢exp⎜ D ⎟ −1⎥
⎝ IS ⎠
⎣ ⎝ nVT ⎠ ⎦
⎛ 7x10−5 A ⎞
⎛ 5x10−6 A ⎞
=
1.05
0.025V
ln
(a) VD = 1.05(0.025V )ln⎜1+
=
0.837V
|
(b)
V
⎟
⎜1+
⎟ = 0.768V
(
)
D
10−18 A ⎠
10−18 A ⎠
⎝
⎝
⎡ ⎛
⎞ ⎤
0
(c) I D = 10−18 A⎢exp⎜
⎟ −1⎥ = 0 A
⎣ ⎝1.05⋅ 0.025V ⎠ ⎦
⎡ ⎛ −0.075V ⎞ ⎤
−19
(d) I D = 10−18 A⎢exp⎜
⎟ −1⎥ = −0.943x10 A
⎣ ⎝ 1.05⋅ 0.025V ⎠ ⎦
⎡ ⎛
⎞ ⎤
−5V
−18
(e) I D = 10−18 A⎢exp⎜
⎟ −1⎥ = −1.00x10 A
⎣ ⎝1.05⋅ 0.025V ⎠ ⎦
3.23
⎡ ⎛V ⎞ ⎤
⎛ I ⎞
VD = nVT ln⎜1+ D ⎟ | I D = I S ⎢exp⎜ D ⎟ −1⎥
⎝ IS ⎠
⎣ ⎝ nVT ⎠ ⎦
⎛ 10−4 A ⎞
⎛ 10−5 A ⎞
(a) VD = 0.025V ln⎜1+ −17 ⎟ = 0.748V | (b) VD = 0.025V ln⎜1+ −17 ⎟ = 0.691V
⎝ 10 A ⎠
⎝ 10 A ⎠
38
⎡ ⎛ 0 ⎞ ⎤
(c) ID = 10−17 A⎢exp⎜
⎟ −1⎥ = 0 A
⎣ ⎝ 0.025V ⎠ ⎦
⎡ ⎛ −0.06V ⎞ ⎤
−17
(d) ID = 10−17 A⎢exp⎜
⎟ −1⎥ = −0.909x10 A
⎣ ⎝ 0.025V ⎠ ⎦
⎡ ⎛ −4V ⎞ ⎤
−17
(e) ID = 10−17 A⎢exp⎜
⎟ −1⎥ = −1.00x10 A
⎣ ⎝ 0.025V ⎠ ⎦
3.24
⎡ ⎛V ⎞ ⎤
⎡ ⎛ 0.675 ⎞ ⎤
−6
I D = I S ⎢exp⎜ D ⎟ −1⎥ = 10−17 A ⎢exp⎜
⎟ −1⎥ = 5.32x10 A = 5.32 μA
⎣ ⎝ 0.025 ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎛ 15.9x10−6 A ⎞
⎞
⎛I
VD = VT ln⎜ D + 1⎟ = (0.025V )ln⎜
+ 1⎟ = 0.703 V
−17
⎠
⎝ IS
⎝ 10 A
⎠
3.25
⎛ I ⎞
⎛
40 A ⎞
VD = nVT ln⎜1+ D ⎟ = 2(0.025V )ln⎜1+ −10 ⎟ = 1.34 V
⎝ 10 A ⎠
⎝ IS ⎠
⎛
100 A ⎞
VD = 2(0.025V )ln⎜1+ −10 ⎟ = 1.38 V
⎝ 10 A ⎠
3.26
ID
2mA
=
= 1.14x10−17 A
⎡ ⎛ V ⎞ ⎤ ⎡ ⎛ 0.82 ⎞ ⎤
⎢exp⎜ D ⎟ −1⎥ ⎢exp⎜
⎟ −1⎥
⎣ ⎝ nVT ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ ⎦
⎡ ⎛ −5 ⎞ ⎤
−17
(b) I D = 1.14x10−17 A ⎢exp⎜
⎟ −1⎥ = −1.14x10 A
0.025
⎠ ⎦
⎣ ⎝
(a) I S =
3.27
ID
300μA
=
= 2.81x10−17 A
⎡ ⎛ V ⎞ ⎤ ⎡ ⎛ 0.75 ⎞ ⎤
⎢exp⎜ D ⎟ −1⎥ ⎢exp⎜
⎟ −1⎥
⎣ ⎝ nVT ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ ⎦
⎡ ⎛ −3 ⎞ ⎤
−17
(b) I D = 2.81x10−17 A ⎢exp⎜
⎟ −1⎥ = −2.81x10 A
⎣ ⎝ 0.025 ⎠ ⎦
(a) I S =
39
3.28
⎛ I ⎞
VD = nVT ln⎜1+ D ⎟ | 10-14 ≤ I S ≤ 10−12
⎝ IS ⎠
⎛ 10−3 A ⎞
VD = (0.025V )ln⎜1+ −14 ⎟ = 0.633 V
⎝ 10 A ⎠
3.29
|
⎛ 10−3 A ⎞
VD = (0.025V )ln⎜1+ −12 ⎟ = 0.518 V
⎝ 10 A ⎠
|
So, 0.518 V ≤ VD ≤ 0.633 V
1.38x10−23 (307)
⎡ ⎛ V
⎞ ⎤
D
=
0.0264V
|
I
=
I
exp
⎢
⎟ −1⎥
⎜
D
S
1.60x10−19
⎣ ⎝ 0.0264n ⎠ ⎦
Varying n and IS by trial-and-error with a spreadsheet:
VT =
n
IS
1.039
7.606E-15
VD
ID-Measured
ID-Calculated
Error
Squared
0.500
0.550
0.600
0.650
0.675
0.700
0.725
0.750
0.775
6.591E-07
3.647E-06
2.158E-05
1.780E-04
3.601E-04
8.963E-04
2.335E-03
6.035E-03
1.316E-02
6.276E-07
3.885E-06
2.404E-05
1.488E-04
3.702E-04
9.211E-04
2.292E-03
5.701E-03
1.418E-02
9.9198E-16
5.6422E-14
6.0672E-12
8.518E-10
1.0261E-10
6.1409E-10
1.8902E-09
1.1156E-07
1.0471E-06
Total Squared Error
3.30
(
1.1622E-06
)
−23
kT 1.38x10 J / K T
=
= 8.63x10−5 T
VT =
q
1.60x10−19 C
For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV
3.31
−23
⎛
10−3 ⎞
kT 1.38x10 (303)
=
=
26.1
mV
|
V
=
0.0261V
ln
1+
= 0.757 V
⎜
(
)
D
−16 ⎟
q
1.60x10−19
⎝ 2.5x10 ⎠
ΔV = (−1.8mV / K )(20K ) = −36.0 mV | VD = 0.757 − 0.036 = 0.721 V
40
3.32
−23
⎛ 10−4 ⎞
kT 1.38x10 (298)
=
= 25.67 mV | (a) VD = (0.02567V )ln⎜1+ −15 ⎟ = 0.650 V
q
1.602x10−19
⎝ 10 ⎠
ΔV = (−2.0mV / K )(25K ) = −50.0 mV
(b) VD = 0.650 − 0.050 = 0.600 V
3.33
−23
⎛ 2.5x10−4 ⎞
kT 1.38x10 (298)
=
=
25.67
mV
|
(a)
V
=
0.02567V
ln
(
) ⎜1+ 10−14 ⎟ = 0.615 V
D
q
1.602x10−19
⎝
⎠
(b) ΔV = (−1.8mV / K )(60K )= −50.0 mV
(c) ΔV = (−1.8mV / K )(−80K )= +144 mV
VD = 0.615 − 0.108 = 0.507 V
VD = 0.615 + 0.144 = 0.758 V
3.34
mV
dvD vD − VG − 3VT 0.7 −1.21− 3(0.0259)
=
=
= −1.96
dT
T
300
K
41
3.35
3
⎡ ⎛ E ⎞⎛ 1 1 ⎞⎤ ⎛ T ⎞3
⎡⎛ E ⎞⎛ T ⎞⎤
I S 2 ⎛ T2 ⎞
= ⎜ ⎟ exp⎢−⎜ G ⎟⎜ − ⎟⎥ = ⎜ 2 ⎟ exp⎢⎜ G ⎟⎜1− 1 ⎟⎥
I S1 ⎝ T1 ⎠
⎣ ⎝ k ⎠⎝ T2 T1 ⎠⎦ ⎝ T1 ⎠
⎣⎝ kT1 ⎠⎝ T2 ⎠⎦
⎡⎛ E ⎞⎛ 1 ⎞⎤
3
T
x= 2
f (x ) = (x ) exp⎢⎜ G ⎟⎜1− ⎟⎥
T1
⎣⎝ kT1 ⎠⎝ x ⎠⎦
Using trial and error with a spreadsheet yields T = 4.27 K, 14.6 K, and 30.7 K to increase the
saturation current by 2X, 10X, and 100X respectively.
x
f(x)
Delta T
1.00000
1.00500
1.01000
1.01500
1.01400
1.01422
1.01922
1.02422
1.02922
1.03422
1.03922
1.04422
1.04922
1.04880
1.10000
1.10239
1.00000
1.27888
1.63167
2.07694
1.97945
2.00051
2.54151
3.22151
4.07433
5.14160
6.47438
8.13522
10.20058
10.00936
90.67434
100.0012
0.00000
1.50000
3.00000
4.50000
4.20000
4.26600
5.76600
7.26600
8.76600
10.26600
11.76600
13.26600
14.76600
14.64000
30.00000
30.71610
3.36
wd = wdo 1+
VR
| (a) wd = 1μm 1+
φj
3.37
(
5
10
= 2.69 μm (b) wd = 1μm 1+
= 3.67 μm
0.8
0.8
)(
)
1016 cm −3 1015 cm −3
N AND
φ j = VT ln
= (0.025V )ln
= 0.633 V
ni2
1020 cm−6
(
)
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1 ⎛
⎞
2εs ⎛ 1
1 ⎞
1
1
+
wdo =
⎜
⎟ φj =
⎜ 16 −3 + 15 −3 ⎟ (0.633V)
−19
q ⎝ N A ND ⎠
10 cm ⎠
1.602x10 C
⎝10 cm
wdo = 0.949 μm
wd = 0.949μm 1+
42
|
wd = wdo 1+
VR
φj
10V
= 3.89 μm
0.633V
|
wd = 0.949μm 1+
100V
= 12.0 μm
0.633V
3.38
(
)(
)
1018 cm −3 1020 cm −3
N AND
φ j = VT ln
= (0.025V )ln
= 1.04 V
ni2
1020 cm−6
(
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1
2εs ⎛ 1
1 ⎞
wdo =
+
⎟ φj =
⎜
q ⎝ N A ND ⎠
1.602x10−19 C
|
wd = 0.0368μm 1+
5
= 0.0887 μm
1.04
3.39
Emax =
3x105
2(φ j + VR )
wd
=
2(φ j + VR )
wdo 1+
VR
=
⎞
1
1
+
⎟ (1.04V)
18
−3
1020 cm−3 ⎠
⎝ 10 cm
VR
wdo = 0.0368 μm
wd = wdo 1+
)⎛⎜
φj
|
wd = 0.0368μm 1+
25
= 0.184 μm
1.04
2φ j
V
1+ R
wdo
φj
φj
2(0.6V )
V
V
= −4
1+ R → VR = 374 V
cm 10 cm
0.6
3.40
2(0.748V )
2φ
kV
= 15.2
|
E= j =
−4
wdo 0.984x10 cm
cm
E
w
φ j + VR = max do =
2
φj
3x105
(
V
0.984x10−4 cm
cm
2 0.748V
)
VR = 291.3 − 0.748 = 291 V
3.41
VZ = 4 V; RZ = 0 Ω since the reverse breakdown slope is infinite.
3.42
Since NA >> ND, the depletion layer is all on the lightly-doped side of the junction. Also, VR
>> φj, so φj can be neglected.
Emax =
qN A x p
εS
=
qN A wd
(
)
(
εS
=
qN A
εS
(
2εS VR
q NA
)
2
3x105 (11.7) 8.854x10−14
Emax
εS
NA =
=
= 2.91 x 1014 / cm3
−19
2qVR
2 1.602x10 1000
2
)
43
3.43
φ j = VT ln
N AND
10151020
=
0.025ln
= 0.864V
ni2
1020
(
)
2(11.7) 8.854x10−14 ⎛ 1
2εS ⎛ 1
1 ⎞
1 ⎞
−4
wdo =
+
+
⎟φ j =
⎜
⎟0.864 = 1.057x10 cm
⎜
q ⎝ N A ND ⎠
1.602x10−19
⎝ 1015 1020 ⎠
C =
"
jo
εS
=
wdo
(
11.7 8.854x10−14
1.057x10−4
) = 9.80x10
-9
F / cm
2
| Cj =
C "jo A
1+
9.80x10-9 (0.05)
=
5
1+
0.864
VR
φj
= 188 pF
3.44
φ j = VT ln
N AND
10181015
=
0.025ln
= 0.748V
ni2
1020
(
2(11.7) 8.854x10−14
2εS ⎛ 1
1 ⎞
wdo =
+
⎟φ j =
⎜
q ⎝ N A ND ⎠
1.602x10−19
C =
"
jo
εS
wdo
=
(
11.7 8.854x10−14
0.984x10−4
3.45
(
) = 10.5x10
-9
(c) CD =
(
0.025V
3.46
(
)
) = 100 pF
(c) CD =
44
(
0.025V
| Cj =
C "jo A
)
) = 0.04 μF
=
10.5x10-9 (0.02)
10
1+
0.748
VR
φj
(
)
(b) Q = I D τ T = 10−4 A 10−10 s = 10 fC
(
)
| Q = I D τ T = 5x10−3 A 10−10 s = 0.50 pC
−8
I D τ T 1A 10 s
(a) CD =
=
= 0.400 μF
VT
0.025V
100mA 10−8 s
1 ⎞
1
0.748 = 0.984x10−4 cm
+
18
15 ⎟
10
10
⎠
⎝
1+
−4
−10
I D τ T 10 A 10 s
(a) CD =
=
= 400 fF
VT
0.025V
25x10−3 A 10−10 s
F / cm
2
)⎛⎜
(
)
(b) Q = I D τ T = 1A 10−8 s = 10.0 nC
(
)
| Q = I D τ T = 100mA 10−8 s = 1.00 nC
= 55.4 pF
3.47
N AND
10191017
=
0.025ln
= 0.921V
ni2
1020
φ j = VT ln
(
)
2(11.7) 8.854x10−14 ⎛ 1
2εS ⎛ 1
1 ⎞
1 ⎞
wdo =
+
+
⎟φ j =
⎜
⎟0.921 = 0.110 μm
⎜
q ⎝ N A ND ⎠
1.602x10−19
⎝ 1019 1017 ⎠
C jo =
εS A
wdo
=
(
)( ) = 9.42 pF / cm
11.7 8.854x10−14 10−4
−4
0.110x10
C jo
| Cj =
2
1+
=
VR
φj
9.42 pF
5
1+
0.921
= 3.72 pF
3.48
N AND
10191016
=
0.025ln
= 0.864V
ni2
1020
φ j = VT ln
(
)⎛⎜
11.7 8.854x10−14 0.25cm2
) = 7750 pF
2(11.7) 8.854x10−14
2εS ⎛ 1
1 ⎞
wdo =
+
⎟φ j =
⎜
q ⎝ N A ND ⎠
1.602x10−19
C jo =
εS A
wdo
=
(
)(
−4
0.334x10
1 ⎞
1
+
⎟0.864 = 0.334μm
⎝ 1019 1016 ⎠
| Cj =
C jo
1+
=
VR
φj
7750 pF
3
1+
0.864
= 3670 pF
3.49
L=
RFC
C
VDC
C=
C jo
1+
(b) C =
(a) C =
VR
φj
39 pF
10V
1+
0.75V
39 pF
1V
1+
0.75V
= 25.5 pF | f o =
= 10.3 pF | f o =
10 μH
10 μH
C
1
2π LC
=
2π
1
2π LC
(
1
=
)
2π
(
1
)
10−5 H 25.5 pF
10−5 H 10.3 pF
= 9.97 MHz
= 15.7 MHz
3.50
⎛
⎛
50 A ⎞
50 A ⎞
(a) VD = (0.025V )ln⎜1+ −7 ⎟ = 0.501 V | (b) VD = (0.025V )ln⎜1+ −15 ⎟ = 0.961 V
⎝ 10 A ⎠
⎝ 10 A ⎠
45
3.51
⎛ 4x10−3 A ⎞
⎛ 4x10−3 A ⎞
(a) VD = (0.025V )ln⎜1+
=
0.025V
ln
=
0.495
V
|
(b)
V
⎟
⎜
(
) 1+ 10−14 A ⎟ = 0.668 V
D
10−11 A ⎠
⎝
⎝
⎠
3.52
Rp = ρ p
RS = R p + Rn
Rn = ρ n
Lp
0.025cm
= (1Ω − cm)
= 2.5Ω
Ap
0.01cm 2
Ln
0.025cm
= (0.01Ω − cm)
= 0.025Ω
An
0.01cm 2
RS = 2.53 Ω
3.53
⎛
⎛ I ⎞
10−3 ⎞
(a) VD' = VT ln⎜1+ D ⎟ = (0.025V )ln⎜1+
= 0.708V
−16 ⎟
⎝ IS ⎠
⎝ 5x10 ⎠
VD = VD' + I D RS = 0.708V + 10−3 A(10Ω)= 0.718 V
(b) VD = VD' + I D RS = 0.708V + 10−3 A(100Ω)= 0.808 V
3.54
10Ω − μm2
ρ c = 10Ω − μm
Ac = 1μm
RC =
=
= 10Ω / contact
Ac
1μm2
5 anode contacts and 14 cathode contacts
2
2
ρc
10Ω
= 2Ω
5
10Ω
= 0.71Ω
Resistance of cathode contacts =
14
Resistance of anode contacts =
3.55
(a) From Fig. 3.21a, the diode is approximately 10.5 μm long x 8 μm wide. Area = 84 μm2.
(b) Area = (10.5x0.13 μm) x (8x0.13μm) = 1.42 μm2.
46
3.56
(a) 5 = 10 4 I D + VD | VD = 0 I D = 0.500mA | I D = 0 VD = 5V
4.5V
= 0.450 mA
10 4 Ω
| VD = 0 I D = −2.00mA | I D = 0 VD = −6V
Forward biased - VD = 0.5 V I D =
(b) − 6 = 3000I D + VD
−2V
= −0.667 mA
3kΩ
| VD = 0 I D = −1.00mA | I D = 0 VD = −3V
In reverse breakdown - VD = −4 V I D =
(c) − 3 = −3000I D + VD
Reverse biased - VD = −3 V I D = 0
iD
2 mA
1 mA
(a) Q-point
(c) Q-point
-6
-5
-4
-3
-2
vD
-1
1
2
5
4
3
6
(b) Q-point
-1 mA
-2 mA
3.57
(a) 10 = 5000I D + VD | VD = 0 I D = 2.00 mA | VD = 5 V I D = 1.00 mA
9.5V
= 1.90 mA
5kΩ
| VD = 0 I D = −2.00 mA | VD = −5 V I D = −1.00 mA
Forward biased - VD = 0.5V I D =
(b) -10 = 5000I D + VD
−6V
= −1.20 mA
5kΩ
| VD = 0 I D = −1.00 mA | I D = 0 VD = −2 V
In reverse breakdown - VD = −4V I D =
(c) − 2 = 2000I D + VD
Reverse biased - VD = −2 V I D = 0
iD
2 mA
(a) Q-point
1 mA
(c) Q-point
-6
-5
-4
-3
-2
v
-1
1
(b) Q-point
2
3
4
5
D
6
-1 mA
-2 mA
47
3.58
*Problem 3.58 - Diode Circuit
V 1 0 DC 5
R 1 2 10K
D1 2 0 DIODE1
.OP
.MODEL DIODE1 D IS=1E-15
.END
SPICE Results
VD = 0.693 V
ID = 0.431 μA
3.59
(a) −10 = 10 4 I D + VD | VD = 0 I D = −1.00 mA | VD = −5 V
I D = −0.500 mA
−10 − (−4)V
= −0.600 mA
10kΩ
| VD = 0 I D = 1.00 mA | VD = 5 V I D = 0.500 mA
In reverse breakdown - VD = −4 V I D =
(b) 10 = 10 4 I D + VD
10 − 0.5V
= 0.950 mA
10kΩ
| VD = 0 I D = −2.00 mA | I D = 0
Forward biased - VD = 0.5 V I D =
(c) − 4 = 2000I D + VD
Reverse biased - VD = −4 V I D = 0
iD
2 mA
1 mA
(b) Q-point
(c) Q-point
-6
-5
-4
-3
-2
vD
-1
1
(a) Q-point
-1 mA
-2 mA
48
2
3
4
5
6
VD = − 4 V
iD (A)
0.002
0.001
-7
-6
-5
-4
-3
-2
v D (V)
-1
1
2
3
4
5
6
7
-0.001
-0.002
49
3.60
R
i
D
+
v
+
V
-
D
-
The load line equation: V = iD R + vD
We need two points to plot the load line.
(a) V = 6 V and R = 4kΩ: For vD = 0, iD = 6V/4 kΩ = 1.5 mA and for iD = 0, vD = 6V.
Plotting this line on the graph yields the Q-pt: (0.5 V, 1.4 mA).
(b) V = -6 V and R = 3kΩ: For vD = 0, iD = -6V/3 kΩ = -2 mA and for iD = 0, vD = -6V.
Plotting this line on the graph yields the Q-pt: (-4 V, -0.67 mA).
(c) V = -3 V and R = 3kΩ: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA)
(d) V = +12 V and R = 8kΩ: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA)
(e) V = -25 V and R = 10kΩ: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA)
i (A)
D
.002
Q-Point
(0.5V,1.45 mA)
(d)
.001
Q-Point
(-3V,0 mA)
-7
-6
-5
-4
-3
Q-Point
(0.5V,1.4 mA)
-2
-1
1
(c)
Q-Point
(-4V,-0.67 mA)
-.001
Load line for (b)
-.002
Q-Point
(-4V,-2.1 mA)
50
(e)
2
Load line for (a)
3
4
5
6
7
v (V)
D
3.61
-9
Using the equations from Table 3.1, (f = 10-10 exp ..., etc.)
VD = 0.7 V requires 12 iterations, VD = 0.5 V requires 22 iterations,
VD = 0.2 V requires 384 iterations - very poor convergence because the second iteration (VD =
9.9988 V) is very bad.
3.62
Using Eqn. (3.28),
⎛i ⎞
V = iD R + VT ln⎜ D ⎟
⎝ IS ⎠
(
)
or 10 = 10 4 iD + 0.025ln 1013 iD .
(
4
13
We want to find the zero of the function f = 10 −10 iD − 0.025ln 10 iD
iD
f
.001
-0.576
.0001
8.48
.0009
0.427
.00094
0.0259 - converged
)
3.63
⎛ I ⎞
0.025
f = 10 −10 4 ID − 0.025ln⎜1+ D ⎟ | f ' = −10 4 −
ID + IS
⎝ IS ⎠
x
f(x)
f'(x)
1.0000E+00
9.2766E-04
9.4258E-04
9.4258E-04
9.4258E-04
-9.991E+03
1.496E-01
3.199E-06
9.992E-16
9.992E-16
-1.000E+04
-1.003E+04
-1.003E+04
-1.003E+04
-1.003E+04
3.64
Create the following m-file:
function fd=current(id)
fd=10-1e4*id-0.025*log(1+id/1e-13);
Then: fzero('current',1) yields ans = 9.4258e-04 + 1.0216e-21i
51
3.65
The one-volt source will forward bias the diode. Load line:
1 = 10 4 I D + VD | I D = 0 VD = 1V | VD = 0 I D = 0.1mA → (50 μA, 0.5 V )
[
]
−9
Mathematical model: f = 1−10 exp(40VD )−1 + VD → (49.9 μA, 0.501 V )
Ideal diode model: ID = 1V/10kΩ = 100μA; (100μA, 0 V)
Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0μA; (40.0μA, 0.6 V)
3.66
Using Thévenin equivalent circuits yields and then combining the sources
I
1.2 k Ω
1.2 V
-
+
V
I
1k Ω
-
+
+
-
-
V
2.2 k Ω
+
+
1.5 V
-
0.3 V
(a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will
assume it is "on". Substituting the ideal diode model for the forward region yields
0.3V
I=
= 0.136 mA . This current is greater than zero, which is consistent with the diode
2.2kΩ
being "on". Thus the Q-pt is (0 V, +0.136 mA).
I
-
V
V
on
-
+
+
I
2.2 k Ω
0.6 V
+
-
2.2 k Ω
+
0.3 V
-
0.3 V
CVD:
Ideal Diode:
(b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it
0.3V − 0.6V
= −136 μA .
is "on". Substituting the CVD model with Von = 0.6 V yields I =
2.2kΩ
This current is negative which is not consistent with the assumption that the diode is "on".
Thus the diode must be off. The resulting Q-pt is: (0 mA, -0.3 V).
-
V +
I=0
2.2 k Ω
+
52
0.3 V
(c) The second estimate is more realistic. 0.3 V is not sufficient to forward bias the diode into
-15
significant conduction. For example, let us assume that IS = 10 A, and assume that the full
0.3 V appears across the diode. Then
⎡ ⎛ 0.3V ⎞ ⎤
iD = 10−15 A⎢exp⎜
⎟ −1⎥ = 163 pA , a very small current.
⎣ ⎝ 0.025V ⎠ ⎦
3.67
The nominal values are:
⎛ R2 ⎞
⎛ 2kΩ ⎞
VA = 3V ⎜
⎟ = 3V ⎜
⎟ = 1.20V
⎝ 2kΩ + 3kΩ ⎠
⎝ R1 + R2 ⎠
and RTHA =
2kΩ(3kΩ)
R1 R2
=
= 1.20kΩ
R1 + R2 2kΩ + 3kΩ
⎛ R4 ⎞
⎛ 2kΩ ⎞
2kΩ(2kΩ)
R3 R4
VC = 3V ⎜
=
= 1.00kΩ
⎟ = 3V ⎜
⎟ = 1.50V and RTHC =
R3 + R4 2kΩ + 2kΩ
⎝ 2kΩ + 2kΩ ⎠
⎝ R3 + R4 ⎠
⎛ 1.50 −1.20 ⎞ V
I Dnom = ⎜
= 136 μA
⎟
⎝1.20 + 1.00 ⎠ kΩ
For maximum current, we make the Thévenin equivalent voltage at the diode anode as large as
possible and that at the cathode as small as possible.
VA =
VC =
3V
3V
=
= 1.65V and
R1
2kΩ(0.9)
1+
R2 1+ 2kΩ(1.1)
3V
3V
=
= 1.06V and
R3
3kΩ(1.1)
1+
R4 1+ 2kΩ(0.9)
RTHA =
2kΩ(0.9)2kΩ(1.1)
R1 R2
=
= 0.990kΩ
R1 + R2 2kΩ(0.9)+ 2kΩ(1.1)
RTHC =
3kΩ(1.1)2kΩ(0.9)
R3 R4
=
= 1.17kΩ
R3 + R4 3kΩ(1.1)+ 2kΩ(0.9)
⎛ 1.65 −1.06 ⎞ V
I Dmax = ⎜
= 274 μA
⎟
⎝ 0.990 + 1.17 ⎠ kΩ
For minimum current, we make the Thévenin equivalent voltage at the diode anode as small as
possible and that at the cathode as large as possible.
VA =
VC =
3V
3V
=
= 1.350V and
R1
2kΩ(1.1)
1+
R2 1+ 2kΩ(0.9)
3V
3V
=
= 1.347V and
R3
3kΩ(0.9)
1+
R4 1+ 2kΩ(1.1)
RTHA =
2kΩ(1.1)2kΩ(0.9)
R1 R2
=
= 0.990kΩ
R1 + R2 2kΩ(1.1)+ 2kΩ(0.9)
RTHC =
3kΩ(0.9)2kΩ(1.1)
R3 R4
=
= 1.21kΩ
R3 + R4 3kΩ(0.9)+ 2kΩ(1.1)
⎛1.350 −1.347 ⎞ V
= 1.39 μA ≅ 0
I Dmin = ⎜
⎟
⎝ 0.990 + 1.21 ⎠ kΩ
53
3.68
SPICE Input
*Problem 3.68
V1 1 0 DC 4
R1 1 2 2K
R2 2 0 2K
R3 1 3 3K
R4 3 0 2K
D1 2 3 DIODE
.MODEL DIODE D IS=1E-15 RS=0
.OP
.END
Results
NAME D1
MODEL
ID
VD
DIODE
1.09E-10
3.00E-01
The diode is essentially off - VD = 0.3 V and ID = 0.109 nA. This result agrees with the CVD
model results.
3.69 (a)
(a) Diode is forward biased :V = 3 − 0 = 3 V | I =
3 − (−7)
= 0.625 mA
16kΩ
5 − (−5)
= 0.625 mA
(b) Diode is forward biased :V = −5 + 0 = −5 V | I =
16kΩ
(c) Diode is reverse biased : I = 0 | V = −5 + 16kΩ(I )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ(I ) = 7 V | VD = −10 V
(b)
(a) Diode is forward biased :V = 3 − 0.7 = 2.3 V | I =
2.3 − (−7)
= 0.581 mA
16kΩ
5 − (−4.3)
(b) Diode is forward biased :V = −5 + 0.7 = −4.3 V | I =
= 0.581 mA
16kΩ
(c) Diode is reverse biased : I = 0 | V = −5 + 16kΩ(I )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ(I ) = 7 V | VD = −10 V
3.70 (a)
3 − (−7)
= 100 μA
100kΩ
5 − (−5)
(b) Diode is forward biased :V = −5 + 0 = −5 V | I =
= 100 μA
100kΩ
(c) Diode is reverse biased : I = 0 A | V = −5 + 100kΩ(I )= −5 V | VD = −10 V
(a) Diode is forward biased :V = 3 − 0 = 3 V | I =
(d ) Diode is reverse biased : I = 0 A | V = 7 −100kΩ(I )= 7 V | VD = −10 V
(b)
54
2.4 − (−7)
= 94.0 μA
100kΩ
5 − (−4.4)
(b) Diode is forward biased :V = −5 + 0.6 = −4.4 V | I =
= 94.0 μA
100kΩ
(c) Diode is reverse biased : I = 0 | V = −5 + 100kΩ(I ) = −5 V | VD = −10 V
(a) Diode is forward biased :V = 3 − 0.6 = 2.4 V | I =
(d ) Diode is reverse biased : I = 0 | V = 7 −100kΩ(I ) = 7 V | VD = −10 V
3.71 (a)
(a) D1 on, D2 on : I D2 =
D1 : (409 μA, 0 V )
0 − (−9)
= 409μA | I D1 = 409μA −
22kΩ
D2 : (270 μA, 0 V )
6−0
= 270μA
43kΩ
6−0
= 140 μA | VD2 = −9 − 0 = −9V
43kΩ
D2 : (0 A, − 9 V )
(b) D1 on, D2 off : I D2 = 0 | I D1 =
D1 : (140 μA, 0 V )
(c) D1 off, D2 on : I D1 = 0 | I D2 =
D1 : (0 A,−3.92 V )
65kΩ
D2 : (230 μA,0 V )
(d ) D1 on, D2 on : I D 2 =
D1 : (140 μA,0 V )
6 − (−9)
0 − (−6)
= 230 μA | VD1 = 6 − 43x103 I D2 = −3.92 V
= 140 μA | I D1 =
43kΩ
D2 : (270 μA,0 V )
9−0
−140 μA = 270 μA
22kΩ
(b)
(a) D1 on, D2 on :
I D2 =
-0.75 − 0.75 − (−9)
= 341μA | I D1 = 341μA −
22kΩ
D1 : (184 μA, 0.75 V) D2 : (341 μA, 0.75 V)
6 − (−0.75)
43kΩ
= 184μA
(b) D1 on, D2 off :
6 − 0.75
= 122μA | VD 2 = −9 − 0.75 = −9.75V
43kΩ
D1 : (122 μA, 0.75 V) D2 : (0 A, − 9.75 V)
I D2 = 0 | I D1 =
55
(c) D1 off, D2 on :
I D1 = 0 | I D2 =
6 − 0.75 − (−9)
= 219μA | VD1 = 6 − 43x103 I D2 = −3.43V
65kΩ
D1 : (0 A, − 3.43 V ) D2 : (219 μA, 0.75 V )
(d) D1 on, D2 on :
I D2 =
0.75 − 0.75 − (−6)
43kΩ
D1 : (235 μA, 0.75 V)
9 − 0.75
− 400μA = 235μA
22kΩ
D2 : (140 μA, 0.75 V)
= 140μA | I D1 =
3.72 (a)
(a) D1 and D2 forward biased
I D2 =
0 − (−9) V
= 600μA
15 kΩ
D1 : (0 V, 200 μA)
I D1 = I D2 −
D2 : (0 V, 600 μA)
6 − (0) V
= 200μA
15 kΩ
(b) D1 forward biased, D2 reverse biased
6−0 V
= 400μA
VD2 = −9 − 0 = −9 V
15 kΩ
D1 : (0 V, 400 μA) D2 : (-9 V, 0 A )
I D1 =
(c) D1 reverse biased, D2 forward biased
6V − (−9V )
= 500μA
VD1 = 6 −15000I D2 = −1.50V
30kΩ
D1 : (−1.50 V, 0 A) D2 : (0 V, 500 μA)
I D2 =
(d) D1 and D2 forward biased
I D2 =
0 − (−6) V
= 400μA
15 kΩ
D1 : (0 V, 200 μA)
I D1 =
9 − (0) V
15 kΩ
D2 : (0 V, 400 μA)
− I D2 = 200μA
(b)
(a) D1 on, D2 on :
I D2 =
-0.75 − 0.75 − (−9)
= 500μA | I D1 = 500μA −
15kΩ
D1 : (50.0 μA, 0.75 V) D2 : (500 μA, 0.75 V)
56
6 − (−0.75)
15kΩ
= 50.0μA
(b) D1 on, D2 off :
6 − 0.75
= 350μA | VD2 = −9 − 0.75 = −9.75V
15kΩ
D1 : (350 μA, 0.75 V) D2 : (0 A, − 9.75 V)
I D2 = 0 | I D1 =
(c) D1 off, D2 on :
6 − 0.75 − (−9)
= 475μA | VD1 = 6 −15x103 I D2 = −1.13V
30kΩ
D1 : (0 A, −1.13 V ) D2 : (475 μA, 0.75 V )
I D1 = 0 | I D2 =
(d) D1 on, D2 on :
I D2 =
0.75 − 0.75 − (−6)
15kΩ
D1 : (150 μA, 0.75 V)
9 − 0.75
− 400μA = 150μA
15kΩ
D2 : (400 μA, 0.75 V)
= 400μA | I D1 =
3.73 Diodes are labeled from left to right
(a) D1 on, D2 off, D3 on : I D2 = 0 | I D1 =
0 − (−5)
10 − 0
= 0.990mA
3.3kΩ + 6.8kΩ
→ I D3 = 1.09mA | VD2 = 5 − (10 − 3300I D1 ) = −1.73V
2.4kΩ
D1 : (0.990 mA, 0 V) D2 : (0 mA, −1.73 V) D3 : (1.09 mA, 0 V)
I D3 + 0.990mA =
(b) D1 on, D2 off, D3 on : I D2 = 0 | I D3 = 0
(10 − 0)V
= 0.495mA | VD2 = 5 − (10 − 8200I D1 )= −0.941V
8.2kΩ + 12kΩ
0 − (−5V )
− I D1 = 0.005mA
I D3 =
10kΩ
D1 : (0.495 mA, 0 V ) D2 : (0 A, − 0.941 V ) D3 : (0.005 mA, 0 V )
I D1 =
57
(c) D1 on, D2 on, D3 on
0 − (−10)
0 − (2)
V = 1.22mA > 0 | I12K =
V = −0.167mA | I D2 = I D1 + I12K = 1.05mA > 0
8.2kΩ
12kΩ
2 − (−5)
V = 0.700mA | I D3 = I10K − I12K = 0.533mA > 0
I10K =
10kΩ
D1 : (1.22 mA, 0 V ) D2 : (1.05 mA, 0 V ) D3 : (0.533 mA, 0 V )
I D1 =
(d) D1 off, D2 off, D3 on : I D1 = 0, I D2 = 0
12 − (−5)
V
= 1.21mA > 0 | VD1 = 0 − (−5 + 4700I D3 ) = −0.667V < 0
4.7 + 4.7 + 4.7 kΩ
VD2 = 5 − (12 − 4700I D3 ) = −1.33V < 0
I D3 =
D1 : (0 A, − 0.667 V ) D2 : (0 A, −1.33 V ) D3 : (1.21 mA, 0 V )
3.74 Diodes are labeled from left to right
(a) D1 on, D2 off, D3 on : I D2 = 0 | I D1 =
−0.6 − (−5)
10 − 0.6 − (−0.6)
3.3kΩ + 6.8kΩ
= 0.990mA
→ I D3 = 0.843mA | VD2 = 5 − (10 − 0.6 − 3300I D1 )= −1.13V
2.4kΩ
D1 : (0.990 mA, 0.600 V) D2 : (0 A, −1.13 V) D3 : (0.843 mA, 0.600V)
I D3 + 0.990mA =
(b) D1 on, D2 off, D3 off : I D2 = 0 | I D3 = 0
10 − 0.6 − (−5)
V = 0.477mA | VD2 = 5 − (10 − 0.6 − 8200I D1 )= −0.490V
8.2kΩ + 12kΩ + 10kΩ
VD3 = 0 − (−5 + 10000I D1 )= +0.230V < 0.6V so the diode is off
I D1 =
D1 : (0.477 mA, 0.600 V) D2 : (0 A, − 0.490 V) D3 : (0 A, 0.230 V)
(c) D1 on, D2 on, D3 on
I D1 =
−0.6 − (−9.4) V
8.2
kΩ
= 1.07mA > 0 | I12K =
58
12
kΩ
= −0.167mA
1.4 − (−5) V
= 0.640mA | I D3 = I10K − I12K = 0.807mA > 0
kΩ
10
D2 : (0.906 mA, 0.600 V) D3 : (0.807 mA, 0.600 V)
I D2 = I D1 + I12K = 0.906mA > 0 | I10K =
D1 : (1.07 mA, 0.600 V)
−0.6 − (1.4) V
(d) D1 off, D2 off, D3 on : I D1 = 0, I D2 = 0
11.4 − (−5) V
= 1.16mA > 0 | VD1 = 0 − (−5 + 4700I D3 ) = −0.452V < 0
4.7 + 4.7 + 4.7 kΩ
VD2 = 5 − (11.4 − 4700I D3 ) = −0.948V < 0
I D3 =
D1 : (0 A, − 0.452 V ) D2 : (0 A, − 0.948 V ) D3 : (1.16 mA, 0.600 V )
3.75
*Problem 3.75(a) (Similar circuits are used for the other three cases.)
V1 1 0 DC 10
V2 4 0 DC 5
V3 6 0 DC -5
R1 2 3 3.3K
R2 3 5 6.8K
R3 5 6 2.4K
D1 1 2 DIODE
D2 4 3 DIODE
D3 0 5 DIODE
.MODEL DIODE D IS=1E-15 RS=0
.OP
.END
NAME
D1
D2
D3
MODEL DIODE DIODE
DIODE
ID
9.90E-04 -1.92E-12 7.98E-04
VD
7.14E-01 -1.02E+00 7.09E-01
NAME
D1
D2
MODEL DIODE
DIODE
ID
4.74E-04 -4.22E-13
VD
6.95E-01 -4.21E-01
NAME
D1
MODEL DIODE
ID
8.79E-03
VD
7.11E-01
D3
DIODE
2.67E-11
2.63E-01
D2
D3
DIODE DIODE
1.05E-03 7.96E-04
7.16E-01 7.09E-01
NAME
D1
D2
D3
MODEL DIODE
DIODE DIODE
ID
-4.28E-13 -8.55E-13 1.15E-03
VD
-4.27E-01 -8.54E-01 7.18E-01
For all cases, the results are very similar to the hand analysis.
3.76
59
I D1 =
10 − (−20)
= 1.50mA | I D2 = 0
10kΩ + 10kΩ
0 − (−10)
I D3 =
= 1.00mA | VD2 = 10 −10 4 I D1 − 0 = −5.00V
10kΩ
D1 : (1.50 mA, 0 V) D2 : (0 A,−5.00 V) D3 : (1.00 mA, 0 V)
3.77
*Problem 3.77
V1 1 0 DC -20
V2 4 0 DC 10
V3 6 0 DC -10
R1 1 2 10K
R2 4 3 10K
R3 5 6 10K
D1 3 2 DIODE
D2 3 5 DIODE
D3 0 5 DIODE
.MODEL DIODE D IS=1E-14 RS=0
.OP
.END
NAME
D1
D2
D3
MODEL DIODE
DIODE
DIODE
ID
1.47E-03 -4.02E-12 9.35E-04
VD
6.65E-01 -4.01E+00 6.53E-01
The simulation results are very close to those given in Ex. 3.8.
3.78
3.9kΩ
= 6.28V | RTH = 11kΩ 3.9kΩ = 2.88kΩ
3.9kΩ + 11kΩ
6.28 − 4
IZ =
= 0.792mA > 0 | (I Z ,VZ )= (0.792 mA,4 V )
2.88kΩ
VTH = 24V
60
3.79
−6.28 = 2880I D + VD | I D = 0,VD = −6.28V | VD = 0, I D =
-6
-5
-4
-3
-2
-1
v
Q-point
−6.28
= −2.18mA
2880
D
-1 mA
-2 mA
i
3.80
IS =
3.81
IS =
D
Q-Point: (-0.8 mA, -4 V)
27 − 9
9V
= 1.20mA → I L < 1.20 mA | RL >
= 7.50 kΩ
15kΩ
1.2mA
27 − 9
= 1.20mA | P = (9V )(1.20mA) = 10.8 mW
15kΩ
3.82
⎛ 1
VS − VZ VZ VS
1⎞
−
=
− VZ ⎜ + ⎟ | PZ = VZ I Z
RS
RL RS
⎝ RS RL ⎠
⎛ 1
30V
1 ⎞
nom
I Znom =
− 9V ⎜
+
⎟ = 0.500 mA | PZ = 9V (0.500mA)= 4.5 mW
15kΩ
15kΩ
10kΩ
⎝
⎠
⎞
⎛
30V (1.05)
1
1
⎟ = 0.796 mA
I Zmax =
− 9V (0.95)⎜⎜
+
⎟
10kΩ(1.05)
15kΩ(0.95)
15kΩ
0.95
(
)
⎠
⎝
IZ =
PZmax = 9V (.95)(0.796mA)= 6.81 mW
I
min
Z
30V (0.95)
⎞
⎛
1
1
⎟ = 0.215 mA
⎜
=
− 9V (1.05)⎜
+
⎟
10kΩ(0.95)
15kΩ(1.05)
15kΩ
1.05
(
)
⎠
⎝
PZmin = 9V (1.05)(0.215mA)= 2.03 mW
3.83
100Ω
24 −15
= 24.0V | RTH = 150Ω 100Ω = 60Ω | I Z =
= 150 mA
150Ω + 100Ω
60
60 −15
= 300 mA | P = 15I Z = 4.50 W
P = 15I Z = 2.25 W | (b) I Z =
150
(a) VTH = 60V
61
3.84
IZ =
⎛ 1
VS − VZ VZ VS
1⎞
−
=
− VZ ⎜ + ⎟
RS
RL RS
⎝ RS RL ⎠
PZ = VZ I Z
(60 −15)V − 15V
= 150 mA | PZnom = 15V (150mA) = 2.25 W
150Ω
100Ω
⎛
⎞
60V (1.1)
1
1
⎟ = 266 mA
=
−15V (0.90)⎜⎜
+
⎟
100Ω(1.1)
150Ω
0.90
150Ω(0.90)
(
)
⎝
⎠
I Znom =
I Zmax
|
PZmax = 15V (0.90)(266mA)= 3.59 W
I Zmin =
60V (0.90)
⎛
⎞
1
1
⎟ = 43.9 mA
−15V (1.1)⎜⎜
+
⎟
100Ω(0.9)
150Ω
1.1
150Ω(1.1)
(
)
⎝
⎠
PZmin = 15V (1.1)(43.9mA)= 0.724 W
3.85
Using MATLAB, create the following m-file with f = 60 Hz:
function f=ctime(t)
f=5*exp(-10*t)-6*cos(2*pi*60*t)+1;
Then: fzero('ctime',1/60) yields ans = 0.01536129698461
and T = (1/60)-0.0153613 = 1.305 ms.
ΔT =
1
120π
ΔT =
1
120π
3.86
IT
5
2Vr
| Vr =
=
= 0.8333V
VP
C 0.1(60)
2(0.8333)
6
= 1.40 ms
⎛ I ⎞
⎛ 48.6 A ⎞
VD = nVT ln⎜1+ D ⎟ = 2(0.025V )ln⎜1+ −9 ⎟ = 1.230 V
⎝ 10 A ⎠
⎝ IS ⎠
62
3.87
⎛ I ⎞
Von = nVT ln⎜1+ D ⎟ | VD = Von + I D RS
⎝ IS ⎠
⎛ 100 A ⎞
VD = 1.6(0.025V )ln⎜1+ −8 ⎟ + 100 A(0.01Ω) = 1.92 V
⎝ 10 A ⎠
⎛100 A ⎞⎛ 1ms ⎞
I ΔT
= 0.92V ⎜
Pjunction ≅ Von I DC = Von P
⎟⎜
⎟ = 2.75 W
2T
⎝ 2 ⎠⎝16.7ms ⎠
2
4⎛ T ⎞ 2
4 ⎛16.7ms ⎞
PR ≅ ⎜ ⎟ I DC
RS = ⎜
⎟(3A) 0.01Ω = 2.00 W
3 ⎝ ΔT ⎠
3 ⎝ 1ms ⎠
Ptotal = 4.76 W
3.88
VDC =
1
T
1⎡
T
∫ v (t )dt = T ⎢⎣(V
P
− Von )T −
0
VDC = 0.975(18V )= 17.6 V
0.05(VP − Von )⎤
TVr ⎤ ⎡
⎥ = 0.975(VP − Von )
⎥ = ⎢(VP − Von )−
2 ⎦ ⎢⎣
2
⎥⎦
3.89
1
PD =
T
2
1 ΔT 2 ⎛
t ⎞
∫ i (t )RS dt = T ∫ I P ⎜⎝1− ΔT ⎟⎠ RS dt
0
0
T
2
D
ΔT
2
⎛
2t
t2 ⎞
I P2 RS ⎛
t2
t3 ⎞
1−
+
dt
=
t
−
+
⎟
⎟
⎜
⎜
∫ ΔT ΔT 2
T ⎝ ΔT 3ΔT 2 ⎠
⎠
0 ⎝
0
2
I R ⎛
ΔT ⎞ 1 2 ⎛ ΔT ⎞
PD = P S ⎜ΔT − ΔT +
⎟ = I P RS ⎜ ⎟
T ⎝
3 ⎠ 3
⎝ T ⎠
I2R
PD = P S
T
ΔT
3.90
Using SPICE with VP = 10 V.
15V
Voltage
10V
5V
0V
-5V
-10V
t
-15V
0s
10ms
20ms
30ms
40ms
50ms
63
3.91
(
)
(a) Vdc = −(VP − Von ) = − 6.3 2 −1 = −7.91V
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V
(e) ΔT =
(b) C =
I T 7.91 1 1
=
= 1.05F
Vr 0.55 0.5 60
(
)
(d) I surge = ωCVP = 2π (60)(1.05) 6.3 2 = 3530 A
2(.25)
2Vr
2T 7.91 2
1
1
=
= 0.628ms | I P = I dc
=
= 841A
ΔT
.5 60 .628ms
ω VP 2π (60) 6.3 2
1
3.92
(
)
VOnom = −(VP − Von )= − 6.3 2 −1 = −7.91V
(
= −(V
) [
]
)= −[6.3(0.9) 2 −1]= −7.02V
VOmax = − VPmax − Von = − 6.3(1.1) 2 −1 = −8.80V
VOmin
min
P
− Von
Circuit3_93b-Transient-8
3.93
*Problem 3.93
VS 1 0 DC 0 AC 0 SIN(0 10 60)
D1 2 1 DIODE
R 2 0 0.25
C 2 0 0.5
.MODEL DIODE D IS=1E-10 RS=0
.OPTIONS RELTOL=1E-6
.TRAN 1US 80MS
.PRINT TRAN V(1) V(2) I(VS)
.PROBE V(1) V(2) I(VS)
.END
+0.000e+000
+10.000m
+20.000m
+30.000m
I P = I dc
64
9 2 1
2T
=
= 923A
ΔT 0.25 60 1.3ms
Time (s)
+50.000m
+10.000
+5.000
+0.000e+000
-5.000
-10.000
V(2)
*REAL(Rectifier)*
SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A
I T 9.00V 1 1
Vdc = −(VP − Von ) = −(10 −1)= −9.00V | Vr =
=
= 1.20V
C 0.25Ω 60s 0.5F
I SC = ωCVP = 2π (60)(0.5)(10)= 1890 A | ΔT =
+40.000m
2(1.2)
1
2Vr
=
= 1.30ms
ω VP 2π (60) 10
1
+60.000m
+70.000m
Circuit3_93b-Transient-11
(Amp)
+0.000e+000
+20.000m
+40.000m
+60.000m
+80.000m
Time (s)
+100.000m
+120.000m
+140.000m
+10.000
+5.000
+0.000e+000
-5.000
-10.000
V(1)
V(2)
*REAL(Rectifier)*
SPICE Graph Results: VDC = -6.55 V, Vr = 0.58 V, IP = 150 A, ISC = 370 A
Note that a significant difference is caused by the diode series resistance.
3.94
(
)
(a) Vdc = −(VP − Von )= − 6.3 2 −1 = −7.91V
(
)
2(.25)
2Vr
2T 7.91 2
1
1
=
= 94.3μs | I P = I dc
=
= 839 A
ΔT
ω VP 2π (400) 6.3 2
.5 400 94.3μs
1
3.95
(
)
(a) Vdc = −(VP − Von )= − 6.3 2 −1 = −7.91V
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V
(e) ΔT =
I T 7.91 1 1
=
= 0.158F
Vr 0.25 0.5 400
(d ) I surge = ωCVP = 2π (400)(0.158) 6.3 2 = 3540 A
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V
(e) ΔT =
(b) C =
1
2Vr
=
ω VP 2π 105
1
( )
I T 7.91 1 1
=
= 633μF
Vr 0.25 0.5 105
( )
(
)
(d ) I surge = ωCVP = 2π 105 (633μF ) 6.3 2 = 3540 A
2(.25)
6.3 2
(b) C =
= 0.377μs | I P = I dc
2T 7.91 2
1
=
= 839 A
5
.5 10 0.377μs
ΔT
3.96
65
(a) C =
1
IT
1
=
= 556 μF
Vr 3000(0.01) 60
3000
(c) Vrms =
= 2120 V
(d ) ΔT =
2
⎞
2T ⎛ 2 ⎞⎛
1
= 1⎜ ⎟⎜
I P = I dc
⎟ = 88.9 A
ΔT ⎝ 60 ⎠⎝ 0.375ms ⎠
(b) PIV ≥ 2VP = 2 ⋅ 3000 = 6000V
1
1
2Vr
=
2(0.01) = 0.375ms
ω VP 2π (60)
(e) I surge = ωCVP = 2π (60)(556μF )(3000)= 629 A
3.97 Assuming Von = 1 V:
V − Von 1
1 ⎛ 1 ⎞⎛ 30 ⎞
3.3 + 1
C= P
T =
= 3.04 V
⎜ ⎟⎜ ⎟ = 6.06 F | PIV = 2VP = 2(3.3 + 1)V = 8.6 V | Vrms =
Vr
R 0.025 ⎝ 60 ⎠⎝ 3.3⎠
2
ΔT =
1
ω
I P = I dc
⎛ 1 ⎞⎛ 3.3V ⎞
1
2T VP − Von
2
=
⎜ s⎟⎜
⎟ = 0.520 ms
RC VP
2π (60) 0.110Ω(6.06F )⎝ 60 ⎠⎝ 4.3V ⎠
⎛ 2 ⎞⎛
⎞
1
2T
= 30⎜ s ⎟⎜
⎟ = 1920 A | I surge = ωCVP = 2π (60 / s)(6.06F )(4.3V ) = 9820 A
ΔT
⎝ 60 ⎠⎝ 0.520ms ⎠
3.98
40V
vO
20V
v1
0V
vS
Time
-20V
0s
5ms
10ms
15ms
20ms
3.99
*Problem 3.99
VS 2 1 DC 0 AC 0 SIN(0 1500 60)
D1 2 3 DIODE
D2 0 2 DIODE
C1 1 0 500U
C2 3 1 500U
RL 3 0 3K
.MODEL DIODE D IS=1E-15 RS=0
.OPTIONS RELTOL=1E-6
.TRAN 0.1MS 100MS
.PRINT TRAN V(2,1) V(3) I(VS)
66
25ms
30ms
VDC = 2(VP - Von) = 2(17 - 1) = 32 V.
.PROBE V(3) V(2,1) I(VS)
.END
4.0kV
3.0kV
vO
2.0kV
1.0kV
vS
0V
-1.0kV
Time
-2.0kV
0s
20ms
40ms
60ms
80ms
100ms
Simulation Results: VDC = 2981 V, Vr = 63 V
The doubler circuit is effectively two half-wave rectifiers connected in series. Each capacitor is
discharged by I = 3000V/3000 = 1 A for 1/60 second. The ripple voltage on each capacitor is
33.3 V. With two capacitors in series, the output ripple should be 66.6 V, which is close to the
simulation result.
3.100
(
)
(a) Vdc = −(VP − Von ) = − 15 2 −1 = −20.2 V (b) C =
(c) PIV ≥ 2VP = 2 ⋅15 2 = 42.4 V
(e) ΔT =
( )
(d ) I surge = ωCVP = 2π (60)(1.35) 15 2 = 10800 A
2(.25)
1
20.2V ⎛ 1 ⎞
1
2Vr
T
=
= 0.407 ms | I P = I dc
=
= 1650 A
⎜ s⎟
ω VP 2π (60) 15 2
ΔT 0.5Ω ⎝ 60 ⎠ 0.407ms
1
3.101
(
)
(a) Vdc = −(VP − Von )= − 9 2 −1 = −11.7 V (b) C =
(c) PIV ≥ 2VP = 2 ⋅ 9 2 = 25.5 V
(e) ΔT =
I ⎛ T ⎞ 20.2V ⎛ 1 ⎞⎛ 1 ⎞
⎜ ⎟=
⎜
⎟⎜
⎟ = 1.35 F
Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝120s ⎠
I ⎛ T ⎞ 11.7V ⎛ 1 ⎞⎛ 1 ⎞
⎜ ⎟=
⎜
⎟⎜
⎟ = 0.780 F
Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝120s ⎠
( )
(d ) I surge = ωCVP = 2π (60)(0.780) 9 2 = 3740 A
⎞
2(.25)
1
1
2Vr
T 11.7V ⎛ 1 ⎞⎛
=
= 0.526 ms | I P = I dc
=
⎜ s⎟⎜
⎟ = 958 A
ω VP 2π (60) 9 2
ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407ms ⎠
1
67
3.102
*Problem 3.102
VS1 1 0 DC 0 AC 0 SIN(0 14.14 400)
VS2 0 2 DC 0 AC 0 SIN(0 14.14 400)
D1 3 1 DIODE
D2 3 2 DIODE
C 3 0 22000U
R303
.MODEL DIODE D IS=1E-10 RS=0
.OPTIONS RELTOL=1E-6
.TRAN 1US 5MS
.PRINT TRAN V(1) V(2) V(3) I(VS1)
.PROBE V(1) V(2) V(3) I(VS1)
.END
20V
10V
vS
0V
-10V
vO
Time
-20V
0s
1.0ms
2.0ms
3.0ms
4.0ms
Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 A
VDC = VP − Von = 10 2 − 0.7 = 13.4 V | Vr =
ΔT =
1
120π
I P = I dc
2Vr
1
=
VP 120π
2(0.254)
14.1
1
13.4 1
= 0.254 V
3 800 22000μF
= 0.504 ms
1
T 13.4V 1
=
s
= 150 A
3Ω 60 0.504 ms
ΔT
Simulation with RS = 0.02 Ω.
Circuit3_102-Transient-15
+0.000e+000
+2.000m
+4.000m
+6.000m
+8.000m
Time (s)
+10.000m
+12.000m
+14.000m
+15.000
+10.000
+5.000
+0.000e+000
-5.000
-10.000
-15.000
V(1)
V(2)
*REAL(Rectifier)*
Simulation Results: VDC = -12.9 V, Vr = 0.20 V, IP = 33.3 A, ISC = 362 A. RS results in a
significant reduction in the values of IP and ISC.
68
5.0ms
3.103
(a) C =
1 ⎛ 1s ⎞⎛ 30 A ⎞
VP − Von 1
T =
⎜
⎟⎜
⎟ = 3.03 F (b) PIV = 2VP = 2(3.3 + 1)V = 8.6 V
R 0.025 ⎝120 ⎠⎝ 3.3V ⎠
Vr
3.3 + 1
(c) Vrms =
2
(e) I P = I dc
= 3.04 V
2(0.025)(3.3)
2Vr
1
=
= 0.520 ms
ω VP 2π (60)
4.3
1
(d) ΔT =
⎞
⎛ 1 ⎞⎛
T
1
= 30 A⎜ s ⎟⎜
⎟ = 962 A | I surge = ωCVP = 2π (60 / s)(3.03F )(4.3V )= 4910 A
ΔT
⎝ 60 ⎠⎝ 0.520ms ⎠
3.104
(a) C =
I T
1
1
=
= 139 μF
Vr 2 3000(0.01) 2 ⋅120
(c) VS =
I P = I dc
3000
2
= 2120 V
(d ) ΔT =
1
2
ω
⎛ 1 ⎞⎛
⎞
1
T
= 1⎜ s ⎟⎜
⎟ = 44.4 A
ΔT ⎝ 60 ⎠⎝ 0.375ms ⎠
(b) PIV ≥ 2VP = 6000 V
Vr
1
=
2(0.01) = 0.375 ms
VP 2π (60)
(e) I surge = ωCVP = 2π (60 / s)(139μF )(3000V )= 157 A
3.105
The circuit is behaving like a half-wave rectifier. The capacitor should charge during the first
1/2 cycle, but it is not. Therefore, diode D1 is not functioning properly. It behaves as an open
circuit.
3.106
(
)
(a) Vdc = −(VP − 2Von )= − 15 2 − 2 = −19.2 V (b) C =
(c) PIV ≥ VP = 15 2 = 21.2 V
( )
(d ) I surge = ωCVP = 2π (60 / s)(1.28F ) 15 2 = 10200 A
⎞
2(.25)
1
2Vr
T 19.2V ⎛ 1s ⎞⎛
1
=
= 0.407 ms | I P = I dc
=
⎜ ⎟⎜
⎟ = 1570 A
ω VP 2π (60) 15 2
ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407ms ⎠
1
(e) ΔT =
3.107
(a) C =
(c) VS =
I P = I dc
I ⎛ T ⎞ 19.2V ⎛ 1 ⎞⎛ 1 ⎞
s⎟ = 1.28 F
⎜ ⎟=
⎜
⎟⎜
Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝ 120 ⎠
⎛ 1 ⎞
I ⎛T ⎞
1A
s⎟ = 278 μF
⎜ ⎟=
⎜
Vr ⎝ 2 ⎠ 3000V (0.01)⎝ 120 ⎠
3000
2
= 2120 V
(d ) ΔT =
(b) PIV ≥ VP = 3000 V
1
1
2Vr
=
2(0.01) = 0.375 ms
ω VP 2π (60)
⎛1 ⎞
1
T
= 1A⎜ s⎟
= 44.4 A
ΔT
⎝ 60 ⎠ 0.375ms
(e) I surge = ωCVP = 2π (60)(278μF )(3000) = 314 A
69
3.108
(a) C =
⎛ 1 ⎞
I ⎛T ⎞
30 A
s⎟ = 3.03 F
⎜ ⎟=
⎜
Vr ⎝ 2 ⎠ (0.025)(3.3V )⎝ 120 ⎠
(c) Vrms =
I P = I dc
5.3
2
= 3.75 V
(d ) ΔT =
⎡ 0.025(3.3)⎤
1
2Vr
⎥ = 0.468 ms
=
2⎢
ω VP 2π (60) ⎢⎣ 5.3 ⎥⎦
1
⎛1 ⎞
T
1
= 30 A⎜ s⎟
= 1070 A
ΔT
⎝ 60 ⎠ 0.468ms
3.109
V1 = VP - Von = 49.3 V
and
(e) I surge = ωCVP = 2π (60 / s)(3.03F )(3.3V )= 3770 A
V2 = -(VP -Von) = -49.3V.
3.110
*Problem 3.110
VS1 1 0 DC 0 AC 0 SIN(0 35 60)
VS2 0 2 DC 0 AC 0 SIN(0 35 60)
D1 1 3 DIODE
D4 2 3 DIODE
D2 4 1 DIODE
D3 4 2 DIODE
C1 3 0 0.1
C2 4 0 0.1
R1 3 0 500
R2 4 0 500
.MODEL DIODE D IS=1E-10 RS=0
.OPTIONS RELTOL=1E-6
.TRAN 10US 50MS
.PRINT TRAN V(3) V(4)
.PROBE V(3) V(4)
.END
3.111
(
)
(a) Vdc = −(VP − 2Von )= − 15 2 − 2 = −19.2 V
(c) PIV ≥ VP = 15 2 = 21.2 V
(e) ΔT =
70
(b) PIV ≥ Vdc + 2Von = (3.3 + 2) = 5.3 V
40V
v1
20V
0V
-20V
v2
Time
-40V
0s
(b) C =
10ms
20ms
30ms
40ms
50ms
I ⎛ T ⎞ 19.2V ⎛ 1 ⎞⎛ 1 ⎞
⎜ ⎟=
⎜
⎟⎜
⎟ = 1.28 F
Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25⎠⎝120 ⎠
(
)
(d ) I surge = ωCVP = 2π (60 / s)(1.28F ) 15V 2 = 10200 A
⎞
2(.25)
1
2Vr
T 19.2V ⎛ 1 ⎞⎛
1
=
= 0.407 ms | I P = I dc
=
⎜ s ⎟⎜
⎟ = 1570 A
ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407ms ⎠
ω VP 2π (60) 15 2
1
3.112
3.3-V, 15-A power supply with Vr ≤ 10 mV. Assume Von = 1 V.
Rectifier Type
Half Wave
Full Wave
Full Wave Bridge
Peak Current
533 A
266 A
266 A
PIV
8.6 V
8. 6 V
5.3 V
Filter
Capacitor
25 F
12.5 F
12.5 F
(i) The large value of C suggests we avoid the half-wave rectifier. This will reduce the cost
and size of the circuit.
(ii) The PIV ratings are all low and do not indicate a preference for one circuit over another.
(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and
also indicate an advantage for these circuits.
(iv) We must choose between use of a center-tapped transformer (full-wave) or two extra
diodes (bridge). At a current of 15 A, the diodes are not expensive and a four-diode bridge
should be easily found. The final choice would be made based upon cost of available
components.
3.113
200-V, 3-A power supply with Vr ≤ 4 V. Assume Von = 1 V.
Rectifier Type
Half Wave
Full Wave
Full Wave Bridge
Peak Current
189 A
94.3 A
94.3 A
PIV
402 V
402 V
202 V
12,500 μF
6250 μF
6250 μF
Filter
Capacitor
(i) The the half-wave rectifier requires a larger value of C which may lead to more cost.
(ii) The PIV ratings are all low enough that they do not indicate a preference for one circuit
over another.
(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and
also indicate an advantage for these circuits.
(iv) We must choose between use of a center-tapped transformer (full-wave) or two extra
diodes (bridge). At a current of 3 A, the diodes are not expensive and a four-diode bridge
should be easily found. The final choice would be made based upon cost of available
components.
71
3.114
3000-V, 1-A power supply with Vr ≤ 120 V. Assume Von = 1 V.
Rectifier Type
Half Wave
Full Wave
Full Wave Bridge
Peak Current
133 A
66.6 A
66.6 A
PIV
6000 V
6000 V
3000 V
Filter
Capacitor
41.7 μF
20.8 μF
20.8 μF
(i) A series string of multiple capacitors will normally be required to achieve the voltage
rating.
(ii) The PIV ratings are high, and the bridge circuit offers an advantage here.
(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers but
neither is prohibitively large.
(iv) We must choose between use of a center-tapped transformer (full wave) or extra diodes
(bridge). With a PIV of 3000 or 6000 volts, multiple diodes may be required to achieve the
require PIV rating.
3.115
5V
5 − VD 5 − 0.6
= 5 mA | I F =
=
= 4.4 mA
1kΩ
1kΩ
1kΩ
⎛
−3 − 0.6
4.4mA ⎞
Ir =
= −3.6 mA | τ S = (7ns) ln⎜1−
⎟ = 5.59 ns
1kΩ
⎝ −3.6mA ⎠
( )
iD 0 + =
3.116
*Problem 3.143 - Diode Switching Delay
V1 1 0 PWL(0 0 0.01N 5 10N 5 10.02N -3
20N -3)
R1 1 2 1K
D1 2 0 DIODE
.TRAN .01NS 20NS
.MODEL DIODE D TT=7NS IS=1E-15
.PROBE V(1) V(2) I(V1)
.OPTIONS RELTOL=1E-6
.OP
.END
10
5
v1
vD
0
-5
Time
-10
0s
Simulation results give
72
S = 4.4 ns.
5ns
10ns
15ns
20ns
3.117
5V
5 − Von 5 − 0.6
= 1 A | IF =
=
= 0.880 A
5Ω
1Ω
5Ω
⎛
−3 − 0.6
0.880 A ⎞
= −0.720 A | τ S = (250ns) ln⎜1−
IR =
⎟ = 200 ns
5Ω
⎝ −0.720 A ⎠
( )
iD 0 + =
3.118
*Problem 3.145(a) - Diode Switching Delay
V1 1 0 DC 1.5 PWL(0 0 .01N 1.5 7.5N 1.5
7.52N -1.5 15N -1.5)
R1 1 2 0.75K
D1 2 0 DIODE
.TRAN .02NS 100NS
.MODEL DIODE D TT=50NS IS=1E-15
CJO=0.5PF
.PROBE V(1) V(2) I(V1)
.OPTIONS RELTOL=1E-6
.OP
.END
For this case, simulation yields
2.0
v1
1.0
vD
0
-1.0
Time
-2.0
0s
5ns
10ns
15ns
20ns
25ns
S = 3 ns.
*Problem 3.145(b) - Diode Switching Delay
V1 1 0 DC 1.5 PWL(0 1.5 7.5N 1.5 7.52N -1.5 15N -1.5)
R1 1 2 0.75K
D1 2 0 DIODE
.TRAN .02NS 100NS
.MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF
.PROBE V(1) V(2) I(V1)
.OPTIONS RELTOL=1E-6
.OP
.END
2.0
v1
1.0
vD
0
-1.0
Time
-2.0
0s
10ns
20ns
30ns
For this case, simulation yields
40ns
S = 15.5 ns.
73
In case (a), the charge in the diode does not have time to reach the steady-state value given by Q
= (1mA)(50ns) = 50 pC. At most, only 1mA(7.5ns) = 7.5 pC can be stored in the diode. Thus is
turns off more rapidly than predicted by the storage time formula. It should turn off in
approximately t = 7.5pC/3mA = 2.5 ns which agrees with the simulation results. In (b), the
diode charge has had time to reach its steady-state value. Eq. (3.103) gives: (50 ns) ln (1-1mA/(3mA)) = 14.4 ns which is close to the simulation result.
3.119
[
]
IC = 1−10−15 exp(40VC )−1 A | For VC = 0, I SC = 1A
VOC =
1 ⎛
1 ⎞
ln⎜1+ −15 ⎟ = 0.864 V
40 ⎝ 10 ⎠
[
[
]]
P = VC IC = VC 1−10−15 exp(40VC )−1
[
]
dP
= 1−10−15 exp(40VC )−1 − 40x10−15 VC exp(40VC ) = 0
dVC
Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts
3.120
(a) For VOC, each of the three diode teminal currents must be zero, and
1
VOC = VC1 + VC 2 + VC 3 = V ln(1.05x1015 )+ ln(1.00x1015 )+ ln(0.95x1015 ) = 2.59 V
40
(b) For ISC, the external currents cannot exceed the smallest of the short circuit current
[
]
of the individual diodes. Thus, ISC = min[1.05A,1.00A,0.95A] = 0.95 A
Note that diode three will be reversed biased in part (b).
Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts
3.121
hc
λ=
E
) = 1.11 μm - far infrared
1.12eV (1.602x10 j / eV )
6.625x10 J − s(3x10 m / s)
= 0.875 μm - near infrared
(b) λ =
1.42eV (1.602x10 j / eV )
(a) λ =
(
6.625x10−34 J − s 3x108 m / s
−19
−34
8
−19
74
CHAPTER 4
4.1
(a) VG > VTN corresponds to the inversion region (b) VG << VTN corresponds to the accumulation
region (c) VG < VTN corresponds to the depletion region
4.2 (a)
(
)
−14
3.9εo 3.9 8.854x10 F / cm
F
nF
C =
=
=
= 6.91x10−8 2 = 69.1 2
−9
Tox
Tox
50x10 m(100cm / m)
cm
cm
εox
"
ox
(b), (c) & (d): Scaling the result from part (a) yields
Cox" = 69.1
nF 50nm
nF
nF 50nm
nF
nF 50nm
nF
= 173 2 | Cox" = 69.1 2
= 346 2 | Cox" = 69.1 2
= 691 2
2
cm 20nm
cm
cm 10nm
cm
cm 20nm
cm
4.3
εs
Cd =
2εs
(0.75V )
qN B
=
(
11.7 8.854x10−14 F / cm
(
2(11.7) 8.854x10−14 F / cm
−19
1.602x10
(10
15
/ cm
4.4 (a)
3
)
)
)(0.75V )
= 10.5x10−9 F / cm 2
(
)
−14
3.9εo ⎛
cm 2 ⎞ 3.9 8.854x10 F / cm
K = μ nC = μ n
= μn
= ⎜ 500
⎟
Tox
Tox
V − sec ⎠ 50x10−9 m(100cm / m)
⎝
'
n
εox
"
ox
F
A
μA
= 34.5 x 10−6 2 = 34.5 2
V − sec
V
V
(b) & (c) Scaling the result from part (a) yields
Kn' = 34.5x10−6
Kn' = 34.5
μA 50nm
2
V 20nm
= 86.3
μA
V
| Kn' = 34.5
2
4.5
(a) Q
(b) Q
"
"
= C (VGS − VTN )=
εox
= C (VGS − VTN )=
εox
"
ox
"
ox
Tox
Tox
(V
GS
(V
GS
− VTN ) =
− VTN )=
μA 50nm
2
V 10nm
= 173
μA
V
2
| Kn' = 34.5
(
3.9 8.854x10−14 F / cm
(
3.9 8.854x10−14 F / cm
5nm
−7
)(2V )= 6.91x10
10x10 m(100cm / m)
−9
V
2
)(1V )= 1.38x10
25x10 m(100cm / m)
−9
μA 50nm
−7
= 345
μA
V2
C
cm 2
C
cm 2
75
4.6
⎛
V ⎞
cm2 ⎞⎛
6 cm
v
=
−
μ
E
=
−
500
a
⎜
⎟⎜2000 ⎟ = −1.00x10
() n n
cm ⎠
s
V − s ⎠⎝
⎝
⎛
V ⎞
cm2 ⎞⎛
cm
(a) vn = −μnE = −⎜400 V − s ⎟⎜⎝4000 cm ⎟⎠ = −1.60x106 s
⎠
⎝
4.7
The carrier velocity must increase as the carriers travel down the channel to compensate for the
decrease in carrier density
4.8
(a) 0 < 0.8V → I D = 0
(b) VGS - VTN = 0.2V , VDS = 0.25V → Saturation region
⎛ 200 μA ⎞⎛ 5μm ⎞
2
K' W ⎛
V ⎞
I D = n ⎜VGS − VTN − DS ⎟VDS = ⎜
⎟(1− 0.8) = 40.0 μA
2 ⎟⎜
2 L⎝
2 ⎠
⎝ 2 V ⎠⎝ 0.5μm ⎠
(c) VGS - VTN = 1.2V , VDS = 0.25V → triode region
⎛
0.25 ⎞
W⎛
μA ⎞⎛ 5μm ⎞⎛
V ⎞
I D = Kn' ⎜VGS − VTN − DS ⎟VDS = ⎜200 2 ⎟⎜
⎟⎜2 − 0.8 −
⎟(0.25) = 538 μA
2 ⎠
2 ⎠
L⎝
V ⎠⎝ 0.5μm ⎠⎝
⎝
(d ) VGS - VTN = 2.2V , VDS = 0.25V → triode region
⎛
0.25 ⎞
W⎛
μA ⎞⎛ 5μm ⎞⎛
V ⎞
I D = Kn' ⎜VGS − VTN − DS ⎟VDS = ⎜200 2 ⎟⎜
⎟⎜3 − 0.8 −
⎟(0.25)= 1.04 mA
2 ⎠
2 ⎠
L⎝
V ⎠⎝ 0.5μm ⎠⎝
⎝
mA
W ⎛
μA ⎞⎛ 5μm ⎞
(e) Kn = Kn' L = ⎜⎝200 V 2 ⎟⎠⎜⎝ 0.5μm ⎟⎠ = 2.00 V 2
4.9
mA
W
μA ⎛ 60μm ⎞
= 200 2 ⎜
⎟ = 4.00 2
L
V
V ⎝ 3μm ⎠
mA
mA
μA ⎛ 3μm ⎞
μA ⎛ 10μm ⎞
(b) Kn = 200 V 2 ⎜⎝ 0.15μm ⎟⎠ = 4.00 V 2 | (c) Kn = 200 V 2 ⎜⎝ 0.25μm ⎟⎠ = 8.00 V 2
(a) K
n
= Kn'
4.10
(a) 0 < 1V → cutoff region, I D = 0
(b) 1V = 1V → cutoff region, I
D
=0
(c) VGS - VTN = 1V , VDS = 0.1V → triode region
⎛
0.1⎞
W⎛
μA ⎞⎛10μm ⎞⎛
V ⎞
I D = Kn' ⎜VGS − VTN − DS ⎟VDS = ⎜250 2 ⎟⎜
⎟⎜2 −1− ⎟(0.1) = +231 μA
2 ⎠
2 ⎠
L⎝
V ⎠⎝ 1μm ⎠⎝
⎝
76
(d) VGS - VTN = 2V , VDS = 0.1V → triode region
⎛
W⎛
μA ⎞⎛10μm ⎞⎛
V ⎞
0.1⎞
I D = Kn' ⎜VGS − VTN − DS ⎟VDS = ⎜250 2 ⎟⎜
⎟⎜3 −1− ⎟(0.1)= +488 μA
2 ⎠
L⎝
2 ⎠
V ⎠⎝ 1μm ⎠⎝
⎝
W ⎛
μA ⎞⎛10μm ⎞
mA
(e) Kn = Kn' L = ⎜⎝250 V 2 ⎟⎠⎜⎝ 1μm ⎟⎠ = 2.50 V 2
4.11
Identify the source, drain, gate and bulk terminals and find the current I in the transistors in Fig.
P-4.3.
W
=
L
+0.2 V
10
1
D
G
+5
(a)
B
V
+
V
S
-
GS
S
B
D
10
1
V
G
+5
W
=
L
I
+
DS
+
VGS
I
-0.2 V
-
DS
+
(b)
(a)
VGS = VG − VS = 5V VDS = VD − VS = 0.2V → Triode region operation
⎛
W⎛
μA ⎞⎛10 ⎞⎛
V ⎞
0.2 ⎞
I = I D = Kn' ⎜VGS − VTN − DS ⎟VDS = ⎜100 2 ⎟⎜ ⎟⎜ 5 − 0.70 −
⎟0.2 = +840 μA
2 ⎠
L⎝
2 ⎠
V ⎠⎝ 1 ⎠⎝
⎝
(b)
VGS = VG − VS = 5 − (−0.2) = 5.2V VDS = VD − VS = 0 − (−0.2)= 0.2V
⎛
0.2 ⎞
μA ⎞⎛ 10 ⎞⎛
I = −I S = −⎜100 2 ⎟⎜ ⎟⎜5.2 − 0.70 −
⎟0.2 = −880 μA
2 ⎠
V ⎠⎝ 1 ⎠⎝
⎝
4.12
⎛
0.5 ⎞
μA ⎞⎛ 10 ⎞⎛
(a) I = ⎜100 2 ⎟⎜ ⎟⎜5 − 0.75 −
⎟0.5 = 2.00 mA
2 ⎠
V ⎠⎝ 1 ⎠⎝
⎝
⎛
0.2 ⎞
μA ⎞⎛10 ⎞⎛
(b) I = ⎜100 2 ⎟⎜ ⎟⎜3 − 0.75 −
⎟0.2 = 430 μA
2 ⎠
V ⎠⎝ 1 ⎠⎝
⎝
4.13
(a) VGS = VG − VS = 5 − (−0.5) = 5.5 V
VDS = VD − VS = 0 − (−0.5)= 0.5 V
⎛
0.5 ⎞
μA ⎞⎛10 ⎞⎛
I = −I S = −⎜100 2 ⎟⎜ ⎟⎜5.5 − 0.75 −
⎟0.5 = −2.25 mA
2 ⎠
V ⎠⎝ 1 ⎠⎝
⎝
(b) V
GS
= VG − VS = 3 − (−1) = 4 V
VDS = VD − VS = 0 − (−1) = 1 V
⎛
1⎞
μA ⎞⎛10 ⎞⎛
I = −I S = −⎜100 2 ⎟⎜ ⎟⎜4 − 0.75 − ⎟1 = −2.75 mA
2⎠
V ⎠⎝ 1 ⎠⎝
⎝
77
4.14
Kn = Kn'
(a) W =
W
L
Kn
4mA /V 2
L
=
(0.5μm)= 20 μm
Kn'
100μA /V 2
(b) W =
800μA /V 2
(0.5μm)= 4 μm
100μA /V 2
4.15
(a) R
on
1
=
(b) R
on
Kn'
=
W
(VGS − VTN )
L
=
1
= 23.0 Ω
⎞
⎛
−6 100
100x10 ⎜
⎟(5 − 0.65)
⎝ 1 ⎠
1
= 35.7 Ω
⎞
⎛
−6 100
100x10 ⎜
⎟(3.3 − 0.50)
⎝ 1 ⎠
4.16
Ron =
1
or
W
1
= '
L Kn (VGS − VTN )Ron
W
(VGS − VTN )
L
W
1
4.71
W
1
7.84
(a)
=
=
| (b)
=
=
−6
−6
L 100x10 (5 − 0.75)(500)
1
L 100x10 (3.3 − 0.75)(500)
1
Kn'
4.17
Ron ≤
0.1V
ID
5
A
= 0.020Ω = 20mΩ | Kn =
=
= 17.0 2
5A
V
(VGS − VTN − 0.5VDS )VDS 5 − 2 − 0.5(0.1) (0.1)
(
Note that this will require
)
W Kn 17.0 170
=
=
=
L Kn'
0.1
1
4.18
Picking two values in saturation :
K
K
2
2
395μA = n (4 − VTN ) and 140μA = n (3 − VTN )
2
2
Taking the ratio of these two equations :
μA
125 2
W
V = 1.25
=
L 100 μA
1
2
V
From the graph, VTN is somewhat less than 2 V. VTN > 0 → enhancement - mode transistor
395 (4 − VTN )
μA
=
2 → VTN = 1.5 V → K n = 125
140 (3 − VTN )
V2
2
78
4.19
Using the parameter values from problem 4.22:
800uA
VGS = 5 V
VGS = 4.5 V
Drain Current (A)
600uA
VGS = 4 V
400uA
VGS = 3.5 V
200uA
VGS = 3 V
VGS = 2.5 V
VGS = 2 V
0
A 0V
1.0V
2.0V
3.0V
4.0V
5.0V
6.0V
Drain-Source Voltage (V)
4.20
(a) For VGS = 0, VGS ≤ VTN and ID = 0 (b) For VGS = 1 V, VGS = VTN and ID = 0
(c ) VGS − VTN
ID =
mA
375 μA ⎛ 5μm ⎞
μA ⎛ 5μm ⎞
2 2
' W
= 375 2 ⎜
⎟ = 3.75 2
⎟(2 −1) V = 1.88 mA | K n = K n
2 ⎜
V
2 V ⎝ 0.5μum ⎠
L
V ⎝ 0.5um ⎠
(d) VGS − VTN
ID =
= 2 -1 =1V and VDS = 3.3 | VDS > (VGS − VTN ) so the saturation region is correct
= 3 -1 = 2V and VDS = 3.3 | VDS > (VGS − VTN ) so the saturation region is correct
375 μA ⎛ 5μm ⎞
2 2
⎟(3 −1) V = 7.50 mA
2 ⎜
2 V ⎝ 0.5μm ⎠
4.21
(a) For VGS = 0, VGS < VTN and ID = 0 (b) For VGS = 1 V, VGS < VTN and ID = 0
(c ) VGS − VTN
ID =
200 μA ⎛10μm ⎞
μA ⎛ 10μm ⎞
mA
2 2
' W
V
=
250
μ
A
|
K
=
K
=
200
2
−1.5
⎜
⎟
⎟ = 2.00 2
(
)
n
n
2
2 ⎜
2 V ⎝ 1um ⎠
L
V ⎝ 1um ⎠
V
(d) VGS − VTN
ID =
= 2 -1.5 = 0.5V and VDS = 4 | VDS > (VGS − VTN ) so the saturation region is correct
= 3 -1.5 =1.5V and VDS = 4 | VDS > (VGS − VTN ) so the saturation region is correct
200 μA ⎛10μm ⎞
2 2
⎟(3 −1.5) V = 2.25 mA
2 ⎜
2 V ⎝ 1μm ⎠
79
4.22
(a)
VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 0.2 V.
VDS < VGS - VTN so the transistor is operating in the triode region.
⎛
V ⎞
0.2 ⎞
W⎛
μA ⎞⎛10 ⎞⎛
⎜VGS − VTN − DS ⎟VDS = ⎜200 2 ⎟⎜ ⎟⎜ 2 − 0.75 −
⎟0.2 = 460 μA
⎝
2 ⎠
2 ⎠
L⎝
V ⎠⎝ 1 ⎠⎝
(b)
VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 2.5 V.
VDS > VGS - VTN so the transistor is operating in the saturation region.
⎛ 200 μA ⎞⎛10 ⎞
K' W
2
2
ID = n
⎟(2 − 0.75) = 1.56 mA
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
(c)
VGS < VTN so the transistor is cutoff with ID = 0.
⎛ 300 ⎞
ID ∝ K n' so (a) ID = ⎜
(d)
⎟460μA = 690μA (b) ID = 2.34 mA (c) ID = 0
⎝ 200 ⎠
ID = K n'
4.23
(a) VGS - VTN = 4 V, VDS = 6 V. VDS > VGS - VTN --> Saturation region
(b) VGS < VTN --> Cutoff region
(c) VGS - VTN = 1 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region
(d) VGS - VTN = 0.5 V, VDS = 0.5 V. VDS = VGS - VTN --> Boundary between triode and saturation
regions
(e) The source and drain of the transistor are now reversed because of the sign change in VDS.
Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b),
VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 - 1 = 1.5 V, and VDS = 0.5 V --> triode region
(f) The source and drain of the transistor are again reversed because of the sign change in VDS.
Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b),
VGS = 3 - (-6) = 9 V, VGS - VTN = 9 - 1 = 8.0 V, and VDS = 6 V --> triode region
D --> 'S'
-
D --> 'S'
G
G
6.0 V
0.5 V
+
+
2V
(e)
+
S --> 'D'
VG'S' = +2.5 V
VD'S' = +0.5 V
80
3V
(f)
+
S --> 'D'
V G'S' = +9.0 V
V D'S' = +6.0 V
4.24
(a) VGS - VTN = 2.6 V, VDS = 3.3 V. VDS > VGS - VTN --> Saturation region
(b) VGS < VTN --> Cutoff region
(c) VGS - VTN = 1.3 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region
(d) VGS - VTN = 0.8 V, VDS = 0.5 V. VDS < VGS - VTN --> triode region
(e) The source and drain of the transistor are now reversed because of the sign change in VDS.
Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b),
VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 – 0.7 = 1.8 V, and VDS = 0.5 V --> triode region
(f) The source and drain of the transistor are again reversed because of the sign change in VDS.
Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b),
VGS = 3 - (-3) = 6 V, VGS - VTN = 6 – 0.7 = 5.3 V, and VDS = 3 V --> triode region
4.25
R
R
2
4
V
D
DD
G
B
+
-
S
R
R
1
4.26
3
+V
DD
D
G
I
B
S
+V
DD
D
D
G
B
G
I
B
S
S
D
D
B
B
G
G
S
(a)
S
(b)
4.27
81
VDS = 3.3V, VGS – VTN = 1.3 V; VDS > VGS - VTN so the transistor is saturated.
(a) gm = K n (VGS − VTN ) = 250
μA ⎛ 20μm ⎞
⎜
⎟(2 − 0.7) = 6.50 mS
V 2 ⎝ 1μm ⎠
(b) gm = K n (VGS − VTN ) = 250
μA ⎛ 20μm ⎞
⎜
⎟(3.3 − 0.7) = 13.0 mS
V 2 ⎝ 1μm ⎠
4.28
(a) gm =
ΔiD
760 −140 μA
=
= 310 μS | As a check, we can use the results from Problem 4.22.
ΔvGS
5−3 V
gm = K n (VGS − VTN ) = 125
(b) gm =
ΔiD
ΔvGS
μA
(4 −1.5)V = 313 μS
V2
μA
390 −15 μA
=
= 188 μS | Checking : gm = 125 2 (3 −1.5)V = 188 μS
V
4−2 V
4.29
VDS > VGS - VTN so the transistor is saturated.
Kn
250 μA
2
2
5 − 0.75) (1+ 0.025(6)) = 2.60 mA
(VGS − VTN ) (1+ λVDS ) =
2 (
2
2 V
K
250 μA
2
2
(b) ID = n (VGS − VTN ) =
5 − 0.75) = 2.26 mA
2 (
2
2 V
(a) ID =
4.30
VDS > VGS - VTN so the transistor is saturated.
Kn
500 μA
2
2
4 −1) (1+ 0.02(5)) = 2.48 mA
(VGS − VTN ) (1+ λVDS ) =
2 (
2
2 V
K
500 μA
2
2
(b) ID = n (VGS − VTN ) =
4 −1) = 2.25 mA
2 (
2
2 V
(a) ID =
4.31
(a) The transistor is saturated by connection.
ID =
12V − VGS 100x10−6 ⎛10 ⎞ A
2
=
⎜ ⎟ 2 (VGS − 0.75V )
5
⎝ 1 ⎠V
10 Ω
2
12.5VGS2 −17.8VGS − 4.97 = 0
VGS = 0.266V , 1.214V ⇒ VGS = 1.214 V since it must exceed 0.75V
ID =
12 −1.214
= 108 μA
10 5
(b) ID =
82
Checking :
100x10−6 ⎛10 ⎞ A
2
⎜ ⎟ 2 (1.214 − 0.75V ) = 108 μA
⎝ 1 ⎠V
2
12V − VGS 1000x10−6 A
2
=
V − 0.75V ) (1+ 0.025VGS )
5
2 ( GS
10 Ω
2
V
Starting with the solution from part (a) and solving iteratively yields VGS = 1.20772 V and ID =
108 μA, essentially no change.
(c)
12V − VGS 100x10−6 ⎛ 25 ⎞ A
2
ID =
=
⎜ ⎟ 2 (VGS − 0.75V )
5
⎝ 1 ⎠V
10 Ω
2
62.5VGS2 − 91.75VGS + 11.16 = 0
VGS = 0.446V , 1.046V ⇒ VGS = 1.046 V since VGS must exceed the threshold voltage.
12 −1.046
ID =
= 110 μA
10 5
100x10−6 ⎛ 25 ⎞ A
2
Checking : ID =
⎜ ⎟ 2 (1.046 − 0.75V ) = 110 μA
⎝ 1 ⎠V
2
4.32
(a) The transistor is saturated by connection.
12V − VGS 100x10−6 ⎛10 ⎞ A
2
ID =
=
⎜ ⎟ 2 (VGS − 0.75V )
4
⎝ 1 ⎠V
5x10 Ω
2
31.25VGS2 − 45.88VGS + 5.58 = 0
VGS = 0.0588V , 1.401V ⇒ VGS = 1.401 V since VGS must exceed the threshold voltage.
12 −1.401
100x10−6 ⎛10 ⎞ A
2
=
212
μ
A
Checking
:
I
=
⎜ ⎟ 2 (1.401− 0.75V ) = 212 μA
D
4
⎝ 1 ⎠V
5x10
2
−6
12V − VGS 1000x10 A
2
=
V − 0.75V ) (1+ 0.02VGS )
(b) ID =
4
2 ( GS
5x10 Ω
2
V
ID =
Starting with the solution from part (a) and solving iteratively yields VGS = 1.3925 V and IDS =
212 μA, essentially no change
4.33
(a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated.
K n' W
K n' W
2
2
I
=
V
−
V
and
I
=
( GS1 TN )
(VGS 2 − VTN ) .
Therefore
D1
D2
2 L
2 L
From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET:
K n' W
K n' W
2
2
I = ID1 = ID 2 or
(VGS1 − VTN ) =
(VGS 2 − VTN )
2 L
2 L
which requires VGS1 = VGS2. Using KVL:
VDD = VDS1 + VDS 2 = VGS1 + VGS 2 = 2VGS 2
V
VGS1 = VGS 2 = DD = 5V
2
'
K W
100 μA 10
2
2
I= n
(VGS1 − VTN ) =
(5 − 0.75) V 2 = 9.03 mA
2
2 V 1
2 L
(b) The current simply scales by a factor of two (see last equation above), and ID = 18.1 mA.
(c) For this case,
83
K n' W
K n' W
2
2
V
−
V
1
+
λ
V
and
I
=
( GS1 TN ) (
(VGS 2 − VTN ) (1 + λVDS2 ).
DS1 )
D2
2 L
2 L
Since VGS = VDS for both transistors
K n' W
K n' W
2
2
ID1 =
(VGS1 − VTN ) (1 + λVGS1 ) and ID 2 =
(VGS 2 − VTN ) (1 + λVGS2 )
2 L
2 L
and ID1 = ID2 = I
K' W
K n' W
2
2
(VGS1 − VTN ) (1 + λVGS1 ) = n (VGS 2 − VTN ) (1 + λVGS2 )
2 L
2 L
which again requires VGS1 = VGS2 = VDD/2 = 5V.
K' W
100 μA 10
2
2
I= n
(VGS1 − VTN ) (1+ λVDS ) =
(5 − 0.75) V 2 (1+ (.04 )5) = 10.8 mA
2
2 L
2 V 1
ID1 =
4.34
(a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (“by
connection”).
K n' ⎛ W ⎞
K n' ⎛ W ⎞
2
2
ID1 =
⎜ ⎟ (VGS1 − VTN ) and ID 2 =
⎜ ⎟ (VGS 2 − VTN ) .
Therefore
2 ⎝ L ⎠1
2 ⎝ L ⎠2
From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET:
K n' ⎛ 10 ⎞
K n' ⎛ 40 ⎞
2
2
V
I = ID1 = ID 2 or
−
V
=
⎜ ⎟( GS1 TN )
⎜ ⎟(VGS 2 − VTN )
2 ⎝1⎠
2 ⎝1⎠
which requires VGS1 = 2VGS2 - VTN. Using KVL:
VDD = VDS1 + VDS 2 = VGS 2 + VGS1 = 3VGS 2 − VTN
V + VTN 10 + 0.75
VGS 2 = DD
=
= 3.583V VGS1 = 6.417
3
3
100 μA 10
K' W
2
2
I= n
(VGS1 − VTN ) =
(6.417 − 0.75) V 2 = 16.1 mA
2
2 V 1
2 L
100 μA 40
2
Checking : I =
(3.583 − 0.75) V 2 = 16.1 mA which agrees.
2
2 V 1
(b) For this case with VGS = VDS for both transistors and ID1 = ID2,
K' ⎛W ⎞
K' ⎛W ⎞
2
2
ID1 = n ⎜ ⎟ (VGS1 − VTN ) (1 + λVGS1 ) and ID 2 = n ⎜ ⎟ (VGS 2 − VTN ) (1 + λVGS2 )
2 ⎝ L ⎠1
2 ⎝ L ⎠2
where VGS2 = VDD – VGS1. Therefore,
K ' ⎛ 40 ⎞
K n' ⎛10 ⎞
2
2
⎜ ⎟(VGS1 − VTN ) (1 + λVGS1 ) = n ⎜ ⎟(10 − VGS1 − VTN ) (1 + λ(10 − VGS1 ))
2 ⎝1⎠
2 ⎝1⎠
VGS1 = 6.3163, VGS2 = 3.6837, ID1 = 20.4 mA, Checking: ID2 = 20.4 mA which agrees.
84
4.35
(a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (“by
connection”).
K n' ⎛ W ⎞
K n' ⎛ W ⎞
2
2
V
−
V
and
I
=
⎜
⎟
⎜ ⎟ (VGS 2 − VTN ) .
(
)
GS1
TN
D2
Therefore
2 ⎝ L ⎠1
2 ⎝ L ⎠2
From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET:
K n' ⎛ 25 ⎞
K n' ⎛12.5 ⎞
2
2
I = ID1 = ID 2 or
⎜ ⎟(VGS1 − VTN ) =
⎜
⎟(VGS 2 − VTN )
2 ⎝1⎠
2 ⎝ 1 ⎠
Solving for VGS2 yields: VGS 2 = 2VGS1 − 2 −1 VTN
ID1 =
(
)
Also, VDD = VDS1 + VDS 2 or VGS1 = 10 − VGS 2
VGS1 =
10 +
( 2 −1)V
TN
1+ 2
= 4.271V VGS 2 = 5.729V
100 μA ⎛ 25 ⎞
K n' W
2
2 2
⎟(4.271− 0.75) V = 15.5 mA
(VGS1 − VTN ) =
2 ⎜
2 V ⎝1⎠
2 L
100 μA ⎛ 12.5 ⎞
2 2
Checking : I =
⎟(5.729 − 0.75) V = 15.5 mA - agrees.
2 ⎜
2 V ⎝ 1 ⎠
(b) For this case with VGS = VDS for both transistors and ID1 = ID2,
K' ⎛W ⎞
K' ⎛W ⎞
2
2
ID1 = n ⎜ ⎟ (VGS1 − VTN ) (1 + λVGS1 ) and ID 2 = n ⎜ ⎟ (VGS 2 − VTN ) (1 + λVGS2 )
2 ⎝ L ⎠1
2 ⎝ L ⎠2
I=
where VGS2 = VDD – VGS1. Therefore,
K n' ⎛ 40 ⎞
K n' ⎛10 ⎞
2
2
⎜ ⎟(VGS1 − VTN ) (1 + λVGS1 ) =
⎜ ⎟(10 − VGS1 − VTN ) (1 + λ(10 − VGS1 ))
2 ⎝1⎠
2 ⎝1⎠
VGS1 = 4.3265 V, VGS2 = 5.6735 V, ID1 = 19.4 mA, Checking: ID2 = 19.4 mA – both agree
4.36
VGS - VTN = 5 - (-2) = 7 V > VDS = 6 V so the transistor is operating in the triode region.
6⎞
−6 ⎛
(a) ID = 250x10 ⎜5 − (−2) − ⎟6 = 6.00 mA
⎝
2⎠
(b) Our triode region model is independent of λ, so ID = 6.00 mA.
4.37
Since VDS = VGS, and VTN < 0 for an NMOS depletion mode device, VGS - VTN will be greater
than VDS and the transistor will be operating in the triode region.
85
4.38
(a) VDS = 6V | VGS − VTN = 0 − (−3) = 3V so the transistor is saturated
2
Kn
250 μA
2
0 − (−3V )] = 1.13 mA
(VGS1 − VTN ) =
2 [
2 V
2
2
250 μA
0 − (−3V )] (1+ 0.025(6)) = 1.29 mA
(b) ID =
2 [
2 V
ID =
4.39
D
G
+10 V
-10 V
100 k Ω
100 k Ω
I DS
S
G
W = 10
L
1
S
(a)
I DS
W = 10
L
1
D
(b)
(a) If the transistor were saturated, then
100x10−6 ⎛ 10 ⎞
2
ID =
⎜ ⎟(−2) = 2.00 mA
⎝1⎠
2
but this would require a power supply of greater than 200 V (2 mA x 100 kΩ). Thus the
transistor must be operating in the triode region.
10V − VDS
VDS ⎞
−3 ⎛
=
10
0
−
−2
−
⎜
⎟VDS
(
)
⎝
10 5 Ω
2 ⎠
10 − VDS = 50VDS (4 − VDS ) and VDS = 0.0504V using the quadratic equation.
⎛ 0.0504 ⎞
10V − VDS
ID = 10−3 ⎜2 −
= 99.5 μA
⎟0.0504 = 99.5 μA Checking :
⎝
2 ⎠
10 5 Ω
(b) For R = 50 kΩ and W/L = 20/1,
⎛
10V − VDS
V ⎞
= 2x10−3⎜ 0 − (−2) − DS ⎟VDS
4
⎝
5x10 Ω
2 ⎠
10 − VDS = 50VDS (4 − VDS ), the same as part (a).
⎛ 0.0504 ⎞
10V − VDS
ID = 2x10−3 ⎜2 −
= 199 μA
⎟0.0504 = 199 μA Checking :
⎝
5x10 4 Ω
2 ⎠
(c) In this circuit, the drain and source terminals of the transistor are reversed because of the
power supply voltage, and the current direction is also reversed. However, now VDS = VGS and
since the transistor is a depletion-mode device, it is still operating in the triode region.
86
⎛
10V − VDS
V ⎞
= 1000x10−6 ⎜VDS − (−2) − DS ⎟VDS
5
⎝
10 Ω
2 ⎠
10 − VDS = 50VDS (4 + VDS ) and VDS = 0.04915V using the quadratic equation.
⎛
0.04915 ⎞
10V − VDS
ID = 10−3 ⎜0.04915 − (−2) −
= 99.5 μA
⎟0.04915 = 99.5 μA Checking :
⎝
10 5 Ω
2 ⎠
(d) In this circuit, the drain and source terminals of the transistor are reversed because of the
power supply voltage, and the current direction is also reversed. However, now VDS = VGS and
since the transistor is a depletion-mode device, it is still operating in the triode region.
⎛
10V − VDS
V ⎞
= 2000x10−6⎜VDS − (−2) − DS ⎟VDS
4
⎝
5x10 Ω
2 ⎠
10 − VDS = 50VDS (4 + VDS ) Same as part (c). VDS = 0.04915V using the quadratic equation.
⎛
0.04915 ⎞
10V − VDS
ID = 10−3 ⎜0.04915 − (−2) −
= 99.5 μA
⎟0.04915 = 99.5 μA Checking :
⎝
10 5 Ω
2 ⎠
4.40 See figures in previous problem but use W/L = 20/1.
25x10−6 ⎛ 20 ⎞
2
=
I
⎜ ⎟(−1) = 250 μA but this would require
(a) If the transistor were saturated, then D
2 ⎝1⎠
a power supply of greater than 25 V. Thus the transistor must be operating in the triode region.
⎛ 20 ⎞⎛
10V − VDS
V ⎞
= 100x10−6 ⎜ ⎟⎜0 − (−1) − DS ⎟VDS
5
⎝ 1 ⎠⎝
10 Ω
2 ⎠
10 − VDS = 100VDS (2 − VDS ) and VDS = 0.05105V using the quadratic equation.
⎛ 0.05105 ⎞
10 − 0.0510
ID = 2.00x10−3 ⎜1−
V = 99.5 μA
⎟0.05105 = 99.5 μA Checking :
⎝
2 ⎠
10 5 Ω
(b) In this circuit, the drain and source terminals of the transistor are reversed because of the
power supply voltage, and the current direction is also reversed. However, now VDS = VGS and
since the transistor is a depletion-mode device, it is still operating in the triode region.
⎛ 20 ⎞⎛
V ⎞
VDS = 10 − (10 5 )(100x10−6 )⎜ ⎟⎜VDS − (−1) − DS ⎟VDS
⎝ 1 ⎠⎝
2 ⎠
⎛ V ⎞
VDS = 10 − 200VDS ⎜1+ DS ⎟ and VDS = 0.04858V using the quadratic equation.
⎝
2 ⎠
⎛ 0.04858 ⎞
10 - 0.04858 V
= 99.5 μA
ID = 2000x10−6 ⎜1+
⎟0.04858 = 99.5 μA Checking :
⎝
Ω
2 ⎠
10 5
87
4.41
(a) VTN = 0.75 + 0.75 1.5 + 0.6 − 0.6 = 1.26V
(
)
VGS − VTN = 2 −1.26 = 0.74V > VDS = 0.2V ⇒ Triode region
⎛10 ⎞⎛
0.2 ⎞
ID = 200x10−6 ⎜ ⎟⎜ 2 −1.26 −
⎟0.2 = 256 μA (compared to 460 μA)
⎝ 1 ⎠⎝
2 ⎠
(b) VGS − VTN = 2 −1.26 = 0.74V < VDS = 2.5V ⇒ Saturation region
200x10−6 ⎛ 10 ⎞
2
ID =
⎜ ⎟(2 −1.26) = 548 μA (compared to 1.56 mA)
⎝1⎠
2
(c) VGS < VTN so the transistor is cut off, and ID = 0.
⎛ 300 ⎞
(d) ID ∝ K n' so (a) ID = ⎜
⎟256μA = 384 μA (b) ID = 822 μA (c) ID = 0
⎝ 200 ⎠
4.42
(a) VTN = 1.5 + 0.5
( 4 + 0.75 −
)
0.75 = 2.16V | VGS < VTN ⇒ Cutoff & ID = 0
(b) ID = 0. The result is independent of VDS .
4.43
(a) VTN = 1+ 0.7
( 3 + 0.6 −
)
0.6 = 1.79V
VGS − VTN = 2.5 −1.79 = 0.71V < VDS = 5V ⇒ Saturation region
100x10−6 ⎛ 8 ⎞
2
⎜ ⎟(0.71) = 202 μA
⎝ 1⎠
2
(b) 0.5 < 0.71 ⇒ Triode region
⎛ 8 ⎞⎛
0.5 ⎞
ID = 100x10−6 ⎜ ⎟⎜ 0.71−
⎟0.5 = 184 μA
⎝ 1 ⎠⎝
2 ⎠
ID =
4.44
0.85 = −1.5 + 1.5 VSB + 0.75 − 0.75
(
)
| Solving for VSB yields VSB = 5.17 V
( 5.17 + 0.75 −
Checking : VTN = −1.5 + 1.5
)
0.75 = 0.85 V
4.45
Using trial and error with a spreadsheet yielded:
VTO = 0.74V γ = 0.84 V 2φ F = 0.87V RMS Error = 51.9 mV
88
4.46
−14
3.9εo ⎛
cm 2 ⎞ 3.9(8.854 x10 F /cm)
(a) K = μ p C = μ p
= μp
= ⎜200
⎟
Tox
Tox
V − sec ⎠ 50x10−9 m(100cm /m)
⎝
F
μA
K 'p = 13.8x10−6
= 13.8 2
V − sec
V
μA 50nm
μA
(b) Scaling the result from Part (a) yields: K n' = 13.8 2
= 34.5 2
V 20nm
V
μA 50nm
μA
= 69.0 2
(c) K n' = 13.8 2
V 10nm
V
μA 50nm
μA
= 138 2
(d) K n' = 13.8 2
V 5nm
V
'
p
εox
"
ox
4.47
The pinchoff points and threshold voltage can be estimated directly from the graph: e. g. VGS =
-3 V curve gives VTP = 2.5 - 3 = - 0.5 V or from the VGS = -5 V curve gives VTP = 4.5 - 5 = 0.5 V.
Alternately, choosing two points in saturation, say ID = 1.25 mA for VGS = -3 V and ID = 4.05
mA for VGS = -5 V:
ID1 (VGS1 − VTP )
=
or
ID 2 (VGS 2 − VTP )
1.25 (−3 − VTP )
=
4.05 (−5 − VTP )
Solving for VTP yields : 0.8VTP = −0.4V and VTP = −0.500V .
Solving for K p : K p =
2ID
(VGS − VTP )
2
=
2(1.25mA)
(−3 + 0.5)
2
= 0.400
400
μA
mA
W
V 2 = 10
|
=
=
μA
V2
L K 'p
1
40 2
V
Kp
4.48 Using the values from the previous problem
PMOS Output Characteristics
0.0045
0.004
0.0035
-2 V
-3 V
-3.5 V
-4 V
-4.5 V
-5 V
0.003
0.0025
0.002
0.0015
0.001
0.0005
0
-6
-5
-4
-3
-2
-1
0
VDS
(IDSAT, VDSAT): (0.45 mA,-1.5 V) (1.25 mA,-2.5 V) (1.8 mA,-3 V) (2.45 mA,-3.5 V) (3.7 mA,-4V)
(4.05 mA,-4 V)
89
4.49
(a) VGS − VTP = −1.1+ 0.75 = −0.35V | VDS = −0.2V → Triode region
(−0.2)⎤⎥ −0.2 = 40.0 μA
40μA ⎛ 20 ⎞⎡
⎢
ID =
−1.1−
−0.75
−
⎜
⎟
(
)
( )
2 ⎥⎦
V 2 ⎝ 1 ⎠⎢⎣
(b) VGS − VTP = −1.3 + 0.75 = −0.55V | VDS = −0.2V → Triode region
(−0.2)⎤⎥ −0.2 = 72.0 μA
40μA ⎛ 20 ⎞⎡
⎢
ID =
−1.3
−
−0.75
−
⎜ ⎟
( ) 2 ⎥( )
V 2 ⎝ 1 ⎠⎢⎣
⎦
[
)]
(
(c) VTP = − 0.75 + .5 1+ .6 − −6 = −0.995V
VGS − VTP = −1.1− (−0.995) = −0.105V | VDS = −0.2V → saturation region
1 ⎛ 40μA ⎞⎛ 20 ⎞
2
ID = ⎜ 2 ⎟⎜ ⎟(−1.1+ 0.995) = 4.41 μA
2 ⎝ V ⎠⎝ 1 ⎠
(d) VGS − VTP = −1.3 + 0.995 = −0.305V | VDS = −0.2V → triode region
10μA ⎛ 10 ⎞⎡
(−0.2)⎤ −0.2 = 32.8 μA
ID =
⎟⎢−1.3 − (−0.995) −
)
⎥(
2 ⎜
2 ⎦
V ⎝ 1 ⎠⎣
4.50
For PMOS : Ron =
1
or
W
1
= '
L K p VGS − VTN Ron
or
W
1
= '
L K p VGS − VTN Ron
W
VGS − VTP
L
W
1
5810
W
1
2330
(a)
=
=
| (b)
=
=
−6
−6
L 40x10 −5 + 0.70 (1)
1
L 100x10 (5 − 0.70)(1)
1
K 'p
4.51
For PMOS : Ron =
1
W
VGS − VTP
L
W
1
2.91
W
1
1.16
(a)
=
=
| (b)
=
=
−6
−6
L 40x10 −5 + 0.70 (2000)
1
L 100x10 (5 − 0.70)(2000)
1
90
K 'p
4.52
For PMOS : Ron =
1
K 'p
W
VGS − VTP
L
1
= 29.4 Ω
−6 ⎛ 200 ⎞
40x10 ⎜
⎟ −5 − (−0.75)
⎝ 1 ⎠
W
1
499
(c)
=
=
−6
L 40x10 −5 − (−0.75)(11.8)
1
(a) Ron =
1
= 11.8 Ω
⎛
−6 200 ⎞
100x10 ⎜
⎟(5 − 0.75)
⎝ 1 ⎠
⎛W ⎞
K' ⎛W ⎞
500
Checking : ⎜ ⎟ = n' ⎜ ⎟ = 2.5(200) =
⎝ L ⎠ p K p ⎝ L ⎠n
1
(b) Ron =
4.53
+18 V
R
R4
2
R
S
VDD
B
G
+
-
S
B
D
R
1
G
R
D
3
(b)
(a)
4.54
(a) For VIN = 0, the NMOS device is on with VGS = 5 and VSB = 0, and the PMOS transistor is
off with VGS = 0, VO = 0, and VSB = 0.
1
Ron =
= 235 Ω
−6
(100x10 )(10)(5 − 0.75)
(b) For VIN = 5V, the NMOS device is off with VGS = 0, and the PMOS transistor is on with
VGS = -5V, VO = 5V, and VSB = 0.
1
Ron =
= 235 Ω
−6
(40x10 )(25)(−5 + 0.75)
4.55
Ron ≤
0.1V
0.5A
ID
A
= 0.2Ω K p =
=
= 0.629 2
0.5A
V
(VGS − VTP − 0.5VDS )VDS [−10V − (−2V ) − 0.5(−0.1V )](−0.1V )
4.56
VTP = −0.75 − 0.5
( 4 + 0.6 −
)
0.6 = −1.44V
VGS - VTP = −1.5 − (−1.44 ) = −0.065 | VDS = −4V ⇒ Saturation region
ID =
2
40x10−6 A ⎛ 25 ⎞
⎟[−1.5 − (−1.44 )] = 1.80 μA
2⎜
V ⎝1⎠
2
91
4.57
VGS − VTP = −1.5 − (−0.75) = −0.75V | VDS = −0.5V
VGS - VTN < VDS ⇒ Triode region | ID = 40x10−6
A ⎛ 40 ⎞⎡
(−0.5)⎤ −0.5 = 400 μA
⎟⎢−1.5 − (−0.75) −
)
⎥(
2⎜
V ⎝ 1 ⎠⎣
2 ⎦
4.58
The PMOS transistor could be either an enhancement-mode or a depletion-mode device
depending upon the specific values of R1, R2 and R4. Thus an enhancement device with VTP < 0
is correct and the symbol is correct.
4.59
If this PMOS transistor is conducting, then its threshold voltage must be greater than zero and it
is a depletion-mode device. The symbol is that of an enhancement-mode device and is incorrect.
4.60
+18 V
R4
R2
R
S
G
VDD
+
D
R1
G
R
D
3
(a)
(b)
4.61
RD
R2
75 k Ω
1.5 M Ω
D
VDD
+
G
S
R1
-3 V
R
S
1 MΩ
92
39 k Ω
S
10 V
4.62
R
D
75 k Ω
R
EQ
600 k Ω
V EQ
VDD
D
G
S
+
-5 V
10 V
RS
4V
39 k Ω
93
4.63
n+
22 λ
Metal
Polysilicon
12 λ
[(2x20)/(12x22)]= 0.152 or 15.2%
4.64
n+
14 λ
Metal
Polysilicon
18 λ
94
[2x10/(18x14)]= 0.079 or 7.9%
4.65
Metal
12 λ
n+
Polysilicon
12 λ
(2x10/122)= 0.139 or 13.9%
4.66
Metal
12 λ
n+
Polysilicon
14 λ
[2x10/(14x12)] = 0.119 or 11.9%
95
4.67
(a)
"
Cox
=
εox
Tox
=
⎛
⎝
(3.9)⎜8.854 x10−14
−6
5x10 cm
F⎞
⎟
cm ⎠
= 6.906x10−8
F
cm 2
⎛
F ⎞
"
CGC = Cox
WL = ⎜6.906x10−8 2 ⎟(20x10−4 cm)(2x10−4 cm)= 27.6 fF
⎝
cm ⎠
F
"
(b) Cox
= 1.73 x 10−7 2 | CGC = 69.1 fF
cm
F
"
= 3.45 x 10−7 2 | CGC = 138 fF
(c) Cox
cm
F
"
= 7.90 x 10−7 2 | CGC = 276 fF
(d) Cox
cm
4.68
"
Cox
=
εox
Tox
=
⎛
⎝
(3.9)⎜8.854 x10−14
F⎞
⎟
cm ⎠
−6
1x10 cm
= 3.46x10−7
F
cm 2
⎛
F ⎞
"
CGC = Cox
WL = ⎜ 3.46x10−7 2 ⎟(5x10−4 cm)(5x10−5 cm)= 8.64 fF
⎝
cm ⎠
4.69
'
COL
⎛
⎝
F⎞
⎟
pF
ε
cm ⎠
= ox L =
0.5x10−4 cm)= 17.3
(
⎛
cm ⎞
cm
Tox
10x10−9 m⎜10 2
⎟
⎝
m⎠
(3.9)⎜8.854 x10−14
4.70
⎛
⎞
−15 F
10μm)(1μm)
⎜1.4 x10
2 ⎟(
⎛
μm ⎠
C WL
F ⎞
⎝
'
+ COLW =
+ ⎜ 4 x10−15
(a) CGS = CGD =
⎟(10μm) = 47 fF
2
2
μm ⎠
⎝
2 "
2
'
WL + COL
W = 14 fF + 40 fF = 49 fF
(b) CGS = COX
3
3
⎛
F ⎞
'
CGD = COL
W = ⎜ 4 x10−15
⎟(10μm) = 40 fF
μm ⎠
⎝
⎛
F ⎞
'
W = ⎜ 4 x10−15
(c ) CGS = CGD = COL
⎟(10μm ) = 40 fF
μm ⎠
⎝
"
OX
96
4.71
⎛
⎝
F⎞
⎟
εox
F
cm ⎠
"
Cox =
=
= 3.453x10−8 2
⎛
⎞
cm
Tox
cm
100x10−9 m⎜10 2
⎟
⎝
m⎠
(3.9)⎜8.854 x10−14
CGC
⎛ −4 cm ⎞ 2
⎛
−8 F ⎞
6
2
= C WL = ⎜ 3.453x10
⎟(50x10 μm )⎜10
⎟ = 17.3 nF
⎝
μm ⎠
cm 2 ⎠
⎝
"
ox
4.72
"
L = 2Λ = 1μm | W =10L = 5μm | Cox
=
εox
Tox
=
3.9(8.854 x10−14 F /cm)
−7
150x10 cm
= 0.23 fF / μm 2
Triode region :
"
0.23 fF / μm 2 )(5μm 2 )
(
Cox
WL
+ CGSOW =
+ (0.02 fF / μm)(5μm) = 0.675 fF
CGS = CGD =
2
2
2 "
WL + CGSOW = 0.867 fF | CGS = CGSOW = 0.10 fF
Saturation region : CGS = Cox
3
Cutoff : CGS = CGD = CGSOW = 0.10 fF
4.73
⎛
F⎞
⎟
cm ⎠
εox
F
"
= 3.453x10−8 2
(a) Cox = T =
⎛ cm ⎞
cm
ox
100x10−9 m⎜102
⎟
m⎠
⎝
⎛
F ⎞
CGC = Cox" WL = ⎜3.453x10−8 2 ⎟ 10x10−4 cm 1x10−4 cm = 3.45 fF
cm ⎠
⎝
⎛
F ⎞
(b) CGC = Cox" WL = ⎜⎝3.453x10−8 cm2 ⎟⎠ 100x10−4 cm 1x10−4 cm = 34.5 fF
(3.9)⎜⎝8.854x10
−14
(
)(
(
)
)(
)
97
4.74
CSB = C j AS + C jsw PS | CDB = C j AD + C jsw PD | AS = 50Λ2 = 12.5μm 2 | PS = 30Λ = 15μm
Cj =
εs
w do
⎛ NA ND ⎞
⎛10 201016 ⎞
2εs ⎛ 1
1 ⎞
=
0.025ln
+
φ
|
φ
=
V
ln
⎜
⎟ j
⎜ 2 ⎟
⎜
⎟ = 0.921V
j
T
20
q ⎝ NA NA ⎠
⎝ 10
⎠
⎝ ni ⎠
| w do =
2(11.7)(8.854 x10−14 )⎛ 1
1 ⎞
-5
w do =
⎜ 20 + 16 ⎟0.921 = 3.45x10 cm
−19
⎝ 10
10 ⎠
1.602x10
Cj =
CSB
(11.7)(8.854 x10−14 F /cm)
−5
= 3.00x10−8 F /cm 2
3.45x10 cm
= (3.00x10−8 F /cm 2 )(12.5x10−8 cm 2 )+ 5x10−4 cm(3.00x10−8 F /cm 2 )(15x10−4 cm)= 26.3 fF
CDB = CSB = 26.3 fF
4.75
KP = K 'n = K n
L
μA ⎛ 0.25μm ⎞
= 175 2 ⎜
⎟ = 8.75U | VTO = VTN = 0.7
W
V ⎝ 5μm ⎠
PHI = 2φ F = 0.8V | L = 0.25U | W = 5U | LAMBDA = 0.02
4.76
(a) VTO = 0.7 | PHI = 2φ F = 0.6 | GAMMA = 0.75
(b) VTO = 0.74
| PHI = 0.87 | GAMMA = 0.84
4.77
KP = K 'n = 50U VTO = VTN = 1 V L = 0.5U W = 2.5U LAMBDA = 0
4.78
KP = K 'n = 10U VTO = VTN = 1 V L = 0.6U W = 1.5U LAMBDA = 0
4.79
KP = K 'p = 10U VTO = VTP = −1 V L = 0.5U
Using the - 3 - V curve, K P = 2
50μA
[-3 - (-1)]
2
= 25
μA
V2
W = 1.25U LAMBDA = 0
4.80
KP = K 'n = 25U VTO = VTN = 1 V L = 0.6U W = 0.6U LAMBDA = 0
98
NMOS i-v Characteristics for Load-Line Problems
800
5V
Drain Current (uA)
600
400
4V
200
3V
2V
0
0
1
2
3
4
5
6
Drain Voltage (V)
99
4.81
For VDS = 0, ID =
4V
= 0.588mA. For ID = 0,VDS = 4V .
6.8kΩ
Also, VGS = 4V. From the graph, the transistor is operating below pinchoff in the triode region
and the Q-point is Q-point: (350 μA, 1.7V)
800
5V
Drain Current (uA)
600
Q-point
(4.82)
400
4V
Q-point
(4.81)
200
3V
2V
0
0
1
2
3
4
5
6
Drain Voltage (V)
4.82
For VDS = 0, ID =
5V
= 0.602mA. For ID = 0,VDS = 5V .
8.3kΩ
For VGS = 5V, the Q-point is (450 μA, 1.25 V). From the graph in Prob. 4.81, the transistor is
operating below pinchoff in the triode region.
4.83
800
5V
Drain Current (uA)
600
Q-point
(4.84)
400
4V
Q-point
(4.83)
200
3V
2V
0
0
1
2
3
Drain Voltage (V)
100
4
5
6
VDD
= 3V | 6 =10 4 ID +V DS | VDS = 0, ID = 0.6mA | ID = 0, VDS = 6V
2
From the graph, Q-pt: (140 μA, 4.6V) in the saturation region.
VGS =
4.84
VDD
= 4V | 8 =10 4 ID +V DS | VDS = 6, ID = 0.2mA | VDS = 0, ID = 0.8mA
2
See graph for Problem 4.83: Q-pt: (390 μA, 4.1 V) in saturation region.
VGS =
4.85
100kΩ
12V = 3.75V | Assume saturation
100kΩ + 220kΩ
−6 ⎞
⎛
⎛ 5⎞
2
3
3 100x10
3.75 = VGS + 24 x10 ID = VGS + 24 x10 ⎜
⎟⎜ ⎟(VGS −1)
2
⎝
⎠⎝ 1 ⎠
(a)
VGG =
6VGS2 −11VGS + 2.25 = 0 → VGS = 1.599V and ID = 89.7μA
VDS = 12 − 36x10 3 ID = 8.77V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct.
Q − point : (89.7 μA, 8.77 V )
(b)
Assume saturation
⎛ 100x10−6 ⎞⎛ 10 ⎞
2
3.75 = VGS + 24 x10 ID = VGS + 24 x10 ⎜
⎟⎜ ⎟(VGS −1)
2
⎠⎝ 1 ⎠
⎝
3
3
12VGS2 − 23VGS + 8.25 = 0 → VGS = 1.439V and ID = 96.4 μA
VDS = 12 − 36x10 3 ID = 8.53V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct.
Q − point : (96.4 μA, 8.53 V )
4.86
10kΩ
12V = 3.75V | Assume saturation
10kΩ + 22kΩ
⎛ 100x10−6 ⎞⎛ 20 ⎞
2
3.75 = VGS + 2.4 x10 3 ID = VGS + 24 x10 3⎜
⎟⎜ ⎟(VGS −1)
2
⎠⎝ 1 ⎠
⎝
VGG =
2.4VGS2 − 3.8VGS + 1.35 = 0 → VGS = 1.882V and ID = 778μA
VDS = 12 − 3.6x10 3 ID = 9.20V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 2.4 x10 3 ID + VGS = 3.75V which is correct.
Q − point : (778 μA, 9.20 V )
101
4.87
1MΩ
12V = 3.75V | Assume saturation
1MΩ + 2.2MΩ
⎛ 100x10−6 ⎞⎛ 5 ⎞
2
3.75 = VGS + 2.4 x10 3 ID = VGS + 2.4 x10 5⎜
⎟⎜ ⎟(VGS −1)
2
⎠⎝ 1 ⎠
⎝
VGG =
60VGS2 −119VGS + 56.25 = 0 → VGS = 1.206V and ID = 10.6μA
VDS = 12 − 3.6x10 5 ID = 8.18V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 2.4 x10 5 ID + VGS = 3.75V which is correct.
Q − point : (10.6 μA, 8.18 V )
4.88
100kΩ
15V = 4.69V | Assume Saturation
100kΩ + 220kΩ
⎛ 100x10−6 ⎞⎛ 5 ⎞
2
4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3⎜
⎟⎜ ⎟(VGS −1)
2
⎠⎝ 1 ⎠
⎝
(a)
VGG =
6VGS2 −11VGS + 1.31 = 0 → VGS = 1.705V and ID = 124 μA
VDS = 15 − 36x10 3 ID = 10.5 V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct.
Q − point : (124 μA, 10.5 V )
(b) VGG = 4.69V | Assume Saturation
⎛ 100x10−6 ⎞⎛ 10 ⎞
2
4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3⎜
⎟⎜ ⎟(VGS −1)
2
⎝
⎠⎝ 1 ⎠
12VGS2 − 23VGS + 7.31 = 0 → VGS = 1.514V and ID = 132 μA
VDS = 15 − 36x10 3 ID = 10.3 V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct.
Q − point : (132 μA, 10.3 V )
4.89
200kΩ
12V = 3.81V | Assume Saturation
200kΩ + 430kΩ
⎛ 100x10−6 ⎞⎛ 5 ⎞
2
3.81 = VGS + 47x10 3 ID = VGS + 47x10 3 ⎜
⎟⎜ ⎟(VGS −1)
2
⎝
⎠⎝ 1 ⎠
(a) VGG =
23.5VGS2 − 45VGS + 15.88 = 0 → VGS = 1.448V and ID = 50.3 μA
VDS = 12 − 71x10 3 ID = 8.43 V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 47x10 3 ID + VGS = 3.81V which is correct.
Q − point : (50.3 μA, 8.43 V )
102
(b) VGG = 3.81V
| Assume Saturation
⎛ 100x10−6 ⎞⎛ 15 ⎞
2
3.81 = VGS + 47x10 ID = VGS + 47x10 ⎜
⎟⎜ ⎟(VGS −1)
2
⎝
⎠⎝ 1 ⎠
3
3
70.5VGS2 −139VGS + 62.88 = 0 → VGS = 1.269V and ID = 54.3 μA
VDS = 12 − 71x10 3 ID = 8.15 V | VDS > VGS − VTN Saturation is correct.
Checking : VGG = 47x10 3 ID + VGS = 3.82V which is correct.
Q − point : (54.3 μA, 8.15 V )
4.90
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 89.6 μA, VGS = 1.60 V and VDS
= 8.77 V 4
(a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 96.3 μA, VGS = 1.44 V and VDS
= 8.53 V 4
4.91
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 124 μA, VGS = 1.71 V and VDS =
10.5 V 4
(a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 132 μA, VGS = 1.51 V and VDS
= 10.2 V 4
4.92
(a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 50.2 μA, VGS = 1.45 V and VDS
= 8.43 V 4
(a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 54.1 μA, VGS = 1.27 V and VDS
= 8.16 V 4
4.93
(300 kΩ, 700 kΩ) or (1.2 MΩ, 2.8 MΩ). We normally desire the current in the gate bias
network to be much less than ID. We also usually like the parallel combination of R1 and R2 to
be as large as possible.
4.94
35x10−6
2
(a) ID =
(4 −1−1700ID ) and using the quadratic equation,
2
ID = 134μA. VDS =10 -134 x10−6 (1700 + 38300) = 4.64V
25x10−6
2
(4 − 0.75 −1700ID ) and using the quadratic equation,
2
ID = 116μA. VDS =10 -116x10−6 (1700 + 38300) = 5.36V
(b)
ID =
103
4.95
(a) Example 4.3
Setting KP = 25U and VTO = 1 yields ID = 34.4 μA, VGS = 2.66 V and VDS = 6.08 V 4
Results agree with hand calculations
(b) Example 4.4
Setting KP = 25U and VTO = 1 yields ID = 99.2 μA, VGS = 3.82 V and VDS = 6.03 V 4
Results are almost identical to hand calculations
4.96
K n' = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 4V and VGS − VTN = 1V
RS =
4
4.1
= 40kΩ ⇒ 39kΩ and VR S = 3.9V | RD =
= 41kΩ ⇒ 43kΩ
100μA
100μA
VGS − VTN =
2ID
2I
W 2
= 1V and K n = D2 = 200μA /V 2 ⇒
=
|
Kn
1V
L 1
VG = VS + VGS = 3.9 + 1+ 0.75 = 5.65V
5.65V =
⎛ 12V ⎞
R1
R R ⎛ 12V ⎞
12V | 5.65V = 1 2 ⎜
⎟ = 530kΩ ⇒ 560kΩ
⎟ | R2 = 250kΩ⎜
⎝ 5.65V ⎠
R1 + R2
R1 + R2 ⎝ R2 ⎠
5.65V =
R1
12V ⇒ R1 = 500kΩ ⇒ 510kΩ
R1 + 560kΩ
R1 = 510kΩ, R2 = 560kΩ, RS = 39kΩ, RD = 43kΩ,
W 2
=
L 1
4.97
K n' = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 3V and VGS − VTN = 1V
3
3
= 12kΩ | RD =
= 12kΩ
0.25mA
0.25mA
2ID
2I
W 5
VGS − VTN =
= 1V and K n = D2 = 500μA /V 2 ⇒
=
Kn
1V
L 1
RS =
VG = VS + VGS = 3 + 1+ 0.75 = 4.75V
4.75V =
⎛ 9V ⎞
R1
R R ⎛ 9V ⎞
9V | 4.75V = 1 2 ⎜ ⎟ | R2 = 250kΩ⎜
⎟ = 473kΩ ⇒ 470kΩ
⎝ 4.75V ⎠
R1 + R2
R1 + R2 ⎝ R2 ⎠
4.75V =
R1
9V ⇒ R1 = 525kΩ ⇒ 510kΩ
R1 + 470kΩ
R1 = 510kΩ, R2 = 470kΩ, RS = 12kΩ, RD = 12kΩ,
104
W 5
=
L 1
4.98
K n' = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 5V and VGS − VTN = 1V
5
5
= 10kΩ | RD =
= 10kΩ
0.5mA
0.5mA
2ID
2I
W 10
VGS − VTN =
= 1V and K n = D2 = 1mA /V 2 ⇒
=
Kn
1V
L
1
RS =
VG = VS + VGS = 5 + 1+ 0.75 = 6.75V
6.75V =
⎛ 15V ⎞
R1
R R ⎛ 15V ⎞
15V | 6.75V = 1 2 ⎜
⎟ = 1.33MΩ ⇒ 1.5MΩ
⎟ | R2 = 600kΩ⎜
⎝ 6.75V ⎠
R1 + R2
R1 + R2 ⎝ R2 ⎠
R1
15V ⇒ R1 = 1.23MΩ ⇒ 1.2MΩ
R1 + 1.5MΩ
W 10
R1 = 1.2MΩ, R2 = 1.5MΩ, RS = 10kΩ, RD = 10kΩ,
=
L
1
6.75V =
4.99
⎛10−3 ⎞
2
Assume Saturation. For IG = 0, VGS = −10 4 ID = −10 4 ⎜
⎟(VGS + 5)
⎝ 2 ⎠
5VGS2 + 51VGS + 125 = 0 ⇒ VGS = −4.10 V and ID = 410 μA
VDS = 15 −15000ID = 8.85 V | VDS > VGS − VTN so saturation is ok.
Q - Point : (410μA, 8.85V)
4.100
⎛ 6x10−4 ⎞
2
Assume Saturation. For IG = 0, VGS = −27x10 3 ID = −27x10 4 ⎜
⎟(VGS + 4 )
⎝ 2 ⎠
8.1VGS2 + 65.8VGS + 129.6 = 0 ⇒ VGS = −3.36 V and ID = 124 μA
VDS = 12 − 78000ID = 2.36 V | VDS > VGS − VTN so saturation is ok.
Q - Point : (124 μA, 2.36V)
4.101
Kn = 1 mA/V 2
RD
V TN = -5 V
+
V
GS
RG
VDD
ID
IG = 0
-
+
5V
-
+
15 V
R
S
105
⎛ 1mA/V 2 ⎞
2
Assume Saturation. IG = 0. 250μA = ⎜
⎟(VGS + 5)
2 ⎠
⎝
0.25mA
4.29V
= −4.29V | RS =
= 17.2kΩ ⇒ 18kΩ
0.5mA
0.25mA
15 − 5 − 4.29 V
VDS = 15 − ID (RD + RS ) ⇒ RD =
= 22.88kΩ ⇒ 24kΩ
0.25
mA
RG is arbitrary but normally fairly large. Choose RG = 510 kΩ.
VGS = −5 +
4.102
+
5V
R
R
D
2
+
VDD
+
5V
+
R1
R
S
15 V
5V
-
Assume Saturation. IG = 0. Assume power supply is split in thirds: VDS = VR D = VRS = 5V
Kn 2
VTN and this will require VGS > 0.
2
⎛ 0.25mA/V 2 ⎞
5V
2
RS =
= 2.5kΩ ⇒ 2.4kΩ and VRS will be 4.8 V | 2mA = ⎜
⎟(VGS + 2)
2
2mA
⎝
⎠
Note that although, this is a depletion - mode device, I D exceeds
VGS = −2 +
2mA
= +2.00V | VG = VS + VGS = 4.8 + 2 = 6.8V
0.125mA
R1
⇒ R1 = 680 kΩ and R2 = 820 kΩ is one convenient possibility.
R1 + R2
Another is R1 = 68 kΩ and R2 = 82 kΩ. Both choices have IR 2 << I D .
6.8 = 15
RD =
106
(15 − 5 − 4.8)V
2mA
= 2.60kΩ ⇒ 2.70 kΩ.
4.103
(a)
The transistor is saturated by connection.
VGS = 12 −10 5 ID and ID =
100x10−6 ⎛ 10 ⎞⎛ A ⎞
2
⎜ ⎟⎜ 2 ⎟(VGS − 0.75V )
⎝ 1 ⎠⎝ V ⎠
2
50VGS2 − 74VGS + 16.13 = 0 ⇒ VGS = 1.214V, − 0.266V ⇒ VGS = 1.214 V since VGS must
100x10−6 ⎛10 ⎞⎛ A ⎞
2
⎜ ⎟⎜ 2 ⎟(1.214 − 0.75V )
⎝ 1 ⎠⎝ V ⎠
2
12 −1.21
ID = 104μA | Checking : ID =
= 108 μA | Q - Point : (108 μA,1.21 V)
10 5
exceed the threshold voltage. | ID =
7
(b) Using KVL, VDS = 10 IG +VGS. But, since IG = 0, VGS = VDS. Also VTN = 0.75 V > 0, so the
transistor is saturated by connection.
+12 V
10
W =
L
1
ID =
10 M Ω
330 k Ω
IDS
I
VDS
+
V GS
-
-
VGS = 12 − 330kΩ(ID + IG ) −10MΩ(IG ) but I G = 0
VGS = 12 − 330kΩ(ID )
+
G
⎛100 μA ⎞⎛10 ⎞
K n' W
2
2
⎟(VGS − 0.75)
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
⎛ 1.00x10−3 A ⎞
2
VGS = 12 − (3.30x10 5 )⎜
V − 0.75)
2 ⎟( GS
2
V ⎠
⎝
165VGS2 − 246.5VGS + 80.81 = 0 yields VGS = 1.008V , 0.486V
VGS must be 1.008 V since 0.486 V is below threshold.
⎛100 μA ⎞ 10
2
ID = ⎜
(1.008 − 0.75) = 33.3 μA and VDS = VGS
2⎟
⎝ 2 V ⎠1
Q-Point: (33.3 μA, 1.01 V)
Checking: ID =(12-1.01)V/330kΩ = 33.3 μA. 4
4.104
(a)
The transistor is saturated by connection.
VGS = 12 −10 5 ID
and
ID =
100x10−6 ⎛ 20 ⎞⎛ A ⎞
2
⎜ ⎟⎜ 2 ⎟(VGS − 0.75V )
⎝ 1 ⎠⎝ V ⎠
2
100VGS2 −149VGS + 44.25 = 0 ⇒ VGS = 1.08V, 0.410V ⇒ VGS = 1.08 V since VGS must
exceed the threshold voltage. ID =
Checking : ID =
100x10−6 ⎛ 20 ⎞⎛ A ⎞
2
⎜ ⎟⎜ 2 ⎟(1.08 − 0.75V ) = 109μA
⎝ 1 ⎠⎝ V ⎠
2
12 −1.08
= 109 μA | Q - Point : (109 μA, 1.08 V)
10 5
7
(b) Using KVL, VDS = 10 IG +VGS. But, since IG = 0, VGS = VDS. Also VTN = 0.75 V > 0, so the
transistor is saturated by connection.
107
+12 V
10
W =
L
1
ID =
10 M Ω
330 k Ω
IDS
I
G
V GS
-
VGS = 12 − 330kΩ(ID + IG ) −10MΩ(IG ) but
IG = 0
VGS = 12 − 330kΩ(ID )
+
VDS
+
⎛100 μA ⎞⎛ 20 ⎞
K n' W
2
2
⎟(VGS − 0.75)
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
-
⎛
A⎞
2
VGS = 12 − (3.30x10 5 )⎜10−3 2 ⎟(VGS − 0.75)
⎝
V ⎠
3.3VGS2 − 4.94VGS + 1.736 = 0 yields VGS = 0.933V, 0.564V
VGS must be 0.933 V since 0.564 V is below threshold.
⎛100 μA ⎞ 20
12 − 0.933
2
ID = ⎜
V = 33.5μA
(0.933 − 0.75) = 33.5 μA Checking :
2⎟
⎝ 2 V ⎠ 1
330kΩ
and VDS = VGS :
Q-Point: (33.5 μA, 0.933 V)
4.105
(a) Assume saturation :
ID =
⎛100 μA ⎞⎛ 10 ⎞
K n' W
2
2
⎟(VGS − 0.75)
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
VGS = 15 − 330kΩ(ID + IG ) −10MΩ(IG ) but I G = 0
VGS = 15 − 330kΩ(ID ) and VGS = VDS so saturation regioin operation is correct
⎛ 10−3 A ⎞
2
VGS = 15 − (3.30x10 5 )⎜
V − 0.75)
2 ⎟( GS
⎝ 2 V ⎠
3.30VGS2 − 4.93VGS + 1.556 = 0 yields VGS = 1.041V, 0.453V
⎛100 μA ⎞⎛ 10 ⎞
15 −1.041
2
ID = ⎜
V = 42.3μA
⎟(1.041− 0.75) = 42.3μA | Checking : ID =
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
330kΩ
Q − Point : (42.3 μA, 1.04 V)
(b) Assume saturation
ID =
⎛100 μA ⎞⎛ 25 ⎞
K n' W
2
2
⎟(VGS − 0.75)
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
VGS = 15 − 330kΩ(ID + IG ) −10MΩ(IG ) but I G = 0
VGS = 15 − 330kΩ(ID ) and VGS = VDS so saturation region operation is correct.
⎛ 2.50x10−3 A ⎞
2
V − 0.75)
VGS = 15 − (3.30x10 5 )⎜
2 ⎟( GS
V ⎠
2
⎝
8.25VGS2 −12.355VGS + 4.341 = 0 yields VGS = 0.9345V, 0.563V
⎛100 μA ⎞⎛ 25 ⎞
15 − 0.9345
2
V = 42.6μA
ID = ⎜
⎟(0.9345 − 0.75) = 42.6μA | Checking : ID =
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
330kΩ
Q - Point :
108
(42.6 μA, 0.935 V)
4.106
(a)
Asssume saturation
ID =
⎛100 μA ⎞⎛10 ⎞
K n' W
2
2
⎟(VGS − 0.75)
(VGS − VTN ) = ⎜
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
2 L
VGS = 12 − 470kΩ(ID + IG ) −10MΩ(IG ) but IG = 0
VGS = 12 − 470kΩ(ID ) and VGS = VDS so saturation region operation is correct.
⎛ 10−3 A ⎞
2
VGS = 12 − (4.70x10 5 )⎜
V − 0.75)
2 ⎟( GS
⎝ 2 V ⎠
4.70VGS2 − 7.03VGS + 2.404 = 0 yields VGS = 0.9666V, 0.529V
⎛100 μA ⎞⎛ 10 ⎞
12 − 0.967
2
ID = ⎜
= 23.5μA
⎟(0.9666 − 0.75) = 23.5μA | Checking : ID =
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
470kΩ
Q - Point : (23.5 μA, 0.967 V)
(b) Since the current in RG is zero, the drain current is independent of RG.
4.107
(a) Create an M-file:
function f=bias(id)
vtn=1+0.5*(sqrt(22e3*id)-sqrt(0.6));
f=id-(25e-6/2)*(6-22e3*id-vtn)^2;
fzero('bias',1e-4) yields ans = 8.8043e-05
(b) Modify the M-file:
function f=bias(id)
vtn=1+0.75*(sqrt(22e3*id)-sqrt(0.6));
f=id-(25e-6/2)*(6-22e3*id-vtn)^2;
fzero('bias',1e-4) yields ans = 8.3233e-05
4.108
Using a spreadsheet similar to Table 4.2 yields: (a) 88.04 μA, (b) 83.23 μA.
4.109
100kΩ
12V = 3.75V | Assume saturation
100kΩ + 220kΩ
2ID
VTN = 1+ 0.6 24 x10 3 ID + 0.6 − 0.6 | VGS = VTN +
5x10−4
3.75 − VGS
ID =
| Solving iteratively yields ID = 73.1μA with VTN = 1.460V
24kΩ
V = 12V − ID (24kΩ + 12kΩ) = 9.37 V Transistor is saturated. Q - Point : (73.1 μA, 9.37 V)
VGG =
(
)
109
4.110
VGG =
100kΩ
12V = 3.75V | Assume saturation
100kΩ + 220kΩ
(a) VTN = 1+ 0.75( 24 x10 3 ID + 0.6 −
0.6
)|
VGS = VTN +
2ID
5x10−4
3.75 − VGS
| Solving iteratively yields ID = 69.7μA with VTN = 1.550V
24kΩ
= 12V − ID (24kΩ + 12kΩ) = 9.49 V The transistor is saturated. Q - Point : (69.7 μA, 9.49 V)
ID =
VDS
(b) VTN = 1+ 0.6( 24 x10 3 ID + 0.6 −
0.6
)|
VGS = VTN +
2ID
5x10−4
3.75 − VGS
| Solving iteratively yields ID = 73.1μA with VTN = 1.460V
24kΩ
V = 12V − ID (24kΩ + 24kΩ) = 8.49 V The transistor is saturated. Q - Point : (73.1 μA, 8.49 V)
ID =
4.111
(a) γ = 0.6
(b) γ = 0.75
(c ) γ = 0.6
VTN = 1.46 V
VTN = 1.55 V
VTN = 1.46 V
ID = 73.1 μA
ID = 69.7 μA
ID = 73.1 μA
VDS = 9.37 V
VDS = 9.49 V
VDS = 8.49 V
These results all agree with the hand calculations.
(They should - they are all solving the same sets of equations.)
4.112
(a) γ = 0
VTN = 1.00 V
ID = 50.2 μA
VDS = 8.43 V
γ = 0.5 VTN = 1.42 V
(b) γ = 0 VTN = 1.00 V
γ = 0.5 VTN = 1.44 V
ID = 42.2 μA
VDS = 9.01 V
ID = 54.1 μA
VDS = 8.16 V
ID = 45.2 μA
VDS = 8.79 V
The γ = 0 values agree with the hand calculations in the original problem. Including body effect
in the simulations reduces the Q-point current by approximately 15%. Although this may sound
large, it is within the error that will be introduced by the use of 5% resistors and typical device
tolerances. So, we normally omit γ in our hand calculations, and then refine the results using
SPICE.
4.113
(a) γ = 0
VTN = 1.00 V
ID = 89.6 μA
VDS = 8.77 V
γ = 0.5 VTN = 1.39 V
(b) γ = 0 VTN = 1.00 V
γ = 0.5 VTN = 1.41 V
ID = 75.5 μA
VDS = 9.28 V
ID = 96.3 μA
VDS = 8.53 V
ID = 80.8 μA
VDS = 9.09 V
The γ = 0 values agree with the hand calculations in the original problem. Including body effect
in the simulations reduces the Q-point current by approximately 15%. Although this may sound
110
large, it is within the error that will be introduced by the use of 5% resistors and typical device
tolerances. So, we normally omit γ in our hand calculations, and then refine the results using
SPICE.
4.114
γ =0
γ = 0.5
VTN = 1.00 V
VTN = 1.35 V
ID = 778 μA
ID = 661 μA
VDS = 9.20 V
VDS = 9.62 V
The γ = 0 values agree with the hand calculations in the original problem. Including body effect
in the simulations reduces the Q-point current by approximately 15%. Although this may sound
large, it is within the error that will be introduced by the use of 5% resistors and typical device
tolerances. So, we normally omit γ in our hand calculations, and then refine the results using
SPICE.
4.115
γ =0
γ = 0.5
VTN = 1.00 V
VTN = 1.45 V
ID = 10.5 μA
ID = 9.18 μA
VDS = 8.03 V
VDS = 8.69 V
The γ = 0 values agree with the hand calculations in the original problem. Including body effect
in the simulations reduces the Q-point current by approximately 15%. Although this may sound
large, it is within the error that will be introduced by the use of 5% resistors and typical device
tolerances. So, we normally omit γ in our hand calculations, and then refine the results using
SPICE.
4.116
(a) Both transistors are saturated by connection and the two drain currents must be equal.
K
K
2
2
ID1 = n1 (VGS1 − VTN1 ) and ID 2 = n 2 (VGS 2 − VTN 2 )
2
2
But since the transistors are identical, ID1 = ID2 requires VGS1 = VGS2 = VDD/2 = 2.5V.
100x10−6 ⎛ 20 ⎞
2
ID1 = ID 2 =
⎜ ⎟(2.5 −1) = 2.25 mA
⎝1⎠
2
(b) For this case, the same arguments hold, and VGS1 = VGS2 = VDD/2 = 5V.
100x10−6 ⎛ 20 ⎞
2
ID1 = ID 2 =
⎜ ⎟(5 −1) =16.0 mA
⎝1⎠
2
.
(c) For this case, the threshold voltages will be different due to the body-effect in the upper
transistor. The drain currents must be the same, but the gate-source voltages will be different:
VGS1 = VTN1 +
VTN1 =1V
2ID
2ID
; VGS 2 = VTN 2 +
; VGS1 + VGS 2 = 5V .
Kn
Kn
(
VTN 2 = 1+ 0.5 VGS1 + 0.6 − 0.6
)
111
Combining these equations yields
(
)
5 - 2VGS1 − 0.5 VGS1 + 0.6 − 0.6 = 0 ⇒ VGS1 = 2.27V ; VGS 2 = 5 − VGS1 = 2.73V
ID 2 = ID1 =
100x10−6 ⎛ 20 ⎞
2
⎜ ⎟(2.27 −1) = 1.61 mA.
⎝1⎠
2
(
)
Checking : VTN 2 = 1+ 0.5 2.27 + 0.6 − 0.6 = 1.46V
ID 2 =
100x10−6 ⎛ 20 ⎞
2
⎜ ⎟(2.73 −1.46) = 1.61 mA.
⎝1⎠
2
4.117
If we assume saturation, we find ID = 234 μA and VDS = 0.65 V, and the transistor is not
saturated. Assuming triode region operation,
VGS = 10 − 2x10 4 ID | VDS = 10 − 4 x10 4 ID
μA ⎛ 2 ⎞⎛
10 − 4 x10 4 ID ⎞
4
ID = 100 2 ⎜ ⎟⎜10 − 2x10 4 ID −1−
⎟(10 − 4 x10 ID )
⎝
⎠
2
V 1 ⎝
⎠
Collecting terms : 16.5x10 4 ID = 40 → ID = 242 μA
VDS = 10 − 4 x10 4 (2.42x10−4 )= 0.320V | Q - Pt : (242 μA, 0.320V )
Checking the operating region : VGS − VTN = 4.16V > VDS
and the triode region assumption is correct. Checking : ID =
10 − 0.32
V = 242μA
40kΩ
4.118
If we assume saturation, we find an inconsistent answer. Assuming triode region operation,
VGS = 10 − 2x10 4 ID | VDS = 10 − 3x10 4 ID
μA ⎛ 4 ⎞⎛
10 − 3x10 4 ID ⎞
4
ID = 100 2 ⎜ ⎟⎜10 − 2x10 4 ID −1−
⎟(10 − 3x10 ID )
2
V ⎝ 1 ⎠⎝
⎠
Collecting terms : 1.5x10 8 ID2 −1.725x10 5 ID + 40 = 0 → ID = 322μA
VDS = 10 − 3x10 4 (3.22x10−4 )= 0.340V | Q - Pt : (322 μA, 3.18 V )
Checking the operating region : VGS − VTN = 2.56V > VDS
and the triode region assumption is correct. Checking : ID =
112
10 − 0.34
V = 322μA
30kΩ
4.119
For (a) and (b), γ = 0. The transistor parameters are identical so 3VGS = 15V or VGS = 5V.
⎛ 20 ⎞
1
2
(a) ID = (100x10−6 )⎜ ⎟(5 − 0.75) = 18.1 mA
⎝1⎠
2
⎛ 50 ⎞
1
2
(b) ID = (100x10−6 )⎜ ⎟(5 − 0.75) = 45.2 mA
⎝1⎠
2
(c) Now we have three different threshold voltages and need an iterative solution. Using
MATLAB:
function f=Prob112(id)
gamma=0.5;
vgs1=.75+sqrt(2*id/2e-3);
vtn2=0.75+gamma*(sqrt(vgs1+0.6)-sqrt(0.6));
vgs2=vtn2+sqrt(2*id/2e-3);
vtn3=0.75+gamma*(sqrt(vgs1+vgs2+0.6)-sqrt(0.6));
vgs3=vtn3+sqrt(2*id/2e-3);
f=15-vgs1-vgs2-vgs3;
fzero('Prob112',1e-4) --> ans = 0.0130
ID = 13.0 mA
4.120
W 20
=
VTN = 0.75 V ID = 18.1 mA VDS = 5.00 V
1
L
W 50
=
VTN = 0.75 V ID = 45.2 mA VDS = 5.00 V
(b) γ = 0
L
1
W 20
=
VTN 3 = 1.95 V ID 3 = 13.0 mA VDS 3 = 5.56 V
(b) γ = 0.5
L
1
VTN 2 = 1.48 V ID 2 = 13.0 mA VDS 2 = 5.09 V VTN1 = 0.75 V ID1 = 13.0 mA
(a) γ = 0
VDS1 = 4.36 V
Results are identical to calculations in Prob. 4.119
4.121
For VGS = 5 V and VDS = 0.5 V, the transistor will be in the triode region.
(5 − 0.5)V = 54.88μA | 54.88x10−6 = 100x10−6⎛ W ⎞⎛5 − 0.75 − 0.5 ⎞0.5 | W = 0.274 = 1
ID =
⎜ ⎟⎜
⎟
⎝ L ⎠⎝
2 ⎠
L
1
3.64
82kΩ
4.122
For VGS = 3.3 V and VDS = 0.25 V, the transistor will be in the triode region.
(3.3 − 0.25)V = 16.94 μA | 16.94 x10−6 = 100x10−6⎛ W ⎞⎛ 3.3 − 0.75 − 0.25 ⎞0.25 | W = 0.280 = 1
ID =
⎜ ⎟⎜
⎟
⎝ L ⎠⎝
2 ⎠
L
1
3.57
180kΩ
113
4.123
(a) The transistor is saturated by connection. For this circuit,
VGS = VDD + ID R = −15 + 75000ID
4 x10−5 ⎛ 1⎞
2
ID =
⎜ ⎟(−15 + 75000ID + 0.75) ⇒ 153 μA
2 ⎝ 1⎠
VGS = −15 + 75000ID = −3.525V
VDS = VGS = −3.525V | Q - point : (153 μA,−3.53 V )
(b) Here the transistor has VGS = -15 V, a large value, so the transistor is most likely operating in
the triode region.
⎛
VDS − (−15)
V ⎞
= 4 x10−5 ⎜ −15 − (−0.75) − DS ⎟VDS ⇒ VDS = −0.347 V and ID = 195 μA.
⎝
75000
2 ⎠
15 − 0.347
V = 195μA
Q - point : (195 μA,-0.347 V )
Checking : ID =
785kΩ
Checking the region of operation: VDS = −0.347V > VGS − VTP = −15 + 0.75 = −14.25V
ID =
Triode region is correct
4.124
Set W=1U L=1U KP=40U VTO=-0.75 GAMMA=0
Results are almost identical to hand calculations for both parts.
4.125
(a) IDP = IDN , and both transistors are saturated by connection. 10 = -VGSP + VGSN
1 ⎛ 100μA ⎞⎛ 20 ⎞
1 ⎛ 40μA ⎞⎛ 20 ⎞
2
2
⎜ 2 ⎟⎜ ⎟(−10 + VGSN + 0.75) = ⎜
⎟⎜ ⎟(VGSN − 0.75)
2
2 ⎝ V ⎠⎝ 1 ⎠
2 ⎝ V ⎠⎝ 1 ⎠
(9.25 − VGSN ) =
2.5 (VGSN − 0.75) → VGSN = 4.04V | VGSP = −5.96V
IDP = IDN = 10.8 mA | VO = VGSN = 4.04V
(b) Everything is the same except the currents scale by 80/20:
IDP = IDN = 43.2 mA
4.126
For (a) and (b), γ = 0. The transistor parameters are identical so -3VGS = 15V or VGS = -5V.
1
40
2
40x10−6 ) (−5 + 0.75) = 14.4 mA
(
2
1
1
75
2
(b) ID = (40x10−6 ) (−5 + 0.75) = 27.1 mA
2
1
(a) ID =
(c) Now we have three different threshold voltages and need an iterative solution. Using
MATLAB:
function f=PMOSStack(id)
gamma=0.5;
114
vsg1=.75+sqrt(2*id/1.6e-3);
vtp2=-0.75-gamma*(sqrt(vsg1+0.6)-sqrt(0.6));
vsg2=-vtp2+sqrt(2*id/1.6e-3);
vtp3=-0.75-gamma*(sqrt(vsg1+vsg2+0.6)-sqrt(0.6));
vsg3=-vtp3+sqrt(2*id/1.6e-3);
f=15-vsg1-vsg2-vsg3;
fzero('PMOSStack',1e-1) --> ans = 0.0104 ID = 10.4 mA.
4.127
(a) W=40U L=1U KP=40U VTO=-0.75 GAMMA=0
(b) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0
(c) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0.5
Results agree with hand calculations.
4.128
4V
= 2mA. For ID = 0,VDS = −4V . (VSD = +4V )
2kΩ
300kΩ
VGS = VEQ = −4V
= −3V (VSG = +3V )
300kΩ + 100kΩ
From the graph, the transistor is operating below pinchoff in the linear region.
For VDS = 0, ID =
5000
VSG = 5 V
Drain Current (
μA)
4000
3000
V
SG
=4V
2000
Q-point
VSG = 3 V
1000
V
SG
=2V
0
-1000
-1
0
1
2
3
4
Source-Drain Voltage (V)
5
6
Q-point: (1.15 mA, 1.7V)
115
PMOS Transistor Output Characteristics
5000
V GS = -5 V
Drain Current ( μA)
4000
3000
VGS = -4 V
2000
V GS = -3 V
1000
V GS = -2 V
0
-1000
-1
0
1
2
3
4
Source-Drain Voltage (V)
116
5
6
4.129
(a) VGG =
⎛ 4 x10−5 ⎞ 20
15V
2
= 7.5V | 7.5 = 10 5 ID - VGS | 7.5 = 10 5 ⎜
⎟ (VGS + 0.75) - VGS
2
⎝ 2 ⎠ 1
4VGS2 + 5.9VGS −1.5 = 0 → VGS = −1.148V and ID = 63.5μA
VDS = −(15 − (100kΩ + 50kΩ)ID ) = −5.47V | Q - point : (59.78 μA,−5.47 V )
(b) For saturation, VDS ≤ VGS − VTP or VSD ≥ VSG + VTP
15 − (100kΩ + R)ID ≥ 7.5 −100kΩID − 0.75 → R ≤ 130 kΩ
4.130
Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 yields results identical to the previous
problem.
4.131
(a) Using MATLAB:
function f=bias4(id)
gamma=0.5;
vbs=1e5*id;
vgs=-7.5+vbs;
vtp=-0.75-gamma*(sqrt(vbs+0.6)-sqrt(0.6));
f=id-(8e-4/2)*(vgs-vtp)^2;
fzero('bias4',4e-5) --> ans = 5.5278e-05 --> ID = 55.3 μA
VDS = -15 + (100kΩ+R) ID
(b) VDS ≤ VGS - VTP | -15 + (100kΩ+R)ID ≤ -1.972 + 1.600 |
R ≤ 164 kΩ
4.132
Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 GAMMA=0.5 yields results identical
to the previous problem.
4.133
(a)
V
DD
R
V
+
GS
-
S
VDS
G
D
+
I
SD
The arrow identifies the transistor as a PMOS device. Since γ = 0, we do not need to worry
about body effect: VTP = VTO. Since VDS = VGS, and VTP < 0, the transistor is saturated.
117
K 'p W
2
ID =
(VGS − VTP ) and - VGS = 12 −105 ID
2 L
⎛ 4 x10−5 ⎞⎛ 10 ⎞
2
-VGS = 12 −10 5 ⎜
⎟⎜ ⎟(VGS − (−0.75))
⎝ 2 ⎠⎝ 1 ⎠
20VGS2 + 29VGS − 0.75 = 0 yields VGS = −1.475V,+0.0255V
We require VGS < VTP = -0.75 V for the transistor to be conducting so
2
4 x10−5 ⎛ 10 ⎞ A
VGS = −1.475V and ID =
⎜ ⎟ 2 (−1.475 − (−0.75)) = 105 μA
2 ⎝ 1 ⎠V
Since VDS = VGS, the Q-point is given by: Q-Point = (105 μA, -1.475 V).
(b) Using MATLAB for the second part (Set gamma = 0 for part (a)):
function f=bias2(id)
gamma=1.0;
vgs=-12+1e5*id;
vsb=-vgs;
vtp=-0.75-gamma*(sqrt(vsb+0.6)-sqrt(0.6));
f=id-5e-5*(vgs-vtp)^2;
fzero('bias2',1e-4) --> ans = 9.5996e-05 and Q-Point = (96.0 μA, 2.40 V).
4.134
VGG
−5 ⎞
⎛
2
15V
40
4
4 4x10
=
= 7.5V | 7.5 = 5x10 I D - VGS | 7.5 = 5x10 ⎜
⎟ (VGS + 0.75) - VGS
2
⎝ 2 ⎠ 1
890VGS2 + 119VGS − 30 = 0 → VGS = −1.166V and I D = 138 μA
(
)
VDS = − 15 − (R + 50kΩ)I D = −5V → R = 22.3 kΩ
4.135
VGG =
I D = 138 μA
⎛ 4x10−5 ⎞ 40
2
15V
= 7.5V | 7.5 = 15 - 5x10 4 I D + VGS | 7.5 = 5x10 4 ⎜
⎟ (VGS − VTP ) - VGS
2
⎝ 2 ⎠ 1
(
VTP = −0.75 − 0.5 5x10 4 I D + 0.6 − 0.6
)
Solving iteratively yields I D = 111 μA with VTP = −1.60V
(
)
VDS = − 15 − (R + 50kΩ)I D = −5V → R = 40.1 kΩ
118
4.136
510kΩ
= 9.81V | 9.81 = 15 -105 I D + VGS
510kΩ + 270kΩ
⎛ 4x10−5 ⎞ 20
2
5.19 = 105⎜
⎟ (VGS + 0.75) − VGS
⎝ 2 ⎠ 1
(a) VGG = 15V
40VGS2 + 59VGS + 17.31 = 0 → VGS = −1.071 V and I D = 41.2 μA
(b) For saturation, VDS ≤ VGS − VTP
−15 + (100kΩ + R)I D ≤ −1.071+ 0.75 → R ≤ 256 kΩ
4.137
510kΩ
= 9.81V | 9.81 = 15 -105 I D + VGS
510kΩ + 270kΩ
⎛ 4x10−5 ⎞ 20
2
5
5.19 = 105 ⎜
⎟ (VGS − VTP ) − VGS | VTP = −0.75 − 0.5 10 I D + 0.6 − 0.6
⎝ 2 ⎠ 1
(a) VGG = 15V
(
)
Solving iteratively yields I D = 35.2 μA with VTP = −1.38 V and VGS = −1.67 V
(b) For saturation, VDS ≤ VGS − VTP
−15 + (100kΩ + R)I D ≤ −1.67 + 1.38 → R ≤ 318 kΩ
4.138
(a) Assume an equal voltage (5V) split between R D , RS and VDS. We need VDS ≤ VGS − VTP
or - 5 ≤ VGS − VTP . Choose VGS − VTP = −2V. Kn =
(V
GS
2I D
− VTP )
2
=
2mA W 12.5
→ =
.
4
L
1
VGS = −2 − 0.75 = −2.75V. VEQ = 5 − VGS = 7.75V.
7.75 = 15
R1
15 R1 R2
15
=
= 100kΩ. R2 = 193.5kΩ → 200kΩ.
R1 + R2 R2 R1 + R2 R2
7.75 = 15
220kΩ
R1
→ R1 = 214kΩ → 220kΩ. VEQ = 15
= 7.86V
220kΩ + 200kΩ
R1 + R2
7.86V − 2.75V
15 − 5 − 5.1
= 5.11kΩ → 5.1kΩ | RD =
= 4.9kΩ → 4.7kΩ
1mA
1mA
Note that R1 is connected between the gate and + 15 V, and R 2 is connected between
RS =
the gate and ground.
(b) For the NMOS case, choose W/L = 5/1. The resistors now have the same values
except R2 is now connected between the gate and +15 V, R1 is connected between
the gate and ground, and RD =
15 − 6 − 5.1
= 3.9kΩ
1mA
119
4.139
(a) Assume an equal voltage (3V) split between R D , RS and VDS. We need VDS ≤ VGS − VTP
or - 3 ≤ VGS − VTP . Choose VGS − VTP = −1V. Kn =
(V
GS
2I D
=
− VTP )
2
1mA W 25
→ = .
1
L 1
VGS = −1− 0.75 = −1.75V. VEQ = 3 − VGS = 4.75V.
4.75 = 9
R1
9 R1 R2
9
=
= 1MΩ. R2 = 1.7 MΩ →1.8 MΩ.
R1 + R2 R2 R1 + R2 R2
4.75 = 9
2 MΩ
R1
→ R1 = 2.01MΩ → 2 MΩ | VEQ = 9
= 4.74V
1.8 MΩ + 2 MΩ
R1 + R2
4.74V −1.75V
9 −3−3
= 5.97kΩ → 6.2kΩ | RD =
= 6.0kΩ → 6.2kΩ
0.5mA
0.5mA
Note that R1 is connected between the gate and + 9 V, and R 2 is connected between
RS =
the gate and ground. R1 = 2 MΩ, R2 = 1.8 MΩ, RS = RD = 6.2kΩ, W / L = 25/1
(b) For the NMOS case, choose W/L
= 40/1. The resistors now have the same values
except R2 is now connected between the gate and + 9 V, and R1 is connected between
the gate and ground.
4.140
⎛ 40 μA ⎞⎛ 10 ⎞ 4
VGS = 10 4 I D | Assume saturation : I D = ⎜
⎟ 10 I D − 4
2 ⎟⎜
⎝ 2 V ⎠⎝ 1 ⎠
(
)
2
Collecting terms : 108 I D2 − 8.5x10 4 I D + 16 = 0 → I D = 281 μA
(
)
VDS = − 15 −10 4 I D = −12.2V | Q - Pt : (281 μA,−12.2 V )
Checking : VGS − VTP = 2 − 4 = −1.19 V | VDS = −12.2 | Saturation is correct.
4.141
(
VGS = 10 4 I D | VTP = 4 − 0.25 VGS + 0.6 − 0.6
)
| ID =
Solving these equations iteratively yields I D = 260 μA
(
)
2
4x10−4
VGS − VTP )
(
2
VDS = − 15 −10 4 I D = −12.4V | Q - Pt : (260 μA,−12.4 V )
120
4.142
Note: The answers are very sensitive to round-off error and are best solved iteratively using
MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will
generally yield poor results.
Saturated by connection with VTP = −1
2
4x10−5 ⎛ 10 ⎞
5
6
11 2
ID =
⎜ ⎟ 3.3x10 I D −12 − (−1) →121− 7.265x10 I D + 1.089x10 I D = 0
2 ⎝1⎠
[
]
I D = 34.6μA, 32.1μA | VDS = 3.3x105 I D −12 = −0.582V ,−1.407V | Q - point : (32.1 μA,−1.41 V )
since the transistor would not be conducting for VGS = −0.582V.
4.143
Note: The answers are very sensitive to round-off error and are best solved iteratively using
MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will
generally yield poor results.
Saturated by connection with VTP = −3
⎛ 4x10−5 ⎞⎛ 30 ⎞
2
5
6
11 2
ID = ⎜
⎟⎜ ⎟ 3.3x10 I D −12 − (−3) → 81− 5.941x10 I D + 1.089x10 I D = 0
⎝ 2 ⎠⎝ 1 ⎠
[
]
I D = 27.07μA | VDS = 3.3x105 I D −12 = −3.067V | Q - point : (27.1 μA,−3.07 V )
4.144
Note: The answers are very sensitive to round-off error and are best solved iteratively using
MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will
generally yield poor results.
(a) Large VGS − Assume triode region.
⎛ 10 ⎞⎛
V ⎞
| I D = 40x10−6 ⎜ ⎟⎜12 − 0.75 − DS ⎟VDS
2 ⎠
⎝ 1 ⎠⎝
VDS = 12 − 3.3x105 I D
Q − po int : (36.1 μA, 80.6 mV) | VDS < VGS − VTN so triode region is correct.
2
10x10−6 ⎛ 25 ⎞
5
Saturated
by
connection
:
I
=
b
⎜ ⎟ 3.3x10 I D −12 + 0.75
()
D
2 ⎝1⎠
(
Q − point : (32.4μA,-1.32V)
(c) V
TP
ID =
(
= −0.75 − 0.5 3.3x105 I D + 0.6 − 0.6
40x10−6 ⎛ 25 ⎞
5
⎜ ⎟ 3.3x10 I D −12 − VTP
2 ⎝1⎠
(
)
2
)
)
(
| VDS = − 12 − 3.3x105 I D
)
Q − point : (28.8 μA,-2.49 V)
121
4.145
(a) Kn = μn
ID =
(b) K
'
n
Tox
(a) C
GC
⎥⎝ 2μm ⎠
⎦
V2
2
864μA ⎛ 4 1 ⎞
| I =
⎜ − ⎟ = 0.972 mA
2 ⎝ 2 2⎠
V
| V =
2
'
'
D
(
⎡ 3.9 8.854x10−14 F / cm
⎢
= C WL =
⎢
20x10−7 cm
⎣
(b) α = 2
"
OX
'
| CGC
=
CGC
1 gm
2π CGC
| gm =
)⎥⎤ 20x10
(
⎥
−4
⎦
)(
)
cm 10−4 cm = 34.5 fF
= 17.3 fF
α
4.147
fT =
)⎥⎤⎛⎜ 20μm ⎞⎟ = 432 μA
2
432μA
4 −1) = 1.94 mA
(
2
= 2Kn
4.146
(
−14
⎡
cm2 ⎢ 3.9 8.854x10 F / cm
= 500
V −s⎢
L
40x10−7 cm
⎣
εox W
∂iD
" W
"
= KP (VGS − VTP ) = μ pCOX
| CGC = COX
WL
∂vGS
L
1 μp
(VGS − VTP ), but (VGS − VTP )< 0 for PMOS transistor.
2π L2
1 μp
Since f T should be positive, f T =
(VGS − VTP )
2π L2
fT =
4.148
1 ⎛μ⎞
⎜ ⎟(VGS − VTN )
2π ⎝ L2 ⎠
⎡
⎤
1 ⎢400cm2 /V − s ⎥
fTN =
(1V )= 6.37 GHz | fTP = 0.4 fTN = 2.55 GHz
2π ⎢ 10−4 cm 2 ⎥
⎢⎣
⎥⎦
⎡
⎤
1 ⎢ 400cm 2 /V − s ⎥
(b) fTN = 2π ⎢ −5 2 ⎥(1V )= 637 GHz | fTP = 0.4 fTN = 255 GHz
⎢⎣ 10 cm
⎥⎦
(a) f
T
122
=
(
)
(
)
4.149
(a) Kn = μn
ID =
Tox
(
−14
⎡
cm2 ⎢ 3.9 8.854x10 F / cm
= 400
V − s⎢
L
80x10−7 cm
⎣
εox W
345μA 2
(2) = 690μA
2
(
)⎥⎤⎛⎜ 2μm ⎞⎟ = 345 μA
3.9 8.854x10−14 F / cm
W
(b) I D = C 2 (VGS − VTN )vsat =
80x10−7 cm
"
ox
⎥⎝ 0.1μm ⎠
⎦
V2
)⎛⎜ 2x10
cm ⎞
7
⎟(2V ) 10 cm / s = 86.3μA
2
⎠
−4
⎝
(
)
4.150
For VGS = 0, ID = 10-22 A. For VTN = 0.5 V and VGS = 0, ID = 10-15 A.
123
CHAPTER 5
5.1
Base Contact = B
n-type Emitter = D
Collector Contact = A
n-type Collector = F
Emitter Contact = C
Active Region = E
5.2
-
iC
v BC
iB
βR =
C
+
B
+
V
+
-
E
vBE
For VBE > 0 and VBC = 0, IC = β F I B or β F =
iE
-
IC 275μA
=
= 68.8
4μA
IB
0.5
αR
=
=1
1− α R 1− 0.5
⎛V ⎞
IC = I S exp⎜ BE ⎟ or I S =
⎝ VT ⎠
IC
275μA
=
= 2.10 fA
⎛ VBE ⎞
⎛ 0.64 ⎞
exp⎜
⎟ exp⎜
⎟
⎝ 0.025 ⎠
⎝ VT ⎠
5.3
i
vBE
i
B
-
E
E
+
B
V
+
-
+
v BC
C
-
iC
For VBC > 0 and VBE = 0, I E = −β R I B or β R = −
βF =
⎛ −275μA ⎞
IE
= −⎜
⎟ = 2.20
IB
⎝ 125μA ⎠
αF
0.975
=
= 39
1− α F 1− 0.975
⎛V ⎞
I E = −I S exp⎜ BC ⎟ or I S =
⎝ VT ⎠
IC
275μA
=
= 3.13 fA
⎛ VBE ⎞
⎛ 0.63 ⎞
exp⎜
⎟ exp⎜
⎟
⎝ 0.025 ⎠
⎝ VT ⎠
123
5.4
Using β =
α
β
and α =
:
1− α
β +1
Table 5.P1
0.167
0.200
0.400
0.667
0.750
3.00
0.909
10.0
0.980
49.0
0.995
200
0.999
1000
0.9998
5000
5.5
(a) For this circuit, VBE = 0 V, VBC = -5 V and I = IC. Substituting these values into the collector
current expression in Eq. (5.13):
⎡
⎛ −5 ⎞⎤ I S ⎡ ⎛ −5 ⎞ ⎤
IC = I S ⎢exp(0)− exp⎜
⎟⎥ − ⎢exp⎜
⎟ −1⎥
⎝ .025 ⎠⎦ β R ⎣ ⎝ .025 ⎠ ⎦
⎣
⎛
⎛ 1⎞
1⎞
I = IC = I S ⎜1+ ⎟ = 10−15 A⎜1+ ⎟ = 2 fA.
⎝ 1⎠
⎝ βR ⎠
(b) For this circuit, the constraints are VBC = -5 V and IE = 0. Substituting these values into the
emitter current expression in Eq. (5.13):
⎡ ⎛V ⎞
⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤
I E = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ −1⎥ = 0 which gives
⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎛V ⎞
⎛V ⎞
βF
1
+
exp⎜ BC ⎟. Substituting this result into IC :
exp⎜ BE ⎟ =
⎝ VT ⎠ 1+ β F 1+ β F
⎝ VT ⎠
⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤
I ⎡
ICBO = S ⎢1− exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ −1⎥.
1+ β F ⎣
⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎡ 1
⎡ 1 1⎤
1⎤
+ ⎥ = 10−15 A⎢
+ ⎥ = 1.01 fA, and
For VBC = -5V , ICBO = I S ⎢
⎣101 1⎦
⎣1+ β F β R ⎦
⎛ 1 ⎞
⎛ 1 ⎞
VBE = VT ln⎜
⎟ = 0.025V ln⎜
⎟ = −0.115 V ≠ 0!
⎝101⎠
⎝1+ β F ⎠
5.6
124
IB
+
150 μA
B
VBC
-
C
IC
(a) - (c)
(b) npn transistor
E
+
V BE
IE
(d) VBE = VBC
I S ⎡ ⎛ VBE ⎞ ⎤
⎢exp⎜
⎟ −1⎥
β R ⎣ ⎝ VT ⎠ ⎦
I ⎡ ⎛V ⎞ ⎤
I E = + S ⎢exp⎜ BE ⎟ −1⎥
β F ⎣ ⎝ VT ⎠ ⎦
⎛ 1
1 ⎞ ⎡ ⎛V ⎞ ⎤
IB = IS ⎜
+ ⎟ ⎢exp⎜ BE ⎟ −1⎥
⎝ β F β R ⎠ ⎣ ⎝ VT ⎠ ⎦
(e) I
IE
=
IB
C
=−
1
1+
and
βF
βR
IE
β
=− R
IC
βF
βF
I = − 400I E and I B = I E − IC = 401 I E
βR E
For the circuit I B = 150μA
150μA
Therefore I E =
= 0.374 μA, and IC = −149.6 μA.
(f ) Using
VBC = VBE
IC = −
401
⎛
⎞
⎛
⎞
⎜
⎟
⎜
⎟
IB
150μA
⎜
⎟
⎜
⎟ = 0.591 V
= (0.025V )ln
= VT ln
⎞
⎛
⎞
⎜ ⎛ 1
⎟
⎜
1
1 ⎟
1
+
+ ⎟⎟
⎟⎟
⎜ IS ⎜
⎜ 2 fA⎜
⎝ 100 0.25 ⎠ ⎠
⎝
⎝ ⎝ βF βR ⎠ ⎠
5.7
VBC
IB
- C IC
+
B
npn transistor
+
V BE
E
-
IE
⎛
1 ⎞⎡ ⎛ VBE ⎞ ⎤
IE
| IC = β F I B
For VBC = 0, I E = I S ⎜1+
⎟⎢exp⎜
⎟ −1⎥ | I B =
βF + 1
⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦
175μA
100
= 1.73 μA | IC =
175μA = 173 μA
101
101
⎛ β
⎛100 175μA ⎞
IE ⎞
= VT ln⎜ F
+ 1⎟ = 0.025V ln⎜
+ 1⎟ = 0.630 V
⎠
⎝ 101 2 fA
⎠
⎝ βF + 1 IS
I E = 175 μA | I B =
175 μA
VBE
125
5.8
- E IE
VBE
IB
npn transistor
+
B
+
V BC
C
-
IC
175 μA
⎛
1 ⎞⎡ ⎛ V ⎞ ⎤
I
For VBE = 0, IC = −I S ⎜1+ ⎟⎢exp⎜ BC ⎟ −1⎥ | I B = − C
| I E = −β R I B
βR + 1
⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦
⎛ −175μA ⎞
0.25
IC = −175 μA | I B = −⎜
175μA = −35 μA
⎟ = 140 μA | I E = −
1.25
⎝ 1.25 ⎠
⎛ β
⎞
⎛ 0.25 175μA ⎞
I
+ 1⎟ = 0.590 V
VBC = VT ln⎜− R C + 1⎟ = 0.025V ln⎜
⎝ 1.25 2 fA
⎠
⎝ βR + 1 IS
⎠
5.9
Using vBC = 0 in Eq. 5.13 and recognizing that i = iC + iB = iE :
⎛
1 ⎞⎡ ⎛ vBE ⎞ ⎤
i = iE = I S ⎜1 +
⎟⎢exp⎜
⎟ − 1⎥ , and the reverse saturation current
⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦
⎛
⎛
1 ⎞
1 ⎞
of the diode connected transistor is I S' = I S ⎜1 +
⎟ = (2 fA)⎜1 +
⎟ = 2.02 fA
⎝ 100 ⎠
⎝ βF ⎠
5.10
Using vBE = 0 in Eq. 5.13 and recognizing that i = −iC :
⎛
1 ⎞⎡ ⎛ v ⎞ ⎤
i = −iC = −I S ⎜1 + ⎟⎢exp⎜ BC ⎟ − 1⎥ , and the reverse saturation current
⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦
⎛
⎛ 1⎞
1⎞
of the diode connected transistor is I S' = I S ⎜1 + ⎟ = (5 fA)⎜1 + ⎟ = 6.67 fA
⎝ 3⎠
⎝ βR ⎠
5.11
⎡ ⎛v ⎞
⎡ ⎛ 0.75 ⎞
⎛ v ⎞⎤
⎛ −3 ⎞⎤
= I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ = 5x10−16 A⎢exp⎜
⎟ − exp⎜
⎟⎥ = 5.34 mA
⎝ 0.025 ⎠⎦
⎝ VT ⎠⎦
⎣ ⎝ 0.025 ⎠
⎣ ⎝ VT ⎠
(b) The current is symmetric: For VBC = 0.75 V and VBE = -3 V, iT = -5.34 mA.
(a) i
T
5.12
⎡ ⎛v ⎞
⎡ ⎛ 0.70 ⎞
⎛ vBC ⎞⎤
⎛ −3 ⎞⎤
−15
BE
a
i
=
I
exp
A
exp
=
10
−
exp
−
exp
⎢
⎥
⎢ ⎜
⎟
⎜
⎟⎥ = 1.45 mA
( ) T S ⎜⎝ V ⎟⎠ ⎜⎝ V ⎟⎠
⎝ 0.025 ⎠⎦
⎣ ⎝ 0.025 ⎠
⎣
T
T ⎦
(b) The current is symmetric: For VBC = 0.70 V and VBE = -3 V, iT = -1.45 mA.
5.13
Base Contact = F
p-type Emitter = D
5.14
126
Collector Contact = G
p-type Collector = A
Emitter Contact = E
Active Region = C
(a) pnp transistor
v
iB
EB
E
E
-
B
v
100 μA
i
+
C
CB
+
iC
(b)-(c)
(d) Using Eq. (5.17) with vEB = 0 and dropping the "-1" terms:
⎛
⎛v ⎞
⎛v ⎞
1 ⎞ ⎛v ⎞
I
iC = −I S ⎜1+ ⎟ exp⎜ CB ⎟
iE = −I S exp⎜ CB ⎟
iB = S exp⎜ CB ⎟
βR
⎝ β R ⎠ ⎝ VT ⎠
⎝ VT ⎠
⎝ VT ⎠
IE
βR
1
IE
=
=
= αR
= −β R
1
IC
βR + 1
IB
1+
βR
IC = −100μA, I E = α R IC = 0.25IC = −25.0μA
αR
I
0.25
1
IB = − E β R =
=
=
I B = +75μA
βR
1− α R 1− 0.25 3
α
0.985
βF = F =
= 65.7
1− α F 1− 0.985
⎛V ⎞
⎛ I ⎞
VEB = 0 and I E = −I S exp⎜ CB ⎟ VCB = VT ln⎜ − E ⎟
⎝ VT ⎠
⎝ Is ⎠
⎛ −25x10−6 A ⎞
VCB = 0.025V ln⎜ −
⎟ = 0.599 V
10−15 A ⎠
⎝
127
5.15
(a) pnp
VCB
IC
+
C
-
IB
B
V
+
V EB
E
+
IE
(b)-(c)
(d) Using Eq. (5.17) with vCB = 0 and droping the "-1" terms:
⎛
⎛v ⎞
1 ⎞ ⎛ vEB ⎞
iE = I S ⎜1 +
iC = −I S exp⎜ EB ⎟
⎟ exp⎜
⎟
⎝ β F ⎠ ⎝ VT ⎠
⎝ VT ⎠
IC
300μA
=
= 2.29 fA
IS =
⎛ VEB ⎞
⎛ 0.640 ⎞
exp⎜
⎟ exp⎜
⎟
⎝ 0.025V ⎠
⎝ VT ⎠
βF =
iB =
⎛v ⎞
exp⎜ EB ⎟
βF
⎝ VT ⎠
IS
αR
IC 300μA
0.2
=
= 75 | β R =
=
= 0.25
4μA
IB
1− α R 1− 0.2
5.16
Using VCB = 0 in Eq. 5.17 and recognizing that i = iE :
⎛
1 ⎞⎡ ⎛ vEB ⎞ ⎤
i = iE = I S ⎜1 +
⎟⎢exp⎜
⎟ − 1⎥ , and the reverse saturation current
⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦
⎛
⎛
1 ⎞
1 ⎞
of the diode connected transistor is I S' = I S ⎜1 +
⎟ = (2 fA)⎜1 +
⎟ = 2.02 fA
⎝ 100 ⎠
⎝ βF ⎠
5.17
v
CB
+
35 μA
i
C
C
B
i
B - v
E
EB
+
iE
(a)-(c)
(b) pnp transistor(d)
⎛ 1
I ⎡ ⎛v ⎞ ⎤
I ⎡ ⎛v ⎞ ⎤
1 ⎞⎡ ⎛ v ⎞ ⎤
+ ⎟⎢exp⎜ EB ⎟ −1⎥
vEB = vCB iC = − S ⎢exp⎜ EB ⎟ −1⎥ iE = + S ⎢exp⎜ EB ⎟ −1⎥ iB = + I S ⎜
β R ⎣ ⎝ VT ⎠ ⎦
β F ⎣ ⎝ VT ⎠ ⎦
⎝ β F β R ⎠⎣ ⎝ VT ⎠ ⎦
1
IE
βR
4
IE
β
4
βF
=
=
=
= 0.0506
= − R = − = −0.0533
1
1
IB
β F + β R 79
IB
βF
75
+
βF
128
βR
4
75
I B = 1.77 μA IC = − I E = −33.2 μA
79
4
⎛ 4 −33.2x10−6 A
⎛ β R IC ⎞
VCB = VEB = 0.025V ln⎜1−
= VT ln⎜1−
⎟
⎜
Is ⎠
2x10−15 A
⎝
⎝
I B = 35 μA
VEB
IE =
(
)⎟⎞ = 0.623 V
⎟
⎠
5.18
C IC
VCB +
IB
B
V EB
E
+
IE
pnp transistor
⎛
1 ⎞ ⎡ ⎛ VEB ⎞ ⎤
For VCB = 0, I E = I S ⎜1+
⎟ ⎢exp⎜
⎟ −1⎥
⎝ β F ⎠ ⎣ ⎝ VT ⎠ ⎦
| IB =
IE
| IC = β F I B
βF + 1
300μA
= 2.97 μA | IC = 100(2.97μA)= 297 μA
101
⎤
⎡⎛ β ⎞ I
⎡⎛ 100 ⎞ 300μA ⎤
= VT ln⎢⎜ F ⎟ E + 1⎥ = 0.025V ln⎢⎜
+ 1⎥( )= 0.626 V
⎟
⎦
⎣⎝ 101⎠ 4 fA
⎣⎝ β F + 1⎠ I S ⎦
I E = 300 μA | I B =
I
VEB
5.19
VEB
IB
+ E IE
B
V CB
I
C
+
IC
pnp transistor
⎛
1 ⎞⎡ ⎛ V ⎞ ⎤
I
| I E = −β R I B
For VEC = 0, IC = −I S ⎜1+ ⎟⎢exp⎜ CB ⎟ −1⎥ | I B = − C
βR + 1
⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦
⎛ −300μA ⎞
IC = −300 μA | I B = −⎜
⎟ = 150 μA | I E = −1(150μA)= −150 μA
⎝ 2 ⎠
⎡ ⎛ β ⎞I
⎤
⎡ 1 ⎛ −300μA ⎞ ⎤
VCB = VT ln⎢−⎜ R ⎟ C + 1⎥ = 0.025V ln⎢− ⎜
⎟ + 1⎥ = 0.603 V
⎣ 2 ⎝ 5 fA ⎠ ⎦
⎣ ⎝ β R + 1⎠ I S ⎦
5.20
⎡ ⎛v ⎞
⎡ ⎛ 0.70 ⎞
⎛ v ⎞⎤
⎛ −3 ⎞⎤
= I S ⎢exp⎜ EB ⎟ − exp⎜ CB ⎟⎥ = 5x10−16 A⎢exp⎜
⎟ − exp⎜
⎟⎥ = 723 μA
⎝ 0.025 ⎠⎦
⎝ VT ⎠⎦
⎣ ⎝ 0.025 ⎠
⎣ ⎝ VT ⎠
(b) The current is symmetric: For VCB = 0.75 V and VEB = -3 V, iT = -723 μA.
(a) i
T
129
5.21
⎡ ⎛v ⎞ ⎤
⎡ ⎛ 0.73V ⎞ ⎤
i
=
I
a
⎢
( ) F S exp⎜⎝ VBE ⎟⎠ −1⎥ = 4x10−15 A⎢exp⎜⎝ 0.025V ⎟⎠ −1⎥ = 19.2 mA
⎣
⎦
⎣
⎦
T
⎡ ⎛v ⎞ ⎤
⎡ ⎛ −3V ⎞ ⎤
iR = I S ⎢exp⎜ BC ⎟ −1⎥ = 4x10−15 A⎢exp⎜
⎟ −1⎥ = −4.00 fA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
19.2mA
i
−4.00 fA
= 240 μA | R =
= −2.00 μA
βF
βR
80
2
⎡ ⎛v ⎞ ⎤
⎡ ⎛ −3V ⎞ ⎤
(b) iF = I S ⎢exp⎜⎝ VBE ⎟⎠ −1⎥ = 4x10−15 A⎢⎣exp⎜⎝ 0.025V ⎟⎠ −1⎥⎦ = −4.00 fA
⎦
⎣
T
⎡ ⎛v ⎞ ⎤
⎡ ⎛ 0.73V ⎞ ⎤
iR = I S ⎢exp⎜ BC ⎟ −1⎥ = 4x10−15 A⎢exp⎜
⎟ −1⎥ = 19.2 mA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
iT = iF − iR = 19.2 mA |
iF
iT = iF − iR = −19,2 mA |
=
iF
βF
−4.00 fA
i
19.2mA
= −0.05 μA | R =
= 9.60 mA
βR
80
2
=
5.22
⎡ ⎛v ⎞ ⎤
⎡ ⎛ 0.68V ⎞ ⎤
= I S ⎢exp⎜ EB ⎟ −1⎥ = 6x10−15 A⎢exp⎜
⎟ −1⎥ = 3.90 mA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛ −3V ⎞ ⎤
iR = I S ⎢exp⎜ CB ⎟ −1⎥ = 6x10−15 A⎢exp⎜
⎟ −1⎥ = −6.00 fA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
(a) i
F
3.90mA
i
−6.00 fA
= 65.0 μA | R =
= −2.00 μA
βF
βR
60
3
⎡ ⎛v ⎞ ⎤
⎡ ⎛ −3V ⎞ ⎤
(b) iF = I S ⎢exp⎜⎝ VEB ⎟⎠ −1⎥ = 6x10−15 A⎢⎣exp⎜⎝ 0.025V ⎟⎠ −1⎥⎦ = −6.00 fA
⎣
⎦
T
⎡ ⎛v ⎞ ⎤
⎡ ⎛ 0.68V ⎞ ⎤
iR = I S ⎢exp⎜ CB ⎟ −1⎥ = 6x10−15 A⎢exp⎜
⎟ −1⎥ = 3.90 mA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
iT = iF − iR = 3.90 mA |
iT = iF − iR = −3.90 mA |
130
iF
iF
=
βF
=
−6.00 fA
i
3.90mA
= −0.100 fA | R =
= 1.30 mA
60
3
βR
5.23
⎡ ⎛v ⎞
⎡ ⎛v ⎞
⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤
⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
iE = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ −1⎥ = I S ⎢exp⎜ BE ⎟ −1− exp⎜ BC ⎟ + 1⎥ + S ⎢exp⎜ BE ⎟ −1⎥
⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦
⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎣ ⎝ VT ⎠
⎡ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
⎛
⎛ v BC ⎞ ⎤
1 ⎞⎡ ⎛ vBE ⎞
BC
S
BE
BC
iE = I S ⎜1+
exp
exp
exp
−1−
exp
+
1
−
I
−1
=
−1
−
I
⎢
⎥
⎢
⎥
⎢
⎥
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎟ −1⎥
S
S ⎢exp⎜
⎝ β F ⎠⎣ ⎝ VT ⎠
⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦ α F ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞
⎡ ⎛v ⎞
⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤
⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
iC = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ −1⎥ = I S ⎢exp⎜ BE ⎟ −1− exp⎜ BC ⎟ + 1⎥ − S ⎢exp⎜ BC ⎟ −1⎥
⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎣ ⎝ VT ⎠
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
⎛
1 ⎞⎡ ⎛ v ⎞ ⎤
iC = I S ⎢exp⎜ BE ⎟ −1⎥ − I S ⎜1+ ⎟⎢exp⎜ BC ⎟ −1⎥ = I S ⎢exp⎜ BE ⎟ −1⎥ − S ⎢exp⎜ BC ⎟ −1⎥
⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦ α R ⎣ ⎝ VT ⎠ ⎦
Defining I ES =
IS
αF
and ICS =
IS
αR
, then we see I S = α F I ES = α R ICS and
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iE = I ES ⎢exp⎜ BE ⎟ −1⎥ − α R I S ⎢exp⎜ BC ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iC = α F I ES ⎢exp⎜ BE ⎟ −1⎥ − ICS ⎢exp⎜ BC ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iB = iE − iC = (1− α F )I ES ⎢exp⎜ BE ⎟ −1⎥ + (1− α R )I S ⎢exp⎜ BC ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
5.24
αF =
βF
βR
100
0.5
I
2 fA
= 0.990 | α R =
=
= 0.333 | I ES = S =
= 2.02 fA
β F + 1 101
β R + 1 1.5
α F 0.990
ICS =
IS
αR
=
=
2 fA
= 6.00 fA | α F I ES = α R ICS = I S
0.333
131
5.25
⎡ ⎛v ⎞
⎡ ⎛v ⎞ ⎤
⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
⎛
1 ⎞⎡ ⎛ v EB ⎞ ⎤
CB
iE = I S ⎢exp⎜ EB ⎟ −1− exp⎜ CB ⎟ + 1⎥ + S ⎢exp⎜ EB ⎟ −1⎥ = I S ⎜1+
⎟⎢exp⎜
⎟ −1⎥ − I S ⎢exp⎜
⎟ −1⎥
⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦
⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞
⎡ ⎛v ⎞ ⎤
⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤
⎛
1 ⎞⎡ ⎛ v ⎞ ⎤
iC = I S ⎢exp⎜ EB ⎟ −1− exp⎜ CB ⎟ + 1⎥ − S ⎢exp⎜ CB ⎟ −1⎥ = I S ⎢exp⎜ EB ⎟ −1⎥ − I S ⎜1+ ⎟⎢exp⎜ CB ⎟ −1⎥
⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iE = I ES ⎢exp⎜ EB ⎟ −1⎥ − α R ICS ⎢exp⎜ CB ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iC = α F I ES ⎢exp⎜ EB ⎟ −1⎥ − ICS ⎢exp⎜ CB ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
⎡ ⎛v ⎞ ⎤
⎡ ⎛v ⎞ ⎤
iB = iE − iC = (1− α F )I ES ⎢exp⎜ EB ⎟ −1⎥ + (1− α R )ICS ⎢exp⎜ CB ⎟ −1⎥
⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
5.26
At IC = 5 mA and VCE = 5 V , I B = 60μA : β F =
IC 5mA
=
= 83.3
I B 60μA
At IC = 7 mA and VCE = 7.5 V , I B = 80μA : β F =
IC 7mA
=
= 87.5
I B 80μA
At IC = 10 mA and VCE = 14 V , I B = 100μA : β F =
IC 10mA
=
= 100
I B 100μA
5.27
See Problem 5.28
5.28
20mA
10mA
0A
-4mA
0V
2V
4V
6V
-I(VCC)
V_VCC
132
8V
10V
5.29
3.0mA
2.0mA
1.0mA
0A
-1.0mA
-2V
0V
2V
4V
6V
8V
10V
-I(VCB)
V_VCB
5.30
See Problem 5.31
5.31
20mA
10mA
0A
-4mA
0V
2V
4V
6V
8V
10V
-I(VEC)
V_VEC
133
5.32
3.0mA
2.0mA
0A
-1.0mA
-2V
0V
2V
4V
6V
8V
10V
-I(VBC)
V_VBC
5.33
The change in vBE for a decade change in iC is ΔVBE = VT ln (10) = 2.30VT .
⎛1.38x10-23 ⎞
kT
The reciprocal of the slope is 2.30VT = 2.30
= 2.30⎜
⎟T (V/dec)
-19
q
⎝1.60x10 ⎠
(a) 39.6 mV/dec (b) 49.5 mV/dec (c) 59.4 mV/dec (d) 69.3 mV/dec
5.34
(a) The break down voltage is equal to that of the emitter-base junction: VZ = 6 V. (b) The break
down voltage is determined by the base-collector junction: VZ = 50 V. (c) The break down
voltage is set by the emitter-base junction: VZ = 6 V.
5.35
(a) The base-emitter junction breaks down with VEB = 6.3 V.
5 − 6.3 − (−5) V
= 2.31 mA
IR =
Ω
1600
(b) The base-emitter junction is forward biased; VBE = 0.7 V
5 − 0.7 − (−5) V
IR =
= 388 μA
24000
Ω
(c) VBE = 0, and the collector-base junction is reversed biased with VBC ≈ -10V which is less
than the breakdown voltage of 75 V. The transistor is operating in cutoff.
⎛
I
1⎞
Using Eq. (5.13), I R = IC = I S (1− 0)− S (0 −1)= I S ⎜1+ ⎟ ≈ 0
βR
⎝ βR ⎠
5.36
VCE = VCB + VBE = VCB + 0.7 ≤ 65.7 V
134
5.37
(a) IB is forced to be negative by the current source, and the largest negative base current
according to the Transport model is
⎛ 1
⎛1
1⎞
1 ⎞
IB = −IS ⎜ + ⎟ = −10−15 A⎜ +
⎟ = −2.02 fA
⎝ 50 0.5 ⎠
⎝ βF βR ⎠
.
(b) IB is forced to be -1 mA by the current source. One or both of the junctions must enter the
breakdown region in order to supply this current. For the case of a normal BJT, the base-emitter
junction will break down and supply the current since it has the lower reverse breakdown
voltage.
5.38
Base-Emitter
Voltage
Base-Collector
Voltage
0.7 V
-5.0 V
-5.0 V
Reverse
Active
Cutoff
0.7 V
Saturation
Forward
Active
5.39
(a) vBE > 0, vBC = 0, forward-active region; vBE = 0, vBC > 0, reverse-active region; vBE > 0, vBC
= 0, forward-active region
(b) vEB < 0, vCB < 0, cutoff region
(c) vEB > 0, vCB < 0, forward-active region
(d) vBE > 0, vBC < 0, forward-active region; vBE > 0, vBC > 0, saturation region
5.40
(a) vBE = 0, vBC < 0 cutoff region
(b) vBC < 0, IE = 0, cutoff region
5.41
(a) vBE > 0, vBC > 0 saturation region
(b) vBE > 0, vBC = 0, forward-active region
(c) vBE = 0, vBC > 0, reverse-active region
135
5.42
Emitter-Base
Voltage
Collector-Base
Voltage
0.7 V
-0.65 V
0.7 V
Saturation
Forward Active
-0.65 V
Reverse Active
Cutoff
5.43
(a) vBE > 0, vBC = 0, forward-active region
(b) vBE = 0, vBC > 0, reverse-active region
5.44
(a) vEB = 0, vCB > 0, reverse-active region
(b) vEB > 0, vCB = 0, forward-active region
5.45
(a) vEB > 0, vCB > 0, saturation region
(b) vEB > 0, vCB = 0, forward-active region
(c) vEB = 0, vCB > 0, reverse-active region
5.46
(a ) pnp transistor with VEB = −3V and VCB = −3V → Cutoff | Using Eq. (5.17) :
10−15 A
I
10−15 A
= 0.5x10-15 = 0.5 fA | I E = − S =
= 13.3x10-18 = 13.3 aA
2
75
βR
βF
⎛ 1
⎛ 1 1⎞
1⎞
+ ⎟ = 10−15 A⎜ + ⎟ = 0.263x10−15 = 0.263 fA
I B = −I S ⎜
⎝ 75 4 ⎠
⎝ βF βR ⎠
IC = +
IS
=
(b) npn transistor with V
BE
= −5V and VBC = −5V → Cutoff | The currents are the same as
in part (a).
5.47
⎡ ⎛ 0.3 ⎞
⎛ −5 ⎞⎤ 10−16 ⎡ ⎛ −5 ⎞ ⎤
iC = 10−16 ⎢exp⎜
−
exp
⎢exp⎜
⎟
⎜
⎟⎥ −
⎟ −1⎥ = 16.3 pA
1 ⎣ ⎝ 0.025⎠ ⎦
⎝ 0.025 ⎠⎦
⎣ ⎝ 0.025 ⎠
⎡ ⎛ 0.3 ⎞
⎛ −5 ⎞⎤ 10−16 ⎡ ⎛ 0.3 ⎞ ⎤
iE = 10−16 ⎢exp⎜
⎢exp⎜
⎟ − exp⎜
⎟⎥ +
⎟ −1⎥ = 17.1 pA
⎝ 0.025 ⎠⎦ 19 ⎣ ⎝ 0.025⎠ ⎦
⎣ ⎝ 0.025 ⎠
10−16 ⎡ ⎛ 0.3 ⎞ ⎤ 10−16 ⎡ ⎛ −5 ⎞ ⎤
iB =
⎢exp⎜
⎢exp⎜
⎟ −1⎥ +
⎟ −1⎥ = 0.857 pA
19 ⎣ ⎝ 0.025 ⎠ ⎦
1 ⎣ ⎝ 0.025 ⎠ ⎦
136
These currents are all very small - for most practical purposes it still appears to be cutoff. Since
VBE > 0 and VBC < 0, the transistor is actually operating in the forward-active region. Note that
I C = β FI B .
5.48
An npn transistor with VBE = 0.7V and VBC = −0.7V → Forward - active region
Using Eq. (5.45) : I E = (β F + 1)I B | β F =
IE
10mA
−1 =
−1 = 65.7
0.15mA
IB
⎛
1 ⎞ ⎛V ⎞
0.01A
= 6.81x10−15 A = 6.81 fA
I E = I S ⎜1+ ⎟ exp⎜ BE ⎟ | I S =
⎛
⎞
⎛
⎞
1
0.7
⎝ β F ⎠ ⎝ VT ⎠
⎜1+
⎟ exp⎜
⎟
⎝ 65.7 ⎠ ⎝ 0.025⎠
5.49
A pnp transistor with VEB = 0.7V and VCB = −0.7V ⇒ Forward - active region
⎛V ⎞
I
2.5mA
2.5mA
Using Eq. (5.44) : β F = C =
= 62.5 | IC = I S exp⎜ EB ⎟ | I S =
= 1.73 fA
⎛ 0.7V ⎞
I B 0.04mA
⎝ VT ⎠
exp⎜
⎟
⎝ 0.025V ⎠
5.50
IE =
−0.7V − (−3.3V )
47kΩ
= 55.3μA | I B =
IE
55.3μA
=
= 0.683μA
81
βF + 1
IC = β F I B = 80(0.683μA)= 54.6μA | Check : I B + IC = I E is ok
5.51
(a) f β =
fT
βF
=
500 MHz
= 6.67 MHz
759
(b) The graph represents the Bode magnitude plot. Thus β (s) =
βF
1+
s
ωβ
=
βFωβ
ωT
=
s + ωβ s + ωβ
ωT
βF
β (s)
s + ωβ
βFωβ
ωT
αF
βF + 1
α (s)=
=
=
=
=
≈
ωT
s
s
s + ωT + ω β s + (β F + 1)ω β
β (s)+ 1
+1
1+
1+
s + ωβ
ωT
(β F + 1)ω β
α ( jω ) =
αF
⎛ ω ⎞2
1+ ⎜ ⎟
⎝ ωT ⎠
137
5.52
vEB > 0
vCB < −4VT
⎛V ⎞ I
⎛V ⎞
iC = I S exp⎜ EB ⎟ + S ≈ I S exp⎜ EB ⎟
⎝ VT ⎠ β R
⎝ VT ⎠
⎛V ⎞ I
⎛V ⎞ I
⎛V ⎞
iE = I S exp⎜ EB ⎟ + S exp⎜ EB ⎟ = S exp⎜ EB ⎟
⎝ VT ⎠ β F
⎝ VT ⎠ α F
⎝ VT ⎠
⎛V ⎞ I
⎛V ⎞
I
I
iB = S exp⎜ EB ⎟ − S ≈ S exp⎜ EB ⎟
βF
⎝ VT ⎠ β R β F
⎝ VT ⎠
iC = β F iB | iC = α F iE
iB
iC
C
B
-
+
i = β i
vEB
C
F B
0.7 V
i
E
E
5.53
An npn transistor with VBE = −0.7V and VBC = +0.7V → Reverse - active region
Using Eq. (5.51) : IC = −(β R + 1)I B | β R = −
⎛V ⎞
I E = −I S exp⎜ BC ⎟ | I E = −35μA | I S = −
⎝ VT ⎠
IC
−75μA
−1 = −
−1 = 0.875
40μA
IB
−35μA
= 2.42x10−17 A = 0.0242 fA = 24.2 aA
⎛ 0.7 ⎞
exp⎜
⎟
⎝ 0.025⎠
5.54
A pnp transistor with VEB = −0.7 V and VCB = +0.7 V → Reverse − active region
⎛V ⎞ I
⎛V ⎞
⎛V ⎞
I
iC = −I S exp⎜ CB ⎟ − S exp⎜ CB ⎟ = − S exp⎜ CB ⎟
αR
⎝ VT ⎠ β R
⎝ VT ⎠
⎝ VT ⎠
⎛V ⎞ I
⎛V ⎞
iE = −I S exp⎜ CB ⎟ − S ≅ −I S exp⎜ CB ⎟
⎝ VT ⎠ β F
⎝ VT ⎠
⎛V ⎞ I
⎛V ⎞
I
I
iB = − S + S exp⎜ CB ⎟ ≅ S exp⎜ CB ⎟
βF βR
⎝ VT ⎠ β R
⎝ VT ⎠
βR = −
5.55
IC = −
(−0.1mA) = 0.667 | I = −
iE
=−
S
iB
0.15mA
−0.7V − (−3.3V )
56kΩ
iE
−10−4 A
=−
= 6.91x10−17 A
⎛ VCB ⎞
⎛ 0.7 ⎞
exp⎜
exp⎜
⎟
⎟
⎝ 0.025⎠
⎝ VT ⎠
= −46.4 μA | I B = −
IC
−46.4μA
=−
= 26.5 μA
1.75
βR + 1
I E = IC + I B = −46.4μA + 26.5μA = −19.9 μA
138
5.56
⎡
⎤
β
1+ FOR ⎥
⎢
⎛ 1 ⎞ (β R + 1)⎥
βR
I
1mA
2
| αR =
β FOR = C =
= 1 | VCESAT = VT ln⎢⎜ ⎟
=
⎢⎝ α R ⎠
⎛β ⎞ ⎥
I B 1mA
βR + 1 3
1− ⎜ FOR ⎟ ⎥
⎢
⎝ βF ⎠ ⎦
⎣
⎡
1 ⎤
⎢ 1+
⎥
⎛ 3 ⎞ (2 + 1)⎥
⎢
= 17.8 mV
VCESAT = 0.025ln ⎜ ⎟
⎢⎝ 2 ⎠
⎛1⎞⎥
1− ⎜ ⎟ ⎥
⎢
⎝ 50 ⎠ ⎦
⎣
⎡
⎤
⎢
⎥
⎡ 1mA + (1− 0.667)1mA ⎤
I B + (1− α R )IC ⎥
⎢
⎥ = 0.724 V
= 0.025V )ln⎢ −15
VBE = VT ln
⎞⎥ (
⎢ ⎛ 1
⎢⎣10 A(0.02 + 1− .0.667)⎥⎦
+ 1− α R ⎟ ⎥
⎢ IS ⎜
⎠⎦
⎣ ⎝ βF
5.57
⎛v ⎞ I
⎛v ⎞
⎛v ⎞ I
⎛v ⎞
I
iC = I S exp⎜ EB ⎟ − S exp⎜ CB ⎟ | iB = S exp⎜ EB ⎟ + S exp⎜ CB ⎟ | Simultaneous
βF
⎝ VT ⎠ α R
⎝ VT ⎠
⎝ VT ⎠ β R
⎝ VT ⎠
i
iB − C
iB + (1− α R )iC
βF
solution yields : vEB = VT ln
| vCB = VT ln
⎤
⎤
⎡1
⎡ 1 ⎤⎡ 1
I S ⎢ + (1− α R )⎥
I S ⎢ ⎥⎢ + (1− α R )⎥
⎦
⎦
⎣β F
⎣α R ⎦⎣β F
⎡
⎤
iC
1+
⎢
⎥
⎛ 1 ⎞ (β R + 1) iB ⎥
i
⎢
vECSAT = vEB − vCB = VT ln ⎜ ⎟
for i B > C
⎢⎝ α R ⎠
⎥
i
βF
1− C
⎢
⎥
β F iB ⎦
⎣
5.58
(a) Substituting iC = 0 in Eq. 5.30 gives
⎛ 1 ⎞
⎛ 1 ⎞
VCESAT = VT ln⎜ ⎟ = (0.025V )ln⎜ ⎟ = 0.0173 V = 17.3 mV
⎝ 0.5⎠
⎝αR ⎠
(b) By symmetry
⎛ 1 ⎞
VECSAT = VT ln⎜ ⎟
⎝αF ⎠
or by using iE = 0 and iC = -iB,
139
VCESAT
βR
1
1−
⎛ 1 ⎞ β +1
⎛ 1 ⎞ β +1
⎛ 1 ⎞α
R
= VT ln⎜ ⎟
= VT ln⎜ ⎟ R
= VT ln⎜ ⎟ R
⎝ α R ⎠ 1+ 1
⎝ α R ⎠ βF + 1
⎝αR ⎠ 1
βF
βF
αF
⎛ 1 ⎞
VCESAT = VT ln (α F ) and VECSAT = VT ln⎜ ⎟
⎝αF ⎠
⎛ 1 ⎞
⎛ 1 ⎞
VECSAT = VT ln⎜ ⎟ = (0.025V )ln⎜
⎟ = 0.000251 V = 0.251 mV
⎝ 0.99 ⎠
⎝αF ⎠
5.59
(a) Substituting iC = 0 in Eq. 5.30 gives
⎛ 1 ⎞
⎛ 1 ⎞
VCESAT = VT ln⎜ ⎟ = (0.025V )ln⎜
⎟ = 27.7 mV
⎝ 0.33⎠
⎝αR ⎠
(b) By symmetry
⎛ 1 ⎞
⎛ 1 ⎞
VECSAT = VT ln⎜ ⎟ = (0.025V )ln⎜
⎟ = 1.28 mV
⎝ 0.95⎠
⎝αF ⎠
5.60
(a) VCESAT
⎡
⎤
β
1+ FOR ⎥
⎢
⎛ 1 ⎞ (β R + 1)⎥
0.9
βR
= VT ln⎢⎜ ⎟
=
= 0.4737
| αR =
⎢⎝ α R ⎠
⎛ β FOR ⎞ ⎥
β R + 1 0.9 + 1
1− ⎜
⎢
⎟⎥
⎝ βF ⎠ ⎦
⎣
⎡
β FOR ⎤
1+
⎢
⎥
⎛ 1 ⎞ (0.9 + 1)⎥
⎢
0.1 = 0.025ln ⎜
→ β FOR = 11.05
⎢⎝ 0.4737 ⎟⎠
⎛ β FOR ⎞ ⎥
1− ⎜
⎢
⎟⎥
⎝ 15 ⎠ ⎦
⎣
1+
0.4737exp(4) =
β FOR
(0.9 + 1) → β
⎛β ⎞
1− ⎜ FOR ⎟
⎝ 15 ⎠
FOR
= 11.05 | I B =
IC
β FOR
=
20 A
= 1.81A
11.05
⎡
β FOR ⎤
1+
⎢⎛
⎥
1 ⎞ (0.9 + 1) ⎥
I
20A
(b) 0.04 = 0.025ln⎢⎜
= 10.1A
→ β FOR = 1.97 | I B = C =
⎟
⎞
⎛
β FOR 1.97
⎢⎝ 0.4737 ⎠ 1− β FOR ⎥
⎟
⎜
⎢⎣
⎝ 15 ⎠ ⎥⎦
140
5.61
With VBE = 0.7 and VBC = 0.5, the transistor is technically in the saturation region, but
calculating the currents using the transport model in Eq. (5.13) yields
⎡ ⎛ 0.7 ⎞
⎛ 0.5 ⎞⎤ 10−16 ⎡ ⎛ 0.5 ⎞ ⎤
iC = 10−16 ⎢exp⎜
⎢exp⎜
⎟ − exp⎜
⎟⎥ −
⎟ −1⎥ = 144.5 μA
1 ⎣ ⎝ 0.025 ⎠ ⎦
⎝ 0.025 ⎠⎦
⎣ ⎝ 0.025 ⎠
⎡ ⎛ 0.7 ⎞
⎛ 0.5 ⎞⎤ 10−16 ⎡ ⎛ 0.7 ⎞ ⎤
−16
iE = 10 ⎢exp⎜
⎢exp⎜
⎟ − exp⎜
⎟⎥ +
⎟ −1⎥ = 148.3 μA
⎝ 0.025 ⎠⎦ 39 ⎣ ⎝ 0.025⎠ ⎦
⎣ ⎝ 0.025 ⎠
iB =
10−16 ⎡ ⎛ 0.7 ⎞ ⎤ 10−16 ⎡ ⎛ 0.5 ⎞ ⎤
⎢exp⎜
⎢exp⎜
⎟ −1⎥ +
⎟ −1⎥ = 3.757 μA
39 ⎣ ⎝ 0.025⎠ ⎦
1 ⎣ ⎝ 0.025 ⎠ ⎦
At 0.5 V, the collector-base junction is not heavily forward biased compared to the base-emitter
junction, and IC = 38.5IB ≅ β F IB . The transistor still acts as if it is operating in the forwardactive region.
5.62
(a) The current source will forward bias the base - emitter junction (VBE ≅ 0.7V ) and
the collector - base junction will then be reverse biased (VBC ≅ −2.3V ). Therefore, the
npn transistor is in the forward - active region.
⎛ 50 175x10−6 A ⎞
⎛ VBE ⎞
⎟ = 0.803 V
IC = β F I B = I S exp⎜
⎟ | VBE = 0.025ln⎜
−16
⎜
⎟
V
10
A
⎝ T ⎠
⎝
⎠
(b) Since I B = 175μA and IC = 0, IC < β F I B , and the transistor is saturated.
(
)
175x10−6 + 0
Using Eq. (5.53) : VBE = 0.025ln
= 0.714 V | Using Eq. (5.54) with iC = 0,
⎡
⎛
.5 ⎞⎤
−16 1
10 ⎢ + ⎜1− ⎟⎥
⎣ 50 ⎝ 1.5 ⎠⎦
⎛ 1 ⎞
⎛ β + 1⎞
⎛ 1.5 ⎞
VCESAT = 0.025ln⎜ ⎟ = 0.025ln⎜ R ⎟ = 0.025ln⎜ ⎟ = 27.5 mV
⎝ 0.5 ⎠
⎝αR ⎠
⎝ βR ⎠
141
5.63
⎡ ⎛v ⎞
⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤
iC = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BC ⎟ −1⎥
⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
⎡ ⎛v ⎞
⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤
iE = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ −1⎥
⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
vBE > 4VT and vBC < −4VT
⎡ ⎛ v ⎞⎤
⎛
1 ⎞⎡ ⎛ vBE ⎞⎤ I S ⎡ ⎛ v BE ⎞⎤
iC ≅ I S ⎢exp⎜ BE ⎟⎥ and iE = I S ⎜1+
⎢exp⎜
⎟⎢exp⎜
⎟⎥ =
⎟⎥ → iC ≅ α F iE
⎝ β F ⎠⎣ ⎝ VT ⎠⎦ α F ⎣ ⎝ VT ⎠⎦
⎣ ⎝ VT ⎠⎦
⎛i ⎞
⎛α i ⎞
vBE ≅ VT ln⎜ C ⎟ = VT ln⎜ F E ⎟
⎝ IS ⎠
⎝ IS ⎠
5.64
I SD =
IS
αF
=
1 fA
= 1.02 fA
0.98
5.65
Both transistors are in the forward - active region. For simplicity, assume VA = ∞.
I = IC1 + I B1 + I B 2 | Since the transistors are identical and have the same VBE ,
IC 2 = IC 1 and I B1 = I B 2 | I = IC1 + 2I B1 = (β F + 2)I B1 | IC 2 = β F I B 2 = β F I B1
IC 2 =
βF
βF + 2
I=
25
25μA | IC 2 = 23.2 μA | See the Current Mirror in Chapter 15.
25 + 2
5.66
CD =
IC
50x10−12
τF =
IC = 2x10−9 IC (F) (a) 4 fF (b) 0.4 pF (c) 40 pF
VT
0.025
5.67
Using Fig. 2.8 with N =
1018
cm2
cm2
,
μ
=
260
and
μ
=
100
n
p
v-s
v-s
cm3
W2
WB2
(a) npn : τ F = B =
=
2Dn 2VT μn
W2
WB2
=
(b) pnp : τ F = B =
2D p 2VT μ p
142
(1x10
−4
cm
)
2
⎛
cm2 ⎞
2(0.025V )⎜260
⎟
v - s⎠
⎝
(1x10
−4
cm
)
= 0.769 ns
2
⎛
cm2 ⎞
2(0.025V )⎜100
⎟
v - s⎠
⎝
= 2.00 ns
5.68
For f >> f β , f T = β ⋅ f = 10(75MHz)= 750 MHz | f β =
fT
βF
=
750 MHz
= 3.75 MHz
200
5.69
f T 900 MHz
f
900 MHz
=
= 180 | For f >> 5 MHz, β (f ) = T =
= 18
5MHz
50 MHz
fβ
f
βF =
5.70
6x1018
cm2
cm2
cm2
→
μ
=
130
using
Fig.
2.8.
D
=
μ
V
=
130
0.025V
=
3.25
(
)
n
n
n T
v−s
v−s
s
cm3
2 ⎞⎛
20 ⎞
⎛
cm 10
1.60x10−19 C 25x10−8 cm 2 ⎜3.25
⎟⎜
⎟
2
s ⎠⎝ cm6 ⎠
qADn ni
⎝
IS =
=
= 5.42x10−20 A
18
N AWB
6x10
0.4x10−4 cm
cm3
NA =
(
)
(
)
5.71
WB = 2Dnτ F | τ F ≤
18
1
1
=
= 31.8 ps
2πf 2π 5x109
(
)
2
5x10
cm
→ μn = 135
using Fig. 2.8.
3
v−s
cm
cm 2
cm2
0.025V
=
3.38
Dn = μnVT = 135
(
)
v−s
s
⎛
cm2 ⎞
−12
WB ≤ 2⎜ 3.38
⎟31.8x10 s = 0.147 μm
s ⎠
⎝
NA =
5.72
⎛ V ⎞
⎛
⎛ 10 ⎞ 265μA
5 ⎞ 240μA
and β FO ⎜1+ ⎟ =
IC = β F I B = β FO ⎜1+ CE ⎟ I B | β FO ⎜1+ ⎟ =
3μA
3μA
⎝ VA ⎠
⎝ VA ⎠
⎝ VA ⎠
⎛ 10 ⎞
⎜1+ ⎟
80
⎝ VA ⎠ 265μA
=
⇒ VA = 43.1 V | β FO =
= 71.7
⎛
⎛
5 ⎞
5 ⎞ 240μA
⎜1+ ⎟
⎜1+
⎟
⎝ 43.1⎠
⎝ VA ⎠
5.73
143
⎡ ⎛ V ⎞ ⎤⎛ V ⎞
⎡ ⎛ 0.72V ⎞ ⎤⎛ 10V ⎞
( a ) IC = I S ⎢exp⎜ BE ⎟ −1⎥⎜1+ CE ⎟ = 10−16 A⎢exp⎜
⎟ −1⎥⎜1+
⎟ = 371 μA
⎣ ⎝ 0.025V ⎠ ⎦⎝ 65V ⎠
⎣ ⎝ VT ⎠ ⎦⎝ VA ⎠
⎡ ⎛V ⎞ ⎤
⎡ ⎛ 0.72V ⎞ ⎤
( b ) IC = I S ⎢exp⎜ BE ⎟ −1⎥ = 10−16 A⎢exp⎜
⎟ −1⎥ = 322 μA
⎣ ⎝ 0.025V ⎠ ⎦
⎣ ⎝ VT ⎠ ⎦
( c ) 1.15 :1 (a) is 15% larger than (b) due to the Early effect.
5.74
⎛ V ⎞
IC = β F I B = β FO ⎜1+ CE ⎟ I B | We need two Q - points from the output characteristics.
⎝ VA ⎠
For example : (10 mA, 14 V) and
(5 mA, 5 V)
⎛ 14 ⎞
⎛
5⎞
10mA = β FO ⎜1+ ⎟0.1mA and 5mA = β FO ⎜1+ ⎟0.06mA yields
⎝ VA ⎠
⎝ VA ⎠
⎛ 14 ⎞
⎛
5⎞
100 = β FO ⎜1+ ⎟ and 83.3 = β FO ⎜1+ ⎟. Solving these two equations yields
⎝ VA ⎠
⎝ VA ⎠
β FO = 72.9 and VA = 37.6 V.
5.75
⎡ ⎛ V ⎞ ⎤
⎡ ⎛ V +V ⎞ ⎤ I
⎛V ⎞
BE
Fig. 5.16(a) : I E = IC + I B = ⎢β FO ⎜1+ CE ⎟ + 1⎥ I B ≅ ⎢β FO ⎜1+ CB
⎟ + 1⎥ s exp⎜ BE ⎟
VA
⎠ ⎦ β FO
⎝ VT ⎠
⎣ ⎝ VA ⎠ ⎦
⎣ ⎝
⎛ V ⎞
⎡ 5 + VBE 1 ⎤
+ ⎥ 5 x10−15 exp⎜ BE ⎟ = 100μA → VBE = 0.589V by iteration
⎢1+
19 ⎦
50
⎣
⎝ 0.025 ⎠
⎛ 5.589 ⎞
100μA
= 4.52 μA | IC = 19⎜1+
IB =
⎟ I B = 95.48 μA
⎡ ⎛ 5.589 ⎞ ⎤
50 ⎠
⎝
⎢19⎜1+
⎟ + 1⎥
50 ⎠ ⎦
⎣ ⎝
⎛V ⎞
100μA
For VA = ∞, I E = I s exp⎜ BE ⎟ | VBE = 0.025ln
= 0.593 V
5 fA
⎝ VT ⎠
(
)
⎛V ⎞
19(100μA)
exp⎜ BE ⎟ → VBE = 0.025ln
= 0.667 V
β FO
5 fA
⎝ VT ⎠
⎛ V ⎞
⎛
5⎞
IC = β FO ⎜1+ CE ⎟ I B = 19⎜1+ ⎟100μA = 2.09 mA | I E = IC + I B = 2.19 mA
⎝ 50 ⎠
⎝ VA ⎠
Fig. 5.16(b) : I B =
Is
VBE is independent of VA in the equation above.
5.76
144
⎛ V ⎞
⎛ 9 + 0.7 ⎞
IC = β F I B | I E = (β F + 1)I B | β F = β FO ⎜1+ CE ⎟ = 50⎜1+
⎟ = 59.7
50 ⎠
⎝
⎝ VA ⎠
IE =
(9 − 0.7)V = 1.01 mA |
8200Ω
5.77
I
gm = C
VT
IE
1.01mA
=
= 16.7 μA | IC = 59.7 I B = 0.996 mA
βF + 1
60.7
(
)
−23
kT 1.38x10 J / K (300K )
| VT =
=
= 25.9 mV
q
1.60x10−19 C
m =
10−5 A
10−4 A
= 0.387 mS (b) g m =
= 3.87 mS
VT
VT
m =
10−3 A
10−2 A
= 38.7 mS (d ) g m =
= 0.387 S
VT
VT
(a) g
(c) g
IB =
(e) The values of g
m
are the same for the pnp.
5.78
IC
10x10−12
CD = τ F =
IC = 3.88x10−10 IC (F) (a) 0.388 fF (b) 0.388 pF (c) 3.88 pF
VT
0.0258
5.79
The following are from the Cadence website and the file psrefman.pdf:
IS = 10fA, BF = 100, BR = 1, VAF = ∞, VAR = ∞, TF = 0, TR = 0,
NF = 1, NE = 1.5, RB = 0, RC = 0, RE = 0, ISE = 0, ISC = 0, ISS = 0,
IKF = ∞, IKR = ∞, CJE = 0, CJC = 0.
These default values apply to both npn and pnp transistors.
5.80
⎡
⎛ i ⎞⎤
⎛ 1mA ⎞
1+ ⎢1+ 4⎜ F ⎟⎥
1+ 1+ 4⎜
⎟
⎝ IKF ⎠⎦
⎝10mA ⎠
⎣
=
= 1.09 → 8.3% reduction
(a) KBQ =
2
2
⎛ 10mA ⎞
1+ 1+ 4⎜
⎟
⎝ 10mA ⎠
i
= 1.62 | iC = F = 0.62iF → 38% reduction
(a) KBQ =
2
1.62
⎛ 50mA ⎞
1+ 1+ 4⎜
⎟
⎝ 10mA ⎠
i
= 2.79 | iC = F = 0.36iF → 64% reduction
(a) KBQ =
2
2.79
NK
5.81
145
12.000
10.000
8.000
6.000
Series1
4.000
2.000
0.000
0.1
1
10
100
1000
Collector Current
5.82
36kΩ
10V = 3.462V | REQ = 36kΩ 68kΩ = 23.54kΩ
36kΩ + 68kΩ
3.462 − 0.7
V
IB =
= 1.618μA | IC = 50I B = 80.9 μA | I E = 51I B = 82.5 μA
23.54 + (50 + 1)33 kΩ
(a) VEQ =
VCE = 10 − 43000IC − 33000I E = 3.797V | Q - point : (80.9 μA,3.80 V)
7.2kΩ
10V = 3.462V | REQ = 7.2kΩ 13.6kΩ = 4.708kΩ
7.2kΩ + 13.6kΩ
3.462 − 0.7
V
IB =
= 8.092μA | IC = 50I B = 404.6μA | I E = 51I B = 412.7 μA
4.708 + (50 + 1)6.6 kΩ
(b) VEQ =
VCE = 10 − 8600IC − 6600I E = 3.7976V | Q - point : (405 μA,3.80 V)
68kΩ
10V = 6.538V | REQ = 36kΩ 68kΩ = 23.54kΩ
36kΩ + 68kΩ
10 − 0.7 − 6.538 V
= 1.618μA | IC = 50I B = 80.9 μA | I E = 51I B = 82.5 μA
IB =
23.54 + (50 + 1)33 kΩ
(c) VEQ =
VEC = 10 − 33000IC − 43000I E = 3.797V | Q - point : (80.9 μA,3.80 V)
13.6kΩ
10V = 6.538V | REQ = 7.2kΩ 13.6kΩ = 4.708kΩ
7.2kΩ + 13.6kΩ
10 − 0.7 − 6.538 V
= 8.092μA | IC = 50I B = 404.6μA | I E = 51I B = 412.7 μA
IB =
4.708 + (50 + 1)6.6 kΩ
(b) VEQ =
VEC = 10 − 6600IC − 8600I E = 3.7976V | Q - point : (405 μA,3.80 V)
5.83
146
36kΩ
10V = 3.462V | REQ = 36kΩ 68kΩ = 23.54kΩ
36kΩ + 68kΩ
V
3.462 − 0.7
= 1.629μA | IC = 75I B = 122.2μA | I E = 76I B = 123.8μA
IB =
23.54 + (75 + 1)22 kΩ
(a) VEQ =
VCE = 10 − 43000IC − 22000I E = 2.022V | Q - point : (122μA,2.02V)
68kΩ
10V = 6.538V | REQ = 36kΩ 68kΩ = 23.54kΩ
36kΩ + 68kΩ
10 − 0.7 − 6.538 V
= 1.629μA | IC = 75I B = 122.2μA | I E = 76I B = 123.8μA
IB =
23.54 + (75 + 1)22 kΩ
(b) VEQ =
VEC = 10 − 22000IC − 43000I E = 2.022V | Q - point : (122μA,2.02V)
5.84
*Problem 5.83(a)
VCC 1 0 10
R1 3 0 36K
R2 1 3 68K
RC 1 2 43K
RE 4 0 33K
Q1 2 3 4 NPN
.MODEL NPN NPN IS=1E-16 BF=50 BR=0.25
.OP
.END
*Problem 5.83(b)
VCC 1 0 10
R1 3 0 36K
R2 1 3 68K
RC 1 2 43K
RE 4 0 33K
Q1 2 3 4 NPN
.MODEL NPN NPN IS=1E-16 BF=50 BR=0.25 VAF=60
.OP
.END
*Problem 5.83(c)
VCC 1 0 10
R1 1 3 36K
R2 3 0 68K
RC 4 0 43K
RE 1 2 33K
Q1 4 3 2 PNP
.MODEL PNP PNP IS=1E-16 BF=50 BR=0.25
.OP
.END
*Problem 5.83(d)
VCC 1 0 10
147
R1 1 3 36K
R2 3 0 68K
RC 4 0 43K
RE 1 2 33K
Q1 4 3 2 PNP
.MODEL PNP PNP IS=1E-16 BF=50 BR=0.25 VAF=60
.OP
.END
5.85
6.2 kΩ
= 3.41V and REQ = 6.2 kΩ 12 kΩ = 4.09 kΩ
6.2kΩ + 12 kΩ
3.41− 0.7
IC = β F I B = 100
= 0.356 mA.
4090 + 101(7500)
VEQ = 10
101
0.356 mA(7.5kΩ)= 5.49V
100
Q − po int : (0.356 mA, 5.49 V )
VCE = 10 − 0.356 mA(5.1kΩ)−
5.86
120 kΩ
= 5.00V and REQ = 120 kΩ 240 kΩ = 80 kΩ
120kΩ + 240 kΩ
5.00 − 0.700
IC = β F I B = 100
= 42.2μA.
80000 + 101(100000)
VEQ = 15
(
)
VCE = 15 − 42.2 x10−6 A 105 Ω −
Q − po int : (42.2 μA, 4.39 V )
5.87
(a) I
E
=
(
)
101
42.2 x10−6 A 1.5 x105 Ω = 4.39V
100
⎛ 101⎞
2V
=⎜
= 1.98kΩ → 2.0 kΩ
⎟1mA = 1.01mA | RE =
1.01mA
α F ⎝ 100 ⎠
IC
RC =
(12 − 5 − 2)V = 5kΩ → 5.1 kΩ
R2 =
(12 − 2.7)V =
| VB = 2 + 0.7 = 2.7V
1.00mA
V
2.7V
= 27kΩ → 27 kΩ
Set R1 = B =
10I B 10(0.01mA)
148
11I B
9.3V
= 84.55kΩ → 82 kΩ
11(0.01mA)
27kΩ
12V = 2.972V | REQ = 27kΩ 82kΩ = 20.31kΩ
27kΩ + 82kΩ
V
2.972 − 0.7
= 10.22μA | IC = 100I B = 1.022mA | I E = 101I B = 1.033mA
IB =
20.31+ (100 + 1)2 kΩ
(b) VEQ =
VCE = 12 − 5100IC − 2000I E = 4.723V | Q - po int : (1.02mA,4.72V )
5.88
⎛ 76 ⎞
= ⎜ ⎟10μA = 10.13μA | Let VRC = VR E = VCE = 6V
α F ⎝ 75 ⎠
6V
= 592kΩ → 620 kΩ
RE =
10.13μA
6V
= 600kΩ → 620 kΩ | VB = 6 + 0.7 = 6.7V
RC =
10μA
V
6.7V
= 5.03MΩ → 5.1 MΩ
Set R1 = B =
⎛
10I B
10μA ⎞
10⎜
⎟
⎝ 75 ⎠
(a) I
E
IC
=
(18 − 6.7)V =
11.3V
= 7.71MΩ → 7.5 MΩ
⎛10μA ⎞
11I B
11⎜
⎟
⎝ 75 ⎠
5.1MΩ
18V = 7.286V | REQ = 5.1MΩ 7.5MΩ = 3.036 MΩ
(b) VEQ =
5.1MΩ + 7.5MΩ
7.286 − 0.7
V
= 0.1313μA | IC = 75I B = 9.848μA | I E = 76I B = 9.980μA
IB =
3036 + (75 + 1)620 kΩ
R2 =
VCE = 18 − 620000IC − 620000I E = 5.707V | Q - point : (9.85 μA, 5.71 V)
5.89
(a) I
RC =
E
(5 − 2 −1)V = 2.35kΩ → 2.4 kΩ
850μA
Set R1 =
R2 =
⎛ 61 ⎞
1V
= ⎜ ⎟850μA = 864.2μA | RE =
= 1.16kΩ →1.2 kΩ
864.2μA
α F ⎝ 60 ⎠
IC
=
VR1
10I B
VR 2
11I B
=
=
| VB = 5 −1− 0.7 = 3.3V
5 − 3.3V
= 12.0kΩ →12 kΩ
⎛ 850μA ⎞
10⎜
⎟
⎝ 60 ⎠
3.3V
= 21.2kΩ → 22 kΩ
⎛ 850μA ⎞
11⎜
⎟
⎝ 60 ⎠
149
22kΩ
5V = 3.24V | REQ = 12kΩ 22kΩ = 7.77kΩ
12kΩ + 22kΩ
5 − 0.7 − 3.24 V
= 13.1μA | IC = 60I B = 786μA | I E = 61I B = 799μA
IB =
7.77 + (60 + 1)1.2 kΩ
(b) VEQ =
VCE = 5 −1200I E − 2400IC = 2.14V | Q - point : (786 μA, 2.14 V)
5.90
11mA
= 0.220mA
50
⎛ 51 ⎞
I
1V
I E = C = ⎜ ⎟11mA = 11.22mA | RE =
= 89.1Ω → 91 Ω
α F ⎝ 50 ⎠
11.22mA
(a ) V
RE
= 1 V ,VRC = 9 V | I B =
9V
= 818Ω → 820 Ω | VB = −15 + 1+ 0.7 = −13.3V
11.0mA
−13.3V − (−15V )
V
Set R1 = R1 =
= 772Ω → 750 Ω
10I B
10(0.220mA)
RC =
R2 =
0 − (13.3V )
11I B
=
13.3V
= 5.50kΩ → 5.6 kΩ
11(0.220mA)
5.6kΩ
(−15V )= −13.2V | REQ = 0.75kΩ 5.6kΩ = 0.661kΩ
0.75kΩ + 5.6kΩ
−13.2 − 0.7 − (−15) V
= 0.207mA | IC = 50I B = 10.3mA | I E = 51I B = 10.6mA
IB =
661+ (50 + 1)(91) Ω
(b) VEQ =
VEC = 0 − 820IC − 91I E − (−15V ) = 5.59V | Q - point : (10.2 mA, 5.59 V)
5.91 Problem numbers on graph
3.3kΩ
VEQ =
10V = 3.056V | REQ = 7.5kΩ 3.3kΩ = 2.292kΩ
3.3kΩ + 7.5kΩ
VCE = 10 − 820IC −1200I E | From characteristics at VCE = 5V : β F ≅
5mA
= 83
60μA
84
IC = 10 - 2034IC
83
Load line points : IC = 0, VCE = 10V and VCE = 0, IC = 4.9mA
VCE = 10 − 820IC −1200
IB =
3.056 − 0.7
= 23μA | From Graph : Q - point : (1.9 mA, 6.0 V)
2292 + (83 + 1)1200
150
10mA
IB = 100
C
o
l
l
e
c
t
o
r 5mA
Q-Point
Prob. 5.92
IB = 92 μA
μA
I = 80 μA
B
IB = 60 μA
C
u
r
r
e
n
t
I = 40 μA
B
IB = 23 μA
IB = 20 μA
Q-Point
Prob. 5.91
0A
0V
5V
VCE
10V
15V
5.92
VEQ =
6.8kΩ
(10)= 6.538V | REQ = 6.8kΩ 3.6kΩ = 2.354kΩ
6.8kΩ + 3.6kΩ
VEC = 10 − 420IC − 330I E | From characteristics at VEC = 5V : β F ≅
VEC = 10 − 420IC − 3300
5mA
= 83
60μA
84
IC = 10 − 754IC
83
Load line points : IC = 0, VEC = 10V and VEC = 0, IC = 13.3mA − off the graph
VEC = 5V , IC = 6.63mA | I B =
10 − 0.7 − 6.538
= 92μA
2354 + (83 + 1)330
From Graph : Q - point : (7.5 mA, 4.3 V)
151
5.93
Writing a loop equation starting at the 9 V supply gives: 9 = 1500(IC + IB ) + 10000IB + VBE
Assuming forward-active region operation, VBE = 0.7 V and IC = βFIB.
9 = 1500(β F I B + I B )+ 10000I B + 0.7
β F (9 − 0.7)
9 − 0.7
and IC = β F I B =
1500(β F + 1)+ 1000
1500(β F + 1)+ 1000
IB =
(a ) I
C
(b) I
(c) I
=
C
=
C
=
(d ) I
C
30(9 − 0.7)V
1.5kW(30 + 1)+ 10kW
100(9 − 0.7)V
1.5kΩ(100 + 1)+ 10kΩ
250(9 − 0.7)V
1.5kΩ(250 + 1)+ 10kΩ
=
(9 − 0.7)V = 5.53 mA
1500Ω
= 4.41 mA | VCE = 9 −1500I E = 2.17V | Q - pt : (4.41mA,2.17V)
= 5.14 mA | VCE = 9 −1500I E = 1.21V | Q - pt : (5.14mA,1.21V)
= 5.37 mA | VCE = 9 −1500I E = 0.913V | Q - pt : (5.37mA,0.913V)
| VCE = 9 −1500I E = 0.705V | Q - pt : (5.53mA,0.705V)
5.94
⎛
I ⎞
V − 0.7
VCE = 9 − (IC + I B )1500 | VCE = 9 − ⎜ IC + C ⎟1500 | I B = CE 4
βF ⎠
10
⎝
5mA
From Fig. P5.26 at 5V : β F =
= 83.3 | VCE = 9 −1518IC
60μA
IC = 0, VCE = 9V | VCE = 0, IC = 5.93mA
VCE = 0.9V , I B = 20μA | VCE = 1.3V , I B = 60μA | VCE = 1.7V , I B = 100μA
From graph : Q - point = (5.0 mA, 1.3 V)
10mA
IB = 100
I = 100
B
μA
μA
VCE = 1.7 V
I = 80 μA
B
Collector Current
IB = 80 μA
VCE = 1.5 V
5mA
IB = 60 μA
Q-Point
IB = 60 μA
VCE = 1.3 V
IB = 40 μA
IB = 40 μA
V CE = 1.1 V
IB = 20 μA
I = 20 μA
B
V CE = 0.9 V
0A
0V
5V
V
CE
152
10V
15V
5.95
(a) VEC = 10 − (IC + I B )RC = 10 − I E RC | I E =
RC =
(10 − 3)V = 689Ω → 680 Ω
10.17mA
| RB =
IC
αF
=
61
βF + 1
IC = 10mA = 10.17mA
60
βF
VEC − VEB (3 − 0.7)V
=
= 13.8kΩ →14 kΩ
0.1667mA
IB
(b) 5 − 0.7 −14000I B − 680(IC + I B )− (−5)= 0
IB =
10 − 0.7
V
= 222.1μA | IC = β F I B = 8.88 mA
14000 + 41(680) Ω
VEC = 10V − (8.88mA)680Ω = 3.96 V | Q - po int : (8.88 mA, 3.96 V )
5.96
VCE = 1.5 − (IC + I B )RC → RC =
RB =
1.5 − 0.9
= 29.4kΩ → 30 kΩ
20μA
20μA +
50
VCE − VBE 0.9 − 0.65
=
= 625kΩ → 620 kΩ
20μA
IB
50
For RC = 30kΩ : VCE = 1.5 − 30kΩ(IC + I B )RC = 1.5 − 30kΩ(126)I B | I B =
VCE − 0.65
620kΩ
V − 0.65
VCE = 1.5 − 30kΩ(126) CE
→ VCE = 0.770V
620kΩ
0.770 − 0.65
IC = 125I B = 125
= 24.2μA | Q - po int : (24.2 μA, 0.770 V )
620kΩ
5.97
12 = RC (IC + I B )+ VZ + VBE = 500(I E )+ 7.7 | I E =
IB =
12 − 7.7
= 8.60mA
500
IE
8.60mA
=
= 85.2μA | IC = β F I B = 8.52mA | VCE = 7.70V
101
βF + 1
Q - point = (8.52 mA, 7.70 V)
5.95
15− 6
= 6.114V | REQ = 100Ω 7800Ω = 98.73Ω
7800+ 100
20mA
VO
20mA 6.14 − 98.7I B − VBE
101.1− VBE
+
=
+
→ I C = 50I B = 50
IB =
51
51
51(4700Ω)
51(4700Ω)
2.398x105
VEQ = 6+ 100
VBE = 0.025ln
IC
10−16
153
Using MATLAB: fzero('IC107',.02) ---> ans =0.0207
function f=IC107(ic)
vbe=0.025*log(ic/1e-16);
f=ic-50*(101.1-vbe)/2.398e5;
VO = 6.14 − 98.7
20.7mA
20.7mA
− .025ln
= 5.276 V
51
10−16
5.99
*Problem 5.98
VCC 1 0 DC 15
R1 1 2 7.8K
RZ 2 4 100
VZ 4 0 DC 6
Q1 1 2 3 NPN
RE 3 0 4.7K
IL 3 0 20MA
.MODEL NPN NPN IS=1E-16 BF=50 BR=0.25
.OP
.END
Output voltages will differ slightly due to different value of VT.
5.100
vO = 7 −100iB − vBE = 7 −100iB − VT ln
vO = 7 −100iB − VT ln iL − VT ln
iC
α i
= 7 −100iB − VT ln F L
IS
IS
αF
IS
⎛
dv
di
V ⎞ 100Ω 0.025V
Ro = − O = −⎜ −100Ω B − T ⎟ =
+
= 3.21Ω
51
0.02 A
diL
diL iL ⎠
⎝
5.101
Since the voltage across the op - amp input must be zero, vO = VZ = 10 V.
Since the input current to the op amp is zero, I E =
I +15 = I Z + IC = I Z + α F I E =
154
VO
= 100 mA
100
15V −10V 60
+ 100mA = 98.5 mA
47kΩ
61
5.102
47Ω
= VZ
47Ω + 47Ω
and vO = 10 V. Since the input current to the op amp is zero,
I
10V ⎛ 41⎞
15V − 5V
+ 109mA = 109 mA
IE = C =
⎜ ⎟ = 109 mA I +15 = I Z + I E =
α F 94Ω ⎝ 40 ⎠
82kΩ
Since the voltage across the op - amp input must be zero, vO
5.103
IC = β F I B = β F
VEQ − VBE
REQ + (β F + 1)RE
| For ICmin : VCC = 0.95(15) = 14.25 V
R1 = 0.95(82kΩ)= 77.9kΩ | R2 = 1.05(120kΩ)= 126kΩ | RE = 1.05(6.8kΩ)= 7.14kΩ
77.9
14.25V = 5.44V | REQ = 77.9kΩ 126kΩ = 48.1kΩ
77.9 + 126
5.44V − 0.7V
= 100
= 616 μA
48.1kΩ + (101)7.14kΩ
VEQ =
ICmin
[
]
VCEmax = 14.25 − ICmin 0.95(6.8kΩ) − I Emin 7.14kΩ
VCEmax = 14.25 − 3.98 − 4.44 = 5.83V | Q - po int : (616 μA, 5.83 V )
For ICmax : VCC = 1.05(15) = 15.75 V
R1 = 1.05(82kΩ) = 86.1kΩ | R2 = 0.95(120kΩ)= 114kΩ | RE = 0.95(6.8kΩ)= 6.46kΩ
86.1
15.75V = 6.78V | REQ = 86.1kΩ 114kΩ = 49.0kΩ
86.1+ 114
6.78V − 0.7V
= 100
= 867μA
49.0kΩ + (101)6.46kΩ
VEQ =
ICmax
[
]
VCEmin = 15.75 − ICmax 1.05(6.8kΩ) − I Emax 6.46kΩ
VCEmin = 15.75 − 6.19 − 5.66V = 3.90V | Q - po int : (867 μA, 3.90 V )
155
5.104
Using the Spreadsheet approach in Fig. 5.40, Eq. set (5.66) becomes:
(
)
(
)
= 6800 (1+ 0.1 (RAND() − 0.5))
1.
VCC = 15 1+ 0.1 (RAND() − 0.5)
3.
R2 = 120000 1+ 0.1 (RAND() − 0.5) | 4.
RE
5.
RC = 6800 1+ 0.1 (RAND() − 0.5)
β F = 100
(
(
|
)
)
| 6.
500 Cases
IC (A)
VCE (V)
Average
Std. Dev.
Min
Max
7.69E-04
4.02E-05
6.62E-04
8.93E-04
4.51
0.40
3.31
5.55
5.105
IC = β F I B = β F
VEQ − VBE
REQ + (β F + 1)RE
2.
R1 = 82000 1+ 0.1 (RAND() − 0.5)
| For ICmin : VCC = 0.95(12) = 11.4 V
R1 = 0.95(18kΩ) = 17.1kΩ | R2 = 1.05(36kΩ) = 37.8kΩ | RE = 1.05(16kΩ)= 16.8kΩ
17.1
11.4V = 3.55V | REQ = 17.1kΩ 37.8kΩ = 11.8kΩ
17.1+ 37.8
3.55V − 0.7V
= 50
= 164 μA
11.8kΩ + (51)16.8kΩ
VEQ =
ICmin
[
]
VCEmax = 11.4 − ICmin 0.95(22kΩ) − I Emin16.8kΩ
VCEmax = 11.4 − 3.43 − 2.81 = 5.16V | Q - po int : (164 μA, 5.16 V )
For ICmax : VCC = 1.05(12) = 12.6 V
R1 = 1.05(18kΩ) = 18.9kΩ | R2 = 0.95(36kΩ)= 34.2kΩ | RE = 0.95(16kΩ)= 15.2kΩ
18.9
12.6V = 4.49V | REQ = 18.9kΩ 34.2kΩ = 12.2kΩ
18.9 + 34.2
4.49V − 0.7V
= 150
= 246μA
12.2kΩ + (151)15.2kΩ
VEQ =
ICmax
[
]
VCEmin = 12.6 − ICmax 1.05(22kΩ) − I Emax 15.2kΩ
VCEmin = 12.6 − 5.68 − 3.77V = 3.15V | Q - po int : (246 μA, 3.15 V )
500 Cases
IC (A)
VCE (V)
Average
Std. Dev.
Min
Max
156
2.02E-04
1.14E-05
1.71E-04
2.35E-04
4.26
0.32
3.43
5.21
The averages are close to the hand calculations that go with Fig. 5.35. The minimum and
maximum values fall within the worst-case analysis as we expect.
5.106
Using the Spreadsheet approach with zero tolerance on the current gain, Eq. set (5.66) becomes:
(
)
1.
VCC = 12 1+ 0.0 (RAND() − 0.5)
|
3.
R2 = 36000 1+ 0.1 (RAND() − 0.5)
| 4.
5.
RC
(
)
= 22000 (1+ 0.1 (RAND() − 0.5))
500 Cases
IC (A)
VCE (V)
Average
Std. Dev.
Min
Max
2.03E-04
1.10E-05
1.74E-04
2.36E-04
4.29
0.32
3.46
5.27
(
)
R = 16000 (1+ 0.1 (RAND() − 0.5))
β = 100 (1+ 0.0(RAND() − 0.5))
R1 = 18000 1+ 0.1 (RAND() − 0.5)
2.
E
| 6.
F
Note that the current gain tolerance has little effect on the results.
5.107
(a) Approximately 22 cases fall outside the interval [170μA,250μA]: 100%
(b) Approximately 125 cases fall inside the interval [3.2V,4.8V]: 100%
22
= 4.4% fail
500
125
= 25% fail
500
5.108
Using the Spreadsheet approach with 50% tolerance on the current gain, a tolerance TP on VCC,
and a tolerance TR on resistor values, Eq. set (5.66) becomes:
1.
VCC = 12 * (1+ 2 * TP * (RAND() − 0.5))
3.
R2 = 36000 * (1+ 2 * TR * (RAND() − 0.5)) |
| 2.
4.
R1 = 18000 * (1+ 2 * TR * (RAND() − 0.5))
RE = 16000 * (1+ 2 * TR * (RAND() − 0.5))
5. RC = 22000 * (1+ 2 TR (RAND() − 0.5)) | 6. β F = 100 * (1+ 1* (RAND() − 0.5))
10,000 case Monte Carlo runs indicate that the specifications cannot be achieved even with ideal
resistors. For TP = 5% and TR = 0%, 18 % of the circuits fail. With TP = 2% and TR = 0%,
1.5% percent fail. The specifications can be met with TP = 1% and TR = 1%.
157
5.109
IC = β F I B = β F
VEQ − VBE
REQ + (β F + 1)RE
| For ICmin : VCC = 0.95(12) = 11.4 V
R1 = 0.8(18kΩ) = 14.4kΩ | R2 = 1.2(36kΩ)= 43.2kΩ | RE = 1.2(16kΩ)= 19.2kΩ
14.4
11.4V = 2.85V | REQ = 14.4kΩ 43.2kΩ = 10.8kΩ
14.4 + 43.2
2.85V − 0.7V
= 50
= 109 μA
10.8kΩ + (51)19.2kΩ
VEQ =
ICmin
[
]
VCEmax = 11.4 − ICmin 0.8(22kΩ) − I Emin19.2kΩ
VCEmax = 11.4 −1.91− 2.13 = 7.36V | Q - point : (109 mA, 7.36 V)
For ICmax : VCC = 1.05(12) = 12.6 V
R1 = 1.2(18kΩ)= 21.6kΩ | R2 = 0.80(36kΩ)= 28.8kΩ | RE = 0.80(16kΩ)= 12.8kΩ
21.6
12.6V = 5.40V | REQ = 21.6kΩ 28.8kΩ = 12.3kΩ
21.6 + 28.8
5.4V − 0.7V
= 150
= 362μA
12.3kΩ + (151)12.8kΩ
VEQ =
ICmax
[
]
VCEmin = 12.6 − ICmax 1.2(22kΩ) − I Emax 12.8kΩ
VCEmin = 12.6 − 9.57 − 4.67V = −1.64V! Saturated!
The forward - active region assumption is violated. See the next problem.
Based upon a Monte Carlo analysis, only about 1% of the circuits actually
have this problem, although VCE will be relatively small in many circuits.
158
5.110
Using the Spreadsheet approach:
1. VCC = 12 * (1+ .1* (RAND() − 0.5))
| 2.
3.
R2 = 36000 * (1+ 0.4 * (RAND() − 0.5)) | 4.
5.
RC = 22000 * (1+ 0.4 * (RAND() − 0.5))
7.
VA = 75 * (1+ 0.66 * (RAND() − 0.5))
R1 = 18000 * (1+ 0.4 * (RAND() − 0.5))
| 6.
RE = 16000 * (1+ 0.4 * (RAND() − 0.5))
β F = 100 * (1+ 1* (RAND() − 0.5))
In order to avoid an iterative solution at each step, assume that VCE does not influence the base
current. Then,
IB =
VEQ − 0.7
and VCE
REQ + (β FO + 1)RE
⎛
R ⎞
VCC − β FO IB ⎜ RC + E ⎟
⎛ V ⎞
αF ⎠
⎝
=
| IC = β FO IB ⎜1+ CE ⎟
⎛
β
R ⎞
⎝ VA ⎠
1+ FO IB ⎜ RC + E ⎟
VA ⎝
αF ⎠
500 Cases
VCE (V)
IC (A)
Average
Std. Dev.
Min**
Max
3.81E+00
1.26E+00
-2.07E-01
6.94E+00
2.049E-04
3.785E-05
1.264E-04
3.229E-04
**Note: In this particular simulation, there were 4 cases in which the transistor was saturated.
159
CHAPTER 6
6.1
(a ) Pavg =
1W
10-5W / gate
= 10 μW / gate (b) I =
= 4 μA / gate
105 gates
2.5V
6.2
(a) Pavg =
100
5x10-6W / gate
=
5
μ
W
/
gate
(b)
I
=
= 2 μA/ gate
2.5V
2x10 7 gates
(c) I total = 2
μA
(2x10 gates)= 40 A
gate
7
6.3
⎛ 2.5 − 0 ⎞ 5
(a ) VH = 2.5 V | VL = 0 V | PVH = I R = 0 mW | PVL = ⎜
⎟ 10 = 62.5 μW
5
⎝ 10 ⎠
2
2
⎛ 3.3 − 0 ⎞ 5
(b) VH = 3.3 V | VL = 0 V | PVH = I R = 0 mW | PVL = ⎜
⎟ 10 = 109 μW
5
⎝ 10 ⎠
2
2
6.4
vO
VH (3.3 V)
vI
V (0V)
L
1.1 V
(V REF)
3.3V
V+
6.5
v
O
V H (3.3 V)
vI
V L (0V)
1.1 V
(V REF)
3.3V
V+
()
Z= A =A
6-1
6.6
V REF
vI
AV
6.7
V H = 3 V | VL = 0 V | VIH = 2 V | VIL = 1 V | AV =
dvO −3V
=
= −3
dv I
1V
6.8
V (3 V)
H
v
O
Slope = +9
1.5 V
v
V (0V)
I
6.9
VOH = 5 V
1.5 V
1.67 V
1.33 V
L
3V
V
+
VIH = VREF = 2 V
NM H = 5 − 2 = 3 V
VOL = 0 V
VIL = VREF = 2 V
NM L = 2 − 0 = 2 V
6.10
We would like to achieve the highest possible noise margins for both states and have them be
symmetrical. Therefore VREF = 3.3/2=1.65 V.
6.11
V H = 3.3 V | VL = 0 V | VIH = 1.8 V | VOL ≅ 0.25 V | VIL = 1.5 V | VIH ≅ 3.0 V
NM H = 3.0 −1.8 = 1.2 V | NM L = 1.5 − 0.25 = 1.25 V
6.12
VH = 2.5 V | VL = 0.20 V
6-2
6.13
V H = −0.80 V | VL = −1.35 V
6.14
VIH = VOH − NM H = −0.8 − 0.5 = −1.3 V | VIL = NM L + VOL = 0.5 + (−2) = −1.5 V
6.15
-13
-4
-9
τP = PDP/P = 10 J/10 W = 10 s = 1 ns
6.16
4 x10-6W / gate
1W
=
μ
= 1.60 μA / gate
4
W
/
gate
(
b
)
I
=
2.5V
2.5 x105 gates
(c) PDP = 2ns (4 μW ) = 8 fJ
(a ) Pavg =
6.17
1μW / gate
100W
= 1 μW / gate (b) I =
= 0.4 μA / gate
8
2.5V
10 gates
(c) PDP = 1ns (1μW ) = 1 fJ
(a) Pavg =
6.18
PDP
250
100
Slope = 2
10
Slope = 1
1
P
1
10
50
100
6-3
6.19
dv (t )
dvc (t )
| v(t ) = RC c + vC (t ) | v(t ) = 1 for t ≥ 0
dt
dt
⎛ t ⎞
⎛ t ⎞
v(t ) = 1 − exp⎜ −
⎟ | 0.9 = 1 − exp⎜ − 90% ⎟ → t90% = − RC ln (0.1)
⎝ RC ⎠
⎝ RC ⎠
⎛ t ⎞
0.1 = 1 − exp⎜ − 10% ⎟ → t10% = − RC ln (0.9 ) | tr = t90% − t10% = RC ln (9) = 2.20 RC
⎝ RC ⎠
⎛ t ⎞
⎛ t ⎞
(b) v(t ) = 0 vC (0 ) = 1 v(t ) = exp⎜ −
⎟ | 0.9 = exp⎜ − 90% ⎟ → t90% = − RC ln (0.9)
⎝ RC ⎠
⎝ RC ⎠
⎛ t ⎞
0.1 = exp⎜ − 10% ⎟ → t10% = − RC ln (0.1) | t f = t10% − t90% = RC ln (9 ) = 2.20 RC
⎝ RC ⎠
(a ) v(t ) = i (t )R + vC (t ) | i (t ) = C
6.20
(a) VH = 2.5V | VL = 0.20V
(b) V10% = VL + 0.1ΔV = 0.20 + 0.23 = 0.43V → t10% ≅ 23 ns for vO
V90% = VL + 0.9ΔV = 0.20 + 2.07 = 2.27V → t90% ≅ 33 ns for vO → t r = 33 − 23 = 10 ns
For fall time : t10% ≅ 2.5 ns for vO t90% ≅ 0.8 ns for vO → t f = 1.7 ns
For v I , t10% ≅ 0 ns t90% ≅ 1 ns t r = 1 ns | t f ≅ 1 ns
(c) τ
PHL
≅ 1.5ns − 0.5ns = 1 ns | τ PLH ≅ 26ns − 21ns = 5 ns (d ) τ P =
1+ 5
ns = 3 ns
2
6.21
(a) VH = −0.78V | VL = −1.36V
(b) V10% = VL + 0.1ΔV = −1.36 + 0.1(0.58) = −1.30V → t10% ≅ 32.5 ns for vO
V90% = VL + 0.9ΔV = −1.36 + 0.9(0.58) = −0.84V → t90% ≅ 42 ns for vO
tr = 42 − 32.5 = 9.5 ns
For fall time : t10% ≅ 11.5 ns for vO t90% ≅ 2 ns for vO → t f = 9.5 ns
For vI , t10% ≅ 0 ns t90% ≅ 1 ns tr = 1 ns | t f ≅ 1 ns
(c ) V50% = − 0.78 − 1.36 = −1.07V | τ PHL ≅ 4 ns | τ PLH
2
6-4
≅ 4 ns (d ) τ P =
4+4
ns = 4 ns
2
6.22
(A + B)(A + C)
AA + AC + BA + BC
A + AC + BA + BC
A(1 + C) + AB + BC
A + AB + BC
A(1+ B) + BC
A + BC
6.23
Z = ABC + ABC + ABC
Z = ABC + + ABC + ABC + ABC
(
) (
)
Z = AB C + C + A + A BC
Z = AB(1) + (1)BC
Z = AB + BC
6.24
A B C Z
0
0
0
0
0
1
0
0
0
0
1
1
1
0
0
1
0
0
1
0
1
1
1
0
1
1
1
0
1
0
1
1
Z = AB+BC
6-5
6.25
Z = ABC + ABC + ABC + ABC
(
)
Z = C (AB + AB + AB + AB)
Z = C (A(B + B)+ A(B + B))
Z = C (A(1) + A(1))
Z = C (A + A)
Z = C AB + AB + AB + AB
Z = C (1)
Z =C
6.26
A B C
0 0 0
Z
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
0
0
1 1 1
Z =C
1
6-6
6.27
A B C D Z
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 0
1 1 1 0 0
1 1 1 1 1
6.28
A B C Z1
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
6.29
(a) Fanout = 2
Z = AB + CD
⎛ ⎞⎛ ⎞
Z = ⎜ AB⎟⎜CD⎟
⎝ ⎠⎝ ⎠
Z = ABCD
Z2
1
1
0
1
0
1
1
1
Z1 = AB = AB
Z 2 = AB + C
(b) Fanout = 1
6-7
6.30
i(t)
0.132 A
For each line : i = C
dv
dt
3.3V
= 132mA = 0.132 A
10− 9 s
For all 64 lines, I = 64(0.132 A) = 8.45 A!
i = 40 x10−12 F
1 ns
t
0
6.31
i(t)
1.32 A
For each line : i = C
3.3V
= 1.32 A
10−10 s
For all 64 lines, I = 64(1.32 A) = 84.5 A!
i = 40 x10−12 F
0.1 ns
t
0
dv
dt
6.32
F ⎞⎛ 7.5mm 0.1cm ⎞
⎛
3.9⎜ 8.854 x10 −14
⎟(1.5μm )
⎟⎜
⎛ ε ox A ⎞
3.9ε o LW
cm ⎠⎝ 2
mm ⎠
⎝
⎟⎟ = 3
= 0.583 pF
C = 3⎜⎜
=3
1μm
tox
⎝ tox ⎠
6.33
CΔV
KCox" WLΔV
| Let W* = αW and L* = αL
ΔT =
=
1
I
⎛W ⎞
μ nCox" ⎜ ⎟(VGS − VTN )2
2
⎝L⎠
C *ΔV *
K (αW )(αL )(αΔV )
ΔT =
=
= α ΔT
*
1 ⎛ αW ⎞
I
2
μn ⎜
⎟(αVGS − αVTN )
2 ⎝ αL ⎠
*
V
ε ⎛W
⎞
2
⎟(VGS − VTN ) = μ n ox ⎜
2 Tox ⎝ L
⎠
αV
ε ⎛ αW ⎞
2
2
P* =
μn ox ⎜
⎟(αVGS − αVTN ) = α P
2
αTox ⎝ αL ⎠
P = VI =
V
⎛W
μ nCox" ⎜
2
⎝L
PDP* = P*ΔT * = (αΔT )α 2 P = α 3 PΔT = α 3 PDP
Power density =
6-8
P
P
P*
α 2P
P
=
| *=
=
A WL A
αW (αL ) A
⎞
2
⎟(VGS − VTN )
⎠
6.34
1
1 ε ⎛W ⎞
⎛W ⎞
μnCox" ⎜ ⎟(VGS − VTN )2 = μ n ox ⎜ ⎟(VGS − VTN )2
2
2 Tox ⎝ L ⎠
⎝L⎠
⎛W ⎞
⎜ ⎟
ε
1
2
I D* = μn ox ⎜ 2 ⎟(VGS − VTN ) = 2 I D
T
L
2
ox ⎜
⎜ ⎟⎟
2 ⎝ 2 ⎠
(b) P* = V (2 I ) = 2VI = 2 P - The power has increased by a factor of two.
ε
ε W L CG
(c) CG = Cox" WL = ox WL | CG* = ox
=
Tox 2 2
Tox
2
2
The capacitance has decreased by a factor of two.
(a) I D =
⎛ W ⎞⎛ L ⎞
K ⎜ ⎟⎜ ⎟(ΔV )
ΔT
C ΔV
⎝ 2 ⎠⎝ 2 ⎠
=
(d ) ΔT * =
=
*
4
I
⎛W ⎞
1 ⎜ 2 ⎟
2
μ n ⎜ ⎟(VGS − VTN )
2 ⎜ L ⎟
⎜ ⎟
⎝ 2 ⎠
*
*
6.35
( a ) Pavg =
6.36
(a) Pavg =
(
1W
6
0.5 2 x10 gates
)
= 1 μW / gate ( b ) I =
1μW / gate
= 0.556 μA / gate
1.8V
20W
1.5μW / gate
= 1.5 μW / gate (b) I =
= 0.833 μA/ gate
⎛2⎞
1.8V
6
⎜ ⎟20x10 gates
⎝ 3⎠
6.37
V
2.5 - 0.2
50 μW
= 20 μA | Let VL = TN = 0.2V | R =
= 115 kΩ
2x10-5
2.5V
3
0.926
1
0.2 ⎞
⎛W ⎞
⎛W ⎞ ⎛
=
M S is in the triode region : 20x10-6 = 60x10-6 ⎜ ⎟ ⎜ 2.5 − 0.6 −
⎟0.2 → ⎜ ⎟ =
1
1.08
2 ⎠
⎝ L ⎠S
⎝ L ⎠S ⎝
I DD =
6-9
6.38
(a ) For MS off, I D = 0 and VH = 2.5V.
⎛
⎛ 3⎞⎛ μA ⎞
2.5 − VL
V ⎞
μA
= Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 180 2
200kΩ
2⎠
V
⎝
⎝ 1 ⎠⎝ V ⎠
⎛
V ⎞
μA ⎞⎛
2.5 − VL = 2x105 ⎜180 2 ⎟⎜2.5 − 0.6 − L ⎟VL → 36VL2 −138.8VL + 5 = 0
2⎠
V ⎠⎝
⎝
2.5 − 0.0364
= 12.3 μA | P = 2.5V (12.3 μA) = 30.8 μW
VL = 0.0364 V | I D =
200kΩ
0.0364 ⎞
μA ⎛
Checking : I D = 180 2 ⎜ 2.5 − 0.6 −
⎟0.0364 = 12.3 μA
2 ⎠
V ⎝
(b) For MS off, I D = 0 and VH = 2.5V.
For VL , I D =
(
)
⎛
⎛ 6 ⎞⎛ μA ⎞
2.5 − VL
V ⎞
μA
= Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 360 2
400kΩ
2⎠
V
⎝
⎝ 1 ⎠⎝ V ⎠
⎛
V ⎞
μA ⎞⎛
2.5 − VL = 4x105 ⎜ 360 2 ⎟⎜2.5 − 0.6 − L ⎟VL →144VL2 − 549.2VL + 5 = 0
2⎠
V ⎠⎝
⎝
2.5 − 0.00913
= 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW
VL = 9.13 mV | I D =
400kΩ
0.00913⎞
μA ⎛
Checking : I D = 360 2 ⎜ 2.5 − 0.6 −
⎟0.00913 = 6.21 μA
2 ⎠
V ⎝
For VL , I D =
(
)
6.39
(a ) For MS off, I D = 0 and VH = 2.5V.
⎛
⎛ 3⎞⎛ μA ⎞
2.5 − VL
V ⎞
μA
= Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 180 2
200kΩ
2⎠
V
⎝
⎝ 1 ⎠⎝ V ⎠
⎛
V ⎞
μA ⎞⎛
2.5 − VL = 2x105 ⎜180 2 ⎟⎜2.5 − 0.8 − L ⎟VL → 36VL2 −124.4VL + 5 = 0
2⎠
V ⎠⎝
⎝
2.5 − 0.0407
= 12.3 μA | P = 2.5V (12.3 μA) = 30.7 μW
VL = 0.0407 V | I D =
200kΩ
0.0407 ⎞
μA ⎛
Checking : I D = 180 2 ⎜ 2.5 − 0.8 −
⎟0.0407 = 12.3 μA
2 ⎠
V ⎝
For VL , I D =
(
)
⎛
μA ⎞⎛
(b) 2.5 − V = (2x10 )⎜⎝180 V
5
L
VL ⎞
2
2.5
−
0.4
−
⎟
⎜
⎟VL → 36VL −153.2VL + 5 = 0
2
2⎠
⎠⎝
2.5 − 0.0329
= 12.3 μA | P = 2.5V (12.3 μA) = 30.8 μW
200kΩ
0.0329 ⎞
μA ⎛
Checking : I D = 180 2 ⎜ 2.5 − 0.4 −
⎟0.0329 = 12.3 μA
2 ⎠
V ⎝
VL = 0.0329 V | I D =
6-10
6.40
(a) V
IL
= VTN +
1
1
1
= 0.6V +
= 0.6 +
= 0.627 V
Kn R
36
3 ⎛ μA ⎞
⎜ 60 ⎟(200kΩ)
1⎝ V 2 ⎠
VOH = VDD −
1
1
2VDD
5
= 2.5 −
= 2.49V | VOL =
=
= 0.215V
3Kn R
2Kn R
72
108
VIH = VTN −
1
V
1
2.5
+ 1.63 DD = 0.6 − + 1.63
= 1.00V
Kn R
36
36
Kn R
NM L = 0.627 − 0.215 = 0.412 V | NM H = 2.49 −1.00 = 1.49 V
1
1
1
= 0.6 +
= 0.607 V
(b) VIL = VTN + K R = 0.6V + 6 ⎛ μA⎞
144
n
⎜ 60 ⎟(400kΩ)
1⎝ V2 ⎠
VOH = VDD −
1
1
2VDD
5
= 2.5 −
= 2.50V | VOL =
=
= 0.108V
3Kn R
2Kn R
288
432
VIH = VTN −
1
V
1
2.5
+ 1.63 DD = 0.6 −
+ 1.63
= 0.807V
Kn R
144
144
Kn R
NM L = 0.607 − 0.108 = 0.499 V | NM H = 2.50 − 0.807 = 1.69 V
6.41
(a ) For MS off, I D = 0 and VH = 2.5V.
⎛
⎛ 6 ⎞⎛ μA ⎞
2.5 − VL
V ⎞
μA
= Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 360 2
400kΩ
2⎠
V
⎝
⎝ 1 ⎠⎝ V ⎠
⎛
V ⎞
μA ⎞⎛
2.5 − VL = 4x105 ⎜ 360 2 ⎟⎜2.5 − 0.6 − L ⎟VL →144VL2 − 549.2VL + 5 = 0
2⎠
V ⎠⎝
⎝
2.5 − 0.00913
= 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW
VL = 9.13 mV | I D =
400kΩ
0.00913⎞
μA ⎛
Checking : I D = 360 2 ⎜ 2.5 − 0.6 −
⎟0.00913 = 6.23 μA
2 ⎠
V ⎝
⎛
V ⎞
(b) 2.5 − VL = 144⎜⎝2.5 − 0.5 − 2L ⎟⎠VL →144VL2 − 578VL + 5 = 0
2.5 − 0.00867
= 6.33 μA | P = 2.5V (6.33 μA)= 15.8 μW
VL = 8.67 mV | I D =
400kΩ
μA ⎛
0.00867 ⎞
Checking : I D = 360 2 ⎜ 2.5 − 0.5 −
⎟0.00867 = 6.23 μA
2 ⎠
V ⎝
For VL , I D =
(
)
6-11
(c) 2.5 − V
L
⎛
V ⎞
= 144⎜2.5 − 0.7 − L ⎟VL →144VL2 − 520.4VL + 5 = 0
2⎠
⎝
2.5 − 0.00963
= 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW
400kΩ
μA ⎛
0.00963 ⎞
Checking : I D = 360 2 ⎜ 2.5 − 0.7 −
⎟0.00963 = 6.22 μA
2 ⎠
V ⎝
VL = 9.63 mV | I D =
In this design, we see that VL is not sensitive to VTN.
6.42
(a) V
IL
= VTN +
1
1
1
= 0.6V +
= 0.6 +
= 0.607V
Kn R
144
6 ⎛ μA ⎞
⎜ 60 ⎟(400kΩ)
1⎝ V2 ⎠
VOH = VDD −
1
1
2VDD
5
= 2.5 −
= 2.50V | VOL =
=
= 0.108V
2Kn R
288
432
3Kn R
VIH = VTN −
1
V
1
2.5
+ 1.63 DD = 0.6 −
+ 1.63
= 0.807V
Kn R
144
144
Kn R
NM L = 0.607 − 0.108 = 0.499 V | NM H = 2.50 − 0.807 = 1.69 V
(b) V
IL
= 0.6 +
1
5
| VOL =
| NM L = VIL-VOL = 0
Kn R
3Kn R
Solving for Kn R yields no solution. Zero noise margin will not occur.
6.43
(a) I
D
=
P
0.25mW
V − VL 2.5 − 0.5
=
= 100μA | R = DD
=
= 20.0 kΩ
ID
VDD
2.5V
1x10-4
Using the values corresponding to Fig. 6.12, K 'p = 100μA/V 2
⎛W ⎞ ⎛
⎛ W ⎞ 1.21
0.5 ⎞
100μA = 100x10-6 ⎜ ⎟ ⎜2.5 − 0.6 −
⎟0.5 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
(
(b) V
IL
)
= VTN +
1
1
1
= 0.6V +
= 0.6 +
= 1.01 V
⎛
Kn R
2.42
μA ⎞
⎜1.21x100 2 ⎟(20kΩ)
V ⎠
⎝
VOH = VDD −
1
1
2VDD
5
= 2.5 −
= 2.29V | VOL =
=
= 0.830V
2Kn R
4.84
7.26
3Kn R
VIH = VTN −
1
V
1
2.5
+ 1.63 DD = 0.6 −
+ 1.63
= 1.84V
Kn R
2.42
2.42
Kn R
NM L = 1.01− 0.83 = 0.18 V | NM H = 2.29 −1.84 = 0.450 V
6-12
6.44
P
0.25mW
V − VL
3.3 − 0.2
=
= 75.76μA | R = DD
=
= 40.9 kΩ
ID
VDD
3.3V
75.76x10-6
⎛W ⎞ ⎛
⎛ W ⎞ 1.52
0.2 ⎞
75.76x10-6 = 100x10-6 ⎜ ⎟ ⎜3.3 − 0.7 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
(a) I
=
D
(
(b) V
IL
= VTN +
)
1
= 0.7V +
Kn R
1
1
= 0.7 +
= 0.861 V
⎛
6.22
μA ⎞
(1.52)⎜⎝100 V 2 ⎟⎠(40.9kΩ)
VOH = VDD −
1
1
2VDD
6.6
= 3.3 −
= 3.22 | VOL =
=
= 0.594V
3Kn R
2Kn R
12.4
18.7
VIH = VTN −
1
V
1
3.3
+ 1.63 DD = 0.7 −
+ 1.63
= 1.73V
Kn R
Kn R
6.22
6.22
NM L = 0.861− 0.594 = 0.267 V | NM H = 3.22 −1.73 = 1.49 V
6.45
VDD − VL 3 − 0.25
=
= 83.3 kΩ
ID
33x10-6
⎛W ⎞ ⎛
⎛ W ⎞ 1.04
0.25 ⎞
33x10−6 = 60x10-6 ⎜ ⎟ ⎜3 − 0.75 −
⎟0.25 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
(a) R =
(
)
(b) SPICE yields VL = 0.249 V with ID = 33.0 μA.
6.46
VDD − VL 2 − 0.15
=
= 185 kΩ
ID
10x10-6
⎛W ⎞ ⎛
⎛W ⎞
0.15 ⎞
1
10−5 = 75x10-6 ⎜ ⎟ ⎜2 − 0.6 −
⎟0.15 → ⎜ ⎟ =
2 ⎠
⎝ L ⎠S ⎝
⎝ L ⎠ S 1.49
(a) R =
(
)
(b) SPICE yields VL = 0.15 V with ID = 10 μA.
6-13
6.47
1
1
= 417 Ω
⎛
⎞
' W
−6 10
Kn (VGS − VTN )
60x10 ⎜ ⎟(5 −1)
L
⎝1⎠
1
1
=
= 1000 Ω
(b) Ron =
⎛ ⎞
' W
−6 10
K p (VSG + VTP )
25x10 ⎜ ⎟(5 −1)
L
⎝1⎠
(c) A resistive connection exists between the source and drain.
1
1
20
W
= '
=
=
(d )
−6
L Kn (VGS − VTN )Ron 60x10 (3 −1)(417) 1
(a) Ron =
=
W
1
1
20
= '
=
=
−6
L K p (VSG + VTP )Ron 25x10 (3 −1)(1000) 1
6.48
(
VH = VDD − VTO + γ
(V
H
(V
SB
+ 2φ F − 2φ F
))→ V
H
(
(
= 3.3 − 0.75 + 0.75 VH + 0.7 − 0.7
))
− 4.88) = 0.5625(VH + 0.7) → VH2 − 6.918VH + 9.706 = 0
2
VH = 4.962V , 1.956V → VH = 1.96 V
(
)
Checking : VTN = 0.75 + 0.75 1.956 + 0.75 − 0.75 = 1.345V | VH = 3.3 −1.345 = 1.96V
6.49
(
VH = VDD − VTO + γ
(V
H
(V
SB
+ 2φ F − 2φ F
))→ V
H
))
2
Checking : VTN = 0.5 + 0.6
6.50
(
VH = VDD − VTO + γ
H
(
− 3.165) = 0.36(VH + 0.6)→ VH2 − 6.69VH + 9.80 = 0
VH = 2.166V , 4.524V → VH = 2.17 V
(V
(
= 3.3 − 0.6 + 0.6 VH + 0.6 − 0.6
(V
SB
( 2.166 + 0.6 −
+ 2φ F − 2φ F
)
0.6 = 1.133V | VH = 3.3 −1.133 = 2.167V
))→ V
H
(
(
= 2.5 − 0.5 + 0.85 VH + 0.6 − 0.6
))
− 2.659) = 0.7225(VH + 0.6)→ VH2 − 6.04VH + 6.634 = 0
2
VH = 1.444V , 4.596V → VH = 1.44 V
(
)
Checking : VTN = 0.5 + 0.85 1.444 + 0.6 − 0.6 = 1.057V | VH = 2.5 −1.057 = 1.443V
6-14
6.51
For γ = 0, VH = VDD − VTN = 3.3 − 0.6 = 2.7V | For VL : I DL = I DS
⎛ ⎞⎛
2
Kn' ⎛ 1 ⎞
VL ⎞
' 4
2
⎜ ⎟(3.3 − VL − 0.6) = Kn ⎜ ⎟⎜ 2.7 − 0.6 + − ⎟VL → 9VL − 39VL + 7.29 = 0
2 ⎝ 2⎠
2⎠
⎝ 1 ⎠⎝
2
60x10−6 ⎛ 1 ⎞
VL = 0.1958V | I DD =
⎜ ⎟(3.3 − 0.1958 − 0.6) = 94.1μA
2 ⎝ 2⎠
P = (3.3V )(94.08μA)= 0.311 mW
⎛ 4 ⎞⎛
0.1958 ⎞
Checking : I DD = 60x10−6⎜ ⎟⎜2.7 − 0.6 −
⎟0.1958 = 94.1μA
2 ⎠
⎝ 1 ⎠⎝
6.52
(a ) For γ = 0, VH = VDD − VTN = 3.3 − 0.8 = 2.5V | For VL : I DL = I DS
⎛ 4 ⎞⎛
2
Kn' 1
V ⎞
3.3 − VL − 0.8) = Kn' ⎜ ⎟⎜2.5 − 0.8 − L ⎟VL → 9VL2 − 32.2VL + 6.25 = 0 | VL = 0.206V
(
2 2
2⎠
⎝ 1 ⎠⎝
2
60x10−6 1
I DD =
3.3 − 0.206 − .8) = 78.9μA | P = 3.3V(78.9μA)= 0.260 mW
(
2
2
⎛ 4 ⎞⎛
0.206 ⎞
Checking : I DD = 60x10−6⎜ ⎟⎜2.5 − 0.8 −
⎟0.206 = 79.0μA
2 ⎠
⎝ 1 ⎠⎝
(b) For γ = 0, VH = VDD − VTN = 3.3 − 0.4 = 2.9V | For VL : I DL = I DS
⎛ 4 ⎞⎛
2
Kn' 1
V ⎞
3.3 − VL − 0.4) = Kn' ⎜ ⎟⎜2.9 − 0.4 − L ⎟VL → 9VL2 − 45.8VL + 8.41 = 0 | VL = 0.191 V
(
2 2
2⎠
⎝ 1 ⎠⎝
2
60x10−6 1
3.3 − 0.191− 0.4) = 110μA | P = 3.3V(6.55μA)= 0.363 mW
(
2
2
⎛ 4 ⎞⎛
0.191⎞
Checking : I DD = 60x10−6⎜ ⎟⎜2.9 − 0.4 −
⎟0.191 = 110μA
2 ⎠
⎝ 1 ⎠⎝
I DD =
6.53
VIL = VTNS = 0.6 V | VOH = VH = VDD − VTNL = 3.3 − 0.6 = 2.7V
At VIH (See Eq. 6.29 Second Edition) VOL =
VDD − VTNL 3.3 − 0.6
1+ 3KR
= 0.540V
4
1+ 3
0.5
(VDD − VOL − VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 − 0.54 − 0.6 2 = 1.41V
V
+ OL +
(
)
2
2
2KRVOL
2(4) 0.54
2
VIH = VTNS
NM H = 2.7 −1.41 = 1.29 V | NM L = 0.60 − 0.54 = 0.06 V
These values are readily confirmed with SPICE.
6-15
4.0
2.0
0
-2.0
6.54
(a ) For γ = 0, VH = VDD − VTN = 3.3 − 0.6 = 2.7V | For VL : I DL = I DS
⎛ 8 ⎞⎛
2
Kn' 1
V ⎞
3.3 − VL − 0.6) = Kn' ⎜ ⎟⎜2.7 − 0.6 − L ⎟VL → 9VL2 − 39VL + 7.29 = 0
(
2 1
2⎠
⎝ 1 ⎠⎝
−6 ⎛ ⎞
2
60x10 1
VL = 0.1958V | I DD =
⎜ ⎟(3.3 − 0.1958 − 0.6) = 188μA
2 ⎝ 1⎠
P = (3.3V )(188μA)= 0.621 mW
⎛ 8 ⎞⎛
0.1958 ⎞
Checking : I DD = 60x10−6⎜ ⎟⎜2.7 − 0.6 −
⎟0.1958 = 188μA - check is ok
2 ⎠
⎝ 1 ⎠⎝
(b) VIL = VTNS = 0.6 V | VOH = VH = VDD − VTNL = 3.3 − 0.6 = 2.7V
At VIH (See Eq. 6.29 Second Edition) VOL =
VDD − VTNL 3.3 − 0.6
1+ 3KR
= 0.540V
4
1+ 3
0.5
(VDD − VOL − VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 − 0.54 − 0.6 2 = 1.41V
V
+ OL +
(
)
2
2KRVOL
2
2(4) 0.54
2
VIH = VTNS
NM H = 2.7 −1.41 = 1.29 V | NM L = 0.60 − 0.54 = 0.06 V
These values are easily checked with SPICE. See Prob. 6.53,
(c) For γ = 0, VH = VDD − VTN = 3.3 − 0.7 = 2.6V | For VL : I DL = I DS
⎛ 8 ⎞⎛
2
Kn' 1
V ⎞
3.3 − VL − 0.7) = Kn' ⎜ ⎟⎜2.6 − 0.7 − L ⎟VL → 9VL2 − 32.2VL + 6.25 = 0
(
2 1
2⎠
⎝ 1 ⎠⎝
2
60x10−6 1
3.3 − 0.200 − 0.7) = 173μA
(
1
2
P = (3.3V )(173μA)= 570 μW
VL = 0.200V | I DD =
⎛ 8 ⎞⎛
0.200 ⎞
Checking : I DD = 60x10−6⎜ ⎟⎜2.6 − 0.7 −
⎟0.200 = 173μA - check is ok
2 ⎠
⎝ 1 ⎠⎝
6-16
6.55
(
(a) VH = VDD − VTO + γ
(V
(V
SB
+ 2φ F − 2φ F
))→ V
H
(
(
= 3.3 − 0.7 + 0.5 VH + 0.6 − 0.6
))
− 2.987) = 0.25(VH + 0.6)→ VH2 − 6.225VH + 8.772 = 0 → VH = 2.156 V
2
H
VL = 0.20V | I D =
⎛W ⎞ ⎛
0.25mW
0.20 ⎞
= 75.76μA | 75.76 = 100⎜ ⎟ ⎜ 2.156 − 0.7 −
⎟0.20
3.3V
2 ⎠
⎝ L ⎠S ⎝
⎛W ⎞
2.79
| VTNL = 0.7 + 0.5 0.2 + 0.6 − 0.6 = 0.760V
⎜ ⎟ =
1
⎝ L ⎠S
⎛W ⎞
2
1
100 ⎛W ⎞
75.76 =
⎜ ⎟ (3.3 − 0.20 − 0.760) → ⎜ ⎟ =
2 ⎝ L ⎠L
⎝ L ⎠ L 3.61
(b) VIL = VTNS = 0.70V | VOH = VH = 2.16V
(
)
Finding VIH (See Eq 6.29 in 2nd Ed.) : VOL =
VDD − VTNL
1+ 3
(
VTNL = 0.7 + 0.5 VOL + 0.6 − 0.6
)|
(W / L)
(W / L)
S
(
L
=
3.3 − VTNL
1 + 3(2.79)(3.61)
(
=
3.3 − VTNL
5.587
5.587VOL = 3.3 − 0.7 + 0.5 VOL + 0.6 − 0.6
))
Using the quadratic equation : VOL = 0.4432V → VTNL = 0.8234V
VIH = VTNS +
2
VOL (W / L)L 1
+
VDD − VOL − VTNL )
(
2
(W / L) 2VOL
S
VIH = 0.7 +
2
0.443
1 ⎛ 1⎞ 1
+
3.3 − 0.443 − 0.823) | VIH = 1.39V
⎜ ⎟
(
2
10.07 ⎝ 2 ⎠ 0.443
NM H = 2.16 − 1.39 = 0.77 V | NM L = 0.7 − 0.443 = 0.26 V
4.0
2.0
0
-2.0
6-17
6.56
(
)
0.4mW
= 160μA | VTNL = 0.6 + 0.5 0.3 + 0.6 − 0.6 = 0.687V
2.5V
⎛ W ⎞ 1.40
2
100x10−6 ⎛W ⎞
160x10−6 =
⎜ ⎟ (2.5 − 0.3 − 0.687) → ⎜ ⎟ =
2
1
⎝ L ⎠L
⎝ L ⎠L
⎛W ⎞ ⎛
⎛W ⎞
0.3 ⎞
6.67
160x10−6 = 100x10−6 ⎜ ⎟ ⎜1.55 − 0.6 −
⎟0.3 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
I DD =
6.57
(a) VDD = 3.3 V VTN =1 V ID = 75 μA VL = 0.2 V VH = VDD - VTN = 3.3 V - 0.6V = 2.7 V
+3.3 V
+
M
V
L
DSL
+
GS
-
ID
V = 3.1 V
= 3.1 V
VL = 0.2V
+
M
V
H
= 2.7 V
S
V
DSS
= 0.2 V
-
I DS = I DL = 75μA
⎛W ⎞ ⎛
V ⎞
I DS = Kn' ⎜ ⎟ ⎜VGSS − VTHS − DSS ⎟VDSS
2 ⎠
⎝ L ⎠S ⎝
2
K ' ⎛W ⎞
I DL = n ⎜ ⎟ (VGSL − VTNL )
2 ⎝ L ⎠L
(
(b) VH = VDD − VTO + γ
(V
H
(V
SB
75μA =
75μA = 100
100
μA
V2
2
+ 2φ F − 2φ F
μA ⎛W ⎞ ⎛
⎛W ⎞ 1.88
0.2 ⎞
⎜ ⎟ ⎜ 2.7 − 0.6 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
V ⎝ L ⎠S ⎝
⎝ L ⎠S
2
⎛W ⎞
⎛W ⎞
2
1
⎜ ⎟ (3.3 − 0.2 − 0.6) → ⎜ ⎟ =
⎝ L ⎠L
⎝ L ⎠ L 4.17
))→ V
H
(
(
= 3.3 − 0.6 + 0.5 VH + 0.6 − 0.6
− 3.087) = 0.25(VH + 0.6)→ VH2 − 6.424VH + 9.381 = 0 → VH = 2.245
2
⎛W ⎞
0.2 ⎞
2.43
2.245
−
0.6
−
0.2
→
⎜
⎟
⎜
⎟
⎜ ⎟ =
2
2 ⎠
1
V ⎝ L ⎠S ⎝
⎝ L ⎠S
2
K ' ⎛W ⎞
I DL = n ⎜ ⎟ (VGSL − VTNL ) | VTNL = 0.6 + 0.5 0.2 + 0.6 − 0.6 = 0.660V
2 ⎝ L ⎠L
75μA = 100
μA ⎛ W ⎞ ⎛
(
75μA =
6-18
100
μA
V2
2
⎛W ⎞
⎛W ⎞
2
1
⎜ ⎟ (3.3 − 0.2 − 0.66) → ⎜ ⎟ =
⎝ L ⎠L
⎝ L ⎠ L 3.97
)
))
6.58
(a ) For γ = 0, V
H
= VDD − VTO = 2 − 0.6 = 1.4V
⎛W ⎞ ⎛
⎛W ⎞ ⎛
⎛W ⎞
V ⎞
0.15⎞
2.30
I DS = Kn' ⎜ ⎟ ⎜VH − VTNS − L ⎟VL | 25x10−6 = 100x10−6 ⎜ ⎟ ⎜1.4 − 0.6 −
⎟0.15 → ⎜ ⎟ =
2 ⎠
1
2⎠
⎝ L ⎠S ⎝
⎝ L ⎠S ⎝
⎝ L ⎠S
⎛W ⎞
2
2
K ' ⎛W ⎞
100x10−6 ⎛W ⎞
1
I DL = n ⎜ ⎟ (VGSL − VTNL ) | 25x10−6 =
⎜ ⎟ (2 − 0.15 − 0.6) → ⎜ ⎟ =
2 ⎝ L ⎠L
2
⎝ L ⎠L
⎝ L ⎠ L 3.13
(b) For γ = 0.6, V
H
[
)]
(
= VDD − VTNL = 2 − 0.6 + 0.6 VH + 0.6 − 0.6 → VH = 1.09V
⎛W ⎞ ⎛
⎛W ⎞ ⎛
⎛W ⎞
V ⎞
0.15 ⎞
4.02
I DS = Kn' ⎜ ⎟ ⎜VH − VTNS − L ⎟VL | 25x10−6 = 100x10−6 ⎜ ⎟ ⎜1.09 − 0.6 −
⎟0.15 → ⎜ ⎟ =
2 ⎠
1
2⎠
⎝ L ⎠S ⎝
⎝ L ⎠S ⎝
⎝ L ⎠S
( 0.15 + 0.6 −
For vO = VL = 0.15V , VTN = 0.6 + 0.6
)
0.6 = 0.655V
⎛W ⎞
2
2
Kn' ⎛W ⎞
100x10−6 ⎛W ⎞
1
−6
V
−
V
|
25x10
=
⎜ ⎟ ( GSL TNL )
⎜ ⎟ (2 − 0.15 − 0.655) → ⎜ ⎟ =
2 ⎝ L ⎠L
2
⎝ L ⎠L
⎝ L ⎠ L 2.86
(c) Using LEVEL=1 KP=100U VTO=0.6 GAMMA=0, the values of ID and VL agree with our
hand calculations. The results also agree for GAMMA=0.6.
I DL =
6.59
⎛W ⎞ ⎛
2
VDSL ⎞
Kn' ⎛W ⎞
I DS = I DL | K ⎜ ⎟ ⎜VGSS − VTNS −
⎟VDSL =
⎜ ⎟ (VGSL − VTNL )
2 ⎠
2 ⎝ L ⎠L
⎝ L ⎠S ⎝
⎛ 4.71⎞⎛
2
Kn' ⎛ 1 ⎞
VO ⎞
Kn' ⎜
⎜
⎟⎜2.5 − 0.6 − ⎟VO =
⎟(2.5 − VO − VTNL )
2⎠
2 ⎝ 1.68 ⎠
⎝ 1 ⎠⎝
'
n
(
VTNL = 0.6 + 0.5 VO + 0.6 − 0.6
)
An iterative solution yields VO = 0.1061 V
6.60
⎛W ⎞ ⎛
2
V ⎞
K ' ⎛W ⎞
I DS = I DL | Kn' ⎜ ⎟ ⎜VH − VTNS − L ⎟VL = n ⎜ ⎟ (2.5 − VL − VTNL )
2⎠
2 ⎝ L ⎠L
⎝ L ⎠S ⎝
which is independent of K'n . Ratioed logic maintains VL and VH independent
of K'n . So VH = 1.55V and VL = 0.20V. However, I DS = I DL ∝ Kn' :
80 μA V 2
= 64.0 μA
P = 2.5V (64μA)= 0.160 mW
100 μA V 2
μA ⎛ 4.71⎞⎛
0.2 ⎞
Checking : I DS = 80 2 ⎜
⎟⎜1.55 − 0.6 −
⎟0.2 = 64.1μA
2 ⎠
V ⎝ 1 ⎠⎝
So, I D = 80μA
6-19
6.61
⎛W ⎞ ⎛
2
V ⎞
K ' ⎛W ⎞
I DS = I DL | Kn' ⎜ ⎟ ⎜VH − VTNS − L ⎟VL = n ⎜ ⎟ (2.5 − VL − VTNL )
2⎠
2 ⎝ L ⎠L
⎝ L ⎠S ⎝
which is independent of K'n . Ratioed logic maintains VL and VH independent
of K'n . So VH = 1.55V and VL = 0.20V. However, I DS = I DL ∝ Kn' :
120 μA V 2
= 96.0 μA
P = 2.5V (96μA)= 0.240 mW
100 μA V 2
0.2 ⎞
μA ⎛ 4.71⎞⎛
Checking : I DS = 120 2 ⎜
⎟⎜1.55 − 0.6 −
⎟0.2 = 96.1μA
2 ⎠
V ⎝ 1 ⎠⎝
So, I D = 80μA
6.62
Noise Margins vs. KR
2.5
2
1.5
NMH
NML
1
0.5
0
0
2
4
6
8
10
12
-0.5
KR
6.63
(a) VH = VDD – VTNL does not depend upon λ. However, VL is dependent upon λ.
(b) SPICE yields VL = 0.20 V, 0.207 V, 0.217 V, and 0.232 V for λ = 0, 0.02/V, 0.05/V, and
0.1/V respectively. The current also increases: IDD = 80.1, 82.8, 86.9 and 93.3 μA, respectively.
6.64
VTNL = 0.6 + 0.5 0.20 + 0.6 − 0.6 = 0.660V
(
)
VGSL − VTNL = 4 − 0.2 − 0.66 = 3.14V | VDSL = 2.5 − 0.2 = 2.30V → Triode region
⎛W ⎞
2.3⎞
1
μA ⎛ W ⎞ ⎛
80μA = 100 2 ⎜ ⎟ ⎜ 4 − 0.2 − 0.66 −
⎟2.3 → ⎜ ⎟ =
2 ⎠
V ⎝ L ⎠L⎝
⎝ L ⎠ L 5.72
⎛W ⎞ 2.22
0.2 ⎞
μA ⎛ W ⎞ ⎛
80μA = 100 2 ⎜ ⎟ ⎜ 2.5 − 0.6 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
V ⎝ L ⎠S ⎝
⎝ L ⎠L
6-20
6.65
(
)
For linear operation at vo = VL : VTNL = 0.8 + 0.5 0.2 + 0.6 − 0.6 = 0.860V
VGSL − VTNL ≥ VDSL : VGG − 0.20 − 0.860 ≥ 2.5 − 0.2 → VGG ≥ 3.36V
(
)
We also require : VGG ≥ 2.5 + VTNL = 2.5 + 0.8 + 0.5 2.5 + 0.6 − 0.6 = 3.79V so VGG ≥ 3.79V
6.66
(
)
If VH = 3.3 V , VTNL = 0.6 + 0.5 3.3 + 0.6 − 0.6 = 1.2V
5 -1.2 = 3.8V > 3.3V so VH = 3.3 V is correct.
⎛ ⎞⎛
2
VL ⎞
Kn' ⎛ 1 ⎞
' 5
I DS = I DL | Kn ⎜ ⎟⎜ 3.3 − 0.6 − ⎟VL =
⎜ ⎟(3.3 − VL − VTNL )
2⎠
2 ⎝ 2⎠
⎝ 1 ⎠⎝
(
VTNL = 0.6 + 0.5 VL + 0.6 − 0.6
I DS =
) An interative solution gives V = 0.1222 V , V
L
TNL
= 0.6376 V
2
100μA ⎛ 1 ⎞
⎜ ⎟(3.3 − 0.1222 − 0.6376) = 161 μA | P = 3.3V (161μA) = 0.532 mW
2 ⎝ 2⎠
6.67
We require VGG ≥ VDD + VTNL so VH = VDD
VTNL = VTO + γ
(V
SB
)
+ 0.6 − 0.6 = 0.6 + 0.6
( 3.3 + 0.6 −
)
0.6 = 1.32V
VGG ≥ 3.3 + 1.32 = 4.62V
6.68
We require VGG ≥ VDD + VTNL so VH = VDD
VTNL = VTO + γ
(V
SB
)
(
)
+ 0.6 − 0.6 = 0.6 + 0.6 3.3 + 0.6 − 0.6 = 1.32V
VGG ≥ 3.3 + 1.32 = 4.62V | Design decision - Choose VGG = 5 V
I DD =
300μW
= 90.9μA
3.3V
⎛W ⎞ ⎛
⎛ W ⎞ 1.75
0.2 ⎞
For MS : 90.9μA = 100μA⎜ ⎟ ⎜3.3 − 0.6 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
(
)
For ML : VTNL = 0.6 + 0.6 0.2 + 0.6 − 0.6 = 0.672V
⎛W ⎞ ⎛
⎛W ⎞
3.3 − 0.2 ⎞
1
90.9μA = 100μA⎜ ⎟ ⎜ 5 − .2 − 0.672 −
⎟(3.3 − 0.2)→ ⎜ ⎟ =
2 ⎠
⎝ L ⎠L⎝
⎝ L ⎠ L 8.79
6.69
We require VTNL ≤ 0 : −1+ γ
( 2.5 + 0.6 −
)
0.6 ≤ 0 → γ ≤ 1.014
6-21
6.70
(a) V
H
⎛W ⎞ ⎛
2
V ⎞
K ' ⎛W ⎞
= VDD | I DS = I DL | Kn' ⎜ ⎟ ⎜VDD − VTNS − L ⎟VL = n ⎜ ⎟ (VTNL )
2⎠
2 ⎝ L ⎠L
⎝ L ⎠S ⎝
For ratioed logic, both VH and VLare independent of K'n . VH = 2.5 V | VL = 0.2 V
⎛ 80 ⎞
However, I D ∝ Kn' | I DS = 80μA⎜
⎟ = 64μA | P = 2.5V (64μA) = 0.160 mW
⎝ 100 ⎠
⎛120 ⎞
(b) VH = 2.5 V VL = 0.2 V I DS = 80μA⎜⎝100 ⎟⎠ = 96μA | P = 2.5V (96μA)= 0.240 mW
6.71
(
)
0.20mW
= 60.1μA VTNL = −1+ 0.5 3.3 + 0.6 − 0.6 = −0.400V → VH = 3.3V
3.3V
⎛W ⎞ ⎛
⎛W ⎞ 1.16
0.20 ⎞
| For VO = VL = 0.2V ,
60.1μA = 100μA⎜ ⎟ ⎜ 3.3 − 0.6 −
⎟0.20 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
⎛ W ⎞ 1.36
2
100μA ⎛ W ⎞
VTNL = −1+ 0.5 0.20 + 0.6 − 0.6 = −0.940V | 60.1μA =
⎜ ⎟ (−0.940) | ⎜ ⎟ =
1
2 ⎝ L ⎠L
⎝ L ⎠L
I DD =
(
)
6.72
(
)
Assume VH = VDD = 3.3V | Checking : VTNL = −2 + 0.5 3.3 + 0.6 − 0.6 = −1.40
P
250μW
=
= 75.8μA
VDD
3.3V
⎛W ⎞ ⎛
⎛W ⎞ 1.46
0.2 ⎞
For MS in the triode region, 75.8μA = 100μA⎜ ⎟ ⎜ 3.3 − 0.6 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
VTNL < 0, so our assumption is correct. | I DD =
(
)
For ML in the saturation region, VTNL = −2 + 0.5 0.2 + 0.6 − 0.6 = −1.94V and
75.8μA =
2
100μA ⎛ W ⎞
⎜ ⎟ (0 − VTNL ) →
2 ⎝ L ⎠L
⎛W ⎞
1
⎜ ⎟ =
⎝ L ⎠ L 2.48
6.73
(a) No, VH does not depend upon λ.
(b) As λ increases, IDD increases in ML, and VL increases.
λ
0
0.02/V
0.05/V
0.1/v
6-22
IDD
78.2 μA
81.4 μA
86.0 μA
93.6 μA
VL
195 mV
203 mV
214 mV
231 mV
6.74
(a) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA.
⎛W ⎞ ⎛
⎛ W ⎞ 1.80
0.25 ⎞
Also, VH = 2.5 V. 80x10-6 = 100x10-6⎜ ⎟ ⎜2.5 − 0.6 −
⎟0.25 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠n ⎝
⎝ L ⎠n
1.80(100)
V +V
K
2.5 − 0.6
= 1.02 V
(b) VIL = VTNS + DD2 TP KR = KS = 1.11 40 = 4.05 VIL = 0.6 +
( )
4.052 + 4.05
K R + KR
L
⎛
⎛
KR ⎞
4.05 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟ = 2.30V
VOH = VDD − (VDD + VTP )⎜⎜1−
5.05 ⎟⎠
KR + 1 ⎠
⎝
⎝
V + VTP 2.5 − 0.6
=
= 0.545V
VIH = VTNS + 2VOL = 0.6 + 2(0.545) = 1.69V
VOL = DD
3KR
3(4.05)
NM H = VOH − VIH = 2.30 −1.69 = 0.610 V
NM L = VIL − VOL = 1.02 − 0.545 = 0.475 V
6.75
(a) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA.
⎛ 2.22 ⎞ ⎛
VL ⎞
Also, VH = 2.5 V. 80x10-6 = 100x10-6⎜
⎟ ⎜2.5 − 0.5 − ⎟VL → VL = 0.189 V
2⎠
⎝ 1 ⎠n ⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.6
(b) VIL = VTNS + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 5.00 VIL = 0.5 + 2 = 0.847 V
5 +5
KR + KR
L
⎛
⎛
KR ⎞
5 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟ = 2.33V
VOH = VDD − (VDD + VTP )⎜⎜1−
5 + 1 ⎟⎠
KR + 1 ⎠
⎝
⎝
V + VTP 2.5 − 0.6
=
= 0.491V
VIH = VTNS + 2VOL = 0.5 + 2(0.491) = 1.48V
VOL = DD
3KR
3(5)
NM H = VOH − VIH = 2.33 −1.48 = 0.849 V
NM L = VIL − VOL = 0.847 − 0.491 = 0.356 V
(c) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA.
⎛ 2.22 ⎞ ⎛
VL ⎞
Also, VH = 2.5 V. 80x10-6 = 100x10-6⎜
⎟ ⎜2.5 − 0.7 − ⎟VL → VL = 0.213 V
2⎠
⎝ 1 ⎠n ⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.6
(d) VIL = VTNS + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 5.00 VIL = 0.7 + 2 = 1.05 V
5 +5
KR + KR
L
⎛
⎛
KR ⎞
5 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟ = 2.33V
VOH = VDD − (VDD + VTP )⎜⎜1−
5 + 1 ⎟⎠
KR + 1 ⎠
⎝
⎝
V + VTP 2.5 − 0.6
=
= 0.490V
VIH = VTNS + 2VOL = 0.7 + 2(0.490)= 1.68V
VOL = DD
3KR
3(5)
NM H = VOH − VIH = 2.33 −1.68 = 0.650 V
NM L = VIL − VOL = 1.05 − 0.490 = 0.560 V
6-23
6.76
(a) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA.
⎛ 2.22 ⎞ ⎛
VL ⎞
Also, VH = 2.5 V. 80x10-6 = 120x10-6⎜
⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.165 V
2⎠
⎝ 1 ⎠n ⎝
V +V
K
2.22 ⎛120 ⎞
2.5 − 0.6
(b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 6 VIL = 0.6 + 2 = 0.893V
6 +6
KR + KR
L
VOH
⎛
= VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
=
3KR
2.5 − 0.6
3(6)
⎛
KR ⎞
6 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟⎟ = 2.36V
6
+
1
KR + 1 ⎠
⎝
⎠
= 0.448V
VIH = VTN + 2VOL = 0.6 + 2(0.448)= 1.50V
NM H = VOH − VIH = 2.36 −1.50 = 0.860 V NM L = VIL − VOL = 0.893 − 0.448 = 0.445 V
(c) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA.
⎛ 2.22 ⎞ ⎛
VL ⎞
Also, VH = 2.5 V. 80x10-6 = 80x10-6 ⎜
⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.254 V
2⎠
⎝ 1 ⎠n ⎝
V +V
K
2.22 ⎛ 80 ⎞
2.5 − 0.6
(d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 4 VIL = 0.6 + 2 = 1.03V
4 +4
KR + KR
L
⎛
VOH = VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
3KR
=
2.5 − 0.6
3(4)
⎛
KR ⎞
4 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟ = 2.30V
4 + 1 ⎟⎠
KR + 1 ⎠
⎝
= 0.549V
VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V
NM H = VOH − VIH = 2.30 −1.70 = 0.600 V
6-24
NM L = VIL − VOL = 1.03 − 0.549 = 0.481 V
6.77
(a) The PMOS load is still saturated, and VH = 2.5 V.
2
40x10-6 ⎛ 1.11⎞
I DD =
⎜
⎟ (2.5 − 0.5) → I DD = 88.8 μA
2 ⎝ 1 ⎠n
⎛ 2.22 ⎞ ⎛
VL ⎞
For MS : 88.8x10-6 = 100x10-6 ⎜
⎟ ⎜ 2.5 − 0.6 − ⎟VL → VL = 0.224 V
2⎠
⎝ 1 ⎠n⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.5
(b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 5 VIL = 0.6 + 2 = 0.965V
5 +5
KR + KR
L
⎛
VOH = VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
=
3KR
2.5 − 0.5
3(5)
⎛
KR ⎞
5 ⎞
⎟⎟ = 2.5 − (2.5 − 0.5)⎜⎜1−
⎟⎟ = 2.33V
5
+
1
KR + 1 ⎠
⎝
⎠
= 0.516V
VIH = VTN + 2VOL = 0.6 + 2(0.516) = 1.63V
NM H = VOH − VIH = 2.33 −1.63 = 0.700 V NM L = VIL − VOL = 0.965 − 0.516 = 0.449 V
(c) The PMOS load is still saturated, and VH = 2.5 V.
2
40x10-6 ⎛ 1.11⎞
I DD =
⎜
⎟ (2.5 − 0.7) → I DD = 71.9 μA
2 ⎝ 1 ⎠n
⎛ 2.22 ⎞ ⎛
VL ⎞
For the NMOS device : 71.9x10-6 = 100x10-6 ⎜
⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.179 V
2⎠
⎝ 1 ⎠n⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.7
(d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 40 ⎟⎠ = 5 VIL = 0.6 + 2 = 0.929V
5 +5
KR + KR
L
⎛
VOH = VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
3KR
=
2.5 − 0.7
3(5)
⎛
KR ⎞
5 ⎞
⎟⎟ = 2.5 − (2.5 − 0.7)⎜⎜1−
⎟ = 2.34V
5 + 1 ⎟⎠
KR + 1 ⎠
⎝
= 0.465V
VIH = VTN + 2VOL = 0.6 + 2(0.465) = 1.53V
NM H = VOH − VIH = 2.34 −1.53 = 0.810 V
NM L = VIL − VOL = 0.929 − 0.465 = 0.464 V
6-25
6.78
(a) The PMOS load is still saturated, and VH = 2.5 V.
2
50x10-6 ⎛ 1.11⎞
I DD =
⎜
⎟ (2.5 − 0.6) → I DD = 100 μA
2 ⎝ 1 ⎠n
⎛ 2.22 ⎞ ⎛
VL ⎞
For MS : 100x10-6 = 100x10-6 ⎜
⎟ ⎜ 2.5 − 0.6 − ⎟VL → VL = 0.254 V
2⎠
⎝ 1 ⎠n⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.6
(b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 50 ⎟⎠ = 4 VIL = 0.6 + 2 = 1.03V
4 +4
KR + KR
L
⎛
VOH = VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
=
3KR
2.5 − 0.6
3(4)
⎛
KR ⎞
4 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟⎟ = 2.30V
4
+
1
KR + 1 ⎠
⎝
⎠
= 0.549V
VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V
NM H = VOH − VIH = 2.30 −1.70 = 0.600 V NM L = VIL − VOL = 1.03 − 0.549 = 0.481 V
(c) The PMOS load is still saturated, and VH = 2.5 V.
2
30x10-6 ⎛ 1.11⎞
I DD =
⎜
⎟ (2.5 − 0.6) → I DD = 60.0 μA
2 ⎝ 1 ⎠n
⎛ 2.22 ⎞ ⎛
VL ⎞
For the NMOS device : 60x10-6 = 100x10-6 ⎜
⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.148 V
2⎠
⎝ 1 ⎠n⎝
V +V
K
2.22 ⎛100 ⎞
2.5 − 0.6
= 0.866V
(d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜⎝ 30 ⎟⎠ = 6.67 VIL = 0.6 +
2
6.67 + 6.67
KR + KR
L
⎛
VOH = VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
3KR
=
2.5 − 0.6
3(6.67)
⎛
KR ⎞
6.67 ⎞
⎟⎟ = 2.5 − (2.5 − 0.6)⎜⎜1−
⎟ = 2.37V
6.67 + 1 ⎟⎠
KR + 1 ⎠
⎝
= 0.425V
VIH = VTN + 2VOL = 0.6 + 2(0.425)= 1.45V
NM H = VOH − VIH = 2.37 −1.45 = 0.920 V
6.79
(a) I DD =
NM L = VIL − VOL = 0.866 − 0.425 = 0.441 V
P 100μW
=
= 55.6μA
VDD
1.8V
For the saturated PMOS load : 55.6x10-6 =
⎛W ⎞
2
25x10-6 ⎛W ⎞
2.63
⎜ ⎟ (1.8 − 0.5) → ⎜ ⎟ =
2 ⎝ L ⎠p
1
⎝ L ⎠p
⎛W ⎞ ⎛
⎛W ⎞ 3.86
0.2 ⎞
For the linear NMOS switch : 55.6x10-6 = 60x10-6 ⎜ ⎟ ⎜1.8 − 0.5 −
⎟0.2 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ n⎝
⎝ L ⎠n
6-26
(b) V
IL
VOH
= VTN +
VDD + VTP
K + KR
⎛
= VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
2
R
=
3KR
1.8 − 0.5
3(3.52)
KR =
KS 3.86 ⎛ 60 ⎞
=
⎜ ⎟ = 3.52
KL 2.63 ⎝ 25 ⎠
(a) I D =
1.8 − 0.5
3.522 + 3.52
= 0.826V
⎛
KR ⎞
3.52 ⎞
⎟⎟ = 1.8 − (1.8 − 0.5)⎜⎜1−
⎟ = 1.65V
KR + 1 ⎠
3.52 + 1 ⎟⎠
⎝
= 0.400V
VIH = VTN + 2VOL = 0.5 + 2(0.400) = 1.30V
NM H = VOH − VIH = 1.60 −1.30 = 0.300 V
6.80
VIL = 0.5 +
NM L = VIL − VOL = 0.826 − 0.400 = 0.426 V
P
200μW
=
= 66.7μA
VDD
3V
For the saturated PMOS load : 66.7x10-6 =
⎛W ⎞
2
25x10-6 ⎛W ⎞
0.926
1
=
⎜ ⎟ (3 − 0.6) → ⎜ ⎟ =
2 ⎝ L ⎠p
1
1.08
⎝ L ⎠p
⎛W ⎞ ⎛
⎛W ⎞ 1.65
0.3 ⎞
For the linear NMOS switch : 66.7x10-6 = 60x10-6 ⎜ ⎟ ⎜ 3 − 0.6 −
⎟0.3 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ n⎝
⎝ L ⎠n
⎛ 60 ⎞
V +V
K
3 − 0.6
= 1.11V
(b) VIL = VTN + DD2 TP KR = KS = 1.65(1.08)⎜⎝ 25 ⎟⎠ = 4.28 VIL = 0.6 +
4.282 + 4.28
KR + KR
L
VOH
⎛
= VDD − (VDD + VTP )⎜⎜1−
⎝
VOL =
VDD + VTP
3KR
=
3 − 0.6
3(4.28)
⎛
KR ⎞
4.28 ⎞
⎟⎟ = 3 − (3 − 0.6)⎜⎜1−
⎟⎟ = 2.76V
4.28
+
1
KR + 1 ⎠
⎝
⎠
= 0.670V
NM H = VOH − VIH = 2.76 −1.94 = 0.821 V
VIH = VTN + 2VOL = 0.6 + 2(0.670)= 1.94V
NM L = VIL − VOL = 1.11− 0.670 = 0.440 V
6.81
With A = 1 = B, the circuit is equivalent to a single 4.44/1 switching device.
⎛ 4.44 ⎞⎛
2
VL ⎞
100μA ⎛1.81⎞
100μA⎜
⎟⎜ 2.5 − 0.6 − ⎟VL =
⎜
⎟(VTNL ) | VTNL = −1+ 0.5 VL + 0.6 − 0.6
2⎠
2 ⎝ 1 ⎠
⎝ 1 ⎠⎝
2
100μA ⎛ 1.81⎞
Solving iteratively → VL = 0.1033V | VTNL = −0.968V (b) I DD =
⎜
⎟(0.968) = 84.8 μA
2 ⎝ 1 ⎠
(
)
6-27
6.82
⎛W ⎞ ⎛
⎛W ⎞
0.1⎞
4.32
80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟0.1 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A⎝
⎝ L ⎠A
(
)
VTNB = 0.6 + 0.5 0.1+ 0.6 − 0.6 = 0.631
⎛W ⎞ ⎛
⎛W ⎞
0.1⎞
4.65
80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.1− 0.631− ⎟0.1 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠A
6.83
We require
Ron
R
R
+ on = on and the total area AT ∝ (WL)A + (WL)B
⎛W ⎞
⎛W ⎞
K
⎜ ⎟
⎜ ⎟
⎝ L ⎠A ⎝ L ⎠B
1
1
1
KW B
KW B
W B2
+
= → WA =
→ AT ∝
+ WB =
WB − K
WB − K
WB − K
WA WB K
d ⎛ W B2 ⎞ W B2 − 2KW B
Finding the minimum :
= 0 → W B = 2K & WA = 2K.
⎜
⎟=
dWB ⎝ W B − K ⎠ (W B − K )2
Setting L = 1,
6.84
+2.5 V
ML
1.81
1
Y
M
MA
A
6-28
2.22
1
B
2.22
1
MD
MC
B
C
2.22
1
D
2.22
1
6.85
+2.5 V
ML
1.81
1
Y
D
MD
8.88
1
C
MC
8.88
1
B
MB
8.88
1
A
MA
8.88
1
6.86
+2.5 V
ML
1.11
1
Y
A
2.22
1
MC
MB
MA
B
2.22
1
C
2.22
1
(a ) With A = B = C = 1, the circuit is equivalent to a single 6.66/1 switching device.
⎛ 6.66 ⎞⎛
2
40μA ⎛1.81⎞
VL ⎞
100μA⎜
⎟⎜ 2.5 − 0.6 − ⎟VL =
⎜
⎟(0.6) → VL = 0.1033V
2⎠
2 ⎝ 1 ⎠
⎝ 1 ⎠⎝
2
40μA ⎛ 1.81⎞
(b) I DD = 2 ⎜⎝ 1 ⎟⎠(0.6) = 13.0 μA
6-29
6.87
+2.5 V
(a)
1.11
1
ML
Y
C
MC
6.66
1
B
MB
6.66
1
A
MA
6.66
1
(b) The PMOS device remains saturated with I
DD
= 80μA.
⎛ 6.66 ⎞⎛
VDSA ⎞
For MA : 80μA = 100μA⎜
⎟⎜ 2.5 − 0.6 −
⎟VDSA → VDSA = 0.0643V
2 ⎠
⎝ 1 ⎠⎝
(
)
For MB :VTNB = 0.6 + 0.5 0.0643 + 0.6 − 0.6 = 0.620
⎛ 6.66 ⎞⎛
VDSB ⎞
80μA = 100μA⎜
⎟⎜ 2.5 − .0643 − 0.620 −
⎟VDSB → VDSB = 0.0674V
2 ⎠
⎝ 1 ⎠⎝
(
)
For MC :VTNC = 0.6 + 0.5 0.0643 + 0.674 + 0.6 − 0.6 = 0.640
⎛ 6.66 ⎞⎛
VDSC ⎞
80μA = 100μA⎜
⎟⎜ 2.5 − .0674 − .0643 − 0.640 −
⎟VDSC → VDSC = 0.0709V
2 ⎠
⎝ 1 ⎠⎝
VL = VDSA + VDSB + VDSC = 0.203 V
(c) Assume equal values of
0.2
= 0.0667V
3
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.43
For MA : 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.6 −
⎟0.0667 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A⎝
⎝ L ⎠A
(
VDS =
)
For MB :VTNB = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.74
80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621−
⎟0.0667 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
(
)
For MC :VTNC = 0.6 + 0.5 0.1334 + 0.6 − 0.6 = 0.641
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
7.09
80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.1334 − 0.641−
⎟0.0667 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
6-30
6.88
VDD
1/1.7
vO
2/14.7/1
A
4.7/1 2/1
B
Ground
6.89
VDD
1/1.7
vO
C
A
4.7/1 2/1
2/14.7/1
Ground
6.90
4.7/1 2/1
B
⎛ W ⎞ 1.81
Y = (A + B)(C + D)(E + F ) | ⎜ ⎟ =
1
⎝ L ⎠L
⎛ W⎞
⎛ 2.22 ⎞ 6.66
| ⎜ ⎟
=3⎜
⎟=
1
⎝ L ⎠ A−F
⎝ 1 ⎠
6.91
(a) The only change to the schematic is to connect the gate of load transistor ML to its drain
instead of its source.
(b) There is no change to the logic function Y = (A + B)(C + D)(E + F )
⎛ W⎞
⎛ W⎞
⎛ 4.71⎞ 14.1
1
| ⎜ ⎟
=3⎜
⎟=
(c) ⎜ ⎟ =
1
⎝ L ⎠ L 1.68
⎝ L ⎠ ABCDEF
⎝ 1 ⎠
6-31
6.92
⎛ W ⎞ 1.11
Y = (A + B)(C + D)E | ⎜ ⎟ =
1
⎝ L ⎠L
⎛ W⎞
⎛ 2.22 ⎞ 6.66
| ⎜ ⎟
=3⎜
⎟=
1
⎝ L ⎠ A− E
⎝ 1 ⎠
6.93
(a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load
device as in Fig. 6.29(b).
(b) The logic function is unchanged: Y = (A + B)(C + D)E
⎛ W⎞
⎛ W⎞
⎛ 4.71⎞ 14.1
1
(c) ⎜ ⎟ =
| ⎜ ⎟
=3⎜
⎟=
1
⎝ L ⎠ L 1.68
⎝ L ⎠ ABCDE
⎝ 1 ⎠
6.94
(a) Y = ACE + ACDF + BF + BDE (b)
⎛ W⎞
1.11 3.33
=
| ACDF path contains 4 devices
⎜ ⎟ =3
1
1
⎝ L ⎠L
⎡ ⎛ 2.22 ⎞⎤ 26.6
⎛ W⎞
⎛ W⎞
1
1
1
1
17.8
=
3
|
+
+
=
→⎜ ⎟ =
⎢4 ⎜
⎜ ⎟
⎟⎥ =
⎛ W⎞
⎛ 26.6 ⎞ ⎛ W ⎞
2.22
1
1
⎝ L ⎠ A, C , D , F
⎝ L ⎠B , E
⎣ ⎝ 1 ⎠⎦
3
⎜ ⎟
⎜
⎟ ⎜ ⎟
1
⎝ L ⎠B ⎝ 1 ⎠ ⎝ L ⎠E
6.95
(a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load
device as in Fig. 6.29(b).
(b) There is no change to the logic function Y = ACDF + ACE + BDE + BF
⎛ W⎞
⎛ W⎞
⎛ 4.71⎞ 18.8
1
(c) ⎜⎝ L ⎟⎠ = 1.68 | ⎜⎝ L ⎟⎠ = 4 ⎜⎝ 1 ⎟⎠ = 1
L
ACDF
RoB + RonD + RonE setting RoB = RonE :
⎛ W ⎞ 12.6
1
1
1
2
1
1
+
+
=
+
=
→⎜ ⎟ =
⎛ W⎞
⎛ W⎞
18.8 ⎛ W ⎞
18.8 4.71 ⎝ L ⎠ B
1
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ L ⎠B
⎝ L ⎠E ⎝ L ⎠B
Checking : RoB + RonF =
6-32
1
1
1
1
+
=
≤
so path BF is ok.
18.8 12.6 7.42 4.71
6.96
+2.5 V
Y
D
E
B
C
A
⎛W ⎞ 1.81
⎜ ⎟ =
1
⎝ L ⎠L
DCA and ECA paths contain three devices
⎛W ⎞
⎛ 2.22 ⎞ 6.66
= 3⎜
⎜ ⎟
⎟=
1
⎝ L ⎠ A,C , D, E
⎝ 1 ⎠
1
1
1
+
=
⎛W ⎞
⎛W ⎞
⎛ 2.22 ⎞
⎜ ⎟
⎜ ⎟
⎜
⎟
⎝ L ⎠ A ⎝ L ⎠B ⎝ 1 ⎠
⎛W ⎞
1
1
1
3.33
+
=
→⎜ ⎟ =
⎛ 6.66 ⎞
⎛W ⎞
⎛ 2.22 ⎞ ⎝ L ⎠
1
B
⎜
⎟
⎜ ⎟
⎜
⎟
⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠
6-33
6.97
+2.5 V
C
E
B
D
A
⎛W ⎞ 1 ⎛1.11⎞
1
⎜ ⎟ = ⎜
⎟=
⎝ L ⎠ L 2 ⎝ 1 ⎠ 1.80
CBA and EDA paths contain three devices
⎛W ⎞
1 ⎛ 2.22 ⎞ 3.33
= (3)⎜
⎜ ⎟
⎟=
1
⎝ L ⎠ A− E 2 ⎝ 1 ⎠
6-34
6.98
+2.5 V
Y
C
E
B
D
A
⎛W ⎞
1
⎜ ⎟ =
⎝ L ⎠ L 1.68
CBA and EDA paths contain three devices
⎛W ⎞
⎛ 4.71⎞ 14.1
= 3⎜
⎜ ⎟
⎟=
1
⎝ L ⎠ A− E
⎝ 1 ⎠
6-35
6.99
+2.5 V
ML
Y
E
D
B
C
A
⎛W ⎞ 1.11
⎜ ⎟ =
1
⎝ L ⎠L
⎛W ⎞
2.22
⎜ ⎟ =
1
⎝ L ⎠E
DCA path contains three devices
⎛W ⎞
⎛ 2.22 ⎞ 6.66
= 3⎜
⎜ ⎟
⎟=
1
⎝ L ⎠ A,C , D ⎝ 1 ⎠
⎛W ⎞
1
1
1
3.33
+
=
→⎜ ⎟ =
⎛ 6.66 ⎞
⎛W ⎞
⎛ 2.22 ⎞ ⎝ L ⎠
1
B
⎜
⎟
⎜ ⎟
⎜
⎟
⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠
6.100
[
]
Y = A(B + D)(C + E )+ (C + E )G + F = (C + E ) A(B + D)+ G + F
⎛ W⎞
1.81 3.62
=
|
⎜ ⎟ =2
1
1
⎝ L ⎠L
⎛ W⎞
⎛ 2.22 ⎞ 4.44
⎜ ⎟ = 2⎜
⎟=
1
⎝ L ⎠F
⎝ 1 ⎠
6-36
⎡ ⎛ 2.22 ⎞⎤ 13.3
⎛ W⎞
=
2
⎢3⎜
⎜ ⎟
⎟⎥ =
1
⎝ L ⎠ A− E
⎣ ⎝ 1 ⎠⎦
|
⎛ W⎞
1
1
1
6.67
+
=
→⎜ ⎟ =
⎛ W⎞
⎛13.3 ⎞
2.22 ⎝ L ⎠G
1
⎜ ⎟
⎜
⎟ 2
1
⎝ L ⎠G ⎝ 1 ⎠
6.101
(a)
+2.5 V
ML
Y
E
D
B
C
A
⎛W ⎞ 1.81
⎜ ⎟ =
1
⎝ L ⎠L
⎛W ⎞
2.22
⎜ ⎟ =
1
⎝ L ⎠E
DCA path contains three devices
⎛W ⎞
⎛ 2.22 ⎞ 6.66
= 3⎜
⎜ ⎟
⎟=
1
⎝ L ⎠ A,C , D ⎝ 1 ⎠
⎛W ⎞
1
1
1
3.33
+
=
→⎜ ⎟ =
⎛ 6.66 ⎞
⎛W ⎞
⎛ 2.22 ⎞ ⎝ L ⎠
1
B
⎜
⎟
⎜ ⎟
⎜
⎟
⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠
(b) Device E remains the same.
0.20
VL 0.20
V
=
= 0.0667V | B : VDS = 2 L = 2
= 0.133V
3
3
3
3
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.43
100μA⎜ ⎟ ⎜ 2.5 − .6 −
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A⎝
⎝ L ⎠A
A, C , D : VDS =
(
)
VTNB = VTNC = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V
⎛W ⎞ ⎛
⎛W ⎞
0.133⎞
3.45
100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 −
⎟0.133 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.74
100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 −
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠C ⎝
⎝ L ⎠C
(
)
VTND = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
7.09
100μA⎜ ⎟ ⎜ 2.5 − 0.133 − 0.641 −
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠D ⎝
⎝ L ⎠D
6-37
6.102
⎛W ⎞
2.22
Device A remains the same. ⎜ ⎟ =
1
⎝ L ⎠A
VL 0.20
=
= 0.100V
2
2
⎛W ⎞ ⎛
⎛W ⎞
0.100 ⎞
4.32
100μA⎜ ⎟ ⎜2.5 − 0.6 −
⎟0.100 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠C, D ⎝
⎝ L ⎠C , D
B,C, D : VDS =
(
)
VTNB = 0.6 + 0.5 0.100 + 0.6 − 0.6 = 0.631V
⎛W ⎞ ⎛
⎛W ⎞
0.100 ⎞
4.65
100μA⎜ ⎟ ⎜ 2.5 − 0.100 − 0.631−
⎟0.100 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
6.103
The load device remains the same.
VL 0.20
V
0.20
=
= 0.0667V | A : VDS = 2 L = 2
= 0.133V
3
3
3
3
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.43
100μA⎜ ⎟ ⎜ 2.5 − .6 −
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
B,C, D : VDS =
(
)
VTNA = VTND = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V
⎛W ⎞ ⎛
⎛W ⎞
0.133⎞
3.45
100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621−
⎟0.133 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A⎝
⎝ L ⎠A
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.74
100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621−
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠D ⎝
⎝ L ⎠D
(
)
VTNC = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
7.09
100μA⎜ ⎟ ⎜ 2.5 − 0.133 − 0.641−
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠C ⎝
⎝ L ⎠C
6-38
6.104
The load device remains the same.
VL 0.20
V 0.20
=
= 0.10V | C, D : VDS = L =
= 0.050V
2
4
2
4
⎛W ⎞ ⎛
⎛W ⎞
0.10 ⎞
4.32
100μA⎜ ⎟ ⎜ 2.5 − .6 −
⎟0.10 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B ⎝
⎝ L ⎠B
A, B : VDS =
(
)
VTNA = VTND = 0.6 + 0.5 0.1+ 0.6 − 0.6 = 0.631V
⎛W ⎞ ⎛
⎛W ⎞
0.10 ⎞
4.65
100μA⎜ ⎟ ⎜ 2.5 − 0.10 − 0.631−
⎟0.10 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A⎝
⎝ L ⎠A
⎛W ⎞ ⎛
⎛W ⎞
0.05⎞
9.17
100μA⎜ ⎟ ⎜ 2.5 − 0.10 − 0.631−
⎟0.05 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠D ⎝
⎝ L ⎠D
(
)
VTNC = 0.6 + 0.5 0.15 + 0.6 − 0.6 = 0.646V
⎛W ⎞ ⎛
⎛W ⎞
0.05⎞
9.53
100μA⎜ ⎟ ⎜ 2.5 − 0.15 − 0.646 −
⎟0.05 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠C ⎝
⎝ L ⎠C
6.105
Device E and the load device remain the same. In the worst case, for paths BCD or ADE
VL 0.20
=
= 0.0667V
3
3
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.43
100μA⎜ ⎟ ⎜2.5 − .6 −
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠B , E ⎝
⎝ L ⎠B , E
A, B,C, D, E : VDS =
(
)
VTND = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
6.74
100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621−
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠D ⎝
⎝ L ⎠D
(
)
VTNA = VTNC = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V
⎛W ⎞ ⎛
⎛W ⎞
0.0667 ⎞
7.09
100μA⎜ ⎟ ⎜2.5 − 0.133 − 0.641−
⎟0.0667 = 80μA → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠ A, C ⎝
⎝ L ⎠ A,C
6-39
6.106
A
0
(a) 0
1
1
B Y
0 1
1 0
0
1
(b) Y = AB + AB = A ⊕ B
0
1
(c) Assuming equal voltage drops (0.10V) across MP and MS :
MP must carry one unit of load current with one - half the drain - source
⎛ W⎞
4.44
voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). → ⎜ ⎟ =
1
⎝ L ⎠P
MS must carry two units of load current with one - half the drain - source
⎛ W⎞
8.88
voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). → ⎜ ⎟ =
1
⎝ L ⎠S
(d) MS will not change. MP will need to be somewhat larger.
(e) Coincidence gate (Exclusive NOR)
6.107
Original design 0.20 mW - 1 mW requires 5 times larger current.
⎛ W⎞
28.8kΩ
2.22 11.1
(a) R =
= 5.76kΩ ⎜ ⎟ = 5
=
5
1
1
⎝ L ⎠S
⎛ W⎞
1
2.98
(b) ⎜ ⎟ = 5
=
1.68
1
⎝ L ⎠L
⎛ W⎞
1
1
(c) ⎜ ⎟ = 5
=
5.72 1.14
⎝ L ⎠L
⎛ W⎞
1.81 9.05
(d) ⎜ ⎟ = 5
=
1
1
⎝ L ⎠L
⎛ W⎞
4.71 23.6
=
⎜ ⎟ =5
1
1
⎝ L ⎠S
⎛ W⎞
2.22 11.1
=
⎜ ⎟ =5
1
1
⎝ L ⎠S
⎛ W⎞
2.22 11.1
=
⎜ ⎟ =5
1
1
⎝ L ⎠S
⎛ W⎞
1.11 5.55 ⎛ W ⎞
2.22 11.1
(e) ⎜ ⎟ = 5
=
=
⎜ ⎟ =5
1
1
1
1
⎝ L ⎠L
⎝ L ⎠S
6.108
⎛ W⎞
1.81 7.24
=
|
⎜ ⎟ =4
1
1
⎝ L ⎠L
⎛ W⎞
⎛ 2.22 ⎞ 8.88
⎜ ⎟ = 4⎜
⎟=
1
⎝ L ⎠F
⎝ 1 ⎠
6-40
⎡ ⎛ 2.22 ⎞⎤ 26.6
⎛ W⎞
= 4 ⎢3 ⎜
⎜ ⎟
⎟⎥ =
1
⎝ L ⎠ A− E
⎣ ⎝ 1 ⎠⎦
⎛ W ⎞ 13.3
1
1
1
|
+
=
→⎜ ⎟ =
⎛ W⎞
⎛ 26.6 ⎞
2.22
1
⎝ L ⎠G
⎜ ⎟
⎜
⎟ 4
1
⎝ L ⎠G ⎝ 1 ⎠
6.109
⎛ W ⎞ 1 ⎛1.81⎞
1
⎜ ⎟ = ⎜
⎟=
⎝ L ⎠ L 4 ⎝ 1 ⎠ 2.21
⎛ W⎞
1 ⎡ ⎛ 2.22 ⎞⎤ 1.67
| ⎜ ⎟
= ⎢3 ⎜
⎟⎥ =
1
⎝ L ⎠ A− F 4 ⎣ ⎝ 1 ⎠⎦
6.110
⎛ W⎞
⎛ 1.81⎞ 5.43
⎛ W⎞
⎛ 6.66 ⎞ 20.0
⎛ W⎞
⎛ 3.33 ⎞ 9.99
(a) ⎜⎝ L ⎟⎠ = 3⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 3⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 3⎜⎝ 1 ⎟⎠ = 1
L
BCD
A
⎛ W ⎞ 1 ⎛1.81⎞
⎛ W⎞
⎛ W⎞
1
1
1 ⎛ 6.66 ⎞ 1.33
1 ⎛ 3.33 ⎞
(b) ⎜⎝ L ⎟⎠ = 5 ⎜⎝ 1 ⎟⎠ = 2.76 | ⎜⎝ L ⎟⎠ = 5 ⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 5 ⎜⎝ 1 ⎟⎠ = 1.50
L
BCD
A
6.111
⎛ W⎞
⎛ W⎞
⎛ W⎞
⎛ 1 ⎞
1
1
1
1 ⎛1.81⎞
1 ⎛ 4.44 ⎞
(a) ⎜⎝ L ⎟⎠ = 10 ⎜⎝ 1 ⎟⎠ = 5.53 | ⎜⎝ L ⎟⎠ = 10 ⎜⎝ 1 ⎟⎠ = 2.25 | ⎜⎝ L ⎟⎠ = 2⎜⎝ 2.25⎟⎠ = 1.13
L
AB
CD
⎞
⎛ W ⎞ 2.5 ⎛1.81⎞ 4.53
⎛ W⎞
⎛
⎛
⎛ 11.1⎞ 22.2
2.5 4.44 11.1
W⎞
(b) ⎜⎝ L ⎟⎠ = 1 ⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 1 ⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 2⎜⎝ 1 ⎟⎠ = 1
L
AB
CD
6.112
⎛ W⎞
⎛ 1.11⎞ 4.44
⎛ W⎞
⎛ 6.66 ⎞ 26.6
(a) ⎜⎝ L ⎟⎠ = 4⎜⎝ 1 ⎟⎠ = 1 | ⎜⎝ L ⎟⎠ = 4⎜⎝ 1 ⎟⎠ = 1
L
ABCDE
⎞
⎛ W ⎞ 1 ⎛1.11⎞
⎛
1
W
1 ⎛ 6.66 ⎞ 2.22
(b) ⎜⎝ L ⎟⎠ = 3 ⎜⎝ 1 ⎟⎠ = 2.70 | ⎜⎝ L ⎟⎠ = 3 ⎜⎝ 1 ⎟⎠ = 1
L
ABCDE
6.113
⎛W ⎞
2
2
1
1 ε ⎛W ⎞
(a) I D = μnCox" ⎜ ⎟(VGS − VTN ) = μn ox ⎜ ⎟(VGS − VTN )
2
2 Tox ⎝ L ⎠
⎝ L⎠
⎛W ⎞
2
1 ε ⎜ ⎟
I*
I D* = μn ox ⎜ 2 ⎟(VGS − VTN ) = 2I D | D = 2
ID
2 Tox ⎜ L ⎟
2 ⎝2⎠
(b) PD* = V (2I )= 2VI = 2PD - Power dissipation has increased by a factor of two.
6-41
6.114
dv
| Assume the transition occurs in ΔT seconds generating
dt
2.5V
a current pulse with constant amplitude I = 10x10−12 F
.
ΔT
2.5x10−11 ΔT
Then I avg =
= 500μA and P = 64(2.5V )I avg = 64(2.5)(0.50mA)= 80 mW
50ns
ΔT
⎛ 3.3 ⎞2
2
(b) P ∝V so P = 80mW ⎜ ⎟ = 139 mW
⎝ 2.5 ⎠
For each line : i = C
6.115
C
C
τ PHL ∝
and τ PLH ∝
KS
KL
C
C"oxWL
L2
| For either case, τ PHL ∝
=
=
μn
KS
" W
μnCox
L
6.116
τP =
PDP 100 fJ
10−13 J
=
= −4 = 1 ns
PD
100μW 10 W
6.117
2.5 + 0.20
= 1.35V
2
= 0.25 + 0.23 = 0.48V
VH = 2.5V | VL = 0.20V | V50% =
V90% = 2.5 − 0.23 = 2.27V | V10%
(a) vI : t r = 22.5 −1.5 = 21 ns | vO : t r = 81− 58 = 23 ns
vI : t f = 62 − 55 = 7 ns
| vO : t r = 12.5 − 6 = 6.5 ns
(b) τ PHL = 2.5 ns | τ PLH = 7 ns (c) τ P =
6.118
(a) T = 301(τ PHL + τ PLH ) = 602
2.5 + 7
= 4.8 ns
2
(τ PHL + τ PLH ) = 602τ
P = 602(0.1ns) = 60.2 ns
2
(b) An even number of inverters has a potential steady state and may not oscillate.
6-42
6.119
t r = 2.2RC = 2.2(28.8kΩ)(0.5 pF )= 31.7 ns
t f ≅ 3.7RonS C =
3.7(0.5 pF )
3.7C
=
= 4.39 ns
Kn (VGS − VTN ) 2.22 10−4 (2.5 − 0.6)
( )
τ PLH = 0.69RC = 0.69(28.8kΩ)(0.5 pF )= 9.94 ns
τ PHL ≅ 1.2RonS C =
1.2(0.5 pF )
1.2C
=
= 1.78 ns
Kn (VGS − VTN ) 2.22 10−4 (2.5 − 0.6)
( )
τP =
9.94 + 1.78
= 5.86 ns
2
τP =
9.94 + 1.00
= 5.47 ns
2
6.120
t r = 2.2RC = 2.2(28.8kΩ)(0.5 pF )= 31.7 ns
t f ≅ 3.7RonS C =
3.7(0.5 pF )
3.7C
=
= 3.09 ns
Kn (VGS − VTN ) 2.22 10−4 (3.3 − 0.6)
( )
τ PLH = 0.69RC = 0.69(28.8kΩ)(0.5 pF )= 9.94 ns
τ PHL ≅ 1.2RonS C =
1.2(0.5 pF )
1.2C
=
= 1.00 ns
Kn (VGS − VTN ) 2.22 10−4 (3.3 − 0.6)
6.121
Resistive Load : τ P =
( )
τ PLH + τ PHL
2
VH = 2.5V
τ PLH = 0.69RC and τ PHL = 1.2RonS C for RonS =
VL = 0.20V
VTNS = 0.6V
C
C
0.526C
=
=
KS
KS (VH − VTNS ) KS (2.5 − 0.6)
⎛
V ⎞
2.5 − 0.20
2.5 − VL
= KS ⎜VGS − VTN − L ⎟VL → KS R =
= 6.39
⎛
R
2⎠
0.20 ⎞
⎝
⎜2.5 − 0.6 −
⎟0.20
2 ⎠
⎝
⎡
⎛ 6.39 ⎞ 0.526⎤
⎛W ⎞
1
987 16.5
μA
2.5ns = (1pF )⎢0.69⎜
=
and R = 6.47 kΩ
⎥ → KS = 987 2 | ⎜ ⎟ =
⎟+
KS ⎦
2
1
V
⎝ L ⎠ S 60
⎝ KS ⎠
⎣
I DDL =
(2.5 − 0.20)V = 356μA
6.47kΩ
| P=
2.5(356μA)
2
= 0.444 mW
6.122
Resistive load inverter – λ has very little effect on the results:
λ = 0 : t r = 3.8ns t f = 1.3ns τ PLH = 10.0ns τ PHL = 1.6ns
λ = 0.04/V : t r = 31.6ns t f = 3.6ns τ PLH = 9.9ns τ PHL = 1.5ns
6.123
6-43
Ignore body effect for simplicity. Equate drain currents to find VL :
⎛
2
V ⎞
1 Kn'
2.5 − VL − 0.6) = 4Kn' ⎜2.5 − 0.6 − 0.6 − L ⎟VL → VL = 0.156V
(
2 2
2⎠
⎝
1
1
RonL =
=
= 19.1kΩ
KL (VGS − VTN ) 0.5 60x10−6 (2.5 − 0.156 − 0.6)
(
RonS =
1
=
)
(
)
1
KS (VGS − VTN ) 4 60x10−6 (2.5 − 0.6 − 0.6)
= 3.21kΩ
t r ≅ 11.9RonLC = 11.9(0.5 pF )(19.1kΩ) = 114 ns
t f ≅ 3.7RonS C = 3.7(0.5 pF )(3.21kΩ) = 5.94 ns
τ PLH ≅ 3.0RonLC = 3.0(0.5 pF )(19.1kΩ)= 28.7 ns
τ PHL ≅ 1.2RonS C = 1.2(0.5 pF )(3.21kΩ)= 1.93 ns
τP =
28.7 + 1.93
= 15.3 ns
2
6.124
Ignore body effect for simplicity. Equate drain currents to find VL :
VTN = 0.6 → VH = 3.3 − 0.6 = 2.7V
(
)
−6
⎛
2
4
VL ⎞
1 60x10
−6
60x10 ⎜2.7 − 0.6 − ⎟VL =
3.3 − VL − 0.6)
(
2⎠
2
1
2
⎝
(
)
9VL2 − 39.0VL + 7.29 = 0 → VL = 0.196V
RonL =
RonS =
(
0.5 60x10
(
4 60x10
−6
)
1
(3.3 − 0.196 − 0.6)
1
−6
)(2.7 − 0.6)
= 13.3kΩ
= 1.98kΩ
t r = 11.9RonLC = 11.9(13.3kΩ)(0.3 pF )= 47.5 ns
τ PLH = 3.0RonLC = 3.0(13.3kΩ)(0.3 pF ) = 12.0 ns
t f = 3.7RonS C = 3.7(1.98kΩ)(0.3 pF )= 2.20 ns
τ PHL = 1.2RonS C = 1.2(1.98kΩ)(0.3 pF )= 0.713 ns
τP =
6-44
12.0 + 0.713
= 6.36 ns
2
6.125
(
)
2 2x10−9 s
3.0RonL + 1.2RonS
τP =
=
C → 3.0RonL + 1.2RonS =
= 4000Ω
2
2
10−12 F
⎛
2
V ⎞
K
KS ⎜VGSS − VTNS − L ⎟VL = L (VGSL − VTNL ) Ignore body effect for simplicity.
2⎠
2
⎝
⎛
2
0.25 ⎞
KL
KS ⎜ 2.5 − 0.6 − 0.6 −
2.5 − 0.25 − 0.6) → KS = 4.63KL
⎟0.25 =
(
2
2 ⎠
⎝
τ PLH + τ PHL
3.0
1.2
+
= 4000Ω → KL = 5.04x10−4 A/V 2
KL (2.5 − .25 − 0.6) 4.63KL (2.5 − 0.6 − 0.6)
⎛W ⎞ 5.04x10−4 8.41
=
⎜ ⎟ =
−5
1
⎝ L ⎠ L 6.0x10
6.126
VH = 2.5V VL = 0.2V
1
RonL =
KL (VGS − VTN )
=
RonS =
1
=
⎛W ⎞
8.41
= 38.9
⎜ ⎟ = 4.63
1
⎝ L ⎠S
(
)
VTNL = 0.6 + 0.5 0.2 + 0.6 − 0.6 = 0.66V
(
6x10
−5
(
)
5.72
= 30.4kΩ
1
= 3.95kΩ
(4 − 0.2 − 0.66)
KS (VGS − VTN ) 2.22 6x10
−5
)
(2.5 − 0.6)
t r ≅ 3.7RonLC = 3.7(0.7 pF )(30.4kΩ) = 78.7 ns
t f ≅ 3.7RonS C = 3.7(0.7 pF )(3.95kΩ) = 10.2 ns
τ PLH ≅ 0.69RonLC = 0.69(0.7 pF )(30.4kΩ)= 14.7 ns
τ PHL ≅ 1.2RonS C = 1.2(0.7 pF )(3.95kΩ)= 3.32 ns
τP =
14.7 + 3.32
= 9.00 ns
2
6-45
6.127
3.0V
2.0V
1.0V
0V
3.0V
2.0V
1.0V
0V
Results: tf = 1.0 ns, tr = 22.3 ns, τPHL = 0.47 ns, τPLH = 4.0 ns, τP = 4.2 ns
6.128
(
)
2 3x10−9 s
3.6RonL + 1.2RonS
τP =
=
C → 3.6RonL + 1.2RonS =
= 6000Ω
2
2
10−12 F
⎛
2
V ⎞
K
KS ⎜VGSS − VTNS − L ⎟VL = L (−VTNL ) Ignore body effect for simplicity.
2⎠
2
⎝
⎛
0.25 ⎞
KL 2
KS ⎜ 3 − 0.6 −
⎟0.25 =
(3) → KS = 7.91KL
2
2 ⎠
⎝
3.6
1.2
+
= 6000Ω → KL = 2.11x10−4 A/V 2
KL (3) 7.91KL (3 − 0.6)
τ PLH + τ PHL
⎛W ⎞ 2.11x10−4 3.52
=
⎜ ⎟ =
−5
1
⎝ L ⎠ L 6.0x10
t r = 8.1RonL
6-46
⎛W ⎞
3.52
= 27.8
⎜ ⎟ = 7.91
1
⎝ L ⎠S
( ) = 12.8 ns
C=
3.52(6x10 )(3)
8.1 10−12
−5
t f = 3.7RonS C =
( ) = 0.924 ns
27.8(6x10 )(3 − 0.6)
3.7 10−12
−5
6.129
(
)
2 10−9 s
3.6RonL + 1.2RonS
τP =
=
C → 3.6RonL + 1.2RonS =
= 10.0kΩ
2
2
0.2x10−12 F
⎛
2
V ⎞
K
KS ⎜VGSS − VTNS − L ⎟VL = L (−VTNL ) Ignore body effect for simplicity.
2⎠
2
⎝
⎛
KL 2
0.20 ⎞
KS ⎜ 3.3 − 0.75 −
⎟0.20 =
(2) → KS = 4.08KL
2
2 ⎠
⎝
τ PLH + τ PHL
1.2
3.6
+
= 10kΩ → KL = 1.92x10−4 A/V 2
KL (2) 4.08KL (3.3 − 0.75)
⎛W ⎞ 1.92x10−4 3.19
=
⎜ ⎟ =
−5
1
⎝ L ⎠ L 6.0x10
I DD =
⎛W ⎞
3.19
= 13.0
⎜ ⎟ = 4.08
1
⎝ L ⎠S
3.3(384μA)
2
KL
1.92x10−4 2
−V
=
2) = 384μA P =
= 0.634 mW
(
(
TNL )
2
2
2
6.130
(a) VTN = 0.6 + 0.5 0.20 + 0.6 − 0.6 = 0.660V
(
)
⎛W ⎞
2
100x10−6 ⎛W ⎞
1
⎜ ⎟ (2.5 − 0.20 − 0.66) → ⎜ ⎟ =
2
⎝ L ⎠L
⎝ L ⎠ L 1.68
80x10−6 =
(b) 80x10
−6
=
⎛W ⎞
2
100x10−6 ⎛W ⎞
1
⎜ ⎟ (2.5 − 0.20 − 0.6) → ⎜ ⎟ =
2
⎝ L ⎠L
⎝ L ⎠ L 1.81
⎛W ⎞ ⎛
⎛W ⎞
2.3 ⎞
1
= 100x10−6⎜ ⎟ ⎜4 − 0.20 − 0.66 −
⎟2.3 → ⎜ ⎟ =
2 ⎠
⎝ L ⎠L ⎝
⎝ L ⎠ L 5.72
⎛W ⎞ ⎛
⎛W ⎞
2.3 ⎞
1
(d ) 80x10−6 = 100x10−6⎜⎝ L ⎟⎠ ⎜⎝4 − 0.20 − 0.6 − 2 ⎟⎠2.3 → ⎜⎝ L ⎟⎠ = 5.89
L
L
(c) 80x10
−6
(e) V
TN
(
)
= −1+ 0.5 0.20 + 0.6 − 0.6 = −0.940V
⎛W ⎞ 1.81
2
100x10 ⎛W ⎞
⎜ ⎟ (−0.940) → ⎜ ⎟ =
2
1
⎝ L ⎠L
⎝ L ⎠L
⎛W ⎞ 1.60
2
100x10−6 ⎛W ⎞
(f ) 80x10−6 = 2 ⎜⎝ L ⎟⎠ (−1) → ⎜⎝ L ⎟⎠ = 1
L
L
80x10−6 =
−6
6-47
6.131
For VDD = -2.5 V, we have VH = -0.20 V with a power dissipation of 0.20 mW. Since these
gates are all ratioed logic design, the ratio of the W/L ratios of the load and switching
transistors does not change. We only need to scale both equally to achieve the power level.
⎛ W ⎞ 100 2.22 5.55
=
(a) RL = 28.8kΩ | ⎜ ⎟ =
1
⎝ L ⎠ S 40 1
⎛W ⎞ 100 1
⎛ W ⎞ 100 4.71 11.8
1.49
(b) ⎜ ⎟ =
=
| ⎜ ⎟ =
=
1
1
⎝ L ⎠ L 60 1.68
⎝ L ⎠ S 60 1
⎛W ⎞ 100 1
⎛ W ⎞ 100
1
5.55
(c) ⎜ ⎟ =
=
| ⎜ ⎟ =
2.22 =
1
⎝ L ⎠ L 60 5.72 2.29
⎝ L ⎠ S 60
⎛W ⎞ 100 1.81 4.53
⎛ W ⎞ 100 2.22 5.55
(d ) ⎜ ⎟ =
=
| ⎜ ⎟ =
=
1
1
⎝ L ⎠ L 60 1
⎝ L ⎠ S 60 1
⎛W ⎞ 100 1.11 2.78
⎛ W ⎞ 100 2.22 5.55
(e) ⎜ ⎟ =
=
| ⎜ ⎟ =
=
1
1
⎝ L ⎠ L 60 1
⎝ L ⎠ S 60 1
6.132
⎛2⎞
⎛
2
−V ⎞
1 ⎛ 25x10−6 ⎞
VL = −2.5 + 0.6 = −1.9 V | ⎜ ⎟ 25x10-6 ⎜ −1.9 − (−0.6)− H ⎟(−VH )= ⎜
⎟ −2.5 − VH − (−0.6)
2 ⎠
4⎝ 2 ⎠
⎝1⎠
⎝
(
)
9VH2 − 24.6VH + 3.61 = 0 → VH = −0.156 V
6.133
Pretend this is an NMOS gate with VDD = 3.3V and VL = 0.33V
[
)]
(
VH = 3.3 − 0.6 + 0.75 VH + 0.7 − 0.7 → VH = 2.08V
(
)
0.1mW
= 30.3μA
3.3V
⎛W ⎞ 0.303
2
40μA ⎛ W ⎞
1
30.3μA =
=
⎜ ⎟ (3.3 − 0.33 − 0.734) → ⎜ ⎟ =
1
3.30
2 ⎝ L ⎠L
⎝ L ⎠L
⎛W ⎞ ⎛
⎛W ⎞ 1.75
0.33⎞
30.3μA = 40μA⎜ ⎟ ⎜ 2.08 − 0.60 −
⎟0.33 → ⎜ ⎟ =
2 ⎠
1
⎝ L ⎠S ⎝
⎝ L ⎠S
VTNL = 0.6 + 0.75 0.33 + 0.7 − 0.7 = 0.734 | I DD =
6-48
(
)
6.134
[
)]
(
VL = −VTPL | VL = − −0.6 − 0.75 2.5 − VL + 0.7 − 0.7 → VL = 1.07V
⎛W ⎞ ⎛
K 'p ⎛W ⎞
2
VDS ⎞
K ⎜ ⎟ ⎜VGS − VTPS −
⎟VDS =
⎜ ⎟ (VGSL − VTPL )
2 ⎠
2 ⎝ L ⎠L
⎝ L ⎠S ⎝
2
VH − 2.5⎞
1 ⎛ 1⎞
3⎛
⎜1.07 − 2.5 − (−0.6)−
⎟(VH − 2.5) = ⎜ ⎟(VH + VTPL )
2 ⎝ 3⎠
1⎝
2 ⎠
'
p
(
)
VTPL = −0.6 − 0.75 VH + 0.7 − 0.7 Solving the last two equations iteratively : VH = 2.30 V
6.135
Y is low only when both A and B are high : Y = AB or Y = AB.
Alternatively, Y is high when either A or B is low : Y = A + B = AB
6.136
Y is high only when both A and B are low : Y = AB or Y = A + B
6.137
0.0V
-0.5V
-1.0V
-1.5V
2 0V
VL = -1.90 and VH = -0.156 agree with the hand calculations in Prob. 6.132
6.138
-0.0V
-1.0V
-2.0V
3 0V
VH = -0.33 and VL -2.08 agree with the design values in Prob. 6.133.
6-49
6.139
0.0V
-0.5V
-1.0V
-1.5V
2 0V
tr = 16.8 ns, tf = 560 μs, τPLH = 11.7 ns, τPHL = 60 ns, τP = 35.9 ns
6.140
-0.0V
-1.0V
-2.0V
-3.0V
0s
0 2us
tr = 46 ns, tf = 1.1
6-50
0 4us
0 6us
0 8us
1 0us
1 2us
s, τPLH = 21 ns, τPHL = 122 ns, τP = 72 ns
1 4us
1 6us
1 8us
2 0us
CHAPTER 7
7.1
(
−14
cm 2 ⎞ (3.9) 8.854x10 F / cm
3.9εo ⎛
K = µ nC = µ n
= µn
= ⎜ 500
⎟
Tox
Tox
V − sec ⎠ 10x10−9 m(100cm / m)
⎝
'
n
εox
"
ox
)
F
A
µA
= 173 x 10−6 2 = 173 2
V − sec
V
V
⎛
⎞
µ
200
µA
µA
K 'p = µ pCox" = p Kn' = ⎜
⎟173 2 = 69.1 2
µn
V
V
⎝ 500 ⎠
Kn' = 173x10−6
7.2
V
DD
vI
(5 V)
B
V
S
D
p+
p+
n+
vo
D
SS
(0 V)
S
n+
n+
B
p+
p-well
PMOS transistor
Ohmic
contact
NMOS transistor
n-type
substrate
Ohmic
contact
7.3
⎛
pA ⎞
A = ⎜ 500 2 ⎟(1cm x 0.5cm)= 250 pA
cm ⎠
⎝
⎛
pA ⎞
(b) I = I S A + 20x106 ⎜⎝100 cm2 ⎟⎠ 2x10−4 cm 5x10−4 cm = 250 + 200 = 450 pA
(a) I = I
S
(
)
(
)(
)
(c) Same as (b)
7.4
F ⎞⎛ 10mm 0.1cm ⎞
⎛
3.9⎜ 8.854 x10 −14
⎟⎜
⎟(1µm )
⎛ ε ox A ⎞
3.9ε o LW
cm ⎠⎝ 2
mm ⎠
⎝
⎟⎟ = 3
=3
C = 3⎜⎜
= 0.518 pF
t ox
1µm
⎝ t ox ⎠
7.5
(a) VH = 2.5 V, VL = 0 V
(b) VH = 1.8 V, VL = 0 V
7-1
7.6
(a) VH = 2.5 V, VL = 0 V
(b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate.
7.7
(a) VH = 3.3 V, VL = 0 V
(b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate.
7.8
(a) VH = 2.5V | VL = 0V | For MN , VGS = 0, so MN is cut off.
For MP , VGS = −2.5, VDS = 0V and VTP = -0.60V. For VDS < VGS - VTP , M P is in the triode region.
(b) For M
N
, VGS = 2.5, VDS = 0V and VTN = 0.60V. For VDS < VGS - VTN , M N is in the triode region.
For MP , VGS = 0, so M P is cut off.
(c) For M
N
, VGS = 1.25, VDS = 1.25 V and VTN = 0.60V. For VDS > VGS - VTN , MN is saturated.
For MP , VGS = −1.25, VDS = -1.25V and VTP = -0.75V. For VDS > VGS - VTP , MP is saturated.
7.9
(a) VH = 3.3V | VL = 0V | For MN , VGS = 0, so MN is cut off.
For MP , VGS = −3.3, VDS = 0V and VTP = -0.75V. For VDS < VGS - VTP , M P is in the triode region.
(b) For M
N
, VGS = 3.3, VDS = 0V and VTN = 0.75V. For VDS < VGS - VTN , MN is in the triode region.
For MP , VGS = 0, so MP is cut off.
(c) For M
N
, VGS = 1.65, VDS = 1.65 V and VTN = 0.75V. For VDS > VGS - VTN , MN is saturated.
For MP , VGS = −1.65, VDS = -1.65V and VTP = -0.75V. For VDS > VGS - VTP , MP is saturated.
7.10
(a) VH = 0 V, VL = -5.2 V
(b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate.
7-2
7.11
For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
(a) Both transistors are saturated with VDS = VGS
K
2
2
Kn
vI − VTN ) = p (v I − VDD − VTP ) so vI − VTN = VDD − vI + VTP
(
2
2
V + VTN + VTP 2.5 + 0.6 − .6
vO = vI = DD
=
= 1.25V
2
2
2
2
K
100µA ⎛ 2 ⎞
(b) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(1.25 − 0.6) = 42.3µA
K
2
2
40µA ⎛ 5 ⎞
Checking I DP = p (vI − VDD − VTp ) =
⎜ ⎟(1.25 − 2.5 + 0.6) = 42.3µA
2
2 ⎝1⎠
(c) For K
n
= 2.5K p ,
2.5K p
K
2
2
vI − VTN ) = p (vI − VDD − VTP ) or 1.58(vI − VTN )= VDD − vI + VTP
(
2
2
V + 1.58VTN + VTP 2.5 + 1.58(0.6)+ (−0.6)
vO = vI = DD
=
= 1.104V
2.58
2.58
2
100µA ⎛ 2 ⎞
(d ) I DN = 2 ⎜⎝ 1 ⎟⎠(1.104 − 0.6) = 25.4µA | Check by finding IDP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(1.104 − 2.5 + 0.6) = 25.3µA
2 ⎝ 1⎠
7.12
(a) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
K
2
2
Kn
vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP
i
()
(
2
2
V + VTN + VTP 3.3 + 0.75 − 0.75
vO = vI = DD
=
= 1.65V
2
2
2
2
K
100µA ⎛ 2 ⎞
(ii) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(1.65 − 0.75) = 81µA
2
2
K
40µA ⎛ 5 ⎞
Checking : I DP = P (vI − VDD + VTP ) =
⎜ ⎟(1.65 − 3.3 + 0.75) = 81µA
2
2 ⎝ 1⎠
For Kn = 2.5 Kp,
7-3
2.5K p
K
2
2
vI − VTN ) = p (v I − VDD − VTP ) so 1.58(vI − VTN )= VDD − vI + VTP
(
2
2
V + 1.58VTN + VTP 3.3 + 1.58(0.75)+ (−0.75)
vO = vI = DD
=
= 1.448V
2.58
2.58
2
100µA ⎛ 2 ⎞
(iv ) I DN = 2 ⎜⎝ 1 ⎟⎠(1.448 − 0.75) = 48.7µA | Check by finding I DP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(1.448 − 3.3 + 0.75) = 48.6µA
2 ⎝1⎠
(iii )
(b) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
K
2
2
Kn
vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP
i
()
(
2
2
V + VTN + VTP 2.5 + 0.6 − 0.6
vO = vI = DD
=
= 1.25V
2
2
2
2
K
100µA ⎛ 2 ⎞
(ii ) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(1.25 − 0.6) = 42.3µA
K
2
2
100µA ⎛ 2 ⎞
Checking : I DP = p (vI − VDD + VTP ) =
⎜ ⎟(1.25 − 2.5 + 0.6) = 42.3µA
2
2 ⎝1⎠
For Kn = 2.5 Kp,
2.5K p
K
2
2
vI − VTN ) = p (v I − VDD − VTP ) and 1.58(v I − VTN ) = VDD − v I + VTP
(
2
2
V + 1.58VTN + VTP 2.5 + 1.58(0.60)+ (−0.60)
vO = vI = DD
=
= 1.104V
2.58
2.58
2
100µA ⎛ 2 ⎞
(iv ) I DN = 2 ⎜⎝ 1 ⎟⎠(1.104 − 0.60) = 25.4µA | Check by finding I DP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(1.104 − 2.5 + 0.60) = 25.3µA
2 ⎝1⎠
(iii )
7-4
7.13
(a) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
K
2
2
Kn
vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP
i
()
(
2
2
V + VTN + VTP 1.8 + 0.5 − 0.5
vO = vI = DD
=
= 0.90V
2
2
2
2
K
100µA ⎛ 2 ⎞
(ii ) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(0.9 − 0.5) = 16.0µA
2
2
K
40µA ⎛ 5 ⎞
Checking : I DP = P (vI − VDD − VTP ) =
⎜ ⎟(0.9 −1.8 + 0.5) = 16.0µA
2
2 ⎝ 1⎠
For Kn = 2.5 Kp,
2.5K p
K
2
2
vI − VTN ) = p (v I − VDD − VTP ) so 1.58(vI − VTN )= VDD − vI + VTP
(
2
2
V + 1.58VTN + VTP 1.8 + 1.58(0.5)+ (−0.5)
vO = vI = DD
=
= 0.810V
2.58
2.58
2
100µA ⎛ 2 ⎞
(iv ) I DN = 2 ⎜⎝ 1 ⎟⎠(0.810 − 0.5) = 96.2µA | Check by finding I DP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(0.8101−1.8 + 0.5) = 96.0µA
2 ⎝1⎠
(iii )
(b) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
K
2
2
Kn
vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP
i
()
(
2
2
V + VTN + VTP 2.5 + 0.75 − 0.65
vO = vI = DD
=
= 1.30V
2
2
2
2
K
100µA ⎛ 2 ⎞
(ii ) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(1.30 − 0.75) = 30.3µA
K
2
2
40µA ⎛ 5 ⎞
Checking : I DP = p (vI − VDD + VTP ) =
⎜ ⎟(1.30 − 2.5 + 0.65) = 30.3µA
2
2 ⎝ 1⎠
7-5
For Kn = 2.5 Kp,
2.5K
K
2
2
(iii ) 2 p (vI − VTN ) = 2p (vI − VDD − VTP ) and 1.58(vI − VTN )= VDD − vI + VTP
V + 1.58VTN + VTP 2.5 + 1.58(0.75)+ (−0.65)
vO = vI = DD
=
= 1.176V
2.58
2.58
2
100µA ⎛ 2 ⎞
(iv ) I DN = 2 ⎜⎝ 1 ⎟⎠(1.176 − 0.75) = 18.2µA | Check by finding I DP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(1.176 − 2.5 + 0.65) = 18.2µA
2 ⎝1⎠
(c) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the
drain currents with Kn = Kp yields:
K
2
2
Kn
vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP
i
()
(
2
2
V + VTN + VTP 2.5 + 0.65 − 0.75
vO = vI = DD
=
= 1.20V
2
2
2
2
K
100µA ⎛ 2 ⎞
(ii ) I DN = 2n (vI − VTN ) = 2 ⎜⎝ 1 ⎟⎠(1.20 − 0.65) = 30.3µA
K
2
2
40µA ⎛ 5 ⎞
Checking : I DP = p (vI − VDD + VTP ) =
⎜ ⎟(1.20 − 2.5 + 0.75) = 30.3µA
2
2 ⎝ 1⎠
For Kn = 2.5 Kp,
2.5K p
K
2
2
vI − VTN ) = p (v I − VDD − VTP ) and 1.58(v I − VTN ) = VDD − v I + VTP
(
2
2
V + 1.58VTN + VTP 2.5 + 1.58(0.65)+ (−0.75)
vO = vI = DD
=
= 1.076V
2.58
2.58
2
100µA ⎛ 2 ⎞
(iv ) I DN = 2 ⎜⎝ 1 ⎟⎠(1.076 − 0.65) = 18.2µA | Check by finding I DP :
2
40µA ⎛ 2 ⎞
I DP =
⎜ ⎟(1.076 − 2.5 + 0.75) = 18.2µA
2 ⎝1⎠
(iii )
7.14
*PROBLEM 7.14 - CMOS INVERTER TRANSFER CHARACTERISTICS
VIN 1 0 DC 0
VDD 3 0 DC 2.5
M1 2 1 0 0 MOSN W=2U L=1U
M2 2 1 3 3 MOSP W=2U L=1U
.DC VIN 0 2.5 .01
*.DC VIN 2.16 2.17 .0001
.MODEL MOSN NMOS KP=10E-5 VTO=0.6 GAMMA=0
.MODEL MOSP PMOS KP=4E-5 VTO=-0.6 GAMMA=0
7-6
.PRINT DC V(2)
.END
Result: vI = 1.104 V
2.5K p
K
2
2
vI − VTN ) = p (VDD − vI + VTP ) and 1.58(vI − VTN )= VDD − vI + VTP
(
2
2
V + 1.58VTN + VTP 2.5 + 1.58(0.6)+ (−0.6)
vO = vI = DD
=
= 1.10 V
2.58
2.58
7.15
(a) VH = 3.3 V. For vO = VL , assume MP is saturated and MN is in the triode region.
⎛ ⎞⎛
K 'p ⎛1⎞
2
VL ⎞
' 4
⎜ ⎟(−3.3 + 0.6) = Kn⎜ ⎟⎜3.3 − 0.6 + ⎟VL
2 ⎝1⎠
2⎠
⎝ 1 ⎠⎝
−5
⎛ 7.29 ⎞ ⎛
4x10
VL ⎞
2
=
2.7
+
⎜
⎟
⎜
⎟VL and rearranging : VL + 5.4VL − 0.729 = 0
−5
2⎠
2 10x10 ⎝ 4 ⎠ ⎝
(
)
VL = 0.132V. Checking the assumptions - For MN , 3.3 - 0.6 > 0.132. Triode region
is correct. For MP , VGS - VTP = -3.3 + 0.6 = -2.7V and VDS = 0.132 - 3.3 = -3.17V.
Saturation region operation is correct.
(b) V
H
= 2.5 V. For vO = VL , assume MP is saturated and MN is in the triode region.
⎛ ⎞⎛
K 'p ⎛1⎞
2
VL ⎞
' 4
⎜ ⎟(−2.5 + 0.6) = Kn⎜ ⎟⎜2.5 − 0.6 + ⎟VL
2 ⎝1⎠
2⎠
⎝ 1 ⎠⎝
−5
⎛ 3.61⎞ ⎛
4x10
VL ⎞
2
=
1.9
+
⎜
⎟
⎜
⎟VL and rearranging : VL + 3.8VL − 0.361 = 0
−5
2⎠
2 10x10 ⎝ 4 ⎠ ⎝
(
)
VL = 0.0928V. Checking the assumptions - For MN , 2.5 - 0.6 > 0.0928. Triode region
is correct. For MP , VGS - VTP = -2.5 + 0.6 = -1.9V and VDS = 0.0928 - 2.5 = -2.41V.
Saturation region operation is correct.
7-7
7.16
For the NMOS device
1.5 mA
+
+5 V
0.6 V
-
⎛
⎞⎛
⎞
⎟0.6 = 1.5x10
(100x10 )⎜⎝WL ⎟⎠ ⎜⎝5 − 0.6 − 0.6
2 ⎠
−6
n
⎛W ⎞ 61.0
⎜ ⎟ =
1
⎝ L ⎠n
For the PMOS device
+5 V
+
2.6 V
60 µA
⎛W ⎞ ⎛
2.6 ⎞
−5
40x10−6 ⎜ ⎟ ⎜ 5 − 0.6 −
⎟2.6 = 6x10
2
L
⎝ ⎠ p⎝
⎠
(
)
⎛W ⎞
1
⎜ ⎟ =
⎝ L ⎠ p 5.37
7-8
−3
7.17
+2.5 V
vO
+2.5 V
Kn = 2000
µA
and K p = 1600
µA
.
V
V2
Therefore the output will be forced below VDD /2.
2
For both transistors, VGS − VTN = 1.9V. Assume that both devices are in the linear region.
⎛ 40 ⎞
⎛
⎛ 20 ⎞
VO − 2.5 ⎞
-5
-5
⎜ ⎟ 4x10 ⎜ −2.5 + 0.6 −
⎟(VO − 2.5)= ⎜ ⎟ 10x10
2
1
1
⎝ ⎠
⎝
⎠
⎝ ⎠
(
)
(
Rearranging : VO2 −14.2VO + 13 = 0 ⇒
⎛
⎞
)⎜⎝2.5 − 0.6 − V2 ⎟⎠V
O
O
VO = 0.9836V | VDSN = 0.984V | VDSP = −1.52V
and the assumed operating regions are correct.
⎛ 40 ⎞
⎛
0.9836 − 2.5 ⎞
I DP = ⎜ ⎟ 4x10-5 ⎜ −2.5 + 0.6 −
⎟(0.9836 − 2.5)= 2.77 mA, and checking
2
⎝1⎠
⎝
⎠
⎛ 20 ⎞
⎛
0.9836 ⎞
I DN = ⎜ ⎟ 10x10-5 ⎜ 2.5 − 0.6 −
⎟0.9836 = 2.77 mA.
2 ⎠
⎝1⎠
⎝
7.18
KR =
(
)
(
)
2(2.5)(2.5 − 0.6 − 0.6) 2.5 − 2.5(0.6)− 0.6
Kn
= 2.5 | VIH =
−
= 1.22V
2.5 −1
Kp
2.5
−1
1+
3
2.5
( )
( )
(
)
(2.5 + 1)(1.22)− 2.5 − 2.5(0.6)+ 0.6 = 0.174V
2(2.5)
2( 2.5 )(2.5 − 0.6 − 0.6) 2.5 − 2.5(0.6)− 0.6
=
−
= 0.902V
2.5 −1
(2.5 −1)( 2.5 + 3)
(2.5 + 1)(0.902)+ 2.5 − 2.5(0.6)+ 0.6 = 2.38V
=
VOL =
VIL
VOH
NMH = VOH
2
− VIH = 2.38 −1.22 = 1.16 V NML = VIL − VOL = 0.902 − 0.174 = 0.728 V
7-9
7.19
VDD − VTN − 3VTP 3.3-0.75-3(-0.75)
=
= 1.20 V
4
4
V + 3VTN + VTP 3.3+3(0.75)-0.75
and NM L = DD
=
= 1.20 V
4
4
2(2.5)(3.3 − 0.75 − 0.75) 3.3 − 2.5(0.75)− 0.75
K
−
= 1.61V
(b) KR = K n = 2.5 | VIH =
2
.
5
−
1
p
(2.5 − 1) 1 + 3(2.5)
(a ) For K
= 1: NM H =
R
(
)
(2.5 + 1)(1.61)− 3.3 − 2.5(0.75)+ 0.75 = 0.242V
2(2.5)
2( 2.5 )(3.3 − 0.75 − 0.75) 3.3 − 2.5(0.75)− 0.75
=
−
= 1.17V
.
5
−
1
2
2
.
5
−
1
2
.
5
+
3
( )(
)
(2.5 + 1)(1.17)+ 3.3 − 2.5(0.75)+ 0.75 = 3.14V
=
VOL =
VIL
VOH
NM H = VOH
2
− VIH = 3.14 − 1.61 = 1.53 V NM L = VIL − VOL = 1.17 − 0.242 = 0.928V
7.20
4.0V
3.0V
2.0V
1.0V
0V
0V
V(M2:d)
0.5V
V(M4:d) V(M6:d)
1.0V
2.0V
1.5V
V_VI
7-10
2.5V
3.0V
3.5V
7.21
(a)
CMOS Noise Margins
(VDD = 3.3 V, VTN = 0.75 V, VTP = -.75 V)
2.5
2
1.5
NMH
NML
1
0.5
0
0
2
4
6
8
10
12
KR
(b)
CMOS Noise Margins
(VDD = 2.0 V, VTN = 0.5 V, VTP = -0.5 V)
1.4
1.2
1
0.8
NML
NMH
0.6
0.4
0.2
0
0
2
4
6
8
10
12
KR
7-11
7.22
(a) t
r
≅ 3.6 RonP C =
t f ≅ 3.6 RonN C =
3.6(0.25 pF )
3.6C
=
= 2.36 ns
K p VGS − VTP 5 4 x10−5 −2.5 + 0.6
(
)
3.6(0.25 pF )
3.6C
=
= 2.36 ns
Kn (VGS − VTN ) 2 10−4 (2.5 − 0.6)
( )
tr
= 0.788 ns
3
t
0.788 + 0.788
τ PHL ≅ 1.2 RonN C = f = 0.788 ns τ P =
= 0.788 ns
2
3
3.6(0.25 pF )
3.6C
(b) tr ≅ 3.6 RonP C = K V − V = 5 4 x10−5 −2.0 + 0.6 = 3.21 ns
p GS
TP
τ PLH = 1.2 RonP C =
(
t f ≅ 3.6 RonN C =
)
3.6(0.25 pF )
3.6C
=
= 3.21 ns
Kn (VGS − VTN ) 2 10−4 (2 − 0.6)
( )
tr
= 1.07 ns
3
t
1.07 + 1.07
τ PHL ≅ 1.2 RonN C = f = 1.07 ns τ P =
= 1.07 ns
2
3
3.6(0.25 pF )
3.6C
(c) tr ≅ 3.6 RonP C = K V − V = 5 4 x10−5 −1.8 + 0.6 = 3.75 ns
p GS
TP
τ PLH = 1.2 RonP C =
(
t f ≅ 3.6 RonN C =
)
3.6(0.25 pF )
3.6C
=
= 3.75 ns
Kn (VGS − VTN ) 2 10−4 (1.8 − 0.6)
( )
tr
= 1.25 ns
3
t
1.25 + 1.25
τ PHL ≅ 1.2 RonN C = f = 1.25 ns τ P =
= 1.25 ns
2
3
τ PLH = 1.2 RonP C =
7.23
t r ≅ 3.6 RonP C =
3.6(0.5 pF )
3.6C
=
= 11.9 ns
K p VGS − VTP 2 4 x10−5 −2.5 + 0.6
t f ≅ 3.6 RonN C =
3.6(0.5 pF )
3.6C
=
= 4.74 ns
Kn (VGS − VTN ) 2 10−4 (2.5 − 0.6)
(
)
( )
tr
= 3.96 ns
3
t
3.96 + 1.58
τ PHL ≅ 1.2 RonN C = f = 1.58 ns τ P =
= 2.77 ns
2
3
τ PLH = 1.2 RonP C =
7-12
7.24
t r ≅ 3.6 RonP C =
3.6(0.15 pF )
3.6C
=
= 1.42 ns
K p VGS − VTP 5 4 x10−5 −2.5 + 0.6
t f ≅ 3.6 RonN C =
3.6(0.15 pF )
3.6C
=
= 1.42 ns
Kn (VGS − VTN ) 2 10−4 (2.5 − 0.6)
(
)
( )
tr
= 0.474 ns
3
t
0.474 + 0.474
τ PHL ≅ 1.2 RonN C = f = 0.474 ns τ P =
= 0.474 ns
2
3
τ PLH = 1.2 RonP C =
7.25
t r ≅ 3.6 RonP C =
3.6(0.2 pF )
3.6C
=
= 1.41 ns
K p VGS − VTP 5 4 x10−5 −3.3 + 0.75
t f ≅ 3.6 RonN C =
3.6(0.2 pF )
3.6C
=
= 1.41 ns
Kn (VGS − VTN ) 2 10−4 (3.3 − 0.75)
(
)
( )
tr
= 0.470 ns
3
t
0.470 + 0.470
τ PHL ≅ 1.2 RonN C = f = 0.470 ns τ P =
= 0.470 ns
2
3
τ PLH = 1.2 RonP C =
7.26
For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL
3ns =
1.2(1pF)
⎛W ⎞
−6
⎜ ⎟ 100 x10 (2.5 − 0.6)
⎝ L ⎠n
(
)
⎛ W ⎞ 2.11
⎛W ⎞
⎛ W ⎞ 5.26
→⎜ ⎟ =
| ⎜ ⎟ = 2.5⎜ ⎟ =
1
1
⎝ L ⎠n
⎝ L ⎠p
⎝ L ⎠n
t r = t f = 3τ PHL = 9.00 ns
7-13
7.27
(a ) For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL
1ns =
1.2(10pF)
⎛W ⎞
−6
⎜ ⎟ 100 x10 (2.5 − 0.6)
⎝ L ⎠n
(
)
⎛ W ⎞ 63.2
⎛W ⎞
⎛ W ⎞ 158
→⎜ ⎟ =
| ⎜ ⎟ = 2.5⎜ ⎟ =
1
1
⎝ L ⎠n
⎝ L ⎠p
⎝ L ⎠n
t r = t f = 3τ PHL = 3.00 ns
(b) 1ns = ⎛W ⎞
1.2(10pF)
(
)
−6
⎜ ⎟ 100 x10 (3.3 − 0.7)
⎝ L ⎠n
⎛W ⎞ 46.2
⎛W ⎞
⎛W ⎞ 115
→⎜ ⎟ =
| ⎜ ⎟ = 2.5⎜ ⎟ =
1
1
⎝ L ⎠n
⎝ L ⎠p
⎝ L ⎠n
t r = t f = 3τ PHL = 3.00 ns
7.28
For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL
0.2ns =
1.2(0.1pF )
⎛W ⎞
−6
⎜ ⎟ 100 x10 (1.5 − 0.5)
⎝ L ⎠n
(
)
⎛ W ⎞ 6.00
⎛W ⎞
⎛ W ⎞ 15.0
→⎜ ⎟ =
| ⎜ ⎟ = 2.5⎜ ⎟ =
1
1
⎝ L ⎠n
⎝ L ⎠p
⎝ L ⎠n
t r = t f = 3τ PHL = 0.600 ns
7.29
For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL
0.4ns =
1.2(0.1pF)
⎛W ⎞
−6
⎜ ⎟ 100 x10 (2.5 − 0.6)
⎝ L ⎠n
(
)
⎛ W ⎞ 1.58
⎛W ⎞
⎛ W ⎞ 3.95
→⎜ ⎟ =
| ⎜ ⎟ = 2.5⎜ ⎟ =
1
1
⎝ L ⎠n
⎝ L ⎠p
⎝ L ⎠n
t r = t f = 3τ PHL = 1.20 ns
7.30
*PROBLEM 7.30 - CMOS INVERTER DELAY
*SIMULATION USES THE MODELS IN APPENDIX B
VIN 1 0 PULSE (0 2.5 0 0.1N 0.1N 10N 20N)
VDD 3 0 DC 2.5
M1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P
M2 2 1 3 3 MOSP W=10U L=2U AS=40P AD=40P
CL 2 0 100FF
.OP
.TRAN 0.1N 20N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
7-14
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PRINT TRAN V(2)
.PROBE V(1) V(2)
.END
Results: tr = 1.7 ns, tf = 2.25 ns, τPHL = 1.1 ns, τPLH = 0.9 ns
τ PHL = 1.2RonnC
τ PLH = 1.2RonpC
1.1x10−9 ⎛ 2 ⎞
−6
=
C1 =
⎜ ⎟ 50x10 (2.5 − 0.91)= 146 fF | Inverter is symmetrical, so
1.29Ronn
1.2 ⎝ 1 ⎠
9x10−10 ⎛ 5 ⎞
146 + 130
τ
−6
C2 = PLH =
fF = 138 fF
⎜ ⎟ 20x10 (2.5 − 0.77) = 130 fF | C =
2
1.2Ronp
1.2 ⎝ 1 ⎠
τ PHL
(
(
)
)
7.31
*PROBLEM 7.31 - FIVE CASCADED INVERTERS
*SIMULATION USES THE MODELS IN APPENDIX B
VDD 1 0 DC 2.5
VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 10N 20N)
*
MN1 3 2 0 0 MOSN W=16U L=2U AS=64P AD=64P
MP1 3 2 1 1 MOSP W=40U L=2U AS=160P AD=160P
*AS=4UM*W - AD=4UM*W
CL1 3 0 100FF
*
MN2 4 3 0 0 MOSN W=16U L=2U AS=64P AD=64P
MP2 4 3 1 1 MOSP W=40U L=2U AS=160P AD=160P
CL2 4 0 100FF
*
MN3 5 4 0 0 MOSN W=16U L=2U AS=64P AD=64P
MP3 5 4 1 1 MOSP W=40U L=2U AS=160P AD=160P
CL3 5 0 100FF
*
MN4 6 5 0 0 MOSN W=16U L=2U AS=64P AD=64P
MP4 6 5 1 1 MOSP W=40U L=2U AS=160P AD=160P
CL4 6 0 100FF
*
MN5 7 6 0 0 MOSN W=16U L=2U AS=64P AD=64P
MP5 7 6 1 1 MOSP W=40U L=2U AS=160P AD=160P
CL5 7 0 100FF
.OP
.TRAN 0.025N 20N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
7-15
.PROBE V(2) V(3) V(5) V(6)
.END
First inverter : t r = 0.83 ns, t f = 1.4 ns, τ PLH = 0.35 ns, τ PHL = 0.42 ns
Fourth inverter : t r = 0.96 ns, t f = 1.0 ns, τ PLH = 0.64 ns, τ PHL = 1.1 ns
τ PHL = 1.2RonnC
C1 =
C2 =
τ PHL
τ PLH = 1.2RonpC
0.42x10−9 ⎛ 8 ⎞
−6
⎜ ⎟ 50x10 (2.5 − 0.91)= 223 fF | The inverter is symmetrical, so
1
1.2
⎝ ⎠
−9 ⎛
0.35x10 20 ⎞
223 + 202
−6
=
fF = 212 fF
⎜ ⎟ 20x10 (2.5 − 0.77)= 202 fF | C =
2
1.2 ⎝ 1 ⎠
(
=
1.2Ronn
τ PLH
1.2Ronp
)
(
)
The average capacitance of 212 fF that is required to fit the results is consistent with the device
capacitances calculated by SPICE. The approximate 3:1 relationship holds between rise/fall
times and the propagation delay times in the first inverter. The first inverter response is faster
than that of the fourth inverter because of the rapid rise and fall times on the input signal. The
first inverter response is closest to our model used for hand calculations. However, the response
of inverter four would be more representative of the actual logic situation.
7.32
(a)
100W
= 1µW / gate
100 x10 6 gates
(b) I =
100W
= 55.6 A
1.8V
7.33
(a)
5W
= 2.5µW / gate
2x106 gates
(b) C =
2.5x10-6
(
2.52 5x106
)
2
P = CVDD
f ;C =
2.5x10-6
(
3.32 5x106
)
= 45.9 fF
= 80.0 fF
7.34
⎛
pA ⎞
A = ⎜ 400 2 ⎟(0.5cm x 1cm)= 200 pA
cm ⎠
⎝
⎛
pA ⎞
(b) I = I S A + 75x106 ⎜⎝150 cm2 ⎟⎠ 2.5x10−4 cm 1x10−4 cm = 200 + 281 = 481pA
(a ) I = I
S
(
)
(
)(
)
(c) Same as (b)
7.35
(
)( )101
2
(a) P = 64CVDD
f = 64 25x10-12 2.52
7.36
7-16
−8
= 1.00 W
(
)( )101
(b) P = 64 25x10-12 3.32
−8
= 1.74 W
Peak current occurs for vO = v I . Assume both transistors are saturated since vO = vI .
2
2
20 ⎛ 40x10−6 ⎞
20 ⎛100x10−6 ⎞
−
V
=
v
⎜
⎟( I
⎜
⎟(v I − VDD − VTP ) →1.58(v I − VTN ) = VDD − v I + VTP
TN )
1⎝ 2 ⎠
2
1⎝
⎠
3.3 + 1.58(0.6)+ (−0.6)
VDD + 1.58VTN + VTP
=
= 1.414 V
2.58
2.58
2
20 ⎛ 100x10−6 ⎞
iD = ⎜
⎟(1.414 − 0.6) = 663 µA
1⎝
2
⎠
(a) v
O
= vI =
Checking the current : iD =
(b) v
O
= vI =
2
20 ⎛ 40x10−6 ⎞
⎜
⎟(1.414 − 3.3 + 0.6) = 662 µA | Within roundoff error.
1⎝ 2 ⎠
2.5 + 1.58(0.6)+ (−0.6)
2.58
= 1.104 V | iD =
2
20 ⎛ 100x10−6 ⎞
⎜
⎟(1.104 − 0.6) = 254 µA
2
1⎝
⎠
2
20 ⎛ 40x10−6 ⎞
Checking the current : iD = ⎜
⎟(1.104 − 2.5 + 0.6) = 253 µA | Within roundoff error.
1⎝ 2 ⎠
7.37
For a symmetrical inverter, the peak current occurs for vO = v I =
VDD
.
2
Assume both transistors are saturated since vO = vI .
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
⎟(vI − VTN ) = ⎜
⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP
⎜
1⎝
2
1⎝ 2 ⎠
⎠
VDD + VTN + VTP 3.3 + (0.7)+ (−0.7)
=
= 1.65 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
= ⎜
⎟(1.65 − 0.7) = 90.3 µA Checking : iDP = ⎜
⎟(1.65 − 3.3 + 0.7) = 90.3 µA
2
1⎝
1⎝ 2 ⎠
⎠
(a) v
O
iDN
VDD + VTN + VTP 2 + (0.5)+ (−0.5)
=
= 1.00 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
= ⎜
⎟(1.00 − 0.5) = 25.0 µA Checking : iDP = ⎜
⎟(1.0 − 2 + 0.5) = 25.0 µA
2
1⎝
1⎝ 2 ⎠
⎠
(b) v
O
iDN
= vI =
= vI =
7-17
7.38
For a symmetrical inverter, the peak current occurs for vO = v I =
VDD
.
2
Assume both transistors are saturated since vO = vI .
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
⎜
⎟(vI − VTN ) = ⎜
⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP
1⎝
1⎝ 2 ⎠
2
⎠
VDD + VTN + VTP 2.5 + (0.7)+ (−0.7)
=
= 1.25 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
= ⎜
⎟(1.25 − 0.7) = 30.3 µA Checking : iDP = ⎜
⎟(1.25 − 2.5 + 0.7) = 30.3 µA
2
1⎝
1⎝ 2 ⎠
⎠
(a) v
O
iDN
VDD + VTN + VTP 2.5 + (0.65)+ (−0.55)
=
= 1.30 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
1.30
−
0.65
= ⎜
=
42.3
µ
A
Checking
:
i
=
⎟(
⎜
⎟(1.3 − 2.5 + 0.55) = 42.3 µA
)
DP
2
1⎝
1⎝ 2 ⎠
⎠
(b) v
O
iDN
= vI =
= vI =
7.39
For the inverter, the peak current occurs for vO = vI . Assume both transistors are saturated since vO = vI .
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
⎜
⎟(vI − VTN ) = ⎜
⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP
2
1⎝
1⎝ 2 ⎠
⎠
VDD + VTN + VTP 2 + (0.55)+ (−0.45)
=
= 1.05 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
= ⎜
=
25.0
µ
A
Checking
:
i
=
1.05
−
0.55
⎟(
⎟(1.05 − 2 + 0.45) = 25.0 µA
⎜
)
DP
2
1⎝
1⎝ 2 ⎠
⎠
(a) v
O
iDN
VDD + VTN + VTP 2 + (0.45)+ (−0.55)
=
= 0.950 V
2
2
2
2
2 ⎛100x10−6 ⎞
5 ⎛ 40x10−6 ⎞
= ⎜
⎟(0.95 − 0.45) = 25.0 µA Checking : iDP = ⎜
⎟(0.95 − 2 + 0.55) = 25.0 µA
2
1⎝
1⎝ 2 ⎠
⎠
(b) v
O
iDN
= vI =
= vI =
7.40
( )
2
(0.25 pF ) 2.5 = 0.313 pJ P = CV 2 f = 0.25 pF 2.52 108 = 156 µW
CVDD
=
(
)
DD
5
5
2
0.25 pF ) 22
(
CVDD
2
= 0.200 pJ P = CVDD
f = (0.25 pF ) 22 108 = 100 µW
(b) PDP ≅ 5 =
5
2
0.25 pF ) 1.82
(
CVDD
2
= 0.163 pJ P = CVDD
f = (0.25 pF ) 1.82 108 = 81.0 µW
(c) PDP ≅ 5 =
5
(a) PDP ≅
2
()
( )
7-18
( )( )
( )( )
( )( )
7.41
( )
2
0.2 pF ) 3.32
(
CVDD
= 0.290 pJ
(a) PDP ≅ 7.5 =
7.5
(b) f max ≅
1
7.5τ P
τ PLH = 1.2RonP C =
1.2(0.2 pF )
1.2C
=
= 0.471 ns
K p VGS − VTP 5 4x10−5 −3.3 + 0.75
τ PHL ≅ 1.2RonN C =
1.2(0.2 pF )
1.2C
=
= 0.471 ns
Kn (VGS − VTN ) 2 10−4 (3.3 − 0.75)
τP =
(
)
( )
0.471+ 0.471
= 0.471 ns
2
f max ≅
( )(
1
1
=
= 283 MHz
7.5τ P 7.5(0.471ns)
)
2
f = (0.2 pF ) 3.32 2.83x108 = 616 µW
(c) P = CVDD
7.42
( )
2
(0.15 pF ) 2.5 = 0.125 pJ
CVDD
PDP
≅
=
a
()
7.5
7.5
2
(b) f max ≅
1
7.5τ P
τ PLH = 1.2RonP C =
1.2(0.15 pF )
1.2C
=
= 0.474ns
K p VGS − VTP 5 4x10−5 −2.5 + 0.6
τ PHL ≅ 1.2RonN C =
1.2(0.15 pF )
1.2C
=
= 0.474 ns
Kn (VGS − VTN ) 2 10−4 (2.5 − 0.6)
τP =
(
)
( )
0.474 + 0.474
= 0.474 ns
2
f max ≅
( )(
1
1
=
= 282 MHz
7.5τ P 7.5(0.474ns)
)
2
f = (0.15 pF ) 2.52 2.82x108 = 264 µW
(c) P = CVDD
7.43
*PROBLEM 7.41 - INVERTER PDP
VDD 1 0 DC 2.5
VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 15N 30N)
*
MN1 3 2 0 0 MOSN W=4U L=1U AS=8P AD=8P
MP1 3 2 1 1 MOSP W=4U L=1U AS=8P AD=8P
C1 3 0 200fF
*AS=2UM*W - AD=2UM*W
*
MN2 4 2 0 0 MOSN W=8U L=1U AS=16P AD=16P
MP2 4 2 1 1 MOSP W=8U L=2U AS=16P AD=16P
C2 4 0 200fF
*
MN3 5 2 0 0 MOSN W=16U L=1U AS=32P AD=32P
MP3 5 2 1 1 MOSP W=16U L=1U AS=32P AD=32P
7-19
C3 5 0 200fF
*
MN4 6 2 0 0 MOSN W=32U L=1U AS=64P AD=64P
MP4 6 2 1 1 MOSP W=32U L=1U AS=64P AD=64P
C4 6 0 200fF
*
MN5 7 2 0 0 MOSN W=64U L=1U AS=128P AD=128P
MP5 7 2 1 1 MOSP W=64U L=1U AS=128P AD=128P
C5 7 0 200fF
*
MN6 8 2 0 0 MOSN W=100U L=1U AS=200P AD=200P
MP6 8 2 1 1 MOSP W=100U L=1U AS=200P AD=200P
C6 8 0 200fF
*
MN7 9 2 0 0 MOSN W=320U L=1U AS=640P AD=640P
MP7 9 2 1 1 MOSP W=2320U L=1U AS=640P AD=640P
C7 9 0 200fF
*
MN9 11 2 0 0 MOSN W=1000U L=1U AS=2000P AD=2000P
MP9 11 2 1 1 MOSP W=1000U L=1U AS=2000P AD=2000P
C9 11 0 200fF
*
.OP
.TRAN 0.025N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) V(10) V(11)
.END
At small device sizes, the power-delay product will be CVDD2 = 1.25 pJ.
7-20
Delay versus Device Size
1.00E+01
1.00E+00
Series1
1.00E-01
1
10
100
1000
W/L of NMOS & PMOS Devices
7.44
∆T =
C∆V
KCox" WL∆V
=
⎛W ⎞
2
I
1
µnCox" ⎜ ⎟(VGS − VTN )
2
⎝ L⎠
| Let W' = αW , L' = αL, Tox' = αTox , V' = αV
K (αW )(αL)(α∆V )
C '∆V '
=
= α∆T
'
2
I
1 ⎛ αW ⎞
µn ⎜
⎟(αVGS − αVTN )
2 ⎝ αL ⎠
⎛W ⎞
2
2
V
V ε ⎛W ⎞
P = VI = µnCox" ⎜ ⎟(VGS − VTN ) = µn ox ⎜ ⎟(VGS − VTN )
2
2 Tox ⎝ L ⎠
⎝ L⎠
2
αV
ε ⎛ αW ⎞
2
P'=
µn ox ⎜
⎟(αVGS − αVTN ) = α P
αTox ⎝ αL ⎠
2
∆T '=
PDP'= P'∆T'= (α∆T )α 2 P = α 3 P∆T = α 3 PDP
7-21
7.45
∆T =
KCox" WL∆V
C∆V
=
| Let W' = αW , L' = αL, Tox' = αTox
⎛
⎞
2
I
1
W
µnCox" ⎜ ⎟(VGS − VTN )
2
⎝ L⎠
K (αW )(αL)(∆V )
C '∆V '
=
= α 2∆T
'
⎛
⎞
2
I
1
αW
µn ⎜
⎟(VGS − VTN )
2 ⎝ αL ⎠
⎛W ⎞
2
2
V
V ε ⎛W ⎞
P = VI = µnCox" ⎜ ⎟(VGS − VTN ) = µn ox ⎜ ⎟(VGS − VTN )
2
2 Tox ⎝ L ⎠
⎝ L⎠
2
ε ⎛ αW ⎞
V
P
P ' = µn ox ⎜
⎟(VGS − VTN ) =
2 αTox ⎝ αL ⎠
α
∆T ' =
(
PDP ' = P ' ∆T ' = α 2∆T
)αP = αP∆T = α PDP
7.46
(Note: Simulation time needs to be extended.)
*PROBLEM 7.46 - FIVE CASCADED INVERTERS
VDD 1 0 DC 2.5
VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N)
*
MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=5U L=1U AS=40P AD=40P
C1 3 0 0.25P
*AS=8UM*W - AD=8UM*W
*
MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP2 4 3 1 1 MOSP W=5U L=1U AS=40P AD=40P
C2 4 0 0.25P
*
MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP3 5 4 1 1 MOSP W=5U L=1U AS=40P AD=40P
C3 5 0 0.25P
*
MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP4 6 5 1 1 MOSP W=5U L=1U AS=40P AD=40P
C4 6 0 0.25P
*
MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP5 7 6 1 1 MOSP W=5U L8U AS=40P AD=40P
C5 7 0 0.25P
.OP
.TRAN 0.025N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
7-22
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(5) V(6)
.END
3.0V
2.0V
1.0V
0V
First inverter : t r = 4.6 ns, t f = 5.4 ns, τ PLH = 2.6 ns, τ PHL = 2.1 ns
Fourth inverter : t r = 5.8 ns, t f = 6.3 ns, τ PLH = 4.2 ns, τ PHL = 4.7 ns
τ PHL = 1.2RonnC =
τ PLH = 1.2RonpC =
0.25x10−12
⎛ 2⎞
−6
⎜ ⎟ 50x10 (2.5 − 0.91)
1
⎝ ⎠
(
)
0.25x10−12
= 1.58ns
⎛ 5⎞
−6
⎜ ⎟ 20x10 (2.5 − 0.77)
⎝ 1⎠
(
)
= 1.45ns
The inverters are slower than the equations predict because of the additional capacitances in
the transistor models. The effective capacitance appears to be approximately 0.4 pF. The delay
of the interior inverter is substantially slower than predicted by the formula because of the
slow rise and fall times of the driving signals.
7-23
7.47
(Note: Simulation time needs to be extended.)
*PROBLEM 7.47(a) - FIVE CASCADED SYMMETRICAL INVERTERS
VDD 1 0 DC 5
VIN 2 0 PULSE (0 5 0 0.1N 0.1N 75N 150N)
*
MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=5U L=1U AS=40P AD=40P
C1 3 0 1P
*AS=8UM*W - AD=8UM*W
*
MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP2 4 3 1 1 MOSP W=5U L=1U AS=40P AD=40P
C2 4 0 1P
*
MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP3 5 4 1 1 MOSP W=5U L=1U AS=40P AD=40P
C3 5 0 1P
*
MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP4 6 5 1 1 MOSP W=5U L=1U AS=40P AD=40P
C4 6 0 1P
*
MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP5 7 6 1 1 MOSP W=5U L=1U AS=40P AD=40P
C5 7 0 1P
.OP
.TRAN 0.025N 150N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(5) V(6)
.END
7-24
First inverter : t r = 17.7 ns, t f = 20.7 ns, τ PLH = 8.1 ns, τ PHL = 9.9 ns
Fourth inverter : t r = 22.3 ns, t f = 24.2 ns, τ PLH = 16.3 ns, τ PHL = 18.3 ns
τ PHL = 1.2 RonnC =
τ PLH = 1.2 RonpC =
1.2 x10−12
⎛ 2⎞
−6
⎜ ⎟ 50 x10 (2.5 − 0.91)
⎝ 1⎠
(
)
1.2 x10−12
= 7.6 ns
⎛ 5⎞
−6
⎜ ⎟ 20 x10 (2.5 − 0.77)
⎝ 1⎠
(
)
= 6.9 ns
3.0V
2.0V
1.0V
0V
-1.0V
0s
V(CL1:2)
20ns
V(CL2:2)
V(CL3:2)
40ns
V(CL4:2)
60ns
V(CL5:2)
V(VI:+)
80ns
100ns
120ns
140ns
160ns
Time
The inverters are slower than the equations predict because of the additional capacitances in
the transistor models. The delay of the interior inverter is substantially slower than predicted
by the formula because of the slow rise and fall times of the driving signals.
*PROBLEM 7.47(b) - FIVE CASCADED MINIMUM SIZE INVERTERS
VDD 1 0 DC 5
VIN 2 0 PULSE (0 5 0 0.1N 0.1N 125N 250N)
*
MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=4U L=2U AS=16P AD=16P
C1 3 0 1P
*AS=4UM*W - AD=4UM*W
*
MN2 4 3 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP2 4 3 1 1 MOSP W=4U L=2U AS=16P AD=16P
C2 4 0 1P
*
MN3 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP3 5 4 1 1 MOSP W=4U L=2U AS=16P AD=16P
C3 5 0 1P
*
MN4 6 5 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP4 6 5 1 1 MOSP W=4U L=2U AS=16P AD=16P
C4 6 0 1P
*
MN5 7 6 0 0 MOSN W=4U L=2U AS=16P AD=16P
7-25
MP5 7 6 1 1 MOSP W=4U L=2U AS=16P AD=16P
C5 7 0 1P
.OP
.TRAN 0.025N 250N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(5) V(6)
.END
3.0V
2.0V
1.0V
0V
-1.0V
0s
V(CL1:2)
V(CL2:2)
50ns
V(CL3:2)
V(CL4:2)
V(CL5:2)
100ns
V(VI:+)
150ns
200ns
250ns
Time
First inverter : t r = 43.7 ns, t f = 20.6 ns, τ PLH = 19.7 ns, τ PHL = 9.7 ns
Fourth inverter : t r = 47.5 ns, t f = 31.4 ns, τ PLH = 30.4 ns, τ PHL = 25.7 ns
τ PHL = 1.2 RonnC =
τ PLH = 1.2 RonpC =
1.2 x10−12
⎛ 2⎞
−6
⎜ ⎟ 50 x10 (2.5 − 0.91)
⎝ 1⎠
(
)
1.2 x10−12
= 7.5 ns
⎛ 2⎞
−6
⎜ ⎟ 20 x10 (2.5 − 0.77)
⎝ 1⎠
(
)
= 17.3 ns
The inverters are slower than the equations predict because of the additional capacitances in
the transistor models. The delay of the interior inverter is substantially slower than predicted
by the formula because of the slow rise and fall times of the driving signals.
7-26
7.48
(a)
V
DD
20
1
D
20
1
C
20
1
B
20
1
vo
2
1
2
1
B
C
2
1
D
2
1
(b ) NMOS : 3 ⎛⎜ 2 ⎞⎟ = 6
⎝1⎠
⎛ 20 ⎞ 60
| PMOS : 3⎜ ⎟ =
1
⎝ 1 ⎠ 1
7-27
7.49
(a)
V
DD
D
C
B
A
vO
D
C
B
A
W
⎛2⎞ 8
= 4⎜ ⎟ =
L
⎝1⎠ 1
W 5
PMOS :
=
L 1
(b ) NMOS : W = 2(4)⎛⎜ 2 ⎞⎟ = 16
L
⎝1⎠ 1
(a )
NMOS :
PMOS :
7-28
W
⎛ 5 ⎞ 10
= 2⎜ ⎟ =
L
⎝1⎠ 1
7.50
V
DD
D
NMOS:
C
PMOS:
1
5
2
1
B
Z = A+B+C+D
A
PMOS:
C
B
D
0.4 ⎛ 2 ⎞ 1
2
| NMOS:
⎜ ⎟=
4 ⎝1⎠ 5
1
7.51
VDD
C
NMOS:
1
3.75
PMOS:
2
1
B
Z = A+B+C
A
B
C
7-29
7.52
VDD
A
B
C
Z = ABC
A
B
C
(W/L)N = 2/1, (W/L)P = 2.5(2/1)/3, = 1.67/1
7.53
VDD
D
C
B
A
Z = ABCD
D
C
B
A
(W/L)N = 2/1, (W/L)P = 2.5(2/1)/4, = 1.25/1
7-30
7.54
Output Z is A multiplied by B.
A B
Z
0 0
0
0 1
0
1 0
0
1 1
1
From the truth table, Z = AB, a two input AND gate.
Assuming complemented variables are available,
VDD
B
M = ( A + B ) = AB
A
NMOS:
B
PMOS:
4
1
5
1
7.55
(The dc input should be 0 V.)
*PROBLEM 7.55 - TWO-INPUT CMOS NOR GATE
VDD 1 0 DC 2.5
VA 2 0 DC 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N)
VB 5 0 DC 0
*
MNA 4 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MPA 4 2 3 1 MOSP W=10U L=1U AS=80P AD=80P
MNB 4 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MPB 3 5 1 1 MOSP W=10U L=1U AS=80P AD=80P
CL 4 0 1PF
*
MNC 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MPC 6 5 7 1 MOSP W=10U L=1U AS=80P AD=80P
MND 6 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MPD 7 2 1 1 MOSP W=10U L=1U AS=80P AD=80P
CL 6 0 1PF
*
7-31
.OP
.DC VDD 0 2.5 0.01
.TRAN 0.1N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(4) V(6)
.END
3.0V
2.0V
1.0V
0V
0V
0.5V
V2(C2)
1.0V
1.5V
2.0V
2.5V
V2(C3)
V_VA
3.0V
2.0V
1.0V
0V
0s
V(VA:+)
10ns
V2(C3)
20ns
V2(C3)
30ns
40ns
50ns
60ns
70ns
80ns
90ns
100ns
Time
The transitions of the two VTCs are separated by approximately 50 mV. The dynamic
characteristics for switching one input with the other constant are essentially identical. The two
transitions are virtually identical because of the ideal step inputs: τPHL = 3.6 ns, τPLH = 3.6 ns, tf =
8.1 ns, tr = 7.9 ns. With the inputs switched together, τPHL and tf are reduced by 50% because the
two NMOS devices are working in parallel.
7.56
The simulation results show only slight changes from those of Problem 7.55.
7.57
*PROBLEM 7.54 - TWO-INPUT CMOS NAND GATE
VDD 1 0 DC 2.5
VA 2 0 DC 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N)
VB 4 0 DC 2.5
*
MNA 3 4 0 0 MOSN W=4U L=1U AS=16P AD=16P
MPA 5 4 1 1 MOSP W=5U L=1U AS=80P AD=80P
7-32
MNB 5 2 3 0 MOSN W=4U L=1U AS=16P AD=16P
MPB 5 2 1 1 MOSP W=5U L=1U AS=80P AD=80P
CL1 5 0 1PF
*
MNC 6 2 0 0 MOSN W=4U L=1U AS=16P AD=16P
MPC 7 2 1 1 MOSP W=5U L=1U AS=80P AD=80P
MND 7 4 6 0 MOSN W=4U L=1U AS=16P AD=16P
MPD 7 4 1 1 MOSP W=5U L=1U AS=80P AD=80P
CL2 7 0 1PF
*
.OP
.DC VA 0 2.5 0.01
.TRAN 0.05N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0
+LAMBDA=.02 TOX=41.5N CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0 LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(5) V(7)
.END
3.0V
2.0V
1.0V
0V
0V
3.0V
0 5V
1 0V
1 5V
2 0V
2 5V
2.0V
1.0V
0V
0s
10ns
20ns
30ns
40ns
50ns
60ns
70ns
80ns
90ns
100ns
The transitions of the two VTCs are separated by approximately 50 mV. The dynamic
characteristics for switching one input with the other constant are essentially identical. The two
transitions are virtually identical because of the ideal step inputs: τPHL = 3.6 ns, τPLH = 3.7 ns, tf =
8.0 ns, tr = 8.1 ns. With the inputs switched together, τPLH and tr are reduced by 50% because the
two PMOS devices are working in parallel
7.58
The simulation results show only slight changes.
7.59
7-33
Worst-case paths are the same as the symmetrical reference inverter:
⎛ 1 ⎞⎛ 15⎞ ⎛ 5⎞
⎛ 1 ⎞⎛ 4 ⎞ ⎛ 2 ⎞
PMOS tree : ⎜ ⎟⎜ ⎟ = ⎜ ⎟ | NMOS tree : ⎜ ⎟⎜ ⎟ = ⎜ ⎟ | For VDD = 2.5V ,
⎝ 3 ⎠⎝ 1 ⎠ ⎝ 1 ⎠
⎝ 2 ⎠⎝ 1 ⎠ ⎝ 1 ⎠
τ PLH = 1.2 RonP C =
1.2(1.25 pF )
1.2C
=
= 3.95ns
K p VGS − VTP 5 4 x10−5 −2.5 + 0.6
τ PHL ≅ 1.2 RonN C =
1.2(1.25 pF )
1.2C
=
= 3.95 ns
Kn (VGS − VTN ) 2 10−4 (2.5 − 0.6)
(
)
( )
Since there is a 2.5 :1 ratio in transistor sizes, τ PLH = τ PHL ,
τ P = τ PHL = 3.95 ns and t r = t f = 3τ PHL = 3τ P = 11.8 ns
7-34
7.60
(a) A depletion-mode design requires the same number of NMOS transistors in the switching
network, but only one load transistor. The depletion-mode design requires 5 transistors total.
The CMOS design requires 8 transistors.
(b) For the CMOS design, first find the worst - case delay of the circuit in Fig. 7.30
and then scale the result to achieve the desired delay. For VDD = 2.5 V,
τ PHL
1.2C
τ PLH =
K p (VDD + VTP
τP =
( )
(
)
1.2(10 )
=
= 23.7 ns
) 1 ⎛⎜ 2 ⎞⎟(40x10 )−2.5 + 0.6
3⎝ 1⎠
1.2 10−12
1.2C
=
=
= 6.32ns
Kn (VDD − VTN ) 1 ⎛ 2 ⎞
−6
⎜ ⎟ 100x10 (2.5 − 0.6)
2 ⎝1⎠
−12
−6
⎛ W⎞
6.32 + 23.7
15.0ns ⎛ 2 ⎞ ⎛ 3.00 ⎞
ns = 15.0ns → ⎜ ⎟ =
⎜ ⎟=⎜
⎟
2
⎝ L ⎠ all 10ns ⎝ 1 ⎠ ⎝ 1 ⎠
1 = 24.0
Relative Area = 8(3.00)()
−−−−−
For the depletion - mode design, first assume that τ P is dominated by τ PLH.
τ PLH ≅ 2τ P | τ PLH = 3.6RonLC
( )
( )(
)
−12
3.6 10
⎛W ⎞
4.50
W
=
=
| Since NMOS is ratioed logic, the
ratios
⎜ ⎟
−8
−6
1
L
⎝ L ⎠ L 2 10 40x10 ()
1
⎛W ⎞
(2.22 /1) 4.50 = 5.52
must maintain the ratio in Fig. 7.29(d) : ⎜ ⎟ =
1
⎝ L ⎠ S (1.81/1) 1
Using this value, τ PHL =
( )
= 1.14ns and
5.52(100x10 )(2.5 - 0.6)
1.2 10-12
-6
20 + 1.14
ns = 10.6ns
2
Rescaling to achieve τ P = 10ns :
⎛W ⎞ 10.6 ⎛ 4.50 ⎞ 4.77 ⎛ W ⎞ 10.6 ⎛ 5.52 ⎞ 5.85 ⎛W ⎞
⎛ W ⎞ 11.7
| ⎜ ⎟ =
| ⎜ ⎟
= 2⎜ ⎟ =
⎜ ⎟ =
⎜
⎜
⎟=
⎟=
1
1
1
⎝ L ⎠ L 10 ⎝ 1 ⎠
⎝ L ⎠ A 10 ⎝ 1 ⎠
⎝ L ⎠ B −D
⎝ L ⎠A
τP =
Relative area = (4.50)()
1 + 3(11.7)()
1 + (5.85)()
1 = 45.5
The CMOS design uses 47% less area and consumes no static power.
7-35
7.61
(a) Y = (A + B)(C + D)(E + F )
W ⎛ 2⎞ 6
= 3⎜ ⎟ =
L
⎝ 1⎠ 1
(b) NMOS:
PMOS:
⎛ 5 ⎞ 10
W
= 2⎜ ⎟ =
L
⎝1⎠ 1
A
C
E
D
F
B
7.62
(a) Y = (A + B)(C + D)E
⎛ 2 ⎞ 18
W
= 3(3)⎜ ⎟ =
L
⎝ 1⎠ 1
(b) NMOS:
A
C
B
D
E
7-36
⎛ W⎞
⎛ 5⎞ 30
PMOS: ⎜ ⎟
= 2(3)⎜ ⎟ =
⎝ L ⎠ A−D
⎝ 1⎠ 1
⎛ W⎞
⎛ 5⎞ 15
⎜ ⎟ = (3)⎜ ⎟ =
⎝ L ⎠E
⎝ 1⎠ 1
7.63
F
G
E
C
E
D
B
C
D
F
G
A
B
A
(a) Y = F + G(C + E )+ A(B + D)(C + E )= F + (C + E )(G + A(B + D))
⎛ W⎞
⎛ 2 ⎞ 12 ⎛ W ⎞
⎛ 2⎞ 4 ⎛ W⎞
1
6
= 3(2)⎜ ⎟ =
| ⎜ ⎟ = (2)⎜ ⎟ = | ⎜ ⎟ =
=
⎜ ⎟
⎝ L ⎠ A−E
⎝ 1⎠ 1
⎝ L ⎠F
⎝ 1 ⎠ 1 ⎝ L ⎠G 1 − 1 1
4 12
⎛ W⎞
⎛ 5 ⎞ 40 ⎛ W ⎞
⎛
⎞
1
20
W
1
26.7
PMOS : ⎜ ⎟
= 4(2)⎜ ⎟ =
| ⎜ ⎟ =
=
| ⎜ ⎟ =2
=
1
1
1
1
⎝ L ⎠ F ,G, B, D
⎝ 1⎠ 1
⎝ L ⎠A 1 − 2
⎝ L ⎠C , E
−
10 40
10 40
(b) NMOS :
7.64
(a) Y = A + B C + D E + F = AB + CD + EF
(
)(
)(
)
(b) NMOS :
⎛ W⎞
⎛2⎞ 4
= 2⎜ ⎟ =
⎜ ⎟
⎝ L ⎠ A−F
⎝1⎠ 1
A
C
E
B
D
F
⎛ W⎞
⎛ 5 ⎞ 15
PMOS : ⎜ ⎟
= 3⎜ ⎟ =
⎝ L ⎠ A−F
⎝ 1⎠ 1
7-37
7.65
(a) Y = ACE + ACDF + BDE + BF = (A + C)(B + DF )+ E(F + DB)
(
)
⎛ W⎞
⎛ 2 ⎞ 18
= 3(3)⎜ ⎟ =
⎜ ⎟
⎝ L ⎠ A−F
⎝ 1⎠ 1
⎛ W⎞
⎛ 5 ⎞ 60
PMOS : ⎜ ⎟
= 4(3)⎜ ⎟ =
⎝ L ⎠ A,C , D, F
⎝ 1⎠ 1
(b) NMOS :
⎛ W⎞
1
40
=
⎜ ⎟ =2
1 1
1
⎝ L ⎠ B, E
−
15 60
Y
C
E
A
F
D
B
7-38
7.66
VDD
15
1
B
15
1
D
F
15
1
A
15
1
15
1
C
E
15
1
Y
A
4
1
C
6
1
F
B
4
1
D
6
1
E
6
1
2
1
7-39
7.67
V
DD
20
1
A
D
20
1
E
F
20
1
G
I
20
1
B
C
20
1
20
1
20
1
H
20
1
20
1
Y
A
6
1
D
4
1
F
6
1
I
B
6
1
C
6
1
7-40
E
4
1
G
6
1
H
6
1
2
1
7.68
Y
A
C
Y
V DD
E
B
D
Gnd
(a)
An Euler path does not appear to exist.
Y
A
C
E
Y
V DD
B
D
Gnd
(b)
An Euler path does not appear to exist.
F
.
7-41
7.69
10011
VDD
D
B
A
C
E
Y
D
C
A
E
B
10001
VDD
D
B
A
C
E
Y
D
C
A
B
7-42
E
11101
V DD
D
B
A
C
E
Y
D
C
A
E
B
00010
V
D
DD
B
A
C
E
Y
D
C
A
B
E
7-43
7.70
V
DD
B
D
F
A
C
E
A
B
C
D
E
F
(a)
V
DD
B
D
F
A
C
E
A
B
C
D
E
F
(b)
7-44
V
DD
B
D
F
A
C
E
A
B
C
D
E
F
(c)
VDD
B
D
F
A
C
E
A
B
C
D
E
F
(d)
7-45
7.71
+
10
1
A
10
1
B
10
1
D
10
1
C
10
1
E
Y
6
1
A
D
4
1
E
4
1
6
1
B
6
1
C
7.72
V DD
B
40
1
C
D
40
1
E
40
1
20
1
A
40
1
Y
24
1
A
B
24
1
D
24
1
C
24
1
E
24
1
7-46
7.73
VDD
15
1
B
5
1
A
D
15
1
7.5
1
C
E
15
1
Y
6
1
A
B
6
1
C
3
1
E
D
6
1
6
1
7.74
V DD
60
1
E
30
1
A
60
1
C
60
1
D
60
1
B
Y
24
1
A
E
12
1
C
24
1
D
24
1
8
1
B
7-47
7.75
S = X ⊕ Y = XY + XY
and
C = XY
V
Y
10
1
Y
DD
10
1
V
X
10
1
X
DD
10
1
10
1
Y
S = (X + Y)(X + Y)
X
4
1
Y
4
1
X
4
1
Y
4
1
10
1
C=X+Y
X
7-48
2
1
2
1
Y
7.76
Let X = X i
Y = Yi
Xi
Yi
Ci-1
Si
Ci
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
1
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
C = Ci-1
(
) (
S = (C + (X + Y )(X + Y ))(C + (X + Y )(X + Y ))
Si = XYC + XYC + XYC + XYC = C XY + XY + C XY + XY
)
i
Ci = XY + XC + YC = XY + C (X + Y )
(
)(
Ci = X + Y C + XY
)
VDD
15
C
Y
C
1
15
1
15
1
Y
15
1
X
15
X
1
15
1
V
DD
Y
X
15
X
1
15
1
Y
15
1
Y
8
1
Y
X
8
1
8
1
Y
Y
1
8
1
Y
X
4
1
10
1
X
10
1
Y
10
1
Ci
X
8
1
8
1
6
1
Y
6
1
X
6
1
8
1
C
X
C
8
1
C
X
1
15
Si
X
10
10
1
C
3
1
Y
6
1
4
1
7-49
7.77
N
M
Output
N
M
Output
A
B
C
D
O3 O2 O1
O0
A
B
C
D
O3 O2 O1
O0
00
00
0000
10
00
0000
00
01
0000
10
01
0010
00
10
0000
10
10
0100
00
11
0000
10
11
0110
01
00
0000
11
00
0000
01
01
0001
11
01
0011
01
10
0010
11
10
0110
01
11
0011
11
11
1001
(
)
O3 = ABCD = A + B + C + D
(
O2 = AC = A + C
)
(
)(
)(
)(
)
O1 = ABC + ABD + BCD + ACD = A + B + C A + B + D B + C + D A + C + D
(
)
O0 = BD = B + D
VDD
D
V
V
DD
DD
C
C
D
B
O2
O4
O3
A
A
B
C
D
D
⎛W ⎞
2 ⎛W ⎞
20
⎜ ⎟ =
⎜ ⎟ =
⎝ L ⎠N 1 ⎝ L ⎠P 1
7-50
B
C
|
⎛W ⎞
2 ⎛W ⎞
10
⎜ ⎟ =
⎜ ⎟ =
⎝ L ⎠N 1 ⎝ L ⎠P 1
|
⎛W ⎞
2 ⎛W ⎞
10
⎜ ⎟ =
⎜ ⎟ =
⎝ L ⎠N 1 ⎝ L ⎠P 1
V
DD
D
D
D
C
C
C
B
B
A
B
A
A
O1
A
C
D
B
C
D
A
B
D
A
B
C
7.78
(a) τ
PHL
= 1.2RonnC =
τ PLH = 1.2RonpC =
(b) τ
PHL
(
1.2 0.4x10−12
PMOS:
)
−6
(
)
(2 / 3)(40x10 )−2.5 + 0.6
−6
(
1.2 0.4x10−12
)
= 9.47ns | τ P =
(2 1) 100x10−6 (2.5 − 0.6)
1
τ PLH + τ PLH
2
= 5.37 ns
= 1.26ns
−12
−6
15
= 1.26ns
(
)
1.2(0.4x10 )
C=
= 3.16ns
(2 1)(40x10 )−2.5 + 0.6
τ PLH = 1.2Ronp
8
1
(2 1)(100x10 )(2.5 − 0.6)
1.2 0.4x10−12
= 1.2RonnC =
NMOS:
| τP =
τ PLH + τ PLH
2
= 2.21 ns
7-51
7.79
(a) τ
PHL
= 1.2RonnC =
τ PLH = 1.2RonpC =
(b) τ
PHL
(
1.2 0.18x10−12
)
(2 5)(100x10 )(2.5 − 0.6)
−6
(
1.2 0.18x10−12
(
)
)
= 1.42ns | τ P =
(2 /1) 40x10−6 −2.5 + 0.6
= 1.2RonnC =
τ PLH = 1.2RonpC =
= 2.84ns
(
1.2 0.18x10−12
)
(2 1)(100x10 )(2.5 − 0.6)
−6
(
1.2 0.4x10−12
(
(2 1) 40x10
−6
)
)
−2.5 + 0.6
τ PLH + τ PLH
2
= 2.13 ns
= 0.568ns
= 3.16ns | τ P =
τ PLH + τ PLH
2
= 1.86 ns
7.80
The worst-case NMOS path contains 2 transistors. The worst-case PMOS path contains 3
transistors.
1.2 250x10−15
τ PHL = 1.2RonnC =
= 1.58 ns
(2 2) 100x10−6 (2.5 − 0.6)
(
τ PLH = 1.2RonpC =
(
)
(
1.2 250x10−15
)
)
(2 / 3)(40x10 )−2.5 + 0.6
−6
= 5.92 ns
7.81
The worst-case NMOS path contains 3 transistors. The worst-case PMOS path contains 3
transistors.
1.2 400x10−15
τ PHL = 1.2RonnC =
= 3.79 ns
(2 3) 100x10−6 (2.5 − 0.6)
(
τ PLH = 1.2RonpC =
(
)
(
1.2 400x10−15
)
)
(2 / 3)(40x10 )−2.5 + 0.6
−6
= 9.47 ns
7.82
The worst-case NMOS path contains 3 transistors (ABE or CBD). The worst-case PMOS path
also contains 3 transistors.
( )
= 9.47 ns
(2 3)(100x10 )(2.5 − 0.6)
1.2(10 )
C=
= 23.7 ns
(2 / 3)(40x10 )−2.5 + 0.6
τ PHL = 1.2RonnC =
1.2 10−12
−6
−12
τ PLH = 1.2Ronp
7-52
−6
| τP =
τ PLH + τ PLH
2
= 16.6 ns
7.83
The worst-case NMOS path contains 3 transistors.
1.2 10−12
τ PHL = 1.2RonnC =
= 9.47 ns
−6
2
3
100x10
2.5
−
0.6
( )
(
)
(
( )
)
7.84
Student PSPICE will only accept 5 inverters.
*PROBLEM 7.84(a) - FIVE CASCADED INVERTERS
VDD 1 0 DC 2.5
VIN 2 0 PULSE (0 2.55 0 0.1N 0.1N 20N 40N)
*
MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P
C1 3 0 200fF
*AS=8UM*W - AD=8UM*W
*
MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP2 4 3 1 1 MOSP W=2U L=1U AS=16P AD=16P
C2 4 0 200fF
*
MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP3 5 4 1 1 MOSP W=2U L=1U AS=16P AD=16P
C3 5 0 200fF
*
MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP4 6 5 1 1 MOSP W=2U L=1U AS=16P AD=16P
C4 6 0 200fF
*
MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP5 7 6 1 1 MOSP W=2U L=1U AS=16P AD=16P
C5 7 0 200fF
*
.OP
.TRAN 0.025N 40N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(4) V(10) V(11)
.END
(a)
7-53
3.0V
2.0V
1.0V
0V
-1.0V
0s
V(CL1:2)
5ns
V(CL2:2)
10ns
V(CL3:2)
15ns
20ns
V(CL4:2)
V(CL5:2)
V(VI:+)
25ns
30ns
35ns
40ns
45ns
50ns
Time
(b)
3.0V
2.0V
1.0V
0V
-1.0V
0s
V(VI:+)
5ns
V(CL1:2)
10ns
V(CL2:2)
15ns
V(CL3:2)
20ns
V(CL4:2)
25ns
V(CL5:2)
30ns
35ns
40ns
45ns
50ns
55ns
60ns
Time
(a) The minimum size inverters yield
P
= 5.8 ns.
(b) The symmetrical inverters yield τP = 3.7 ns.
Note that these results are larger than the delay equation estimate because of the slope of the
waveforms.
7-54
7.85
(a)
(b)
V DD
CLK
V DD
Z = AB
CLK
A
B
Z=A+B
B
A
CLK
CLK
7-55
7.86
(a)
(b)
V DD
V DD
CLK
CLK
A
A
B
B
Z = A + B = AB
Z = A + B = AB
CLK
CLK
7.87
(a)
(b)
VDD
CLK
A
V DD
CLK
Z
A
B
C
B
Z = A+B+C
C
CLK
7-56
CLK
Z = ABC
7.88
(a)
(b)
V DD
CLK
A
VDD
Z = A B C = A+B+C
CLK
Z
B
A
B
C
C
Z = A + B + C = ABC
CLK
CLK
7.89
VDD
CLK
A
B
C
D
E
F
Z = (AB+CD+EF)
CLK
7-57
7.90
VDD
CLK
A
B
C
D
E
Z = (ABC+DE+F)
F
CLK
7.91
Charge sharing occurs. Assuming C2 and C3 are discharged (the worst case)
(a ) V
=
(c) V
=
C1VDD + C2 (0)
2C2VDD 2
2
= VDD | Node B drops to VDD .
C1 + C2
2C2 + C2 3
3
⎛2
⎞
2
V
3C
⎜
⎟
2
DD
C
+
C
V
+
C
0
( 1 2 )3 DD 3 ( )
V
1
⎝3
⎠
=
= DD | Node B drops to VDD .
(b) VB =
2C2 + C2 + C2
C1 + C2 + C3
2
2
B
B
R≥
=
C1VDD
RC2VDD
R
=
=
VDD ≥ VIH → R(VDD − VIH )≥ 2VIH
C1 + C2 + C3 RC2 + C2 + C2 R + 2
2VIH
2VIH
=
VDD − VIH NM H
Using VDD = 2.5V , VTN = 0.6V , VTP = −0.6V in Eq. (8.9) :
VIH =
R≥
7-58
5(2.5)+ 3(0.6)+ 5(−0.6)
8
2(1.41)
= 1.41V | NM H =
2VIH
=
= 3.05 | C1 ≥ 83.05C2
NM H
0.925
2.5 − 0.6 − 3(−0.6)
4
= 0.925V
7.92
Z = A0 + A1 + A2
V
DD
Clock
A0
A1
A2
Z
C
1
7.93
V DD
CLK
A
B
C
D
Z=(A+B)(C+D)
CLK
7-59
7.94
V DD
CLK
A
C
B
D
CLK
7-60
Z = AB + CD
7.95
V DD
CLK
Z
A
B
C
D
E
F
CLK
7.96
V DD
CLK
Z
A
C
E
B
D
F
CLK
7.97
N opt = ln
1
CL
= ln (4000)= 8.29 → N = 8 | β = (4000)8 = 2.82.
Co
The relative sizes of the 8 inverters are : 1, 2.82, 7.95, 22.4, 63.2, 178, 503, 1420.
Each inverter has a delay of 2.82τ o. The total delay is 8(2.82τ o )= 22.6τ o
7-61
7.98
N opt = ln
1
⎛ 10 pF ⎞
CL
4 = 3.16.
= ln⎜
β
=
100
=
4.61
→
N
=
4
|
(
)
⎟
Co
100
fF
⎝
⎠
The relative sizes of the 4 inverters are : 1, 3.16, 10.0, 31.6
Each inverter has a delay of 3.16τ o. The total delay is 4(3.16τ o )= 12.6τ o
Note, N = 5 yields β = 2.51 and the total delay is 5(2.51τ o ) = 12.6τ o .
However, the area will be significantly larger. See Prob. 7.100.
7.99
N opt
1
⎛ 40 pF ⎞
CL
= ln
= ln⎜
⎟ = 6.69 → N = 6 | β = (800)6 = 2.82.
Co
⎝ 50 fF ⎠
The relative sizes of the 8 inverters are: 1, 2.82, 7.95, 22.4, 63.2, 178, 503, 1420.
Each inverter has a delay of 2.82τ o. The total delay is 8(2.82τ o )= 22.6τ o .
Note, N = 7 yields β = 2.60 and the total delay is 7(2.60τ o ) = 18.2τ o.
However, the area will be significantly larger. See Prob. 7.100.
7.100
AT = Ao (1+ β + β 2 + ...+ β N −1 )= Ao
β N −1
β −1
1000 −1
= 462Ao
3.1623 −1
1000 −1
For N = 7, β = 10001/7 = 2.6827 | A = Ao
= 594 Ao
2.6827 −1
Since the two values of N give similar delays, N = 6 would be used
For N = 6, β = 10001/6 = 3.1623 | A = Ao
because of it requires significantly less area.
7.101
⎡⎛ 20 ⎞
⎤−1
−6
(a) Ronn = K V − V = ⎢⎜⎝ 1 ⎟⎠ 100x10 (2.5 − 0.6)⎥ = 263 Ω
⎦
n ( GS
TN ) ⎣
1
(b) R
onp
=
1
K p VGS − VTP
(
)
−1
⎡⎛ 20 ⎞
⎤
= ⎢⎜ ⎟ 40x10−6 −2.5 + 0.6 ⎥ = 658 Ω
⎣⎝ 1 ⎠
⎦
(
)
(c) A resistive channel exists connecting the source and drain.
7-62
7.102
Gon = Gonn + Gonp = Kn (VGSN − VTN )* ( (VGSN − VTN )> 0) + K p (VTP − VGSP )* ( (VTP − VGSP )> 0)
VGSN = 2.5 − VI
[
(
VSBN = VI
VGSP = −VI
VTN = 0.6 + 0.5 VI + 0.6 − 0.6
(a) R
on
)]
VBSN = 2.5 − VI
[
(
VTP = −0.6 − 0.5 2.5 − VI + 0.6 − 0.6
)]
will be the largest for VI = 1 V
VTN = 0.845V
VTP = −0.937V
( )
(
)
10 −4
10
10 (2.5 −1− 0.845)+
4x10−5 (−0.937 + 1) → Ron = 1470Ω
1
1
(b) Ron will be the largest for the VI at which the NMOS transistor just cuts off :
Gon =
[
)]
(
2.5 − VI = 0.6 + 0.5 VI + 0.6 − 0.6 → VI = 1.554V
Gon =
(
VTN = 0.947V VTP = −0.834V
)
10
4x10−5 (−0.834 + 1.554)→ Ron = 3470Ω
1
7.103
Gon = Gonn + Gonp = Kn (VGSN − VTN )* ( (VGSN − VTN )> 0) + K p (VTP − VGSP )* ( (VTP − VGSP ) > 0)
VGSN = 2.5 − VI
[
VSBN = VI
VGSP = −VI
(
VTN = 0.75 + 0.5 VI + 0.6 − 0.6
)]
VBSN = 2.5 − VI
[
(
VTP = −0.75 − 0.5 2.5 − VI + 0.6 − 0.6
)]
The worst cases occur approximately at the point where the PMOS or NMOS transistors just cut off.
[
(
)]
The NMOS transistor cuts off for 2.5 − VI = 0.75 + 0.5 VI + 0.6 − 0.6 → VI = 1.426V
VTP = −1.01V
⎛W ⎞
⎛W ⎞
1
240
= ⎜ ⎟ 4x10−5 (−1.01+ 1.426)→ ⎜ ⎟ =
250 ⎝ L ⎠ P
1
⎝ L ⎠P
(
)
[
(
)]
The PMOS transistor cuts off for VI = 0.75 + 0.5 2.5 − VI + 0.6 − 0.6 → VI = 1.074V
VTN = −1.01V
⎛W ⎞
⎛W ⎞
1
96.2
= ⎜ ⎟ 10−4 (2.5 −1.074 −1.01)→ ⎜ ⎟ =
250 ⎝ L ⎠ N
1
⎝ L ⎠P
( )
7.104
(a ) The output of the first NMOS transistor will be
[
)]
(
VI = 2.5 − VTN = 2.5 − 0.75 + 0.55 VI + 0.6 − 0.6 → VI = 1.399V | VTN = 1.10V
The output of the other gates reaches this same value. All three nodes = 1.40 V.
For the PMOS transistors, the node voltages will all be 2.5 V.
(b) The node voltages would all be
+ 2.5 V.
7-63
7.105
*Figure 7.38(b) - CMOS Latchup
VDD 1 0
RC 1 2 25
RL 3 4 2000
RN 2 3 2000
RP 4 0 500
Q1 3 4 0 NBJT
Q2 4 3 2 PBJT
.DC VDD 0 5 .01
.MODEL NBJT NPN BF=60 BR=.25 IS=1E-15
.MODEL PBJT PNP BF=60 BR=.25 IS=1E-15
.PROBE I(VDD) V(1) V(2) V(3) V(4)
.OPTIONS ABSTOL=1E-12 RELTOL=1E-6 VNTOL=1E-6
.END
Simulation results from B2SPICE
Circuit 7_91-DC Transfer-1
+0.000e+000
(V)
+1.000
+2.000
+3.000
+2.000
+1.500
+1.000
+500.000m
+0.000e+000
V(1)
V(2)
V(3)
7.106
*Figure 7.39(b) - CMOS Latchup
VDD 1 0
RC 1 2 25
RL 3 4 2000
RN 2 3 200
RP 4 0 50
Q1 3 4 0 NBJT
Q2 4 3 2 PBJT
.DC VDD 0 5 .01
.MODEL NBJT NPN BF=60 BR=.25 IS=1E-15
.MODEL PBJT PNP BF=60 BR=.25 IS=1E-15
.PROBE I(VDD) V(1) V(2) V(3) V(4)
7-64
VDD
+4.000
.OPTIONS ABSTOL=1E-12 RELTOL=1E-6 VNTOL=1E-6
.END
Simulation results from B2SPICE – Latchup does not occur!
Circuit 7_92-DC Transfer-2
+0.000e+000
(V)
+1.000
+2.000
VDD
+3.000
+4.000
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(1)
V(2)
7.107
V DD
v
B
n+
V(3)
I
S
V SS
D
p+
p+
vo
D
S
n+
n+
B
p+
p-well
PMOS transistor
Ohmic
contact
NMOS transistor
Ohmic
contact
n-type substrate
7-65
7.108
(a)
VIH =
VIH =
2K R (VDD − VTN + VTP ) (VDD − K RVTN + VTP )
−
(K R −1)
(K R −1) 1+ 3K R
2K R (VDD − VTN + VTP ) − (VDD − K RVTN + VTP ) 1+ 3K R 0
=
0
(K R −1) 1+ 3K R
3
+ VTN 1+ 3K R
2 1+ 3K R
3 (K R −1)
1+ 3K R +
2 1+ 3K R
2(VDD − VTN + VTP ) − (VDD − K RVTN + VTP )
lim VIH = lim
K R →1
K R →1
3
2(VDD − VTN + VTP ) − (VDD − VTN + VTP ) + 2VTN
5V + 3VTN + 5VTP
4
lim VIH =
= DD
K R →1
2
8
2V − VDD − VTN − VTP VDD − VTN + VTP
VOL = IH
=
2
8
(b)
VIL =
lim VIL =
K R →1
2 K R (VDD − VTN + VTP )
2
(VDD − K RVTN + VTP )
(K R − 1)
(K R − 1) K R + 3
K R (VDD − VTN + VTP ) − (VDD − K RVTN + VTP )
(K R − 1) K R + 3
−
KR + 3
=
0
0
2
1
+ VTN K R + 3
(VDD − VTN + VTP ) − (VDD − K RVTN + VTP )
2 KR
2 KR + 3
lim VIL = lim
K R →1
K R →1
1 (K R − 1)
KR + 3 +
2 KR + 3
lim VIL =
K R →1
(VDD − VTN + VTP ) − (VDD − VTN + VTP )
1
+ 2VTN 3V + 5V + 3V
TN
TP
4
= DD
8
2
2V + VDD − VTN − VTP 7VDD + VTN − VTP
VOH = IH
=
2
8
3V + 5VTN + 3VTP VDD − VTN + VTP VDD + 3VTN + VTP
NM L = VIL − VOL = DD
−
=
8
8
4
7V + VTN − VTP 5VDD + 3VTN + 5VTP VDD − VTN − 3VTP
NM H = VOH − VIH = DD
−
=
8
8
4
7-66
7.109
(a)
For VDD = 5V, VTN = 1V, VTP = −1V
τP =
∆τ P
τ
0.322C
dτ P
0.322C
|
=−
=− P
2
dK n
Kn
Kn
Kn
≈ SτKPn
τP
| SτKPn =
K n dτ P
= −1
τ P dK n
∆K n
∆K
= − n = −(−0.25) = +0.25 | A 25% decrease in K n will cause
Kn
Kn
a 25% increase in propagation delay.
(b)
Assuming a symmetrical inverter with VDD = 5V and VTN = 0.75V,
τP =
⎡ ⎛ V −V
⎞
2VTN ⎤
TN
−1⎟ +
⎥
⎢ln⎜ 4 DD
VDD
K n (VDD − VTN ) ⎣ ⎝
⎠ VDD − VTN ⎦
τP =
⎡ ⎛ 5 − 0.75 ⎞ 2(0.75) ⎤ 0.289C
C
−1⎟ +
⎢ln⎜ 4
⎥=
⎠ 5 − 0.75 ⎦
K n (5 − 0.75) ⎣ ⎝
5
Kn
C
⎡
⎤
−4
⎢
⎥
dτ P
τP
C
2VDD
VDD
⎢
⎥
=
+
+
2
⎛
⎞
⎢
dVTN (VDD − VTN ) K n (VDD − VTN )
VDD − VTN
V − VTN ) ⎥
−1⎟ ( DD
⎢⎜ 4
⎥
VDD
⎠
⎣⎝
⎦
⎡
⎤
⎛ −4 ⎞
⎜ ⎟
⎢
2(5) ⎥ 0.120C
0.289C
C
dτ P
⎝ 5⎠
⎢
⎥=
=
+
+
dVTN (5 − 0.75)K n K n (5 − 0.75) ⎢⎛ 5 − 0.75 ⎞ (5 − 0.75)2 ⎥
Kn
−1⎟
⎜4
⎢⎣⎝
⎥
⎠
5
⎦
V dτ P
0.75K n ⎛ 0.120C ⎞
SτVPTN = TN
=
⎜
⎟ = 0.311
τ P dVTN 0.289C ⎝ K n ⎠
⎛ 0.1 ⎞
∆VTN
∆τ P
≅ SτKPn
= 0.311⎜
⎟ = 0.0415 = 4.15%.
⎝ 0.75 ⎠
τP
VTN
A 13% increase in VTN causes an 4.2% increase in τ P .
7.110
*PROBLEM 7.110 - INVERTER DELAY VS RISETIME
VDD 1 0 DC 2.5
V1 2 0 PULSE (0 2.5 0 0.1N 0.1N 50N 100N)
MN1 3 2 0 0 MOSN W=1U L=1U AS=4P AD=4P
MP1 3 2 1 1 MOSP W=1U L=1U AS=4P AD=4P
C1 3 0 1PF
*
V2 4 0 PULSE (0 2.5 0 0.2N 0.2N 50N 100N)
MN3 5 4 0 0 MOSN W=1U L=1U AS=4P AD=4P
MP3 5 4 1 1 MOSP W=1U L=1U AS=4P AD=4P
C3 5 0 1PF
*
7-67
V3 6 0 PULSE (0 2.5 0 0.5N 0.5N 50N 100N)
MN5 7 6 0 0 MOSN W=1U L=1U AS=4P AD=4P
MP5 7 6 1 1 MOSP W=1U L=1U AS=4P AD=4P
C5 7 0 1PF
*
V4 8 0 PULSE (0 2.5 0 1N 1N 50N 100N)
MN7 9 8 0 0 MOSN W=1U L=1U AS=4P AD=4P
MP7 9 8 1 1 MOSP W=1U L=1U AS=4P AD=4P
C7 9 0 1PF
*
V5 10 0 PULSE (0 2.5 0 2N 2N 50N 100N)
MN9 11 10 0 0 MOSN W=1U L=1U AS=4P AD=4P
MP9 11 10 1 1 MOSP W=1U L=1U AS=4P AD=4P
C9 11 0 1PF
*
.OP
.TRAN 0.025N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) V(10) V(11)
.END
tr
0.1 ns
0.2 ns
0.5 ns
1 ns
5 ns
7-68
τP
14.6 ns
14.7 ns
14.7 ns
14.8 ns
15.8 ns
CHAPTER 8
8.1
( )( )
(c) 256Mb = 2 (2 )(2 )= 2
( )
3
(a) 256Mb = 28 210 210 = 268,435,456 bits (b) 1Gb = 210 = 1,073,741,824 bits
8
10
10
28
( )
| 128kb = 2 7 210 = 217 |
228
= 211 = 2048 blocks
17
2
8.2
I≤
8.3
pA
1mA
= 3.73
28
bit
2 bits
(a) P = CV
(b) P = CV
2
DD
f = 64(10pF)(3.3) (1GHz)= 6.97 W
2
DD
f = 64(10 pF )(2.5) (3GHz)= 12 W
2
2
8.4
⎛ 230 ⎞
2⎛
1 ⎞
2
P = CVDD
f = ⎜ ⎟(100fF)(2.5V ) ⎜
⎟ = 28.0 mW
⎝ 0.012s ⎠
⎝ 2 ⎠
8.5
"1" = VDD = 3 V | "0":
⎛
3 − VO 2
VO ⎞
−6
=
100x10
3
−
0.75
−
⎜
⎟VO → VO = 0.667µV | "0"= 0.667 µV
2⎠
1
1010
⎝
(
)
8-1
8.6
*PROBLEM 8.6 - 6-T Cell
VDD 1 0 DC 3
MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=10U L=2U AS=40P AD=40P
MN2 2 3 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP2 2 3 1 1 MOSP W=10U L=2U AS=40P AD=40P
MN3 3 0 0 0 MOSN W=4U L=2U AS=16P AD=16P
MN4 2 0 0 0 MOSN W=4U L=2U AS=16P AD=16P
.IC V(3)=1.55V V(2)=1.45V V(1)=3
.OP
.TRAN 0.025N 10N UIC
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PRINT TRAN V(2) V(3)
.PROBE V(2) V(3)
.END
4.0V
3.0V
2.0V
1.0V
0V
Time
-1.0V
0s
8-2
2ns
4ns
6ns
8ns
10ns
Result: t = 1.5 ns
8.7
(a)
3V
+3 V
(b)
3V
0.7 V
0.7 V
1.5 V
2.3 V
1.5 V
2.3 V
First Case : Both transistors are in the linear region
⎛1⎞⎛
0.7 ⎞
I DS = 25x10−6 ⎜ ⎟⎜ 2.3 − 0.7 −
⎟0.7 = 21.88µA
2 ⎠
⎝1⎠⎝
⎛ W ⎞⎛
⎛W ⎞
0.8 ⎞
1
21.88µA = 25x10−6 ⎜ ⎟⎜3 − 0.7 − 0.7 −
⎟0.8 → ⎜ ⎟ ≤ 0.911 =
2 ⎠
1.10
⎝ L ⎠⎝
⎝ L⎠
Second Case : Both transistors are in the linear region
⎛ ⎞⎛
⎛ ⎞⎛
⎞
⎛ ⎞
−0.7)⎞
(
−6 1
⎜
⎟(−0.7)= 25x10−6 ⎜ W ⎟⎜3 −1.5 − 0.7 − 0.8 ⎟0.8 → ⎜ W ⎟ ≤ 1.09
10x10 ⎜ ⎟⎜ 0.7 − 3 − (−0.7)−
2 ⎠
1
2 ⎟⎠
⎝1⎠⎝
⎝ L ⎠⎝
⎝ L⎠
⎛W ⎞
1
So ⎜ ⎟ ≤
⎝ L ⎠ 1.10
8-3
8.8
*Problem 8.8 - WRITING THE CMOS SRAM
VWL 6 0 DC 0 PULSE(0 3 1NS 1NS 1NS 100NS)
VDD 3 0 DC 3
VBL1 4 0 DC 0
VBL2 5 0 DC 3
CBL1 4 0 500FF
CBL2 5 0 500FF
*Storage Cell
MCN1 2 1 0 0 MOSN W=1U L=1U AS=4P AD=4P
MCP1 2 1 3 3 MOSP W=1U L=1U AS=4P AD=4P
MCN2 1 2 0 0 MOSN W=1U L=1U AS=4P AD=4P
MCP2 1 2 3 3 MOSP W=1U L=1U AS=4P AD=4P
MA1 4 6 2 0 MOSN W=1U L=1U AS=4P AD=4P
MA2 5 6 1 0 MOSN W=1U L=1U AS=4P AD=4P
*
.OP
.TRAN 0.01NS 20NS
.NODESET V(1)=3 V(2)=0
.MODEL MOSN NMOS KP=2.5E-5 VTO=.70 GAMMA=0.5
+LAMBDA=.05 TOX=20N
+CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10
.MODEL MOSP PMOS KP=1.0E-5 VTO=-.70 GAMMA=0.75
+LAMBDA=.05 TOX=20N
+CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
4.0V
3.0V
2.0V
1.0V
0V
Time
-1.0V
0s
2ns
4ns
6ns
8ns
10ns
Small voltage transients occur on both cell storage nodes which die out in 5 - 7 ns.
8-4
8.9
(a ) The transistor will fully discharge CC : VC 0 = 0 V
(
= 0.6 + 0.5( 2.5 + 0.6 −
)
0.6 )= 1.09V | V
VC1 = 2.5 − VTN = 2.5 − 0.6 − 0.5 VC1 + 0.6 − 0.6 → VC1 = 1.55V | "1"= 1.55 V
For VC1 = 2.5V ,VTN
W/L
≥ 2.5 + 1.09 = 3.59 V
8.10
For "0" = 0V, the bias across the source-substrate junction is 0 V, so the leakage current would
be 0 and the "0" state is undisturbed. For a "1" corresponding to a positive voltage, a reverse
bias across the source-substrate junction, and the diode leakage current will tend to destroy the
"1" state.
iL
C
n
+
n
"OFF"
+
8.11
(
CDG
)
VG
C
VC = 2.5 − VTN = 2.5 − 0.7 − 0.5 VC + 0.6 − 0.6 → VC = 1.47V
∆VC =
1
sCC
1
1
+
sCC sCGS
∆VW / L =
GS
∆VW / L
−2.5
=
= −1.43 V
75 fF
CC
1+
1+
100 fF
CGS
CC
8.12
QI = 60 fF (0V )+ 7.5 pF (2.5V ) | QF = 7.56 pF (VF ) | QF = QI → VF =
VF = 2.48 V | ∆V =
7.5 pF
0.06
2.5V − 2.5V = −2.5
= −19.8 mV
7.56 pF
1.56
8.13
(a) "1"= +3.3 V
(b) "1"= +2.5 V
7.5 pF
2.5V
7.56 pF
(
= 0.7 + 0.5( 2.5 − V
)
0.6 )→ V
| VC = −VTP = 0.7 + 0.5 3.3 − VC + 0.6 − 0.6 → VC = 1.14V | "0"= 1.14 V
| VC = −VTP
C
+ 0.6 −
C
= 1.03V | "0"= 1.03 V
8-5
8.14
Note that the simulation results in Fig. 9.28 assume that the word line is also driven higher
than 3 V. For this case:
(a) V
(b) V
C
TN
(
)
= 5 − VTN = 5 − 0.7 − 0.5 VC + 0.6 − 0.6 → VC = 3.66 V
(
)
= 0.7 − 0.5 1.3 + 0.6 − 0.6 = 1.00V
VGS − VTN = 5 −1.3 −1.00 = 2.7V | VDS = 3.7 −1.3 = 2.4V → linear region
⎛1⎞⎛
2.4 ⎞
iDS = 60x10−6 ⎜ ⎟⎜ 5 −1.3 −1.00 −
⎟2.4 = 216 µA which agrees with the text.
2 ⎠
⎝1⎠⎝
8.15
(a) "0"= +0 V | V = 3 − V = 3 − 0.7 − 0.5( V + 0.6 − 0.6 )→ V = 1.90V | "1"= 1.90 V
(b) A "0" will have 0 V across the drain - substrate junction, so no leakage occurs.
C
TN
C
C
A "1" will have a reverse bias of 1.9 V across the junction, so the junction leakage will tend
to destroy the "1" level. (Note that this discussion ignores subthreshold leakage through the
FET which has not been discussed in the text.)
8.16
(a) "1"= +5 V | V = −V = 0.8 + 0.65( 5 − V
(b) V = −0.8 − 0.65( 5 + 0.6 − 0.6 )= −1.83V
C
TP
TP
C
)
+ 0.6 − 0.6 → VC = 1.60V | "0"= +1.60 V
| VW/L ≤ −1.83V
8.17
⎛1⎞⎛
0.6 ⎞
Original : iD = 60x10−6 ⎜ ⎟⎜ 3 −1.3 −1−
⎟0.6 = 14.4µA
2 ⎠
⎝1⎠⎝
⎛ 1⎞⎛
2.4 ⎞
New :
iD = 60x10−6 ⎜ ⎟⎜5 −1.3 −1−
⎟2.4 = 216µA.
2 ⎠
⎝ 1⎠⎝
2.4
5 −1.3 −1−1.2 1.5
=
= 3.75 | VDS ratio :
=4
Gate drive terms:
0.6
3 −1.3 −1− 0.3 0.4
Improved gate drive yields a 3.75 times improvement, although it is reduced by the larger
VDS term. Improved drain - source voltage yields a 4 times improvement. 4 x 3.75 = 15.
8.18
2⎛
1 ⎞
2
f = (0.5) 230 (100fF)(2.5V ) ⎜
P = CVDD
⎟ = 33.6 mW
⎝ 0.01s ⎠
( )
8-6
8.19
*Problem 8.19 - 4-T Refresh SRAM
VWL 3 0 DC 0 PULSE(0 3 1NS 1NS 1NS 6NS)
VBL 4 0 DC 3
VBLB 5 0 DC 3
CC1 1 0 50FF
CC2 2 0 50FF
*Storage Cell
MCN1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P
MCN2 1 2 0 0 MOSN W=4U L=2U AS=16P AD=16P
MA1 4 3 2 0 MOSN W=4U L=2U AS=16P AD=16P
MA2 5 3 1 0 MOSN W=4U L=2U AS=16P AD=16P
.IC V(1)=0 V(2)=1 V(3)=0 V(4)=3 V(5)=3
.OP
.TRAN 0.01NS 20NS UIC
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.PROBE V(1) V(2) V(3) V(4) V(5)
.END
4.0V
Note the very slow recovery due to
relatively high threshold and gamma values
relative to the power supply voltage.
3.0V
vWL
2.0V
D
1.0V
D
0V
Time
-1.0V
0s
5ns
10ns
15ns
20ns
8.20
*Problem 8.20 4-T READ ACCESS
VPC 7 0 DC 3 PULSE(3 0 1NS .5NS .5NS 100NS)
VWL 6 0 DC 0 PULSE(0 3 2NS .5NS .5NS 100NS)
VDD 3 0 DC 3
CBL1 4 0 1PF
CBL2 5 0 1PF
*Storage Cell
MCN1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P
MCN2 1 2 0 0 MOSN W=4U L=2U AS=16P AD=16P
8-7
MA1 4 6 2 0 MOSN W=4U L=2U AS=16P AD=16P
MA2 5 6 1 0 MOSN W=4U L=2U AS=16P AD=16P
CC1 1 0 50FF
CC2 2 0 50FF
*
*Sense Amplifier
MSN1 4 5 0 0 MOSN W=4U L=2U AS=16P AD=16P
MSP1 4 5 3 3 MOSP W=4U L=2U AS=16P AD=16P
MSN2 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P
MSP2 5 4 3 3 MOSP W=4U L=2U AS=16P AD=16P
MRS 5 7 4 0 MOSN W=4U L=2U AS=16P AD=16P
*
.OP
.TRAN 0.01NS 40NS UIC
.IC V(1)=1.5 V(2)=0 V(3)=3 V(4)=1.7 V(5)=1.7 V(6)=0 V(7)=3
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7)
.END
vPC
3.0V
vWL
BL
2.0V
D
1.0V
BL
D
0V
Time
0s
8-8
10ns
20ns
30ns
40ns
8.21
+3.3 V
10
1
Each inverter will have vI = v O .
vO
vI
5
1
Equating inverter drain currents :
⎛100x10-6 ⎞⎛ 5 ⎞
⎛ 40x10-6 ⎞⎛ 10 ⎞
2
2
−
0.7
=
v
⎟⎜ ⎟( I
⎟⎜ ⎟(3.3 − v I − 0.7)
⎜
⎜
)
2
⎠⎝ 1 ⎠
⎝
⎝ 2 ⎠⎝ 1 ⎠
→ vO = vI = 1.597 V
Sense amp current = 2iDS = 402 µA
P = 1024(3.3V)(402 µA)= 1.36 W
8.22
V
G
C
C
C
C
DG
BL
GS
500 fF
BL
500 fF
The precharge transistor is operating in the linear region with VDS = 0
1
CGS = Cox" WL + CGSOW
2
−14
1 3.9 8.854x10 F / cm
CGS =
10−3 cm 10−4 cm + 4x10−11 F / cm 10−3 cm = 48.6 fF
2
2x10−6 cm
1
∆VG
-3
sC BL
∆VG (s) → ∆V =
=
= 0.266 V
∆V(s) =
1
1
C BL
500
+
+1
+1
CGS
sCGS sC BL
48.6
(
)
(
)(
)(
)(
)
This value provides a good estimate of the drop observed in Fig. 8.25.
8-9
8.23
The precharge transistor is operating in the linear region with VDS = 0
1
CGS = Cox" WL + CGSOW
2
−14
1 3.9 8.854x10 F / cm
CGS =
10−3 cm 10−4 cm + 4x10−11 F / cm 10−3 cm = 48.6 fF
−6
2
2x10 cm
Using CBL = 500 fF as in the previous problem,
(
∆V(s) =
)
(
1
sC BL
1
1
+
sCGS sC BL
)(
∆VG (s)→ ∆V =
)(
)(
)
∆VG
3
=
= 0.266 V
C BL
500
+1
+1
48.6
CGS
This value provides a good estimate of the observed change in Fig. 8.29.
The source - substrate diode will clamp the voltage to ∆V ≤ 0.7 V.
8.24
The bitline will charge to an initial voltage of VBL = 3 - VTN
(
)
VTN = 0.7 + 0.5 3 − VTN + 0.6 − 0.6 → VTN = 1.10V | VBL = 1.90V
The initial charge QI on C BL : QI = 10−12 F (1.9V )= 1.9 pC
After charge sharing : VBL =
1.9 pC
= 1.81V. The voltage will be restored to 1.90V
1.05 pF
by the transistor. The total charge delivered through the transistor is
9.45x10−14 C
∆Q = 0.09V (1.05 pF )= 0.0945 pC | ∆vO =
= 0.945 V
10−13 F
0.945
This sense amplifier provides a voltage gain of AV =
= 10.5
0.09
8.25
*PROBLEM 8.24 Charge Transfer Sense Amplifier
VSW 5 0 DC 0 PULSE(0 3 2NS .5NS .5NS 100NS)
VGG 3 0 DC 3
CL 2 0 100FF
CBL 4 0 1PF
CC 6 0 50FF
M1 2 3 4 0 MOSN W=100U L=2U
M2 4 5 6 0 MOSN W=8U L=2U
.IC V(2)=3 V(4)=1.9
.TRAN 0.02NS 100NS UIC
.MODEL MOSN NMOS KP=25U VTO=0.7 GAMMA=0.5 PHI=0.6
.PROBE V(2) V(3) V(4) V(5) V(6)
.END
8-10
3.0V
vSW
vO
2.0V
vBL
1.0V
0V
Time
0s
20ns
40ns
60ns
80ns
100ns
8-11
8.26
*PROBLEM 8.26 - Cross-Coupled Latch
VDD 1 0 DC 3.3
MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P
MN2 2 3 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP2 2 3 1 1 MOSP W=2U L=1U AS=16P AD=16P
CBL1 3 0 1PF
CBL2 2 0 1PF
.IC V(3)=1V V(2)=1.25V V(1)=5
.OP
.TRAN 0.05N 50N UIC
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PRINT TRAN V(2) V(3)
.PROBE V(2) V(3)
.END
4.0
3.0
2.0
1.0
0
0s
Time
5ns
V2(CBL1)
V2(CBL2)
10ns
15ns
20ns
Time
8-12
25ns
30ns
35ns
40ns
8.27
*PROBLEM 8.27 - Cross-Coupled Latch
VDD 1 0 DC 3
VSW 4 0 DC 0 PULSE(3 0 5NS 1NS 1NS 100NS)
MPC 3 4 2 0 MOSN W=20U L=2U AS=80P AD=80P
MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=8U L=2U AS=32P AD=32P
MN2 2 3 0 0 MOSN W=4U L=2U AS=16P AD=16P
*MN2 2 3 0 0 MOSN W=4.4U L=2U AS=17.6P AD=17.6P
MP2 2 3 1 1 MOSP W=8U L=2U AS=32P AD=32P
CBL1 3 0 400FF
CBL2 2 0 400FF
.OP
.TRAN 0.05N 50N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(4)
.END
3.0V
vPC
2.0V
D1 and D 2
1.0V
0V
Time
0s
10ns
20ns
30ns
40ns
50ns
The latch is perfectly balanced in Part (a) and the voltage levels remain symmetrical even after
the PC transistor turns off. This would not happen in the real case because of small asymmetries
and noise in the latch. Even a small capacitive imbalance will cause the latch to assume a
preferred state. Try setting CBL2 = 425 fF in Part (a) for example. The asymmetry in the latch in
Part (b) causes it to switch to a preferred state.
8-13
8.28
*PROBLEM 8.28 - Clocked NMOS Sense Amplifier
VPC 2 0 DC 0 PULSE(3 0 1NS .5NS .5NS 250NS)
VWL 6 0 DC 0 PULSE(0 3 2NS .5NS .5NS 250NS)
VLC 9 0 DC 0 PULSE(0 3 3NS .5NS .5NS 250NS)
VDD 3 0 DC 3
CBL1 5 0 2PF
CBL2 4 0 2PF
*Storage Cell
MA1 5 6 1 0 MOSN W=2U L=2U AS=8P AD=8P
CC 1 0 100FF
*Dummy Cell
MA2 4 6 7 0 MOSN W=2U L=2U AS=8P AD=8P
CD 7 0 50FF
*Sense Amplifier
MPC 5 2 4 0 MOSN W=10U L=2U AS=40P AD=40P
ML1 3 2 4 0 MOSN W=10U L=2U AS=40P AD=40P
ML2 3 2 5 0 MOSN W=10U L=2U AS=40P AD=40P
MS1 5 4 8 0 MOSN W=50U L=2U AS=200P AD=200P
MS2 4 5 8 0 MOSN W=50U L=2U AS=200P AD=200P
MLC 8 9 0 0 MOSN W=50U L=2U AS=200P AD=200P
*
.OP
.TRAN 0.01NS 250NS
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9)
.END
3.0V
2.0V
1.0V
0V
Time
0s
8-14
50ns
100ns
125ns
With only a 3 V power supply, the maximum bit-line differential is only 1.14 V which is
achieved in 120 ns. (This is relatively slow due to the discharge of large bitline capacitances and
the relatively large threshold voltage of the NMOS transistors.)
8-15
8.29
*PROBLEM 8.29 - Clocked NMOS Sense Amplifier
VPC 2 0 DC 0 PULSE(5 0 1NS .5NS .5NS 250NS)
VWL 6 0 DC 0 PULSE(0 5 2NS .5NS .5NS 250NS)
VLC 9 0 DC 0 PULSE(0 5 3NS .5NS .5NS 250NS)
VDD 3 0 DC 5
CBL1 5 0 2PF
CBL2 4 0 2PF
*Storage Cell
MA1 5 6 1 0 MOSN W=2U L=2U AS=8P AD=8P
CC 1 0 100FF
*Dummy Cell
MA2 4 6 7 0 MOSN W=2U L=2U AS=8P AD=8P
CD 7 0 50FF
*Sense Amplifier
MPC 5 2 4 0 MOSN W=10U L=2U AS=40P AD=40P
ML1 3 2 4 0 MOSN W=10U L=2U AS=40P AD=40P
ML2 3 2 5 0 MOSN W=10U L=2U AS=40P AD=40P
MS1 5 4 8 0 MOSN W=50U L=2U AS=200P AD=200P
MS2 4 5 8 0 MOSN W=50U L=2U AS=200P AD=200P
MLC 8 9 0 0 MOSN W=50U L=2U AS=200P AD=200P
*
.OP
.TRAN 0.01NS 250NS
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9)
.END
6.0V
4.0V
2.0V
0V
Time
0s
5ns
10ns
15ns
20ns
25ns
30ns
With the 5 V power supply, the maximum bit-line differential is 1.75 V. A 1.5 V differential is
achieved in approximately 15 ns, which is much faster than the 3 V case.
8.30
8-16
*PROBLEM 8.30 - Cascaded Inverter Pair
VDD 1 0 DC 3
VI 2 0 DC 0
MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P
MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P
MP2 4 3 1 1 MOSP W=2U L=1U AS=16P AD=16P
.OP
.DC VI 0 3 0.001
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(2) V(3) V(4)
.END
3.0V
2.0V
1.0V
vO
0V
0V
0.5V
1.0V
1.5V
2.0V
2.5V
3.0V
Results: 0 V, 1.429 V, 3 V
8.31
(a) The array requires: (12 transistors/row) 212 rows + (1 load transistor/row) 212 rows
( )
(
)
(
)
= 13 212 = 53,248 transistors. The 24 inverters require an additional 48 transistors.
N = 53,296 transistors | (b) The number is the same.
8-17
8.32
(a) NMOS Pass Transistor Tree : 2 x 20 + 21 + 22 + 23 + 2 4 + 25 + 26 = 254 Transistors
(
)
2 logic inverters per level = 14 inverters = 28 transistors.
Total = 282 Transistors
(b) An estimate : 128 data bits requires 128 7 - input gates for data selectors;
1 - 128 input NOR gate; 14 address bit inverters
Total = 128(8)+ 1(129)+ 14(2) = 1181 transistors without looking closely at the logic detail.
A number of additional inverters may be needed, and the 128 input gate can likely
be replaced with a smaller NOR tree.
8.33
A0
Clock
A1
A2
Clock
V
DD
0
V
DD
1
V
DD
2
V
DD
3
V
DD
4
V
DD
5
V
DD
6
V
DD
7
NMOS Transistor
8-18
8.34
(a) The output of the first NMOS transistor will be
[
)]
(
V1 = 5 − VTN = 5 − 0.75 + 0.55 V1 + 0.6 − 0.6 → V1 = 3.55V | VTN = 1.45V
The output of the other gates reaches this same value. All three nodes = 3.55V.
(b) The node voltages will all be
+ 5 V.
8.35
Charge sharing occurs. Assuming C2 and C3 are discharged (the worst case)
(a) V
B
=
(c) V
B
=
C1VDD + C2 (0)
2C2VDD 2
2
= VDD | Node B drops to VDD .
C1 + C2
2C2 + C2 3
3
⎛2
⎞
2
V
3C
⎜
⎟
2
DD
C
+
C
V
+
C
0
( 1 2 )3 DD 3 ( )
V
1
⎝3
⎠
V
=
=
= DD | Node B drops to VDD .
b
() B
2C2 + C2 + C2
C1 + C2 + C3
2
2
=
C1VDD
RC2VDD
R
C
=
=
VDD ≥ VIH where R = 1
C1 + C2 + C3 RC2 + C2 + C2 R + 2
C2
R(VDD − VIH ) ≥ 2VIH
or R ≥
2VIH
VDD − VIH
Using VDD = 5V , VTN = 0.7V , VTP = −0.7V in Eq. (7.9) :
VIH =
5(5)+ 3(0.7)+ 5(−0.7)
8
= 2.95V | R ≥
2(2.95)
2VIH
=
= 2.88 | C1 ≥ 2.88C2
VDD − VIH
2.05
8.36
Z = A0 + A1 + A2
VDD
Clock
A0
A1
A
Z
2
C1
8-19
8.37
B7
B6
B5
B4
B3
B2
B1
B0
W0
W1
W2
1
0
1
0
1
1
1
0
0
1
0
0
0
0
0
0
1
1
0
1
0
0
0
0
W3
W4
0
0
0
0
1
0
0
0
1
1
0
1
1
1
1
0
W5
0
1
0
0
0
0
0
0
8.38
VDD
Clock
W
L
W
5
W
L
4 W
L
W
L
*PROBLEM 8.38 - Simplified ROM Cross-Section
VCLK 1 0 DC 0 PULSE(0 5 2.5NS 1NS 1NS 25NS)
VW5 3 0 DC 0 PULSE(0 5 4.5NS 1NS 1NS 25NS)
VDD 5 0 DC 5
MPC 4 1 5 5 MOSP W=4U L=2U AS=16P AD=16P
MNC 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P
MW5 4 3 2 0 MOSN W=4U L=2U AS=16P AD=16P
MWW 4 0 2 0 MOSN W=16U L=2U AS=64P AD=64P
.OP
.TRAN 0.01NS 15NS
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(1) V(2) V(3) V(4) V(5)
.END
8-20
6.0V
vO
4.0V
vCLK
2.0V
vW5
0V
Time
0s
2ns
4ns
6ns
8ns
10ns
12ns
8.39
W1
W2
B5
0
1
B4
0
0
B3
1
0
B2
0
1
B1
1
0
B0
0
1
W3
0
1
1
1
0
1
B2
B1
B0
W0
W1
W2
1
1
1
0
1
0
1
0
1
W3
0
1
0
8.40
Note that the input lines are active low.
8-21
8.41
+5 V
4
1
4
1
(2.64V)
Q
Q
0.7V
M
+5 V
R
2
1
2
1
OFF
Regenerative switching of the cell will take place when the voltage at Q is pulled low enough
by transistor MR that the voltage at Q rises above the NMOS transistor threshold voltage.
Equating drain currents for this condition yields the value of VQ . It appears that
the NMOS transistor will be in the linear region, and the PMOS transistor will be saturated.
⎛ ⎞⎛
2
4 x10−5 ⎛ 4 ⎞
0.7 ⎞
−4 2
For VDD = 5V,
⎜ ⎟(5 − VQ − 0.7) = 10 ⎜ ⎟⎜VQ − 0.7 −
⎟0.7 → VQ = 2.64V which agrees with
2 ⎠
2 ⎝1⎠
⎝ 1 ⎠⎝
the assumptions. Now, MR must be large enough to force VQ = 2.64V . MR and the PMOS
load transistor are both in the linear region.
⎛ 4 ⎞⎛
⎛ ⎞ ⎛
⎛W ⎞ 1.16
2.36 ⎞
2.64 ⎞
−4 W
4 x10−5 ⎜ ⎟⎜ 5 − 0.7 − 0.7 −
⎟2.36 ≤ 10 ⎜ ⎟ ⎜ 5 − 0.7 −
⎟2.64 → ⎜ ⎟ ≥
2 ⎠
2 ⎠
1
⎝ 1 ⎠⎝
⎝ L ⎠R ⎝
⎝ L ⎠R
8.42
The inputs are active in the low voltage state. V1 low sets the latch and V2 low resets the
latch. V1 = S V2 = R .
8-22
8.43
*PROBLEM 8.43 - D-Latch
VDD 7 0 DC 2.5
VI 1 0 DC 2.5
VCLK 2 0 DC 0 PULSE(0 2.5 3NS 1NS 1NS 5NS)
VNCLK 3 0 DC 0 PULSE(2.5 0 3NS 1NS 1NS 5NS)
MTN1 1 2 4 0 MOSN W=2U L=1U AS=16P AD=16P
MTP1 1 3 4 7 MOSP W=2U L=1U AS=16P AD=16P
MIN1 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P
MIP1 5 4 7 7 MOSP W=2U L=1U AS=16P AD=16P
MIN2 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P
MIP2 6 5 7 7 MOSP W=2U L=1U AS=16P AD=16P
MTN2 6 3 4 0 MOSN W=2U L=1U AS=16P AD=16P
MTP2 6 2 4 7 MOSP W=2U L=1U AS=16P AD=16P
.OP
.TRAN 0.01N 12N
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
3.0V
2.0V
1.0V
0V
-1.0V
0s
1ns
2ns
3ns
4ns
5ns
6ns
7ns
8ns
9ns
10ns
11ns
12ns
8-23
8.44
*PROBLEM 8.44 - Master-Slave Flip-Flop
VDD 10 0 DC 5
VI 1 0 DC 5 PWL(0 5 17.4NS 5 17.6NS 0 30NS 0)
VCLK 2 0 DC 0 PULSE(0 5 0NS 2.5NS 2.5NS 7.5NS)
VNCLK 3 0 DC 0 PULSE(5 0 0NS 2.5NS 2.5NS 7.5NS)
MTN1 1 2 4 0 MOSN W=4U L=2U AS=16P AD=16P
MTP1 1 3 4 10 MOSP W=4U L=2U AS=16P AD=16P
MIN1 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P
MIP1 5 4 10 10 MOSP W=4U L=2U AS=16P AD=16P
MIN2 6 5 0 0 MOSN W=4U L=2U AS=16P AD=16P
MIP2 6 5 10 10 MOSP W=4U L=2U AS=16P AD=16P
MTN2 6 3 4 0 MOSN W=4U L=2U AS=16P AD=16P
MTP2 6 2 4 10 MOSP W=4U L=2U AS=16P AD=16P
*
MTN3 6 3 7 0 MOSN W=4U L=2U AS=16P AD=16P
MTP3 6 2 7 10 MOSP W=4U L=2U AS=16P AD=16P
MIN3 8 7 0 0 MOSN W=4U L=2U AS=16P AD=16P
MIP3 8 7 10 10 MOSP W=4U L=2U AS=16P AD=16P
MIN4 9 8 0 0 MOSN W=4U L=2U AS=16P AD=16P
MIP4 9 8 10 10 MOSP W=4U L=2U AS=16P AD=16P
MTN4 9 2 7 0 MOSN W=4U L=2U AS=16P AD=16P
MTP4 9 3 7 10 MOSP W=4U L=2U AS=16P AD=16P
.IC V(1)=5 V(2)=0 V(3)=5 V(4)=0 V(5)=5 V(6)=0 V(7)=0
+ V(8)=5 V(9)=0 V(10)=5
.TRAN 0.05N 20N UIC
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5
+LAMBDA=.05 TOX=41.5N
+CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P
.PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(9)
.END
The flip-flop operates normally. Data is transferred to the master following the first clock
transition and to the slave after the second clock transition. The maximum rise and fall times
are highly dependent upon the position of the data transition edge. It is interesting to
experiment with the data delay to see the effect. At some point the flip-flop will fail.
8-24
6.0V
Data
4.0V
CLK
CLK
2.0V
Master
Data
Slave
0V
Time
0s
5ns
10ns
15ns
20ns
8-25
CHAPTER 9
9.1
Since VREF = −1.25V , and v I = −1.6V , Q1 is off and Q2 is conducting.
vC1 = 0 V and vC 2 = −α F I EE RC ≅ −I EE RC = −(2mA)(350Ω) = −0.700 V
9.2
⎛ ∆V ⎞
IC 2
0.995α F I EE
= exp⎜ BE ⎟ ⇒ ∆VBE = 0.025ln
= 0.132V
IC1
0.005α F I EE
⎝ VT ⎠
(a) v I = VREF + ∆VBE = −1.25 + 0.132 = −1.12 V
v I = VREF + ∆VBE = −1.25 − 0.132 = −1.38 V
(b) v I = VREF + ∆VBE = −2.00 + 0.132 = −1.87 V
v I = VREF + ∆VBE = −2.00 − 0.132 = −2.13 V
9.3
Since VREF = −2V , and v I = −1.6V , Q2 is off and Q1 is conducting.
vC 2 = 0 V and vC1 = −α F I EE RC ≅ −I EE RC = −(2.5mA)(700Ω)= −1.75 V
Note that Q1 is beginning to enter the saturation region of operation, but VBC = +0.15 V is not
really enough to turn on the collector-base diode. (See Problems 9.5 or 5.61.)
9.4
vI = VREF + 0.3V ⇒ Q1 on; Q2 off. IC1 = α F I EE ≅ I EE = 0.3mA | IC 2 = 0
vC1 = 0 − IC1 (R1 + RC )= −0.3mA(3.33kΩ + 2 kΩ) = −1.60 V
vC 2 = 0 − IC1 R1 = −0.3mA(3.33kΩ) = −0.999 V
9-1
9.5
With VBE = 0.7 and VBC = 0.3, the transistor is technically in the saturation region, but
calculating the currents using the transport model in Eq. (5.13) yields
βF =
0.98
0.2
αF
αR
=
= 49 | β R =
=
= 0.25
1− α F 1− 0.98
1− α R 1− 0.2
⎡ ⎛ 0.7 ⎞
⎛ 0.3 ⎞⎤ 10−15 ⎡ ⎛ 0.3 ⎞ ⎤
iC = 10−15 ⎢exp⎜
⎟ − exp⎜
⎟⎥ −
⎟ −1⎥ = 1.446 mA
⎢exp⎜
⎝ 0.025 ⎠⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦
⎣ ⎝ 0.025 ⎠
⎡ ⎛ 0.7 ⎞
⎛ 0.3 ⎞⎤ 10−15 ⎡ ⎛ 0.7 ⎞ ⎤
iE = 10−15 ⎢exp⎜
−
exp
⎟
⎜
⎟⎥ +
⎟ −1⎥ = 1.476 mA
⎢exp⎜
⎝ 0.025 ⎠⎦ 49 ⎣ ⎝ 0.025 ⎠ ⎦
⎣ ⎝ 0.025 ⎠
10−15 ⎡ ⎛ 0.7 ⎞ ⎤ 10−15 ⎡ ⎛ 0.3 ⎞ ⎤
iB =
⎟ −1⎥ +
⎟ −1⎥ = 29.52 µA
⎢exp⎜
⎢exp⎜
49 ⎣ ⎝ 0.025 ⎠ ⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦
At 0.3 V, the collector-base junction is not heavily forward-biased compared to the baseemitter junction, and IC = 48.99IB ≅ β F IB . The transistor still acts as if it is operating in the
forward-active region.
9.6
(a) For Q2 off, VH = 0 V. For Q2 on, IC ≈ IE and
−0.2 − 0.7 − (−2)
= 100 µA VL ≅ −4000I E = −0.400 V
1.1x10 4
(b) Yes, these voltages are symmetrically positioned above and below VREF, i. e. VREF ± 0.2 V,
and the current will be fully switched. See Parts (d) and (e).
0 − 0.7 − (−2)
0.4V
= 118 µA R =
= 3.39 kΩ
(c) For v I = 0, IC ≅ I E =
4
118µA
1.1x10
IE =
(d) Q2 is cutoff. Q1 is saturated with VBC = +0.4 V.
(e) Q1 is cutoff. Q2 is saturated with VBC = +0.2 V.
(f) 0.2 V and 0.4 V are not large enough to heavily saturate Q1 or Q2. Although the transistors
are technically operating in the saturation region, the transistors still behave as if they are in
the forward-active region. (See problem 10.5).
9-2
9.7
(a) For v
−0.2 − 0.7 − (−2)
= 100 µA | VL ≅ −4000I E = −0.400 V
1.1x10 4
0 − 0.7 − (−2)
0.4V
= 118 µA | R =
= 3.39 kΩ
For vI = VH = 0V , IC ≅ I E =
4
118µA
1.1x10
⎛100µA + 118µA ⎞
P = 2V ⎜
⎟ = 218 µW
2
⎝
⎠
R
11kΩ
R
4kΩ
R
3.39kΩ
(b) REE' = 5EE = 5 = 2.20 kΩ | RC1' = 5C1 = 5 = 800 Ω | RC' 2 = 5C2 = 5 = 678 Ω
I
= VL , I E =
9.8
VH = 0 − VBE = −0.7 V | VL = −(5mA)(200Ω)− 0.7 = −1.70 V
VREF =
VH + VL
= −1.2 V | ∆V = (5mA)(200Ω)= 1.00 V
2
9.9
VH = 0 − VBE = −0.7 V | VL = −(1mA)(600Ω)− 0.7 = −1.30 V
VREF =
VH + VL
= −1.0 V | ∆V = (1mA)(600Ω) = 0.600 V
2
9.10
I EE = 4(0.3mA) = 1.2 mA | I3 = I 4 = 4(0.1mA)= 0.4 mA | RC =
2kΩ
= 500Ω
4
9.11
(a) R
C
=
∆V
0.8V
=
= 2.67 kΩ | VH = 0 − VBE = −0.7 V | VL = −0.8 − VBE = −1.5 V
I EE 0.3mA
VH + VL
= −1.10 V
2
⎡
⎡
⎛ ∆V ⎞⎤ 0.8
⎛ 0.8
⎞⎤
∆V
(b) NM H = NM L = 2 − VT ⎢1+ ln⎜⎝ V −1⎟⎠⎥ = 2 − 0.025V ⎢⎣1+ ln⎜⎝ 0.025 −1⎟⎠⎥⎦ = 0.289 V
⎣
⎦
T
(c) For Q1: VCB = -0.8 - (-0.7) = -0.1 V which represents a slight forward bias, but it is not
enough to turn on the diode. For Q2: VCB = -0.8 - (-1.10) = +0.3 V which represents a reverse
bias. Both values are satisfactory for operation of the logic gate.
VREF =
9-3
9.12
(b) For Q1 on and Q2 off, IC1 = α F I EE ≅ I EE = 0.3mA | IC2 = 0
VL = 0 − IC1 (R1 + RC )− 0.7V = −0.3mA(3.33kΩ + 2kΩ)− 0.7V = −2.30 V
VH = 0 − IC1 R1 − 0.7V = −0.3mA(3.33kΩ)− 0.7 = −1.70 V
∆V = VH − VL = 0.600 V
VH + VL
= −2.0V = VREF | Yes, the input and output voltage levels are
2
compatible with each other and are symmetrically placed around VREF.
(c)
R1
RC
RC
Q
v
Q
3
4
v O2
O1
vI
Q
1
Q
2
V REF
0.1 mA
0.1 mA
IEE
- 5.2 V
9.13
(a) See Prob. 9.12
(b) ∆V = α
0.4V
= 267 Ω
1.5mA
≅ −I EE R1 − VBE = −1.5mA(800)− 0.7V = −1.90 V
I RC ≅ I EE RC | RC =
F EE
VH = 0 − α F I EE R1 − VBE
VL = 0 − α F I EE (R1 + RC )− VBE ≅ −I EE (R1 + RC )− VBE = −1.5mA(1067)− 0.7V = −2.30 V
VREF =
9-4
VH + VL −1.90 − 2.30
=
= −2.10 V
2
2
9.14
∆V = ∆VBE + ∆iB 4 RC | Let the Fanout = N; β F = 30. Then there will be N base
currents that must be supplied from emitter - follower transistor Q4 : ∆i E4 = N
∆VBE = VT ln
∆iB 4 =
I EE
βF + 1
⎡
⎤
⎛ ∆I ⎞
I EE
IC 4 + ∆IC 4
I + ∆I E 4
⎥
= VT ln E 4
= VT ln⎜1+ E 4 ⎟ = 0.025ln⎢1+ N
IC 4
IE 4
IE 4 ⎠
⎝
⎢⎣
(β F + 1)I E 4 ⎥⎦
∆i E4
I EE
=N
2
βF + 1
(β F + 1)
⎛
0.3mA ⎞
0.3mA
⎟+ N
∆V = ∆VBE + ∆iB 4 RC = 0.025ln⎜⎜1+ N
2kΩ | ∆V ≤ 0.025
2
⎟
31
0.1mA
(
)
⎠
⎝
(31)
⎛ 3N ⎞ 0.6N
0.025 = 0.025ln⎜1+
⎟+
31 ⎠ (31)2
⎝
| Using MATLAB or HP - Solver : N ≤ 11.01 → N = 11
9.15
'
'
= 8RC1 = 8(1.85kΩ) = 14.8 kΩ | RC2
= 8RC2 = 8(2kΩ)= 16.0 kΩ
RC1
'
REE
= 8REE = 8(11.7kΩ)= 93.6 kΩ | R' = 8R = 8(42kΩ)= 336 kΩ
9.16
RC1 1850Ω
R
2000Ω
'
=
= 231 Ω | RC2
= C2 =
= 250 Ω
8
8
8
8
R
11.7kΩ
R 42kΩ
'
REE
= EE =
= 1.46 kΩ | R' = =
= 5.25 kΩ
8
8
8
8
'
(b) RC1
= 5RC1 = 5(1.85kΩ) = 9.25 kΩ | RC' 2 = 5RC 2 = 5(2kΩ)= 10.0 kΩ
'
=
(a) RC1
'
REE
= 5REE = 5(11.7kΩ) = 58.5 kΩ | R' = 5R = 5(42kΩ)= 210 kΩ
9-5
9.17
∆V = α F I EE RC ≅ I EE RC = 0.2mA(2kΩ) = 0.400 V
VH = 0 − α F I EE R1 − VBE ≅ −I EE R1 − VBE = −0.2mA(2kΩ)− 0.7V = −1.10 V
VL = 0 − α F I EE (R1 + RC )− VBE ≅ −I EE (R1 + RC )− VBE = −0.2mA(4kΩ)− 0.7V = −1.50 V
VH + VL −1.10 −1.50
=
= −1.30 V
2
2
⎡
⎡
⎛ ∆V ⎞⎤ 0.4
⎛ 0.4
⎞⎤
∆V
− VT ⎢1+ ln⎜
−1⎟⎥ =
− 0.025V ⎢1+ ln⎜
−1⎟⎥ = 0.107 V
NM L = NM H =
2
⎝ 0.025 ⎠⎦
⎠⎦ 2
⎝ VT
⎣
⎣
VREF =
IE3 + IE 4 =
[V − (−2)]+ [V − (−2)] = (4 −1.10 −1.50)V = 28.0 µA
H
L
50kΩ
R
P = 28µA(2V )+ 0.2mA(5.2V )= 1.10 mW
9.18
NM H =
⎡
⎛ ∆V ⎞⎤
∆V
− VT ⎢1+ ln⎜
−1⎟⎥ |
2
⎝ VT
⎠⎦
⎣
0.1V =
∆V
− 0.025V 1+ ln (40∆V −1)
2
[
]
Solving by trial - and - error, HP - Solver, or MATLAB : ∆V = 0.383 V
function f=dv15(v)
f=4-20*v+1+log(40*v-1);
fzero('dv15',0.5) yields ans = 0.3831
9.19
(a) The change in vBE will be neglected :
∆v BE = VT ln
0.8IC
= −5.6mV
IC
V H = 0 − VBE = 0 − 0.7 = −0.7 V - no change
VL = 0 − α F IEE RC − VBE ≅ −IEE RC − VBE = −0.3mA(1.2)(2kΩ) − 0.7V = −1.42 V
VL has dropped by 0.12V. | ∆V = 0.3mA(1.2)(2kΩ) = 0.72 V
⎡
⎛ ∆V ⎞⎤ 0.72
⎡
⎛ 0.72
⎞⎤
∆V
− VT ⎢1+ ln⎜
−1⎟⎥ =
− 0.025V ⎢1+ ln⎜
−1⎟⎥ = 0.252 V
⎝ 0.025 ⎠⎦
2
2
⎣
⎝ VT
⎠⎦
⎣
(b) At node A : VH = 0 − VBE = 0 − 0.7 = −0.7 V - no change
NM H = NM L =
VL = 0 − α F IEE RC − VBE ≅ −IEE RC − VBE = −
−1.0 − 0.7 − (−5.2)
V (1.2)(2kΩ) − 0.7V = −1.30 V
1.2(11.7kΩ)
VL also has not changed! | Similar results hold at node B because the voltages
are set by resistor ratios.
NM H = NM L =
9-6
⎡
⎛ ∆V ⎞⎤ 0.6
⎡
⎛ 0.6
⎞⎤
∆V
− VT ⎢1+ ln⎜
−1⎟⎥ =
− 0.025V ⎢1+ ln⎜
−1⎟⎥ = 0.197 V, unchanged
⎝ 0.025 ⎠⎦
2
⎣
⎝ VT
⎠⎦ 2
⎣
9.20
(a) V
H
= −0.7V | ∆V = 0.8V | VL = −0.8 − 0.7 = −1.5V | VREF =
VH + VL
= −1.1V
2
−1.1− 0.7 − (−5.2) V
0.8V
= 11.3 kΩ | RC 2 =
= 2.67 kΩ
mA
0.3
0.3mA
−0.7 − 0.7 − (−5.2) V
0.8V
= 0.336mA | RC1 =
= 2.38 kΩ
I E1 =
kΩ
11.3
0.336mA
⎡
⎞⎤
⎛ 0.8
0.8
NM
=
NM
=
−
0.025V
1+
ln
−1
b
⎢
⎟⎥ = 0.289 V
⎜
() H
L
2
⎝ 0.025 ⎠⎦
⎣
REE =
(c) V
CB1
= −0.8 − (−0.7)= −0.1V | VCB2 = −0.8 − (−1.1)= +0.3V
The collector - base junction of Q2 is reverse - biased by 0.3 V. Although the
collector - base junction of Q1 is forward - biased by 0.1 V, this is not large
enough to cause a problem. Therefore the voltages are acceptable.
9.21
NM H =
⎡
⎛ ∆V ⎞⎤
∆V
− VT ⎢1+ ln⎜
−1⎟⎥
2
⎠⎦
⎝ VT
⎣
For room temperature, VT = 0.025V : 0.1V =
For - 55C, VT = 0.0188V : 0.1V =
⎡
⎛ ∆V
⎞⎤
∆V
− 0.025V ⎢1+ ln⎜
−1⎟⎥ → ∆V = 0.383V
2
⎝ 0.025 ⎠⎦
⎣
⎡
⎛ ∆V
⎞⎤
∆V
− 0.0188V ⎢1+ ln⎜
−1⎟⎥ → ∆V = 0.346V
2
⎝ 0.0188 ⎠⎦
⎣
For + 75C, VT = 0.0300V : 0.1V =
⎡
⎛ ∆V
⎞⎤
∆V
− 0.0300V ⎢1+ ln⎜
−1⎟⎥ → ∆V = 0.413V
2
⎝ 0.0300 ⎠⎦
⎣
∆V = 0.413 V
9-7
9.22
In the original circuit : VH = −2mA(2kΩ)− 0.7V = −1.1V | ∆V = 2mA(2kΩ)= 0.4V
VL = −1.1V − ∆V = −1.5V. VH and VL are symmetrically placed about VREF .
−1.3 − 0.7 − (−5.2) V
= 16.0 kΩ. R1 and R C2 remain unchanged.
0.2
mA
−1.1− 0.7 − (−5.2) V
For Q1 on and and Q2 off : I EE =
= 0.2125mA
16.0
kΩ
VL1 = −(0.2125mA)(2kΩ + RC1)− 0.7V | VL1 = −1.5V → RC1 = 1.77 kΩ.
REE =
Note that there are only 3 variables (R1 , RC1 and R C2 ) and four voltage levels.
Thus we cannot force them all to the desired level. For this design,
VH 2 = −(0.2125mA)(2kΩ)− 0.7V = −1.125V rather than the desired -1.10V
9.23
60kΩ
(−5.2V )= −3.0V | REQ = 60kΩ 44kΩ = 25.38kΩ
60kΩ + 44kΩ
−3.0 − 0.7 − (−5.2) V
−3.0 − 0.7 − (−5.2) V
=
| I EE = β F I BS = β F
25.38 + (β F + 1)30 kΩ
25.38 + (β F + 1)30 kΩ
VEQ =
I BS
For large β F , I EE =
−3.0 − 0.7 − (−5.2) V
requires VCBS ≥ 0V | VCBS
= 50.0 µA | Active region operation
30
kΩ
= VREF − VBE2 − VBS = VREF − 0.7V − (−3V )
VREF − 0.7V − (−3V )≥ 0 → VREF ≥ −2.30 V
9.24
The base of QS must not be higher than VL − 0.7 = −0.2mA(4kΩ)− 0.7 − 0.7 = −2.2V
Design choice - Choose VB = −3 V. Assume β F = 50. | I B =
RE =
VB − VEE −3 − (−5.2)
=
= 10.8 kΩ → RE = 11 kΩ
IE
0.204mA
Choose I R 2 = 20µA.
R2 =
I R1 = I R 2 − I B = 16µA. R1 =
9-8
0 − (−3)
20µA
= 150 kΩ → R2 = 150 kΩ
−3 − (−5.2)
16µA
200µA
= 4µA
50
= 138 kΩ → R1 = 136 kΩ
9.25
*PROBLEM 9.25 - ECL INVERTER VTC
VIN 2 0 DC -1.3
VREF 4 0 -1.0
VEE 8 0 -5.2
Q1 1 2 3 NBJT
Q2 5 4 3 NBJT
Q3 0 1 6 NBJT
Q4 0 5 7 NBJT
REE 3 8 11.7K
RC1 0 1 1.85K
RC2 0 5 2K
R3 6 8 42K
R4 7 8 42K
.DC VIN -1.3 -0.7 .01
.TEMP -55 25 85
.MODEL NBJT NPN BF=40 BR=0.25 VA=50
.PROBE V(2) V(1) V(5) V(6) V(7)
.PRINT DC V(2) V(6) V(7)
.END
T
VT
VH
VL
∆V
VREF
VIH
VOH
VIL
VOL
NMH
NML
-55C
0.0188 V
-0.846 V
-1.40 V
0.554 V
-1.00 V
-0.918 V
-0.865 V
-1.08 V
-1.38 V
0.053 V
0.300 V
+25C
0.0257 V
-0.724 V
-1.30 V
0.576 V
-1.00 V
-0.895 V
-0.750 V
-1.10 V
-1.27 V
0.145 V
0.170 V
+85C
0.0309 V
-0.629 V
-1.22 V
0.591 V
-1.00 V
-0.880 V
-0.660 V
-1.12 V
-1.19 V
0.220 V
0.070 V
VIH , VOH , VOL , and VIL were calculated from Eqns. 9.27 - 9.30.
With a fixed reference voltage, the noise margins change with temperature and can become
zero for a large enough temperature change.
9-9
9.26
RC
Q
Q
A
B
C
D
V
2
E
4
REF
Y =A+B+C+D+E
I EE
R
-V
EE
9.27
R
Q
C
3
Q
B
A
C
D
2
V
REF
Y=A+B+C+D
I EE
R
-V
9.28
(a ) For Q
4
on, IC 4 = α F I E 4 ≅ I E 4 =
EE
−0.7 − (−2.5)
840
VL = 1.0V − (2.14mA)(390Ω)− 0.7V = −0.540 V
= 2.14 mA
For Q4 off, and neglecting the base current in Q5 , VH = 1.0 − 0.7 = +0.300 V
(b) For v
= 0.3V, IC 2 = α F I E 2 ≅ I E 2 =
0.3 − 0.7 − (−2.5)
= 2.50 mA
840
∆V
0.84V
∆V = 0.30 − (−0.54)= 0.84V | R =
=
= 336 Ω
IC 2 2.50mA
9-10
A
9.29
(a) For Q
4
on, IC 4 = α F I E 4 ≅ I E 4 =
−0.7 − (−3.2)
840
VL = 1.3V − (2.98mA)(390Ω)− 0.7V = −0.56 V
= 2.98 mA
For Q4 off, and neglecting the base current in Q5 , VH ≅ 1.3 − 0.7 = +0.60 V
(b) For v
= 0.6V , IC 2 = α F I E 2 ≅ I E 2 =
0.6 − 0.7 − (−3.2)
= 3.69 mA
840
∆V
1.16V
∆V = 0.60 − (−0.56)= 1.16V | R =
=
= 314 Ω
IC 2 3.69mA
A
9.30
V CC
390
Ω
A
Q
B
2
Q
3
Q
Q
4
5
Y=A+B
840 Ω
600 Ω
-V EE
(a)
(b) The NOR output is taken from the collectors of Q2/Q3, and the 390Ω resistor, Q5, and the
600-Ω resistor are removed.
9.31
vOmin = − I EE RL = −(2.5mA)(1.2kΩ) = −3.00 V | I E = I EE +
vO
4 − 0.7
= 2.5mA +
= 5.25 mA
RL
1.2kΩ
VBC = 4 − 5 = −1 V , so the transistor is in the forward - active region.
IB =
IE
5.25mA
=
= 0.103 mA and IC = β F I B = 5.15 mA.
β F + 1 50 + 1
9.32
(a-b) See Problem 9.33 (c) IEE ≥ −
(VI − 0.7)V
1kΩ
=
3.7V
= 3.7 mA
1kΩ
9-11
9.33
Simulation Results from B2SPICE
Circuit 9.32-Transient-2
Circuit 9.32-Transient-1
+0.000e+000
(V)
+500.000u
+1.000m
Time (s)
+1.500m
+0.000e+000
(V)
+500.000u
+1.000m
Time (s)
+1.500m
+2.000m
+2.000m
+3.000
+3.000
+2.000
+2.000
+1.000
+1.000
+0.000e+000
+0.000e+000
-1.000
-1.000
-2.000
-2.000
-3.000
-3.000
-4.000
IEE = 4 mA
V(1)
IEE = 2 mA
V(3)
V(1)
V(3)
9.34
(a) vO = v I − 0.7V = (−1.7 + sin2000πt ) V
2.7V
= 0.13 mA with no safety margin.
20kΩ
The transistor will cut off at the bottom of the input waveform for IEE = 0.13 mA.
(b) vOmin = −2.7V
| - IEE RL ≤ −2.7V → IEE ≥
9.35
Simulation results from B2SPICE
Circuit 9.35-Transient-1
(V)
+0.000e+000
+500.000u
+1.000m
Time (s)
+1.500m
+2.000m
+0.000e+000
-500.000m
-1.000
-1.500
-2.000
-2.500
-3.000
V(1)
V(3)
9.36
(a ) The transistor cuts off for vOmin = -I EE RL = -(0.5mA)(1kΩ) = -0.5V. So vI ≥ -0.5 + 0.7 = +0.2V.
For v O > 1.5 V, the transistor enters the saturation region of operation.
Therefore : 0.2 V ≤ v I ≤ 1.5 V.
(b) v
min
O
= −1.5 − 0.7 = −2.2 V. We need - I EE RL ≤ −2.2V → I EE ≥
2.2V
= 2.2 mA
1kΩ
9.37
vOmin = −10 − 0.7 = −10.7 V . We need - IEE RL ≤ −10.7V → IEE ≥
9-12
10.7V
= 10.7 mA
1kΩ
9.38
Assuming Q1 off and using voltage division, −12 = −15
12 − (−15)
12
+
= 60 mA !
2000
500
IE =
9.39
(a) v
min
O
(b) I
⎛ 4.7kΩ ⎞
= −10 − 0.7 = −10.7 V. We need -15V ⎜
⎟ ≤ −10.7V
⎝ 4.7kΩ + RE ⎠
(15 −10.7)(4.7kΩ) = 1.89 kΩ
RE ≤
10.7
| IE =
−0.7 −0.7 − (−15)
+
= 7.43 mA |
1890
4700
=
E
2000
⇒ RE = 500 Ω
2000 + RE
vI − 0.7 vI − 0.7 − VEE
+
RE
RL
(c) I
E
=
−10 − 0.7 −10 − 0.7 − (−15)
+
= 0 mA
1890
4700
9.40
(a) See the solution to Problem 9.41.
(b) v
(c) v
O
= v I − 0.7V = (−2.2 + 1.5sin 2000πt ) V
max
I
IE =
= −1.5 + 1.5 = 0V | vOmax = −0.7V |
vO vO − VEE
+
RL
RE
| I Emax =
−0.7 −0.7 − (−6)
+
= 3.93 mA
1300
4700
−3.7 −3.7 − (−6)
+
= 0.982 mA
1300
4700
⎛ 4.7kΩ ⎞
4.7kΩ(6 − 3.7)
= 2920 Ω
(e) We need - 6V ⎜⎝ 4.7kΩ + R ⎟⎠ ≤ −3.7V → RE ≤
3.7
E
(d ) v
min
O
= −2.2 −1.5 = −3.7V | I Emin =
9.41 Simulation results from B2SPICE
Circuit 9.37-Transient-2
+0.000e+000
+500.000u
+1.000m
+1.500m
Time (s)
+2.000m
+2.500m
+3.000m
+0.000e+000
-1.000
-2.000
-3.000
-4.000
V(1)
V(3)
V(5)
9-13
9.42
The outputs act as a "wired - or" connection.
For v I = −0.7V, vO1 = vO 2 = −0.7 V | IE 3 = 0 | IE 4 = 0.1mA + 0.1mA = 0.200 mA
For v I = −1.3V, vO1 = vO 2 = −0.7 V | IE 3 = 0.1mA + 0.1mA = 0.200 mA | IE 4 = 0
9.43
Y = A+ B | Z = A+ B
9.44
A
Y
OR
B
C
NOR
D
9.45
For Fig. 9.21, P ≅ 0.5mA(5.2V)= 2.6mW = 2600µW. For 20µW, the power must
be reduced by 130X. The currents must be reduced by 130X and the resistors must
increase by this factor to keep the logic swing the same : R C = 130(2kΩ)= 260kΩ.
Using Eq. (9.54), τ P = 0.69(260kΩ)(2 pF ) = 359 ns - rather slow!
9.46
RC 2kΩ
=
= 1kΩ | ∆V = 0.3mA(1kΩ) = 0.3V | VH = 0 − 0.7 = −0.7V
2
2
−0.7 −1.0
V = −0.850 V | P ≅ 0.5mA(5.2V )= 2.6mW
VL = VH − 0.3V = −1.0V | VREF =
2
τ P = 0.69(1kΩ)(2 pF )= 1.38ns | PDP = 2.6mW (1.38ns)= 3.59 pJ
RC' =
9.47
∆V = 0.15mA(2kΩ) = 0.3V | VH = 0 − 0.7 = −0.7V
−0.7 −1.0
V = −0.850 V | P ≅ 0.25mA(5.2V )= 1.30mW
2
τ P = 0.69(2kΩ)(2 pF ) = 2.76ns | PDP = 1.30mW (2.76ns)= 3.59 pJ
VL = VH − 0.3V = −1.0V | VREF =
9-14
9.48
∆V
→ ∆V = 2 0 − 0.7 − (−1) = 0.6V
2
VL = VH − ∆V = 0 − .6 = −0.600 V. Ignoring the base currents, the average power is
⎡ −1.7 − −3.3 V
−1.0 − (−3.3) V ⎤
(
)
⎢
⎥ 3.3V = 5.67 mW
P≈
+
⎢
⎥
1.6kΩ
3.2kΩ
⎣
⎦
(a) At the outputs : V
H
(
= 0 V | VREF = VH − 0.7 −
) (
(
)
)
∆V
0.6
∆V
0.6
=
= 600 Ω | RC1 =
=
= 505 Ω
I EE 2 −1− 0.7 − (−3.3)
I EE1 −0.7 − 0.7 − (−3.3)
RC 2 =
1600
(b) Y = A + B + C Y = A + B + C
(c) 5 versus 6 transistors
1600
9.49
At the outputs : VH = 0 V | VL = VH − ∆V = 0 − .4 = −0.400 V.
At the base of QD : VH → VBD = 0 − 0.7 = −0.7V | VL → VBD = −0.4 − 0.7 = −1.10V
−0.7 −1.1
= −0.90V | VEE ≤ VREF − 0.7 − 0.6 = −0.9 − 0.7 − 0.6 = −2.20 V
2
−0.9 − (−2.2) V
For VEE = −2.20V : RB =
= 1.30 kΩ
1
mA
−0.9 − 0.7 − (−2.2) + −0.7 − 0.7 − (−2.2) V
= 700 Ω
RE =
2
1 mA
0.4V
0.4V
= 350 Ω | RC 2 =
= 467 Ω
RC1 =
−0.7 − 0.7 − (−2.2)
−0.9 − 0.7 − (−2.2)
A
A
700
700
VREF =
[
][
]
9-15
9.50
*PROBLEM 9.50 - ECL DELAY
VIN 1 0 PULSE(-0.6 0 0 .01NS .01NS 15NS)
VB 8 0 -0.6
VREF 6 0 -1.0
VEE 7 0 -3.3
QA 0 1 2 NBJT
QB 0 8 2 NBJT
QC 0 8 2 NBJT
QD 4 2 3 NBJT
QE 5 6 3 NBJT
RB 2 7 3.2K
RE 3 7 1.6K
RC1 0 4 505
RC2 0 5 600
.OP
.TRAN 0.1N 30N
.MODEL NBJT NPN BF=40 BR=0.25
+IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF
+RB=100 RC=5 RE=1
.PROBE V(2) V(1) V(4) V(5) V(6)
.END
200mV
0V
-200mV
-400mV
-600mV
vI
Time
-800mV
0s
5ns
10ns
15ns
Result: τP = 0.95 ns
9-16
20ns
25ns
30ns
9.51
One approach is to scale all the resistor values. To reduce the power from
2.7 mW to 1.0 mW, the resistor values should all be increased a factor of 2.7.
RC1 = 2.7(1.85kΩ)= 5.00 kΩ | RC 2 = 2.7(2kΩ)= 5.40 kΩ
REE = 2.7(11.7kΩ) = 31.6 kΩ | R = 2.7(42kΩ) = 113 kΩ
9.52
Voltage levels remain unchanged : VREF = −1 V , VH = −0.7 V , VL = −1.3 V , I EE = 0.3 mA
−1− 0.7 − (−2) V
0.6V
0.6V
= 1 kΩ | RC1 =
=
= 1 kΩ
0.6mA
0.3
mA
−0.7 − 0.7 − (−2)
A
1 kΩ
−1− (−2) V 0.3 + 0.6
+
mA = 0.650mA | P = 0.65mA(2V )= 1.30 mW (-28%)
I ≅2
2
10
kΩ
Note that this gate will now have quite asymmetrical delays at the two outputs
REE =
since the two collector resistors differ by a factor of two in value.
9.53
The circuit is the pnp version of the ECL gate in Fig. P9.48. | Y = ABC
9.54
0.7 + 1.3
1mW
= +1.0V | I =
= 333µA
2
3V
(1 + 0.7)+ (0.7 + 0.7) = 1.55V
The average voltage at the emitter of QD is
2
(3 −1.55)V = 4.84 kΩ | R = (3 −1)V = 60.1 kΩ | R = 0.6V = 2.23 kΩ
RE =
B
C
0.9(333µA)
0.1(333µA)
(3 −1.7)V
VL = 0 | VH = VL + ∆V = +0.6V | VREF =
4.84kΩ
9.55
*Problem 9.55(a) - PNP ECL GATE DELAY
VI 4 0 PULSE(0.6 0 0 .01NS .01NS 25NS)
VB 7 0 DC 0.6
VREF 6 0 DC 1.0
VEE 1 0 DC 3
QA 0 4 3 PBJT
QB 0 7 3 PBJT
QC 0 7 3 PBJT
QD 0 3 2 PBJT
QE 5 6 2 PBJT
RB 1 3 60.1K
RE 1 2 4.84K
9-17
RC 5 0 2.23K
.OP
.TRAN 0.1N 50N
.MODEL PBJT PNP BF=40 BR=0.25 IS=5E-16
+TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF
+RB=100 RC=5 RE=1
.PROBE V(4) V(3) V(5)
.END
800mV
600mV
vI
400mV
vO
200mV
0V
Time
-200mV
0s
10ns
20ns
30ns
40ns
50ns
Result: τP = 6.0ns. This delay is dominated by a slow charge up at the base of QD.
*Problem 9.55(b) - Prob. 9.4
VIN 1 0 PULSE( -2.3 -1.7 0 .01NS .01NS 15NS)
VREF 6 0 -2.0
IEE 2 0 0.0003
Q1 3 1 2 NBJT
Q2 4 6 2 NBJT
R1 0 5 3.33K
RC1 5 3 2K
RC2 5 4 2K
.OP
.TRAN 0.1N 30N
.MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1
.PROBE V(1) V(3) V(4)
.END
Result: τP = 2.4 ns.
*Problem 9.55(c) - Fig. P9.16
VIN 1 0 PULSE( -1.5 -1.1 0 .01NS .01NS 15NS)
VREF 6 0 DC -1.30
IEE 2 0 DC 0.0002
9-18
Q1 3 1 2 NBJT
Q2 4 6 2 NBJT
Q3 0 3 7 NBJT
Q4 0 4 8 NBJT
R1 0 5 2K
RC1 5 3 2K
RC2 5 4 2K
RE1 7 9 50K
RE2 8 9 50K
VEE 9 0 DC -2
.OP
.TRAN 0.1N 30N
.MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1
.PROBE V(1) V(3) V(4) V(7) V(8)
.END
Result: τP = 3.0 ns.
9.56
Applying the transport model,
⎡ ⎛V ⎞
⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤
IC = IS ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ − 1⎥
⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎣ ⎝ VT ⎠
IS ⎡ ⎛ VBE ⎞ ⎤ IS ⎡ ⎛ VBC ⎞ ⎤
⎢exp⎜
⎢exp⎜
⎟ − 1⎥ +
⎟ − 1⎥
β F ⎣ ⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦
⎡ ⎛ 0.2 ⎞
⎛ −4.8 ⎞⎤ 10−15 ⎡ ⎛ −4.8 ⎞ ⎤
−15
IC = 10 ⎢exp⎜
⎟ − exp⎜
⎟⎥ −
⎟ − 1⎥ = 2.98 pA
⎢exp⎜
⎝ 0.025 ⎠⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦
⎣ ⎝ 0.025 ⎠
IB =
IB =
10−15 ⎡ ⎛ 0.2 ⎞ ⎤ 10−15 ⎡ ⎛ −4.8 ⎞ ⎤
⎟ − 1⎥ +
⎟ − 1⎥ = 74.5 fA
⎢exp⎜
⎢exp⎜
40 ⎣ ⎝ 0.025 ⎠ ⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦
Although the transistor is technically in the forward-active region, (and operating with IC = βF
IB), it is esentially off - its terminal currents are zero for most practical purposes.
9.57
⎛ 1 ⎞
For IC = 0, VCESAT = VT ln⎜ ⎟
⎝αR ⎠
VCESAT
1+
IC
(β R + 1)I B
⎛ 1 ⎞
= VT ln⎜ ⎟
⎝αR ⎠
IC
βF I B
⎛ β + 1⎞
⎛ 1.25 ⎞
= VT ln⎜ R ⎟ = 0.025ln⎜
⎟ = 0.402 V
⎝ 0.25 ⎠
⎝ βR ⎠
1−
9-19
9.58
(a) For the Transport model with VBE = VBC, the transport current iT = 0:
I ⎡ ⎛V ⎞ ⎤
I ⎡ ⎛V ⎞ ⎤ I
β
40
IC = S ⎢exp⎜ BC ⎟ − 1⎥ and IE = S ⎢exp⎜ BE ⎟ − 1⎥ ⇒ C = F =
= 160
β R ⎣ ⎝ VT ⎠ ⎦
β F ⎣ ⎝ VT ⎠ ⎦ IE β R 0.25
(b)
v BE = VB − 0.6 | v BC = VB − 0.8 = v BE − 0.2
⎤
⎡
v
v ⎤ I ⎡
v
iE = IS ⎢exp BE − exp BC ⎥ + S ⎢exp BE −1⎥
VT
VT ⎦ β F ⎣
VT
⎦
⎣
⎡
v
v
−0.2 ⎤ IS ⎡
v BE ⎤
iE = IS ⎢exp BE − exp BE exp
−1⎥
⎢exp
⎥+
VT
VT
VT
VT ⎦ β F ⎣
⎦
⎣
⎛
⎡
v ⎤ I ⎡
v ⎤ I ⎡
v ⎤
1 ⎞⎡
v ⎤
iE ≅ IS ⎢exp BE ⎥ + S ⎢exp BE ⎥ = IS ⎜1+ ⎟⎢exp BE ⎥ = S ⎢exp BE ⎥
VT ⎦ β F ⎣
VT ⎦
VT ⎦ α F ⎣
VT ⎦
⎣
⎝ β F ⎠⎣
−−−−−
⎡
v
v
−0.2 ⎤ IS ⎡
v BE − 0.2 ⎤
iC = IS ⎢exp BE − exp BE exp
−1⎥
⎥ − ⎢exp
VT ⎦ β R ⎣
VT
VT
VT
⎣
⎦
⎡
i
v ⎤
40
iC ≅ IS ⎢exp BE ⎥ | C = α F =
= 0.976
41
VT ⎦
iE
⎣
i
(c ) C = −1 → iB = iE − iC = 2iE | Both junctions will be forward - biased. Neglect
iE
⎛
I
I
v
I
v
v
v ⎞
v
the -1 terms : S exp BE + S exp BC = 2IS ⎜ exp BE − exp BC ⎟ + 2 S exp BE
βF
VT β R
VT
VT
VT ⎠
βF
VT
⎝
v BE − v BC
1
1
0.25 = 27.2mV
= VT ln
= 0.025V ln
1
1
2+
2+
βF
40
2+
βR
(v B − v I ) − (v B − 0.8) = 27.7mV
9-20
2+
| v I = 0.773 V
9.59
(a) For the default value of β F = 40,
IC = β F I B ⇒ Forward − active region | VBE ≅ VT ln
(b) I
IC
10−3 A
= 0.025V ln −15 = 0.691 V
IS
10 A
< β F I B ⇒ saturation region; VBE is given by Eqn. 5.45
C
VBE
⎛ 1 ⎞
IB + ⎜
⎟ IC
I B + (1− α R )IC
⎝ β R + 1⎠
= VT ln
= V ln
⎡ 1 ⎛ 1 ⎞⎤
⎡1
⎤ T
I S ⎢ + (1− α R )⎥
IS ⎢ + ⎜
⎟⎥
⎣β F
⎦
⎣ β F ⎝ β R + 1⎠⎦
VBE
⎛ 1 ⎞ −3
25x10−6 + ⎜
⎟10
⎝ 0.25 + 1⎠
= 0.025V ln
= 0.691 V
⎡
⎛ 1 ⎞⎤
−15 1
10 ⎢ + ⎜
⎟⎥
⎣80 ⎝ 0.25 + 1⎠⎦
(c) I
C
< β F I B ⇒ saturation region | VBE
9.60
βR
αR =
βR + 1
(a) V
CESAT
(b) I
IC
B
=
=
1
|
2
⎛ 1 ⎞ −3
10−3 + ⎜
⎟10
⎝ 0.25 + 1⎠
= 0.025V ln
= 0.710 V
⎡
⎛ 1 ⎞⎤
−15 1
10 ⎢ + ⎜
⎟⎥
⎣40 ⎝ 0.25 + 1⎠⎦
IC 1mA
=
= 40
I B 25µA
⎛ 1 ⎞
= VT ln⎜ ⎟
⎝αR ⎠
1+
IC
(β R + 1)I B
1−
IC
βF I B
1mA
= 25 | VCESAT
40µA
⎡
40 ⎤
⎢⎛ 3 ⎞ 1+ ⎥
2 ⎥ = 114 mV
= (0.025V )ln ⎢⎜ ⎟
40
2
⎝
⎠
⎢ 1− ⎥
⎣
60 ⎦
⎡
25 ⎤
⎢⎛ 3 ⎞ 1+ ⎥
2 ⎥ = 88.7 mV
= (0.025V )ln ⎢⎜ ⎟
25
2
⎢⎝ ⎠ 1− ⎥
⎣
60 ⎦
9-21
9.61
βR
2
1mA
I
| C =
= 40
βR + 1 3
IB 25µA
IC
⎡
40 ⎤
1+
⎛ 1 ⎞ (β R + 1)IB
⎢⎛ 3 ⎞ 1+ 3 ⎥
= (0.025V )ln⎢⎜ ⎟
= 117 mV
(a) VCESAT = VT ln⎜ ⎟
40 ⎥
⎝
⎠
2
⎝ α R ⎠ 1− IC
⎢ 1− ⎥
⎣
50 ⎦
β F IB
αR =
(b) VCESAT
9.62
βR
=
⎡
40 ⎤
⎢⎛ 3⎞ 1+ 3 ⎥
= (0.025V )ln⎢⎜ ⎟
= 89.5 mV
40 ⎥
⎝
⎠
2
⎢ 1−
⎥
⎣
100 ⎦
0.25
= 0.2
β R + 1 1.25
⎞
⎞
⎛V
⎛V
⎛ 0.2V ⎞
⎛ 0.1V ⎞
⎟ = 2980
⎟ = 54.6
(a) Γ = exp⎜ CESAT ⎟ = exp⎜
(b) Γ = exp⎜ CESAT ⎟ = exp⎜
⎝ 0.025V ⎠
⎝ 0.025V ⎠
⎝ VT ⎠
⎝ VT ⎠
⎤
⎤
⎡
⎡
40
40
⎡
⎡
β ⎤
β ⎤
1+ F ⎥
1+ F ⎥
⎢1+ 0.25 2980 ⎥
⎢1+ 0.25 54.6 ⎥
⎢
⎢
I
I
I
I
βRΓ
( )⎥ = IC
βRΓ
( )⎥ = IC
⎥= C ⎢
⎥= C ⎢
IB ≥ C ⎢
IB ≥ C ⎢
1
1
1
1
β F ⎢1−
β F ⎢1−
⎥ 37.9
⎥ 9.25
⎥ 40 ⎢ 1−
⎥ 40 ⎢ 1−
⎥
⎢⎣
⎢
⎢⎣ α R Γ ⎥⎦
⎢
⎥
0.2(2980) ⎦
0.2(54.6) ⎥⎦
⎣ α RΓ ⎦
⎣
I
1 ⎛ 5 − 0.2 ⎞
I
1 ⎛ 5 − 0.1⎞
IB ≥ C =
IB ≥ C =
⎟ = 63.3 µA
⎟ = 265 µA
⎜
⎜
β FOR 37.9 ⎝ 2kΩ ⎠
β FOR 9.25 ⎝ 2kΩ ⎠
αR =
9.63
αR =
=
⎛V
⎞
⎛ 0.1V ⎞
0.25
= 0.2 Γ = exp⎜ CESAT ⎟ = exp⎜
⎟ = 54.6
⎝ 0.025V ⎠
β R + 1 1.25
⎝ VT ⎠
βR
=
⎡
⎤
40
⎡
β ⎤
1+ F ⎥
⎢1+ 0.25 54.6 ⎥
⎢
I
I
βRΓ
( )⎥ = IC
⎥= C ⎢
IB ≥ C ⎢
1
1
β F ⎢1−
⎥ 9.25
⎥ 40 ⎢ 1−
⎢⎣
⎢⎣ α R Γ ⎥⎦
0.2(54.6) ⎥⎦
I
1 ⎛ 5 − 0.1⎞
IB ≥ C =
⎜
⎟ = 147 µA
β FOR 9.25 ⎝ 3.6kΩ ⎠
9-22
9.64
⎛ 1 ⎞
⎛ β + 1⎞
⎛ 1.25 ⎞
For IC = 0, VCESAT = VT ln⎜ ⎟ = VT ln⎜ R ⎟ = 0.025ln⎜
⎟ = 40.2 mV
⎝ 0.25 ⎠
⎝αR ⎠
⎝ βR ⎠
⎛ 1 ⎞
⎛ β + 1⎞
⎛ 41 ⎞
For IE = 0, VECSAT = VT ln⎜ ⎟ = VT ln⎜ F ⎟ = 0.025ln⎜ ⎟ = 0.617 mV
⎝ 40 ⎠
⎝αF ⎠
⎝ βF ⎠
9.65
αF =
τS =
βF
βF + 1
=
(
40
0.25
βR
= 0.976 α R =
=
= 0.200
β R + 1 1.25
41
) = 3.40ns
0.976 0.4ns + 0.2(12ns)
1− 0.976(0.2)
t S = (3.40ns)ln
2mA − (−0.5mA)
2.5mA
− (−0.5mA)
40
iCMAX ≅
5V
= 2.5 mA
2kΩ
= 5.07 ns
9.66
V H = VCC = 3.0 V VL = VCESAT = 0.15
VIL = 0.7 − VCESAT = 0.7 − 0.04V = 0.66V VIH ≅ VBESAT 2 = 0.8 V
3 − 0.7 − 0.8 V
= 375µA | IB 2 = 1.25IB1 = 469µA
v I = 3V : IB1 =
kΩ
4
3 − 0.8 − 0.15 V
3 − 0.15
= −513µA | IC 2SAT =
A = 1.43mA
v I = 0.15V : IIL = −
kΩ
4
2000
1.43mA + N (513µA) ≤ 40(469µA) → N ≤ 33.8 → N ≤ 33.
9.67
vI = VH : I =
5 − 0.7 − 0.8 5 − 0.15
+
= 4.13mA | P = 5(4.13mA) = 20.6 mW
(0.8)4kΩ (0.8)2kΩ
vI = VL : I =
5 − 0.8 − 0.15
= 0.844mA | P = 5(0.844mA)= 4.22 mW
(1.2)4kΩ
Pmax = 20.6 mW | Pmin = 4.22 mW
9-23
9.68
⎛ 1 ⎞
Using Eqs. 9.44 and 9.47 : VCC − iC RC = VT ln⎜ ⎟
⎝αR ⎠
1+
iC
1.25(1.09mA)
⎛1⎞
5 − 2000iC = 0.025ln⎜ ⎟
⎝ .2 ⎠ 1−
iC
40(1.09mA)
1+
iC
(β R + 1)iB
1−
iC
β F iB
→ iC = 2.4659 mA
vCESAT = 5 − 2000iC = 0.0682V
9.69
V H = 2.5 V | VL = VCESAT = 0.15 V
VIL = 0.7 − VCESAT = 0.55V | VOL ≅ VL = 0.15V
VIH ≅ VBESAT 2 = 0.8 V | VOH ≅ V H = 2.5 V
NM L = 0.55 − 0.15 = 0.40 V NM H = 2.5 − 0.8 = 1.7 V
9.70
For vI = VH , we require VCC = VBE2SAT + VBC1 + I B1 RB = 0.8 + 0.7 + ∆V = 1.5V + ∆V
where ∆V is the voltage across the base resistor. ∆V must be large enough to absorb
VBE process variations and to establish the base current. 0.5 V should be sufficient.
Thus VCC = 2.0 V or more is acceptable.
9.71
The VTC transitions are set by the values of vBE and vBESAT and are not changed by the
power supply voltage. (b) VIL = 0.66 V and VIH = 0.80 V. But VOH ≅ VH = 3 V and VOL ≅ VL
= 0.15 V. (c) NMH = 3 - 0.8 = 2.2 V | NML = 0.66-0.15 = 0.51 V.
9.72
We need to reduce the currents by a factor of 11.2. Thus, RB = 11.2 (4kΩ) = 44.8 kΩ and
RC = 11.2 (2kΩ) = 22.4 kΩ
9.73 (a)
*Problem 9.73 - Prototype TTL Inverter +Delay
VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 +50N 0)
VCC 5 0 DC 5
Q1 3 2 1 NBJT
Q2 4 3 0 NBJT
RB 5 2 4K
RC 5 4 2K
*RB 5 2 45.2K
9-24
*RC 5 4 22.6K
.OP
.TRAN .1N 80N
.MODEL NBJT NPN BF=40 BR=0.25 +IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF +RB=100 RC=5 RE=1
.PROBE V(1) V(2) V(3) V(4)
.END
6.0V
vI
4.0V
2.0V
vO
0V
Time
(a)
-2.0V
0s
20ns
Results: (a) τP = 2.9 ns
40ns
60ns
80ns
(b) τP = 15.8 ns.
9.74
(a) VH = 5V | VL = VCE2SAT = 0.15V
5 − 0.15 − 0.6
= −1.06 mA
4000
= −I S ≅ 0 where IS is the diode saturation current.
vI = VL = 0.15V , I IN = −
vI = VH = 5V , I IN
5 − 0.8 − 0.6
5 - 0.15
(b) I = 4000 = 0.90mA; 2000 + N (1.06mA)≤ 40(0.9mA); N ≤ 31.
(c) -1.06 mA compared to -1.01 mA and 0 mA compared to 0.22 mA.
B
9.75
If we assume that the diode on-voltage is 0.7 V to match the base-emitter voltage of the BJT,
then the VTC will be the same as that in Fig. 9.35. Both VTCs will be the same.
9-25
9.76
*Figure 9.76 - Prototype TTL Inverter
VTC's
VI 1 0 DC 0
VCC 5 0 DC 5
*DTL
D1A 6 1 D1
D2A 6 7 D1
RBA 5 6 4K
RCA 5 8 2K
Q2A 8 7 0 NBJT
*TTL
Q1B 3 2 1 NBJT
Q2B 4 3 0 NBJT
RBB 5 2 4K
RCB 5 4 2K
.DC VI 0 5 .01
.MODEL NBJT NPN BF=40 BR=0.25
IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF
RB=100 RC=5 RE=1
.MODEL D1 D IS=5E-16 TT=0.15NS
CJO=1PF
.PROBE V(1) V(2) V(3) V(4) V(6) V(7)
V(8)
.END
5.0V
4.0V
vO
3.0V
2.0V
1.0V
0V
0V
1.0V
2.0V
3.0V
4.0V
5.0V
The TTL transition is sharper (more abrupt)
and is shifted by approximately 50 mV.
9.77
*Figure 9.77 - Prototype Inverter Delays
VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 5 50N 0)
VCC 5 0 DC 5
*DTL
D1A 6 1 D1
D2A 6 7 D1
RBA 5 6 4K
RCA 5 8 2K
Q2A 8 7 0 NBJT
*TTL
Q1B 3 2 1 NBJT
Q2B 4 3 0 NBJT
RBB 5 2 4K
RCB 5 4 2K
.OP
.TRAN 0.1N 100N
.MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1
.MODEL D1 D IS=5E-16 TT=0.15NS CJO=1PF
9-26
vI
.PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(8)
.END
6.0V
vI
4.0V
vO TTL
2.0V
vO DTL
0V
Time
-2.0V
0s
20ns
40ns
60ns
80ns
100ns
The fall time of the output of the TTL gate is somewhat slower than the DTL gate since
transistor Q1 must come out of saturation. However, the rise time of the DTL gate is extremely
slow because there is no reverse base current to remove the charge from the transistor base.
9.78
*Figure 9.78 - DTL Inverter Delays
VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 50N 0)
VCC 5 0 DC 5
*DTLA
D1A 6 1 D1
D2A 6 7 D1
RBA 5 6 4K
RCA 5 8 2K
Q2A 8 7 0 NBJT
*DTL-B
D1B 2 1 D1
D2B 2 3 D1
Q2B 4 3 0 NBJT
RBB 5 2 4K
RCB 5 4 2K
RB1 3 0 1K
.OP
.TRAN 0.1N 100N
.MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1
.MODEL D1 D IS=5E-16 TT=0.15NS CJO=1PF
.PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(8)
.END
9-27
©R. C. Jaeger - February 7, 2007
6.0V
vO (b)
4.0V
vI
2.0V
vO (a)
0V
Time
-2.0V
0s
20ns
40ns
60ns
80ns
100ns
Without the 1-kΩ resistor, the rise time of the DTL gate is extremely slow because there is no
reverse base current to remove the charge from the transistor base. The resistor provides an
initial reverse base current of -0.7 mA to turn off the transistor and significantly reduces the rise
time and propagation delay.
9.79
See problem 9.80.
9.80
*Figure 9.79 - Inverter VTC
VI 1 0 DC 0
VCC 6 0 DC 3.3
Q1 3 2 1 NBJT
Q2 5 3 4 NBJT
Q3 5 4 0 NBJT
R1 6 2 4K
R2 6 5 2K
R3 4 0 3K
.OP
.DC VI 0 3.3 0.01
.MODEL NBJT NPN BF=40 BR=0.25
IS=1E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF
RB=100 RC=5 RE=1
.PROBE V(1) V(2) V(3) V(4) V(5)
.END
Circuit 9.108-DC Transfer-1
VO
(V)
+0.00e+000
+500.00m
+1.00
+1.50
VI
+2.00
+2.50
+3.00
+3.50
+3.00
+2.50
+2.00
+1.50
+1.00
+500.00m
+0.00e+000
V(2)
V(7)
The first break point occurs when the input reaches a voltage large enough to just start turning
on Q2, at approximately VCESAT1 + VBE2 = 0.04V + 0.6 V = 0.64V. The second breakpoint
begins when the input reaches VCESAT1 + VBE2 +VBE3 = 0.04V + 0.7 + 0.6 V = 1.34V. Note
9-28
that the shallow slope is set by the ratio of R2/R3 = 2/3, and also note that Q3 cannot saturate.
From the B2SPICE simulation, VH = 3.3 V, VL ≅ VBE3 + VCESAT2 = 0.82V, VIH = 1.38 V, VOL =
0.84 V, VIL = 1.38 V, VOH = 2.82 V. NMH = 2.82 – 1.38 = 1.54 V. NML = 1.38 - 0.84 = 0.54
V.
9.81
VCC = 5 V
0.875 mA
2k Ω
4kΩ
iR
N(0.875 mA)
vH < 5 V
N
iC = 0
VL
0.15 V
0.875 mA
Q2
Q1
+
"Off"
0.19 V
-
From the analysis in the text, we see that the fanout is limited by the VH condition.
5 − 0.7 − 0.8 V
iB1 =
= 0.875mA | iE1 = −β R iB1 = −0.875mA
kΩ
4
5 − 2000(N )(0.875x10−3 )≥ 1.5 → N ≤ 2 → Fanout = 2
9.82
V CC = 5 V
1.2(4 k Ω)
1.2(2k Ω)
1.2(2k Ω)
1.5 V
iB1
+ 0.7 V
VH
iIH = β RiB1
Q1
N
+
(βR + 1)i B1
Q2
0.15 V
+
0.8 V
-
9-29
From the analysis in the text, we see that the fanout is limited by the VH condition.
5 − 0.7 − 0.8 V
iB1 =
= 0.729mA | iE1 = −β R iB1 = −0.25(0.729mA) = 0.182mA
4 (1.2) kΩ
5 − 2000(1.2)(N )(0.182x10−3 )≥ 1.5 → N ≤ 8.01 → Fanout = 8
iB1 =
5 − 0.7 − 0.8 V
= 1.09mA | iE1 = −β R iB1 = −0.25(1.09mA) = 0.273mA
4 (0.8) kΩ
5 − 2000(0.8)(N )(0.273x10−3 )≥ 1.5 → N ≤ 8.01 → Fanout = 8
The result is independent of the tolerance if the resistors track each other.
Note that Eq. 9.83 also yields N = 8 if more digits are used in the calculation.
9.83
From the analysis in the text, we see that the fanout is limited by the VH condition.
iB1 =
5 − 0.7 − 0.8
RB
iE1 = −β R iB1 = −0.25iB1
5 − 0.7 − 0.8
5 − 2000(N )(0.25)
≥ 1.5 → RB ≥ 5 kΩ
RB
9.84
(a) Q4 is in the forward - active region with I E = (β F + 1)I B
5 − 0.7 − 0.6
= 234 mA
1600
5 − 0.8 − 0.6 5 − 0.6 − 0.15
(b) Q4 saturates; I E = I B + IC = 1600 + 130 = 34.9 mA
I E = 101
9.85
(a) PD = 5V (234mA) = 1.17 W (b) PD = 5V (34.9mA) = 0.175 W
9-30
9.86
*Figure 9.86 - TTL Output Current
VCC 5 0 DC 5
RB 5 3 1.6K
RS 5 4 130
Q1 4 3 2 NBJT
D1 2 1 D1
IL 1 0 DC 0
.DC IL 0 30MA 0.01MA
.MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS
+CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1
.MODEL D1 D IS=5E-16
.PROBE V(1) V(2) V(3) V(4)
.END
5.0V
vO
4.0V
3.0V
2.0V
iL
1.0V
0A
5mA
10mA
15mA
20mA
25mA
30mA
9-31
9.87
*Problem 9.87 - Modified TTL Inverter VTC
VI 1 0 DC 0
VCC 9 0 DC 5
Q1 2 8 1 NBJT
Q2 4 3 0 NBJT
Q3 6 2 3 NBJT
Q4 7 6 5 NBJT
D1 5 4 DN
RB 9 8 4K
RC 9 6 1.6K
RS 9 7 130
RL 4 0 100K
Q5 10 11 0 NBJT
RB5 3 11 3K
RC5 3 10 1K
.DC VI 0 5 .01
.MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.25NS TR=25NS
+CJC=0.6PF CJE=.6PF CJS=1.25PF RB=100 RC=5 RE=1
.MODEL DN D
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
4.0V
vO
3.0V
2.0V
1.0V
0V
0V
1.0V
2.0V
vI
3.0V
4.0V
5.0V
In the modified TTL circuit, Q3 cannot start conducting until its base reaches at least VBE5 + VBE6
= 1.2 V.
9-32
9.88
+ 5V
20 k Ω
8k Ω
+
Q
0.7 V
-
Q
1
+
3
0.8 V
-
Q
+
5k Ω
2
0.8 V
-
(a) v I = VH : Q4 off − IB 4 = 0 = IC 4 | Q2 saturated with IC 4 = 0
(5 − 0.7 − 0.8 − 0.8)V = 135 µA | I = −β I = −0.25 135 µA = −33.8 µA
IB1 =
(
)
E1
R B1
20kΩ
IC1 = −169 µA | IC 3 =
(5 − 0.15 − 0.8)V
8kΩ
= 506 µA
IE 3 = 506µA + 169µA = 675 µA | IB 2 = 675µA −
(b) v I
IB1 =
= VL : Q2 ,Q3 off ; Q4 on
(5 − 0.8 − 0.15)V
20kΩ
0.8V
= 515 µA
5kΩ
= 203 µA = IE1 | IC1 = 0
9-33
9.89
See Problem 9.90.
9.90
4.0V
*Problem 9.90 - Low Power TTL Inverter
VTC versus Temperature
VI 1 0 DC 0
VCC 9 0 DC 5
Q1 2 8 1 NBJT
Q2 4 3 0 NBJT
Q3 6 2 3 NBJT
Q4 7 6 5 NBJT
D1 5 4 DN
RB 9 8 20K
RC 9 6 8K
RS 9 7 650
RL 4 0 100K
RE 3 0 5K
.DC VI 0 5 .01
.TEMP -55 25 85
.MODEL NBJT NPN BF=40 BR=0.25 IS=1E17 TF =0.25NS TR=25NS
+CJC=0.6PF CJE=.6PF CJS=1.25PF RB=100
RC=5 RE=1
.MODEL DN D
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
vO
3.0V
2.0V
-55 o C
1.0V
+85 o C
0V
0V
9.91
+5V
RC
RS
130 Ω
1.6 k Ω
Q4
V H = 5 − 0.7 − 0.7 −
N x 0.17 mA
9-34
N (IIH )
R
βF + 1 C
N (0.17mA)
(1600)
40 + 1
V H ≥ 2.4V → N ≤ 180
V H = 3.6 −
+25 o C
1.0V
2.0V
vI
3.0V
4.0V
5.0V
9.92
For small β R , fanout is limited by the vO = VL case (v I = V H ).
iB 3 = (β R + 1)iB1 = 1.05
(5 − 0.7 − 0.8 − 0.8)V
= 567µA
5kΩ
(5 − 0.15 − 0.8)V + 567µA − 0.8V = 1.95mA
iB 2 = iE 3 − iR E =
2kΩ
1.25kΩ
(5 − 0.8 − 0.15)V = 0.810mA | α = β R = 0.05 = 0.0476
iIL = −iE1 = −iB1 =
R
5kΩ
1+ β R 1.05
1
1−
⎛ 0.15V ⎞
0.0476(403.4 )
Using Eq. 9.61, Γ = exp⎜
= 9.52
⎟ = 403.4 → β FOR = 20
1
20
⎝ 0.025V ⎠
1+
0.05 403.4
N (0.810mA) ≤ 9.52(1.95mA) → N = 22
9.93
For the vO = VL case, the equations are given in Problem 9.92. For vO = VH,
V H = 5 − IB 4 RC − 0.7 − 0.7 ≥ 2.4V
⎛ 5 − 0.7 − 0.8 − 0.8 ⎞
⎛ 2.7 ⎞
IIH = β R ⎜
⎟
⎟ = βR ⎜
⎝
⎠
⎝ 4000 ⎠
4000
NIIH
Nβ R ⎛ 2.7 ⎞
IB 4 =
=
⎜
⎟
β F + 1 β F + 1 ⎝ 4000 ⎠
N≤
1.2(β F + 1)(4000)
2.7β R (1600)
100
Fanout
80
60
40
20
0
0
1
2
3
4
Inverse Current Gain
5
9-35
function [N,X]=P993
br=0;
bf=40;
g=exp(.15/.025);
for i=1:50
br=br+.1;
ar=br/(1+br);
ib3=(1+br)*675;
ib2=1730+ib3;
bfor=40*(1-1/(ar*g))/(1+bf/(br*g));
N1=fix(bfor*ib2/1013);
N2=1.2*(bf+1)*4000/(2.7*br*1600);
N(i)=min(N1,N2);
X(i)=0.1*i;
end
»[Y,X]=p993;
»plot(X,Y)
9.94
+2 V
+2 V
2 kΩ
+
2kΩ
+
VH
iIN
Q1
0.7 V
-
2kΩ
"Saturated"
+
0.8 V
0.8 V
Q3
-
0.8 V
-
-
Q2
+
Q1
0.15 V
(a) VH = 2 − VECSAT 2 = 2 − 0.15 = 1.85 V | VL = VCESAT 3 = 0.15 V
(2 − 0.7 − 0.8)V = 62.5 µA
(b) iIH : iB 2 ≅ 0 | iIH = 0.25
2kΩ
(2 − 0.8 − 0.15)V − (2 − 0.8 − 0.8 − 0.15)V = −650 µA
iIL = −iB1 = −
2kΩ
2kΩ
(c ) Assume β FOR ≤ 28.3
For the pnp transistor : N (62.5µA) ≤ 28.3
(2 − 0.8 − 0.8 − 0.15)V
→ N = 56
2kΩ
(2 − 0.7 − 0.8)V → N = 13
For the npn transistor : N (650µA) ≤ 28.3(1.25)
2kΩ
9-36
9.95
(a) VL = VCESAT 3 = 0.15 V | VH = 2 − VBE 2 = 2 − 0.7 = 1.3 V
⎛ 2 − 0.8 − 0.15 2 − 0.15 − 0.15 ⎞
= 0.15V : iIL = −(iB1 + iC1 ) = −⎜
+
⎟ = −247 µA
⎝ 10000
⎠
12000
⎛ 2 − 0.7 − 0.8 ⎞
v I = 1.3V : iIL = β R iB1 = 0.25⎜
⎟ = 12.5 µA
⎝
⎠
10 4
2 − 0.8 − 0.15 2 − 0.15 − 0.15
+
= 1.875mA
(c ) Using β FOR = 28.3 : iL = iB 2 + iC 2 =
6000
1000
⎛ 2 − 0.7 − 0.8 ⎞
2 − 0.8
iB 3 =
+ 1.25⎜
⎟ = 162.5µA
⎝ 10000 ⎠
12000
(b) v I
28.3(0.1625 mA) ≥ N (0.247 mA) + 1.875mA → N = 11
9.96
(a) Y = ABC
(c ) v I
(b) VL = VCESAT 3 = 0.15 V
| V H = 3.3 − VBE1 − VD = 3.3 −1.4 = 1.9 V
= 1.9V, input diode is off and iIH = 0. v I = 0.15V, iIL = −
2.0V
3.3 − 0.7 − 0.15
= −408 µA
6000
vO
1.5V
1.0V
0.5V
vI
0V
0V
1.0V
2.0V
3.0V
4.0V
The VTC starts to decrease immediately because Q2 is ready to conduct due to the 0.7-V drop
across the input diode. When the input has increased to approximately 0.7 V, Q3 begins to
conduct and the output drops rapidly. The VTC is much sloppier than that of the
corresponding TTL gate. For this particular circuit VIL = 0 and VIH = 0.8 V. Based upon our
definitions, NML = 0. However, the initial slope can be reduced by changing the ratio RC/R2 so
that VIL = 0.7 V.
9-37
9.97
(a) If either input A is low or input B is low, VB2 will be low. Q2 will be off, Q3 will be on and
Y will be low. Therefore Y = A + B → Y = AB .
+5 V
+5 V
4kΩ
4 kΩ
+
0.8 V
+0.3 V
0.15 V
Q2
+4.3 V
+0.45 V
-
VB
Q1
+
10 k Ω
10 k Ω
-5 V
5 kΩ
-5 V
(b) VH = 5 − IB 5 R2 − VBE 5 ≈ 5 − VBE 5 = 5 − 0.7 = 4.30 V
VL = 5 − α F IE 3 R2 − IB 5 R2 − VBE 5 ≈ 5 − IE 3 R2 − VBE 5
+0.7 − 0.7 − (−5)
= 1.00 mA | VL = 5 − 0.001(4000) − 0.7 = +0.300 V
5000
⎛ 5 − 0.7 − VB ⎞ VB − (−5) VB − 0.7 − (−5) 1
+
→ VB = 1.97V
⎟=
(c ) v I = 4.3V : 1.25⎜
⎝ 4000 ⎠
10000
5000
41
5 − 0.7 − 1.97
IB1 =
= 582 µA | IE1 = −0.25 IB1 = −146 µA | IIH = 146 µA
4000
0.3 + 0.15 − (−5)
5 − 0.8 − 0.3
v I = 0.3V : IB1 =
= 975 µA | IC1 = −
= −545 µA
10000
4000
IE1 = IB1 + IC1 = 430 µA | IIL = −430 µA
IE 3 =
9.98
1.5 V
1kΩ
800 Ω
0.70 V
0.25 V
9-38
1kΩ
800 Ω
VH
Off
Q
- 0.45 V +
1.5 V
+
0.45 V
-
Q
1
Off
+
0.7 V
-
1
(a) VH = VCC = 1.5 V | VL = "VCESAT1" = 0.7 − 0.45 = 0.25 V
(b) For v I = 1.5V , the input diode is off, and IIH = 0.
1.5 − 0.45 − 0.25
= 1.00 mA
800
(c ) Note that Q1 operates as if it were in the forward - active region :
For v I = 0.25V , IIL =
β F IB1 ≥ NIIL + IR
2
9.99
vBE + vD2 − vD1 = vCE
⎛ 1.5 − 0.45 − 0.70 ⎞
1.5 − 0.25
| 40⎜
→ N = 16
⎟ ≥ N (0.001) +
⎠
⎝
800
1000
| For vD2 ≅ vD1 , vCE = v BE = 0.7 V
iC = iCC + iD1 | iB = iBB − iD1 | iC = β F iB
iCC + iD1 = β F (iBB − iD1) → iD1 =
β F iBB − iCC 20(0.25)−1
=
mA = 0.191 mA
βF + 1
21
iD2 = iBB − iD1 = 0.25 − 0.191 = 0.059 mA | iC = 20iB = 20iD2 = 1.18 mA
9.100
In this circuit as drawn, the collector - base junction of Q1 is
bypassed by a Schottky diode. Q1 will be "off" with VBC = +0.45 V .
iB1 =
5 − 0.45 − 0.7
= 963 µA | IIN = 0 | iB 2 = iB1 = 963 µA
4000
9.101
4.0V
4.0V
vO
vI
3.0V
3.0V
2.0V
2.0V
1.0V
1.0V
vO
Time
0V
0V
1.0V
2.0V
vI
3.0V
4.0V
5.0V
0V
0s
5ns
10ns
15ns
20ns
25ns
30ns
Result: τP = 3.0 ns
*Problem 9.101 - Schottky TTL Inverter VTC
VI 1 0 DC 3.5 PWL(0 3.5 0.2N 0.25 15N 0.25 15.2N 3.5 30N 3.5)
9-39
VCC 9 0 DC 5
Q1 2 8 1 NBJT
D1 2 8 DS
Q2 4 3 0 NBJT
D2 3 4 DS
Q3 6 2 3 NBJT
D3 2 6 DS
Q4 7 5 4 NBJT
Q5 7 6 5 NBJT
D5 6 7 DS
RB 9 8 2.8K
RC 9 6 900
RS 9 7 50
R5 5 0 3.5K
RL 4 0 100K
Q6 10 11 0 NBJT
D6 11 10 DS
R2 3 11 500
R6 3 10 250
.OP
.DC VI 0 5 .01
.TRAN .025N 30N
.MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.15NS TR=15NS
+CJC=1PF CJE=.5PF CJS=1PF RB=100 RC=10 RE=1
.MODEL DS D IS=1E-12
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
9-40
9.102
0.54 k Ω
RC
Y
C
VREF
+
0.7 V
0.75 k Ω
RE
-3 V
Y = A + B + C | VH = 0 V | VL = −540IC = −540
VREF − 0.7 − (−3)
= −0.72(VREF + 2.3)
750
V H + VL
= VREF − 0.7 + 0.4 + 0.7 = VREF + 0.4
2
0 − 0.72(VREF + 2.3)
= VREF + 0.4 → VREF = −0.903V | VL = −0.72(−.903 + 2.3) = −1.01 V
2
9.103
RC
3.3 k Ω
Y
V REF- 0.4V
V REF
V REF- 0.7V
2.4 k Ω
RE
-3 V
Y = A + B + C | V H = 0 V | VL = VREF − 0.4
0 + (VREF − 0.4 )
V H + VL
= VREF |
= VREF → VREF = −0.40 V | VL = −0.80V
2
2
9-41
9.104
The circuit can be modeled by a normal BJT with a Schottky diode in parallel with the
collector base junction. If iC and iB are defined to be the collector- and base-currents of the
BJT,
iC + iB =
5 − 0.7
1.075mA
= 1.075mA | iC ≅ β F iB → iB =
= 26.9 µA | iC = 1.05 mA
4000
40
+5 V
i=0
4 kΩ
iC
iB
Q1
9.105
If we assume 50% of the gates are switching (an over estimate),
50W
= 2µW /gate | τ P = 1ns | PDP = (2µW )(1ns) = 2 fJ
(a) P =
0.5(50x10 6 )
(b) PDP = (0.1mW )(0.1ns) = 10 fJ
| The result in Part (a) is off the graph!
9.106
If we assume 50% of the gates are switching (an over estimate),
100W
= 1µW /gate | τ P = 0.25ns | PDP = (1µW )(0.25ns) = 0.25 fJ
(a) P =
0.5(200x10 6 )
(b) PDP = (0.1mW )(0.1ns) = 10 fJ
9.107
| The result in Part (a) is off the graph!
PDP 0.5 pJ
=
= 1.67 ns
P
0.3mW
(b) P =
PDP
(a) PDP = (0.7ns)(40mW ) = 28 pJ
| τP =
PDP 28 pJ
=
= 2.8 ns
P
10mW
(a) τ P =
τP
=
0.5 pJ
= 0.5 mW
1ns
9.109
(b) P =
9-42
PDP
τP
=
28 pJ
= 140 mW
0.2ns
9.110
Results from B2SPICE: VH = 4.48 V, VL = 0.54 V, τPHL = 3.3 ns, τPLH = 4.4 ns.
Circuit 9_127-DC Transfer-2
+0.000e+000
(V)
+500.000m
+1.000
+10.000n
+20.000n
+1.500
+2.000
+30.000n
+40.000n
VI
+2.500
+3.000
+3.500
+4.000
+4.500
+60.000n
+70.000n
+80.000n
+90.000n
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(3)
Circuit 9_127-Transient-2
(V)
+0.000e+000
Time (s)
+50.000n
+100.000n
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(3)
9.111
V
M4
B
A
v
M3
DD
Q4
vO
I
B
M2
R
Q
3
A
M1
9-43
9.112
VDD
M
4
M
3
B
A
Q4
vO
M2
B
B
M
6
M
A
1
A
M7
Q3
M
5
9.113
Results from B2SPICE with R = 4 kΩ: VH = 5 V, VL = 0 V, τPHL = 1.9 ns, τPLH = 4.1 ns.
Circuit 9_130-DC Transfer-1
+0.000e+000
(V)
+500.000m
+1.000
+1.500
+2.000
+10.000n
+20.000n
+30.000n
+40.000n
+2.500
VI
+3.000
+3.500
+4.000
+4.500
+60.000n
+70.000n
+80.000n
+90.000n
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(3)
Circuit 9_130-Transient-1
+0.000e+000
(V)
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(3)
9-44
+50.000n
Time (s)
+100.000n
9.114
V
DD
V
DD
M
A
3
A
M3
M
4
B
Q
4
M4
B
Q
M2
A
M
B
R
1
A
B
M
A
M1
R
2
v
4
vO
M
5
M6
O
M
B
6
M5
A
B
(a)
(b)
9.115
Results from B2SPICE: VH = 4.7 V, VL = 0.34 V, τPHL = 7.0 ns, τPLH = 14 ns.
Circuit 9_133-DC Transfer-4
+0.000e+000
(V)
+500.000m
+1.000
+1.500
+2.000
+10.000n
+20.000n
+30.000n
+40.000n
+2.500
VI
+3.000
+3.500
+4.000
+4.500
+60.000n
+70.000n
+80.000n
+90.000n
+5.000
+4.000
+3.000
+2.000
+1.000
+0.000e+000
V(5)
Circuit 9_133-Transient-4
+0.000e+000
(V)
+50.000n
Time (s)
+100.000n
+5.000
+4.000
+3.000
+2.000
+1.000
V(5)
9-45
CHAPTER 10
10.1
A/C temperature
Automobile
coolant temperature
gasoline level
oil pressure
sound intensity
inside temperature
Battery charge level
Battery voltage
Fluid level
Computer display
hue
contrast
brightness
Electrical variables
voltage amplitude
voltage phase
current amplitude
current phase
power
power factor
spectrum
Fan speed
Humidity
Lawn mower speed
Light intensity
Oven temperature
Refrigerator temperature
Sewing machine speed
Stereo volume
Stove temperature
Time
TV picture brightness
TV sound level
Wind velocity
10.2
(a) 20 log (120) = 41.6 dB | 20 log (60) = 35.6 dB | 20 log (50000) = 94.0 dB
20 log(100000) = 100 dB | 20 log(0.90) = −0.915 dB
(b) 20 log (600) = 55.6 dB | 20 log (3000) = 69.5 dB | 20 log (106 ) = 120 dB
20 log(200000) = 106 dB | 20 log(0.95) = −0.446 dB
(c) 10 log (2x109 ) = 93.0 dB | 10 log (4x105 ) = 56.0 dB
10 log (6x108 ) = 87.8 dB | 10 log(1010 ) = 100 dB
10.3 (a)
4
vO
2
0
vS
-2
-4
0
0.5
1
1.5
2
2.5
3
3.5
4
x 10 -3
10-1
(b) 500 Hz :
(c) 500 Hz :
(d) 500 Hz :
(e) Yes
1∠0 o | 1500 Hz : 0.333∠0 o | 2500 Hz : 0.200∠0 o
2∠30 o | 1500 Hz : 1∠30 o | 2500 Hz : 1∠30 o
2∠30 o | 1500 Hz : 3∠30 o | 2500 Hz : 5∠30 o
10.4
Vs = 0.0025V | PO = 40W | Vo = 2PO RL = 2(40)(8) = 25.3V
25.3
= 10100 | 20 log (10100)= 80.1 dB
.0025
0.0025V
V
25.3V
Is =
= 45.45nA | I o = o =
= 3.162 A
5kΩ + 50kΩ
8Ω
8Ω
3.162 A
Ai =
= 6.96 x 10 7 | 20 log 3.48 x 10 7 = 157 dB
45.45nA
40W
Ap =
= 7.04 x 1011 | 10 log 7.04 x 1011 = 118 dB
.0025V (45.45nA)
Av =
(
)
(
)
2
10.5
Vs = 0.01V | PO = 20mW | Vo = 2PO RL = 2(.02)(8) = 0.566V
0.566
= 56.6 | 20 log (56.6) = 35.0 dB
.01
0.01V
V
0.566V
= 192nA | Io = o =
= 70.8mA
Is =
2kΩ + 50kΩ
8Ω
8Ω
70.8mA
Ai =
= 3.68 x 10 5 | 20 log (3.68 x 10 5 )= 111 dB
192nA
0.02W
= 2.08 x 10 7 | 10 log (2.08 x 10 7 )= 73.2 dB
Ap =
.01V (192nA)
2
Av =
10.6
(a) v
Rth
v th
+
-
vo =
(b) v
vo =
th
= voc = 0.768 2 = 1.09 V
⎛ 0.768 − 0.721⎞
RL
v −v
v th → Rth = RL th o = 430⎜
⎟ = 28.0 Ω
0.721
Rth + RL
vo
⎠
⎝
th
= voc = 0.760 2 = 1.08 V
⎛ 0.760 − 0.740 ⎞
RL
v −v
v th → Rth = RL th o = 1040⎜
⎟ = 28.1 Ω
0.740
Rth + RL
vo
⎠
⎝
(c) 1.09 V and 1.08 V → 9% error and 8% error
10-2
G4 laptop – 1 V, 28 Ω.
10.7
10.8
(a) Vo = 2RL PO = 2(8)(20) = 17.9V
Pi =
Vi2
12
1V
17.9V
=
= 25.0µW | I i =
= 49.9µA | I o =
= 2.24 A
20000Ω + 32Ω
8Ω
2Ri 40066
Av =
Vo 17.9V
20W
2.24 A
=
= 17.9 | AP =
= 8.00x105 | Ai =
= 4.49x10 4
1V
25µW
49.9µA
Vi
(b) V = 17.9 V ; recommend ± 20 - V supplies
o
10.9
2P
R
The 24-Ω case represents a good trade off between voltage and current.
V = 2PR
| I=
R (Ω)
V (V)
I (mA)
8
1.27
158
24
2.19
91.3
1000
14.1
14.1
10.10
In the dc steady state, the internal circuit voltages cannot exceed the power supply limits.
(a) +15 V (b) -9 V
10.11
(a) For VB = 0.6V , VO = +8V | Av =
Av = 32 dB ∠AV = 180
o
dvO
dvI
=
v I =0.6V
12 − 4
= −40
0.5 − 0.7
| VM ≤ 0.100 V for linear operation
(b) vI (t )= (0.6 + 0.1sin1000t ) V
vO (t )= (8 − 4sin1000t ) V
10-3
10.12
(a) For VB = 0.5V , VO = +12V |
dvO
is different for positive and negative values of
dvI
VM sin1000t. Thus, the gain is different for positive and negative signal excursions
and the output will always be a distorted sine wave. This is not a useful choice of
bias point for the amplifier.
(b) For VB = 1.1V , VO = +2V and
dvO
= 0. The gain is zero for this bias point.
dvI
Thus this is also not a useful choice of bias point for the amplifier.
10.13
(a) For V
B
= 0.8V , VO = +3V | Av =
dvO
dv I
=
v I =0.8V
4−2
= −10
0.7 − 0.9
Av = 20dB ∠AV = 180 o | VM ≤ 0.100 V for linear operation
(b) For V
B
= 0.2V , VO = +14V | Av =
dvO
dv I
=0
v I =0.8V
The output signal will be distorted regardless of the value of VM .
10.14
14
vO
12
10
8
6
4
Time
2
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
The amplifier is operating in a linear region. vO = 8 - 4 sin 1000t volts There are only two
spectral components: 8 V at dc and -4 V at 159 Hz
10-4
10.15
For sin 1000t ≥ 0, vO = 12 − 4 sin 1000t
For sin 1000t < 0, vO = 12 −1 sin 1000t
14
vO
12
10
8
Time
6
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
Using the MATLAB FFT capability with a fundamental frequency of 1000/2
Hz:
t=linspace(0,2*pi/1000,1000);
y=12-4*sin(1000*t).*(sin(1000*t)>=0)-sin(1000*t).*(sin(1000*t)<0);
z=fftshift(fft(y))/1000;
yields the following series:
v O (t ) = 11.05 − 2.50sin(1000t ) + 0.638cos(2000t ) + 0.127cos(4000t ) + 0.0546cos(6000t )Note:
It is worth plotting this function to see if it is correct.
The Fourier coefficients may also be calculated directly using MATLAB. For example, for the
cosine terms:
Define a function:
function y=four(t)
y=cos(fn*1000*t).*(12-4*sin(1000*t).*(sin(1000*t)>=0)sin(1000*t).*(sin(1000*t)<0));
global fn
fn=0; quad('four',0,pi/500)*1000/pi
10.16
vI = 0.004sin 2000πt V | Av =
THD = 100
5V
= 1250 | Third and fifth harmonics are present.
0.004V
0.252 + 0.102
% = 5.39 %
5
10.17
At the input signal frequency : vi = 0.25sin1200πt V | vo = 4sin1200πt V
4V
0.42 + 0.22
Av =
= 16 | Second and third harmonics are present. THD = 100
% = 11.2 %
0.25V
4
10-5
10.18 (a)
w=200*pi*linspace(0,0.001,512);
v=6+4*sin(w).*(sin(w)>=0)+2*sin(w).*sin(w)<0;
s=fft(v)/512;
vmag=sqrt(s.*conj(s));
vmag(1:3)
ans = 6.6354 1.4985 0.2129
dc component: 6.64 V
fundamental:
1.50 V
2nd harmonic:
0.213 V
(b) Define m-files for a(n) and b(n)
function f=an(t);
global n;
ftemp=6+(4*sin(2000*pi*t)).*(t<=0.0005)+(2*sin(2000*pi*t)).*(t>0.0005);
f=ftemp.*cos(2000*n*pi*t);
function f=bn(t);
global n;
ftemp=6+(4*sin(2000*pi*t)).*(t<=0.0005)+(2*sin(2000*pi*t)).*(t>0.0005);
f=ftemp.*sin(2000*n*pi*t);
Then n=0; quad(‘an’,0,0.001)
Then n=1; quad(‘bn’,0,0.001)
ans: 13.27
ans 3.000 etc
f = 6.63 + 3sin(2000 t) – 0.4244cos(4000 t) –0.0849cos(8000 t)
10-6
10.19
t=linspace(0,0.002,512);
y=max(-1,min(1,1.5*sin(1400*pi*t)));
plot(t,y)
1.5
1
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x 10-3
w=1400*pi*linspace(0,0.001428571,512);
y=max(-1,min(1,1.5*sin(w)));
s=fft(y)/512;
ymag=sqrt(s.*conj(s));
y2=ymag.*ymag;
sumy2=sum(y2)-2*y2(1)-2*y2(2);
thd=100*sqrt(sumy2)/(2*y2(2))
thd = 18.3929 %
10.20
g11 =
i1
: v1 = 10 4 i1 + 101 i1 (240kΩ) → g11 = 4.124 x 10−8 S = 4.12 x 10−8 S
v1 i = 0
2
g12 =
g21 =
g22 =
i1
i2
v1 =0
v2
v1
i2 = 0
: i1 = −
v2
i2
i
+
240kΩ
(i2 + 100i1)→ g12 = −9.90 x 10−3
240kΩ + 10kΩ
: v2 = 101i1 (240kΩ) | i1 = g11v1 → g21 = 1.00
: i2 =
v1 =0
v2
v
v
+ 2 + 100 2 → g22 = 99.0 Ω
240kΩ 10kΩ
10kΩ
i
1
10 k Ω
v1 = 0
-
100 i 1
240 k Ω
i
2
+
+
v
v
2
-
i2 = 0
1
1
-
+
10 k Ω
100 i 1
240 k Ω
v
2
-
10-7
10.21
g11 =
1
24.25 MΩ
g12 = −9.897x10−3
g21 = 0.9996
g22 = 98.97 Ω
10.22
g11 =
g21 =
g22 =
i1
v1
=
i2=0
v2
v1
i2=0
v2
i2
v1 = 0
1
= 0.286 mS
2.5kΩ +1kΩ
: v 2 = v1
=−
v1 = 0
1kΩ
= −0.286
2.5kΩ +1kΩ
: v 2 = (i 2 − 0.04 v a )60kΩ + i 2 (1kΩ 2.5kΩ) | v a = −i 2 (1kΩ 2.5kΩ)
60 k Ω
0.04 v
2.5 k Ω
+
i1
i2
1kΩ
2.5kΩ
+ v1
(−0.04 )(60kΩ) → g21 = −1710
2.5kΩ +1kΩ
2.5kΩ +1kΩ
g22 = 1.78 x 10 6 Ω
i1
| g12 =
a
i2
i2
-
va
v2
1k Ω
v1 = 0
+
-
10.23
g11 =
i1
: v1 = 103 i1 → g11 = 1.00 x 10−3 S
v1 i =0
2
g12 =
g21 =
g22 =
i1
i2
v1 =0
v2
v1
i2 =0
v2
i2
: i1 = − i2 → g12 = −1.00
: v2 = v1 + 0.1v1(20kΩ) | v1 = 2001v1 → g21 = 2001
: i2 =
v1 =0
v2
→ g 22 = 20 kΩ
20kΩ
20 k Ω
20 k Ω
0.1v
i1
0.1v
1
i2
i
2
i1
-
10-8
0.1v
1k Ω
1
i =0
2
=0
i2
2
+
v1
1
i
1k Ω
v1 = 0
i=0
0.1v
1
=0
+
v2
-
10.24
g11 =
1
1.000 kΩ
g12 = −1.000
g21 = 2001
g22 = 20.00 kΩ
10.25
Rin
RL
A
(a) VO = VS
Rin + RS RL + Rout
| A = 10
54
20
= 501.2
16
10 6
Av = 3
501.2)
= 485.5 | AvdB = 20 log(485.5) = 53.7 dB
6 (
10 + 10
0.5 + 16
485.5VS
IO
16
= 3.041 x 10 7 | A idB = 20 log(3.041 x 10 7 )= 150 dB
Ai = =
VS
IS
10 3 + 10 6
485.5VS
VO IO 485.5VS 16
=
= 485.5(3.041 x 10 7 ) = 1.478 x 1010
AP =
V
VS IS
VS 3 S 6
10 + 10
A PdB = 10 log(1.478 x 1010 )= 102 dB
VO2
5.657
→ VO = 5.657 V | VS =
= 11.65 mV
(b) 1 =
2(16)
485.5
(c) R out and R L see the same current.
IO2
1
R = 0.5 = 31.25 mW | PR in
2
16
P = 31.25mW + 67.7 pW = 31.3 mW
PR out =
IO2
I2 1
RL → O =
2
2 16
⎛ 0.01165 ⎞ 2 10 6
IS2
= Rin = ⎜ 3
= 67.7 pW
⎟
⎝10 + 10 6 ⎠ 2
2
P=
10.26
⎛ 54 ⎞ 16
Rin
1000
RL
(a) VO = VS R + R A R + R = VS 1000 + 1000 ⎜⎜10 20 ⎟⎟ 16 + 16 = 125 VS
in
S
L
out
⎝
⎠
VO2
5.66
1=
→ VO = 5.66V | VS =
= 45.3 mV
125
2(16)
(b) Since R out and R L have the same current and same value, the power dissipated in R out is 1 W.
⎛ 45.3mV ⎞2
1
The power lost in R in is ⎜
= 0.257µW | PD = 0.257µW + 1.00W = 1.00 W
⎟
⎝ 2 ⎠ 2(1000)
10-9
10.27
io =
2(0.1W )
2Po
=
= 91.3 mA | v = io (28 + 24 )Ω = 4.75 V
RL
24Ω
iv
i2R
(91.3mA) (28Ω) = 117 mW
PD = o = 217mW | PL = o out =
2
2
2
2
10.28
io =
2(0.1W )
2Po
=
= 14.1 mA | v = io (28 + 1000)Ω = 14.5 V
RL
1000Ω
iv
i2R
(14.1mA) (28Ω) = 2.78 mW
PD = o = 103 mW | PL = o out =
2
2
2
2
10.29
Rin = ∞ | Rout = 0 Ω | vO = AvS = −2000v S | vO = 0.01V (−2000)= −20 V
P=
VO2
202
12.5W
=
= 12.5 W | AP =
=∞
2RL 2(16)
0
10.30
77
20
20kΩ
2kΩ
A
→ A = −7805
20kΩ + 1kΩ 2kΩ + 0.1kΩ
A = −7800 | AdB = 20 log(7800)= 77.8 dB
Av = −10
= −7079 | − 7079 =
10.31
i1 = is
RS
RS + Rin
io = βi1
Rout
Rout + RL
Ai = β
Rout
RS
Rout + RL RS + Rin
Rin = 0
Rout = ∞
10.32
IO = I S
RS
Rout
β
RS + Rin Rout + RL
10.33
| 200 =
300kΩ
200kΩ
β
→ β = 243
200kΩ + 10kΩ 300kΩ + 47kΩ
10−6 A(5000)] 4
[
IO2
125mW
= ∞ | P = RL =
10 Ω = 125 mW | AP =
=∞
2
2
0
2
Rin = 0 Ω | Rout
10-10
10.34
VO = VS
Rin
RL
Rin
A
A
Rin + RS Rin + Rout RL + Rout
5000
100
5000
−1200)
−1200)
= +1.82 x 105
(
(
5000 + 500
100 + 500
5000 + 1000
5
AvdB = 20 log 1.82 x 10 = 105 dB
Av =
(
)
(
)
IO 1.82x105VS 1
=
= +1.09 x 10 7 | AidB = 20 log 1.36 x 10 7 = 141 dB
VS
IS
100
6000
5
1.82x10 VS +1.09 x 10 7 I S
= +1.98 x 1012 | APdB = 10 log 1.98 x 1012 = 123 dB
AP =
VS I S
Ai =
(
)
(
)
10.35
VO
⎛ VO ⎞ 2 RS + Rin
VO IO
RL
AP =
=
=⎜ ⎟
VS
VS IS V
RL
⎝ VS ⎠
S
RS + Rin
⎤
⎡⎛ ⎞ 2
⎛ VO ⎞ 2
⎛ R + Rin ⎞
VO RS + Rin ⎥
⎢
= 10log⎜ ⎟ + 10log⎜ S
APdB = 10log ⎜ ⎟
⎟
RL ⎥⎦
⎢⎣⎝ VS ⎠
⎝ VS ⎠
⎝ RL ⎠
⎛V ⎞
⎛ RL ⎞
⎛ RL ⎞
APdB = 20log⎜ O ⎟ −10log⎜
⎟ = AvdB −10log⎜
⎟
⎝ VS ⎠
⎝ RS + Rin ⎠
⎝ RS + Rin ⎠
VO
⎛ I ⎞2 R
VO IO
IO RL IO
L
AP =
=
=⎜ O⎟
VS IS IS (RS + Rin )IS ⎝ IS ⎠ RS + Rin
2
⎡⎛ ⎞ 2
⎤
⎛ IO ⎞
⎛ RL ⎞
IO
RL
⎥ = 10log⎜ ⎟ + 10log⎜
AidB = 10log⎢⎜ ⎟
⎟
⎢⎣⎝ IS ⎠ RS + Rin ⎥⎦
⎝ IS ⎠
⎝ RS + Rin ⎠
⎛I ⎞
⎛ R
⎞
⎛ R
⎞
L
L
APdB = 20log⎜ O ⎟ + 10log⎜
⎟ = AidB + 10log⎜
⎟
⎝ IS ⎠
⎝ RS + Rin ⎠
⎝ RS + Rin ⎠
Note : APdB =
AvdB + AidB
2
10-11
10.36
(a) A (s)=
i
(s
3x109 s2
2
)(
+ 51s + 50 s2 + 13000s + 3x10 7
)
=
3x109 s2
(s + 1)(s + 50)(s + 3000)(s + 10000)
Zeros : s = 0, s = 0 | Poles : s = -1, s = -50, s = -3000, s = -10000
(b) A (s)= s
(
)
105 s2 + 51s + 50
+ 1000s + 50000s + 20000s + 13000s + 3x10 7
Using MATLAB to find the poles:
v
5
4
3
2
| Zeros : s = -1, s = -50
Av=[1 1000 50000 20000 13000 3e7];
roots(Av)
ans = -947.24 -52.13 -9.170 4.27 + j6.93 4.27 - j6.93
10.37
⎛
⎞
⎜
⎟
R2 ⎜ 1 ⎟
R2
1.5kΩ
Vo
| Amid =
=
=
= 0.600 | Amid-dB = −4.44 dB
s ⎟
Vs R1 + R2 ⎜
R1 + R2 1kΩ + 1.5kΩ
⎜1+
⎟
⎝ ωH ⎠
1
1
=
= 26.5 kHz
fH =
2π R1 R2 C 2π 1kΩ 1.5kΩ 0.01µF
(
)
(
)
10.38
⎛
⎞
⎜
⎟
R2 ⎜ 1 ⎟
R2
100kΩ
Vo
| Amid =
=
=
= 0.909 | Amid-dB = −0.828 dB
s ⎟
Vs R1 + R2 ⎜
R1 + R2 10kΩ + 100kΩ
⎜1+
⎟
⎝ ωH ⎠
1
1
fH =
=
= 1.75 kHz
2π R1 R2 C 2π 10kΩ 100kΩ 0.01µF
(
)
(
)
10.39
−0.5
(a) Amid ≥ 10 20 = 0.944 |
C=
(
1
)
2π R1 R2 f H
(b) R
2
10-12
=
R2
≥ 0.944 → R2 ≥ 9440Ω | Choose R2 = 10 kΩ
560 + R2
(
1
)
2π 560Ω 10kΩ 20kHz
= 10 kΩ | C = 15.0 nF
= 0.015 µF
10.40
Vo
R2 ⎛ s ⎞
R2
20kΩ
=
=
= 0.667 | Amid-dB = −3.52 dB
⎜
⎟ | Amid =
Vs R1 + R2 ⎝ s + ω L ⎠
R1 + R2 10kΩ + 20kΩ
1
1
=
= 531 Hz
2π (R1 + R2 )C 2π (10kΩ + 20kΩ)0.01µF
fL =
10.41
Vo
R2 ⎛ s ⎞
R2
78kΩ
=
=
= 0.886 | Amid-dB = −1.05 dB
⎜
⎟ | Amid =
Vs R1 + R2 ⎝ s + ω L ⎠
R1 + R2 10kΩ + 78kΩ
1
1
=
= 181 Hz
2π (R1 + R2 )C 2π (10kΩ + 78kΩ)0.01µF
fL =
10.42
(a) A
mid
C=
≥ 10
−0.5
20
R2
≥ 0.944 → R2 ≥ 5560Ω | Choose R 2 = 7.5 kΩ
330Ω + R2
= 0.944 |
1
1
=
= 1.02 nF
2π (R1 + R2 )f L 2π (330Ω + 7500Ω)20kHz
(b) R
2
10.43
Av =
= 7.5 kΩ | C = 1.0 nF | Checking : f L = 20.3 kHz
2π x 10 7 s
(s + 20π )(s + 2π x 10 )
4
=
1000s
| A mid = +1000 = 60 dB
⎛
⎞
s
(s + 20π )⎜⎝1+ 2π x 104 ⎟⎠
20π
2π x 10 4
= 10 Hz | f H =
= 10 kHz | BW = 10kHz −10Hz = 9.99 kHz
2π
2π
Bandpass Amplifier
fL =
10.44
10 4 s
| High - pass Amplifier | A mid = +10 4 = 80 dB
s + 200π
200π
fL =
= 100 Hz | f H = ∞ | BW = ∞
2π
Av =
10.45
2π x 106
=
s + 200π
10 4
→ Low - pass Amplifier | A mid = +10 4 = 80 dB
s
1+
200π
200π
= 100 Hz | BW = 100Hz − 0Hz = 100 Hz
f L = 0 Hz | f H =
2π
Av =
10-13
10.46
Av (s) =
7
s
ωo
s2 + s
ωo
5
10 s
10 s
= 102 2
= Amid
5
14
s + 10 s + 10
s + 105 s + 1014
2
Bandpass Amplifier | Amid = 100 = 40 dB | f o =
Q
Q
+ ω 2o
10
ω
= 1.592 MHz | Q = o5 = 100
2π
10
7
1.592 MHz
= 15.92 kHz | For a high Q circuit :
100
BW
= 1.592 MHz −15.92kHz = 1.584 MHz
fL ≈ fo −
2
BW
fH ≈ fo +
= 1.592 MHz + 15.92kHz = 1.600 MHz
2
BW =
10.47
s 2 + 1012
s2 + ω o2
=
A
| Notch Filter
mid
ω
s2 + 10 4 s + 1012
s2 + s o + ω o2
Q
⎞
⎛
ω
ω
s o
s o
⎟
⎜
Q
Q
Note : Av (s) = −20⎜1−
is a bandpass function.
⎟ where
ωo
2
2
⎜⎜ s2 + s ω o + ω o2 ⎟⎟
s + s + ωo
Q
Q
⎠
⎝
Av (s) = −20
Amid = −20 = 20 dB but Av = 0 at f o | f o =
The width of the null = BW =
ω o 10 6
ω
=
= 159.2 kHz | Q = o4 = 100
2π 2π
10
159.2kHz
= 1.592 kHz | For a high Q circuit :
100
BW
= 159.2Hz −1.592kHz = 157.6 kHz
2
BW
= 159.2Hz + 1.592kHz = 160.8 kHz
f H ≈ fo −
2
fL ≈ fo −
10-14
10.48
Av (s) =
4 π 2 x 1014 s2
(s + 20π )(s + 50π )(s + 2π x 10 5 )(s + 2π x 10 6 )
Av (s) =
10 3 s2
⎛
⎞⎛
⎞
s
s
1+
⎟
⎜
⎟
(s + 20π )(s + 50π )⎜1+
⎝ 2π x 10 5 ⎠⎝ 2π x 10 6 ⎠
| Amid = 1000 = 60 dB
Zeros : s = 0, s = 0 | Poles : s = -20π , s = -50π , s = -2π x 10 5 , s = -2π x 10 6
10 3
⎛
⎞⎛
⎞
s
s
1+
⎜1+
5 ⎟⎜
6⎟
⎝ 2π x 10 ⎠⎝ 2π x 10 ⎠
Since the two high frequency poles are separated in frequency by a decade,
For s >> 50π , Av (s) ≈
2π x 10 5
= 100 kHz | However, we are not that lucky at low frequencies.
fH ≈
2π
10 3 s2
10 3 ω L2
10 3
For s << 2π x 10 5, Av (s) ≅
| Av ( jω L ) =
=
2
2
2
(s + 20π )(s + 50π )
ω 2 + (20π ) ω 2 + (50π )
(
[
]
L
ω L4 − (20π ) + (50π ) ω L2 − (20π ) (50π ) = 0 → ω L = ±178 | f L =
2
2
2
2
)(
L
)
178
= 28.3 Hz
2π
10-15
10.49
Using MATLAB: n=[2e7*pi 0]; d=[1 (20*pi+2e4*pi) 40e4*pi^2]; bode(n,d)
70
60
Gain dB
50
40
30
20
10
0
10 0
10 1
10 2
10 3
10 4
Frequency (rad/sec)
10 5
10
6
100
50
Phase deg
0
-50
-100
-150
-200
10 0
10 1
10.50
Using MATLAB:
10 2
10 3
10 4
Frequency (rad/sec)
10 5
10 6
n=[1e4 0]; d=[1 200*pi]; bode(n,d)
Bode Diagrams
80
70
60
50
40
100
80
60
40
20
0 1
10
10
2
10
Frequency (rad/sec)
10-16
3
10
4
10.51
Using MATLAB:
n=[2e6*pi]; d=[1 200*pi]; bode(n,d)
Bode Diagrams
80
75
70
65
60
55
0
-20
-40
-60
-80
-100 2
10
10
3
10
4
Frequency (rad/sec)
10.52
Using MATLAB:
n=[1e7 0]; d=[1 1e5 1e14]; bode(n,d)
100
Gain dB
50
0
-50
-100
10
6
10 7
Frequency (rad/sec)
10 8
Phase deg
100
50
0
-50
-100
10 6
10 7
Frequency (rad/sec)
10 8
10-17
10.53
Using MATLAB:
n=[-20 0 -2e13]; d=[1 1e4 1e12]; bode(n,d)
Gain dB
30
20
10
0
10 5
10 6
Frequency (rad/sec)
10 7
10 6
Frequency (rad/sec)
10
Phase deg
200
150
100
10 5
7
10.54
Using MATLAB:
n=[4e14*pi^2 0 0];
p1=[1 20*pi]; p2=[1 50*pi]; p3=[1 2e5*pi]; p4=[1 2e6*pi];
d=conv(conv(p1,p2),conv(p3,p4));
bode(n,d)
60
Gain dB
40
20
0
-20
-40
-60
10 -1
10 0
10 1
10
2
10 4
10 5
10 3
Frequency (rad/sec)
10
6
10 7
10
8
10 0
10 1
10
2
10 4
10 5
10 3
Frequency (rad/sec)
10
6
10 7
10
8
200
Phase deg
100
0
-100
-200
10 -1
10-18
10.55
Using MATLAB:
n=[2e7*pi 0]; d=conv([1 20*pi],[1 2e4*pi]); w=2*pi*[5 500 50000];
a=freqs(n,d,w); am=abs(a); ap=angle(a)*180/pi
Magnitudes: 447.2135 998.5526 196.1161
Phases:
63.4063 -1.7166 -78.6786
(a) vO = 0.447sin(10πt + 63.4 o ) V
(b) vO = 0.999sin(1000πt −1.72 o ) V
(c ) vO = 0.196sin(105 πt − 78.7 o ) V
10.56
Av (s) =
10 4 s
s + 200π
Av (jω ) =
| Av ( jω )=
10 4 f
10 4 jω
10 4 jf
=
jω + 200π jf + 100
| ∠Av (jω )= 90 o − tan −1
f 2 + 1002
(a) 1 Hz : A (jω ) = 100 | ∠A (jω )= 89.4 |
(b) 50 Hz : A (jω ) = 4472 | ∠A (jω )= 63.4
(c) 5 kHz : A (jω ) = 9998 | ∠A (jω )= 1.15
o
v
v
o
v
v
o
v
v
f
100
(
)
vO = 0.03 sin 2πt + 89.4 o V
(
)
= 3.00 sin (10 πt + 1.15 ) V
| vO = 1.34 sin 100πt + 63.4 o V
| vO
4
o
10.57
Using MATLAB:
n=[1e4 0]; d=[1 200*pi]; w=2*pi*[2 2000 200000];
a=freqs(n,d,w); am=abs(a); ap=angle(a)*180/pi
Magnitudes: 2.0000e+02 9.9875e+03 1.0000e+04
Phases:
88.8542 2.8624 0.0286
(a) vO = 2.00sin 4πt + 88.9o mV
(
)
(b) v = 0.999sin(4000πt + 2.86 )V
(c) v = 0.100sin(4x10 πt + 0.0286 )V
o
O
5
o
O
10-19
10.58
Using MATLAB:
n=[-1e7 0]; d=[1 1e5 1e14];
w=2*pi*[1.59e6 1e6 5e6];
a=freqs(n,d,w);
am=abs(a)
ap=angle(a)*180/pi
Magnitudes: 98.1550 1.0381 0.3542
Phases:
-168.9767 -90.5948 90.2029
(a) vO = 0.393sin 3.18x106 πt −169o V
(
)
(b) v = 4.15sin(2x10 πt − 90.6 )mV
(c) v = 1.42sin(10 πt + 90.2 )mV
6
o
O
7
o
O
10.59
Using MATLAB:
n=[-20 0 -2e13]; d=[1 1e4 1e12];
w=2*pi*[1.59e5 5e4 2e5];
a=freqs(n,d,w);
am=abs(a)
ap=angle(a)*180/pi
Magnitudes: 3.8242 19.9999 19.9953
Phases:
101.0233 179.8003 -178.7570
(a) vO = 0.956sin 3.18x105 πt + 101o mV
(
)
(b) v = 5.00sin(10 πt + 180)V
(c) v = 5.00sin(4x10 πt −179 )V
5
O
5
o
O
10.60
Using MATLAB:
n=[4e14*pi^2 0 0];
p1=[1 20*pi]; p2=[1 50*pi]; p3=[1 2e5*pi]; p4=[1 2e6*pi];
d=conv(conv(p1,p2),conv(p3,p4));
w=2*pi*[5 500 50000];
a=freqs(n,d,w);
am=abs(a)
ap=angle(a)*180/pi
Magnitudes: 87.7058 998.5400 893.3111
Phases:
142.1219 3.6930 -29.3873
(a) vO = 175sin 10πt + 142o mV
(
)
(b) v = 2.00sin(1000πt + 3.69 )V
(c) v = 1.79sin(10 πt − 29.4 )V
o
O
5
O
10-20
o
10.61
26
(a) Amid = +10 20 = +20 | Av =
1+
(b) A
v
=−
20
s
(
2π x 5 x 106
=
)
20
2x108 π
=
s
s +10 7 π
1+ 7
10 π
2x10 π
s +10 7 π
8
10.62
(a) A
mid
= +10
(b) Av (s)= −
40
20
= +100 | Av (s) = 100
s
⎛
⎞
s
(s + 400π )⎜⎝1+ 2π x 105 ⎟⎠
=
2π x 10 7 s
(s + 400π )(s + 2π x 10 )
5
2π x 10 7 s
(s + 400π )(s + 2π x 10 )
5
10-21
10.63
⎛
⎞2
⎜
⎟
⎛ 50000π ⎞
1
Av (s) = −1000⎜
⎟ = −1000⎜
⎟ | A mid = −1000
s
⎝ s + 50000π ⎠
⎜1+
⎟
⎝ 50000π ⎠
⎛
dB
dB ⎞
f H = 0.64 f1 = 0.644(25kHz) = 16.1 kHz | 2⎜ −20
⎟ = −40
⎝
dec
dec ⎠
2
Using MATLAB:
n=1000*(50000*pi)^2;
d=[1 2*50000*pi (50000*pi)^2 ];
bode(n,d)
60
Gain dB
50
40
30
- 40 dB/dec
20
10
0
10 4
10 5
Frequency (rad/sec)
10
6
0
Phase deg
-50
-100
-150
-200
10 4
10-22
10 5
Frequency (rad/sec)
10 6
10.64
⎞3 A
⎛
⎛ ω1 ⎞ 3
ω
1
⎟ = o
Av (s) = Ao ⎜
⎟ | Av ( jω H ) = Ao ⎜⎜
⎟
2
2
2
⎝ s + ω1 ⎠
⎝ ω H + ω1 ⎠
| BW = f H
1
⎛ f H ⎞2
⎛
dB ⎞
dB
2 = 1+ ⎜ ⎟ → f H = f1 2 3 −1 = 25kHz(0.5098) = 12.8 kHz | 3⎜ -20
⎟ = −60
⎝
dec ⎠
dec
⎝ f1 ⎠
1
3
Using MATLAB:
n=2000*(50000*pi)^3;
d=[1 3*50000*pi 3*(50000*pi)^2 (50000*pi)^3];
bode(n,d)
Bode Diagrams
70
60
50
40
30
-60 dB/dec
20
10
200
150
100
50
0
-50
-100
10 4
10 5
10 6
Frequency (rad/sec)
10-23
10.65
(a) To avoid distortion of the waveform, the phase shift must be proportional to frequency. 30o
at 1500 Hz and 50o at 2500 Hz.
o
o
o
(b) vO = 3.16sin 1000πt + 10 + 1.05sin 3000πt + 30 + 0.632sin 5000πt + 50
(
)
(
)
(
)
(c) Using MATLAB:
t=linspace(0,.004);
a=10^(10/20);
vs=sin(1000*pi*t)+0.333*sin(3000*pi*t)+0.200*sin(5000*pi*t);
vo=A*(sin(1000*pi*t+pi/18)+0.333*sin(3000*pi*t+3*pi/18)+0.2*sin(5000*pi*t+5*pi/18));
plot(t,A*vs,t,vo)
3
2
1
0
-1
-2
-3
0
0.5
1
1.5
2
2.5
3
3.5
4
x 10
10-24
-3
CHAPTER 11
11.1
v O = vS
iS =
v
1MΩ
1kΩ
1000)
| Av = O = 990 or 59.9 dB
(
vS
1MΩ + 5kΩ
1kΩ + 0.5Ω
vS
990vS
and iO =
1MΩ + 5kΩ
1kΩ
(
| Ai =
iO 990 6
=
10 = 9.9x105 or 120 dB
iS 1000
)
AP = Av Ai = 990 9.9x105 = 9.8x108 or 89.9 dB | v S =
vO 5V
=
= 5.05 mV
AV 990
11.2
vO = vS
vs =
5kΩ
1kΩ
v
31.6)
and A v = O = 7.91 or 18.0 dB
(
vS
5kΩ + 5kΩ
1kΩ + 1kΩ
vO 10V
=
= 1.27 V Since Rout has the same value as R L , the power
Av 7.91
dissipated in Rout is also 0.5W. The power dissipated in R id will be
⎛ VS ⎞2
⎜ ⎟
V2 ⎝ 2 ⎠
V2
V
= S where VS = O and VO = 2(0.5W )(1000Ω) = 31.6V
PI = id =
2Rid
2Rid
8Rid
7.91
42
PI =
= 0.4mW. The total power dissipated in the amplifier is
8(5000)
P = 500mW + 0.4mW = 500 mW.
11.3
0.99mV ≥ 1mV
11.4
Io =
2(100W )
50Ω
Rid
⇒ R id ≥ 4.95 MΩ
Rid + 50kΩ
= 2 A and
I2o Rout
≤ 5W or Rout ≤ 2.5Ω
2
11.5
v id =
11.6
vid =
vO 10V
10V
= 5 = 0.1 mV |
≤ 10−6V requires A ≥ 10 7 or 140 dB
A
A 10
vO 15V
15V
15µV
= 6 = 15 µV |
≤ 10−6 V requires A ≥ 15x106 or 144 dB | i+ =
= 15 pA
A
1MΩ
A 10
11-1
11.7
(a) A
=−
R2
220kΩ
=−
= −46.8 | 20log(46.8) = 33.4 dB | Rin = R1 = 4.7 kΩ | Rout = 0 Ω
R1
4.7kΩ
(b) A
v
=−
R2
2.2 MΩ
=−
= −46.8 | 20log(46.8)= 33.4 dB | Rin = R1 = 47 kΩ | Rout = 0 Ω
R1
47kΩ
(a) A
=−
R2
120kΩ
=−
= −10.0 | 20log(10.0)= 20.0 dB | Rin = R1 = 12 kΩ | Rout = 0 Ω
R1
12kΩ
(b) A
=−
R2
330kΩ
=−
= −2.36 | 20log(2.36)= 7.46 dB | Rin = R1 = 140 kΩ | Rout = 0 Ω
R1
140kΩ
(c) A
=−
R2
240kΩ
=−
= −55.8 | 20log(55.8)= 34.9 dB | Rin = R1 = 4.3 kΩ | Rout = 0 Ω
R1
4.3kΩ
v
11.8
v
v
v
11.9
Av = −
Is =
R2
8200Ω
=−
= −10.9 | VO = −10.9(0.05V )= −0.547V
750Ω
R1
|
vo (t )= −0.547sin (4638t ) V
Vs 0.05V
=
= 66.7µA | is (t )= 66.7sin (4638t ) µA
R1 750Ω
11.10
110kΩ
(a) A = − 22kΩ = −5 v = A v = 0 (b) V = A V = -5(0.22V )= −1.10 V
(b) v = [−5(0.15V )]sin 2500πt = −0.75sin 2500πt V (d ) v = (−1.10 + 0.75sin 2500πt ) V
v
o
v s
O
v S
o
(e) I
S
(f ) i
(g ) v
O
-
11.11
o
=
⎡0.15V ⎤
0.22V
= 10.0 µA is = ⎢
⎥ sin 2500πt = 6.82sin 2500πt µA iS = (10.0 - 6.82sin 2500πt ) µA
22kΩ
⎣ 22kΩ ⎦
= -iS
IO = −10.0 µA
io = −6.82sin 2500πt µA
iO = (-10.0 + 6.82sin 2500πt ) µA
=0
vO = −iTH R
11.12
40
R
Rin = R1 = 1.5 kΩ | Av = − 2 = −10 20 = −100 → R2 = 100R1 = 150kΩ
R1
The resistors exist as standard values.
11-2
11.13
26
Rin = R1 = 30 kΩ | Av = −
R2
= −10 20 = −20 → R2 = 20R1 = 600kΩ
R1
Using the closest values from Appendix A, R1 → 30.1 kΩ and R2 → 604 kΩ
The values for the final design are Av = −
604 kΩ
= −20.1 and Rin = 30.1 kΩ
30.1 kΩ
11.14
12
R
Rin = R1 = 100 kΩ | Av = − 2 = −10 20 = −15.8 → R2 = 15.8R1 = 1.58 MΩ
R1
Using the closest 1% values from Appendix A, R1 →100 kΩ and R2 →1.00 MΩ + 576kΩ
1.576MΩ
= −15.8 and Rin = 100 kΩ
100 kΩ
If we used 5% values, we could select 100 kΩ and 1.6 MΩ.
The values for the final design are Av = −
11.15
Av = 1+
R2
750kΩ
= 1+
= 92.5 | 20 log(92.5) = 39.3 dB | Rin = ∞ | Rout = 0 Ω
R1
8.2kΩ
11.16
(a) A
= 1+
R2
120kΩ
= 1+
= 6.00 | 20 log(6.00)= 15.6 dB | Rin = ∞ | Rout = 0 Ω
R1
24kΩ
(b) A
= 1+
R2
300kΩ
= 1+
= 21.0 | 20 log(21.0)= 26.4 dB | Rin = ∞ | Rout = 0 Ω
15kΩ
R1
(c) A
= 1+
R2
360kΩ
= 1+
= 84.7 | 20 log(84.7) = 38.6 dB | Rin = ∞ | Rout = 0 Ω
4.3kΩ
R1
v
v
v
11.17
Av = 1+
R2
8200Ω
= 1+
= 10.0 | VO = 10.0(0.05V )= 0.500V | vo (t )= 0.500sin 9125t V
910Ω
R1
11-3
11.18
110kΩ
(a) A = 1+ 22kΩ = +6 v = 6v = 0 (b) V = A V
(b) v = [6(0.18V )]sin 3250πt = 1.08sin 3250πt V
(d ) v = (1.98 −1.08sin 3250πt ) V (e) i = 0
v
o
s
O
v S
= 6(0.33V )= +1.98 V
o
o
(f ) i
O
S
=+
vO
R1 + R2
IO =
1.98V
= 15.0 µA io = 8.18sin 3250πt µA iO = (15.0 + 8.18sin 3250πt ) µA
132kΩ
(g ) v = (0.33 − 0.18sin 3250πt ) V
-
11.19
(a) A
nom
v
= 1+
R2
47kΩ
= 1+
= 262
R1
0.18kΩ
| 20 log(262) = 48.4 dB
Rin = 10kΩ + ∞ = ∞ | Rout = 0 Ω
(b) A
max
v
= 1+
47kΩ(1.1)
0.18kΩ(0.9)
= 320 | Avmin = 1+
Avmax − Avnom 320 − 262
=
= 0.22 |
262
Avnom
(c) Tolerances :
+ 22%, −18%
47kΩ(0.9)
0.18kΩ(1.1)
= 215
Avmin − Avnom 215 − 262
=
= −0.18
262
Avnom
(d )
320
= 1.49 :1
215
(e) function count=c;
c=0;
for i=1:500,
r1=180*(1+0.2*(rand-0.5));
r2=47000*(1+0.2*(rand-0.5));
a=1+r2/r1;
anom=1+47000/180;
if (a>=0.95*anom & a<=1.05*anom), c=c+1; end;
end
c
Executing this function twenty times yields 44% .
11.20
32
R
Av = 1+ − 2 = 10 20 = 39.8 → R2 = 38.8R1
R1
There are many possible pairs of values in Appendix A. Two choices are
R1 → 3.09 kΩ and R2 → 121 kΩ or R1 → 6.19 kΩ and R2 → 243 kΩ
The values for the final design are Av = 1+
11.21
11-4
121 kΩ
= 40.2
3.09 kΩ
26
R
Av = 1+ − 2 = 10 20 = 20 → R2 = 19R1
R1
There are many possible pairs of values in Appendix A. Two choices are
R1 → 1.05 kΩ and R2 → 200 kΩ or R1 → 10.5 kΩ and R2 → 200 kΩ
The values for the final design are Av = 1+
200 kΩ
= 20.0
1.05 kΩ
11.22
6
R
Av = 1+ − 2 = 10 20 = 2.00 → R2 = R1. Any value could theoretically be used, but
R1
we don' t want a heavy load on a real op amp, so we should choose the resistors to be
at least in the 10 - kΩ range or so. A third 100 - kΩ resistor is added in shunt with the input
as in Fig. P11.23(b). R1 → 10.0 kΩ, R2 →10.0 kΩ, and R3 →100 kΩ
The values for the final design are Av = 1+
10.0 kΩ
= 2.00 and Rin = 100 kΩ
10.0 kΩ
11.23
R2
100kΩ
=−
= −5.00 | Rin = R1 = 20 kΩ
R1
20kΩ
(a) A
=−
(b) A
= 1+
(c) A
=−
v
v
v
R2
100kΩ
= 1+
= +6.00 | Rin = 47kΩ ∞ = 47 kΩ
20kΩ
R1
R2
0
=−
= 0 | Rin = R1 = 36 kΩ
R1
36kΩ
(This is not a very useful circuit except possibly as an "electronic ground".)
11.24
vO = −
R3
R
51kΩ
51kΩ
v1 − 3 v2 = −
v1 −
v2 = −51v1 − 25.5v2
1kΩ
R1
R2
2kΩ
[
]
[
]
vO (t )= −51(0.01) sin 3770t + −25.5(0.04) sin10000t
vO (t )= (−0.510sin 3770t −1.02sin10000t ) V and v- (t )≡ 0
11-5
11.25
⎡ R
⎤
⎡ R
⎤
⎡ R
⎤
⎡
⎤
v ⎥ + ()
v ⎥ + (0)⎢−
v ⎥ = −0.3750sin 4000πt V
1 ⎢−
1 ⎢−
(a) v = (0)⎢⎣− 2R v ⎥⎦ + ()
⎣ 4R ⎦
⎣ 8R ⎦
⎣ 16R ⎦
O
S
⎡ R
S
⎤
⎡ R
R
S
⎤
⎡ R
s
⎤
⎡
⎤
1 ⎢−
v ⎥ + (0)⎢−
v ⎥ + ()
v ⎥ + ()
v ⎥ = −0.6875sin 4000πt V
1 ⎢−
1 ⎢−
(b) v = ()
⎣ 2R ⎦
⎣ 4R ⎦
⎣ 8R ⎦
⎣ 16R ⎦
O
S
(c) v
S
R
S
S
= − Asin 4000πt V where the amplitude A is in the table below:
O
0000
0
0100
-0.250 V
1000
-0.5000 V
1100
-0.7500 V
0001
-0.0625 V
0101
-0.3125 V
1001
-0.5625 V
1101
-0.8125 V
0010
-0.1250 V
0110
-0.3750 V
1010
-0.6250 V
1110
-0.8750 V
0011
-0.1875V
0111
-0.4375 V
1011
-0.6875 V
1111
-0.9375 V
11.26
Ron =
1
| The worst - case condition for the switches occurs
⎛W ⎞
K ⎜ ⎟(VGS − VTN − VDS )
⎝ L⎠
'
n
for the one with VDS ≠ 0 and VSB = 0. If VREF = 3V and Ron = 0.01(10kΩ)= 100Ω
VD = 3V and VS = 3V
(
10kΩ
= 2.97V | VDS = 0.03V
10kΩ + 100Ω
)
VTN = 1 + 0.5 2.97 + 0.6 − .6 = 1.56V
⎛W ⎞
⎛ 455 ⎞
1
=⎜
⎜ ⎟=
⎟
−5
⎝ L ⎠ 5x10 (100) (5 − 2.97)−1.56 − 0.03 ⎝ 1 ⎠
[
]
⎛W ⎞
⎛ 50 ⎞
1
When the grounded transistor is on, VDS = 0 | ⎜ ⎟ =
=⎜ ⎟
−5
⎝ L ⎠ 5x10 (100)[5 −1− 0] ⎝ 1 ⎠
11.27
(a ) Using Eq. 11.32, A
v
(b) R
in2
(c) v
O
11-6
=
v2
i2
=−
R2
10R
=−
= −10
R
R1
= 11R = 110 kΩ
v1 = 0
| Rin1 =
v1
i1 v
= −10(v1 − v2 )= −30 + 15cos8300πt V
= R = 10 kΩ
2 =0
(d ) v
O
= −30 + 30cos8300πt V
11.28
V1 = 3.2 V | V2 = 3.1 V | VO = −
(3.20 − 2.82)V = 3.80 µA
R2
10R
V1 − V2 )= −
(
(3.2 − 3.1)= −1.00 V
R
R1
10R
= 2.82 V | V- =V += 2.82 V
R + 10R
100kΩ
V −V
3.2 − 2.82 V
V − V 3.1− 2.82 V
I1 = 1 − =
= 3.80 µA | I2 = 2 + =
= 2.80 µA
100
100 kΩ
100kΩ
100kΩ
kΩ
V2
3.1V
Checking : I2 =
=
= 2.82 µA which agrees within roundoff error.
R + 10R 110kΩ
IO =
| V+ = 3.1V
11.29
⎛ R ⎞⎛ R ⎞ ⎛ 10kΩ ⎞⎛ 100kΩ ⎞
Av = −⎜ 4 ⎟⎜1+ 2 ⎟ = −⎜
⎟⎜1+
⎟ = −51
2kΩ ⎠
⎝ R3 ⎠⎝ R1 ⎠ ⎝ 10kΩ ⎠⎝
[
]
vO = Av (v1 − v2 ) = −51(0.1) sin 2000πt + (−51)(2 − 2.1)= (5.1− 5.1sin 2000π ) V
11.30
⎛ R ⎞⎛ R ⎞ ⎛ 20kΩ ⎞⎛ 100kΩ ⎞
Av = −⎜ 4 ⎟⎜1+ 2 ⎟ = −⎜
⎟⎜1+
⎟ = −12
20kΩ ⎠
⎝ R3 ⎠⎝ R1 ⎠ ⎝ 10kΩ ⎠⎝
vO = Av (v1 − v 2 ) = −12(4 − 0.1sin 4000πt − 3.5) = (−6 + 1.2sin 4000π ) V
11.31
1
vO (t ) = −
RC
t
1
∫ v S (τ ) dτ + vC (0 )= − 10kΩ(0.005µF ) ∫ 0.1sin2000πτ dτ + 0 = 0.318cos2000πt V
0
0
t
+
11.32
vO
vS
1 ms
2 ms
t
5V
t
-5 V
0
11.33
(a) A
v
1 ms
2 ms
⎛R ⎞ 1
= −⎜ 2 ⎟
⎝ R1 ⎠ 1+ s
ωH
(b) A (0) = − R
R2
v
1
=−
| Av (0) = −
R2
10kΩ
1
1
=−
= −5 | f H =
=
= 15.9 kHz
R1
2kΩ
2πR2C 2π (10kΩ)(0.001µF )
56kΩ
1
1
= −20.7 | f H =
=
= 28.4 kHz
2.7kΩ
2πR2C 2π (56kΩ)(100 pF )
11-7
11.34
20
(a) Rin = R1 → R1 = 10 kΩ | Av (0)= −10 20 = −10 → R2 = 10R1 = 100 kΩ
C=
1
1
=
= 79.6 pF
2πR2 f H 2π (100kΩ)(20kHz)
(b) If the input resistance specification is most critical,
R1 = 10 kΩ | R2 = 10R1 = 100 kΩ | C = 82 pF → f H = 19.4 kHz
If the cutoff frequency specification is most critical,
C = 82 pF → f H = 19.9 kHz | R1 = 9.76 kΩ | R2 = 10R1 = 97.6 kΩ
11.35
sCVs = −
Vo
R
| T(s) =
11.36
vO (t )= −RC
11.37
Av (s)= −
dvS (t )
dt
Z2 (s)
Z1 (s)
Vo
= −sRC
Vs
= −(100kΩ)(0.02µF )
Z2 = R2
Z1 (s)=
d
2cos3000πt = +37.7sin 3000πt V
dt
R1
sCR1 + 1
Av (s)= −
R2 ⎛
s ⎞
1
⎜1+ ⎟ for ω L =
R1C
R1 ⎝ ω L ⎠
11.38
(a) A
v
=−
R2
120kΩ
=−
= −6.00 | Rin = R1 = 20 kΩ
R1
20kΩ
Note that the 100 - kΩ resistor does not affect the circuit because v- = 0.
(b) A
= 1+
(c) A
=−
v
v
R2
120kΩ
= 1+
= +9.00 | Rin = 75kΩ ∞ = 75 kΩ
15kΩ
R1
R2
0
=−
= 0 | Rin = R1 = 160 kΩ
R1
36kΩ
(This is not a very useful circuit except possibly as an "electronic ground".)
11-8
11.39
Equating currents at the inverting input node (virtual ground):
20kΩ
vS
vO
=−
10kΩ
100kΩ + 20kΩ 100kΩ 100kΩ + 20kΩ
(
)
116.7kΩ(120kΩ)
vO
=−
= −70.0
vS
10kΩ(20kΩ)
Av =
Rin = 10 kΩ
Rout = 0
11.40
The inverting terminal of the op amp represents a virtual ground (0 V).
V − (−V ) 0 − (−10)
(a) I = I = R = 10 = 1 A
(b) Saturation requires V ≥ V − V where V = V
2()
1
2I
V −V =
=
=2.83V → V ≥ 2.83 V
−
O
EE
D
DS
GS
TN
DS
DD
− V− = VDD and
D
GS
TN
(c) P
R
DD
0.25
Kn
= I 2 R = ()
1 10 = 10W. So the resistor must dissipate 10 W.
2
A 15 - W resistor would provide a reasonable safety margin.
11.41
The inverting terminal of the op amp represents a virtual ground (0 V).
⎛ β ⎞⎡V− − (−VEE )⎤ ⎛ 30 ⎞⎡ 0 − (−15)⎤
⎥ = ⎜ ⎟⎢
⎥ = 0.484 A
(a) IO = IC = α F I E = ⎜ F ⎟⎢
R
30
31
⎝
⎠
⎝ 1+ β F ⎠⎢⎣
⎥⎦
⎢⎣
⎥⎦
(b) VO = V− + VBE = 0 + VBE = VBE = VT ln
IC
0.484
= 0.025V ln −13 = 0.730V
IS
10
(c) Forward - active region operation requires VCE ≥ VBE but VCE = VCC − V− = VCC
Therefore VCC ≥ 0.730 V.
(d ) PR = I 2 R = (0.484) 30 = 7.03W. So the resistor must dissipate 7.03 W.
2
A 10W resistor would provide a reasonable safety margin.
PD = ICVCE + I BVBE = 0.484(15)+
0.484
0.730 = 7.27 W.
30
11-9
11.42
Z2 (s)
(a) A (s)= − Z
v
1
(b)
(s)
Z2 =
1
sC
Z1 (s) = R
Av (s )= −
1
, an inverting integrator.
sRC
VS − V+ VO − V+
V
R1
+
= sCV+ and V+ = V− = VO
= O
R
KR
R1 + KR1 1+ K
Combining these expressions yields : A V (s)=
Vo
1+ K
=+
, a noninverting integrator.
sRC
Vs
11.43
R1
V
V − V+ Vo − V+
= o | s
+
= sCV+ | V+ = V−
R1 + 9R1 10
R
10R
⎛
11 ⎞ V
V
100
Vs = ⎜ sCR + ⎟ o | o =
- - The circuit becomes a low - pass filter.
⎝
Vs 10sCR + 11
10 ⎠ 10
-------Note : The pole moves to the right half plane if the tolerances go the other way.
V
99
For example, if KR1 = 10R1 and KR = 9R, o =
Vs 9sCR −1
V− = Vo
11.44
(a) Applying ideal op-amp assumption 1, the voltage at the top end of R is v1 and the voltage at
the bottom end of R is v2. Applying op-amp assumption 2, the current iO must also equal the
current in R, so
v − v2
iO = 1
R
+
v O1
vx
v1
i
+ x
R
(b) i X =
0V − 0V
v
= 0 | Rout = X = ∞
R
iX
v O1 = A(0 − (v O1 + v X ))
v2
A
v
| v1 = v O1 + v X = X
1+ A
1+ A
v − v2
vX
v
iX = 1
=
| R out = X = (1+ A)R
R
iX
(1+ A)R
v O1 = − v X
v O2
+
11-10
v O2 = A(0 − v O2 ) ⇒ v O2 = 0
11.45
v +v
2
v =
-
O
2
i
v
R
1
R
1
3
i
v
5
+
R
v
R
2
4
R
2
v +v
2
v+ =
O
2
v
O
v
+
i
Z
O
O
L
(a)
iO =
v5 − vO
R
| v 5 = v− − i R3 = v− −
v− =
v2 + vO
2
and v 5 = v 2 − v1 + v O | i O =
v
X
i
2
R
R
1
v1 − v−
R3 = 2v− − v1 since R1 = R3
R1
v 2 − v1 + v O − v O v 2 − v1
=
R
R
3
v
5
i
v
+
R
R
2
X
4
R
v
X
2
v
X
+
i
R = R = R = R
1
2
3
4
v
X
(b)
iX =
vX − v5 vX − vX
v
=
= 0 | Rout = X = ∞
R
R
iX
X
+
-
(an ideal current source)
11-11
11.46
(a) Using voltage division since i
+
= 0, v 2 = v 4 + 6
4.99kΩ
4.99kΩ + 5.00kΩ
4.99kΩ
v 2 = v 4 + (6 − v 4 )
= 0.5005v 4 + 2.997V
4.99kΩ + 5.00kΩ
4 − v1
Since v id = 0, v 1 = v 2 and v 5 = v 1 −
(5.01kΩ)
5kΩ
Solving for v 5 yields v 5 = 1.992V + 1.002v 4
io =
v5 − v4
= 199µA + 2x10−7 v 4
10kΩ
v 4 is unknown; let us assume 2x10-7 v 4 << 199x10−6 A
which requires v 4 << 995V . So for v 4 < 100V,which should almost
always be true in transistor circuits, i O = 199µA.
For ZL = 10 kΩ, v4 = 1.99 V, v2 = 3.99 V, v1 = 3.99 V, v5 = 3.99 V
Note that v5 - v4 = 2 V = (6V - 4V)
5.01 k Ω
v1
i
R
v
4V
1
i
R
1
3
v
5k Ω
5
+
R
R
2
6V
v
2
5k Ω
v2
4
v4
10 k Ω
4.99 k Ω
v4
+
i
Z
11-12
L
O
11.46 (b)
v
R
i
5.01 k Ω
1
i
R
1
3
v
5k Ω
5
+
R
R
2
5k Ω
v
10 k Ω
4
R
4.99 k Ω
2
v
X
+
i
v
X
Rout =
vX
v − v5
and i X = X
iX
10kΩ
X
+
-
So we need to find iX, and hence v5, in terms of vX
5.00kΩ
= 0.5005v X
4.99kΩ + 5.00kΩ
v
v 5 = v 1 + i (5.01kΩ)= v 1 + 1 (5.01kΩ) = 2.002 v 1 = 1.002 v x
5kΩ
0.002 v X
v − v 5 v X −1.002 v X
iX = X
=
=−
and Rout = −5 MΩ ! A negative output resistance!
10kΩ
10kΩ
10kΩ
v1 = v2 = vX
11.47
IL = IS +
VO − I L RL
R1
VO = I L RL + I S R2
⎡
⎛ R ⎞⎤
VS = I S R2 + I L RL = I S ⎢ R2 + RL ⎜1+ 2 ⎟⎥
⎝ R1 ⎠⎦
⎣
⎛ R⎞
I L = I S ⎜1+ 2 ⎟
⎝ R1 ⎠
⎛ R⎞
Rin = R2 + RL ⎜1+ 2 ⎟
⎝ R1 ⎠
Apply test voltage vx to the output with I S = 0. Then v+ = vx ,
vO = vx , and ix = i− +
vx − vx
= 0 + 0 = 0. Rout = ∞
R1
For finite gain : Rout =
v X − vO =
vX
iX
where i X =
v X − vO
R1
and vO = v X
A
1+ A
vX
and Rout = R1 (1+ A)
1+ A
11.48
11-13
Using the result from Prob. 11.47,
⎛ R⎞
⎛
1kΩ ⎞
Rin = R2 + RL ⎜1+ 2 ⎟ = 1kΩ + 3.6kΩ⎜1+
⎟ = 4.96 kΩ
⎝ 10kΩ ⎠
⎝ R1 ⎠
Rout = ∞
11.49
Applying op-amp assumption 1 to the circuit below, the voltage at the top of R2 is vO2, and
applying op-amp assumption 2,
vS
v
= − O2
R1
R2
or v O2 = − v S
R2
R1
Since the op-amp input currents are zero, and
i=
vS
,
R1
⎛R R ⎞
v O1 = −iR2 − iR3 = −⎜ 2 + 3 ⎟ v S
⎝ R1 R1 ⎠
Alternatively, the voltage at the bottom of R2 is zero, so
⎛ R ⎞
⎛ R ⎞⎛ R ⎞
⎛R R ⎞
v O1 = ⎜1+ 3 ⎟ v O2 = ⎜1+ 3 ⎟⎜− 2 ⎟v S = −⎜ 2 + 3 ⎟v S
⎝ R2 ⎠
⎝ R2 ⎠⎝ R1 ⎠
⎝ R1 R1 ⎠
+
+
i
v
O1
-
R
3
i
R
R
v
1
2
i
+
+
S
v
-
11.50
Rin =
vS
iS
11-14
iS =
vS − vO
15kΩ
⎛ 30kΩ ⎞
vO = ⎜1+
⎟v S = 16vS
2kΩ ⎠
⎝
iS =
O2
vS −16vS −15vS
=
15kΩ
15kΩ
Rin = -1kΩ
11.51
An n-bit DAC requires (n+1) resistors. Ten bits requires 11 resistors.
210 R 210
=
or 1024 :1
R
1
A wide range of resistor values is required but it could be done.
For R = 1 kΩ, 1024R = 1.024 MΩ.
11.52
⎛ R
⎛ 1 1⎞
R⎞
+
VO = −VREF ⎜
⎟ = −3.0⎜ + ⎟ = −1.1250 V
⎝ 4R 8R ⎠
⎝ 4 8⎠
⎛ R
⎛1 1 ⎞
R ⎞
+
VO = −VREF ⎜
⎟ = −3.0⎜ + ⎟ = −1.6875 V
⎝ 2R 16R ⎠
⎝ 2 16 ⎠
0000
0.0000 V
1000
-1.5000 V
0001
-0.1875 V
1001
-1.6875 V
0010
-0.3750 V
1010
-1.875 V
0011
-0.5625 V
1011
-2.0625 V
0100
-0.7500 V
1100
-2.2500 V
0101
0.9375 V
1101
-2.4375 V
0110
-1.1250 V
1110
-2.6250 V
0111
-1.3125 V
1111
-2.8125 V
11.53
Taking successive Thévenin equivalent circuits at each ladder node yields:
0001
0010
0100
1000
VTH
RTH
vO
VREF/16
VREF/8
VREF/4
VREF/2
R
R
R
R
-0.3125 V
-0.6250V
-1.250 V
-2.500 V
11.54
The ADC code is equivalent to 2-1 + 2−3 + 2−5 + 2−7 + 2−9 + 2−10 + 2−13 x VFS = 0.66711426 x VFS
(
)
The ADC input may be anywhere in the range :
⎛
⎞
1
⎜0.66711426 ± LSB⎟ x VFS = 3.4154625V ± 0.15625mV | 3.415469V ≤ VX ≤ 3.415781V
2
⎝
⎠
11.55
11-15
(a) 1 LSB = 2
2V
= 1.90735 µV
bits
20
1.63V 20
2 = 854589.4 bits → 85458910 = 110100001010001111012
2V
0.997003V
(c) 2V 220 = 522716.7 bits → 52271710 = 011111111001110111012
(b)
11.56
The time corresponding to 1 LSB is TLSB =
0.2s
= 190.7 ns | 0.1TLSB = 19.1 ns
2 20 bits
11.57
VO (ω ) =
1
RC
TT
∫ v (t )dt =
0
For θ = 0, VO (ω ) =
1
RC
TT
∫ VM cos(ωt + θ )dt =
0
VM sin ωTT VM TT
=
RC ω
RC
VM ωt +θ
V
cos x dx = M [sin(ωTT + θ ) − sin θ ]
∫
ωRC θ
ωRC
⎛ sin ωTT ⎞
⎟
⎜
⎝ ωTT ⎠
11.58 Combine amplifiers A & B; Combine amplifiers B & C
11.59
R
+
v
1
=0
OUT
-10 v
-
v
1
-
+
v
2
O
-
10 k Ω
+
+
v
=0
OUT
-5 v
2
-
24 k Ω
R
+
+50 v
1
24 k Ω
v
1
O
-
Av = 50, Rin = 24 kΩ, Rout = 0
11.60
⎛ 240kΩ ⎞⎛ 50kΩ ⎞
Av = ⎜−
⎟⎜−
⎟ = +50 | Rin = RinA = 24 kΩ | Rout = RoutB = 0
⎝ 24kΩ ⎠⎝ 10kΩ ⎠
11.61
⎛ 390kΩ ⎞⎛ 100kΩ ⎞
Av = ⎜1+
⎟⎜1+
⎟ = +48.0 | Rin = RinA = ∞ | Rout = RoutB = 0
⎝
47kΩ ⎠⎝
24kΩ ⎠
11.62
11-16
⎛ 120kΩ ⎞⎛ 120kΩ ⎞
= ⎜1+
⎟⎜−
⎟ = −42.0 | Rin = RinA = ∞ | Rout = RoutB = 0
20kΩ ⎠⎝ 20kΩ ⎠
⎝
⎛ 120kΩ ⎞⎛ 120kΩ ⎞
(b) Av = ⎜⎝− 20kΩ ⎟⎠⎜⎝1+ 20kΩ ⎟⎠ = −42.0 | Rin = RinA = 20 kΩ | Rout = RoutB = 0
(a) A
v
11.63
+
+
v
-20 v
1
-
v
1
-20 v
2
-
2k Ω
+
+
-
-8000 v
v
-20 v
3
2k Ω
+
2
kΩ
v
3
O
-
+
1
v
O
1
-
v
2
2k Ω
-
⎛ 40kΩ ⎞⎛ 40kΩ ⎞⎛ 40kΩ ⎞
3
Av = ⎜−
⎟⎜ −
⎟⎜−
⎟ = (−20) = −8000 | Rin = RinA = 2 kΩ | Rout = RoutC = 0
⎝ 2kΩ ⎠⎝ 2kΩ ⎠⎝ 2kΩ ⎠
11.64
⎛ 40kΩ ⎞⎛ 40kΩ ⎞⎛ 40kΩ ⎞ ⎛ 40kΩ ⎞3
Av = ⎜−
⎟⎜ −
⎟⎜−
⎟ = ⎜−
⎟ = −1080 | Rin = RinA = 3.9 kΩ | Rout = RoutC = 0
⎝ 3.9kΩ ⎠⎝ 3.9kΩ ⎠⎝ 3.9kΩ ⎠ ⎝ 3.9kΩ ⎠
11.65
40
⎛ 40kΩ ⎞⎛ 40kΩ ⎞⎛ 40kΩ ⎞ ⎛ 40kΩ ⎞3
20
|
A
=
−10
= -100
−
−
=
−
Av = ⎜−
⎟⎜
⎟⎜
⎟ ⎜
⎟
v
R ⎠⎝
R ⎠⎝
R ⎠ ⎝
R ⎠
⎝
40kΩ
= 8.62 kΩ | Rin = RinA = 8.62 kΩ | Rout = RoutC = 0
R= 3
100
11.66
(a) A
v
⎛ 470kΩ ⎞⎛ 150kΩ ⎞⎛ 270kΩ ⎞
= ⎜−
⎟⎜−
⎟⎜ −
⎟ = −1500 | Rin = RinA = 47 kΩ | Rout = RoutC = 0
⎝ 47kΩ ⎠⎝ 15kΩ ⎠⎝ 18kΩ ⎠
(b) Moving from left to right,
0.010 V, 0 V, -10(.01) = -0.100 V, 0 V, -10(-0.100) = +1.00 V, 0 V, -15(1.00) = -15.0 V,
The eighth node is the ground node at 0 V.
11-17
11.67
⎛ 470kΩ ⎞⎛ 150kΩ ⎞⎛ 270kΩ ⎞
Avnom = ⎜−
⎟⎜ −
⎟⎜−
⎟ = −1500
⎝ 47kΩ ⎠⎝ 15kΩ ⎠⎝ 18kΩ ⎠
⎡ 470kΩ ⎛ 1.05 ⎞⎤ ⎡ 150kΩ ⎛ 1.05 ⎞⎤ ⎡ 270kΩ ⎛ 1.05 ⎞⎤
Avmax = ⎢−
⎟⎥ ⎢−
⎟⎥ ⎢−
⎟⎥ = −2025
⎜
⎜
⎜
⎣ 47kΩ ⎝ 0.95 ⎠⎦ ⎣ 15kΩ ⎝ 0.95⎠⎦ ⎣ 18kΩ ⎝ 0.95 ⎠⎦
⎡ 470kΩ ⎛ 0.95⎞⎤ ⎡ 150kΩ ⎛ 0.95⎞⎤ ⎡ 270kΩ ⎛ 0.95⎞⎤
Avmin = ⎢−
⎟⎥ ⎢−
⎟⎥ ⎢−
⎟⎥ = −1011
⎜
⎜
⎜
⎣ 47kΩ ⎝ 1.05 ⎠⎦ ⎣ 15kΩ ⎝ 1.05 ⎠⎦ ⎣ 18kΩ ⎝ 1.05 ⎠⎦
Rinnom = RinA = 47 kΩ
Rinmax = 47kΩ(1.05)= 49.4 kΩ
Rinmin = 47kΩ(0.95)= 44.7 kΩ
nom
max
min
Rout
= Rout
= Rout
= RoutC = 0
11.68
⎛ 39kΩ ⎞⎛ 39kΩ ⎞⎛ 39kΩ ⎞
Av = ⎜1+
⎟⎜1+
⎟⎜1+
⎟ = +2744 | Rin = 1MΩ ∞ = 1 MΩ | Rout = RoutC = 0
3kΩ ⎠⎝
3kΩ ⎠
3kΩ ⎠⎝
⎝
2.00 mV , 2.00 mV , 28.0 mV , 28.0 mV , 392 mV , 392 mV , 5.49 V , 0 V (Ground node)
11.69
⎛ 39kΩ ⎞⎛ 39kΩ ⎞⎛ 39kΩ ⎞
Av = ⎜1+
⎟⎜1+
⎟⎜1+
⎟ = +64000 | Rin = 2 MΩ ∞ = 2 MΩ | Rout = RoutC = 0
1kΩ ⎠⎝
1kΩ ⎠
1kΩ ⎠⎝
⎝
2.00 mV , 2.00 mV , 80.0 mV , 80.0 mV , 3.20V , 3.20 V , 128 V (not realistic), 0 V
11.70
3
nom
v
⎛ 39kΩ ⎞
= ⎜1+
⎟ = +2744
3kΩ ⎠
⎝
max
v
⎡⎛ 39kΩ ⎞⎛ 1.02 ⎞⎤
= ⎢⎜1+
⎟⎜
⎟⎥ = +3094
3kΩ ⎠⎝ 0.98 ⎠⎦
⎣⎝
A
3
A
Rinnom = 1MΩ ∞ = 1 MΩ
3
max
v
A
⎡⎛ 39kΩ ⎞⎛ 0.98 ⎞⎤
= ⎢⎜1+
⎟⎜
⎟⎥ = +2434
3kΩ ⎠⎝ 1.02 ⎠⎦
⎣⎝
Rinmax = 1MΩ(1.02) = 1.02 MΩ
Rinmin = 1MΩ(0.98)= 980 kΩ
nom
max
min
Rout
= Rout
= Rout
= RoutC = 0
11.71
⎛ 150kΩ ⎞⎛ 420kΩ ⎞⎛ 100kΩ ⎞
Av = ⎜1+
⎟⎜ −
⎟⎜1+
⎟ = −1320 | Rin = 75kΩ ∞ = 75 kΩ | Rout = RoutC = 0
20kΩ ⎠
15kΩ ⎠⎝ 21kΩ ⎠⎝
⎝
5.00 mV , 5.00 mV , 55.0 mV , 0 V, -1.10 V , -1.10V , - 6.60 V , 0 V (Ground node)
11-18
11.72
⎛ 150kΩ ⎞⎛ 420kΩ ⎞⎛ 100kΩ ⎞
Avnom = ⎜1+
⎟⎜ −
⎟⎜1+
⎟ = −1320
20kΩ ⎠
15kΩ ⎠⎝ 21kΩ ⎠⎝
⎝
⎡⎛ 150kΩ ⎞⎛ 1.01 ⎞⎤ ⎡⎛ 420kΩ ⎞⎛ 1.01 ⎞⎤ ⎡⎛ 100kΩ ⎞⎛ 1.01 ⎞⎤
Avmax = ⎢⎜1+
⎟⎜
⎟⎥ ⎢⎜−
⎟⎜
⎟⎥ ⎢⎜1+
⎟⎜
⎟⎥ = −1402
20kΩ ⎠⎝ 0.99 ⎠⎦
15kΩ ⎠⎝ 0.99 ⎠⎦ ⎣⎝ 21kΩ ⎠⎝ 0.99 ⎠⎦ ⎣⎝
⎣⎝
⎡⎛ 150kΩ ⎞⎛ 0.99 ⎞⎤ ⎡⎛ 420kΩ ⎞⎛ 0.99 ⎞⎤ ⎡⎛ 100kΩ ⎞⎛ 0.99 ⎞⎤
Avmin = ⎢⎜1+
⎟⎜
⎟⎥ ⎢⎜−
⎟⎜
⎟⎥ ⎢⎜1+
⎟⎜
⎟⎥ = −1243
20kΩ ⎠⎝ 1.01 ⎠⎦
15kΩ ⎠⎝ 1.01 ⎠⎦ ⎣⎝ 21kΩ ⎠⎝ 1.01 ⎠⎦ ⎣⎝
⎣⎝
Rinmax = 75kΩ(1.01)= 75.8 kΩ
Rinnom = RinA = 75 kΩ
Rinmin = 75kΩ(0.99) = 74.3 kΩ
nom
max
min
Rout
= Rout
= Rout
= RoutC = 0
11.73
(a) C
2
= C = 0.005µF | C1 = 2C = 0.01µF | R =
1
For Q =
(b) For f
2
o
: Av (s) =
1
2ωO C
=
1
2 (40000π )0.005µF
= 1.13 kΩ
1
ω 2o
|
A
0
=
1
|
A
j
ω
=
→ ω H = ω o! | f H = 20 kHz
(
)
(
)
o
s2 + 2ω o s + ω 2o
2
= 40kHz : C ' →
C
: C1' = 0.005 µF | C2' = 0.0025 µF | R = 1.13 kΩ
2
11.74
(a)
r1=1130; r2=1130; c1=1e-8; c2=5e-9;
wo=1/sqrt(r1*r2*c1*c2)
n=wo*wo;
d=[1 2/(r1*c1) wo*wo];
bode(n,d)
0
Gain dB
-10
-20
-30
-40
10 4
10 5
Frequency (rad/sec)
10 6
0
Phase deg
-50
-100
-150
-200
10
4
10 5
Frequency (rad/sec)
10 6
11-19
(b)
*PROBLEM 11.74 - Low-pass Filter
R_R1
$N_0002 $N_0001 1.13k
C_C1
$N_0002 $N_0003 0.01UF
V_V1
$N_0004 0 18V
V_V2
0 $N_0005 18V
R_R2
$N_0006 $N_0002 1.13k
V_VS
$N_0006 0 DC 0V AC 1V
C_C2
0 $N_0001 0.005UF
X_U1
$N_0001 $N_0003 $N_0004 $N_0005 $N_0003 uA741
.AC DEC 40 1 1MEG
.PRINT AC IM(VS) IP(VS) VDB(6) VP(6)
.PROBE I(VS) V(6)
.END
1
-0d
2
20
-100d
0
-200d
-20
-300d
>>
(c) The magnitude response is very similar to the ideal case. However, note bumps in the phase
plot and the excess phase shift as one approaches the fT of the amplifier. In this case, it is not
causing a problem, but for higher gain filters the situation would be different. See Probs. 11.82
& 11.84.
11-20
11.75
Using Eq. 11.47 : V1 (s)=
G1VS (s)(sC2 + G2 )
| IS = G1 (VS − V1 )
s 2C1C2 + sC2 (G1 + G2 )+ G1G2
⎛
⎞
G1(sC2 + G2 )
s 2C1C2 + sC2G2
⎟ = G1VS 2
IS = G1VS ⎜⎜1− 2
⎟
s C1C2 + sC2 (G1 + G2 )+ G1G2
⎝ s C1C2 + sC2 (G1 + G2 )+ G1G2 ⎠
s C1C2 + sC2 (G1 + G2 )+ G1G2
V
ZS (s)= S = R1
= R1
IS
s2C1C2 + sC2G2
2
s2 + s
ωo
Q
+ ω 2o
⎛
1 ⎞
s⎜ s +
⎟
R2C1 ⎠
⎝
1
ωo 1 ⎛ 1 1 ⎞
|
= ⎜ + ⎟
Q C1 ⎝ R1 R2 ⎠
R1 R2C1C2
ω 2o =
11.76
r1=2260; r2=2260; c1=2e-8; c2=1e-8;
wsq=1/(r1*r2*c1*c2);
n=r1*[1 2/(r1*c1) wsq];
d=[1 1/(r1*c1) 0];
bode(n,d)
200
Gain dB
150
100
50
0
10
-1
10 0
10 1
10 2
10 3
Frequency (rad/sec)
10 4
10 5
10
6
-1
10 0
10 1
10 2
10 3
Frequency (rad/sec)
10 4
10 5
10
6
0
Phase deg
-50
-100
-150
-200
10
11-21
11.77
G1VS = (sC1 + G1 + G2 )V1 − s(KC1 + G2 )V2
0 = −G2 V1 + (sC2 + G2 )V2
VO = KV2 |
ωo =
VO
K
= 2
VS s R1 R2C1C2 + s R1C1 (1− K )+ C2 (R1 + R2 ) + 1
1
[
| Q=
R1 R2C1C2
]
ω 3o
⎡
⎤
1
⎢1− K +
⎥
⎢ R2C2
⎥
R
R
C
1
2
1⎦
⎣
(
For R1 = R2 = R and C1 = C2 = C, ω o =
)
1
RC
11.78
ωo =
1
R1R2C1C2
SRω1o = −
| SRω1o =
R1 ∂ω o R1
=
ω o ∂R1 ω o
Q=
ω 2o
3− K
SKQ =
K
3− K
3
⎛ 1⎞
1
R ⎛ 1 ωo ⎞
1
−
⎜− ⎟(R1 ) 2 = 1 ⎜ −
⎟=−
2
ω o ⎝ 2 R1 ⎠
R2C1C2 ⎝ 2 ⎠
1
1
| By symmetry, SCω1o = −
2
2
11.79
For C1 = C = C2 : ω o =
SQω o =
1
C R1R2
| Q=
1 R2
1
| ωo =
2 R1
2R1CQ
Q ∂ω o Q ⎛
1 ⎞ Q ⎛ ωo ⎞
ω
=
⎟=
⎜−
⎜ − ⎟ = −1 | SQ o = −1
ω o ∂Q ω o ⎝ 2R1CQ 2 ⎠ ω o ⎝ Q ⎠
11.80
As noted in Design Example 11.8, the maximally flat response corresponds to Q =
1
2
.
ω 2o
1
: AV (s)= 2
| A(0)= 1 | A(jω o ) =
→ ω H = ω o | ∴ f H = 1 kHz
For Q =
2
s + 2ω o s + ω o
2
2
1
R1 = R = R2 and C1 = 2C2 = 2C yields Q =
RC =
1
2π 2 (1000)
1
2
| ωo =
1
2R2C 2
=
1
2RC
= 1.125 x 10−4 | For C = 0.001 µF, R = 112.5 kΩ. The
nearest 5% value is 110 kΩ. The nearest 1% value is 113 kΩ. Using 1% values,
C1 = 0.002 µF | C2 = 0.001 µF | R1 = R2 = 113 kΩ
11-22
11.81
Using the R1 = R2 = R and C1 = C2 = C case, A HP (s) = K
ωo =
1
RC
| Q=
1
3−K
| Q =1→ K = 2
s2
s2 + s
| A HP (s) = 2
ωo
Q
+ +ω o2
s2
s2 + sω o + +ω o2
We need to find the relationship between ω L and ω o .
ω L2
2
=2
2
(ω
2
o
−ω
) + (ω ω )
2 2
L
2
L
→ ω o4 − 2ω o2ω L2 + ω L4 + ω o2ω L2 = 2ω L4 | ω L4 + ω o2ω L2 − ω o4 = 0
o
−ω o2 ± ω o4 + 4ω o4
5 −1
→ ω L2 = ω o2
| ω L = 0.7862ω o | 2π (20kHz) = 0.7862ω o
2
2
1
ω o = 1.599x10 5 | RC =
= 6.256x10−6 | For C = 270 pF, R = 23.17 kΩ
ω L2 =
ωo
The nearest 1% resistor value is 23.2 kΩ.
Final design : C1 = C2 = 270 pF | R1 = R2 = 23.2 kΩ
11.82
1
(a) Q = 2
BW =
1 200kΩ 10
R2
=
=
Rth 2 1kΩ
2
fo
= 7.23 kHz
Q
Av (f o ) =
| fo =
1
(
2π 10 2x10
3
5
)(2.2x10 )
−10
2
= 51.2 kHz
1 R2
= 100 or 40 dB
2 Rth
3.3kΩ
= 3.3 | Rth = 3.3(1kΩ) = 3.3 kΩ | R2 = 3.3(200kΩ)= 660 kΩ
1kΩ
220 pF
= 66.7 pF
C1 = C2 =
3.3
2f
220 pF
(c) KF = f o = 2 | C1 = C2 = 2 = 110 pF | Rth = 1 kΩ | R2 = 200 kΩ
o
(b) K
M
=
11-23
11.83
(a) BW =
f o 1000Hz
1 R2
R
=
= 200 Hz | C1 = C2 = C | Q =
= 5 → 2 = 100
Q
Rth
5
2 Rth
Choose Rth = 1 kΩ → R2 = 100 kΩ | C =
ωo =
ωo
1
1
=
= 0.0159 µF
Rth R2 2π (10 3 )(10 4 )
1
: Checking the nearest standard values :
10Rth C
C = 0.015 µF → R th = 1.06kΩ - not good; C = 0.01 µF → Rth = 1.6 kΩ | R2 = 160 kΩ
Choose R1 = 1.6 kΩ with R3 = ∞.
(b) C ' =
11-24
C
= 0.004 µF | R1 = 1.6 kΩ | R2 = 160 kΩ | R3 = ∞
2.25
11.84
(a)
r1=1000; r2=2e5; c=2.2e-10;
wo=1/sqrt(r1*r2*c*c); q=sqrt(r2/r1)/2;
n=[-2*q*wo 0]; d=[1 wo/q wo*wo];
w=logspace(4,7,300);
bode(n,d,w)
40
Gain dB
30
20
10
0
-10
10 4
10
5
10
5
10 6
Frequency (rad/sec)
10 7
-50
Phase deg
-100
-150
-200
-250
-300
10 4
10
6
10
7
Frequency (rad/sec)
(b) SPICE Simulation Results
1
40
2
0d
0
-200d
-40
>>
-400d
SPICE yields: fo = 38.9 kHz, Q = 8.1, Center frequency gain = 38.8 dB. These values are off
due to the finite bandwidth of the op-amp and its excess phase shift at the center frequency of the
filter. The center frequency is substantially shifted.
11-25
11.85
(a) BW
=
ωo
Q
=
1
rad
rad
| ω L = 0.833
| ω H = 1.167
3
s
s
1
rad
rad
1
| ω 'o = ω o = 1
| Q' =
= 4.65
s
s
0.215
1
1 R2
1
C1 = C2 = C : ω o =
| Q=
|
= 2Qω o
2 Rth
Rth C
C Rth R2
BW ' = BW 2 2 −1 = 0.215
⎛
⎞2
⎞2
⎞2
⎛
⎛
2
⎛ Rth ⎞ ⎜ −2Qsω o ⎟ ⎛ Rth ⎞ ⎜ −6s ⎟
⎜ −6s ⎟
b
A
s
=
| For R3 = ∞, ABP (s) = ⎜
⎟ =⎜ ⎟ ⎜
( ) BP ( ) ⎜ ⎟ ⎜
⎟
s ⎟
⎝ R1 ⎠ ⎜⎜ s2 + s ω o + 1⎟⎟ ⎝ R1 ⎠ ⎜ s 2 + s + 1⎟
⎜ s 2 + + 1⎟
⎝
⎝
Q
3 ⎠
3 ⎠
⎝
⎠
2
11.86
n=conv([-6 0],[-6,0]); d=conv([1 1/3 1],[1 1/3 1]); bode(n,d)
60
50
Gain dB
40
30
20
10
0
-10
10
-1
10 0
Frequency (rad/sec)
10 1
-1
10 0
Frequency (rad/sec)
10
200
150
Phase deg
100
50
0
-50
-100
-150
-200
10
11-26
1
11.87
Using normalized frequency and R3 = ∞ : 5 kHz → ω o = 1 and 6 kHz → ω o = 1.1
⎛ −10s ⎞⎛
⎞
−12s
120s 2
ABP (s) = ⎜ 2
⎟⎜
⎟=
⎝ s + 0.2s + 1⎠⎝ s 2 + 0.24s + 1.44 ⎠ s 4 + 0.44s3 + 2.484s2 + 0.528s + 1.44
At the new center frequency s = jω o , − 0.44ω 3o + 0.528ω o = 0 → ω o = 1.095
and A(jω o ) = 1429 = 63.1 dB
The bandwidth points can be found using MATLAB:
w=linspace(.9,1.5,250);
[m,p,w]=bode([120 0 0],[1 .44 2.484 .528 1.44],w);
20*log10(max(m))
ans = 63.098
((20*log10(a))>60.098).*(w.');
From this last vector one can easily find: ωo = 1.095 or fo = 5.48 kHz, ωL = 0.970 or fo = 4.85
kHz, ωH = 1.237 or fo = 6.19 kHz, BW = 1.34 kHz, Q = 4.09
11-27
11.88
w1=2*pi*5000; q1=5; w2=2*pi*6000; q2=5;
n1=[-2*q1*w1 0]; d1=[1 w1/q1 w1*w1];
n2=[-2*q2*w2 0]; d2=[1 w2/q2 w2*w2];
n=conv(n1,n2); d=conv(d1,d2);
w=logspace(4,5,100);
bode(n,d,w)
70
60
Gain dB
50
40
30
20
10
0
10 4
Frequency (rad/sec)
10 5
200
150
Phase deg
100
50
0
-50
-100
-150
-200
10
4
Frequency (rad/sec)
10 5
11.89
Using ABP = 20 dB at the center frequency : Rin = R1 = 10 kΩ | 10 = KQ =
R2 = 100 kΩ | K =
10
1
1
= 2 | R = KR1 = 20 kΩ | C =
=
= 0.0133 µF.
ω o R 2π (600Hz)20kΩ
Q
11.90 Q is independent of C in the Tow-Thomas biquad.
11-28
R2
R1
SQ
C = 0.
11.91
vO
T/2
T
3T/2
2T
0
t
-5 V
11.92
The waveform going into the low - pass filter is the same as that in Prob. 11.91
⎛ 8.2kΩ ⎞
except the amplitude will be VM = 1V ⎜−
⎟ = −3.037 V.
⎝ 2.7kΩ ⎠
1T
⎛ 10kΩ ⎞ 2 2 (−3.037)
The average value of the waveform is V = ⎜ −
= +0.759 V.
⎟
T
⎝ 10kΩ ⎠
11.93
The Fourier series converges very rapidly since only the even terms exist for n ≥ 2 and the terms
decrease as 1/n2. Thus the RMS value will be dominated by the first term (n = 1).
⎛ ω ⎞2
π
1
ω 120π
2
Require :
≤
0.01
|
50
π
≤
1+
|
ω
≥
=
= 2.40 Hz
(
)
⎜
⎟
o
2
157 157
⎝ωo ⎠
⎛ ω ⎞2
1+ ⎜ ⎟
⎝ωo ⎠
11.94
vO
0.5 V
t
0
T/2
0
11.95
T
3T/2
2T
vO
1.0 V
t
0
0
T/2
T
3T/2
2T
11-29
11.96
*Figure P11.94 - RECTIFIER
VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0)
R1 1 2 10K
R2 4 5 10K
R3 5 6 10K
R4 2 4 10K
R5 1 5 20K
D1 3 2 DIODE
D2 4 3 DIODE
EOP1 3 0 0 2 1E5
EOP2 6 0 0 5 1E5
.MODEL DIODE D IS=1E-12A
.TRAN .01M 8M
.PRINT TRAN V(6)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
11.97
*Figure P11.95 - RECTIFIER
VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0)
R1 0 2 10K
R2 4 5 10K
R3 5 6 20K
R4 2 4 10K
D1 3 2 DIODE
D2 4 3 DIODE
EOP1 3 0 1 2 1E5
EOP2 6 0 1 5 1E5
.MODEL DIODE D IS=1E-12A
.TRAN .01M 8M
.PRINT TRAN V(6)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
11.98
−V
V
V
V1
= I S exp o1 | VO1 = −VT ln 41
| VO 2 = −VT ln 42
10kΩ
VT
10 I S
10 I S
⎛
V
V ⎞
VV
VO3 = −(VO1 + VO2 )= VT ⎜ ln 41 + ln 42 ⎟ = VT ln 18 2 2
10 I S ⎠
10 I S
⎝ 10 I S
⎛ VV ⎞
VV
V
VO = −10 4 I D = −10 4 I S exp D = −10 4 I S exp⎜ ln 18 2 2 ⎟ = − 14 2
VT
10 I S
⎝ 10 I S ⎠
11-30
11.99
Simplify the circuit by taking a Thevenin equivalent of the 5V source and two 10kΩ
10kΩ
resistors : VTH = 5V
= 2.5V | RTH = 10kΩ 10kΩ = 5kΩ
10kΩ + 10kΩ
100kΩ
5kΩ
+5
= 2.62V
VO = 5V − Using superposition : V+ = 2.5
100kΩ + 5kΩ
100kΩ + 5kΩ
100kΩ
= 2.38V VN = 2.62 − 2.38 = 0.24V
VO = 0V : V+ = 2.5
100kΩ + 5kΩ
11.100
4.3kΩ
= −0.993 V
4.3kΩ + 39kΩ
4.3kΩ
= 0.993 V
VO = 10V : V+ = 10
4.3kΩ + 39kΩ
VN = 0.993 − (−0.993)= 1.99 V
VO = −10V : V+ = −10
11.101
4.3kΩ
= 0.487 V
4.3kΩ + 39kΩ
4.3kΩ
For VO = −4.3 − 0.6 = −4.9V : V+ = −4.9
= −0.487 V
4.3kΩ + 39kΩ
VN = 0.487 − (−0.487) = 0.974 V
For VO = 4.3 + 0.6 = 4.9V : V+ = 4.9
11.102
R4
R2
R2
+5 V
+5 V
+
+5 V
v
VTH
O
R3
+
R TH
O
R
R
C
v
C
11-31
For VO = 0 : V+ = VTH
0.05
R2
R4
= 1−
= 0.975V | RTH = R3 R4 | VTH = 5
2
RTH + R2
R3 + R4
For VO = 5 : V+ = VTH
RTH
0.05
R2
+5
= 1+
= 1.025V
2
RTH + R2
RTH + R2
Subtracting : 5
RTH
R
RTH
= 0.05V →
= 0.01 → 2 = 99
RTH + R2
RTH + R2
RTH
R2
97.5
R
RTH + R2 0.975
=
→ VTH 2 = 97.5 → VTH =
= 0.985V
RTH
0.01
99
RTH
RTH + R2
VTH
0.985 = 5
R
R4
→ 3 = 4.077 | Choosing R4 = 2kΩ → R3 = 8.154kΩ
R3 + R4
R4
RTH = 8.154kΩ 2kΩ = 1.606kΩ | R2 = 99(1.606kΩ) = 159kΩ
Choosing standard values : R2 = 160 kΩ | R3 = 8.2kΩ | R4 = 2 kΩ
11.103
24kΩ
3.4kΩ
+ 12
= 6.74 V
3.4kΩ + 24kΩ
3.4kΩ + 24kΩ
24kΩ
For vO = 0V : V+ = 6
= 5.26 V
3.4kΩ + 24kΩ
⎛ t ⎞
v (t )= VF − (VF − VI )exp⎜ −
⎟
⎝ RC ⎠
⎛ T ⎞
6.74
= 50.7 µs
6.74 = 12 − (12 − 5.26)exp⎜− 1 ⎟ → T1 = 6200 3.3x10−8 ln
5.26
⎝ RC ⎠
⎛ T ⎞
6.74
5.26 = 0 − (0 − 6.74)exp⎜− 2 ⎟ → T2 = 6200 3.3x10−8 ln
= 50.7 µs
5.26
⎝ RC ⎠
For vO = +12V : V+ = 6
(
(
f =
)
)
1
= 9.86 kHz
50.7µs + 50.7µs
11.104
f = 0. The circuit does not oscillate. VO = 0 is a stable state.
11-32
11.105
R1
1
1+β
=
| T = 2RC ln
= 2RC ln 3 = 2.197RC
R1 + R2 2
1− β
During steady - state oscillation, the maximum output current from the op- amp is
(a) Let R1 = R2
| β=
5 − (−2.5)
5
5 7.5
+
| Let R = R1 = R2 |
+
≤ 1mA → R ≥ 10kΩ
R1 + R2
2R R
R
0.001s
= 4.55x10−4 s | Selecting C = 0.015µF, R = 30.3kΩ → 30 kΩ (5% values)
RC =
2.197
1
Final values : R = R1 = R2 = 30 kΩ | C = 0.015 µF | f =
= 1.01 kHz
2.197(30 kΩ)(0.015µF )
1
1
1
(b) β = R | β max = 30kΩ 0.95 = 0.525 | β min = 30kΩ 1.05 = 0.475
( )
( )
1+ 2
1+
1+
R1
30kΩ(1.05)
30kΩ(0.95)
I=
1+ 0.525
= 1.213x10−3 s → f min = 825 Hz
1− 0.525
1+ 0.475
= 7.949x10−4 s → f min = 1.26 kHz
Tmin = 2(30kΩ)(0.95)(0.015µF )(0.90)ln
1− 0.475
4.75
= 2.375V
(c ) For vO = +4.75V : V+ = 4.75β =
2
−5.25
For vO = -5.25V : V+ = −5.25β =
= −2.625V
2
⎛ t ⎞
v (t ) = VF − (VF − VI )exp⎜ −
⎟
⎝ RC ⎠
⎛ T ⎞
7.375
= 1.133RC
2.375 = 4.75 − (4.75 − (−2.625))exp⎜ − 1 ⎟ → T1 = RC ln
⎝ RC ⎠
2.375
⎛ T ⎞
7.625
−2.625 = −5.25 − (−5.25 − 2.375)exp⎜ − 2 ⎟ → T2 = RC ln
= 1.066RC
⎝ RC ⎠
2.625
Tmax = 2(30kΩ)(1.05)(0.015µF )(1.1)ln
T = 2.199RC = 2.199(30kΩ)(0.015µF ) = 9.896x10 -4 s → f = 1.01 kHz − Very little change
11-33
11.106
For a triangular waverform with peak amplitude VS and ω o = 2000π :
∞
8VS
nπ
sin
sin nω ot
2 2
2
n=1 n π
v (t )= ∑
For the low - pass filter : Av (s)= −
1
1+
s
3000π
| Av ( jf ) =
1
⎛ f ⎞2
1+ ⎜
⎟
⎝ 1500 ⎠
Av (j1000) = 0.832 | Av (j3000) = 0.447 | Av (j5000) = 0.287
8VS
1 5 2
π = 7.41 V
0.832 8
π
This series contains only odd harmonics : For n = 2, V2000 = 0.
⎛ 5 ⎞1
For f = 3 kHz, n = 3 : V3000 = 0.447⎜
⎟ = 0.298 V
⎝ .832 ⎠ 32
⎛ 5 ⎞1
For f = 5 kHz, n = 5 : V5000 = 0.287⎜
⎟ = 69.0 mV
⎝ .832 ⎠ 52
For a 5 - V fundamental :
2
0.832 = 5 → VS =
11.107
Von
VCC
T = RC ln
1− β
1+
VCC
VEE
Tr = RC ln
V
1− on
VEE
1+ β
11-34
0.6
10 = 841 µs
(51kΩ)(0.033µF )ln 1− 0.357
⎛ 10 ⎞
1+ 0.357⎜ ⎟
⎝ 10 ⎠
= (51kΩ)(0.033µF )ln
= 416 µs
0.6
1−
10
15kΩ
| β=
= 0.357 |
15kΩ + 27kΩ
1+
11.108
VCC
V
V
1+ on
1+ β CC
VEE
VCC T
VEE
| Tr = RC ln
| ln
= ln
Von
V
1− β
Tr
1−
1− on
VEE
VEE
⎛5⎞
0.6
1+ β ⎜ ⎟
1+
⎛ 1.12 ⎞
⎛ 1.12 ⎞ ⎛1+ β ⎞ 2
10µs
1+ β
⎝5⎠
5
ln
=
ln
| ln⎜
| ⎜
⎟
⎟ = 2ln
⎟=⎜
0.6
⎝
⎠
1−
1−
1− β
β
0.88
β
0.88
5µs
⎝
⎠
⎝
⎠
1−
5
R
MATLAB gives β = 0.6998 → 1 = 2.33 | R2 = 13 kΩ | R2 = 30.3kΩ → 30 kΩ
R2
Von
VCC
T = RC ln
1− β
1+
1+ β
10−5
= 7.595µs | C =150 pF | R = 50.6kΩ → 51 kΩ
RC =
0.6
1+
5
ln
1− 0.6998
11-35
CHAPTER 12
12.1
86
(a) A = 10 20 = 2.00x104 | Av-ideal = 1+
A
Av =
=
1+ Aβ
FGE =
150kΩ
= 13.5
12kΩ
2.00x10 4
= 13.49
⎛
⎞
4 12kΩ
1+ 2.00x10 ⎜
⎟
⎝162kΩ ⎠
1
13.5 −13.49
= 6.75x10−4 or 0.0675% | Note : FGE ≅
= 6.75x10−4
Aβ
13.5
150kΩ
(b) Av-ideal = 1+ 1.2kΩ = 126 | Av =
2.00x10 4
= 125
⎛ 1.2kΩ ⎞
4
1+ 2.00x10 ⎜
⎟
⎝ 151.2kΩ ⎠
⎛ 1.2kΩ ⎞
1
= 6.30x10−3
Aβ = 2.00x10 4 ⎜
⎟ = 159 >> 1 | FGE ≅
A
151.2kΩ
β
⎝
⎠
12.2
(a) A = 10
100
20
= 105 | Av-ideal = 1+
47kΩ
= 9.393
5.6kΩ
105
= 9.392
⎛
⎞
5 5.6kΩ
1+ 10 ⎜
⎟
⎝ 52.6kΩ ⎠
1⎛ 1 ⎞
1
−4
GE = ⎜
FGE =
= 9.39x10−5
⎟ = 8.82x10
1+ Aβ
β ⎝1+ Aβ ⎠
Av =
A
=
1+ Aβ
94
(b) A = 10 20 = 5.01x104 | Av-ideal = 1+
47kΩ
= 9.393
5.6kΩ
5.01x10 4
= 9.391
⎛
⎞
4 5.6kΩ
1+ 5.01x10 ⎜
⎟
⎝ 52.6kΩ ⎠
1⎛ 1 ⎞
1
−3
GE = ⎜
FGE =
= 1.87x10−4
⎟ = 1.76x10
β ⎝1+ Aβ ⎠
1+ Aβ
A
Av =
=
1+ Aβ
12-1
12.3
92
220kΩ
= −10
22kΩ
⎡
⎛ 22kΩ ⎞ ⎤
3.98x10 4 ⎜
⎢
⎟ ⎥
R2 ⎛ Aβ ⎞
⎝ 242kΩ ⎠ ⎥
⎢
Av = − ⎜
= −9.997
⎟ = −10
⎞⎥
⎛
⎢
R1 ⎝1+ Aβ ⎠
4 22kΩ
⎟⎥
⎢1+ 3.98x10 ⎜
⎝ 242kΩ ⎠⎦
⎣
(a) A = 10 20 = 3.98x104 | Av-ideal = −
1
10 − 9.997
= 2.76x10−4 or 0.0276% | Note : FGE ≅
= 2.76x10−4
Aβ
10
⎡
⎛ 1.1kΩ ⎞ ⎤
3.98x10 4 ⎜
⎢
⎟ ⎥
220kΩ
R2 ⎛ Aβ ⎞
⎝ 221.1kΩ ⎠ ⎥
⎢
= −199.0
(b) Av-ideal = − 1.1kΩ = −200 | Av = − R ⎜⎝1+ Aβ ⎟⎠ = −200 ⎢
⎛ 1.1kΩ ⎞ ⎥
4
1
⎢1+ 3.98x10 ⎜ 221.1kΩ ⎟ ⎥
⎠⎦
⎝
⎣
⎛ 1.1kΩ ⎞
1
= 5.05x10−3
Aβ = 3.98x10 4 ⎜
⎟ = 198 >> 1 | FGE ≅
A
β
221.1kΩ
⎠
⎝
FGE =
12.4
94
20
47kΩ
= −10
4.7kΩ
⎡
⎛ 4.7kΩ ⎞ ⎤
5.01x10 4 ⎜
⎢
⎟ ⎥
R2 ⎛ Aβ ⎞
⎝ 51.7kΩ ⎠ ⎥
⎢
= −9.998
Av = − ⎜
⎟ = −10
⎛
⎞⎥
⎢
R1 ⎝1+ Aβ ⎠
4 4.7kΩ
⎟⎥
⎢1+ 5.01x10 ⎜
⎝ 51.7kΩ ⎠⎦
⎣
R ⎛ 1 ⎞
1
−10
GE = − 2 ⎜
= -2.20x10-3 | FGE =
= 2.20x10−4
⎟=
⎛
⎞
1+ Aβ
R1 ⎝1+ Aβ ⎠
4.7kΩ
1+ 5.01x10 4 ⎜
⎟
⎝ 51.7kΩ ⎠
(a) A = 10
100
= 5.01x10 4 | Av-ideal = −
47kΩ
= −10
4.7kΩ
⎡
⎛ 4.7kΩ ⎞ ⎤
105 ⎜
⎢
⎟ ⎥
R2 ⎛ Aβ ⎞
⎝ 51.7kΩ ⎠ ⎥
⎢
= −9.999
Av = − ⎜
⎟ = −10
⎛
⎞⎥
⎢
R1 ⎝1+ Aβ ⎠
5 4.7kΩ
⎢1+ 10 ⎜ 51.7kΩ ⎟⎥
⎝
⎠⎦
⎣
R ⎛ 1 ⎞
1
−10
GE = − 2 ⎜
= -1.10x10-3 | FGE =
= 1.10x10−4
⎟=
⎛ 4.7kΩ ⎞
R1 ⎝1+ Aβ ⎠
1+ Aβ
1+ 105 ⎜
⎟
⎝ 51.7kΩ ⎠
(b) A = 10 20 = 105 | Av-ideal = −
12-2
12.5
46
ACL = 10 20 = 200 →
FGE ≅
R2
= 200 | β ≅
R1
1
1
=
R 201
1+ 2
R1
1
201
< 0.001 | A >
= 2.01x105 or 106 dB
Aβ
0.001
12.6
FGE = 1−
1
A
=
≤ 10−4 requires A ≥ 10,000 (80 dB)
1+ A 1+ A
12.7
32
1
1
= Av = 10 20 = 39.8 | FGE ≅
< 0.002
Aβ
β
12.8
Av = 1+
R2
R1
Avmin = 1+
| Avnom = 1+
| A>
39.8
= 1.99x10 4 or 86 dB
0.002
99R1 (1+ 0.0001)
99R1
= 100 | Avmax = 1+
= 100.02
R1
R1(1− 0.0001)
99R1 (1− 0.0001)
R1 (1+ 0.0001)
= 98.98 |
1
< .0001 | A > 106 or 120 dB
Aβ
12.9
Driving the output of the circuit in Fig. 12.3 with a current source of value iX:
vX
v − Av id
iX = iO + i2 | i2 =
; iO = X
; v id = −i 2 R1
R1 + R2
RO
1+ Aβ
v + Ai 2 R1
R1
iO = X
= vX
where β =
RO
RO
R1 + R2
1+ Aβ
vX
v
RO
i X = vX
+
and R out = X =
(R1 + R2 )
RO
R1 + R2
i X 1+ Aβ
12.10
Assuming i− << i2 :
i1 ≅ i2 =
β=
R1
1
vX
=
| i− = −i X = −
=−
R1 + R2 48
Rid (1+ Aβ )
0.1V (1+ 47)
vO
=
= 100 µA and i− << i 2
R1 + R2
48kΩ
0.1V
= −48.0 pA
⎛
⎞
6
5 1
10 Ω⎜1+ 10
⎟
48 ⎠
⎝
12-3
12.11
86
(a) A = 10 20 = 2.00x104
⎛ 12kΩ ⎞
| 1+ Aβ = 1+ 2.00x10 4 ⎜
⎟ = 1482
⎝ 162kΩ ⎠
A
2.00x10 4
=
= 13.49 | Rin = 250kΩ(1+ Aβ ) = 371 MΩ
1482
1+ Aβ
250Ω
150kΩ
=
= 169 mΩ | Note that Av-ideal = 1+
= 13.5
12kΩ
(1+ Aβ )
Av =
Rout
(b) A
v-ideal = 1+
Av =
⎛ 1.2kΩ ⎞
150kΩ
= 126 | 1+ Aβ = 1+ 2.00x10 4 ⎜
⎟ = 159.7
1.2kΩ
⎝ 151.2kΩ ⎠
A
2.00x10 4
250Ω
=
= 124 | Rin = 250kΩ(1+ Aβ ) = 39.9 MΩ | Rout =
= 156 Ω
160.7
1+ Aβ
(1+ Aβ )
12.12
(a) A = 10
100
20
= 105
⎛
⎞
5.6kΩ
| 1+ Aβ = 1+ 105 ⎜
⎟ = 10647
⎝ 5.6kΩ + 47kΩ ⎠
A
105
=
= 9.392 | Rin = 400kΩ(1+ Aβ )= 10.6 MΩ
1+ Aβ 10647
200Ω
47kΩ
=
= 18.8 mΩ | Note that Av-ideal = 1+
= 9.393
5.6kΩ
(1+ Aβ )
Av =
Rout
(b) A = 10
= 5.01x10 4
⎛
⎞
5.6kΩ
| 1+ Aβ = 1+ 5.01x10 4⎜
⎟ = 5335
⎝ 5.6kΩ + 47kΩ ⎠
A
5.01x10 4
=
= 9.391 | Rin = 400kΩ(1+ Aβ )= 2.13 MΩ
1+ Aβ
5335
200Ω
47kΩ
=
= 37.5 mΩ | Note that Av-ideal = 1+
= 9.393
5.6kΩ
(1+ Aβ )
Av =
Rout
94
20
12-4
12.13
100
(a) A = 10 20 = 105
⎛
⎞
5.6kΩ
47kΩ
| 1+ Aβ = 1+ 105⎜
= −8.393
⎟ = 10647 | Av-ideal = −
5.6kΩ
⎝ 5.6kΩ + 47kΩ ⎠
Av = −
R2
R2 ⎛ Aβ ⎞
10646
= −8.392 | Rin = R1 + Rid
= 5.60 kΩ
⎜
⎟ = −8.393
R1 ⎝1+ Aβ ⎠
1+ A
10647
Rout =
200Ω
= 18.8 mΩ
(1+ Aβ )
94
20
(b) A = 10
= 5.01x10 4
⎛
⎞
5.6kΩ
| 1+ Aβ = 1+ 5.01x10 4⎜
⎟ = 5335
⎝ 5.6kΩ + 47kΩ ⎠
Av = −
R2
R2 ⎛ Aβ ⎞
5334
= −8.391 | Rin = R1 + Rid
= 5.60 kΩ
⎜
⎟ = −8.393
R1 ⎝1+ Aβ ⎠
1+ A
5335
Rout =
200Ω
= 37.5 mΩ
(1+ Aβ )
12.14
(a) A = 10
94
20
= 5.01x10 4
⎛
⎞
4.7kΩ
47kΩ
| 1+ Aβ = 1+ 5.01x10 4 ⎜
= −10
⎟ = 4555 | Av-ideal = −
4.7kΩ
⎝ 4.7kΩ + 47kΩ ⎠
Av = −
R2
R2 ⎛ Aβ ⎞
4554
= −9.998 | Rin = R1 + Rid
= 4.70 kΩ
⎜
⎟ = −10
R1 ⎝1+ Aβ ⎠
1+ A
4555
Rout =
100Ω
= 21.6 mΩ
(1+ Aβ )
100
(b) A = 10 20 = 105
⎛
⎞
4.7kΩ
47kΩ
| 1+ Aβ = 1+ 105 ⎜
= −10
⎟ = 9092 | Av-ideal = −
4.7kΩ
⎝ 4.7kΩ + 47kΩ ⎠
Av = −
R2
R2 ⎛ Aβ ⎞
9091
= −9.999 | Rin = R1 + Rid
= 4.70 kΩ
⎜
⎟ = −10
R1 ⎝1+ Aβ ⎠
1+ A
9092
Rout =
100Ω
= 11.0 mΩ
(1+ Aβ )
12.15
⎛
⎞
2R1
Setting v2 = 0, Rin1 = Rid (1+ Aβ )= (500kΩ)⎜1+ 4x10 4
⎟ = 785 MΩ
2R1 + 49R1 ⎠
⎝
RO
75
By symmetry, Rin2 = Rin1 = 785 MΩ | Rout = Rout3 =
=
= 3.75 mΩ
⎛
(1+ Aβ ) ⎜1+ 4x104 ⎞⎟
2 ⎠
⎝
12-5
12.16
Op-amp parameters: Rid = 500 kΩ, Ro = 35 Ω, A = 50,000
Amplifier Requirements: AV = 200, Rin ≥ 200 MΩ, Rout ≤ 0.2 Ω.
We must immediately discard the inverting amplifier case. Rin = R1 requires R1 ≥ 200 MΩ
which can be achieved, but then R2 must be 200 R1 ≥ 40 GΩ which is out of the question (see
values in Appendix A). So, working with the non-inverting amplifier:
Av = 200 ⇒ β =
R out =
R1
1
50000
=
and Aβ =
= 250 >> 1
R1 + R2 200
200
Ro
35
≅
= 0.14Ω meets the specification
1+ Aβ 250
R in = Rid (1+ Aβ )≅ 500kΩ(250)= 125MΩ does not meet the requirements.
So the specifications cannot be met using a single-stage amplifier built using the op-amp that
was specified in the problem.
12.17
The non-inverting amplifier is the only one that can hope to achieve both the required gain and
with such a high value of input resistance:
Av =
1
β
= 200 and Aβ =
10 4
= 50
200
Rin = Rid (1+ Aβ ) = 1MΩ(51) = 51 MΩ - too small
R out =
Ro
100Ω
=
= 1.96Ω - too large
(1+ Aβ ) 51
If the gain specification is met, the input and output resistance specifications will not be met.
12.18
⎛ R2 ⎞⎛ Aβ ⎞
The open-circuit voltage is v th = v s ⎜− ⎟⎜
⎟ . Checking the loop-gain:
⎝ R1 ⎠⎝ 1+ Aβ ⎠
⎛ R ⎞
⎛
⎞
⎛110kΩ ⎞
6.8kΩ
Aβ = (5x10 4 )⎜
⎟ = 2910 >> 1 so v th = v S ⎜− 2 ⎟ = −v S ⎜
⎟ = −16.2v S
⎝ 6.8kΩ + 110kΩ ⎠
⎝ 6.8kΩ ⎠
⎝ R1 ⎠
Rth = Rout =
12-6
Ro
R
250Ω
≅ o =
= 85.9 mΩ
1+ Aβ Aβ 2910
12.19
A
v S . Checking the loop gain :
1+ Aβ
⎛ R2 ⎞
⎛
⎞
⎛
56kΩ ⎞
0.39kΩ
Aβ = 10 4 ⎜
⎟ = 69.2 >> 1 so vth ≅ vS ⎜1+ ⎟ = v S ⎜1+
⎟ = 145 vS
⎝ 0.39kΩ + 56kΩ ⎠
⎝ 0.39kΩ ⎠
⎝ R1 ⎠
The open circuit voltage is vth =
or more exactly : vth = v S
A
10 4
Ro
200Ω
= vS
= 143 vS | Rth = Rout =
=
= 2.85 Ω
1+ 69.2
1+ Aβ 70.2
1+ Aβ
12.20
Applying the definintion of fractional gain error:
R ⎡ R2 (1± ε)⎤ Aβ
⎥
− 2 − ⎢−
⎡ (1± ε)⎤ Aβ
R1 ⎢⎣ R1(1 m ε)⎥⎦ 1+ Aβ
Aβ
⎥
FGE =
= 1− ⎢
≅ 1− (1± 2ε)
R
1+ Aβ
⎢⎣ (1 m ε)⎥⎦ 1+ Aβ
− 2
R1
FGE ≅ 1−
Aβ
1
Aβ
Aβ
m 2ε
=
m 2ε
1+ Aβ
1+ Aβ 1+ Aβ
1+ Aβ
For Aβ >> 1, FGE ≅
1
m 2ε =
Aβ
1
1
2x10
1000
(
5
)
106
1
m 2ε which must be ≤ 0.01. A = 10 20 = 2.00x105
Aβ
m 2ε ≤ 0.01 | Taking the positive sign, 2ε ≤ 0.005 and ε ≤ 0.25%
12.21
Using the results from Prob. 12.20,
54
Av = 10 20 = 501 | For Aβ =
4x10 4
1
= 79.8 >> 1, so FGE ≅
m 2ε which must be ≤ 0.02
Aβ
501
1
1
m 2ε =
m 2ε ≤ 0.02 | Taking the positive sign, 2ε ≤ 0.00747 and ε ≤ 0.374%
79.8
Aβ
12.22
V+ = 3V
99kΩ
3 - 2.722 2.722 − VO
= 2.722 V |
=
10.1kΩ + 99kΩ
9.9kΩ
101kΩ
| VO = −0.111 V | VOideal = 0 V
12-7
12.23
99kΩ
3.95 - 3.675 3.675 − VO
= 3.675 V |
=
| VO = 0.869 V
101kΩ
10.1kΩ + 99kΩ
9.9kΩ
For matched resistors, VOideal = −10(V1 − V2 ) = −10(3.95 − 4.05)= +1.00 V
V+ = 4.05V
⎛ 1− 0.869 ⎞
⎟ = 13.1%
1
⎝
⎠
ε = 100%⎜
12.24
v1 + v 2
= 10sin120πt V and v id = v1 − v 2 = 0.50sin5000πt V
2
v − v− v ic − v + 0.09258
99kΩ
= 0.90742v ic | i = ic
=
=
v ic
(b) v + = v ic
9.9kΩ
10.1kΩ + 99kΩ
9.9kΩ
9.9kΩ
0.09258
v O = v− − i(101kΩ) = v + − i(101kΩ) = 0.90742v ic −
v ic (101kΩ)
9.9kΩ
v
v O = −0.037v ic and Acm = O = −0.037 | The value of Adm = −10 is not
v ic
(a) v ic =
affected by the small tolerances.
(c) CMRR =
Adm
= 270 - a paltry 48.6 dB
Acm
(d) vO = Admv id + Acmv ic = −0.370sin(120πt ) − 5.00sin(5000πt ) V
12.25
5 + 5.01
= 5.005V. The maximum equivalent input error is
2
5.005
VIC
=
= 0.500 mV , but the sign is unknown. Therefore the meter reading
CMRR 10 4
may be anywhere in the range 9.50 mV ≤ Vmeter ≤ 10.5 mV.
VIC =
12-8
12.26
30kΩ
10kΩ
= 3.75 V
10kΩ + 30kΩ
20kΩ
V + V 11.25 + 3.75
V1 - V2 = 15V
= 7.50 V | VCM = 1 2 =
= 7.50V.
2
10kΩ + 30kΩ
2
VCM
VCM
The maximum equivalent input error is
. We need
< 10−4 (V1 - V2 )
CMRR
CMRR
4
10 (7.5V )
CMRR >
= 10 4 or 80 dB.
7.5V
10.2kΩ
10kΩ
(b) V1 = 15V 20.2kΩ = 7.574 V | V2 = 15V 20.2kΩ = 7.426 V
200Ω
V +V
= 0.1485 V | VCM = 1 2 = 7.50V.
V1 - V2 = 15V
2
20.2kΩ
10 4 (7.5V )
VCM
−4
< 10 (V1 - V2 ) or CMRR >
= 5.05x105 or 114 dB.
We need
CMRR
0.1485V
(a) V = 15V 10kΩ + 30kΩ = 11.25 V
1
| V2 = 15V
12.27
One worst-case tolerance assignment is given below. The second is found by reversing the pairs
of resistor values.
i
v
9.995 k Ω
i
10.005 k
Ω
v
O
i
c
+
10.005 k
Ω
9.995 k Ω
v −v
v − v+
0.50025
9.995
= 0.49975v ic | i = ic − = ic
=
v ic
9.995kΩ 9.995kΩ 9.995kΩ
10.005 + 9.995
0.50025
vO = v− − i(10.005kΩ)= v + − i(10.005kΩ)= 0.49975v ic −
v ic (10.005kΩ)
9.995kΩ
v
vO = −0.001vic and Acm = O = −0.001 | The value of Adm = 1 is not affected by the
vic
v+ = v ic
small tolerances. CMRR =
Adm
= 1000 | CMRRdB = 60 dB
Acm
12-9
12.28
⎛
⎞
2kΩ
Setting v2 = 0, Rin1 = Rid (1+ Aβ ) 2Ric = (1MΩ)⎜1+ 7.5x10 4
⎟ 2(500MΩ) = 852 MΩ
⎝
2kΩ + 24kΩ ⎠
Ro
100
By symmetry, Rin 2 = Rin1 = 852 MΩ | Rout = Rout 3 =
=
= 2.67 mΩ
(1+ Aβ ) ⎛⎜1+ 75000 ⎞⎟
⎝
2 ⎠
12.29
V2 − V3
(4.9kΩ)= 4.500
200Ω
9.99kΩ
V −V
V4 = V3 − 2 3 (4.9kΩ)= 5.500 | V6 = V4
= 2.7473V
200Ω
10.01kΩ + 9.99kΩ
V1 - V5
V −V
= 5 O | V5 = V6 | VO = 0.991 V
9.99kΩ 10.01kΩ
V −V V - V
V −V
V4
i1 = - 2 3 - 1 5 = -75.4 µA | i2 = 2 3 = -375 µA
200Ω 9.99kΩ
200Ω 10.01kΩ + 9.99kΩ
V −V
i3 = 5 O = +175 µA
10.01kΩ
The common - mode and differential - mode inputs to the differential subtractor are
V +V
Vcm = 1 4 = 5V & Vdm = V1 − V4 = −1V with V1 = −0.5V and V4 = +0.5V
2
The subtractor outputs for the common - mode and differential - mode inputs are :
V2 = Va = 4.99V | V3 = Vb = 5.01V | V1 = V2 +
9.99kΩ
V1 - V5
V5 − VOcm
CM : V5 = V6 = 5
= 2.4975V |
=
| VOcm = −0.0100 V
9.99kΩ 10.01kΩ
10.01kΩ + 9.99kΩ
9.99kΩ
V1 - V5
V − V dm
= 0.24975V |
= 5 O
| VOdm = 1.00 V
DM : V5 = V6 = 0.5
9.99kΩ 10.01kΩ
10.01kΩ + 9.99kΩ
1.00
−0.01
A
Adm =
= −50.0 | Acm =
= −0.002 | CMRR = dm = 25000 or 88.0 dB
Acm
−0.02
5
12-10
12.30
V2 − V3
(4.9kΩ)= 3.00V
200Ω
V −V
9.99kΩ
V4 = V3 − 2 3 (4.9kΩ)= 3.00V | V6 = V4
= 1.4985V
200Ω
10.01kΩ + 9.99kΩ
V −V
V1 - V5
= 5 O | V5 = V6 | VO = −6.01 mV
9.99kΩ 10.01kΩ
V −V V - V
V −V
V4
i1 = - 2 3 - 1 5 = -150 µA | i2 = 2 3 = -150 µA
200Ω 9.99kΩ
200Ω 10.01kΩ + 9.99kΩ
V −V
i3 = 5 O = +149 µA
10.01kΩ
3+ 3
VO = Admvid + Acmvic | − 6.01x10−3 = -50(3 − 3)+ Acm
| Acm = −2.00x10−3
2
A
50
CMRR = dm =
= 2.50x10 4 or 88.0 dB | It has been assumed that
−3
Acm 2.00x10
V2 = Va = 3.00V | V3 = Vb = 3.00V | V1 = V2 +
the value of Adm is not affected by the small tolerances. See solution to Prob. 12.29.
12.31
⎛ R⎞
⎛ R⎞
⎛ R⎞
VO = (VOS − I B1 R1)⎜1+ 2 ⎟ − I B2 R3 ⎜− 2 ⎟ = (VOS − I B1 R1 )⎜1+ 2 ⎟ + I B 2 R2
⎝ R3 ⎠
⎝ R3 ⎠
⎝ R3 ⎠
⎛ 106 ⎞
VO = ±0.001−10−7105 ⎜1+ 5 ⎟ + 0.95x10−7106 = ±.011− .015V. Worst case VO = −0.026 V.
⎝ 10 ⎠
(
)
Ideal output = 0 V. Error = -26 mV.Yes, R1 should be R2 R3 = 90.9 kΩ.
I B2
1M Ω
R
100 k Ω
R
1M Ω
2
-I
V
O
3
+
+
B2
R
+
3
R
100 k Ω
R
2
VO
3
+
+
V
I
OS
B1
R
1
100 k Ω
R
V
1
OS
I
B1
R
1
100 k Ω
12-11
12.32
⎛ R ⎞
⎛ R ⎞
⎛ R ⎞
VO = (VOS − IB1R1 )⎜1+ 2 ⎟ − IB 2 R3 ⎜− 2 ⎟ = (VOS − IB1R1 )⎜1+ 2 ⎟ + IB 2 R2
⎝ R3 ⎠
⎝ R3 ⎠
⎝ R3 ⎠
⎛ 510kΩ ⎞
−7
VO = ±0.01− 2x10−7 (10 5 ) ⎜1+
⎟ + 2.5x10 (510kΩ) = ±0.061+ 0.0055V
⎝ 100kΩ ⎠
[
]
Worst case VO = +0.066.5 mV. Ideal output = 0 V. Error = 66.5 mV. Yes, R1 should be R2||R3 =
90.9 kΩ.
12.33
vO = Av (vid + VOS ) | Av =
dvO 10 − (−5) V
=
= +7500
dvid
2 − 0 mV
When vO = 0, vid = −VOS and so VOS = − 0.667 mV.
12.34
Av
7,500
-6
-4
4
-2
6
2
vid (mV)
12.35
For I B2 = 0 : Since v + must = v- = VOS , the current through C is iC (t ) =
VOS
R
1 t
1 tV
V
iC (t ) dt = VOS + ∫0 OS dt = VOS + OS t
∫
0
R
RC
C
C
For VOS = 0, iC (t ) = I B2 since v- = v+ = 0.
vO (t )= VOS +
vO (t )=
1
C
1
∫ i (t )dt = C ∫
I B2
t | Summing these two results yields
0
C
V
I
Eq. (11.103) : vO (t ) = VOS + OS t + B2 t | Note that vC (0) = 0 for both cases.
RC
C
12-12
t
0 C
t
I B2 dt =
12.36
R
R2
or 2 = 99 | For bias current compensation, R1 R2 = 10kΩ
R1
R1
R2
1MΩ
R1R2
=
= 10kΩ | R2 = 10kΩ(1+ 99) = 1.00 MΩ and R1 =
= 10.1kΩ
R1 + R2 1+ R2
99
R1
The nearest 5% values would be 1 MΩ and 10 kΩ.
40dB = 100 = 1+
12.37
(a ) Ideal
⎛ 100kΩ ⎞
VO = −0.005V ⎜1+
⎟ = −0.460V
⎝ 1.1kΩ ⎠
10 4
= −0.546V
4 1.1kΩ
1+ 10
101.1kΩ
(b) V = (−0.005V − 0.001V )1+ Aβ = (−0.005V − 0.001V )
A
O
(c) Error
=
-0.460 - (-0.546)
= −0.187 or −18.7%
-0.460
12.38
Inverting Amplifier : vO = Av v S = −6.2vS as long as v O ≤ 10V as constrained
by the op - amp power supply voltages
1 = -6.2V, feedback loop is working and V- = 0
(a) VO = −6.2()
(b) VO = −6.2(−3) = +18V; VO saturates at VO = +10V
The feedback loop is "broken" since the open - loop gain is now 0.
(The output voltage does not change when the input changes so A = 0)
6.2kΩ
1kΩ
+ 10
= −1.19V
By superposition, V− = −3
7.2kΩ
7.2kΩ
12.39
vO
10 V
Gain = -6.2
5V
vS
-2 V
-1 V
1V
2V
-5 V
-10 V
12-13
12.40
Inverting Amplifier : vO = Av v S = −10vS as long as v O ≤ 10V as constrained
by the op - amp power supply voltages
(a) VO = −10(0.5) = -5.00 V, the feedback loop is working, and V- = 0
(b) VO = −10(1.2) = -12.0V; VO saturates at VO = +10V
The feedback loop is "broken" since the open - loop gain is now 0.
(The output voltage does not change when the input changes so A = 0)
10kΩ
1kΩ
−10
= 0.182 V
By superposition, V− = 1.2
11kΩ
11kΩ
12.41
vO
10 V
5V
Gain = -10.0
vS
-2 V
-1 V
1V
2V
-5 V
-10 V
12.42
Noninverting Amplifier : vO = Av v S = +40vS as long as v O ≤ 15V
(a) VO = 40(0.25V ) = +10V, the feedback loop is working, and VID = 0
(b) VO = 40(0.5V ) = 20V; VO saturates at VO = +15V
The feedback loop is "broken" since the open - loop gain is now 0.
(The output voltage does not change when the input changes so A = 0)
1kΩ
= 0.125V.
VID = V+ − V− = 0.5V −15
1kΩ + 39kΩ
12-14
12.43
vO
15 V
7.5 V
Gain = +40.0
vS
-1 V
-0.5 V
0.5 V
1V
-7.5 V
-15 V
12.44
Noninverting Amplifier : vO = Av v S = +43.9vS as long as v O ≤ 15V
(a) VO = 43.9(0.25V ) = +11.0V, feedback loop is working, and VID = 0
(b) VO = 43.9(0.5V ) = 22.0V; VO saturates at VO = +15V
The feedback loop is "broken" since the open - loop gain is now 0.
(The output voltage does not change when the input changes so A = 0)
0.91kΩ
= 0.158 V.
VID = V+ − V− = 0.5V −15
0.91kΩ + 39kΩ
12-15
12.45
v
S
i
+
O
i
L
v
O
i2
10 k Ω
R
R
2
1
iO = iL + i2 and iO ≤ 1.5mA. The output voltage requirement gives i L ≤
which leaves 0.500mA as the maximum value of i2 . i2 =
The closed - loop gain of 40 db (A v = 100) requires
10V
= 1.00mA
10kΩ
10V
gives (R1 + R2 )≥ 20kΩ.
R1 + R2
R2
= 99.
R1
The closest ratio from the resistor tables appears to be
R2
= 100 which is within 1%
R1
of the desired ratio. (This is close enough since we are using 5% resistors.)
There are many many choices that meet both
R2
= 100 and (R1 + R2 )≥ 20kΩ.
R1
However, the choice, R1 = 200Ω and R 2 = 20kΩ is not acceptable because its
minimum value does not meet the requirements : 20.2kΩ(1− 0.05) = 19.2kΩ.
The smallest acceptable pair is R1 = 220Ω and R 2 = 22kΩ.
12-16
12.46
R
v
R
2
1
i
2
S
v
5k Ω
L
R
O
+
i
O
i
L
15V
= 3 mA so i 2 ≤ 1 mA
5kΩ
v
15
i 2 = O ≤ 1 mA requires R 2 ≥
= 15kΩ
.001
R2
i O = i L + i 2 ≤ 4mA and i L =
To account for the resistor tolerance, 0.95R2 ≥ 15kΩ requires R 2 ≥ 15.8kΩ . For AV = 46 dB =
200, R2 = 200 R1, and one acceptable resistor pair would be R1 = 1 kΩ and R2 = 20 kΩ. Many
acceptable choices exist. An input resistance constraint might set a lower limit on R1.
12.47
The maximum base current is limited to 5 mA, so the maximum emitter current is limited to
I E = (β F + 1)I B = 51(5mA)= 255 mA. Since I E =
10V
10V
, R≥
= 39.2 Ω.
R
255mA
12.48
R
v
R
2
1
i
2
S
v
5k Ω
L
R
O
+
i
O
i
L
10V
10V
= 2.5 mA and iL =
= 2 mA so i2 ≤ 0.5 mA
4kΩ
5kΩ
10V
R
R2 ≥
= 20kΩ Av = 46dB ⇒ 2 = 200
R1
0.5mA
iO = iL + i2 ≤
One possible choice would be R2 = 20 kΩ and R1 = 100 Ω. However, the op-amp would not be
able to supply enough output current if tolerances are take into account. Better choices would be
R2 = 22 kΩ and R1 = 110 Ω or R2 = 200 kΩ and R1 = 1 kΩ which would give the amplifier a
much higher input resistance.
(b) V =
vo max
200
=
10V
= 50 mV (c) Rin = R1 = 110 Ω and 1 kΩ for the two designs given above.
200
12.49
12-17
Using the expressions in Table 12.1:
⎛ 105 ⎞
⎜
240kΩ 11 ⎟
24kΩ
1
R2 ⎛ Aβ ⎞
⎜
⎟ = −10.0
=
| A v1 = − ⎜
First stage : β =
⎟=−
24kΩ ⎜ 105 ⎟
R1 ⎝ 1+ Aβ ⎠
24kΩ + 240kΩ 11
⎜1+
⎟
⎝
11 ⎠
⎛
R2 ⎞
240kΩ
Ro
100
Rin1 = R1 + ⎜ Rid
= 24.0 kΩ | Rout1 =
=
= 11.0 mΩ
⎟ = 24kΩ + 500kΩ
5
1+ Aβ ⎠
1+ Aβ
1+ 10
105
⎝
1+
11
⎛ 105 ⎞
10kΩ
1
50kΩ ⎜ 6 ⎟
⎟ = −5.00
⎜
=
| Av2 = −
Second stage : β =
10kΩ + 50kΩ 6
10kΩ ⎜ 105 ⎟
⎟
⎜1+
⎝
6 ⎠
Rin2 = 10kΩ + 500kΩ
240kΩ
100
= 10.0 kΩ | Rout2 =
= 6.00 mΩ
5
1+ 10
105
1+
6
Overall amplifier :
10kΩ
(−5.00)= +50.0 | Rin = 24.0 kΩ | Rout = 6.00 mΩ
10.0kΩ + 11.0mΩ
For all practical purposes, the numbers the same. R out = 6.00 mΩ is a good
Av = −10.0
approximation of 0 Ω, and Av = −10.0(−5.00)= +50.0
12.50
Use the expressions in Table 12.1.
β1 =
47kΩ
A
105
105
= 0.010755 | Av1 =
=
=
= +9.30
47kΩ + 390kΩ
1+ Aβ1 1+ 105 (0.010755) 10756
24kΩ
A
105
105
β2 =
= 0.19355 | Av1 =
=
=
= +5.17
24kΩ + 100kΩ
1+ Aβ2 1+ 105 (0.19355) 19356
Av = (9.30)(5.17)= 48.1
Rin = Rid1 (1+ Aβ )= 250kΩ(10756)= 2.69 GΩ | Rout =
12-18
Ro2
200
=
= 10.3 mΩ
1+ Aβ2 19356
12.51
Use the expressions in Table 12.1.
β1 = β2 =
Av2 = −
20kΩ
1
A
105
105
=
| Av1 =
=
=
= +7.00
20kΩ + 120kΩ 7
1+ Aβ1 1+ 105 7 14287
(
⎛ 14286 ⎞
R2 ⎛ Aβ2 ⎞
⎜
⎟ = −6⎜
⎟ = −6.00 |
R1 ⎝ 1+ Aβ2 ⎠
⎝ 14287 ⎠
)
Av = (7.00)(−6.00) = −42.0
Rin = Rid1 (1+ Aβ1 )= 250kΩ(14287)= 3.57 GΩ | Rout =
Ro2
200
=
= 14.0 mΩ
1+ Aβ2 14287
12.52
Use the expressions in Table 12.1. The three individual amplifier stages are the same.
5.01x10 4
92
2kΩ
1
R ⎛ Aβ ⎞
40kΩ
21
β=
=
| A = 10 20 = 5.01x10 4 | Av = − 2 ⎜
= −20.0
⎟=−
R1 ⎝ 1+ Aβ ⎠
2kΩ + 40kΩ 21
2kΩ
5.01x10 4
1+
21
⎛
⎞
R2
40kΩ
Ro
300Ω
= 2.00 kΩ | Rout =
=
= 126 mΩ
Rin = R1 + ⎜ Rid
⎟ = 2kΩ + 500kΩ
4
1+ A ⎠
1+ Aβ
1+ 5.01x10
5.01x10 4
⎝
1+
21
⎞⎛
⎞
⎛
2kΩ
2kΩ
For the overall amplifier : Av = ⎜−20.0
⎟⎜−20.0
⎟(−20.0) = −8000
2kΩ + 126mΩ ⎠
2kΩ + 126mΩ ⎠⎝
⎝
Rin = 2.00 kΩ | Rout = 126 mΩ
12.53
50 2 < 5000 < 50 3 | Three stages will be required to keep the gain of each stage ≤ 50.
However, the input and output resistance requirements may further constrain the
85
20
gains and must be checked as well. A =10 = 1.778 x 10 4
Ro
100Ω
1
For Rout =
:
≤ 0.1Ω → Aβ ≥ 999 → β ≥ 0.0562 → ≤ 17.8.
1+ Aβ 1 + Aβ
β
1
For Rin = Rid (1+ Aβ ) 2Ric : 1MΩ(1+ Aβ ) 2GΩ ≥ 10MΩ → Aβ ≥ 9 → ≤ 1976
β
17.8(50)(50) > 5000 so three stages is still sufficient.
12-19
12.54
VS = VO
Z1
= VO
Z1 + Z2
R1
R1 +
R2
SC
= VO
(SCR + 1)R
(SCR + 1)R + R
2
1
2
1
2
1
SC
V ⎛ R ⎞ SC R1 R2 + 1
Av (s)= O = ⎜1+ 2 ⎟
VS ⎝ R1 ⎠ SCR2 + 1
R2 +
(
)
12.55
Av (s) = −
Avnom = −
f Hnom =
f Hmin =
1
R2
R
1
| Av (0)= − 2 | f H =
R1 sCR2 + 1
R1
2πCR2
330kΩ(1.1)
330kΩ(0.9)
330kΩ
= −33 | Avmax = −
= −40.3 | Avmax = −
= −27.0
10kΩ
10kΩ(0.9)
10kΩ(1.1)
1
( )3.3x10
2π 10
−10
5
= 4.83kHz | f Hmax =
1
( )(1.2)3.3x10 (1.1)
2π 10
−10
5
1
( )(0.5)3.3x10 (0.9)
2π 10
−10
5
= 10.7kHz
= 3.65kHz
12.56
−60db/decade requires 3 poles - 3 x (-20db/decade). Using three identical
fH3
= 1.96(20kHz) = 39.2kHz.
amplifiers : Av = 3 1000 = 10 and f H1 =
1
R2 = 10R1 | f H =
1
| R2C =
= 4.06x10−6 s
3
2πR2C
2π 39.2x10
Try C = 270 pF. R2 =
12-20
2 3 −1
1
(
)
1
= 15.0 kΩ, R1 = 1.5kΩ
2πf H C
12.57
s + ωB
Ro
Ro
=
= Ro
ω
1+ A(s)β 1+
s + ω B + ωT β
T
β
s + ωB
s
s
1+
1+
s + ωB
Ro
Ro
ωB
ωB
= Ro
=
≅
s
s + ω B (1+ Aoβ ) (1+ Aoβ ) 1+
(1+ Aoβ ) 1+ s
ω B (1+ Aoβ )
βωT
A(s) =
Z out
ωT
s + ωB
| ωT = Aoω B | Z out =
R OUT
RO
Inductive
region
RO
1+A o β
ω
ωB
βω T
12.58
Using MATLAB:
b=1/11; ro=100; wt=2*pi*1e6; wb=wt/1e5;
n=ro*[1 wb]; d=[1 b*wt];w=logspace(0,7);
r=freqs(n,d,w);
mag=abs(r); phase=angle(r)*180/pi;
subplot(212);semilogx(w,phase)
subplot(211);loglog(w,mag)
10 2
1
Magnitud
e
10
10 0
10 -1
10 -2
10 0
10
1
10 2
10 3
10
4
10
5
10 6
10
7
100
80
Phase
60
40
20
0
10
0
10 1
10
2
10 3
10 4
Frequency (rad/s)
10 5
10
6
10 7
12.59
12-21
⎛
⎜
⎛
R2 ⎞
R2
⎟⎟ = R1 + ⎜ Rid
Z in = R1 + ⎜⎜ Rid
ωT
1+ A(s)⎠
⎜
⎝
1+
⎜
s + ωB
⎝
Rid R2
⎞
⎟
⎛
s + ωB ) ⎞
⎟ = R1 + ⎜ Rid R2 (
⎟
s + ω B + ωT ⎠
⎟
⎝
⎟
⎠
(s + ω B )
Rid R2 (s + ω B )
s + ω B + ωT
= R1 +
(s + ω B )
Rid (s + ω B + ωT ) + R2 (s + ω B )
Rid + R2
s + ω B + ωT
⎛
⎛
s ⎞
s ⎞
R2
Rid R2ω B ⎜1+
Rid
⎜1+
⎟
⎟
1+ Ao )
(
⎝ ωB ⎠
⎝ ωB ⎠
Z in = R1 +
= R1 +
R2
s
Rid + R2
Rid ω B (1+ Ao ) + R2ω B + s(Rid + R2 )
Rid +
1+
(1+ Ao ) ω B (1+ Ao ) Rid + R2
(1+ Ao )
Z in = R1 +
⎛
s ⎞
1+
⎜
⎟
⎛
R2 ⎞
⎝ ωB ⎠
⎟
Z in = R1 + ⎜⎜ Rid
Rid + R2
s
(1+ Ao )⎟⎠ 1+
⎝
ω B (1+ Ao ) R + R2
id
(1+ Ao )
12-22
12.60
Using MATLAB:
n1=1e6; d1=[1 2000];
n2=1e6; d2=[1 4000];
n3=1e12; d3=[1 6000 8e6];
w=logspace(2,5);
[m1,p1,w]=bode(n1,d1,w);
[m2,p2,w]=bode(n2,d2,w);
[m3,p3,w]=bode(n3,d3,w);
subplot(211)
loglog(w,m1,w,m2,w,m3)
subplot(212)
semilogx(w,p1,w,p2,w,p3)
.
10 6
Magnitude
10 5
AV
10
4
10
3
AV1
10
2
AV2
10
1
10
0
10 2
10 4
10 3
10 5
Frequency (rad/s)
0
AV2
Phase
-50
A V1
-100
AV
-150
-200
10 2
10 3
Frequency (rad/s)
10 4
10 5
12-23
12.61
Av = −
Z 2 Aβ
Z1 1 + Aβ
Av (s) = −
Av (s) = −
| β=
Z1
Z1 + Z 2
| A=
ωT
R2
| Z1 = R1 | Z 2 =
s +ωo
sCR2 + 1
R2
ωT
R1
⎛ R
⎞
⎛ R ⎞
s2 R2C + s⎜1 + 2 + R2C (ω o + ωT )⎟ + ω o ⎜1 + 2 ⎟ + ωT
⎝ R1
⎠
⎝ R1 ⎠
3.653 x1013
s2 + 3.142 x10 7 s + 1.916 x1012
Using MATLAB: bode(-3.653e13,[1 3.142e7 1.916e12])
30
20
10
Gain dB
0
-10
-20
-30
-40
-50
10
3
10
4
5
10 6
10
Frequency (rad/sec)
10 7
10 8
200
150
Phase deg
100
50
0
-50
-100
-150
-200 3
10
10
4
10
5
10
Frequency (rad/sec)
12-24
6
10
7
10
8
12.62
V (s)
V (s)
1
which is the transfer function of an integrator
(a) S = −sCVO (s) | Av (s) = O = −
R
VS (s)
sRC
(b) Generalizing Eq. (11.122) :
Av (s) = −
Z 2 A(s)β
1
Z1
| Z2 =
| Z1 = R1 | β =
Z1 1+ A(s)β
Z1 + Z 2
sC
ωT
sRC
ωT
sRC
R1
sCR
1
s + ω B sRC + 1
β=
=
| A(s)β =
| Av (s) = −
sRC
1
s + ω B sRC + 1
sCR + 1
sRC 1+ ωT
R1 +
s + ω B sRC + 1
sC
sRCωT
ωT
1
Av (s) = −
=−
sRC (s + ω B )(sRC + 1) + sRCωT
(s + ω B )(sRC + 1) + sRCωT
ωT
ωT
ωT
Av (s) = −
RC
RC
RC
≅−
=−
⎛
⎛
⎞
⎛
1 ⎞ ωB
ω
1 ⎞
B
s + s⎜ω B + ωT +
⎟+
s + ωT )⎜ s +
s + ωT )⎜ s +
(
(
⎟
⎟
⎝
RC ⎠ RC
Ao RC ⎠
⎝ ωT RC ⎠
⎝
2
using dominant root factorization where it is assumed ωT >> ω B and ωT >>
|A (ω)|
1
.
RC
v
A
o
Integrator region
1
A RC
o
ω
T
ω
12-25
12.63
wrc=1/(1e4*470e-12); wt=2*pi*5e6; wb=2*pi*50;
n=wt*wrc; d=[1 wt+wb+wrc wb*wrc];
bode(n,d)
100
Gain dB
50
0
-50
-100
10 -1
10 0
10 1
10 2
10 3
10 4 10
Frequency (rad/sec)
5
10
6
10
7
10
8
10 0
10
1
10 2
10 3
10 4 10
Frequency (rad/sec)
5
10 6
10
7
10 8
0
Phase deg
-50
-100
-150
-200
10 -1
12.64
2kΩ
1
105
=
| Aβ =
= 4760 >> 1
2kΩ + 40kΩ 21
21
40 kΩ
3x106 Hz
R
= −20 | f H = βf T =
= 143kHz
( a ) Av = − 2 = −
R1
2 kΩ
21
β=
( b ) Av = (−20) = −8000 (78dB) | f H 3 = 0.51 f H = 72.9 kHz
3
12-26
12.65
The table below follows the approach used in Table 12.6. The only change is the required gain
85
is Av = 10 20 = 1.778 x 10 4 .
Cascade of Identical Non-Inverting
Amplifiers
# of Stages
AV(0)
Gain per
Stage
1/β
fH
Single
Stage
β x fT
fH
N Stages
RIN
ROUT
1
2
3
4
5
6
7
8
9
10
11
12
2.00E+04
1.41E+02
2.71E+01
1.19E+01
7.25E+00
5.21E+00
4.12E+00
3.45E+00
3.01E+00
2.69E+00
2.46E+00
2.28E+00
5.00E+01
7.07E+03
3.68E+04
8.41E+04
1.38E+05
1.92E+05
2.43E+05
2.90E+05
3.33E+05
3.71E+05
4.06E+05
4.38E+05
5.000E+01
4.551E+03
1.878E+04
3.658E+04
5.320E+04
6.717E+04
7.839E+04
8.724E+04
9.415E+04
9.951E+04
1.037E+05
1.068E+05
6.00E+09
7.08E+11
3.69E+12
8.41E+12
1.38E+13
1.92E+13
2.43E+13
2.90E+13
3.33E+13
3.71E+13
4.06E+13
4.38E+13
8.33E+00
7.06E-02
1.36E-02
5.95E-03
3.62E-03
2.60E-03
2.06E-03
1.72E-03
1.50E-03
1.35E-03
1.23E-03
1.14E-03
We see from the spreadsheet that a cascade of seven identical stages is required to achieve the
bandwidth specification. Fortuitously, it also meets the input and output resistance specs. For
the non-inverting amplifier cascade:
R
R
AV = 1+ 2 = 4.12 → 2 = 3.12
R1
R1
A similar spreadsheet for the cascade of identical inverting amplifiers indicates that it is
impossible to meet the bandwidth requirement.
12-27
12.66
R2
R
= 4.12 → 2 = 3.12 | Exploring the 5% resistor
R1
R1
R
tables, we find R2 = 62kΩ and R1 = 20kΩ yields 2 = 3.10 as a reasonable pair.
R1
(a) From Problem 11.99, Av = 1 +
The nominal gain of the cascade is then Av = (4.10) =1.948 x 10 4 .
7
Av = 86db ± 1dB ⇒ 1.778 x 10 4 ≤ AV ≤ 2.239 x 10 4 and the gain is well within this range.
Many amplifiers will probably fail due to tolerances with 5% resistors. A Monte Carlo
analysis would tell us. If we resort to 1% resistors to limit the tolerance spread,
R2 = 30.9 kΩ and R1 = 10.0 kΩ is one of many possible pairs.
1
5 x10 6
| f H1 = βf T =
= 1.22 MHz
(b) For R2 = 62kΩ and R1 = 20kΩ, β =
4.1
4.1
1
f H = 1.22 MHz 2 7 − 1 = 394 kHz
12-28
12.67/12.68
One possibility: Use a cascade of two non-inverting amplifiers, and shunt the input of the first
amplifier to define the input resistance.
60db → Av = 1000 | A single - stage amplifier with a gain of 1000 would have a bandwidth
of only 5 kHz using this op - amp. Two stages should be sufficient if Rin and R out can also
be met. A design with f H 2 >> f H1 will be tried.
First stage : Non - inverting with bandwidth of 20 kHz
f
20kHz
1
β1 = H1 =
= 0.004 | Av1 = = 250 → Av 2 = 4 → β 2 = 0.25 → f H 2 = 1.25 MHz
fT
5 MHz
β1
Since f H 2 >> f H1, f H = f H1 = 20kHz. Ao = 85 dB = 17800
100
Ro2
=
= 0.0225Ω which is ok.
Checking Rout =
1 + Aoβ 2 1 + 17800(0.25)
Choosing resistors from the Appendix, a possible set is
Amplifier 1: R1 = 1.2kΩ, R 2 = 300 kΩ and shunt the input with R 3 = 27kΩ
Amplifier 2 : R1 = 3.3kΩ, R 2 = 10 kΩ
⎞
⎛
⎜
17800
17800 ⎟
= 997.5 = 60.0 dB
Checking gain : Av =
⎜
17800
17800 ⎟
⎟
⎜1 +
1+
251 ⎝
4.03 ⎠
vS
+
+
vO
27 k Ω
300 k Ω
10 k Ω
1.2 k Ω
3.3 k Ω
12-29
12.69
function sg=Prob103(tol);
sg=0;
for j=1:10
ao=100000; ft=1e6*sqrt(2^(1/6)-1);
for i=1:500,
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g1=ao/(1+ao*beta); b1=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g2=ao/(1+ao*beta); b2=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g3=ao/(1+ao*beta); b3=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g4=ao/(1+ao*beta); b4=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g5=ao/(1+ao*beta); b5=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g6=ao/(1+ao*beta); b6=beta*ft;
gain(i)=g1*g2*g3*g4*g5*g6; bw(i)=(b1+b2+b3+b4+b5+b6)/6;
end;
sg=sg+sum(gain<1e5 | bw<5e4);
end;
end
(a) For 5000 test cases with tol = 0.05, 33.5% of the amplifiers failed to meet either the gain or
bandwith requirement.
(b) For 10000 test cases with tol = 0.015, 0.1% of the amplifiers failed to meet either the gain or
bandwith requirement.
12-30
12.70
1
1
Z
1
⎛ 1 ⎞N
1
1
fH
2 N − 1 (2 − 1)2
fT
N
a
=
G
→
β
=
|
f
=
β
f
|
f
=
2
−
1
|
=
=
for Z =
( )⎜ ⎟
H1
T
H
1
1
1
Z
fT
G
N
⎝β ⎠
GN
GN
GN
1
1
⎡1
⎤
−
G Z ⎢ (2 Z − 1) 2 2 Z ln2⎥ − (2 Z − 1)2 G Z ln G
d ⎛ fH ⎞
⎣2
⎦
For 2 Z = e Z ln 2 and G Z = e Z ln G :
⎜ ⎟=
2Z
dz ⎝ f T ⎠
G
1
1
⎡
⎤
−
d ⎛ fH ⎞
Z 1
Z
Z
Z
Z
Z
Z
Setting
⎜ ⎟ = 0 → G ⎢ (2 − 1) 2 2 ln2⎥ − (2 − 1)2 G ln G = 0 → 2 ln2 = 2(2 − 1)ln G
2
dz ⎝ f T ⎠
⎣
⎦
⎛
2ln G
2ln G ⎞
2Z = −
→ Z ln2 = ln⎜ −
⎟
⎝ ln2 − 2ln G ⎠
ln2 − 2ln G
⎛
⎞
ln G
ln⎜−
⎟
ln2
⎝ ln G − ln 2 ⎠
→ N opt =
Z=
⎛
⎞
ln G
ln2
ln⎜
⎟
⎝ ln G − ln 2 ⎠
ln2
= 22.7 which agrees with the spreadsheet in Table 12.8
(b) N opt = ⎛
⎞
ln10 5
ln⎜
⎟
5
⎝ ln10 − ln 2 ⎠
f Hopt = 106.0 kHz
12.71
22kΩ
1
5x10 4
=
Aβ =
= 7240 >> 1
22kΩ + 130kΩ 6.91
6.91
R
130kΩ
(a) Avnom = 1+ 2 = 1+
= 6.91
22kΩ
R1
β nom =
Avmax = 1+
130kΩ(1.05)
130kΩ(0.95)
R2
R
= 1+
= 7.53 | Avmin = 1+ 2 = 1+
= 6.35
R1
R1
22kΩ(0.95)
22kΩ(1.05)
f Hnom = β nom f T =
106 Hz
106 Hz
106 Hz
= 145 kHz | f Hmax =
= 157 kHz | f Hmin =
= 133 kHz
6.91
6.36
7.53
12-31
12.72
function [gain,bw]=Prob1272a
ao=50000; ft=1e6;
for i=1:500,
r1=22000*(1+0.1*(rand-0.5));
r2=130000*(1+0.1*(rand-0.5));
beta=r1/(r1+r2);
gain(i)=ao/(1+ao*beta); bw(i)=beta*ft;
end;
end
[gain,bw]=prob1272a;
mean(gain) ans = 6.9140 std(gain)
ans = 0.2339
mean(bw)
ans = 1.4478e+05 std(bw)
ans = 4.8969e+03
Three sigma limits: 6.21 ≤ Av ≤ 7.62
130 kHz ≤ BW ≤ 159 kHz
function [gain,bw]=Prob1272b
for i=1:500,
ao=100000*(1+1.0*(rand-0.5));
ft=2e6*(1+1.0*(rand-0.5));
r1=22000*(1+0.1*(rand-0.5));
r2=130000*(1+0.1*(rand-0.5));
beta=r1/(r1+r2);
gain(i)=ao/(1+ao*beta); bw(i)=beta*ft;
end;
end
[gain,bw]=prob1272b;
mean(gain) ans = 6.9201 std(gain)
ans = 0.2414
mean(bw)
ans = 2.8925e+05 std(bw)
ans = 8.5536e+04
Note that the bandwidth is essentially a uniform distribution.
3σ: 6.20 ≤ Av ≤ 7.64 98.9% of the values fall between: 146 kHz ≤ BW ≤ 439 kHz
12.73
(
)
SR ≥ VOω = (15V )(2π ) 2x10 4 Hz = 1.89x106
12.74
f =
SR
10V
1
= −6
= 159 kHz
2πVo 10 s 20πV
12.75
12-32
V
V
or 1.89
s
µs
The negative transition requires the largest slew rate : SR =
V
∆V 20V
=
= 10
2µs
∆t
µs
12.76
(a) For the circuit in Fig. 12.26 : Rid = 250 kΩ | R = 1 kΩ − an arbitrary choice
ωB =
(
2π 5 x106
8 x10
4
) = 125π rad/s |
C=
1
ωB R
=
1
(125π )1000
= 2.55 µF | Ro = 50Ω | Ao = 80,000
(b) Add a resistor from each input terminal to ground of value 2 R
IC
= 1 GΩ.
See Prob. 11.111 for schematics.
12.77 Two possibilities:
105 nA
1 mV
+
v
1
250 k Ω
50 Ω
1k Ω
200
MΩ
- +
+
+
v
v
1
80,000v
2
2
v
o
2.55 µF
200 M Ω
95 nA
1 mV
- +
+
50 Ω
1k Ω
105 nA
125 k Ω
v
1
+
+
100 M Ω
v
1
v
80,000v
2
2
vo
2.55 µF
125 k Ω
95 nA
12-33
12.78
100 Ω
+
v
v1
400 k Ω
ω1 : C =
1
1 Ω
+
v
1.592
v2
2
µF
1.592
80,000v
µF
1
1
=
= 1.592 µF − setting R1 arbitrarily to 100 Ω.
ω1R1 2π (10 3 )(100)
ω 2 : R2 =
1
1
=
= 1 Ω − Using the same value of C.
5
ω 2C 2π (10 )(1.592 µF )
12.79
*PROBLEM 12.79 - Six-Stage Amplifier
VS 1 0 AC 1
XA1 1 2 0 AMP
XA2 2 3 0 AMP
XA3 3 4 0 AMP
XA4 4 5 0 AMP
XA5 5 6 0 AMP
XA6 6 7 0 AMP
.SUBCKT AMP 1 2 7
RID 1 3 1E9
RO 6 2 50
E2 6 7 5 7 1E5
E1 4 7 1 3 1
R 4 5 1K
C 5 7 15.915UF
R2 2 3 130K
R1 3 7 22K
.ENDS
.TF V(7) VS
.AC DEC 40 1 1MEG
.PRINT AC V(1) V(2) V(3) V(4) V(5) V(6) V(7)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7)
.END
12-34
75 Ω
+
v3
+
3
vo
12.80
Student PSPICE cannot handle 6 copies of the uA741 op amp, but since all the stages are the
same, we can square the output from a 3-stage version or cube the output from a two-stage
model.
200
100
0
-100
100Hz
300Hz
1.0KHz
DB(V(I1:+))+ DB(V(I1:+))
3.0KHz
10KHz
30KHz
100KHz
300KHz
1.0MHz
3.0MHz
10MHz
Frequency
From the SPICE graph, BW = 54.3 kHz.
12-35
12.81
*PROBLEM 12.81 - Six Stage Amplifier
VS 1 0 AC 1
XA1 1 2 0 AMP
XA2 2 3 0 AMP
XA3 3 4 0 AMP
XA4 4 5 0 AMP
XA5 5 6 0 AMP
XA6 6 7 0 AMP
.SUBCKT AMP 1 2 8
RID 1 3 1E9
RO 7 2 50
E2 7 8 6 8 1E5
*Two dummy loops provide separate control of Gain & BW tolerances
G1 8 4 1 3 .001
R11 4 8 RG 1000
E1 5 8 4 8 1
RC 5 6 1000
C 6 8 CC 15.915UF
*
R2 2 3 RR 130K
R1 3 8 RR 22K
.ENDS
.MODEL RR RES (R=1 DEV=5%)
.MODEL RG RES (R=1 DEV=50%)
.MODEL CC CAP (C=1 DEV=50%)
.AC DEC 20 1E3 1E6
.PROBE V(7)
.PRINT AC V(7)
.MC 1000 AC V(7) MAX OUTPUT(EVERY 20)
*.MC 1000 AC V(7) MAX OUTPUT(RUNS 77 573 597 777)
.END
Maximum gain = 103 dB; Minimum gain = 98.5 dB
Maximum Bandwidth = 65 kHz; Minimum bandwidth = 38 kHz (These are approximate.)
12.82
1
= 7.96 pF
2π (1000Ω)(20 MHz)
Rid = 1010 Ω
C=
12.83
A ≥ 118 dB
CMRR ≥ 80 dB
I B ≤ 8.8 nA
IOS ≤ 2.2 nA
12-36
Ro not specified
PSRR ≥ 100 dB VOS ≤ 1.5 mV
Power Supplies ± 18 V maximum
Nominal values only : Rid = 1010 Ω
12.84
A = 4 x106
SR = 12.5 V/µs nominal
f T = 20 MHz
(a ) Use the expressions in Tables 12.1 and 12.2.
100
47 kΩ
1
15kΩ
1
18 kΩ
1
=
| βB =
=
| βC =
=
47kΩ + 470kΩ 11
15kΩ + 150kΩ 11
18 kΩ + 270 kΩ 16
5
10
105
R ⎛ Aβ ⎞
R ⎛ Aβ ⎞
470kΩ 11
150 kΩ 11
AvA = − 2 ⎜
= −10.0 AvB = − 2 ⎜
= −10.0
⎟=−
⎟=−
5
R1 ⎝ 1 + Aβ ⎠
47kΩ
R1 ⎝1 + Aβ ⎠
15kΩ
10
105
1+
1+
11
11
5
10
R2 ⎛ Aβ ⎞
270 kΩ 16
AvC = − ⎜
= −15.0 Av = −10.02 (−15)= −1500
⎟=−
5
R1 ⎝ 1 + Aβ ⎠
18kΩ
10
1+
16
⎛
⎞
R2
470 kΩ
Rin = RinA = R1 + ⎜ Rid
= 47.0 kΩ
⎟ = 47 kΩ + 1MΩ
1 + A⎠
1 + 105
⎝
A = 10 20 = 105 | β A =
Ro
250Ω
=
= 40.0 mΩ
1 + Aβ
105
1+
16
2 MHz
2 MHz
f HA = f HB = β A fT =
= 182kHz
f HC = βC fT =
= 125kHz
11
16
For the overall amplifier : Av = −1500 | Rin = 47.0 kΩ | Rout = 40.0 mΩ
Rout = RoutC =
Using the definition of bandwidth :
10
10
⎛ f H ⎞2
1+ ⎜
⎟
⎝182 kHz ⎠
(b) v
S
⎛ f H ⎞2
1+ ⎜
⎟
⎝182 kHz ⎠
15
⎛ f H ⎞2
1+ ⎜
⎟
⎝ 125kHz ⎠
=
1500
2
→ f H = 79.9 kHz
= 50.0 mV | vOA = −10vS = −500 mV | vOB = −10vOA = +5.00 V |
vOA
v
= +5.00 µV | v -B = OB5 = −50.0 µV
5
-10
-10
The output of the third amplifier is saturated at -18 V, and the inverting input is no longer near
270kΩ
18 kΩ
ground potential. Using superposition, v -C = 5V
+ (−18V )
= +3.56 V
18kΩ + 270kΩ
18kΩ + 270kΩ
The remaining three nodes are V+ = +18 V V− = −18 V and Vgnd = 0 V .
vO = −15vOB = −75.0V > 18V → vO = −18 V | v -A =
12-37
12.85
100
47kΩ
1
15kΩ
1
=
| βB =
=
47kΩ + 470kΩ 11
15kΩ + 150kΩ 11
⎛ 470kΩ ⎞⎛ 150kΩ ⎞⎛ 270kΩ ⎞
18kΩ
1
βC =
=
Aβ >> 1 Avnom ≅ ⎜−
⎟⎜ −
⎟⎜−
⎟ = −1500
18kΩ + 270kΩ 16
⎝ 47kΩ ⎠⎝ 15kΩ ⎠⎝ 18kΩ ⎠
A = 10 20 = 105 | β A =
max
v
A
⎛ 1.05 ⎞3
≅ −1500⎜
⎟ = −2025
⎝ 0.95 ⎠
min
v
A
⎛ 0.95 ⎞3
≅ −1500⎜
⎟ = −1111
⎝ 1.05 ⎠
max
Rinmax ≅ RinA
= 47.0 kΩ(1.05)= 49.4 kΩ
nom
Rinnom ≅ RinA
= 47.0 kΩ
min
Rinmin ≅ RinA
= 47.0 kΩ(0.95)= 44.7 kΩ
Ro
250Ω
=
= 40.0 mΩ
1+ Aβ
105
1+
16
R
250Ω
Ro
250Ω
max
max
min
min
o
≅ RoutC
=
=
= 43.9 mΩ
Rout
≅ RoutC
=
=
= 36.4 mΩ
Rout
min
5
max
1+ Aβ
1+ Aβ
10
105
1+
1+
17.6
14.6
2 MHz
2 MHz
= 182kHz
f HC = βC fT =
= 125kHz
f HA = f HB = β A fT =
11
16
Using the definition of bandwidth :
nom
nom
≅ RoutC
=
Rout
10
10
2
⎛ fH ⎞
1+ ⎜
⎟
⎝182kHz ⎠
15
2
⎛ fH ⎞
1+ ⎜
⎟
⎝182kHz ⎠
2
⎛ fH ⎞
1+ ⎜
⎟
⎝ 125kHz ⎠
=
1500
2
→ f H = 79.8 kHz
2 MHz
2 MHz
= 200kHz
f HC = βC fT =
= 137kHz
10.0
14.6
9.048
13.57
1111
=
→ f Hmax = 87.5 kHz
2
2
2
⎛ f max ⎞
⎛ f max ⎞
1+ ⎜ H
⎟ 1+ ⎜ H
⎟
⎝ 200kHz ⎠
⎝ 137kHz ⎠
max
max
f HA
= f HB
= β A fT =
9.048
⎛ f max ⎞2
1+ ⎜ H
⎟
⎝ 200kHz ⎠
2 MHz
2 MHz
= 167kHz
f HC = βC fT =
= 114kHz
12.0
17.6
11.05
16.58
2025
=
→ f Hmax = 73.0 kHz
2
2
2
⎛ f max ⎞
⎛ f max ⎞
H
H
1+ ⎜
⎟ 1+ ⎜
⎟
⎝167kHz ⎠
⎝ 114kHz ⎠
max
max
f HA
= f HB
= β A fT =
11.05
2
⎛ f max ⎞
1+ ⎜ H
⎟
⎝167kHz ⎠
12-38
12.86
(a ) Use the expressions in Tables 12.1 and 12.2. Three identical gain blocks.
A = 10
106
20
= 2.00x105 | β =
3kΩ
1
A
=
| Aβ = 14286 | Av1 =
= 14.0
3kΩ + 39kΩ 14
1+ Aβ
Av = 14.03 = 2744
Rin = RinA = 1MΩ Rid (1+ Aβ )= 1MΩ 500kΩ(1+ 14286)= 1.00 MΩ
Rout = RoutC =
Ro
300Ω
=
= 21.0 mΩ
1+ Aβ 1+ 14286
1
5MHz
= 357kHz
f H = f H1 2 3 −1 = 182kHz
f H1 = β A f T =
14
For the overall amplifier : Av = +2740 | Rin = 1.00 MΩ | Rout = 21.0 mΩ | f H = 182kHz
(b) v
I
= 5.00 mV | vOA = 14vI = 70.0 mV | vOB = 14vOA = 980 mV |
vOA
v
= +5.00 mV | v-B = OB = 70.0 mV
14
14
The output of the third amplifier is saturated at 12 V, and the inverting and non - inverting inputs
vO = 14vOB = 13.7V > 12V → vO = 12.0 V | v- A =
are no longer equal. v-C = 12V
3kΩ
= 0.857 V. V+ = +12 V
3kΩ + 39kΩ
V− = −12 V Vgnd = 0 V
12.87
(a ) Use the expressions in Tables 12.1 and 12.2. Three identical gain blocks.
A = 10
106
20
= 2.00x105 | β =
1.5kΩ
1
A
=
| Aβ = 7407 | Av1 =
= 27.0
1.5kΩ + 39kΩ 27
1+ Aβ
Av = 27.03 = 19700
Rin = RinA = 1.5MΩ Rid (1+ Aβ ) = 1.5MΩ 500kΩ(1+ 14286) = 1.50 MΩ
Rout = RoutC =
Ro
300Ω
=
= 38.8 mΩ
1+ Aβ 1+ 7407
1
5MHz
= 185kHz
f H = f H1 2 3 −1 = 94.3kHz
f H1 = β A f T =
27
For the overall amplifier : Av = +19700 | Rin = 1.50 MΩ | Rout = 38.8 mΩ | f H = 94.3 kHz
(b) v
I
= 5.00 mV | vOA = 27vI = 135 mV | vOB = 27vOA = 3.65 V |
vOA
v
= +5.00 mV | v-B = OB = 135 mV
27
27
The output of the third amplifier is saturated at 12 V, and the inverting and non - inverting inputs
vO = 27vOB = 98.4V > 12V → vO = 12.0 V | v- A =
are no longer equal. v-C = 12V
1.5kΩ
= 0.444 V. V+ = +12 V
1.5kΩ + 39kΩ
V− = −12 V Vgnd = 0 V
12-39
12.88
Three identical stages : β nom =
3kΩ
1
=
3kΩ + 39 kΩ 14
Aβ =
2 x105
= 14286 >> 1
14
A
⎛ R2 ⎞3 ⎛ 39 kΩ ⎞3
= ⎜1 + ⎟ = ⎜1 +
⎟ = 2740
3kΩ ⎠
⎝ R1 ⎠ ⎝
Avmax
⎡ 39 kΩ(1.02)⎤
⎡ 39 kΩ(0.98)⎤
⎥ = 3070 | Avmin = ⎢1 +
⎥ = 2460
= ⎢1 +
3kΩ(1.02) ⎥⎦
⎢⎣ 3kΩ(0.98) ⎥⎦
⎣⎢
nom
v
3
3
Rinnom = 1 MΩ | Rinmax = 1 MΩ(1.02) = 1.02 MΩ | Rinmin = 1 MΩ(0.98) = 980 kΩ
300
300
min
= 21.8 MΩ | Rout
=
= 20.2 MΩ
5
2 x10
2 x105
1+
1+
14.53
13.49
5 MHz
5 MHz
5 MHz
f Hnom = β nom f T =
= 357 kHz | f Hmax =
= 371 kHz | f Hmin =
= 344 kHz
14
13.49
14.53
nom
=
Rout
300
max
= 21.0 MΩ | Rout
=
1 + 14286
12-40
12.89
(a ) Use the expressions in Tables 12.1 and 12.2.
80
10kΩ
1
2kΩ
1
10kΩ
1
=
| βB =
=
| βC =
=
10kΩ + 39kΩ 4.9
2kΩ + 200kΩ 101
10kΩ + 39kΩ 4.9
4
10
A
10000
R2 ⎛ Aβ ⎞
200kΩ 101
AvC = AvA =
=
= +4.90 AvB = − ⎜
= −99.0
⎟=−
10000
R1 ⎝ 1+ Aβ ⎠
1+ Aβ
2kΩ
10 4
1+
1+
4.9
101
⎛
⎞
10000
Av = 4.902 (−99.0)= −2380 Rin = RinA = Rid (1+ Aβ )= 300kΩ⎜1+
⎟ = 613 MΩ
4.9 ⎠
⎝
A = 10 20 = 10 4 | β A =
Ro
250Ω
=
= 98.0 mΩ
1+ Aβ
10 4
1+
4.9
3MHz
3MHz
f HC = f HA = β A fT =
= 612kHz
f HC = βC fT =
= 29.7kHz
4.9
101
For the overall amplifier : Av = −2380 | Rin = 613 MΩ | Rout = 98.0 mΩ
Rout = RoutC =
Using the definition of bandwidth :
4.9
99
2
⎛ fH ⎞
1+ ⎜
⎟
⎝ 612kHz ⎠
4.9
2
⎛ fH ⎞
1+ ⎜
⎟
⎝ 29.7kHz ⎠
2
⎛ fH ⎞
1+ ⎜
⎟
⎝ 612kHz ⎠
=
2380
2
→ f H = 29.6 kHz
Note that the bandwidth is controlled by amplifier B because it's bandwidth is much smaller
than the others.
(b) v
I
= 00.0 mV | vOA = 4.9vOS = +49.0 mV | vOB = −100vOA + 101(10mV ) = −3.89V |
vO = 4.9(vOB + .010) = −19.0V < -15V → vO = −15 V | v -A =
v -C = −
vOA
v
= 10.0 mV | v -B = OB4 = +389 µV
4.9
-10
15V
= −3.06 V The remaining three nodes are V+ = +15 V
4.9
V− = −15 V and Vgnd = 0 V.
12-41
12.90
10kΩ
1
2kΩ
1
A = 10 = 10 4 | βC = β A =
=
| βB =
=
10kΩ + 39kΩ 4.9
2kΩ + 200kΩ 101
2
⎛ 39kΩ ⎞ ⎛ 200kΩ ⎞
Aβ >> 1 Avnom ≅ ⎜1+
⎟ ⎜−
⎟ = −2401
⎝ 10kΩ ⎠ ⎝ 2kΩ ⎠
80
20
2
max
v
A
⎛ 39kΩ ⎛ 1.10 ⎞⎞ ⎛ 200kΩ ⎛ 1.10 ⎞⎞
≅ ⎜1+
⎟⎟ ⎜ −
⎟⎟ = −4064
⎜
⎜
⎝ 10kΩ ⎝ 0.90 ⎠⎠ ⎝ 2kΩ ⎝ 0.90 ⎠⎠
⎛ 39kΩ ⎛ 0.90 ⎞⎞2⎛ 200kΩ ⎛ 0.90 ⎞⎞
A ≅ ⎜1+
⎟⎟ ⎜ −
⎟⎟ = −1437
⎜
⎜
⎝ 10kΩ ⎝ 1.10 ⎠⎠ ⎝ 2kΩ ⎝ 1.10 ⎠⎠
⎡
⎛ 1 ⎞⎤
Rinnom = Rid (1+ Aβ ) = 300kΩ ⎢1+ 10 4 ⎜ ⎟⎥ = 612 MΩ
⎝ 4.9 ⎠⎦
⎣
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
4
min
Rinmax = 300kΩ ⎢1+ 10 4 ⎜
⎟⎥ = 716 MΩ Rin = 300kΩ ⎢1+ 10 ⎜
⎟⎥ = 521 MΩ
⎝ 4.19 ⎠⎦
⎝ 5.77 ⎠⎦
⎣
⎣
min
v
Ro
200Ω
=
= 98.0 mΩ
1+ Aβ
10 4
1+
4.9
Ro
200Ω
Ro
200Ω
max
max
min
min
Rout
≅ RoutC
=
=
= 115 mΩ
Rout
≅ RoutC
=
=
= 83.4 mΩ
min
4
max
1+ Aβ
1+ Aβ
10
10 4
1+
1+
5.77
4.19
3MHz
3MHz
f HA = f HB = β A fT =
= 612 kHz
f HC = β B fT =
= 29.7kHz
4.9
101
The bandwidth is controlled by the narrow bandwidth stage.
nom
nom
≅ RoutC
=
Rout
f Hnom = 29.7 kHz
12-42
f Hmin = β Bmin f T =
3MHz
= 24.3 kHz
123
f Hmax = β Bmax f T =
3MHz
= 36.2 kHz
82.8
CHAPTER 13
13.1
Assuming linear operation : vBE = 0.700 + 0.005sin 2000πt V
⎡⎛ 5mV ⎞
⎤
vce = ⎢⎜
⎟(−1.65V )⎥ sin 2000πt = −1.03sin 2000πt V
⎣⎝ 8mV ⎠
⎦
vCE = 5.00 −1.03sin 2000πt V ; 10 - 3300IC ≥ 0.700 → IC ≤ 2.82 mA
13.2
Assuming linear region operation :
vGS = 3.50 + 0.25sin 2000πt V
⎡ 0.25V
⎤
vds = ⎢
−2V )⎥ sin 2000πt = −1.00sin 2000πt V
(
⎣0.50V
⎦
vDS = 4.80 −1.00sin 2000πt V
vDS ≥ vGS − VTN →10 − 3300I D −1.00sin 2000πt ≥ 3.50 + 0.25sin 2000πt −1
For sin 2000πt = 1, I D ≤
10 −1− 3.5 − 0.25 + 1 V
= 1.89 mA
Ω
3300
13.3
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the ac component of the signal at the collector to the output vO.
C3 is a bypass capacitor. (b) The signal voltage at the top of resistor R4 will be zero.
13.4
(a) C1 is a bypass capacitor. C2 is a coupling capacitor that couples the ac component of vI into
the amplifier. C3 is a coupling capacitor that couples the ac component of the signal at the
collector to output vO. (b) The signal voltage at the base will be vb = 0.
13.5
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the
collector to output vO. (b) The signal voltage at the emitter will be ve = 0.
13.6
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the
drain to output vO. (b) The signal voltage at the source will be vs = 0.
13.7
13-1
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the ac component of the signal at the drain to output vO.
13.8
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the
drain to output vO. (b) The signal voltage at the source will be vs = 0.
13.9
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the
collector to output vO. (b) The signal voltage at the emitter will be ve = 0.
13.10
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the signal from the emitter of Q1 back to the node joining R1 and
R2. C3 is a coupling capacitor that couples the ac component of the signal at the emitter to the
output vO. (b) The signal voltage at the collector will be zero.
13.11
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the
drain to the output vO. (b) The signal voltage at the top of R4 will be zero.
13.12
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the ac component of the signal at the drain to output vO.
13.13
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the ac component of the signal at the drain to the output vO.
13.14
dc voltage sources produce constant values of output voltage. Hence no signal voltage can
appear at the terminals of the source. The signal component of the voltage is forced to be zero,
and a direct path to ground is provided for signal currents.
13-2
13.15
NPN Common-Emitter Amplifier
+18 V
270 k Ω
REQ
REQ
Q1
360kΩ
= 5.84 V
360kΩ + 750kΩ
= R1 R2 = 360kΩ 750kΩ = 243 kΩ
VEQ = 18V
(
)
5.84 = 243x103 I B + 0.7 + 91 228.2x103 I B
243 k Ω
I B = 0.245 µA | IC = 90I B = 22.0 µA
VEQ
5.84 V
228.2 k Ω
VCE = 18 − 2.7x105 IC − 2.28x105 I E = 6.99 V
Q − point : (22.0 µA, 6.99 V )
13.16
SPICE results: (a) (22.5 µA, 6.71 V) (b) (22.6 µA, 6.69 V) The discrepancies between the
results in Probs. 13.15 and 13.16 arise because VBE = 0.575 V with IS = 5 fA. Very little
changes occurs with the addition of VA.
13.17
NPN Common-Emitter Amplifier
+12 V
5kΩ
= −4 V
5kΩ + 10kΩ
= R1 R2 = 5kΩ 10kΩ = 3.33 kΩ
VEQ = −12V + 24V
6k Ω
REQ
R EQ
−4 = 3300I B + 0.7 + 76(4000)I B −12
Q1
3.33 k Ω
I B = 23.8 µA | IC = 75I B = 1.78 mA
VEQ
-4 V
4k Ω
-12 V
VCE = 12 − 6000IC − 4000I E − (−12) = 6.08 V
Q − point : (1.78 mA, 6.08 V )
13-3
13.18
*Problem 13.17 - NPN Common-Emitter Amplifier
VCC 7 0 DC 12
VEE 8 0 DC -12
R1 3 8 5K
R2 7 3 10K
RE 4 8 4K
RC 7 5 6K
Q1 5 3 4 NBJT
.OP
.MODEL NBJT NPN IS=1E-15 BF=75 VA=75
.END
Results: IC
1.78E-03
VCE
6.14E+00
VBE
7.28E-01
13.19
PNP Common-Base Amplifier
-7.5 V
33 k Ω
7.5 = 3000IB + 66(68000)IB + 0.7
Q1
IB = 1.51 µA
| IC = 98.4 µA
3k Ω
68 k Ω
+7.5 V
13-4
VEC = 15 − 33000IC − 68000IE = 4.96 V
Q − point : (98.4 µA,4.96 V )
13.20
*Problem 13.19 - PNP Common-Base Amplifier
VEE 1 0 DC 7.5
VCC 5 0 DC -7.5
RC 5 4 33K
RB 3 0 3K
RE 1 2 68K
Q1 4 3 2 PBJT
.OP
.MODEL PBJT PNP IS=1E-16 BF=65 VA=75
.END
Results:
IC
-9.83E-05
VCE
-4.97E+00 VBE
-7.13E-01
13.21
PNP Common-Emitter Amplifier
+9 V
62kΩ
= 6.80V
62kΩ + 20kΩ
= 62kΩ 20kΩ = 15.1kΩ
VEQ = 9V
REQ
3.9 k Ω
9 = 3900(136)IB + 0.7 + 15100IB + 6.80
R EQ
Q1
VEQ
IB = 2.75 µA | IC = 371 µA | IE = 374 µA
15.1 k Ω
+6.80 V
13 k Ω
VEC = 9 − 3900IE −13000IC = 2.72 V
Q − point : (371 µA, 2.72 V )
13.22
*Problem 13.21 - PNP Common-Emitter Amplifier
VCC 4 0 DC 9
RC 1 0 13K
R2 2 0 62K
R1 4 2 20K
RE 4 3 3.9K
Q1 1 2 3 PBJT
.OP
.MODEL PBJT PNP IS=1E-15 BF=135 VA=75
.END
Results:
IC
-3.73E-04
VCE
-2.68E+00 VBE
-6.88E-01
13-5
13.23
NMOS Common-Source Amplifier
1MΩ
= 4.05V
1MΩ + 2.7MΩ
= 1MΩ 2.7MΩ = 730kΩ
VEQ = 15V
+15 V
REQ
82 k Ω
R EQ
M1
VEQ
ID =
0.25mA
2
(VGS −1)
2
VGS = 4.05 − 27000ID
ID = 0.125mA(3.05 − 27000IDS ) → ID = 82.2 µA
2
730 k Ω
27 k Ω
VDS = 15 − 82000ID − 27000ID = 6.04V
4.05 V
Q − point : (82.2 µA, 6.04 V )
13.24
*Problem 13.23 - NMOS Common-Source Amplifier
VDD 4 0 DC 15
RD 4 3 82K
R2 4 2 2.7MEG
R1 2 0 1MEG
R4 1 0 27K
M1 3 2 1 1 NFET
.OP
.MODEL NFET NMOS KP=250U VTO=1
.END
Results:
ID
8.29E-05 VDS
5.96E+00 VGS
1.81E+00
13.25
Depletion-mode NMOS Common-Gate Amplifier
+15 V
4.3 k Ω
M1
3.9 k Ω
5x10−4
2
(VGS + 2) | VGS = −3900ID
2
5x10−4
2
VGS = −3900
(VGS + 2) → VGS = −0.990V
2
V
ID = − GS = 254 µA
3900
VDS = 15 − 4300ID − 3900ID = 12.9V
ID =
Q − point : (254 µA, 12.9 V )
13-6
13.26
*Problem 13.25 - Depletion-mode NMOS Common-Gate Amplifier
VDD 3 0 DC 15
RD 3 2 4.3K
R1 1 0 3.9K
M1 2 0 1 1 NDMOS
.OP
.MODEL NDMOS NMOS KP=500U VTO=-2
.END
Results:
ID
2.54E-04 VDS
1.29E+01 VGS
-9.92E-01
13.27
PMOS Common-Source Amplifier
VEQ = 18V
22 k Ω
REQ
R EQ
M1
VEQ
1.65 M Ω
24 k Ω
9.00 V
3.3MΩ
= 9.00V
3.3MΩ + 3.3MΩ
= 3.3MΩ 3.3MΩ = 1.65MΩ
+18 V
18 = 22000ID − VGS + 9
⎛ 4 x10−4 ⎞
2
9 = 22000⎜
⎟(VGS + 1) − VGS
⎝ 2 ⎠
VGS = −2.24V | ID = 307 µA
VDS = 18 − 22000ID − 24000ID = −3.88 V
Q − point : (307 µA, 3.88 V )
13.28
*Problem 13.27 - PMOS Common-Source Amplifier
VDD 4 0 DC 18
RD 1 0 24K
R2 4 2 3.3MEG
R1 2 0 3.3MEG
R4 4 3 22K
M1 1 2 3 3 PFET
.OP
.MODEL PFET PMOS KP=400U VTO=-1
.END
Results:
ID
-3.07E-04 VDS
-3.86E+00 VGS
-2.24E+00
13-7
13.29
NPN Common-Collector Amplifier
+9 V
9 = 86000IB + 101(82000)IB + 0.7
Q1
IB = 0.992 µA | IC = 99.2 µA
86 k Ω
82 k Ω
VCE = 18 − 82000IE = 18 − 82000(100µA) = 9.80 V
-9 V
Q − point : (99.2 µA,9.80 V )
13.30
*Problem 13.29 - NPN Common-Collector Amplifier
VCC 5 0 DC 9
VEE 8 0 DC -9
R1 3 6 43K
R2 6 0 43K
RE 4 8 82K
Q1 5 3 4 NBJT
.OP
.MODEL NBJT NPN IS=1E-16 BF=100 VA=75
.END
Results:
IC
9.93E-05
VCE
9.79E+00 VBE
13.31
PMOS Common-Gate Amplifier
7.12E-01
12 + VGS 200x10−6
2
=
(VGS −1)
33000
2
VGS = −0.84V | ID = 338 µA
ID =
VDS = −(12 − 33000ID − 22000ID + 12)
M1
+12 V
33 k Ω
22 k Ω
-12 V
VDS = −5.41 V
Q − point : (338 µA, - 5.41 V )
13-8
13.32
*Problem 13.31 - PMOS Common-Gate Amplifier
VDD 4 0 DC 12
VSS 1 0 DC -12
RD 2 1 22K
R1 4 3 33K
M1 2 0 3 3 PFET
.OP
.MODEL PFET PMOS KP=200U VTO=+1
.END
Results:
ID
-3.38E-04
VDS
-5.40E+00 VGS
-8.39E-01
13.33
RS = 1 kΩ, R4 = 1 kΩ
+18 V
3.9 k Ω IG = 0, so VG = 0. Assume active region operation.
2
4x10−4
VGS + 5) | VGS = −2000I D
(
2
M
2
4x10−4
VGS = −2000
VGS + 5) → VGS = −2.50V
(
2
V
I D = − GS = 1.25 mA
2000
2 k Ω VDS = 18 − 3900I D − 2000I D = 10.6V
ID =
RG
10 M Ω
Q − point : (1.25 mA, 10.6 V )
13.34 SPICE results: The Q-point is the same (1.25 mA, 10.6 V) .
13-9
13.35
+15 V
7.5 k Ω
M
RG
2
225x10−6
0 − (−3) = 1.01 mA
IG = 0, so VG = 0 and VGS = 0. I D =
2
VDS = 15 − 7500I D = 7.41V
Q − point : (1.01 mA, 7.41 V )
[
2.2 M Ω
]
13.36 SPICE results: The Q-point is the same (1.01 mA, 7.41 V).
13.37
(a)
+
Q
R
C
R
RI
vi
vo
3
-
R1
R2
RE
(b)
vi
+
+
RI
RB
rπ
ro
v
-
gmv
RE
13-10
RC
R
3
vo
-
(c) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a
coupling capacitor that couples the ac component of the signal at the collector to the output vO.
C3 is a bypass capacitor.
13.38(a)
Figure P13.4
Q1
RI
v
+
R
RE
i
R
C
vo
3
-
(a)
(c )
ro
-
RI
v
R
i
g v
m
v
E
rπ
+
RC
R3
vo
+
(b)
C1 − Bypass
C2 − Coupling
-
C3 − Coupling
13.38(b)
Figure P13.5
RI
+
Q1
vi
R1
R
R
C
v
3
o
R2
(c )
-
(a)
C1 − Coupling
RI
vi
(b)
R1
R2
rπ
+
+
v
vo
-
gmv
ro
R
C
R
3
-
C2 − Bypass
C3 − Coupling
13-11
13.39(a)
Figure P13.9
+
Q
R
RI
R
vi
R
C
v
3
o
R
1
(c )
-
2
(a)
C1 − Coupling
RI
vi
R
R
1
(b)
rπ
2
+
+
v
v
g v
-
m
r
o
R
C
R
3
Q1
RI
R
1
vi
+
R
RE
2
(a)
R
v
3
o
-
(c )
+
RI
R1
vi
13-12
C2 − Bypass
-
C3 − Coupling
13.39(b)
Figure P13.10
(b)
o
rπ
v
-
C1 − Coupling
ro
g v
m
C2 − Coupling
+
R2
RE
R
v
3
-
o
C3 − Coupling
13.40(a)
Figure P13.6
R
I
+
M
1
R
R
D
vi
R1
v
3
o
R
(c )
-
2
(a)
C1 − Coupling
+
R
I
vi
+
v
R1
R
2
-
(b)
r
gm v
R
o
D
R
vo
3
C2 − Bypass
-
C3 − Coupling
13.40(b)
Figure P13.7
M
1
+
RI
vi
R
D
R1
R
v
3
o
-
(a)
(c )
ro
-
RI
vi
(b)
R
1
v
+
+
gmv
RD
R
3
vo
C1 − Coupling
-
C2 − Coupling
13-13
13.41(a)
Figure P13.8
+
M
1
RD
RI
vi
R1
R
v
3
o
R2
(c )
-
(a)
C1 − Coupling
+
RI
vi
+
v
R1
R2
gmv
-
(b)
ro
R
RD
vo
3
C2 − Bypass
-
C3 − Coupling
13.42(a)
Figure P13.12
M1
+
RI
vi
RD
R
1
R
vo
3
-
(a)
(c )
r
o
-
RI
vi
(b)
R
v
1
+
+
gmv
RD
R
3
v
-
o
C1 − Coupling
C2 − Coupling
13.43
RI: Thévinen equivalent source resistance; R1: base bias voltage divider; R2: base bias voltage
divider; RE and R4: emitter bias resistors - determine the emitter current; RC: collector bias
resistor - sets the collector-emitter voltage; R3: load resistor
13.44
RI: Thévinen equivalent source resistance; R1: gate bias voltage divider; R2: gate bias voltage
divider; R4: source bias resistor - sets source current; RD: drain bias resistor - sets drain-source
voltage; R3: load resistor
13-14
13.45
RI: Thévinen equivalent source resistance; R1: base bias voltage divider; R2: base bias voltage
divider; RE: emitter bias resistor - determines the emitter current; RC: collector bias resistor - sets
the collector-emitter voltage; R3: load resistor
13.46
(a) r
=
(b) I
= 0 rd =
d
D
⎛ ⎛ 0.6 ⎞ ⎞
0.025
| I D = 10−14 ⎜ exp⎜
= 94.4 Ω
⎟ − 1⎟ = 264.9µA | rd =
264.9µA + 10 fA
⎝ ⎝ 0.025 ⎠ ⎠
VT
ID + IS
VD < VT ln
0.025
0.025
= 2.50 TΩ (c)
> 1015 → I D + I S < 2.5 x10−17 A
ID + IS
10 fA
ID + IS
2.5x10−17
= 0.025ln
= −0.150 V
IS
10−14
13.47
kT 1.38x10−23
V
=
T = 8.63x10−5 T | rd ≅ T = 1000VT
VT =
−19
ID
q 1.60x10
T
VT
rd
13.48
⎛ 0.005 ⎞
75K
6.47 mV
6.47 Ω
(a) exp⎜⎝ 0.025⎟⎠ −1 = 0.221
100K
8.63 mV
8.63 Ω
|
⎛ 0.005 ⎞
exp⎜−
⎟ −1 = −0.181
⎝ 0.025 ⎠
⎛ 0.010 ⎞
(b) exp⎜⎝ 0.025 ⎟⎠ −1 = 0.492 |
⎛ 0.010 ⎞
exp⎜−
⎟ −1 = −0.330
⎝ 0.025 ⎠
200K
17.3 mV
17.3 Ω
300K
25.9 mV
25.9 Ω
400K
34.5 mV
34.5 Ω
0.005
= 0.200 → +10.7% error
0.025
| −
0.005
= −0.200 → −9.37% error
0.025
0.010
= 0.400 → +23.0% error
0.025
| −
0.010
= −0.400 → −17.5% error
0.025
13.49
g m 0.03
=
= 0.750 mA = 750 µA
40
40
g m 50x10−6
c
I
=
( ) C 40 = 40 = 1.25 µA
(a) IC =
(b) IC =
g m 250x10−6
=
= 6.25 µA
40
40
13-15
13.50
IC =
β oVT
rπ
=
75(0.025V )
10 4 Ω
= 187.5 µA | Q - point : (188 µA,VCE ≥ 0.7 V )
(
)
g m = 40IC = 40 1.875x10−4 = 7.50 mS | ro =
13.51
IC =
β oVT
rπ
=
125(0.025V )
= 1.56 µA | Q - point : (1.56 µA,VCE ≥ 0.7 V )
2x106 Ω
(
)
g m = 40IC = 40 1.56x10−6 = 62.4 µS | ro =
13.52
IC =
β oVT
rπ
=
100(0.025V )
250kΩ
( )
IC =
β oVT
rπ
=
75(0.025V )
106 Ω
(
VA + VCE VA
75V
≅
=
= 48.1 MΩ
IC
IC 1.56 µA
= 10 µA | Q - point : (10 µA,VCE ≥ 0.7 V )
g m = 40IC = 40 10−5 = 0.400 mS | ro =
13.53
VA + VCE VA 100V
≅
=
= 10 MΩ
IC
IC 10µA
= 1.875 µA | Q - point : (1.88 µA,VCE ≥ 0.7 V )
)
g m = 40IC = 40 1.875x10−6 = 75.0 µS | ro =
13-16
VA + VCE VA
100V
≅
=
= 533 kΩ
IC
IC 187.5 µA
VA + VCE VA
100V
≅
=
= 53.3 MΩ
IC
IC 1.875 µA
13.54
V + VCE
ro = A
; solving for VA : VA = IC ro − VCE
IC
Using the values from row 1: VA = 0.002(40000)−10 = 70 V
Using the values from the second row : β o = g mrπ = 0.12(500)= 60 and β F = β o = 60.
Row 1: g m = 40IC = 40(0.002) = 0.08 S | rπ =
µ f = g mro = 0.08(40000) = 3200
Row 2 : IC =
βo
gm
=
60
= 750 Ω
0.08
g m 0.12
V + VCE
80
=
= 3 mA | ro = A
=
= 26.7 kΩ
40
IC
40
0.003
µ f = g mro = 0.12(26700) = 3200
Row 3 : g m =
ro =
βo
rπ
=
60
g m 1.25 x 10-4
-4
=
1.25
x
10
S
|
I
=
=
= 3.13 µA
C
40
40
4.8 x 105
VA + VCE
80
=
= 25.6 MΩ | µ f = g mro = 1.25 x 10-4 25.6 x 106 = 3200
IC
3.13 x 10-6
13.55
(
⎛ 0.005 ⎞
(a) exp⎜⎝ 0.025⎟⎠ −1 = 0.221
|
)
0.005
= 0.200 → +10.7% error
0.025
⎛ 0.005 ⎞
0.005
= −0.200 → −9.37% error
exp⎜−
⎟ −1 = −0.181 | −
0.025
⎝ 0.025 ⎠
⎛ 0.0075⎞
0.0075
(b) exp⎜⎝ 0.025 ⎟⎠ −1 = 0.350 | 0.025 = 0.300 → +16.7% error
⎛ 0.0075 ⎞
0.0075
= −0.300 → −13.6% error
exp⎜−
⎟ −1 = −0.259 | −
0.025
⎝ 0.025 ⎠
⎛ 0.0025 ⎞
0.0025
(c) exp⎜⎝ 0.025 ⎟⎠ −1 = 0.105 | 0.025 = 0.100 → +5.17% error
⎛ 0.0025 ⎞
0.0025
= −0.100 → −4.84% error
exp⎜−
⎟ −1 = −0.0952 | −
0.025
⎝ 0.025 ⎠
13.56
(a) β F =
IC 350µA
∆I
600µA −125µA
≅
≅ 90 | β o = C ≅
≅ 120
4µA
6µA − 2µA
IB
∆I B
(b) β F ≅
750µA
900µA − 600µA
≅ 95 | β o ≅
≅ 75
8µA
4µA
13.57(a)
13-17
t=linspace(0,.004,1024);
ic=.001*exp(40*.005*sin(2000*pi*t));
IC=fft(ic);
z=abs(IC(1:26)/1024);
z(1)
ans = 0.001
plot(t,ic)
x10-3
1.2
1
0.8
0
1
2
3
4
x10-3
z([5 9 13])
ans = 0.0001 0.0000 0.0000
13.57(b)
t=linspace(0,.004,1024);
ic=.001*exp(40*.005*sin(2000*pi*t));
IC=fft(ic);
z=abs(IC(1:26)/1024);
z(1)
ans = 0.0023
plot(t,ic)
-3
8 x10
6
4
2
0
0
1
z([5 9 13])
ans = 0.0016 0.0007 0.0002
13-18
2
3
4
x10-3
13.58 (a)
NAME
Q1
MODEL
NBJT
IB
2.21E-05
IC
1.78E-03
VBE
7.28E-01
VBC
-5.41E+00
VCE
6.14E+00
BETADC
8.04E+01
GM
6.87E-02
RPI
1.17E+03
RX
0.00E+00
RO
4.52E+04
BETAAC
8.04E+01
T = 27 o C | VT = 8.625x10−5 (300) = 25.9mV
IC 1.78mA
=
= 68.7 mS
VT 25.9mV
⎛ V ⎞
⎛ 6.14 ⎞
β o = β FO ⎜1+ CE ⎟ = 75⎜1+
⎟ = 81.1
⎝
75 ⎠
⎝ VA ⎠
β
81.1
= 1180 Ω
rπ = o =
gm 0.0687
gm =
ro =
VA + VCE 75 + 6.14
=
= 45.6 kΩ
IC
1.78mA
13 - 19
13.58(b)
MODEL
PBJT
IB
-2.69E-06
IC
-3.73E-04
VBE
-6.88E-01
VBC
1.99E+00
VCE
-2.68E+00
BETADC
1.39E+02
GM
1.44E-02
RPI
9.61E+03
RX
0.00E+00
RO
2.06E+05
BETAAC
1.39E+02
T = 27 o C | VT = 8.625x10−5 (300) = 25.9mV
IC 0.373mA
=
= 14.4 mS
VT 25.9mV
⎛ V ⎞
⎛ 2.68 ⎞
β o = β FO ⎜1+ CE ⎟ = 135⎜1+
⎟ = 140
⎝
75 ⎠
⎝ VA ⎠
gm =
rπ =
ro =
βo
gm
=
140
= 9.72 kΩ
0.0144
VA + VCE 75 + 2.68
=
= 208 kΩ
IC
0.373mA
Note: The SPICE model actually is using VCB instead of VCE in the current gain calculations.
⎛
⎝
(a) β o = 75⎜1+
13-20
5.41⎞
⎟ = 80.4
75 ⎠
⎛
⎝
(b) β o = 135⎜1+
1.99 ⎞
⎟ = 139
75 ⎠
13.59
ix
+
rπ
y 11
v
-
gm v
re
vx
ro
ie
αo ie
1
rπ
v
v 1− α o
For the T - model : ix = x − α o x =
vx
re
re
re
For the hybrid pi model: y11 =
y11 =
re =
ix 1− α o
=
=
vx
re
1−
βo
βo + 1
re
=
1
→ r = (β o + 1)re
(β o + 1)re π
rπ
βo
α αV V
=
= o= o T = T
IE
(β o + 1) gm (β o + 1) gm IC
13.60
g m = 40(50µA) = 2.00mS | rπ =
100
75V + 10V
= 50kΩ | ro =
= 1.70 MΩ
2.00mS
50µA
RBB = RB rπ = 100kΩ 50kΩ = 33.3kΩ
⎛
⎞
33.3kΩ
Av = −⎜
⎟(2mS ) 1.70 MΩ 100kΩ 100kΩ = −95.0
⎝ 33.3kΩ + 0.75kΩ ⎠
(
)
13.61
For β o = 100, see Prob. 13.60.
60
= 30kΩ | RBB = RB rπ = 100kΩ 30kΩ = 23.1kΩ
2.00mS
⎛
⎞
23.1kΩ
Av = −⎜
⎟(2mS ) 1.70 MΩ 100kΩ 100kΩ = −94.1
⎝ 23.1kΩ + 0.75kΩ ⎠
−95.0 ≤ Av ≤ −94.1 − only a small variation
rπ =
(
)
13.62
75
50 + 7.5 V
= 750Ω | ro =
= 23.0kΩ
2.5 mA
0.1S
= RB rπ = 4.7kΩ 750Ω = 647Ω
g m = 40(2.5mA)= 0.100S | rπ =
RBB
⎛ 647Ω ⎞
Av = ⎜
⎟(−0.100S ) 23.0kΩ 4.3kΩ 10kΩ = −247
⎝ 50Ω + 647Ω ⎠
(
)
13-21
13.63
g m = 40(1µA)= 40µS | rπ =
40
50 + 1.5 V
= 1MΩ | ro =
= 51.5MΩ
µA
40µS
1
RBB = RB rπ = 5MΩ 1MΩ = 833kΩ
⎛
⎞
833kΩ
Av = ⎜
⎟(−40µS ) 51.5MΩ 1.5MΩ 3.3MΩ = −40.0
⎝10kΩ + 833kΩ ⎠
(
)
13.64
SPICE Results: IC = 248 µA, VCE = 3.30 V, AV = -15.1 dB -- IC differs by 1.2% - AV is off by
0.5%
13.65
[10(V )] = [10(9)]
N
CC
N
≥ 20000 → N ≥
log(20000)
log(90)
= 2.20 → N = 3
(or quite possibly 2 by droping a larger fraction of the supply voltage across RC )
13.66
*Problem 13.21 - PNP Common-Emitter Amplifier - Figure P13.5
VCC 7 0 DC 9
VI 1 0 AC 1
RI 1 2 1K
C1 2 3 100U
R1 7 3 20K
R2 3 0 62K
RE 7 4 3.9K
C 7 4 100U
RC 5 0 13K
C3 5 6 100U
R3 6 0 100K
Q1 5 3 4 PBJT
.OP
.MODEL PBJT PNP IS=1E-15 BF=135 VA=75
.AC DEC 10 100Hz 10000Hz
.PRINT AC VM(6) VDB(6) VP(6)
.END
Results: IC = 373 µA, VEC = 2.68V, AV = -134
Hand calculations in Prob. 13.21 yielded (371 µA, 2.72V)
135
= 9.12kΩ | ro = ∞
14.8mS
RB = R1 R2 = 20kΩ 62kΩ = 15.1kΩ | RBB = 15.1kΩ 9.12kΩ = 5.69kΩ
gm = 40(371µA) = 14.8mS | rπ =
⎛ 5.69kΩ ⎞
Av = −⎜
⎟(14.8mS ) ∞ 13kΩ 100kΩ = −145
⎝ 1kΩ + 5.69kΩ ⎠
(
13-22
)
SPICE AV result is somewhat lower because ro is included.
13.67
Av ≅ −10VCC = −10(12) = −120
13.68
Av ≅ −10(VCC + VEE ) = −10(15 + 15) = −300
13.69
Av = −10(VCC + VEE )= −10(1.5 + 1.5)= −30; This estimate says no.
However, if we look a bit deeper,
Av = −40(IC RC ) = −40VRC ,and we let VRC =
then we can achieve A v = −40(1.5)= −60.
(V
CC
+ VEE )
2
= 1.5V ,
So, with careful design, we can probably achieve a gain of 50.
13.70
Using our rule - of - thumb estimate,
1 = −10
Av ≅ −10VCC = −10(1.5)= −15 | Av = −10()
Note that this result assumes that IC varies with VCC.
13.71
5V
(a) i = 10kΩ = 0.5 mA, but i ≤ 0.2I for small - signal operation.
(b) V ≥ V + i R + I R = 0.7 + 5 + 25 = 30.7 V
c
CC
c
BE
c
L
C
C
So IC ≥ 5ic = 2.5 mA.
L
13.72
Av = 40dB = 100 | v o = 100v be = 100(0.005V ) = 0.500V .
13-23
13.73
For common - emitter stage : Av = 50dB → Av = −316
15V
= 47.5mV which is far too big for small- signal operation.
316
The will be significant distortion of the sine wave.
vbe =
13.74
20kΩ
18 = −4.61V | REQ = 20kΩ 62kΩ = 15.1kΩ
62kΩ + 20kΩ
−4.61− 0.7 − (−9)
= 6.76µA | IC = 135I B = 913µA
IB =
15.1kΩ + 136(3.9kΩ)
(a) V
EQ
= −9 +
VCE = 9 −13000IC − 3900I E − (−9) = 2.54V
g m = 40IC = 0.0365S | rπ =
135
= 3.70kΩ | ro = ∞ | R L = 13kΩ 100kΩ = 11.5kΩ
gm
⎛ 2.97kΩ ⎞
Av = −⎜
⎟(0.0365)(11.5kΩ)= −314
⎝ 1kΩ + 2.97kΩ ⎠
(b) For V
CC
= 18V , the answers are the same : IC = 913µA | VEC = 2.54V | Av = −314
13.75
Using the information from Row 1:
1
λ
(
)(
)
= I D ro − VDS = 8x10−4 4x10 4 − 6 = 26V ; λ = 0.0385V -1
gm2
From Row 2 : K n =
=
2ID (1+ λVDS )
(2x10 ) = 3.25x10
⎛
6⎞
2(5x10 )⎜1+ ⎟
⎝ 26 ⎠
−4 2
−4
−5
A
V2
⎛
6⎞
Row 1: gm = 2K n ID (1+ λVDS ) = 2(3.25x10−4 )(8x10−4 )⎜1+ ⎟ = 8x10−4 S
⎝ 26 ⎠
µ f = gm ro = 8x10−4 (4 x10 4 )= 32
2ID
= 0.2
0.2(VGS − VTN ) = 0.2
K n (1+ λVDS )
13-24
2(8x10−4 )
= 0.40V
6⎞
−4 ⎛
3.25x10 ⎜1+ ⎟
⎝ 26 ⎠
1
Row 2 : ro = λ
+ VDS
ID
=
26 + 6
= 640kΩ | µ f = gm ro = 2x10−4 (6.4 x10 5 )= 128
−5
5x10
2(5x10−5 )
= 0.10V
6⎞
−4 ⎛
3.25x10 ⎜1+ ⎟
⎝ 26 ⎠
⎛
6⎞
Row 3 : gm = 2K n ID (1+ λVDS ) = 2(3.25x10−4 )(10−2 )⎜1+ ⎟ = 2.83x10−3 S
⎝ 26 ⎠
2ID
0.2(VGS − VTN ) = 0.2
= 0.2
K n (1+ λVDS )
1
ro = λ
+ VDS
=
ID
26 + 6
= 3.2kΩ | µ f = gm ro = 2.83x10−3 (3.2x10 3 )= 9.06
0.01
0.2(VGS − VTN ) = 0.2
2ID
= 0.2
K n (1+ λVDS )
2(10−2 )
= 1.41V
6⎞
−4 ⎛
3.25x10 ⎜1+ ⎟
⎝ 26 ⎠
MOSFET Small-Signal Parameters
ID
(S)
gm
(Ω)
ro
µf
Small-Signal Limit
vgs (V)
0.8 mA
0.0008
40,000
32
0.40
50 µA
0.0002
640,000
128
0.10
10 mA
0.00283
3200
9.06
1.41
13.76
⎛1
⎞ 2Kn (1+ λVDS ) ⎛ 1 ⎞ 2Kn ⎛ 1 ⎞ 2Kn' ⎛ W ⎞
µ f = ⎜ + VDS ⎟
≅⎜ ⎟
=⎜ ⎟
⎜ ⎟
ID
⎝λ
⎝ λ ⎠ ID
⎝ λ ⎠ ID ⎝ L ⎠
⎠
2 2x10-4
2 I
W
50
= (µ f λ ) D ' = 250(0.02)
=
L
1
2Kn
2 5x10-5
[
VGS − VTN
]
(
(
(
)
)
)
2 2x10-4
2I D
≅
=
= 0.160 V
Kn
50 5x10-5
13-25
13.77
−4
⎛1
⎞ 2K n (1+ λVDS ) ⎛ 1 ⎞ 2K n
⎛ 1 ⎞ 2(2.5x10 )
µ f = ⎜ + VDS ⎟
≅⎜ ⎟
| ⎜ ⎟
≤ 1 → ID ≥ 1.25 A
⎝λ
⎠
⎝ λ ⎠ ID
⎝ .02 ⎠
ID
ID
13.78
gm =
2I D
VGS − VTN
| ID =
(.005)(0.5) = 1.25 mA
2
|
2(1.25mA) 250
W
2I D
=
=
=
2
L K ' (V − V ) 4x10−5 (0.5)2
1
n
GS
TN
13.79
2
(1+ 0.2) −1 = 0.44 | 2(0.2) = 0.40 →10% error
(1+ 0.4)
2
−1 = 0.96 | 2(0.4) = 0.80 → 20% error
13.80
From the results of Problem 13.24:
ID = 8.29E-05, VGS = 1.81E+00, VDS=5.96E+00, GM = 2.04E-04, GDS = 0.00E+00
gm =
2(82.9µA)
2I D
1
=
= 205 µS | λ = 0 → ro = ∞ | g ds = = 0
VGS − VTN
1.81−1
ro
13.81
From the results of Problem 13.28:
ID = 3.07E-04, VGS = -2.24E+00, VDS=-3.86E+00, GM = 4.96E-04, GDS = 0.00E+00
2(307µA)
2I D
1
=
= 495 µS | λ = 0 → ro = ∞ | g ds = = 0
ro
VGS − VTP −2.24 + 1
gm =
13.82
Rout =
ID =
RD ro
V
9V
50V + 9V
7.8V
| Using VDS = DD | RD =
| ro =
→ Rout =
RD + ro
2
ID
ID
ID
7.8V
= 156µA → RD = 57.6kΩ | ro = 378kΩ | Q - point : (156 µA, 9 V )
50kΩ
13.83
Virtually any Q-point is possible. RIN is set by RG which can be any value desired since there is
no gate current. (Note this is not the case with a BJT for which base current must be
considered.)
13-26
13.84
Note that iG ≅ 0 for this device.
Load line : 400 = 133000iP + v PK and vGK = −1.5V
Two points (i P , v PK ) : (3mA, 0V) and (0mA, 400V) → Q - pt : (1.4 mA, 215 V)
ro =
250V − 200V
2.3mA − 0.7mA
= 55.6kΩ | g m =
= 1.6mS | µ f = 89.0
2.15mA −1.25mA
−1V − (−2V )
(
)
(
)
Av = −g m RP ro = −1.6mS 133kΩ 55.6kΩ = −62.7
13.85
gm2
(0.5S) = 5 A!
BJT : IC = gmVT = 0.5S (0.025V ) = 12.5mA | MOSFET : ID =
=
2K n 2(25mA /V 2 )
2
The BJT can achieve the required transconductance at a 400 times lower current than the
MOSFET. For a given power supply voltage, the BJT will therefore use 400 times less power.
60
= 120Ω (versus ∞ for the FET).
Note, however, that rπ is small for the BJT: rπ =
0.5
13.86
Since a relatively high input resistance is required at a relatively high current, a FET should be
used. If a BJT were selected, it would be very difficult to achieve the required input resistance
because its value of rπ is low:
β V 100(.025V )
rπ = o T =
= 250 Ω
IC
10mA
13.87
⎛1
⎞ 2Kn (1+ λVDS )
40(VA + VCE )= ⎜ + VDS ⎟
ID
⎝λ
⎠
40(35) = 60
2(0.025)(1.2)
ID
→ I D = 111 µA | µ f = 40(35) = 1400
13.88
V
I
µ f ≅ A = 40VA = 40(50)= 2000 | g m = C = 40IC = 40 2x10−4 = 8.00 mS
VT
VT
(
)
(
)
2 2x10−4
2
2
2I D
µf ≅
=
= 200 | g m =
=
= 0.800 mS
VGS − VTN
0.5
λ(VGS − VTN ) 0.02(0.5)
13-27
13.89
Either transistor could be used. For a BJT operating in the common-emitter configuration or an
FET operating in the common-source configuration:
100(0.025V )
= 33.3 mA - A fairly high current
rπ
75Ω
For the FET : Rin = RG and setting RG = 75 Ω is satisfactory, particularly if a depletion mode FET is available.
The input of a CE circuit operating at a much lower current could also be swamped by the
addition of a 75-Ω resistor in parallel with its input. (Note that common-base and common-gate
amplifiers from Chapter 14 could also be used.)
For the BJT : Rin ≅ rπ | BJT : IC =
β oVT
=
13.90
26
(a) Av = 10 20 = 20.0. Either a BJT or MOSFET can achieve the required gain. However, based
upon the material in Chapter 13, an FET should be chosen since the input voltage of 0.25 V is 50
time larger than the permissible value for vbe (0.005 V) for the BJT. For the FET, a value of VGS
– VTN = 1.25 V will satisfy the small-signal limit with vgs = 0.25 V. (The generalized commonemitter stage with emitter degeneration can also satisfy the requirements.) (b) The FET is also
best for this case, since the amplifier will see input signals much greater than the 5-mV limit of
the BJT.
13.91
Av ≅ −
(12) = −12 or 21.6 dB
VDD
=−
VGS − VTN
1
13.92
15
7.5V
vd = = 7.5V peak | 15dB → Av = −5.62 | vgs =
= 1.34V | VGS − VTN ≥ 5(1.34)= 6.70V
2
5.62
Yes, it is possible although the required value of (VGS − VTN ) is getting rather large.
13.93
Av ≅
VDD
VGS − VTN
|
9
≥ 30 → VGS − VTN ≤ 0.300 V
VGS − VTN
13.94
For VDS =
ID =
VDD
VDD
, Av = −
2
VGS − VTN
| 30 =
15
VGS − VTN
| VGS − VTN = 0.5 V
2
1mA
VGS − VTN ) = 125 µA | Q - point : (125 µA,7.5 V )
(
2
13-28
13.95
0.1
= 0.5V
0.2
Av = 35dB → Av = −56.2. Using the rule - of - thumb estimate to select VDD :
vgs ≤ 0.2(VGS − VTN ) requires (VGS − VTN ) ≥
Av = −
and VDD = 56.2(0.5V )= 28 V
VDD
VGS − VTN
13.96
0.5
= 2.5V
0.2
Av = 20dB → Av = −10. Using the rule - of - thumb estimate to select VDD :
vgs ≤ 0.2(VGS − VTN ) requires (VGS − VTN ) ≥
Av = −
and VDD = 10(2.5) = 25 V
VDD
VGS − VTN
13.97
⎛ VDD ⎞ N
⎛ 10 ⎞ N
We desire ⎜
⎟ ≥ 1000 | ⎜
⎟ ≥ 1000
⎝ VGS − VTN ⎠
⎝ VGS − VTN ⎠
For VGS − VTN = 1 V , N = 3 meets the requirements, but with no safety margin.
For VGS − VTN = 0.75 V , N = 3 easily meets the requirements.
13.98
For the bias network : VEQ = 10V
430kΩ
= 4.343V | REQ = 430kΩ 560kΩ = 243kΩ
430kΩ + 560kΩ
2
5x10−4
VGS −1) | VGS = 4.343 − 2x10 4 I D → VGS = 1.72 V | I D = 131 µA
(
2
= 10 − 63kΩ(131µA)= 1.75V ≥ VGS − VTN so active region assumption is ok.
ID =
VDS
⎛ 1
⎞
+ 1.75⎟
⎜
⎝ 0.0133
⎠
g m = 2 5x10−4 (131µA) = 362µS | ro =
= 586kΩ
131µA
243kΩ
Av = −
(362µS ) 586kΩ 43kΩ 100kΩ = −10.3
243kΩ + 1kΩ
(
)
(
)
13.99
⎛
µA ⎞
50 + 5V
g m = 2⎜ 500 2 ⎟(100µA)(1+ 0.02(5)) = 332µS | ro =
= 550kΩ
100µA
V ⎠
⎝
⎛
⎞
6.8 MΩ
Av = −⎜
⎟(332µS ) 550kΩ 50kΩ 120kΩ = −10.9
⎝ 6.8 MΩ + 0.1MΩ ⎠
(
)
13-29
13.100
(
)
(
)
g mmax = 2 700µA/V 2 (100µA) = 374µS | g mmin = 2 300µA/V 2 (100µA) = 245µS
⎛
⎞
6.8 MΩ
Av = −⎜
⎟(g m ) 550kΩ 50kΩ 120kΩ = (−32.7kΩ)(g m )
⎝ 6.8 MΩ + 0.1MΩ ⎠
(
)
Avmax = −12.2 | Avmin = −8.01
13.101
(
)
g m = 2 100µA/V 2 (10µA)(1+ 0.02(5)) = 46.9µS | ro =
50 + 5V
= 5.50 MΩ
10µA
⎛
⎞
10 MΩ
Av = −⎜
⎟(46.9µS ) 5.50 MΩ 560kΩ 2.2 MΩ = −19.2
⎝ 10 MΩ + 0.1MΩ ⎠
(
13.102
)
(
)
g m = 2Kn I DS (1+ λVDS ) = 2(0.001)(0.002) 1+ 0.015(7.5) = 2.11x10−3 S
1
ro = λ
+ VDS
I DS
(
1
+ 7.5
0.015
=
= 37.1kΩ
0.002
)
(
)
Avt = −g m ro RD R3 = −2.11x10−3 37.1kΩ 3.9kΩ 270kΩ = −7.35
Av =
10 MΩ
Avt = −7.34
10kΩ + 10 MΩ
13-30
13.103
*Problem 13.103 - NMOS Common-Source Amplifier - Figure P13.6
VDD 7 0 DC 15
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VI 1 0 AC 0
VI 1 0 AC 1
RI 1 2 1K
C1 2 3 100UF
R1 3 0 1MEG
R2 7 3 2.7MEG
R4 4 0 27K
C2 4 0 100UF
RD 7 5 82K
C3 5 6 100UF
R3 6 0 470K
M1 5 3 4 4 NFET
.OP
.MODEL NFET NMOS KP=250U VTO=1
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI)
*.PRINT AC IM(C3) IP(C3)
.END
Results: ID = 8.29E-05 VGS = 1.81E+00 VDS = 5.96E+00
VM(6) = 1.420E+01 VP(6) = -1.800E+02 IM(VI) = 1.369E-06 IP(VI) = -1.800E+02
VM(3) = 9.986E-01 VP(3) = 1.248E-04 IM(C3) = 1.220E-05 IP(C3) = -1.800E+02
Av = −14.6 | Rin =
VM (3) 0.9986
1
1
=
= 729 kΩ | Rout =
=
= 82.0 kΩ
IM (C3) 12.20µA
IM (VI ) 1.369µA
13-31
13.104
*Problem 13.104 - PMOS Common-Source Amplifier - Figure P13.8
VDD 7 0 DC 18
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VI 1 0 AC 0
*
VI 1 0 AC 1
RI 1 2 1K
C1 2 3 100U
R2 7 3 3.3MEG
R1 3 0 3.3MEG
R4 7 4 22K
C2 7 4 100U
RD 5 0 24K
C3 5 6 100U
R3 6 0 470K
M1 5 3 4 4 PFET
.OP
.MODEL PFET PMOS KP=400U VTO=-1
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: ID = 3.07E-04
VGS = -2.24E+00
VDS = -3.86E+00
VM(6) = 12.76E+01 VP(6) = -1.799E+02 IM(VI) = 6.057E-07 IP(VI) = -1.800E+02
VM(3) = 9.994E-01 VP(3) = 5.523E-05 IM(C3) = 4.167E-05
Av = −12.8 | Rin =
13-32
IP(C3) = -1.800E+02
VM (3) 0.9994V
1
1
=
= 1.65 MΩ | Rout =
=
= 24.0 kΩ
IM (C3) 41.67µA
IM (VI ) 0.6057µA
13.105
*Problem 13.105 - Depletion-mode NMOS Common-Source Amplifier - Figure P13.11
VDD 7 0 DC 18
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VI 1 0 AC 0
*
VI 1 0 AC 1
RI 1 2 10K
C1 2 3 100UF
RG 3 0 10MEG
R1 4 0 2K
C3 4 0 100UF
RD 7 5 3.9K
C2 5 6 100UF
R3 6 0 36K
J1 5 3 4 NFET
.OP
.MODEL NFET NMOS KP=400U VTO=-5
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI) VM(3) VP(3)
*.PRINT AC IM(C2) IP(C2)
.END
Results: ID = 1.25E-03
VGS = -2.50E+00
VDS = -1.06E+01
VM(6) = 3.515E+00 VP(6) = -1.799E+02 IM(VI) = 9.991E-08 IP(VI) = -1.800E+02
VM(3) = 9.990E-01 VP(3) = 9.106E-06 IM(C3) = 2.564E-04 IP(C3) = -1.800E+02
Av = −3.52 | Rin =
VM (3) 0.9990V
1
1
=
= 10.0 MΩ | Rout =
=
= 3.90 kΩ
IM (C3) 256.4 µA
IM (VI ) 99.91nA
13.106
SPICE Results: Q-point: (1.01 mA, 7.41 V), Av = 15.1 dB, Rin = 2.20 MΩ, Rout = 7.50 kΩ
13.107
g m = 40(50µA) = 2.00mS | rπ =
100
75V + 10V
= 50kΩ | ro =
= 1.70 MΩ
50µA
2.00mS
Rin = RB rπ = 100kΩ 50kΩ = 33.3kΩ | Rout = 1.7 MΩ 100kΩ = 94.4kΩ
13-33
13.108
Rin = RB rπ | rπmin =
60
100
= 30kΩ | rπmax =
= 50kΩ
40(50µA)
40(50µA)
Rinmin = RB rπ = 100kΩ 30kΩ = 23.1kΩ | Rinmax = RB rπ = 100kΩ 50kΩ = 33.3kΩ
Rout = RC ro = 100kΩ
75 + 10
= 100kΩ 1.7 MΩ = 94.4kΩ independent of βo
50µA
13.109
40(0.025V )
50 + 1.5 V
= 1.00 MΩ | ro =
= 51.5MΩ
rπ =
1
1µA
µA
Rin = RB rπ = 5MΩ 1MΩ = 833 kΩ | Rout = RC ro = 1.5MΩ 51.5MΩ = 1.46 MΩ
13.110
75(0.025V )
50 + 7.5 V
rπ =
= 750Ω | ro =
= 23.0kΩ
2.5mA
2.5 mA
Rin = RB rπ = 4.7kΩ 0.75kΩ = 647 Ω | Rout = RC ro = 4.3kΩ 23kΩ = 3.62kΩ
13.111
From Prob. 13.98 : Q - Point = (131µA, 1.75V)
Rin = R1 R2 = 430kΩ 560kΩ = 243 kΩ | Rout = 43kΩ ro
⎛ 1
⎞
+ 1.75⎟V
⎜
⎝ 0.0133
⎠
ro =
= 587kΩ | Rout = 43kΩ 587kΩ = 40.1kΩ
0.131mA
13.112
Rin = RG = 6.8 MΩ | Rout = 50kΩ ro
ro =
(50 + 5)V = 550kΩ
0.1mA
| Rout = 50kΩ 550kΩ = 45.8kΩ
13.113
Rin = RG = 6.8 MΩ which is independent of Kn | Rout = RD ro
⎛ 1
⎞
+ 5⎟V
⎜
⎝ 0.02 ⎠
ro =
= 550kΩ | Rout = 50kΩ 550kΩ = 45.8 kΩ, also independent of Kn
0.1mA
13-34
13.114
Rin = RG = 10 MΩ | Rout = RD ro
⎛ 1
⎞
+ 5⎟V
⎜
⎝ 0.02 ⎠
| ro =
= 5.50 MΩ
10µA
Rout = 560kΩ 5.50 MΩ = 508 kΩ
13.115
⎛ 1
⎞
+ 7.5⎟V
⎜
⎝ 0.015
⎠
| ro =
= 37.1 kΩ
2mA
Rin = RG = 1MΩ | Rout = RD ro
Rout = 3.9kΩ 37.1kΩ = 3.53 kΩ
13.116
g m = 40(50µA) = 2.00mS | rπ =
100
75V + 10V
= 50kΩ | ro =
= 1.70 MΩ
2.00mS
50µA
RBB = RB rπ = 100kΩ 50kΩ = 33.3kΩ
⎛
⎞
33.3kΩ
vth = −vi ⎜
⎟(2mS ) 1.70 MΩ 100kΩ = −185vi
⎝ 33.3kΩ + 0.75kΩ ⎠
(
)
Rth = 1.70 MΩ 100kΩ = 94.4 kΩ
13.117
75
50V + 7.5V
= 750Ω | ro =
= 23.0kΩ
100mS
2.5mA
= RB rπ = 4.7kΩ 0.75kΩ = 647Ω
g m = 40(2.5mA) = 100mS | rπ =
RBB
⎛ 647Ω ⎞
vth = −vi ⎜
⎟(100mS ) 23.0kΩ 4.3kΩ = −336v i
⎝ 647Ω + 50Ω ⎠
(
)
Rth = 23.0kΩ 4.3kΩ = 3.62 kΩ
13.118
(
)
g m = 2 500µA/V 2 (100µA)(1+ 0.02(5)) = 332µS | ro =
(50 + 5)V = 550kΩ
100µA
⎛
⎞
6.8 MΩ
vth = −vi ⎜
⎟(332µS ) 550kΩ 50kΩ = −15.0vi
⎝ 6.8 MΩ + 0.1MΩ ⎠
(
)
Rth = 550kΩ 50kΩ = 45.8 kΩ
13-35
13.119
(
)
g m = 2 100µA/V 2 (10µA)(1+ 0.02(5)) = 46.9µS | ro =
(50 + 5)V = 5.50 MΩ
10µA
⎛
⎞
10 MΩ
vth = −vi ⎜
⎟(46.9µS ) 5.50 MΩ 560kΩ = −23.6vi
⎝10 MΩ + 0.1MΩ ⎠
(
)
Rth = 5.50 MΩ 560kΩ = 508 kΩ
13.120
SPICE Results: Q-point: (242 µA, 3.61 V), Av = 31.1 dB, Rin = 14.8 kΩ, Rout = 9.81 kΩ
13.121
0 = 10000I B + 0.7 + 66(1615)I B − 5
I B = 36.9 µA | IC = 65I B = 2.40 mA
VCE = 5 −1000IC −1615I E − (−5)= 3.66 V
Q − point : (2.40 mA, 3.66 V )
65
50V + 3.66V
= 677 Ω | ro =
= 22.4 kΩ
96.0mS
2.40mA
Rin = RB rπ (1+ g m RE )= 10kΩ 677Ω 1+ 0.096(15) = 1.42 kΩ
g m = 40(2.40mA)= 96.0mS | rπ =
[
]
⎛
⎞ 96.0mS
1.42kΩ
1kΩ 220kΩ = −31.8
Av = −⎜
⎟
⎝ 0.33kΩ + 1.42kΩ ⎠ 1+ 0.096(15)
(
[
)
]
Rout = 1kΩ 22.4kΩ 1+ 0.096(15) = 982 Ω
13.122
SPICE Results: Q-point: (2.39 µA, 3.69 V), Av = 29.7 dB, Rin = 1.49 kΩ, Rout = 977 Ω
The results agree closely with the hand calculations in Prob. 13.121. The small disagreements
arise from

m
atu
us
in

13.123
0 = 106 I B + 0.7 + 66(161.5kΩ)I B − 5
I B = 0.369 µA | IC = 65I B = 24.0 µA
VCE = 5 −1000IC −1615I E − (−5)= 3.66 V
g m = 40(24.0µA) = 0.959mS | rπ =
Q − point : (24.0 µA, 3.66 V )
65
50V + 3.66V
= 67.8 kΩ | ro =
= 2.24 MΩ
0.959mS
24.0µA
[
]
Rin = RB rπ (1+ g m RE )= 1MΩ 67.8kΩ 1+ 0.959mS (1.5kΩ) = 142 kΩ
⎛
⎞
0.959mS
142kΩ
Av = −⎜
100kΩ 220kΩ = −27.0
⎟
⎝ 0.33kΩ + 142kΩ ⎠ 1+ 0.959mS (1.5kΩ)
(
[
]
Rout = 100kΩ 2.24 MΩ 1+ 0.959mS (1.5kΩ) = 98.2 kΩ
13-36
)
13.124
SPICE Results: Q-point: (24.6 µA, 3.52 V), Av = 28.4 dB, Rin = 148 kΩ, Rout = 98.1 kΩ
The results agree closely with the hand calculations in Prob. 13.123. The small disagreements
arise from


m
atu
us
in

13.125
SPICE Results: Q-point: (251 µA, 4.45 V), Av = 16.3 dB, Rin = 1.00 MΩ, Rout = 28.7 kΩ
The results agree closely with the hand calculations in Ex. 13.10.
13.126
*Problem 13.126 - NMOS Common-Source Amplifier - Figure P13.98
VDD 7 0 DC 10
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VI 1 0 AC 0
*
VI 1 0 AC 1
RI 1 2 1K
C1 2 3 100U
R1 3 0 430K
R2 7 3 560K
R4 4 0 20K
C3 4 0 100U
RD 7 5 43K
C2 5 6 100U
R3 6 0 100K
M1 5 3 4 4 NFET
.OP
.MODEL NFET NMOS KP=500U VTO=1 LAMBDA=0.0133
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) VM(3) VP(3) IM(VI) IP(VI)
*.PRINT AC IM(C2) IP(C2)
.END
Results: ID = 1.31E-04 VDS = 1.73E+00
VM(6) = 1.044E+01 VP(6) = -1.800E+02 IM(VI) = 4.094E-06 IP(VI) = -1.800E+02
VM(3) = 9.959E-01 VP(3) = 3.734E-04 IM(C3) = 2.496E-05 IP(C3) = -1.800E+02
Av = −10.4 | Rin =
VM (3) 0.9959V
1
1
=
= 243 kΩ | Rout =
=
= 40.1 kΩ
IM (C3) 24.96µA
IM (VI ) 4.094 µA
13-37
13.127
I B = 3.71µA IC = 241µA I E = 245µA VCE = 3.67V
PR B = I B2 RB = (3.71µA) (100kΩ)= 1.38 µW | PRC = IC2 RC = (241µA) (10kΩ)= 0.581 mW
2
2
PR E = I E2 RE = (245µA) (16kΩ) = 0.960 mW
2
PBJT = ICVCE + I BVBE = (241µA)(3.67V )+ (3.71µA)(0.7V )= 0.887 mW
PS = 5V (241µA)+ (−5V )(−245µA) = 2.43 mW | PS = PR B + PRC + PR E + PQ
13.128
I D = 250µA VDS = 4.75V
PJFET = I DVDS = (250µA)(4.75V )= 1.19 mW | PR D = I D2 RD = (250µA) (27 kΩ)= 1.69 mW
2
PR 4 = I D2 R4 = (250µA) (2 kΩ) = 0.125 mW | PRG = 0
2
PS = 12V (250µA) = 3.00 mW | PJFET + PR D + PR 4 + PRG = 3.00 mW
13.129
Using the values from Prob. 13.17 :
I B = 23.8µA IC = 1.78mA I E = 1.81mA VCE = 6.08V VB = −4.09V
2
2
V12 (−4.09 − ( −12)) V
V22 (12 − ( −4.09)) V
PR1 =
=
= 12.5 mW | PR 2 =
=
= 25.9 mW
R1
5000Ω
R2
10000Ω
2
2
PRC = IC2 RC = (1.78 mA) (6 kΩ) = 19.0 mW | PR E = I E2 RE = (1.81mA) (4 kΩ)= 13.0 mW
2
2
PBJT = ICVCE + I BVBE = (1.78 mA)(6.08V )+ (23.8µA)(0.7V ) = 10.8 mW
⎡
⎡
12 − ( −4.09)V ⎤
−4.09 − ( −12)V ⎤
PS = 12V ⎢1.78 mA +
⎥ + 12V ⎢1.81mA +
⎥ = 81.3 mW
10000Ω ⎦
5000Ω
⎦
⎣
⎣
PS = PR1 + PR 2 + PRC + PR E + PBJT = 81.2 mW
13.130
Using the values from Prob. 13.19 :
I B = 1.51µA IC = 98.4µA I E = 99.9µA VCE = 4.96V
PR B = I B2 RB = (1.51µA) 3kΩ = 6.84 nW | PRC = IC2 RC = (98.4µA) (33kΩ)= 0.320 mW
2
2
PR E = I E2 RE = (99.9µA) (68 kΩ)= 0.679 mW
2
PBJT = ICVCE + I BVBE = (98.4µA)(4.96V )+ (1.51µA)(0.7V ) = 0.489 mW
PS = 7.5(98.4µA)+ 7.5V (99.9µA)= 1.49 mW | PS = PR B + PRC + PR E + PBJT = 1.49 mW
13-38
13.131
Using the values from Prob. 13.23: I D = 82.2µA VDS = 6.04V
PFET = I DVDS = (82.2µA)(6.04V ) = 0.497 mW | PR D = I D2 RD = (82.2µA) (82kΩ) = 0.554 mW
2
PR 4 = I D2 R4 = (82.2µA) (27kΩ)= 0.182 mW | I2 =
2
15V
= 4.05µA
3.7 MΩ
PR1 = I22 R1 = (4.05µA) (1MΩ)= 16.4 µW | PR 2 = I22 R2 = (4.05µA) (2.7 MΩ) = 44.3 µW
2
2
PS = 15V (82.2µA + 4.05µA)= 1.29 mW | PFET + PR D + PR 4 + PR1 + PR 2 = 1.29 mW
13.132
Using the values from Prob. 13.27 : I D = 307µA VSD = 3.88V
PFET = I DVSD = (307µA)(3.88V ) = 1.19 mW | PR D = I D2 RD = (307µA) (24kΩ)= 2.26 mW
2
PR 4 = I D2 R4 = (307µA) (22kΩ) = 2.07 mW | I2 =
2
18V
= 2.73µA
6.6 MΩ
PR1 = I22 R1 = (2.73µA) (3.3MΩ) = 24.6 µW | PR 2 = I22 R2 = (2.73µA) (3.3MΩ)= 24.6 µW
2
2
PS = 18V (307µA + 2.73µA)= 5.58 mW | PFET + PR D + PR 4 + PR1 + PR 2 = 5.57 mW
13.133
Using the values from Prob. 13.33: I D = 1.25mA VDS = 10.6V
PFET = I DVDS = (1.25mA)(10.6V ) = 13.3 mW | PR D = I D2 RD = (1.25mA) (3.9kΩ) = 6.09 mW
2
PR S = I D2 RS = (1.25mA) (1kΩ)= 1.56 mW
| PR 4 = I D2 R4 = (1.25mA) (1kΩ) = 1.56 mW
2
PRG = IG2 RG = 0
13.134
V
IC = CC
3RC
2
PS = 18V (1.25mA) = 22.5 mW | PFET + PR D + PR S + PR1 = 22.5 mW
| ic ≤ 0.2IC | v c = ic RC ≤ 0.2
VCC
V
RC = CC
3RC
15
13.135
2
500x10−6
0 − (−1.5) = 563 µA
2
vgs ≤ 0.2(VGS − VTN )= 0.2 0 − (−1.5) = 0.3 V
(
id ≤ 0.4I D | I D =
(
)
)
vds = id RD ≤ 0.4(563µA)(15kΩ)= 3.38 V
To insure saturation : vDS ≥ vGS − VTN = vgs − VTN = 0.3 − (−1.5) = 1.8 V
VDD ≥ 1.8 + 3.38 + (563µA)(15kΩ)= 13.6 V
13.136
13-39
Assuming
VCC
>> VCESAT :
2
2
2
2
⎛ VCC ⎞ VCC
VCC
1 ⎛ VCC ⎞ 1 VCC
| Pdc = VCC ⎜
| Pac = ⎜
=
vo ≤
⎟=
⎟
2 ⎝ 2 ⎠ RL 8RL
2
⎝ 2RL ⎠ 2RL
13.137
The Q - point from problem 13.21 is (371µA, 2.72V).
(
)
(
2
VCC
8R
| ε = 100% 2 L = 25 %
VCC
2RL
)
vo ≤ g m vbe ro RC R3 = 40(371µA)(5mV ) ∞ 13kΩ 100kΩ = 0.854 V
13.138
The Q - point from Problem 13.23 is (82.2µA,6.04V ).
(
)
(
)
id ≤ 0.4I D = 32.9µA | vds ≤ 0.4I D ro RD R3 = 32.9µA ∞ 82kΩ 470kΩ = 2.30V
13.139
The Q - point from Problem 13.27 is (307µA,3.88V ).
(
)
(
)
id ≤ 0.4I D = 307µA | vo ≤ 0.4I D ro RL R3 = 307µA ∞ 24kΩ 470kΩ = 2.80V
Checking the bias point : VR D = 307µA(24kΩ) = 7.37V | 2.80 < 7.37 & 2.80 < 3.88 -1
13.140
The Q - point from problem 13.17 is (1.78mA, 6.08V ).
(
)
(
)
ic ≤ 0.2IC = 0.356 mA | vc ≤ 0.2IC ro RC R3 = 0.356mA ∞ 6kΩ 100kΩ = 2.02V
13.141
The Q - point from problem 13.33 is (1.25 mA, 10.6 V ).
(
)
(
Neglecting ro ,
)
id ≤ 0.4I D = 0.500mA | vd ≤ 0.4I D RD R3 = 0.500mA 3.9kΩ 36kΩ = 1.76 V
13.142
The Q - point from problem 13.35 is (1.01 mA, 7.41 V ).
(
)
(
)
id ≤ 0.4I D = 0.404mA | vd ≤ 0.4I D RD R3 = 0.404mA 7.5kΩ 220kΩ = 2.93 V
13-40
13.143
VCE = 20 − 20000IC : Two points on the load line (0mA, 20V ) , (1mA, 0V )
At IB = 2µA, the maximum swing is approximately 2.5 V limited by VRC .
For IB = 5µA,the maximum swing is approximately - 8.5 V limited by VCE .
IC
I B= 10 µA
1000 µA
I = 8 µA
B
750 µA
I = 6 µA
B
500 µA
I B = 5 µA
I = 4 µA
B
250 µA
8.5 V
9V
I = 2 µA
B
V
0
CE
0
10 V
2.5 V
20 V
13-41
CHAPTER 14
14.1
(a) Common-collector Amplifier (npn) (emitter-follower)
RI
Q1
R
vi
R
1
2
+
R
R
E
3
vo
-
(b) Not a useful circuit because the signal is injected into the drain of the transistor.
M
1
RI
+
vi
R3
RD
R1
vo
-
(c) Common-emitter Amplifier (pnp)
R
1
R
2
RI
Q
1
1kΩ
vi
RC
R
+
3
vo
100 k Ω
-
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-1
14.1 (d) Common-source Amplifier (NMOS)
+
RI
M
1k Ω
R
vi
1
R
1
R
3
RD
vo
470 k Ω
-
2
(e) Common-gate Amplifier (PMOS)
RI
R
vi
+
M1
RD
1
R3
vo
-
(f) Common-collector Amplifier (emitter-follower) (npn)
RI
Q
vi
R
1
1
+
R
2
RE
R3
v
o
-
14-2
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.1(h) Common-base Amplifier (pnp)
+
R3
RC
v
Q1
o
-
R
I
v
RE
i
(i) Not a useful circuit since the signal is being taken out of the base terminal.
Q1
+
vo
-
RB
R3
RI
vi
RE
(j) Common-source Amplifier (PMOS)
RI
M1
1k Ω
vi
R
1
R
2
+
RD
R3
470 k Ω
v
o
-
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-3
14.1(k) Common-gate Amplifier (Depletion-mode NMOS)
M1
RI
vi
+
R
RS
vo
3
-
RD
(l) Not a useful circuit because the signal is injected into the drain of the transistor.
M
RI
1
+
R3
vi
RD
vo
RS
-
(n) Common-emitter Amplifier (npn)
+
RI
R3
Q1
vo
R
vi
B
(o) Common-drain Amplifier (Source-follower) (NMOS)
RI
M
vi
R
G
+
R3
vo
-
14-4
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.2
V DD
RS
CC1
RI
1k Ω
vi
CC2
CB
+
RG
RL
RD
-V SS
vo
100 k Ω
14.3
VDD
RS
RI
vi
CC2
CC1
RL
1k Ω
RG
+
vo
100 k Ω
CB
RD
-V SS
14.4
V DD
RS
RI
CC1
1k Ω
CB
CC2
vi
+
RG
RL
RD
-V SS
vo
100 k Ω
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-5
14.5
+15 V
15 kΩ
RC
CC2
RI
+
RL
CC1
v
Q1
150 kΩ
3.3 k Ω
O
-
RB
vI
220 kΩ
CB
RE
22 kΩ
-15 V
14.6
+15 V
15 kΩ
RC
+15 V
RI
RI
CB
CC1
CC1
Q1
Q1
3.3 k Ω
vI
3.3 k Ω
RB
RB
CC2
220 kΩ
RE
RL
22 kΩ
(a)
150 kΩ
vI
+
CC2
220 kΩ
RE
-
22 kΩ
(b)
-15 V
+
RL
vO
vO
-
150 kΩ
-15 V
14.7
+15 V
+15 V
15 kΩ
RC
15 kΩ
RC
CC2
CC2
vO
CB
Q1
150 kΩ
RB
220 kΩ
CC1
(a)
14-6
-
150 kΩ
CC1
RI
-15 V
vI
22 kΩ
(b)
RI
3.3 kΩ
RE
vI
22 kΩ
vO
Q1
3.3 kΩ
RE
+
RL
+
RL
-15 V
© R. C. Jaeger & T. N. Blalock - February 20, 2007
-
14.8
(a) Neglecting R
out
Av = Avt
: Avt = −
(0.005S )(2000Ω) = −3.77
g m RL
=−
1+ g m RS
1+ (0.005S )(330Ω)
⎛
⎞
RG
2 MΩ
= -3.77⎜
⎟ = -3.64
RI + RG
⎝ 75kΩ + 2 MΩ ⎠
[
Rin = RG = 2 MΩ
|
]
Rout = ro (1+ g m RS ) = 10kΩ 1+ (0.005S )(330Ω) = 26.5 kΩ >> 2kΩ
Ai = −RG
(b) A
v
gm
0.005S
= −2 MΩ
= −3770
1+ g m RS
1+ (0.005S )(330Ω)
⎛ RG ⎞
⎛
⎞
2 MΩ
= −g m RL ro ⎜
⎟ = −(0.005S ) 2kΩ 10kΩ ⎜
⎟ = −8.03
⎝ 75kΩ + 2 MΩ ⎠
⎝ RI + RG ⎠
(
)
(
)
Rin = RG = 2 MΩ | Rout = ro = 10.0 kΩ | Ai = −RG g m = −2 MΩ(0.005S ) = −10000
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-7
14.9
(a) g
m
75
= 3750Ω
.02
rπ + (β o + 1)RE = 15kΩ 3750Ω + 76(300Ω) = 9.58 kΩ
= 0.02S | rπ =
Rin = RB
[
]
[
]
⎛ Rin ⎞
⎛ Rin ⎞
β o RL
Neglecting Rout : Av = Avt ⎜
⎟=−
⎜
⎟
rπ + (β o + 1)RE ⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
Av = −
Rout
75(12kΩ)
⎛
⎞
9.58kΩ
⎜
⎟ = −32.2
3750Ω + 76(300Ω)⎝ 500Ω + 9.58 kΩ ⎠
⎡
⎤
⎛
⎞
75(300Ω)
β o RE
⎢
⎥ = 596 kΩ >> 12kΩ
≅ ro ⎜1+
⎟ = 100kΩ 1+
⎢
⎥
15kΩ
500Ω
+
3750Ω
+
300Ω
⎝ Rth + rπ + RE ⎠
⎣
⎦
(
)
Ai = −β o
RB
15kΩ
= −75
= −27.1
RB + rπ + (β o + 1)RE
15kΩ + 3750Ω + 76(300Ω)
(b) g
75
= 3750Ω
.02
rπ + (β o + 1)RE = 15kΩ 3750Ω + 76(620Ω) = 11.6 kΩ
m
= 0.02S | rπ =
Rin = RB
[
]
[
]
⎛ Rin ⎞
⎛ Rin ⎞
β o RL
Neglecting Rout : Av = Avt ⎜
⎟=−
⎜
⎟
rπ + (β o + 1)RE ⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
Av = −
75(12kΩ)
⎛
⎞
11.6kΩ
⎜
⎟ = −17.0
3750Ω + 76(620Ω)⎝ 500Ω + 11.6kΩ ⎠
⎤
⎡
⎛
⎞
75(620Ω)
β o RE
⎥ = 1060 kΩ >> 12kΩ
⎢
=
100kΩ
1+
Rout ≅ ro ⎜1+
⎟
⎥
⎢
⎝ Rth + rπ + RE ⎠
15kΩ
500Ω
+
3750Ω
+
620Ω
⎦
⎣
(
Ai = −β o
)
RB
15kΩ
= −75
= −17.1
15kΩ + 3750Ω + 76(620Ω)
RB + rπ + (β o + 1)RE
14.10
(a ) For large β
o
: Av ≅ −
8.2kΩ 47kΩ
RL
=−
= −6.91 |
RE
330Ω + 680Ω
parallel with the 330Ω resistor. Then Av ≅ −
(b) Place a bypass capacitor in
8.2kΩ 47kΩ
RL
=−
= −10.3 |
RE
680Ω
bypass capacitor in parallel with the 680Ω resistor. Then Av ≅ −
(c) Place a
8.2kΩ 47kΩ
= −21.2
330Ω
(d) Place a bypass capacitor from the emitter to ground. (e) Av ≅ −10(VCC + VEE )= −240.
14-8
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.11
Rin = rπ + (β o + 1)RE = 250 kΩ
Avt = −
75RL
β o RL
=−
= −10 → RL = 33.3 kΩ → 33 kΩ
250kΩ
rπ + (β o + 1)RE
Assuming (β o + 1)RE >> rπ , RE ≅
250 kΩ 250 kΩ
=
= 3.29 kΩ → 3.3 kΩ
76
βo + 1
As indicated above, the nearest 5% values would be 33 kΩ and 3.3 kΩ.
14.12
Rin = rπ = 250 kΩ | rπ =
βo
gm
=
β oVT
IC
| IC =
β oVT
rπ
=
75(0.025V )
250kΩ
= 7.50 µA
⎞
⎛ r
75RL
β R
Av = −g m RL ⎜ π ⎟ = − o L = −
= −10 | RL = 33.3 kΩ → 33 kΩ
Rth + rπ
100Ω + 250kΩ
⎝ Rth + rπ ⎠
The closest 5% value is RL = 33 kΩ.
14.13
⎡ix − gm v gs⎤ ⎡ go
−go ⎤⎡v d ⎤
⎢
⎥=⎢
⎥⎢ ⎥ | v gs = −v s
⎣ +gm v gs ⎦ ⎣−go go + GS ⎦⎣ v s ⎦
⎡ix ⎤ ⎡ go
−(gm + go ) ⎤⎡v d ⎤
⎥⎢ ⎥
⎢ ⎥=⎢
⎣0 ⎦ ⎣−go gm + go + GS ⎦⎣ v s ⎦
vd
gm vgs
+ vgs
R th
r
o
-
R
ix
5
ix (gm + go + GS )
=
ix
∆
go G S
⎛ g
⎛
v
G ⎞
r ⎞
= d = RS ⎜1+ m + S ⎟ = RS ⎜1+ µ f + o ⎟
ix
RS ⎠
⎝ go go ⎠
⎝
∆ = goGS | v d = (gm + go + GS )
Rout
Rout = ro + (1+ µ f )RS ≅ ro + µ f RS = ro (1+ gm RS )
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-9
14.14
(15 − 0.7 )V
= 1.29 µA | I C = 129 µA | VCE = 30 − 39000 I C − 100000 I E
1MΩ + (100 + 1)100kΩ
100(0.025V )
Active region is correct. | r =
= 19.4kΩ | r - no V specified - neglect
IB =
π
129 µA
o
= 11.9 V
A
RL = 500kΩ 39kΩ = 36.2kΩ | Rin = RB rπ = 1MΩ 19.4kΩ = 19.0kΩ
⎛ Rin ⎞
19.0kΩ
⎞
⎟⎟ = 40(0.129mA)(36.2kΩ )⎛⎜
Av = − g m RL ⎜⎜
⎟ = −182
⎝ 500Ω + 19.0kΩ ⎠
⎝ RI + Rin ⎠
Rout = RC ro = 39 kΩ
⎛ Rin ⎞
19.0kΩ
0.005V
⎛
⎞
⎟⎟ = vi ⎜
vbe = vi ⎜⎜
vi ≤
= 5.13 mV
⎟ = 0.974vi
R
R
500
Ω
+
19
.
0
k
Ω
0
.
974
+
⎝
⎠
I
in
⎠
⎝
Av ≅ −10(VCC + VEE ) = −10(30 ) = −300. | A closer estimate is - 40VR C = -40(5.03) = −201
14.15
62kΩ
= 6.80V | REQ = 20kΩ 62kΩ = 15.1kΩ
20kΩ + 62kΩ
(9 − 0.7 − 6.80)V = 4.82µA | I = 361 µA | V = 9 − 3900I − 8200I = 4.61 V
IB =
C
EC
E
C
15.1kΩ + (75 + 1)3.9kΩ
VEQ = 9
Active region is correct. | rπ =
75(0.025V )
361µA
= 5.19kΩ | VA not specified, choose ro = ∞
Rin = 15.1kΩ 5.19kΩ = 3.86 kΩ | Rout = ro 8.2kΩ = 8.2 kΩ | g m = 40IC = 12.6 mS
RL = ro 8.2kΩ 100kΩ = 8.2kΩ 100kΩ = 7.58kΩ
⎛ Rin ⎞
⎛ 3.86kΩ ⎞
Av = −g m RL ⎜
⎟ = −(12.6mS )(7.58kΩ)⎜
⎟ = −75.9
⎝1kΩ + 3.86kΩ ⎠
⎝ RI + Rin ⎠
Rout
8.2kΩ
RB
15.1kΩ
−β o )
=
−75)
= −4.23
Ai =
(
(
RB + rπ
Rout + R3 15.1kΩ + 5.19kΩ
8.2kΩ + 100kΩ
vbe = vi
Rin
3.86kΩ
5.00mV
= vi
= 0.794v i | vi =
= 6.30 mV
1kΩ + 3.86kΩ
0.794
RI + Rin
Av ≅ −10VCC = −10(9)= −90. | The voltage gain is slightly below the rule - of - thumb estimate.
14-10
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.16
VEQ = 15
500kΩ
= 3.95V | REQ = 500kΩ 1.4 MΩ = 368kΩ
1.4 MΩ + 500kΩ
2I D
+ 27000I D → I D = 79.7µA
250x10−6
= 15 − I D (75kΩ + 27kΩ) = 6.87 V | Active region operation is correct.
3.95 = VGS + 27000I D = 1+
VDS
(
)(
)
g m = 2 250x10−6 79.7x10−6 = 0.200mS | Assume λ = 0, ro = ∞.
RL = ro 75kΩ 470kΩ ≅75kΩ 470kΩ = 64.7kΩ
Rin = RG = R1 R2 = 368 kΩ | Rout = ro 75kΩ ≅ 75kΩ
Av = −g m RL
⎛ 368kΩ ⎞
Rin
= −(0.200mS )(64.7kΩ)⎜
⎟ = −12.9
RI + Rin
⎝ 1kΩ + 368kΩ ⎠
Ai = RG (−g m )
vgs = vi
RD
75kΩ
= 368kΩ(−0.200mS )
= −10.1
RD + R3
75kΩ + 470kΩ
Rin
368kΩ
= vi
= 0.997v i | VGS − VTN =
RI + Rin
1kΩ + 368kΩ
vgs ≤ 0.2(VGS − VTN )→ vi ≤ 0.2
2(79.7µA)
250µA/V 2
= 0.798V
0.798V
VDD
15
= 0.160 V | Av ≅ −
=−
= −18.8
0.997
0.798
VGS − VTN
VDD
. We have VR L = 79.7µA(75kΩ)= 5.98V = 0.399VDD
2
The estimate also doesn't account for the presence of R3 .
The rule - of - thumb estimate assumes VR L =
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-11
14.17
2.2 MΩ
= 11.0V | RG = REQ = 2.2 MΩ 2.2 MΩ = 1.10 MΩ
2.2 MΩ + 2.2 MΩ
Assume active region operation.
VEQ = 22
22 = 22000I D − VGS + 11 | 11 = 22000I D + 1+
[
]
2(400x10 )(391x10 ) = 0.559mS
2I D
→ I D = 391 µA
400x10−6
VDS = − 22 − I D (22kΩ + 18kΩ) = −6.36 V | Active region operation is correct.
gm =
−6
−6
| Assume λ = 0, ro = ∞
RL = ro 18kΩ 470kΩ ≅18kΩ 470kΩ = 17.3kΩ | Rin = 1.10 MΩ | Rout = ro 18kΩ = 18kΩ
⎛ Rin ⎞
1.1MΩ
Av = −g m RL ⎜
= −9.48
⎟ = −(0.559mS )(17.3kΩ)
22kΩ + 1.1MΩ
⎝ RI + Rin ⎠
RD
18kΩ
Ai = −g m Rin
= −0.559mS (1.1MΩ)
= −22.7
RD + R3
18kΩ + 470kΩ
Rin
1.1MΩ
= vi
= 0.980vi | VGS − VTN =
RI + Rin
22kΩ + 1.1MΩ
⎛1.40V ⎞
vgs ≤ 0.2(VGS − VTN ) | vi ≤ 0.2⎜
⎟ = 0.286 V
⎝ 0.980 ⎠
vgs = vi
2(391µA)
400µA/V 2
14.18
2
2
Kn
4x10−4
V
=
−5) = 5.00 mA
(
(
TN )
2
2
= 16 −1800I D = 7.00V | Active region operation is correct.
VGS = 0 → I D =
VDS
(
)(
)
g m = 2 4x10 4 5x10−3 = 2.00mS | Assume λ = 0, ro = ∞.
⎛ RG ⎞
⎛ 10 MΩ ⎞
Av = -g m RL ⎜
⎟ = −(2.00mS ) 1.8kΩ 36kΩ ⎜
⎟ = −3.43
⎝ 10 MΩ + 5kΩ ⎠
⎝ RI + RG ⎠
(
)
Rin = 10.0 MΩ | Rout = RD ro = 1.80 kΩ
⎛ RD ⎞
⎛
⎞
1.8kΩ
Ai = −g m RG ⎜
⎟ = −2.00mS (10 MΩ)⎜
⎟ = −952
⎝ 1.8kΩ + 36kΩ ⎠
⎝ RD + R3 ⎠
⎛ 10 MΩ ⎞
vgs = vi ⎜
⎟ ≤ 0.2VGS − VTN → vi ≤ 1 V
⎝10 MΩ + 5kΩ ⎠
14-12
© R. C. Jaeger & T. N. Blalock - February 20, 2007
= 1.40V
14.19
(12 − 0.7)V
= 14.9µA | I = 1.19 mA | V = 24 − 9100I = 13.2 V
20kΩ + (80 + 1)9.1kΩ
80(0.025V )
(100 + 13.2)V = 95.1kΩ
Active region is correct. | r =
= 1.68kΩ | r =
IB =
C
CE
π
E
o
1.19mA
1.19mA
RL = ro 1MΩ = 95.1kΩ 1MΩ = 86.9kΩ | Rin = RB rπ = 20kΩ 1.68kΩ = 1.55 kΩ | Rout = ro = 95.1 kΩ
⎛ Rin ⎞
1.55kΩ
= −3560
Av = -g m RL ⎜
⎟ = −40(1.19mA)(86.9kΩ)
250Ω + 1.55kΩ
⎝ RI + Rin ⎠
⎛ RB ⎞⎛ ro ⎞
⎛
⎞⎛
⎞
20kΩ
95.1kΩ
Ai = −β o ⎜
⎟⎜
⎟ = −80⎜
⎟⎜
⎟ = −6.41
⎝ 20kΩ + 1.68kΩ ⎠⎝ 95.1kΩ + 1MΩ ⎠
⎝ RB + rπ ⎠⎝ ro + R3 ⎠
⎛ Rin ⎞
1.55kΩ
5.00mV
= 0.861v i | vi ≤
= 5.81 mV
vbe = vi ⎜
⎟ = vi
250Ω + 1.55kΩ
0.861
⎝ RI + Rin ⎠
14.20
80
= 200Ω | Assume VA = ∞, ro = ∞.
0.4S
Rin = RB rπ + (β o + 1)RL = 47kΩ 200Ω + 81(1kΩ) = 29.8 kΩ
rπ =
[
Rout =
Av = +
]
(
)
[
]
47kΩ 10kΩ + 200Ω
Rth + rπ
=
= 104 Ω
βo + 1
81
(β + 1)R
+ (β + 1)R
o
rπ
L
o
⎛ Rin ⎞
⎛
⎞
81(1kΩ)
29.8kΩ
⎟=
⎜
⎜
⎟ = 0.747
200Ω + 81(1kΩ)⎝10kΩ + 29.8kΩ ⎠
L ⎝ RI + Rin ⎠
⎡
⎤⎛ r ⎞
47kΩ
RB
⎢
⎥⎜ o ⎟ = 81
= 29.7
Ai = +(β o + 1)
47kΩ + 200Ω + 81(1kΩ)
⎢⎣ RB + rπ + (β o + 1)RL ⎥⎦⎝ ro + RL ⎠
14.21
Assume λ = 0, ro = ∞. | Rin = RG = 2 MΩ | Rout =
Av = +
1
= 100 Ω
gm
⎞
0.01(2kΩ) ⎛
g m RL ⎛ Rin ⎞
2 MΩ
⎜
⎟=
⎜
⎟ = 0.907
1+ g m RL ⎝ RI + Rin ⎠ 1+ 0.01(2kΩ)⎝100kΩ + 2 MΩ ⎠
⎛ r ⎞
Ai = +g m RG ⎜ o ⎟ = 0.01(2 MΩ)= 2x10 4
⎝ ro + RL ⎠
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-13
14.22
Defining v1 as the source node:
(a) 2kΩ 100kΩ = 1.96kΩ
(v − v ) + 3.54x10 v − v = v
( ) 1960
10
1
i
−3
6
−3
1 MΩ
1
1
i
+
−3
3.541x10 vi = 4.051x10 v1
v1 = 0.874v i | Av = 0.874
Rin =
+
vi
v
RG
g v
gs
m gs
-
-
v
vi
v
= −6 i
= 7.94 MΩ
ii 10 (vi − v1 )
1
2kΩ
100 k Ω
Driving the output with current source ix :
Rout : ix =
Rout =
v1
v
+ 1 + 3.54x10−3 v1
6
10 2000
(b) R
v1
= 247 Ω |
ix
in
=∞
14.23
100kΩ
= −6.00V | REQ = 100kΩ 100kΩ = 50.0kΩ
100kΩ + 100kΩ
(−0.7 + 6)V
= 8.25µA | IC = 1.03 mA | VCE = 24 − 2000IC − 4700I E = 17.1 V
IB =
50.0kΩ + (126)(4.7kΩ)
VEQ = −12 + 12
Active region is correct. | rπ =
125(0.025V )
= 3.03kΩ | ro =
(50 + 17.1)V = 65.1kΩ
1.03mA
1.03mA
RB = R1 R2 = 100kΩ 100kΩ = 50.0kΩ | RL = R3 RE ro = 24kΩ 4.7kΩ 65.1kΩ = 3.71kΩ
[
]
[
]
Rin = RB rπ + (β o + 1)RL = 50.0kΩ 3.03kΩ + (126)3.71kΩ = 45.2 kΩ
Av = +
(β + 1)R
+ (β + 1)R
o
rπ
L
o
⎞
⎛ R
⎛
⎞
126(3.71kΩ)
50.0kΩ
in
⎟=
⎜
⎜
⎟ = 0.984
R
+
R
500Ω
+
50.0kΩ
3.03kΩ
+
126
3.71kΩ
⎝
⎠
⎠
⎝
(
)
I
in
L
⎞ ⎛
⎤
⎞⎛
⎛ R
⎞⎡
rπ
3.03kΩ
50.0kΩ
in
⎟=⎜
⎢
⎥ = 6.34x103 vi
vbe = vi ⎜
⎟⎜⎜
⎟
⎟
R
+
R
500Ω
+
50.0kΩ
r
+
β
+
1
R
3.03kΩ
+
126
3.71kΩ
⎠⎢⎣
⎝ I
( o ) L⎠ ⎝
(
)⎥⎦
in ⎠⎝ π
vi ≤
0.005V
= 0.784 V | R out = RE
6.34x10−3
14-14
(R
B
)
RI + rπ
βo + 1
= 4.7kΩ
(50.0kΩ 500Ω)+ 3.03kΩ = 27.8 Ω
© R. C. Jaeger & T. N. Blalock - February 20, 2007
126
14.24
VGS = 5V | I D =
2
5x10−4
5 −1.5) = 3.06 mA | VDS = 5 − (−5)= 10V - Pinchoff region
(
2
(
)
[
]
operation is correct. | g m = 2 5x10−4 (3.06mA) 1+ 0.02(10) = 1.92mS
1
+ 10
V
ro = 0.02
= 19.6 kΩ − Cannot neglect! | RL = 19.6kΩ 100kΩ = 16.4kΩ
3.06 mA
1
Rin = RG = 1 MΩ | Rout =
ro = 507 Ω
gm
⎛ 1.92mS (16.4kΩ) ⎞
Rin ⎛ g m RL ⎞
1MΩ
⎟ = 0.960
⎜
Av = +
=
+
⎜
⎟
RI + Rin ⎝ 1+ g m RL ⎠
10kΩ + 1MΩ ⎜⎝1+ 1.92mS (16.4kΩ)⎟⎠
⎤
⎛
⎞⎡
⎛ Rin ⎞⎛
106 Ω
1
1 ⎞
⎥ = 0.0305v i
⎢
v gs = v i ⎜
⎟
⎟⎜
⎟ = vi⎜ 4
6
⎝ RI + Rin ⎠⎝ 1+ g m RL ⎠
⎝ 10 Ω + 10 Ω ⎠⎢⎣1+ 1.92mS (16.4kΩ)⎥⎦
vi ≤
VDS
0.2(5 −1.5)
= 23.0 V But, vDS must exceed vGS − VTN ≅ VGS − VTN = 4V for pinchoff.
0.0305
= 10 − vo = 10 − 0.970vi ≥ 4 → vi ≤ 6.19 V − Limited by the Q - point voltages
14.25
(9 − 0.7)V
= 187 nA | I = 18.7 µA | V
1MΩ + (100 + 1)430kΩ
100(0.025V )
= 134kΩ |
Active region is correct. | r =
IB =
C
π
CE
18.7µA
= 18 − 430000I E = 9.89 V
ro =
(60 + 9.89)V = 3.74 MΩ - neglected
18.7µA
In the ac model, R1 appears in parallel with rπ . The circuit appears to be using a transistor with
rπ' = 500kΩ rπ = 106kΩ
and β o' = g mrπ' = 40(18.7µA)106kΩ = 79.0
(
)
RL = 500kΩ 430kΩ 500kΩ = 158kΩ | Rin = rπ' + β o' + 1 RL = 106kΩ + 79.0(158kΩ)= 12.6 MΩ
(β + 1)R ⎛⎜ R
A =
r + (β + 1)R ⎝ R + R
'
o
v
'
π
Rout
'
o
in
L
I
⎞
⎛
⎞
80.0(158kΩ)
12.6 MΩ
⎟=
⎜
⎟ = +0.992
106kΩ + 80.0(158kΩ)⎝ 500Ω + 12.6 MΩ ⎠
in ⎠
RI + rπ'
500Ω + 106kΩ
= RE R2 '
= 430kΩ 500kΩ
= 1.34 kΩ
79.0
βo + 1
vbe = vi
vi ≤
L
rπ'
(
)
RI + rπ + β + 1 RL
'
'
o
= vi
106kΩ
= 8.32x10−3 vi
500Ω + 106kΩ + 80.0(158kΩ)
0.005V
= 0.601 V
8.32x10−3
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-15
14.26
vi ≤ 0.005(1+ g m RL ) | RL = RE R3 ≅ RE
⎛ I R ⎞
vi ≤ 0.005(1+ g m RL ) = 0.005(1+ g m RE ) = 0.005⎜1+ C E ⎟
VT ⎠
⎝
⎛
⎛ I R ⎞
I R ⎞
vi ≤ 0.005⎜1+ α F E E ⎟ ≅ 0.005⎜1+ E E ⎟
VT ⎠
VT ⎠
⎝
⎝
⎛ VR ⎞
⎛
VR E ⎞
vi ≤ 0.005⎜1+ E ⎟ = 0.005⎜1+
⎟ = 0.005 + 0.2VR E
VT ⎠
⎝
⎝ 0.025 ⎠
14.27
β o = g mrπ = 3.54mS (1MΩ) = 3540 | RL = 2kΩ 100kΩ = 1.96kΩ
(β + 1)R
r + (β + 1)R
= r + (β + 1)R
Av =
Rin
o
L
π
o
L
π
o
L
Rout = 2kΩ
(3540 + 1)(1.96kΩ) = 0.874
1MΩ + (3540 + 1)(1.96kΩ)
= 1MΩ + (3540 + 1)(1.96kΩ)= 7.94 MΩ
=
rπ
106
= 2kΩ
= 247 Ω
(β o + 1)
(3541)
14.28
(a) v
be
(b) A
v
= v i − vo | 0.005 ≤ 5 − v o → Av =
=
(β + 1)R
+ (β + 1)R
o
rπ
E
o
E
=
vo 4.995
≥
= 0.999
5
vi
1
1
1
1
=
=
=
βo
rπ
αo
rπ
V
1+
1+
1+
1+ T
g m RE
I E RE
(β o + 1)RE
(β o + 1) βo RE
V
0.025V
1
≥ 0.999 → T ≤ 0.001 → I E RE ≥
= 25.0 V
VT
0.001
I E RE
1+
I E RE
14-16
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.29
vbe = vi − vo = (1− Av )vi | 0.005 ≤ (1− Av )7.5 → Av =
vo 7.5 − 0.005
≥
= 0.999333
vi
7.5
1
500RE
1
| RL = RE 500Ω =
| Av =
⎛
V
500 + RE
500 + RE ⎞
V
1+ T
1+ T ⎜
⎟
I E RL
I E RE ⎝ 500 ⎠
1
V ⎛ 500 + RE ⎞
−4
≥ 0.999333 → T ⎜
⎟ ≤ 6.67x10
⎛
⎞
I
R
500
500 + RE
V
⎠
E E ⎝
1+ T ⎜
⎟
I E RE ⎝ 500 ⎠
From Prob. 14.28, Av =
500I E RE
0.025V
≥
= 37.5V | VCC ≥ I E RE + 0.7 + 7.5
500 + RE 6.67x10−4
Some design possibilities are listed in the table below.
RE
IE
VCC
VCC IE
100 Ω
450 mA
53 V
24 W
250 Ω
225 mA
64 V
16 W
360 Ω
179mA
73V
13 W
500 Ω
150 mA
83 V
12 W
750 Ω
125 mA
102 V
13 W
1000 Ω
113mA
120 V
14 W
2000 Ω
93.8 mA
196 V
18 W
Using a result near the minimum-power case in the table: RE = 510 Ω  E= 149 mA and VCC = 85
V.
149mA
= 2.92 mA | Set I R1 = 5I B = 14.6mA ≅ 15mA
For β F = 50 : I B ≅
51
V + VBE 149mA(510Ω)+ 0.7
R1 = E
=
= 5.07kΩ → 5.1 kΩ | I R 2 = I R1 + I B ≅ 18mA
I R1
15mA
85 − VBE − VBE 8.3V
=
= 462Ω → 470 Ω
I R2
18mA
It is obviously very difficult to achieve the required level of linearity!
R2 =
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-17
14.30
(a) g m = 40(12.5µA)= 0.5mS
Rin = R4
Av =
rπ
α
0.99
= R4 o = 100kΩ
= 1.94 kΩ
βo + 1
gm
0.5mS
⎛ R4 ⎞
⎛ 100kΩ ⎞
0.5mS (100kΩ)
⎜
⎟=
⎜
⎟ = 48.7
1+ g m RI R4 ⎝ RI + R4 ⎠ 1+ 0.5mS 50Ω 100kΩ ⎝ 50Ω + 100kΩ ⎠
g m RL
(
[
)
(
)]
)
60V
1+ 0.5mS (50Ω) = 4.92 MΩ
12.5µA
⎛ R4 ⎞
⎛ 100kΩ ⎞
Ai = α o ⎜
⎟ = 0.990⎜
⎟ = 0.990
⎝ 50Ω + 100kΩ ⎠
⎝ RI + R4 ⎠
(
[
Rout = ro 1 + g m RI R4 =
(b) A
=
v
]
0.5mS (100kΩ)
⎛
⎞
100kΩ
⎜
⎟ = 23.6 | Rin = 1.94kΩ - no change
1+ 0.5mS 2.2kΩ 100kΩ ⎝ 2.2kΩ + 100kΩ ⎠
(
[
Rout = ro 1 + g m
)
60V
(R R )]= 12.5
[1+ 0.5mS (2.2Ω)]= 10.1 MΩ
µA
4
I
14.31
(a) R
in
Av =
1
1
= 3kΩ
= 1.20 kΩ | Rout = ∞ (assume λ = 0)
gm
0.5mS
= R4
⎛ R ⎞
0.5mS (60kΩ) ⎛ 3kΩ ⎞
4
⎜
⎟=
⎜
⎟ = +28.8
1+ g m RI R4 ⎝ RI + R4 ⎠ 1+ 0.5mS 50Ω 3kΩ ⎝ 50Ω + 3kΩ ⎠
g m RL
(
Ai = 1
(b) A
v
R4
R4 +
=
1
gm
)
=
(
)
3kΩ
= 0.600
3kΩ + 2kΩ
0.5mS (60kΩ)
⎛ 3kΩ ⎞
1
= 1.43 kΩ
⎜
⎟ = +5.81 | Rin = 5kΩ
0.5mS
1+ 0.5mS 5kΩ 3kΩ ⎝ 5kΩ + 3kΩ ⎠
Rout = ∞ | Ai =
(
)
5kΩ
= 0.714
5kΩ + 2kΩ
14.32
The voltage gain is approximately 0. The signal is injected into the collector and taken out of the
emitter. This is not a useful amplifier circuit.
14-18
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.33
IB =
(12 − 0.7)V
= 2.64µA |
100kΩ + (50 + 1)82kΩ
IC = 132 µA
VCE = 24 − 82000I E − 39000IC = 7.81 V | Active region operation is correct.
g m = 40IC = 5.28mS | rπ =
βo
gm
= 9.47kΩ | ro =
(50 + 7.81)V = 438kΩ - neglected
132µA
RI RE = 0.5kΩ 82kΩ = 497Ω | RL = RC R3 = 39kΩ 100kΩ = 28.1kΩ
Av =
⎛ RE ⎞ 5.28mS (28.1kΩ) ⎛
⎞
82kΩ
⎜
⎟=
⎜
⎟ = 40.7
1+ g m RI RE ⎝ RI + RE ⎠ 1+ 5.28mS (497Ω)⎝ 500Ω + 82kΩ ⎠
g m RL
(
Rin = 82kΩ
)
rπ
500Ω + 185Ω
R + Rin
= 185 Ω | Ai = Av I
= 40.7
= 0.279
100kΩ
βo + 1
R3
Rout = RC = 39.0 kΩ | veb = vi
Rin
≤ 5.00mV | 0.270vi ≤ 5.00mV | vi ≤ 18.5 mV
RI + Rin
14.34
IB =
(9 − 0.7)V
= 194nA |
1000kΩ + (50 + 1)820kΩ
IC = 9.69 µA
VCE = 18 − 820000I E − 390000IC = 6.12 V | Active region is correct.
g m = 40IC = 0.388mS | rπ =
βo
gm
= 129kΩ | ro =
(50 + 6.12)V = 5.79 MΩ - neglected
9.69µA
RI RE = 5kΩ 820kΩ = 4.97kΩ | RL = RC R3 = 390kΩ 1MΩ = 281kΩ
Av =
⎛ RE ⎞
0.388mS (281kΩ) ⎛ 820kΩ ⎞
⎜
⎟=
⎜
⎟ = 37.0
1+ g m RI RE ⎝ RI + RE ⎠ 1+ 0.388mS (4.97kΩ)⎝ 5kΩ + 820kΩ ⎠
g m RL
(
Rin = 820kΩ
)
rπ
5kΩ + 2.52kΩ
R + Rin
= 2.52 kΩ | Ai = Av I
= 37.0
= 0.278
1MΩ
βo + 1
R3
Rout ≅ RC = 390 kΩ | veb = vi
Rin
≤ 5.00mV | 0.335vi ≤ 5.00mV | vi ≤ 14.9 mV
RI + Rin
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-19
14.35
(5x10 )(V
= −3900
−4
VGS = −3900I D
(5x10 )(V
2
−4
ID =
gm =
2
2(254µA)
2 − 0.992
+ 2) | VGS = −0.975(VGS + 2) → VGS = −0.9915V
2
GS
2
+ 2) = 254µA | VDS = 16 − 23.9kΩI D = 9.92V - Pinched off.
2
GS
= 0.504mS | Rin = 3.9kΩ
1
= 1.32kΩ | Rout ≅ RD = 20kΩ
gm
RL = 20kΩ 51kΩ = 14.4kΩ
Av =
⎛ RS ⎞
0.504mS (14.4kΩ) ⎛ 3.9kΩ ⎞
⎜
⎟=
⎟ = 4.12
⎜
1+ g m RI RS ⎝ RI + RS ⎠ 1+ 0.504mS (0.796kΩ)⎝1kΩ + 3.9kΩ ⎠
g m RL
(
)
Ai = Av
1kΩ + 1.32kΩ
RI + Rin
= 4.12
= 0.187
51kΩ
R3
vgs = vi
1.32kΩ
1.32kΩ
≤ 0.2(VGS + 2) | v i
≤ 0.2(−0.992 + 2)→ vi ≤ 0.354 V
1kΩ + 1.32kΩ
1kΩ + 1.32kΩ
14.36
(2x10 )(V
=
−4
ID
GS
+ 1) |
2
2
15 + VGS
= 10−4 (VGS + 1) → VGS = −2.363V
68kΩ
2
15 + VGS
= 186µA | VDS = − 30 − (68kΩ + 43kΩ)I D = −9.35V |
ID =
68kΩ
2(186µA)
1
= 0.274mS | Rin = 68kΩ
= 3.46kΩ
Pinchoff region is correct. | g m =
2.36 −1
gm
[
]
Rout ≅ RD = 43kΩ | RL = 43kΩ 200kΩ = 35.4kΩ
Rin
3.46kΩ
g m RL =
(0.274mS )(35.4kΩ)= 9.05
RI + Rin
0.250kΩ + 3.46kΩ
⎛ 3.71kΩ ⎞
R + Rin
3.46kΩ
Ai = Av I
= 9.05⎜
≤ 0.2(VSG −1)
⎟ = 0.168 | vgs = vi
R3
0.250kΩ + 3.46kΩ
⎝ 200kΩ ⎠
3.46kΩ
vi
≤ 0.2(2.36 −1)→ v i ≤ 0.292 V
0.250kΩ + 3.46kΩ
Av =
14-20
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.37
VGS = −10 + (33kΩ)I D | VGS
(3.3x10 )(2x10 )(V
= −10 +
4
(2x10 )(V
−4
VGS = −2.507 V & I D =
VDS
GS
−4
2
GS
+ 1)
2
+ 1) = 227µA
2
2
= − 20 − I D (33kΩ + 24kΩ) = −7.06 V - Active region operation is correct.
[
]
2(2x10 )(2.27x10 ) = 3.01x10
−4
gm =
−4
−4
S | RI RS = 0.5kΩ 33kΩ = 493Ω
Assume λ = 0, ro = ∞ | RL = RD R3 = 24kΩ 100kΩ = 19.4kΩ
Av =
⎛ RS ⎞ 0.301mS (19.4kΩ) ⎛
⎞
33kΩ
⎜
⎟=
⎜
⎟ = 5.01
1+ g m RI RS ⎝ RI + RS ⎠ 1+ 0.301mS (493Ω)⎝ 500Ω + 33kΩ ⎠
g m RL
(
Ai = 1
)
⎛ RD ⎞
⎛
⎞
33kΩ
24kΩ
⎜
⎟=
⎜
⎟ = 0.176
1 ⎝ RD + R3 ⎠ 33kΩ + 3.32kΩ ⎝ 24kΩ + 100kΩ ⎠
RS +
gm
Rin = RS
vgs = vi
RS
1
= 3.02 kΩ | Rout = RD = 24kΩ
gm
RIN
3.02kΩ
≤ 0.2VGS + 1 | vi
≤ 0.2(1.51)→ vi ≤ 0.352 V
0.5kΩ + 3.02kΩ
RI + RIN
14.38
For Rth <<
1
, Av ≅ g m RL All of the input voltage appears across the gate - source
gm
terminals of the transistor.
1
R
For Rth >>
, Av ≅ L For large Rth , all of the Thevenin equivalent source
Rth
gm
current,
14.39
Rin =
vth
, goes into the transistor source terminal.
Rth
75(0.025V )
rπ + 1.5kΩ
1.88kΩ + 1.5kΩ
| rπ =
= 1.88kΩ | Rin =
= 44.5 Ω
βo + 1
1mA
76
14.40
g m = 2(1.25mA)(1mA) = 1.58mS | Rin =
1
= 633 Ω
gm
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-21
14.41
⎡
15V − 0.7V
β o RE ⎤
a
R
=
r
1+
⎢
( ) out o⎣ r + R ⎥⎦ | I E = 143kΩ = 100µA | For β F = 100, IC = 99.0µA
π
E
rπ =
Rout
100(0.025V )
= 25.3kΩ | ro ≅
50V
= 505kΩ
99.0µA
99.0µA
⎡
100(143kΩ) ⎤
⎥ = 43.4 MΩ
⎢
= 505kΩ 1+
⎢⎣ 25.3kΩ + 143kΩ⎥⎦
(b) 0 V
100(0.025V )
15V − 0.7V
= 953µA | For β F = 100, IC = 944µA | rπ =
= 2.65kΩ
15kΩ
944µA
⎡
100(15kΩ) ⎤
50V
⎥ = 4.56 MΩ | VCB ≥ 0 V
ro ≅
= 53.0kΩ | Rout = 53.0kΩ ⎢1+
944µA
⎢⎣ 15kΩ + 2.65kΩ⎥⎦
(c) I
E
=
14.42
⎛V + V ⎞
⎛ 50 + 10.7 ⎞
Rout = (β o + 1)ro = (β o + 1) ⎜ A CE ⎟ = 126 ⎜
⎟ = 154 MΩ
⎝ 49.6µA ⎠
⎝ IC ⎠
14.43
43
Rin = 350Ω | Av = 10 20 = 141 | Low Rin , large gain
A common - base amplifier can achieve these speocfications.
1
1
→ IC ≅
= 71.4 µA
gm
40(350)
Rin ≅
A common emitter amplifier operating at a higher current is an alternate choice.
100
Rin ≅ rπ → IC ≅
= 7.14 mA
40(350)
For both cases, Av ≅ 10VCC → VCC = 14 V
14.44
Rin = 0.3 MΩ | Av = 10
46
20
= 200 | Fairly large Rin , large gain
A common - emitter amplifier operating at a low current can achieve both a
large gain and input resistance. Av ≅ 20VCC → VCC = 10V
Achieving this gain with an FET is much more difficult :
Av ≅
VDD
V
= DD → VDD ≅ 50V which is unreasonably large.
VGS −V TN 0.25V
14-22
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.45
26
Rin = 10 MΩ | Av = 10 20 = 20 | Large Rin , moderate gain
These requirements are readily met by a common - source amplifier.
VDD
15V
For example, Av ≅
=
= 30.
VGS −V TN 0.5V
A common - emitter stage operating at a low collector current with
⎛
⎞
10 MΩ
an unbypassed emitter resistor ⎜ R E ≅
= 100kΩ⎟is a second possibility,
100
⎝
⎠
but the circuit will require careful design.
14.46
58
20
Rin = 50kΩ | Av = 10 = 792 | A bipolar transistor would be required
for such a large gain. However, this is a large fraction of the BJT amplification
factor [i. e. (40/V)(75V) = 3000] and will be very difficult to achieve with
the information thus far (the active load discussed later is a possibility).
Using our rule - of - thumb for the common - emitter amplifier,
Av ≅ 10VCC → VCC = 80 V which is too large. Thus, it is not possible is the best answer.
14.47
An inverting amplifier with a gain of 40 dB is most easily achieved with a common - emitter
stage : Av ≅ 10VCC → VCC = 10 V. The input resistance can be achieved by shunting the
input with a 5 - Ω resistor. Setting rπ = 5 Ω would require IC ≅
100(0.025V )
= 0.5A and would
5Ω
waste a large amount of power to achieve the required input resistance.
It would be better to operate the transistor at a much lower current and "swamp" the input resistance
by shunting the input with a 5 - Ω resistor
14.48
A non - inverting amplifier with a gain of 20 and an input resistance of 5 kΩ should
be readily achievable with either a common - base or common - gate amplifier with
proper choice of operating point. The gain of 10 is easily achieved with either the
VDD
1
FET or BJT design estimate : Av ≅
or Av ≅ 10VCC . Rin ≅
= 5 kΩ is within
gm
VGS − VTN
easy reach of either device. The gain and input resistance can also be easily met
with either a common - emitter, or common - source stage with a resistor shunt at the input.
14.49
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-23
0 - dB gain corresponds to a follower ( Av = 1).
For an emitter - follower, R in ≅ (β o + 1)RL ≅ 101(20kΩ) = 2.02 MΩ. So an BJT
cannot meet the input resistance requirement. A source follower provides a gain of
approximately 1 and can easily achieve the required input resistance.
14.50
A gain of 0.97 and an input resistance of 400kΩ should be achievable with
g m RL
either a source - follower or an emitter - follower. For the FET, Av ≅
= 0.97
1+ g m RL
requires g m RL = 33.3 :
2I D RL
= 33.3 → I D RL = 8.3V for a design with VGS − VTN = 0.5V.
VGS − VTN
The BJT can achieve the required gain with a much lower power supply and
can still meet the Rin requirement : Rin ≅ β o RL ≅ 100(5kΩ) = 500kΩ.
Av ≅
g m RL
= 0.97 | g m RL = 33.3 → IC RE = 33.3(0.025)= 0.833 V.
1+ g m RL
The requirements can be met with careful bias circuit design and specification
of a BJT with minimum current gain of at least 100.
14.51
66
Av = 10 20 = 2,000. This value of voltage gain approaches the amplification factor
of the BJTs : Av ≤ µ f = 40VA = 40(75)= 3000. Such a large gain
requirement cannot be met with single - transistor BJT amplifiers using the
resistive loaded amplifiers in this chapter (Remember the 10VCC limit). FETs
typically have much lower values of µf and are at an even worse disadvantage.
None of the single - transistor amplifier configurations can meet the gain requirements.
14-24
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.52
Such a large output resistance will require either a CE or CB stage or a CS or CG stage.
For a BJT, Rout ≤ β oro
β oVA
100(75V )
= 7.5µA using typical
109 Ω
values for β o and VA . We need to also see how much voltage is required.
IC
= 109 Ω or IC =
75
> 109 Ω → IC RE > 2.5 V whihc is reasonable.
We also need ro (1+ g m RE )> 109 Ω or 40(IC RE )
7.5µA
For a MOSFET, Rout = ro (1+ g m RS ) ≅ g mro RS =
RS =
109 (0.01/V )(0.25V )
2
= 1.25 MΩ
2RS
= 109 Ω
λ(VGS − VTN )
For typical values,
Using this value to estimate the required voltage,
2
2Kn ⎛ RS ⎞ 2(0.001)⎛1.25x106 ⎞
RS
9
= 10 Ω → I D = 18 ⎜ ⎟ =
g mro RS = 2Kn I D
⎟ = 31.3µA and
⎜
λI D
10 ⎝ λ ⎠
1018 ⎝ 0.01 ⎠
2
VR S = 39 V which is getting large. So the BJT appears to be the best choice.
14.53
Rout =
Rth + rπ
βo + 1
| Assuming Rth ≅ RI and rπ = 0, Rout ≥
RI
250
=
= 1.66 Ω
β o + 1 151
14.54
Rin = rπ + (β o + 1)RE ≅ rπ + β o RE = rπ (1+ g m RE ) | rπ' = rπ (1+ g m RE )
g m' =
ro' =
ic
gm
βo
βo
βo
=
≅
=
=
vi rπ + (β o + 1)RE rπ + β o RE rπ (1+ g m RE ) 1+ g m RE
ic
vc
⎛
βR ⎞ ⎛ β R ⎞
= ro⎜1+ o E ⎟ ≅ ro⎜1+ o E ⎟ = ro (1+ g m RE ) for rπ >> RE
rπ ⎠
⎝ rπ + RE ⎠
⎝
v i =0
⎛
⎛ gm ⎞
gm ⎞
'
' '
⎟rπ (1+ g m RE ) = β o | µ f = g mro = ⎜
⎟ro (1+ g m RE ) = µ f
⎝1+ g m RE ⎠
⎝1+ g m RE ⎠
β o' = g m' rπ' = ⎜
14.55
*Problem 14.55 - Common-Emitter Amplifier 5mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.005 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-25
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
*Problem 14.55 - Common-Emitter Amplifier 10mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.01 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
*Problem 14.55 - Common-Emitter Amplifier 15mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.015 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
Results:
1 kHz
2 kHz
3 kHz
vi
14-26
THD
© R. C. Jaeger & T. N. Blalock - February 20, 2007
5 mV
5.8 mV
10 mV
15 mV
12.4 mV
20.6 mV
0.335 mV (5.7%)
0.043 mV
(0.74%
1.54 mV (12.5%) 0.258 mV (2.1%)
4.32 mV (21%) 1.18 mV (5.4%)
5.9%
12.8%
22%
14.56
*Problem 14.56 - Output Resistance
VCC 2 0 DC 10
IB1 0 1 DC 10U
Q1 2 1 0 NBJT
IB2 0 3 DC 10U
RE 4 0 10K
Q2 2 3 4 NBJT
.OP
.DC VCC 10 20 .025
.MODEL NBJT NPN IS=1E-16 BF=60 VA=20
.PRINT DC IC(Q1) IC(Q2)
.PROBE IC(Q1) IC(Q2)
.END
Results: A small value of Early voltage has been used deliberately to accentuate the results.
Note that the transistors have significantly different values of
F because of the collectoremitter voltage differences and low value of VA.
NAME
Q1
Q2
MODEL
NBJT
NBJT
IB
1.00E-05 1.00E-05
IC
8.77E-04 6.72E-04
VBE
7.61E-01 7.61E-01
VBC
-9.24E+00 -2.41E+00
VCE
1.00E+01 3.18E+00
BETADC 8.77E+01 6.72E+01
GM
3.39E-02 2.60E-02
RPI
2.59E+03 2.59E+03
RO
3.33E+04 3.33E+04
From SPICE : Rout1 =
(20 −10)
V
(20 −10) V = 43.5 kΩ
= 34.1 kΩ | Rout 2 =
(1.17 − 0.877) mA
(903 − 673) µA
For circuit 1: Rout1 = ro1 = 33.3 kΩ
⎛
⎞
β o RE
For circuit 2 : Rout 2 = ro2⎜1+
⎟ + (Rth + rπ ) RE (See Eq. 14.28)
⎝ Rth + rπ + RE ⎠
But Rth = ∞ → Rout 2 = ro2 + RE = 33.3kΩ + 10kΩ = 43.3 kΩ
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-27
14.57
⎛
⎛
100(270Ω) ⎞
β o RI ⎞
⎜
⎟ = 384 kΩ
Rth ≅ ro ⎜1+
⎟ = 250kΩ⎜1+
⎟
270Ω
+
50kΩ
⎝ RI + rπ ⎠
⎝
⎠
vi
vth = isc Rth =
RI +
αo
α o Rth =
gm
vi
0.990
270Ω +
0.002
0.990(384 kΩ) → vth = 479v i
14.58
vth = Gm Rth = −
[
]
gm
ro (1+ g m RS ) vi = −µ f v i = −(0.5mS )(250kΩ)vi = −125vi
1+ g m RS
Rth = ro (1+ g m RS ) = ro + µ f RS = 250kΩ + 125(18kΩ)= 2.50 MΩ
14.59
(β
vth = v i
o
+ 1)ro
RI + rπ + (β o + 1)ro
= vi
1
rπ
RI
+
+1
(β o + 1)ro (βo + 1)ro
= vi
1
g m RI
βo
+
+1
(β o + 1)µ f (β o + 1)µ f
RI + rπ
R +r
RI
α
ro ≅ I π =
+ o
βo + 1
βo + 1 βo + 1 gm
Rth ≅
14.60
(a) g
i1
i2
g12 =
(b) g
21
21
=
v2
v1
v1 =0
=
i2 =0
g m RE
≅1
1+ g m RE
⎛ 1 ⎞ ie ⎛ 1 ⎞ R E
1 ⎛ g m RE ⎞ 1
=⎜
=
⎟ ≅⎜
⎟
⎜
⎟≅
βo + 1 ⎝1+ g m RE ⎠ βo
⎝ β o + 1⎠ i 2 ⎝ β o + 1⎠ R + 1
E
gm
= 0.960
| g12 = 9.51x10−3 | g21 >> g12
14.61
(a) g
g12 =
21
i1
i2
=
| v2 =
i2 =0
(
g m RD ro
(
)
1+ g m RD ro
v1 | g21 =
)
(
g m RD ro
(
)
1+ g m RD ro
)
≅1
| i1 =
v1 =0
(b) g21 =
14-28
v2
v1
(
)
0.2mS 50kΩ 450kΩ
(
)
1+ 0.2mS 50kΩ 450kΩ
= 0.947 | g12 = 0 | g21 >> g12
© R. C. Jaeger & T. N. Blalock - February 20, 2007
≅ vi
14.62
(a) g
i1
i2
g12 =
(b) g
21
21
=
v2
v1
(
)
(
| v2 = g m RC ro v1 | g21 = g m RC ro
i2 =0
=−
v1 =0
RC
ro + RC
R r r + RC
g21
= gm C o o
= g mro = µ f >> 1
g12
ro + RC RC
|
(
)
)
= 3mS 18kΩ 800kΩ = 52.8
g12 = −
18kΩ
= 0.0220
18kΩ + 800kΩ
i1
i2
=−
14.63
(a) g
(b) g
21
21
=
v2
v1
(
= +g m RD ro
i2 =0
)|
(
g12 =
v1 =0
RD
| g21 = −g mro g12 ≅ −µ f g12
RD + ro
)
= +0.5mS 100kΩ 500kΩ = 41.7 | g12 = −
100kΩ
= −0.167 | g21 >> g12
100kΩ + 500kΩ
14.64
(a) g
i1 ≅ −
(b) g
21
=
v2
v1
| v2 ≅ −
i2 = 0
g m RC
g R
v1 | g21 = − m C
1+ g m RE
1+ g m RE
| g12 =
i1
i2
v1 = 0
⎛ RE ⎞
⎛ RE ⎞
⎛ RE ⎞
i2 RC
v2
RC
⎟≅−
⎟ | g12 = −
⎟
⎜
⎜
⎜
ro (1+ g m RE )⎝ RE + rπ ⎠
ro (1+ g m RE )⎝ RE + rπ ⎠
ro (1+ g m RE )⎝ RE + rπ ⎠
21
=−
2mS (130kΩ)
1+ 2mS (12kΩ)
= -10.4 | g12 = −
⎞
⎛
12kΩ
-3
⎟ = -1.01x10
⎜
1MΩ 1+ 2mS (12kΩ) ⎝12kΩ + 50kΩ ⎠
[
130kΩ
]
g21 >> g12
14.65
(a) g
g12 =
(b) g
21
i1
i2
=
v2
v1
v1 = 0
21 = −
i2 = 0
g m RD
g R
v1 | g21 ≅ − m D
1+ g m RS
1+ g m RS
| i1 = 0 |
g12 = 0 |
| v2 ≅ −
0.75mS (130kΩ)
1+ 0.75mS (12kΩ)
= −9.75
g12
=0
g21
| g21 >> g12
| g12 = 0
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-29
14.66
gm + go
vx
g o + GL
⎛ g + go ⎞
⎛ GL − g m ⎞
i
| x = g m + g o ⎜1− m
⎟ = g m + g o⎜
⎟
vx
⎝ g o + GL ⎠
⎝ g o + GL ⎠
At the output node v o : g mvx = (g o + GL )vo − g ovx | vo =
ix = g mvx + g o (vx − v o )
⎛G − g ⎞
ix
gm + go
m
= g m + g o⎜ L
⎟ = GL
vx
g o + GL
⎝ g o + GL ⎠
RL
v
1
1 ⎛ RL ⎞
ro
| Rin = x =
≅
⎜1+ ⎟
1
ix g m
ro ⎠
gm ⎝
1+
1+
µf
14.67
IC = 100
5 − 0.7
( )
10 + 101 10
4
3
= 3.87mA | g m = 40IC = 0.155S | rπ =
100
= 645Ω
gm
RL = 1kΩ 20kΩ = 952Ω | RE = 1kΩ 20kΩ = 952Ω
Av1 = −
Av2 =
100(952Ω)
β o RL
=−
= −0.984
rπ + (β o + 1)RE
645Ω + 101(952Ω)
(β + 1)R
+ (β + 1)R
o
rπ
E
o
E
=
101(952Ω)
645Ω + 101(952Ω)
= 0.993
The small - signal requirement limits the output signal to :
vbe = vi − vo2 = vi (1− 0.993) = 0.007vi | v i ≤
vo1 ≤ 0.984(0.714V ) = 0.703V
0.005
= 0.714V
0.007
We also need to check VCB : VC = 5 − 3.87mA(1kΩ)= 1.13V and VB = −10 4 I B = −0.387V.
The total collector - base voltage of the transistor is therefore : VCB = 1.52V − 0.984v i − vi .
We require VCB ≥ 0 for forward - active region operation. Therefore : vi ≤ 0.766 V.
The small - signal limit is the most restrictive.
14-30
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.68
100kΩ
= 0V | REQ = 100kΩ 100kΩ = 50kΩ
100kΩ + 100kΩ
0 − 0.7 − (−15)
= 22.3µA | IC = 2.78 mA | I E = 2.81 mA
IB =
50kΩ + 126(4.7kΩ)
(a) VEQ = −15 + 30
VCE = 30 − 2000IC − 4700I E = 11.4 V
rπ =
125(0.025V )
= 1.12kΩ | ro =
(50 + 11.4)V = 22.1kΩ
2.78mA
2.78mA
RB = R1 R2 = 100kΩ 100kΩ = 50kΩ | RL = R3 RE ro = 24kΩ 4.7kΩ 22.1kΩ = 3.34kΩ
[
]
[
]
Rin = RB rπ + (β o + 1)RL = 50kΩ 1.12kΩ + (126)3.34kΩ = 44.7 kΩ
Av = +
(β + 1)R
+ (β + 1)R
o
rπ
Rout = RE ro
L
o
(R
B
⎞
⎛ R
⎛
⎞
126(3.34kΩ)
44.7kΩ
in
⎟=
⎜
⎜
⎟ = 0.984
1.12kΩ + 126(3.34kΩ)⎝ 0.600kΩ + 44.7 kΩ ⎠
L ⎝ RI + Rin ⎠
)
RI + rπ
= 4.7kΩ 22.1kΩ
(44.7kΩ 600Ω)+ 1.12kΩ = 13.5 Ω
126
βo + 1
(b) SPICE Results: Q-point: (2.81 mA, 11.1 V), Av = 0.984, Rin = 45.5 kΩ Rout = 13.0 Ω
14.69
(10 − 0.7)V = 1.43 µA | I = 114 µA | V
1MΩ + (80 + 1)68kΩ
81(0.025V )
Active region is correct. | r =
= 17.8 kΩ
IB =
C
π
114µA
CE
= 20 − 39000IC − 68000I E = 7.71 V
| ro =
75 + 7.71
= 726 kΩ
114µA
RL = 500kΩ 39kΩ 726kΩ = 34.5kΩ | Rin = RB rπ = 1MΩ 17.8kΩ = 17.5 kΩ
⎛ Rin ⎞
⎞
⎛
17.5kΩ
Av = −g m RL ⎜
⎟ = 40(0.114mA)(34.5kΩ)⎜
⎟ = −153
⎝ 500Ω + 17.5kΩ ⎠
⎝ RI + Rin ⎠
Rout = RC ro = 39kΩ 726kΩ = 37.0 kΩ
⎛ Rin ⎞
⎞
⎛
17.5kΩ
vbe = vi ⎜
⎟ = vi ⎜
⎟ = 0.972vi
⎝ 500Ω + 17.5kΩ ⎠
⎝ RI + Rin ⎠
vi ≤
0.005V
= 5.14 mV
0.972
Av ≅ −10(VCC + VEE ) = −10(20)= −200. | A closer estimate is - 40VR C = -40(4.47) = −178
(b) SPICE Results: Q-point: (116 µA, 7.53 V), Av = -150, Rin = 19.6 kΩ, Rout = 37.0 kΩ
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-31
14.70
62kΩ
= 9.07V | REQ = 62kΩ 20kΩ = 15.1kΩ
62kΩ + 20kΩ
12 − 0.7 − 9.07
IB =
= 4.19µA | IC = 314 µA | I E = 319 µA
15.1kΩ + 76(6.8kΩ)
(a) VEQ = 12
VEC = 12 −16000IC − 6800I E = 4.81 V
75(0.025)
60 + 4.81
= 206kΩ | rπ =
= 5.97kΩ
−6
314x10
314x10−6
RL = ro 16kΩ 100kΩ = 206kΩ 16kΩ 100kΩ = 12.9kΩ
ro =
Rin = 15.1kΩ 5.97kΩ = 4.28 kΩ | Rout = ro 16kΩ = 14.8 kΩ | g m = 40IC = 12.6 mS
⎛ Rin ⎞
⎛ 4.28kΩ ⎞
Av = −g m RL ⎜
⎟ = −(12.6mS )(12.9kΩ)⎜
⎟ = −132
⎝1kΩ + 4.28kΩ ⎠
⎝ RI + Rin ⎠
(b) SPICE Results: Q-point: (309 µA, 4.93 V), Av = -127, Rin = 4.65 kΩ Rout = 14.9 kΩ
14.71
500kΩ
= 4.73V | REQ = R1 R2 = 500kΩ 1.4 MΩ = 368kΩ
500kΩ + 1.4 MΩ
Assume Active Region Operation
(a) V
EQ
= 18
2I D
+ 27000I D → I D = 113 µA
5x10−4
= 18 − (27000 + 75000)I D = 6.47 V > 3.73 V - Active region is correct.
4.73 = VGS = 27000I D → 4.73 = 1+
VDS
50 + 6.47
= 500kΩ | g m = 2 500x10−6 113x10−6 1+ 0.02(6.47) = 357µS
−6
113x10
RL = ro RD 470kΩ = 500kΩ 75kΩ 470kΩ = 57.3kΩ
ro =
(
)(
)[
]
Rin = R1 R2 = 500kΩ 1.4 MΩ = 368kΩ | Rout = ro 75kΩ = 65.2 kΩ
⎛ Rin ⎞
⎛ 368kΩ ⎞
Av = −g m RL ⎜
⎟ = −(0.357mS )(57.3kΩ)⎜
⎟ = −20.4
⎝ 1kΩ + 368kΩ ⎠
⎝ RI + Rin ⎠
(b) SPICE Results: Q-point: (115 µA, 6.30 V), Av = -20.5, Rin = 368 kΩ Rout = 65.1 kΩ
14-32
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.72
(2.5x10 )(V
=
−4
ID
2
10 + VGS
= 1.25x10−4 (VGS + 1) → VGS = −2.358V
33kΩ
+ 1) |
2
GS
2
10 + VGS
= 232µA | VDS = −10 + 0.232mA(24kΩ)− 2.36 = −6.79V | Pinchoff region is correct.
ID =
33kΩ
(
)(
)[
]
g m = 2 2.5x10−4 2.32x10−4 1+ 0.02(6.79) = 0.363mS | RI RS = 0.5kΩ 33kΩ = 493Ω
ro=
50 + 6.79
1
=245kΩ RL = ro RD R3 = 245kΩ 24kΩ 100kΩ = 17.9kΩ | Rin = 33kΩ
= 2.69 kΩ
−4
gm
2.32x10
Rout = ro RD = 21.9 kΩ Av =
⎛ RS ⎞
⎞
0.363mS (17.9kΩ) ⎛
33kΩ
⎟=
⎜
⎜
⎟ = 5.42
1+ g m RI RS ⎝ RI + RS ⎠ 1+ 0.363mS (0.493kΩ)⎝ 500Ω + 33kΩ ⎠
g m RL
(
)
(b) SPICE Results: Q-point: (234 µA, -6.67V), Av = +5.56, Rin = 2.69 kΩ, Rout= 18.1 kΩ
14.73
IC = 100
rπ =
5 − 0.7
= 12.5µA | VCE = 5 − IC (330kΩ)− (−5)= 5.87V
500kΩ + 500kΩ + (101)330kΩ
100
60 + 5.87
= 200kΩ | ro =
= 5.27 MΩ | RL = 500kΩ 330kΩ 500kΩ ro = 139kΩ
40(12.5µA)
12.5x10−6
Absorb R1 into the transistor : rπ' = rπ R1 = 143kΩ | β o' = g mrπ' = 71.4
(
)
Rin = rπ' + β o' + 1 RL = 143kΩ + 71.4(139kΩ)= 10.1 MΩ
(R
I
Rout = 330kΩ 500kΩ ro
)
RB + rπ'
β o' + 1
(β + 1)R
A =+
(R R )+ r + (β + 1)R
'
o
v
L
'
I
B
π
'
o
L
=
= 1.97 kΩ
72.4(139kΩ)
1kΩ + 143kΩ + 72.4(139kΩ)
= 0.986
(b) SPICE Results: Q-point: (12.7 µA, 5.78 V), Av = +0.986, Rin = 10.7 MΩ, Rout= 2.00 kΩ
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-33
14.74
IB =
(12 − 0.7)V = 2.69 µA |
100kΩ + (51)82kΩ
Active region is correct.
g m = 40(135µA) = 5.40 mS | rπ =
IC = 135 µA | VCE = 24 − 39000IC − 82000I E = 7.58 V
50(0.025V )
135µA
= 9.26 kΩ | ro =
50 + 7.58
= 427 kΩ
135µA
Rth = RI RE = 0.5kΩ 82kΩ = 497Ω | RL = ro (1+ g m Rth ) RC R3 = 1.57 MΩ 39k Ω 100kΩ = 27.6 kΩ
Av =
⎛ RE ⎞ 5.40mS (27.6kΩ) ⎛
⎞
82kΩ
⎟=
⎜
⎜
⎟ = 40.2
1+ g m RI RE ⎝ RI + RE ⎠ 1+ 5.40mS (497Ω)⎝ 500Ω + 82kΩ ⎠
g m RL
Rin = 82kΩ
(
)
rπ
= 181 Ω | Rout = ro (1+ g m Rth ) RC = 38.1 kΩ
βo + 1
(b) SPICE Results: Q-point: (132 µA, 7.79V), Av = 39.0, Rin = 204 Ω Rout = 38.0 kΩ
14.75
2.2 MΩ
= 9.00V | REQ = R1 R2 = 2.2mΩ 2.2 MΩ = 1.10 MΩ
2.2 MΩ + 2.2 MΩ
Assume Active Region Operation
VEQ = 18
2I D
+ 110000I D → I D = 67.5 µA
4x10−4
= 18 − (110000 + 90000)I D = 4.50 V > 0.575 V - Active region is correct.
18 - 9 = 110000I D − VGS → 9 = 1+
VDS
50 + 4.50
= 807kΩ | g m = 2 400x10−6 67.5x10−6 1+ 0.02(4.50) = 243µS
−6
67.5x10
RL = ro RD 470kΩ = 807kΩ 90kΩ 470kΩ = 69.7kΩ
ro =
(
)(
)[
]
Rin = R1 R2 = 2.2 MΩ 2.2 MΩ = 1.1 MΩ | Rout = ro 90kΩ = 81.0 kΩ
⎛ Rin ⎞
⎛ 1.1MΩ ⎞
Av = −g m RL ⎜
⎟ = −(0.243mS )(69.7kΩ)⎜
⎟ = −16.9
⎝ 1kΩ + 1.1MΩ ⎠
⎝ RI + Rin ⎠
(b) SPICE Results: Q-point: (66.7 µA, 4.47V), Av = -16.8, Rin = 1.10 MΩ Rout = 81.0 kΩ
14-34
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.76
5x10−4
(a ) Assume Active Region operation. V = −51000I I = 2 (V + 2) → I
V = 15 − (20000 + 51000)I = 12.7 V > 0.36 V - Active Region is correct.
GS
DS
D
D
(
Rout
D
= 32.2 µA
D
)(
)[
]
g m = 2 5x10−4 32.2x10−6 1+ 0.02(12.70) = 0.201mS | Rin = 51kΩ
ro =
2
GS
1
= 4.53kΩ
gm
50 + 12.7
= 1.95MΩ | Rth = RI RS = 981Ω
32.2x10−6
= ro (1+ g m Rth ) RD = 19.8 kΩ | RL = Rout 10kΩ = 6.65kΩ
Av =
⎛ R ⎞
0.201mS (6.65kΩ) ⎛ 51kΩ ⎞
S
⎟=
⎜
⎜
⎟ = 1.10
1+ g m RI RS ⎝ RI + RS ⎠ 1+ 0.201mS (0.981kΩ)⎝1kΩ + 51kΩ ⎠
g m RL
(
)
(b) SPICE Results: Q-point: (32.9 µA, 12.7 V), Av = +1.10, Rin = 4.50 kΩ, Rout= 19.8 kΩ
14.77 The power supply should be +16 V.
(a ) Assume Active Region operation. Since there is no negative feedback (RS = 0),
we should include the effect of channel - length modulation. VGS = 0
2
4x10−4
−5) (1+ 0.02VDS ) and VDS = 16 −1800I D → I D = 5.59 mA
(
2
= 16 −1800I D = 5.93 V > 5 V - Active region is correct.
ID =
VDS
50 + 5.93
= 10.0kΩ
5.59x10−3
⎛ RG ⎞
⎛ 10 MΩ ⎞
Av = -g m RL ⎜
⎟ = −(2.24mS ) 10.0kΩ 1.8kΩ 36kΩ ⎜
⎟ = −3.27
⎝ 10 MΩ + 5kΩ ⎠
⎝ RI + RG ⎠
(
)
[
]
g m = 2 4x10−4 (5.59mA) 1+ 0.02(5.93) = 2.24mS | ro =
(
)
Rin = 10.0 MΩ | Rout = RD ro = 1.52 kΩ
(b) SPICE Results: Q-point: (5.59 mA, -5.93V), Av = -3.27, Rin = 10.0 MΩ, Rout= 1.52 kΩ
14.78
(10 − 0.7)V
= 14.0µA | I = 1.12 mA | V = 20 − 7800I = 11.3 V
33kΩ + (80 + 1)7.8kΩ
80(0.025V )
(100 + 11.3)V = 99.4kΩ
= 1.79kΩ | r =
Active region is correct. | r =
IB =
C
π
CE
E
o
1.12mA
1.12mA
RL = ro 1MΩ = 95.1kΩ 1MΩ = 86.8kΩ | Rin = RB rπ = 33kΩ 1.79kΩ = 1.70 kΩ | Rout = ro = 99.4 kΩ
⎛ Rin ⎞
1.70kΩ
= −3390
Av = -g m RL ⎜
⎟ = −40(1.12mA)(86.8kΩ)
250Ω + 1.70kΩ
⎝ RI + Rin ⎠
(b) Results: Q-point: (1.12 mA, 11.2 V), Av = -3440, Rin = 1.93 kΩ Rout= 98.9 kΩ
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-35
14.79
Bias forces Active Region operation : VDS = 6 − (−6)= 12 V VGS = 6 V
2
2
Kn
4x10−4
ID =
VGS − VTN ) (1+ λVDS ) =
6 −1) 1+ 0.02(12) = 6.20 mA
(
(
2
2
[
(
)(
)[
]
]
g m = 2 4x10−4 6.2x10−3 1+ 0.02(12) = 2.48 mS
1
+ 12
V
0.02
= 10.0kΩ − Cannot neglect! | RL = 10.0kΩ 100kΩ = 9.09kΩ
ro =
6.20 mA
1
Rin = RG = 2 MΩ | Rout =
ro = 388 Ω
gm
⎛ 2.48mS (9.09kΩ) ⎞
2 MΩ
Rin ⎛ g m RL ⎞
⎟ = 0.953
⎜
Av = +
⎟=+
⎜
10kΩ + 2 MΩ ⎜⎝1+ 2.48mS (9.09kΩ)⎟⎠
RI + Rin ⎝ 1+ g m RL ⎠
(b) SPICE Results: Q-point: (6.20 mA, 12.0V), Av = 0.953, Rin = 2.00 MΩ, Rout= 388 Ω
14.80
(a) V
EQ
= 15
500kΩ
= 3.95 V | REQ = 500kΩ 1.4 MΩ = 368 kΩ
1.4 MΩ + 500kΩ
2I D
+ 27000I D → I D = 85.1 µA
400x10−6
= 15 − I D (75kΩ + 27kΩ) = 6.32V | Active region operation is correct.
Assume active region | 3.95 = VGS + 27000I D = 1+
VDS
(
)(
)
g m = 2 400x10−6 85.1x10−6 = 0.261 mS | ro =
C1 ≥ 10
50 + 6.32
= 662 kΩ | RG = R1 R2 = 368 kΩ
85.1µA
1
10
=
| C1 ≥ 0.0108 µF → 0.01 µF
2πf (RI + RG ) 2π (400Hz)(1kΩ + 368kΩ)
1
10
=
| C2 ≥ 1.19 µF →1.2 µF
⎛
⎛
⎞
⎞
1
1
2πf ⎜ RS
⎟
⎟ 2π (400Hz)⎜27kΩ
0.261mS ⎠
gm ⎠
⎝
⎝
1
10
C3 ≥ 10
=
| C3 ≥ 7.40 nF → 8200 pF
2π (400Hz)(67.4kΩ + 470kΩ)
2πf RD ro + R3
C2 ≥ 10
[( ) ]
(b) C
2
14-36
=
1
⎛
1⎞
2πf ⎜ RS
⎟
gm ⎠
⎝
=
1
⎛
⎞
1
2π (4000Hz)⎜27kΩ
⎟
0.261mS ⎠
⎝
= 0.0119 µF → 0.012 µF
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.81
62kΩ
= 9.07V | REQ = 20kΩ 62kΩ = 15.1kΩ
20kΩ + 62kΩ
(12 − 0.7 − 9.07)V = 7.16µA | I = 537 µA | V = 12 − 3900I − 8200I = 5.47 V
IB =
C
EC
E
C
15.1kΩ + (75 + 1)3.9kΩ
(a) V
EQ
= 12
Active region is correct.
rπ =
75(0.025V )
537µA
= 3.49kΩ | ro =
60 + 5.47
= 122kΩ | g m = 40IC = 21.5 mS
537µA
Rin = R1 R2 rπ = 15.1kΩ 3.49kΩ = 2.83 kΩ | Rout = ro RC = 122kΩ 8.2kΩ = 7.68 kΩ
C1 ≥ 10
1
10
=
| C1 ≥ 4.16 µF → 0.01 µF
2πf (RI + Rin ) 2π (100Hz)(1kΩ + 2.83kΩ)
1
10
| C2 ≥ 273 µF → 270 µF
⎛R R +r ⎞
⎛1kΩ 15.1kΩ + 3.49kΩ ⎞
I
EQ
π
⎟ 2π (100Hz)⎜
⎟
2πf ⎜⎜
⎟
⎟
⎜
β
+
1
76
o
⎠
⎠
⎝
⎝
1
10
=
| C3 ≥ 0.148 µF → 0.15 µF
C3 ≥ 10
2πf [Rout + R3 ] 2π (100Hz)(7.68kΩ + 100kΩ)
C2 ≥ 10
(b) C
2
=
=
1
⎛R R +r ⎞
I
EQ
π
⎟
2πf ⎜⎜
⎟
⎝ βo + 1 ⎠
=
1
⎛ 1kΩ 15.1kΩ + 3.49kΩ ⎞
⎟
2π (1000Hz)⎜⎜
⎟
76
⎝
⎠
= 2.73 µF → 2.7 µF
14.82
IC = 100
5 − 0.7
( )
10 4 + 101 103
= 3.87mA | g m = 40IC = 0.155S | rπ =
100
= 645Ω | ro = ∞
gm
The capacitors will be negligible at a frequency 10 times the individual break frequencies :
[
(
)]
Req1 = 10kΩ rπ + (β o + 1) 1kΩ 20kΩ = 9.06kΩ | f1 =
10
= 87.8 Hz
2π (9.06kΩ)2µF
⎡
rπ ⎤
10
⎥ = 20.0kΩ | f2 =
= 7.96 Hz
Req2 = 20kΩ + ⎢1kΩ
2π (20.0kΩ)10µF
⎢⎣
(β o + 1)⎥⎦
Req3 = Rout + 20kΩ = 21.0kΩ | f 3 =
10
= 7.56 Hz
2π (21.0kΩ)10µF
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-37
14.83
80 ⎛ 5 − 0.7 ⎞
80
= 319µA | g m = 40IC = 12.8mS | rπ =
= 6.26kΩ | ro = ∞
⎜
3⎟
81 ⎝ 13.3x10 ⎠
gm
⎡
10
rπ ⎤
⎥ = 152Ω | C1 =
= 0.209 µF → 0.20 µF
Req1 = 75Ω + ⎢13.3kΩ
2π (50kHz)152Ω
⎢⎣
(β o + 1)⎥⎦
(a) I
C
= αF IE =
Req2 = Rout + 100kΩ = 8.25kΩ + 100kΩ = 108kΩ | C2 =
⎛ 50kHz ⎞
(b) C = 0.209µF⎜⎝100Hz ⎟⎠ = 105 µF →100 µF
1
10
= 295 pF → 270 pF
2π (50kHz)108kΩ
⎛ 50kHz ⎞
| C2 = 295 pF ⎜
⎟ = 0.148 µF → 0.15 µF
⎝ 100Hz ⎠
14.84
51kΩ
= −4.87V | REQ = 51kΩ 100kΩ = 33.8kΩ
51kΩ + 100kΩ
−4.87 − 0.7 − (−15) V
= 18.5µA | IC = 1.85 mA
Assume active region operation | I B =
33.8kΩ + (101)(4.7kΩ)
VEQ = −15 + 15
[
]
VCE = 30 − 4700I E = 21.2 V | g m = 40(1.85mA)= 7.40mS
Active region is correct. | rπ =
100(0.025V )
= 1.35kΩ | ro =
(50 + 21.2)V = 38.5kΩ
1.85mA
1.85mA
RB = R1 R2 = 51kΩ 100kΩ = 33.8kΩ | RL = R3 RE ro = 24kΩ 4.7kΩ 38.5kΩ = 3.57kΩ
[
(R
=R
]
[
]
Rin = RB rπ + (β o + 1)RL = 33.8kΩ 1.35kΩ + (101)3.57kΩ = 30.9 kΩ
Rout
B
E
)
RI + rπ
βo + 1
= 4.7kΩ
(33.8kΩ 500Ω)+ 1.35kΩ = 18.2 Ω
101
C1 =
10
10
=
= 1.01µF →1.0 µF
2πf (RI + Rin ) 2π (50Hz)(500Ω + 30.9kΩ)
C2 =
10
10
=
= 1.33µF →1.5 µF
2πf (R3 + Rout ) 2π (50Hz)(24kΩ + 18.2Ω)
14-38
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.85
Note : RS ≡ R1 = 3.9kΩ | Assume active region operation.
(5x10 )(V
= −3900
−4
VGS = −3900I D
(5x10 )(V
−4
ID =
2
2(254µA)
gm =
2 − 0.992
[
GS
GS
2
+ 2) | VGS = −0.975(VGS + 2) → VGS = −0.9915V
2
2
+ 2) = 254µA | VDS = 15 − 23.9kΩI D = 8.92V - Active region is correct.
2
= 0.504mS | ro =
(
Rout = RD ro 1+ g m RS RI
50 + 8.92
1
= 232kΩ | Rin = 3.9kΩ
= 1.32kΩ
−6
gm
254x10
)]= 20kΩ 232kΩ[1+ 0.504mS (3.9kΩ 1kΩ)]= 18.8kΩ
C1 =
10
10
=
= 1.72µF → 1.8 µF
2πf (RI + Rin ) 2π (400Hz)(1kΩ + 1.32kΩ)
C2 =
10
10
=
= 0.0335µF → 0.033 µF
2πf (R3 + Rout ) 2π (400Hz)(100kΩ + 18.8kΩ)
14.86
(a) Use C2 to set the lower cutoff frequency to 1 kHz. C1 and C3 remain negligible at 1 kHz.
C2 = 0.056 µF, C1 = 1800 pF, C3 = 0.015 µF
(b) SPICE Results: fL = 925 Hz
14.87
(a ) Use C3 to set the lower cutoff frequency to 2 kHz. C1 remains negligible at 2 kHz.
C1 = 8200 pF, C3 = 820 pF
(b) Use C to set the lower cutoff frequency to 1 kHz. C
1
2
and C3 remain negligible at 1 kHz.
C1 = 0.042 µF, C2 = 1800 pF, C3 = 0.015 µF
(c) SPICE Results: For (a) fL = 1.96 kHz. For (b) fL = 1.02 kHz
14.88
Av =
g m RL
2I D
≥ 0.95 → g m RL ≥ 19 | g m = 2Kn I D | VGS − VTN =
= 0.5V
1+ g m RL
Kn
(0.5) (0.03) = 3.75mA
=
2
ID
2
| RL ≥
19
2(0.03)(0.00375)
= 1.27kΩ
RL = RS 3kΩ → RS ≥ 2.19kΩ | VSS = VGS + (3.75mA)RS | Possible designs :
2.4kΩ, 11.5V ; 2.7kΩ, 12.6V ; 3.0kΩ, 13.75V - Making a choice which uses
a nearly minimum value of supply voltage gives : VSS = 12 V , RS = 2.4 kΩ.
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-39
14.89
For a common - emitter amplifier with RE = 0, Rin ≅ rπ =
β oVT
IC
| IC =
100(0.025V )
75Ω
= 33.3 mA
14.90
Using Eqn. 14.73 : 50 =
50 = g m RL
RE
4.3V
→ RE = 8.65kΩ | IC ≅
= 497µA
RE
1+ 40(4.3)
2(50)
Rin
g R
| Assuming RI = 50Ω, 50 = m L → RL =
= 5.03kΩ
RI + Rin
2
40(497µA)
RL = RC 100kΩ → RC = 5.30kΩ | VEC = 5 + 0.7 − IC RC = 3.07V - Active region is ok.
C1 >>
C2 >>
1
2π (500kHz)(50Ω + 50Ω)
= 3.18nF → C1 = 0.033 µF
1
= 3.02 pF → C2 = 33 pF
2π (500kHz)(105kΩ)
14.91
The base voltage should remain half way between the positive and negative power supply
voltages. If VEE = +10V and VCC = 0V, then VB should = 5 V which can be obtained using a
resistive voltage divider from the +10V supply. We now have the standard four-resistor bias
circuit. The base current is 327 µA/80 = 4.08 µA.
C1
C2
+
75 Ω
v
13 k Ω
8.2 k Ω
RE
RC
100 k Ω
S
120 k Ω
+ 10 V
R1
110 k Ω
CB
v
O
-
R2
Setting the current in R1 to 10I B = 40µA, R1 =
5V
= 125kΩ →120kΩ.
40µA
5V
= 114kΩ →110kΩ.
44µA
Note that the base terminal must now be bypassed with a capacitor.
The current in R2 = 11I B = 44µA, and R 2 =
14-40
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.92
Using Eqn. 14.73 : 75 =
50 = g m RL
RE
8.3V
→ RE = 25.0kΩ | IC ≅
= 332µA
RE
1+ 40(8.3)
2(50)
Rin
g R
| 50 = m L → RL =
= 7.52kΩ | RL = RC 100kΩ → RC = 8.13kΩ
RI + Rin
2
40(332µA)
VEC = 9 + 0.7 − IC RC = 6.98V - Active region is ok.
1
C1 >>
2π (500kHz)(75Ω + 75Ω)
C2 >>
= 2.12nF → C2 = 0.022 µF
1
= 2.95 pF → C2 = 30 pF
2π (500kHz)(108kΩ)
14.93
Rin ≅
(b) I
1
0.01
→ g m = 2Kn I D = 0.1S | I D =
|
gm
2Kn
D
=
(a) I
D
=
0.01
=1 A
2(0.005)
0.01
= 10.0 mA | The second FET achieves the desired input resistance
2(0.5)
at much lower current and hence much lower power for a given supply voltage.
14.94
1 VT kT
=
=
∝T
g m IC qIC
At − 40 o C = 233K,
⎛ 233k ⎞
⎛ 323K ⎞
1
1
o
= 50Ω⎜
= 50Ω⎜
⎟ = 38.8 Ω. | At + 50 C = 323K,
⎟ = 53.8 Ω.
gm
gm
⎝ 300K ⎠
⎝ 300K ⎠
Another approach : At 27 o C = 300K, IC =
At − 40 o C = 233K,
300k 6k
= .
50q
q
1 233
1 323
=
= 38.8 Ω. | At + 50 o C = 323K,
=
= 53.8 Ω.
gm
6
gm
6
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-41
14.95
This analysis assumes that the source and load resistors are fixed, and that only
⎛ Rin ⎞
the amplifier parameters are changing. Av = g m RL ⎜
⎟
⎝ RI + Rin ⎠
1
g R
1
≅
| Av ≅ m L
Since RE >> 75Ω, RE RI ≅ 75Ω and Rin ≅ RE
1+ g m RI
gm gm
To achieve Avmax , RL → RLmax , g m → g mmax which requires
(5.25 − 0.7)V = 364µA
13kΩ(0.95)
IC → ICmax = 0.988
g m = 40(364µA) = 14.6mS | Avmax =
| RLmax = 8.2kΩ(1.05)100kΩ = 7.93kΩ
14.6mS (7.93kΩ)
1+ 0.0146(75)
= 55.3
min
min
To achieve A min
which requires
v , R L → RL , g m → g m
IC → ICmin = 0.988
(4.75 − 0.7)V = 293µA
13kΩ(1.05)
g m = 40(293µA) = 11.7mS | Avmin =
| RLmin = 8.2kΩ(0.95)100kΩ = 7.23kΩ
11.7mS (7.23kΩ)
1+ 0.0117(75)
= 45.1 | 45.1 ≤ Av ≤ 55.3
The range is only slightly larger than that observed in the Monte Carlo analysis in Table 14.15.
14.96
*Problem 14.96 - Common-Base Amplifier - Monte Carlo Analysis
*Generate Voltage Sources with 5% Tolerances
IEE 0 8 DC 5
REE 8 0 RTOL 1
EEE 6 0 8 0 1
*
ICC 0 9 DC 5
RCC 9 0 RTOL 1
ECC 7 0 9 0 -1
*
VS 1 0 AC 1
RS 1 2 75
C1 2 3 47U
RE 3 6 RTOL 13K
Q1 4 0 3 PBJT
RC 4 7 RTOL 8.2K
C2 4 5 4.7U
R3 5 0 100K
.OP
.AC LIN 1 10KHZ 10KHZ
.PRINT AC VM(5) VP(5)
.MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%)
14-42
© R. C. Jaeger & T. N. Blalock - February 20, 2007
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 AC VM(5) YMAX
.END
Results: Mean value Av = 47.5; 3σ limits: 42.5 ≤ Av ≤ 52.5. However, the worst-case values
observed in the analysis are Avmin = 43.2 and Avmax = 51.9. The mean is 5% lower than the design
value. The width of the distribution is approximately the same as that in Table 14.15.
14.97
(a ) This analysis assumes that the source and load resistors are fixed, and that only
the amplifier parameters are changing. Av =
Rth = RE RI and
RE
=
RI + RE
Since R E >> RI , Rth and
g m RL ⎛ RE ⎞
⎜
⎟
1+ g m Rth ⎝ RI + RE ⎠
1
where RE = 13.3kΩ and RI = 75Ω
RI
1+
RE
RE
are essentially constant. To achieve Avmax , RL → RLmax , g m → g mmax
RI + RE
which requires IC → ICmax = 0.988
(5.10 − 0.7)V = 330µA | R = 8.25kΩ 1.01 100kΩ = 7.69kΩ
( )
13.3kΩ(0.99)
13.2mS (7.69kΩ)⎡ 13.3(0.99) ⎤
⎢
⎥ = 50.7
=
1+ 0.0132(75) ⎢⎣ 75 + 13.3(0.99)⎥⎦
max
L
g m = 40(330µA) = 13.2mS | Avmax
To achieve Avmin , RL → RLmin , g m → g mmin which requires
IC → ICmin = 0.988
(4.90 − 0.7)V = 309µA
13.3kΩ(1.01)
| RLmin = 8.25kΩ(0.99)100kΩ = 7.55kΩ
12.4mS (7.55kΩ)⎡ 13.3(1.01) ⎤
⎢
⎥ = 48.2
1+ 0.0124(75) ⎢⎣ 75 + 13.3(1.01)⎥⎦
(b) Using a Spreadsheet similar to Table 14.16: Mean value Av = 49.6; 3σ limits: 48.2 ≤ Av ≤
50.9. The worst-case values observed in the analysis are Avmin = 48.4 and Avmax = 50.8.
g m = 40(309µA) = 12.4mS | Avmax =
14.98
*Problem 14.98 - Common-Base Amplifier - Monte Carlo Analysis
*Generate Voltage Sources with 2% Tolerances
IEE 0 8 DC 5
REE 8 0 RTOL 1
EEE 6 0 8 0 1
*
ICC 0 9 DC 5
RCC 9 0 RTOL 1
ECC 7 0 9 0 -1
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-43
*
VS 1 0 AC 1
RS 1 2 75
C1 2 3 100U
RE 3 6 RR 13.3K
Q1 4 0 3 PBJT
RC 4 7 RR 8.25K
C2 4 5 1U
R3 5 0 100K
.OP
.AC LIN 1 10KHZ 10KHZ
.PRINT AC VM(5) VP(5) IM(VS) IP(VS)
.MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%)
.MODEL RTOL RES (R=1 DEV 2%)
.MODEL RR RES (R=1 DEV 1%)
.MC 1000 AC VM(5) YMAX
*.MC 1000 AC IM(VS) YMAX
.END
Results: Mean value: Av = 47.2; 3σ limits: 45.7 ≤ Av ≤ 48.5
Mean value: Rin = 83.4 Ω; 3σ limits: 79.5 Ω ≤ Av ≤ 87.6 Ω
14.99
Rin = RE
1
RE
RE
=
=
=
g m 1+ g m RE 1+ 40IC RE
I E RE = 2.5 − 0.7 = 1.8V | 75 =
RE
RE
=
80
1+ 39.5I E RE
1+ 40 I E RE
81
RE
→ RE = 5.41kΩ
1+ 39.5(1.8)
80 1.8V
= 329µA | Rth = 75Ω 5.41kΩ = 74.0Ω | g m = 40(329µA)= 13.2mS
81 5.41kΩ
g m RL ⎛ RE ⎞
Av =
⎟ → RL = 7.59kΩ
⎜
1+ g m Rth ⎝ RI + RE ⎠
IC =
7.59kΩ = RC 100kΩ → RC = 8.21kΩ | VC = −2.5V + IC RC = +0.201V | Oops! We are violating
our definition of the forward - active region. If we use the nearest 5% values, RE = 5.6kΩ and
RC = 8.2kΩ, IC = 318µA and VC = +0.108V. The transistor is just entering saturation.
14-44
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.100
Using a Spreadsheet similar to Table 14.15:
Mean value: Av = 0.960; 3σ limits: 0.942 ≤ Av ≤ 0.979
Mean value: ID = 4.91 mA; 3σ limits: 4.27 mA ≤ ID ≤ 5.55 mA
Mean value: VDS = 7.03 V; 3σ limits: 4.52 V ≤ VDS ≤ 9.54 V
*Problem 14.102 - Common-Drain Amplifier - Fig. 14.34
*Generate Voltage Sources with 5% Tolerances
IDD 0 7 DC 5
RDD 7 0 RTOL 1
EDD 5 0 7 0 1
*
ISS 0 8 DC 20
RSS 8 0 RTOL 1
ESS 6 0 8 0 -1
*
VGG 1 0 DC 0 AC 1
C1 1 2 4.7U
RG 2 0 RTOL 22MEG
RS 3 6 RTOL 3.6K
C2 3 4 68U
R3 4 0 3K
M1 5 2 3 3 NMOSFET
.OP
.AC LIN 1 10KHZ 10KHZ
.DC VGG 0 0 1
.MODEL NMOSFET NMOS (VTO=1.5 DEV 33.33%) (KP=20M DEV 50%) LAMBDA=0.02
.MODEL RTOL RES (R=1 DEV 5%)
.PRINT AC VM(4) VP(4) IM(VGG) IP(VGG)
.MC 1000 DC ID(M1) YMAX
*.MC 1000 DC VDS(M1) YMAX
*.MC 1000 AC IM(VGG) YMAX
*.MC 1000 AC VM(4) YMAX
.END
Results: Mean value: ID = 4.97 mA; 3σ limits: 4.32 mA ≤ ID ≤ 5.62 mA
Mean value: VDS = 7.19 V; 3σ limits: 6.18 V ≤ VDS ≤ 8.20 V
Mean value: Rin = 22.0 MΩ; 3σ limits: 20.3 Ω ≤ Rin ≤ 24.0 Ω
Mean value: Av = 0.956; 3σ limits: 0.936 ≤ Av ≤ 0.976
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-45
14.101
ac equivalent circuit
Rin3
Q
Rin2
R
RIN
Q
RI
M1
R
1 MΩ
vi
G
RS1
R B2
D1
620 Ω
17.2 k Ω
200 Ω
OUT
2
R C2
R
10 k Ω
3
R
B3
R
4.7 k Ω
RE2
E3
)(
3.3 k Ω
4.7 k Ω
)
(
)
RL1 = 620Ω 17.2kΩ rπ 2 + (β o2 + 1)1.6kΩ = 598Ω 2.39kΩ + (151)1.6kΩ = 597Ω
Avt1 = −
0.01(597)
g m1 RL1
=−
= −1.99 | RL2 = 3.54kΩ (Eq. 15.7)
1+ g m1 RS 1
1+ (0.01)200
Avt 2 = −
150(3.54kΩ)
β o2 RL2
=−
= −2.18
rπ 2 + (β o2 + 1)1.6kΩ
2.39kΩ + (151)1.6kΩ
⎛
⎞
1MΩ
Avt 3 = +0.950 | Av = −⎜
⎟(1.99)(−2.18)(0.950)= +4.08
⎝ 10kΩ + 1MΩ ⎠
⎛R +r ⎞
⎛
⎞
β o2 RE 2
Rout = (3300Ω) ⎜ th3 π 3 ⎟ | Rth3 = RI 3 Ro2 =RI 3 ro2 ⎜1+
⎟ ≅ RI 3 = 4.31kΩ
⎝ β o3 + 1 ⎠
⎝ Rth2 + rπ 2 + RE 2 ⎠
⎛ 4.31+ 1.00 ⎞
kΩ⎟ = 64.3Ω
Rout = (3.30kΩ) ⎜
81
⎝
⎠
14-46
L
51.8 k Ω
The Q - points and small - signal parameter values have already been found in the text.
⎛
⎞
Rin
The bypass capacitors do not affect Rin : Rin = RG = 1 MΩ | Av = ⎜
⎟ Avt1 Avt 2 Avt 3
⎝ 10kΩ + Rin ⎠
(
R
© R. C. Jaeger & T. N. Blalock - February 20, 2007
250 Ω
14.102
M1 : ID =
0.01
2
(VGS + 2) | VGS = −9000ID → VGS = −1.80V
2
0.01
2
(−1.8 + 2) = 200 µA | VDS = 15 − (15kΩ + 9kΩ)ID = 10.2 V
2
(50 + 10.2)V = 301kΩ
gm = 2(10mA /V 2 )(0.2mA) = 2mS | ro =
0.2mA
ID =
43kΩ
= 3.18V | REQ 2 = 160kΩ 43kΩ = 33.9kΩ
43kΩ + 160kΩ
⎞
⎛
3.18 - 0.7
151
=1.35 mA | VCE 2 = 15 − ⎜ 4.7kΩ +
1.6kΩ⎟IC = 6.49 V
IC 2 = 150
⎠
⎝
33.9kΩ +151(1.6kΩ)
150
Q2 : VEQ 2 = 15
gm 2 = 40(1.35mA) = 54.0mS | rπ 2 =
150
(80 + 6.49)V = 64.1kΩ
= 2.78kΩ | ro2 =
1.35mA
54.0mS
120kΩ
= 8.53V | REQ 3 = 120kΩ 91kΩ = 51.8kΩ
120kΩ + 91kΩ
⎞
⎛ 81
8.53 - 0.7
= 2.72 mA | VCE 3 = 15 − ⎜ 2.2kΩ⎟IC = 8.93 V
IC 3 = 80
⎠
⎝ 80
51.8kΩ + 81(2.2kΩ)
Q3 : VEQ 3 = 15
gm 3 = 40(2.72mA) = 109mS | ro3 =
(60 + 8.93)V
2.72mA
⎛
⎞
RG
Av = ⎜
⎟ Avt1 Avt 2 Avt 3
⎝10kΩ + RG ⎠
(
= 25.3kΩ | rπ 3 =
80
= 734Ω
109mS
)
Avt1 = −(2mS ) 301kΩ 15kΩ 33.9kΩ 2.78kΩ = −4.36
⎡
⎤
Avt 2 = (−54.0mS )⎢64.1kΩ 4.7kΩ 51.8kΩ 734 + 81(2.2kΩ 250Ω) ⎥ = −180
⎣
⎦
(
Avt 3 =
)
81(2.2kΩ 250Ω)
⎛
⎞
1MΩ
= 0.961 | Av = −4.36(−180)(0.961)⎜
⎟ = 747
⎝10kΩ + 1MΩ ⎠
734 + 81(2.2kΩ 250Ω)
v be 3 = v b 3 (1− Avt 3 ) = 0.99Avt1 Avt 2v s (1− Avt 3 ) ≤ 5mV | v s ≤
0.005
= 165µV
0.99(4.36)(180)(1− 0.961)
The gain is actually reduced rather than improved. The signal range increased since the gain was
reduced.
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-47
14.103
*Problem 14.102/14.103 - Multistage Amplifier – Figure P14.102
VCC 12 0 DC 15
VI 1 0 AC 1
*For output resistance
*VI 1 0 AC 0
*VO 11 0 AC 1
RS 1 2 10K
C1 2 3 22U
RG 3 0 1MEG
M1 5 3 4 4 NMOSFET
RS1 4 0 9K
C2 4 0 22U
RD 12 5 15K
C3 5 6 22U
R1 12 6 160K
R2 6 0 43K
Q2 8 6 7 NBJT1
RC 12 8 4.7K
RE2 7 0 1.6K
C4 7 0 22U
C5 8 9 22U
R3 12 9 91K
R4 9 0 120K
Q3 12 9 10 NBJT2
RE3 10 0 2.2K
C6 10 11 22U
RL 11 0 250
.MODEL NMOSFET NMOS VTO=-2 KP=.01 LAMBDA=0.02
.MODEL NBJT1 NPN IS=1E-16 BF=150 VA=80
.MODEL NBJT2 NPN IS=1E-16 BF=80 VA=60
.OPTIONS TNOM=17.2
.OP
.AC LIN 1 2KHZ 2KHZ
.PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(11) VP(11) IM(C6) IP(C6)
.END
VM(3)
1
Results : Av = VM(11) = +879 | Rin =
= 1.00 MΩ | Rout =
= 51.8 Ω
IM(VI)
IM(C6)
14-48
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.104
⎛
⎞
Rin
The bypass capacitors do not affect Rin : Rin = RG = 1 MΩ | Av = ⎜
⎟ Avt1 Avt 2 Avt 3
⎝ 10kΩ + Rin ⎠
RL1 = (15kΩ 160kΩ 43kΩ) (rπ 2 + (β o2 + 1)1.6kΩ) = 10.4kΩ [2.78kΩ + 151(1.6kΩ)] = 9.98kΩ
Avt1 = −
2(9.98kΩ)
gm1RL1
=−
= −1.05
1+ gm1 RS1
1+ 2(9kΩ)
[
]
RL 2 = (4.7kΩ 91kΩ 120kΩ) rπ 3 + (β o3 + 1)(2.2kΩ 250Ω) = 4.31kΩ [734Ω + 81(225Ω)] = 3.51kΩ
Avt 2 = −
Avt 3 =
150(3.51kΩ)
β o2 RL 2
=−
= −2.17
rπ 2 + (β o2 + 1)1.6kΩ
2.39kΩ + (151)1.6kΩ
81(2.2kΩ 250Ω)
⎞
⎛
1MΩ
= 0.961 | Av = −⎜
⎟(−1.05)(−2.17)(0.961) = +2.17
⎝10kΩ + 1MΩ ⎠
734 + 81(2.2kΩ 250Ω)
⎛R + r ⎞
⎛
⎞
β o2 RE 2
Rout = (3300Ω) ⎜ th 3 π 3 ⎟ | Rth 3 = RI 3 Ro2 =RI 3 ro2 ⎜1+
⎟ ≅ RI 3 = 4.31kΩ
⎝ β o3 + 1 ⎠
⎝ Rth 2 + rπ 2 + RE 2 ⎠
⎛ 4.31+ 1.00 ⎞
Rout = (3.30kΩ) ⎜
kΩ⎟ = 64.3Ω
⎠
⎝
81
14.105
*Problem 14.105 - Use the listing from Problem 14.103, but remove C2 and C4.
Result: Av = VM(11) = +2.20
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-49
14.106
100kΩ
= 1.63V | REQ1 = 100kΩ 820kΩ = 89.1kΩ
100kΩ + 820kΩ
⎞
⎛
1.63 - 0.7
101
= 319 µA | VCE1 = 15 − ⎜18kΩ +
2kΩ⎟IC = 8.61 V
IC1 = 100
⎠
⎝
89.1kΩ +101(2kΩ)
100
Q1 : VEQ1 = 15
gm1 = 40(319 µA) = 12.8mS | rπ 1 =
100
(70 + 8.61)V = 246kΩ
= 7.81kΩ | ro1 =
319 µA
12.8mS
43kΩ
= 3.18V | REQ 2 = 160kΩ 43kΩ = 33.9kΩ
43kΩ + 160kΩ
⎞
⎛
3.18 - 0.7
101
=1.27 mA | VCE 2 = 15 − ⎜ 4.7kΩ +
1.6kΩ⎟IC = 6.98 V
IC 2 = 100
⎠
⎝
33.9kΩ +101(1.6kΩ)
100
Q2 : VEQ 2 = 15
gm 2 = 40(1.27mA) = 50.8mS | rπ 2 =
100
(70 + 6.98)V = 60.6kΩ
= 1.97kΩ | ro2 =
1.27mA
50.8mS
1.2MΩ
= 8.53V | REQ 3 = 1.2MΩ 910kΩ = 518kΩ
1.2MΩ + 910kΩ
2ID 3
+ 3000ID 3 → ID 3 = 1.87mA | VGS 3 − VTN 3 = 1.93V
8.53 = VGS 3 + 3000ID 3 =1 +
0.001
M 3 : VEQ 3 = 15
VDS 3 = 15 − 3000ID 3 = 9.39V | gm 3 = 2(0.001)(0.00187) = 1.93mS
⎛ Rin ⎞
Av = ⎜
⎟ Avt1 Avt 2 Avt 3 | Rin = 820kΩ 100kΩ rπ 1 = 820kΩ 100kΩ 7.81kΩ = 7.18kΩ
⎝ RI + Rin ⎠
Avt1 = −gm1 (RC1 RB 2 rπ 2 ) = −12.8mS (18kΩ 33.9kΩ 1.97kΩ) = −21.6
Avt 2 = −gm 2 (RC 2 RG 3 ) = −50.8mS (4.7kΩ 518kΩ) = −237
Avt 3 =
gm 3 (RE 3 RL )
=
1.93mS (3.0kΩ 250Ω)
1+ gm 3 (RE 3 RL ) 1+ 1.93mS (3.0kΩ 250Ω)
= 0.308
⎞
⎛
7.18kΩ
Av = ⎜
⎟(−21.6)(−237)(0.308) = 659
⎝10kΩ + 7.18kΩ ⎠
⎛ Rin ⎞
v gs3 = v g 3 (1− Avt 3 ) = ⎜
⎟ Avt1 Avt 2v i (1− Av 3 ) ≤ 0.2(VGS 3 − VTN 3 )
⎝ RI + Rin ⎠
vi ≤
0.2(1.93)
= 261µV
0.418(21.6)(237)(1− 0.308)
The gain is reduced rather than improved. The signal range increased since the gain was
reduced.
14-50
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.107
*Problem 14.107 - Multistage Amplifier – Figure P14.106
VCC 12 0 DC 15
VI 1 0 AC 1
RI 1 2 10K
*For output resistance
*VI 1 0 AC 0
*VO 11 0 AC 1
C1 2 3 22U
R1 12 3 820K
R2 3 0 100K
Q1 5 3 4 NBJT
RE1 4 0 2K
C2 4 0 22U
RC1 12 5 18K
C3 5 6 22U
R3 12 6 160K
R4 6 0 43K
Q2 8 6 7 NBJT
RC 12 8 4.7K
RE2 7 0 1.6K
C4 7 0 22U
C5 8 9 22U
R5 12 9 910K
R6 9 0 1.2MEG
M3 12 9 10 10 NMOSFET
RE3 10 0 3K
C6 10 11 22U
RL 11 0 250
.OP
.AC LIN 1 3KHZ 3KHZ
.MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(11) VP(11) IM(C6) IP(C6)
.END
VM(3)
1
Results : Av = VM(11) = +711 | Rin =
= 8.29 kΩ | Rout =
= 401 Ω
IM(VI)
IM(C6)
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-51
14.108
Rin = R1 R2 [rπ 1 + (β o1 + 1)RE1 ] = 100kΩ 820kΩ [7.81kΩ + 101(2kΩ)] = 62.66kΩ
Avt1 = −
β o1 (RC1 RB 2 RI 2 )
| RI 2 = rπ 2 + (β o2 + 1)RE 2 = 1.97kΩ + 101(1.6kΩ) = 164kΩ
rπ 1 + (β o1 + 1)RE1
R3 R4 = 43kΩ 160kΩ = 33.9kΩ | Avt1 = −
100(18kΩ 33.9kΩ 164kΩ)
7.81kΩ + 101(2kΩ)
RG 3 = R5 R6 = 1.2MΩ 910kΩ = 518kΩ | Avt 2 = −
Avt 3 =
gm 3 (RE 3 RL )
=
1.93mS (3.0kΩ 250Ω)
1+ gm 3 (RE 3 RL ) 1+ 1.93mS (3.0kΩ 250Ω)
= −6.69
100(4.7kΩ 518kΩ)
β o2 (RC 2 RG 3 )
=−
= −2.85
rπ 2 + (β o2 + 1)RE1
1.97kΩ + 101(1.6kΩ)
= 0.308
⎛
⎞
⎛
⎞
Rin
62.6kΩ
Av = ⎜
⎟(−6.69)(−2.85)(0.308) = 5.05
⎟ Avt1 Avt 2 Avt 3 = ⎜
⎝ 10kΩ + 62.6kΩ ⎠
⎝10kΩ + Rin ⎠
14-52
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.109
Note that the equivalent circuits are the same for Q1 and Q2 .
180kΩ
15V = 5.63V | REQ = 180kΩ 300kΩ = 113 kΩ
180kΩ + 300kΩ
5.63 − 0.7 V
IB =
= 2.31µA | IC = 100I B1 = 232µA | I E = 101I B1 = 234µA
113 + 101(20) kΩ
VEQ =
VCE = 15 − 2x10 4 I E − 2x10 4 IC = 5.71V
rπ =
100(0.025V )
232µA
= 10.8kΩ | ro =
RI
v1
232µA
v2
Q 20 k Ω
20 k Ω
Q
1
2kΩ
vi
(70 + 5.71)V = 326kΩ
2
300 k Ω
100 k Ω
16.7 k Ω
180 k Ω
R
2kΩ
113 k Ω
180 k Ω
R
I1
L
300 k Ω
=17.0 k Ω
⎛ Rin
⎞
Av = ⎜
⎟ Avt1 Avt 2
⎝ 2kΩ + Rin ⎠
Avt1 =
100(17kΩ 10.8kΩ)
β o1 (RI1 rπ 2 )
v2
=−
=−
= −3.10
v1
rπ 1 + (β o1 + 1)R5
10.8kΩ + (101)2kΩ
Avt 2 =
vo
= −gm 2 RL | RL = 100kΩ 20kΩ = 16.7kΩ | Avt 2 = −40(232µA)(16.7kΩ) = −155
v2
Rin = RB1 (rπ 1 + (β o1 + 1)R5 ) = 300kΩ 180kΩ [10.8kΩ + (101)2kΩ] = 73.6kΩ
⎛ 73.6kΩ ⎞
Av = ⎜
⎟(−3.10)(−155) = +468 | Rout = 20kΩ ro2 = 20kΩ 326kΩ = 18.8kΩ
⎝ 2kΩ + 73.6kΩ ⎠
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-53
14.110
The ac equivalent circuit from Problem 14.109 becomes:
v1
Q
2k Ω
v2
20 k Ω
Q
1
2
100 k Ω
300 k Ω
vi
vo
20 k Ω
16.7 k Ω
180 k Ω
RL
20 k Ω
(
113 k Ω
)
180 k Ω
300 k Ω
20 k Ω
RI1 =17.0 k Ω
[
]
Avt1 =
100 17kΩ (10.8kΩ + (101)20kΩ)
β o RI1 [rπ 2 + (β o2 + 1)R6 ]
v2
=−
=−
= −0.830
v b1
rπ 1 + (β o1 + 1)R5
10.8kΩ + (101)20kΩ
Avt 2 =
100(16.7kΩ)
β o2 RL
vo
=−
=−
= −0.822
v2
rπ 2 + (β o2 + 1)R6
10.8kΩ + (101)20kΩ
Rin
Avt1 Avt 2 | Rin = 113kΩ (rπ 1 + (β o1 + 1)R5 ) = 113kΩ (10.8kΩ + (101)20kΩ) = 107kΩ
RI + Rin
107kΩ
Av =
(−0.830)(−0.822) = +0.670
2kΩ + 107kΩ
The voltage gain is completely lost. | Rout ≅ 20kΩ
Av =
14-54
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.111
*Problem 14.111 - Multistage Amplifier – Figure P14.109
VCC 11 0 DC 15
VI 1 0 AC 1
RI 1 2 2K
*For output resistance
*VI 1 0 AC 0
*VO 10 0 AC 1
C1 2 3 10U
R1 3 0 180K
R2 11 3 300K
Q1 6 3 4 NBJT
RE1 4 5 2K
RE2 5 0 18K
C2 5 0 10U
RC1 11 6 20K
C3 6 7 10U
R3 7 0 180K
R4 11 7 300K
Q2 9 7 8 NBJT
RC2 11 9 20K
RE3 8 0 20K
C4 8 0 10U
C5 9 10 10U
RL 10 0 100K
.OP
.AC LIN 1 5KHZ 5KHZ
.MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(10) VP(10) IM(C5) IP(C5)
.END
VM(3)
1
Results : Av = VM(10) = +454 | Rin =
= 74.7 kΩ | Rout =
= 18.8 kΩ
IM(VI)
IM(C5)
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-55
14.112
⎛ Rin
⎞
Av = ⎜
⎟ Avt1 Avt 2
⎝ 2kΩ + Rin ⎠
v
Avt1 = 2 = −gm1 (RI1 rπ 2 ) = −40(232µA)(17kΩ 10.8kΩ) = −61.3
v1
v
Avt 2 = o = −gm 2 RL | RL = 100kΩ 20kΩ = 16.7kΩ | Avt 2 = −40(232µA)(16.7kΩ) = −155
v2
Rin = RB1 rπ 1 = 300kΩ 180kΩ 10.8kΩ = 10.0kΩ
⎛ 10.0kΩ ⎞
Av = ⎜
⎟(−61.3)(−155) = +7920 | Rout = 20kΩ ro2 = 20kΩ 326kΩ = 18.8kΩ
⎝ 2kΩ + 10.0kΩ ⎠
14.113
0.05
2
(VGS + 2) and VGS = −1800ID
M1: Assume saturation: ID =
2
0.05
2
VGS = −1800
(VGS + 2) or 45VGS2 + 181VGS + 180 = 0
2
VGS = −2.22V,−1.80V | VGS = −1.80 V and ID =1 mA
VDS = 20 −15000(0.001) −1800(0.001) = 3.2 V > VGS − VTN
0.05
2
(VGS + 2) and VGS = −2500ID
M2: Assume saturation: ID =
2
0.05
2
VGS = −2500
(VGS + 2) or 62.5VGS2 + 251VGS + 250 = 0
2
VGS = −1.83 V and ID = 0.723 mA
VDS = 20 − 2500(0.723mA) =18.2 V > VGS − VTN
gm1 = 2(0.05)(0.001) = 10.0mS | gm 2 = 2(0.05)(7.23x10−4 ) = 8.50mS
Rin = 1800
1
1
= 1800 100 = 94.7Ω | Rout = 2500
= 2500 118 = 113Ω
gm1
gm 2
Avt1 = +gm1 (15kΩ 1MΩ) = 0.01S (14.8kΩ) = 148
Avt 2 = +
gm 2 (2.5kΩ 10kΩ)
1+ gm 2 (2.5kΩ 10kΩ)
=+
8.5x10−3 (2.5kΩ 10kΩ)
1+ 8.5x10−3 (2.5kΩ 10kΩ)
= 0.944
Av = Avt1 Avt 2 = +140
14-56
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14.114
100kΩ
= 0V | REQ = 100kΩ 100kΩ = 50kΩ
100kΩ + 100kΩ
0 − 0.7 − (−15)
= 22.3µA | IC = 2.78 mA | I E = 2.81 mA
IB =
50kΩ + 126(4.7kΩ)
VEQ = −15 + 30
VCE = 30 − 2000IC − 4700I E = 11.4 V | rπ =
ro =
(50 + 11.4)V = 22.1kΩ
125(0.025V )
2.78mA
= 1.12kΩ
| RB = R1 R2 = 100kΩ 100kΩ = 50kΩ
2.78mA
Rin = RB rπ + (β o + 1)RL = 50kΩ 1.12kΩ + (126)3.34kΩ = 44.7 kΩ
[
(R
B
Rout = RE ro
)
]
RI + rπ
βo + 1
[
]
(44.7kΩ 600Ω)+ 1.12kΩ = 13.5 Ω
= 4.7kΩ 22.1kΩ
126
R1S = RI + Rin = 45.3kΩ R2S = R3 + Rout = 24.0kΩ R3S ≅ RC = 2kΩ
⎤
1 ⎡
1
1
⎢ −5
⎥ = 0.492 Hz SPICE result : f L = 0.39Hz
fL ≅
+
2π ⎢⎣10 (45.3kΩ) 47x10−6 (24.0kΩ)⎥⎦
Note that C3 is not in the signal path and doesn't contribute to f L .
14.115 Use C3 = 2.2 µF
(10 − 0.7)V = 1.43 µA | I = 114 µA | V = 20 − 39000I − 68000I = 7.71 V
IB =
C
CE
C
E
1MΩ + (80 + 1)68kΩ
Active region is correct. | rπ =
81(0.025V )
114µA
= 17.8 kΩ | ro =
75 + 7.71
= 726 kΩ
114µA
RL = 500kΩ 39kΩ 726kΩ = 34.5kΩ | Rin = RB rπ = 1MΩ 17.8kΩ = 17.5 kΩ
⎛ Rin ⎞
⎛
⎞
17.5kΩ
Av = −g m RL ⎜
⎟ = 40(0.114mA)(34.5kΩ)⎜
⎟ = −153
⎝ 500Ω + 17.5kΩ ⎠
⎝ RI + Rin ⎠
Rout = RC ro = 39kΩ 726kΩ = 37.0 kΩ R1S = RI + Rin = 18.0kΩ R3S = Rout + R3 = 537kΩ
R2S = RE
(R
B
)
RI + rπ
βo + 1
= 68kΩ
(1MΩ 500Ω)+ 17.8kΩ = 225Ω
81
⎤
1
1 ⎡
1
1
⎥ = 19.2 Hz
⎢
+
+
fL ≅
2π ⎢⎣2.2µF (18.0kΩ) 47µF (225Ω) 2.2µF (537kΩ)⎥⎦
© R. C. Jaeger & T. N. Blalock - February 20, 2007
SPICE result : f L = 18 Hz
14-57
14.116
62kΩ
= 9.07V | REQ = 62kΩ 20kΩ = 15.1kΩ
62kΩ + 20kΩ
12 − 0.7 − 9.07
IB =
= 4.19µA | IC = 314 µA | I E = 319 µA
15.1kΩ + 76(6.8kΩ)
(a) VEQ = 12
VEC = 12 −16000IC − 6800I E = 4.81 V
75(0.025)
60 + 4.81
=
206kΩ
|
r
=
= 5.97kΩ
π
314x10−6
314x10−6
Rin = 15.1kΩ 5.97kΩ = 4.28 kΩ | Rout = ro 16kΩ = 14.8 kΩ | g m = 40IC = 12.6 mS
ro =
R1S = RI + Rin = 5.28kΩ R3S = Rout + R3 = 115kΩ
R2S = RE
fL ≅
(R
B
)
RI + rπ
βo + 1
= 6.8kΩ
(15.1kΩ 1kΩ)+ 5.97kΩ = 90.0Ω
76
⎤
1
1 ⎡
1
1
⎢
⎥ = 51.5 Hz
+
+
2π ⎢⎣2.2µF (5.28kΩ) 47µF (90.0Ω) 10µF (115kΩ)⎥⎦
SPICE result : f L = 43.8 Hz
14.117
500kΩ
= 4.73V | REQ = R1 R2 = 500kΩ 1.4 MΩ = 368kΩ
500kΩ + 1.4 MΩ
Assume Active Region Operation
VEQ = 18
2I D
+ 27000I D → I D = 113 µA
5x10−4
= 18 − (27000 + 75000)I D = 6.47 V > 3.73 V - Active region is correct.
4.73 = VGS = 27000I D → 4.73 = 1+
VDS
50 + 6.47
= 500kΩ | g m = 2 500x10−6 113x10−6 1+ 0.02(6.47) = 357µS
−6
113x10
RL = ro RD 470kΩ = 500kΩ 75kΩ 470kΩ = 57.3kΩ
ro =
(
)(
)[
]
Rin = R1 R2 = 500kΩ 1.4 MΩ = 368kΩ | Rout = ro 75kΩ = 65.2 kΩ
R1S = RI + Rin = 369kΩ R3S = Rout + R3 = 535kΩ R2S = RS
1
1
= 27kΩ
= 2.54kΩ
gm
0.357mS
⎤
1 ⎡
1
1
1
⎢
⎥ = 1.56 Hz
fL ≅
+
+
2π ⎢⎣2.2µF (369kΩ) 47µF (2.54kΩ) 10µF (535kΩ)⎥⎦
14-58
© R. C. Jaeger & T. N. Blalock - February 20, 2007
SPICE result : f L = 1.22 Hz
14.118
2.5x10−4
2
2
10 + VGS
VGS + 1) |
= 1.25x10−4 (VGS + 1) → VGS = −2.358V
ID =
(
2
33kΩ
10 + VGS
ID =
= 232µA | VDS = −10 + 0.232mA(24kΩ)− 2.36 = −6.79V | Pinchoff region is correct.
33kΩ
50 + 6.79
= 245kΩ
g m = 2 2.5x10−4 2.32x10−4 1+ 0.02(6.79) = 0.363mS ro =
2.32x10−4
1
Rin = 33kΩ
= 2.69 kΩ | R1S = RI + Rin = 3.19kΩ | R2S ≅ RD + R3 = 124kΩ
gm
⎤
1
1 ⎡
1
⎥ = 5.94 Hz SPICE result : f L = 5.00 Hz
⎢
fL ≅
+
2π ⎢⎣10µF (2.69kΩ) 47µF (124kΩ)⎥⎦
(
)
(
)(
)[
]
14.119
(12 − 0.7)V = 2.69 µA | I = 135 µA | V
100kΩ + (51)82kΩ
50(0.025V )
Active region is correct. r =
= 9.26 kΩ
IB =
C
π
[
(
R1S = RB rπ + (β o + 1) RE RI
CE
135µA
= 24 − 39000IC − 82000I E = 7.58 V
| ro =
50 + 7.58
= 427 kΩ
135µA
)]= 100kΩ [9.26kΩ + (51)(82kΩ 500Ω)]= 25.7kΩ
rπ
9.26kΩ
= 500 + 82kΩ
= 681Ω | R3S = RC + R3 = 139kΩ
51
βo + 1
⎤
1 ⎡
1
1
1
⎢
⎥ = 6.40 Hz SPICE result : f L = 5.72 Hz
fL ≅
+
+
2π ⎢⎣4.7µF (25.7kΩ) 47µF (681Ω) 10µF (139kΩ)⎥⎦
R2S = RI + RE
14.120
2
5x10−4
VGS + 2) → I D = 32.2 µA
(
2
= 15 − (20000 + 51000)I D = 12.7 V > 0.36 V - Active Region is correct.
Assume Active Region operation. VGS = −51000I D I D =
VDS
(
)(
)[
]
g m = 2 5x10−4 32.2x10−6 1+ 0.02(12.70) = 0.201mS | Rin = 51kΩ
1
= 4.53kΩ
gm
50 + 12.7
= 1.95MΩ R1S = RI + Rin = 5.53kΩ
R2 S ≅ RD + R3 = 30kΩ
32.2x10−6
⎤
1
1 ⎡
1
⎢
⎥ = 13.2 Hz SPICE result : f L = 12.8 Hz
fL ≅
+
2π ⎢⎣2.2µF (5.53kΩ) 47µF (30kΩ)⎥⎦
ro =
© R. C. Jaeger & T. N. Blalock - February 20, 2007
14-59
14.121 The power supply should be +16 V.
Assume Active Region operation. Since there is no negative feedback (RS = 0),
we should include the effect of channel - length modulation. VGS = 0
2
4x10−4
−5) (1+ 0.02VDS ) and VDS = 16 −1800I D → I D = 5.59 mA
(
2
= 16 −1800I D = 5.93 V > 5 V - Active region is correct.
ID =
VDS
50 + 5.93
= 10.0kΩ Rin = 10.0 MΩ | Rout = RD ro = 1.52 kΩ
5.59x10−3
R1S = RI + Rin = 10.0 MΩ R2 S = Rout + R3 = 37.5kΩ
⎤
1 ⎡
1
1
⎢
⎥ = 0.497 Hz SPICE result : f L = 0.427 Hz
+
fL ≅
2π ⎢⎣2.2µF (10.0 MΩ) 10µF (37.5kΩ)⎥⎦
ro =
14.122 Use C1 = C2 = C3 = 1 µF
0.05
2
(VGS + 2) and VGS = −1800ID
M1: Assume saturation: ID =
2
2
0.05
VGS + 2) or 45VGS2 + 181VGS + 180 = 0
VGS = −1800
(
2
VGS = −2.22V ,−1.80V | VGS = −1.80 V and I D = 1 mA
VDS = 20 −15000(0.001)−1800(0.001) = 3.2 V > VGS − VTN
0.05
2
(VGS + 2) and VGS = −2500ID
M2: Assume saturation: ID =
2
2
0.05
VGS + 2) or 62.5VGS2 + 251VGS + 250 = 0
VGS = −2500
(
2
VGS = −1.83 V and I D = 0.723 mA
VDS = 20 − 2500(0.723mA) = 18.2 V > VGS − VTN
(
)
g m1 = 2(0.05)(0.001) = 10.0mS | g m2 = 2(0.05) 7.23x10−4 = 8.50mS
R1S = Rin = 1800
1
= 1800 100 = 94.7Ω | R2 S = 15kΩ + 1MΩ = 1.02 MΩ
g m1
1
= 2500 118 = 113Ω | R3 S = 10kΩ + Rout = 10.1kΩ
g m2
⎤
1 ⎡
1
1
1
⎥ = 1.42 kHz SPICE result : f L = 1.68 kHz
⎢
fL ≅
+
+
2π ⎢⎣1µF (113Ω) 1µF (1.02 MΩ) 1µF (10.1kΩ)⎥⎦
Rout = 2500
14-60
© R. C. Jaeger & T. N. Blalock - February 20, 2007
CHAPTER 15
15.1
(a) I
C
= αF IE =
1 β F 12 − VBE 1 ⎛100 ⎞⎛ 12 − 0.7 ⎞
= ⎜
= 20.7 µA | VC = 12 − 3.3x105 IC = 5.17V
⎟⎜
5⎟
2 β F + 1 REE
2 ⎝ 101 ⎠⎝ 2.7x10 ⎠
VCE = VC − (−0.7V ) = 5.87V | Q − Point = (20.7µA, 5.87V )
(b) A
= −g m RC = −40(20.7µA)(330kΩ)= −273
dd
Rid = 2rπ = 2
(c) A
cc
=−
Add = −
IC
=2
100(0.025V )
E
= 243 kΩ | Rod = 2RC = 660 kΩ
g m RC
−137
= −137 | Acd = Acc | CMRR =
= 227 or 47.1 dB (very low)
2
−0.604
2
15.2
20.7µA
100(330kΩ)
β o RC
=−
= −0.604
rπ + (β o + 1)2REE
122kΩ + 2(101)270kΩ
rπ + (β o + 1)2REE
Ric =
(a) I
β oVT
=
122kΩ + 2(101)270kΩ
2
= 27.3 MΩ
1 ⎛ 1.5 − 0.7 ⎞ V
60
= ⎜
= 5.33µA | IC = α F I E =
I E = 5.25µA
3 ⎟
2 ⎝ 75x10 ⎠ Ω
61
VCE = 1.5 −105 IC − (−0.7)= 1.68V | Q - Pt : (5.25µA, 1.68V )
(b) g
m
Acc = −
= 40IC = 0.210mS | rπ =
60
= 286kΩ | Add = −g m RC = −0.210mS (100kΩ)= −21.0
gm
60(100kΩ)
β o RC
=−
= −0.636
rπ + (β o + 1)2REE
286kΩ + 61(150kΩ)
For differential output : CMRR =
−21.0
=∞
0
−21.0
2
For single - ended output : CMRR =
= 16.5, a paltry 24.4 dB!
−0.636
Rid = 2rπ = 572kΩ | Ric =
rπ + (β o + 1)2REE
Rod = 2RC = 200 kΩ | Roc =
2
=
286 + 61(150)
2
kΩ = 4.72 MΩ
RC
= 50kΩ
2
15.3
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-1
*Problem 15.3
VCC 2 0 DC 12
VEE 1 0 DC -12
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RC1 2 3 330K
RC2 2 7 330K
Q1 3 4 5 NBJT
Q2 7 6 5 NBJT
REE 5 1 270K
.MODEL NBJT NPN BF=100 VA=60 IS=1FA
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
1
Results : Add = VM(3,7) = −241 | Rid =
= 269 kΩ | Acc = -0.602 | Ric = 23.2 MΩ
IM(VID1)
Problem15.45(b)-Transient-7
+0.000e+000
+1.000m
+2.000m
Time (s)
+3.000m
+4.000m
+6.000
+4.000
+2.000
+0.000e+000
-2.000
-4.000
-6.000
V(IVOUT)
Simulation results from B2SPICE.
Problem15.3(b)-Fourier-Table
FREQ
mag
phase
norm_mag norm_phase
+0.000
+49.786n +0.000
+0.00
+0.000
+1.000k
+5.766
+180.000 +1.000
+0.000
+2.000k
+99.572n +93.600
+17.268n -86.400
+3.000k
+80.305m -180.000 +13.927m -360.000
+4.000k
+99.572n +97.200
+17.268n -82.800
+5.000k
+1.161m +179.993 +201.326u -7.528m
+6.000k
+99.572n +100.800 +17.268n -79.200
+7.000k
+13.351u -179.005 +2.315u
-359.005
+8.000k
+99.572n +104.400 +17.268n -75.600
Using the Fourier analysis capability of SPICE, THD = 1.39%
15-2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.4
(a) I
E
=
18V − 0.7V
(
)
2 4.7 x10 Ω
4
= 184 µA | IC = α F I E =
100
I E = 182 µA
101
VCE = 18 − 105 IC − (−0.7)= 0.92 V | Q - point : (182 µA, 0.92 V )
Note that RC is quite large and the common - mode input range is poor.
More realistic choices might be 47 kΩ or 51 kΩ
100
(b) g m = 40 IC = 7.28 mS | rπ = g = 13.7 kΩ | Add = −g m RC = −7.28mS (100kΩ)= −728
m
Acc = −
100(100 kΩ)
β o RC
=−
= −1.05
rπ + (β o + 1)2 REE
13.7 kΩ + 101(94 kΩ)
For differential output : CMRR =
−33.7
=∞
0
−728
For single - ended output : CMRR = 2 = 346, a paltry 50.8 dB!
−1.05
Rid = 2rπ = 27.4 kΩ | Ric =
rπ + (β o + 1)2 REE
Rod = 2 RC = 200 kΩ | Roc =
2
=
13.7 + 101(94)
2
kΩ = 4.75 MΩ
RC
= 50 kΩ
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-3
15.5
(a) I
C
1 ⎛ β ⎞⎛ 12 − VBE ⎞ 1 ⎛ 100 ⎞⎛ 12 − 0.7 ⎞
= α F I E = ⎜ F ⎟⎜
⎟= ⎜
⎟⎜
⎟ = 20.7 µA
2 ⎝ β F + 1⎠⎝ REE ⎠ 2 ⎝ 101 ⎠⎝ 2.7 x105 ⎠
VC1 = VC 2 = 12 − 2.4 x105 IC = 7.03V | VCE = VC − (−0.7V )= 7.73V
Q − Point = (20.7µA, 7.73V ) | rπ =
Acc = −
100(0.025V )
20.7 µA
= 121kΩ
100(240 kΩ)
β o RC
5.000 + 5.000
=
= −0.439 | vic =
= 5.00V
2
rπ + (β o + 1)2 REE 121kΩ + (101)540kΩ
vic = 5 V , vC1 = vC 2 = 7.03 + Accvic = 7.03 − 0.439(5) = 4.84 V
Note that the BJT's are just beyond the edge of saturation!
1 β F ⎛ 5V − VBE − (−12V )⎞ 1 ⎛ 100 ⎞⎛17V − 0.7V ⎞
⎟= ⎜
b
I
=
α
I
=
( ) C F E 2 β + 1⎜⎜
⎟ 2 ⎝ 101⎟⎠⎜⎝ 2.7 x105 ⎟⎠ = 29.9 µA
R
F
EE
⎝
⎠
VC1 = VC 2 = 12 − 2.4 x105 IC = 4.82 V | Part (a) has a small error of 0.02 mV
(c) The common - mode signal voltage applied to the base - emitter junction is
vbe = vic
rπ
121kΩ
=5
= 11.1 mV > 5mV .
rπ + (β o + 1)2 REE
121kΩ + (101)540 kΩ
A common - mode input voltage of 5 volts exceeds the small - signal limit.
15.6
We should first check the feasibility of the design using the Rule- of - Thumb estimates similar to
those developed in Chapter 13 (Eq. (13.55)). The required Add = 794 (58 db).
(This sounds fairly large - a significant fraction of the BJT amplification factor µf .)
Even assuming we choose to drop all of the positive power supply voltage across RC
(which provides no common - mode input range) : Add = g m RC = 40IC RL ≤ 40VCC = 40(9)= 360.
Thus, a gain of 794 is not feasible with this topology!
15-4
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.7
We should first check the feasibility of the design using the Rule- of - Thumb estimates
similar to those developed in Chapter 13 (Eq. (13.55)). The required Add = 200 (46 db).
For symmetric supplies, Add ≅ 10(VCC + VEE ) = 240. Thus, a gain of 200 appears feasible.
Rid = 2rπ = 1MΩ → rπ = 500kΩ | IC =
IE =
IC
αF
=
β oVT
rπ
=
100(0.025V )
500kΩ
= 5.00 µA
101
V − VBE (12 − 0.7)V
IC = 5.05µA | REE = EE
=
= 1.12 MΩ
100
2I E
2(5.05µA)
Add = −g m RC = −200 (46dB) | RC =
200
200
=
= 1.00 MΩ
g m 40 5x10−6
(
)
Checking the collector voltage : VC = 12 − (990kΩ)(5µA)= 7V | Picking the closest 5%
valuesfrom the table in the Appendix : REE = 1.1 MΩ and RC = 1 MΩ are the final design
values. These values give IC = 5.09µA and Add = −204 (46.2dB).
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-5
15.8
(a) I
= αF IE = αF
C
I EE 100 ⎛ 400µA ⎞
4
=
⎜
⎟ = 198µA | VCE = 12 − 3.9x10 IC − (−0.7)= 4.98V
2 101 ⎝ 2 ⎠
Q - point : (198µA,4.98V )
(b)
g m = 40IC = 7.92mS | rπ =
Acc = −
100
= 12.6kΩ | Add = −g m RC = −7.92mS (39kΩ)= −309
gm
100(39kΩ)
β o RC
=−
= −0.0965
rπ + (β o + 1)2REE
12.6kΩ + 101(400kΩ)
For a differential output : CMRR =
−309
=∞
0
−309
2
= 1600 or 64.1 dB
For a single - ended output : CMRR =
−0.0965
Rid = 2rπ = 25.2 kΩ | Ric =
rπ + (β o + 1)2REE
12.6kΩ + 101(400kΩ)
kΩ = 20.2 MΩ
2
2
( Note that this value is approaching the βoro limit and hence is not really correct.)
Rod = 2RC = 78.0 kΩ | Roc =
=
RC
= 19.5 kΩ
2
50 + 4.98
= 278kΩ
198x10−6
Add = −g m RC ro = −7.92mS 39kΩ 278kΩ = −271
(c) r
o
=
(
Acc ≅ −
)
(
)
100(39kΩ)
β o RC
=−
= −0.0965
rπ + (β o + 1)2REE
12.6kΩ + 101(400kΩ)
−271
=∞
0
For differential output : CMRR =
−271
2
For single - ended output : CMRR =
= 1400 or 62.9 dB
−0.0965
Rid = 2rπ = 25.2 kΩ | Ric =
(
)
Rod = 2 RC ro = 68.4 kΩ | Roc =
15-6
(
rπ + (β o + 1) 2REE ro
2
CB
RC Rout
(
2
)≅ R
) = 12.6kΩ + 101(164kΩ)kΩ = 8.29 MΩ
C
2
2
(
)
CB
= 19.5 kΩ since Rout
≅ µ f 2REE rπ = 24.4 MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.9
I EE 75 400µA
=
= 197µA | VC1 = VC 2 = 12 − 3.9x10 4 IC = 4.32V
76 2
2
= 4.32 − (−0.7)= 5.02V | Q - point : (197µA,5.02V )
IC = α F I E = α F
VCE
g m = 40IC = 7.88mS | rπ =
Acc = −
75
= 9.52kΩ | Add = −g m RC = −7.88mS (39kΩ) = −307
gm
75(39kΩ)
β o RC
=−
= −0.0962
rπ + (β o + 1)2REE
9.52kΩ + 76(400kΩ)
2.005 + 1.995
= 2.00V
2
0.01V
v
vC1 = VC1 + Add id + Accv ic = 4.32V − 307
− 0.0962(2V ) = 2.593 V
2
2
0.01V
v
− 0.0962(2V )= 5.663 V
vC 2 = VC 2 − Add id + Accv ic = 4.32V + 307
2
2
vOD = 2.593 − 5.663 = −3.07 V
vid = 2.005 −1.995 = 0.01V | v ic =
VCB = VC1 + AccVIC − VIC ≥ 0 | VIC ≤
15.10
Rid = 2rπ =
2(100)(0.025V )
2β oVT
I
101
→ IC
= 1.00µA | I EE = 2 C = 2
(1µA)= 2.02 µA
100
IC
5MΩ
αF
CMRR = g m REE = 105 → REE =
15.11
(a ) I
C
4.32
= 3.94 V
1+ 0.0962
= αF IE = αF
105
= 2.5 GΩ !
40(1.00µA)
I EE 100 ⎛ 20µA ⎞
5
=
⎜
⎟ = 9.90µA | VC 2 = 10 − 9.1x10 IC = 0.991V
2
101 ⎝ 2 ⎠
g m = 40IC = 0.396mS | Add = −g m RC = −0.396mS (910kΩ) = −360 | REE = ∞ → Acc ≅ 0
vid
+ Accv ic | For vs = 0 : vC 2 = VC2 = 0.991
2
For v s = 2mV : vC 2 = 0.991V + 360(0.001V )− 0(0.001V )= +1.35 V
vC 2 = VC 2 − Add
(b) For v
BC
≥ 0, vs − (0.991−180vs )≥ 0 → vs ≤
0.991V
= 5.48 mV
181
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-7
15.12
(a ) I
C
= αF IE = αF
I EE 120 ⎛ 200µA ⎞
5
=
⎜
⎟ = 99.2µA | VO = 12 −1.10x10 IC = 1.09 V
2
121 ⎝ 2 ⎠
g m = 40IC = 3.97mS | Add = −g m RC = −3.97mS (110kΩ)= −437 | REE = ∞ → Acc ≅ 0
For vs = 0 : VO = 1.09 V and v o = − Add
vid
=0
2
0.001V
= 0.219 V
For vs = 1 mV : VO = 1.09 V and vo = −(−437)
2
⎛
⎞
A
(b) For vCB ≥ 0, vs − ⎜⎝VO − 2dd vs ⎟⎠ ≥ 0 → vs ≤ 4.96 mV
15.13
*Problem 15.13 - Figure P15.11
VCC 2 0 DC 12
VEE 1 0 DC -12
V1 3 7 AC 1
V2 5 7 AC 0
VIC 7 0 DC 0
RC 2 6 110K
Q1 2 3 4 NBJT
Q2 6 5 4 NBJT
IEE 4 1 DC 200U
.MODEL NBJT NPN VA=60V BF=120
.OP
.AC LIN 1 1KHz 1KHZ
.PRINT AC IM(V1) IP(V1) VM(6) VP(6)
.TF V(6) VIC
.END
1
Results : Add = VM(6) = −193 | Rid =
= 82.0 kΩ | Acc = +0.0123 | Ric = 45.8 MΩ
IM(V1)
Circuit 15.55-Transient-3
+0.000e+000
+1.000m
+2.000m
+3.000m
+4.000m
+5.000m
Time (s)
+6.000m
+7.000m
+8.000m
+9.000m
+13.000
+12.000
+11.000
+10.000
+9.000
+8.000
+7.000
+6.000
+5.000
*REAL(Vout)*
Simulation results from B2SPICE.
The amplifier is over driven causing the output to be distorted.
Using the Fourier analysis capability of SPICE, THD = 16.9%
15.14
15-8
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
150 ⎡ 15 − 0.7 ⎤
⎢
⎥ = 47.4µA | VEC = 0.7 − −15 + 2x105 IC = 6.22V
IC = α F I E =
151 ⎢⎣2(150kΩ)⎥⎦
(
)
Q - points : (47.4µA, 6.22V ) | g m = 40IC = 1.90mS | rπ =
Add = −g m RC = −1.90mS (200kΩ) = −380
Acc = −
150
= 79.0kΩ
gm
150(200kΩ)
β o RC
=−
= −0.661
rπ + (β o + 1)2REE
79.0kΩ + 151(300kΩ)
Rid = 2rπ = 158kΩ | Ric =
rπ + (β o + 1)2REE
79.0kΩ + 151(300kΩ)
= 22.7 MΩ
2
For a differential output : Adm = Add = −380 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
2
=
Add
= −190 | Acm = Acc = −0.661
2
−190
= 287 or 49.2dB
−0.661
15.15
IC = α F I E =
100 ⎡ 10 − 0.7 ⎤
⎢
⎥ = 10.7µA | VC1 = VC 2 = −10 + 5.6x105 IC = −4.01V
101 ⎢⎣2(430kΩ)⎥⎦
VEC = 0.7 − (−4.01) = 4.71V | g m = 40IC = 0.428mS | rπ =
Add = −g m RC = −0.428mS (560kΩ)= −240
Acc = −
100
= 234kΩ
gm
100(560kΩ)
β o RC
=−
= −0.643
rπ + (β o + 1)2REE
234kΩ + 101(860kΩ)
1+ 0.99
= 0.995V
2
v
0.01V
− 0.643(0.995V )= −5.850 V
vC1 = VC1 + Add id + Accv ic = −4.01V − 240
2
2
0.01V
v
vC 2 = VC 2 − Add id + Accv ic = −4.01V + 240
− 0.643(0.995V ) = −3.450 V
2
2
−5.850 − 3.450
= −4.65
vOD = −5.850 − (−3.450)= −2.40 V | Note : Add vid = −2.40V and vOC =
2
Also note : vOC = VC + Accv ic = −4.01− 0.643(0.995V ) = −4.65V
vid = 1− 0.99 = 0.01V | v ic =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-9
15.16
*Problem 15.16 – Figure P15.14
VCC 2 0 DC 10
VEE 1 0 DC -10
V1 4 8 AC 1
V2 6 8 AC 0
VIC 8 0 DC 0
RC1 5 1 560K
RC2 7 1 560K
Q1 5 4 3 PBJT
Q2 7 6 3 PBJT
REE 2 3 430K
.MODEL PBJT PNP VA=60V BF=100
.OP
.AC LIN 1 5KHz 5KHZ
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(7) VIC
.END
1
Add = VM(5,7) = −213 | Rid =
= 511 kΩ
IM(V1)
213
= 332 → 50.4dB
Acc = −0.642 | Ric = 37.5 MΩ | CMRR =
0.642
15-10
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.17
IC = α F
I EE 80 ⎛10µA ⎞
5
= ⎜
⎟ = 4.94µA | VEC = 0.7 − −3 + 3.9x10 IC = 1.77V
2
81 ⎝ 2 ⎠
(
)
Q - points : (4.94µA, 1.77V ) | g m = 40IC = 0.198mS | rπ =
Add = −g m RC = −0.198mS (390kΩ)= −77.2
Acc = −
80
= 404kΩ
gm
80(390kΩ)
β o RC
=−
= −0.0385
rπ + (β o + 1)2REE
404kΩ + 81(10 MΩ)
Rid = 2rπ = 808kΩ | Ric =
Note that Ric is similar to
rπ + (β o + 1)2REE
2
βoro
2
For example, if VA were 80V ,
=
808kΩ + 81(10 MΩ)
2
= 405 MΩ
so that Ric = 405 MΩ will not be fully acheived.
β oro
2
≅
80 ⎛ 80 ⎞
⎜
⎟ = 648 MΩ
2 ⎝ 4.94µA ⎠
For a differential output : Adm = Add = −77.2 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −38.6 | Acm = Acc = −0.661
2
−38.6
= 1000 or 60.0dB | VBC ≥ 0 requires VIC ≥ VC = −1.07V and
−0.0385
Without detailed knowledge of the circuit for IEE , we can only estimate that VIC should not exceed
VIC + 0.7 ≤ VCC − 0.7V which allows 0.7V for biasing I EE → −1.07V ≤ VIC ≤ +1.6V.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-11
15.18
IC = α F
I EE 120 ⎛1mA ⎞
4
=
⎜
⎟ = 496µA | VC1 = VC 2 = −22 + 1.5x10 IC = −14.6V
2 121 ⎝ 2 ⎠
Checking VEC = 0.7 − (−14.6)= 15.3V | g m = 40IC = 19.8mS | rπ =
Add = −g m RC = −19.8mS (15kΩ)= −297
Acc = −
120
= 6.06kΩ
gm
120(15kΩ)
β o RC
=−
= −0.0149
rπ + (β o + 1)2REE
6.06kΩ + 121(1MΩ)
0.01+ 0
= 0.005V
2
0.01V
v
− 0.0149(0.005V )= −16.09 V
vC1 = VC1 + Add id + Accv ic = −14.6V − 297
2
2
0.01V
v
− 0.0149(0.005V )= −13.12 V
vC 2 = VC 2 − Add id + Accv ic = −14.6V + 297
2
2
vOD = −16.09 − (−13.12)= −2.97 V | Note : Add v id = −2.97V
vid = 0.01− 0 = 0.01V | v ic =
−16.09 −13.12
= −14.6 and
2
= VC + Accvic = −14.60 − 0.0149(0.005V )= −14.6V
Also note : vOC =
vOC
15.19
IC = α F I E =
100 ⎡15V − 0.7V ⎤
100
⎢
⎥ = 70.8µA | g m = 40IC = 2.83mS | rπ =
= 35.3kΩ
101 ⎢⎣ 2(100kΩ) ⎥⎦
gm
⎛
⎛
∆R ⎞
∆R ⎞
| vod = vc1 − vc2 = ic1⎜ R +
⎟ − ic2 ⎜ R −
⎟
2 ⎠
2 ⎠
⎝
⎝
vid ⎞⎛
∆R ⎞
v ⎛
∆R ⎞ ⎛
vod = −gm id ⎜ R +
⎟ − ⎜−gm ⎟⎜ R −
⎟ = −g m R vid | Add = −g m R = −283
2 ⎠
2 ⎠ ⎝
2⎝
2 ⎠⎝
⎛
⎛
v
∆R ⎞
∆R ⎞
Acd = od | vod = vc1 − vc2 = ic1⎜ R +
⎟ − ic2 ⎜ R −
⎟
2 ⎠
2 ⎠
vic
⎝
⎝
Add =
vod
vid
For a common - mode input, ic1 = ic2 =
vod = −
Acd = −
rπ + (β o + 1)2REE
vic
⎡⎛
∆R ⎞ ⎛
∆R ⎞⎤
β o∆R
vic⎢⎜ R +
v ic
⎟ −⎜R −
⎟⎥ = vod = −
2 ⎠ ⎝
2 ⎠⎦
rπ + (β o + 1)2REE
rπ + (β o + 1)2REE ⎣⎝
βo
100(100kΩ)
∆R
βo R
= −0.01
= −.00494
R rπ + (β o + 1)2REE
35.3kΩ + (101)200kΩ
CMRR =
15-12
βo
−283
= 57300 or 95.2 dB
−0.00494
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.20
*Problem 15.20 – Figure P15.19
VCC 2 0 DC 15
VEE 1 0 DC -15
V1 4 8 AC 0.5
V2 6 8 AC -0.5
VIC 8 0 DC 0
RC1 2 5 100.5K
RC2 2 7 99.5K
Q1 5 4 3 NBJT
Q2 7 6 3 NBJT
REE 3 1 100K
.MODEL NBJT NPN BF=100
.OP
.AC LIN 1 100 100
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(5,7) VIC
.END
Results: Add = VM (5,7) = −274 | A cd = −0.00494 | CMRR = 55500 or 94.9 dB
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-13
15.21
For a differential - mode input :
⎛v
⎞⎛
⎛ v
⎞⎛
∆g ⎞
∆g ⎞
vod = −⎜ id − ve ⎟⎜ g m + m ⎟ R + ⎜− id − ve ⎟⎜ g m − m ⎟ R = −g m R vid + ∆g m Rve
2 ⎠
2 ⎠
⎝2
⎠⎝
⎝ 2
⎠⎝
⎛
⎞
∆g
vod = −g m R⎜ vid + m ve ⎟ | At the emitter node :
gm ⎠
⎝
⎛ vid
⎞⎛
⎞ ⎛ v
⎞⎛
⎞
∆g m
∆g
+ g π ⎟ + ⎜− id − ve ⎟⎜ g m − m + g π ⎟ − GEE ve = 0
⎜ − ve ⎟⎜ g m +
2
2
⎝2
⎠⎝
⎠ ⎝ 2
⎠⎝
⎠
1 ∆g m
1 ∆g m
β o REE
ve =
vid ≅
vid << v id | vod ≅ −g m R vid | Add ≅ −g m R = −300
2 g m rπ + (β o + 1)2REE
4 gm
For a common - mode input :
⎛
⎛
∆g ⎞
∆g ⎞
∆g
vod = −(vic − ve )⎜ g m + m ⎟ R + (vic − ve )⎜ g m − m ⎟ R = −∆g m R(vic − v e )= − m g m R(vic − ve )
2 ⎠
2 ⎠
gm
⎝
⎝
⎛
⎞
∆g
∆g
At the emitter node : (vic − ve )⎜ g m + m + g π + g m − m + g π ⎟ − GEE ve = 0
2
2
⎝
⎠
ve =
(β + 1)2R
+ (β + 1)2R
o
rπ
Acd =
EE
o
vic | vic − v e =
EE
rπ
vic
rπ + (β o + 1)2REE
βo R
gm R
gm R
v od
∆g
∆g
∆g
=− m
≅− m
=− m
= −0.00499
vic
g m rπ + (β o + 1)2REE
g m 1+ 2g m REE
g m 1+ 2g m REE
−1
⎡ ∆g ⎤
CMRR ≅ 2g m REE ⎢ m ⎥ = 60000 or 95.6 dB
⎣ gm ⎦
Note for Problems 15.22 - 15.37
The MATLAB m-file listed below 'FET Bias' can be used to help find the drain currents in the
FET circuits in Problems 15.22 - 15.37. Use fzero('FET Bias',0) to find .ID
function f=bias(id)
kn=4e-4; vto=1; gamma=0.0;
rss=62e3; vss=15;
vsb=2*id*rss;
vtn=vto+gamma*(sqrt(vsb+0.6)-sqrt(0.6));
f=vss-vtn-sqrt(2*id/kn)-vsb;
15-14
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.22
This solution made use of the m-file above. The solution to Problem 15.23 gives an example of
direct hand calculation.
2
µA
12 − VGS
K
12 − VGS
⇒ 2 n (VGS − VTN ) =
and for Kn = 400 2 and VTN = 1V
2I S =
220kΩ
2
220kΩ
V
2
2
12 − VGS = 88 VGS − 2VGS + 1 or 88VGS −175VGS + 76 = 0 and VGS = 1.348V
(
)
1 ⎛12 −1.35 ⎞
5
ID = IS = ⎜
⎟ = 24.2µA. VD = 12 − 3.3x10 (I D ) = 4.01V
2 ⎝ 220kΩ ⎠
VDS = 4.01− (−VGS )= 5.36V (> VGS − VTN ). Q - Point = (24.2µA, 5.36V )
(
)
2 24.2x10−6
2I D
gm =
=
= 1.39mS | Add = −g m RD = −1.39ms(330kΩ)= −45.9
VGS − VTN
0.348
Acc = −
1.39ms(330kΩ)
g m RD
=
= −0.738
1+ 2g m RSS 1+ 2(1.39ms)(220kΩ)
For a differential output : Adm = Add = −45.9 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −23.0 | Acm = Acc = −0.738
2
23.0
= 31.2 | CMRRdb = 29.8 dB | Rid = ∞ | Ric = ∞
0.738
15.23
VSS − VGS = 2I D RSS | VGS = VTN +
(
2I D
2I D
| VSS = 2I D RSS + VTN +
Kn
Kn
)
2I D
→ I D = 107µA | VGS − VTN = 0.731V
4x10−4
= 15 − (62kΩ)I D − (−VGS )= 10.1V > 0.731V - Active | Q - pt : (107µA, 10.1V )
15 = 2I SS 62x103 + 1+
VDS
gm =
2(107µA)
2I D
=
= 0.293mS | Add = −g m RD = −(0.293mS )(62kΩ)= −18.2
VGS − VTN
0.731V
Acc = −
(0.293mS )(62kΩ) = −0.487
g m RD
=−
1+ 2g m RSS
1+ 2(0.293mS )(62kΩ)
For a differential output : Adm = Add = −18.2 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −9.10 | Acm = Acc = −0.487
2
9.10
= 18.7 | CMRRdb = 25.4 dB | Rid = ∞ | Ric = ∞
0.487
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-15
15.24
*Problem 15.24 - Figure P15.22
VCC 2 0 DC 12
VEE 1 0 DC -12
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RD1 2 3 330K
RD2 2 7 330K
M1 3 4 5 5 NFET
M2 7 6 5 5 NFET
REE 5 1 220K
.MODEL NFET NMOS KP=400U VTO=1
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
Results : Add = VM(3,7) = −45.9 | Acc = −0.738 | CMRR = 31.2 | Rid = ∞ | Ric = ∞
(b) Results from B2SPICE
Problem15.67(b)-Transient-0
+0.000e+000
(V)
+1.000m
+2.000m
Time (s)
+3.000m
+4.000m
+1.000
+500.000m
+0.000e+000
-500.000m
-1.000
V(IVOUT)
Problem 15.24(b)-Fourier-Table
FREQ
mag
phase
+0.000
+1.000k
+2.000k
+3.000k
+4.000k
+5.000k
-4.011
+586.074m
+82.747u
+90.858u
+14.251n
+9.695n
norm_mag
norm_phase
+0.000
+0.000
+0.000
+180.000 +1.000
+0.000e+000
-90.000
+141.188u -270.000
+179.996
+155.029u -3.577m
+93.958
+24.316n -86.042
+53.059
+16.541n -126.941
THD = 0.021% is very low due to the Level-1 square law model used in the simulation
15.25
15-16
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
First, we should check our rule - of - thumb. Since we have symmetric power supplies,
Add ≅ VDD + VSS = 10 or 20 dB. We should be ok.
Rod = 2RD = 5kΩ → RD = 2.5kΩ | Selecting closest 5% value : RD = 2.4kΩ
20
Add = −g m RD = 10 20 = 10 | g m =
(4.17x10 ) = 348µA
=
2(25x10 )
2
−3
ID
−3
RSS =
10
= 4.17mS = 2Kn I D
2400
| VGS = VTN +
2I DS
= 1.16V
Kn
VSS − VGS
5 −1.16
=
= 5.52kΩ | Selecting closest 5% value : RSS = 5.6kΩ
2I D
2(348µA)
15.26
(a ) This solution made use of the m - file listed above Prob. 15.22.
VSS − VGS = 2I S RSS = 2I D RSS | VGS = VTN +
VTN = VTO + γ
(V
SB
)
+ 0.6 − 0.6 = VTO + γ
2I D
2I D
| VSS = 2I D RSS + VTN +
Kn
Kn
( 2I
D
RSS + 0.6 − 0.6
µA
| VTO = 1V | γ = 0.75 V yields
V2
= 3.01V | VGS = 3.69V
Solving iteratively with RSS = 62kΩ | Kn = 400
I D = 91.3µA | VGS − VTN = 0.676V | VTN
)
VDS = 15 − (62kΩ)I D − (−VGS )= 12.9V > 0.676V - Saturated | Q - pt : (91.3µA, 12.9V )
(b) g
m
Acc = −
=
2(91.3µA)
2I D
=
= 0.270mS | Add = −g m RD = −(0.270mS )(62kΩ)= −16.7
VGS − VTN
0.676V
(0.270mS )(62kΩ) = −0.486 assuming η = 0
g m RD
=−
1+ 2g m (1+ η)RSS
1+ 2(0.270mS )(62kΩ)
For a differential output : Adm = Add = −16.7 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −8.35 | Acm = Acc = −0.486
2
8.35
= 17.2 | CMRRdb = 24.7 dB | Rid = ∞ | Ric = ∞
0.486
(c) For γ = 0, V
TN
I D = 107 µA
= VTO | VSS = 2I D RSS + VTO +
2I D
2I D
| 15 = 124000I D + 1+
Kn
4x10−4
VDS = 30 − 62000I D −128000I S = 10.1 V Saturated | Q - pt : (107 µA, 10.1 V )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-17
15.27
*Problem 15.27 - Figure P15.26
VCC 2 0 DC 15
VEE 1 0 DC -15
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RD1 2 3 62K
RD2 2 7 62K
M1 3 4 5 1 NFET
M2 7 6 5 1 NFET
REE 5 1 62K
.MODEL NFET NMOS KP=400U VTO=1 PHI=0.6 GAMMA=0.75
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
Results: Add = VM(3,7) = −16.8 | Acc = −0.439 | CMRRdB = 25.6 dB | Rid = ∞ | Ric = ∞
(b)
Problem15.70(b)-Transient-1
(V)
+0.000e+000
+1.000m
+2.000m
Time (s)
+3.000m
+4.000m
+400.000m
+200.000m
+0.000e+000
-200.000m
-400.000m
V(IVOUT)
Problem15.27(b)-Fourier-Table
FREQ
+0.000
+1.000k
+2.000k
+3.000k
+4.000k
+5.000k
15-18
mag
phase
-345.019n
+444.350m
+35.476n
+15.056u
+23.789n
+22.342n
+0.000
+180.0
-30.192
-179.929
-50.644
-57.083
THD = 0.0034%
norm_mag
norm_phase
+0.000
+0.000
+1.000
+0.000
+79.838n -210.192
+33.884u -359.929
+53.536n -230.643
+50.280n -237.083
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.28
This solution made use of the m - file listed above Prob. 15.22.
(a) V
2I D
2I D
| VSS = 2I D RSS + VTN +
Kn
Kn
− VGS = 2I S RSS = 2I D RSS | VGS = VTN +
SS
VTN = VTO + γ
(V
SB
)
+ 0.6 − 0.6 = VTO + γ
( 2I
D
RSS + 0.6 − 0.6
µA
| VTO = 1V | γ = 0.75 V yields
V2
= 2.74V | VGS = 3.05V
Solving iteratively with RSS = 220kΩ | Kn = 400
I D = 20.3µA | VGS − VTN = 0.319V | VTN
)
VDS = 12 − (330kΩ)I D − (−VGS )= 8.35V > 0.319V - Active region | Q - pt : (20.3µA, 8.35V )
(b) g
m
=
2(20.3µA)
2I D
=
= 0.127mS | Add = −g m RD = −(0.127mS )(330kΩ)= −41.9
VGS − VTN
0.319V
(0.127mS )(330kΩ) = −0.737 assuming η = 0
g m RD
=−
1+ 2g m (1+ η)RSS
1+ 2(0.127mS )(220kΩ)
Acc = −
For a differential output : Adm = Add = −41.9 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −21.0 | Acm = Acc = −0.737
2
21.0
= 28.4 | CMRRdb = 29.1 dB | Rid = ∞ | Ric = ∞
0.737
(c) For γ = 0, V
TN
I D = 24.2 µA
= VTO | VSS = 2I D RSS + VTO +
2I D
2I D
| 12 = 440000I D + 1+
Kn
4x10−4
VDS = 24 − 330000I D − 440000I S = 5.37 V Saturated | Q - pt : (25.2 µA, 5.37 V )
15.29
(
2 2x10−5
I SS
2I D
(a) I D = 2 = 20µA | VGS = VTN + K = 1+ 4x10−4
n
) = 1.316V | V
GS
− VTN = 0.316V
VDS = 9 − (300kΩ)I D − (−VGS )= 4.32V > 0.316V - Active region | Q - pt : (20µA, 4.32V )
(b) g
m =
Acc = −
2(20µA)
2I D
=
= 0.127mS | Add = −g m RD = −(0.127mS )(300kΩ) = −38.0
VGS − VTN 0.316V
(0.127mS )(300kΩ) = −0.120
g m RD
=−
1+ 2g m RSS
1+ 2(0.127mS )(1.25MΩ)
For a differential output : Adm = Add = −38.0 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −19.0 | Acm = Acc = −0.120
2
19.0
= 158 | CMRRdb = 44.0 dB | Rid = ∞ | Ric = ∞
0.120
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-19
15.30
(
2 1.5x10−4
I SS
2I D
(a) I D = 2 = 150µA | VGS = VTN + K = 1+ 4x10−4
n
) = 1.866V
| VGS − VTN = 0.866V
VDS = 15 − (75kΩ)I D − (−VGS ) = 5.62V > 0.866V - Active region | Q - pt : (150µA, 5.62V )
(b) g
2(150µA)
2I D
=
= 0.346mS | Add = −g m RD = −(0.346mS )(75kΩ)= −26.0
VGS − VTN
0.866V
=
m
Acc = −
(0.346mS )(75kΩ) = −0.232
g m RD
=−
1+ 2g m RSS
1+ 2(0.346mS )(160kΩ)
For a differential output : Adm = Add = −26.0 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −13.0 | Acm = Acc = −0.232
2
13.0
= 56.0 | CMRRdb = 35.0 dB | Rid = ∞ | Ric = ∞
0.232
15.31
(
)
2 2x10−5
I SS
2I D
(a) I D = 2 = 20µA | VGS = VTN + K = VTN + 4x10−4 = VTN + 0.316V
n
VTN = VTO + γ
(V
SB
)
(
0.6 )→ V
+ 0.6 − 0.6 = 1+ 0.75 9 − VGS + 0.6 − 0.6
(
VGS − 0.316 = 1+ 0.75 9 − VGS + 0.6 −
GS
)
= 2.71V | VTN = 2.39V
VDS = 9 − (300kΩ)I DS − (−VGS ) = 5.71V > 0.316V - Active region | Q - pt : (20µA, 5.71V )
(b) g
m
Acc = −
=
2(20µA)
2I D
=
= 0.127mS | Add = −g m RD = −(0.127mS )(300kΩ)= −38.1
VGS − VTN 0.316V
(0.127mS )(300kΩ) = −0.120 assuming η = 0
g m RD
=−
1+ 2g m (1+ η)RSS
1+ 2(0.127mS )(1.25MΩ)
For a differential output : Adm = Add = −38.1 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
15-20
Add
= −19.0 | Acm = Acc = −0.120
2
19.0
= 158 | CMRRdb = 44.0 dB | Rid = ∞ | Ric = ∞
0.120
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.32
(
)
2 1.5x10−4
I SS
2I D
(a) I D = 2 = 150µA | VGS = VTN + K = VTN + 4x10−4 = VTN + 0.866V
n
VTN = VTO + γ
(V
SB
)
( 15 − V
0.6 )→ V
+ 0.6 − 0.6 = VTO + γ
(
VGS − 0.866 = 1+ 0.75 15 − VGS + 0.6 −
)
GS
+ 0.6 − 0.6
GS
= 3.86V | VTN = 2.99V
VDS = 15 − (75kΩ)I DS − (−VGS )= 7.61V > 0.866V - Active region | Q - pt : (150µA, 7.61V )
(b) g
m
Acc = −
=
2(150µA)
2I D
=
= 0.346mS | Add = −g m RD = −(0.346mS )(75kΩ)= −26.0
VGS − VTN
0.866V
(0.346mS )(75kΩ) = −0.233 assuming η = 0
g m RD
=−
1+ 2g m (1+ η)RSS
1+ 2(0.346mS )(160kΩ)
For a differential output : Adm = Add = −26.0 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −13.0 | Acm = Acc = −0.233
2
13.0
= 55.8 | CMRRdb = 34.9 dB | Rid = ∞ | Ric = ∞
0.233
15.33
30
Add = −g m RD = 10 20 = 31.6 | First, we should check our rule - of - thumb.
Since we have symmetric power supplies, Add ≅ (VDD + VSS ) = 15, within a factor of about 2.
We should be ok if we reduce the value of VGS − VTN . (Remember, our rule - of - thumb
used VGS − VTN = 1V.) | g m RD = 31.6 =
2I D RD
VGS − VTN
Maximum common - mode range requires minimum I D RD ⇒ minimum VGS − VTN
Choosing VGS − VTN = 0.25V to insure strong inversion operation,
I D RD =
0.25(31.6)
2
(0.25) (0.005) = 156µA
2I D
→ ID =
Kn
2
2
= 3.95V | 0.25V =
I SS = 2I D = 312 µA | RD =
3.95V
= 25.3kΩ → 27kΩ, the nearest 5% value.
156µA
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-21
15.34
(a) I
D
Kp
2
1 ⎛ 18 − VGS ⎞
18 + VGS
µA
= ⎜
VGS − VTP ) =
and for K p = 200 2 and VTP = −1V
⎟⇒
(
2
112kΩ
2 ⎝ 56kΩ ⎠
V
18 + VGS = 11.2(VGS + 1) → VGS = −2.19V | VGS − VTP = −1.19V | I D = 142µA
2
{
[
]}
) = 0.238mS
VDS = − −VGS − (91kΩ)I D −18 = −7.27V ≤ −1.19V − Active region | Q - Point = (142µA, 7.27V )
(b) g
m
Acc = −
(
)(
= 2 2x10−4 1.42x10−4
| Add = −g m RD = −0.238mS (91kΩ)= −21.7
0.238mS (91kΩ)
g m RD
=
= −0.785
1+ 2g m RSS 1+ 2(0.238mS )(56kΩ)
For a differential output : Adm = Add = −21.7 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −10.9 | Acm = Acc = −0.785
2
10.9
= 13.9 | CMRRdb = 22.9 dB | Rid = ∞ | Ric = ∞
0.785
15.35
*Problem 15.35 – Figure P15.34
VCC 2 0 DC 18
VEE 1 0 DC -18
VIC 8 0 DC 0
V1 4 8 AC 0.5
V2 6 8 AC -0.5
RD1 5 1 91K
RD2 7 1 91K
M1 5 4 3 3 PFET
M2 7 6 3 3 PFET
REE 2 3 56K
.MODEL PFET PMOS KP=200U VTO=-1
.OP
.AC LIN 1 3KHZ 3KHZ
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(7) VIC
.END
Results : Add = VM(5,7) = −21.6 | Acc = −0.783 | CMRR = 13.8 | Rid = ∞ | Rid = ∞
15-22
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.36
(
)
2 2x10−5
I SS
2I D
(a) I D = 2 = 20µA | VGS = VTP − K = VTP − 2x10−4 = VTP − 0.447V
p
{
( V + 0.6 − 0.6 )= −1− 0.6( 10 + V + 0.6 − 0.6 )}
= −0.447 − {
1+ 0.6( 10.6 + V − 0.6 )}→ V = −2.67V | V = −2.23V
VTP = − VTO − γ
VGS
BS
GS
GS
GS
[
]
2(2x10 )(2x10 ) = 89.4µS
TP
VDS = VGS + −10 + (300kΩ)I D = −6.67V ≤ −0.447V - Active region | Q - pt : (20µA,-6.67V )
(b) g
m
Acc = −
=
−5
−4
| Add = −g m RD = −(89.4µS )(300kΩ)= −26.8
(89.4µS )(300kΩ) = −0.119 assuming η = 0
g m RD
=−
1+ 2g m (1+ η)RSS
1+ 2(89.4µS )(1.25MΩ)
For a differential output : Adm = Add = −26.8 | Acm = 0 | CMRR = ∞
For a single - ended output : Adm =
CMRR =
Add
= −13.4 | Acm = Acc = −0.119
2
13.4
= 113 | CMRRdb = 41.0 dB | Rid = ∞ | Ric = ∞
0.119
15.37
(a) I
D
=
I SS
= 10µA | VO = −12 + (820kΩ)I D = −3.80V | For vI = 0, vO = VO = −3.80 V
2
( )
2 10−5
2I D
VGS = VTP +
= 1−
= 0.86 V | VGS − VP = −0.14V
KP
10−3
VDS ≤ −0.14V for pinchoff. So VD ≤ −1 V for pinchoff.
g m = 2(1mA)(10µA) = 141µS | Add = −g m RD = −(141µS )(820kΩ)= −116
Acc = 0 for RSS and ro = ∞ | vO = VO −
(b) 2 ≤ 0.2V
v1
GS
Add
116.0
v1 = −3.80 +
(0.02)= −2.64V
2
2
− VP = 0.2(0.14) = 28.0mV | v1 ≤ 56 mV based upon the small - signal limit
Also vO ≤ −1 V for pinchoff. -1 ≤ −3.80 +
116.0
v1 → v1 = 48.3 mV | So, v1 ≤ 48.3 mV
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-23
15.38
+22 V
RC
RC
200 k Ω
Q
200 k Ω
Q
1
R1 2 k Ω
+
-
RC
Q
1
+
-
vic
1
vid
R 1 +2R EE
2R EE
200 k Ω
402 k Ω
400 k Ω
2
R1
2kΩ
-22 V
Common-mode
half circuit
dc half-circuit
Differential-mode
half circuit
(a)
150 22 − 0.7
= 52.6 µA | VCE = 22 − (200kΩ)IC − (−0.7) = 12.2 V
151 402kΩ
150(0.025V )
= 71.3kΩ
Q − Point = (52.6µA, 12.2V ) for both transistors | rπ =
52.6µA
(b) IC = α F IE =
Acc = −
Add = −
β o RC
rπ + (β o + 1)(R1 + 2REE )
=−
150(200kΩ)
= −0.494
71.3kΩ + (151)402kΩ
150(200kΩ)
β o RC
=−
= −80.4
rπ + (β o + 1)R1
71.3kΩ + (151)2kΩ
Rid = 2[rπ + (β o + 1)R1 ] = 2[71.3kΩ + (151)2kΩ] = 747 kΩ
Note also : Ric = 0.5[rπ + (β o + 1)(R1 + 2REE )] = 0.5[71.3kΩ + (151)(402kΩ)] = 30 MΩ
and single - ended CMRR =
15-24
40.2
= 81.4, a paltry 38.2 dB
0.494
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.39
*Problem 15.39 – Figure P15.38
VCC 2 0 DC 22
VEE 1 0 DC -22
VIC 10 0 DC 0
V1 4 10 AC 0.5
V2 8 10 AC -0.5
RC1 2 5 200K
RC2 2 9 200K
Q1 5 4 3 NBJT
Q2 9 8 7 NBJT
RE1 3 6 2K
RE2 7 6 2K
REE 6 1 200K
.MODEL NBJT NPN BF=150
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(V1) IP(V1) VM(5,9) VP(5,9)
.TF V(9) VIC
.END
Results : Add = VM(5,9) = −79.9 | Acc = −0.494 | Rid =
1
= 751 kΩ
IM(V1)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-25
15.40
+V
CC
100 k Ω
100 k Ω
100 k Ω
500 k Ω
Q
Q
1
+
-
EE
-V
v
+
v
ic
1
2.5 k Ω
id
2
100 µA
I
Q
1
REE
600 k Ω
REE
600 k Ω
EE
dc half-circuit
Common-mode
half circuit
(a)
Differential-mode
half circuit
100
(100µA)= 99.0 µA VCE = 20 −105 IC − (−0.7)= 10.8 V
101
100(0.025V )
Q − Po int = (99.0µA, 10.8V ) for both transistors | rπ =
= 25.3kΩ
99.0 µA
(b) I
C
= αF IE =
Acc = −
β o RL'
100kΩ
=−
= −0.165
rπ + (β o + 1)REE
25.3kΩ + 101(600kΩ)
Add = −
β o RL
| RL = 100kΩ 500kΩ =83.3kΩ | R5 = 600kΩ 2.5kΩ =2.49kΩ
rπ + (β o + 1)R5
Add = −
100(83.3kΩ)
25.3kΩ + 101(2.49kΩ)
[
] [
⎛ 30.1 ⎞
Note : Single - ended CMRR = 0.5⎜
⎟ = 91.2 and
⎝ 0.165 ⎠
[
]
[
]
Ric = 0.5 rπ + (β o + 1)R5 = 0.5 25.3kΩ + 101(600kΩ) = 30.3 MΩ
15-26
]
= −30.1 | Rid = 2 rπ + (β o + 1)R5 = 2 25.3kΩ + 101(2.49kΩ) = 554kΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.41
*Problem 15.41 – Figure P15.40
VCC 2 0 DC 20
VEE 1 0 DC -20
VIC 9 0 DC 0
V1 4 9 AC 0.5
V2 7 9 AC -0.5
RC1 2 5 100K
RC2 2 8 100K
RL 5 8 1MEG
Q1 5 4 3 NBJT
Q2 8 7 6 NBJT
REE 3 6 5K
IEE1 3 1 67.8U
RE1 3 1 600K
IEE2 6 1 67.8U
RE2 6 1 600K
.MODEL NBJT NPN BF=100
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(V1) IP(V1) VM(5,8) VP(5,8)
.TF V(8) VIC
.END
Results : Add = VM(5,8) = −30.0 | Acc = −0.165 | Rid =
1
= 555 kΩ
IM(V1)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-27
15.42
+V
CC
R
R
D
R
D
M
M
1
Q
D
Q
1
+
-
v
C
= αF
+
-
2R
2
(b) I
Q
ic
EE
-V
1
1
I
dc half-circuit
M
1
1
v
id
2
Differential-mode
half circuit
EE
Common-mode
half circuit
EE
2
2x10−4
I EE 100 ⎛ 100µA ⎞
=
VGS − (−4) → VGS = −3.29V
⎜
⎟ = 49.5 µA | 49.5µA =
2 101 ⎝ 2 ⎠
2
[
]
VCE = −VGS = 3.29V | VDS = 15 − 7.5x10 4 IC − VCE − (−0.7)= 8.70 V
BJT Q − Points = (49.5µA, 3.29V ) | JFET Q − Points = (49.5µA, 8.70V )
rπ =
100(0.025V )
49.5 µA
= 50.5kΩ | Acc = − −
β o RL
rπ + (β o + 1)(2REE )
=−
100(75kΩ)
50.5kΩ + 101(1.2 MΩ)
= −0.0619
Add = −g m1 RD = −40(49.5µA)(75kΩ)= −149 | Rid = 2rπ = 101kΩ
⎛ 149 ⎞
Note : Single - ended CMRR = 0.5⎜
⎟ = 1200 or 61.6 dB and
⎝ 0.0619 ⎠
[
]
[
]
Ric = 0.5 rπ + (β o + 1)2REE = 0.5 50.5kΩ + 101(1.2 MΩ) = 60.6 MΩ
(c) From (a) V
CE
15-28
= −VGS = 3.29V | VBE = 0.7V | VCE ≥ VBE → Active − region operation
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.43
+V DD = 6 V
I1
200 µA
M1
M1
M3
M1
v1
v ic
I2
100 µA
M3
M3
v id
2
RD
RD
30 k Ω
30 k Ω
RD
30 k Ω
- V SS = -6 V
(a)
Differential-mode
half circuit
Common-mode
half circuit
dc half-circuit
(b) I D1 = I2 = 100µA | VGS1 = VTN +
2I D1
10−4
= 1+ 2 −3 = 1.447V | VGS1 − VTN = 0.447V
Kn
10
I D3 = I1 − I D1 = 200µA −100µA = 100µA | VDS1 = −VGS 3 = 1+ 2
{
[
VGS 3 − VTP = −0.632V | VDS 3 = − VS1 − VGS 3 − −6 + (30kΩ)I D3
10−4
= 1.632 V
5x10−4
]}= −(−1.45 + 1.63 + 6 − 3)= −3.18V
Both M1 and M3 are saturated. Q - points : M1 : (100µA, 1.63V ) M3 : (100µA, − 3.18V )
( )( )
Add = −g m RD = − 2 10−4 10−3 (30kΩ) = −13.4 | For ro = ∞, Acc = 0 | Rid = ∞
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-29
15.44
(a)
+V DD = 1.5 V
I1
200 µA
V S3
M1
-1.45 V
+0.18 V
M1
M3
100 µA
M3
M3
V S1
I2
M1
-0.5 V
v ic
vid
2
R
RD
D
RD
10 k Ω
10 k Ω
10 k Ω
- V SS = 1.5 V
(b) I
Differential-mode
half circuit
Common-mode
half circuit
dc half-circuit
D1 = I 2 = 100µA | VGS1 = VTN +
2I D1
10−4
= 1+ 2 −3 = 1.447V | VGS1 − VTN = 0.447V
Kn
10
I D3 = I1 − I D1 = 200µA −100µA = 100µA | VDS1 = −VGS 3 = 1+ 2
{
[
VGS 3 − VTP = −0.632V | VDS 3 = − VS1 − VGS 3 − −1.5 + (30kΩ)I D3
10−4
= 1.632 V
5x10−4
]}= −(−1.45 + 1.63 + 1.5 − 3)= −1.32V
These voltages can barely be supported by the 1.5- V negative power supply.
VS1 = −VGS1 = −1.45V which is more negative than the -1.5 - V supply. Also,
VS 2 = VS1 − VGS 3 = −1.45V + 1.63 = +0.18V , but for I D3 = 100µA, VD3 = −1.5 + 10−4 (10kΩ)= −0.5V.
M3 is just beyond pinchoff, and current source I2 has a very small voltage across it.
15-30
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.45
(a) I
C1
I1 100 ⎛ 50µA ⎞
=
⎜
⎟ = 24.8 µA | VCE2 = 12 − VEB3 − (−VBE2 )= 12 V
2 101 ⎝ 2 ⎠
= αF
12V
= 500 µA
24kΩ
Q - points : (24.8 µA, 12V ) (24.8 µA, 12V ) (500 µA, 12V )
For VO = 0 → VEC3 = 12 V | IC 3 =
(b) R
C1
rπ 2 =
rπ 3 =
=
0.7V
VEB3
0.7V
= 28.2kΩ | RC 2 =
=
= 35.4kΩ
24.8µA
IC 2 − I B3 24.8µA − 5µA
100(0.025V )
24.8µA
100(0.025V )
500µA
= 101kΩ | ro2 =
= 5kΩ | ro3 =
(
60 + 12
= 2.90 MΩ
24.8µA
60 + 12
= 144kΩ
500µA
) ( )
g m2
ro2 RC 2 rπ 3 g m3 ro3 R
2
40(24.8µA)
Adm =
2.90 MΩ 35.4kΩ 5kΩ (40)(500µA) 144kΩ 24kΩ = 893
2
Rid = 2rπ 2 = 202 kΩ | (c) Rout = ro3 R = 20.6 kΩ
Adm =
(
(d ) R
ic ≅
(β
o
+ 1)ro2
2
)
=
(101)2.90 MΩ = 147 MΩ
2
(
|
(e) v
2
)
is the non - inverting input
15.46
vic ≥ −VEE + 0.75V + VBE1 = −12 + 0.7 + 0.75 = −10.6V
vic ≤ VCC − VEB3 = 12 − 0.7 = 11.3V | −10.6 V ≤ vic ≤ 11.3 V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-31
15.47
Note that the parameters of the transistors and values of RC have been carefully adjusted to
permit open-loop operation and achieve VO = 0.
*Problem 15.47 - Two Stage Amplifier – Figure P15.45
VCC 1 0 DC 12
VEE 2 0 DC -12
RC1 1 5 28.2K
RC2 1 7 33.9K
Q1 5 4 3 NBJT
Q2 7 6 3 NBJT
I1 3 2 DC 50U
Q3 8 7 1 PBJT
R 8 2 24K
V1 4 10 AC 0.5
V2 6 10 AC -0.5
VIC 10 0 DC 0
.MODEL NBJT NPN BF=100 VA=60
.MODEL PBJT PNP BF=100 VA=60 IS=0.288F
.OPTIONS TNOM=17.2
.OP
.AC LIN 1 1KHZ 1KHZ
.TF V(8) VIC
.PRINT AC VM(8) VP(8) IM(V1) IP(V1)
.END
Adm = VM(8) = 1030 | Acm = −6.07 x 10−3 | CMRRdB = 105 dB
1
Results:
Rid =
= 239 kΩ | Rout = 20.6 kΩ
IM(V1)
15-32
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.48
IC1 = β F1 I B1 = β F1 I B
[
[(β + 1)I ]≅ β
]
IC 2 = β F 2 I B2 = β F 2 (β F1 + 1)I B ≅ β F1β F 2 I B
IC = IC1 + IC 2 = β F 1 I B1 + β F 2
F1
B
β I
F1 F 2 B
Assume β o2 = β o1 = β F 2 = β F1
g m1 = 40IC1 = β F1 I B
rπ 1 =
ro1 ≅
β o1
g m1
rπ 2 =
β o2
g m2
VA
V
= A
IC1 β F1 I B
vbe1 = v be
g m2 = 40IC2 ≅ β F 2β F1 I B g m2 = β o2 g m1 = β o1g m1
=
β o1
β o1g m1
ro2 ≅
rπ 1
rπ 1 + (β o1 + 1)rπ 2
1
rπ 1 = β o1rπ 2
g m1
=
VA
VA
≅
IC2 β F1β F 2 I B
≅ vbe
ro1 ≅ β o2ro2 = β o1ro2
β o1rπ 2
v
v
≅ be → vbe2 ≅ be
2
β o1rπ 2 + (β o1 + 1)rπ 2 2
g m2
g
vbe → Gm ≅ m2
2
2
RiB = rπ 1 + (β o1 + 1)rπ 2 = β o1rπ 2 + (β o1 + 1)rπ 2 ≅ 2β o1rπ 2
Gmvbe = g m1vbe1 + g m2vbe2 ≅
⎡
(β o2 + 1)
1
ic = v ce2⎢ +
⎢ ro2 r 1+ g r r
o1
m1 π 2 π 1
⎣
RiC ≅ ro2
[
(
(
(
ro1 1+ g m1 rπ 2 rπ 1
β o2
⎤
⎡
β o2
⎥≅v ⎢ 1 +
ce2
⎥
⎢ ro2 r 1+ g r r
o1
m1 π 2 π 1
⎦
⎣
))
)]≅ r
[
[
(
]= r
ro1 1+ g m1 (rπ 2 )
o2
Note that βo = Gm RiB = β o1β o2 ≅ β o2
β o2
µf = Gm RiC =
o2
2
ro1
β o2
⎤
⎥
⎥
⎦
)]
2
= ro2 2ro2 = ro2
3
µf 2
3
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-33
15.49
+V CC
RC
RC
Q
4
Q3
Q
vO
Q
1
v1
2
I2
v2
I1
-V EE
(a) I
C1
I1 100 ⎛ 50µA ⎞
=
⎜
⎟ = 24.8 µA | VCE 2 = VCC − VEB 3 − VEB 4 − (−VBE 2 )= 12 − 0.7 = 11.3 V
2 101 ⎝ 2 ⎠
= αF
For VO = 0 → VEC 4 = 12 V and VEC 3 = 12 − 0.7 = 11.3 V | For balance, VCE1 = VCE 2 = 11.3 V
⎛
1 ⎞ 12V
I
IC 4 + IC 3 = IC 4 + α F 3 I B 4 = IC 4 + α F 3 C 4 = IC 4 ⎜1+
| IC 4 = 495 µA | IC 3 = 4.95 µA
⎟=
βF
⎝ β F + 1⎠ 24kΩ
Q - points : (24.8 µA, 11.3V ) (24.8 µA, 11.3V ) (4.95 µA, −11.3V ) (495 µA, 12V )
(e) v
145 MΩ |
(b) R
C2
rπ 2 =
=
is the non - inverting input
2
VEB 3 + VEB 4
1.4V
1.4V
=
= 56.6kΩ | RC1 =
= 56.5kΩ
IC 2 − I B 3
24.8µA − 0.0495µA
24.8µA
100(0.025V )
24.8µA
= 101kΩ | ro2 =
100(0.025V )
60 + 11.3
= 2.88 MΩ | rπ 3 =
= 505kΩ
24.8µA
4.95µA
100(0.025V )
60 + 11.3
60 + 11.3
= 14.4 MΩ | rπ 4 =
= 5.05kΩ | ro 4 =
= 144kΩ
4.95µA
495µA
495µA
⎛ g ⎞⎛ 2r
⎞
g
g
Darlington
Adm = m2 ro2 RC 2 RinDarlington g mDarlington Rout
R = m2 ro2 RC 2 2β orπ 4 ⎜ m 4 ⎟⎜ o 4 R⎟
2
2
⎝ 2 ⎠⎝ 3
⎠
ro3 =
(
Adm =
)
(
)
(
)
40(24.8µA)
2
40 495µA
(2.88 MΩ 56.6kΩ 101kΩ)( )(2 )(96kΩ 24kΩ)= 9180
2ro4
R = 19.2 kΩ
3
(β + 1)r (101)2.88 MΩ = 145 MΩ | e v is the non - inverting input
(d ) Ric ≅ o 2 o2 =
() 2
2
Rid = 2rπ 2 = 202 kΩ | (c) Rout =
15-34
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.50
15V
= 300µA | VC 2 = 15 − 2400I E 3 − VEB 3 = 15 − 0.729 − 0.7 = 13.6V
50kΩ
⎛ 200µA ⎞ 80 ⎛ 200µA ⎞
IC 3 300µA
I C1 = I C 2 = α F ⎜
=
= 3.75µA
⎟= ⎜
⎟ = 98.8µA | I B 3 =
βF 3
80
⎝ 2 ⎠ 81 ⎝ 2 ⎠
(a) For V
O
= 0, IC 3 =
VCE1 = VCE 2 = 13.6 − (−0.7)= 14.3V | VEC 3 = 15 − 2400I E 3 − VO = 14.3V
Q - points : (98.8µA, 14.3V ) (98.8µA, 14.3V ) (300µA, 14.3V )
80(0.025V )
15 −13.6 V
= 15.1kΩ | rπ 3 =
= 6.67kΩ
0.3mA
(98.8 − 3.75) µA
RC =
⎛g ⎞
vc2
= −⎜ m1 ⎟ RC rπ 3 + (β o3 + 1)RE
vid
⎝ 2 ⎠
⎛ 40(98.8µA)⎞
⎟ 15.1kΩ 6.67kΩ + 81(2.4kΩ) = −27.7
Av1 = −⎜⎜
⎟
2
⎝
⎠
[ [
(b) A
v1 =
[
]]
(
)]
Av2 =
80(50kΩ)
vo
β o3 RL
=−
=−
= −19.9
vc2
rπ 3 + (β o3 + 1)RE
6.67kΩ + 81(2.4kΩ)
Av =
vc2 vo
= −27.7(−19.9)= 551
vid v c2
Rid = 2rπ 1 = 2
(c) R
out
(d ) R
β o1VT
IC1
=2
80(0.025V )
98.8µA
= 40.5kΩ | ro3 =
70 + 14.3
= 281kΩ
0.3mA
⎤
⎡
⎛
⎞
80(2.4kΩ)
β o RE
⎥ =49.0kΩ
= 50kΩ ro3 ⎜1+
⎟ =50kΩ 281kΩ⎢1+
⎝ RC + rπ 3 + RE ⎠
⎢⎣ 15.1kΩ + 6.67kΩ + 2.4kΩ⎥⎦
ic =
(β
o1
+ 1)ro1
2
=
81 ⎛ 70 + 14.3⎞
⎜
⎟ = 34.6 MΩ |
2 ⎝ 98.8µA ⎠
(e) v
2
is the non - inverting (+) input.
15.51
vic ≥ −VEE + 0.75V + VBE1 = −15 + 0.7 + 0.75 = −13.6V | vic ≤ VCC − I E 3 RE − VEB 3
⎛ 81 ⎞⎛ 15V ⎞
From Prob. 15.92, I E 3 RE = ⎜ ⎟⎜
⎟(2.4kΩ) = 0.729V
⎝ 80 ⎠⎝ 50kΩ ⎠
vic ≤ 15 − 0.729 − 0.7 = 13.6V | −13.6 V ≤ vic ≤ 13.6 V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-35
15.52
15V
= 300µA VEC 3 = 15 − VO = 15 − 0 = 15V
50kΩ
⎛ 200µA ⎞ 80 ⎛ 200µA ⎞
IC 3 300µA
I C1 = I C 2 = α F ⎜
=
= 3.75µA
⎟= ⎜
⎟ = 98.8µA I B 3 =
βF 3
80
⎝ 2 ⎠ 81 ⎝ 2 ⎠
(a) For V
O
= 0, IC 3 =
VCE1 = VCE 2 = 15 − VEB 3 − (−VBE1 ) = 15.0V
Q - points : (98.8µA, 15.0V ) (98.8µA, 15.0V ) (300µA, 15.0V )
RC 2 =
rπ 2 =
rπ 3 =
V
0.7
0.7 V
= 7.37kΩ | For balance, RC1 =
= 7.09kΩ
98.8 µA
(98.8 − 3.75) µA
80(0.025V )
98.8µA
80(0.025V )
0.3mA
15.53
For VO = 0, IC 3 =
= 20.2kΩ | ro2 =
70V + 15V
= 860kΩ
98.8µA
= 6.67kΩ | ro3 =
70V + 15V
= 283kΩ
0.3mA
15V
I
300µA
81
= 300µA | I B3 = C 3 =
= 3.75µA | I E3 =
IC 3 = 304µA
50kΩ
80
80
βF 3
⎛ 200µA ⎞ 80 ⎛ 200µA ⎞
0.7V + I E3 RE 0.7V + (304µA)RE
IC1 = IC 2 = α F ⎜
=
⎟= ⎜
⎟ = 98.8µA | RC =
IC1 − I B3
98.8µA − 3.75µA
⎝ 2 ⎠ 81 ⎝ 2 ⎠
VC2 = 15 − 2400I E3 − VEB 3 = 15 − 0.729 − 0.7 = 13.6V
(
)
(
)
(
)
g m1
RC Rin3 = −20(98.8µA) RC Rin3 = -1.976x10-3 RC Rin3 | Rin3 = rπ 3 + (β o + 1)RE
2
⎛
⎞
β o RL
β o RE
Avt2 = −
| RL = R ro3 ⎜1+
⎟ | Av = Avt1 Avt2
rπ 3 + (β o + 1)RE
⎝ Rth + rπ 3 + RE ⎠
Avt1 = −
Voltage Gain
1000
100
10
1
0
10000
20000
30000
Emitter Resistance
15-36
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.54
[
]
First, use our gain estimates to check feasibility of the design : Av ≅ 10(VCC + VEE ) = 32,400
2
There is plenty of margin available.
Rout = R ro3 |
10−3 =
1
1 1 1
IC 3
1 1⎛ I R ⎞
= +
= +
= + ⎜ C3
⎟ | VO = 0V
Rout R ro3 R VA + VEC3 R R ⎝ VA + VEC3 ⎠
100(0.025V )
1⎛
9 ⎞
9
= 308Ω
⎜1+
⎟ → R = 1.11 kΩ | IC3 = = 8.11mA | rπ 3 =
8.11mA
R ⎝ 70 + 9 ⎠
R
Avt2 = −g m3 Rout = −40(8.11mA)(1kΩ)= −324 | Avt1 =
Avt1 = −
(
Av 2000
=
= −6.165
Avt2 −324
)
g m2
I Rr
I R
0.7
RC rπ 3 = −20 C 2 C π 3 = −20 C 2 C ≅ −20
neglecting IB3
RC
RC
2
RC + rπ 3
+1
+1
rπ 3
rπ 3
0.7
0.7V
0.7V 8.11mA
= −6.165 → RC = 391Ω | IC1 =
+ I B3 =
+
= 1.87mA
RC
391Ω
391Ω
100
+1
308
Selecting the closest 5% values : R = 1.1 kΩ , RC = 390 Ω , I1 = 3.74 mA
−20
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-37
15.55
(a) For VO = 0, IC 3 = I2 = 300µA VC 2 = 15 − 2400I E 3 − VEB3 = 15 − 0.729 − 0.7 = 13.6V
⎛ 200µA ⎞ 80 ⎛ 200µA ⎞
IC 3 300µA
IC1 = IC 2 = α F ⎜
=
= 3.75µA
⎟= ⎜
⎟ = 98.8µA | I B3 =
βF 3
80
⎝ 2 ⎠ 81 ⎝ 2 ⎠
VCE1 = VCE 2 = 13.6 − (−0.7)= 14.3V | VEC 3 = 15 − 2400I E3 − VO = 14.3V
Q - points : (98.8µA, 14.3V ) (98.8µA, 14.3V ) (300µA, 14.3V )
RC 2 =
15 −13.6 V
15 −13.6 V
= 14.7kΩ | For balance, RC1 =
= 14.2kΩ
98.8 µA
(98.8 − 3.75) µA
80(0.025V )
rπ 3 =
0.3mA
= 6.67kΩ | ro2 =
80V
= 810kΩ
98.8µA
⎛g ⎞
v c2
= −⎜ m1 ⎟ RC 2ro2 rπ 3 + (β o3 + 1)RE
v id
⎝ 2 ⎠
⎛ 40(98.8µA)⎞
⎟ 15.1kΩ 1.62 MΩ 6.67kΩ + 81(2.4kΩ) = −27.5
Avt1 = −⎜⎜
⎟
2
⎝
⎠
⎛
⎞
vo
β o3 RL
β o RE
70 + 14.3
⎜
⎟ | ro3 =
Avt2 =
=−
| RL = ro3⎜1+
= 281kΩ
⎟
300µA
vc2
rπ 3 + (β o3 + 1)RE
⎝ RC 2ro2 + rπ 3 + RE ⎠
⎛
⎞
80(2.4kΩ)
80(2.53MΩ)
⎟ = 2.53MΩ | Avt2 = −
= −1010
RL = 281kΩ⎜⎜1+
⎟
6.67kΩ + 81(2.4kΩ)
⎝ 14.8kΩ + 6.67kΩ + 2.4kΩ ⎠
(b) A
vt1
{
=
]}
[
[
)]
(
Av = Avt1 Avt2 = −27.5(−1010)= 27800 | Rid = 2rπ 1 = 2
β o1VT
IC1
=2
80(0.025V )
98.8µA
= 40.5kΩ
Rout = RL = 2.51 MΩ
15.56
The amplifier has an offset voltage of approximately 3.92 mV. Use this value to force the output
to nearly zero. A transfer function analysis then yields
Av = +28,627 Rout = 2.868 MΩ
Rin = +50.051 kΩ
These values are similar to the hand calculations in Prob. 15.55. Rin and Rout are larger because
the hand calculations did not adjust the value of current gain based upon the Early voltage.
15.57
⎛ 200µA ⎞ 100 ⎛ 200µA ⎞
IC1 = IC 2 = α F ⎜
⎟=
⎜
⎟ = 99.0µA | VCE1 = VCE2 = 15 − VEB 3 − (−VBE1 ) = 15V
⎝ 2 ⎠ 101 ⎝ 2 ⎠
For VO = 0, IC 3 = I2 = 300µA | VEC3 = 15 − VO = 15V
Q - points : (99.0µA, 15.0V ) (99.0µA, 15.0V ) (300µA, 15.0V )
15-38
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.58
{
]}
[
g m1
RC 2ro2 rπ 3 + (β o3 + 1)RE
2
⎡
⎤
β o Rout
βo
β
R
o
E
⎥
Avt2 = −
=−
ro3⎢1+
rπ 3 + (β o + 1)RE
rπ 3 + (β o + 1)RE ⎢⎣
RC 2ro2 + rπ 3 + RE ⎥⎦
Av = Avt1 Avt2 | Avt1 = −
(
)
I1 80 200µA
=
= 98.8µA | For VO = 0, IC3 = I2 = 300µA
2 81 2
300µA
81
=
= 3.75µA | I E3 =
IC 3 = 303.8µA | VEC 3 = 15 − I E 3 RE
80
80
IC1 = IC 2 = α F
IC3
I B3 =
βF 3
RC2 =
rπ 3 =
0.7V + I E3 RE 0.7V + (303.8µA)RE
=
IC1 − I B3
98.8µA − 3.75µA
80(0.025V )
300µA
= 6.67kΩ | ro3 =
g m1 = 40(98.8µA)= 3.95mS | ro2 =
70 + 15 − (303.8µA)RE
300µA
70 + 14.3 − (303.8µA)RE
98.8µA
Voltage Gain
30000
20000
10000
0
0
10000
20000
30000
40000
Emitter Resistance
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-39
15.59
(
)
2 2.5x10-4
500µA
2I D2
(a) I D2 = 2 = 250µA | VS = VSG = −VTP + K = 1+ 5x10-3 = 1.32V
p
VD = −15 + 0.7 = −14.3V | VDS = VD − VS = −14.3 −1.32 = −15.6V | Q - pt : (250µA, −15.6V )
IC3 = 500µA | VCE 3 = VC 3 - VE3 = 0 - (-15) = 15V | Q - pt : (500µA, 15V )
VBE
=
I D2 − I B3
RD =
(b) g
0.7V
= 2.87kΩ
500µA
250µA −
80
= 2(0.005)(0.00025) = 1.58x10−3 S | rπ 3 =
80(0.025V )
= 4kΩ
0.5mA
v v
g
1.58mS
Av = d2 o = Avt1 Avt2 | Avt1 = − m2 RD rπ 3 = −
2.87kΩ 4kΩ = −1.30
2
vid v d2
2
⎛ 75V + 15V
⎞
Avt2 = −g m3 ro3 R2 = −40(0.5mA)⎜
2 MΩ⎟ = −0.02 180kΩ 2 MΩ = −3300
⎝ 0.5mA
⎠
m2
(
(
)
)
(
)
(
)
Av = −1.30(−3300) = 4300 | Rin = ∞ | Rout = ro3 R2 = 180kΩ 2 MΩ = 165kΩ
(c) v
2
15-40
is the non - inverting (+) input
(d ) v
1
is the inverting (-) input
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.60
Note that the parameters of the transistors and values of RD have been carefully adjusted to
permit open-loop operation and achieve VO = 0.
*Problem 15.60 – Figure P15.59
VCC 7 0 DC 15
VEE 8 0 DC -15
V1 1 9 AC 0.5
V2 3 9 AC -0.5
VIC 9 0 DC 0
I1 7 2 DC 493.2U
R1 7 2 2MEG
M1 4 1 2 2 PFET
M2 5 3 2 2 PFET
RD1 4 8 2.863K
RD2 5 8 2.863K
Q3 6 5 8 NBJT
I2 7 6 DC 492.5U
R2 7 6 2MEG
.MODEL PFET PMOS KP=5M VTO=-1
.MODEL NBJT NPN BF=80 VA=75 IS=0.2881FA
.OP
.AC LIN 1 1000 1000
.TF V(6) VIC
.PRINT AC IM(V1) IP(V1) VM(6) VP(6)
.OPTIONS TNOM=17.2
.END
Adm = VM(6) = 4630 | Acm = −1.46 | CMRRdB = 70.0 dB
1
Results: R =
= ∞ | Rout = 164 kΩ
id
IM(V1)
15.61
v v
g
Av = d2 o = Avt1 Avt2 | Avt1 = − m2 RD rπ 3
vid v d2
2
(
)|
IC3 = 100µA | rπ 3 =
80(0.025V )
100µA
= 20kΩ
500µA
= 250µA | g m2 = 2(0.005)(0.00025) = 1.58x10−3 S
2
VBE
0.7V
1.58mS
RD =
=
= 2.81kΩ | Avt1 = −
2.81kΩ 20kΩ = −1.95
100µA
I D2 − I B3
2
250µA −
80
⎛ 75V + 5V
⎞
Avt2 = −g m3 ro3 R2 = −40(100µA)⎜
10 MΩ⎟ = −0.02 800kΩ 10 MΩ = −2960
⎝ 100µA
⎠
I D2 =
(
(
)
(
)
)
Av = −1.95(−2960)= 5770
15.62
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-41
VGS 2 = VTP −
2I D
5x10 4
= −1−
= −1.316V
KP
5x103
For PMOS active region : VDS 2 ≤ VGS 2 − VTP = −0.316V
vic ≥ −VEE + VBE3 − VDS 2 + VGS 2 = −15 + 0.7 + 0.316 −1.316 = −15.3V
vic ≤ VCC − 0.75 + VGS 2 = 15 − 0.75 −1.316 = 12.9V | −15.3 V ≤ vic ≤ 12.9 V
15.63
(
)
2 2.5x10-4
500µA
2I D2
(a) I D2 = 2 = 250µA | VS = −VGS = −VTP + K = −1+ 5x10-3 = 1.32V
p
VD = −5 + 0.7 + 0.7 = −3.6V | VDS = VD − VS = −3.6 −1.32 = −4.92V | Q - pt : (250µA, - 4.92V )
IC3 + IC 4 = 500µA | IC3 + β F I E3 = (β F + 2)IC 3 = 500µA → IC 3 = 6.10µA | IC 4 = 494µA
For VO = 0, VCE 4 = 5V and VCE3 = 5 - 0.7 = 4.30V
Q - pts : (250µA, - 4.92V ) (250µA, - 4.92V ) (6.10µA, 4.30V ) (494µA, 5.00V )
VBE3 + VBE 4
=
I D2 − I B 3
1.4V
= 5.60kΩ | Based upon results for the Darlington
6.10µA
250µA −
80
80(0.025V )
g
circuit in Prob. 15.48 : Avt1 = m1 RD 2rπ 3 | rπ 3 =
= 328kΩ
2
6.10µA
1.58mS
g m1 = 2(0.005)(0.00025) = 1.58mS | Avt1 = −
5.60kΩ 656kΩ = −4.39
2
⎤
⎞
40(494µA)⎡2 ⎛ 75V + 5V ⎞
g ⎛2
Avt2 = − m4 ⎜ ro4 R2 ⎟ = −
⎢ ⎜
⎟ 1MΩ⎥ = −9.88mS 108kΩ 1MΩ = −963
2 ⎝3
2
⎠
⎣ 3 ⎝ 494µA ⎠
⎦
⎛2 ⎞
Av = −4.39(−963)= 4230 | Rid = ∞ | Rout = ⎜ ro4 ⎟ R2 = 108kΩ 1MΩ = 97.5 kΩ
⎝3 ⎠
RD =
(
)
(
)
(
15.64
*Problem 15.64 – Figure P15.63
*Vos (the dc value of V2) has been carefully adjusted to set Vo ≈ 0
VCC 8 0 DC 5
VEE 9 0 DC -5
V1 1 10 AC 0.5
V2 3 10 DC 1.21M AC -0.5
VIC 10 0 DC 0
I1 8 2 DC 496.3U
R1 8 2 1MEG
M1 4 1 2 2 PFET
M2 5 3 2 2 PFET
RD1 4 9 5.6K
RD2 5 9 5.6K
15-42
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
)
Q3 7 5 6 NBJT
Q4 7 6 9 NBJT
I2 8 7 DC 495U
R2 8 7 1MEG
.OP
.MODEL PFET PMOS KP=5M VTO=-1
.MODEL NBJT NPN BF=80 VA=75
.AC LIN 1 1000 1000
.TF V(7) VIC
.PRINT AC IM(V1) IP(V1) VM(7) VP(7)
.END
Adm = VM(7) = 4080 | Acm = −2.58 | CMRRdB = 64.0 dB
1
Results: R =
= ∞ | Rout = 96.2 kΩ
id
IM(V1)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-43
15.65
+V
R
C
R
CC
C
Q
3
v
1
v
Q
Q
1
Q4
2
2
Q5
vO = 0
I
I1
R
I
2
L
3
2kΩ
-V EE
The new output stage can be treated as an improved single transistor using
the results from Prob. 15.48. Using the results from Ex. 15.5:
Rin4-5 = 2β orπ 4 + β o2 RL = 2(100)(505Ω)+ 1002 (2kΩ)= 20.1MΩ | Av2 becomes
(
)
(
)
Av2 = −g m3 ro3 Rin4−5 = −22mS 161kΩ 20.1MΩ = −3510 | The gain of the
g m4 RL
40 4.95x10−3 (2kΩ)
g m4 RL
2
emitter follower becomes Av3 ≅
=
=
0 = 0.995
g m4 RL 2 + g m4 RL 2 + 40 4.95x10−3 (2kΩ)
1+
2
Av = −3.50(−3510)(0.995)= 12200.
(
(
CMRR and Rid do not change : CMRR = 63.5 dB and Rid = 101 kΩ
Rout =
15-44
2
r
2
161kΩ
+ o32 =
+
= 26.2 Ω
−3
g m4 β o 40 4.95x10
10 4
(
)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
)
)
15.66
+V
R
C
R
CC
C
Q
3
v
1
v
Q
Q
1
Q4
2
2
Q5
vO = 0
I
I1
R
I
2
L
3
2kΩ
-V EE
The new output stage can be treated as an improved single transistor using the equation
set from prob. 15.48 and the results from Ex. 15.4
Rin4-5 = 2β orπ 4 + β o2 RL = 2(100)(505Ω)+ 1002 (2kΩ) = 20.1MΩ | Av2 becomes
(
)
(
)
Av2 = −g m3 ro3 Rin4−5 = −22mS 161kΩ 20.1MΩ = −3510 | The gain of the
g m4 RL
40 4.95x10−3 (2kΩ)
g m4 RL
2
emitter follower becomes Av3 ≅
=
=
0 = 0.995
g m4 RL 2 + g m4 RL 2 + 40 4.95x10−3 (2kΩ)
1+
2
Av = −3.50(−3510)(0.995)= 12200.
(
(
)
)
CMRR and Rid do not change : CMRR = 63.5 dB and Rid = 101 kΩ
Rout =
2
r
2
161kΩ
+ o32 =
+
= 26.2 Ω
−3
g m4 β o 40 4.95x10
10 4
(
)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-45
15.67
I1 100 ⎛ 100µA ⎞
=
⎜
⎟ = 49.5µA | VEC 2 = +0.7V − (−15V + 0.7V )= 15.0V
2 101 ⎝ 2 ⎠
⎛100 ⎞
For VO = 0, IC 4 = α F I3 = ⎜
⎟1.00mA = 990µA | VEC 4 = 0 − (−15V )= 15.0V
⎝ 101 ⎠
1mA
IC3 = I2 + I B 4 = 350µA +
= 360µA | VCE3 = VO − 0.7V − (−15) = 14.3
101
Q − pts : (49.5µA, 15.0V ) (49.5µA, 15.0V ) (360µA, 14.3V ) (990µA, 15.0V )
(a) I
C1
(b) R
C
= IC 2 = α F
=
0.7V
0.7V
=
= 15.3kΩ
IC 2 − I B3 (49.5 − 3.60)µA
100(0.025V )
50V
= 1.01MΩ | rπ 3 =
= 6.94kΩ
49.5µA
360µA
50 + 14.3
g
g
ro3 =
= 179kΩ | Av = Avt1 Avt2 Avt3 = m1 RC 2ro2 rπ 3 (g m3ro3 )()
1 = m1 RC 2ro2 rπ 3 µ f 3
2
2
360µA
ro2 =
(
)
(
40(49.5µA)
Av =
(15.3kΩ 2.02 MΩ 6.94kΩ)(40)(64.3)= 12100
2
Rid = 2rπ 1 = 2
(c) R
out
(d ) R
ic
15-46
=
=
100(0.025V )
49.5µA
ro3 + rπ 4
β o4 + 1
(β
o1
| rπ 4 =
+ 1)ro1
2
= 101 kΩ
100(0.025V )
| ro1 =
990µA
= 2.53kΩ | Rout =
179kΩ + 2.53kΩ
= 1.80 kΩ
101
101(1.31MΩ)
50V + 15V
1.31MΩ | Ric =
= 66.3 MΩ (e) v2
49.5µA
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
)
15.68
*Problem 15.68 – Figure P15.67
*RC and Vos (see V2) have been carefully adjusted to set Vo ≈ 0
VCC 7 0 DC 15
VEE 8 0 DC -15
V1 1 9 DC 0.117M AC 0.5
V2 3 9 AC -0.5
VIC 9 0 DC 0
I1 7 2 DC 100U
Q1 4 1 2 PBJT
Q2 5 3 2 PBJT
RC1 4 8 15.8K
RC2 5 8 15.8K
Q3 6 5 8 NBJT
I2 7 6 DC 350U
Q4 8 6 10 PBJT
I3 7 10 DC 1M
.MODEL PBJT PNP BF=100 VA=50
.MODEL NBJT NPN BF=100 VA=50
.NODESET V(10)=0
.OP
.AC LIN 1 1000 1000
.TF V(10) VIC
.PRINT AC IM(V1) IP(V1) VM(10) VP(10)
.END
Adm = VM(10) = 13800 | Acm = −0.0804 | CMRRdB = 105 dB
1
Results: R =
= 133 kΩ | Rout = 1.37 kΩ
id
IM(V1)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-47
15.69
(a) Working backwards from the output :V = V − V = 12 − 0 = 12.0V | I = I = 5.00mA
2(0.005)
2I
V =V +
= 0.75 +
= 2.16V | V = −(V − V )= −(12 − 2.16)= −9.84V
K
0.005
DS 4
D4
GS 4
DD
DS 3
TN
D4
O
DD
3
GS 4
n
I D3 = I2 = 2.00mA | VGS 3 = VTP −
2(0.002)
2I D 4
= −0.75V −
= −2.16V
Kn
0.002
VD2 = VDD + VGS 3 = 12V − 2.16V = 9.84V | I D1 = I D2 =
(
2 2.5x10−4
) = 1.07V
I1
= 250µA
2
| VDS1 = VDS 2 = 9.84V − (−1.07V )= 10.9V
5x10
Q − pts : (250µA,10.9V) (250µA,10.9V) (2.00mA,-9.84V) (5.00mA,12.0V)
VGS 2 = 0.75V +
(b) A
dm
=
−3
µf 4
g r
g m1
g
2.16V
RD (g m3ro3 ) m4 o4 = m1 RD µ f 3
| RD =
= 8.64kΩ
2
1+ g m4ro4
2
1+ µ f 4
0.25mA
(
g m1 = 2 5x10−3
)(
Adm =
15-48
)[
]
( )( )[
]
2(5x10 )(5x10 )[1+ 0.02(12)] = 7.87mS
g m3 = 2 2x10−3
g m4 =
1
+ 9.84
−4
0.015
2.5x10 1+ 0.02(10.9) = 1.75mS | ro3 =
= 38.3kΩ
2mA
1
+ 12
2x10−3 1+ 0.015(9.84) = 3.03mS | ro4 = 0.02
= 12.4kΩ
5mA
−3
−3
| µ f 3 = g m3ro3 = 116 | µ f 4 = 97.6
1.75ms
1
97.6
8.64kΩ)(116)
= 868 | Rid = ∞ | Rout =
= 127Ω
(
2
g m4
1+ 97.6
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.70
*Problem 15.70 – Figure P15.69
*The values of RD have been adjusted to bring the offset voltage to ≈ 0
VCC 8 0 DC 12
VEE 9 0 DC -12
V1 1 10 AC 1
V2 3 10 AC 1
VIC 10 0 DC 0
I1 2 9 DC 500U
M1 4 1 2 2 NFET
M2 5 3 2 2 NFET
RD1 8 4 8.28K
RD2 8 5 8.28K
M3 6 5 8 8 PFET
M4 8 6 7 7 NFET
I2 6 9 DC 2M
I3 7 9 DC 5M
.MODEL PFET PMOS KP=2M VTO=-0.75 LAMBDA=0.015
.MODEL NFET NMOS KP=5M VTO=0.75 LAMBDA=0.02
.OP
.AC LIN 1 1000 1000
.TF V(7) VIC
.PRINT AC VM(7) VP(7) IM(V1) IP(V1)
.END
Adm = VM(7) = 802 | Acm = −4.74 x10 -7 ≅ 0 | CMRRdB = ∞
1
Results:
Rid =
= 10 30 ≅ ∞ | Rout = 126 Ω
IM(V1)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-49
15.71
(a) Working backwards from the output :VDS 4 = −(VO − VSS )= 0 + (−5)= −5.00V
I D 4 = I3 = 2.00mA | VGS 4 = VTP −
2(0.002)
2I D 4
= −0.7 −
= −2.11V
Kp
0.002
VDS 3 = VO + VGS 4 − (−VSS ) = 0 − 2.11+ 5 = 2.89V
(
)
2 5x10−4
2I D 4
= 0.75V +
= 1.15V
I D3 = I2 = 500µA | VGS 3 = VTN +
Kn
5x10−3
VD2 = −VSS + VGS 3 = −5 + 1.15V = −3.85V | I D1 = I D2 =
(
2 3x10−4
) = −1.25V
I1
= 300µA
2
[
]
| VDS 2 = VDS 2 = − 1.25 − (3.85) = −5.10V
2x10
Q − pts : (300µA,−5.10V ) (300µA,−5.10V ) (500µA,2.89V ) (2.00mA,5.00V )
VGS1 = VGS 2 = −0.7V −
(b) A
dm
=
−3
µf 4
g m1
g
1.15V
g r
RD (g m3ro3 ) m4 o4 = m1 RD µ f 3
| RD =
= 3.83kΩ
2
1+ g m4ro4
2
1+ µ f 4
0.3mA
(
)(
g m1 = 2 2x10−3 3x10−4
)[
Adm =
]
( )( )[
]
2(2x10 )(2x10 )[1+ 0.015(15)] = 2.93mS
g m3 = 2 5x10−3 5x10−4
g m4 =
1
+ 2.89
0.02
1+ 0.015(5.10) = 1.14mS | ro3 =
= 106kΩ
0.5mA
1
+ 5.00
1+ 0.02(2.89) = 2.30mS | ro4 = 0.015
= 35.8kΩ
2mA
−3
−3
| µ f 3 = g m3ro3 = 244 | µ f 4 = 105
1.14ms
1
105
3.83kΩ)(244)
= 528 | Rid = ∞ | Rout =
= 341Ω
(
2
g m4
1+ 105
15.72
The amplifier has an offset voltage of approximately –6.69 mV. This value is used to force the
output to nearly zero. A transfer function analysis then yields
Av = +517 Rout = 339 Ω
Rin = +1.00 x 1020 Ω
These values are similar to the hand calculations in Prob. 15.71.
15-50
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.73
(a) Working backwards from the output :VDS 4 = −(VO − VSS )= 0 + (−5)= −5.00V
I D 4 = I3 = 2.00mA | VGS 4 = VTP −
2(0.002)
2I D 4
= −0.7 −
= −2.11V
Kp
0.002
VCE3 = VO + VGS 4 − (−VSS )= 0 − 2.11+ 5 = 2.89V
IC3 = I2 = 500µA | VD2 = −VSS + VBE3 = −5 + 0.7V = −4.30V | I D1 = I D2 =
(
2 3x10−4
) = −1.25V
[
I1
= 300µA
2
]
| VDS1 = VDS 2 = − 1.25 − (−4.30) = −5.55V
2x10
Q − pts : (300µA,−5.55V ) (300µA,−5.55V ) (500µA,2.89V ) (2.00mA,5.00V )
VGS1 = VGS 2 = −0.7V −
(b) A
dm =
RD =
−3
⎛ g r ⎞ g
µf 4
g m1
RD rπ 3 (g m3ro3 )⎜ m4 o4 ⎟ = m1 RD rπ 3 µ f 3
2
1+ µ f 4
⎝ 1+ g m4ro4 ⎠ 2
(
)
(
)
150(0.025V )
0.7V
= 2.33kΩ | rπ 3 =
= 7.53kΩ
0.3mA
500µA
1
+ 2.89
g m1 = 2 2x10−3 3x10−4 1+ 0.015(5.10) = 1.14mS | ro3 = 0.02
= 106kΩ
0.5mA
1
+ 5.00
= 35.8kΩ
g m4 = 2 2x10−3 2x10−3 1+ 0.015(15) = 2.93mS | ro4 = 0.015
2mA
µ f 3 = g m3ro3 = 40(70) = 2800 | µ f 4 = 105
Adm =
(
)(
)[
(
)(
)[
(
]
]
)
1.14ms
1
105
2.33kΩ 7.53kΩ (2800)
= 2810 | Rid = ∞ | Rout =
= 341Ω
2
g m4
1+ 105
15.74
The amplifier has an offset voltage of approximately 48.69 mV. This value is used to force the
output to nearly zero. A transfer function analysis then yields
Av = +2810
Rout = 339 Ω
Rin = +1.00 x 1020 Ω
These values are similar to the hand calculations in Prob. 15.73
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-51
15.75
(a) Working backwards from the output with VO = 0 : VDS 4 = VCC − VO = 5 − 0 = 5V
ID 4 = I3 = 2mA | VGS 4 = VTN +
2(0.002)
2ID 4
= 0.7 +
= 1.59V | IC 3 = I2 = 500µA
Kn
0.005
VEC 3 = 5 − VGS 4 = 5 −1.59 = 3.41V | VCE 2 = 5 − VEB 3 − (−VBE 2 ) = 5 − 0.7 + 0.7 = 5.00V
IC1 = IC 2 = α F
VEB 3
0.7V
I1 100 200µA
=
= 99.0µA | RC =
=
= 7.45kΩ
2 101 2
IC 2 − IB 3 (99.0 − 5.00)µA
VCE1 = 5 − IC1RC − (−VBE 2 ) = 5 − 99.0µA(7.45kΩ) + 0.7 = 4.96V
Q = pts : (99.0µA, 4.96V ) (99.0µA, 5.00V ) (500µA, 3.41V ) (2.00mA, 5.00V )
(b) Using current division at the collector of Q2 :
Adm =
rπ 3 =
⎛ gm 4 RL ⎞ gm1
gm1 ⎛ RC ⎞
g R
RC rπ 3 )µ f 3 m 4 L
(
⎜
⎟β o3 ro3 ⎜
⎟=
2 ⎝ RC + rπ 3 ⎠
1+ gm 4 RL
⎝ 1+ gm 4 RL ⎠ 2
100(0.025V )
= 5.00kΩ | gm 4 = 2(0.005)(0.002) = 4.47mS
500µA
Adm =
40(99.0µA)
4.47mS (2kΩ)
7.45kΩ 5.00kΩ)(40)(50 + 3.41)
= 11400
(
2
1+ 4.47mS (2kΩ)
Rid = 2rπ 1 = 2
100(0.025V )
1
= 50.5 kΩ | Rout =
= 224 Ω
gm 4
99.0µA
(c)
*Problem 15.75 – Figure P15.75
*The values of RC have been adjusted to set Vo ≈ 0.
VCC 8 0 DC 5
VEE 9 0 DC -5
VIC 10 0 DC 0
V1 1 10 AC 0.5
V2 3 10 AC -0.5
I1 2 9 DC 200U
Q1 4 1 2 NBJT
Q2 5 3 2 NBJT
RC1 8 4 8.00K
RC2 8 5 8.00K
Q3 6 5 8 PBJT
I2 6 9 DC 500U
M4 8 6 7 7 NFET
I3 7 9 DC 2M
RL 7 0 2K
.MODEL NBJT NPN BF=100 VA=50
.MODEL PBJT PNP BF=100 VA=50
.MODEl NFET NMOS KP=5M VTO=0.70
.OP
.AC LIN 1 2KHZ 2KHZ
15-52
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
.PRINT AC VM(7) VP(7) IM(V1) IP(V1)
.TF V(7) VIC
.END
Adm = VM(7) = 11200 | Acm = −0.0957 | CMRRdB = 101 dB
1
Results: R =
= 56.4 kΩ ≅ ∞ | Rout = 201 Ω
id
IM(V1)
15.76
(a) I
C1
I1 100 10µA
I
50 50µA
=
= 4.95 µA | IC 3 = IC 4 = α F 2 =
= 24.5 µA
2 101 2
2 51 2
⎛
24.5 µA ⎞
− (IC 2 − I B3 )RC − (−VBE 2 )= 3V − ⎜4.95µA −
⎟300kΩ − (−0.7)= 2.36V
50 ⎠
⎝
= IC2 = α F
VCE2 = VCC
50
(250µA)= 245µA | VEC 5 = 3.00 V | VC 4 = −0.7V
51
⎛
24.5 µA ⎞
VC1 = 3 − ⎜ 4.95µA −
⎟300kΩ = 1.66V | VEC 3 = VEC 4 = 1.66 + 0.7 − (−0.7)= 3.06V
50 ⎠
⎝
For VO = 0 : IC5 = α F I3
Q - pts : (4.95µA,2.36V ) (4.95µA,2.36V ) (24.5µA,3.06V ) (24.5µA,3.06V ) (245µA,3.00V )
(b) A
dm
rπ 3 =
ro1 =
(
= g m1 RC1 rπ 3 ro1
50(0.025V )
24.5µA
) (
[
g m3
RC 2 2ro4 rπ 5 + (β o5 + 1)RL
2
= 51.0kΩ | ro4 =
⎛ (β + 1)R ⎞
o5
L
⎟
⎜
⎜r + β +1 R ⎟
⎝ π 5 ( o5 ) L ⎠
])
50(0.025V )
70
= 2.86 MΩ | rπ 5 =
= 5.10kΩ
245µA
24.5µA
(β o5 + 1)RL = 51(5kΩ) = 0.980
50
= 10.1MΩ |
4.95µA
rπ 5 + (β o5 + 1)RL 5.10kΩ + 51(5kΩ)
(
)
Adm = 40(4.95µA) 300kΩ 51.0kΩ 10.1MΩ •
(40)(24.5µA) 78kΩ 5.72 MΩ
(
2
Rid = 2rπ 1 = 2
100(0.025V )
4.95µA
= 1.01 MΩ | Rout =
[5.10kΩ + 51(5kΩ)])0.980 = 235
RC 2 + rπ 5 78kΩ 5.72 MΩ + 5.10kΩ
=
= 1.59 kΩ
β o5 + 1
51
(c) v is the non - inverting input - v is the inverting input
(d) A = (10V )(10V )= 30 = 900 | r << R is substantailly reducing the gain
A
B
2
v
CC
CC
π3
C
Also, the input resistance of the emitter follower is low, Rin5 = 107kΩ ≈ RC 2 , and
is reducing the gain by an additional factor of almost 2.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-53
15.77
I1 100 ⎛ 100µA ⎞
I2 50 ⎛ 200µA ⎞
=
⎜
⎟ = 49.5 µA | IC 3 = IC 4 = α F = ⎜
⎟ = 98.0 µA
2 101 ⎝ 2 ⎠
2 51 ⎝ 2 ⎠
⎛
98.0 µA ⎞
VCE2 = VCC − (IC 2 − I B3 )RC − (−VBE2 )= 18V − ⎜49.5µA −
⎟120kΩ − (−0.7)= 13.0V
50 ⎠
⎝
50
For VO = 0 : IC 5 = α F I3 (750µA) = 735µA | VEC 4 = 18 V | VC 4 = −0.7V
51
⎛
98.0 µA ⎞
VC1 = 18 − ⎜ 49.5µA −
⎟120kΩ = 12.3V | VEC 3 = VEC 4 = 12.3 + 0.7 − (−0.7) = 13.7V
50 ⎠
⎝
(a) I
C1
= IC 2 = α F
Q - pts : (49.5µA,13.0V ) (49.5µA,13.0V ) (98.0µA,13.7V ) (98.0µA,13.7V ) (735µA,18.0V )
(
(b) A
dm = g m1 RC1 rπ 3 ro2
rπ 3 =
ro2 =
50(0.025V )
98.0µA
(β o5 + 1)RL
g m3
RC2 2ro4 rπ 5 + (β o5 + 1)RL
2
rπ 5 + (β o5 + 1)RL
) (
])
[
= 12.8kΩ | ro4 =
50(0.025V )
70V
= 714kΩ | rπ 5 =
= 1.70kΩ
98.0µA
735µA
(β o5 + 1)RL = 51(2kΩ) = 0.984
50V
= 1.01MΩ |
49.5µA
rπ 5 + (β o5 + 1)RL 1.70kΩ + 51(2kΩ)
(
)
Adm = 40(49.5µA) 120kΩ 12.8kΩ 1.01MΩ •
(40)(98.0µA) 170kΩ 1.43MΩ
Rid = 2rπ 1 = 2
2
100(0.025V )
49.5µA
(c) For positive v
IC
(
= 101 kΩ | Rout =
[1.70kΩ + 51(2kΩ)])0.984 = 2700
RC 2 + rπ 5 170 + 1.70
=
kΩ = 3.37 kΩ
51
β o5 + 1
, vIC ≤ VC1 = 12.3V | For negative v IC , the characteristics of I1 will
deterimine vIC . For the ideal current source, the negative limit of v IC is not defined.
(d) The actual voltage at the collector of Q
4
would be
⎛
735µA ⎞
VC 4 = −18V + (IC 4 + I B 5 )RC2 = −18V + ⎜ 98.0µA +
⎟170kΩ = 1.16V and VO = +1.86V
50 ⎠
⎝
should be - 0.7V. The value of offset voltage required to bring the output back to zero is
∆V 1.86V
VOS = O =
= 0.689 mV.
2700
Adm
15-54
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.78
(a) I
I1 100 ⎛ 200µA ⎞
100 ⎛ 12V ⎞
=
⎜
⎟ = 99.0µA | For VO = 0 : IC 3 = α F I E3 =
⎜
⎟ = 990µA
2 101 ⎝ 2 ⎠
101 ⎝ 12kΩ ⎠
C1 = I C 2 = α F
VCE3 = 12V − 0V = 12.0V | VCE 2 = +0.7 − (−0.7)= 1.4V
RC =
12V − 0.7V
0.7V
=
= 104 kΩ | VCE1 = 12 − 99.0µA(104kΩ)− (−0.7) = 2.40V
IC 2 + I B 3
(99.0 + 9.90)µA
Q - points : (99.0µA,2,40V ) (99.0µA,1.40V ) (990µA,12V )
(
[
g
(b) Adm = 2m2 RC 2ro2 rπ 3 + (β o3 + 1)R
ro2 =
⎛ (β + 1)R ⎞
o3
⎜
⎟
⎜r + β +1 R⎟
⎝ π 3 ( o3 ) ⎠
])
100(0.025V )
70V
= 707kΩ | rπ 3 =
= 2.53kΩ
990µA
99.0µA
Adm =
40(99.0µA)
2
Rid = 2rπ 1 = 2
(104kΩ 1.41MΩ [2.53kΩ + (101)12kΩ])2.53kΩ( +)(101)12kΩ = 177
101 12kΩ
100(0.025V )
99.0µA
15.79
= 50.5 kΩ | Rout =
RC 2ro2 + rπ 3
β o3 + 1
=
96.9kΩ + 2.53kΩ
= 984 Ω
101
100(0.025V )
= 16.7µA. For VO = 0, VC1 = 0.7V
300kΩ
12 − 0.7 11.3V 11.3V
R +r
677kΩ
≅
=
= 677kΩ | Rout ≅ C π 3 ≥
= 6.7kΩ
RC =
IC1 + I B3
IC1
16.7µA
101
β o3 + 1
300kΩ = 2rπ 1 → IC1 = 2
The Rout specification cannot be meet if the Rid specification is met and vice - versa.
Either Rid must be reduced or Rout must be increased, or both must be changed.
15.80
100(0.025V )
= 5.00µA | For VO = 0, VC1 = 0.7V
1MΩ
9 − 0.7 8.3V
8.3V
R +r
1.66 MΩ
≅
=
= 1.66 MΩ | Rout ≅ C π 3 ≥
= 16.4kΩ
RC =
IC1 + I B3
IC1
5.00µA
101
β o3 + 1
1MΩ = 2rπ 1 → IC1 = 2
The Rout specification cannot be meet if the Rid specification is met and vice - versa.
Either Rid must be reduced or Rout must be increased, or both must be changed.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-55
15.81
(a) For V
O
= 0, IC 6 = α F I E 6 = α F I3 =
IC5 = α F I E 5 = α F
βF 4
IC3
αF3
IC 6
βF 6
=
100
(5mA)= 4.95mA
101
4.95mA
49.0µA
= 49.0µA | IC 4 + IC 3 = I2 + I B5 = 500µA +
= 500µA
100
101
+ IC 3 = (β F 3 + 2)IC 3 = 500µA → IC3 = 9.62µA | IC 4 = 500µA − IC3 = 490µA
IC1 = IC 2 = α F
I1 100 ⎛ 50µA ⎞
=
⎜
⎟ = 24.8µA | VCE6 = 18 − 0 = 18V | VCE 5 = VCE 6 − VBE6 = 17.3V
2 101 ⎝ 2 ⎠
VEC 4 = 18 − VBE5 − VBE6 = 16.6V | VEC 3 = VEC 4 − VEB 4 = 15.9V
VCE1 = VCE2 = 18 − VEB 4 − VEB3 − (−VEB2 )= 17.3V
Q − pts : (24.8µA,17.3V) (24.8µA,17.3V) (9.62µA,15.9V) (490µA,16.6V)
(49.0µA,17.3V ) (4.95mA,18.0V )
| RC =
1.4V
1.4V
=
= 56.9kΩ
IC 2 − I B 3 ⎛
9.62 ⎞
⎜24.8 +
⎟µA
50 ⎠
⎝
(b) Using the properties of the Darlington configuration from Prob. 15.48 :
Adm =
⎞ β o5β o6 RL
g ⎛2
g m2
RC Rin3 m4 ⎜ ro4 Rin5 ⎟
2
2 ⎝3
⎠ 2rπ 5 + β o5β o6 RL
(
)
50(0.025V )
Rin3 ≅ 2β o3rπ 4 = 2(50)
= 255kΩ
490µA
100(0.025V )
+ 100(100)(2kΩ)= 20.1MΩ
Rin5 ≅ 2β o5rπ 6 + β o5β o6 RL = 2(100)
4.95mA
40(24.8µA)
µ ⎛ 20 MΩ ⎞
70V + 16.6V
Rin5 >> ro4 =
= 177kΩ | Adm =
56.9kΩ 255kΩ f 4 ⎜
⎟
2
3 ⎝ 20.1MΩ ⎠
490µA
(
µf 4
3
Rout
=
40(70 + 11.6)
3
= 1155 | Adm = 26500 or (88.5dB) | Rid = 2rπ 1 = 2
100(0.025V )
2
2
118kΩ + 2
ro4 + 2rπ 5
R + 2rπ 5 3
49.0µA
3
= th5
=
=
= 18.1 Ω
β o5β o6
β o5β o6
100(100)
15-56
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
)
100(0.025V )
24.8µA
= 202 kΩ
15.82
(a) For V
O
= 0, IC 6 = α F I E 6 = α F I3 =
IC5 = α F I E 5 = α F
βF 4
IC3
αF3
IC 6
βF 6
=
100
(5mA)= 4.95mA
101
4.95mA
49.0µA
= 49.0µA | IC 4 + IC 3 = I2 + I B5 = 500µA +
= 500µA
100
101
+ IC 3 = (β F 3 + 2)IC 3 = 500µA → IC3 = 9.62µA | IC 4 = 500µA − IC3 = 490µA
IC1 = IC 2 = α F
I1 100 ⎛ 50µA ⎞
=
⎜
⎟ = 24.8µA | VCE6 = 22 − 0 = 22V | VCE5 = VCE6 − VBE6 = 21.3V
2 101 ⎝ 2 ⎠
VEC 4 = 22 − VBE5 − VBE6 = 20.6V | VEC 3 = VEC 4 − VEB 4 = 19.9V
VCE1 = VCE2 = 22 − VEB 4 − VEB3 − (−VEB2 ) = 21.3V
Q − pts : (24.8µA,21.3V ) (24.8µA,21.3V ) (9.62µA,19.9V ) (490µA,20.6V )
(49.0µA,21.3V ) (4.95mA,22.0V )
| RC =
1.4V
1.4V
=
= 56.9kΩ
IC2 − I B 3 ⎛
9.62 ⎞
⎜24.8 +
⎟µA
50 ⎠
⎝
(b) Using the properties of the Darlington configuration from Prob. 15.48 :
⎞⎤⎡ β o5β o6 RL ⎤
⎡g
⎤⎡ g ⎛ 2
Adm = AV1 AV 2 AV 3 = ⎢ m2 RC Rin3 ⎥⎢ m4 ⎜ ro4 Rin5 ⎟⎥⎢
⎥
⎣ 2
⎦⎣ 2 ⎝ 3
⎠⎦⎣ 2rπ 5 + β o5β o6 RL ⎦
(
)
50(0.025V )
= 255kΩ
Rin3 ≅ 2β o3rπ 4 = 2(50)
490µA
100(0.025V )
+ 100(100)(2kΩ)= 20.1MΩ
Rin5 ≅ 2β o5rπ 6 + β o5β o6 RL = 2(100)
4.95mA
40(24.8µA)
µ ⎛ 20 MΩ ⎞
70V + 20.6V
Rin5 >> ro4 =
= 185kΩ | Adm =
56.9kΩ 255kΩ f 4 ⎜
⎟
2
3 ⎝ 20.1MΩ ⎠
490µA
(
µf 4
3
Rout
=
40(70 + 20.6)
3
)
= 1208 | Adm = 27700 (88.9dB) | Rid = 2rπ 1 = 2
100(0.025V )
2
2
185kΩ + 2
ro4 + 2rπ 5
R + 2rπ 5 3
49.0µA
3
= th5
=
=
= 22.5 Ω
β o5β o6
β o5β o6
100(100)
100(0.025V )
24.8µA
= 202 kΩ
15.83
Since the transistor parameters are the same, VGS1 = −VGS 2 =
I D2 = I D1 =
2.2V
= 1.1V
2
2
6x10−4
1.1− 0.75) = 36.8 µA
(
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-57
15.84
2I D1 ⎛
2I D2 ⎞
⎜
⎟ where I D2 = I D1
− VTP −
2.2V = VGS1 − VGS 2 = VTN +
Kn ⎜⎝
K p ⎟⎠
⎛
2 ⎞
0.7
2
⎟ |
2.2 = 0.7 + 0.8 + I D1 ⎜⎜
+
→ I D2 = I D1 = 29.7µA
I D1 =
−4
−4 ⎟
128.5
4x10 ⎠
⎝ 6x10
15.85
Since the values of IS and IE are the same, VBE1 = VEB 2
I
1.30
1.30 = VBE1 + VEB 2 = 2VT ln C | IC = 10−15 exp
= 196 µA
IS
2(0.025V )
15.86
IC
IC
IC2
+ VT ln
= VT ln
1.30 = VBE1 + VEB2 = VT ln
I S1
IS 2
I S1 I S 2
IC =
15-58
1.30
= 391 µA
(4x10 )(10 )exp 0.025
−15
−15
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.87
vO
Slope = 1
+0.7 V
vS
-0.7 V
15.88
*Problem 15.88 – Figure P15.87
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 1.3
VS 1 0 DC 0
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 1K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.DC VS -10 +8.7 0.01
.PROBE
.END
10V
vO
0V
vS
-10V
-10V
-5V
0V
5V
10V
15.89
Since the base currents are zero (β F = ∞), VBE1 + VEB 2 = (250µA)(5kΩ) = 1.25V
1.25V = VT ln
IC
I
I2
+VT ln C = VT ln C
IS1
IS 2
IS1IS 2
| IC =
1.25
= 22.8 µA
(10 )(10 )exp 0.025
−15
−16
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-59
15.90
VGS1 + VGS 2 = (0.5mA)(4kΩ)= 2.00V | 2.00V = VTN +
⎛
2 ⎞
2
⎟ |
2.00 = 0.75 + 0.75 + I D1 ⎜⎜
+
−4
2x10−4 ⎟⎠
⎝ 5x10
2I D1 ⎛
2I D2 ⎞
⎟ | I D2 = I D1
− ⎜⎜VTP −
Kn ⎝
K p ⎟⎠
I D1 =
0.5
→ I D2 = I D1 = 9.38 µA
163.3
15.91
I SS ≥
5V 5V
=
= 5.00mA | iS = I SS + iL
RL 1kΩ
5V
5V
= I SS + 5.00mA | iSmin = I SS −
= I S − 5.00mA
1kΩ
1kΩ
= 5.00mA, iSmax = 10.0mA | iSmin = 0 | iD = 0.005(1+ sin 2000πt ) A
iSmax = I SS +
For I SS
Power delivered from the supplies : P (t )= 10V (iD )+ 10V (I SS )= 0.05(2 + sin 2000πt ) W
1
T
Pav =
T
∫ 0.05(2 + sin 2000πt )dt = 100mW
0
⎛ 5 ⎞2 1
12.5mW
Signal power developed in RL : Pac = ⎜ ⎟
= 12.5mW | η = 100%
= 12.5%
100mW
⎝ 2 ⎠ 1kΩ
15.92
+5V
T
t
0
-5V
15-60
2
2
1 ⎡(+5V ) T (−5V ) T ⎤
⎥ = 10.0mW
Pac = ⎢
+
5kΩ 2 ⎥⎦
T ⎢⎣ 5kΩ 2
1⎡
5V T
−5V T ⎤
− 5V
Pav = ⎢5V
= 10.0mW
T ⎣ 5kΩ 2
5kΩ 2 ⎥⎦
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
|
η = 100%
15.93
+10 V
t
0
T
4
T
2
T
-10 V
2
1
Pac =
T
T
1
T
T
Pav =
∫
0
v 2 (t )
4
dt =
T
R
∫ 10i(t )dt =
0
T
4
∫
20
T
0
⎛ 40t ⎞
⎜
⎟
6400
⎝ T ⎠
dt = 3
T R
R
T
2
∫ i(t )dt =
0
40
T
T
4
∫
0
T
4
∫ t dt =
0
2
100
3R
1600
40t
dt = 2
T R
TR
T
4
∫
0
100
50
tdt =
| η = 100% 3R = 66.7%
50
R
R
15.94
*Problem 15.94(a) VBB = 0 V
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 0
VS 1 0 DC 0 SIN(0 4 2000)
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 2K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.TRAN 1U 2M
.FOUR 2000 V(4)
.PROBE
.END
*Problem 15.94(b) VBB = 1.3 V
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 1.3
VS 1 0 DC 0 SIN(0 4 2000)
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 2K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.TRAN 1U 2M
.FOUR 2000 V(4)
.PROBE
.END
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-61
(a)
HARMONIC FREQUENCY
FOURIER
NORMALIZED
NORMALIZED
NO
(HZ)
COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
2.000E+03
4.000E+03
6.000E+03
8.000E+03
1.000E+04
1.200E+04
1.400E+04
1.600E+04
1.800E+04
3.056E+00
2.693E-02
2.112E-01
3.473E-02
7.718E-02
4.064E-02
3.179E-02
4.109E-02
2.127E-02
1.000E+00
8.811E-03
6.910E-02
1.136E-02
2.525E-02
1.330E-02
1.040E-02
1.345E-02
6.960E-03
-4.347E-01
-1.300E+02
-1.744E+02
-1.550E+02
-1.678E+02
-1.679E+02
-1.580E+02
-1.736E+02
-1.568E+02
PHASE
PHASE (DEG)
0.000E+00
-1.296E+02
-1.740E+02
-1.545E+02
-1.674E+02
-1.675E+02
-1.576E+02
-1.731E+02
-1.564E+02
TOTAL HARMONIC DISTORTION = 7.831458E+00 PERCENT with VBB = 0
(b)
HARMONIC FREQUENCY
FOURIER
NORMALIZED
NORMALIZED
NO
(HZ)
COMPONENT COMPONENT (DEG)
1
2
3
4
5
6
7
8
9
2.000E+03
4.000E+03
6.000E+03
8.000E+03
1.000E+04
1.200E+04
1.400E+04
1.600E+04
1.800E+04
3.853E+00
1.221E-02
1.537E-02
1.504E-02
1.501E-02
1.531E-02
1.435E-02
1.467E-02
1.382E-02
1.000E+00
3.169E-03
3.990E-03
3.903E-03
3.897E-03
3.973E-03
3.726E-03
3.807E-03
3.587E-03
2.544E-01
6.765E+01
9.046E+01
5.520E+01
5.500E+01
4.231E+01
3.680E+01
2.823E+01
2.087E+01
PHASE
PHASE (DEG)
0.000E+00
6.740E+01
9.020E+01
5.495E+01
5.475E+01
4.206E+01
3.654E+01
2.798E+01
2.062E+01
TOTAL HARMONIC DISTORTION = 1.064939E+00 PERCENT with VBB = 1.3 V
15-62
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.95
VBE 2 0.7V
=
= 70.0 mA.
R
10Ω
0.07
+ 0.7 + 0.7 + 250(0.07) = 19.6 V
v S = 1000iB + VBE1 + VBE 2 + 250iE = 1000
101
The current begins to limit at iE =
15.96
*Problem 15.96 – Figure 15.38
VCC 3 0 DC 50
VS 1 0 DC 1
R1 1 2 1K
Q1 3 2 4 NBJT
Q2 2 4 5 NBJT
R 4 5 10
RL 5 0 250
.MODEL NBJT NPN IS=1FA BF=100
.OP
.DC VS 1 50 .05
.PROBE
.END
120mA
Slope = 1k
80mA
Ω
iL
40mA
Slope = 250
Ω
0A
0V
10V
20V
vS
30V
40V
50V
The results agree well with hand calculations.
15.97
I2 RG = VGS 4 − VGS 5 |
(0.25mA)(7kΩ)= VTN 4 +
2I D 4 ⎛
2I D5 ⎞
⎟ | I D5 = I D 4
− ⎜⎜VTP5 −
0.005 ⎝
0.002 ⎟⎠
⎛
2
2 ⎞
⎟ → I D5 = I D 4 = 23.5 µA
1.75 − 0.75 − 0.75 = I D 4 ⎜⎜
+
0.002 ⎟⎠
⎝ 0.005
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-63
15.98
+15 V
0.2 V
50 Ω
Q
+
4
+
0.7 V
2.4 k Ω
+
1.2 V
0.5 V
-
For VBE 4 = 0.7V, VEB 5 = I2 RB - VBE 4
-
2k Ω
Q
VEB 5 = 1.2 − 0.7 = 0.5V and Q 5 is off.
VEQ = 15 − 0.2 − 500µA(50Ω) = 14.8V
5
REQ = 50Ω
500 µ A
IC 4 = 100
-15 V
15.99
Rout =
(14.8 − 0.7)V = 6.98 mA
50Ω + 101(2kΩ)
1 ⎛ rπ ⎞ 1 ⎛ β o ⎞⎛ VT ⎞ 1 ⎛ VT ⎞ 1 ⎛ 0.025V ⎞
⎜
⎟ = 25.0 mΩ
⎜
⎟=
⎜
⎟⎜ ⎟ =
⎜ ⎟=
n 2 ⎝ β o + 1⎠ 100 ⎝ β o + 1⎠⎝ IC ⎠ 100 ⎝ IE ⎠ 100 ⎝ 10mA ⎠
15.100
IC = 100I B = 100
9 − 0.7
V
= 97.9 µA | Looking back into the
200kΩ + 101(82kΩ) Ω
transformer : Rth =
1 ⎛ rπ ⎞
1 100(0.025V )
rπ
=
= 253Ω
⎟ |
2⎜
β o + 1 101 97.9µA
n ⎝ β o + 1⎠
Desire to match the Thevenin equivalent resistance to RL :
(β + 1)n R
+ (β + 1)n R
2
vth =
o
rπ
L
2
o
L
vs =
(101)253 v
25.6kΩ + (101)253
relationships : vth = i1 Rth + nvo | i1 =
vo =
vth
R
n + th
nRL
| vo =
s
1
253Ω = 10Ω → n = 5.03
n2
= 0.500v s | Using the ideal transformer
1
1 vo
i2 =
n
n RL
| v th =
1 vo
Rth + nvo
n RL
0.500v s
= 0.0497vs | vo = 0.0497sin 2000πt
253Ω
5.03 +
5.03(10Ω)
⎛ 0.0497 ⎞2 1
Po = ⎜
= 0.124 mW
⎟
⎝
2 ⎠ 10
15-64
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.101
2 MΩ
= −6V | REQ = 2 MΩ 2 MΩ = 1MΩ
2 MΩ + 2 MΩ
−6 − 0.7 − (−12) V
100(0.025V )
IO = 100I B = 100
= 22.8 µA | rπ =
= 110kΩ
22.8µA
1MΩ + 101(220kΩ) Ω
(a) V
EQ
= −12V
(50 + 6.98)V = 2.50 MΩ
101
22.8µA)(220kΩ)= 6.98V | ro =
(
100
22.8µA
⎞
⎛
⎛
⎞
100(220kΩ)
β o RE
⎟ = 43.9 MΩ
⎜
= ro ⎜1+
⎟ = 2.50 MΩ⎜1+
⎟
1MΩ
+
110kΩ
+
220kΩ
⎝ Rth + rπ + RE ⎠
⎠
⎝
VCE = 12 − I E (220kΩ)= 12 −
Rout
(b) Using the CVD model for the diode,
2 MΩ
VEQ = −12 + (12V − 0.7V )
+ 0.7 = −5.65V | REQ = 2 MΩ 2 MΩ = 1MΩ
2 MΩ + 2 MΩ
−5.65 − 0.7 − (−12) V
100(0.025V )
IO = 100I B = 100
= 24.3 µA | rπ =
= 103kΩ
24.3µA
1MΩ + 101(220kΩ) Ω
(50 + 6.60)V = 2.33MΩ
101
24.3µA)(220kΩ)= 6.60V | ro =
(
100
24.3µA
⎞
⎛
⎛
⎞
100(220kΩ)
β o RE
⎟ = 41.1 MΩ
= ro ⎜1+
⎟ = 2.33MΩ⎜⎜1+
⎟
⎝ Rth + rπ + RE ⎠
⎝ 1MΩ + 103kΩ + 220kΩ ⎠
VCE = 12 − I E (220kΩ)= 12 −
Rout
15.102
The dc analysis is the same as Problem 15.101. However, the bypass capacitor provides as
ac ground at the base of the transistor so that Rth = 0.
⎛
⎛
100(220kΩ) ⎞
β o RE ⎞
⎟ = 169 MΩ
⎜
Rout = ro ⎜1+
⎟ = 2.50 MΩ⎜1+
⎟
110kΩ
+
220kΩ
⎝ rπ + RE ⎠
⎠
⎝
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-65
15.103
430kΩ
= −5.53V | REQ = 270kΩ 430kΩ = 166kΩ
270kΩ + 430kΩ
−5.53 − 0.7 − (−9)
150(0.025V )
V = 144 µA | rπ =
= 26.0kΩ
IO = 100I B = 150
144µA
166kΩ + 151(18kΩ)
(a) V
EQ
= −9V
(75 + 6.39)V = 565kΩ
151
144µA)(18kΩ)= 6.39V | ro =
(
144µA
150
⎛
⎞
⎛
⎞
150(18kΩ)
β o RE
⎟ = 7.83 MΩ
= ro⎜1+
⎟ = 565kΩ⎜⎜1+
⎟
⎝ Rth + rπ + RE ⎠
⎝ 166kΩ + 26.0kΩ + 18kΩ ⎠
VCE = 9 − I E (18kΩ) = 9 −
Rout
(b) Using the CVD model for the diode,
⎛
⎞
270kΩ
VEQ = −9 + (9V − 0.7V )⎜
⎟ + 0.7 = −5.10V | REQ = 270kΩ 430kΩ = 166kΩ
⎝ 270kΩ + 430kΩ ⎠
IO = 100I B = 150
−5.10 − 0.7 − (−9)
V = 166 µA | rπ =
166kΩ + 151(18kΩ)
150(0.025V )
166µA
= 22.6kΩ
(75 + 5.99)V = 488kΩ
151
166µA)(18kΩ)= 5.99V | ro =
(
166µA
150
⎛
⎞
⎛
⎞
150(18kΩ)
β o RE
⎟ = 6.87 MΩ
Rout = ro⎜1+
⎟ = 488kΩ⎜⎜1+
⎟
166kΩ
+
22.6kΩ
+
18kΩ
⎝ Rth + rπ + RE ⎠
⎝
⎠
200kΩ
(c) VEQ = −5V 100kΩ + 200kΩ = −3.33V | REQ = 100kΩ 200kΩ = 66.7kΩ
−3.33 − 0.7 − (−5)
100(0.025V )
IO = 100I B = 100
V = 61.3 µA | rπ =
= 40.8kΩ
61.3µA
66.7kΩ + 101(15kΩ)
VCE = 9 − I E (18kΩ) = 9 −
(75 + 4.07)V = 1.29 MΩ
101
61.3µA)(15kΩ)= 4.07V | ro =
(
61.3µA
100
⎛
⎞
⎛
⎞
100(15kΩ)
β o RE
⎜
⎟
=
1.29
MΩ
1+
= ro⎜1+
⎟
⎜ 66.7kΩ + 40.8kΩ + 15kΩ ⎟ = 17.1 MΩ
⎝ Rth + rπ + RE ⎠
⎝
⎠
VCE = 5 − I E (15kΩ)= 5 −
Rout
(d ) Using the CVD model for the diode,
⎛
⎞
100kΩ
VEQ = −5 + (5V − 0.7V )⎜
⎟ + 0.7 = −2.87V | REQ = 100kΩ 200kΩ = 66.7kΩ
⎝100kΩ + 200kΩ ⎠
IO = 100I B = 100
−2.87 − 0.7 − (−5)
V = 90.4 µA | rπ =
66.7kΩ + 101(15kΩ)
100(0.025V )
90.4µA
= 27.7kΩ
(75 + 3.63)V = 870kΩ
101
90.4µA)(15kΩ)= 3.63V | ro =
(
100
90.4µA
⎛
⎞
⎛
⎞
100(15kΩ)
β o RE
⎟ = 12.8 MΩ
= ro⎜1+
⎟ = 870kΩ⎜⎜1+
⎟
⎝ Rth + rπ + RE ⎠
⎝ 66.7kΩ + 27.7kΩ + 15kΩ ⎠
VCE = 5 − I E (15kΩ)= 5 −
Rout
15-66
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.104
2MΩ
= −6V | REQ = 2MΩ 2MΩ = 1MΩ
2MΩ + 2MΩ
−6 − 0.7 − (−12) V
100(0.025V )
IO = 100IB = 100
= 22.8 µA | rπ =
= 110kΩ
1MΩ + 101(220kΩ) Ω
22.8µA
(a) VEQ = −12V
101
(50 + 6.98)V = 2.50MΩ
(22.8µA)(220kΩ) = 6.98V | ro =
100
22.8µA
⎞
⎞
⎛
⎛
100(220kΩ)
β o RE
Rout = ro⎜1+
⎟ = 2.50MΩ⎜1+
⎟ = 43.9 MΩ
⎝ Rth + rπ + RE ⎠
⎝ 1MΩ + 110kΩ + 220kΩ ⎠
(b) Using the CVD model for the diode (diode - connected transistor),
VCE = 12 − IE (220kΩ) = 12 −
2 MΩ
VEQ = −12 + (12V − 0.7V )
+ 0.7 = −5.65V | REQ = 2 MΩ 2 MΩ = 1MΩ
2 MΩ + 2 MΩ
−5.65 − 0.7 − (−12) V
100(0.025V )
IO = 100I B = 100
= 24.3 µA | rπ =
= 103kΩ
24.3µA
1MΩ + 101(220kΩ) Ω
(50 + 6.60)V = 2.33MΩ
101
24.3µA)(220kΩ)= 6.60V | ro =
(
100
24.3µA
⎞
⎛
⎛
⎞
100(220kΩ)
β o RE
⎟
⎜
= ro ⎜1+
=
2.33MΩ
1+
⎟
⎜ 1MΩ + 103kΩ + 220kΩ ⎟ = 41.1 MΩ
⎝ Rth + rπ + RE ⎠
⎠
⎝
VCE = 12 − I E (220kΩ) = 12 −
Rout
15.105
IC
R
1
IC
RB
Q
R
Q
V
2
RE
-12 V
B
RE
-12 V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-67
A spread sheet will be used to assist in this design using β F = β o = 100 & VA = 70V
The maximum current in the two bias resistors is 0.2mA. To allow
some room for tolerances, choose I1 ≅ 0.15mA. Neglecting the transistor base current,
12V
V
V
= 80kΩ | R2 = B (R1 + R2 ) = B 80kΩ | RB = R1 R2
0.15mA
12
12
(VB − 0.7) or R = 1 ⎛⎜100(VB − 0.7) − R ⎞⎟ | V = 12 − I R
IC = 100
E
B⎟
CE
E E
101 ⎜⎝
RB + 101RE
IC
⎠
⎛
⎞
β o RE
70 + VCE
ro =
| Rout = ro ⎜1+
⎟
IC
⎝ RB + rπ + RE ⎠
Now, a spreadsheet MATLAB, MATHCAD, etc. can be used to explore the design space
with VB as the primary design variable.
R1 + R2 =
VB
R2
R1
RB
0.500
0.600
0.700
0.800
0.900
1.000
1.100
1.200
1.300
1.400
1.500
1.600
1.700
1.800
1.900
2.000
3.33E+03
4.00E+03
4.67E+03
5.33E+03
6.00E+03
6.67E+03
7.33E+03
8.00E+03
8.67E+03
9.33E+03
1.00E+04
1.07E+04
1.13E+04
1.20E+04
1.27E+04
1.33E+04
7.67E+04
7.60E+04
7.53E+04
7.47E+04
7.40E+04
7.33E+04
7.27E+04
7.20E+04
7.13E+04
7.07E+04
7.00E+04
6.93E+04
6.87E+04
6.80E+04
6.73E+04
6.67E+04
3.19E+03
3.80E+03
4.39E+03
4.98E+03
5.55E+03
6.11E+03
6.66E+03
7.20E+03
7.73E+03
8.24E+03
8.75E+03
9.24E+03
9.73E+03
1.02E+04
1.07E+04
1.11E+04
-2.30E+02 -1.74E+05
-1.37E+02 -8.00E+04
-4.35E+01 1.41E+04
4.97E+01 1.08E+05
1.43E+02 2.03E+05
2.37E+02 2.97E+05
3.30E+02 3.91E+05
4.24E+02 4.86E+05
5.18E+02 5.81E+05
6.11E+02 6.76E+05
7.05E+02 7.71E+05
8.00E+02 8.66E+05
8.94E+02 9.61E+05
9.88E+02 1.06E+06
1.08E+03 1.15E+06
1.18E+03 1.25E+06
5.57E+05
9.73E+04
5.13E+03
1.80E+05
5.56E+05
1.09E+06
1.75E+06
2.52E+06
3.38E+06
4.31E+06
5.32E+06
6.38E+06
7.50E+06
8.68E+06
9.90E+06
1.12E+07
7.50E+04
6.80E+04
5.73E+03
1.02E+04
1.50E+02
1.00E+03
5.85E+05
8.86E+06
RE
ro
Rout
Two
possible
solutions
0.916
1.800
6.20E+03
1.20E+04
2.10E+05
1.07E+06
IO
1.04E-03
9.89E-04
The first solution is the lowest value of VB that was found to meet the output specification using
the nearest 5% values. The second is one in which the values were found to be very close to
existing standard 5% resistor values, but it uses twice the value of VB and has a smaller output
voltage compliance range.
15-68
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.106
330kΩ
10V = 3.27V | REQ = 330kΩ 680kΩ = 222kΩ
330kΩ + 680kΩ
⎛ 5x10−4 ⎞
2
Assume active region operation : I D = ⎜
⎟(VGS −1)
⎝ 2 ⎠
⎛ 5x10−4 ⎞
2
VGS = 3.27 − (33kΩ)I D = 3.27 − 33x103 ⎜
⎟(VGS −1) ⇒ VGS = 1.467V | IO = I D = 54.5 µA
⎝ 2 ⎠
VEQ =
VDS = 10 − (33kΩ)I D = 8.20V | Active region operation is correct.
ro =
100 + 8.20 V
= 1.99 MΩ | g m = 2 5x10−4 54.5x10−6 1+ (0.01)8.20 = 0.243 mS
54.5 µA
(
[
(
)]
)(
[
)[
]
]
Rout = ro 1+ g m 33x103 = 1.99 MΩ 1+ 0.243mS (33kΩ) = 17.9 MΩ
15.107
68kΩ
3V = 0.760V | REQ = 68kΩ 200kΩ = 50.8kΩ
68kΩ + 200kΩ
< VTN , (0.76V < 1V) so the transistor is off, and IO = ID = 0.
VEQ =
VEQ
The circuit designer made an error and failed to check the final design.
15.108
100kΩ
6V = 2V | REQ = 100kΩ 200kΩ = 66.7kΩ
100kΩ + 200kΩ
⎛ 5x10−4 ⎞
2
Assume active region operation : 2 = VGS + (16kΩ)I D | I D = ⎜
⎟(VGS −1)
⎝ 2 ⎠
VEQ =
4VGS2 − 7VGS + 2 = 0 ⇒ VGS = 1.390V | IO = I D = 38.1 µA
VDS = VO − (16kΩ)I D = 6 − (16kΩ)(38.1µA)= 5.39V | Active region is correct.
ro =
100 + 5.39 V
= 2.77 MΩ | g m ≅ 2 5x10−4 38.1x10−6 1+ 0.01(5.39) = 0.200 mS
38.1 µA
(
[
(
)]
[
)(
)[
]
]
Rout = ro 1+ g m 16x103 = 2.77 MΩ 1+ 0.200mS (16kΩ) = 11.6 MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-69
15.109
200kΩ
= 10V | REQ = 200kΩ 100kΩ = 66.7kΩ
200kΩ + 100kΩ
(15 − 0.7 −10)V = 88.6 µA | r = 75(0.025V ) = 21.2kΩ
IO = 75I B = 75
π
88.6µA
66.7kΩ + 76(47kΩ)
VEQ = 15V
(50 + 10.7)V = 685kΩ
76
88.6 µA)(47kΩ) = 10.7V | ro =
(
75
88.6µA
⎛
⎞
⎛
⎞
75(47kΩ)
β o RE
⎟ = 18.6 MΩ
= ro ⎜1+
⎟ = 685kΩ⎜⎜1+
⎟
⎝ Rth + rπ + RE ⎠
⎝ 66.7kΩ + 21.2kΩ + 47kΩ ⎠
VEC = 15 − I E RE = 15 −
Rout
15.110
33kΩ
= 3.84V | REQ = 33kΩ 10kΩ = 7.67kΩ
33kΩ + 10kΩ
75(0.025V )
5 − 0.7 − 3.84
IO = 75IB = 75
V = 284 µA | rπ =
= 6.60kΩ
284 µA
7.67kΩ + 76(1.5kΩ)
VEQ = 5V
76
(60 + 4.57)V = 227kΩ
(284 µA)(1.5kΩ) = 4.57V | ro =
75
284 µA
⎞
⎞
⎛
⎛
75(1.5kΩ)
β o RE
= ro⎜1+
⎟ = 227kΩ⎜1+
⎟ = 1.85 MΩ
⎝ Rth + rπ + RE ⎠
⎝ 7.67kΩ + 6.60kΩ + 1.5kΩ ⎠
VEC = 5 − IE RE = 5 −
Rout
15.111
300kΩ
= 7.50V | REQ = 300kΩ 100kΩ = 75.0kΩ
300kΩ + 100kΩ
90(0.025V )
10 − 0.7 − 7.50
V = 94.6 µA | rπ =
= 23.4kΩ
IO = 90I B = 90
94.6µA
75.0kΩ + 91(18kΩ)
VEQ = 10V
(75 + 8.28)V = 880kΩ
91
94.6µA)(18kΩ)= 8.28V | ro =
(
90
94.6µA
⎛
⎞
⎛
⎞
90(18kΩ)
β o RE
⎟ = 13.1 MΩ
= ro⎜1+
⎟ = 880kΩ⎜⎜1+
⎟
⎝ Rth + rπ + RE ⎠
⎝ 75.0kΩ + 23.4kΩ + 18kΩ ⎠
VEC = 10 − I E RE = 10 −
Rout
15-70
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.112
200kΩ
6V = 4V | REQ = 200kΩ 100kΩ = 66.7kΩ | 6 −162 −16e3* x103 I D + VGS = 4
200kΩ + 100kΩ
2
7.5x10−4
Assume active region operation : I D =
VGS + 0.75)
(
2
3
2 −16x10 I D + VGS = 0 ⇒ VGS = −1.13V | IO = I D = 54.3 µA
VEQ =
(
)
VDS = − 6 −16x103 I D = −5.13V | Active region is correct.
ro =
100 + 5.13 V
= 1.94 MΩ | g m = 2 7.5x10−4 54.3x10−6 1+ 0.01(5.13) = 0.292mS
54.3 µA
(
)(
[
)[
]
]
Rout = ro (1+ g m RS )= 1.94 MΩ 1+ 0.292mS (16kΩ) = 11.0 MΩ
15.113
VEQ = 9V
2 MΩ
= 6V | REQ = 2 MΩ 1MΩ = 667kΩ | 9 −105 I D + VGS = 6
2 MΩ + 1MΩ
Assume active region operation : VGS = VTP −
2I D
Kp
⎛
2I D ⎞
⎟ ⇒ IO = I D = 17.0 µA
1.2x105 I D = 3 + ⎜⎜ −0.75 −
7.5x10−4 ⎟⎠
⎝
(
)
VDS = − 9 −1.2x105 I D = −6.96V | Active region is correct.
ro =
100 + 6.96 V
= 6.29 MΩ | g m = 2 7.5x10−4 17.0x10−6 1+ 0.01(6.96) = 0.165mS
17.0 µA
Rout = ro (1+ g m
(
)(
)[
R )= 6.29 MΩ[1+ 0.165mS (1.2x10 )]= 131 MΩ
]
5
S
15.114
VEQ =
200kΩ
4V = 3.05V | REQ = 200kΩ 62kΩ = 47.3kΩ
200kΩ + 62kΩ
Assume active region operation : 4 − 43x103 I D + VGS = 3.05 | I D =
2
7.5x10−4
VGS + 0.75)
(
2
0.95 − 43x103 I D + VGS = 0 ⇒ VGS = −0.6034V | IO = I D = 8.06 µA
(
)
VDS = − 4 − 43x103 I D = −3.65V | Active region is correct.
ro =
100 + 3.65 V
= 12.9 MΩ | g m = 2 7.5x10−4 8.06x10−6 1+ 0.01(3.65) = 112 µS
8.06 µA
(
[
)(
)[
]
]
Rout = ro (1+ g m RS )= 12.9 MΩ 1+ 112µS (43kΩ) = 75.0 MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-71
15.115
50V ⎛
⎞
⎜1+ 2 2x10−4 1.75x10−4 RS ⎟
−4 ⎝
⎠
1.75x10
Note that including λ in the g m expression will increase Rout above this estimate.
(
Estimating Rout ≅ ro (1+ g m RS )≅
)(
)
Hence neglecting λ represents a conservative simplification.
(
)
286kΩ 1+ 2.65x10−4 RS ≥ 2.5MΩ ⇒ RS ≥ 29.2kΩ | Choose RS = 33kΩ
⎛
⎞
2 1.75x10−4 ⎟
⎜
VG = VDD − I D RS + VGS = 12 −1.75x10 3.3x10 + −1−
= 3.90V
⎜
2x10−4 ⎟
⎝
⎠
12V
R4
12 = 3.90 | I2 ≤ 25µA | Assign I2 = 20µA | R3 + R4 =
= 600kΩ
R3 + R4
20µA
−4
(
4
)
(
)
3.90
(R3 + R4 ) = 195kΩ ⇒ R4 = 200kΩ | R3 = 430kΩ
12
15.116
68kΩ
(a) VEQ = −12V 68kΩ + 33kΩ = −8.08V | REQ = 68kΩ 33kΩ = 22.2kΩ | VB = −8.08 − (I B1 + I B2 )RTH
⎛ V − 0.7 − (−12) V − 0.7 − (−12)⎞
B
B
⎟22.2kΩ → VB = −8.11V
VB = −8.08 − ⎜⎜
+
126(100kΩ) ⎟⎠
⎝ 126(20kΩ)
125 ⎛⎜ VB − 0.7 − (−12)⎞⎟
125 ⎛⎜ VB − 0.7 − (−12)⎞⎟
µ
A
|
I
=
α
I
=
IC1 = α F I E1 =
=
158
C2
F E2
⎟
⎟ = 31.7 µA
126 ⎜⎝
126 ⎜⎝
20kΩ
100kΩ
⎠
⎠
R4 =
VCE = 0 − (−8.11− 0.7)= 8.87V | ro1 =
rπ 1 =
125(0.025V )
158 µA
= 19.8kΩ | rπ 2 =
(50 + 8.11)V = 368kΩ |
158µA
125(0.025V )
31.7µA
[
[
= 98.6kΩ
]
R th1 = 22.2kΩ 98.6kΩ + (126)(100kΩ) = 22.2kΩ
⎛
⎞
⎛
⎞
125(20kΩ)
β o RE
⎜
⎟
Rout1 = ro1⎜1+
=
368kΩ
1+
⎟
⎜ 22.2kΩ + 19.8kΩ + 20kΩ ⎟ = 15.2 MΩ
⎝ Rth + rπ 1 + RE ⎠
⎝
⎠
ro2 =
(50 + 8.11)V = 1.83MΩ
31.7µA
[
[
]
| Rth2 = RTH rπ 1 + (β o + 1)(20kΩ)
]
Rth2 = 22.2kΩ 19.8kΩ + (126)(20kΩ) = 22.0kΩ
⎛
⎞
⎛
⎞
125(100kΩ)
β o RE
⎜
⎟
Rout2 = ro2⎜1+
=
1.83MΩ
1+
⎟
⎜ 22.0kΩ + 98.6kΩ + 100kΩ ⎟ = 106 MΩ
⎝ Rth + rπ 2 + RE ⎠
⎝
⎠
15-72
]
Rth1 = REQ rπ 2 + (β o + 1)(100kΩ)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.116 cont.
(b) Using the CVD model for diode Q for dc calculations,
(12V − 0.7) 68kΩ = −7.61V | R = 68kΩ 33kΩ = 22.2kΩ | V = −8.08 − I + I R
V =
( )
(
)
68kΩ + 33kΩ
⎛ V − 0.7 − (−12) V − 0.7 − (−12)⎞
⎟22.2kΩ → V = −7.65V
V = −7.61− ⎜⎜
+
⎟
126
20kΩ
126
100kΩ
(
)
(
)
⎠
⎝
125 ⎛ V − 0.7 − (−12)⎞
125 ⎛ V − 0.7 − (−12)⎞
⎜
⎟ = 181 µA | I = α I =
⎜
⎟ = 36.2 µA
I =α I =
3
EQ
EQ
B
B
EQ
B
F
E1
126 ⎜⎝
B
20kΩ
⎟
⎠
C2
VCE = 0 − (−7.65 − 0.7)= 8.35V | ro1 =
rπ 1 =
B2
B
B
C1
B1
125(0.025V )
181 µA
= 17.3kΩ | rπ 2 =
[
F E2
(50 + 8.35)V = 322kΩ |
181µA
125(0.025V )
36.2µA
126 ⎜⎝
B
100kΩ
[
⎟
⎠
]
Rth1 = REQ rπ 2 + (β o + 1)(100kΩ)
= 86.3kΩ
]
Rth1 = 22.2kΩ 86.3kΩ + (126)(100kΩ) = 22.2kΩ
⎞
⎛
⎛
⎞
125(20kΩ)
β o RE
⎟
⎜
Rout1 = ro1⎜1+
=
322kΩ
1+
⎟
⎜ 22.2kΩ + 17.3kΩ + 20kΩ ⎟ = 13.9 MΩ
⎝ Rth + rπ 1 + RE ⎠
⎠
⎝
ro2 =
(50 + 8.35)V = 1.61MΩ
36.2µA
[
[
]
| Rth2 = RTH rπ 1 + (β o + 1)(20kΩ)
]
Rth2 = 22.2kΩ 17.3kΩ + (126)(20kΩ) = 22.0kΩ
⎛
⎞
⎛
⎞
125(100kΩ)
β o RE
⎜
⎟ = 98.2 MΩ
Rout2 = ro2 ⎜1+
⎟ = 1.61MΩ⎜1+
⎟
22.0kΩ
+
86.3kΩ
+
100kΩ
⎝ Rth + rπ 2 + RE ⎠
⎝
⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-73
15.117
20kΩ
= −4.07V | REQ = 20kΩ 39kΩ = 13.2kΩ
20kΩ + 39kΩ
⎛ 1 VB + 0.7 1 VB + 0.7 1 VB + 0.7 ⎞
I B = −⎜
+
+
⎟ | VB = −4.07V + 13200I B → VB = −3.95V
76 16kΩ
76 8.2kΩ ⎠
⎝ 76 33kΩ
⎛
β o (33kΩ) ⎞
75 ⎛ 0 − 0.7V − (−3.95V )⎞
⎟ = 97.2µA | Rout1 = ro1⎜1+
⎟
IC1 = ⎜⎜
⎜ R + r + 33kΩ ⎟
⎟
76 ⎝
33kΩ
th1
π
1
⎝
⎠
⎠
VEQ = −12V
[
][
]
Rth1 = 13.2kΩ rπ 2 + (β o2 + 1)16kΩ rπ 3 + (β o3 + 1)8.2kΩ ≅ 13.2kΩ
Rout1 =
IC2 =
⎞
75(33kΩ)
60 + 8.75 ⎛
⎜1+
⎟ = 27.4 MΩ
97.2µA ⎜⎝ 13.2kΩ + 19.3kΩ + 33kΩ ⎟⎠
⎛
β o (16kΩ) ⎞
75 ⎛ 0 − 0.7V − (−3.95V )⎞
⎜
⎟ = 201µA | Rout2 = ro2 ⎜1+
⎟
⎜ R + r + 16kΩ ⎟
⎟
76 ⎜⎝
16kΩ
th2
π2
⎝
⎠
⎠
[
][
]
Rth2 = 13.2kΩ rπ 1 + (β o1 + 1)33kΩ rπ 3 + (β o3 + 1)8.2kΩ ≅ 13.2kΩ
⎞
75(16kΩ)
60 + 8.75 ⎛
⎜
⎟ = 11.0 MΩ
Rout2 =
1+
201µA ⎜⎝ 13.2kΩ + 9.33kΩ + 16kΩ ⎟⎠
⎛
β o (8.2kΩ) ⎞
75 ⎛ 0 − 0.7V − (−3.95V )⎞
⎜
⎜
⎟
⎟
IC3 = ⎜
⎟ = 391µA | Rout3 = ro3 ⎜1+ R + r + 8.2kΩ ⎟
76 ⎝
8.2kΩ
th2
π2
⎝
⎠
⎠
[
][
]
Rth3 = 13.2kΩ rπ 1 + (β o1 + 1)33kΩ rπ 2 + (β o2 + 1)16kΩ ≅ 13.2kΩ
⎞
75(8.2kΩ)
60 + 8.75 ⎛
⎜
⎟ = 4.30 MΩ
Rout3 =
1+
391µA ⎜⎝ 13.2kΩ + 4.80kΩ + 8.2kΩ ⎟⎠
15-74
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.118
2 MΩ
= 6.00V | REQ = 2 MΩ 2 MΩ = 1.00 MΩ
2 MΩ + 2 MΩ
2I D1
Assume saturation : 12 −105 I D1 + VGS1 = 6 | VGS1 = −1−
2.5x10−4
I D1 = 44.1 µA, VGS1 = −1.59V , VDS1 = −(6 − VGS1 )= −7.59V
VEQ = 12V
Rout1 = ro1(1+ g m1 R1 )=
50V + 7.59V ⎛
⎞
⎜1+ 100kΩ 2(250µA)(44.1µA) 1+ 0.02(7.69) ⎟ = 22.2 MΩ
⎝
⎠
44.1µA
[
]
2I D2
2.5x10−4
I D2 = 10.0 µA, VGS 2 = −1.28 V , VDS 2 = −(6 − VGS 2 ) = −7.28V
12 − 4.7x105 I D2 + VGS 2 = 6 | VGS 2 = −1−
Rout2 = ro2 (1+ g m2 R2 )=
50V + 7.28V ⎛
⎞
⎜1+ 470kΩ 2(250µA)(10.0µA) 1+ 0.02(7.28) ⎟ = 210 MΩ
⎝
⎠
10.0µA
[
]
15.119
*Problem 15.119 – Figure P15.118
VCC 1 0 DC 12
R1 1 2 100K
R4 1 3 2MEG
R3 3 0 2MEG
R2 1 4 470K
M1 5 3 2 2 PFET
M2 6 3 4 4 PFET
VD1 5 0 DC 0
VD2 6 0 DC 0
.MODEL PFET PMOS VTO=-1 KP=250U LAMBDA=0.02
.OP
*.TF I(VD1) VD1
.TF I(VD2) VD2
.END
Results: IO1 = 44.4 µA, ROUT1 = 22.1 MΩ, IO1 = 10.1 µA, ROUT1 = 209 MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-75
15.120
For large A, IO ≅ α F
VREF 120 5V
=
= 99.2 µA
121 50kΩ
R
vx
+
v
-A v e
rπ
ix
gmv
ve
ro
R
For the small - signal model above,
vx = ve + (ix − g mv )ro | v = (− Av e )− ve = −ve (1+ A) | ix = Gve + g π (1+ A)ve | Combining :
Rout =
ro =
Rout
vx 1+ µ f (1+ A) 1
=
+
≅ ro (1+ β o ) for g π (1+ A)>> G and µ f (1+ A)>> 1
ix G + g π (1+ A) g o
50V + 10V
= 605kΩ | Rout = 605kΩ(121)= 73.2 MΩ
99.2µA
cannot exceed β oro because of the loss of base current through rπ .
15.121
ROUT is limited to βoro of the BJT. We need to increase the effective current gain of the
transistor which can be done by replacing Q1 with a Darlington configuration of two transistors.
+V CC
I
o
+
V
REF
A
Q
2
Q1
R
-V
EE
2 2
Now ROUT can approach the βoro product of the Darlington which is Rout ≅ β o ro2 . See Prob.
3
15.48
15-76
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.122
For large A, IO ≅
VREF
5V
=
= 100µA
50kΩ
R
vx
+
v
-A v s
gm v
-
ix
ro
vs
R
For the small - signal model above,
v x = v s + (ix − gm v )ro | v = (−Av s ) − v s = −v s (1+ A) | v s = ix R | Combining :
Rout =
vx
50V + 10V
= R + ro [1+ gm R(1+ A)] | ro =
= 600kΩ
100µA
ix
gm = 2(8x10−4 )(10−4 )[1+ 0.02(10)] = 0.438mS
[
]
Rout = 50kΩ + 600kΩ 1+ 0.438mS (50kΩ)(1+ 5x10 4 ) = 6.57x1011 Ω !!
15.123
91kΩ
= 9.03V | REQ = 91kΩ 30kΩ = 22.6kΩ
91kΩ + 30kΩ
12 − 0.7 − 9.03
V
= 9.34 µA
IC 3 = 85I B 3 = 85
22.6kΩ + 86(240kΩ) Ω
(a) V
EQ
= 12V
86
(9.34µA)(240kΩ)− 0.7 = 9.03V
85
85 ⎛ 9.34µA ⎞
I
IC1 = IC 2 = α F C 3 = ⎜
⎟ = 4.62µA | VEC1 = VEC 2 = 0.7 − −12 + 1.2 MΩ(4.62µA) = 7.16V
86 ⎝ 2 ⎠
2
VEC 3 = 12 − I E RE − 0.7 = 12 −
[
]
Q − po int s : (4.62µA,7.62V ) (4.62µA,7.62V ) (9.34µA,9.03V )
(b) r
π3
=
85(0.025V )
9.34µA
= 228kΩ | ro3 =
(70 + 9.03)V = 8.46 MΩ
9.34µA
⎛
⎞
⎛
⎞
85(240kΩ)
β o RE
⎟ = 360 MΩ
Rout 3 = ro3⎜1+
⎟ = 8.46 MΩ⎜⎜1+
⎟
⎝ Rth + rπ 3 + RE ⎠
⎝ 22.6kΩ + 228kΩ + 240kΩ ⎠
g m RC
= 20(4.62µA)(1.2 MΩ) = +111 (40.9dB)
2
CMRR = g m1 Rout 3 = 40(4.62µA)(360 MΩ)= 6.65x10 4 (96.5dB)
For a single - ended output, Av =
(c) The answers are the same as parts (a) and (b).
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-77
15.124
(a) V
EQ
= −15V
100kΩ
= −9.93V | REQ = 100kΩ 51kΩ = 33.8kΩ | Assume saturation :
100kΩ + 51kΩ
−9.93 = −15 + 7500I D 3 + VGS 3 | VGS 3 = 1+
I D1 = I D 2 =
2I D3
| I D3 = 363 µA, VGS 3 = 2.35V
4x10−4
2(363µA)
I D3
= 182µA | VGS 1 = 1+
= 1.95V
2
2(400µA)
VDS 3 = −VGS 1 − 7500I D 3 − (−15)= 10.3V
VDS1 = VDS 2 = VD1 − VS 1 = 15 − 36000I D1 − (−VGS 1)= 10.4V
(182µA, 10.4V ) (182µA, 10.4V ) (363µA, 10.3V )
50V + 10.3V
= 166kΩ
363µA
⎛
⎞
Rout 3 = ro3 (1+ g m3 RS )= 166kΩ⎜1+ 2 4x10−4 3.63x10−4 1+ 0.02(10.3) (7.5kΩ)⎟ = 903 kΩ
⎝
⎠
(b) r
o3
=
(
(
)
(
)(
Add = −g m RD ro2 = − 2 4x10−4 1.82x10−4
For a single - ended output, Acd ≅ −
CMRR =
)(
)[
]
)[1+ 0.02(10.4)](36kΩ 332kΩ)= 0.419mS (325kΩ)= −13.6
RD
36kΩ
=−
= −0.199
2Rout 3
2(903kΩ)
13.6 / 2
= 342 or 50.7 dB
0.0199
The approximate CMRR estimate is CMRR ≅ g m1 Rout 3 = 0.419mS (903kΩ) = 378 (51.6 dB)
15.125
ro1
since the collector current of the
2
current source is twice that of the input transistors. For a single- ended output,
Assuming all devices are identical, Rout = β o1
Add = −
β µ
g m1 RC
RC
R
g β r
| Acc = −
= − C | CMRR = m1 o1 o1 = o1 f 1
⎛ r ⎞
2
β o1ro1
2
2
2⎜β o1 o1 ⎟
2⎠
⎝
Using our default paramters: CMRR ≅ 20β o1VA1 = 20(100)(70)= 140,000 (103dB)
(Note that this analysis neglects the contribution of the output resistance ro of the input pair. If
this resistance is included, a theoretical cancellation occurs and Acc = 0! Of course the output
β o ro
resistance expression Rout =
is not precise, but an improvement over the CMRR expression
2
above is possible.)
15-78
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.126
Rout = ro (1+ g m RS ) ≅ µF RS = g mro RS ≅ 2Kn I D
VR S = I D RS =
( )
λ I D1.5 Rout
2Kn
=
1
R
λI D S
( ) (5x10 ) = 3.16 V
2(5x10 )
0.02 10−4
1.5
6
−4
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-79
15.127
*Problem 15.127 - Fig. 15.49(a) - BJT Current Source Monte Carlo Analysis
*Generate a Voltage Source with 5% Tolerances
IEE 0 5 DC 1
REE 5 0 RTOL 15
EEE 1 0 5 0 -1
*
VO 4 0 AC 1
RE 1 2 RTOL 18.4K
R1 1 3 RTOL 113K
R2 3 0 RTOL 263K
Q1 4 3 2 NBJT
.OP
.DC VO 0 0 .01
.AC LIN 1 1000 1000
.PRINT AC IM(VO) IP(VO)
.MODEL NBJT NPN BF=150 VA=75
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 DC I(VO) YMAX
*.MC 1000 AC IM(VO) YMAX
.END
Results - 3σ limits: IO = 199 µA ± 32.5 µA, ROUT = 11.8 MΩ ± 2.6 MΩ
*Problem 15.127 - Fig. 15.49(b) - MOSFET Current Source
*Generate a Voltage Source with 5% Tolerance
IEE 0 5 DC 1
REE 5 0 RTOL 15
EEE 1 0 5 0 -1
*
VO 4 0 AC 1
RS 1 2 RTOL 18K
R3 1 3 RTOL 240K
R4 3 0 RTOL 510K
M1 4 3 2 2 NFET
.OP
.DC VO 0 0 .01
.AC LIN 1 1000 1000
.PRINT AC IM(VO) IP(VO)
.MODEL NFET NMOS KP=9.95M VTO=1 LAMBDA=0.01
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 DC I(VO) YMAX
*.MC 1000 AC IM(VO) YMAX
.END
Results - 3σ limits: IO = 201 µA ± 34.7 µA, ROUT = 21.7 MΩ ± 3.6 MΩ
15-80
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.128
4.02kΩ(1+ 0.15)(1− 0.03) ≤ R ≤ 4.02kΩ(1+ 0.15)(1+ 0.03) | 4.48kΩ ≤ R ≤ 4.76kΩ
15.129
⎛V ⎞
⎛V ⎞ I
⎛V − V ⎞
I
IC1 = I S1 exp⎜ BE1 ⎟ | IC2 = I S 2 exp⎜ BE 2 ⎟ | C2 = S 2 exp⎜ BE 2 BE1 ⎟ | ∆VBE = VBE2 − VBE1
VT
⎝ VT ⎠
⎝ VT ⎠ IC1 I S1
⎝
⎠
⎛ ∆I ⎞
⎛ ∆I ⎞
I +I
∆I S = I S1 − I S 2 | I S = S1 S 2 | I S1 = I S ⎜1+ S ⎟ | I S 2 = I S ⎜1− S ⎟
2
⎝ 2I S ⎠
⎝ 2I S ⎠
⎡
⎛ ∆I ⎞ ⎤
⎢ I S ⎜1+ S ⎟ ⎥
⎤
⎡
⎛ IC 2 I S1 ⎞
⎢ 1 ⎝ 2I S ⎠ ⎥ = 0.025ln⎢ (1.05)⎥ = 2.50 mV
I
=
I
:
∆V
=
V
ln
a
=
0.025ln
⎜
⎟
( ) C 2 C1
BE
T
⎢() ⎛ ∆I S ⎞ ⎥
⎝ IC1 I S 2 ⎠
⎢⎣(0.95)⎥⎦
⎟⎥
⎢ I S ⎜1−
⎝ 2I S ⎠ ⎦
⎣
⎡ (1.10)⎤
(b) ∆VBE = 0.025ln⎢⎢ 0.90 ⎥⎥ = 5.02 mV
)⎦
⎣(
⎛ ∆I S ⎞
⎜1+
⎟
⎛V − V ⎞
⎛ 0.001⎞
IS ⎠
∆I
I S1 ⎝
(c) I = ⎛ ∆I ⎞ = exp⎜⎝ BE2V BE1 ⎟⎠ = exp⎜⎝ 0.025⎟⎠ = 1.04 → I S = 0.02 or 2%
S2
T
S
S
⎜1−
⎟
IS ⎠
⎝
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-81
15.130
(a) For v1 = 0 = v2 , VBE1 = VBE 2 and the collector currents are the same.
So, VOS = 0. Only the base currents will be mismatched.
⎛ VBE ⎞⎛ VCE ⎞
⎛ VBE ⎞⎛ VCE ⎞
1
+
|
I
I
=
I
1−
0.025
exp
=
I
1+
0.025
exp
b
⎜
⎟
⎜
⎟
( ) C1 S (
) ⎝ V ⎠⎝ V ⎠ C 2 S (
) ⎜⎝ V ⎟⎠⎜⎝1 + V ⎟⎠
T
A
T
A
⎛ V ⎞⎛ V ⎞
⎛ V ⎞⎛ V ⎞
I +I
∆IC = IC 2 − IC1 = 0.05I S exp⎜ BE ⎟⎜1 + CE ⎟ | IC = C1 C 2 = I S exp⎜ BE ⎟⎜1 + CE ⎟
VA ⎠
2
VA ⎠
⎝ VT ⎠⎝
⎝ VT ⎠⎝
∆I
∆I
VOS = C = VT C = 0.025V (0.05)= 1.25 mV
gm
IC
⎞
⎞
⎛ VBE ⎞⎛
⎛ VBE ⎞⎛
VCE
VCE
⎜
⎟
⎜
c
1
+
1
+
I
=
I
exp
=
I
exp
|
I
( ) C1 S ⎜⎝ V ⎟⎠⎜ V 1+ 0.025 ⎟ C 2 S ⎜⎝ V ⎟⎠⎜ V 1− 0.025 ⎟⎟
)⎠
)⎠
T ⎝
T ⎝
A(
A(
⎛ V ⎞⎛
⎛ V ⎞⎛
V
V ⎞
V ⎞
∆IC = IC 2 − IC1 ≅ I S exp⎜ BE ⎟⎜1-1.025 CE −1+ 0.975 CE ⎟ = I S exp⎜ BE ⎟⎜0.05 CE ⎟
VA
VA ⎠
VA ⎠
⎝ VT ⎠⎝
⎝ VT ⎠⎝
⎛ VCE ⎞
⎜
⎟
⎛ VBE ⎞⎛ VCE ⎞
IC1 + IC 2
∆IC
∆IC
⎝ VA ⎠
IC =
= I S exp⎜
= VT
= 0.025V (0.05)
⎟⎜1 +
⎟ | VOS =
⎛ VCE ⎞
2
VA ⎠
gm
IC
⎝ VT ⎠⎝
⎜1 +
⎟
VA ⎠
⎝
V
For CE = 0.1, VOS = 114 µV
VA
(d ) V
OD
VOS =
[
]
= IC (RC + 0.025RC )− (RC − 0.025RC ) = 0.05IC RC
VOD
V
= VT OD = 0.025V (0.05) = 1.25 mV
g m RC
IC RC
15.131
⎛V ⎞
⎛ V + 0.002 ⎞ IS1
⎛ 0.002 ⎞
IS1 exp⎜ BE1 ⎟ = IS 2 exp⎜ BE1
= exp⎜
⎟ = 1.08 | IS1 = 1.08IS 2
⎟ |
⎝ 0.025 ⎠
VT
⎝ VT ⎠
⎝
⎠ IS 2
I +I
∆IS 0.08
∆IS = IS1 − IS 2 = 0.08IS 2 | IS = S1 S 2 = 1.04IS 2 |
=
= 7.7%
2
IS 1.04
⎛ 10V ⎞
⎛ 10V ⎞
β F1 = 100(1+ 0.025)⎜1+
⎟ = 123 | β F 2 = 100(1− 0.025)⎜1+
⎟ = 117
⎝ 50V ⎠
⎝ 50V ⎠
100µA
100µA
IB1 =
= 0.813 µA | IB 2 =
= 0.855 µA
123
117
Note: IOS = -42.0 nA.
15-82
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.132
(a) ID =
(250)(1± 0.05) µA
2
[2 − (1± 0.025)]
2
| IDmax =
(250)(1+ 0.05) µA
2
[1+ 0.025]
V
V
2
2
(250)(1− 0.05) µA 1− 0.025 2 = 113µA
IDmin =
[
]
2
V2
138µA +113µA
∆ID
= 125.5µA | ∆ID = 138µA −113µA = 25µA |
= 19.8%
ID =
ID
2
(b) IDmax =
(250)(1+ 0.05) µA
2
=138µA
[3 + 0.025]
2
=1.20mA
V
2
(250)(1− 0.05) µA 3 − 0.025 2 = 1.05mA
IDmin =
[
]
2
V2
1.20mA +1.05mA
∆ID
= 1.125mA | ∆ID = 1.20mA −1.05mA = 0.150mA |
= 13.3%
ID =
ID
2
15.133
VGS1 = VTN +
VGS 2 = VTN +
2
1
2IDS1
2IDS1
2IDS1 ⎛ ∆ (W /L)⎞
= VTN +
≅ VTN +
⎜1−
⎟
⎛W ⎞
⎛W ⎞
⎛W ⎞
∆ (W /L)
4 (W /L) ⎠
K n' ⎜ ⎟
K n' ⎜ ⎟ 1+
K n' ⎜ ⎟ ⎝
⎝ L ⎠1
⎝L⎠
⎝L⎠
2(W /L)
1
2IDS 2
2IDS 2
2IDS 2 ⎛ ∆ (W /L)⎞
= VTN +
≅ VTN +
⎜1+
⎟
∆ (W /L)
4 (W /L) ⎠
' ⎛W ⎞
' ⎛W ⎞
' ⎛W ⎞ ⎝
Kn ⎜ ⎟
K n ⎜ ⎟ 1−
Kn ⎜ ⎟
⎝ L ⎠2
⎝L⎠
⎝L⎠
2(W /L)
IDS 2 = IDS1 : VGS 2 − VGS1 =
⎛ ∆ (W /L)⎞
2IDS 2 ⎛ ∆ (W /L)⎞
⎜
⎟ = (VGS − VTN )⎜
⎟
2(W /L) ⎠
' ⎛ W ⎞ ⎝ 2(W /L ) ⎠
⎝
Kn⎜ ⎟
⎝L⎠
⎛ ∆ (W /L)⎞
⎛ 0.10 ⎞
⎟ = (0.5)⎜
⎟ = 25 mV
⎝ 2 ⎠
⎝ 2(W /L ) ⎠
(a) ∆VGS = (VGS − VTN )⎜
(b)
∆ (W /L)
∆VGS
0.003
=2
=2
= 1.2 % |
0.5
(W /L)
(VGS − VTN )
(c )
∆ (W /L )
∆VGS
0.001
=2
=2
= 0.4 %
0.5
(W /L)
(VGS − VTN )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-83
15.134
⎡⎛ W ⎞ ⎛ W ⎞ ⎤ K '
2
∆ID = ID 2 − ID1 = ⎢⎜ ⎟ − ⎜ ⎟ ⎥ n (VGS − VTN ) (1+ λVDS )
⎣⎝ L ⎠ 2 ⎝ L ⎠1⎦ 2
⎛ 0.05 ⎞
∆I
∆I
∆ID = 0.05ID | VOS = D = (VGS − VTN ) D = 0.75V ⎜
⎟ = 18.8 mV
⎝ 2 ⎠
gm
2ID
K
2
2
(b) ∆ID = ID 2 − ID1 = n (1+ λVDS ) (VGS − VTN + 0.025VTN ) − (VGS − VTN − 0.025VTN )
2
ID
K
∆ID = n (1+ λVDS )(0.1VTN )(VGS − VTN ) = (0.1VTN )
2
(VGS − VTN )
(a) Assume active region operation :
[
]
∆I
∆ID
= (VGS − VTN ) D = 0.05 VTN | If VTN = 1V, VOS = 50 mV
gm
2ID
K
2
(c ) ∆ID = ID 2 − ID1 = n (VGS − VTN ) [(1+ λVDS + 0.025 λVDS ) − (1+ λVDS − 0.025 λVDS )]
2
⎛ 0.05 ⎞⎛ λVDS ⎞
0.05 λVDS
∆I
∆I
∆ID = ID 2 − ID1 = ID
| VOS = D = (VGS − VTN ) D = 0.75V ⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 1+ λVDS ⎠
1+ λVDS
gm
2ID
VOS =
If λVDS = 0.1, VOS = 1.71 mV
(d) VOD = ID [(RD + 0.025RD ) − (RD − 0.025RD )] = 0.05ID RD
VOS =
VOD
V
0.05ID
= OD =
(VGS − VTN ) = 0.025(0.75V ) = 18.8 mV
Avt gm RD
2ID
15.135
(a) I
OX
⎛W ⎞
⎜ ⎟
1+ λVDSX
⎝ L ⎠X
=
I
⎛ W ⎞ REF 1+ λVDS1
⎜ ⎟
⎝ L ⎠1
1
| RoutX
(
(
= λ
+ VDSX
IOX
)
)
2 30x10−6
2I D1
VDS1 = VGS1 = VTN +
= 0.75 +
= 1.52V
Kn
4 25x10−6
1
+ 10
10
0.015
IO2 = (30µA)
= 84.3µA | Rout2 =
= 909kΩ
84.3µA
4
1+ 0.015(1.52)
1+ 0.015(10)
1
+8
20
0.015
= 164µA | Rout3 =
= 455kΩ
IO3 = (30µA)
164µA
4
1+ 0.015(1.52)
1+ 0.015(8)
1
+ 12
40
0.015
= 346µA | Rout 4 =
= 227kΩ
IO 4 = (30µA)
346µA
4
1+ 0.015(1.52)
1+ 0.015(12)
15-84
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.135 cont.
⎛W ⎞
⎜ ⎟
L
1+ λV
(b) IOX = ⎝⎛W⎠⎞X I REF 1+ λVDSX
DS1
⎜ ⎟
⎝ L ⎠1
1
| RoutX = λ
(
(
+ VDSX
IOX
)
)
2 50x10−6
2I D1
VDS1 = VGS1 = VTN +
= 0.75 +
= 1.75V
Kn
4 25x10−6
1
+ 10
10
IO2 = (50µA)
= 140µA | Rout2 = 0.015
= 548kΩ
140µA
4
1+ 0.015(1.75)
1+ 0.015(10)
1
+8
20
IO3 = (50µA)
= 273µA | Rout3 = 0.015
= 274kΩ
273µA
4
1+ 0.015(1.75)
1+ 0.015(8)
1
+ 12
1+ 0.015(12)
40
= 575µA | Rout 4 = 0.015
= 137kΩ
IO 4 = (50µA)
575µA
4
1+ 0.015(1.75)
(c) I
O2
IO 4 =
=
40
(30µA)= 300 µA | Rout 4 = ∞
4
15.136
(a) I
OX
⎛W ⎞
⎜ ⎟
1+ λVDSX
⎝ L ⎠X
=
I REF
⎛W ⎞
1+ λVDS1
⎜ ⎟
⎝ L ⎠1
(
(
)
)
2 30x10−6
2I D1
| VDS1 = VGS1 = VTN +
= 0.75 +
= 1.73V
Kn
2.5 25x10−6
1+ 0.015(10)
1+ 0.015(8)
10
20
30µA)
= 135 µA | IO3 =
30µA)
= 262 µA
(
(
2.5
2.5
1+ 0.015(1.73)
1+ 0.015(1.73)
IO2 =
IO 4 =
(b) V
1+ 0.015(12)
40
30µA)
= 552 µA
(
2.5
1+ 0.015(1.73)
DS1
IO3 =
10
20
30µA) = 75 µA | Rout2 = ∞ | IO3 = (30µA)= 150 µA | Rout3 = ∞
(
4
4
= 0.75 +
(
) = 1.27V
6(25x10 )
2 20x10−6
−6
| IO2 =
1+ 0.015(10)
10
20µA)
= 37.6 µA
(
6
1+ 0.015(1.27)
1+ 0.015(8)
1+ 0.015(12)
20
40
20µA)
= 73.3 µA | IO 4 = (20µA)
= 154 µA
(
6
6
1+ 0.015(1.27)
1+ 0.015(1.27)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-85
15.137
(a) For λ = 0, I
O2
⎛ 10 ⎞
⎛ 20 ⎞
⎛ 40 ⎞
= 30µA⎜ ⎟ = 75µA, IO3 = 30µA⎜ ⎟ = 150µA, IO2 = 30µA⎜ ⎟ = 300µA
⎝4⎠
⎝4⎠
⎝4⎠
(b) From Prob. 16.8, I
O2
= 84.3µA, IO3 = 164µA, IO 4 = 346µA
∆IO2 84.3 − 75
∆IO3 164 −150
∆IO 4 346 − 300
=
= 0.124 LSB,
=
= 0.187 LSB,
=
= 0.613 LSB
IO2
IO2
IO2
75
75
75
15.138
*Problem 15.135(a) - NMOS Current Source Array
IREF 0 1 DC 30U
VD2 2 0 DC 10 AC 1
VD3 3 0 DC 8 AC 1
VD4 4 0 DC 12 AC 1
M1 1 1 0 0 NFET W=4U L=1U
M2 2 1 0 0 NFET W=10U L=1U
M3 3 1 0 0 NFET W=20U L=1U
M4 4 1 0 0 NFET W=40U L=1U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.015
.OP
.AC LIN 1 1000 1000
.PRINT AC IM(VD2) IM(VD3) IM(VD4) IP(VD2) IP(VD3) IP(VD4)
.END
The results are identical to the hand calculations.
15.139
2
5 + VGS 1
15x10-6 ⎛ 2 ⎞
5 + VGS 1
I D1 =
|
→ VGS 1 = −2.985V
⎜ ⎟(VGS 1 + 0.9) (1− 0.01VGS 1 )=
R
2 ⎝ 1⎠
3x10 4
I REF =
5 − 2.985
= 67.2µA
3x10 4
1
+ VDS 2
2
15x10-6 ⎛ 8 ⎞
100 + 5
λ
IO2 =
=
= 383kΩ
⎜ ⎟(−2.985 + 0.9) 1− 0.01(−5) = 274µA | Rout 2 =
274µA
2 ⎝ 1⎠
IO 2
2
15x10-6 ⎛16 ⎞
100 + 10
= 192kΩ
IO3 =
⎜ ⎟(−2.985 + 0.9) 1− 0.01(−10) = 574µA | Rout 3 =
574µA
2 ⎝1⎠
[
[
15-86
]
]
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.140
5 + VGS1
15x10-6 ⎛ 3.3 ⎞
5 + VGS 1
2
|
→ VGS 1 = −2.655V
⎜ ⎟(VGS 1 + 0.9)1 (1− 0.01VGS 1 )=
R
2 ⎝ 1 ⎠
3x10 4
2
5 − 2.655
15x10-6 ⎛ 8 ⎞
=
78.2
µ
A
|
I
=
I REF =
⎜ ⎟(−2.655 + 0.9) 1− 0.01(−5) = 194µA
O2
4
2 ⎝1⎠
3x10
-6 ⎛
2
15x10 16 ⎞
IO3 =
⎜ ⎟(−2.655 + 0.9) 1− 0.01(−10) = 407µA
2 ⎝1⎠
5+V
15x10-6 ⎛ 4 ⎞
5+V
(b) I D1 = R GS1 | 2 ⎜⎝ 1 ⎟⎠(VGS1 + 0.9)12 (1− 0.01VGS1)= 5x10GS41 → VGS1 = −2.241V
2
5 − 2.241
15x10-6 ⎛ 8 ⎞
=
55.2
µ
A
|
I
=
I REF =
⎜ ⎟(−2.241+ 0.9) 1− 0.01(−5) = 113µA
O2
4
2 ⎝1⎠
3x10
2
15x10-6 ⎛16 ⎞
IO3 =
⎜ ⎟(−2.241+ 0.9) 1− 0.01(−10) = 237µA
2 ⎝1⎠
(a) I D1 =
[
[
]
[
]
]
[
]
15.141
*Problem 15.141 - PMOS Current Source Array
RREF 0 1 30K
VSS 4 0 DC 5
VD2 2 0 DC 0 AC 1
VD3 3 0 DC -5 AC 1
M1 1 1 4 4 PFET W=2U L=1U
M2 2 1 4 4 PFET W=8U L=1U
M3 3 1 4 4 PFET W=16U L=1U
.MODEL PFET PMOS KP=15U VTO=-0.9 LAMBDA=0.01
.OP
.AC LIN 1 1000 1000
.PRINT AC IM(VD2) IM(VD3) IP(VD2) IP(VD3)
.END
The results are identical to the hand calculations.
15.142
K ' ⎛W ⎞
2
2
15x10-6 ⎛ 8 ⎞
I D2 = p ⎜ ⎟(VGS 2 − VTP ) 1+ λ VDS 2 | 55x10-6 =
⎜ ⎟(VGS1 + 0.9) 1+ 0.01−5
2 ⎝ L⎠
2 ⎝ 1⎠
⎛W ⎞
⎛ 8⎞
⎜ ⎟ 1+ λ VDS 2
⎜ ⎟ 1+ 0.01(5)
I D2 ⎝ L ⎠2
55µA
⎝ 1 ⎠2
VGS1 = −1.834V |
=
|
=
→ I REF = 13.3µA
⎛ 2⎞
I REF
I REF ⎛W ⎞
⎜ ⎟ 1+ λ VDS1
⎜ ⎟ 1+ 0.01(1.834)
⎝ L ⎠1
⎝ 1 ⎠1
5 + VGS1
5 −1.834
I REF =
| R=
= 238 kΩ
R
13.3µA
[
]
[
[
]
[
]
[
[
]
]
]
15.143
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-87
(a) I
REF
=
⎛ V ⎞⎛ 14.3 VBE ⎞
12 − 0.7
= 151 µA | I REF = IC1 + (1+ 5 + 8.3)I B | I REF = I S exp⎜ BE ⎟⎜1+
+
⎟
4
7.5x10
⎝ VT ⎠⎝ β FO VA ⎠
V
5
1+ CE 2
1+
⎛ VBE ⎞⎛ VCE2 ⎞
VA
60
IO2 = 5I S exp⎜
| IO2 = 5(151µA)
= 631 µA
⎟⎜1+
⎟ = 5I REF
14.3 VBE
14.3 0.7
VA ⎠
⎝ VT ⎠⎝
1+
+
1+
+
β FO VA
50
60
VA + VCE2
60 + 5
=
= 103 kΩ
IC 2
6.31x10−4
V
3
1+ CE 3
1+
VA
60
IO3 = 8.3I REF
= 8.3(151µA)
= 1.02 mA
14.3 VBE
14.3 0.7
1+
+
1+
+
β FO VA
50
60
Rout2 = ro2 =
Rout3 = ro3 =
VA + VCE3
60 + 3
=
= 61.8 kΩ
IC 3
1.02x10−3
(b) Since all areas are scaled equally, the current ratios stay the same, and there is no change
from part (a). This ignores the slight change in VBE of Q1 due to its area change.
I
12 − 0.7 − 0.7
= 141 µA | I REF = IC1 + (1+ 5 + 8.3) B
4
β FO + 1
7.5x10
⎛ V ⎞⎛
2V ⎞
14.3
I REF = I S exp⎜ BE ⎟⎜⎜1+
+ BE ⎟⎟
⎝ VT ⎠⎝ β FO (β FO + 1) VA ⎠
V
1+ CE 2
⎛ VBE ⎞⎛ VCE2 ⎞
VA
IO2 = 5I S exp⎜
⎟⎜1+
⎟ = 5I REF
14.3
2V
VA ⎠
⎝ VT ⎠⎝
1+
+ BE
β FO (β FO + 1) VA
(c) I
REF
=
5
V + VCE2 60V + 5V
60
IO2 = 5(141µA)
= 745 µA | Rout2 = ro2 = A
=
= 87.2 kΩ
14.3 1.4
IC 2
745µA
1+
+
50(51) 60
1+
VCE3
3
1+
VA
60
IO3 = 8.3I REF
= 8.3(141µA)
= 1.20 mA
14.3
2VBE
14.3 1.4
1+
+
1+
+
50(51) 60
β FO (β FO + 1) VA
1+
Rout3 = ro3 =
15-88
VA + VCE3 60V + 3V
=
= 52.5 kΩ
IC 3
1.20mA
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.144
*Problem 15.144 – Figure P15.143(a) - NPN Current Source Array
RREF 2 1 75K
VCC 2 0 DC 12
VC2 3 0 DC 5 AC 1
VC3 4 0 DC 3 AC 1
Q1 1 1 0 NBJT 1
Q2 3 1 0 NBJT 5
Q3 4 1 0 NBJT 8.3
.MODEL NBJT NPN BF=50 VA=60
.OP
.AC LIN 1 1000 1000
.PRINT AC IM(VC2) IM(VC3) IP(VC2) IP(VC3)
.END
*Problem 15.144 – Figure 15.143(b) - Buffered NPN Current Source Array
RREF 2 5 75K
VCC 2 0 DC 12
VC2 3 0 DC 5 AC 1
VC3 4 0 DC 3 AC 1
Q1 5 1 0 NBJT 1
Q2 3 1 0 NBJT 5
Q3 4 1 0 NBJT 8.3
Q4 2 5 1 NBJT 1
.MODEL NBJT NPN BF=50 VA=60
.OP
.AC LIN 1 1000 1000
.PRINT AC IM(VC2) IM(VC3) IP(VC2) IP(VC3)
.END
The results are almost identical to the hand calculations.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-89
15.145
⎛ VBE ⎞⎛ 14.3 VBE ⎞
=
I
+
1+
5
+
8.3
I
=
I
+
14.3I
=
I
exp
+
⎜
⎟⎜1+
⎟
(
)
REF
C1
B
C1
B
S
⎝ VT ⎠⎝ β FO VA ⎠
⎛ V ⎞⎛ 14.3 0.7 ⎞
⎛ VBE ⎞
+
I REF ≅ I S exp⎜ BE ⎟⎜1+
⎟
⎟ = 1.298I S exp⎜
50
60 ⎠
⎝ VT ⎠⎝
⎝ VT ⎠
V
1+ CE2
⎛ VBE ⎞⎛ VCE3 ⎞
150µA(1.298)
VA
IO3 = 8.3I S exp⎜
| I REF =
= 22.3µA
⎟⎜1+
⎟ = 8.3I REF
⎛
VA ⎠
1.298
3⎞
⎝ VT ⎠⎝
8.3⎜1+ ⎟
⎝ 60 ⎠
V
5
1+ CE2
1+
12 − 0.7
12 − 0.7
VA
I REF =
| R=
= 507 kΩ | IO2 = 5I REF
= 5(22.3µA) 60 = 93.1 µA
R
22.3µA
1.298
1.298
⎛ VBE ⎞⎛
14.3
14.3I B
2VBE ⎞
IB
⎜
b
1+
I
=
I
+
1+
5
+
8.3
=
I
+
=
I
exp
+
( ) REF C1 (
)β + 1 C1 β + 1 S ⎜⎝ V ⎟⎠⎜ β β + 1 V ⎟⎟
) A⎠
FO
FO
T ⎝
FO ( FO
⎛ V ⎞⎛
⎛V ⎞
14.3 1.4 ⎞
⎟ = 1.029I S exp⎜ BE ⎟
+
I REF ≅ I S exp⎜ BE ⎟⎜⎜1+
⎝ VT ⎠⎝ 50(51) 60 ⎟⎠
⎝ VT ⎠
(a) I
V
1+ CE2
⎛ VBE ⎞⎛ VCE3 ⎞
VA
IO3 = 8.3I S exp⎜
⎟⎜1+
⎟ = 8.3I REF
VA ⎠
1.029
⎝ VT ⎠⎝
I REF
150µA(1.029)
= 17.7µA
⎛
3⎞
8.3⎜1+ ⎟
⎝ 60 ⎠
V
5
1+ CE2
1+
12 − 0.7 − 0.7
12 −1.4
VA
=
| R=
= 599 kΩ | IO2 = 5I REF
= 5(17.7µA) 60 = 93.2 µA
R
17.7µA
1.029
1.029
15-90
| I REF =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.146
(a) I
REF
=
⎛ VBE ⎞⎛ 7.65 VBE ⎞
⎛ 5 8.3 ⎞
12 − 0.7
= 151 µA | I REF = IC1 + ⎜1+ +
+
⎟⎜1+
⎟
⎟ I B | I REF = I S exp⎜
3
75x10
⎝ 2 2 ⎠
⎝ VT ⎠⎝ β FO VA ⎠
V
5
1+ CE2
1+
⎛
⎞
⎛
⎞
VCE2
5
VBE
VA
75
IO2 = I S exp⎜
| IO2 = 2.5(151µA)
= 376 µA
⎟⎜1+
⎟ = 2.5I REF
7.65 VBE
7.65 0.7
VA ⎠
2
⎝ VT ⎠⎝
1+
+
1+
+
β FO VA
125 75
VCE3
3
1+
8.3
VA
75
IO3 =
I REF
= 4.15(151µA)
= 609 µA
7.65 VBE
7.65 0.7
2
1+
+
1+
+
β FO VA
125 75
⎛ 5 8.3 ⎞ I
12 − 0.7 − 0.7
(b) I REF = 75x104 = 141 µA | I REF = IC1 + ⎜⎝1+ 2 + 2 ⎟⎠ β B+ 1
FO
⎛ V ⎞⎛
2V ⎞
7.65
I REF = I S exp⎜ BE ⎟⎜⎜1+
+ BE ⎟⎟
⎝ VT ⎠⎝ β FO (β FO + 1) VA ⎠
V
1+ CE2
⎛
⎞
⎛
⎞
5
V
V
VA
IO2 = I S exp⎜ BE ⎟⎜1+ CE2 ⎟ = 2.5I REF
7.65
2V
VA ⎠
2
⎝ VT ⎠⎝
1+
+ BE
β FO (β FO + 1) VA
1+
5
75
IO2 = 2.5(141µA)
= 370 µA
7.65
1.4
1+
+
125(126) 75
1+
VCE3
3
1+
8.3
VA
75
I REF
= 4.15(141µA)
= 596 µA
IO3 =
7.65
2VBE
6.65
1.4
2
1+
+
1+
+
β FO (β FO + 1) VA
125(126) 75
1+
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-91
15.147
⎛ 5 8.3⎞ I B
12 − 0.7 − 0.7
=
106
µ
A
|
I
=
I
+
1+ +
⎜
⎟
REF
C1
100x103
⎝ 3 3 ⎠ β FO + 1
⎛ V ⎞⎛
2V ⎞
5.43
I REF = I S exp⎜ BE ⎟⎜⎜1+
+ BE ⎟⎟
⎝ VT ⎠⎝ β FO (β FO + 1) VA ⎠
V
1+ CE 2
⎛
⎞
⎛
⎞
V
5
5
V
VA
IO2 = I S exp⎜ BE ⎟⎜1+ CE 2 ⎟ = I REF
5.43
2V
VA ⎠ 3
3
⎝ VT ⎠⎝
1+
+ BE
β FO (β FO + 1) VA
I REF =
5
IO2 = (106µA)
3
1+
5
75
= 185 µA
5.43
1.4
+
100(101) 75
1+
VCE 3
3
1+
8.3
8.3
VA
I REF
=
IO3 =
(106µA) 5.4375 1.4 = 299 µA
5.43
2VBE
3
3
1+
+
1+
+
100(101) 75
β FO (β FO + 1) VA
1+
15-92
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.148
(a) I
REF
=
⎛ VBE ⎞⎛ 14.3 VBE ⎞
12 − 0.7
=
80.7
µ
A
|
I
=
I
+
1+
5
+
8.3
I
|
I
=
I
exp
+
⎟
⎜
⎟⎜1+
(
)
REF
C1
B
REF
S
1.4x105
⎝ VT ⎠⎝ β FO VA ⎠
VCE 2
5
1+
⎛ V ⎞⎛ V ⎞
V
75
A
IO2 = 5I S exp⎜ BE ⎟⎜1+ CE2 ⎟ = 5I REF
| IO2 = 5(80.7µA)
= 383 µA
14.3 VBE
14.3 0.7
VA ⎠
⎝ VT ⎠⎝
1+
+
1+
+
125 75
β FO VA
1+
VCE 3
3
1+
VA
75
IO3 = 8.3I REF
= 8.3(80.7µA)
= 620 µA
14.3 VBE
14.3 0.7
1+
+
1+
+
125 75
β FO VA
⎛ VBE ⎞⎛
IB
14.3I B
2VBE ⎞
14.3
⎜
(b) I REF = IC1 + (1+ 5 + 8.3)β + 1 = IC1 + β + 1 = I S exp⎜⎝ V ⎟⎠⎜1+ β β + 1 + V ⎟⎟
) A⎠
FO
FO
T ⎝
FO ( FO
⎛ V ⎞⎛
⎛V ⎞
1.4 ⎞
14.3
⎟ = 1.029I S exp⎜ BE ⎟
+
I REF = I S exp⎜ BE ⎟⎜⎜1+
⎝ VT ⎠⎝ 125(126) 75 ⎟⎠
⎝ VT ⎠
V
1+ CE 2
⎛ V ⎞⎛ V ⎞
620µA(1.02)
VA
| I REF =
= 73.3µA
IO3 = 8.3I S exp⎜ BE ⎟⎜1+ CE3 ⎟ = 8.3I REF
⎛
VA ⎠
1.02
3⎞
⎝ VT ⎠⎝
8.3⎜1+ ⎟
⎝ 75 ⎠
1+
12 − 0.7 − 0.7
12 −1.4
| R=
= 145 kΩ
R
73.3µA
V
5
1+ CE 2
1+
VA
= 5(73.3µA) 75 = 383 µA - Correct.
Checking : IO2 = 5I REF
1.02
1.02
I REF =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-93
15.149
Use βFO = 50 and VA = 60 V.
(a) I
REF
=
⎛ V ⎞⎛ 14.3 VBE ⎞
12 − 0.7
= 113 µA | I REF = IC1 + (1+ 5 + 8.3)I B | I REF = I S exp⎜ BE ⎟⎜1+
+
⎟
5
10
⎝ VT ⎠⎝ β FO VA ⎠
V
5
1+ CE 2
1+
⎛ VBE ⎞⎛ VCE2 ⎞
VA
60
IO2 = 5I S exp⎜
| IO2 = 5(113µA)
= 472 µA
⎟⎜1+
⎟ = 5I REF
14.3 VBE
14.3 0.7
VA ⎠
⎝ VT ⎠⎝
1+
+
1+
+
β FO VA
50
60
VCE 3
3
1+
VA
60
IO3 = 8.3I REF
= 8.3(113µA)
= 759 µA
14.3 VBE
14.3 0.7
1+
+
1+
+
β FO VA
50
60
1+
6
(b) IO2 = 5(113µA) 14.360 0.7 = 479 µA | No change in IO3.
1+
+
50
60
11− 0.7
(c) I REF = 105 = 103 µA | IO2 and IO3 are proportional to I REF
103µA
103µA
IO2 = 472µA
= 430 µA | IO3 = 472µA
= 430 µA
113µA
113µA
1+
103µA
= 692 µA
113µA
60 + 5
∆V
1V
=
= 138kΩ | ∆IO2 =
=
= 7.25µA
472µA
ro2 138kΩ
IO3 ∝ I REF | IO2 = 759µA
(d ) R
out2
= ro2 =
VA + VCE2
IC 2
IO2−6V − IO2−5V = 479µA − 472µA = 7µA - Agrees within the calculation precision.
15.150
I REF =
⎛ V ⎞
15 − 0.7
= 238µA | I REF = 2IC 1 + (2 + 1+ 6 + 9)I B = 2β FO ⎜1+ EC1 ⎟ I B + 18I B
4
VA ⎠
6x10
⎝
238µA
= 2.00µA
⎛ 0.7 ⎞
18 + 2(50)⎜1+
⎟
⎝ 60 ⎠
⎛ V ⎞
⎛ 15 ⎞
60 + 15
IO2 = β FO ⎜1+ EC 2 ⎟ I B = 50⎜1+ ⎟(2.00µA) = 125µA | Rout 2 = ro2 =
= 600 kΩ
VA ⎠
1.25x10−4
⎝ 60 ⎠
⎝
⎛ V ⎞
⎛
9⎞
60 + 9
= 100 kΩ
IO3 = 6β FO ⎜1+ EC 3 ⎟ I B = 300⎜1+ ⎟(2.00µA) = 690µA | Rout 3 = ro3 =
VA ⎠
6.90x10−4
⎝ 60 ⎠
⎝
⎛ V ⎞
⎛ 27 ⎞
60 + 27
= 66.4 kΩ
IO 4 = 9β FO ⎜1+ EC 4 ⎟ I B = 450⎜1+ ⎟(2.00µA)= 1.31mA | Rout 4 = ro 4 =
VA ⎠
1.31x10−3
⎝ 60 ⎠
⎝
IB =
15-94
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.151
⎛ V ⎞
15 − 0.7
= 238µA | I REF = 2IC 1 + (2 + 1+ 6 + 9)I B = 2β FO ⎜1+ EC1 ⎟ I B + 18I B
VA ⎠
R
⎝
⎡
⎛ VEC 3 ⎞
⎛ 0.7 ⎞⎤
65µA
= I B ⎢18 + 2(50)⎜1+
= 0.189µA
⎟IB | IB =
⎟⎥ = 119I B | IO3 = 6β FO ⎜1+
⎛
VA ⎠
9⎞
⎝ 60 ⎠⎦
⎝
⎣
300⎜1+ ⎟
⎝ 60 ⎠
I REF =
I REF
15 − 0.7
= 63.8 kΩ
22.4µA
⎛ V ⎞
⎛ 15 ⎞
IO2 = β FO ⎜1+ EC 2 ⎟ I B = 50⎜1+ ⎟(0.189µA)= 11.8 µA
VA ⎠
⎝ 60 ⎠
⎝
⎛ V ⎞
⎛ 27 ⎞
IO 4 = 9β FO ⎜1+ EC 4 ⎟ I B = 450⎜1+ ⎟(0.189µA)= 123 µA
VA ⎠
⎝ 60 ⎠
⎝
I REF = 119I B = 22.4µA | R =
15.152
15 V
2A
Q
1
9A
Q
Q
A
Q
R
6A
A
Q
3
2
4
5
I O2
I O4
I O3
+6 V
-12 V
(2 + 1+ 6 + 9)I B
15V − 0.7V − 0.7V
= 544kΩ | I REF = IC1 +
25µA
β FO + 1
⎛ V ⎞
18I B
25µA
I REF = 2β FO ⎜1+ EC1 ⎟ I B +
| IB =
= 0.2435µA
⎛
VA ⎠
β FO + 1
1.4 ⎞ 18
⎝
2(50)⎜1+
⎟+
⎝ 60 ⎠ 51
⎛ V ⎞
⎛ 15 ⎞
IO2 = β FO ⎜1+ EC 2 ⎟ I B = 50⎜1+ ⎟ I B = 15.2 µA
VA ⎠
⎝ 60 ⎠
⎝
⎛ V ⎞
⎛
9⎞
IO3 = 6β FO ⎜1+ EC 3 ⎟ I B = 300⎜1+ ⎟ I B = 84.0 µA
VA ⎠
⎝ 60 ⎠
⎝
⎛ V ⎞
⎛ 27 ⎞
50
IO 4 = 9β FO ⎜1+ EC 4 ⎟ I B = 450⎜1+ ⎟ I B = 159 µA | IC 5 = α F I E 5 = 18I B = 4.30 µA
VA ⎠
51
⎝ 60 ⎠
⎝
R=
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-95
15.153
VBE2 + I E 2 R2 = VBE3 + I E 3 R3 → I E3 =
R2
V −V
I E 2 + BE2 BE 3
R3
R3
In order to have equal base - emitter voltages, the two transistors must operate at the
equal collector − current densities :
I E2 I E 3 5I E 2
=
=
→ n = 10
2 A nA
nA
15.154
⎞ I
I REF
I
I
76 ⎛ I
= REF | I E1 = C1 = ⎜ REF ⎟ = REF
7 1.093
α F 75 ⎝1.093 ⎠ 1.079
1+
75
I
12V - 0.7V
12 = I REF (10kΩ)+ 0.7V + REF (10kΩ) | I REF =
= 586µA | I E1 = 543µA
1.079
10kΩ(1.927)
I REF = IC1 + 7I B | IC1 ≅
75
(543µA)= 1.07mA | VE1 = 543µA(10kΩ)= 5.43V
76
75
1
1
| IO3 = α F (4I E1 )= 4 (543µA) = 2.14mA |
=
= 46.6Ω
76
g m1 40(536µA)
I E2 = 2I E1 | IO2 = α F (2I E1 )= 2
I E3 = 4I E1
⎛ 1
⎞
Rth = R ⎜
+ R1 ⎟ = 10kΩ (46.6Ω + 10kΩ) = 5.01kΩ | Rth2 = Rth rπ 3 + (β o3 + 1)(2.5kΩ)
⎝ g m1
⎠
[
rπ 2 =
75(0.025V )
1.07mA
= 1.75kΩ | rπ 3 =
75(0.025V )
[
2.14mA
= 0.876kΩ | ro2 =
]
60 + (10 − 5.43)
]
Rth2 = 5.01kΩ 0.876kΩ + (76)(2.5kΩ) = 4.88kΩ
1.07mA
= 60.4kΩ
⎛
⎞
⎛
⎞
75(5kΩ)
β o R2
⎟ = 2.01 MΩ
Rout2 = ro2⎜1+
⎟ = 60.4kΩ⎜⎜1+
⎟
⎝ Rth2 + rπ 2 + R2 ⎠
⎝ 4.88kΩ + 1.75kΩ + 5kΩ ⎠
ro3 =
(60 + 10 − 5.43)V = 30.2kΩ
2.14mA
[
[
]
| Rth2 = Rth rπ 2 + (β o + 1)(5kΩ)
]
Rth3 = 5.01kΩ 1.75kΩ + (76)(5kΩ) = 4.95kΩ
⎛
⎞
⎛
⎞
75(2.5kΩ)
β o R3
⎜
⎟
=
30.2kΩ
1+
Rout3 = ro3⎜1+
⎟
⎜ 4.95kΩ + 0.876kΩ + 2.5kΩ ⎟ = 710 kΩ
⎝ Rth + rπ 3 + R3 ⎠
⎝
⎠
15.155
For VBE2 = VBE 3, IO3 R3 = IO2 R2 | R3 =
15-96
IO2
I
I
2
R2 = 3(5kΩ) = 15 kΩ | O2 = O3 | n =
IO3
2 A nA
3
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.156
⎞ I
I REF
I
I
76 ⎛ I
= REF | I E1 = C1 = ⎜ REF ⎟ = REF
13 1.173
α F 75 ⎝1.173 ⎠ 1.158
1+
75
I
12V - 0.7V
12 = I REF (10kΩ)+ 0.7V + REF (20kΩ) | I REF =
= 414µA | I E1 = 357µA
1.079
10kΩ(2.73)
I REF = IC1 + 13I B | IC1 ≅
75
(357µA)= 1.41 mA | VE1 = 357µA(20kΩ)= 7.14V
76
75
1
1
| IO3 = α F (8I E1 )= 8 (357µA)= 2.82 mA |
=
= 71.0Ω
76
g m1 40(352µA)
I E2 = 4I E1 | IO2 = α F (4I E1 )= 4
I E3 = 8I E1
⎛ 1
⎞
Rth = R ⎜
+ R1 ⎟ = 10kΩ (71.0Ω + 20kΩ)= 6.68kΩ | Rth2 = Rth rπ 3 + (β o3 + 1)(2.5kΩ)
⎝ g m1
⎠
[
rπ 2 =
75(0.025V )
1.41mA
= 1.33kΩ | rπ 3 =
75(0.025V )
[
2.82mA
= 0.665kΩ | ro2 =
]
60 + (10 − 7.14)
]
Rth2 = 6.68kΩ 0.665kΩ + (76)(2.5kΩ) = 6.45kΩ
1.41mA
= 44.6kΩ
⎛
⎞
⎛
⎞
75(5kΩ)
β o R2
⎜
⎟ = 1.35 MΩ
Rout2 = ro2⎜1+
⎟ = 44.2kΩ⎜1+
⎟
6.45kΩ
+
1.33kΩ
+
5kΩ
⎝ Rth2 + rπ 2 + R2 ⎠
⎝
⎠
ro3 =
(60 + 10 − 7.14)V = 22.3kΩ
2.82mA
[
[
]
| Rth2 = Rth rπ 2 + (β o + 1)(5kΩ)
]
Rth3 = 6.68kΩ 1.33kΩ + (76)(5kΩ) = 6.57kΩ
⎛
⎞
⎛
⎞
75(2.5kΩ)
β o R3
⎜
⎟ = 452 kΩ
Rout3 = ro3⎜1+
⎟ = 22.3kΩ⎜1+
⎟
6.57kΩ
+
0.665kΩ
+
2.5kΩ
⎝ Rth + rπ 3 + R3 ⎠
⎝
⎠
15.157
IC1 = IREF = 15µA | IB =
IC1
2ID
| VCE1 = VBE1 + VGS 3 = 0.7 + VTN +
⎛ V ⎞
Kn
β FO ⎜1+ CE1 ⎟
VA ⎠
⎝
15µA
4IB
| VCE1 = 1.45 +
| Solving iteratively yields IB = 0.147µA
⎛ VCE1 ⎞
50x10−6
100⎜1+
⎟
VA ⎠
⎝
⎛ V ⎞
⎛
5⎞
(75 + 5)V = 5.10 MΩ
IO 2 = β FO ⎜1+ CE 2 ⎟ IB = 100⎜1+ ⎟(0.147µA) = 15.7 µA | Rout = ro2 =
⎝ 75 ⎠
15.7µA
VA ⎠
⎝
⎛
I
5⎞
Note : If one assumes IB ≅ C1 = 0.15µA, then IO 2 = 100⎜1+ ⎟(0.15µA) = 16.0 µA, a very
⎝ 75 ⎠
β FO
IB =
similar result and a valid approximation, since VCE1 << VA .
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-97
15.158
Results from SPICE were IC1 = 15.0 µA, IO = 15.7 µA and Rout = 5.06 MΩ.
ELTOL
VNTOL

an

 


results are very close to our hand calculations. Remember, SPICE uses
ro3 =
 These
VA + VCB (75 + 4.3)V
≅
= 5.05 MΩ .
15.7µA
IC
15.159
(a) R
in
=
vref
iref
| v ref = (iref − g m1vbe1 )ro1 = iref ro1 − µ f 1vbe1
(
)
⎡ g r r
m3 π 1 π 2
vbe1 = v ref ⎢
⎢1+ g m3 rπ 1 rπ 2
⎣
(
⎛
rπ 1 ⎞
⎤
⎟
⎜ g m3
⎞
⎛
2 ⎟ = v ⎜ g m3rπ 1 ⎟
⎥≅v ⎜
ref
r
⎥ ref ⎜
⎝ 2 + g m3rπ 1 ⎠
⎦
1+ g m3 π 1 ⎟
⎝
2 ⎠
)
−1
1
ro1
1
1 ⎛ g m3rπ 1 ⎞
Rin =
≅
=
⎛ g m3rπ 1 ⎞ g m1 ⎛ g m3rπ 1 ⎞ g m1 ⎜⎝ 2 + g m3rπ 1 ⎟⎠
1+ µ f 1⎜
⎟
⎟
⎜
⎝ 2 + g m3rπ 1 ⎠
⎝ 2 + g m3rπ 1 ⎠
(
)(
)
From Prob. 15.157, I D = 0.294µA, and g m3 = 2 50x10−6 0.294x10−6 = 5.42x10−6 S
g m1 = 40(15µA)= 0.600mS
Rin ≅
rπ 1 =
⎡2.905⎤
1
⎢
⎥ = 5.35 kΩ
0.600mS ⎣0.905⎦
100
= 167kΩ g m3rπ 1 = 0.905
0.600mS
(b) SPICE yields Rin = 5.35 kΩ
15.160
⎛I I ⎞
10V − 0.7V − 0.7V
= 860µA ≅ IC1 | IO2 R2 = VT ln⎜ C1 S 2 ⎟
10kΩ
⎝ IO2 I S1 ⎠
⎛ 860µA 2 ⎞
10 4 IO2 = 0.025ln⎜
⎟ → IO2 = 12.4 µA | VE 2 ≅ 12.4 µA(10kΩ)= 0.124V
⎝ IO2 1 ⎠
60 + 5 − 0.124
ro2 =
= 5.23 MΩ | Rth is small and the voltage across R2 is also small.
12.4µA
I REF =
[
]
Rout2 ≅ ro2 (1+ g m2 R2 )= 5.23MΩ 1+ 40(12.4 µA)(10kΩ) = 31.2 MΩ
⎛ 860µA 12 ⎞
5x103 IO3 = 0.025ln⎜
⎟ → IO3 = 29.3 µA | VE 3 ≅ 29.3 µA(5kΩ)= 0.147V
⎝ IO2 1 ⎠
60 + 5 − 0.147
ro3 =
= 2.21 MΩ | Rout3 ≅ ro3 (1+ g m3 R3 )= 2.21MΩ 1+ 40(0.147) = 15.2 MΩ
29.3µA
[
15-98
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
]
15.161
(10 − 0.7 − 0.7)V = 115kΩ | R = VT ln⎛ IREF IS 2 ⎞ = 0.025V ln⎛ 75µA 2 ⎞ = 17.0 kΩ
R=
⎜
⎟
⎜
⎟
2
Io2 ⎝ IO 2 IS1 ⎠
75µA
5µA
⎝ 5µA 1 ⎠
⎛I
⎛ 75µA n ⎞
I ⎞
Io3 R3 = VT ln⎜ REF S 3 ⎟ | (10µA)(2kΩ) = 0.025V ln⎜
⎟ → n = 0.297
⎝ 10µA 1 ⎠
⎝ IO 3 IS1 ⎠
15.162
*Problem 15.162 - Buffered NPN Widlar Current Source Array
RREF 1 3 115K
VCC 1 0 DC 10
VC2 2 0 DC 5 AC 1
Q1 3 4 0 NBJT 1
Q2 2 4 5 NBJT 2
R2 5 0 17K
Q3 2 4 6 NBJT 0.297
R3 6 0 2K
Q4 1 3 4 NBJT 1
.MODEL NBJT NPN BF=100 VA=60
.OP
.AC LIN 1 1000 1000
.PRINT AC IC(Q2) IC(Q3)
.END
Results: IC1 = 75.7 µA, IO2 = 5.18 µA, IO3 = 10.5 µA, ROUT2 = 53.4 MΩ ROUT3 = 11.1 MΩ
15.163
I REF =
IO2 =
ro2 =
5V − 0.7V − 0.7V − (−5V )
40kΩ
= 215µA ≅ IC1 | IO2 =
VT ⎛ IC1 I S 2 ⎞
ln⎜
⎟
R2 ⎝ IO2 I S1 ⎠
0.025V ⎛ 215µA 10 ⎞
ln⎜
⎟ → IO2 = 22.7 µA | VE 2 ≅ 22.7µA(5kΩ)= 0.114V
5kΩ
⎝ IO2 1 ⎠
70 + 5 − 0.114
= 3.30 MΩ | Rth is small and the voltage across R2 is also small.
22.7µA
[
]
Rout2 ≅ ro2 (1+ g m2 R2 )= ro2 (1+ 40IC 2 R2 ) = 3.30 MΩ 1+ 40(0.114) = 18.3 MΩ
IO3 =
ro3 =
0.025V ⎛ 215µA 20 ⎞
ln⎜
⎟ → IO3 = 45.5 µA | VE 3 ≅ 45.4µA(2.5kΩ)= 0.114V
2.5kΩ ⎝ IO3 1 ⎠
70 + 5 − 0.114
= 1.65MΩ | Rout3 ≅ ro3 (1+ g m3 R3 )= 1.65MΩ 1+ 40(0.114) = 9.17 MΩ
45.5µA
[
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
]
15-99
15.164
5 − 0.7 − 0.7 − (−5) V
= 172 kΩ
R=
50µA
I ⎞ 0.025V ⎛ 50µA 10 ⎞
V ⎛I
R2 = T ln⎜ REF S 2 ⎟ =
ln⎜
⎟ = 9.78 kΩ
I o2 ⎝ IO2 I S1 ⎠ 10µA ⎝ 10µA 1 ⎠
⎛I
⎛ 50µA n ⎞
I ⎞
I o3 R3 = VT ln⎜ REF S 3 ⎟ | (10µA)(2kΩ)= 0.025V ln⎜
⎟ → n = 0.445
⎝ 10µA 1 ⎠
⎝ IO3 I S1 ⎠
[
]
15.165
For the current mirror at the bottom: IC 2 = β F IB 2 | IE 3 = IC1 + IB1 + IB 2 = nβ F IB 2 + nIB 2 + IB 2
IC 2 =
βF
β F 1+ n (β F + 1)
I
IE 3 | IO = α F IE 3 =
IC 2 | IC 2 = IREF − IB 3 = IREF − O
1+ n (β F + 1)
βF + 1
βF
βF
n+
IO =
1
1+ n(β F + 1) ⎛
β F (β F + 1)
I ⎞
n
IREF ≅
I ≅ nIREF for β F >> n
⎜ IREF − O ⎟ → IO =
n REF
n
1
βF + 1 ⎝
βF ⎠
1+
1+
+
β F β F (β F + 1)
βF
(a) IO ≅
1
50µA = 49.6 µA | Rout ≅
β o ro
=
125 40 − 0.7 − (−5) V
= 55.8 MΩ
49.6
µA
2
1
2
125
β r 125 44.3 V
3
50µA = 146 µA | Rout ≅ o o =
= 19.0 MΩ
(b) IO ≅
3
146
2
µ
A
2
1+
125
(c ) VCS = IO Rout = 146µA(19.0MΩ) = 2770V (d) VCB 3 = VEE − 0.7V − 0.7V ≥ 0 → VEE ≥ 1.40V
15-100
1+
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.166
i1
+
βo i 1
r π3
v1
r o3
vb
ix
1
gm1
i
i
vx
ix
ve
r o2
+
-
-
⎡ix ⎤ ⎡gm1 + gπ 3
−gπ 3 ⎤⎡v e ⎤
| ⎢ ⎥=⎢
⎥⎢ ⎥
⎣0 ⎦ ⎣gm1 − gπ 3 gπ 3 + go2 ⎦⎣v b ⎦
IC1 ≅ IC 2 ≅ IC 3 so the small - signal parameters are matched
⎛
β
1⎞
∆ = 2gm1gπ 3 + gm1go2 + gπ 3 go2 = gm1gπ 3 ⎜⎜2 + o + ⎟⎟ ≅ 2gm1gπ 3 for µ f >> β o >> 1
µf µf ⎠
⎝
g + go2
g −g
i ⎛ β ⎞
i
= x ⎜⎜1+ o ⎟⎟ ≅ x
| v b = −ix m1 π 3
v e = ix π 3
∆
2gm1 ⎝ µ f ⎠ 2gm1
∆
g + go2
i ⎛
1⎞ i
i
v b − v e = −ix m1
= − x ⎜⎜1+ ⎟⎟ ≅ x rπ 3 | i1 = gπ 3 (v b − v e ) = x
∆
2gπ 3 ⎝ µ f ⎠ 2
2
i
βr
βr
i
v
1
v x = v e + (ix − β oi1)ro3 = x + ix ro3 + β o ro3 x | Rout = x =
+ ro3 + o o3 ≅ o o3
2gm1
2
ix 2gm1
2
2
Rout =
vx
ix
15.l67
Rout ≅
β oro
2
≅
β oVA
2IO
=
β oVA
2nI REF
| For Prob. 15.165, Rout ≅
125(40)
2n (50µA)
=
50
MΩ
n
15.168
VCB3 = VC 3 − VBE3 − VBE 2 − (−VEE )≥ 0 → VC 3 ≥ −VEE + VBE 3 + VBE2
IC 3
I
I
I
I
β +1
+ VT ln C1 | IC1 + C1 + C1 = C 3 ≅ F
I
| From Prob.
IS3
I S1
β F nβ F α F
βF C3
VBE3 + VBE1 = VT ln
15.165 : IC3 ≅
n
1+
n
βF
5
I REF =
1+
5
125
15µA = 72.1µA | IC1 =
βF + 1
I = 72.0µA
1
1
β F C3
+
1+
β F nβ F
1
⎛ 72.1µA
72.0µA ⎞
+ ln
VBE3 + VBE 2 = 0.025V ⎜ ln
⎟ = 1.16V | VC 3 ≥ −VEE + 1.16 V
3 fA
15 fA ⎠
⎝
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-101
15.169
(a) Assuming balanced drain voltages, I
VGS1 = VTN +
2I D1
2I REF
= 0.75 +
Kn1
4 5Kn'
5 − 0.75 −
I REF =
( )
2I REF
( )
4 5K
'
n
D3
= I D1 = I REF
| VGS 3 = 0.75 +
2I REF
− 0.75 −
(
4 20Kn'
30kΩ
)
|
⎛W ⎞
⎜ ⎟
⎝ L ⎠1 I REF
=
⎛W ⎞
4
⎜ ⎟
⎝ L ⎠2
2I REF
(
4 20Kn'
(30kΩ)I
REF
| I REF =
5 − VGS1 − VGS 3
30kΩ
)
= 3.5 −1.5
I REF
10Kn'
2
Using Kn' = 25x10-6 and rearranging : 9x108 I REF
− 2.19x105 I REF + 12.25 = 0
I REF
= 21.8 µA. Drain voltage balance on M1 and M2
4
⎛ W ⎞ 80
2I REF
2I REF
| VTN +
= VTN +
|
⎜ ⎟ =
⎛W ⎞ '
⎝ L ⎠4 1
4 20Kn'
⎜ ⎟ Kn
⎝ L ⎠4
I REF = 87.2 µA and IO =
requires VGS 4 = VGS 3
(
)
(b) This part requires an iterative solution or the use of a computer solver.
Assuming VDS balance between M1 and M2 , λ ≠ 0 will not affect the
current mirror ratio, but it will change VGS and hence I REF slightly.
One iterative approach :
Guess VGS1 | Then I D1 =
2
5Kn'
VGS1 − VTN ) (1+ λVGS1 )
(
2
Since I D3 = I D1 , VGS 3 = VTN +
[
2I DS1
]
20K 1+ λ (5 − VGS1 )
'
n
5 − VGS1 − VGS 3
I
and I D1 = REF .
30kΩ
4
If the second value of I D1 does not agree with the first, then try a new VGS1.
I REF =
A spreadsheet yields : VGS1 = 1.336V , VGS 3 = 1.381V , I REF = 87.5µA, IO = 21.7µA.
Note: There is essentially no change from the first answer!
15-102
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.169 cont.
(c)
*Problem 15.169 - NMOS Wilson Source
RREF 1 0 30K
VSS 4 0 DC -5
M1 3 3 4 4 NFET W=5U L=1U
M2 2 3 4 4 NFET W=20U L=1U
M3 0 1 3 3 NFET W=20U L=1U
M4 1 1 2 2 NFET W=80U L=1U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.015
.OP
.END
15.170
IO = I REF
Rout =
1
λ2
⎛W ⎞
⎜ ⎟
⎝ L ⎠1
⎛W ⎞
⎜ ⎟
⎝ L ⎠2
| Rout = µ f 2ro3 = µ f 2
⎛W ⎞
2⎜ ⎟ Kn'
1
⎝ L ⎠2
I REF
λ3 I REF
1
1
=
λ3 IO λ2
2Kn 1
I D2 λ3 IO
3
⎡⎛ W ⎞ ⎤2
⎛W ⎞
⎢⎜ ⎟ ⎥
⎜ ⎟
1 ⎢⎝ L ⎠2 ⎥
⎝ L ⎠2
=
⎛ W ⎞ λ2 λ3 ⎢ I REF ⎥
⎜ ⎟
⎥
⎢
⎝ L ⎠1
⎦
⎣
2Kn'
⎛W ⎞
⎜ ⎟
⎝ L ⎠1
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-103
15.171
ix
M3
+
-
vx
+
ro2
v
-
gm2 v
ix = go2v x + gm 2v1
ix = go2v x + gm 2
vx
2
1
g m1
⎛ 1 ⎞
gm 3 ⎜ ⎟
v
⎝ gm1 ⎠
| v1 = v x
= x
⎛ 1 ⎞ 2
1+ gm 3 ⎜ ⎟
⎝ gm1 ⎠
| Rin =
| gm1 = gm 3 | v1 =
vx
2
2
vx
2
= ro
≅
ix
gm 2 gm 2
15.172
⎛W ⎞
⎜ ⎟
⎝ L ⎠1 150µA
Since λ = 0, ID 3 = ID1 = IREF
=
= 37.5µA
⎛W ⎞
4
⎜ ⎟
⎝ L ⎠2
VDS 3 ≥ VGS 3 − VTN 3 | VD 3 − (−10 + VGS1 ) ≥ VGS 3 − VTN 3 | VD 3 ≥ −10 + VGS1 + VGS 3 − VTN 3
VD 3 ≥ −10 + VTN1 +
2ID1
2ID 3
+ VTN 3 +
− VTN 3
K n1
Kn 3
2(37.5µA)V 2
2(37.5µA)V 2
VD 3 ≥ −10V + 0.75V +
+
= −8.09 V
5(25µA)
20(25µA)
15.173
⎛W ⎞ ⎛W ⎞
1
1
Rout ≅ µ f 2 ro3 | ⎜ ⎟ = ⎜ ⎟ ⇒ IO = IREF = 50µA | ro3 ≅
=
= 1.60MΩ
⎝ L ⎠ 2 ⎝ L ⎠1
λID 3 0.0125(50µA)
µf 2 =
Rout 250MΩ
1 2K n
=
= 156 | Using Eq. 13.71, µ f 2 =
ro3 1.60MΩ
λ ID 2
⎛W ⎞
2 I
2
5x10−5
3.80
=
⎜ ⎟ = (λµ f 2 ) D 2' = [0.0125(156)]
−5
⎝ L ⎠2
2K n
1
2(2.5x10 )
15.174
15-104
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
The circuit is the same as Fig. 16.20 with the addition of RREF in parallel with ro2. We require
RREF >> ro2 in order not to reduce the gain of the feedback loop. A current source with a source
resistor which achieves ROUT = ro(1+gmRs) should be sufficient. A cascode or Wilson source will
also work.
15.175
M1 and M2 are voltage balanced.
(a) Rout = µ f 4ro2 | All Kn are the same : IO = I REF = 17.5µA
VGS1 = 0.75 +
(
2 17.5x10−6
−6
75x10
) = 1.43V
| ∆V3 = VGS 3 − VTN 3 = 1.43V − 0.75V = 0.680
1
1
+ 1.43
+ (5 −1.43)
= 4.65MΩ | ro4 = 0.0125
= 4.78 MΩ
ro2 = 0.0125
17.5µA
17.5µA
(
)(
)(
)
g m4 = 2 75x10−6 17.5x10−6 1+ 0.0125(5 −1.43) = 5.24x10−5 S
µ f 4 = 5.24x10−5 S (4.78 MΩ) = 250 | Rout = 1.16 GΩ
min
(b) VCS = IO Rout = 20.3 kV! (c) VDD
= VGS1 + ∆V4 = 1.43 + 0.680 = 2.11V
15.176
*Problem 15.176 - NMOS Cascode Source
IREF 0 1 DC 17.5U
VDD 2 0 DC 5
M1 3 3 0 0 NFET W=3U L=1U
M2 4 3 0 0 NFET W=3U L=1U
M3 1 1 3 3 NFET W=3U L=1U
M4 2 1 4 4 NFET W=3U L=1U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.0125
.OP
.TF I(VDD) VDD
.END
Results: IO = 17.5 µA ROUT = 1.17 GΩThe same as the hand analysis.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-105
15.177
⎛W ⎞
⎜ ⎟
⎝L⎠
(a) M1 and M 2 are voltage balanced, so IO = ⎛ W ⎞2 IREF = 1.05IREF = 18.4µA, a 5% error.
⎜ ⎟
⎝ L ⎠1
(b) To first order, IO does not depend upon W/L of M 3 and M 4 .
The mismatch will create
a small VDS mismatch between M1 and M 2 , but this error will be negligible.
An estimate of this effect is ∆Io = go2∆VDS where
⎛
2ID
2ID 3 ⎞ ⎛
2ID 4 ⎞
2ID
2ID
∆VDS = VGS 3 − VGS 4 = ⎜VTN +
−
= 0.026
⎟ − ⎜VTN +
⎟=
Kn 3 ⎠ ⎝
Kn 4 ⎠
Kn3
0.95K n 3
Kn3
⎝
∆VDS = 0.026
2(17.5)
= 0.0218V | ∆Io = 0.0125(17.5x10−6 )(0.0218) = 3.89 nA
50
15.178
⎛W ⎞ ⎛W ⎞
1
1
Rout = µ f 4 ro2 | ⎜ ⎟ = ⎜ ⎟ ⇒ IO = IREF = 50µA | ro2 ≅
=
= 1.60MΩ
⎝ L ⎠ 2 ⎝ L ⎠1
λID 2 0.0125(50µA)
µf 4 =
Rout 250MΩ
1 2K n 4
=
= 156 | Using Eq. 13.71, µ f 4 =
ro2 1.60MΩ
λ ID 4
⎛W ⎞
2 I
2
5x10−5
3.80
=
⎜ ⎟ = (λµ f 4 ) DS 4' = [0.0125(156)]
−5
⎝ L ⎠4
2K n
1
2(2.5x10 )
15-106
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.179
(a − a) Rout = µ f 4 ro2 | All K n are the same : IO = IREF = 25µA
VGS1 = 0.75 +
2(25x10−6 )
75x10−6
= 1.57V | ∆V3 = VGS 3 − VTN 3 = 1.57V − 0.75V = 0.817V
1
1
+ 1.57
+ (5 −1.57)
= 3.26MΩ | ro4 = 0.0125
= 3.38MΩ
ro2 = 0.0125
25µA
25µA
gm 4 = 2(75x10−6 )(25x10−6 )(1+ 0.0125(5 −1.57)) = 6.25x10−5 S
µ f 4 = 6.25x10−5 S (3.38MΩ) = 211 | Rout = 689 MΩ
min
(a − b) VCS = IO Rout = 17.2 kV! (a - c) VDD
= VGS1 + ∆V4 = 1.57 + 0.817 = 2.39 V
(b − a) Rout = µ f 4 ro2 | All K n are the same : IO = IREF = 25µA
VGS1 = 0.75 +
2(50x10−6 )
75x10−6
= 1.91V | ∆V3 = VGS 3 − VTN 3 = 1.91V − 0.75V = 1.16V
1
1
+ 1.91
+ (5 −1.91)
= 1.64 MΩ | ro4 = 0.0125
= 1.66MΩ
ro2 = 0.0125
50µA
50µA
gm 4 = 2(75x10−6 )(50x10−6 )(1+ 0.0125(5 −1.91)) = 8.83x10−5 S
µ f 4 = 8.83x10−5 S (1.66MΩ) = 147 | Rout = 240 MΩ
min
(b − b) VCS = IO Rout = 12.0 kV! (b - c) VDD
= VGS1 + ∆V4 = 1.91+ 1.16 = 3.07 V
15.180
1
1
2
2
R=
+
=
=
= 39.0 kΩ
−6
gm 3 gm1 gm1
2(75x10 )(17.5x10−6 )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-107
15.181
(a ) I
= IC 3 + I B3 + I B 4 = I E3 + I B 4 = IC1 +
REF
I REF
IC1 =
1+
2
βF
+
1
βF + 1
2IC1
βF
+
IC 2
2I
I
= IC1 + C1 + C1
βF + 1
βF βF + 1
| IO = IC 4 = α F IC 2 = α F IC1 =
βF
βF + 1
I REF
1+
2
βF
+
1
βF + 1
⎛
⎞
⎜
110(50)
17.5µA ⎟
βr
110
IO =
= 163 MΩ
⎜
⎟ = 16.9 µA | Rout = o o ≅
2
1 ⎟
2
111 ⎜
2(16.9µA)
1+
+
⎝ 110 111 ⎠
(b) V
CS
= IO Rout = 16.9µA(163MΩ) = 2750 V
(c) V
CC
≥ 2VBE = 1.40 V
15.182
*Figure 15.182 - NPN Cascode Current Source
IREF 0 1 17.5U
VCC 2 0 DC 5
Q1 3 3 0 NBJT 1
Q2 4 3 0 NBJT 1
Q3 1 1 3 NBJT 1
Q4 2 1 4 NBJT 1
.MODEL NBJT NPN BF=110 VA=50
.OP
.TF I(VCC) VCC
.END
Results: IO = 16.9 µA Rout = 164 MΩ  the same as the hand analysis
15.183
⎛ 1
⎞
1
+
rπ 2 ⎟
R =⎜
⎝ g m3 g m1 ⎠
15-108
[r + (β
π4
o4
]
+ 1)ro2 ≅
1
1
2
2
+
≅
=
= 2.86 kΩ
g m3 g m1 g m1 40(17.5µA)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.184
(a) Assuming β
IO =
o
= 80 and VA = 60V :
VT ⎛ I REF AE2 ⎞ 0.025V ⎛ 80µA ⎞
20⎟ → IO = 127 µA
⎟=
⎜ ln
⎜ ln
R2 ⎝
IO AE1 ⎠ 500Ω ⎝
IO
⎠
100(0.025)
60V + 10 − 0.0635
= 551kΩ | rπ 2 =
= 19.7kΩ | g m1 = 40(80µA) = 3.2mS
0.127mA
0.127mA
⎞
⎛
⎟
⎜
⎞
⎛
100(0.5kΩ)
β o R2
⎟ = 551kΩ⎜1+
⎟
Rout = ro2 ⎜1+
⎜ 0.313kΩ + 19.7kΩ + 0.5kΩ ⎟ = 1.89 MΩ
1
⎟
⎜
⎠
⎝
+ rπ 2 + R2 ⎟
⎜
g
⎠
⎝
m1
⎛
⎞
0.025V
80µA
(b) IO = 500Ω ⎜⎝ ln I 20(1.05)⎟⎠ → IO = 129 µA
O
⎛
A ⎞ 0.025V ⎛ 80µA ⎞
V
I
(c) IO = RT ⎜⎝ ln IREF AE 2 ⎟⎠ = 500Ω ⎜⎝ ln I 14⎟⎠ → IO = 114 µA
2
O
E1
O
ro2 =
ro2 =
Rout
100(0.025)
60V + 10 − 0.057
= 614kΩ | rπ 2 =
= 21.9kΩ | g m1 = 40(80µA)= 3.2mS
0.114mA
0.114mA
⎛
⎞
100(0.5kΩ)
⎟ = 1.97 MΩ
= 614kΩ⎜⎜1+
⎟
0.313kΩ
+
21.9kΩ
+
0.5kΩ
⎝
⎠
15.185
Assuming βo = 80 and VA = 60V :
V ⎛ I
A ⎞ 0.025V ⎛ 35µA ⎞
(a) IO = RT ⎜⎝ ln IREF AE 2 ⎟⎠ = 935Ω ⎜⎝ ln I 20⎟⎠ → IO = 64 µA
2
O
E1
O
100(0.025)
60V + 10 − 0.0598
= 1.09 MΩ | rπ 2 =
= 39.1kΩ | g m1 = 40(35µA) = 1.40mS
0.064mA
0.064mA
⎞
⎛
⎟
⎜
⎞
⎛
100(0.935kΩ)
β o R2
⎟
⎜
⎟ = 3.59 MΩ
⎜
= 1.09 MΩ⎜1+
Rout = ro2 1+
⎟
1
0.714kΩ
+
39.1kΩ
+
0.935kΩ
⎟
⎜
⎠
⎝
+ rπ 2 + R2 ⎟
⎜
⎠
⎝ g m1
V ⎛ I
A ⎞ 0.025V ⎛ 35µA ⎞
(b) IO = RT ⎜⎝ ln IREF AE 2 ⎟⎠ = 935Ω ⎜⎝ ln I 10⎟⎠ → IO = 51.3 µA
2
O
E1
O
ro2 =
ro2 =
Rout
100(0.025)
60V + 10 − 0.0480
= 1.36 MΩ | rπ 2 =
= 48.7kΩ | g m1 = 40(35µA)= 1.40mS
51.3µA
51.3µA
⎞
⎛
⎟
⎜
⎞
⎛
100(0.935kΩ)
β o R2
⎟ = 1.36 MΩ⎜1+
⎜
⎟
= ro2 1+
⎜ 0.714kΩ + 48.7kΩ + 0.935kΩ ⎟ = 3.89 MΩ
1
⎟
⎜
⎠
⎝
+ rπ 2 + R2 ⎟
⎜
⎠
⎝ g m1
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-109
15.186
VT ⎛ IREF
⎜ ln
IO ⎝ IO
V ⎛ I
(b) R2 = T ⎜ln REF
IO ⎝ IO
(a) R2 =
(c ) R2 =
VT
IO
AE 2 ⎞ 0.025V ⎛ 73µA ⎞
20⎟ = 4.77 kΩ
⎟=
⎜ln
AE1 ⎠ 22µA ⎝ 22µA ⎠
AE 2 ⎞ 0.025V ⎛ 73µA ⎞
20⎟ = 24.3 kΩ
⎟=
⎜ln
AE1 ⎠ 5.7µA ⎝ 5.7µA ⎠
⎛ IREF AE 2 ⎞ 0.025V ⎛ 73µA ⎞
10⎟ = 21.3 kΩ
⎟=
⎜ ln
⎜ln
⎝ IO AE1 ⎠ 5.7µA ⎝ 5.7µA ⎠
15.187
VT ⎛ IREF
⎜ ln
IO ⎝ IO
V ⎛ I
(b) R2 = T ⎜ln REF
IO ⎝ IO
(a) R2 =
AE 2 ⎞ 0.025V ⎛ 62µA ⎞
10⎟ = 8.22 kΩ
⎟=
⎜ln
AE1 ⎠ 12µA ⎝ 12µA ⎠
AE 2 ⎞ 0.025V ⎛ 62µA ⎞
10⎟ = 8.22 kΩ
⎟=
⎜ln
AE1 ⎠ 512µA ⎝ 512µA ⎠
15.188
*Problem 15.188 - NPN Widlar Current Source
IREF 2 1 50U
VCC 2 0 DC 10
Q1 1 1 0 NBJT 1
Q2 2 1 3 NBJT 20
R2 3 0 4K
.MODEL NBJT NPN BF=110
.OP
.DC IREF 50U 5M 50U
.PROBE IC(Q2)
.END
50uA
I O2
40uA
30uA
I REF
20uA
0A
15-110
1.0mA
2.0mA
3.0mA
4.0mA
5.0mA
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.189
⎛V
⎞ V
V
I
IO = α F IE 2 = α F ⎜ BE1 + IB1 ⎟ ≅ BE1 = T ln C1
R2 IS1
⎝ R2
⎠ R2
V − VBE 2 − VBE1
V − VBE 2 − VBE1
IC1 = EE
− IB 2 ≅ EE
R1
R1
0.025V
15 −1.4
V
0.7V
ln 4 −15 = 318 µA | Note : BE1 ≅
= 318 µA
(a) IC 2 ≅
R2
2.2kΩ 10 (10 )
2.2kΩ
(b) IC 2 ≅
0.025V
3.3 −1.4
V
0.7V
ln 4 −15 = 295 µA | Note : BE1 ≅
= 318 µA
R2
2.2kΩ 10 (10 )
2.2kΩ
(c ) IC 2 ≅
VT VCC − VEB1 − VEB 2 0.025V
5 −1.4
0.7V
ln
=
ln 4 −15 = 66.5 µA |
= 70 µA
R2
IS1R1
10kΩ 10 (10 )
10kΩ
15.190
VEE − VBE 2 − VBE1
3.3 −1.4
− IB 2 ≅
= 317kΩ
6µA
IC1
⎛ 6µA ⎞
I
VT ln C1 0.0258V ⎜ln
⎟
⎛V
⎞
IS1
⎝ 0.1 fA ⎠
=
= 21.2kΩ
IO = α F IE 2 = α F ⎜ BE1 + IB1 ⎟ | R2 =
131
IO
6µA
⎝ R2
⎠
−I
(30µA) −
130
α F B1 130
(a) Choose IC1 = 0.2IO
| R1 =
(b) This is the same circuit implemented with pnp transistors.
VCC − VEB 2 − VEB1
3.3 −1.4
− IB 2 ≅
= 317kΩ
6µA
IC1
⎛ 6µA ⎞
I
VT ln C1 0.0258V ⎜ln
⎟
⎛V
⎞
IS1
⎝ 0.1 fA ⎠
EB1
+ IB1 ⎟ | R2 =
=
= 21.2kΩ
IO = α F IE 2 = α F ⎜
IO
6µA
131
⎝ R2
⎠
−I
(30µA) −
130
α F B1 130
Choose IC1 = 0.2IO | R1 =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-111
15.191
i
+
−β o i
r π4
v1
r o4
R th
+
-
ix
i
vx
ix
-
+
r o2
ve
-
v x = v1 + v e = (ix − β oi1)ro4 + (ix − 2i)ro2 | i1 = −i | v x = (ix + β oi)ro4 + (ix − 2i)ro2
i=
(ix − 2i)ro2
rπ 4 + Rth
i = ix
where Rth =
µf 2
β o4 + 2µ f 2
≅
1
1
(i − 2i)ro2
+
<< rπ 4 | i ≅ x
rπ 4
gm 3 gm 2
| i = ix
ro2
rπ 4 + 2ro2
⎞
⎛
βr
ix
for 2µ f 2 >> β o | v x = ix ⎜ ro2 + ro4 + o o4 − ro2 ⎟
⎠
⎝
2
2
⎛β
⎞ βr
Rout ≅ ro4 ⎜ o + 1⎟ ≅ o o4
⎝2
⎠
2
for Rth << rπ 4 and 2µf2 >> β o
15.192
An iterative solution is required :
1. Choose VGS 2 . Then I D2 =
[
]
2
Kn2
VGS 2 − VTN 2 ) 1 + λ (VDD − I D2 R1 )
(
2
2
Kn2
VGS 2 − VTN 2 ) (1 + λVDD )
(
or I D2 = 2
and
2
Kn2
1+
(VGS 2 − VTN 2 ) (λR1)
2
2I D1
2. VGS1 = VTN1 +
Kn1 1 + λ(VDD − I D1 R2 )
[
3. I D2 =
I D1 =
VGS 2
R2
]
VDD − VGS1 − VGS 2
Compare to I D2 in step 1 and choose new VGS 2
R1
−−−−−
I D2
2
2.5x10-4
VGS 2 − 0.75) (1.17)
(
V
2
=
| I D1 = GS 2
-4
2
15kΩ
2.5x10
1+
VGS 2 − 0.75) (0.017 *10000)
(
2
VGS1 = 0.75 +
[
2I D1
]
2.5x10 1 + 0.017(10 −15000I D1)
-4
| I D2 =
10 − VGS1 − VGS 2
10kΩ
Iteration yields VGS 2 = 2.744V , IO = I D2 = 536 µA, and I D1 = 183 µA.
15-112
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.193
Choose ID 2 = 0.2IO = 15µA | The effect of λ1 can be ignored because of the presence of
2(75x10−6 )
2ID1
= 0.75 +
= 1.53V | Then,
resistor R2 : VGS1 = VTN1 +
K n1
250x10−6
Kn 2
250µA
2
2
(VGS 2 − VTN 2 ) [1 + λ(VGS 2 + VGS1 )] | 15µA =
(VGS 2 − 0.75) [1 + λ(VGS 2 + 1.53)]
2
2
V
1.09V
An iterative solution gives VGS 2 = 1.089V | R2 = GS 2 =
= 14.5 kΩ
IO
75µA
V − VGS1 − VGS 2 6 −1.53 −1.09
R1 = DD
=
= 225 kΩ
ID 2
15µA
ID 2 =
15.194
An iterative solution is required :
1. Choose VGS1. Then I D1 =
K p1
2
(V
GS1
[
2
K p1
2
VGS1 − VTP1 ) (1 + λVDD )
(
or I D2 = 2
and
K p1
2
1+
(VGS1 − VTP1) (λR1)
2
2I D2
2. VGS 2 = VTP2 −
KP2 1 + λ(VDD − I D2 R2 )
[
3. I D1 =
]
− VTP1) 1 + λ (VDD − I D1 R1 )
I D2 =
VSG1
R2
]
VDD − VSG1 − VSG2
Compare to I D2 in step 1 and choose new VGS1
R1
−−−−−
2
10-4
VGS1 + 0.75) (1.10)
(
2
I D1 =
-4
| I D2 =
2
10
VGS1 + 0.75) (0.02 *10000)
(
2
2I D2
VGS 2 = −0.75 −
-4
10 1 + 0.02(5 −18000I D2 )
1+
[
]
VGS1
18kΩ
| I D1 =
5 − VSG1 − VSG2
10kΩ
Iteration yields VGS1 = −1.984V , IO = I D2 = 110 µA, and I D1 = 82.4 µA.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-113
15.195
Choose ID1 = 0.2IO = 25µA | The effect of λ2 can be ignored because of the presence of
resistor R2 : VGS 2 = VTP 2 −
2(125x10−6 )
2ID 2
= −0.75 −
= −2.33V | Then,
KP 2
100x10−6
K p1
100µA
2
2
(VGS1 − VTP1 ) 1 + λ(VGS1 + VGS 2 ) | 25µA =
(VGS1 + 0.75) 1 + λ(VGS1 + 2.33)
2
2
VGS 1.43V
An iterative solution gives VGS1 = −1.432V | R2 = −
=
= 11.4 kΩ
IO 125µA
V + VGS1 + VGS 2 9 −1.43 − 2.33
=
= 210 kΩ
R1 = DD
25µA
ID1
[
ID1 =
]
15.196
IC1 = IC 3 = 3IC 4 | IC 4 = IC 2 =
IC 2 =
[
VBE1 − VBE 2 VT ⎛ IC1
I ⎞ V ⎛ I I ⎞
= ⎜ln
− ln C 2 ⎟ = T ⎜ln C1 S 2 ⎟
R
R ⎝ IS1
IS 2 ⎠ R ⎝ IC 2 IS1 ⎠
0.025V ⎛ 3IC 2 20A ⎞
⎜ln
⎟ = 46.5 µA | IC1 = 3IC 2 = 140 µA
2.2kΩ ⎝ IC 2 A ⎠
15.197
*Problem 15.197 - BJT reference current cell P15.196
VCC 1 0 DC 1.5 AC 1
VEE 5 0 DC -1.5
Q4 2 2 1 PBJT 1
Q3 3 2 1 PBJT 3
Q1 3 3 5 NBJT 1
Q2 2 3 4 NBJT 20
R 4 5 2.2K
.MODEL NBJT NPN BF=100 VA=50
.MODEL PBJT PNP BF=100 VA=50
.OP
.AC LIN 1 1000 1000
.PRINT AC IC(Q1) IC(Q2)
.END
Results: IC1 = 140 µA
IC2 = 47.8 µA
15.198
IC1 = IC 3 = 3IC 4 | IC 4 = IC 2 =
IC 4 = IC 2 =
VBE1 − VBE 2 VT
=
R
R
S IVC1CC = 2. 92 x10
−2
⎛ IC1
I ⎞ V
− ln C 2 ⎟ = T
⎜ln
IS 2 ⎠ R
⎝ IS1
S IVC2CC = 9 . 92 x 10 −3
⎛ IC1 IS 2 ⎞
⎜ln
⎟
⎝ IC 2 IS1 ⎠
0.025V ⎛ 3IC 2 8A ⎞
⎜ ln
⎟ = 19.9 µA | IC1 = 3IC 2 = 59.6 µA | IC 3 = IC1 = 59.6 µA
4kΩ ⎝ IC 2 A ⎠
15.199
15-114
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
]
VBE1 − VBE 2 must be greater than 0. | VBE1 − VBE 2 = VT ( ) > 0
ln
nA
1
IC1IS 2
> 0 or 3
>1 | n >
IC 2 IS1
A
3
15.200
VBE1 − VBE 2 VT ⎛ IC1
I ⎞ V ⎛I I ⎞
= ⎜ ln
− ln C 2 ⎟ = T ln⎜ C1 S 2 ⎟
R
R ⎝ IS1
IS 2 ⎠ R ⎝ IC 2 IS1 ⎠
−23
kT 1.38x10 (323)
V ⎛ I I ⎞ 27.9mV
=
= 0.0279V | R = T ln⎜ C1 S 2 ⎟ =
ln(3⋅ 5) = 2.16 kΩ
VT =
−19
q
1.6x10
IC 2 ⎝ IC 2 IS1 ⎠
35µA
−23
kT 1.38x10 (273)
VT ⎛ IC1 IS 2 ⎞ 23.6mV
=
0.0236V
|
R
=
ln⎜
ln(3⋅10) = 2.29 kΩ
(b) VT = =
⎟=
q
1.6x10−19
IC 2 ⎝ IC 2 IS1 ⎠
35µA
(a) IC1 = IC 3 = 3IC 4
| IC 4 = IC 2 =
15.201
The M 3 - M 4 current mirror forces ID1 = ID 2 .
VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN +
ID 2 =
2ID1
2ID 2
− VTN −
'
10K n
20K n'
| ID 2 R =
2ID 2 ⎛
1 ⎞
1−
⎟
' ⎜
10K n ⎝
2⎠
0.293 1
0.293
1
=
⇒ ID 2 = 26.4 µA
'
R
5K n 5100 5(25x10−6 )
15.202
(a) The M 3 - M 4 current mirror forces ID1 = ID 2 .
VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN1 +
ID 2 =
1
0.293 1
0.293
=
⇒ ID 2 = 6.86 µA
4
'
5K n
R
10
5(25x10−6 )
(b) VTN1 = VTO
ID 2 R = VTO +
ID 2 =
2ID1
2ID 2
2ID 2 ⎛
1 ⎞
− VTN 2 −
| ID 2 R =
1−
⎟
'
'
' ⎜
10K n
20K n
10K n ⎝
2⎠
| VTN 2 = VTO + γ
( 2φ
F
)
(
+ VSB − 2φ F = VTO + 0.5 0.6 + ID 2 R − 0.6
(
)
)
2ID1
2ID 2
− VTO − 0.5 0.6 + ID 2 R − 0.6 −
'
10K n
20K n'
0.293 2IDS 2 0.5
−
10 4 10K n' 10 4
( 0.6 + 10 I
4
D2
)
− 0.6 → ID 2 = 3.96 µA
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-115
15.203 This problem should refer to Prob. 15.202 (a) and (b)
*Problem 15.203(a) - MOS reference current cell
VDD 1 0 DC 5 AC 1
VSS 5 0 DC -5
M1 3 3 5 5 NFET W=10U L=1U
M2 2 3 4 5 NFET W=20U L=1U
M3 3 2 1 1 PFET W=10U L=1U
M4 2 2 1 1 PFET W=10U L=1U
R 4 5 10K
.MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0 LAMBDA=0.017
.MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0 LAMBDA=0.017
*.MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0 LAMBDA=0
*.MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0 LAMBDA=0
*Problem 15.203(b) - MOS reference current cell
*.MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0.5 LAMBDA=0.017
*.MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0.75 LAMBDA=0.017
.OP
.AC LIN 1 1000 1000
.PRINT AC ID(M1) ID(M2)
.END
Results: (a) ID2 = 13.9 µA ID2 = 12.3 µA
D1
D2
SVI DD
= 7.64 x10−2 SVI DD
= 6.23x10−2
The currents differ considerably from the hand calculations.
D1
D2
= 7.75x10−2 SVI DD
= 6.31x10−2
Results: (b) ID1 = 8.19 µA ID2 = 7.24 µA SVI DD
The currents differ considerably from the hand calculations. The currents are quite sensitive to
the value of λ. The hand calculations used λ = 0. If the simulations are run with λ = 0, then the
results are identical to the hand calculations.
15.204
V ⎛ I 5A ⎞
0.025V ⎛ 2IC 2 5A ⎞
IC 2 = T ln⎜ C1
ln⎜
⎟ | IC1 = IC 3 = 2IC 4 = 2IC 2 | IC 2 =
⎟ = 5.23 µA
R ⎝ A IC 2 ⎠
11kΩ ⎝ A IC 2 ⎠
V ⎛ I 3A ⎞ 0.025V ⎛15.7µA ⎞
IC 7 = 5IC 4 = 5(5.23µA) = 26.2 µA | IC 8 = T ln⎜ C 4
ln⎜
⎟=
⎟ → IC 8 = 6.00 µA
R8 ⎝ A IC 8 ⎠
4kΩ ⎝ IC 8 ⎠
V ⎛ I A ⎞ 0.025V ⎛ 10.4µA ⎞
ln⎜
IC 5 = 2.5IC1 = 5IC 2 = 26.2 µA | IC 6 = T ln⎜ C1
⎟=
⎟ → IC 6 = 5.42 µA
R6 ⎝ A IC 6 ⎠
3kΩ ⎝ IC 6 ⎠
15-116
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.205
V ⎛ I 10A ⎞
0.025V ⎛ IC 2 10A ⎞
IC 2 = T ln⎜ C1
ln⎜
⎟ | IC1 = IC 3 = IC 4 = IC 2 | IC 2 =
⎟ = 5.23 µA
R ⎝ A IC 2 ⎠
11kΩ ⎝ A IC 2 ⎠
V ⎛ I 3A ⎞ 0.025V ⎛15.7µA ⎞
IC 7 = 5IC 4 = 5(5.23µA) = 26.2 µA | IC 8 = T ln⎜ C 4
ln⎜
⎟=
⎟ → IC 8 = 6.00 µA
R8 ⎝ A IC 8 ⎠
4kΩ ⎝ IC 8 ⎠
V ⎛ I A ⎞ 0.025V ⎛ 5.23µA ⎞
ln⎜
IC 5 = 2.5IC1 = 13.1 µA | IC 6 = T ln⎜ C1
⎟=
⎟ → IC 6 = 3.45 µA
R6 ⎝ A IC 6 ⎠
3kΩ ⎝ IC 6 ⎠
15.206
VT ⎛ IC1 IS 2 ⎞ 0.025V ⎛ 2IC 2 7IS1 ⎞
⎜ln
⎟=
⎜ln
⎟ = 15.3 µA
R ⎝ IC 2 IS1 ⎠ 4.3kΩ ⎝ IC 2 IS1 ⎠
= IC 5 = IC 6 = IC1 = 30.6 µA | IC 3 = IC 7 = IC 2 = 15.3 µA
(a) IC1 = 2IC 2 set by Q 4 and Q 3
IC1 = 2IC 2 = 30.6 µA | IC 4
(b) No change.
| IC 2 =
The currents are independent of the areas of Q5 , Q6 , and Q7 .
15.207
*Problem 15.207 - NPN Cascode Current Source
VCC 1 0 DC 5 AC 1
Q4 2 2 1 PBJT 2
Q3 3 2 1 PBJT 1
Q5 4 3 2 PBJT 1
Q1 6 6 0 NBJT 1
Q2 5 6 7 NBJT 7
Q6 4 4 6 NBJT 1
Q7 3 4 5 NBJT 1
R 7 0 4.3K
.MODEL NBJT NPN BF=100 VA=50
.MODEL PBJT PNP BF=50 VA=50
.OP
.AC LIN 1 1000 1000
.PRINT AC IC(Q7) IC(Q5)
.END
Results: IC2 = 15.2 µA IC1 = 28.5 µA - Similar to hand calculations.
C1
C2
= 1.81x10−3 SVICC
= 7.07x10−4
SVICC
15.208
VT ⎛ IC1 IS 2 ⎞ 0.025V ⎛ IC 2 7IS1 ⎞
⎜ ln
⎟=
⎜ ln
⎟ = 11.3 µA
R ⎝ IC 2 IS1 ⎠ 4.3kΩ ⎝ IC 2 IS1 ⎠
= IC 5 = IC 6 = IC1 = 11.3 µA | IC 3 = IC 7 = IC 2 = 11.3 µA
(a) IC1 = IC 2 set by Q 4 and Q 3
IC1 = IC 2 = 11.3 µA | IC 4
(b) No change.
| IC 2 =
The currents are independent of the areas of Q5 , Q6, and Q7 .
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-117
15.209
(a) The M 3 - M 4 current mirror forces ID1 = 1.5ID 2 .
VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN +
2ID1
2ID 2
− VTN −
'
10K n
30K n'
⎛
⎞
1 ⎛ 3ID 2
2ID 2 ⎞
2ID 2
3ID 2
1 ⎜
⎟
ID 2 = ⎜
−
−
⎟=
30K n' ⎠ 3300 ⎜⎝ 10(25x10−6 )
R ⎝ 10K n'
30(25x10−6 )⎟⎠
57.9
ID 2 =
⇒ ID 2 = 308 µA ID1 = 462 µA
3300
ID 4 = ID 5 = ID 6 = ID1 = 462 µA | ID 3 = ID 7 = ID 2 = 308 µA
(b) With λ
= 0, the currents do not depend upon the W/L ratios of M5 , M 6 or M 7
as long as all transistors remain in the active region. For λ ≠ 0, there will be a
weak dependence, since the drain - source voltages of M2 and M 3 will change slightly.
15.210
*Problem 15.210 - MOS reference current cell
VDD 1 0 DC 15 AC 1
M3 3 2 1 1 PFET W=10U L=1U
M4 2 2 1 1 PFET W=15U L=1U
M5 4 3 2 2 PFET W=10U L=1U
M6 4 4 6 6 NFET W=10U L=1U
M7 3 4 5 5 NFET W=10U L=1U
M1 6 6 0 0 NFET W=10U L=1U
M2 5 6 7 7 NFET W=30U L=1U
R 7 0 3.3K
*.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0
*.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017
.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017
.OP
.AC LIN 1 1000 1000
.PRINT AC ID(M1) ID(M2)
.END
Results: ID2 = 265 µA ID1 = 377 µA These differ from the hand calculations due to the nonzero value of λ. Simulation with λ = 0 gives results very close to the hand calculations.
D2
D2
= 9.82 x 10−4 SVI DD
= 6.99 x 10−4
SVI DD
15-118
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.211
The M 3 - M 4 current mirror forces I D1 = I D2.
⎛
2I D1 ⎞ ⎛
2I D2 ⎞
⎟
⎜
⎟
VGS1 − VGS 2 = I D2 R | I D2 R = ⎜⎜VTN +
−
V
+
TN
10Kn' ⎟⎠ ⎜⎝
30Kn' ⎟⎠
⎝
⎛
I D2
1 ⎛ 2I D2
2I D2 ⎞
I D2
1 ⎜
⎜
⎟
−
−
I D2 = ⎜
=
'
' ⎟
−
6
R ⎝ 10Kn
30Kn ⎠ 3300 ⎜ 5 25x10
15 25x10−6
⎝
34.1
I D2 =
⇒ I D2 = 107 µA I D1 = 107 µA
3300
I D 4 = I D5 = I D6 = I D1 = 107 µA | I D3 = I D 7 = I D2 = 107 µA
(
)
(
⎞
⎟
⎟
⎠
)
15.212
1
g m3
-
r o4
v3
+
r ο3
icc
Current mirror model:
1
Assuming VDS << since it is unknown :
gm4 v3
i cc
λ
⎛ 50 50 ⎞
ro4
≅ 2(5x10−4 )(10−4 )⎜ -4
= 0.316mS (250kΩ) = 79.1
-4 ⎟
2
⎝ 10 10 ⎠
Acd is determined by the mirror ratio error :
icc
g − g + go3
1
= icc m 3 m 4
= icc
for gm 3 = gm 4
icc + gm 4 v 3 = icc - gm 4
gm 3 + go3
gm 3 + go3
µf 3 +1
Add = gm1 (ro2 ro4 ) ≅ gm1
This error current goes through ro4 to produce the output voltage since the
common - mode output resistance at the drain of M2 is very large : v od = icc
icc = v ic
gm1
1
≅ v ic
2RSS
1+ 2gm1RSS
CMRR =
| Acd =
ro4
µf 3 + 1
⎤ 500kΩ
1 ⎛ ro4 ⎞ ⎡
1
= 6.28 x 10−5
⎥
⎜
⎟=⎢
µ f 3 + 1 ⎝ 2RSS ⎠ ⎣2(79.1) + 1⎦ 50MΩ
79.1
= 1.26 x 10 6 (122 dB)
6.28x10 -5
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-119
15.213
*Problem 15.213 - MOS Amplifier with Active Load
VDD 1 0 DC 10
VSS 5 0 DC -10
V1 6 8 DC 0 AC 0.5
V2 7 8 DC 0 AC -0.5
VIC 8 0 DC 0
M3 2 2 1 1 PFET W=50U L=1U
M4 3 2 1 1 PFET W=50U L=1U
M1 2 6 4 4 NFET W=20U L=1U
M2 3 7 4 4 NFET W=20U L=1U
ISS 4 5 DC 199.5U
RSS 4 5 25MEG
.MODEL NFET NMOS KP=25U VTO=1 LAMBDA=0.02
.MODEL PFET PMOS KP=10U VTO=-1 LAMBDA=0.02
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(3) VP(3)
.TF V(3) VIC
.END
Results: Adm = 98.8 Acd = 6.16 x10-5. The results are similar to hand calculations. The
discrepancies result from not including VDS in the hand calculations for both gm and ro.
15-120
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.214
-
1
r o4
v3
g m3
+
gm4 v3
r ο3
i cc
icc
Current mirror model:
Assuming VDS <<
1
λ
since it is unknown :
Add = gm1 (ro2 ro4 ) ≅ gm1
⎛ 66.7
ro4
66.7 ⎞
= 0.707mS (66.7kΩ) = 47.2
≅ 2(5x10−4 )(5x10−4 )⎜
−4
−4 ⎟
2
⎝ 5x10 5x10 ⎠
Acd is determined by the mirror ratio error :
icc
g − g + go3
1
icc + gm 4 v 3 = icc - gm 4
= icc m 3 m 4
= icc
for gm 3 = gm 4
gm 3 + go3
gm 3 + go3
µf 3 +1
This error current goes through ro4 to produce the output voltage since the
common - mode output resistance at the drain of M2 is very large : v od = icc
icc = v ic
gm1
1
≅ v ic
2RSS
1+ 2gm1RSS
CMRR =
| Acd =
ro4
µf 3 + 1
⎤ 133kΩ
1 ⎛ ro4 ⎞ ⎡
1
= 6.97 x 10−5
=
⎢
⎥
⎜
⎟
µ f 3 + 1 ⎝ 2RSS ⎠ ⎣2(47.2) + 1⎦ 20MΩ
47.2
= 6.77 x 10 5 (117 dB)
-5
6.97x10
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-121
15.215
*Problem 15.214 - MOS Amplifier with Active Load
VDD 1 0 DC 12
VSS 5 0 DC -12
V1 6 8 DC 0 AC 0.5
V2 7 8 DC 0 AC -0.5
VIC 8 0 DC 0
M3 2 2 1 1 PFET W=50U L=1U
M4 3 2 1 1 PFET W=50U L=1U
M1 2 6 4 4 NFET W=20U L=1U
M2 3 7 4 4 NFET W=20U L=1U
ISS 4 5 DC 1M
RSS 4 5 10MEG
.MODEL NFET NMOS KP=25U VTO=1 LAMBDA=0.02
.MODEL PFET PMOS KP=10U VTO=-1 LAMBDA=0.02
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(3) VP(3)
.TF V(3) VIC
.END
Results: Adm = 56.2 Acd = 6.82 x10-5. The results are similar to hand calculations. The
discrepancies result from not including VDS in the hand calculations for gm and ro.
15-122
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.216
Assuming VCE << VA since it is unknown :
⎛ 60 60 ⎞
r
Add = g m2 ro2 ro4 ≅ g m2 o4 ≅ 40 10−4 ⎜ -4
= 4.00mS (300kΩ)= 1200
-4 ⎟
2
⎝ 10 10 ⎠
Acd : This circuit is often the input stage of a feedback amplfier and the feedback
(
)
( )
applies an offset voltage Vos that forces vC1 = vC 2 . In this case, the induced collector
current imbalance exactly matches the current mirror imbalance and Acd = 0.
The case for VC1 ≠ VC2 is a more difficult problem!
The mirror ratio causes a mismatch in the collector currents and therefore
∆g m
∆g
g m2 = g m − m | The common - mode voltage that is
2
2
cm
vπ is multiplied by g m1 and g m2. The common g mvπcm term
a mismatch in g m : g m1 = g m +
developed across rπ 1 and rπ 2
( )
is canceled out by the current mirror, but the mismatch terms add at the output of the
current mirror. The output voltage is given approximately by
(
)
vo = ∆g m2vπcm ro4 ro2 = ∆g m2vπcm
vπcm = v ic
Acd =
rπ 2
(
rπ 2 + (β o + 1) 2R1 ro2
)
ro2 ∆g m2 cm µ f 2
=
vπ
2
g m2
2
≅ v ic
(
1
g m2 2R1 ro2
)
≅
vic
v
= ic for 2R1 >> ro2
g m2ro2 µ f 2
vo 1 ∆g m2
≅
v ic 2 g m2
The collector current imbalance can be found as follows : Assume that VEC 4 = VEC 3 + ∆V
0.7 + ∆V
⎛ ∆V ⎞
VA 60V
VA
and equal Early voltages : IC 4 = IC 2 ⇒ IC1
= IC1⎜1−
=
= 0.48V
⎟ | ∆V ≅
2 0.7
β F 125
VA ⎠
⎝
1+
+
β F VA
⎛ V ⎞
⎛ V − ∆V ⎞
∆V
V
| IC 0 = I S exp BE ≅ IC
∆IC = IC1 − IC2 = IC 0 ⎜1+ C1 ⎟ − IC0 ⎜1+ C1
⎟ = IC 0
VT
VA ⎠
VA
⎝ VA ⎠
⎝
1+
∆IC = IC 0
Acd =
∆V
∆V
≅ IC
VA
VA
|
∆IC ∆V
1
≅
=
IC
VA β F
|
∆g m2 ∆IC
1
=
=
g m2
IC
βF
| Acd =
1
2β F
1
1200
= 4x10-3 | CMRR =
= 3x105 (110 dB)
-3
2(125)
4x10
Note that VOS ≅
∆V
Add
| ∆V ≅
60
0.48
= 0.48V | VOS ≅
= 0.400 mV
125
1200
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-123
15.216 cont.
For VIC : VCB1 = VCC − VEB3 − VIC ≥ 0 | VCC ≥ VIC + VEB3 = 1.5 + 0.7 = 2.2 V
Assume that the current source needs VCS = 0.7 V across it to operate properly.
VIC − VBE − (−VEE ) ≥ VCS | VEE ≥ VCS + VBE − VIC = 0.7 + 0.7 − (−1.5) = 2.9V
We need ± 2.9 - V supplies or approximately ± 3V.
These results can be easily checked with SPICE - See Problem 15.217
15.217
*Figure 15.83 - BJT Differential Amplifier with Active Load
VCC 1 0 DC 5
VEE 5 0 DC -5
Q4 3 2 1 PBJT 1
Q3 2 2 1 PBJT 1
Q1 2 6 4 NBJT 1
Q2 3 7 4 NBJT 1
*Apply offset voltage to balance collector voltages
V1 6 8 DC 0.4107M AC 0.5
*V1 6 8 DC 0 AC 0.5
V2 7 8 DC 0 AC -0.5
VIC 8 0 DC 0
I1 4 5 199.8U
R1 4 5 25MEG
.MODEL NBJT NPN BF=125 VA=60
.MODEL PBJT PNP BF=125 VA=60
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(3) VP(3) VM(4) VP(4)
.TF V(3) VIC
.END
Results: Adm = 1200 Acd = 5.11x10-6. CMRR = 167 dB. The results are similar to hand
calculations. Note that a very high CMRR is achieved when the circuit is brought back to
balance, as is the case in operational amplifier input stages with feedback applied. For the case
with no offset voltage applied, Acd = 3.73x10-3, and ∆V = 0.49 V. These agree well with the
analysis in Prob. 15.216. The value of the required offset voltage is also very similar to the hand
calculations.
15-124
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.218
(a ) Assuming VCE << VA since it is unknown :
(
)
Add = g m2 ro2 ro4 ≅ g m2
⎛ 75
ro4
75 ⎞
≅ 40 5x10−5 ⎜
= 2.00mS (750kΩ)= 1500
−5
−5 ⎟
2
⎝ 5x10 5x10 ⎠
(
)
Acd : This circuit is often the input stage of a feedback amplfier and the feedback
applies an offset voltage Vos that forces vC1 = vC 2 . In this case, the induced collector
current imbalance exactly matches the current mirror imbalance and Acd = 0.
The case for VC1 ≠ VC2 is a more difficult problem!
The mirror ratio causes a mismatch in the collector currents and therefore
∆g m
∆g
g m2 = g m − m | The common - mode voltage that is
2
2
cm
vπ is multiplied by g m1 and g m2 . The common g mvπcm term
a mismatch in g m : g m1 = g m +
developed across rπ 1 and rπ 2
( )
is canceled out by the current mirror, but the mismatch terms add at the output of the
current mirror. The output voltage is given approximately by
(
)
vo = ∆g m2vπcm ro4 ro2 = ∆g m2vπcm
vπcm = v ic
Acd =
rπ 2
(
rπ 2 + (β o + 1) 2R1 ro2
)
ro2 ∆g m2 cm µ f 2
=
vπ
2
g m2
2
≅ v ic
(
1
g m2 2R1 ro2
)
≅
vic
v
= ic for 2R1 >> ro2
g m2ro2 µ f 2
vo 1 ∆g m2
≅
v ic 2 g m2
The collector current imbalance can be found as follows: Assume that VEC 4 = VEC 3 + ∆V
0.7 + ∆V
⎛ ∆V ⎞
75V
V
VA
= IC 2 ⇒ IC1
= IC1⎜1−
= 0.60V
| ∆V ≅ A =
⎟
2 0.7
β F 125
⎝ VA ⎠
1+
+
β F VA
1+
and equal Early voltages: IC 4
⎛ V ⎞
⎛ V − ∆V ⎞
V
∆V
∆IC = IC1 − IC 2 = IC 0 ⎜1+ C1 ⎟ − IC 0 ⎜1+ C1
| IC 0 = IS exp BE ≅ IC
⎟ = IC 0
VA ⎠
VA
VT
⎝ VA ⎠
⎝
∆V
∆V
∆IC ∆V
1
∆gm 2 ∆IC
1
1
∆IC = IC 0
≅ IC
|
≅
=
|
=
=
| Acd =
IC
gm 2
IC
βF
VA
VA
VA β F
2β F
1
1500
= 4 x10 -3 | CMRR =
= 3.75x10 5 (112 dB)
-3
2(125)
4 x10
15.218 continued on next page
Acd =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-125
15.218 cont.
Note that VOS ≅
∆V
Add
| ∆V ≅
75
0.60
= 0.60V | VOS ≅
= 0.400 mV
125
1500
For VIC : VCB1 = VCC − VEB 3 − VIC ≥ 0 | VCC ≥ VIC + VEB 3 = 1.5 + 0.7 = 2.2 V
Assume that the current source needs VCS = 0.7 V across it to operate properly.
VIC − VBE − (−VEE ) ≥ VCS | VEE ≥ VCS + VBE − VIC = 0.7 + 0.7 − (−1.5) = 2.9V
We need ± 2.9 - V supplies or approximately ± 3V.
(
)
(b) Add = g m2 ro2 ro4 ≅ g m2
⎛ 100
ro4
100 ⎞
≅ 40 5x10−5 ⎜
= 2.00mS (1.00 MΩ) = 2000
−5
−5 ⎟
2
⎝ 5x10 5x10 ⎠
(
)
0.7 + ∆V
⎛ ∆V ⎞
v
1 ∆g m2
VA 100V
VA
| IC 4 = IC 2 ⇒ IC1
= IC1⎜1−
=
= 0.80V
Acd = o ≅
⎟ | ∆V ≅
2 0.7
VA ⎠
125
v ic 2 g m2
βF
⎝
1+
+
β F VA
⎛ V ⎞
⎛ V − ∆V ⎞
V
∆V
| IC0 = I S exp BE ≅ IC
∆IC = IC1 − IC2 = IC 0 ⎜1+ C1 ⎟ − IC0 ⎜1+ C1
⎟ = IC 0
VA ⎠
VA
VT
⎝ VA ⎠
⎝
∆IC ∆V
1
∆g m2 ∆IC
1
1
∆V
∆V
∆IC = IC 0
≅ IC
|
≅
=
|
=
=
| Acd =
VA β F
2β F
VA
VA
IC
g m2
IC
βF
1+
Acd =
1
2000
= 4x10-3 | CMRR =
= 5.00x105 (114 dB)
-3
2(125)
4x10
For VIC : VCB1 = VCC − VEB 3 − VIC ≥ 0 | VCC ≥ VIC + VEB 3 = 1.5 + 0.7 = 2.2 V
Assume that the current source needs VCS = 0.7 V across it to operate properly.
VIC − VBE − (−VEE )≥ VCS | VEE ≥ VCS + VBE − VIC = 0.7 + 0.7 − (−1.5) = 2.9V
We need ± 2.9 - V supplies or approximately ± 3V.
15.219
Results: For VA = 75 V, Adm = 1470 and Acd = 6.92x10-3. CMRR = 106 dB. The results are
similar to hand calculations. Note that a very high CMRR is achieved when the circuit is
brought back to balance (with VOS = 0.728mV), as is the case in operational amplifier input
stages with feedback applied. For the case with the offset voltage applied, Acd = 2.71x10-7.
15-126
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.220
(a) I
D5
= I D 4 = I D3 = I D2 = I D1 =
I SS
= 100 µA
2
( )
−4
2 10
2I D1
= 0.75 +
= 1.20V
VGS 2 = VGS1 = VTN +
Kn1
40 2.5x10−5
(
)
( )
( )
2 10−4
2I D 4
VGS 5 = VGS 4 = VGS 3 = VTP −
= −0.75 −
= −1.25V | VDS 3 = VDS 4 + VGS 5 = −2.50V
K p4
80 10−5
VDS1 = VDS 2 = 10 + VGS 4 + VGS 5 − (−VGS1 )= 8.70V | VDS 5 = VDS 4 = VGS 4 = −1.25V
Q − Pts : (100µA,8.70V ) (100µA,8.70V ) (100µA,-2.50V ) (100µA,-1.25V ) (100µA,-1.25V )
(b) A
dd
⎛ 2
⎞
= g m2 ⎜ ro2 µ f 3ro5 ⎟ ≅ g m2ro2
⎝ 3
⎠
⎛ 1
⎞
+ 8.7 ⎟
⎜
Add = 2(40) 2.5x10−5 10−4 1+ 0.017(8.7) ⎜ 0.017 −4
⎟ = 0.479mS (675kΩ) = 323
10
⎜
⎟
⎝
⎠
(Note that the loop - gain of the Wilson source is reduced by the presence of Rout1 ≅ 2ro1.)
⎛ 1
⎞
+ 1.25⎟
⎜
(c) Add = g m1 ro1 ro3 | ro3 = ⎜ 0.01710−4 ⎟ = 601kΩ
⎜
⎟
⎝
⎠
(
)( )[
(
]
)
(
)
Add = 0.479mS 675kΩ 601kΩ = 152 - The Wilson source yields a 2X improvement.
15.221
M
1
2
3
4
5
ID(µA) 101 99.0 101 99.0 99.0
VDS 8.69 7.33 -2.48 -1.24 -2.60
(V)
Results: Adm = 313. The gain and Q-points are similar to hand calculations.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-127
15.222
ID 2 = ID1 =
I1
I
= 125 µA | ID 3 = ID 4 = ID 5 = ID 6 = ID 7 = I2 − 1 = 125 µA
2
2
2(1.25x10−4 )
2ID
= 0.75 +
= 1.25V
For the NMOS transistors VGS = VTN +
Kn
40(2.5x10−5 )
2(1.25x10−4 )
2ID
For the PMOS transistors VGS = VTP −
= −0.75 −
= −1.54V
Kp
40(10−5 )
VDS1 = VDS 2 = −VGS 3 = 1.54V | VDS 7 = VDS 6 = VGS 6 = 1.25V | VDS 5 = VGS 6 + VGS 7 = 2.50V
VDS 4 = VDS 3 = (−5 + VDS 5 ) − (−VGS1 − VGS 3 ) = −5 + 2.50 − (−1.25 + 1.54) = −2.79V
Q − Pts : (125µA,1.54V) (125µA,1.54V) (125µA,-2.79V) (125µA,-2.79V)
(125µA,2.50V) (125µA,1.25V) (125µA,1.25V)
M3
v id
M1
2
RL
(b) Add = gm 2 (µ f 4 ro2 µ f 5 ro7 )≅
µ f 2µ f 4
2
⎞
⎛ 1
+ 1.54 ⎟
⎜
gm 2 = 2(40)(2.5x10−5 )(1.25x10−4 )[1+ 0.017(1.54 )] = 0.507mS | ro3 = ⎜ 0.017 −4 ⎟ = 483kΩ
⎜ 1.25x10 ⎟
⎠
⎝
⎞
⎛ 1
+ 2.79 ⎟
⎜
gm 4 = 2(40)(10−5 )(1.25x10−4 )[1+ 0.017(2.79)] = 0.324mS | ro4 = ⎜ 0.017 −4 ⎟ = 493kΩ
⎜ 1.25x10 ⎟
⎠
⎝
Add ≅
µ f 2µ f 4
15-128
2
=
(0.507mS)(483kΩ)(0.324mS )(493kΩ) = 19600
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.223
*Figure P15.222 - CMOS Folded Cascode Amplifier with Active Load
VDD 1 0 DC 5
VSS 10 0 DC -5
*An offset voltage must be applied to bring output to -2.5V
V1 4 11 DC -5.085M AC 0.5
V2 5 11 DC 0 AC -0.5
VIC 11 0 DC 0
M1 2 4 6 6 NFET W=40U L=1U
M2 3 5 6 6 NFET W=40U L=1U
M3 8 6 2 2 PFET W=40U L=1U
M4 7 6 3 3 PFET W=40U L=1U
M5 8 9 10 10 NFET W=40U L=1U
M6 9 9 10 10 NFET W=40U L=1U
M7 7 8 9 9 NFET W=40U L=1U
I2A 1 2 DC 250U
I2B 1 3 DC 250U
I1 6 10 DC 250U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017
.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(7) VP(7) VM(6) VP(6) VM(8) VP(8)
.TF V(7) VIC
.END
8
Results: Add = 23700, Acd = 1.81x10-4. Rout = 47.7 MΩ, CMRR = 1.31 x 10 . The values of Add
and Rout are similar to hand calculations. Acd and the CMRR are limited by the small residual
mismatches in device parameters.
From Problem 15.222, g m2 = 0.507mS | ro3 = 483kΩ | g m4 = 0.324mS | ro4 = 493kΩ
Rout = µ f 4ro2 µ f 5ro7 ≅
µ f 4ro2
2
=
0.324mS (493kΩ)(493kΩ)
2
= 39.4 MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-129
15.224
M
M
1
M
2
+5
V
3
25
1
25
1
I2
25
1
I2
R
I1
M
M
1
10
1
-5V
2
10
1
Using a reference current of 250µA and 1:1 current mirrors with − VGSP = VGSN :
−VGSP = VGSN = VTN +
2(2.5x10−4 )
2ID
10 − 2.16 − 2.16 V
= 0.75 +
= 2.16V | R =
= 22.7kΩ
−6
Kn
2.5x10−4
A
10(25x10 )
15.225
⎛ R ⎞
VBE 2
R2 = VBE 2 ⎜1+ 2 ⎟
R1
⎝ R1 ⎠
⎛
VBE 2 ⎞
⎜ 200µA − R ⎟
1 ⎟
| VBE 2 = 0.025ln⎜
10
fA
⎜
⎟
⎜
⎟
⎝
⎠
For β F = ∞, IB = 0. | VBE 3 + VEB 4 = VBE 2 +
⎛ 200µA − I1 ⎞
VBE 2
VBE 2 = 0.025ln⎜
⎟ | I1 =
R1
⎝ 10 fA ⎠
⎛
VBE 2 ⎞
⎜ 200µA − 20kΩ ⎟
0.589
= 171µA
VBE 2 = 0.025ln⎜
⎟ → VBE 2 = 0.589V | IC 2 = 200µA −
20kΩ
10
fA
⎜
⎟
⎝
⎠
⎛ 20kΩ ⎞
1
Since IS 4 = IS 3 , VBE 3 = VEB 4 = (0.589)⎜1+
⎟ = 0.589V and IC 4 = IC 3 = 171 µA.
⎝ 20kΩ ⎠
2
15-130
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.226
(a) β F = ∞ : VBE 3 + VEB 4 = VBE1 + VEB 2
VT ln
IO
IO
I2
I2
+VT ln
- V ln
- V ln
=0
AE 3
AE 4 T
AE1 T
A
ISON
ISOP
ISON
ISOP E 2
AEO
AEO
AEO
AEO
2
⎡
IO2 AE1 AE 2
IO2 AEO
ISON ISOP AE1 AE 2 ⎤
AE 3 AE 4
VT ln⎢
=
0
→
= 1 | IO = I2
⎥
2 2
2
I2 AEO
I2 AE 3 AE 4
AE1 AE 2
⎣ ISON ISOP AE 3 AE 4
⎦
(b) IO = 300µA
AE 3 AE 4
= 100 µA
3AE 3 3AE 4
15.227
(a) I D8 = I REF = 250µA | I D10 = I D9 = 2I D8 = 500µA
I D1 = I D2 = I D3 = I D 4 =
2(250µA)
I D9
= 250µA | VDS 8 = VGS 8 = 0.75 +
= 2.16V
2
10 25x10−6
(
I D5 = I D11 = I D10 = 500µA | VGS11 = 0.75 +
V
VGS 6 = −VGS 7 = GS11 = 1.789V | I D 7 = I D6
2
−VDS 4 = −VDS 3 = −VGS 3 = VGS 2 = VTN +
2(500µA)
)
= 3.58V
(
)
10(25x10 )
=
(1.789 − 0.75) = 135µA
5 25x10−6
−6
2
2
2(250µA)
2I D2
= 0.75 +
= 1.75V
Kn2
20 25x10−6
(
VDS1 = VDS 2 = 5 − VSD 4 − (−VGS 2 )= 5.00V | VDS10 = −VDS 5 = 5 −
)
VGS11
= 3.21V
2
VDS 6 = −VDS 7 = 5.00V | VDS 9 = 5 − VGS 2 = 5 −1.75 = 3.25V
ID (µA)
VDS (V)
SPICE
ID (µA)
VDS (V)
1
2
3
4
5
6
7
8
9
10
11
250
250
250
250
500
135
135
250
500
500
500
5.00
5.00
-1.75
-1.75
-3.21
5.00
-5.00
2.16
3.25
3.21
3.58
255
4.97
255
4.99
255
-1.74
255
-1.73
509
-3.24
139
5.01
139
-5.00
250
2.14
509
3.28
509
3.23
509
3.52
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-131
(b) Adm = Avt1 Avt 2 Avt 3 = [gm 2 (ro2
][
]
ro4 ) gm 5 (ro5 ro12 ) [1]
gm 2 = 2(20)(25x10−6 )(2.50x10−4 )[1+ 0.017(5.00)] = 0.521mS
1
58.8 + 5.00 V
58.8 + 1.75 V
= 58.8V | ro2 =
= 255kΩ | ro4 =
= 242kΩ | AV1 = 64.7
−4
λ 0.017
A
2.5x10
2.5x10−4 A
58.8 + 3.21 V
gm 5 = 2(100)(10−5 )(5.00x10−4 )[1+ 0.017(3.21)] = 1.03mS | ro12 = ro5 =
= 124kΩ
5.00x10−4 A
Avt 2 = 63.9 | Adm = Avt1 Avt 2 Avt 3 = [64.7] [63.9] [1] = 4130
1
=
(c ) The amplification factor is inversely proportional to the square root of current.
Thus,
⎡ 64.7⎤ ⎡ 63.9 ⎤
Adm = Avt1 Avt 2 Avt 3 ≅ ⎢
[1] = 2065
⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦
SPICE Results: Adm = 4000, Acm = 0.509, Rout = 1.81 kΩ.
15.228
Adm ≅
µ f 2µ f 5
4
| For the MOSFET, µ f 2 ∝
But, ID 2 ∝ IREF and ID 5 ∝ IREF → Adm ∝
(a)
Adm = 16000
100µA
= 6400
250µA
(b)
1
ID
∴ Adm ∝
1
ID 2
1
ID 5
1
IREF
Adm = 16000
100µA
= 80000
20µA
15.229
Adm = Av1 Av 2 Av 3 = gm 2 (ro2 ro4 ) gm 5 (ro5 ro12 ) [1]
[
][
]
ID10 = IREF =100µA | ID12 = ID11 = 2ID10 = 200µA | ID1 = ID 2 = ID 3 = ID 4 =
ID 5 = ID12 = 200µA | VDS 4 = VGS 3 = −VGS 2 = −VTN −
ID11
= 100µA
2
2(100µA)
2IDS 2
= −0.75 −
= −1.38V
Kn2
20(25x10−6 )
VDS 2 = 10 + VDS 4 − (−VGS 2 ) = 10.0V | gm 2 = 2(20)(25x10−6 )(100µA)[1+ 0.017(10)] = 0.342mS
1
λ
=
1
58.8 + 10 V
58.8 + 1.38 V
= 58.8V | ro2 =
= 688kΩ | ro4 =
= 602kΩ | Av1 = 110
−4
0.017
10
A
10−4
A
VDS12 = −VDS 5 = 10 −
VGSGG
2
| VGSGG = 0.75 +
2(200µA)
= 2.54V | VDS12 = −VDS 5 = 8.73V
5(25x10−6 )
gm 5 = 2(100)(10−5 )(2x10−4 )[1+ 0.017(8.73)] = 0.678mS | ro12 = ro5 =
Av 2 = 115 | Adm = Av1 Av 2 Av 3 = [110] [115] [1] = 12600
58.8 + 8.73 V
= 338kΩ
2x10−4 A
Adm = 10900 if (1 + λVDS ) is neglected in the gm calculation.
15-132
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.230
*Figure P15.229 - CMOS Amplifier with Active Load
VDD 8 0 DC 10
VSS 14 0 DC -10
*An offset voltage is used to set Vo to approximately zero volts.
V2 1 15 DC 0.3506M AC 0.5
V1 2 15 DC 0 AC -0.5
VIC 15 0 DC 0
M1 3 1 5 14 NFET W=20U L=1U
M2 4 2 5 14 NFET W=20U L=1U
M3 3 3 8 8 PFET W=50U L=1U
M4 4 3 8 8 PFET W=50U L=1U
M5 6 4 8 8 PFET W=100U L=1U
*The offset can be adjusted to zero by correcting the value of W/L
*M5 6 4 8 8 PFET W=89.5U L=1U
M6 8 6 13 14 NFET W=10U L=1U
M7 14 7 13 8 PFET W=25U L=1U
MGG 6 6 7 14 NFET W=5U L=1U
M10 9 9 14 14 NFET W=10U L=1U
M11 5 9 14 14 NFET W=20U L=1U
M12 7 9 14 14 NFET W=20U L=1U
IREF 0 9 DC 100U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017
.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017
*.MODEL NFET NMOS KP=25U VTO=0.75 GAMMA=0.6 LAMBDA=0.017
*.MODEL PFET PMOS KP=10U VTO=-0.75 GAMMA=0.75 LAMBDA=0.017
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(13) VP(13) VM(4) VP(4)
.TF V(13) VIC
.END
Results: Adm = 11200, Acm = 0.604, Rout = 3.10 kΩ.
(a)
1
2
3
4
5
6
7
GG 10
11
12
IDS (µA) 112 112 112 112 223 44.2 44.2 223 100 223 223
VDS (V) 9.96 9.99 -1.41 -1.37 -8.70 10.0 -10.0 -2.60 1.63 8.63 8.70
(b)
IDS (µA) 110 110 110 110 219
0
0
219 100 219 219
VDS (V) 11.2 11.2 -1.40 -1.37 -8.85 10.0 -9.97 -3.79 1.63 7.41 7.35
Note that the body effect has increased the threshold voltages of M6 and M7 to the point that they
are no longer conducting. VTN6 = 2.24 V and VTP7 = -2.61V. The W/L ratio of MGG needs to be
redesigned to solve this problem.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-133
15.231
ID10 = IREF = 250µA | ID11 = 2ID10 = 500µA | ID12 = 4ID10 = 1000µA
ID1 = ID 2 = ID 3 = ID 4 =
2(250µA)
ID11
= 250µA | VDS10 = VGS10 = 0.75 +
= 2.16V
2
10(25x10−6 )
ID 5 = IDGG = ID12 = 1000µA | VGSGG = 0.75 +
2(1000µA)
= 4.75V
5(25x10−6 )
10(25x10
V
VGS 6 = −VGS 7 = GSGG = 2.375V | ID 7 = ID 6 =
2
2
VDS 4 = VDS 3 = VGS 3 = −VGS 2 = −VTN −
−6
)(2.375 − 0.75)
2
= 330µA
2(250µA)
2ID 2
= −0.75 −
= −1.75V
Kn2
20(25x10−6 )
VDS1 = VDS 2 = 7.5 + VDS 4 − (−VGS 2 ) = 7.50V | VDS12 = −VDS 5 = 7.5 +
VGSGG
= 5.13V
2
VDS 6 = −VDS 7 = 7.5V | VDS11 = 7.5 − VGS 2 = 7.5 −1.75 = 5.75V
M
IDS (µA)
VDS (V)
SPICE
IDS (µA)
VDS (V)
1
2
3
4
5
6
7
GG
10
11
12
250 250 250 250 1000
7.50 7.50 -1.75 -1.75 -5.13
330
7.50
330 1000 250 500 1000
-7.50 4.75 2.16 5.75 5.13
264 266 -264 -266 1050
7.46 7.09 -1.76 -2.14 -5.20
359
7.54
-359 1050 250 530 1050
-7.46 4.69 2.14 5.78 5.11
[
][
]
Adm = Av1 Av 2 Av 3 = gm 2 (ro2 ro4 ) gm 5 (ro5 ro12 ) [1]
gm 2 = 2(20)(25x10−6 )(250µA)[1+ 0.017(7.50)] = 0.531mS
1
58.8 + 7.50 V
58.8 + 1.75 V
= 58.8V | ro2 =
= 265kΩ | ro4 =
= 242kΩ | Av1 = 67.2
−4
A
λ 0.017
2.5x10
2.5x10−4 A
58.8 + 5.13 V
= 63.9kΩ
gm 5 = 2(100)(10−5 )(10−3 )[1+ 0.017(5.13)] = 1.48mS | ro12 = ro5 =
10−3
A
Av 2 = 47.3 | Adm = Av1 Av 2 Av 3 = [67.2] [47.3] [1] = 3180
1
=
SPICE Results:
Adm = 2950, Acm = 0.03, Rout = 1.10 kΩ.
15-134
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.232
(a) For saturation of M11 :
VDS11 = 0 − VGS 2 − (−VSS ) = VSS − VGS 2 ≥
VGS 2 = VTN +
2(200µA)
2IDS11
=
= 0.894
K n11
20(25x10−6 )
2(100µA)
2IDS 2
= 0.75 +
= 1.38V | VSS −1.38 ≥ 0.894 → VSS ≥ 2.27V
Kn 2
20(25x10−6 )
For saturation of M12 :
VDS12 = 0 −
2(200µA)
VGSGG
V
2ID12
− (−VSS ) = VSS − GSGG ≥
=
= 0.894
2
2
K n12
20(25x10−6 )
VGSGG = 0.75 +
2(200µA)
2.54
= 2.54V | VSS −
≥ 0.894 → VSS ≥ 2.16V
−6
2
5(25x10 )
For saturation of M1 and M 2 :
VDS1 = VDD + VGS 3 − (−VGS1 ) = VDD + VGS 3 +VGS1 ≥
2(100µA)
2ID1
=
= 0.633
K n1
20(25x10−6 )
VGS 3 = −VGS1 : VDD ≥ 0.633V
For saturation of M 5 :
−VDS 5 = VDD −
2(200µA)
2.54
2ID 5
≥
=
= 0.633V → VDD ≥ 1.90V
K p5
2
100(10x10−6 )
M 6 and M 7 are always saturated : e.g. VDS 6 ≥ VGS 6
The minimum supply voltages are : VDD ≥ 1.90V VSS ≥ 2.27V
For the symmetrical supply case, VDD = VSS ≥ 2.27V
(b) The values of VDD and VSS in part (a) do not provide any significant common - mode
input voltage range. For saturation of M11 with VIC = −5V,
VDS11 = VIC − VGS 2 − (−VSS ) = VSS −1.38 − 5 ≥ 0.894 → VSS ≥ 7.27V
For saturation of M1 and M 2 :
VDS1 = VDD − VSG 3 − (VIC − VGS1) = VDD − 5 −1.38 +1.38 ≥ 0.633 → VDD ≥ 5.63V
For an output range of 5V, saturation of M12 requires
V
2.54
≥ 0.894 → VSS ≥ 7.25V
VDS12 = −5 − GSGG − (−VSS ) = VSS − 5 −
2
2
For Saturation of M 5 :
2.54
− 5 ≥ 0.633V → VDD ≥ 6.90V
VSD 5 = VDD −
2
The minimum supply voltages are : VDD ≥ 6.90V VSS ≥ 7.25V
For the symmetrical supply case, VDD = VSS ≥ 7.25V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-135
15.233
ID 9 = ID10 = ID12 = ID11 = IREF = 250µA | ID 6 = ID 7 = ID 8 = 3ID12 = 750µA
ID1 = ID 2 = ID13 = ID 5 = ID 3 = ID 4 =
ID 9
= 125µA
2
VDS10 = VDS12 = VDS11 = VGS12 = VTN +
VGS1 = 0.75 +
2(250µA)
2ID12
= 0.75 +
= 2.75V
K n12
5(25x10−6 )
2(125µA)
= 1.25V | VDS 9 = −VGS1 − (−10 + VGS12 ) = 6V
40(25x10−6 )
VDS 7 = VO − (−10 + VGS12 + VGS11 − VGS 7 ) = 0 + 10 − 2.75 − 2.75 + 0.75 +
2(750µA)
= 7.25V
15(25x10−6 )
VDS 8 = 10 − VDS 7 = 2.75V
VDS 6 = −(10 − VO ) = −10V + 0 = −10V | VDS 5 = VDS13 = VDS 3 = VDS 4 = −0.75 −
2(125µA)
= −1.31V
80(10x10−6 )
M
1
2
3
4
5
6
7
8
9
10
11
12
13
ID (µA)
125
125
125
125
125
750
750
750
250
250
250
250
125
VDS (V)
8.63
8.63
-1.31
-1.31
-1.31
-10
7.25
2.75
6.00
2.75
2.75
2.75
-1.31
K n' ⎛ W ⎞
10−5 ⎛ W ⎞
2
2
⎜ ⎟ (−VDS 4 − VDS 5 + 0.75) = 750µA
(b) ⎜ ⎟ (VGS 6 − VTP ) =
2 ⎝ L ⎠6
2 ⎝ L ⎠6
⎛W ⎞
10−5 ⎛ W ⎞
2
⎜ ⎟ (−2.62 + 0.75) = 750µA → ⎜ ⎟ = 42.9
⎝ L ⎠6
2 ⎝ L ⎠6
Add = Av1 Av 2 = (gm 2 ro2 )(gm 6 ro6 ) = µ f 2µ f 6 | µ f 2 ≅
µf 6 ≅
1
λp
15-136
1
λn
2(40)(25x10−6 )
2K n 2
1
=
= 235
ID 2
125x10−6
0.017
2(42.9)(10x10−6 )
2K p 6
1
=
= 62.9 | Add = 235(62.9) = 14800
ID 6
750x10−6
0.017
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.234
*Figure P15.233 - CMOS Amplifier with Active Load
VDD 8 0 DC 10
VSS 14 0 DC -10
*Connect feedback to determine Vos
*V1 1 13 DC 0
*The offset voltage must be used to set Vo to approximately zero voltages.
V1 1 15 DC 0.4423M AC 0.5
V2 2 15 DC 0 AC -0.5
VIC 15 0 DC 0
M1 3 1 5 5 NFET W=40U L=1U
M2 4 2 5 5 NFET W=40U L=1U
M3 6 7 8 8 PFET W=80U L=1U
M4 7 7 8 8 PFET W=80U L=1U
M5 4 3 7 7 PFET W=80U L=1U
M6 13 4 8 8 PFET W=42.9U L=1U
*The offset can be adjusted to zero by correcting the value of W/L
*M6 13 4 8 8 PFET W=37.25U L=1U
M7 13 9 12 12 NFET W=15U L=1U
M8 12 10 14 14 NFET W=15U L=1U
M9 5 9 11 11 NFET W=5U L=1U
M10 11 10 14 14 NFET W=5U L=1U
M11 9 9 10 10 NFET W=5U L=1U
M12 10 10 14 14 NFET W=5U L=1U
M13 3 3 6 6 PFET W=80U L=1U
IREF 0 9 DC 250U
.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017
.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(13) VP(13) VM(4) VP(4)
.TF V(13) VIC
.END
Results: Vos = 0.4423 mV, Adm = 22500, Acm = 0.2305, CMRR = 99.9 dB, ROUT = 90.3 MΩ. The
values of Add and Rout are similar to hand calculations. Acd and the CMRR are limited by the
offset induced mismatches in the devices. With the W/L of M6 corrected, Vos ≈ 0, Adm = 20800,
A = 9.28 x 10-3. R = 90.3 MΩ, CMRR = 127 dB.
cm
out
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-137
15.235
+V
25
1
50
1
M8
50
1
M 10
M9
I
I REF
I
M6
12.5
v
10
1
1
M1
1
M2
50
1
2
DD
50
1
2
v
M 11
vO
M7
1
M5
40
1
25
1
M4
M3
20
1
20
1
-V
SS
The W/L ratios have been scaled to keep the Q-points and gain the same. Note that the output
stage should remain a source follower pair and is not mirrored.
15-138
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.236
A
A
Q
Q
10
Q
9
I
I
VCC
5A
I1
8
2
Q
REF
A
5A
v
Q 12
Q
1
2
v
1
3
A
O
A
R
L
Q
7
2
Q
Q
v
Q 11
5A
Q
7
5
AE5
Q
4
A
-V EE
Note that the output stage should remain complementary emitter − followers.
I
The gain of the first stage is approximately Av1 = gm1rπ 5 = C1 β o5 , the mirror
IC 5
image amplifier with an npn transistor for Q5 will have the highest gain. The
voltage gain of the rest of the amplifier is the same.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-139
15.237
VEB 7 + VEB 8 = VEB 6 + VEB 4 = 2VT ln
⎡ 250µA ⎤
IC 4
I
= 2VT ln C14 = 0.05ln⎢
⎥ = 1.142V
IS 4
2IS 4
⎣ 2(15 fA)⎦
VEB 7 + VEB 8 = VT ln
I
IC 7
I
60 IC 8 IC 8
+ VT ln C 8 IC 7 = α F IB 8 = α F C 8 =
=
IS 7
IS 8
β F 61 60 61
VEB 7 + VEB 8 = VT ln
IC 8
IC 8
IC2 8
+ VT ln
→ 0.025ln
= 1.142
61(15 fA)
4 (15 fA)
61(15 fA)(4 )(15 fA)
IC 8 = 1.946 mA | IC16 ≅ IC 8 | AE16 =
IC16
1946µA
AE12 =
(1) = 7.78
250µA
IC12
⎛ 75µA ⎞
2(0.5583V )
VBE 6 = VEB10 = 0.025ln⎜
= 574 Ω
⎟ = 0.558V | RBB =
1.946 mA
⎝ 15 fA ⎠
⎤
⎡ (β + 1)r
o
π8
Adm = Av1 Av 2 Av 3 ≅ gm 2 ro2 [rπ 7 + (β o + 1)rπ 8 ] ⎢
gm 8 (ro8 ro16 )⎥ [1]
⎦
⎣ rπ 7 + (β o + 1)rπ 8
[ {
}]
[ (
(β o + 1)rπ 8 ⎛ µ f 8 ⎞⎤ 1
⎜
⎟⎥ [ ]
π 7 + (β o + 1)rπ 8 ⎝ 2 ⎠⎦
125µA
(40)(60 + 4.3) = 3.03 x 10 5
60
Adm = Av1 Av 2 Av 3 ≅ gm 2 ro2 rπ 7 + (β o + 1)rπ 8
[
Rid = 2rπ 1 = 2
15-140
µf 8
]4
Adm ≅ gm 2 (ro2 2rπ 7 )
≅
⎡
)]⎢⎣ r
µ
IC 2
β o7 f 8 =
31.8µA
IC 7
2
2
150(0.025V )
= 60 kΩ
125µA
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.238
*Figure P15.237 - Bipolar Op-Amp
VCC 1 0 DC 5
VEE 14 0 DC -5
V1 6 15 DC -74.17U AC 0.5
V2 7 15 DC 0 AC -0.5
VIC 15 0 DC 0
Q1 4 6 8 NBJT 1
Q2 5 7 8 NBJT 1
Q3 4 4 2 PBJT 1
Q4 5 4 3 PBJT 1
Q5 2 3 1 PBJT 1
Q6 3 3 1 PBJT 1
Q7 14 5 10 PBJT 1
Q8 11 10 1 PBJT 4
Q9 1 11 12 NBJT 1
Q10 14 13 12 PBJT 1
Q12 9 9 14 NBJT 1
Q14 8 9 14 NBJT 1
Q16 13 9 14 NBJT 7.78
IB 0 9 250U
RBB 11 13 574
.MODEL NBJT NPN BF=150 VA=60 IS=15F
.MODEL PBJT PNP BF=60 VA=60 IS=15F
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(12) VP(12)
.TF V(12) VIC
.END
SPICE Results: Vos = -74.17µV, Adm = 2.83 x 105, Acm = 0.507, CMRR = 115 dB,
Rid = 81.6 kΩ, Rout = 523 Ω.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-141
15.239
(a) We require forward - active region operation of all transistors.
For Q14 : VCB14 = 0 − VBE1 − (−VEE + VBE14 ) ≥ 0 → VEE ≥ 1.4V
For Q1 : VCB1 = VCC − VEB 6 − VEB 4 ≥ 0 → VCC ≥ 1.4V
For Q16 : VCB16 = 0 − VEB10 − (−VEE + VBE14 ) ≥ 0 → VEE ≥ 1.4V
For Q 8 : VBC 8 = VCC − VEB 8 − VBE 6 ≥ 0 → VCC ≥ 1.4V
So VCC ≥ 1.4V and VEE ≥ 1.4V
(b) We require forward - active region operation of all transistors with VIC present.
For Q14 : VCB14 = VIC − VBE1 − (−VEE + VBE14 ) = −1− 0.7 − (−VEE + 0.7) =≥ 0 → VEE ≥ 2.4V
For Q1 : VCB1 = VCC − VEB 6 − VEB 4 − VIC = VCC − 0.7 − 0.7 −1 ≥ 0 → VCC ≥ 2.4V
For an output range of ± 1V,
For Q16 : VCB16 = VO − VEB10 − (−VEE + VBE14 ) = −1 ≥ 0 → VEE ≥ 1.4V
For Q 8 : VBC 8 = VCC − VEB 8 − (VO + VBE 6 ) = VCC − 0.7 − (1+ 0.7) ≥ 0 → VCC ≥ 2.4V
So VCC ≥ 2.4V and VEE ≥ 2.4V
15.240
(a) I
C 22
= IC 20 =
VCC − VEB22 − VBE20 − (−VEE ) 3V − 0.7 − 0.7 − (−3V )
=
= 46.0µA
R1
100kΩ
IC23 = 3IC 22 = 138 µA | IC 24 = IC 22 = 46.0 µA |
⎛ 46.0µA ⎞
V ⎛ I ⎞ 0.025V ⎛ 46.0µA ⎞
ln⎜
I1 = T ln⎜ C 20 ⎟ =
⎟ = 6.25µAln⎜
⎟ =→ I1 = 9.72µA
4kΩ
R ⎝ I1 ⎠
⎝ I1 ⎠
⎝ I1 ⎠
(b) I
22V − 0.7 − 0.7 − (−22V )
= 426µA
100kΩ
IC23 = 3IC 22 = 128 µA | IC 24 = IC 22 = 426 µA
⎛ 426µA ⎞
V ⎛I ⎞
I1 = T ln⎜ C 20 ⎟ = 6.25µAln⎜
⎟ → I1 = 19.3µA
R ⎝ I1 ⎠
⎝ I1 ⎠
C 22
= IC20 =
(c) The input bias current and input resistance of the amplifier are directly
dependent upon I1 , whereas the gain of the interior amplifier stages is
approximately independent of bias current.
15-142
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.241
250µA
= 83.3 µA | I3 = IREF = 83.3 µA
3
V − VEB 22 − VBE 20 − (−VEE )
12 − 0.7V − 0.7 − (−12) V
IREF = CC
| R1 =
= 271 kΩ
R1
83.3
µA
V ⎛ I ⎞ 0.025V ⎛ 83.3µA ⎞
ln⎜
R2 = T ln⎜ REF ⎟ =
⎟ = 255 Ω
I1 ⎝ I1 ⎠ 50µA ⎝ 50µA ⎠
I2 = 3IREF → IREF =
15.242
IREF = I3 = 300 µA | I2 = 3IREF = 900 µA
VCC − VEB 22 − VBE 20 − (−VEE )
15 − 0.7V − 0.7 − (−15) V
| R1 =
= 95.3 kΩ
R1
300
µA
V ⎛ I ⎞ 0.025V ⎛ 300µA ⎞
ln⎜
R2 = T ln⎜ REF ⎟ =
⎟ = 462 Ω
I1 ⎝ I1 ⎠ 75µA ⎝ 75µA ⎠
IREF =
15.243
For forward - active region operation of Q3 , VBC 3 ≥ 0
VEE ≥ VIC + VBE1 + VBE 3 + VBE 7 + VBE 5 + VR1
For forward - active region operation of Q1, VCB1 ≥ 0
VCC − VEB 9 ≥ VIC
For the output stage, VCC ≥ VBE15 + VI 3 = 0.7 + 0.7 = 1.4 V
−VEB16 − VEB12 ≥ VBE11 + VR 5 − (−VEE ) → VEE ≥ 3VBE + VR1 ≅ 2.1V
(a) VIC = 0, VEE ≥ 4VBE + VR
1
≅ 2.8V | VCC − VEB 9 ≥ 0 → VCC ≥ 0.7V
Combining these results yields : VCC ≥ 1.4V and VEE ≥ 2.8V
(b) VIC = ±1, VEE ≥ 1+ 4VBE + VR
1
≅ 3.8V | VCC − VEB 9 ≥ 1 → VCC ≥ 1.7V
If also account for the output stage, VCC ≥ VO + VBE15 + VI 3 = 1 + 0.7 + 0.7 = 2.4 V
Combining these results yields : VCC ≥ 2.4V and VEE ≥ 3.8V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-143
15.244
50µA
(7.32µA) = 20.3µA
18µA
50µA
Using Eq. 16.139 : io = −20(20.3µA)v id = (−0.406 mS )v id | IC 4 =
(7.25µA) = 20.1µA
18µA
⎛ 20.3µA(1kΩ)⎞
60V
Rout 6 ≅ ro6⎜1+
= 2.84 MΩ
⎟ = 1.81ro6 | Rth = 1.81ro6 2ro4 = 0.952ro4 = 0.95
0.025V ⎠
20.1µA
⎝
The input stage current is proportional to I1 : IC 2 =
io = (−4.06 x 10 -4 )v id | Rth = 2.84 MΩ
As a check, we know that g m ∝ IC and ro ∝
⎛ 50µA ⎞
-4
io = 1.46x10−4 v id ⎜
⎟ = −4.06 x 10 v id
µ
A
18
⎝
⎠
which underestimates Rth .
15.245
(a)
R2 =
β o2 ro2
2
=
1
. Using the results from Fig. 16.60,
IC
⎛ 18µA ⎞
and Rth = 6.54 MΩ⎜
⎟ = 2.35MΩ
⎝ 50µA ⎠
50 60 + 15.7
= 2.84 MΩ | The cascode source uses up an extra VEB .
2 0.666mA
(b) [y 22 ] = Rout11 R2 = 407kΩ 2.84 MΩ = 356kΩ
(c ) Adm = 256(6.70mS)(356kΩ) = 6.11 x 10 5
−1
| The other y - parameters are unchanged.
15.246
+VCC
A
A
3A
Q 23A
Q 22A
Q 24
I
A
3
3A
Q 22B
Q 23B
I
R
2
1
I
1
Q
Q 20
A
21
A
R
2
-V
EE
15-144
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.247
gm10 = 40(19.8µA) = 0.792mS | rπ 10 =
150(0.025V )
60V
= 189kΩ | ro10 =
= 3.03MΩ
19.8µA
19.8µA
150(0.025V )
60V
= 5.63kΩ | ro11 =
= 90.1kΩ
0.666mA
0.666mA
*Problem 15.247 - Small Signal Parameters.
V1 1 0 DC 0
V2 4 0 AC 1
RPI10 1 2 189K
RO10 2 0 3.03MEG
GM10 0 2 1 2 0.792M
RE10 2 0 50K
RPI11 2 3 5.63K
RO11 4 3 90.1K
GM11 4 3 2 3 26.6M
RE11 3 0 100
R2 4 0 115K
.TF I(V2) V1
.AC LIN 1 1000 1000
.PRINT AC IM(V2) IP(V2) IM(V1) IP(V1)
.END
gm11 = 40(0.666mA) = 26.6mS | rπ 11 =
-1
Results : y11-1 = 2.38 MΩ | y12 = 3.27 x 10−10 S ≅ 0 | y 21 = 6.66 mS | y 22
= 81.9 kΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-145
15.248
IC11
= 50µA | IC 3 = IC 6 = IC 4 = 50µA
2
I
IC1 = IC 3 = IC 7 = 50µA | IC 2 = IC 6 = IC 8 = 50µA | IC 9 = 2 C 8
(a) Assume large β F :
IC11 = IREF = 100µA | IC 4 = IC 5 =
βF
VCE1 = VCE 2 = 15 − (−0.7) = 15.7V | VEC 4 = VEC 5 = 0.7V
VCE 7 = VCE 8 = 0.7 + 0.7 = 1.4V | VCE 9 = 15 − (−15 + 0.7) = 29.3V | VCE10 = 0.7V
VEC 3 = VEC 2 = (0 − 0.7) − (−15 + 1.4) = 12.9 | VCE11 = 0 − 0.7 − 0.7 − (−15) = 13.6V
Q
1
2
3
IC (µA) 100 100 -50
VCE (V) 15.7 15.7 -12.9
4
5
6
7
8
9
10
11
-50
-0.7
-50
-0.7
-50
-12.9
50
1.4
50
1.4
--29.3
100
0.7
100
13.6
(b) Transistor Q11 replicates the reference current. This current divides in two and controls two
matched current mirrors formed of Q4-Q3 and Q5-Q6. The currents of Q1 and Q7, and Q2 and Q8
are equal to the output current of Q3 and Q4.
(c) v1 is the inverting input; v2 is the non-inverting input.
g m5 = g m6 | g m2 = 2g m6 | ro8 = ro6 | io = g m6ve6
ve6 = v id
Differential-mode
Half Circuit
Q2
v id
2
+
-
Q6
1
g m5
15-146
io
2
1
g m2
1
= v | io = g m6 vid
⎛ 1 1 ⎞ 2 id
2
1+ g m2 ⎜
⎟
⎝ g m5 g m6 ⎠
1
1
1
Gm = g m6 = g m2 = (40)(100µA)= 1.00 mS
2
4
4
⎡
⎛
⎞⎤
1
6
= ro8 ro6⎢1+ g m6⎜
Rout = ro8 Rout
⎟⎥
⎝ g m5 + g m2 ⎠⎦
⎣
⎛ 61.4V ⎞
⎛ 72.9V ⎞
Rout = ro8 1.33r06 = ⎜
⎟ 1.33⎜
⎟ = 752 kΩ
⎝ 50µA ⎠
⎝ 50µA ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15.249
(a) Assume large β F :
IC 3 = IC 4 = β F
IC 8 = IREF = 100µA | IC10 = IC 9 = IB 8 =
IC 8
βF
IC10 IC 8
I
=
= 50µAIC1 = IC 3 = IC 5 = 50µA | IC 2 = IC 4 = IC 6 = 50µA | IC 7 = 2 C 5
2
2
βF
VCE1 = VCE 2 = 15 − (−0.7) = 15.7V | VCE 5 = VCE 6 = 0.7 + 0.7 = 1.4V
VEC 3 = VEC 4 = 0 − 0.7 − (−15 + 1.4) = 12.9V | VCE 7 = 15 − (−15 + 0.7) = 29.3V
VEC 8 = 0.7 + 0.7 = 1.4V | VCE 9 = 0.7V | VCE10 = 0 − 0.7 − 0.7 − (−15) = 13.6V
Q
1
2
3
4
IC (µA) 50
50
-50
-50
VCE (V) 15.7 15.7 -12.9 -12.9
5
6
7
8
9
10
50
1.4
50
1.4
--29.3
-100
1.4
--0.7
--13.6
(b) Transistors Q9 and Q10 form a current mirror that replicates the base current of transistor Q8.
The output current divides in two and forms the base currents of Q3 and Q4. Since Q3 and Q4
match Q8, the collector currents of Q1-Q6 will all be equal to IREF/2.
(c) v1 is the inverting input; v2 is the non-inverting input.
io
= g m4ve4 | io = 2g m4ve4
2
g m2
v
v
1
ve4 = id
= vid | io = g m4 id
⎛ 1 ⎞ 4
2
2
1+ g m2⎜
⎟
⎝ g m4 ⎠
g m2 = g m4 | ro6 = ro4 |
Q2
v id
2
Differential-mode
Half Circuit
io
+
-
Q4
2
1
1
1
Gm = g m4 = g m2 = (40)(50µA)= 1.00 mS
2
2
2
⎡
⎛ 1 ⎞⎤
4
= ro6 ro4 ⎢1+ g m4 ⎜
Rout = ro6 ROUT
⎟⎥
⎝ g m2 ⎠⎦
⎣
⎛ 61.4V ⎞ ⎛ 72.9V ⎞
Rout = ro6 2ro4 = ⎜
⎟ 2⎜
⎟ = 864 kΩ
⎝ 50µA ⎠ ⎝ 50µA ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
15-147
CHAPTER 16
16.1
Av (s) = 50
s2
s2
| Amid = 50 | FL (s)=
| Poles : - 2,-30 | Zeros : 0,0
(s + 2)(s + 30)
(s + 2)(s + 30)
Yes, s = −30 | Av (s) ≈ 50
s
rad
| ω L ≅ 30
s
(s + 30)
| fL =
ω L 30
≅
= 4.77 Hz
2π 2π
2
2
1
302 + 22 − 2(0) − 2(0) = 4.79 Hz
2π
50ω 2
| MATLAB : f L = −4.80 Hz
Av (jω ) =
ω 2 + 22 ω 2 + 302
fL =
16.2
Av (s) =
400s2
s2
s2
=
200
|
A
=
200
|
F
s
=
(
)
mid
L
2s2 + 1400s + 100,000
(s + 619)(s + 80.8)
(s + 619)(s + 80.8)
Poles : - 619,-80.8
Av (s) ≈ 200
rad
s
| Zeros : 0, 0 | Yes, a 5 :1 split is sufficient | s = −619
s
rad
| ω L ≅ 619
s
(s + 619)
| fL ≅
619
= 98.5 Hz
2π
2
2
1
80.82 + 6192 − 2(0) − 2(0) = 99.4 Hz
2π
200ω 2
| MATLAB : 100 Hz
Av (jω ) =
ω 2 + 80.82 ω 2 + 6192
fL ≅
16.3
Av (s) = −150
s (s + 15)
(s + 12)(s + 20)
Poles : -12, - 20
fL ≅
| Amid = −150 | FL (s)=
s(s + 15)
(s + 12)(s + 20)
rad
rad
| Zeros : 0, -15
| No, the poles and zeros are closely spaced.
s
s
2
2
1
122 + 202 − 2(0) − 2(15) = 1.54 Hz
2π
Av (jω ) =
150ω ω 2 + 152
ω 2 + 122 ω 2 + 202
| MATLAB : f L = 2.72 Hz | ω L = 17.1
rad
s
Note that ωL =16.1 rad/s does not satisfy the assumption used to obtain Eq. (16.15), and the
estimate using Eq. (16.15) is rather poor.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-1
16.4
(
)( )( )
3x1011 10−4 10−5
9x1011
300
Av (s) = 2
=
=
5
9
⎛ s
⎞⎛ s
⎞
⎛ s
⎞⎛ s
⎞
3s + 3.3x10 s + 3x10
⎜ 4 + 1⎟⎜ 5 + 1⎟
⎜ 4 + 1⎟⎜ 5 + 1⎟
⎝ 10
⎠⎝10
⎠
⎝ 10
⎠⎝10
⎠
1
Amid = 300 | FH (s)=
⎛ s
⎞⎛ s
⎞
⎜ 4 + 1⎟⎜ 5 + 1⎟
⎝ 10
⎠⎝10
⎠
300
rad
| ω H ≅ 10 4
s
s
+1
4
10
⎞−1
⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎟
⎜ 4 ⎟ + ⎜ 5 ⎟ − 2⎜ ⎟ − 2⎜ ⎟ ⎟ = 1.58 kHz
⎝ 10 ⎠ ⎝ 10 ⎠
⎝ ∞⎠
⎝ ∞⎠
⎠
Poles : -10 4 ,-105
⎛
1 ⎜
fH ≅
2π ⎜
⎝
Av (jω ) =
rad
s
| Yes : Av (s) ≈
3x1011
( )
ω + 10
2
16.5
4
2
( )
ω + 10
2
⎛
5
2
| fH ≅
10 4
= 1.59kHz
2π
| MATLAB : f H = 1.58 kHz
⎞
⎛
s ⎞
⎜1+
⎟
⎝ 2x109 ⎠
Av (s) =
= 300
⎛
⎛
s ⎞⎛
s ⎞
s ⎞
s ⎞⎛
10 7 ⎜1+ 7 ⎟⎜1+ 9 ⎟
⎜1+ 7 ⎟⎜1+ 9 ⎟
⎝ 10 ⎠⎝ 10 ⎠
⎝ 10 ⎠⎝ 10 ⎠
⎛
s ⎞
⎜1+
⎟
⎝ 3x109 ⎠
Amid = 300 | FH (s)=
| Poles : -10 7 , -109 Zeros : - 3x109 , ∞
⎛
⎞
⎛
⎞
s
s
⎜1+ 7 ⎟⎜1+ 9 ⎟
⎝ 10 ⎠⎝ 10 ⎠
s
⎟
(3x10 )⎜⎝1+ 3x10
⎠
9
9
300
rad
Yes : Av (s ) ≈
| ω H ≅ 10 7
⎛
⎞
s
s
⎜1+ 7 ⎟
⎝ 10 ⎠
10 4
| fH ≅
= 1.59 MHz
2π
−1
⎛
2
2
2
2⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
1 ⎜
1
1
1
1
− 2⎜ ⎟ ⎟ = 1.59 MHz
fH =
⎜ 7 ⎟ + ⎜ 9 ⎟ − 2⎜
9⎟
⎜
2π ⎝ 10 ⎠ ⎝ 10 ⎠
⎝ 3x10 ⎠
⎝ ∞⎠ ⎟
⎠
⎝
Av (jω ) =
17-2
(
2x109 ω 2 + 3x109
( )
ω + 10
2
7
2
)
2
( )
ω + 10
2
9
2
| MATLAB : f H = 1.59 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.6
⎛
⎛
s ⎞
s ⎞
2x109 5x105 ⎜1+
1+
⎟
⎜
⎟
⎝ 5x105 ⎠
⎝ 5x105 ⎠
Av (s) =
= 3333
⎛
⎛
s ⎞⎛
s ⎞
s ⎞
s ⎞⎛
5
6
1.5x10 2x10 ⎜1+
1+
1+
1+
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝ 1.3x105 ⎠⎝ 2x106 ⎠
⎝ 1.5x105 ⎠⎝ 2x106 ⎠
⎛
s ⎞
⎜1+
⎟
rad
⎝ 5x105 ⎠
Amid = 3333 | FH (s)=
| Poles : -1.5x105 , - 2x106
⎛
s
s ⎞⎛
s ⎞
1+
⎜1+
⎟
⎜
⎟
⎝ 1.5x105 ⎠⎝ 2x106 ⎠
rad
Zeros : - 5x105
, ∞ | No, the poles and zeros are closely spaced and will interact.
s
−1
⎛
2
2
2
2⎞
⎛ 1 ⎞
⎛1⎞
1 ⎜ ⎛ 1 ⎞ ⎛ 1 ⎞
fH ≅
+⎜
− 2⎜
− 2⎜ ⎟ ⎟ = 26.3 kHz
⎜
⎟
⎟
⎟
5
6
5
2π ⎜ ⎝ 1.5x10 ⎠ ⎝ 2x10 ⎠
⎝ 5x10 ⎠
⎝ ∞⎠ ⎟
⎠
⎝
(
(
)(
)(
)
)
(
)
+ (2x10 )
2x109 ω 2 + 5x105
Av (jω ) =
(
ω + 1.3x10
2
5
)
2
ω
2
2
6
2
| MATLAB : 26.3 kHz
16.7
6x108
s2
1
s ⎞
s ⎞⎛
1000(2000) (s + 1)(s + 2) ⎛
⎜1+
⎟⎜1+
⎟
⎝ 1000 ⎠⎝ 2000 ⎠
⎤
⎡
⎥
⎢
⎡
⎤
s2
1
⎥ = 200F (s)F (s)
⎢
⎥
Av (s) = 300 ⎢
L
H
⎛
⎞
⎛
⎞
⎢
s
s
+
1
s
+
2
s
⎢⎣ (
)( )⎥⎦ ⎢⎜1+ ⎟⎜1+ ⎟⎥⎥
⎣⎝ 500 ⎠⎝ 1000 ⎠⎦
Av (s) =
rad
| Zeros : 0, 0, ∞, ∞ | No. | No.
s
6x108 ω 2
Poles; -1, - 2, -1000, - 2000
Av (jω ) =
fL =
1
2π
12 + ω 2 22 + ω 2 10002 + ω 2 20002 + ω 2
1 + (2) − 2(0) − 2(0)
()
1
fH =
2π
03/09/2007
2
2
2
2
= 0.356 Hz | MATLAB : 0.380 Hz
⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2
⎜
⎟ +⎜
⎟ − 2⎜ ⎟ − 2⎜ ⎟ = 142 Hz | MATLAB : 133 Hz
⎝1000 ⎠ ⎝ 2000 ⎠
⎝ ∞⎠
⎝ ∞⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-3
16.8
⎛
s ⎞
1+
⎜
⎟
s (s + 1)
10 (200)
⎝ 200 ⎠
Av (s) =
2
2
(100) (300) (s + 3)(s + 5)(s + 7)⎛⎜1+ s ⎞⎟ ⎛⎜1+ s ⎞⎟
⎝ 100 ⎠ ⎝ 300 ⎠
⎤
⎡
⎛
s ⎞
⎥
⎢
⎜1+
⎟
⎤
⎡
s2 (s + 1)
200 ⎠
⎥
⎢
⎝
5
⎥
= 6.67x105 FL (s )FH (s)
Av (s) = 6.67x10 ⎢
2
⎥
⎢
⎞⎛
⎞
⎢⎣(s + 3)(s + 5)(s + 7)⎥⎦ ⎛
⎢⎜1+ s ⎟ ⎜1+ s ⎟ ⎥
⎢⎣⎝ 100 ⎠ ⎝ 300 ⎠ ⎥⎦
10
2
No dominant pole at either low or high frequencies.
Av (jω ) =
fL =
1
2π
1010 ω 2 ω 2 + 12 ω 2 + 2002
(
)
ω 2 + 32 ω 2 + 52 ω 2 + 72 ω 2 + 1002 ω 2 + 3002
1 − 2(0)
(3) + (5) + (7) − 2()
2
2
2
2
2
= 1.43 Hz | MATLAB : 1.62 Hz
−1
⎛
2
2
2
2
2⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
1 ⎜
1
1
1
1
1
+⎜
+⎜
− 2⎜
− 2⎜ ⎟ ⎟ = 12.5 Hz | MATLAB : 10.6 Hz
fH =
⎜
⎟
⎟
⎟
⎟
2π ⎜ ⎝ 100 ⎠ ⎝ 100 ⎠ ⎝ 300 ⎠
⎝ 200 ⎠
⎝ ∞⎠ ⎟
⎠
⎝
17-4
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.9
Low frequency:
0.1 µF
2kΩ
0.1 µF
Rout
v
2.43 M Ω
Rin
43 k Ω
i
1 MΩ
10 µF
13 k Ω
Mid-band:
2kΩ
Rout
v
43 k Ω
Rin
i
1 MΩ
2.43 M Ω
(b) Avt =
(
vd
= −g m RL = −g m Rout R3
vg
)|
Amid =
2(0.2mA)
Rin
2I D
Avt | g m =
=
= 0.400mS
RI + Rin
VGS − VTN
1V
Rin = 2.43MΩ | Rout = RD ro ≅ RD = 43kΩ assuming λ = 0 since it is not specified.
⎛ 2.43MΩ ⎞
Amid = −⎜
⎟(0.400mS ) 43kΩ 1MΩ = −16.5
⎝ 2kΩ + 2.43MΩ ⎠
1
rad
1
rad
ω1 = −7
= 4.11
| ω2 =
= 9.59
−7
s
s
10 F (2.43MΩ + 2kΩ)
10 F (43kΩ + 1MΩ)
(
(
ω3 =
)
)
(
)
1
1
rad
=
= 47.7
−5
⎛
⎞
s
10 F 13kΩ 2.5kΩ
1
10−5 F ⎜13kΩ
⎟
gm ⎠
⎝
(
(
)
ω3 is dominant : f L ≅
)(
)
| ωZ =
1
(10 F )(13kΩ)
−5
= 7.69
rad
s
ω3
= 7.59 Hz
2π
2
2
2
2
1
4.11) + (9.59) + (47.7) − 2(7.69) = 7.58 Hz
(
2π
= 0.2mA(56kΩ)+ 5V = 16.2 V
Using Eq. (16.15) yields : f L ≅
(c) VDD = I D (RD + RS )+ VDS
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-5
16.10
0.1 µF
5kΩ
0.1 µF
Rout
v
Rin
220 k Ω
43 k Ω
i
430 k Ω 560 k Ω
10 µF
13 k Ω
Low frequency:
5k Ω
Rout
220 k Ω
43 k Ω
Rin
vi
243 k Ω
Mid-band:
Avt =
(
vd
= −g m RL = −g m Rout R3
vg
)|
Amid =
2(0.2mA)
Rin
2I D
Avt | g m =
=
= 0.400mS
RI + Rin
VGS − VTN
1V
Rin = 243kΩ | Rout = RD ro ≅ RD = 43kΩ assuming λ = 0 since it is not specified.
⎛ 243kΩ ⎞
Amid = −⎜
⎟(0.400mS ) 43kΩ 220kΩ = −14.1
⎝ 5kΩ + 243kΩ ⎠
(
ω1 =
ω3 =
1
(10 F )(243kΩ + 5kΩ)
−7
rad
s
| ω2 =
1
(10 F )(43kΩ + 220kΩ)
−7
1
1
rad
=
= 47.7
−5
⎛
⎞
s
10 F 13kΩ 2.5kΩ
1
10−5 F ⎜13kΩ
⎟
gm ⎠
⎝
(
(
)
Using Eq. (17.16) : f L ≅
17-6
= 40.3
)
1
2π
)(
)
| ωZ =
(40.3) + (38.0) + (47.7) − 2(7.69)
2
2
2
2
= 38.0
1
rad
s
(10 F )(13kΩ)
−5
= 11.5 Hz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
= 7.69
rad
s
16.11
(a ) Assume that ω
C3 =
is dominant : f L ≅ ω3 = 2π (50) = 314
(
3
)
= 1.5 µF | ω3 =
1
(10 F )(2.43MΩ + 2kΩ)
−7
(c) Assume that ω
⎛
1
ω3⎜ RS
⎝
1⎞
⎟
gm ⎠
=
3
1
= 4.11
rad
s
| ω2 =
(
2
)
1
(10 F )(43kΩ + 1MΩ)
−7
2
2
(
1
)
1
2π
rad
s
| ω2 =
= 318
2
= 49.3 Hz
rad
1
rad
| ωZ =
= 51.3
s
s
(1.5µF )(13kΩ)
1
(10 F )(43kΩ + 220kΩ)
−7
2
rad
s
2I D
0.4mA
=
= 0.4mS
1V
VGS − VTN
(40.3) + (38.0) + (318) − 2(51.3)
2
= 9.59
rad
s
= 1.52 µF where g m =
1.5 µF 13kΩ 2.5kΩ
= 40.3
rad
1
rad
| ωZ =
= 51.3
s
s
(1.5µF )(13kΩ)
(4.11) + (9.59) + (318) − 2(51.3)
314 13kΩ 2.5kΩ
Using Eq. (16.15) yields : f L ≅
03/09/2007
1
2π
1
(10 F )(243kΩ + 5kΩ)
−7
)
= 318
is dominant : f L ≅ ω3 = 2π (50)= 314
Choose C3 = 1.5 µF | ω3 =
ω1 =
(
1
1.5 µF 13kΩ 2.5kΩ
Using Eq. (17.16) yields : f L ≅
C3 =
rad
s
1
1
2I D
0.4mA
=
= 1.52 µF where g m =
=
= 0.4mS
⎛
⎞
VGS − VTN
1V
314 13kΩ 2.5kΩ
1
ω3⎜ RS
⎟
gm ⎠
⎝
(b) Choose C
ω1 =
3
2
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
= 38.0
rad
s
= 50.1 Hz
17-7
16.12
4.7 µF
C1
RI
C2
1 µF
100 Ω
Rin
vi
Rout
RD
1.3 k Ω
R
100 k Ω
4.3 k Ω
S
R3
Low Frequency:
(b) Av (s) = Amid
(c )Amid
s2
| ω1 =
(s + ω1)(s + ω 2 )
1
1
| 2 zeros at ω = 0
⎛
C2 (RD + R3 )
1⎞
C1⎜ RI + RS
⎟
gm ⎠
⎝
⎛ Rin ⎞
⎛ Rin ⎞
⎛ Rin ⎞
=⎜
⎟ Avt = ⎜
⎟ gm RL = ⎜
⎟ gm (Rout R3 ) | gm = 5mS
⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
Rin = RS
| ω2 =
1
= 173Ω | RL = Rout R3 | Rout = RD ro = RD = 4.3kΩ (assuming ro = ∞)
gm
⎛
⎞
173Ω
Amid = ⎜
⎟(0.005)(4.3kΩ 100kΩ) = +13.1 → 22.3 dB
⎝ 100Ω + 173Ω ⎠
1
rad
1
rad
ω1 =
= 779
| ω 2 = −6
= 9.59
−6
4.7x10 (100 + 173)
s
10 (4.3kΩ + 100kΩ)
s
ω1 is dominant : f L ≅
ω1
= 124 Hz
2π
16.13
(a) Assume ω1
Rin = RS
17-8
rad
1
| C1 =
ω1 (RI + Rin )
s
1
1
= 1300Ω 200Ω = 173Ω | C1 =
= 0.583 µF
3
gm
6.28x10 (100 + 173)
(b) Choose
ω2 =
is dominant : ω L ≅ ω1 = 2π (1000Hz) = 6280
C1 = 0.56 µF | ω1 =
1
rad
= 6540
0.56x10 (100 + 173)
s
−6
ω
1
rad
= 10.3
| ω1 is dominant : f L ≅ 1 = 1040 Hz
2π
10 (22kΩ + 75kΩ)
s
−6
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.14
4.7 µF
200 Ω
v
1 µF
Rin
Rout
4.3 k Ω
i
2.2 k Ω
51 k Ω
Low frequency:
200 Ω
Rin
vi
Rout
4300 Ω
2.2 k Ω
51 k Ω
Mid-band:
s2
| ω =
mid
(s + ω1)(s + ω2 ) 1
1
(b) A (s)= A
v
(c) A
mid
1
| ω2 =
| 2 zeros at ω = 0
⎛
C2 (RC + R3 )
1⎞
C1⎜ RS + RE
⎟
gm ⎠
⎝
⎛ Rin ⎞
⎛ Rin ⎞
⎛ Rin ⎞
=⎜
⎟ Avt = ⎜
⎟ g m RL = ⎜
⎟ g m Rout R3 | g m = 40(1mA)= 0.04S
⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
⎝ RI + Rin ⎠
(
)
1
= 24.9Ω | RL = Rout R3 | Rout = RC ro = RC = 2.2kΩ
gm
⎛
⎞
24.9Ω
Amid = ⎜
⎟(0.04) 2.2kΩ 51kΩ = +9.34 →19.4 dB
⎝ 200Ω + 24.9Ω ⎠
1
rad
1
rad
ω1 =
= 946
| ω2 = −6
= 18.8
−6
s
s
4.7x10 (200 + 24.9)
10 (2.2kΩ + 51kΩ)
Rin = RE
(
ω1 is dominant : f L ≅
(d) g
m
)
ω1
= 151 Hz
2π
= 40(10µA)= 0.0004S | Rin = 2.49kΩ | Rout = RC ro = RC = 220kΩ
⎛
⎞
2.49kΩ
Amid = ⎜
⎟(0.0004) 220kΩ 510kΩ = +56.9 → 35.1 dB
⎝ 200Ω + 2.49kΩ ⎠
1
rad
1
rad
ω1 =
= 79.1
| ω2 = −6
= 1.37
−6
s
s
4.7x10 (200 + 2.49kΩ)
10 (220kΩ + 510kΩ)
(
ω1 is dominant : f L ≅
03/09/2007
)
ω1
= 12.6 Hz
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-9
16.15
(a ) Assume ω
1
is dominant : ω L ≅ ω1 = 2π (500Hz)= 3140
rad
s
C1 =
1
1
| Rin = RE
| g m = 40(1mA) = 0.04S | Rin = 4300Ω 25Ω = 24.9Ω
gm
ω1(RI + Rin )
C1 =
1
= 1.42 µF
3.14x10 (200 + 24.9)
3
(b) Choose
ω2 =
C1 = 1.5 µF | ω1 =
1
rad
= 2960
s
1.5x10 (200 + 24.9)
1
rad
= 18.8
s
10 (2.2kΩ + 51kΩ)
−6
(c) Assume ω
1
−6
| ω1 is dominant : f L ≅
is dominant : ω L ≅ ω1 = 2π (500Hz)= 3140
ω1
= 472 Hz
2π
rad
s
C1 =
1
1
| Rin = RE
| g m = 40(10µA) = 0.0004S | Rin = 430kΩ 2500Ω = 2.49kΩ
gm
ω1(RI + Rin )
C1 =
1
= 0.118 µF | Choose C1 = 0.12 µF
3.14x10 (200 + 2490)
ω1 =
1
rad
= 2100
s
0.12x10 (200 + 2490)
3
−6
ω1 is dominant : f L ≅
17-10
| ω2 =
1
rad
= 18.8
s
10 (2.2kΩ + 51kΩ)
−6
ω1
= 493 Hz
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.16
(a) gm = 40IC = 40(0.175mA) = 7.00mS
| rπ =
βo
gm
=
100
= 14.3kΩ | ro = ∞ (VA not given)
7.00mS
Rin = R1 R2 rπ = 100kΩ 300kΩ 14.3kΩ = 12.0kΩ | RL = RC R3 = 43kΩ 100kΩ = 30.1kΩ
Amid =
Rin
12.0kΩ
gm RL =
(7.00mS)(30.1kΩ) = −194
RI + Rin
1kΩ + 12.0kΩ
SCTC : R1S = RI + Rin = 1kΩ + 12.0kΩ = 13.0kΩ | Rth = R1 R2 RI = 100kΩ 300kΩ 1kΩ = 987Ω
R2S = RE
fL ≅
Rth + rπ
987Ω + 14.3kΩ
= 15kΩ
= 150Ω | R3S = RC + R3 = 43kΩ + 100kΩ = 143kΩ
βo + 1
101
⎤ (38.5 + 667 + 69.9)
1
1 ⎡
1
1
+
+
= 123 Hz
⎢
⎥=
2π
2π ⎣2x10−6 (13.0kΩ) 10x10−6 (150Ω) 1x10−7 (143kΩ)⎦
(b) Note that the Q-point assumed in part (a) is not quite correct.
SPICE yields: (144 µA, 3.67 V), Amid = 43.9 dB, fL = 91 Hz
(c ) VEQ = VCC
IC = β F
R1
100kΩ
= 12
= 3V | REQ = R1 R2 = 100kΩ 300kΩ = 75.0kΩ
R1 + R2
100kΩ + 300kΩ
VEQ − VBE
REQ + (β F + 1)RE
= 100
3 − 0.7
= 145µA
75kΩ + (101)15kΩ
VCE = VCC − IC RC − IE RE = 12 − (0.145mA)(43kΩ) −
101
(0.145mA)(15kΩ) = 3.57 V
100
These values agree with the SPICE results listed above in part (b).
16.17
(a ) Use the values from Section 16.3.1, and assume ω3 is dominant.
ω L = 2π (2500Hz)= 15700
C3 =
rad
s
| ω3 = ω L − ω1 − ω2 = 15700 − 225 + 96.1 = 15390
rad
s
1
1
=
= 2.86 µF
ω3 R3S 15390(22.7Ω)
(b) Choose
C3 = 2.7 µF | ω3 =
1
2.7x10
−6
(22.7Ω)
= 16320
rad
s
rad
rad
225 + 96.1+ 16320
| ω1 = 225
| fL ≅
= 2650 Hz
s
s
2π
or if ω L must be no more than 2500 Hz, choose C3 = 3.3 µF
ω2 = 96.1
ω3 =
1
3.3x10
03/09/2007
−6
(22.7Ω)
= 13350
rad
s
| fL ≅
225 + 96.1+ 13350
= 2180 Hz
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-11
16.18
1k Ω
43 k Ω
43 k Ω
Rin
v
i
100k Ω 300k Ω
75 k Ω
21.5 k Ω
(a) Mid-band:
1k Ω
1 µF
5 µF
43 k Ω
vi
43 k Ω
75 k Ω
13 k Ω
22 µF
Low frequency:
(b) g
m
= 40IC = 40(0.164mA)= 6.56mS | rπ =
βo
gm
=
100
= 15.2kΩ | ro = ∞ (VA not given)
6.56mS
Rin = R1 R2 rπ = 100kΩ 300kΩ 15.2kΩ = 12.6kΩ | RL = RC R3 = 43kΩ 43kΩ = 21.5kΩ
Amid =
Rin
12.6kΩ
g m RL =
(6.56mS )(21.5kΩ)= −131
RI + Rin
1kΩ + 12.6kΩ
SCTC : R1S = RI + Rin = 1kΩ + 12.6kΩ = 13.6kΩ | Rth = R1 R2 RI = 100kΩ 300kΩ 1kΩ = 987Ω
Rth + rπ
987Ω + 15.2kΩ
= 13kΩ
= 158Ω | R3S = RC + R3 = 43kΩ + 43kΩ = 86kΩ
βo + 1
101
⎤ (14.7 + 288 + 11.6)
1
1 ⎡
1
1
⎢
⎥=
fL ≅
+
+
= 50.0 Hz
2π
2π ⎢⎣5x10−6 (13.6kΩ) 22x10−6 (158Ω) 1x10−6 (86kΩ)⎥⎦
⎛ 101 ⎞
(c) VCC = I E RE + VEC + IC RC = 0.164mA⎜⎝100 ⎟⎠(13kΩ)+ 2.79V + 0.164mA(43kΩ)= 12.0 V
R2S = RE
17-12
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.19
SCTC : R1S = RI + RG = 1kΩ + 1MΩ = 1.00 MΩ | ω1 =
R2S = RS
1
1
1
rad
= 6.8kΩ
= 607Ω | ω2 =
= 165
gm
1.5mS
s
607Ω(10µF )
R3S = RD + R3 = 22kΩ + 68kΩ = 90kΩ | ω3 =
fL ≅
1
rad
= 10.0
s
1.00 MΩ(0.1µF )
(10.0 + 165 + 111) = 45.5 Hz
1
rad
= 111
s
90kΩ(0.1µF )
2π
16.20
SCTC : R1S = RI + RG = 1kΩ + 500kΩ = 501kΩ | ω1 =
R2S = RS
1
1
1
rad
= 10kΩ
= 1.18kΩ | ω2 =
= 84.8
gm
0.75mS
s
1.18kΩ(10µF )
R3S = RD + R3 = 43kΩ + 10kΩ = 53kΩ | ω3 =
fL ≅
1
rad
= 20.0
s
501kΩ(0.1µF )
(20.0 + 84.8 + 189) = 46.8 Hz
03/09/2007
1
rad
= 189
s
53kΩ(0.1µF )
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-13
16.21
C1
R
I 2k Ω
v
C
4.7 µF
R
3
12 k Ω
i
C
R4
892 k Ω
2
3
1 µF
R7
22 k Ω
100 k Ω
0.1 µF
(a) Low Frequency:
2kΩ
vi
Rin
12 k Ω
22 k Ω
100 k Ω
Mid-band:
2(0.1mA)
1
= 0.200mS |
= 5000Ω
1V
gm
1
Rin = RS
= 12kΩ 5kΩ = 3.53kΩ | RL = 22kΩ 100kΩ = 18.0kΩ
gm
Rin
3.53kΩ
Amid =
gm RL =
(0.200mS)(18kΩ) = 2.30 (7.24dB)
RI + Rin
2kΩ + 3.53kΩ
1
1
rad
ω1 =
=
= 38.5
| ω 2 = doesn't matter since ig = 0!
−6
C1(RI + Rin ) 4.7x10 (2kΩ + 3.53kΩ)
s
gm =
ω3 =
1
1
rad
1
= −7
= 82.0
| fL ≅
(38.5 + 82.0) = 19.2Hz
C3 (R3 + R7 ) 10 (100kΩ + 22kΩ)
s
2π
17-14
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.22
2kΩ
2kΩ
4.7 µF
10 µF
vi
Rin
v
75 k Ω
13 k Ω
in
11.5 k Ω
100 k Ω
(a) Low frequency
(b) R
75 k Ω
i
Mid-band
= R1 R2 rπ + (β o + 1)RL | RL = 13kΩ 100kΩ = 11.5kΩ | rπ =
[
]
[
100
= 10.0kΩ
40(0.25mA)
]
Rin = R1 R2 rπ + (β o + 1)RL = 100kΩ 300kΩ 10.0kΩ + (101)11.5kΩ = 70.5kΩ
Amid =
101(11.5kΩ)
Rin (β o + 1)RL
= 0.972
= 0.963 | RB = R1 R2 = 75kΩ
RI + Rin
Rin
2 + 10.0 + 101(11.5) kΩ
[
[
]
[
]
]
R1S = RI + RB rπ + (β o + 1)RL = 2kΩ + 75kΩ 10.0kΩ + (101)11.5kΩ = 72.5kΩ
ω1 =
1
rad
= 2.94
−6
s
(72.5kΩ)4.7x10
R3S = R7 + RE
ω3 =
1
( )
10−5 105
03/09/2007
RB RI + rπ
(β
=1
o
+ 1)
rad
s
= 100kΩ + 13kΩ
fL ≅
1.95kΩ + 10.0kΩ
= 100kΩ
101
(2.94 + 1) = 0.627Hz
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-15
16.23
2k Ω
4.7 µF
v
892 k Ω
0.1 µF
i
100 k Ω
12 k Ω
(a) Low Frequency:
2k Ω
Rin
v
i
892 k Ω
10.7 k Ω
Mid-band:
(b) g
m
Amid =
=
2(0.1mA)
0.75V
= 0.267mS | Rin = R1 R2 = 892kΩ | RL = 12kΩ 100kΩ = 10.7kΩ
(0.267mS )(10.7kΩ) = +0.739
Rin
g m RL
= 0.998
RI + Rin 1+ g m RL
1+ (0.267mS )(10.7kΩ)
(-2.62 dB)
ω1 =
1
1
rad
=
= 0.238
−6
s
C1(RI + Rin ) 4.7x10 (2kΩ + 892kΩ)
ω3 =
1
1
rad
=
= 97.2
⎤
⎡
⎤
⎡
⎛
⎛
⎞
s
1
1⎞
−7
C3⎢ R7 + ⎜ RS
⎟⎥ 10 ⎢100kΩ + ⎜12kΩ
⎟⎥
0.267mS ⎠⎥⎦
g m ⎠⎥⎦
⎢⎣
⎢⎣
⎝
⎝
1
(0.238 + 97.2)= 15.5 Hz
2π
(c) VDD = VDS + I S RS = 8.8V + 0.1mA(12kΩ)= 12.0 V
fL ≅
17-16
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.24
3
1
rad
= 2π (500) = 3140
s
i=1 RisC i
SCTC requires : ω L ≅ ∑
ω1 =
1
rad
= 4.11
s
(10 F )(2.43MΩ + 1kΩ)
−7
| ω2 =
1
rad
= 9.59
s
(10 F )(43kΩ + 1MΩ)
−7
ω1 + ω 2 << ω L | ω 3 will be dominant → ω 3 ≅ ω L
ω3 =
1
| gm =
2(0.2mA)
2ID
1
=
= 0.400mS |
= 2.50kΩ
VGS − VTN
1V
gm
⎛
1⎞
C3⎜ RS
⎟
gm ⎠
⎝
1
C3 =
= 0.152 µF → 0.15 µF from Appendix C
3140(13kΩ 2.5kΩ)
16.25
3
SCTC requires : ω L ≅ ∑
i=1
ω2 =
1
C2 (RC + R3 )
=
1
rad
= 2π (100) = 628
RisCi
s
1
(10 F )(2.2kΩ + 51kΩ)
−6
= 18.8
rad
<< ω L | ω1 will be dominant → ω L ≅ ω1
s
1
1
1
|
=
= 25Ω
⎞
⎛
g m 40 10−3
1
C1⎜ RI + RE
⎟
gm ⎠
⎝
1
C1 =
= 7.08 µF → 6.8 µF nearest value in Appendix C
628 200Ω + 4.3kΩ 25Ω
ω1 =
( )
(
)
Note : We might want to choose 8.2 µF to insure that f L ≤ 100 Hz.
3
1
(b) SCTC requires : ω ≅ ∑ R C
L
i=1
ω2 =
1
is
(10 F )(220kΩ + 510kΩ)
−6
= 2π (100) = 628
i
= 1.37
rad
s
rad
s
| ω1 will be dominant → ω L ≅ ω1
1
1
1
|
=
= 2.5kΩ
⎞
⎛
g m 40 10−5
1
C1⎜ RI + RE
⎟
gm ⎠
⎝
1
C1 =
= 0.592 µF → 0.56 µF nearest value in Appendix C
628 200Ω + 430kΩ 2.5kΩ
ω1 =
( )
(
)
Note : We might want to use 0.68 µF to insure that f L ≤ 100 Hz.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-17
16.26
gm = 40IC = 40(0.164mA) = 6.56mS | rπ =
SCTC requires :
3
1
i=1
is
∑R C
= 2π (20) = 126
i
βo
gm
=
100
= 15.2kΩ | ro = ∞ (VA not given)
6.56mS
rad
s
R1S = RI + (RB rπ ) = 1kΩ + (75kΩ 15.2kΩ) = 13.6kΩ | ω1 =
R3S = RC + R3 = 43kΩ + 43kΩ = 86kΩ | ω 3 =
ω 2 = 126 −14.7 −11.6 = 99.7
C2 ≅
1
= 14.7
5x10 (13.6kΩ)
−6
1
= 11.6
1x10 (86kΩ)
−6
(RB RI )+ rπ = 13kΩ 987Ω + 15.2kΩ = 158Ω
rad
| R2S = RE
βo + 1
s
101
1
= 63.5 µF → 68 µF from Appendix C
99.7(158)
16.27
SCTC requires :
3
1
i=1
is
∑R C
= 2π (1) = 6.28
i
rad
s
However, R3S = R3 + R7 = 22kΩ + 100kΩ = 122kΩ
1
rad
rad
ω3 =
= 82.0
> 6.28
| The design goal cannot be met.
−7
1x10 (122kΩ)
s
s
It is not possible to force f L below the limit set by C3 .
16.28
3
1
rad
= 2π (10) = 62.8
| RG = R1 R2 = 892kΩ
R
s
C
is i
i=1
SCTC requires : ω L ≅ ∑
ω1 =
1
1
rad
=
= 0.238
| ω L >> ω1 → ω 3 is dominant
−6
C1(RI + RG ) 4.7x10 (2kΩ + 892kΩ)
s
1
1
0.75V
|
=
= 3.75kΩ
⎡
⎛
gm 2(0.1mA)
1 ⎞⎤
C3⎢R7 + ⎜ RS
⎟⎥
⎢⎣
⎝ gm ⎠⎥⎦
1
rad
= 0.155
→ 0.15 µF using Appendix C
C3 =
s
62.8 100kΩ + (12kΩ 3.75kΩ)
ωL ≅ ω3 =
[
17-18
]
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.29
3
SCTC requires : ω L ≅ ∑
i=1
1
rad
= 2π (5) = 31.4
RisCi
s
RL = 13kΩ 100kΩ = 11.5kΩ | rπ =
[
]
100
= 10.0kΩ
40(0.25mA)
[
]
R1S = RI + RB rπ + (β o + 1)RL = 2kΩ + 75kΩ 10.0kΩ + (101)11.5kΩ = 72.5kΩ
ω1 =
1
rad
= 2.94
−6
s
(72.5kΩ)4.7x10
R3S = R7 + RE
C3 =
(R
B
)
RI + rπ
(β
o
+ 1)
| ω3 = 31.4 − 2.94 = 28.5
= 100kΩ + 13kΩ
rad
s
1.95kΩ + 10.0kΩ
= 100kΩ
101
1
= 0.351 µF → 0.39 µF using the values from Appendix C.
28.5(100kΩ)
16.30
fT =
1 ⎛ gm ⎞
g
⎜⎜
⎟⎟ | Cπ = m − Cµ | gm = 40IC
2π ⎝ C π + C µ ⎠
2πf T
IC
fT
Cπ
Cµ
1/2πrxCµ
10 µA
50 MHz
0.773 pF
0.5 pF
1.59 GHz
100 µA
300 MHz
0.75 pF
1.37 pF
580 MHz
50 µA
1 GHz
2.93 pF
0.25 pF
3.19 GHz
10 mA
6.06 GHz
10 pF
0.500 pF
1.59 GHz
1 µA
3.18 MHz
1 pF
1 pF
795 MHz
1.18 mA
5 GHz
1 pF
0.5 pF
1.59 GHz
16.31
Cπ = gm τ F | Cπ =
gm
ωT
− Cµ | VCB = 5 − 0.7 = 4.3V | Cµ =
Cµo
1+
VCB
=
φ jc
2 pF
= 0.832 pF
4.3V
1+
0.9V
40(2x10−3 )
Cπ 24.6x10−12
Cπ =
− 0.832 pF = 24.6 pF | τ F =
=
= 0.308ns = 308 ps
gm 40(2x10−3 )
2π (5x10 8 )
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-19
16.32
⎞
gm
1 ⎛
⎜
⎟ | gm = 2K n ID
2π ⎝ CGS + CGD ⎠
fT =
ID
fT
CGS
CGD
10 µA
15.9 MHz
1.5 pF
0.5 pF
250 µA
79.6 MHz
1.5 pF
0.5 pF
2.47 mA
250 MHz
1.5 pF
0.5 pF
16.33
(a) fT =
3 µn (VGS − VTN ) 3 600(0.25V ) cm 2
=
= 22.5 GHz
2
2 10−4 2 V − s
L2
(b) fT =
3 µn (VGS − VTN ) 3 250(0.25V ) cm 2
=
= 9.38 GHz
2
2 10−4 2 V − s
L2
( )
( )
(c) NMOS : f
T
3 µn (VGS − VTN ) 3 600(0.25V ) cm2
=
= 2.25 THz
2
2 10−5 2 V − s
L2
=
PMOS : f T =
( )
3 µn (VGS − VTN ) 3 250(0.25V ) cm2
=
= 938 GHz
2
2 10−5 2 V − s
L2
( )
(d ) NMOS : fT =
PMOS : f T =
16.34
(a) r
π
=
3 µn (VGS − VTN ) 3 600(0.25V ) cm2
=
= 36.0 THz
2
2 2.5x10−6 2 V − s
L2
)
(
)
3 µn (VGS − VTN ) 3 250(0.25V ) cm2
=
= 15.0 GHz
2
2 2.5x10−6 2 V − s
L2
125(0.025V )
1mA
(
= 3.13kΩ | Rin = 7.5kΩ (rx + rπ ) = 2.44kΩ | RL = 4.3kΩ 100kΩ = 4.12kΩ
⎛ 2.44kΩ ⎞
Rin
g m RL = −⎜
⎟(40mS )(4.12kΩ)= −117
RI + Rin
⎝1kΩ + 2.44kΩ ⎠
⎛ 2.21kΩ ⎞
(b) Rin = 7.5kΩ rπ = 2.21kΩ | Amid = −⎜⎝1kΩ + 2.21kΩ ⎟⎠(40mS )(4.12kΩ)= −113
( )
g m = 40 10−3 = 40mS | Amid = −
17-20
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.35
125(0.025V )
rπ =
= 3.13kΩ | g m = 40(1mA) = 40mS | RL = 3kΩ 47kΩ = 2.82kΩ
1mA
Rin = RB rx + rπ + (β o + 1)RL = 100kΩ 0.35kΩ + 3.13kΩ + (126)2.82kΩ = 78.2kΩ
[
(a) A
mid
(b) R
in
Amid
]
= Amid
[
[
]
⎤
⎤
β o + 1)RL
126(2820)
(
78.2kΩ ⎡
Rin ⎡
⎥
⎥ = 0.978
⎢
⎢
=
=
RI + Rin ⎢⎣ rx + rπ + (β o + 1)RL ⎥⎦ 1kΩ + 78.2kΩ ⎢⎣350 + 3130 + 126(2820)⎥⎦
]
[
]
= RB rπ + (β o + 1)RL = 100kΩ 3.13kΩ + (126)2.82kΩ = 78.2kΩ
⎤
126(2820)
78.2kΩ ⎡
⎥ = 0.978
⎢
=
1kΩ + 78.2kΩ ⎢⎣ 350 + 3130 + 126(2820)⎥⎦
16.36
125(0.025V )
rπ =
= 31.25kΩ | g m = 40(0.1mA)= 4mS
0.1mA
r +r
200Ω + 31.25kΩ
= 248Ω | RL = 22kΩ 75kΩ = 17.0kΩ
Rin = RE x π = 43kΩ
126
βo + 1
Rin
248Ω
g m RL =
(4mS )(17.0kΩ)= 48.5
100Ω + 248Ω
RI + Rin
(a) A
=
(b) R
= RE
mid
in
03/09/2007
247Ω
rπ
31.25kΩ
= 43kΩ
= 247Ω | Amid =
(4mS )(17.0kΩ)= 48.4
100Ω + 247Ω
126
βo + 1
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-21
16.37
(a) s
2
s=
+ 5100s + 500000 | s1 ≅ −
(
−5100 ± 51002 − 4 5x105
5100
5x105
= −5100 | s2 ≅ −
= −98.0
1
5100
) = −5100 ± 4900 → −100,
2
2
2
2
(b) 2s + 700s + 30000 = 2 s + 350s + 15000
(
s1 ≅ −
− 5000 | 2% error
)
350
15000
= −350 | s2 ≅ −
= −42.9
1
350
−350 ± 3502 − 4(15000)
−350 ± 250
→ −50, − 300 | 14% error
2
2
3300
3x105
2
3s
+
3300s
+
300000
|
s
≅
−
=
−1100
|
s
≅
−
= −90.9
c
()
1
2
3
3300
s=
s=
=
(
−3300 ± 33002 − 4(3) 3x105
) = −3300 ± 2700 → −100,
6
6
2
2
(d ) 0.5s + 300s + 40000 = 0.5 s + 600s + 80000
(
s1 ≅ −
s=
−1000 | 11% error
)
600
80000
= −600 | s2 ≅ −
= −133
1
600
−600 ± 6002 − 4(80000)
2
=
−600 ± 200
→ −200, − 400 | 34%, 50% error
2
16.38
s3 + 1110s2 + 111000s + 1000000
1110
111000
1000000
= −1110 | s2 ≅ −
= −100 | s3 ≅ −
= −9.01
1
1110
111000
Factoring the polynomial : s3 + 1110s2 + 111000s + 1000000 = (s + 10)(s + 100)(s + 1000)
s1 ≅ −
s = −1000, −100, −10 | 11% error in s1 , 10% error in s3
In MATLAB: roots([1 1110 111000 1000000])
16.39
f (s) = s6 + 142s 5 + 4757s 4 + 58230s3 + 256950s2 + 398000s + 300000
f ' (s)= 6s5 + 710s 4 + 19028s 3 + 174690s2 + 513900s + 398000
s
i +1
=s −
i
()
f (s )
f si
'
i
| Using a spreadsheet, four real roots are found : -100, - 20, -15, - 5
Using MATLAB: roots([1 142 4757 58230 256950 398000 300000])
ans = -100, -20, -15, -5, -1+i, -1-i
17-22
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.40
r
RI
1kΩ
v
π
Cµ
300 Ω
Rin
r
RB
π
=
100(0.025)
0.001
Ω
2500
+
+
i
7.5 k Ω
(a) r
x
v
1
RC
2
gm v1 -
-
4.3 k Ω
R3
100 k Ω
( ) − 0.75 pF = 12.0 pF
2π (5x10 )
= 4.3kΩ 100kΩ = 4.12kΩ | g = 40(10 )= 40mS
= 2500Ω | Cµ = 0.75 pF | Cπ =
Rin = 7.5kΩ (rx + rπ ) = 2.03kΩ | RL
v
Cπ
40 10−3
8
−3
m
⎛ Rin ⎞⎛ rπ ⎞
⎛ 2.03kΩ ⎞⎛
⎞
2500Ω
Amid = −⎜
⎟⎜
⎟ g m RL = −⎜
⎟⎜
⎟(40mS )(4.12kΩ)= −98.6
⎝1kΩ + 2.03kΩ ⎠⎝ 300Ω + 2500Ω ⎠
⎝ RI + Rin ⎠⎝ rx + rπ ⎠
1
ωH =
| rπo = rπ rx + RB RI = 2500 300 + 7500 1000 = 803 Ω
rπoCT
⎡
4120⎤
1
= 1.42 MHz
CT = 12.0 + 0.75 ⎢1+ 40 10−3 (4120)+
⎥ = 140 pF | f H =
803 ⎦
⎣
2π (803) 1.4x10−10
[ (
)]
( )
(b) GBW = 98.6(1.42 MHz)= 140 MHz
03/09/2007
[
(
)]
(
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
)
17-23
16.41
+12 V
RC
R2
2.15 k Ω
15 k Ω
R1
RE
5k Ω
650 Ω
(a)
5kΩ
= 3V | REQ = 5kΩ 15kΩ = 3.75kΩ
5kΩ + 15kΩ
VEQ − VBE
(3 − 0.7)V
IC = β F IB = β F
= 100
= 3.31 mA
3.75kΩ + (101)(0.65kΩ)
REQ + (β F + 1)RE
VEQ = 12V
⎛
⎞
101
VCE = VCC − IC RC − IE RE = 12 − (3.31mA)⎜ 2.15kΩ +
0.65kΩ⎟ = 2.71 V | Q - Point : (3.31 mA, 2.71 V )
⎝
⎠
100
(b)
r
RI
1kΩ
v
x
Cµ
300 Ω
Rin
RB
+
+
v
rπ
i
1
2
gm v 1 -
-
3.75 k Ω
RC
v
Cπ
R
3
2.15 k Ω 100 k Ω
Note: As designers, we are free to change the amplifier design, but we typically cannot change
the characteristics of the source and load resistances.
40(3.31x10−3 )
100(0.025)
rπ =
= 755Ω | Cµ = 0.75 pF | Cπ =
− 0.75 pF = 41.4 pF
3.31mA
2π (5x10 8 )
Rin = 3.75kΩ (rx + rπ ) = 823Ω | RL = 2.15kΩ 100kΩ = 2.11kΩ
gm = 40(3.31x10−3 )= 132mS
Amid = −
ωH =
⎛
⎞⎛
⎞
Rin ⎛ rπ ⎞
823Ω
755Ω
⎟⎜
⎟(132mS )(2.11kΩ) = −90.0
⎜
⎟gm RL = −⎜
⎝1000Ω + 823Ω ⎠⎝ 300Ω + 755Ω ⎠
RI + Rin ⎝ rx + rπ ⎠
1
rπoCT
[
]
[
]
| rπo = rπ rx + (RB RI ) = 755Ω 300 + (3.75kΩ 1kΩ) = 260Ω
⎡
2.11kΩ ⎤
1
= 312 pF | f H =
CT = 41.4 + 0.75 ⎢1+ (132mS )(2.11kΩ) +
= 1.96 MHz
⎥
⎣
0.260kΩ⎦
2π (260Ω)(3.12x10−10 F )
(c ) GBW
17-24
= 90.0(1.96MHz) = 176 MHz | GBW ≤
1 ⎛ 1 ⎞
1
= 707 MHz
⎜⎜
⎟⎟ =
2π ⎝ rx Cµ ⎠ 2π (300Ω)(0.75 pF )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.42
+12 V
RC
R2
215 k Ω
1.5 M Ω
R1
RE
500 k Ω
65 k Ω
(a)
r
RI
50 k Ω
RB
375 k Ω
Cµ
300 Ω
Rin
vi
x
rπ
+
+
v
-
1
v
Cπ
2
gm v1 -
(b)
RC
R3
215 k Ω 5 M Ω
500kΩ
= 3V | REQ = 500kΩ 1.5MΩ = 3.75kΩ
500kΩ + 1.5MΩ
VEQ − VBE
(3 − 0.7)V
= 100
= 33.1 µA
IC = β F I B = β F
375kΩ + (101)(65kΩ)
REQ + (β F + 1)RE
VEQ = 12V
⎛
⎞
101
VCE = VCC − IC RC − I E RE = 12 − (33.1µA)⎜ 215kΩ +
65kΩ⎟ = 2.71 V | Q - Point : (33.1 µA, 2.71 V )
100
⎝
⎠
Note: As designers, we are free to change the amplifier design, but we typically cannot change
the characteristics of the source and load resistances. However, the problem statement indicated
changing all resistors.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-25
rπ =
100(0.025)
33.1µA
= 75.5kΩ | Cµ = 0.75 pF | Cπ =
(
40 33.1x10−6
(
2π 5x10
8
)
)− 0.75 pF = −0.329 pF
The constant fT model is breaking down at low currents. Set Cπ = 0
Rin = 375kΩ (rx + rπ ) = 63.1kΩ | RL = 215kΩ 5MΩ = 211kΩ
Amid = −
ωH =
(
- not possible.
)
| g m = 40 33.1x10−6 = 1.32mS
⎛
⎞⎛
⎞
75.5kΩ
Rin ⎛ rπ ⎞
63.1kΩ
⎜
⎟ g m RL = −⎜
⎟⎜
⎟(1.32mS )(211kΩ) = −155.0
RI + Rin ⎝ rx + rπ ⎠
⎝ 50kΩ + 63.1kΩ ⎠⎝ 300Ω + 75.5kΩ ⎠
1
rπoCT
[ (
| rπo = rπ rx + RB RI
)]= 75.5kΩ [300 + (375kΩ 50kΩ)]= 28.0kΩ
⎡
211kΩ ⎤
1
CT = 0 + 0.75 ⎢1+ (1.32mS )(211kΩ)+
= 24.9 kHz
⎥ = 228 pF | f H =
8.86kΩ⎦
⎣
2π (28.0kΩ) 2.28x10−10 F
(
(c) GBW = 155(24.9kHz)= 3.86 MHz |
17-26
GBW ≤
)
1 ⎛ 1 ⎞
1
= 707 MHz
⎜⎜
⎟⎟ =
2π ⎝ rx Cµ ⎠ 2π (300Ω)(0.75 pF )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.43
Rin = R1 R2 = 4.3MΩ 5.6 MΩ = 2.43MΩ | RL = 43kΩ 470kΩ = 39.4kΩ
gm =
2(0.2mA)
2I D
=
= 0.400mS |
VGS − VTN
1
Amid = −
fH =
Rin
2.43MΩ
g m RL = −
0.400mS (39.4kΩ)= −15.7
2kΩ + 2.43MΩ
RI + Rin
1
2πrπoCT
| rπo = R1 R2 RI = 2.00kΩ
⎡
39.4kΩ⎤
CT = 2.5 pF + 2.5 pF ⎢1+ (0.400mS )(39.4kΩ)+
⎥ = 93.7 pF
2kΩ ⎦
⎣
fH =
(
1
2π (2kΩ) 93.7x10−12 F
)
= 849 kHz
16.44
*Problem 16.44 - Common-Source Amplifier
VDD 7 0 DC 0
VS 1 0 AC 1
RS 1 2 2K
C1 2 3 0.1UF
R1 3 0 4.3MEG
R2 3 7 5.6MEG
RD 7 5 43K
R4 4 0 13K
C3 4 0 10UF
C2 5 6 0.1UF
R3 6 0 1MEG
*Small-Signal FET Model
GM 5 4 3 4 0.4MS
CGS 3 4 2.5PF
CGD 3 5 2.5PF
*
.AC DEC 20 1 10MEG
.PRINT AC VM(6)
.PROBE
.END
Results: Amid = -15.7, fL = 8.52 Hz, fH = 866 MHz
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-27
16.45
V DD = 12 V
RD
R2
10 k Ω
560 k Ω
R1
RS
430 k Ω
3.9 k Ω
(a)
430kΩ
= 5.21V | REQ = 430kΩ 560kΩ = 243kΩ
430kΩ + 560kΩ
K
2
Assume active region operation : ID = n (VGS − VTN ) | VEQ = VGS + ID RS
2
⎛ 0.5mA ⎞
2
5.21 = VGS + 3.9kΩ⎜
⎟(VGS −1) → VGS = 2.629V and ID = 663µA
⎝ 2 ⎠
VEQ = 12V
VDS = VDD − ID RD − IS RS = 12 − (663µA)(13kΩ + 3.9kΩ) = 0.795 V
The transistor is not in pinch off! Reduce RD to 10 kΩ.
VDS = VDD − ID RD − IS RS = 12 − (663µA)(10kΩ + 3.9kΩ) = 2.78 V - Active region is correct.
RI
1kΩ
CGD
RG
vi
243 k Ω
+
+
Rin
v
1
-
CGS
v
RD
R3
10 k Ω
100 k Ω
2
g mv 1 -
(b)
Rin = R1 R2 = 430kΩ 560kΩ = 243kΩ | RL = 10kΩ 100kΩ = 9.09kΩ
gm =
2(0.663mA)
2ID
=
= 1.33mS
VGS − VTN
1
Amid = −
fH =
Rin
243kΩ
gm RL = −
(1.33mS )(9.09kΩ) = −12.0
1kΩ + 243kΩ
RI + Rin
1
2πrπoCT
| rπo = R1 R2 RI = 243kΩ 1kΩ = 0.996kΩ
⎡
9.09kΩ ⎤
CT = 2.5 pF + 2.5 pF ⎢1+ (1.33mS)(9.09kΩ) +
= 58.1pF
⎣
0.996kΩ⎥⎦
1
= 2.75 MHz
fH =
2π (0.996kΩ)(58.1x10−11 F )
(c ) GBW
17-28
= 12.0(2.75 MHz) = 33 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.46
r
RI
1kΩ
x
Cµ
300 Ω
Rin
RB
vi
+
+
v
rπ
1
g m = 40IC = 40(0.164mA) = 6.56mS | rπ =
βo
gm
2
gm v 1 -
-
75 k Ω
v
Cπ
=
RC
R3
43 k Ω
43 k Ω
100
= 15.2kΩ | ro = ∞ (VA not given )
6.56mS
Rin = R1 R2 rπ = 100kΩ 300kΩ 15.2kΩ = 12.6kΩ | RL = RC R3 = 43kΩ 43kΩ = 21.5kΩ
Amid = −
Cπ =
ωH =
gm
ωT
Rin
12.6kΩ
g m RL = −
(6.56mS )(21.5kΩ)= −131
1kΩ + 12.6kΩ
RI + Rin
− Cµ =
1
rπoCT
6.56mS
(
2π 5x108 Hz
)
− 0.75 = 1.34 pF
(
)
| rπo = rπ rx + R1 R2 RI = 15.2 kΩ (300 + 987)= 1.19 kΩ
⎛
⎡
R ⎞
21.5kΩ⎤
CT = Cπ + Cµ ⎜1+ g m RL + L ⎟ = 1.34 pF + 0.75 pF ⎢1+ 6.56mS (21.5kΩ)+
⎥ = 121pF
1.19kΩ ⎦
rπo ⎠
⎣
⎝
1
= 1.10 MHz
fH ≅
2π (1.19kΩ) 1.21x10−10 F
(
)
16.47
*Problem 16.47 - Common-Emitter Amplifier
VCC 7 0 DC 0
VS 1 0 AC 1
RS 1 2 1K
C1 2 3 5UF
R1 3 0 300K
R2 3 7 100K
RC 5 0 43K
R4 7 4 13K
C2 7 4 22UF
C3 5 6 1UF
R3 6 0 43K
*Small-signal Model for the BJT
GM 5 4 8 4 6.56MS
RX 3 8 0.3K
RPI 8 4 15.24K
CPI 8 4 1.34PF
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-29
CU 8 5 0.75PF
*
.AC DEC 100 1 10MEG
.PRINT AC VM(6)
.PROBE
.END
Results: Amid = -128, fL = 47 Hz, fH = 1.10 MHz
16.48
(a ) See Eqs. (16.88 -16.96).
⎤ ⎡V (s)⎤
⎡I (s)⎤ ⎡s(C + C )+ g
−sC
⎥⎢
⎥
⎢
⎥=⎢
s(C + C )+ g ⎥⎦ ⎢⎣V (s)⎥⎦
⎣ 0 ⎦ ⎢⎣ −(sC − g )
∆ = s [C (C + C )+ C C ]+ s[C g + C (g + g + g )+ C g ]+ g g
π
S
µ
µ
2
(b) ω
π
P1
ω P2 ≅
≅
µ
L
πo
µ
µ
m
µ
L
1
L
π
L
2
L
µ
m
g L g πo
=
Cπ g L + Cµ (g m + g πo + g L )+ CL g πo
Cπ g L + Cµ (g m + g πo + g L )+ CL g πo
Cπ (Cµ + CL )+ Cµ CL
=
πo
L
L
πo
L
πo
1
⎡
R⎤
rπo ⎢Cπ + Cµ (1+ g m RL )+ (Cµ + CL ) L ⎥
rπo ⎦
⎣
gm
⎛ C ⎞
Cπ ⎜⎜1+ L ⎟⎟ + CL
⎝ Cµ ⎠
(c) The three capacitors form a loop, and there are only two independent voltages
among the three capacitors.
16.49
CT = Cπ + Cµ (1+ g m RL )= 20 pF + 1pF 1+ 40(1mA)(1kΩ) = 61 pF
[
fT =
]
1 ⎛ g m ⎞ 1 ⎡ 40(1mA) ⎤
⎥ = 303 MHz
⎢
⎜
⎟=
2π ⎜⎝ Cπ + Cµ ⎟⎠ 2π ⎢⎣ 20 pF + 1pF ⎥⎦
17-30
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.50
(
)
1+ A 1+ A
=
= sC(1+ A) | Cin = C (1+ A)= 10−10 F 1+ 105 = 10 µF
1
Z (s)
sC
Z (s)
s + 10
s + 10
1
105
(b) Zin = Y = 1+ A s = 106 = 105 s + 10 + 106 ≅ 105 s + 106
( ) 1+
in
s + 10
Using MATLAB : Zin (j2000π )= (4.95 + j6.28) Ω
(a) Y
in
(
=
)
(
)
Zin j105 π = (8.98 + j28.6) kΩ | Zin j2πx106 = (97.5 + j15.5) kΩ
16.51
10 Ao
s + 10
⎛ 1 ⎞ A(s)
10 Ao
⎟
⎜
1+
⎛ 1 ⎞
⎝ RC ⎠ 1+ A(s)
10 Ao
s + 10
| A(s) =
| Av (s)= ⎜
(a) Av (s)=
⎟
1
1
s + 10
RC ⎠
⎝
s+
s+
⎛ 10 Ao ⎞
RC 1+ A(s)
RC⎜1+
⎟
⎝ s + 10 ⎠
⎛ 10 Ao ⎞
⎟
⎜
⎛ 1 ⎞
10 Ao
⎝ RC ⎠
Av (s) = ⎜
=
⎟
⎡ 1
⎤ 10
⎝ RC ⎠ s2 + s 1+ A 10 + s + 10
+ 10(1+ Ao )⎥ +
( o ) RC s2 + s⎢ RC
⎣
⎦ RC
⎛ 106 ⎞
⎛ 106 ⎞
⎟
⎟
⎜
⎜
1
⎝ RC ⎠
⎝ RC ⎠
Av (s) =
≅
; ωL = 5
⎛
⎡ 1
⎤ 10
1 ⎞
10 RC
s2 + s⎢
s + 106 ⎜ s + 5
+ 106⎥ +
⎟
⎣ RC
⎦ RC
⎝ 10 RC ⎠
7
⎛ 10 ⎞
⎛ 106 ⎞
⎟
⎟
⎜
⎜
1
⎝ RC ⎠
⎝ RC ⎠
≅
; ωL = 6
(b) Av (s)=
⎛
⎡ 1
⎤ 10
1 ⎞
10 RC
s2 + s⎢
s + 10 7 ⎜ s + 6
+ 10 7 ⎥ +
⎟
⎣ RC
⎦ RC
⎝ 10 RC ⎠
⎛10 Ao ⎞
⎟
⎜
1
⎝ RC ⎠
(c) lim Av (s)= 10 A s = sRC
Ao →∞
o
[
]
(
)
(
03/09/2007
)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-31
16.52
250 Ω
r
x
C
rπ
vi
2500
T
Ω
⎡
2500⎤
rπo = 2500Ω 250Ω = 227Ω | CT = 15 + 1⎢1+ 0.04(2500)+
⎥ = 127 pF
227 ⎦
⎣
(a) At 1 kHz, Z
=
C
1
( )
j (2π ) 10 (127 pF )
3
Using MATLAB : Z = 250 +
(b) At 50 kHz, Z
C
=
(c) At 1 MHz, Z
C
=
)
j (2π ) 5x10 (127 pF )
4
Using MATLAB : Z = 250 +
1
( )
)
2500ZC
= (2750 − j4.99) Ω | SPICE : (2750 − j4.56) Ω
2500 + ZC
1
(
(
= − j 1.25x106
= − j2.51x10 4 Ω
2500ZC
= (2730 − j247) Ω | SPICE : (2730 − j226) Ω
2500 + ZC
j (2π ) 106 (127 pF )
= − j (12.53)
2500ZC
= (752 − j1000) Ω | SPICE : (836 − j1040) Ω
2500 + ZC
(d) *Problem 16.52 - Common-Emitter Amplifier
IS 0 1 AC 1
RX 1 2 0.25K
RPI 2 0 2.5K
CPI 2 0 15PF
CU 2 3 1PF
GM 3 0 2 0 40MS
RL 3 0 2.5K
.AC LIN 1 1KHZ 1KHZ
*.AC LIN 1 50KHZ 50KHZ
*.AC LIN 1 1MEG 1MEG
.PRINT AC VR(1) VI(1) VM(1) VP(1)
.END
Using MATLAB : Z = 250 +
Note that the CT approximation does not provide as good an estimate of Zin at high frequencies
(note the discrepancy at 1 MHz).
16.53
Amid = 39.2 dB, fL = 0 Hz, fH = 5.53 MHz
16.54
17-32
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
(a) g
m
= 40IC = 40(1mA) = 40.0mS | rπ =
[ (
rπo = rπ rx + RB RI
Cπ =
gm
ωT
− Cµ =
β oVT
IC
=
125(0.025)
1mA
= 3.13kΩ | ro = ∞ (VA not given )
)]= 3.13kΩ (500 + (7.5kΩ 1kΩ))= 959Ω
40.0mS
(
2π 5x10 Hz
8
)
− 0.75 pF = 12.0 pF | f H ≅
| RL = RC R3 = 4.3kΩ 100kΩ = 4.12kΩ
1
2π rπoCT
⎛
⎡
R ⎞
4.12kΩ ⎤
CT = Cπ + Cµ ⎜1+ g m RL + L ⎟ = 12.0 pF + 0.75 pF ⎢1+ 40.0mS (4.12kΩ)+
⎥ = 140 pF
0.959kΩ⎦
rπo ⎠
⎣
⎝
fH ≅
(b) r
1
(
[r + (R
2π (959Ω) 1.40x10−10 F
πo
= rπ
x
B
RI
)
= 1.19 MHz
)]= 3.13kΩ (0 + (7.5kΩ 1kΩ))= 688Ω
⎡
4.12kΩ ⎤
CT = 12.0 pF + 0.75 pF ⎢1+ 40.0mS (4.12kΩ)+
⎥ = 141pF
0.688kΩ⎦
⎣
fH ≅
1
(
2π (688Ω) 1.41x10−10 F
03/09/2007
)
= 1.64 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-33
16.55
r
RI
1kΩ
x
Cµ
400 Ω
Rin
RB
vi
v
rπ
m
= 40IC = 40(0.1mA)= 4.00mS | rπ =
[ (
rπo = rπ rx + RB RI
1
β oVT
IC
100(0.025)
0.1mA
R3
43 k Ω
100 k Ω
2
gm v1 -
=
RC
v
Cπ
-
75 k Ω
(a) g
+
+
= 25kΩ | ro = ∞ (VA not given )
)]= 25kΩ (400 + (75kΩ 1kΩ))= 1.31kΩ
Rin = R1 R2 (rx + rπ ) = 100kΩ 300kΩ 25.4kΩ = 19.0kΩ | RL = RC R3 = 43kΩ 100kΩ = 30.1kΩ
Amid = −
Cπ =
gm
ωT
Rin
19.0kΩ
g m RL = −
(4.00mS )(30.1kΩ)= −114
1kΩ + 19.0kΩ
RI + Rin
− Cµ =
4.00mS
(
2π 5x10 Hz
8
)
− 0.75 pF = 0.523 pF | f H ≅
1
2π rπoCT
⎛
⎡
R ⎞
30.1kΩ⎤
CT = Cπ + Cµ ⎜1+ g m RL + L ⎟ = 0.523 pF + 0.75 pF ⎢1+ 4.00mS (30.1kΩ)+
⎥ = 109 pF
rπo ⎠
1.31kΩ ⎦
⎣
⎝
fH ≅
1
(
2π (1.31kΩ) 1.09x10−10 F
)
= 1.12 MHz
(b) GBW = 114(1.12 MHz)= 128 MHz
|
1
1
=
= 531 MHz
2π rx Cµ 2π (400Ω)(0.75 pF )
16.56
Rin = R1 R2 = 4.3MΩ 5.6 MΩ = 2.43MΩ | RL = 43kΩ 470kΩ = 39.4kΩ
gm =
2(0.2mA)
2I D
=
= 0.400mS |
VGS − VTN
1
Amid = −
fH =
Rin
2.43MΩ
g m RL = −
0.400mS (39.4kΩ) = −15.7
RI + Rin
2kΩ + 2.43MΩ
1
2πrπoCT
| rπo = R1 R2 RI = 2.00kΩ
⎡
39.7kΩ⎤
CT = 5 pF + 2 pF ⎢1+ (0.400mS )(39.7kΩ)+
⎥ = 78.5 pF
2kΩ ⎦
⎣
fH =
(
1
2π (2kΩ) 78.5x10−12 F
17-34
)
= 1.01 MHz | GBW = 15.7(1.01 MHz)= 15.9 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.57
Problem 16.57 should refer to Fig. 16.34, ot Fig. 1631.
1
1
1
=
| CT =
= 48.5 pF
fH =
2πrπoCT 2π (656Ω)CT
2π (656Ω)(5MHz)
⎛
⎡
R⎤
1 ⎞ C − Cπ
48.5 pF −19.9 pF
CT = Cπ + Cµ ⎢1+ g m RL + L ⎥ | RL ⎜ g m + ⎟ = T
−1 =
−1 = 56.2
rπo ⎦
Cµ
rπo ⎠
0.5 pF
⎣
⎝
RL =
56.2
⎛
1 ⎞
⎜.064S +
⎟
656Ω ⎠
⎝
= 858Ω | RL = RC 100kΩ → RC = 865Ω
100(858Ω)
= −31.9 | GBW = 31.9(5MHz)= 160 MHz
882Ω + 250Ω + 1560Ω
The nearest 5% value is RC = 820 Ω | RL = 820Ω 100kΩ = 813Ω
Amid = −
Amid = −
fH =
100(813Ω)
⎡
813⎤
= −30.2 | CT = 19.9 + 0.5⎢1+ 0.064(813)+
⎥ = 47.0 pF
882Ω + 250Ω + 1560Ω
656 ⎦
⎣
1
1
=
= 5.16 MHz | GBW = 156 MHz
2πrπoCT 2π (656Ω)(47.0 pF )
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-35
16.58
+12 V
43 k Ω
100 Ω
43 k Ω 100 k Ω
75 k Ω
30.1 k Ω
Rin
vi
75 k Ω
3V
3kΩ
13 k Ω
IC = 100
3 − 0.7
= 0.166mA | VCE = 12 − 43kΩIC −13kΩI E = 2.70 V
75kΩ + 101(13kΩ)
g m = 40(0.166mA)= 6.64mS
rπ 0 =
100
= 15.1kΩ
6.64mS
Cπ =
gm
ωT
− Cµ = 3.02 pF
Short-Circuit Time Constants
[
]
R1S = 100Ω + 75kΩ 300Ω + 15.1kΩ + 101(3kΩ) = 60.8kΩ
⎛
15.1kΩ + 99.9Ω ⎞
R2S = 10kΩ ⎜ 3kΩ +
⎟ = 2.40kΩ
101
⎠
⎝
R3S = 43kΩ + 100kΩ = 143kΩ
⎤
1
1 ⎡
1
1
⎢
⎥ = 43.9Hz
+
+
fL ≈
2π ⎢⎣(60.8kΩ)(1µF ) (2.40kΩ)(2.2µF ) (143kΩ)(0.1µF )⎥⎦
Open-Circuit Time Constants
[
(
)]
Using the results from Table 16.2 on page 1037 : rπ 0 = 15.1kΩ 300 + 100 75kΩ = 390 Ω
⎡ (6.64mS )(30.1kΩ) 30.1kΩ⎤
3.02 pF
⎥
CTB =
+ 0.5 pF ⎢1+
+
1+ (6.64mS )(3kΩ)
⎢⎣ 1+ (6.64mS )(3kΩ) 390Ω ⎥⎦
CTB = 44.0 pF | f H =
(b) A
17-36
mid
1
= 9.27 MHz
2π (390Ω)(44.0 pF )
⎛ 60.7kΩ ⎞ (6.64mS )(30.1kΩ)
= −⎜
= −9.54 GBW = 9.54(9.27 MHz − 43.9Hz)= 88.0 MHz
⎟
⎝ 60.8kΩ ⎠ 1+ (6.64mS )(3kΩ)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.59
Using the results from Table 16.2 on page 1037 and Prob. 16.58
1
= 54.4
rπ 0 = 15.1kΩ 300 + 100 75kΩ = 390 Ω CTB =
2π (7.5MHz)(390Ω)
[
CTB =
(
)]
⎡ (6.64mS )(30.1kΩ) 30.1kΩ⎤
3.02 pF
⎥ = 54.4 pF
+ 0.5 pF ⎢1+
+
390Ω ⎥⎦
1+ (6.64mS )RE
1+ (6.64mS )RE
⎢⎣
Using MATLAB : RE = 862 Ω
Amid = 0.999
−100(30.1kΩ)
99.9Ω + 300Ω + 15.1kΩ + 101(862Ω)
= −29.3 | GBW = 220 MHz
The closest 5% resistor values are RE = 820 Ω and R6 = 12 kΩ
⎡
(6.64mS )(30.1kΩ) + 30.1kΩ⎤⎥ = 55.1 pF
3.02 pF
+ 0.5 pF ⎢1+
CTB =
1+ (6.64mS )0.82kΩ
⎢⎣ 1+ (6.64mS )0.82kΩ 390Ω ⎥⎦
fH =
1
= 7.41MHz
2π (390Ω)(55.1pF )
Amid = 0.999
−100(30.1kΩ)
= −30.6 | GBW = 227 MHz
99.9Ω + 300Ω + 15.1kΩ + 101(0.82kΩ)
Note: The Q-point will actually change somewhat and this has been neglected.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-37
16.60
⎡
⎤
⎛I ⎞
3 − 0.7
⎢
⎥
(a) IC = 100⎢7.5kΩ + 101 1.3kΩ ⎥ = 1.66mA | VCE = 12 − 4.3kΩ(IC )−1.3kΩ⎜⎝ αC ⎟⎠ = 2.69V
( )⎦
F
⎣
2.69V ≥ 0.7V Active region operation is correct. | rπ =
g m = 40(1.66 mA)= 66.4mS | Cπ =
66.4mS
(
2π 2x108
)
100(0.025)
1.66 mA
= 1.51 kΩ
−1 = 51.8 pF | rx = 300Ω | Cµ = 1.0 pF
+12 V
4.3 k Ω
250 Ω
4.3 k Ω 47 k Ω
7.5 k Ω
3.94 k Ω
Rin
vi
7.5 k Ω
3V
200 Ω
1.3 k Ω
[
]
[
]
Rin = R1 R2 rx + rπ + (β o + 1)RE1 = 10kΩ 30kΩ 0.350kΩ + 1.51kΩ + (101)200Ω = 5.60 kΩ
Rth = 7.5kΩ 250Ω = 242Ω | RL = 4.3kΩ 47kΩ = 3.94kΩ
Amid = −
⎤
⎡100(3.94kΩ)⎤
5.60kΩ
Rin ⎡
β o RL
⎥=−
⎥ = −16.9
⎢
⎢
350Ω + 5.60kΩ ⎢⎣ 22.0kΩ ⎥⎦
RI + Rin ⎢⎣rx + rπ + (β o + 1)RE1 ⎥⎦
(b) Using the Short-Circuit Time Constants:
R1S = 250Ω + 7.5kΩ 350Ω + 1.51kΩ + 101(200Ω) = 5.85kΩ
[
]
R2S = 4.3kΩ + 43kΩ = 47.3kΩ
⎛
1.51kΩ + 350 + 242Ω ⎞
R3S = 1.1kΩ ⎜ 200Ω +
⎟ = 184Ω
101
⎝
⎠
⎤
1
1 ⎡
1
1
⎢
⎥ = 193Hz
+
+
fL ≅
2π ⎢⎣(5.85kΩ)(5µF ) (47.3kΩ)(1µF ) (184Ω)(4.7µF )⎥⎦
(c) Using the Open-Circuit Time Constants:
Using the results from Table 16.2 on page 1037 : rπ 0 ≅ Rth + rx = 242Ω + 350Ω = 592Ω
⎡
(66.4mS )(3.94kΩ) + 3.94kΩ⎤⎥ = 29.6 pF
51.8 pF
+ 1pF ⎢1+
CTB =
1+ (66.4mS )(200Ω)
⎢⎣ 1+ (66.4mS )(200Ω) 592Ω ⎥⎦
CTB = 29.6 pF
17-38
fL =
1
= 9.08 MHz
2π (592Ω)(29.6 pF )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.61
Using the results in Table 16.2 on page 1037 and the values from Prob. 16.60 :
rπ 0 ≅ Rth + rx = 242Ω + 350Ω = 592Ω | CTB =
CTB =
1
= 22.4 pF
2π (592Ω)(12 MHz)
⎡ (66.4mS )(3.94kΩ) 3.94kΩ⎤
51.8
⎥ = 22.4 pF
+ 1pF ⎢1+
+
592Ω ⎥⎦
1+ (66.4mS )RE
1+ (66.4mS )RE
⎢⎣
Using MATLAB : RE = 305 Ω
The closest 5% resistor values are RE = 300 Ω and R6 = 1 kΩ
⎡ (66.4mS )(3.94kΩ) 3.94kΩ⎤
51.8 pF
⎥ = 22.6 pF
CTB =
+ 1pF ⎢1+
+
592Ω ⎥⎦
1+ (66.4mS )300Ω
⎢⎣ 1+ (66.4mS )300Ω
fH =
1
= 11.9 MHz
2π (592Ω)(22.6 pF )
[
]
[
]
Rin = R1 R2 rx + rπ + (β o + 1)RE1 = 10kΩ 30kΩ 0.300kΩ + 1.51kΩ + (101)300Ω = 6.08 kΩ
Rth = 7.5kΩ 250Ω = 242Ω | RL = 4.3kΩ 47kΩ = 3.94kΩ
Amid = −
⎡100(3.94kΩ)⎤
⎤
Rin ⎡
β o RL
6.08kΩ
⎢
⎢
⎥=−
⎥ = −11.8
RI + Rin ⎢⎣rx + rπ + (β o + 1)RE1 ⎥⎦
250Ω + 6.08kΩ ⎢⎣ 32.1kΩ ⎥⎦
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-39
16.62
1k Ω
1k Ω
R 2S
R 1S
10 k Ω
1k Ω
1k Ω
1k Ω
R 2O
R 1O
1k Ω
10 k Ω
10 k Ω
1k Ω
(a) SCTC:
R1S = 10kΩ 1kΩ = 909Ω | R2S = 1kΩ 1kΩ = 500Ω | ω L =
(b) OCTC:
R1O = 10kΩ 2kΩ = 1.67kΩ | R2O = 1kΩ 11kΩ = 917Ω | ω H =
1
1
rad
+
= 1300
−6
−5
s
909(10 ) 500(10 )
1
rad
= 92.3
−5
s
1670(10 )+ 917(10 )
−6
(c) There are two poles. The SCTC technique assumes both are at low frequency and yields the
largest pole. The OCTC assumes both are at high frequency and yields the smallest pole.
(d )
⎡(sC1 + G1 + G2 )
⎢
−G2
⎣
⎤⎡V1 ⎤
−G2
⎥⎢ ⎥ = 0
(sC2 + G2 + G3 )⎦⎣V2 ⎦
∆ = s 2C1C2 + s[C2 (G1 + G2 ) + C1 (G2 + G3 )]+ G1G2 + G2G3 + G1G3
∆ = s 210−11 + s(1.30x10−8 )+ 1.20x10−6
∆ = s 2 + 1300s + 1.20x10 5 → s = −1200,−100
17-40
rad
s
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.63
100 Ω R
in
vi
1.3 k Ω
4.3 k Ω
100 k Ω
RL = 4.3kΩ 100kΩ = 4.12kΩ | CGS = 3.0 pF | CGD = 0.6 pF
Rin = RS
1
1
= 1.3kΩ
= 173Ω
gm
5mS
Rin
173Ω
g m RL =
(5ms)(4.12kΩ)= +13.1
RI + Rin
100Ω + 173Ω
⎛
⎞
1
1 ⎛ 1 ⎞ 1
⎜
⎟ = 64.4 MHz
fH ≅
=
⎜
⎟
2π ⎝ CGD RL ⎠ 2π ⎜⎝ 0.6 pF (4.12kΩ)⎟⎠
Amid =
16.64
*Problem 16.12 - Common-Gate Amplifier - ac small-signal model
VI 1 0 AC 1
RI 1 2 100
C1 2 3 4.7UF
RS 3 0 1.3K
RD 4 0 4.3K
C2 4 5 1UF
R3 5 0 100K
G1 4 3 0 3 5mS
.OP
.AC DEC 100 0.01 100MEG
.PRINT AC VM(5) VP(5)
.END
Results: Amid = +13.3, fL = 123 Hz, fH = 64.4 MHz
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-41
16.65
200 Ω
vi
Rin
2.2k Ω
4.3 k Ω
2.11 k Ω
gm = 40(1 mA) = 0.04S | rx = 300Ω | rπ =
Cπ =
40(10−3 )
2π (5x10 8 )
Rin = RE
Amid
51k Ω
100(0.025)
= 2.50kΩ | Cµ = 0.6 pF
1 mA
− 0.6 = 12.1 pF | Rth = 4.3kΩ 200Ω = 191Ω | RL = 2.2kΩ 51kΩ = 2.11kΩ
(rx + rπ ) = 4.3kΩ (0.3kΩ + 2.50kΩ) = 27.6Ω
βo + 1
101
100(2.11kΩ)
Rin ⎛ β o RL ⎞
27.6Ω
=
= +9.14
⎜
⎟=
RI + Rin ⎝ rx + rπ ⎠ 200Ω + 27.6Ω 2.80kΩ
1
ωH =
191
fH =
17-42
⎡
0.04 (2110) ⎤
12.1pF ⎛ 300 ⎞
⎥ + 0.6 pF (2110Ω)
⎜1+
⎟ + 0.6 pF (300Ω)⎢1+
1+ 0.04 (191) ⎝ 191 ⎠
⎣ 1+ 0.04 (191)⎦
⎞
1 ⎛
1
= 40.9MHz
⎜
−10
−9
−9 ⎟
2π ⎝ 6.876x10 + 1.938x10 + 1.266x10 ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.66
First, estimate the required SPICE parameters:
0.333
⎛
CJC
2.8 ⎞
| CJC = CJC = 0.6 pF ⎜1+
≅ 1.01pF
Cµ =
⎟
ME
⎝ 0.75 ⎠
⎛
VCB ⎞
⎟
⎜1+
⎝ PHIE ⎠
C
1 Cµ
1
0.6 pF
τF = π =
−
= 9 −
= 303 ps
gm ωT gm 10 π 40(1mA)
*Figure P16.14 - Common-Base Amplifier
VCC 6 0 DC 5
VEE 7 0 DC -5
VI 1 0 AC 1
RI 1 2 200
C1 2 3 4.7UF
RE 3 7 4.3K
Q1 4 0 3 NBJT
RC 4 6 2.2K
C2 4 5 1UF
R3 5 0 51K
.MODEL NBJT NPN BF=100 RB=300 CJC=1.01PF TF=303PS
.OP
.AC DEC 50 1 50MEG
.PRINT AC VM(5)
.PROBE
.END
Results: Amid = 19.1 dB, fL = 149 Hz, fH = 43.8 MHz
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-43
16.67
200 Ω
Rin
vi
2.2k Ω
4.3 k Ω
IC = α F IE =
51k Ω
2.11 k Ω
100 ⎡−0.7 − (−10)⎤
⎢
⎥ = 2.14 mA | VCE = 10 − (2.14mA)(2.2kΩ) − (−0.7) = 5.99 V
101 ⎣ 4300 ⎦
gm = 40(2.14mA) = 85.6mS | rx = 300Ω | rπ =
Cπ =
85.6mS
− 0.6 = 26.7 pF | Rth = 4.3kΩ 200Ω = 191Ω | RL = 2.2kΩ 51kΩ = 2.11kΩ
2π (5x10 8 )
Rin = RE
Amid =
(rx + rπ ) = 4.3kΩ (0.3kΩ + 1.17kΩ) = 14.5Ω
101
1
191
17-44
βo + 1
100(2.11kΩ)
14.5Ω
Rin ⎛ β o RL ⎞
= +9.70
⎜
⎟=
RI + Rin ⎝ rx + rπ ⎠ 200Ω + 14.5Ω 1.47kΩ
ωH =
fH =
100(0.025)
= 1.17kΩ | Cµ = 0.6 pF
2.14 mA
⎡
⎛ 300 ⎞
0.0856(2110) ⎤
26.7 pF
⎥ + 0.6 pF (2110Ω)
⎜1+
⎟ + 0.6 pF (300Ω)⎢1+
1+ 0.0856(191) ⎝ 191 ⎠
⎣ 1+ 0.0856(191)⎦
⎞
1 ⎛
1
= 39.1 MHz
⎜
−10
−9
−9 ⎟
2π ⎝ 7.556x10 + 2.054 x10 + 1.266x10 ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.68
2kΩ
v
i
Rin
12 k Ω
22 k Ω
100 k Ω
Rth = 12kΩ 2kΩ = 1.71kΩ | RL = 22kΩ 100kΩ = 18.0kΩ | CGS = 3.0 pF | CGD = 0.6 pF
gm =
2(0.1mA)
1V
= 0.200mS | Rin = 12kΩ
1
1
= 12kΩ
= 3.53kΩ
gm
0.200mS
Rin
3.53kΩ
g m RL =
(0.200ms)(18.0kΩ)= +2.30
2kΩ + 3.53kΩ
RI + Rin
⎛
⎞
⎞
⎛
⎜
⎟
⎟
⎜
1
1 ⎜
1
1
⎜
⎟ = 10.9 MHz
⎟=
fH =
⎜
⎟
3.0 pF
2π ⎜ CGS
⎟ 2π
+ CGD RL ⎟
+ 0.6 pF (18.0kΩ)⎟
⎜
⎜
⎠
⎝ Gth + g m
⎝ (0.5848 + 0.200)mS
⎠
Amid =
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-45
16.69
(a) First find VTN and K n based upon Problem 16.21
1.5MΩ
12V = 4.87V | VGG − VGS = 0.1mA(12kΩ)→ VGS = 3.67V
1.5MΩ + 2.2 MΩ
2(0.1mA)
2I D
=
= 0.2mS
VGS − VTN = 1V → VTN = 2.67V | Kn =
2
2
1
V
−
V
(
)
VGG =
GS
TN
1.5MΩ
18V = 7.30V
1.5MΩ + 2.2 MΩ
= 1.5MΩ 2.2 MΩ = 892kΩ | VGG − VGS = I D RS
Now, find the Q - point for VDD = 18V : VGG =
RGG
⎛ 0.2mS ⎞
2
7.30 - VGS = (12kΩ)⎜
⎟(VGS − 2.67) → VGS = 4.73V | I D = 0.254 mA | VDS = 9.37V ok
⎝ 2 ⎠
2kΩ
v
i
Rin
12 k Ω
22 k Ω
100 k Ω
Rth = 12kΩ 2kΩ = 1.71kΩ | RL = 22kΩ 100kΩ = 18.0kΩ | CGS = 3.0 pF | CGD = 0.6 pF
gm =
2(0.254mA)
(4.26 − 2.67)V
= 0.320mS | Rin = 12kΩ
1
1
= 12kΩ
= 2.48kΩ
gm
0.320mS
Rin
2.48kΩ
g m RL =
(0.320ms)(18.0kΩ)= +3.19
2kΩ + 2.48kΩ
RI + Rin
⎛
⎞
⎞
⎛
⎜
⎟
⎟
⎜
1
1 ⎜
1
1
⎜
⎟ = 11.3 MHz
⎟=
fH =
⎜
⎟
3.0
pF
2
π
2π ⎜ CGS
⎟
+ CGD RL ⎟
+ 0.6 pF (18.0kΩ)⎟
⎜
⎜
⎠
⎝ Gth + g m
⎝ (0.5848 + 0.32)mS
⎠
Amid =
Note that the contribution of the input pole cannot be neglected
because of the low fT of the MOSFET.
R1S = RI + Rin = 4.48kΩ | R3S = R7 + Rout ≅ 100kΩ + 22kΩ = 122kΩ
1 ⎛ 1
1 ⎞
fL =
+
⎟ = 20.6 Hz
⎜
2π ⎝ R1S C1 R3S C3 ⎠
Note that the is no signal current in C2 , so it does not contribute to f L .
17-46
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.70
g m = 40(0.25 mA) = 10.0mS | rx = 300Ω | rπ =
Cµ = 0.6 pF
| Cπ =
0.01
(
2π 5x108
)
100(0.025)
0.25 mA
= 10.0kΩ
− 0.6 = 2.58 pF | RB = 100kΩ 300kΩ = 75.0kΩ
RL = 13kΩ 100kΩ = 11.5kΩ | Rth = 75kΩ 2kΩ = 1.95kΩ
[
]
[
]
Rin = RB rx + rπ + (β o + 1)RL = 75.0kΩ 300Ω + 10.0kΩ + (101)11.5kΩ = 70.5kΩ
⎛ Rin ⎞
101(11.5kΩ)
(β o + 1)RL = 0.972
Amid = ⎜
= 0.964
⎟
0.300 + 10.0 + 101(11.5) kΩ
⎝ RI + Rin ⎠ rx + rπ + (β o + 1)RL
[
fH ≅
1
2π
1
⎤
⎡
2.58 pF
⎥
⎢
1950
+
300
+
0.6
pF
(
)⎢1+ 10mS 11.5kΩ
⎥⎦
(
)
⎣
]
=
1
1
= 114 MHz
2π (2250)(0.622 pF )
(b) Calculating the required SPICE parameters:
⎛ 11.8 ⎞ 0.333
CJC
Cµ =
|
CJC
=
0.6
pF
≅ 1.54 pF
⎟
⎜1+
ME
⎝ 0.75 ⎠
⎛
VCB ⎞
⎟
⎜1+
⎝ PHIE ⎠
C
1 Cµ
1
0.6 pF
τF = π =
−
= 9 −
= 260 ps | TF = 260 ps
gm ωT gm 10 π 40(0.25mA)
*Problem 16.70 - Common-Collector Amplifier
VCC 6 0 DC 15
VS 1 0 AC 1
RS 1 2 2K
C1 2 3 4.7UF
R1 3 0 100K
R2 6 3 300K
Q1 6 3 4 NBJT
R4 4 0 13K
C3 4 5 10UF
R7 5 0 100K
.MODEL NBJT NPN BF=100 TF=260PS CJC=1.54PF RB=300
.OP
.AC DEC 100 0.1 200MEG
.PRINT AC VM(5) VP(5)
.END
Results: Amid = 0.962, fL = 0.52 Hz, fH = 110 MHz
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-47
16.71
100kΩ
= 2.25V | RB = 100kΩ 300kΩ = 75.0kΩ
100kΩ + 300kΩ
(2.25 − 0.7)V = 0.251mA
IC = 100
75.0kΩ + 101(13kΩ)
VBB = 9V
gm = 40(0.251mA) = 10.0mS | rx = 300Ω | rπ =
Cµ = 0.6 pF
| Cπ =
100(0.025)
= 9.96kΩ
0.251mA
0.01
− 0.6 = 2.58 pF
2π (5x10 8 )
RL = 13kΩ 100kΩ = 11.5kΩ | Rth = 75kΩ 2kΩ = 1.95kΩ
Rin = RB [rx + rπ + (β o + 1)RL ] = 75.0kΩ [300Ω + 9.96kΩ + (101)11.5kΩ] = 70.5kΩ
⎛ Rin ⎞
101(11.5kΩ)
(β o + 1)RL
Amid = ⎜
= 0.972
= 0.964
⎟
⎝ RI + Rin ⎠ rx + rπ + (β o + 1)RL
[0.300 + 9.96 + 101(11.5)]kΩ
fH ≅
17-48
1
2π
1
⎤
2.58 pF
+ 0.6 pF ⎥
⎦
⎣1+ 10mS (11.5kΩ)
⎡
(1950 + 300)⎢
=
1
1
= 114 MHz
2π (2250)(0.622 pF )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.72
First, find the value of Kn required for I D = 0.1 mA
1.5MΩ
10V = 4.05V | VGG − VGS = 0.1mA(12kΩ) → VGS = 2.85V
1.5MΩ + 2.2 MΩ
2(0.1mA)
2I D
mA
=
= 0.356 2
VGS − VTN = 0.75V → VTN = 2.10V | Kn =
2
2
V
(VGS − VTN ) (0.75)
VGG =
gm =
2(0.1mA)
0.75V
Amid =
= 0.267mS | Rin = R1 R2 = 892kΩ | RL = 12kΩ 100kΩ = 10.7kΩ
(0.267mS )(10.7kΩ) = +0.739
g m RL
Rin
= 0.998
RI + Rin 1+ g m RL
1+ (0.267mS )(10.7kΩ)
From Table 16.2 on page 1037 : f H =
f P1 =
1
= 0.379Hz
2π (894kΩ)(4.7µF )
(-2.62 dB)
1
⎡
⎤
3 pF
+ 0.6 pF ⎥
2π 2kΩ 892kΩ ⎢
⎢⎣1+ (0.267mS )(10.7kΩ)
⎥⎦
(
fP2 =
)
1
⎡
⎛
⎞⎤
1
2π ⎢100kΩ + ⎜12kΩ
⎟⎥(0.1µF )
0.267mS ⎠⎥⎦
⎢⎣
⎝
= 57.9 MHz
= 15.5 Hz
f L ≅ 15.5 Hz
Note that a low frequency RHP zero makes the calculation of fH a very poor estimate for the FET
case. See the analysis in Prob. 17.73 which shows ωz = gm/CGS.
*Problem 16.72 - Common-Drain Amplifier
VDD 6 0 DC 10
VS 1 0 AC 1
RS 1 2 2K
C1 2 3 4.7UF
R1 3 0 1.5MEG
R2 6 3 2.2MEG
M1 6 3 4 4 NFET
R4 4 0 12K
C3 4 5 0.1UF
R7 5 0 100K
.MODEL NFET NMOS VTO=2.10 KP=0.356MA CGSO=30NF CGDO=6NF
.OP
.AC DEC 100 1 500MEG
.PRINT AC VM(5) VP(5)
.END
Results: Amid = 0.740, fL = 15.5 Hz, fH = 195MHz - Note that there is peaking in the response.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-49
16.73
First find VDD , VTN and Kn from Prob. 17.22 : VDD = VDS + I D RS = 10 V
1.5MΩ
10V = 4.05V | VGG − VGS = 0.1mA(12kΩ)→ VGS = 2.85V
VGG =
1.5MΩ + 2.2 MΩ
2(0.1mA)
2I D
mA
=
= 0.356 2
VGS − VTN = 0.75V → VTN = 2.10V | Kn =
2
2
V
(VGS − VTN ) (0.75)
Now find the new Q - point with VDD = 20 V.
1.5MΩ
20V = 8.11V | RGG = 1.5MΩ 2.2 MΩ = 892kΩ | VGG − VGS = I D RS
1.5MΩ + 2.2 MΩ
⎛ 0.356mA ⎞
2
8.11- VGS = (12kΩ)⎜
⎟(VGS − 2.10) → VGS = 3.56V | I D = 0.379 mA | VDS = 15.5 V ok
2
⎝ 2V
⎠
VGG =
gm =
2(0.379mA)
(3.56 − 2.10)V
Amid =
= 0.519mS | Rin = R1 R2 = 892kΩ | RL = 12kΩ 100kΩ = 10.7kΩ
(0.519mS )(10.7kΩ) = +0.846
g m RL
Rin
= 0.998
RI + Rin 1+ g m RL
1+ (0.519mS )(10.7kΩ)
From Table 16.2 on page 1037 : f H =
17-50
(-1.46 dB)
1
⎤
⎡
3 pF
2π 2kΩ 892kΩ ⎢
+ 0.6 pF ⎥
⎥⎦
⎢⎣1+ (0.519mS )(10.7kΩ)
(
)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
= 75.4 MHz
16.74
i1
ix
vx
Cµ
rπ
+
gm v
v
-
Cπ
RL
Ix = sCµVx + I1 | Vx =
⎛
⎞
I1
I1
+ ⎜ I1 + gm
⎟ RL
(sCπ + gπ ) ⎝
(sCπ + gπ )⎠
Vx sCπ rπ RL + RL + rπ + β o RL sCπ rπ RL + rπ + (β o + 1)RL
=
=
I1
sCπ rπ + 1
sCπ rπ + 1
1
1
sCπ rπ
sCπ
+
+
1 r + (β o + 1)RL rπ + (β o + 1)RL (1+ gm RL ) rπ + (β o + 1)RL
Y1 = = π
≅
for β o >> 1
Cπ rπ RL
Cπ RL
Z1
s
+1
s
+1
rπ + (β o + 1)RL
(1+ gm RL )
Z1 =
⎞
⎞
1 ⎛ 1
Cπ RL
1 ⎛ 1
<< 1 → ω <<
⎜ + gm ⎟ but
⎜ + gm ⎟ > ωT
Cπ ⎝ RL
Cπ ⎝ RL
(1+ gm RL )
⎠
⎠
Cπ
1
So, for ω << ωT , Y1 ≅ s
+
(1+ gm RL ) rπ + (β o + 1)RL
ω
Cin = Cµ +
Cπ
and Rin = rπ + (β o + 1)RL
(1+ gm RL )
ω H is determined by the input capacitance Cin and the source resistance Rth + rx .
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-51
16.75
+
ix
v
C GD
vx
C GS
gm v
vs
RL
Ix = sCGDVx + sCGS (Vx − VS ) | Vx = V + (sCGSV + gmV )RL | V =
Ix = sCGDVx + sCGS
Vx
(1+ gm RL + sCGS RL )
| Note : VS =
Vx
(1+ gm RL + sCGS RL )
(sCGS + gm )RL
(1+ gm RL + sCGS RL )
Vx
⎤
⎡
⎥
⎢
CGS RL
Ix
1
CGS
C R
C
g
⎥ |
= s⎢CGD +
≈ GS L = GS & m > ωT
R
C
Vx
1+ gm RL 1+ s GS L ⎥
1+ gm RL
gm RL
gm
CGS
⎢
⎢⎣
1+ gm RL ⎥⎦
CGS
g
Assuming ω << ωT : CIN ≈ CGD +
| Note the zero in VS at ω z = − m
1+ gm RL
CGS
16.76
Av = −141 (43dB) | f H = 6x106 Hz | fT ≥ 2(141) 6x106 = 1.69 GHz
(
GBW ≤
1
rx Cµ
| rx Cµ ≤
1
(
2π 1.69x109 Hz
)
)
= 94.2 ps
16.77
Av = 100 (40dB) | f H = 40x106 Hz | f T ≥ 2(100) 4x10 7 = 8.00 GHz
(
GBW ≤
17-52
1
rx Cµ
| rx Cµ ≤
(
1
2π 8x109 Hz
)
)
= 19.9 ps
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.78
Amid = g m RL | g m =
100
100
β
= 1.00mS | rπ = o ≅
= 100kΩ
100kΩ
g m 1.00mS
Assume rπ >> rx | rπo = rπ rx ≅ rx
1
ωH =
⎡
⎛
R ⎞⎤
rx ⎢Cπ + Cµ ⎜1+ g m RL + L ⎟⎥
rx ⎠⎦
⎝
⎣
rx Cµ (1+ g m RL )+ RLCµ =
Cµ =
0.884 pF
1+ 1.01x10−3 rx
1
ωH
≅
1
⎛
R ⎞
rx Cµ ⎜1+ g m RL + L ⎟
rx ⎠
⎝
| rx Cµ (1+ 100)+ 105 Cµ =
(
)
2π 25x106 =
| Amid =
(
1
2π 1.8x10
6
)
= 8.84x10−8
(0.75pF, 177Ω) (0.5pF,760Ω)
16.79
1
RLCGD
1
rx Cµ (1+ g m RL )+ RLCµ
| Cµ cannot exceed 0.884 pF for an ideal transistor with rx = 0.
Other more realistic possibilities (Cu ,rx ):
ωH ≅
=
Rin
g R
g m RL ≅ m L
RI + Rin
1+ g m RI
⎛ 1
⎞
⎛ 1
⎞
| RL = Amid ⎜ + RI ⎟ = 20⎜ + 100⎟
⎝ gm
⎠
⎝ gm
⎠
1
→ g m = 164mS
⎞
⎛ 1
−12
20⎜ + 100Ω⎟3x10 F
⎠
⎝ gm
⎛ 1
⎞
(164mS ) = 672 mA
g2
RL = 20⎜
+ 100⎟ = 2.12 kΩ | I D = m =
⎛ mS ⎞
2Kn
⎝ 0.164
⎠
2⎜20
⎟
⎝ V ⎠
2
Note that we cannot supply I D through RL since I D RL = 1420V >> VDD .
RI
C
C2
1
vi
L
R
RL
S
One Possibility:
+V DD
This is really not a realistic design. The current and power are far too high. We need to find an
FET with a much higher Kn.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-53
16.80
1
1
=
⎡
⎡
⎤
⎡
⎡
R ⎤⎤
R ⎤
Rth ⎢CGS + CGD ⎢1+ g m RL + L ⎥⎥ 100⎢12 pF + 5 pF ⎢1+ g m RL + L ⎥⎥
100 ⎦⎦
Rth ⎦⎦
⎣
⎣
⎣
⎣
⎡
⎤
1
RL ⎢
1
6
⎥1
2π 25x10 =
| g m RL +
=
−17
⎡
⎡
⎥5
100 ⎢ 2π 25x106 (100)10−12
R ⎤⎤
⎣
⎦
100⎢17 pF + 5 pF ⎢g m RL + L ⎥⎥
100
⎣
⎦
⎣
⎦
ωH =
(
)
gm =
(
)
9.33
g2
g2
− 0.01 → RL ≤ 933Ω | I D = m = m = 20g m2
2Kn 0.05
RL
For strong inversion (for the square - law model to be valid), we desire
(V
GS
− VTN ) ≥ 0.25V → I D ≥
2
0.025
0.25) = 781 µA.
(
2
(0.015)
=
2
g m must exceed 0.01S. Choose g m = 0.015S. I D
RL =
.05
= 4.5 mA.
9.33
= 1.87 kΩ | g m RL = 28.0
g m − .01
16.81
fH ≤
1
1
=
→ f H ≤ 8.33 MHz
2πRLCµ 2π 12kΩ 47kΩ (2 pF )
(
)
16.82
i
r x1
b
+
rπ1
v1
-
c u1
Q
c π1
g
m1
v
RL
i
b
1
Q2
Cπ & C µ
1
1
r
x2
+rπ
1
g
2
Cπ2 & C µ2
Use the open-circuit time-constant approach:
17-54
1
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
m1
Q
2
R
L
⎡ (β o + 1)(rx 2 + rπ 2 ) ⎤⎛
⎞
βo
Rµ1O : v x ≅ ix rx1 − ix rx1⎢
RL ⎟ − (−ix RL )
⎥⎜−
⎣ rπ 1 + (β o + 1)(rx 2 + rπ 2 )⎦⎝ rx 2 + rπ 2 ⎠
⎡
⎞⎤
⎛
β o rπ 2
Rµ1O ≅ ix ⎢RL + rx1⎜1+
gm 2 RL ⎟⎥ assuming rx 2 << rπ 2 .
⎠⎦
⎝ rπ 1 + β o rπ 2
⎣
⎛ 10
⎞
10
v
β
rπ 1 ≅ 10rπ 2 | β o = 100 | gm1 ≅ o =
| Rµ1O = x = RL + rx1⎜1+ gm 2 RL ⎟
⎝ 11
⎠
rπ 1 rπ 2
ix
Rπ 1O : Split ix and use superposition with rx 2 << rπ 2 :
⎡
ix
rπ 1
(β o + 1)(rx 2 + rπ 2 ) ⎤ + ix ≅ i r
v x ≅ ix rx1⎢1−
+
⎥
x x1
rπ 1 + β o rπ 2 gπ 2 + 10gπ 2
⎣ rπ 1 + (β o + 1)(rx 2 + rπ 2 )⎦ gπ 2 + gm1
v 10rx1 + rπ 2
Rπ 1O = x ≅
ix
11
Rπ 2O : The circuit is the same as that used for the CT calculation.
(a)
⎛
⎛
1 ⎞
rπ 2 ⎞
Rπ 2O = rπ 2 ⎜ rx 2 +
⎟
⎟ = rπ 2 ⎜ rx 2 +
⎝
gm1 ⎠
10 ⎠
⎝
Rµ 2O : The circuit is the same as that used for the CT calculation except the
additional ib = ix /2 is returned back to the output :
⎛
R
RL ⎞
Rµ 2O = Rπ 2O + Rπ 2O gm 2 RL + L = Rπ 2O ⎜1+ gm 2 RL +
⎟
2
2Rπ 2O ⎠
⎝
1
ωH =
⎡
⎛ 10
⎛
⎛10r + r ⎞
R ⎞
RL ⎞⎤
Cπ 1⎜ x1 π 2 ⎟ + Cµ1rx1⎜1+ gm 2 RL + L ⎟ + Rπ 2O ⎢Cπ 2 + Cµ 2 ⎜1+ gm 2 RL +
⎟⎥
⎝
⎠
11
rx1 ⎠
2Rπ 2O ⎠⎦
⎝ 11
⎝
⎣
100(0.025V )
r
50V
= 2.50kΩ | Use RL = o 2 =
= 25.0kΩ
1mA
2 2mA
40(10−4 )
40(10−3 )
− 0.5 pF = 1.62 pF | Cπ 2 =
− 0.5 pF = 20.7 pF
Cπ 1 =
6x10 8 π
6x10 8 π
⎛
⎛
r ⎞
2.50kΩ ⎞
Rπ 2O = rπ 2 ⎜ rx 2 + π 2 ⎟ = 2.50kΩ ⎜ 300 +
⎟ = 451Ω
⎝
⎝
10 ⎠
10 ⎠
rπ 2 =
−1
⎧
⎛ 3kΩ + 2.5kΩ ⎞
⎡
25kΩ ⎤⎫
⎟ + 0.5 pF (300Ω)⎢1+ 40mS (25kΩ) +
⎪
⎪1.62 pF ⎜
⎣
⎝
⎠
11
300Ω⎥⎦⎪
1 ⎪
fH =
⎬ = 393 kHz
⎨
⎡
⎛
2π ⎪
25kΩ ⎞⎤
⎪
+451Ω⎢20.7 pF + 0.5 pF ⎜1+ 40mS (25kΩ) +
⎟⎥
⎪⎭
⎪⎩
⎝
902Ω ⎠⎦
⎣
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-55
(b)
The circuit is almost the same except for two important changes: Cµ1 sees only rx1,
and the ib = ix /2 is not returned to the output for Cµ 2 .
ωH =
1
⎡
⎛
R ⎞⎤
10r + r
Cπ 1 x1 π 2 + Cµ1rx1 + Rπ 2O ⎢Cπ 2 + Cµ 2⎜1+ gm 2 RL + L ⎟⎥
11
Rπ 2O ⎠⎦
⎝
⎣
−1
⎫
⎧
⎛ 3kΩ + 2.5kΩ ⎞
⎟ + 0.5 pF (300Ω)
⎪
⎪1.62 pF ⎜
⎝
⎠
11
⎪
1 ⎪
fH =
⎬ = 640kHz
⎨
⎡
⎛
2π ⎪
25kΩ ⎞⎤⎪
+451Ω⎢20.7 pF + 0.5 pF ⎜1+ 40mS (25kΩ) +
⎟⎥
⎪⎩
⎝
451Ω ⎠⎦⎪⎭
⎣
(c )
The C - C/C - E cascade offers significantly better bandwidth than the
Darlington configuration because Cµ1 is not subject to Miller multiplication.
(d)
Improved bandwidth is one reason for the use of the C - C/C - E cascade
in the 741 op - amp.
17-56
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.83
C
R
C
u
+
v
rπ
cm
v
be
ro
Cπ
-
g v
m be
2R EE
CEE
2
Assume RL represents a differential-mode load between the collectors. Degradation of the
CMRR starts with the zero in the common-mode gain transfer function.
⎡⎛
C ⎞
G ⎤
(sCπ + g m + g π )vcm = ⎢s⎜⎝Cπ + 2EE ⎟⎠ + g m + g π + g o + 2EE ⎥ve − g ovc
⎣
⎦
(sC
µ
− g m )vcm = −(g m + g o )ve + (sCµ + g o + GC )v cm
Keeping the dominant terms, the numerator poynomial is approximately
⎤
⎛
C ⎞ ⎡⎛
C ⎞
G
N ≅ s 2Cµ ⎜ Cπ + EE ⎟ + s⎢⎜ Cµ − EE ⎟ g m + Cπ g o ⎥ + g o g π − g m EE
2 ⎠ ⎣⎝
2 ⎠
2
⎝
⎦
Using dominant root factorization :
⎛ 1
1 ⎞
GEE
−
⎟
⎜
gogπ − g m
⎛ 1
1 ⎞
1
⎝ β oro 2REE ⎠
2
ωZ ≅
=⎜
−
≅
⎟
⎛
⎛
CEE ⎞
CEE ⎞ Cπ
CEE ⎞
⎝ β oro 2REE ⎠ ⎛
⎜ Cµ −
⎜Cµ −
⎜Cµ −
⎟ g m + Cπ g o
⎟+
⎟
2 ⎠
2 ⎠ µf
2 ⎠
⎝
⎝
⎝
The dominant terms in the determinant are
⎡⎛
⎤
⎛
⎛
C ⎞
C ⎞
C ⎞
∆ = s 2Cµ ⎜ Cπ + EE ⎟ + sCµ g m + s⎜ Cπ + EE ⎟GC + g mGC = ⎢s⎜Cπ + EE ⎟ + g m ⎥[sCµ + GC ]
2 ⎠
2 ⎠
2 ⎠
⎝
⎝
⎣⎝
⎦
The high pole is in the vicinity of ωT . The dominant pole is ω Pcm =
1
.
RC Cµ
⎛ 1
1 ⎞
At dc, the common - mode gain is approximately Acm = RC ⎜
−
⎟
⎝ β oro 2REE ⎠
For zero base resistance rx , the dominant pole in the differential mode response is
⎛
1
R ⎞
ω Pdm =
and Adm = 0.5g m ⎜ RC L ⎟ ω Pdm and ω Pcm approximately cancel.
⎛
2⎠
R ⎞
⎝
⎜ RC L ⎟Cµ
2⎠
⎝
The CMRR reaches - 6 dB when C µ shorts the base to the collector.
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-57
50 + 10.1
= 611kΩ
10−4
⎡
⎛ 1
1 ⎞
1 ⎤
1
⎥ = −2.02x10−4 (-73.9dB)
⎢
−
−
Acm = RC ⎜
=
6kΩ
⎟
⎝ β oro 2REE ⎠
⎢⎣100(611kΩ) 20 MΩ⎥⎦
⎛
100kΩ ⎞
Adm = −0.5 4x10−3 ⎜6kΩ
⎟ = −10.7 CMRRdB = 94.5 dB
2 ⎠
⎝
⎛ 1
1 ⎞
−
⎟
⎜
1 ⎝ β oro 2REE ⎠
1 ⎛ -3.36x10-8 ⎞
=
fZ ≅
⎜
⎟ = 26.8 kHz
2π ⎛
2π ⎝ -0.2 pF ⎠
CEE ⎞
⎜ Cµ −
⎟
2 ⎠
⎝
g m = 40IC = 4x10−3 ro =
(
ω Pdm =
)
1
= 88.4 MHz
RLCµ
ω Pdm =
1
= 99.0 MHz
⎛
RL ⎞
⎜ RC
⎟Cµ
2⎠
⎝
See the PSPICE graph in Prob. 16.84 below.
17-58
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.84
The three curves from top to bottom are: CMRR, -20log(Acm), 20log(Adm/2).
frequencies (near 100 MHz), the pole of Adm approximately cancels the pole in Acm.
At high
16.85
ωH =
gm1
CGS1 + CGS 2 + CGD 2 (1+ gm1ro2 + gm 2 ro2 )
⎛ 5⎞
50V
gm1 = gm 2 = 2(25x10−6 )⎜ ⎟(10−4 ) = 158 µS | ro2 ≅
= 500kΩ
⎝ 1⎠
0.1mA
CGS1 = 3pF | CGS 2 = 3pF | CGD1 = 0.5 pF | CGD 2 = 0.5 pF
1
158µS
fH =
= 294 kHz
2π 3pF + 3pF + 0.5 pF [1+ 2(0.158mS )500kΩ]
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-59
16.86
ωH =
50V
gm1
| ID 2 = 5ID1 = 1.00mA | ro2 =
= 50kΩ
1mA
CGS1 + CGS 2 + CGD 2 (1+ gm1ro2 + gm 2 ro2 )
⎛ 5⎞
⎛ 25 ⎞
gm1 = 2(25x10−6 )⎜ ⎟(2x10−4 ) = 224 µS | gm 2 = 2(25x10−6 )⎜ ⎟(1x10−3 ) = 1.12mS
⎝ 1⎠
⎝1⎠
CGS & CGD ∝W : CGS1 = 3pF | CGS 2 = 15 pF | CGD1 = 1pF | CGD 2 = 5 pF
1
0.224mS
fH =
= 99.3 kHz
2π 3pF + 15 pF + 5 pF [1+ (1.12mS + 0.224mS )50kΩ]
16.87
The most probable answer that will be produced is
60V
gm1
ωH =
| IC 2 ≅ 10IC1 = 1.00mA | ro2 =
= 60kΩ
1.00mA
Cπ 1 + Cπ 2 + Cµ 2 [1+ (gm1 + gm 2 )ro2 ]
Cπ 1 =
fH =
40(10−4 )
1.2x10 9 π
− 0.5 pF = 0.561pF | Cπ 2 =
40(10−3 )
1.2x10 9 π
− 0.5 pF = 10.1pF
40(10−4 )
1
= 478 kHz
2π 0.561pF + 10.1pF + 0.5 pF [1+ 40(1.1mA)60kΩ]
However, C should be approximately proportional to emitter area:
Cµ 2 = 10Cµ1 = 5.00 pF | Cπ 2 =
fH =
17-60
40(10−3 )
1.2x10 9 π
40(10−4 )
− 5.00 pF = 5.10 pF
1
= 48.2 kHz
2π 0.561pF + 5.10 pF + 5.00 pF [1+ 40(1.1mA)60kΩ]
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.88
With the addition of rx, we must re-evaluate the open-circuit time constants.
c µ1
r o1
c π1
g m1 v1
r x1
+
v1
r x2
c µ2
+
r π1
v2
r π2
-
r o2
c π2
g m2 v2
-
Assume: rx << ro
⎛
1 ⎞ 1+ gm1rx 2
Cπ 2 & Cµ 2 are part of a common - emitter stage with rπo2 = rπ 2 ⎜ rx 2 +
⎟≅
gm1 ⎠
gm1
⎝
⎞
⎛
⎟
⎜
gm1
rx1
1
⎟ | Rπ 1o ≅
Cπ 1 : Rπ−11o = gπ 1 + gx1⎜1+
≅
1 ⎟
1+ gm1rx1 gm1
⎜ g +g +
⎟
⎜
x1
o1
rx 2 + rπ 2 ⎠
⎝
rx1
Cµ1 : Rµ1o =
with R = ro1 (rx 2 + rπ 2 ) | Rµ1o ≅ rx1
1
1
1+
+
β o gm R
−1
⎫
⎧
1+ gm1rx 2
rx1
ω H = ⎨Cπ 1
+ Cµ1rx1 +
Cπ 2 + Cµ 2 (1+ gm 2 ro2 )]⎬
[
gm1
⎭
⎩ 1+ gm1rx1
1
The last term will be dominant : ω H ≅
1+ gm1rx 2
Cµ 2 (1+ gm 2 ro2 )
gm1
The most probable answer that will be generated is
50V
IC 2 ≅ 4IC1 = 1.00mA | ro2 =
= 50kΩ | gm 2 = 40(0.001) = 40mS
1.00mA
40(2.5x10−4 )
40(10−3 )
Cπ 1 =
−
0.3pF
=
2.88
pF
|
C
=
− 0.3pF = 9.73pF
π2
10 9 π
10 9 π
1
1
fH ≅
= 964 kHz
2π 1+ 0.01S (175Ω)
0.3pF [1+ 40mS (50kΩ)]
0.01S
However, C should be approximately proportional to emitter area:
1
1
Cµ 2 = 4Cµ1 = 1.2 pF | f H ≅
= 241 kHz
2π 1+ 0.01S (175Ω)
1.2 pF [1+ 40mS (50kΩ)]
0.01S
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-61
16.89
ωH =
gm1
| IC 2 ≅ IC1 = 100µA
Cπ 1 + Cπ 2 + Cµ 2 [1+ (gm1 + gm 2 )ro2 ]
40(10−4 )
60V
ro2 =
= 600kΩ | Cπ 2 = Cπ 1 =
− 2 pF = 10.7 pF
100µA
10 8 π
⎤
40(10−4 )
1 ⎡
⎥ = 66.2 kHz
⎢
fH =
2π ⎢⎣10.7 pF + 10.7 pF + 2 pF (1+ 2(40)(0.100mA)600kΩ)⎥⎦
16.90
This analysis utilizes the open-circuit time-constant approach.
I
C3
I
O
+
M3
REF
v1
C2
gm v1
-
+
M2
ro
M1
ix
g m1 vx
C1
1
gm
vx
-
C1:
C1 = CGS1 + CGS 2 | C2 = CGD 2 + CGS 3 | C1 = CGD 3
R1O : v x = (ix + gm v1 )
1
gm
| v1 = −µ f v x − v x | R1O =
vx
1
=
ix gm (µ f + 2)
ix
+
ix
vx
-
+ v1
vg
gm vx
-
+
vx
ro
gm v1
i
ro
i
1
gm
i
ro
gm1 vx
C2:
17-62
1
gm
i
gm1 vx
C3:
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
vs
R2O : v x = (ix − i)ro − (gm v x − ix )
1
gm
| i = gm v x − ix
1
2ro +
⎞
⎛
1
v
2
gm
2v x = ix ⎜ ro + ⎟ − ro (gm v x − ix ) | R2O = x =
≅
gm ⎠
gm
ix
µf + 2
⎝
R3O : v x = (ix − gm v1 )ro +
R3O =
ωH ≅
fH ≅
ix
i
− (ix + i)ro | i = ix | v1 = −2ix ro − x
gm
gm
vx
1
= 2µ f ro + 4ro +
≅ 2(µ f + 2)ro ≅ 2µ f ro
gm
ix
1
1
1
≅
=
2
2CGS
CGS + CGD
+2
+ 2µ f roCGD 2µ f roCGD 2gm ro CGD
gm µ f
gm
1
2π
1
⎛ 50 ⎞ 2 −12
2 2(2.5x10 )(2.5x10 )⎜
⎟ (10 )
⎝ 2.5x10−4 ⎠
−4
= 5.63 kHz
−4
Note: R3O neglects any attached load resistance. If a load exists, essentially all of ix will go
through the load RL, and the frequency response will significantly improve. For that case, R3O ≈
RL + ro ≈ ro.
16.91
(a) rπ =
40(15x10−6 )
100(0.025)
=167kΩ
|
C
=
0.5
pF
|
C
=
− 0.5 pF = 0.773 pF
µ
π
15x10−6
2π (75x10 6 )
rπo = rπ rx = 167kΩ 500Ω = 499 Ω | gm = 40(15x10−6 )= 0.6mS | ω H =
1
rπoCT
|
⎡
430kΩ⎤
1
= 568 kHz
CT = 0.773 + 0.5 ⎢1+ 0.6mS(430kΩ) +
= 561pF | f H =
⎥
⎣
499Ω ⎦
2π (499)(5.61x10−10 )
(b) rπ =
40(5x10−5 )
100(0.025)
=
50.0kΩ
|
C
=
0.5
pF
|
C
=
− 0.5 pF = 3.74 pF
µ
π
5x10−5
2π (75x10 6 )
rπo = rπ rx = 50kΩ 500Ω = 495 Ω | gm = 40(5x10−5 )= 2.00mS | ω H =
1
rπoCT
|
⎡
140kΩ⎤
1
= 285 pF | f H =
= 1.13 MHz
CT = 3.74 + 0.5 ⎢1+ 2.0mS(140kΩ) +
⎥
⎣
495Ω ⎦
2π (495)(2.85x10−10 )
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-63
16.92
(a) Cµ = 1 pF
| Cπ =
40(125x10−6 )
2π (100x10
6
)
−1pF = 6.96 pF | rx = 500Ω | gm = 40(125x10−6 )= 5.00mS
⎡ 5.00mS(62kΩ)
62kΩ ⎤
1
+
= 1.11 MHz
CT = 6.96 + 1.0 ⎢2 +
⎥ = 288 pF | f H =
0.500kΩ⎦
2
2π (500)(2.88x10−10 )
⎣
(b) Cπ =
40(1x10−3 )
2π (100x10
6
)
−1pF = 62.7 pF | rx = 500Ω | gm = 40(1x10−3 )= 40.0mS
⎡ 40.0mS(7.5kΩ)
7.5kΩ ⎤
1
+
= 1.39 MHz
CT = 62.7 + 1.0 ⎢2 +
⎥ = 230 pF | f H =
2
0.500kΩ⎦
2π (500)(2.30x10−10 )
⎣
(a) Cµ = 1 pF
| Cπ =
40(100x10−6 )
2π (100x10
6
)
16.93
−1pF = 5.37 pF | rx = 500Ω | gm = 40(100x10−6 )= 4.00mS
100(0.25)
= 25kΩ | rπo = 500Ω 25kΩ = 490Ω
10−4
1
= 2.01 MHz
fH =
2π [(490Ω)(5.37 + 2)pF + (500 + 75kΩ)1pF ]
rπ =
(b) Cµ = 1 pF
| Cπ =
40(1x10−3 )
2π (100x10 6 )
−1pF = 62.7 pF | rx = 500Ω | gm = 40(1x10−4 )= 40.0mS
100(0.25)
= 2.5kΩ | rπo = 500Ω 2.5kΩ = 417Ω
10−3
1
= 4.55 MHz
fH =
2π [(417Ω)(62.7 + 2)pF + (500 + 7.5kΩ)1pF ]
rπ =
17-64
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.94
For Amid , refer to Section 14.8 : RB3 =
51.8kΩ
1kΩ
= 25.9kΩ, RE3 = 1.65kΩ, rπ 3 =
= 500Ω
2
2
[
(
)]
Avt1 = doesn' t change | RL2 = 4.7kΩ 25.9kΩ 500Ω + 81 1.65kΩ 250Ω = 3.26kΩ
Avt2 = −62.8mS (3.26kΩ)= −205 | Avt3 =
Av =
81(217)
500 + 81(217)
= 0.972
1MΩ
(−4.78)(−205)(0.972)= +943 or 59.5 dB
1.01MΩ
For f L , refer to Section 16.10.5
(
)
[
)]
(
R 5S = 4.7kΩ 54.2kΩ + 25.9kΩ 500Ω + 81 1.65kΩ 250Ω = 15.0kΩ
3.71kΩ + 0.5kΩ
= 300Ω
81
⎞
1 ⎛
1
1
⎜99 + 319 + 372 + 2340 +
⎟ = 533 Hz
fL =
+
2π ⎜⎝
15.0kΩ(1µF ) 300Ω(22µF )⎟⎠
Rth3 = 25.9kΩ 4.7Ω 54.2kΩ = 3.71kΩ | R6S = 250 + 1.65kΩ
For f H , refer to Section 16.10.5
[
)]
(
R L2 = 54.2kΩ 4.7kΩ 25.9kΩ 500Ω + 81 1.65kΩ 250Ω = 3.07kΩ
⎡
⎛
3.07kΩ ⎞⎤
−7
Rπo2CT 2 = (610Ω) ⎢39 pF + 1pF ⎜1+ 67.8mS (3.07kΩ)+
⎟⎥ = 1.54x10 s
0.610kΩ
⎝
⎠⎦
⎣
C π + Cµ =
gm
ωT
∝ IC → Cπ 3 = 2(51pF )−1pF = 101pF | Rth3 = 25.9kΩ 4.7Ω 54.2kΩ = 3.71kΩ
3.71kΩ + 250Ω
101pF + (3.71kΩ + 250Ω)1pF = 1.52x10−8 s
1+ 159.4mS (0.217kΩ)
fH =
(
1
2π 1.07x10 s + 1.54x10−7 s + 1.52x10−8 s
03/09/2007
−7
)
= 576 kHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-65
16.95
For Amid , refer to Section 14.8 : R1 = 39kΩ, R2 = 11kΩ, RE21 = 750Ω, RC 2 = 2.35kΩ
2.39kΩ
rπ 2 =
= 1.40kΩ | RI1 = 620Ω 39kΩ 11kΩ = 578Ω | Avt1 = −10mS 578Ω 1.20Ω = −3.91
2
RI 2 = 2.35Ω 51.8kΩ = 2.25kΩ | Avt2 = −62.8mS 2.25kΩ 19.8kΩ = −127 | Avt3 doesn't change
(
(
Av =
)
)
1MΩ
(−3.91)(−127)(0.95)= +467 or 53.4 dB
1.01MΩ
For f L , refer to Section 16.10.5
⎛17.2kΩ 2.39kΩ ⎞
17.2kΩ
R3S = 620Ω 12.2kΩ + ⎜
620Ω 12.2kΩ = 552Ω
⎟ = 1.64kΩ | Rth =
2 ⎠
2
⎝ 2
⎛R r ⎞
552 + 1195
R4S = 750
= 11.4Ω | R5S = ⎜ C2 o2 ⎟ + RB 3 Rin3 = 16.3kΩ
151
⎝ 2 2⎠
(
)
(
Rth3 = 51.8kΩ 2.35Ω 27.1kΩ = 2.08kΩ | R6S = 250 + 3.3kΩ
fL =
)
2.08kΩ + 1.00kΩ
= 288Ω
81
1
(99 + 319 + 610 + 3987 + 61.4 + 158)= 833 Hz
2π
For f H , refer to Section 16.10.5
RL1 = 598Ω
⎡
⎛
2.39kΩ
399Ω ⎞⎤
−8
= 399Ω | RthCT1 = (9.9kΩ) ⎢5 pF + 1pF ⎜1+ 0.01S (399Ω)+
⎟⎥ = 9.92x10 s
2
9900Ω ⎠⎦
⎝
⎣
Rth2 = 598Ω
12.2kΩ
2.39kΩ
= 544Ω | rπo2 = R4S =
(544Ω + 250Ω)= 596Ω
2
2
RL2 = 51.8kΩ
Cπ + Cµ =
gm
ωT
4.7kΩ
Rin2 = 2.25kΩ 19.8kΩ = 2.02kΩ
2
∝ IC → Cπ 2 = 2(40 pF )−1pF = 79 pF
⎡
⎛
2.02kΩ ⎞⎤
−7
rπo2CT 2 = (544)⎢79 pF + 1pF ⎜1+ 136mS (2.02kΩ)+
⎟⎥ = 1.95x10 s
0.544kΩ
⎠
⎝
⎣
⎦
Rth3 =
54.2kΩ 4.7kΩ
51.8kΩ = 2.08kΩ
2
2
2.08kΩ + 250Ω
50 pF + (2.08kΩ + 250Ω)1pF = 8.31x10−9 s
1+ 79.6mS (0.232k )
fH =
(
1
2π 9.92x10 s + 1.95x10−7 s + 8.31x10−9 s
17-66
−8
)
= 526 kHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.96
fo =
gm =
1
1
=
= 22.5 MHz
−5
2π LCGD 2π 10 (5x10−12 )
2ID
0.02
=
= 0.01S
2
VGS − VTN
1
+ 10
ro = 0.0167
= 6.99kΩ
0.01
Av = −gm (ro RL ) = −0.01S(6.99kΩ 10kΩ) = −41.1
BW =
1
1
22.5
=
= 7.75 MHz Q =
= 2.90
2πRP CGD 2π (4.11kΩ)(5 pF )
7.75
16.97
(a) f o =
(b) ro =
1
2π
(C + C )L
µ
→C =
1
1
− Cµ =
− 2 pF = 20.1pF
2
2
(2πf o ) L
2π (10.7x10 6 Hz) 10−5 H
[
]
75V + 10V
1
10.7
= 8.50kΩ | BW =
= 847kHz | Q =
= 12.6
10mA
2π (8.5kΩ)(22.1pF )
0.847
(c ) Q = 100
| BW =
fo
1
1
= 107kHz | ro =
=
= 67.3kΩ
Q
ω o (C + Cµ ) 2π (107kHz)(22.1pF )
67.3kΩ
= 7.918 | n = 2.81
8.50kΩ
C
2 pF
= 0.253pF | C = 22.1− 0.253 = 21.9 pF
(d ) Cµ' = µ2 =
7.918
n
n2 =
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-67
16.98
220 pF
VC
C
6 µH
D
20(220)
20 pF
20 pF
(a) CD =
= 20 pF | C =
pF = 18.3pF
20 + 220
VC
0
1+
1+
0.9
0.9
1
1
=
= 15.2MHz
fo =
2π LC 2π (6µH )(18.3pF )
CD =
5.75(220)
20 pF
= 5.75 pF | C =
pF = 5.60 pF
5.75 + 220
10
1+
0.9
1
1
=
= 27.5MHz
fo =
2π LC 2π (6µH )(5.60 pF )
(b) CD =
17-68
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.99(a)
4
C1
C2
1 pF
1
rπ
5 pF
+
v
RL
-
gm v
5 µΗ
20
pF
R EQ
20
pF
C EQ
C EQ
n 2 R EQ
n2
8.24
pF
3.33
kΩ
CEQ = Cπ + Cµ [1+ gm RL ] = 5 pF + 1pF [1+ 40(1mA)(5kΩ)] = 206 pF
CP = 20 pF +
REQ = rπ
CEQ
206 pF
= 20 pF +
= 28.2 pF | f o =
2
52
n
2π
RL
(1+ gm RL )(ωRL Cµ )
2
= 2.5kΩ
1
= 13.4 MHz
(5µH )(28.2 pF )
5000
(1+ 200)[2π (13.4 MHz)(5kΩ)(1pF )]
2
=2.5kΩ 140Ω =133Ω
1
13.4
= 1.70 MHz | Q =
= 7.88
2π (3.33kΩ)(28.2 pF )
1.70
Note the huge error that would be caused by using only rπ as the input resistance term.
RP = n 2 REQ = 25(133Ω) = 3.33kΩ | BW =
20 pF
vs
2
CA
v1
SC A
vs
2
n 2 R EQ
20
pF
5 µH
C EQ
n2
⎛ 1⎞
v1 = j2π (13.4 MHz)(20 pF )⎜ ⎟(3.33kΩ)v i = j2.80v s
⎝ 2⎠
j2.80v s
v o = (−gm RL )
= −40(10−3 )(5kΩ) j0.560v i | Av = 112∠ − 90 o
5
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-69
+
j
2.80
vs
5
v
RL
gm v
-
16.99(b)
10 pF
R EQ
10 pF
C EQ
RL
v
5 µH
vs
+
133 Ω
206 pF
gm v
-
CT = 5 pF + 1pF [1+ 40(1mA)(5kΩ)] = 206 pF | CP = 20 pF + 206 pF = 226 pF
fo =
1
2π
REQ = rπ
(5µH )(226 pF )
= 4.74 MHz
RL
(1+ gm RL )(ωRL Cµ )
2
= 2.5kΩ
REQ = 2.5kΩ 1.12kΩ =774Ω | BW =
5000
(1+ 200)[2π (4.74 MHz)(5kΩ)(1pF )]
2
1
4.74
= 910 kHz | Q =
= 5.21
2π (774Ω)(226 pF )
0.910
v o = j2π (4.74 MHz)(10 pF )(774Ω)(−gm RL )v i
v o = j2π (4.74 MHz)(10 pF )(774Ω)[−0.04mS(5kΩ)]v i | Av = 46.1∠ − 90 o
17-70
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.100 (a)
Referring to Eqns. (16.180-16.182) in Jaeger and Blalock,
1
1
=
= 10.1 MHz
fo =
2π L(CGD + C ) 2π (10µH )(25 pF )
ro ≅
1
1
=
= 2500Ω
λID (0.02 /V )(20mA)
BW =
Q=
gm ≅ 2K n ID = 2(0.005)(0.02 = 14.1 mS
1
1
=
= 2.55 MHz
2π (ro )(CGD + C ) 2π (2.5kΩ)(25 pF )
10.1
= 3.96 | Amid = −gm ro = −µ f = −35.4
2.55
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-71
16.100(b)
V1
C
CGD
r
V
CGS
S
o
C
+
2
VO
1
L
g V
m
-
S
⎡(sCGD − gm )Vs⎤ ⎡s(C1 + CGD + C2 ) + go
⎢
⎥=⎢
−sC2
0
⎣
⎦ ⎢⎣
−sC2 ⎤⎡V ⎤
jωC2 ( jωCGD − gm )
V
1
1 ⎥⎢ ⎥ | AV ( jω ) = o =
sC2 + ⎥⎣Vo ⎦
Vs
∆
sL ⎦
Let CEQ = C1 + CGD
⎡
C + C2 1 ⎤
g
go
∆ (s) = CEQ C2 ⎢s2 + s o +
+ EQ
⎥
CEQ sCEQ C2 L
CEQ C2 L ⎦
⎣
⎡ C + C2 1
⎤
CEQ + C2 1
g
go
∆ ( jω ) = CEQ C2 ⎢ EQ
− ω 2 + jω o +
⎥ | ω o2 =
CEQ jωCEQ C2 L ⎦
CEQ C2 L
⎣ CEQ C2 L
g
C
ω o m − jω o2 GD
CEQ
CEQ
C ⎞
gm ro ⎛
Av ( jω o ) = −
=−
1− jω o GD ⎟
⎜
g
go
1 ⎝
gm ⎠
ωo o −
1− 2
CEQ ω oCEQ C2 L
ω o LC2
µf
Av ( jω o ) = −
1−
But ω1 =
CEQ
CEQ + C2
⎛
⎛ CEQ ⎞⎛
ω ⎞
ω ⎞
⎟⎜1− j o ⎟
⎜1− j o ⎟ = −µ f ⎜1+
ω1 ⎠
C2 ⎠⎝
ω1 ⎠
⎝
⎝
⎛ C ⎞
gm
> ωT . So, Av ( jω o ) ≅ −µ f ⎜1+ EQ ⎟ for ω1 << ωT
CGD
C2 ⎠
⎝
Referring to Eqs. (17.192 -17.193) :
BW =
ωo
Q
=
go
g
1 ⎛
1 ⎞
− 2 o
=
⎟=
⎜1− 2
CEQ ω o CEQ C2 L roCEQ ⎝ ω o LC2 ⎠
1
⎛ C ⎞
roCEQ ⎜1+ EQ ⎟
C2 ⎠
⎝
⎛ C ⎞
Q = ω o roCEQ ⎜1+ EQ ⎟
C2 ⎠
⎝
17-72
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
C2CEQ
45(40)
=
pF = 21.2 pF | f o =
CEQ + C2 45 + 40
2π
ro ≅
1
(10µH )(21.2 pF )
= 10.9MHz
⎛ 45 ⎞
1
= 2.50kΩ | Q = 2π (10.9MHz)(2.50kΩ)(45 pF )⎜1+ ⎟ = 16.4
⎝ 40 ⎠
0.02(20mA)
10.9MHz
= 666 kHz
16.4
⎛ C ⎞
⎛ 45 pF ⎞
= −µ f ⎜1+ EQ ⎟ = − 2(0.005)(0.02)(2500)⎜1+
⎟ = −75.1
C2 ⎠
⎝ 40 pF ⎠
⎝
BW =
Amid
⎛ 1 ⎞ g
⎛ 1 ⎞ 2(0.005)(0.02)
Checking the approximation : ⎜ ⎟ m = ⎜ ⎟
= 450 MHz >> ω o
⎝ 2π ⎠ CGD ⎝ 2π ⎠
5 pF
16.101
CEQ =
fo =
(C1 + 5 pF )C2
C1 + 5 pF + C2
1
2π
= 25 pF | C2 = 50 pF | C1 = 45 pF
(10µH )(25 pF )
= 10.1 MHz | ro =
Using the results from Prob. 17.113: BW =
Q=
10.1
= 15.9 | Amid
0.635
03/09/2007
1
= 2.50kΩ
0.02(20mA)
1
= 635 kHz
⎛ 50 pF ⎞
2π (2.50kΩ)(50 pF )⎜1+
⎟
⎝ 50 pF ⎠
⎛ C ⎞
⎛ 50 pF ⎞
= −µ f ⎜1+ 1 ⎟ = − 2(0.005)(0.02)(2500)⎜1+
⎟ = −70.7
⎝ 50 pF ⎠
⎝ C2 ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-73
16.102
*Problem 16.102(a) - Fig. P16.100(a)
VS 1 0 AC 1
CGD 1 2 5PF
GM 2 0 1 0 14.1MS
RO 2 0 2.5K
C1 2 0 20PF
L1 2 0 10UH
.AC LIN 400 8MEG 12MEG
.PRINT AC VM(2) VP(2)
.PROBE V(2)
.END
Results: Amid = 35.3, fo = 10.1 MHz, BW = 2.50 MHz
*Problem 16.102(b) - Fig. P16.100(b)
VS 1 0 AC 1
CGD 1 2 5PF
GM 2 0 1 0 14.1MS
RO 2 0 2.5K
C1 2 0 40PF
C2 2 3 40PF
L1 3 0 10UH
.AC LIN 400 8MEG 12MEG
.PRINT AC VM(2) VP(2) VM(3) VP(3)
.PROBE V(2) V(3)
.END
Results: Amid = 75.1, fo = 10.1 MHz, BW = 670 kHz
*Problem 16.102(c) - Problem 16.101
VS 1 0 AC 1
CGD 1 2 5PF
GM 2 0 1 0 14.1MS
RO 2 0 2.5K
C1 2 0 45PF
C2 2 3 50PF
L1 3 0 10UH
.AC LIN 400 8MEG 12MEG
.PRINT AC VM(2) VP(2) VM(3) VP(3)
.PROBE V(2) V(3)
.END
Results: Amid = 70.7, fo = 10.1 MHz, BW = 640 kHz
17-74
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
16.103
⎛ g ⎞
a
C
≅
C
+
C
1+
g
R
=
C
+
C
⎜
( ) EQ GS1 GD1( m1 L ) GS1 GD1⎝1+ g m1 ⎟⎠
m2
I D2 = I D1 =
2(0.005)
2
0.01
0 − (−1) = 5.00mA | VGS 2 = −4 +
= −3V
0.01
2
[
]
VDS1 = VSG2 = +3V > 1V → Saturation region is ok. | g m2 = g m1 = 2(0.01)(0.005) = 10.0mS
CEQ = 20 pF + 5 pF [1+ 1]= 30 pF | CP = C1 + C3 + CEQ = 20 pF + 20 pF + 30 pF = 70 pF
Require C2 + CGD = CP → C2 = 70 pF − 5 pF = 65 pF
(b) f
o
=
RP = RG
1
2π LCP
(1+ g
=
1
2π
(10µH )(70 pF )
RL1
RL1 )(ωRL1CGD1 )
2
m1
= 100kΩ
RP = 100kΩ 140kΩ =58.3kΩ | BW1 =
= 6.02 MHz | RL1 =
1
g m2
100
(1+ 1)[2π (6.02 MHz)(100)(5 pF )]
2
1
1
=
= 39.0kHz
2πRP CP 2π (58.3kΩ)(70 pF )
1
2
BW2 ≅ BW1 2 −1 = 25.1kHz | Note that this is an approximation since R P = 100kΩ
6.02 MHz
= 240 | vo = (ωC3vi )(RP )(−g m RD )
25.1kHz
= 2π (6.02 MHz)(20 pF )(58.3kΩ)(−10.0mS )(100kΩ)= 4.41x10 4
at the output and 58.3 kΩ at the input. | Q =
Amid
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-75
16.104
*Problem 16.103 - Synchronously-Tuned Cascode Amplifier
VDD 5 0 DC 12
VS 1 0 AC 1
C3 1 2 20PF
L1 2 0 10UH
C1 2 0 20PF
RG 2 0 100K
M1 3 2 0 0 NFET1
CGS1 2 0 20PF
CGD1 2 3 5PF
M2 4 0 3 3 NFET2
CGS2 3 0 20PF
CGD2 4 3 5PF
L2 4 5 10UH
C2 4 5 65PF
RD 4 5 100K
.MODEL NFET1 NMOS VTO=-1 KP=10M
.MODEL NFET2 NMOS VTO=-4 KP=10M
.OP
.AC LIN 200 5.5MEG 6.5MEG
.PRINT AC VM(2) VP(2) VM(3) VP(3) VM(4) VP(4)
.PROBE
.END
Results: Amid = 279, fo = 6.10 MHz, Q = 24
The amplifier is actually stagger-tuned. Note that the loop of capacitors around M1 messes up
the hand results based upon the CEQ approximation. The CEQ approximation itself may not be
accurate enough for precise synchronous tuning. Plot a graph of V(2) and V(3) to show the
problem. Evidence of the problem is also provided by the huge error in the mid-band gain.
16.105
From Prob. 16.103 : CP 2 =
1
(2πf o ) L
2
=
1
2
⎛
⎞
1.02
⎜ 2π
⎟L
⎜
LCP1 ⎟⎠
⎝
=
CP1
(1.02)
2
=
70 pF
(1.02)
2
= 67.3 pF
C2 = CP2 − CGD2 = 67.3 pF − 5 pF = 62.3 pF | R p2 = 100kΩ
BW1 =
1
1
= 39.0kHz | BW2 =
= 23.7kHz
5
2π (58.3kΩ)(70 pF )
2π 10 Ω (67.3 pF )
(
)
BW2 39.0kHz
23.7kHz
BW1
+ 0.02 f o1 +
=
+ 0.02(6.02 MHz)+
= 152kHz
2
2
2
2
f + 1.02 f o
6.08 MHz
= 6.08 MHz | Q =
= 40
The new f o ≅ o1
2
152kHz
BW ≅
16.106
17-76
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
*Problem 16.105 - Stagger-Tuned Cascode Amplifier
VDD 5 0 DC 12
VS 1 0 AC 1
C3 1 2 20PF
L1 2 0 10UH
C1 2 0 20PF
RG 2 0 100K
M1 3 2 0 0 NFET1
CGS1 2 0 20PF
CGD1 2 3 5PF
M2 4 0 3 3 NFET2
CGS2 3 0 20PF
CGD2 4 3 5PF
L2 4 5 10UH
C2 4 5 62.3PF
RD 4 5 100K
.MODEL NFET1 NMOS VTO=-1 KP=10M
.MODEL NFET2 NMOS VTO=-4 KP=10M
.OP
.AC LIN 200 5.5MEG 6.5MEG
.PRINT AC VM(2) VP(2) VM(3) VP(3) VM(4) VP(4)
.PROBE
.END
Results: Amid = 512, fo = 6.19 MHz, BW = 0.19 MHz, Q = 33
16.107
+
Yin
C
µ
v
r
π
C
-
π
Yin = gπ + sCπ + Y1 |
I = sCµ (V − Vo ) = sCµ
Y1 ( jω ) = jωCµ
RL
gm v
(sC
µ
+
Y1
C
µ
v
RL
-
− gm )V = (sCµ + GL )Vo | Vo =
gm v
(sC
(sC
µ
µ
− gm )
V
+ GL )
gm + GL
g + GL
1+ gm RL
I
V | Y1 = = sCµ m
= sCµ
V
(sCµ + GL )
(sCµ + GL )
(sCµ RL + 1)
1− jωCµ RL
2
1+ gm RL
= jωCµ (1+ gm RL )
| For (ωCµ RL ) << 1,
2
(jωCµ RL + 1)
(ωCµ RL ) + 1
Y1 ( jω ) ≅ jωCµ (1+ gm RL ) +
(1+ gm RL ) ωC
RL
(
RL )
2
µ
From the results we see that the input capacitance is correctly modeled by the total Miller
input capacitance, but the input resistance is not correctly modeled by just rπ:
03/09/2007
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-77
RL
Cin = Cπ + Cµ (1+ gm RL ) | Rin = rπ
(1+ gm RL )(ωCµ RL )
2
(b) Cin = CGS + CGD (1+ gm RL ) = 6 pF + 2 pF [1+ 5mS (10kΩ)] = 108 pF
Rin =
RL
(1+ gm RL )(ωCGD RL )
2
Note also that X C in =
=
10kΩ
[1+ 5mS(10kΩ)][2π (5x10 6 )(2 pF )(10kΩ)]
2
1
2π (5x10 )(108 pF )
6
= 497Ω !
= 295 Ω | Both values are far less than infinity.
Although the CT approximation gives an excellent estimate for the dominant pole of the
common-emitter amplifier, it does not do a good job of representing the input admittance at
high frequencies. An improved estimate is needed for several of the problems to come.
+
Yin
C
µ
v
r
π
C
π
-
Yin = gπ + sCπ + Y1 |
I = sCµ (V − Vo ) = sCµ
Y1 ( jω ) = jωCµ
+
RL
Y1
gm v
(sC
µ
C
µ
v
RL
-
− gm )V = (sCµ + GL )Vo | Vo =
gm v
(sC
(sC
µ
µ
− gm )
V
+ GL )
gm + GL
g + GL
1+ gm RL
I
V | Y1 = = sCµ m
= sCµ
V
(sCµ + GL )
(sCµ + GL )
(sCµ RL + 1)
1− jωCµ RL
2
1+ gm RL
= jωCµ (1+ gm RL )
| For (ωCµ RL ) << 1,
2
(jωCµ RL + 1)
(ωCµ RL ) + 1
Y1 ( jω ) ≅ jωCµ (1+ gm RL ) +
(1+ gm RL ) ωC
RL
(
RL )
2
µ
From the results we see that the input capacitance is correctly modeled by the total Miller
input capacitance, but the input resistance is not correctly modeled by just rπ:
Cin = Cπ + Cµ (1+ gm RL ) | Rin = rπ
17-78
RL
(1+ gm RL )(ωCµ RL )
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
CHAPTER 17
17.1
(a) T = Aβ = ∞
(b) A = 10
80
20
1
Av =
|
β
=5
FGE = 0
|
= 10000 | T = 10000(0.2)= 2000
A
10000
100% 100%
=
= 5.00 | FGE =
=
= 0.05%
1+ Aβ 1+ 2000
1+ Aβ 2001
A
10
100%
(c) T = 10(0.2)= 2 | Av = 1+ Aβ = 1+ 2 = 3.33 | FGE = 1+ 2 = 33.3%
Av =
17.2
(a) β = R + R
R1
1
1kΩ
1
=
101kΩ 101
=
2
(b) T = Aβ = 10
86
20
⎛ 1 ⎞
A
2x10 4
=
= 100
⎜
⎟ = 198.6 | Av =
1+ Aβ
200
⎝ 101⎠
17.3
80
⎛ 1 ⎞
R1
1kΩ
1
=
=
| T (s) = Aβ = 10 20 ⎜
⎟ = 99.0
⎝ 101⎠
R1 + R2 101kΩ 101
⎛100kΩ ⎞⎛ 99 ⎞
R Aβ
Av = − 2
= −⎜
⎟⎜
⎟ = −99
⎝ 1kΩ ⎠⎝ 100 ⎠
R1 1+ Aβ
(a) β (s) =
17.4
β (s)=
Z1 (s)
Z1 (s)+ Z2 (s)
=
R
R+
1
sC
s
=
s+
1
RC
80
=
s
| A = 10 20 = 10 4
s + 5000
10 4 s
⎛ 1 ⎞ s + 5000
⎛ 1 ⎞⎛ 1 ⎞
10 s
Z Aβ
T (s)= Aβ =
| Av = − 2
= −⎜
= −⎜
⎟
⎟⎜
⎟
4
s + 5000
Z1 1+ Aβ
10 s
⎝ sRC ⎠
⎝ RC ⎠⎝ s + 0.5 ⎠
1+
s + 5000
Instead of a pole at the origin, the integrator has a low - pass response
with a pole a ω = 0.5 rad/s.
4
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-1
17.5
AV =
A
A
1
1
=
| From Chapter 12, GE =
=
1+ Aβ 1+ A
1+ Aβ 1+ A
1
≤ 10−4 → A ≥ 9999 | A ≥ 80 dB
1+ A
17.6
2.0V
Output Voltage
0V
No feedback
Feedback
-2.0V
-2.0V
-1.0V
0V
VS
Input
1.0V
2.0V
Voltage
*Problem 17.6 – Figure P17.6- Class-B Amplifiers
VCC 3 0 DC 10
VEE 4 0 DC -10
VI 1 0 DC 0
Q1 3 1 2 NBJT
Q2 4 1 2 PBJT
RL1 2 0 2K
RID 1 7 100K
E1 5 0 1 7 5000
RO 5 6 100
Q3 3 6 7 NBJT
Q4 4 6 7 PBJT
RL2 7 0 2K
.MODEL NBJT NPN
.MODEL PBJT PNP
.OP
.DC VS -10 10 .01
.PROBE V(1) V(2) V(7)
.END
17.7
1
1
≅
1+ Aβ Aβ
1
200
200
= 200 | GE ≅
≤ 0.002 → A ≥
= 105 | A ≥ 100 dB
β
A
0.002
Av =
A
1+ Aβ
| From Chapter 12, GE =
17.8
17-2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
SAA v =
A ∂A v
Av ∂A
Av =
A
1+ Aβ
∂A v (1+ Aβ )1− Aβ
1
=
=
2
2
∂A
(1+ Aβ )
(1+ Aβ )
SAA v =
∂A v
Av
1
1+ 10 (0.01)
5
= SAA v
∂A
A
=
=
SAA v =
1
A
1
1
=
≈
2
A 1+ Aβ
(
) 1+ Aβ Aβ
1+ Aβ
1
1001
1
10% = 9.99x10−3%
1001
17.9
(a) Series-series feedback (b) Shunt-series feedback
(c) Shunt-shunt feedback (d) Series-shunt feedback
17.10
(a) Series-shunt feedback (b) Shunt-series feedback
(c) Series-series feedback (d) Shunt-shunt feedback
17.11
(a) Shunt-series and series-series feedback (b) Shunt-shunt and series-shunt feedback
17.12
(a) Series-shunt and series-series feedback (b) Shunt-series and shunt-shunt and feedback
17.13
86
20
20000
io
| io = ii (40kΩ)
→ Ai = 8.00 x 10 5
ii
1kΩ
With resistive feedback, the closed - loop gain cannot exceed the open - loop gain.
(a) Av = 10
= 20000 | Ai =
Therefore, Ai ≤ 8.00 x 10 5.
(b) Atr =
io
io
Ai
8x10 5
=
=
| Atr ≤
= 20 S
v i ii (40kΩ) (40kΩ)
4 x10 4 Ω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-3
17.14
90
A = 10 20 = 31600
(a) R
= Rid (1+ Aβ ) | For β = 1, Rin = 40kΩ(1+ 31600) = 1.26 GΩ
(b) R
in
=
(c) R
out
= Ro (1+ Aβ ) | For β = 1, Rout = 1kΩ(1+ 31600)= 31.6 MΩ
in
(d ) R
out
Rid
40kΩ
| For β = 1, Rin =
= 1.27 Ω
(1+ Aβ )
(1+ 31600)
=
Ro
1kΩ
| For β = 1, Rout =
= 31.6 mΩ
(1+ Aβ )
(1+ 31600)
17.15
The circuit topology is identical to Fig. 17.8.
h11F = 5kΩ 45kΩ = 4.50kΩ | h22F = (45kΩ + 5kΩ) = (50.0kΩ)
-1
β = h12F =
v1
v2
=
i2 = 0
5kΩ
1
=
5kΩ + 45kΩ 10
| RL
-1
1
= 5kΩ 50kΩ = 4.55kΩ
h22F
4.55kΩ
20kΩ
4000)
= 2570
(
1kΩ + 4.55kΩ
1kΩ + 20kΩ + 4.5kΩ
A
2570
2570
=
=
= 9.96
Av =
⎛
⎞
1 + Aβ
258
1
1 + 2570⎜ ⎟
⎝10 ⎠
A=
Rin = RinA (1 + Aβ )= (1kΩ + 20kΩ + 4.5kΩ)(258)= 6.58 MΩ
Rout =
17-4
A
5kΩ 50kΩ 1kΩ
Rout
=
= 3.18 Ω
1 + Aβ
258
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.16
(a)
+
A
RI
+
RL
v
o
vI
R
2
R1
Feedback Network
(b) h
A
11
=
i2
i1
h21A =
(c) A =
v 2 =0
-1
i 1 =0
=−
-1
20kΩ(4000)
1kΩ
v 2 =0
v1
v2
h12A =
= 15kΩ | h11F = 4.3kΩ 39kΩ = 3.87kΩ | h11T = 18.9kΩ
= (1kΩ) = (1kΩ)
i2
v2
h22A =
v1
i1
= 0 | h12F =
i 1 =0
v1
v2
−h21A
(R + h )(h
I
T
11
T
22
+ GL
)
| h22F = (39kΩ + 4.3kΩ) = (43.3kΩ)
-1
= −80,000 | h21F =
=
i 2 =0
=
i2
i1
=−
v 2 =0
-1
| h22T = +1.02mS
4.3kΩ
= −0.0993
39kΩ + 4.3kΩ
4.3kΩ
= 0.0993
39kΩ + 4.3kΩ
−(−80000)
= 2680
⎛ 1
1
1 ⎞
(1kΩ + 20kΩ + 3.87kΩ)⎜⎝ 5.6kΩ + 1kΩ + 43.3kΩ ⎟⎠
β = 0.0993
(d ) Av =
(e) h
A
21
2680
= 10.0
1+ 2680(0.0993)
>> h21F | h12F >> h12A
| Note : (R in = 6.64 MΩ, Rout = 3.11 Ω)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-5
17.17
+
+
A
v
v
o
_
S
R
2
R
1
1k Ω
+
vs
40 k Ω
v1
1kΩ
8.5 k Ω
-
v1
i1 v
2=0
v1
v2
=
i2 = 0
o
_
316v 1
F
= 1kΩ 7.5kΩ = 882Ω | h22
=
β = h12F =
+
v
882 Ω
h11F =
R2 7.5 k Ω
R1
i2
v2
=
i1 = 0
1
1
=
1kΩ + 7.5kΩ 8.5kΩ
1kΩ
1
=
1kΩ + 7.5kΩ 8.5
100(.025V )
100
= 12.6kΩ
(200µA) =198µA | rπ =
198µA
101
v
40kΩ
(β o + 1)8.5kΩ
(101)8.5kΩ
A= o =
= 309
= 304
(316)
v s 40kΩ + 0.882kΩ
RO + rπ + (β o + 1)8.5kΩ
1kΩ + 12.6kΩ + (101)8.5kΩ
IC = α F IE =
Av =
A
=
1+ T
304
304
=
= 8.27
⎛ 1 ⎞ 36.8
1+ 304⎜ ⎟
⎝ 8.5 ⎠
Rin = RinA (1+ T ) = 40.9kΩ(36.8) = 1.51 MΩ | Rout =
17-6
A
out
R
=
1+ T
8.5kΩ
12.6kΩ + 1kΩ
101
= 3.60 Ω
36.8
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.18
+
+
vO
_
v ref
R
R
v ref
A-Circuit
v
h11F = 1
i1 v
=0
2=0
(
i
F
| h22
= 2
v2
1
=
R
i =0
1
=1
i2 = 0
(β o + 1)R = g
m1
(ro2
π 5 + (β o + 1)R
)r
A = gm1 ro2 ro4 [rπ 5 + (β o + 1)R]
v
| h12F = 1
v2
ro2 ro4
(β o + 1)R
ro4 )+ rπ 5 + (β o + 1)R
50 + 1.4
50 + 11.3
= 514kΩ | ro4 =
= 613kΩ | ro2 ro4 = 280kΩ
−4
10
10−4
100(.025)
12
IC 5 ≅ IE 5 ≅ 4 = 1.2mA | rπ 5 =
= 2.08kΩ
10
1.2mA
(101)10kΩ
= 876
A = 40(10−4 )(280kΩ)
280kΩ + 2.08kΩ + (101)10kΩ
ro2 =
A
876
109
=
=
= 0.999
1+ T 1+ 876(1) 110
100(0.025)
Rin = Rid (1+ T ) = 2rπ 1 (1+ T ) = 2
(877) = 43.9 MΩ
10−4
r +r r
R π 5 o2 o4 10kΩ 2.08kΩ + 280kΩ
βo + 1
101
Rout =
=
= 2.49 Ω
1+ T
877
io
100 ⎛ 1 ⎞
v
α v
io = α oie = α o o |
= o o =
⎟(0.999) = 98.9 µS
⎜
R
v ref
R v ref 101 ⎝10 4 ⎠
Av =
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-7
17.19
+
RI
A
1 kΩ
+
vo
-
vI
R2
91 k Ω
R1
10 k Ω
RL
2 kΩ
Feedback Network
h11F =
v1
i1 v
= 10kΩ 91kΩ = 9.01kΩ | h22F =
2 =0
β = h12F =
A=
=
i2 = 0
=
i1 = 0
1
2kΩ (1kΩ + 7.5kΩ)
=
1
1.96kΩ
10kΩ
1
=
10kΩ + 91kΩ 0.0990
( )
vo
25kΩ
1.96kΩ
=
10 4
= 4730
1kΩ + 1.96kΩ
vi 1kΩ + 25kΩ + 9.01kΩ
Av =
A
4730
=
= 10.1 | Rin = RinA (1+ T ) = 34.0kΩ(469) = 16.0 MΩ
1+ Aβ 1+ 4730(0.990)
Rout =
17-8
v1
v2
i2
v2
A
1.96kΩ 1kΩ
Rout
=
= 1.41 Ω
1+ T
469
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.20
(a) S
(b)
=
R in
A
∂Rin
=S
Rin
(c) S
R out
A
=
S AR out = −
(d)
∂Rout
Rout
A ∂Rin
A
Aβ
| Rin = RinA (1+ Aβ ) | S AR in = A
RinAβ =
≈1
Rin ∂A
Rin (1+ Aβ )
(1+ Aβ )
R in
A
∂A
A
=
A ∂Rout
Rout ∂A
5x10 4 (0.01)
10% = 9.98%
1+ 5x10 4 (0.01)
| Rout =
A
A
Rout
∂Rout
βRout
|
=−
(1+ Aβ ) ∂A (1+ Aβ )2
A
A(1+ Aβ ) βRout
Aβ
=−
≅ −1
2
A
Rout
1+
A
β
(
)
1+
A
β
(
)
=S
R out
A
∂A
A
=−
105 (0.01)
10% = −9.99%
1+ 105 (0.01)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-9
17.21
+
i
v1
i
100 k Ω
36 k Ω
1kΩ
+
4000 v
-
20 k Ω
5kΩ
1
36 k Ω
36 k Ω
i1
v1 v
y11F =
=
2 =0
1
36kΩ
| y22F =
i2
v2
=
v1 = 0
1
36kΩ
| y12F =
i1
v2
(20kΩ 36kΩ 100kΩ)= 11.4kΩ | (5kΩ 36kΩ)= 4.39kΩ
A=
=−
v1 = 0
1
36kΩ
vo
4.39kΩ
= −4000
(11.4kΩ)= −3.71x107
1kΩ + 4.39kΩ
ii
Atr =
Rin =
A
=
1+ Aβ
−3.71x10 7
= −36.0 kΩ
⎛
1 ⎞
7
1+ −3.71x10 ⎜−
⎟
⎝ 36x103 ⎠
(
)
(20kΩ 36kΩ 100kΩ)
⎛
1 ⎞
1+ −3.71x10 ⎜ −
⎟
⎝ 36x103 ⎠
17-10
(
7
)
= 11.1 Ω | Rout =
(36kΩ 5kΩ 1kΩ)
⎛
1 ⎞
1+ −3.71x10 ⎜ −
⎟
⎝ 36x103 ⎠
(
7
)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
= 0.790 Ω
17.22
A
+
+
RL
RI
iI
vo
10 k Ω
100 k Ω
RF
10 k Ω
Feedback Network
y11A =
y11F =
i1
v1 v
=
2 =0
i1
v1 v
=
2 =0
1
20kΩ
| y22A =
1
10kΩ
| y22F =
i2
v2
i2
v2
=
v1 =0
=
v1 =0
1
1kΩ
| y21A =
1
10kΩ
i2
v1 v
| y12F =
=−
(−4000) = 4S
2 =0
i1
v2
=−
v1 =0
1kΩ
1
10kΩ
| y12A =
| y21F =
i2
v1 v
i1
v2
=0
v1 =0
=−
2 =0
1
10kΩ
1
1
1
1
T
+
= 0.150mS | y22
=
+
= 1.10mS
20kΩ 10kΩ
1kΩ 10kΩ
− y21A
−4
A=
=
= −2.08x10 7 Ω | β = y12F = −10−4
−5
−3
−3
−4
T
T
10 + 0.150x10 1.1x10 + 10
GI + y11 y22 + GL
y11T =
(
Atr =
)(
) (
)(
)
A
−2.08x10 Ω
=
= −10.0 kΩ | Aβ = 2080
1+ Aβ 1+ −2.08x10 7 Ω −10−4 S
Note : Rin =
7
(
)(
100kΩ 10kΩ 20kΩ
1+ 2080
)
= 1.92 Ω | Rout =
10kΩ 1kΩ 10kΩ
1+ 2080
= 0.400 Ω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-11
17.23
+12 V
500 µA
Q
Q2
1kΩ
Q
Q
2
4.7 k Ω
1
1
v
i
36 k Ω
36 k Ω
1kΩ
1k Ω
IC1 = 500µA − IB 2 | IE 2 = IB1 +
IC1 = 500µA −
IE 2 = 37
37IB1 + 700µA
101
36000IB1 + 0.7
I
= 37IB1 + 700µA | IB 2 = E 2
1000
101
= 493µA − 0.366IB1 → IC1 = 491.2µA
IC1
100
+ 700µA = 881.7µA | IC 2 =
IE 2 = 873µA
100
101
Q
Q
i
1
2
1k Ω
36 k Ω
i
0.001v
1kΩ
36 k Ω
i
R
EQ
17-12
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
4.7k Ω
i1
v1 v
y11F =
rπ 1 =
=
2=0
1
36kΩ
F
| y 22
=
i2
v2
=
v1 = 0
1
36kΩ 1kΩ
| y12F =
i1
v2
=−
v1 = 0
1
36kΩ
100(0.025)
100(0.025)
50 + 1.6
= 5.09kΩ | rπ 2 =
= 2.86kΩ | ro1 =
= 105kΩ
491µA
873µA
493x10−6
(
)
RE = 1kΩ 36kΩ 4.7kΩ = 807Ω
A=
rπ 2 + (β o + 1)RE
vo
= 1kΩ 36kΩ rπ 1 gm1 ro1 (rπ 2 + (β o + 1)RE )
ii
ro1 + rπ 2 + (β o + 1)RE
A=
(β o + 1)REQ
vo
= − 1kΩ 36kΩ 5.09kΩ gm1 ro1 (rπ 2 + (β o + 1)REQ )
ii
rπ 2 + (β o + 1)REQ
) [
(
]
) [
(
]
(1kΩ 36kΩ r )= (1kΩ 36kΩ 5.09kΩ)= 817Ω
π1
[r (r
o1
π2
| gm = 40(491µA) = 19.6mS
][
]
+ (β o + 1)REQ ) = 105kΩ (2.86kΩ + (101)806Ω) = 46.8kΩ
REQ = 1kΩ 36kΩ 4.7kΩ = 806Ω |
(β o + 1)REQ
rπ 2 + (β o + 1)REQ
=
(101)806Ω
= 0.966
2.86kΩ + (101)806Ω
⎛
1 ⎞
A = −(817Ω)(19.6mS )(46.8kΩ)(0.966) = −724 kΩ | Aβ = −724 kΩ⎜ −
⎟ = 20.1
⎝ 36kΩ ⎠
A
−724kΩ
Atr =
=
= −36.0 kΩ
1+ Aβ 1+ 20.1
Note : Rin =
Rin =
(1kΩ 36kΩ 5.09kΩ) = 38.7Ω
1+ 20.1
⎛
r +r ⎞ 1
| Rin = ⎜ REQ π 2 o1 ⎟
= 21.8Ω
β o + 1 ⎠ 1+ Aβ
⎝
(1kΩ 36kΩ 5.09kΩ) = 817Ω = 82.2 Ω
1+ Aβ
9.94
⎛
rπ 2 + ro1 ⎞ ⎛
2.86kΩ + 105kΩ ⎞
⎟ ⎜806Ω
⎟
⎜1kΩ 36kΩ 4.7kΩ
101 ⎠ ⎝
101
⎠
⎝
Rout =
=
= 46.2 Ω
1+ Aβ
9.94
v
vo
ii = 10−3 v i → Av = o =
= −32.4
v i 1000ii
Note that this amplifier can be analyzed as a shunt-series feedback amplifier. This is good for
student practice - see Problem 17.32.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-13
17.24
*Problem 17.24 – Figure P17.23
VCC 5 0 DC 12
IDC 5 4 DC 500UA
II 0 2 AC 1
*IX 0 7 AC 1
RS 2 0 1K
C1 2 3 82UF
Q1 4 3 0 NBJT
Q2 5 4 6 NBJT
RF 3 6 36K
RE 6 0 1K
C2 6 7 47UF
RL 7 0 4.7K
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.OP
.AC DEC 100 1E2 1E7
.PRINT AC VM(7) VP(7) VM(2) VP(2)
.END
Results: Atr = -34.4 kΩ, Rin = 36.8 Ω, Rout = 17.6 Ω -- Note that these values are highly
sensitive to the precise value of rπ1.
17.25
0.002v
+
i
gs
v gs
i
100 k Ω
y11F =
i1
v1 v
(
-
1 MΩ
= 10−6 S | y22F =
2 =0
)
40 k Ω
i2
v2
10 k Ω
= 10−6 S | y12F =
v1 =0
vgs = ii 100kΩ 1MΩ = (90.9kΩ)ii
A=
10 k Ω
(
i1
v2
) (
1MΩ
= −10−6 S
v1 =0
vo
= −(2mS )(4.44kΩ)(90.9kΩ) = −8.08x105
ii
A
−8.08x105
−8.08x105
Atr =
=
=
= −446 kΩ
1+ Aβ 1+ −8.08x105 −10−6
1.81
Rin
(
)( )
(100kΩ 1MΩ) = 90.9kΩ = 50.2 kΩ
=
Rout =
(1+ Aβ )
1.81
(40kΩ 10kΩ 10kΩ 1MΩ) = 4.44kΩ = 2.45kΩ
(1+ Aβ )
1.81
17.26
17-14
)
| vo = − 2x10−3 v gs 40kΩ 10kΩ 10kΩ 1MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
i
i
gmv1
+
RI
Cπ
rπ
v1
Cµ
R
y12F = −sCµ | A =
ZinA =
RL
vo
rπo
g mrπo RL
=−
gm )
=−
(
ii
sCµ RL + 1
s(Cπ + Cµ )rπo + 1
s(Cπ + Cµ )rπo + 1 [sCµ RL + 1]
[
rπo
ZinA
| Zin =
=
1+ Aβ
s(Cπ + Cµ )rπo + 1
| Zin =
[s(C
π
Zin =
Zin ≅
]
rπo
s(Cπ + Cµ )rπo + 1
1−
[s(C
π
where rπo = rπ RI
Cµ
L
g mrπo RL
]
(−sC )
+ Cµ )rπo + 1 [sCµ RL + 1]
rπo (sCµ RL + 1)
µ
]
+ Cµ )rπo + 1 [sCµ RL + 1]+ sCµ g mrπo RL
rπo (sCµ RL + 1)
rπo (sCµ RL + 1)
= 2
⎡
s (Cπ + Cµ )Cµ rπo RL + srπoCT + 1
R ⎤
s2 (Cπ + Cµ )Cµ rπo RL + srπo ⎢Cπ + Cµ (1+ g m RL )+ L ⎥ + 1
rπo ⎦
⎣
rπo (sCµ RL + 1)
⎡
R ⎤
srπo ⎢Cπ + Cµ (1+ g m RL )+ L ⎥ + 1
rπo ⎦
⎣
=
rπo (sCµ RL + 1)
srπoCT + 1
for ω << ωT
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-15
17.27
+
M3
vo
iref
i
o
_
+
M
3
i2
1
i
r o1
g m2
2
ro1
i
ref
i1
v1 v
y11F =
= g o1 | y22F =
2 =0
i2
v2
1
v
r
g m2
A = o = ro1
= o1
1
iref
2
1+ g m3
g m2
g m3
Rin =
ro1
=
1+ Aβ
ro1
1+
µf 2
2
io = i2 = g m2vo = g m2
17-16
=
= g m2 | y12F =
v1 =0
A
| Atr =
=
1+ Aβ
i1
v2
= g m2
v1 =0
ro1
2
1+
ro1
(g m2 )
2
=
ro1
2 + µf 1
36kΩ
2
= 973Ω | Note : Rin ≅
72
gm
1+
2
ro1
i
µ f 1 + 2 ref
|
1
g
µf 1
io
=
= 0.973
iref µ f 1 + 2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
m2
v
o
_
17.28
y11F =
i1
v1 v
= go1
F
| y 22
=
2=0
io = α oi2 = α o gm 2v o
| A=
i2
v2
= gm 2
| y12F =
v1 = 0
(β o + 1)
i1
v2
β
=
β+2
v1 = 0
gm 2 ≅ gm 2
1
gm 2
vo
(β o + 1) ≅ β o r = r
= ro1
≅ ro1
1
iref
µ f + 2β o + 1 µ f o1 π 1
ro1 + rπ 3 + (β o + 1)
gm 2
+
Q
3
vo
i
ref
_
io
+
β
i
β+2 2
r
i2
g
o1
ro1
1
1
ro1
m2
g m2
vo
_
i
ref
(β o + 1)
µ f + 2β o + 1
βo + 1
r
1
≅ ro1
≅ o1 =
(β o + 1) g
µ f β o + 2µ f + 2β o + 2 µ f gm1
1+ ro1
( m2)
µ f + 2β o + 1
Atr =
A
=
1+ Aβ
Atr =
1
= 20.0 Ω
50mS
| AI =
io
v
g
= α o gm 2 o = α o m 2 ≅ 1
iref
iref
gm1
⎡
1 ⎤
RinA = ro1 ⎢rπ 3 + (β o + 1)
⎥ ≅ ro1 2rπ 3 ≅ 2rπ 3
gm 2 ⎦
⎣
RinA
Rin =
=
1+ Aβ
Rin =
ro1 2rπ 3
ro1 2rπ 3
r 2r
2r
2
≅
≅ o1 π 3 ≅ π 3 ≅
µ (β + 1)
(β o + 1) g
β o + 1 β o + 1 gm 3
1+ ro1
( m 2 ) 1+ f o
µ f + 2β o + 1
µ f + 2β o + 1
ro1 2rπ 3 40kΩ 4kΩ
=
= 36.0 Ω
βo + 1
101
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-17
17.29
*Problem 17.29 - BJT Wilson Source
*Current gain = 100
VCC 0 3 DC -6
IREF 0 1 DC 100UA
Q1 1 2 0 NBJT
Q2 2 2 0 NBJT
Q3 3 1 2 NBJT
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.OP
.TF I(VCC) IREF
β o ro3 100(55.3V )
=
= 27.7 MΩ | SPICE : 29.9 MΩ
.END
2
2(100µA)
*Problem 17.29 - BJT Wilson Source
*Current gain = 10K
VCC 0 3 DC -6
IREF 0 1 DC 100UA
Q1 1 2 0 NBJT
Q2 2 2 0 NBJT
Q3 3 1 2 NBJT
.MODEL NBJT NPN BF=10K VA=50 IS=1E-15
.OP
.TF I(VCC) IREF
.END
SPICE : 799 MΩ
*Problem 17.29 - BJT Wilson Source
*Current gain = 1MEG
VCC 0 3 DC -6
IREF 0 1 DC 100UA
Q1 1 2 0 NBJT
Q2 2 2 0 NBJT
Q3 3 1 2 NBJT
.MODEL NBJT NPN BF=1MEG VA=50 IS=1E-15
.OP
.TF I(VCC) IREF
55.3V
µ f 1ro3 = 40(51.4)
= 1.14 GΩ | SPICE : 1.08 GΩ
.END
100µA
17-18
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.30
+
is
v
100 k Ω
11 k Ω
20 k Ω
-
io
1k Ω
1
+
_
5kΩ
4000 v 1
909 Ω
10 k Ω
1kΩ
g11F =
i1
1
v
=
| g22F = 2
v1 i = 0 11kΩ
i2
2
(
= 1kΩ 10kΩ = 909Ω | g12F =
v1 = 0
=−
v1 = 0
1kΩ
1
=−
10kΩ + 1kΩ
11
)
A=
i2
4000
= − 100kΩ 11kΩ 20kΩ
= −384.84x103
ii
(5 + 1+ 0.909)kΩ
Ai =
A
=
1+ Aβ
Rin =
i1
i2
−3.84x103
= −10.97
⎛ 1⎞
3
1+ −3.84x10 ⎜ − ⎟
⎝ 11⎠
(
)
(100kΩ 11kΩ 20kΩ) = 6.63kΩ = 18.9 Ω
(1+ Aβ )
351
Rout = (5kΩ + 1kΩ + 0.909kΩ)(1+ Aβ )= (6.91kΩ)(351)= 2.43MΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-19
17.31
+
A
RI
iI
+
RL
vo
10 k Ω
150 k Ω
R1
-
20 k Ω
R2
2kΩ
Feedback Network
i1
1
v
=
| g22F = 2
v1 i = 0 22kΩ
i2
g11F =
2
g11T =
v1 = 0
i1
i2
=−
v1 = 0
1
1
1
v
T
+
=
| g22
= 1.82kΩ + 1kΩ = 2.82kΩ | g21A = 2
20kΩ 22kΩ 10.5kΩ
v1
A=−
g
T
11
A
=
1+ Aβ
Rin =
A
21
(G + g )(g
I
Ai =
= 2kΩ 20kΩ = 1.82kΩ | g12F =
T
22
+ RL
)
=−
(150kΩ 10.5kΩ) = 9.81kΩ = 35.2 Ω
1+ 278
Rout = (10kΩ + 2.82kΩ)(1+ Aβ )= (12.8kΩ)(279) = 3.57 MΩ
17-20
= 4000
i2 = 0
4000
1
= −3060 | β = g12F = −
⎛ 1
11
1 ⎞
+
⎟(2.82kΩ + 10kΩ)
⎜
⎝150kΩ 10.5kΩ ⎠
−3060
= −11.0 | Aβ = 278
⎛ 1⎞
1+ (−3060)⎜ − ⎟
⎝ 11⎠
(1+ Aβ )
2kΩ
1
=−
20kΩ + 2kΩ
11
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.32
io
+12 V
Q
1
500 µA
Q
Q
RI
2
ii
1
4.7 k Ω
Q2
1k Ω
36 k Ω
36 k Ω
1k Ω
1k Ω
IC1 = 500µA − IB 2 | IE 2 = IB1 +
IC1 = 500µA −
IE 2 = 37
37IB1 + 700µA
101
36000IB1 + 0.7
I
= 37IB1 + 700µA | IB 2 = E 2
101
1000
= 493µA − 0.366IB1 → IC1 = 491.2µA
IC1
100
+ 700µA = 881.7µA | IC 2 =
IE 2 = 873µA
100
101
Q
Q
io
2
4.7k Ω
1
36 k Ω
0.001v i
g11F =
1kΩ
i1
1
v
F
=
| g22
= 2
v1 i = 0 37 kΩ
i2
2
1kΩ 37kΩ = 974 Ω | rπ 1 =
1kΩ
1k Ω
36 k Ω
= 36kΩ 1kΩ = 973 Ω | g12F =
v1 = 0
i1
i2
=−
v1 = 0
1kΩ
1
=−
1kΩ + 36kΩ
37
100(0.025)
100(0.025)
= 5.09 kΩ | rπ 2 =
= 2.86 kΩ
491µA
873µA
A=
⎛
⎞
4700Ω
io
974Ω
=−
⎟ = −1340
(−100)(101)⎜
⎝ 973Ω + 4700Ω ⎠
974Ω + 5090Ω
ii
Ai =
A
=
1+ Aβ
−1340
−1340
=
= −36.0
⎛ 1⎞
37.2
1+ (−1340)⎜ − ⎟
⎝ 37 ⎠
| 1 + Aβ = 37.2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-21
v o 973io
i
=
= 0.973 o = −35.0
v i 1000ii
ii
Av =
(1kΩ 37kΩ r ) = (1kΩ 37kΩ 5.09kΩ) = 22.0 Ω
50 + 1.6
= 105kΩ
1+ Aβ
37.2
493x10−6
⎛
rπ 2 + ro1 ⎞ ⎛
5.09kΩ + 105kΩ ⎞
⎜1kΩ 36kΩ 4.7kΩ
⎟ ⎜1kΩ 36kΩ 4.7kΩ
⎟
101
101 ⎠ ⎝
⎝
⎠
=
=
= 12.5 Ω
1+ Aβ
37.2
π1
Rin =
Rout
| ro1 =
17.33
+
2kΩ
+
-
vi
20 k Ω
v
i
1k Ω
1
-
+
_
4000 v
o
5kΩ
1
5k Ω
5kΩ
5kΩ
v1
v
= 5 kΩ | z22F = 2
i1 i =0
i2
z11F =
2
A=
= 5 kΩ | β = z12F =
i1 =0
v1
= 5 kΩ
i2 i =0
1
⎛
⎞
io
4000
20kΩ
=
⎜
⎟ = 0.269
vs 2kΩ + 20kΩ + 5kΩ ⎝ 5kΩ + 1kΩ + 5kΩ ⎠
Atr =
io
A
0.269
v
i
=
=
= 0.200 mS | o = −5000 o = −1.00 | 1+ Aβ = 1350
vs 1+ Aβ 1+ 0.269(5000)
vs
vs
Rin = (2kΩ + 20kΩ + 5kΩ)(1+ Aβ ) = (27kΩ)(1350)= 36.5 MΩ
Rout = (5kΩ + 1kΩ + 5kΩ)(1+ Aβ )= (11kΩ)(1350) = 14.9 MΩ
17-22
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.34
RI
vi
2 kΩ
A
+
+
RL
+
vo
5 kΩ
-
R1
5 kΩ
Feedback Network
v1
v
= 5 kΩ | z22F = 2
i1 i =0
i2
z11F =
2
= 5 kΩ | β = z12F =
i1 =0
v1
= 5 kΩ
i2 i =0
1
T
z11T = 5kΩ + 20kΩ = 25 kΩ | z22
= 5kΩ + 1kΩ = 6 kΩ | z21A =
A=
z21A
(R + z )(z
I
Atc =
T
11
T
22
+ RL
)
=
v2
i1
= 20kΩ(4000) = 80 MΩ
i2 =0
80 MΩ
= 0.269 S
(2kΩ + 25kΩ)(6kΩ + 5kΩ)
A
0.269
=
= 2.00x10−4 S | Aβ = 1345
1+ Aβ 1+ 0.269(5kΩ)
(
)
Note : Rin = RI + z11T (1+ Aβ ) = (27kΩ)(1346) = 36.3 MΩ
(
)
T
+ RL (1+ Aβ )= (11kΩ)(1346) = 14.8 MΩ
Rout = z22
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-23
17.35
RI
+
-
vi
io
i1
β i
rπ
o1
io
RE
RI
+
-
vi
RE
+
vo
-
i2
i1
rπ
β o i1
RE
+
-
i2 RE
By carefully drawing the circuit, it can be represented as a series-series feedback amplifier. In
particular, rπ and the current generator are connected within the feedback network.
i
βo
A= 2 =
| β = z12F = RE
v s RS + rπ + RE
βo
io
A
βo
RS + rπ + RE
=
=
=
βo
v s 1+ Aβ 1+
RE RS + rπ + (β o + 1)RE
RS + rπ + RE
Atc =
Av =
v o io RE
βo
(β o + 1)RE =
(β o + 1)RE
=
=
v s v s α o RS + rπ + (β o + 1)RE
βo
RS + rπ + (β o + 1)RE
⎛
⎞
βo
Rin = RinA (1+ Aβ ) = (RS + rπ + RE )⎜1+
RE ⎟ = RS + rπ + (β o + 1)RE
⎝ RS + rπ + RE ⎠
Both answers agree with our previous direct derivations.
17-24
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.36
Av =
Av =
Atc =
+
-
β o RL
β o RL
=
RS + rπ + (β o + 1)RE RS + rπ + RE + β o RE
βo
RS + rπ + RE
1+
βo
RS + rπ + RE
βo
RS + rπ + RE
R
I
vi
RL =
RE
Atc
RL
1+ Atcβ
| β = RE
i
i1
o
β i
rπ
o1
io
RE
+
vo
-
+
-
R
i1
R
rπ
I
vi
E
i2
β i
o 1
R
E
+
-
i R
2 E
By carefully drawing the circuit, it can be represented as a series-series feedback amplifier. In
particular, rπ and the current generator are connected within the feedback network.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-25
17.37
i
o
Q
io
3
Q3
i
ref
iref
r
o1
+
v
r o1
g
g11F =
v
1
-
g
F
| g22
=
2
Ai =
m2
m2
m1 1
i1
1
=
v1 i = 0 ro1
io = iref ro1
1
g
1
v2
i2
βo
ro1 + rπ 3 + (β o3 + 1)
=
v1 = 0
1
gm 2
1
gm 2
≈ iref
| g12F =
β o ro1
ro1 + 2rπ 3 +
i1
i2
1
gm 2
=
v1 = 0
gm1
≅1
gm 2
| A=
β oµ f
io
=
≈ βo
iref µ f + 2β o + 1
A
βo
=
= α o ≈ 1 which is correct.
1+ Aβ 1+ β o (1)
⎛
1 ⎞
ro1 ⎜ rπ 3 + (β o3 + 1)
⎟
gm 2 ⎠ 2rπ
⎝
2
≈
=
which is correct.
Rin =
1+ β o
β o gm
⎞
⎛
1
β o3
⎜
⎛
⎞
βo
gm 2 ⎟
⎟ = (1+ β o )ro ⎜⎜1+
⎜
Rout = (1+ β o )ro 3 1+
⎟ ≈ β o ro − not correct!
µ f + β o + 1⎟⎠
⎜ r +r + 1 ⎟
⎝
⎟
⎜
o1
π3
gm 2 ⎠
⎝
17-26
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.38
5
Q3
Q
R
I
+
-
1
Q
RL3
2
1
RL2
RL1
v
i
3
2
4
RF
R
E1
R
E2
The amplifier is not a two-port. It has five separate terminals. It can be analyzed correctly as a
series-shunt configuration with the output defined at terminal 4. In the series-shunt
configuration, RL is absorbed into the amplifier thereby making it a two-port.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-27
17.39
R = r o2
M4
M4
Amplifier
ii
RI
M3
M3
R = r o2
Feedback
Network
(a)
(b)
(c) Yes, see figure.
g r
1 ro 3
≅
A(i) = ro 3 m 4 o2
1+ gm 4 ro2 ro 2 ro 2
| β = gm 3 ro2 | Aβ = gm 3 ro 3 = µ f 3
A
A
Rout
= µ f 4 ro 2 | Rout = Rout
(1+ Aβ ) = µ f 4 (1+ µ f 3 )ro2
Also, RinA = ro 3 | Rin =
RinA
r
1
= o3 ≅
(1+ Aβ ) 1+ µ f 3 gm 3
(d)
ix
+
vgs4
-
g
m4
v
r
+
-
vx
o4
gs4
+
r
v
o3
g m3 v gs3
gs3
ro2
-
v x = (ix − gm 4 v gs4 )ro4 + ix ro2 | v gs4 = (−µ f 3v gs3 − v gs3 )= −ix ro 2 (µ f 3 + 1)
v x = ix ro 4 + ix ro 2 + gm 4 ro 4 (1+ µ f 3 )ix ro2 | Rout = µ f 4 (1+ µ f 3 )ro 2
17-28
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
(e)
⎡ 50 + 8⎤
⎢
µf = gm ro ≅ 2(0.75 x10 )(10 ) −4 ⎥ = (0.387mS )(580kΩ) = 194
⎢⎣ (10 )⎥⎦
−3
−4
Rout = µ f 4 (1 + µ f 3 )ro2 = 194 (195)(580kΩ) = 21.9 GΩ
(f) SPICE yields 28.0 GΩ
values from SPICE.

with the formula above using the parameter
17.40
+
A
R1
R2
R
2
+
vx
R1
vo
_
Note: The loading effects of the feedback network must be carefully included.
R1 R2 = 1kΩ 9.1kΩ = 901Ω | For vS = 0, IC = α F I = 198µA | rπ =
T=
T=
100(0.025V )
(β o + 1)(R1 + R2 ) ⎛ R1 ⎞
vo
RID
=
A)
⎟
⎜
(
vx RID + 901Ω
RO + rπ + (β o + 1)(R1 + R2 )⎝ R1 + R2 ⎠
198µA
= 12.6kΩ
⎛
⎞
(101)(10.1kΩ)
vo
40kΩ
1kΩ
=
316)
⎜
⎟ = 30.2
(
vx 40kΩ + 901Ω
1kΩ + 12.6kΩ + (101)(10.1kΩ)⎝1kΩ + 9.1kΩ ⎠
17.41
v
(β o + 1)R
T = o = gm 2 (ro2 ro4 )
| gm1 = 40(10−4 )= 4.00mS
vx
(ro2 ro4 )+ rπ 3 + (β o + 1)R
ro2 =
100(0.025)
50 + 1.4
50 + 11.3
= 514kΩ | ro4 =
= 613kΩ | rπ 3 =
= 2.08kΩ
−4
−4
10
10
(12V /10kΩ)
T = (4 x10−3 )(280kΩ)
(101)10kΩ
280kΩ + 2.08kΩ + 101(10kΩ)
= 876 (58.9 dB)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-29
17.42
Q3
ix
i
+
i
v1
r o1
T=
g
-
1
m2
(β o + 1)µ f
io
ro
(101)(2000) = 91.8
= (β o + 1)
=
=
1
ix
µ f + 2β o + 1 2000 + 200 + 1
ro + rπ + (β o + 1)
gm
17.43
Q
Q
v
x
Q
2
Q
1
36 k Ω
+
-
R
1k Ω
1k Ω
1k Ω||36k Ω||4.7k Ω
EQ
vx
36 k Ω
1k Ω
36 k Ω
y11F =
rπ 1 =
T=
i1
v1 v
=
2=0
1
36kΩ
F
| y 22
=
i2
v2
=
v1 = 0
1
36kΩ 1kΩ
i1
v2
=−
v1 = 0
1
36kΩ
100(0.025)
100(0.025)
50 + 1.6
= 5.09kΩ | rπ 2 =
= 2.86kΩ | ro1 =
= 105kΩ
491µA
873µA
493x10−6
rπ 2 + (β o + 1)REQ
vo
1
=
1kΩ 36kΩ rπ 1 gm1 ro1 (rπ 2 + (β o + 1)REQ )
v s 36kΩ
ro1 + rπ 2 + (β o + 1)REQ
) [
(
(1kΩ 36kΩ r ) = (1kΩ 36kΩ 5.09kΩ) = 0.0227
π1
36 kΩ
[r (r
o1
| y12F =
π2
36kΩ
][
]
| gm = 40(491µA) = 19.6mS
]
+ (β o + 1)REQ ) = 105 kΩ (2.86 kΩ + (101)806Ω) = 46.8kΩ
rπ 2 + (β o + 1)REQ
2.86kΩ + (101)806Ω
=
= 0.430
ro1 + rπ 2 + (β o + 1)REQ 105 kΩ + 2.86 kΩ + (101)806Ω
T = (0.0227)(19.6mS )(46.8kΩ)(0.430) = 8.95
17-30
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
2
1
REQ
+
806 Ω
-
vo
17.44
C
v1
v2
1
vx
vx
+
R
R
1
2
C
⎡sC1Vx ⎤ ⎡sC1 + G1 + G2
⎢
⎥=⎢
−G2
⎣ 0 ⎦ ⎣
2
−G2 ⎤⎡V1 ⎤
⎥⎢ ⎥
sC2 + G2 ⎦⎣V2 ⎦
V2 =
sC1G2
Vx
∆
∆ = s 2C1C2 + s[C1G2 + C2 (G1 + G2 )]+ G1G2
T=
V2
=
Vx
s
R2C2
⎡ 1
⎤
1
1
⎥+
s 2 + s⎢
+
⎢⎣ R2C2 (R1 R2 )C1 ⎥⎦ R1R2C1C2
17.45
R
v1
vx
1
v2
vo
+
C1
K
C2
R
2
⎡G1Vx ⎤ ⎡s(C1 + C2 ) + G1
⎢
⎥=⎢
−sC2
⎣ 0 ⎦ ⎣
−sC 2 ⎤⎡V1 ⎤
⎥⎢ ⎥
sC2 + G2⎦⎣V2 ⎦
Vo = KV2
∆ = s 2C1C2 + s[C1G2 + C2 (G1 + G2 )]+ G1G2
T=
Vo
=K
Vx
s
1
R1C1
⎡ 1
⎤
1
1
⎥+
s2 + s⎢
+
⎢⎣ R2C2 (R1 R2 )C1 ⎥⎦ R1R2C1C2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-31
17.46
*Problem 17.46 - Fig. P17.17
VCC 4 0 DC 10
IS 5 0 DC 200U
VS 1 0 DC 0
Q1 4 3 5 NBJT
RID 1 8 40K
RO 2 3 1K
E1 2 0 1 8 316.2
R2 6 5 7.5K
R1 8 0 1K
VB 7 8 DC 0
VX 7 6 AC 0
IX 0 7 AC 1
*VX 7 6 AC 1
*IX 0 7 AC 0
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.OP
.AC LIN 1 10 10
.PRINT AC IM(VX) IP(VX) IM(VB) IP(VB) VM(7) VP(7) VM(6) VP(6)
.END
Results: I(VX) = 0.9759 A, I(VB) = 0.0241 A, V(7) = 3.078 mV, V(6) = -0.9969 V
−.9969V
0.9759A
= 324 | Ti =
= 40.5x10 4
3.078mV
0.0241A
324 (40.5) −1
T T −1
R2 1+ Tv 1+ 324
T= v i
=
= 35.8 |
=
=
= 7.83
2 + Tv + Ti 2 + 324 + 40.5
R1 1+ Ti 1+ 40.5
Tv = −
17-32
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.47
*Problem 17.47 - Fig. P17.18 BJT Op-amp
VCC 8 0 DC 12
VEE 9 0 DC -12
IS 2 9 DC 200U
VS 1 0 DC 0
Q1 4 1 2 NBJT
Q2 5 3 2 NBJT
Q3 4 4 8 PBJT
Q4 5 4 8 PBJT
Q5 8 5 6 NBJT
R 6 9 10K
VB 7 3 DC 0
VX 7 6 AC 0
IX 0 7 AC 1
*VX 7 6 AC 1
*IX 0 7 AC 0
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.MODEL PBJT PNP BF=100 VA=50 IS=1E-15
.OP
.AC LIN 1 10 10
.PRINT AC IM(VX) IP(VX) IM(VB) IP(VB) VM(7) VP(7) VM(6) VP(6)
Results: I(VX) = 1.000 A, I(VB) = 3.703 x 10-5 A, V(7) = 1.188 mV, V(6) = -1.000 V
Tv = −
−.9988V
1.000A
= 841 | Ti =
= 2.70x10 4
1.188mV
37.03µA
841(2.70x10 4 )−1
Tv Ti −1
R2 1+ Tv
1+ 841
T=
=
= 816 |
=
=
= 0.0312
4
R1 1+ Ti 2.70x10 4
2 + Tv + Ti 2 + 841+ 2.70x10
17.48
The circuit description is the same as Problem 17.46 except for the change in the values of R1
and R2.
R2 6 5 300K
R1 8 0 40K
Results: I(VX) = 0.9548 A, I(VB) = 45.19 mA, V(7) = 3.011 mV, V(6) = -0.9970 V
−.9970V
0.9548A
= 331 | Ti =
= 21.1
3.011mV
45.19mA
331(21.1) −1
R2 1+ Tv 1+ 331
T T −1
=
= 19.7 |
=
=
= 15.0
T= v i
R1 1+ Ti 1+ 21.1
2 + Tv + Ti 2 + 331+ 21.1
Tv = −
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-33
17.49
*Problem 17.49 - Fig. 17.2(b)
IS 0 1 DC 0
RS 1 0 1K
RID 1 0 15K
RO 2 3 1K
E1 3 4 1 0 5000
RL 2 0 4.7K
R2 4 0 1K
R1 4 6 36K
VB 7 1 DC 0
VX 7 6 AC 0
IX 0 7 AC 1
*VX 7 6 AC 1
*IX 0 7 AC 0
.OP
.AC LIN 1 10 10
.PRINT AC IM(VX) IP(VX) IM(VB) IP(VB) VM(7) VP(7) VM(6) VP(6)
.END
Results: I(VX) = 0.9500 A, I(VB) = 49.97 mA, V(7) = 1.271 mV, V(6) = -0.9987 V
−.9987V
0.9500A
= 786 | Ti =
= 19.0
1.271mV
49.97mA
786(19.0) −1
R2 1+ Tv 1+ 786
T T −1
=
= 18.5 |
=
=
= 39.4
T= v i
R1 1+ Ti 1+ 19.0
2 + Tv + Ti 2 + 786 + 19.0
Tv = −
17-34
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.50
*Problem 17.50
VCC 5 0 DC 6
IREF 0 1 DC 100UA
Q1 1 4 0 NBJT
Q2 4 4 0 NBJT
Q3 5 3 4 NBJT
IX 0 2 DC 0
VX1 2 1 DC 0
VX2 2 3 DC 0
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.OP
.TF I(VX1) IX
---> 0.9910
*.TF I(VX2) IX
---> 9.043 x 10-3
.END
*Problem 17.50
VCC 5 0 DC 6
IREF 0 1 DC 100UA
Q1 1 4 0 NBJT
Q2 4 4 0 NBJT
Q3 5 3 4 NBJT
IX 0 2 DC 0
VX1 2 1 DC 0
VX2 2 3 DC 0
.MODEL NBJT NPN BF=100 VA=50 IS=1E-15
.OP
.TF V(1) VX1
---> -0.9990
*.TF V(2) VX1
---> 1.012 x 10-3
.END
−0.9990
0.9910
Tv = −
= 987 | Ti =
= 110
−3
1.012x10
9.043x10−3
987(110) −1
R2 1+ 987
= 98.8 |
=
= 8.90
T=
2 + 987 + 110
R1 1+ 110
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-35
17.51
Since the output resistance of the amplifier is zero (R2 = 0), the simplified method can be used:
T = Tv.
*Problem 17.51
VS 1 0 DC 0
C1 1 2 0.005UF
C2 2 3 0.005UF
R1 2 7 2K
R2 3 0 2K
E1 4 0 3 0 1
R 4 5 31.83
C 5 0 1NF
E2 6 0 5 0 2
VX 7 6 AC 1
.OP
.AC DEC 100 1 1E6
.PROBE V(6) V(7)
.END
0 dB
T = V(6)/V(7)
-20 dB
Frequency
-40 dB
1.0 Hz
17-36
100 Hz
10 KHz
1.0 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.52
(a ) With gain A = 0, RinD = RI + Rid + R1 R2 + Ro RL
RinD
[
]
= 1kΩ + 20kΩ + 5kΩ [45kΩ + 1kΩ 5kΩ]= 25.5kΩ
With the input open - circuited, the current in Rid is zero, and so TOC = 0.
With the input set to zero, The Thevenin equivalent looking back into R1 is
⎤
⎡
⎛ 5kΩ ⎞⎤⎡
5kΩ
⎢
⎥ = 328 and
Vth = ⎢4000⎜
⎟⎥
⎝ 5kΩ + 1kΩ ⎠⎦⎢⎣ 5kΩ + 45kΩ + 5kΩ 1kΩ ⎥⎦
⎣
[
]
(
[
)
]
Rth = R1 R2 + Ro RL = 5kΩ 45kΩ + 5kΩ 1kΩ = 4.51kΩ
TSC = 328
⎛ 1+ 257 ⎞
20kΩ
= 257 | Rin = 25.5kΩ⎜
⎟ = 6.58 MΩ
4.51kΩ + 20kΩ + 1kΩ
⎝ 1+ 0 ⎠
With gain v s and A = 0,
[ (
)]
[
)]
(
RoutD = RL Ro R2 + R1 (Rid + RI ) = 5kΩ 1kΩ 45kΩ + 5kΩ (20kΩ + 1kΩ) = 819Ω
With the output shorted, TSC = 0. With the output open - circuited, TOC = 257
⎛ 1+ 0 ⎞
Rout = 819Ω⎜
⎟ = 3.17 Ω | Rin and Rout agree with both Prob. 17.15 and SPICE.
⎝1+ 257 ⎠
(b) With gain A = 0,
[
]
[
]
RinD = RI Rid R1 + R2 (Ro + RL ) = 100kΩ 20kΩ 10kΩ + 1kΩ (1kΩ + 5kΩ) = 6.57kΩ
With the input short - circuited, TSC = 0.
With the input open - circuited, TOC =
TOC = −
4000
[ (
RL + Ro + R2 R1 + RI Rid
)]
R2
(
R2 + R1 + RI Rid
⎡
1kΩ
⎢
5kΩ + 1kΩ + 1kΩ 10kΩ + 100kΩ 20kΩ ⎢⎣1kΩ + 10kΩ + 100kΩ 20kΩ
4000
[
⎛ 1+ 0 ⎞
Rin = 6.57kΩ⎜
⎟ = 18.9 Ω
⎝1+ 346 ⎠
)]
(
|
(
(R
)
I
Rid
)
⎤
⎥ 100kΩ 20kΩ = −346
⎥
⎦
)
(
)
Since the circuit is series feedback at the output,
we look into the circuit between ground and the bottom of R L . With gain A = 0,
[ (
)]
[
(
)]
RoutD = RL + Ro + R2 R1 + RI Rid = 5kΩ + 1kΩ + 1kΩ 10kΩ + 100kΩ 20kΩ = 6.96kΩ
With the output open, io = 0, and TOC = 0. With the output short - circuited, TSC is
TSC = −
4000
[ (
RL + Ro + R2 R1 + RI Rid
R2
)]R + R + (R
2
1
I
Rid
(R
)
I
)
Rid = −346
⎛ 1+ 346 ⎞
Rout = 6.96kΩ⎜
⎟ = 2.42 MΩ | Rin and Rout agree with both Prob. 17.30 and SPICE.
⎝ 1+ 0 ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-37
(b) cont.
We can instead choose to look at the shunt output. With gain A = 0,
{
[ (
RoutD = RL Ro + R2 R1 + RI Rid
)]}= 5kΩ {1kΩ + 1kΩ [10kΩ + (100kΩ 20kΩ)]}= 1.40kΩ
4000
With the output shorted, TSC = −
[ (
Ro + R2 R1 + RI Rid
)]
R2
(
R2 + R1 + RI Rid
(R
)
I
)
Rid = −1227
With the output open - circuited,
TOC = −
4000
[ (
RL + Ro + R2 R1 + RI Rid
)]
R2
(
R2 + R1 + RI Rid
(R
)
I
)
Rid = −346
⎛ 1+ 1227 ⎞
Rout = 1.40kΩ⎜
⎟ = 4.954 kΩ and removing the 5kΩ resistor yields Rout = 544kΩ.
⎝ 1+ 346 ⎠
The error is due to loss of significance in the calculations.
If we remove RL from across the output, RoutD = 1.94kΩ, TSC = −1227, and TOC = 0.
⎛ 1+ 1227 ⎞
Rout = 1.94kΩ⎜
⎟ = 2.38 MΩ | Rout again agrees with both Prob. 17.30 and SPICE.
⎝ 1+ 0 ⎠
(c) With gain A = 0,
R = R + R + R (R
inD
I
1
id
o
+ RL )= 2kΩ + 20kΩ + 5kΩ (1kΩ + 5kΩ)= 24.7kΩ
When the input is open - circuited, zero current exists in Rid and TOC = 0.
With the input short - circuited,
⎞
⎛
4000
R1
TSC =
⎟ Rid
⎜
RL + Ro + R1 (Rid + RI ) ⎝ R1 + Rid + RI ⎠
[
TSC =
]
⎞
⎛
5kΩ
⎟20kΩ = 1471
⎜
5kΩ + 1kΩ + 5kΩ (20kΩ + 2kΩ) ⎝ 5kΩ + 20kΩ + 2kΩ ⎠
4000
[
]
⎛1+ 1471⎞
Rin = 24.7kΩ⎜
⎟ = 36.3 MΩ
⎝ 1+ 0 ⎠
With gain A = 0, (remember, this is series feedback and we look into the bottom of R L )
[
]
[
]
RoutD = RL + Ro + R1 (Rid + RI ) = 5kΩ + 1kΩ + 5kΩ (20kΩ + 2kΩ) = 10.1kΩ
With the output open - circuited, TOC = 0. With the output shorted,
⎛1+ 1471⎞
TSC = 1730, the same as above, and Rout = 10.1kΩ⎜
⎟ = 14.9 MΩ
⎝ 1+ 0 ⎠
Rin and Rout agree with both Prob. 17.33 and SPICE.
17-38
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
(d) With gain A = 0,
[ (
)]
[
(
)]
RinD = RI Rid RF + RL Ro = 100kΩ 20kΩ 36kΩ + 5kΩ 1kΩ = 11.5kΩ
With the input short - circuited, TSC = 0.
When the input is open - circuited and using a Norton equivalent,
(
)
)
(
RL Ro
⎛ 4000 ⎞
TOC = ⎜
⎟
⎝ Ro ⎠ RL Ro + RF + RI Rid
(R R )
)
⎛ 4000 ⎞
(5kΩ 1kΩ)
=⎜
⎟
(100kΩ 20kΩ)= 1040
⎝ 1kΩ ⎠ (5kΩ 1kΩ)+ 36kΩ + (100kΩ 20kΩ)
(
TOC
⎛ 1+ 0 ⎞
Rin = 11.5kΩ⎜
⎟ = 11.0 Ω
⎝1+ 1040 ⎠
With gain A = 0,
[ (
)]
I
id
[
(
)]
RoutD = RL Ro RF + RI Rid = 5kΩ 1kΩ 36kΩ + 100kΩ 20kΩ = 820Ω
With the output short - circuited, TSC = 0. With the output open,
⎛ 1+ 0 ⎞
TOC = 1040, the same as above, and Rout = 820Ω⎜
⎟ = 0.788 Ω
⎝ 1+ 1040 ⎠
Rin and Rout agree with both Prob. 17.21 and SPICE.
17.53
IC1 = 500µA − IB 2 | IE 2 = IB1 +
IC1 = 500µA −
37IB1 + 700µA
101
36000IB1 + 0.7
I
= 37IB1 + 700µA | IB 2 = E 2
1000
101
= 493µA − 0.366IB1 → IC1 = 491.2µA
IC1
100
+ 700µA = 881.7µA | IC 2 =
IE 2 = 873µA | gm1 = 40(491.2µA) = 19.6mS
101
100
100(0.025)
100(0.025)
50 + 1.6
= 5.09kΩ | rπ 2 =
= 2.86kΩ | ro1 =
= 105kΩ
rπ 1 =
493x10−6
491µA
873µA
IE 2 = 37
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-39
After replacing vi and R I with their Norton equivalent, and with g m1 = 0,
⎡
⎛ r + r ⎞⎤
RinD = RI rπ 1 ⎢RF + RE RL ⎜ π 2 o1 ⎟⎥
⎝ β o 2 + 1 ⎠⎦
⎣
⎡
⎛ 2.86kΩ + 105kΩ ⎞⎤
RinD = 1kΩ 5.09kΩ ⎢36kΩ + 1kΩ 4.7kΩ ⎜
⎟⎥ = 817Ω
⎝
⎠⎦
101
⎣
With the input shorted, TSC = 0. With the input open and starting at the output of Q1,
⎫⎪⎡ R r
⎧⎪
(β o 2 + 1) RE RL (RF + RI rπ 1 )
( I π1) ⎤⎥
TOC = (−gm1ro1 )⎨
⎬⎢
⎪⎩ ro1 + rπ 2 + (β o2 + 1) RE RL (RF + RI rπ 1 ) ⎪⎭⎢⎣ RF + (RI rπ 1 )⎥⎦
⎫⎪⎡
⎧⎪
⎤
(101) 1kΩ 4.7kΩ (36kΩ + 1kΩ 5.09kΩ)
1kΩ 5.09kΩ
⎥
TOC = (−2062)⎨
⎬⎢
⎪⎩105kΩ + 2.86kΩ + (101) 1kΩ 4.7kΩ (36kΩ + 1kΩ 5.09kΩ) ⎪⎭⎢⎣ 36kΩ + (1kΩ 5.09kΩ)⎥⎦
[
]
[
[
[
]
]
]
TOC = (−2062)(0.403)(0.0227) = −18.9
⎛ 1+ 0 ⎞
Rin = 817Ω⎜
⎟ = 41.2Ω
⎝ 1+ 18.9 ⎠
r +r
2.86kΩ + 105kΩ
RoutD = RE RL (RF + RI rπ 1 ) π 2 o1 = 1kΩ 4.7kΩ (36kΩ + 1kΩ 5.09kΩ)
= 460Ω
(β o 2 + 1)
(101)
⎛ 1+ 0 ⎞
TSC = 0, TOC = 18.9, Rout = 460Ω⎜
⎟ = 23.1Ω | The results agree with SPICE.
⎝1+ 18.9 ⎠
17-40
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.54
⎡
100(0.025V )
r + Ro ⎤
With gain A = 0, RinD = Rid + R1 ⎢R2 + π
= 12.5kΩ
⎥ | rπ =
βo + 1 ⎦
200µA
⎣
⎡
12.5kΩ + 1kΩ⎤
RinD = 40kΩ + 1kΩ ⎢7.5kΩ +
⎥⎦ = 40.9kΩ
⎣
101
With the input open - circuited, the current in Rid is zero, and so TOC = 0.
With the input set to zero, the load on the emitter follower is
RLEQ = R2 + (R1 Rid ) = 7.5kΩ + (1kΩ 40kΩ) = 8.48kΩ
TSC = A
(β o + 1)RLEQ
Ro + rπ + (β o + 1)RLEQ
⎛ 0.976kΩ ⎞
101(8.48kΩ)
(R1 Rid ) = 316
⎜
⎟ = 35.8
R2 + (R1 Rid )
1kΩ + 12.5kΩ + 101(8.48kΩ) ⎝ 8.48kΩ ⎠
⎛1+ 35.8 ⎞
Rin = 40.9kΩ⎜
⎟ = 1.51 MΩ
⎝ 1+ 0 ⎠
⎛r + R ⎞
o
With gain A = 0, RoutD = R2 + (R1 Rid ) ⎜ π
⎟ = 8.48kΩ 134Ω = 132Ω
⎝ βo + 1 ⎠
With the output shorted, TSC = 0. With the output open - circuited, TOC = 32.0.
⎛ 1+ 0 ⎞
Rout = 132Ω⎜
⎟ = 3.59 Ω | Rin and Rout agree with both Prob. 18.18 and SPICE.
⎝1+ 35.8 ⎠
[
]
17.55
1/g m3
Q3
Q4
Q4
i
i
Q5
i
Q1
v ref
Q2
r o1
R
Q5
Q2
i
R || 1.33r π2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-41
RinD = 4rπ 1 = 100kΩ (See analysis below **.)
⎡ (β o 5 + 1)(R 1.33rπ 2 ) ⎤
⎥
With the input shorted, TSC = gm 2 ro 2 ro 4 (β o 5 + 1)(R 1.33rπ 2 ) Ω⎢
⎢⎣ rπ 5 + (β o 5 + 1)(R 1.33rπ 2 )⎥⎦
[
Avef =
]
(β o 5 + 1)(R 1.33rπ 2 )
(101)(10kΩ 33.3kΩ)
=
= 0.997
rπ 5 + (β o 5 + 1)(R 1.33rπ 2 ) 2.08kΩ + (101)(10kΩ 33.3kΩ)
[
]
TSC = 40(100µA) 500kΩ 500kΩ (101)(10kΩ 33.3kΩ) (0.997) = 757
With the input open, TOC ≅ −
2i
2
2
ro2 ro 4 )Avef ≅ −
ro2 ro 4 )Avef ≅ − (ro 2 ro 4 )(0.997) ≅ −0.997
(
(
1
vi
ro1
ro1 +
gm 3
⎛ 1+ 757 ⎞
Rin = 100kΩ⎜
⎟ = 38.0MΩ
⎝1+ 0.997 ⎠
RoutD = R 1.33rπ 1
rπ 5 + (ro2 ro 4 )
β o5 + 1
= 10kΩ 33.3kΩ
2.08k + (500kΩ 500kΩ)
101
= 757. With the output shorted, TSC = 0
= 1.88kΩ
With the output open, TOC
⎛ 1+ 0 ⎞
Rin = 1.88kΩ⎜
⎟ = 2.48 Ω.
⎝1+ 757 ⎠
The values of R in and R out agree well with simulation when the effect of imbalance
due to offset voltage is considered. (Try a buffered current mirror in SPICE.)
-v i/4
vi
+
-
+
rπ
v
+µf vi /2
+
ro
v
1
-
2
ro
g m v1
gm v 2
rπ
-
ve
** The input resistance to the differential pair with active load is 4rπ rather than the 2rπ that one
might expect. Because of the high gain (µf /2) to the output node, a significant current is fed back
through ro2 . At the emitter node :
⎛µ
⎞
⎛ v
⎞
gπ (v i − v e ) + gm (v i − v e ) + gm (0 − v e ) + gπ (0 − v e ) + go⎜ f v i − v e ⎟ + go ⎜− i − v e ⎟ = 0
⎝ 4
⎠
⎝2
⎠
⎛3
g ⎞
⎜ gm + gπ + o ⎟
⎛3
g ⎞
v
⎝2
2⎠ 3
≅
⎜ gm + gπ − o ⎟v i = (2gm + 2gπ + 2go )v e | e =
⎝2
4⎠
v i (2gm + 2gπ + 2go ) 4
vi
, so Rin ≅ 4rπ . If an input is applied to the
4
µ
4
right side instead of the left, the sign changes on the go f v i term, and Rin ≅ rπ .
2
3
The voltage across rπ of the input transistor is
17-42
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.56
RinD = 100kΩ 1MΩ + (10kΩ 10kΩ 40kΩ) = 91.0kΩ | TSC = 0
[
]
1+ 0
100kΩ
= 0.808 | Rin = 91.0kΩ
= 50.3 kΩ
TOC ≅ gm (10kΩ 10kΩ 40kΩ)
1+ 0.808
100kΩ + 1MΩ
RoutD = 10kΩ 10kΩ 40kΩ 1.1MΩ = 4.43kΩ | TSC = 0
1+ 0
100kΩ
= 0.808 | Rout = 4.43kΩ
= 2.45 kΩ
TOC ≅ gm (10kΩ 10kΩ 40kΩ)
1+ 0.808
100kΩ + 1MΩ
17.57
gm 4 (ro 2 ro 4 )
µf 4
RoutD = µ f 4 ro 2 | TOC = 0 | TSC = (gm 3 ro 3 )
≅ µf 3
≅ µf 3
1+ gm 4 (ro 2 ro 4 )
2 + µf 4
1+ µ f 3
= µ f 4 ro2 (µ f 3 + 1)
1+ 0
gm 4 (ro 2 ro 4 )
1
1+ 0
| TSC = 0 | TOC = µ f 3
≅ µ f 3 | Rin = ro 3
=
1+ gm 4 (ro 2 ro 4 )
1+ µ f 3 gm 3
Rout = µ f 4 ro 2
RinD = ro 3
17.58
⎡
⎡
⎤
β o (ro 2 rπ 3 ) ⎤
β o RE
βr
β2
⎥ ≅ ro 4 o π 3 = ro4 o
Using Wq. (14.28), RoutD = ro ⎢1 +
⎥ = ro 4 ⎢1 +
ro 3
µf
⎢⎣ ro 3 + rπ 4 + (ro 2 rπ 3 )⎥⎦
⎣ Rth + rπ + RE ⎦
TOC = (gm 3 ro 3 )
ro 2 rπ 3
r
≅ (gm 3 ro3 ) π 3 = β o
ro 3 + rπ 4 + ro 2 rπ 3
ro 3
(β o + 1)(ro2 rπ 3 )
TSC ≅ (gm 3 ro 3 )
ro3 + rπ 4 + (β o + 1)(ro 2
rπ 3 )
≅ µ f | Rout
β o2 1+ µ f
≅ ro4
≅ β o ro 4
µ f 1+ β o
We cannot exceed the β o ro 4 limit as long as the base current of Q 4 reaches ground!
[
]
RinD = ro 3 rπ 4 + (β o + 1)(ro 2 rπ 3 ) ≅ ro 3 β o rπ 3 ≅ ro 3 | TSC = 0
TOC = µ f 3
gm 4 (ro 2 ro 4 )
1+ gm 4 (ro 2 ro 4 )
≅ µ f 3 | Rin = ro 3
1+ 0
1
≅
1+ µ f 3 gm 3
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-43
17.59
2π x1010 s
2π x10 6 )
(
10 4 s
(a) A(s) =
=
⎛
s ⎞
s ⎞
3 ⎛
(s + 2π x10 )⎜⎝1+ 2π x106 ⎟⎠ (s + 2π x103 )⎜⎝1+ 2π x10
6⎟
⎠
A(s) represents a band - pass amplifier with two widely - spaced poles
Open − loop : Ao = 10 4 or 80 dB
(b)
| f L = 1 kHz
2π x1010 s
(s + 2π x103 )(s + 2π x106 )
Av (s) =
1+
2π x1010 s
(0.01)
(s + 2π x103 )(s + 2π x106 )
=
| f H = 1 MHz
6.28 x1010 s
s 2 + 1.01(2π x10 8 )s + 4π 2 x10 9
Using dominant - root factorization :
fH =
(c )
1.01(2π x10 8 )
2π
= 101 MHz, f L =
1 ⎛ 4 π 2 x10 9 ⎞
⎟ = 9.90 Hz
⎜
2π ⎜⎝ 1.01(2π x10 8 )⎟⎠
2π x1010 s
(s + 2π x103 )(s + 2π x106 )
Av (s) =
1+
2π x1010 s
(0.025)
(s + 2π x103 )(s + 2π x106 )
=
6.28 x1010 s
s 2 + 2.51(2π x10 8 )s + 4 π 2 x10 9
Using dominant - root factorization :
fH =
17-44
2.51(2π x10 8 )
2π
1 ⎛ 4 π 2 x10 9 ⎞
⎟ = 3.98 Hz
⎜
= 251 MHz, f L =
2π ⎜⎝ 2.51(2π x10 8 )⎟⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.60
2x1014 π 2
2π x10 3 )(2π x10 5 )
(
5x10 5
=
(a) A(s) =
⎛
s ⎞ ⎛
s ⎞
s ⎞⎛
s ⎞⎛
1+
1+
⎟
⎜1+
⎜1+
3 ⎟⎜
5⎟
3 ⎟⎜
⎝ 2π x10 ⎠⎝ 2π x10 ⎠ ⎝ 2π x10 ⎠⎝ 2π x10 5 ⎠
A(s) represents a low - pass amplifier with two widely - spaced poles
Open − loop : Ao = 5x10 5 = 114dB
| f H ≅ f1 = 1000 Hz
| fL = 0
(b)A common mistake would be the following :
Closed − loop : f H = 1000Hz[1+ 5x10 5 (0.01)]= 5MHz
Oops! - This exceeds f 2 = 100 kHz! This is a two - pole low - pass amplifier.
2x1014 π 2
(s + 2π x103 )(s + 2π x105 )
Av (s) =
1+
2x1014 π 2
(0.01)
(s + 2π x103 )(s + 2π x105 )
=
2x1014 π 2
s2 + 1.01(2π x10 5 )s + 2x1012 π 2
Using dominant - root factorization : f1 = 101 kHz, f 2 = 4.95 MHz
So the closed - loop values are f H = 101 kHz and f L = 0.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-45
17.61
4 π 2 x1018 s2
2π x10 6 )(2π x10 7 )
(
(a) A(s) =
⎛
s ⎞⎛
s ⎞
1+
⎟
(s + 200π )(s + 2000π )⎜1+
6 ⎟⎜
⎝ 2π x10 ⎠⎝ 2π x10 7 ⎠
10 5 s2
⎛
s ⎞⎛
s ⎞
1+
⎟
(s + 200π )(s + 2000π )⎜1+
6 ⎟⎜
⎝ 2π x10 ⎠⎝ 2π x10 7 ⎠
A(s) represents a band - pass amplifier with four widely - spaced poles
A(s) =
Open − loop : Ao = 10 5 or 100 dB
(b)
| f L ≅ 1 kHz | f H ≅ 1 MHz
4 π 2 x1018 s2
(s + 200π )(s + 2000π )(s + 2π x10 6 )(s + 2π x10 7 )
Av (s) =
1+
4 π 2 x1018 s2
(0.01)
(s + 200π )(s + 2000π )(s + 2π x10 6 )(s + 2π x10 7 )
Using MATLAB : D(s) = s4 + 6.9122x10 7 s3 + 3.9518x1017 s 2 + 2.7288x1018 s + 1.5585x10 21
The closed loop amplifier has complex roots:
10 8
1
−3.45
±
j62.7
=
−0.637
±
j9.98
Hz
and
(
) (
)
(−0.346 ± j6.28) = (−0.0637 ± j1) x108 Hz
2π
2π
Note that the amplifier is stable since all the poles are in the left half plane. See Problem 18.67.
(c )
2π x1010 s
(s + 2π x103 )(s + 2π x106 )
Av (s) =
1+
2π x1010 s
(0.025)
(s + 2π x103 )(s + 2π x106 )
Using MATLAB : D(s) = s4 + 6.9122x10 7 s3 + 9.909x1016 s2 + 2.7288x1018 s + 1.5585x10 21
The closed loop amplifier has complex roots:
10 8
1
−13.8
±
j124.7
=
−2.20
±
j19.9
Hz
and
(
) (
)
(−0.346 ± j3.13) = (−0.637 ± j4.98)x10 7 Hz
2π
2π
Note that the amplifier is stable since all the poles are in the left half plane.
17-46
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.62
C2
C1
1 µF
1 µF
100 k Ω
ii
ω1 =
10 k Ω
1MΩ
1M Ω
1
rad
= 0.909
10 (100kΩ + 1MΩ)
s
| ω2 =
−6
Separate, widely - spaced, poles → f LA = f 2 =
10 k Ω
1
rad
= 58.5
s
10 10kΩ + 25kΩ 10kΩ 1MΩ
−6
(
)
58.5
= 9.31 Hz
2π
CGD
3pF
+
100 k Ω
1 MΩ
C
GS
R
L
v1
-
4.15 k Ω
gmv1
10 pF
25k Ω||1M Ω||10k Ω||10k Ω
ω AH =
f HA =
1
=
rπoCT
⎡
⎛
⎢⎣
⎝
4.15kΩ ⎞⎤
⎠⎥⎦
(100kΩ 1MΩ)⎢10 pF + 3pF ⎜1+ 2mS(4.15kΩ) + 100kΩ 1MΩ ⎟⎥
1
1
= 46.1 kHz
2π (90.9kΩ)(38.0 pF )
v gs = is (100kΩ 1MΩ) = (90.9kΩ)is
A=
1
| v o = −(2x10−3 )v gs (25kΩ 10kΩ 10kΩ 1MΩ)
vo
= −(2mS )(4.15kΩ)(90.9kΩ) = −7.55x10 5 Ω | y12F = −10−5 S
is
1+ Aβ = 1+ (−7.55x10 5 Ω)(−10−6 S)= 1.76
fL =
9.31
= 5.29 Hz
1.76
fH = 46.1kHz(1.76) = 81.0 kHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-47
17.63
C2
1 µF
C2
R
i
Q
1 µF
S
RF
R
C
2kΩ
R
F
100 k Ω
R
+
L
v
5kΩ
-
i
5k Ω
o
100 k Ω
From the Exercise : gm = 40.3 mS | rπ = 3.72 kΩ | ro = 50.8 kΩ | 1 + Aβ = 2.19
100(0.025)
50 + 1.4
50 + 11.3
= 514kΩ | ro4 =
= 613kΩ | rπ 5 =
= 2.08kΩ
−4
−4
10
10
0.0012
40.3mS
Cπ 1 =
− 0.75 pF = 12.1pF
2π (500MHz)
ro2 =
Using the open - circuit time constant approach with C1 = C2 = 1 µF :
R1O = 5kΩ + 100kΩ rπ 5kΩ + 100kΩ 3.72kΩ = 8.59kΩ
R2O = 5kΩ + 50.8kΩ 2kΩ 100kΩ = 6,89kΩ
fL =
⎤
1 ⎡
1
fL
1
F
+
= 19.0 Hz
⎢
⎥ = 41.6 Hz | f L =
2π ⎣1µF (8.59kΩ) 1µF (6.89kΩ)⎦
1 + Aβ
Q
R
R
S
F
i
R
C
2k Ω
i
5kΩ
+
R
F
100 k Ω
R
L
5k Ω
100 k Ω
rπo = 3.72kΩ 100kΩ 5kΩ = 2.09kΩ | RL = 50.8kΩ 2kΩ 100kΩ 5kΩ = 1.37kΩ
⎡
1.37kΩ ⎤
CT = 12.1pF + 0.75 pf ⎢1+ 40.3mS (1.37kΩ) +
= 54.8 pF
⎣
2.09kΩ⎥⎦
1
1
=
= 1.39 MHz | f HF = f H (1+ Aβ ) = 3.04 MHz
fH =
2πrπoCT 2π (1.37kΩ)(54.8 pF )
17-48
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
v
o
-
17.64
A(s) =
2π x 10 7
s + 2000π
| A=
⎛ 2π x 10 7
25 kΩ
⎜
1kΩ + 25 kΩ + 9.01kΩ ⎝ s + 2000π
⎞ 1.96 kΩ
2.97 x 10 7
=
⎟
⎠ 1.96 kΩ + 1kΩ s + 2000π
2.97 x 10 7
2.97 x10 7
10.1
s
+
2000
π
AV (s) =
=
=
7
6
s
2.97 x 10
s + 2.95 x10 1+
1+
(0.0990)
2.95 x10 6
s + 2000π
17.65
SAωoH =
F
Ao ∂ω FH
ω FH ∂Ao
| ω FH = ω AH (1+ Aβ ) | SAωoH =
F
2.95 x10 6
| fH =
= 470 kHz
2π
Ao
Aoβ
ω AH β =
≅ +1
ω (1+ Aoβ )
(1+ Aoβ )
A
H
10 5 (0.01)
∂ω FH
ω ∂Ao
=
S
=
10% = 9.99%
A
ω FH
Ao 1+ 10 5 (0.01)
F
H
o
17.66
(a) At high frequencies with β = 0.01, Aβ =
(2 x10 π )
ω + (2πx10 )
12
Aβ =
ω
(2 x10
(2 x10 π )
(s + 2000π )(s + 2πx10 ) s(s + 2πx10 )
14
π 2 )(0.01)
12
≅
5
2
5
2
5 2
2
= 1 | Using MATLAB, ω = 4.42 x10 6
⎛ 4.42 x10 6 ⎞
∠Aβ = −90 − tan ⎜
= 171.9 o | ΦM = 8.1o
5 ⎟
⎝ 2πx10 ⎠
−1
(b) At high frequencies with β = 0.025,
(5 x10 π )
ω + (2πx10 )
12
Aβ =
ω
2
(2 x10
(5 x10 π )
Aβ =
≅
(s + 2000π )(s + 2πx10 ) s(s + 2πx10 )
14
π 2 )(0.025)
12
5
2
5
2
5 2
= 1 | Using MATLAB, ω = 7.01x10 6
⎛ 7.01x10 6 ⎞
∠Aβ = −90 − tan−1⎜
= 174.9 o | Φ M = 5.1o
5 ⎟
⎝ 2πx10 ⎠
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-49
17.67
s(2π x1010 )(0.01)
2π x10 8
≅
(a) At high frequencies with β = 0.01, Aβ =
(s + 2000π )(s + 2π x10 6 ) (s + 2πx10 6 )
Aβ =
2π x10 8
ω 2 + (2πx10
)
6 2
= 1 | ω ≅ 2π x10 8
⎛ 2π x10 8 ⎞
∠Aβ = −tan−1⎜
= −89.4 o | ΦM = 90.6 o
6 ⎟
⎝ 2πx10 ⎠
(b) At high frequencies with β = 0.025,
Aβ =
5π x10 8
ω ω 2 + (2πx10
)
6 2
=1 |
Aβ =
s(2π x1010 )(0.025)
(s + 2000π )(s + 2π x10 6 )
≅
5π x10 8
(s + 2πx106 )
ω ≅ 5π x10 8
⎛ 5π x10 8 ⎞
∠Aβ = −tan−1⎜
= −89.8 o | ΦM = 90.2 o
6 ⎟
⎝ 2π x10 ⎠
17.68
(a) At high frequencies with β = 0.01, Aβ =
Aβ ≅
Aβ =
s2 (4 π 2 x1018 )(0.01)
s 2 (s + 2π x10 6 )(s + 2π x10 7 )
≅
ω 2 + (2πx10
)
(s + 200π )(s + 2000π )(s + 2π x106 )(s + 2π x107 )
4π 2 x1016
(s + 2π x106 )(s + 2π x107 )
4π 2 x1016
6 2
s2 (4 π 2 x1018 )(0.01)
ω 2 + (2πx10
)
6 2
= 1 | Using MATLAB, ω ≅ 6.2673x10 8
8⎞
⎛ 6.2673x10 8 ⎞
⎛
−1 6.2673x10
o
o
∠Aβ = −tan−1⎜
−
tan
⎜
⎟
⎟ = −173.7 | Φ M = 6.3
6
7
⎝ 2πx10 ⎠
⎝ 2πx10 ⎠
(b) At high frequencies with β = 0.025, Aβ =
Aβ ≅
Aβ =
s 2 (4 π 2 x1018 )(0.025)
s 2 (s + 2π x10 6 )(s + 2π x10 7 )
≅
ω + (2πx10
)
6 2
(s + 2π x10 )(s + 2π x10 )
ω + (2πx10
2
(s + 200π )(s + 2000π )(s + 2π x106 )(s + 2π x107 )
π 2 x1017
6
π 2 x1017
2
s2 (4 π 2 x1018 )(0.025)
)
6 2
7
= 1 | Using MATLAB, ω ≅ 9.9246x10 8
8⎞
⎛ 9.9246x10 8 ⎞
⎛
−1 9.9246x10
o
o
∠Aβ = −tan−1⎜
−
tan
⎜
⎟
⎟ = −176.0 | ΦM = 4.0
6
7
⎝ 2πx10 ⎠
⎝ 2πx10 ⎠
17-50
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.69
(a) T (s) =
4 x 1019 π 3
β | ∠T ( jω ) = −tan−1
(s + 2π x 10 )(s + 2π x 10 )
5 2
4
f
f
− 2tan−1 5 = −180 o
4
10
10
f
= −90 o → f = 10 5 Hz. Using this as a starting point
10 5
for iteration, we find f = 110 kHz or ω = 2.2 x 10 5 π
For f >> 10 4 , − 2tan−1
4 x 1019 π 3
(b) A(j2.2x10 5 π ) =
= 2048
(2.2x10 π ) + (2π x 10 ) [(2.2x10 π ) + (2π x 10 ) ]
2
5
4 2
5
2
5 2
The amplifier will oscillate for closed - loop gains ≤ 2048 (66.2 dB).
17.70
1
sCL
⎛ 10 ⎞
⎛ 10 7 ⎞
⎛ 10 7 ⎞
1
1
T (s) = Aβ = ⎜
=
=
⎜
⎜
⎟
⎟
⎟
⎝ s + 50 ⎠ R + 1
⎝ s + 50 ⎠ sCL RO + 1 ⎝ s + 50 ⎠ 500 sCL + 1
O
sCL
7
Assume that the unity - gain occurs at ω1 >> 50 : ∠T ( jω1 ) = ∠A + ∠β = −90 o − tan−1 (500ω1CL )
−90 o − tan−1 (500ω1CL ) = −180 o + 60 o | tan−1 (500ω1CL ) = 30 o | 500ω1CL = 0.5774
T ( jω1 ) = 1 |
CL =
10 7
ω1 1+ (500ω1CL )
2
tan(30 o )
500(8.66 x10 6 )
17.71
(a) T = Aβ =
=
10 7
[
]
ω1 1+ tan(30 o )
2
= 1 → ω1 = 8.66 x 10 6
= 133 pF
⎛ 1⎞
⎜
⎟ | Yes, it is a second - order system and will
s + 2x103 π s + 2x10 4 π ⎝ 4 ⎠
4x1013 π 2
(
)(
)
have some phase margin, although Φ M may be vanishingly small.
(b) For ω >> 2π x 10 , T (jω ) ≈
4
(
)
∠T j9.935x106 = − tan −1
1x1013 π 2
ω
2
and T (jω ) = 1 for ω = 9.935x106
rad
s
6
9.935x10
9.935x106
− tan −1
= 179.6 o → Φ M = 0.4 o | A very
2000π
20000π
small phase margin.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-51
17.72
(a) T = Aβ =
⎛ 1⎞
2x1014 π 2
⎜ ⎟ | Yes, it is a second - order system and will
3
5
(s + 2x10 π )(s + 2x10 π )⎝ 5 ⎠
have some phase margin, although Φ M may be vanishingly small.
(b) For ω >> 2π x 10 5 ,
T ( jω ) ≈
∠T (j1.987x10 7 )= −tan−1
4 x1013 π 2
ω
2
and T ( jω ) = 1 for ω = 1.987x10 7
rad
s
7
1.987x10
1.987x10 7
− tan−1
= 178.2 o → ΦM = 1.83o | A very
5
2000π
2x10 π
small phase margin.
17.73
(a) The following command line will generate the complete bode plot:
w=logspace(3,8,400); bode((2e14*pi^2/5),conv([1 2000*pi],[1 2e5*pi]),w)
The following command line will generate the bode plot between 107 and 108rad/s:
w=logspace(7,8,100); bode((2e14*pi^2/5),conv([1 2000*pi],[1 2e5*pi]),w)
The second plot agrees with the results calculated in the previous problem.
20
Gain dB
10
0
-10
-20
-30
10
7
Frequency (rad/sec)
10 8
-176
Phase deg
-177
-178
-179
-180
10 7
Frequency (rad/sec)
10 8
(b) w=logspace(7,8,100); bode((2e14*pi^2),conv([1 2000*pi],[1 2e5*pi]),w)
yields a phase margin of only 0.75 degrees
17-52
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.74
C
vi
R
A(s)
+
β=
R
R+
1
sC
=
2π x 10 6 sRC
2π x 10 6
sRC
| T=
| For ωRC >> 1, T ≅
sRC + 1
s + 20π (sRC + 1)
s + 20π
and T = 1 for ω = 2π x 10 6. Given RC = 10 -8 (10 5 )= 10−3 , β =
s
s + 1000
6
2π x 10 6
−1 2π x 10
∠T = 90 − tan
− tan
= −90.0 o | ΦM = 90.0 o
20π
1000
o
−1
17.75
s10 5 (10−8 )
sRC
s
β=
=
=
=
5
−8
1
sRC + 1 s10 (10 )+ 1 s +1000
R+
sC
10 5
4 π 2 x 1013
A(s) =
=
⎛
⎞ (s + 2000π )(s + 200000π )
s
s ⎞⎛
⎜1 +
⎟⎜1 +
⎟
⎝ 2000π ⎠⎝ 200000π ⎠
R
T=
s
4 π 2 x 1013
4 π 2 x 1013
| At high frequencies, T ≅
s2
(s + 2000π )(s + 200000π ) (s + 1000)
and the integrator will have a positive phase margin, although ΦM may be very small.
For ω >> 2π x 10 5, T ( jω1) ≈
∠T = 90 o − tan−1
7
4 π 2 x 1013
ω
2
1
= 1 ⇒ ω = 1.987 x 10 7
rad
>> 2π x 10 5
s
1.987 x 10
1.987 x 10 7
1.987 x 10 7
− tan−1
− tan−1
2000π
200000π
1000
| ΦM = 1.83o
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-53
17.76
Gain dB
100
50
0
-50
101
0
102
103
104
10 5
106
102
103
104
10 5
106
Phase deg
-50
-100
-150
-200
101
Frequency (rad/sec)
(a) β =
R1
1
sCC
R1 +
1
R2 +
sCC
R2
For CC
=
R1
R1 +
R2
sCC R2 + 1
=
R1
sCC R2 + 1
R1 + R2 sCC (R1 R2 )+ 1
2x1011 π 2
1
= 0, T =
(s + 200π )(s + 20000π ) 21
The graphs above were generated using
bode(2e11*pi^2/21,[1 2.02e4*pi 4e6*pi^2])
Blowing up the last decade:
w=linspace(1e5,1e6); bode(2e11*pi^2/21,[1 2.02e4*pi 4e6*pi^2],w)
and the phase margin is approximately 12o
Gain dB
20
0
-20
Phase deg
-40
105
-140
Frequency (rad/sec)
106
Frequency (rad/sec)
106
-150
-160
-170
-180
105
Setting the zero to cancel the second pole,
17-54
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
1
R1
sCC R2 + 1
CC R2
β (s) =
=
1
R1 + R2 sCC (R1 R2 )+ 1 s +
CC (R1 R2 )
s+
T=
2x1011 π 2
2x1011 π 2
(s + 20000π ) =
(s + 200π )(s + 20000π ) s + 1.319x10 6 s2 + 1.320x10 6 s + 8.288x10 6
Using MATLAB:
bode(2e11*pi^2,[1 1.320e6 8.288e8])
and then
w=linspace(1e6,1e7); bode(2e11*pi^2,[1 1.320e6 8.288e8],w)
Gain dB
20
0
-20
-40
106
Frequency (rad/sec)
107
Frequency (rad/sec)
107
Phase deg
-120
-140
-160
-180
106
The phase margin is now approximately 50o
17.77
num=4e19*pi^3;
p=conv([1 2e5*pi],[1 2e5*pi]);
den=conv([1 2e4*pi],p];
bode(num,den)
5
Results: Frequency = 6.9 x10 rad/s and approximately 66 dB
which agree with the hand calculations in Problem 17.69.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-55
17.78
Bode Diagrams
120
100
80
60
40
20
0
-50
-100
-150
-200 3
10
10
4
10
5
10
6
10
7
Frequency (rad/sec)
Expanded view:
Bode Diagrams
70
60
50
40
30
20
-145
-150
-155
-160
-165
-170
-175
-180
10 6
10 7
Frequency (rad/sec)
For a gain of 100 (40 dB), the phase margin is approximately 8o. For a gain of 40 (32 dB), the
phase margin is approximately 5o. The gain margin is infinite in both cases since the phase shift
never reaches 180o. These values agree with the calculations in Problem 17.59. (b) The
amplifier will not oscillate.
17-56
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.79
Bode Diagrams
80
70
60
50
40
100
50
0
-50
-100 2
10
10
3
10
4
10
5
10
6
10
7
10
8
Frequency (rad/sec)
The phase shift ranges from +90o at low frequencies to –90o at high frequencies. The phase
margin for both gains of 100 and 40 is 90o at the high frequency intersection and 270o at the low
end. These values agree with the calculations in Problem 17.60. (b) The amplifier will not
oscillate.
Expanded view at high frequencies:
Bode Diagrams
80
70
60
50
40
30
20
10
0
-20
-40
-60
-80
-100
6
10
10
7
10
8
10
9
10
10
Frequency (rad/sec)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-57
17.80
Bode Diagrams
100
80
60
40
20
0
200
100
0
-100
-200
1
10
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
Frequency (rad/sec)
Bode Diagrams
80
60
40
20
0
-20
-140
-150
-160
-170
-180
8
10
10
9
10
10
Frequency (rad/sec)
The phase margin is approximately 6o and 4o for gains of 40 dB and 32 dB respectively. The
gain margin is infinite in both cases. These values agree with the calculations in Problem 17.61.
(b) The amplifier will not oscillate.
17-58
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.81
num=2e11*pi^2
den=conv([1 200*pi],[1 20000*pi]);
bode(num/100,den)
w=logspace(5,6,100); bode(num/100,den,w)
Results: Yes, the amplifier is stable with a phase margin of approximately 26o.
Bode Diagrams
60
40
20
0
-20
-40
0
-50
-100
-150
-200 2
10
10 3
10 4
10 5
10 6
Frequency (rad/sec)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-59
17.82
Aoω o
2π x 10 6
ωT
ωT
s + ωo
Av (s) =
=
≅
=
Aω
s + (1+ Ao )ω o s + ωT s + 2π x 10 6
1+ o o
s + ωo
5x10 4 s
s2 + 7x10 4 s + 5x10 8
1011 πs
T (s) = Av (s)β (s) =
(s + 2π x 106 )(s2 + 7x104 s + 5x108 )
From problem 18.45 : β (s) =
Using MATLAB:
num=[pi*1e11 0]
den=conv([1 3e5 1e10],[1 5e6])
bode(num,den)
It is also instructive to use: nyquist(num,den)
One finds that |T(jω)| < 1 for all ω, so the phase margin is undefined. The filter is stable. Note
that this is a positive feedback system so the point of interest is +1. The gain of the filter is
approximately -3 dB. So the filter has a gain margin of 3 dB.
Bode Diagrams
0
-20
-40
-60
-80
-100
100
50
0
-50
-100
-150
-200 2
10
10 3
10 4
10 5
10 6
10 7
10 8
Frequency (rad/sec)
17-60
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.83
⎛ 10 7
⎞⎛
⎞
10 5 s
1012 s
=
T (s) = K (s)β (s) = ⎜
6 ⎟⎜ 2
5
10 ⎟
6
2
5
10
⎝ s + 5 x 10 ⎠⎝ s + 3x10 s + 10 ⎠ (s + 5 x 10 )(s + 3x10 s + 10 )
Using MATLAB:
num=[1e12 0]
den=conv([1 3e5 1e10],[1 5e6])
bode(num,den)
It is also instructive to use: nyquist(num,den)
One finds that |T(jω)| < 1 for all ω, so the phase margin is undefined. The filter is stable. Note
that this is a positive feedback system so the point of interest is +1. The gain of the filter is
approximately -3 dB. So the filter has a gain margin of 3 dB.
Bode Diagrams
0
-20
-40
-60
-80
-100
100
50
0
-50
-100
-150
-200 2
10
10 3
10 4
10 5
10 6
10 7
10 8
Frequency (rad/sec)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-61
17.84
sRC
s
2π x 10 6 s
=
| T=
sRC + 1 s + 1000
(s + 20π )(s + 1000)
Using MATLAB:
num=[2e6*pi 0]; den=conv([1 20*pi],[1 1000]);
w=logspace(0,8,400);
bode(num,den,w)
The phase margin is 90o which agrees with Problem 17.74.
β=
Bode Diagrams
80
60
40
20
0
-20
-40
100
50
0
-50
-100 0
10
10 2
10 4
10 6
10 8
Frequency (rad/sec)
17-62
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.85
β=
sRC
s
10 5
4 π 2 x 1013
=
| A(s) =
=
⎞ (s + 2000π )(s + 200000π )
⎛
s
s ⎞⎛
sRC + 1 s +1000
⎟⎜1 +
⎟
⎜1 +
⎝ 2000π ⎠⎝ 200000π ⎠
4 π 2 x 1013 s
T=
(s + 2000π )(s + 200000π )(s + 1000)
Using MATLAB:
num=[4e13*pi^2 0];
den=conv([1 1000],conv([1 2000*pi],[1 200000*pi]));
w=logspace(7,8,100); bode(num,den,w)
The phase margin is 1.8o which agrees with Problem 17.75.
Bode Diagrams
20
10
0
-10
-20
-30
-176
-177
-178
-179
-180 7
10
10
8
Frequency (rad/sec)
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-63
17.86
R1
R1
1
1
sCS R1 + 1
β (s) =
=
=
−6
R1
+ R2 R1 + R2 sCS (R1 R2 )+ 1 9.30(1.89x10 s + 1)
sCS R1 + 1
10 7
T (s) = Av (s)β (s)
s + 50
5.69x1011
For ω >> 50, T ( jω ) ≅
→ T ( jω ) = 1 for ω = 6.68x10 5
2
ω ω 2 + (5.29x10 5 )
β (s) =
5.69x10 4
s + 5.29x10 5
ΦM = 180 − tan−1
Av (s) =
6.68x10 5
6.68x10 5
− tan−1
= 38.4 o
5
50
5.29x10
17.87
2(0.001)(125x10−6 )
gm1
=
= 10.6 MHz |
(a) fT =
2πCC
2π (7.5x10−12 )
Since I 2 > I1, the slew rate
I1 250x10−6
V
V
is approximately symmetrical. | SR =
=
= 33.3x10 6 = 33.3
−12
s
CC 7.5x10
µs
2(0.001)(250x10−6 )
gm1
=
= 11.3 MHz | Since I 2 > I1, the slew rate
(b) fT =
2πCC
2π (10−11 )
is asymmetrical. | SR+ =
I1 500µA
V
250µA
V
=
= 50
| SR− =
= 25
10 pF
CC 10 pF
µs
µs
17.88
2(0.001)(250x10−6 )
gm1
fT =
=
= 11.3 MHz | Since I 2 > I1, the slew rate
2πCC
2π (10−11 )
is symmetrical. | SR =
17-64
V
I1 500x10−6
V
=
= 50x10 6 = 50
−11
CC
10
µs
s
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.89
*Problem 17.89 - CMOS Op-amp
VDD 8 0 DC 10
VSS 9 0 -10
I1 1 9 250U
I2 6 9 500U
I3 7 9 2M
V1 4 0 DC -2.23M AC 0.5
V2 2 0 AC -0.5
M1 3 2 1 1 NFET W=20U L=1U
M2 5 4 1 1 NFET W=20U L=1U
M3 3 3 8 8 PFET W=40U L=1U
M4 5 3 8 8 PFET W=40U L=1U
M5 6 5 8 8 PFET W=160U L=1U
M6 8 6 7 7 NFET W=60U L=1U
CC 5 6 7.5PF
*CC 5 10 7.5PF
*RZ 10 6 1K
.MODEL NFET NMOS KP=2.5E-5 VTO=0.70 GAMMA=0.5
+LAMBDA=0.05 TOX=20N
+CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10
.MODEL PFET PMOS KP=1.0E-5 VTO=-0.70 GAMMA=0.75
+LAMBDA=0.05 TOX=20N
+CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10
.OP
.TF V(7) V1
.AC DEC 100 1 20MEG
.PRINT AC VM(7) VP(7)
.PROBE
.END
Results: 8.1 MHz, -110 degrees; 8.0 MHz, -92 degrees
17.90
gm
40IC1
=
2πCC 2πCC
I1
2
40(25µA)
= 13.3 MHz
2π (12 pF )
(a) fT
=
SR =
I1 25µA
MV
V
=
= 2.09
= 2.09
since I2 > I1. The slew rate is symmetrical.
s
CC 12 pF
µs
(b) fT
=
| IC1 =
| fT =
40(100µA)
I
100µA
MV
V
= 53.1 MHz | SR = 1 =
= 8.33
= 8.33
s
2π (12 pF )
CC 12 pF
µs
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-65
17.91
40(250µA)
I1
| fT =
= 159 MHz
2
2π (10 pF )
fT =
gm
40IC1
=
2πCC 2πCC
SR =
I1 500µA
MV
V
=
= 50
= 50
since I2 > I1 . The slew rate is symmetrical.
s
CC 10 pF
µs
17.92
SR =
| IC1 =
V
I1 40µA
V
=
= 8x10 6 = 8
s
CC 5 pF
µs
*Problems 17.92 - Bipolar Op-amp
VCC 8 0 DC 10
VEE 9 0 -10
I1 1 9 40U
I2 6 9 400U
I3 7 9 500U
V1 4 0 DC 0 PWL (0 0 5U 0 5.1U 5 10U 5 10.2U -5 15U -5 15.2U 5 20U 5)
VF 2 7 DC -0.0045
Q1 3 2 1 NBJT
Q2 5 4 1 NBJT
Q3 3 10 8 PBJT
Q4 5 10 8 PBJT
Q11 0 3 10 PBJT
Q5 6 5 8 PBJT
Q6 8 6 7 NBJT
CC 5 6 5PF
.MODEL NBJT NPN BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF
.MODEL PBJT PNP BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF
.OP
.TRAN .05U 20U
.PROBE V(4) V(5) V(6) V(7)
.END
Results: -8V/µs, +6V/µs
17.93
SR =
V
I1 100µA
V
=
= 12.5x10 6 = 12.5
CC
8 pF
µs
s
17-66
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.94
fT =
gm
40IC1
=
2πCC 2πCC
| IC 1 =
I1
2
| fT =
40(50µA)
= 21.2 MHz
2π (15 pF )
*Problem 17.94 - Bipolar Op-amp
VCC 8 0 DC 10
VEE 9 0 -10
I1 1 9 100U
I2 6 9 500U
I3 7 9 500U
V1 4 0 DC 2.10M AC 0.5
V2 2 0 DC 0 AC -0.5
Q1 3 2 1 NBJT
Q2 5 4 1 NBJT
Q3 3 10 8 PBJT
Q4 5 10 8 PBJT
Q11 0 3 10 PBJT
Q5 6 5 8 PBJT
Q6 8 6 7 NBJT
CC 5 6 15PF
.MODEL NBJT NPN BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF
.MODEL PBJT PNP BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF
.OP
.TF V(7) V1
.AC DEC 100 1 20MEG
.PRINT AC VM(7) VP(7)
.PROBE
.END
Spice Results: (a) 16.2MHz (b) 16.3 MHz - 15 pF does not represent the effective value of
CC.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-67
17.95
(a) The SPICE results agree with those given in Ex. 11.10: Ao = 67.2 dB, dB, fH = 4.5 kHz, fT =
10.5 MHz, and with a phase margin of 74o
100
-0
-100
-200
-300
1.0Hz
DB(V(R3:2))
10Hz
P(V(R3:2))
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
100MHz
Frequency
(b) Note, λ should be 0.02V-1 for the MOSFETs.
(b) The SPICE results give Ao = 86.5 dB, fT = 8.5 MHz and with a phase margin of 68.4o. SR+ =
17.6 V/µsec, and SR- = -15.6 V/µsec.
12V
8V
4V
0V
-4V
-8V
-12V
0s
1us
2us
3us
4us
5us
6us
7us
V(M3:d)
Time
17-68
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
8us
9us
10us
17.96
⎛ 2R ⎞
Vo1
1
=−
Vo 2 = ⎜1+
⎟V = 2V+
⎝ 2R ⎠ +
sRC
Vo 2
G
(V+ − Vo1) + sCV+ + (V+ − Vo2 )GF = 0 Combining these yields
2
V
G
1
and T (s) = Av1 Av 2 =
Av 2 = o 2 =
⎞
⎛G
⎛
Vo1
1 R⎞
sC + ⎜ − GF ⎟
sRC⎜ sRC + − ⎟
⎠
⎝2
2 RF ⎠
⎝
Av1 =
∠T ( jω o ) = 0 → RF = 2R and T ( jω o ) = 1 → ω o =
1
RC
17.97
Define V1 as the output of the inverting amplifier and V2 as the output of the right - hand
non - inverting amplifier.
⎛
⎛
⎞
⎞
⎞
⎛
⎜
⎜
⎟
R ⎟⎛ R2 ⎞
R
Z
R
sCR
R
V1 = −V2 2 = −V2
= −V2
| V2 = V1⎜
1+ 2 ⎟⎜
1+
⎜
⎟
1 ⎟⎜⎝ R1 ⎟⎠
1
1 ⎝ R1 ⎠
Z1
SCR + 1
⎜R+
⎜R+
⎟
⎟
R+
⎝
⎝
sC ⎠
sC
sC ⎠
2
2
⎡ ⎛
⎛ jωCR ⎞ 3 ⎛ R2 ⎞
⎞ 3 ⎛ R2 ⎞ ⎤
sCR
V2⎢1+ ⎜
⎟ ⎜1+ ⎟ ⎥ = 0 | ⎜
⎟ ⎜1+ ⎟ = −1
⎝ jωCR + 1⎠ ⎝ R1 ⎠
⎢⎣ ⎝ SCR + 1⎠ ⎝ R1 ⎠ ⎥⎦
3[90 o − tan−1 (ωCR)]= 180 o
|
tan−1 (ωCR) = 30 o
| ωCR = tan(30 o )= 0.5774
3
⎤
2⎡
⎞3 ⎛
o
⎛ R2 ⎞ 2⎛
R2 ⎞ ⎢ tan(30 ) ⎥
R
ωCR
⎜
⎟
= ⎜1+ ⎟
= 1 → 2 = 8 −1 = 1.83
⎜1+ ⎟
2
2
o
⎥
⎢
⎜
⎟
R1
⎝ R1 ⎠ ⎝ 1+ (ωCR) ⎠ ⎝ R1 ⎠ 1+ tan (30 )
⎦
⎣
17.98
10 k Ω
10 k Ω
15 k Ω
6.2 k Ω
fo =
1
2π (5kΩ)(500 pF )
= 63.7 kHz |
vO =
3(0.7V )
= 6.85 V
⎛ 15kΩ ⎞⎛ 10kΩ ⎞ 10kΩ
⎜2 −
⎟⎜1+
⎟−
⎝ 10kΩ ⎠⎝ 6.2kΩ ⎠ 10kΩ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-69
17.99
*Problem 17.99 - Wien-Bridge Oscillator
C1 1 0 500PF IC=1
RA 1 0 5K
C2 1 2 500PF
RB 2 3 5K
E1 3 0 1 6 1E6
R1 6 0 10K
R2 5 6 15K
R3 3 5 6.2K
R4 4 5 10K
D1 3 4 DMOD
D2 4 3 DMOD
.MODEL DMOD D
.TRAN 10U 10M UIC
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
Results: f = 60.0 kHz, amplitude = 6.8 V
17.100
- +
v
47 k Ω
i
68 k Ω
0.7 V
B
15 k Ω
i
v
O
+
Using Eq. (17.135), f o =
1
( )
2π 3 (5000) 10−9
= 18.4 kHz
Using Eq. (17.136), the total feedback resistance should be R1 = 12R = 60kΩ.
The current in R1 is I =
VB = I (47kΩ) =
VO
V
V
= O = O . The voltage at VB is
R1 12R 60kΩ
47kΩ
V
V − VB VO − 0.7 − VB
VO | In the diode network, I = O = O
+
60kΩ
15kΩ
68kΩ
60kΩ
13
13
VO
VO
60 + 60 − VO = 0.7 → V = 10.7 V
O
15kΩ 68kΩ 60kΩ 68kΩ
17-70
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.101
*Problem 17.101 - Phase Shift Oscillator
C1 1 6 1000PF IC=1
RA 1 0 5K
C2 1 2 1000PF
RB 2 0 5K
C3 2 3 1000PF
E1 3 0 0 6 1E6
R2 6 5 47K
R3 5 3 15K
R4 5 4 68K
D1 3 4 DMOD
D2 4 3 DMOD
.MODEL DMOD D
.TRAN 10U 20M UIC
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
Results: f = 17.5 kHz, amplitude = 11.5 V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-71
17.102
Note that the presence of rπ makes the analysis more complex than the FET case. C4 is a
coupling capacitor, and its impedance is neglected in the analysis. C5 = C1 + Cπ. However, the
effect of rπ can usually be neglected in the fo calculation as shown below.
⎡
1
⎢s(C5 + Cµ )+ gπ + sL
⎢ − sC + g + g
⎣ ( 5
m
π)
⎤⎡ ⎤ ⎡ ⎤
⎥⎢Vb ⎥ = ⎢0⎥
s(C2 + C5 ) + gm + gπ + GE ⎥⎦⎣Ve ⎦ ⎣0⎦
−(sC5 + gπ )
∆ (s) = s2 [C5C2 + Cµ (C2 + C5 )]+ s[C2 gπ + Cµ (gm + gπ + GE ) + C5GE ]
+
gm + gπ + GE
(C + C5 )
+ gπ GE + 2
sL
L
vb
+
g m (v b -v e )
v
C
1
Cπ
-
rπ
L
ve
Cµ
RE
C2
∆ ( jω o ) = 0 | ω o2 =
⎤
⎤
1 ⎡1
1 ⎡1
1
gm
CC
⎢ +
⎥=
⎢ +
⎥ | CTC = Cµ + 2 5
CTC ⎣ L rπ RE (C2 + C5 )⎦ CTC ⎣ L β o RE (C2 + C5 )⎦
C 2 + C5
gm + gπ + GE
ω oL
⎤
⎤
⎡
⎡
⎥
⎥
⎢
⎢
C2
C5
C2
C5
⎥ = 1 | ω o2 L⎢Cµ +
⎥ =1
ω o2 L⎢Cµ +
+
+
rπ
RE ⎥
βo
gm RE ⎥
⎢
⎢
β o + 1+
1+ gm RE +
β o + 1+
1+ gm RE +
⎢⎣
⎢⎣
RE
rπ ⎥⎦
gm RE
β o ⎥⎦
⎤
100 pF (20 pF )
1 ⎡ 1
10mS
= 16.7 pF | ω o2 =
+
⎢
⎥
(a) CTC =
100 pF + 20 pF
16.7 pF ⎣5µH 100(1kΩ)(100 pF + 20 pF )⎦
1
ω o2 =
[2x105 + 833]→ f o = 17.5 MHz
16.7 pF
1
1
Note that the correction term is negligible : ω o ≅
(b) CTC = 1 1 1
LCTC
+
+
C 2 C5 C3
ω o [C2 gπ + Cµ (gm + gπ + GE ) + C5GE ]=
17-72
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
1
1
1
= 3.85 pF | f o ≅
=
= 36.3 MHz
1
1
1
2
π
LC
π
5
µ
H
3.85
pF
2
(
)
(
)
TC
+
+
100 pF 20 pF 5 pF
1
1
max
CTC
=
= 12.5 pF | f o ≅
= 20.1 MHz
1
1
1
π
5
µ
H
12.5
pF
2
(
)
(
)
+
+
100 pF 20 pF 50 pF
⎤
⎤
⎡
⎡
⎥ 1 ⎢
⎥
⎢
C2
C5
C2
C5
⎥≅
⎥=1
⎢
+
+
(c ) ω o2 L⎢
βo
gm RE ⎥ CTC ⎢
gm RE ⎥
⎢β + 1+ β o
1+
g
R
+
β
+
1+
1+
g
R
+
m E
m E
⎢⎣ o
⎢⎣ o
gm RE
β o ⎥⎦
gm RE
β o ⎥⎦
⎤
⎡
⎥
⎢
1 ⎢ 100 pF
20 pF
⎥ = 1 | MATLAB yields gm = 0.211 mS
+
gm (1kΩ) ⎥
16.7 pF ⎢101+ 100
1+ gm (1kΩ) +
⎢⎣
gm (1kΩ)
100 ⎥⎦
min
=
CTC
IC = (0.211mS)(0.025V ) = 5.28 µA
17.103
Assuming the effect of rπ is negligible:
(a) f o ≅
CEQ =
(b) Cπ
fo ≅
1
2π
1
CEQ L
| CEQ =
1
1
+
C4 C +
µ
| Cπ =
1
1
1
1
+
C1 + Cπ C2
1
1
+
0.01uF 3pF +
=
1
2π
40(5mA)
− 3pF = 60.7 pF
10 9 π
= 47.4 pF | f o ≅
1
1
1
1
+
(20 + 60.7)pF 100 pF
40(10mA)
− 3pF = 124 pF | CEQ =
10 9 π
1
2π
1
= 5.17MHz
47.4 pF (20µH )
1
1
+
0.01uF 3pF +
= 61.6 pF
1
1
1
1
+
20 pF + 124 pF 100 pF
1
= 4.53MHz
61.6 pF (20µH )
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-73
17.104
CTC =
1
1
1
=
= 21.1pF | CTC =
2
2
1
1
ω o L 4x10 7 π (3µH )
+
C1 C2
(
)
| gmR ≥
C1
2I D R
C
→
≥ 1
VGS − VTN C2
C2
2
2
Kn
1.25mA
VGS − VTN ) =
VGS + 4) | − 4 ≤ VGS Suppose we pick VGS = -2 V to provide
(
(
2
2
2
2V
= 800Ω → 820 Ω
good value of g m : VGS = −2V → I D = 0.625mA(−2 + 4) = 2.50mA | R =
2.5mA
2(2.5mA)820Ω
2I D R
C1
≤
=
= 2.05 | C1 ≤ 2.05C2 | Select C1 ≅ C2 ≅ 42 pF | Choosing
C2 VGS − VTN
−2 − (−4)
ID =
C1 = 47 pF from Appendix C, C2 =
If 47pF and 39pF are used : f o =
1
1
1
−
21.1pF 47 pF
= 38.3 pF which is close to 39 pF.
1
2π
(21.3 pF )(3µH )
= 19.9 MHz
In order to obtain an exact frequency of oscillation, a 33-pF capacitor in parallel with a small
variable capacitor could be used. Note that including the FET capacitances would modify the
design values.
17.105
L
C
CGD
C2
C3
C1
L
C
GS
CGS
C
1
1
= 4 pF +
= 31.27 pF
1
1
1
1
+
+
C2 C1 + C3 + CGS
50 pF 50 pF + 0 + 10 pF
1
1
fo =
=
= 9.00MHz
−5
2π LCTC 2π (10 H )(31.27x10−12 F )
CTC = CGD +
gm ro ≥
17-74
2
CGD
C1 + C3 + CGS 50 pF + 0 + 10 pF
=
= 1.20 which is easily met.
50 pF
C2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
1
C
3
17.106
(a) CTC = CGD +
fo =
1
1
1
+
C2 C1 + C3 + CGS
1
=
2π LCTC 2π
max
| CTC
= 4 pF +
1
(10µH )(32.3pF )
1
= 32.3pF
1
1
+
50 pF 50 pF + 5 pF + 10 pF
= 8.87 MHz
1
1
= 38.4 pF | f o =
= 8.12 MHz
1
1
2π (10µH )(38.4 pF )
+
50 pF 50 pF + 50 pF + 10 pF
C +C +C
50 pF + 5 pF + 10 pF
= 1.30 and
(b) gm ro ≥ 1 3 GS | gm ro ≥
C2
50 pF
min
CTC
= 4 pF +
gm ro ≥
50 pF + 50 pF + 10 pF
= 2.20 | ∴ gm ro ≥ 2.20 which is easily met.
50 pF
17.107
(a) CD =
C jo
1+
CD =
f omin
VTUNE
| CD =
φj
20 pF
1
= 10.7 pF | CTC =
= 40.0 pF
1
1
2V
+
1+
75 pF + 10.7 pF 75 pF
0.8V
20 pF
1
= 3.92 pF | CTC =
= 38.5 pF
1
1
20V
+
1+
75 pF + 3.92 pF 75 pF
0.8V
1
1
=
= 7.96 MHz | f omax =
= 8.11 MHz
2π (10µH )(40.0 pF )
2π (10µH )(38.5 pF )
(b) In this circuit, R = ro : µ f
= gm ro ≥
C1 + CD
C2
| µf ≥
78.9 pF
= 1.05
75 pF
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-75
17.108
CTC =
1
1
1
+
470 pF 220 pF
The required g m R ≥
= 150 pF | f o =
1
2π
(10µH )(150 pF )
= 4.11 MHz
C2 220 pF
=
= 0.468 is met :
C1 470 pF
2
2
Kn
1.25mA
VGS − VTN ) =
VGS + 4) and VGS = −820I D → VGS = −2.016V , I D = 2.46mA
(
(
2
2
2(2.46mA)(820Ω)
2I D R
≅
= 1.02
gm R =
VGS − VTN
−2.02 − (−4)
This analysis is borne out by the SPICE simulation below.
VDD 3 0 DC 10
R 1 0 820
C1 1 0 470PF IC=2
C2 2 1 220PF IC=0
*C1 1 0 220PF IC=2
*C2 2 1 470PF IC=0
L 2 0 10UH
M1 3 2 1 NFET
.MODEL NFET NMOS VTO=-4 KP=1.25MA
.OP
.TRAN 10N 30U UIC
.PROBE
.END
C 470 pF
(b) For this case, the required g m R ≥ C2 = 220 pF = 2.14 is not met,
1
ID =
and the circuit fails to oscillate.
SPICE simulation confirms that the circuit does not oscillate.
17-76
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.109
CTC = 3pF +
1
= 30.3pF | f o =
1
1
2π
+
50 pF + 10 pF 50 pF
1
(10µH )(30.3pF )
= 9.15 MHz
*Problem 17.109 NMOS Colpitts Oscillator
VDD 3 0 DC 12
LRFC 3 2 20MH
C1 1 0 50PF
C2 2 0 50PF
L 2 1 10UH
M1 2 1 0 0 NFET
CGS 1 0 10PF
CGD 1 2 4PF
.MODEL NFET NMOS VTO=1 KP=10MA LAMBDA=0.02
.OP
.TRAN 50N 40U UIC
.PROBE
.END
Results: f = 7.5 MHz, amplitude = 80 V peak-peak. There is little to set the amplitude in this
circuit, and the frequency of oscillation is significantly in error. Also, µf of the transistor greatly
exceeds the gain required for oscillation and the waveform at the drain is highly nonlinear. The
voltage at the gate is filtered by the LC network and is more sinusoidal in character. A diode
from ground to gate could be employed to help limit the amplitude of the oscillation.
17.110
fo =
2π
1
(10µH + 10µH )(20 pF )
= 7.96 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-77
17.111
1
1
20 pF
| CTC =
| C = 220 pF | CD =
| L = L1 + L2 = 20µH
1
1
2π LCTC
V
+
1+ TUNE
C CD
0.8V
20 pF
1
(a) CD =
= 10.7 pF | CTC =
= 10.2 pF
1
1
2V
+
1+
220 pF 10.7 pF
0.8V
fo =
fo =
1
2π 20µH (10.2 pF )
= 11.1 MHz
20 pF
1
= 3.92 pF | CTC =
= 3.85 pF
1
1
20V
+
1+
220PF 3.92 pF
0.8V
1
L
fo =
= 18.1 MHz (b) µ f ≥ 1 = 1.00
L2
2π 20µH (3.85 pF )
CD =
17.112
ωS =
1
RQ
| L=
ωS
LCS
(a) L =
40(25000)
1
1
= 15.915 mH | CS = 2 =
= 15.916 fF
2
7
7
2x10 π
ω S L (2x10 π ) 15.915mH
1
1
= 15.890 fF | f P =
= 10.008 MHz
1
1
π
15.915mH
15.890
fF
2
(
)
+
15.915 fF 10 pF
1
1
(c) CP =
= 15.907 fF | f P =
= 10.003 MHz
1
1
2π 15.915mH (15.907 fF )
+
15.915 fF 32 pF
(b) CP =
17-78
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17.113
C2
15 mH
5pF
C1
20fF
(a) CTC =
fP =
40(2.14mA)
5 − 0.7
= 2.14mA | Cπ =
− 5 pF = 49.5 pF
100kΩ + 101(1kΩ)
2π (2.5x10 8 Hz)
1
1
+
20 fF 5 pF +
1
1
+
100 pF 470 pF + 49.5 pF
2π 15mH (19.996 fF )
max
CTC
=
= 19.996 fF
1
1
17.114
max
CTC
100pF
1
1
= 19.995 fF | f P =
= 9.190 MHz
1
1
1
2π 15mH (19.995 fF )
+
+
20 fF 470 pF 100 pF
(b) IC = 100
CTC =
Cπ
470pF
Cµ
1
= 9.190 MHz
= 19.60 fF | f P =
1
= 9.28 MHz
1
1
1
1
2
π
15mH
19.60
fF
(
)
+
+
+
20 fF 1pF 100 pF 470 pF
1
1
=
= 19.98 fF | f P =
= 9.19 MHz
1
1
1
1
2
π
15mH
19.98
fF
(
)
+
+
+
20 fF 35 pF 100 pF 470 pF
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
17-79
17.115
*Problem 17.115 BJT Colpitts Crystal Oscillator
VCC 1 0 DC 5
VEE 4 0 DC -5
Q1 1 2 3 NBJT
RE 3 4 1K
RB 2 0 100K
C1 3 0 100PF
C2 2 3 470PF
LC 2 6 15M
CC 6 5 20FF IC=5
RC 5 0 50
.MODEL NBJT NPN BF=100 VA=50 TF=1N CJC=5PF
.OP
.TRAN 2N 20U UIC
.PROBE
.END
In the period of time used in the simulation results, node 6 at the interior of the crystal oscillates
vigorously, but the oscillation is not coupled well to the other nodes.
17-80
©Richard C. Jaeger and Travis N. Blalock - 3/10/07
Microelectronic Circuit Design
Fourth Edition - Part I
Solutions to Exercises
CHAPTER 1
Page 11
5.12V
5.12V
VLSB = 10
=
= 5.00 mV
2 bits 1024bits
11000100012 = 29 + 28 + 2 4 + 20 = 78510
or
€
€
(
5.12V
= 2.560V
2
VO = 785(5.00mV ) = 3.925 V
VMSB =
)
VO = 2−1 + 2−2 + 2−6 + 2−10 5.12V = 3.925 V
Page 12
5.0V
5.00V
1.2V
VLSB = 8
=
= 19.53 mV
N=
256bits = 61.44bits
5.00V
2 bits 256bits
61 = 32 + 16 + 8 + 4 + 1 = 25 + 2 4 + 23 + 22 + 20 = 001111012
Page 12
The dc component is VA = 4V.
The signal consists of the remaining portion of vA: va = (5 sin 2000πt + 3 cos 1000 πt) Volts.
Page 23
vo = −5cos 2000πt + 25o = − −5sin 2000πt + 25o − 90 o = 5sin 2000πt − 65o
) [
(
Vo = 5∠ − 65
€
Page 25
R
Av = − 2
R1
o
|
Vi = 0.001∠0
−5=−
o
)]
(
(
)
5∠ − 65o
Av =
= 5000∠ − 65o
o
0.001∠0
R2
→ R2 = 50 kΩ
10kΩ
€
1
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 26
vs = 0.5sin (2000πt ) + sin (4000πt ) + 1.5sin (6000πt )
[
]
The three spectral component frequencies are f1 = 1000 Hz
f2 = 2000 Hz
f3 = 3000 Hz
(a ) The gain of the band - pass filter is zero at both f and f . At f , V = 10(1V ) = 10V ,
and v = 10.0sin (4000πt ) volts.
(b) The gain of the low - pass filter is zero at both f and f . At f , V = 6(0.5V ) = 3V ,
and v = 3.00sin (2000πt ) volts.
1
3
2
o
o
2
3
2
o
o
Page 27
€
39kΩ(1− 0.1) ≤ R ≤ 39kΩ(1+ 0.1)
or
3.6kΩ(1− 0.01) ≤ R ≤ 3.6kΩ(1+ 0.01)
€
Page 29
VI2
P=
R1 + R2
P
max
=
P nom =
(1.1x15)
35.1 kΩ ≤ R ≤ 42.9 kΩ
or
152
= 4.17 mW
54kΩ
2
0.95x54kΩ
3.56 kΩ ≤ R ≤ 3.64 kΩ
= 5.31 mW
P
min
=
(0.9x15)
2
1.05x54kΩ
= 3.21 mW
Page 33
€
 10−3

R = 10kΩ 1+ o (−55 − 25) oC = 9.20 kΩ
C


 10−3

R = 10kΩ 1+ o (85 − 25) oC = 10.6 kΩ
C


€
2
©R. C. Jaeger and T. N. Blalock
08/08/10
CHAPTER 2
Page 47
ni =
(2.31x10
30
K cm
−3
−6
)


−0.66eV

(300K ) exp 8.62x10−5 eV / K 300K  = 2.27x1013 cm3
(
)

3
(
)
Page 47
€
ni =
(1.08x10
31
ni =
(1.08x10
31
K cm
−6
K cm
−6
−3
−3
)


−1.12eV

(50K ) exp 8.62x10−5 eV / K 50K  = 4.34x10−39 cm3
( )

)


−1.12eV

(325K ) exp 8.62x10−5 eV / K 325K  = 4.01x1010 cm3
( )

3
(
)
3
(
)
3
 −2 m 
cm 3
10
L =
10
 → L = 6.13x10 m
−39 
cm 
4.34x10 
3
Page 49
€
v p = µ p E = 500
E=
€
cm2  V 
3 cm
10  = 5.00x10
V − s  cm 
s
vn = −µn E = −1350
cm2 
V 
6 cm
1000  = −1.35x10
V − s
cm 
s
V
1
V
V
=
= 5.00x103
−4
L 2x10 cm
cm
Page 49
(a) From Fig. 2.5 : The drift velocity for Ge saturates at 6x106 cm / sec.
At low fields the slope is constant. Choose E = 100 V/cm
v
v
4.3x105 cm / s
cm2
2.1x105 cm / s
cm2
µn = n =
= 4300
µp = p =
= 2100
E
100V / cm
s
E
100V / cm
s
7
(b) The velocity peaks at 2x10 cm / sec
µn =
vn 8.5x105 cm / s
cm2
=
= 8500
E
100V / cm
s
€
3
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 51


3
−1.12
 = 5.40x1024 / cm 6 | ni = 2.32x1012 / cm3
ni2 = 1.08x1031 (400) exp
−5
8.62x10
400

( )
ρ=
1
1
=
= 1450 Ω − cm
−19
12
σ 1.60x10
2.32x10 (1350) + 2.32x1012 (500)
[(
)
(
]
)


3
−1.12
 = 1.88x10−77 / cm6 | ni = 4.34x10−39 / cm3
ni2 = 1.08x1031 (50) exp
−5
 8.62x10 (50) 
ρ=
1
1
=
= 1.69x1053 Ω − cm
σ 1.60x10−19 4.34x10−39 (6500) + 4.34x10−39 (2000)
[(
)
(
)
]
Page 55
€


3
−1.12
 = 5.40x1024 / cm 6
ni2 = 1.08x1031 (400) exp
−5
 8.62x10 (400) 
p = N A − N D = 1016 − 2x1015 = 8x1015
holes
cm3
n=
ni2 5.40x1024
electrons
=
= 6.75x108
15
p
8x10
cm3
−−−
Antimony (Sb) is a Column - V element, so it is a donor impurity.
p=
ni2
1020
holes
=
= 5.00x103
16
n 2x10
cm3
n = N D = 2x1016
electrons
cm 3
n > p → n - type silicon
Page 56
€
Reading from the graph for NT = 1016/cm3, 1230 cm2/V-s and 405 cm2/V-s.
Reading from the graph for NT = 1017/cm3, 800 cm2/V-s and 230 cm2/V-s.
4
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 58
(
1
= 0.001 Ω − cm
σ
)
σ = 1000 = 1.60x10−19 µn n → un n = 6.25x1021 / cm3 = 6.25x1019 (100) | ρ =
−−−
(a) N D = 2x1016 / cm 3
µn = 92 +
| N A = 0 / cm3
1270
0.91
 2x10 
1+ 

17
1.3x10 
16
| N T = 2x1016 / cm 3
= 1170 cm 2 /V − s | µ p = 48 +
|
447
0.76
 2x10 
1+ 

16
 6.3x10 
16
= 363 cm2 /V − s
(b) N T = N D + N A = 2x1016 / cm 3 + 3x1016 / cm3 = 5x1016 / cm3
µn = 92 +
€
1270
0.91
 5x1016 
1+ 

17
1.3x10 
= 990 cm 2 /V − s | µ p = 48 +
447
0.76
 5x1016 
1+ 

16
 6.3x10 
= 291 cm2 /V − s
Page 59
Boron (B) is a Column - III element, so it is an acceptor impurity.
p = N A − N D = 4x1018
holes
cm3
| n=
ni2
1020
electrons
=
= 25
18
p 4x10
cm3
| p - type material
4x1018
and mobilities from Fig. 2.8 : µn = 150 cm 2 /V − s | µp = 70 cm2 /V − s
3
cm
1
1
ρ≅
=
= 0.022 Ω − cm
qµ p p 1.60x10−19 4x1018 (70)
NT =
(
)
--Indium (In) is a Column - III element, so it is an acceptor impurity.
holes
p = N A − N D = 7x10
cm3
19
ni2
1020
electrons
| n=
=
= 1.43
| p - type material
19
p 7x10
cm3
7x1019
and mobilities from Fig. 2.8 : µn = 100 cm 2 /V − s | µp = 50 cm2 /V − s
3
cm
1
1
ρ= =
= 1.79 mΩ − cm
−19
σ 1.60x10 7x1019 (50)
NT =
(
)
€
5
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 60
−23
1.38x10−23 (300)
kT 1.38x10 (50)
VT =
=
= 4.31 mV | VT =
= 25.8 mV
q
1.602x10−19
1.602x10−19
VT =
1.38x10−23 (400)
1.602x10−19
= 34.5mV
−−−
Dn =

kT
cm2 
cm 2
µn = 25.8mV 1362
 = 35.1
q
V − s
s

| Dp =

kT
cm2 
cm2
µ p = 25.8mV  495
 = 12.8
q
V − s
s

−−−
jn = qDn
 cm2  1016  10 4 µm 
dn
A
= 1.60x10−19 C 20
 3

 = 320 2
dx
s  cm − µm  cm 
cm

j p = −qD p
 cm2  1016  10 4 µm 
dp
A
= −1.60x10−19 C 4
 3

 = −64 2
dx
cm
 s  cm − µm  cm 
€
6
©R. C. Jaeger and T. N. Blalock
08/08/10
CHAPTER 3
Page 79
 2x1018 1020
N N 
A D
φ j = VT ln 2  = 0.025V ln

1020
 ni 

( )  = 1.05 V


(
)
2(11.7) 8.85x10−14  1
2εs  1
1 
1 
wdo =
+
+
1.05) = 2.63x10−6 cm = 0.0263 µm

φ j =

−19
18
20 (
q  N A ND 
1.60x10
10 
 2x10
€
Page 80
qN A x p
1 0
Emax = ∫ −qN A dx =
εs −x p
εs
Emax =
Emax
1
=−
εs
(
)(
11.7(8.854x10 F / cm)
1.6x10−19 C 1017 / cm3 1.13x10−5 cm
−14
xn
∫ qN
D
dx =
0
qN D x n
εs
For the values in Ex.3.2 :
) = 175 kV
cm
−−−
2(1.05V )
kV
cm
2.63x10 cm
−1
 2x1018 
x p = 0.0263µm1+
 = 0.0258 µm
1020 

Emax =
−6
= 798
−1

1020 
−4
xn = 0.0263µm1+
 = 5.16x10 µm
18
 2x10 
Page 83
1.381x10 (300)
kT
300
n
=1
= 0.0259 | T =
= 291 K
−19
q
1.03
1.602x10
−23
€
Page 85
€
  0.55  
  0.70  
−15
iD = 40x10−15 Aexp
 −1 = 143 µA iD = 40x10 Aexp
 −1 = 57.9 mA
  0.025 
  0.025  

6x10−3 A 
VD = (0.025V ) ln1+
 = 0.643 V
−15
 40x10 A 
−−−
  −0.04  
  −2.0  
−15
iD = 5x10−15 Aexp
 −1 = −3.99 fA | iD = 5x10 Aexp
 −1 = −5.00 fA
  0.025  
  0.025  
€
7
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 87
 40x10−6 A 
 I 
D
a
V
=
V
ln
1+
=
0.025V
ln
1+
 = 0.593 V
( ) BE T  I 
−15
 2x10 A 
S
 400x10−6 A 
 I 
VBE = VT ln1+ D  = 0.025V ln1+
 = 0.651 V ΔVBE = 57.6 mV
2x10−15 A 
 IS 

 40x10−6 A 
 I 
(b) VBE = VT ln1+ ID  = 0.0258V ln1+ 2x10−15 A  = 0.612 V


S
−6
 400x10 A 
 I 
VBE = VT ln1+ D  = 0.0258V ln1+
 = 0.671 V ΔVBE = 59.4 mV
2x10−15 A 
 IS 

€
Page 89
i 
ln D  = ln (iD ) − ln ( I S ) | Assume iD is constant, and EG = EGO
 IS 
 E 
dvD k  iD  kT 1 dI S v d
1 dI S
= ln  −
= − VT
| I S = Kni2 = KBT 3 exp − GO 
dT q  I S  q I S dT T
I S dT
 kT 
 E 
 E  E 
dI S
= 3KBT 2 exp− GO  + KBT 3 exp − GO  + GO2 
dT
 kT 
 kT  kT 
1 dI S 3 EGO 3 qVGO 3 VGO
dvD v d
1 dI S v d − VGO − 3VT
= + 2= +
= +
|
= − VT
=
2
I S dT T kT
T kT
T VT T
dT
T
I S dT
T
Page 90
€
wd = 0.113µm 1+
2(V + φ j )
2(10.979V )
10V
kV
= 0.378 µm | Emax =
=
= 581
−4
0.979V
wd
cm
0.378x10 cm
−−−
I S = 10 fA 1+
€
10V
= 36.7 fA
0.8V
Page 93
11.7 8.854x10−14 F / cm
nF
nF
C jo =
= 91.7 2 | C j(0V ) = 91.7 2 10−2 cm 1.25x10−2 cm = 11.5 pF
−4
0.113x10 cm
cm
cm
11.5 pF
C j (5V ) =
= 4.64 pF
5V
1+
0.979V
−−−
(
)
(
)(
)
10−5 A −8
8x10−4 A −8
5x10−2 A −8
CD =
10 s = 4.00 pF | CD =
10 s = 320 pF | CD =
10 s = 0.02 µF
0.025V
0.025V
0.025V
8
€
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 98
5V
= 1 mA; I D = 0, VD = 5V
5kΩ
The intersection of the two curves occurs at the Q - pt : (0.88 mA, 0.6 V)
Two points on the load line : VD = 0, I D =
Page 100
€
10 = 10 4 I D + VD
 I 
 I 
| VD = VT ln1+ D  | 10 = 10 4 I D + 0.025ln1+ D 
 IS 
 IS 
Page 101
fzero(@(id) (10-10000*id-0.025*log(1+id/1e-13), 5e-4)
€
ans = 9.4258e-004
Page 102
fzero(@(id) (10-10000*(1e-14)*(exp(40*vd)-1)-vd), 0.5)
ans = 0.6316
Page 109
From the answer, the diodes are on,on,off.
10V - VB VB − 0.6V − (−20V ) VB − 0.6V − (−10V )
=
+
= 0 → VB = 1.87 V
2.5kΩ
10kΩ
10kΩ
1.87 − 0.6 − (−20V )
1.87 − 0.6 − (−10V )
I D1 =
= 2.13 mA | I D2 =
= 1.13 mA | VD3 = −(1.87 − 0.6) = −1.27 V
10kΩ
10kΩ
I D1 > 0, I D2 > 0, VD3 < 0. These results are consistent with the assumptions.
I1 = I D1 + I D2
€
€
Page 111
5kΩ
1kΩ
Rmin =
= 1.67 kΩ | VO = 20V
= 3.33 V (VZ is off) | VO = 5 V (VZ is conducting)
20
5kΩ + 1kΩ
−1
5
Page 112
VL − 20V VL − 5V
V
6.25V − 5V
+
+ L = 0 → VL = 6.25 V | I Z =
= 12.5mA
1kΩ
0.1kΩ 5kΩ
0.1kΩ
2
PZ = 5V (12.5mA) + 100Ω(12.5mA) = 78.1 mW
−−−
Line Regulation =
€
0.1kΩ
mV
= 19.6
0.1kΩ + 5kΩ
V
Load Regulation = 0.1kΩ 5kΩ = 98.0 Ω
9
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 118
Vdc = VP − Von = 6.3 2 −1 = 7.91 V
ΔT =
I dc =
7.91V
= 15.8 A
0.5Ω
 0.527V 
1
V
1
2 r =
2
 = 0.912 ms
ω VP 2π (60)  8.91V 
Vr =
I dc
15.8 A 1
T=
s = 0.527 V
C
0.5F 60
θ c = 120π (0.912ms)
180 o
= 19.7 o
π
−−−
Vdc = VP − Von = 10 2 −1 = 13.1 V
ΔT =
I dc =
13.1V
= 6.57 A
2Ω
 0.1V 
1
V
1
2 r =
2
 = 0.316 ms
ω VP 2π (60) 14.1V 
C=
I dc
6.57 A 1
T=
s = 1.10 F
Vr
0.1F 60
θ c = 120π (0.316ms)
180 o
= 6.82 o
π
Page 120
€
−23
kT  I D  1.38x10 J / K (300K )  48.6 A 
At 300K : VD =
ln1+  =
ln1+ −15  = 0.994 V
q  IS 
1.60x10−19 C
 10 A 
−23
kT  I D  1.38x10 J / K (273K + 50K )  48.6 A 
At 50 C : VD =
ln1+  =
ln1+ −15  = 1.07 V
q  IS 
1.60x10−19 C
 10 A 
 48.6 A 
kT  I D 
Note : For VT = 0.025V , VD =
ln1+  = 0.025V ln1+ −15  = 0.961 V
q  IS 
 10 A 
o
Page 127
€
Vrms =
Vdc + Von
2
=
15 + 1
2
= 11.3 V | C =
I dc
2A
1
T=
s = 0.222F = 222,000 µF
Vr
60
0.01(15V )
I SC = ωCVP = 2π (60Hz )(0.222F )(16V ) = 1340 A
ΔT =
(0.01)(15) = 0.363 ms | I = I 2T = 2 A 2s
1
V
1
2 r =
2
= 184 A
P
dc
ω VP 2π (60)
16
ΔT
60(0.363ms)
€
10
©R. C. Jaeger and T. N. Blalock
08/08/10
CHAPTER 4
Page 153
(
)
14
εox
cm2 3.9 8.854x10 F / cm
C
µA
K = µn
= 500
= 69.1x10−6 2
= 69.1 2
−7
Tox
V −s
25x10 cm
V −s
V
--W
µA  20µm 
µA
µA  60µm 
µA
Kn = Kn'
= 50 2 
 = 1000 2 Kn = 50 2 
 = 1000 2
L
V  1µm 
V
V  3µm 
V
µA  10µm 
µA
Kn = 50 2 
 = 2000 2
V  0.25µm 
V
--For VGS = 0V and 1V , VGS < VTN and the transistor is off. Therefore I D = 0.
'
n
VGS - VTN = 2 -1.5 = 0.5V and VDS = 0.1 V → Triode region operation
W
V 
µA  10µm 
0.1
2
I D = Kn' VGS − VTN − DS VDS = 25 2 
 2 −1.5 − 0.1V = 11.3 µA
L
2 
2
V  1µm 
VGS - VTN = 3 -1.5 = 1.5V and VDS = 0.1 V → Triode region operation
W
V 
µA  10µm 
0.1
2
I D = Kn' VGS − VTN − DS VDS = 25 2 
 3 −1.5 − 0.1V = 36.3 µA
L
2 
2
V  1µm 
Kn' = 25
µA  100µm 
µA
 = 250 2
2 
V  1µm 
V
Page 154
€
1
1
= 4.00 kΩ | Ron =
= 1.00 kΩ
µA
µA
' W
Kn (VGS − VTN ) 250 2 (2 −1)
250 2 (4 −1)
L
V
V
1
1
VGS = VTN +
= 1V +
= 3.00 V
µA
Kn Ron
250 2 (2000Ω)
V
Ron =
1
=
€
11
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 157
VGS - VTN = 5 -1 = 4 V and VDS = 10 V → Saturation region operation
2
2
Kn
1 mA
VGS − VTN ) =
5 −1) V 2 = 8.00 mA
(
2 (
2
2V
W
W 1mA 25
Kn = Kn'
→ =
=
W = 25L = 8.75 µm
L
L 40µA 1
ID =
−−−
Assuming saturation region operation,
2
2
Kn
1 mA
VGS − VTN ) =
2.5 −1) V 2 = 1.13 mA
(
2 (
2
2V
mA
g m = Kn (VGS − VTN ) = 1 2 (2.5 −1)V = 1.50 mS
V
2I D 2(1.125mA)
Checking : g m =
= 1.50 mS
VGS − VTN (2.5 −1)V
ID =
€
Page 158
VGS - VTN = 5 -1 = 4 V and VDS = 10 V → Saturation region operation
2
2
Kn
1 mA
VGS − VTN ) (1+ λVDS ) =
5 −1) V 2 1+ 0.02(10) = 9.60 mA
(
2 (
2
2V
2
2
K
1 mA
I D = n (VGS − VTN ) (1+ λVDS ) =
5 −1) V 2 1+ 0(10) = 8.00 mA
2 (
2
2V
--VGS - VTN = 4 -1 = 3 V and VDS = 5 V → Saturation region operation
[
ID =
[
]
]
2
2 2
Kn
25 µA
VGS − VTN ) (1+ λVDS ) =
4
−1
V 1+ 0.01(5) = 118 µA
(
(
)
2
2 V2
2
2
K
25 µA
I D = n (VGS − VTN ) (1+ λVDS ) =
5 −1) V 2 1+ 0.01(10) = 220 µA
2 (
2
2 V
[
ID =
[
]
]
Page 159
€
Assuming pinchoff region operation, I D =
VGS = VTN +
2
2
Kn
50 µA
0 − VTN ) =
+2V ) = 100 µA
(
2 (
2
2 V
2(100 µA)
2I D
= 2V +
= 4.00 V
Kn
50µA /V 2
−−−
Assuming pinchoff region operation, I D =
€
2
2
Kn
50 µA
VGS − VTN ) =
1−
−2V
= 225 µA
(
(
)
2
2 V2
[
12
]
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 161
( V + 0.6V − 0.6V ) = 1+ 0.75(
= 1+ 0.75( 1.5 + 0.6 − 0.6 ) = 1.51 V | V
VTN = VTO + γ
VTN
)
0 + 0.6 − 0.6 = 1 V
SB
TN
(
)
= 1+ 0.75 3 + 0.6 − 0.6 = 1.84 V
---
(a ) V
is less than the threshold voltage, so the transistor is cut off and ID = 0.
( b) V
- VTN = 2 -1 = 1 V and VDS = 0.5 V → Triode region operation
GS
GS

V 
mA 
0.5 
2
I D = KnVGS − VTN − DS VDS = 1 2  2 −1−
0.5V = 375 µA
2 
2 
V 

(c) VGS - VTN = 2 -1 = 1 V and VDS = 2 V → Saturation region operation
2
2
Kn
mA
VGS − VTN ) (1+ λVDS ) = 0.5 2 (2 −1) V 2 1+ 0.02(2) = 520 µA
(
2
V
[
ID =
€
]
Page 163
(a ) VGS is greater than the threshold voltage, so the transistor is cut off and I D = 0.
( b) V
GS
- VTN = -2 + 1 = 1 V and VDS = 0.5 V → Triode region operation

V 
mA 
−0.5 
2
I D = KnVGS − VTN − DS VDS = 0.4 2  −2 + 1−
(−0.5)V = 150 µA
2
2
V




(c) V
GS
ID =
- VTN = −2 + 1 = 1 V and VDS = 2 V → Saturation region operation
2
2
Kn
0.4 mA
VGS − VTN ) (1+ λVDS ) =
−2 + 1) V 2 1+ 0.02(2) = 208 µA
(
2 (
2
2 V
[
]
€
13
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 167

µF 
CGC = 200 2  5x10−6 m 0.5x10−6 m = 0.500 fF
m 

(
)(
)

CGC
pF 
−6
+ CGSOW = 0.25 fF +  300
 5x10 m = 1.75 fF
2
m 


2
pF 
−6
Saturation region : CGS = CGC + CGSOW = 0.333 fF +  300
 5x10 m = 1.83 fF
3
m 


pF 
−6
CGD = CGSOW =  300
 5x10 m = 1.50 fF
m

Triode region : CGD = CGS =
(
€
(
)
(
)
)
Page 169
KP = Kn = 150U | LAMBDA = λ = 0.0133 | VTO = VTN = 1 | PHI = 2φ F = 0.6
W = W = 1.5U | L = L = 0.25U
Page 170
€
 1µm 2
Circuits/cm ∝ α = 
 = 16
 0.25µm 
2
2
Power - Delay Product ∝
€
1
1
1
= 3=
→ 64 times improvement
3
64
α
4
Page 171
ε W /α 
v DS 
*
*
iD* = µn ox
 vGS − VTN −
v DS = α iD | P = VDD iD = VDD (α iD ) = α P
Tox / α L / α 
2 
P*
αP
P
=
= α3
*
A
A (W / α )( L / α )
−−−
(a ) fT =
1 500cm2 /V − s
(1V ) = 7.96 GHz |
2
−4
2π
10 cm
(
)
(b) fT =
1 500cm2 /V − s
(1V ) = 127 GHz
2π 0.25x10−4 cm 2
(
)
−−−

 5 V  −5
V
V
V  −4
≥ 105
→ L = 105
 10 cm = 10 V | L = 10
 10 cm = 1 V
L
cm
 cm 
 cm 
(
)
(
)
€
14
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 172
(a ) From the graph for VGS = 0.25 V , I D ≅ 10−18 A |
(b) From the graph for V = V − 0.5 V , I ≅ 3x10
(c) I = (10 transistors)(3x10 A/transistors) = 3 µA
GS
9
TN
D
−15
A
-15
Page 176
2
€
Active area = 10Λ(12Λ) = 120Λ2 = 120(0.125µm ) = 1.88 µm 2
L = 2Λ = 0.250 µm | W = 10Λ = 1.25 µm
2
Gate area = 2Λ(10Λ) = 20Λ2 = 20(0.125µm) = 0.313 µm 2
2
Transistor area = (10Λ + 2Λ + 2Λ)(12Λ + 2Λ + 2Λ) = 224Λ2 = 224(0.125µm ) = 3.50 µm 2
(10 µm)
N=
4
3.50µm
€
2
2
= 28.6 x 106 transistors
Page 180
Assume saturation region operation and λ = 0. Then I D is indpendent of VDS , and I D = 50 µA.
VDS = VDD − I D RD = 10 − 50kΩ(50µA) = 7.50 V. VDS ≥ VGS − VTN , so our assumption was correct.
Q - Point = (50.0 µA, 7.50 V )
--270kΩ
10V = 2.647 V | REQ = 270kΩ 750kΩ | Assume saturation region.
270kΩ + 750kΩ
2 2
25x10−6 A
ID =
2.647
−1
V = 33.9 µA | VDS = VDD − I D RD = 10 −100kΩ(33.9µA) = 6.61 V.
(
)
2
V2
VDS ≥ VGS − VTN , so our assumption was correct.
Q - Point = (33.9 µA, 6.61 V )
VEQ =
--2
30x10−6 A
3 −1) V 2 = 60.0 µA
2 (
2
V
= VDD − I D RD = 10 −100kΩ(60.0µA) = 4.00 V. VDS ≥ VGS − VTN , so our assumption was correct.
VGS does not change : VGS = 3.00 V | I D =
VDS
Q - Point = (60.0 µA, 4.00 V )
€
15
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 181
2
25x10−6 A
3 −1.5) V 2 = 28.1 µA
2 (
2
V
= VDD − I D RD = 10 −100kΩ(28.1µA) = 7.19 V. VDS ≥ VGS − VTN , so our assumption was correct.
VGS does not change : VGS = 3.00 V | I D =
VDS
Q - Point = (28.1 µA, 7.19 V )
--VDS = 10 −
VDS =
( ) (3 −1) (1+ 0.01V ) → V
2
25x10−6 105
10 − 5
V = 4.76 V
1.05
2
DS
ID =
DS
= 10 − 5(1+ 0.01VDS )
2
25x10−6
3 −1) 1+ 0.01(4.76) = 52.4 µA
(
2
[
]
Page 182
€
10V
= 150µA.
66.7kΩ
= 3 - V curve at I D = 50 µA, VDS = 6.7 V.
For I D = 0, VDS = 10 V. For VDS = 0, I D =
The load line intersects the VGS
€
16
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 185 Upper group
 2

2V
V + VGS 
− 2VTN  + VTN2 − EQ = 0
Kn RS
 Kn RS

2
GS
2
 1

 1

2V
| VGS = −
− VTN  ± 
− VTN  − VTN2 + EQ
Kn RS
 Kn RS

 Kn RS

2
 1  2VTN
2V
1
1
VGS = VTN −
± 
+ VTN2 − VTN2 + EQ = VTN +
 −
Kn RS
Kn RS
Kn RS
 Kn RS  Kn RS
( 1+ 2K R (V
n
S
EQ
)
− VTN ) −1
−−−
Assume saturation region operation.
ID =
1
2Kn RS2
(
)
2
1+ 2Kn RS (VEQ − VTN ) −1 =
1
2(30µA)(39kΩ)
2
(
)
2
1+ 2(30µA)(39kΩ)(4 −1) −1 = 36.8 µA
VDS = 10 −114kΩ(36.8µA) = 5.81 V | Saturation region is correct. | Q - Point : (36.8 µA, 5.81 V )
−−−
Assume saturation region operation. I D =
1
2(25µA)(39kΩ)
2
(
)
2
1+ 2(25µA)(39kΩ)(4 −1.5) −1 = 26.7 µA
VDS = 10 −114kΩ(26.7µA) = 6.96 V | Saturation region is correct. | Q - Point : (26.7 µA, 6.96 V )
Assume saturation region operation. I D =
1
2(25µA)(62kΩ)
2
(
)
2
1+ 2(25µA)(62kΩ)(4 −1) −1 = 36.8 µA
VDS = 10 −137kΩ(25.4µA) = 6.52 V | Saturation region is correct. | Q - Point : (25.4 µA, 6.52 V )
€
17
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 185 Lower group
R1
1MΩ
VEQ =
VDD =
10V = 4 V | Assume saturation region operation.
R1 + R2
1MΩ + 1.5MΩ
ID =
1
2Kn RS2
(
)
2
1+ 2Kn RS (VEQ − VTN ) −1 =
1
2(25µA)(1.8kΩ)
2
(
)
2
1+ 2(25µA)(1.8kΩ)(4 −1) −1 = 99.5 µA
VDS = 10 − 40.8kΩ(99.5µA) = 5.94 V | Saturation region is correct. | Q - Point : (99.5 µA, 5.94 V )
VEQ =
ID =
R1
1.5MΩ
VDD =
10V = 6 V | Assume saturation region operation.
R1 + R2
1.5MΩ + 1MΩ
1
2(25µA)(22kΩ)
2
(
)
2
1+ 2(25µA)(22kΩ)(6 −1) −1 = 99.2 µA
VDS = 10 − 40kΩ(99.2µA) = 6.03 V | Saturation region is correct. | Q - Point : (99.2 µA, 6.03 V )
−−−
I Bias =
VDD
V
10V
→ R1 + R2 = DD =
= 5 MΩ
R1 + R2
I Bias 2µA
VEQ =
V
R1
4V
VDD → R1 = ( R1 + R2 ) EQ = 5MΩ
= 2 MΩ
R1 + R2
VDD
10V
R2 = 5MΩ − R1 = 3 MΩ | REQ = R1 R2 = 1.2 MΩ
Page 187
€
(
VGS = 6 − 22000I D VSB = 22000I D VTN = 1+ 0.75 VSB + 0.6 − 0.6
)
ID =
2
25µA
VGS − VTN )
(
2
Spreadsheet iteration yields ID = 83.2 µA.
Page 183
€
[
(
)
]
Equation 4.55 becomes 6 - 1 + 0.75 VSB + 0.6 − 0.6 + 2.83 − VSB = 0.
VSB2 − 6.065VSB + 7.231 = 0 → VSB = 1.63V.
RS =
1.63V
= 16.3 kΩ →16 kΩ
10−4 A
RD =
(10 − 6 −1.63)V = 23.7kΩ → 24 kΩ
10−4 A
€
18
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 189
Assume saturation region operation.
10 − 6 =
(
25x10−6 62x10 4
2
) (V
GS
2
+ 1) − VGS → VGS2 + 0.710VGS − 4.161 = 0 → VGS = −2.426V , + 1.716V
2
25x10
−2.426 + 1) = 25.4 µA | VDS = − 10 −137kΩ(25.4µA) = −6.52 V
(
2
Saturation region is correct. | Q - Point : (25.4 µA, - 6.52 V )
−6
ID =
€
[
]
Page 195
(a) VDS = 3 V , VGS − VP = −2 − (−3.5) = +1.5 V. The transistor is pinched off.
2
  −2V 
 V 2
I D = I DSS 1− GS  = 1mA 1− 
 = 184 µA | Pinchoff requires VDS ≥ VGS − VP = +1.5 V
 VP 
  −3.5V 
(b)
VDS = 6 V , VGS − VP = −1− (−3.5) = +2.5 V. The transistor is pinched off.
2
2
  −1V 
 VGS 
I D = I DSS 1−
 = 1mA 1− 
 = 510 µA | Pinchoff requires VDS ≥ VGS − VP = +2.5 V
 VP 
  −3.5V 
(c)
VDS = 0.5 V , VGS − VP = −2 − (−3.5) = +1.5 V. The transistor is in the triode region.
ID =
2(1mA) 
2I DSS 
VDS 
0.5
V
−
V
−
V
=
−2
+
3.5
−



0.5 = 51.0 µA
GS
P
DS
2
2 
2
VP2 


−3.5
( )
Pinchoff requires VDS ≥ VGS − VP = +1.5 V
−−−
(a )
VDS = 0.5 V , VGS − VP = −2 − (−4) = +2 V. The transistor is in the triode region.
ID =
(b)
2(0.2mA) 
2I DSS 
VDS 
0.5
V
−
V
−
V
=
−2
+
4
−



0.5 = 21.9 µA
GS
P
DS
2
2 
2 
VP2 
(−4) 
VDS = 6 V , VGS − VP = −1− (−4) = +3 V. The transistor is pinched off.
2
  −1V 
 V 2
I D = I DSS 1− GS  = 0.2mA 1− 
 = 113 µA
 VP 
  −4V 
€
19
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 197
(a) VDS = −3 V , VGS − VP = 3 − 4 = −1 V. The transistor is pinched off.
 VGS 2
 3V 2
I D = I DSS 1−
 = 2.5mA 1−  = 156 µA | Pinchoff requires VDS ≤ VGS − VP = −1 V
 4V 
 VP 
( b)
VDS = −6 V , VGS − VP = 1− (4) = −3 V. The transistor is pinched off.
 VGS 2
 1V 2
I D = I DSS 1−
 = 2.5mA 1−  = 1.41 mA | Pinchoff requires VDS ≤ VGS − VP = −3 V
 4V 
 VP 
(c)
VDS = −0.5 V , VGS − VP = 2 − (4) = −2 V. The transistor is in the triode region.
ID =
2(2.5mA) 
2I DSS 
VDS 
−0.5 
V
−
V
−
V
=
2
−
4
−



(−0.5) = 273 µA
GS
P
DS
2 
2 
VP2 
42

Pinchoff requires VDS ≤ VGS − VP = −2 V
Page 198
€
BETA =
I DSS 2.5mA
=
= 0.625 mA | VTO = VP = −2 V | LAMBDA = λ = 0.025 V -1
2
2
VP
(−2)
−−−
BETA =
I DSS 5mA
= 2 = 1.25 mA | VTO = VP = 2 V | LAMBDA = λ = 0.02 V -1
2
VP
2
€
20
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 200
VTO = VP = −5 V | BETA =
I DSS 5mA
=
= 0.2 mA | LAMBDA = λ = 0.02 V -1
VP2 (−5)2
−−−
VGS = VG − VS = −IG RG − I S RS = 0 − I D RS = −I D RS
2
2
V

 V 
2
K
K
K
VGS = − n (VGS − VTN ) RS = − n VTN2 RS  GS −1 = −I DSS RS 1− GS  for I DSS = n VTN2 and VP = VTN
2
2
2
 VTN 
 VP 
2
2
 V 
 V 
2
0.4mA
VGS = −
−5) (1kΩ)1− GS  = 51+ GS  → VGS2 −15VGS + 25 = 0 and the rest is identical.
(
2
5 
 −5 

--Assuming pinchoff, I D = 1.91 mA and VDS = 9 −1.91mA(2kΩ + 1kΩ) = 3.27 V.
VGS − VP = 3.09 V , VDS > VGS − VP , and pinchoff is correct.
--2
 V 
 VGS 2
−3
3
2
GS
Assuming pinchoff, VGS = −I DSS RS 1−
 = − 5x10 2x10 1+
 → VGS + 12.5VGS + 25 = 0
5 

 VP 
(
)(
)
2
2
 V 
 −2.5 
GS
VGS = −2.5 V | I D = I DSS 1−
 = 5mA1−
 = 1.25 mA
−5 

 VP 
VDS = 12 −1.25mA(2kΩ + 2kΩ) = 7.00 V.
VGS − VP = −2.5 − (−5) = +2.5 V , VDS > VGS − VP , and pinchoff is correct. Q - Point : (1.25 mA, 7.00 V )
---
(a ) V
G
= −IG RG = −10nA(680kΩ) = −6.80 mV.
2
2
 VGS 
 VGS 
−3
3
2
VGS = VG − VS = −IG RG − I DSS RS 1−
 = −.00680 − 5x10 10 1+
 → VGS −15VGS + 25 = 0
V
5



P 
(
)( )
The value of VG is insignificant with respect to the constant term of 25. So the answers are
the same to 3 significant digits.
(b) V
G
= −IG RG = −1µA(680kΩ) = 0.680 V and now cannot be neglected.
2
 V 2
 VGS 
−3
3
2
GS
VGS = VG − VS = −IG RG − I DSS RS 1−
 = −0.680 − 5x10 10 1+
 → VGS −15VGS + 28.4 = 0
V
5



P 
(
)( )
2
 V 
 −2.226 2
GS
VGS = −2.226 V | I D = I DSS 1−
 = 5mA1−
 = 1.54 mA
−5 

 VP 
VDS = 12 −1.54mA(2kΩ + 1kΩ) = 7.38 V | Q - Point : (1.54 mA, 7.38 V )
€
21
©R. C. Jaeger and T. N. Blalock
08/09/10
CHAPTER 5
Page 223
(a )
βF =
αF
0.970
0.993
0.250
=
= 32.3 | β F =
= 142 | β F =
= 0.333
1− α F 1− 0.970
1− 0.993
1− .250
(b)
αF =
βF
40
200
3
=
= 0.976 | α F =
= 0.995 | α F = = 0.750
β F + 1 41
201
4
Page 225
€
  0.700 
 −9.30  10−15 A   −9.30  
iC = 10−15 Aexp
−
exp
exp


 −
 −1 = 1.45 mA
0.5   0.025  
 0.025 
  0.025 
  0.700 
 −9.30  10−15 A   0.700  
iE = 10−15 Aexp
−
exp
exp


 +
 −1 = 1.46 mA
 0.025  100   0.025  
  0.025 
10−15 A   0.700   10−15 A   −9.30  
iB =
exp
exp
 −1 +
 −1 = 14.5 µA
100   0.025  
0.5   0.025  
Page 227
€
  0.750 
 0.700  10−16 A   0.700  
iC = 10−16 Aexp
exp
 − exp
 −
 −1 = 563 µA
0.4   0.025  
 0.025 
  0.025 
  0.750 
 0.700  10−16 A   0.750  
iE = 10−16 Aexp
−
exp
exp


 +
 −1 = 938 µA
75   0.025  
 0.025 
  0.025 
iB =
10−16 A   0.750   10−16 A   0.700  
exp
exp
 −1 +
 −1 = 376 µA
75   0.025  
0.4   0.025  
−−−
  0.750 
 −2 
iT = 10−15 Aexp
 − exp
 = 10.7 mA
 0.025
  0.025 
  0.750 
 −4.25 
iT = 10−16 Aexp
 − exp
 = 1.07 mA
 0.025 
  0.025 
€
22
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 230
 10−4 A 
I

VBE = VT ln C + 1 = 0.025V ln −16 + 1 = 0.691 V
 IS

 10 A 
 10−3 A 
I

VBE = VT ln C + 1 = 0.025V ln −16 + 1 = 0.748 V
 IS

 10 A 
Page 231
€
€
npn :VBE > 0, VBC < 0 → Forward − Active Region | pnp :VEB > 0, VCB > 0 → Saturation Region
Page 233
αF
0.95
βF =
=
= 19
1− α F 0.05
VBE = 0, VBC
I E = I S = 0.100 fA
IB = −
IS
10−16 A
=−
= −0.300 fA
βR
0.333
VBE << 0, VBC << 0 :
IC =
IS
= 3x10−16 A = 0.300 fA
βR
IE =
€
αR
0.25 1
=
=
1− α R 0.75 3


1
1 
<< 0 : IC = I S 1+  = 10−16 A1+
 = 0.400 fA
 0.333 
 βR 
βR =
I S 10−16 A
=
= 5.26 aA
βF
19.0
IB = −
IS IS
10−16 A 10−16 A
−
=−
−
= −0.305 fA
βF βR
19.0
1/ 3
Page 235
(a ) The currents do not depend upon VCC as long as the collector - base junction is reverse
biased by more than 0.1 V. (Later when Early votlage VA is discussed, one should revisit
this problem.)
IE
100µA
=
= 1.96 µA | IC = β F I B = 50I B = 98.0 µA
βF + 1
51
I

 98.0µA 
= VT ln C + 1 = 0.025V ln −16 + 1 = 0.690 V
 10 A 
 IS

(b) I
VBE
E
= 100 µA | I B =
€
23
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 236
(a ) The currents do not depend upon VCC as long as the collector - base junction is reverse
biased by more than 0.1 V. (Later when Early voltage VA is discussed, one should revisit
this problem.)
(b) Forward - active region : I
B
= 100 µA | I E = (β F + 1) I B = 5.10 mA | IC = β F I B = 5.00 mA
I

 5.00mA 
VBE = VT ln C + 1 = 0.025V ln −16 + 1 = 0.789 V | Checking : VBC = −5 + 0.789 = −4.21
 10 A

 IS

−−−
Forward - active region with VCB ≥ 0 requires VCC ≥ VBE or VCC ≥ 0.764 V
€
Page 238
(a) Resistor R is changed.
−0.7V − (−9V )
IE =
5.6kΩ
= 1.48 mA | I B =
IE
I
= E = 29.1 µA | IC = β F I B = 50I B = 1.45 mA
β F + 1 51
VCE = VC − VE = (9 − 4300IC ) − (−0.7) = 3.47 V | Q - Po int : (1.45 mA, 3.47 V )
(b) I
E
=
−0.7V − (−9V ) 8.3V
βF + 1
51
IC = 100µA = 102µA | R =
=
= 81.4 kΩ
βF
50
102µA
102µA
The nearest 5% value is 82 kΩ.
€
Page 239
−0.7V − (−9V )
I
I
IE =
= 1.48 mA | I B = E = E = 29.1 µA | IC = β F I B = 50I B = 1.45 mA
5.6kΩ
β F + 1 50
−−−
I

βF + 1
51
IC =
IC = 1.02IC VBE = VT ln C + 1 = 0.025ln 2x1015 IC + 1
βF
50
 IS


 V  
VBE + 8200 1.02 5x10−16 exp BE  −1 = 9 → VBE = 0.7079 V using a calculator solver
 0.025 

 0.7079 
or spreadsheet. IC = 5x10−16 exp
 = 992 µA | VCE = 9 − 4300IC − (−0.708) = 5.44 V
 0.025 
−−−
(
IE =
(
I SD =
)
)
I SBJT 2x10−14 A
=
= 21.0 fA
αF
0.95
€
24
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 242
Resistor R is changed :
−IC =
−0.7V − (−9V )
5.6kΩ
= 1.48 mA | I B =
−IC
−I
= C = 0.741 mA | − I E = β R I B = (1) I B = 0.741 mA
βR + 1
2
Page 244
€
VCESAT

1mA 
1+


 1 
2(40µA) 

= (0.025V ) ln  
= 99.7 mV
 0.5 
1mA 
1−


50(40µA) 

−−−
VBESAT
VBCSAT




0.1mA + (1− 0.5)1mA

= (0.025V ) ln
= 0.694 mV

 −15  1
 10 A + 1− 0.5 
 50




1mA


0.1mA −
50

 = 0.627 mV | V
= (0.025V ) ln
CESAT = VBESAT − VBCSAT = 67.7mV

 −15  1  1
10 A  + 1− 0.5
 0.5 50


Page 247
€
(a )
Dn =
kT
µn = 0.025V 500cm2 /V − s = 12.5cm2 / s
q
(
)
(
)(
(
)(
)(
)
−19
2
−4
2
20
6
qADn ni2 1.6x10 C 50µm 10 cm / µm 12.5cm / s 10 / cm
= 10−18 A = 1 aA
( b) I S = N W =
18
3
10 / cm (1µm)
AB
€
)
Page 250
1.38x10−23 J / K (373K )
I
10 A
VT =
= 32.2mV | CD = C τ F =
4x10−9 s = 1.24 µF
−19
VT
0.0322V
1.60x10 C
−−−
f
300 MHz
fβ = T =
= 2.40 MHz
βF
125
(
)
(
)
€
25
©R. C. Jaeger and T. N. Blalock
08/09/10
Page 251
 0.7  10 
 10 
1.74mA
IC = 10−15 Aexp
= 19.3 µA
1+  = 1.74 mA | β F = 751+  = 90.0 | I B =
90.0
 0.025  50 
 50 
 0.7 
1.45mA
IC = 10−15 Aexp
= 19.3 µA
 = 1.45 mA | β F = 75 | I B =
75
 0.025 
€
€
Page 253
40 −4
40 −3
gm =
10 A = 4.00 mS | g m =
10 A = 40.0 mS
V
V
CD = 4.00mS (25 ps) = 0.100 pF | CD = 40.0mS (25 ps) = 1.00 pF
(
)
(
Page 256
−23
kT 1.38x10 (300)
VT =
=
= 25.9 mV | IS =
q
1.60x10−19
)
IC
350µA
=
= 1.39 fA
 VBE 
 0.68 
exp
 exp

 0.0259 
 VT 
BF = 80 | VAF = 70 V
Page 260
€
18kΩ
12V = 4.00 V | REQ = 18kΩ 36kΩ = 12 kΩ
18kΩ + 36kΩ
4.00 − 0.7 V
IB =
= 2.687 µA | IC = 75I B = 202 µA | I E = 76I B = 204 µA
12 + (75 + 1)16 kΩ
VEQ =
VCE = 12 − 22000IC −16000I E = 4.29 V | Q - po int : (202 µA, 4.29 V )
−−−
180kΩ
12V = 4.00 V | REQ = 180kΩ 360kΩ = 120 kΩ
180kΩ + 360kΩ
4.00 − 0.7
V
IB =
= 2.470 µA | IC = 75I B = 185 µA | I E = 76I B = 188 µA
120 + (75 + 1)16 kΩ
VEQ =
VCE = 12 − 22000IC −16000I E = 4.29 V | Q - po int : (185 µA, 4.93 V )
€
26
©R. C. Jaeger and T. N. Blalock
08/08/10
Page 261
I
50I B
I2 = C =
= 10I B
5
5
−−−
18kΩ
12V = 4.00 V | REQ = 18kΩ 36kΩ = 12 kΩ
18kΩ + 36kΩ
4.00 − 0.7
V
IB =
= 0.4111 µA | IC = 500I B = 205.6 µA | I E = 76I B = 206.0 µA
12 + (500 + 1)16 kΩ
VEQ =
VCE = 12 − 22000IC −16000I E = 4.18 V | Q - point : (206 µA, 4.18 V )
€
Page 262
The voltages all remain the same, and the currents are reduced by a factor of 10. Hence all
the resistors are just scaled up by a factor of 10.
120 kΩ →1.2 MΩ 82 kΩ → 820 kΩ 6.8 kΩ → 68 kΩ
Page 264
€


76
VCE = 0.7 V at the edge of saturation. 12V -  RC + 16kΩ(205µA) ≥ 0.7V → RC ≤ 38.9 kΩ
75


−−−
VBESAT = 4 −12kΩ(24µA) −16kΩ(184µA) = 0.768 V
VCESAT = 12 − 56kΩ(160µA) −16kΩ(184µA) = 0.096 V
€
Page 265
9 − 0.7
V
IB =
= 95.4 µA | IC = 50I B = 4.77 mA | I E = 51I B = 4.87 mA
36 + (50 + 1)1 kΩ
VCE = 9 −1000( IC + I B ) = 4.13 V | Q - point : (4.77 mA, 4.13 V )
€
Page 266
VBE (V)
0.70000
0.67156
0.67178
0.67178
IC (A)
V'BE (V)
2.0155E-04
2.0328E-04
2.0327E-04
2.0327E-04
0.67156
0.67178
0.67178
0.67178
27
©R. C. Jaeger and T. N. Blalock
08/09/10
Microelectronic Circuit Design
Fourth Edition - Part II
Solutions to Exercises
CHAPTER 6
Page 292
NM L = 0.8V − 0.4V = 0.4 V | NM H = 3.6V − 2.0V = 1.6 V
€
Page 294
V10% = VL + 0.1 (ΔV ) = −2.6V + 0.1 −0.6 − (−2.6) = −2.4 V
[
]
[
]
Checking : V10% = VH − 0.9 (ΔV ) = −0.6V − 0.9 −0.6 − (−2.6) = −2.4 V
[
]
V90% = VH − 0.1 (ΔV ) = −0.6V − 0.1 −0.6 − (−2.6) = −0.8 V
[
]
Checking : V90% = VL + 0.9 (ΔV ) = −2.6V + 0.9 −0.6 − (−2.6) = −0.8 V
V50% =
€
VH + VL −0.6 − 2.6
=
= −1.6 V | t r = t 4 − t3 = 3 ns | t f = t2 − t1 = 5 ns
2
2
Page 295
At P = 1 mW : PDP = 1mW (1ns) = 1 pJ
At P = 3 mW : PDP = 3mW (1ns) = 3 pJ
At P = 20 mW : PDP = 20mW (2ns) = 40 pJ
€
Page 297
Z = ( A + B)( B + C) = AB + AC + BB + BC = AB + BB + AC + BB + BC
Z = AB + B + AC + B + BC = B( A + 1) + AC + B(C + 1) = B + AC + B
Z = B + B + AC = B + AC
€
1
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 300
P
0.4mW
V − VL 2.5V − 0.2V
I DD =
=
= 160 µA | R = DD
=
= 14.4 kΩ
VDD
2.5V
I DD
160µA
W 
A W  
0.2 
4.44
2
1.6x10−4 A = 10−4 2   2.5 − 0.6 −
0.2 V →   =
2 
1
V  L S 
 L S
€
Page 301
V − VL 3.3V − 0.1V
I DD = DD
=
= 31.4µA
R
102kΩ
W 
A W  
0.1
2.09
31.4x10−6 A = 6x10−5 2    3.3 − 0.75 − 0.1 V 2 →   =
2 
1
V  L S 
 L S
Page 303
€
Ron
2.5V → Ron = 1.84 kΩ
Ron + 28.8kΩ
W 
W 
1
2.98
→  =
  =


1
0.15
 L S
 L S
10−4  2.5 − 0.60 −
(1.84kΩ)
2 

0.15V =
−−−
Ron =
1
 1.03 
0.2 
6x10−5 
 3.3 − 0.75 −

2 
 1 
= 6.61 kΩ | VL =
6.61kΩ
3.3V = 0.201 V
6.61kΩ + 102kΩ
−−−
 1  V2 1 V2
=
=V

=
 Kn R  A Ω V
Page 305
€
1.03 
6.30
5
Kn R = 6x10−5 
 1.02x10 =
V
 1 
(
)
NM H = 3.3 − 0.75 +
(
)
1
3.3
−1.63
= 1.45 V
6.30
2(6.30)
NM L = 0.75 +
2(3.3)
1
−
= 0.318 V
6.30
3(6.30)
€
2
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 309
Using MATLAB :
fzero(@(vh) ((vh -1.9 - 0.5* sqrt(0.6))^2 - 0.25(vh + 0.6)), 1) | ans = 1.5535
fzero(@(vh) ((vh -1.9 - 0.5* sqrt(0.6))^2 - 0.25(vh + 0.6)), 4) | ans = 3.2710
−−−
[
)]
(
VH = 5 − 0.75 + 0.5 VH + 0.6 − 0.6 → VH = 3.61 V
fzero(@(vh) (5- 0.75 - 0.5* (sqrt(vh + 0.6) - sqrt(0.6)) - vh), 1) | ans = 3.6112
−−−
(a )
80x10−6 A = 100x10−6
W 
A W  
0.15 
6.10
2
 1.55 − 0.60 −
0.15 V →   =
2
2 
1
V  L S 
 L S
(
)
VTNL = 0.6 + 0.5 .15 + 0.6 − 0.6 = 0.646 V
 W  0.551
2 2
100x10−6 A W 
1
=
 (2.5 − 0.15 − 0.646) V →   =
2
2
1
1.82
V  L L
 L L
W 
A W  
0.1
8.89
(b) 80x10−6 A = 100x10−6 2   1.55 − 0.60 − 0.1 V 2 →   =
2 
1
V  L S 
 L S
80x10−6 A =
(
)
VTNL = 0.6 + 0.5 .1+ 0.6 − 0.6 = 0.631 V
 W  0.511
2 2
100x10 A W 
1
2.5
−
0.1−
0.631
V
→
=


  =
(
)
2
2
1
1.96
V  L L
 L L
−6
80x10−6 A =
€
Page 312
The high logic level is unchanged : VH = 2.11
W 
A W  
0.1
9.16
60x10−6 A = 50x10−6 2    2.11− 0.75 − 0.1 V 2 →   =
2 
1
V  L S 
 L S
(
)
VTNL = 0.75 + 0.5 .1+ 0.6 − 0.6 = 0.781 V
60x10−6 A =
 W  0.410
2 2
50x10−6 A  W 
1
=
 (3.3 − 0.1− 0.781) V →   =
2
2
1
2.44
V  L L
 L L
€
3
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 314
Using MATLAB :
fzero(@(vh) ((vh -1.9 - 0.5* sqrt(0.6))^2 - 0.25(vh + 0.6)), 1) | ans = 1.5535
−−−
γ = 0 → VTN = 0.6V | VH = 2.5 - 0.6 = 1.9 V | I DD = 0 for vO = VH
 
2
VL 
100x10−6  2 
−6 10
100x10  1.9 − 0.6 − VL =
 (2.5 − VL − 0.6)
2
2
 1 
 1
10 
0.235
6VL2 −116.8VL + 3.61 = 0 → VL = 0.235V | I DD = 100x10−6  1.9 − 0.6 −
0.235 = 278 µA
2 
 1 
2
100x10−6  2 
Checking : I DD =
 (2.5 − 0.235 − 0.6) = 277 µA
2
 1
€
Page 319
VTNL = −1.5 + 0.5 0.2 + 0.6 − 0.6 = −1.44V
(
)
W  
 W  1.17
0.2 
60.6x10−6 = 100x10−6    3.3 − 0.6 −
0.2 →   =
2 
1
 L S 
 L S
 W  0.585
2
100x10−6  W 
1
60.6x10−6 =
=
  (0 −1.44) →   =
2
1
1.71
 L L
 L L
Page 320
€
 2.22 
0.2 
I DS = 100x10−6 
 2.5 − 0.6 −
0.2 = 79.9 µA which checks.
2 
 1 
€
Page 321
The PMOS transistor is still saturated so I DL = 144 µA, and VH = 2.5 V.
 5
V 
144x10−6 = 100x10−6   2.5 − 0.6 − L VL → VL = 0.158 V
2
 1 
Page 326
€
W 2.22
=
in parallel with transistors A and B.
L
1
 W  1.81
The W/L ratio of the load transistor remains unchanged :   =
1
 L L
Place a third transistor with
€
4
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 327
Place a third transistor in series with transistors A and B.
W 
2.22 6.66
The new W/L ratios of transistors A, B and C are  
=3
=
.
1
1
 L  ABC
 W  1.81
The W/L ratio of the load transistor remains unchanged :   =
1
 L L
Page 333
€
M L1 is saturated for all three voltages. I DD =
2
40x10−6 1.11

 −2.5 − (−0.6) = 80.1 µA
2  1 L
[
]
−−−
The voltages can be estimated using the on - resistance method.
For the 11000 case, RonA =
132mV − 64.4mV
64.4mV
= 844 Ω RonB =
= 804 Ω
80.1µA
80.1µA
For the 00101 case, RonE =
64.4mV
= 804 Ω.
80.1µA
For the 01110 case, RonC =
203mV −132mV
132mV − 64.4mV
= 886 Ω RonD =
= 844 Ω
80.1µA
80.1µA
The voltage across a given conducting device is I D Ron . Small variations in Ron are ignored.
€
ABCDE
Y (mV)
2 (mV)
3 (mV)
IDD (uA)
ABCDE
Y (mV)
2 (mV)
3 (mV)
IDD (uA)
00000
00001
00010
00011
00100
00101
00110
00111
01000
01001
01010
01011
01100
01101
01110
01111
2.5 V
2.5 V
2.5 V
2.5 V
2.5 V
130
2.5 V
130
2.5 V
2.5 V
2.5 V
2.5 V
2.5 V
130
200
114
0
0
0
0
0
0
2.5 V
64
0
0
0
0
0
0
64
21
0
0
0
0
2.5 V
64
2.5 V
64
0
0
0
0
2.5 V
64
130
43
0
0
0
0
0
80.1
0
80.1
0
0
0
0
0
80.1
80.1
80.1
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
2.5 V
2.5 V
2.5 V
200
2.5 V
130
2.5 V
100
130
130
130
110
130
66
110
65
2.5 V
2.5 V
2.5 V
130
2.5 V
130
2.5 V
83
64
64
64
43
64
32
64
32
0
0
2.5 V
64
2.5 V
64
2.5 V
64
0
0
64
22
64
32
87
32
0
0
0
80.1
0
80.1
0
80.1
80.1
80.1
80.1
80.1
80.1
80.1
80.1
80.1
5
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 334
2.5V (80µA)
Pav =
= 0.100 mW
2
Page 335
2
(
)
F (2.5V ) (3.2x10 Hz ) = 2x10
PD = 10-12 F (2.5V ) 32x106 Hz = 2x10−4W = 200 µW or 0.200 mW
€
€
PD = 10-12
2
9
W = 0.02 W or 20 mW
−4
Page 336
The inverter in Fig. 6.38(a) was designed for a power dissipation of 0.2 mW.
To reduce the power by a factor of two, we must reduce the W/L ratios by a factor of 2.
W  1  1 
W 
1
1  4.71 2.36
|   = 
  = 
=
=
1
 L  L 2  1.68  3.36
 L S 2  1 
−−−
4mW
, we must increase the W/L ratios by a factor of 20.
0.2mW
W 
 2.22  44.4
|   = 20
=
1
 L S
 1 
To increase the power by a factor of
W 
 1.81 36.2
  = 20
=
1
 L L
 1 
−−−
To reduce the power by a factor of three, we must reduce the W/L ratios by a factor of 3.
W  1  1.81 0.603
W 
W 
1
1  3.33 1.11
1  6.66  2.22
=
|   = 
|  
= 
  = 
=
=
=
1
1.66
1
1
 L L 3  1 
 L A 3 1 
 L  BCD 3  1 
€
6
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 339
t r = 2.2RC = 2.2 28.8x103 Ω 2x10−13 F = 12.7 ns
(
)(
(
)
)(
)
τ PLH = 0.69RC = 0.69 28.8x103 Ω 2x10−13 F = 3.97 ns
−−
 t 
 VH + VL 
vO (t ) = VF − (VF − VI ) exp −
 | vO (τ PHL ) = VH − 0.5
 = 2.5 −1.15 = 1.35 V
 RC 
 2 
 τ 
1.35 = 0.2 − (0.2 − 2.5) exp − PHL  → τ PLH = −RC ln 0.5 = 0.69RC
 RC 
vO (t1 ) = VH − 0.1(VH + VL ) = 2.5 + 0.23 = 2.27 V
 t 
2.27 = 0.2 − (0.2 − 2.5) exp − 1  → t1 = −RC ln 0.9
 RC 
vO (t2 ) = VL + 0.1(VH + VL ) = 0.2 + 0.23 = 0.43 V
 t 
0.43 = 0.2 − (0.2 − 2.5) exp − 2  → t2 = −RC ln 0.1
 RC 
t f = t2 − t1 = −RC ln 0.1+ RC ln 0.9 = RC ln 9 = 2.2RC
€
Page 343
t f = 3.7 2.37x103 Ω 2.5x10−13 F = 2.19 ns | τ PHL = 1.2 2.37x103 Ω 2.5x10−13 F = 0.711 ns
tr
(
)(
)
= 2.2(28.8x10 Ω)(2.5x10 F ) = 15.8 ns
τP =
3
−13
(
| τ PLH
)(
)
= 0.69(28.8x10 Ω)(2.5x10 F ) = 4.97 ns
3
−13
0.711 ns + 4.97 ns
= 2.84 ns
2
Page 346
€
(
)
T = 2Nτ P 0 = 2(401) 10−9 s = 802 ns | f =
1
1
=
= 1.25 MHz
T 802ns
€
7
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 347
For our Psuedo NMOS inverter with VL = 0.2 V ,
τ PHL = 1.2RonS C = 1.2
Cox" WL
L2
= 1.2
W
µn (VGS − VTN )
µnCox"
VGS − VTN )
(
L
−9
τ PHL
2
2
(250x10 m) (100cm m) = 0.606 ps
= 1.2
(500cm V − s)(3.3 − 0.825)V
(250x10 m) (100cm m) = 2.63 ps
L
= 1.2R C = 1.2
= 1.2
0.4µ (V − V )
(125cm V − s)(3.1− 0.825)V
2
−9
2
τ PLH
onL
p
τP =
GS
2
2
2
TN
0.606 ps + 2.63 ps
= 1.62 ps
2
Page 349
€
The PMOS transistor is saturated for vO = VL.
Pav =
2.5V (1.71mA)
2
I DD =
2
40x10−6  23.7 

 −2.5 − (−0.6) = 1.71 mA
2  1 L
[
]
2
1 
= 2.14 mW | PD = 5x10−12 F (2.5V − 0.2V ) 
 = 13.2 mW
 2x10−9 s 
−−−
 20 pF  2ns 
We must increase the power by a factor of 

 = 8,
 5 pF  1ns 
so the W/L ratios must also be increased by a factor of 8.
W 
 23.7  190
W 
 47.4  379
2
1 
|   = 8
| PD = 20x10−12 F (2.5V − 0.2V )  −9  = 106 mW
  = 8
=
=
1
 L L
 1  1
 L S
 1 
 10 s 
€
8
©R. C. Jaeger and T. N. Blalock
08/12/10
CHAPTER 7
Page 370
(a )
 20 
µA
K p = 40x10−6   = 800 2
V
1
( b) V
(c) V
€
 20 
µA
mA
| Kn = 100x10−6   = 2000 2 = 2.00 2
V
V
1
(
)
TN
= 0.6 + 0.5 2.5 + 0.6 − 0.6 = 1.09 V
TP
= −0.6 − 0.75 2.5 + 0.7 − 0.7 = −1.31 V
(
)
Page 372
(a ) For vI = 1 V , VGSN − VTN = 1− 0.6 = 0.4V and VGSP − VTP = −1.5 + 0.6 = −0.9V
MN is saturated for vO ≥ 0.4 V. MP is in the triode region for vO ≥ 1.6 V. ∴ 1.6 V ≤ v O ≤ 2.5 V
(b) M
(c) M
P
is saturated for vO ≤ 1.6 V. ∴ 0.4 V ≤ v O ≤ 1.6 V
N
is in the triode region for vO ≤ 0.4 V. MP is saturated for v O ≤ 1.6 V. ∴ 0 ≤ v O ≤ 0.4 V
−−−
W 
 10  25
Kn  W 
  =
  = 2.5  =
 L P K p  L  N
1 1
€
Page 373
Both transistors are saturated since VGS = VDS .
K
2
2
Kn
VGSN − VTN ) = p (VGSP − VTP )
(
2
2
VGSN = −VGSP → vI = VDD − v I → v I =
Kn = K p
VTN = VTP
VDD
2
10K p
2
K
2
VGSN − VTN ) = p (VGSP − VTP ) → 10 (VGSN − VTN ) = −VGSP + VTP
(
2
2
10 (vI − 0.6) = 4 − v I − 0.6 → vI = 1.273 V
Kp
2
10K p
2
VGSN − VTN ) =
VGSP − VTP ) → (VGSN − VTN ) = 10 (−VGSP + VTP )
(
(
2
2
vI − 0.6 = 10 (4 − v I − 0.6) → vI = 2.37 V
€
9
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 375
W 
Kn 
 L  N Kn
KR =
=
= 2.5
W 
Kp
K p 
 L P
VIH =
2KR (VDD − VTN + VTP )
VOL
(V
DD
− KRVTN + VTP )
K −1
( K −1) 1+ 3K
2(2.5)(2.5 − 0.6 − 0.6) (2.5 − 2.5(0.6) − 0.6)
=
−
= 1.22V
2.5
−1
2.5
−1
1+
3
2.5
(
)
( )
( K + 1)V − V − K V − V = (2.5 + 1)1.22 − 2.5 − 2.5(0.6) + 0.6 = 0.174V
=
2K
2(2.5)
2 K (V − V + V ) (V − K V + V )
=
−
K −1
( K −1) K + 3
2 2.5 (2.5 − 0.6 − 0.6) (2.5 − 2.5(0.6) − 0.6)
=
−
= 0.902V
2.5
−1
2.5
−1
2.5
+
3
(
)
( K + 1)V + V − K V − V = (2.5 + 1)0.902 + 2.5 − 2.5(0.6) + 0.6 = 2.38V
=
R
VIH
−
R
R
R
IH
DD
R TN
TP
R
VIL
R
DD
R
VIL
VOH
R
NM H = VOH
TN
TP
DD
DD
TP
R
R
IL
R TN
R TN
TP
2
2
− VIH = 2.38 −1.22 = 1.16 V | NM L = VIL − VOL = 0.902 − 0.174 = 0.728 V
Page 376
€
Symmetrical Inverter : τ P = 1.2RonnC = 1.2
10−12 F
( )
2 10−4 (2.5 − 0.6)
Ω = 3.16 ns
Page 377
€
Symmetrical Inverter : Ronn =
τP
10−9 s
=
= 167Ω
1.2C 1.2 5x10−12 F
(
)
W 
1
1
31.5
=
=
  =
'
−4
1
 L  N Ronn Kn (VGS − VTN ) 167 10 (2.5 − 0.6)
( )
|
W 
W 
78.8
  = 2.5  =
1
 L P
 L N
€
10
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 379
The inverters need to be increased in size by a factor of
W 
 3.77  4.22
  = 1.12
=
1
 L N
 1 
|
280ps
= 1.12.
250ps
W 
 9.43  10.6
  = 1.12
=
1
 L P
 1 
−−−
W 
 3.77  3.3 − 0.75  3.43
  =

=
1
 L  N  1  3.3 − 0.5 
W   9.43  3.3 − 0.75  8.59
|   =

=
1
 L  P  1  3.3 − 0.5 
Page 380
€
τ PHL = 2.4RonnC =
2.4C
2.4C
C
=
= 1.26
Kn
Kn (VGS − VTN ) Kn (2.5 − 0.6)
τ PLH = 2.4RonpC =
2.4C
2.4C
C
=
= 1.26
Kp
K p (VGS − VTN ) K p (2.5 − 0.6)
−−−
€
τ PHL = 2.4RonnC =
2.4C
2.4C
C
=
= 0.94
Kn
Kn (VGS − VTN ) Kn (3.3 − 0.75)
τ PLH = 2.4RonpC =
2.4C
2.4C
C
=
= 0.94
Kp
K p (VGS − VTN ) K p (3.3 − 0.75)
Page 381
The inverter in Fig. 7.12 is a symmetrical design, so the maximum current occurs
2
VDD
10-4  2 
for vO = v I =
. Both transistors are saturated : iDN =
 (1.25 − 0.6) = 42.3 µA
2
2  1
2
4x10-5  5 
Checking : iDP =
 (1.25 − 0.6) = 42.3 µA
2 1
Page 382
€
10
CV 2
(a) PDP ≅ 5DD =
F (2.5V )
2
10
CVDD
b
PDP
≅
=
()
5
F (3.3V )
2
10
CVDD
c
PDP
≅
=
()
5
F (1.8V )
−13
−13
−13
€
2
5
2
5
5
= 0.13 pJ = 130 fJ
= 0.22 pJ = 220 fJ
2
= 0.065 pJ = 65 fJ
11
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 388
Remove the NMOS and PMOS transistors connected to input E, and ground the source of
the NMOS transistor connected to input D. The are now 4 NMOS transistors in series, and
W 
 2 8
W 
5
|   =
  = 4  =
 L N
 1 1
 L P 1
€
€
Page 392
There are two NMOS transistors in series in the AB and CD NMOS paths, and three PMOS
transistors in the ACE and BDE PMOS paths. Therefore :
W 
 2 4
W 
W 
 5  15
2
= 2  =
|  
=
|   = 3  =
 
 L  N − ABCD
 1 1
 L  N −E 1
 L P
 1 1
Page 396
(a) The logic network for F = AB + C is
B
C
A
(
)
2
(
)
2
P = CVDD
f = 50x10−12 F (5V ) 10 7 Hz = 12.5 mW
€
12
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 400
1
 50 pF  2
β =
 = 31.6
 50 fF 
−−−
τ P = 31.6τ o + 31.6τ o = 63.2τ o
1
z = e ln z
( )
| z ln z = e ln z
1
ln z
=e
−−−
1
 50 pF  7
β =
 = 2.683
 50 fF 
1, 2.68 , 2.6832 = 7.20, 2.6833 = 19.3, 2.6834 = 51.8, 2.6835 = 139, 2.6836 = 373
A6 = (1+ 3.16 + 10 + 31.6 + 100 + 316) Ao = 462 Ao
A7 = (1+ 2.68 + 7.20 + 19.3 + 51.8 + 139 + 373) Ao = 594 Ao
€
Page 401
From the figure, 10/1 devices give a maximum Ron of 4 kΩ. The W/L ratios must be 4 times
W   10  40
larger in order to reduce the maximum Ron to 1 kΩ. ∴   = 4  =
 L  1 1
€
13
©R. C. Jaeger and T. N. Blalock
08/12/10
CHAPTER 8
Page 419
(a )
€
NS =
28 • 220
= 211 = 2048 segments |
2 7 • 210
( b)
NS =
230
= 211 = 2048 segments
29 • 210
Page 422
(a) N = 28 • 220 = 228 = 268,435,456
( b)
I DD =
0.05W
15.2mA
= 15.2 mA | Current/cell = 28
= 56.4 pA
3.3V
2 cells
−−−
Reverse the direction of the substrate arrows, and connect the substrates of the PMOS
transistors to VDD .
€
Page 426
MA1 : At t = 0 + , VGS − VTN = 4 V and VDS = 2.5V , so transistor MA1 is operating in the triode region.
 1
2.5 
i1 = 60x10−6  5 −1−
2.5 = 413 µA
2 
 1
MA2 : At t = 0 + , VGS = VDS , so transistor MA2 is operating in the saturation region.
2
60x10−6  1
VTN 2 = 1+ 0.6 2.5 + 0.6 − 0.6 = 1.592V
i2 =
 (5 − 2.5 −1.592) = 24.8 µA
2  1
(
€
)
Page 428
MA1 : At t = 0 + , VGS = VDS , so transistor MA1 is operating in the saturation region.
2
60x10−6 1
i1 =
 (5 −1) = 480 µA
2  1
MA2 : At t = 0 + , VGS = VDS , so transistor MA2 is operating in the saturation region.
2
60x10−6 1
i1 =
 (5 −1) = 480 µA
2  1
€
14
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 431
(a ) At t = 0+ , VGS − VTN = 3 - 0.7 = 2.3 V and VDS = 1.9 V , so transistor MA is operating in
the triode region.
 1
1.9 
i1 = 60x10−6   3 − 0 − 0.7 − 1.9 = 154 µA
2 
 1
(b) From Table 6.10 :
t f = 3.7RonC = 3.7
50x10−15 F
= 1.34 ns
60x10−6 (3 − 0.7)
−−−
VC = VBL − VTN
[
)]
(
| VC = 3 − 0.7 + 0.5 VC + 0.6 − 0.6 → VC = 1.89 V | VC = 3 − 0.7 = 2.3 V
−−−
n=
−15
CV 25x10 F (1.89V )
=
= 2.95 x 105 electrons
−19
q
1.60x10 C
Page 432
€
€
VC − VBL 1.9 − 0.95
V − VBL 0 − 0.95
=
V = 19.0 mV | ΔV = C
=
V = −19.0 mV
CBL
49CC
CBL
49CC
+1
+1
+1
+1
CC
CC
CC
CC
(a )
ΔV =
(b)
τ = Ron
CC
25 fF
= 5kΩ
= 0.123 ns
CC
1
+1
+1
CBL
49
or
τ ≅ RonCC = 5kΩ(25 fF ) = 0.125 ns
Page 434
At t = 0 + , VGS − VTN = (3 - 0) - 0.7 = 2.3 V and VDS = 1.5 V , so transistor MA2 is operating in
the triode region.
 2 
1.5 
iD = 60x10−6   3 − 0.7 − 1.5 = 279 µA
2
 1 
€
15
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 436
In setting the drain currents equal, we see that the change in W/L cancels out, and
the voltages remain the same.
 5
2
1
∴iD = 60x10-6  (1.33 − 0.7) = 59.5 µA | PD = 2(59.5µA)(3V ) = 0.357 mW
2
 1
5
As a check, the current should scale with W/L : iD = (23.5µA) = 58.8 µA
2
−−−
 2
 2
2
2
1
1
Equating drain currents :
25x10-6  (2.5 − VO − 0.6) = 60x10-6  (VO − 0.6)
2
2
 1
 1
(
)
(
)
(
)
1.4VO2 + 0.92VO − 2.746 = 0 → VO = 1.11V
 2
2
1
iD = 25x10-6  (2.5 −1.11− 0.6) = 15.6 µA | PD = 2(15.6µA)(2.5V ) = 78.0 µW
2
 1
 2
2
1
Checking :
60x10-6  (1.11− 0.6) = 15.6 µA
2
 1
(
)
(
)
Page 488
€
€
€
Ron =
1
= 23.8 kΩ | τ = 23.8kΩ(25 fF ) = 0.595 ns
60x10 (3 −1.3 −1)
−6
Page 440
For all possible input combinations there will be two inverters and 3 output lines in the low state.
PD = 5(0.2mW ) = 1.0 mW
Page 442
W 
2  1.81 1.63
  =

=
1
 L  L 2.22  1 
€
16
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 444
For a 0 - V input, all transistors will be on and the input nodes will all discharge to 0 V.
For the 3 - V input, the nodes will all charge to 3 V as long as VTN ≤ 2 V.
(
)
VTN = 0.7 + 0.5 3 + 0.6 − 0.6 = 1.26 V. Thus the nodes will all be a 3 V.
2 ≥ 0.7 + γ
(
)
3 + 0.6 − 0.6 → γ ≤ 1.158
−−−
The output will drop below VDD / 2. For the PMOS device, VGS − VTP = 3 −1.9 − 0.7 = 0.4V.
The PMOS transistor will be saturated. For the NMOS device, VGS − VTP = 1.9 − 0.7 = 1.2V.
Assume linear region operation.
 
2
40x10-6  5
VO 
-6 2
 (−1.1+ 0.7) = 100x10  1.9 − 0.7 − VO
2  1
2
 1 
VO2 − 2.4VO + 0.16 = 0 → VO = 68.6 mV
€
17
©R. C. Jaeger and T. N. Blalock
08/12/10
CHAPTER 9
Page 462
 0.2V 
 0.3V 
 0.4V 
iC 2
iC 2
iC2
3
5
6
= exp
= exp
= exp
 = 2.98 x 10 |
 = 1.63 x 10 |
 = 8.89 x 10
iC1
0.025V
i
0.025V
i
0.025V






C1
C1
€
€
Page 464
The current must be reduced by 5 while the voltages remain the same.
300µA
I EE =
= 60 µA | RC = 5(2kΩ) = 10 kΩ
5
Page 465
I
IB = E
βF + 1
| IB3 =
92.9µA
107µA
= 4.42 µA | I B 4 =
= 5.10 µA
21
21
I B3 RC = 4.42µA(2kΩ) = 8.84 mV << 0.7 V | I B 4 RC = 5.10µA(2kΩ) = 10.2 mV << 0.7 V
€
Page 467
VH = 0 − 0.7 = −0.7 V | VL = 0 − 0.2mA(2kΩ) − 0.7V = −1.1 V
VREF =
−0.7V + (−1.10V )
2
= −0.9 V | ΔV = -0.7V - (1.1V ) − 0.4 V
Page 469
€
NM H = NM L =

 0.4

0.4V
− 0.025V 1+ ln
−1 = 0.107 V
2
 0.025 

€
18
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 471
P = 3.3V (0.3mA + 0.2mA) = 1.65 mW | P = 3.3V (0.357mA + 0.2mA) = 1.84 mW
NM H = NM L =

 0.6

0.6V
− 0.025V 1+ ln
−1 = 0.20 V
2
 0.025 

−−−
From the graph, the VTC slope is -1 for VIL = −1.08 V , VOH = −0.71 V and
VIH = −0.91 V , VOL = −1.28 V. NM H = −0.71− (−0.91) = 0.20 V. NM L = −1.08 − (−1.28) = 0.20 V
−−−
The voltages remain the same. Thus the currents must be reduced by a factor of 3,
and the resistor values must be increase by a factor of 3.
--REE =
−1.7V − (−5.2V )
0.20mA
= 17.5 kΩ | I E =
−1.4V − (−5.2V )
18 kΩ
= 0.211 mA | RC1 =
0.4V
= 1.90 kΩ
0.211mA
Page 472
€
For all inputs low : I EE =
−1− 0.7 − (−5.2) V
= 299µA
11.7
kΩ
ΔV
= VH − VREF = −0.7 − (−1) = 0.3 V | ΔV = 0.6 V |
2
For an inputs high : I EE =
RC 2 =
0.6V
= 2.00 kΩ
299µA
−0.7 − 0.7 − (−5.2) V
0.6V
= 325µA | RC1 =
= 1.85 kΩ
11.7
kΩ
325µA
Based upon analysis above, RC =
0.6V
= 1.85 kΩ
325µA
Page 473
€
For all inputs low : I EE =
−1− 0.7 − (−5.2) V
= 299µA
11.7
kΩ
ΔV
= VH − VREF = −0.7 − (−1) = 0.3 V | ΔV = 0.6 V
2
€
|
RC =
0.6V
= 2.00 kΩ
299µA
Page 474
VE − (−VEE ) 0 − 0.7 − (−5.2V ) V
RE =
=
= 15.0 kΩ
0.3mA
0.3
mA
€
19
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 475
(a )
€
For I E = 0, vO = −5.2 V. ( b) For I E = 0, vO = − - 5.2V
10kΩ
= −2.08 V
10kΩ + 15kΩ
Page 476
The transistor's power dissipation is

50 
P = VCB IC + VBE I E = 5V 2.55mA  + 0.7V (2.55mA) = 14.3 mW
51 

The total power dissipation in the circuit is

50 
P = VCC IC + VEE I E = 5V 2.55mA  + 5V (2.55mA) = 25.3 mW
51 

For vO = −3.7V , I E =
−3.7 − (−5)
1300
−0.7 − (−5)
−
3.7
= 260 µA.
5000
0.7
= 3.17 mA
1300
5000
The transistor's power dissipation is

50 
P = VCB IC + VBE I E = 5V 3.17mA  + 0.7V (3.17mA) = 17.8 mW
51 

--10kΩ
(a ) − 4V = −5.2V 10kΩ + R → RE = 3.00 kΩ
E
At the Q - point, I E =
( b)
IE =
−
−4 − (−5.2)
4 − (−5.2)
5.2V
4
4
= 1.73 mA | I E =
−
= 0 | IE =
+
= 3.47 mA
3kΩ
3000
10000
3000
10000
Page 478
Increase the value of each resistor by a factor of 10.
€
€
Page 481
ΔV
0.6V
RC =
=
= 1.2 kΩ | τ P = 0.69(1.2kΩ)(2 pF ) = 1.66 ns
I EE 0.5mA
P = 5.2V (0.5 + 0.1+ 0.1)mA = 3.64 mW | PDP = 6.0 pJ
€
20
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 483
RC2 = RC1 =
0 − VL
0.4V
=
= 800 Ω | P = I EEVEE = 0.5mA(2.8V ) = 1.40 mW
I EE
0.5mA
PDP = 1.4mW (50 ps) = 70 fJ
−−−
I EF =
I EE
= 250 µA | P = I EEVEE = 0.25mA(2.8V ) = 0.70 mW
2
−−−
VH = 0 | VL = −0.2 V | VBias = 0.1 V | VBH = −0.7 | VBL = −0.9 V | VBiasB = −0.8 V
VAH = −1.4 | VAL = −1.6 V | VBiasA = −1.5 V
−VEE = VAH − 0.7V − 0.7V = −1.4 − 0.7 − 0.7 = −2.8 V | VEE = 2.8 V
−−−
VH = 0 V , VL = −0.4 V , : The C - level bias is VBiasC =
VH + VL
= −0.2 V
2
Using the level shifter in Fig. 9.27,
VBH = −0.7 | VBL = −1.1 V | VBiasB = −0.9 V
VAH = −1.4 | VAL = −1.8 V | VBiasA = −1.6 V
−VEE = VemitterA − 0.7V = −0.7 = −2.8 V | VEE = 2.8 V
€
Page 488
For vO = VH , IC = 0, and P = 0.
P = VDD I DD = 5V (2.43mA) = 12.1 mA
Increase R by a factor of 10 : R = 10(2kΩ) = 20kΩ.
Page 490
€
 0.1 
Γ = exp
 = 48.2
 0.0258 

20 
1+

10 A  0.1(48.2) 
10 A
| IB ≥
= 3.34 A | β FOR =
= 3.00
11 
20 
3.34 A
 1−

48.2 

−−−
αR =
0.2
1
=
0.2 + 1 6

20 
1+

10 A  0.2(54.6) 
| IB ≥
= 1.59 A

6
20 
 1−

54.6 

Page 491
€
21
©R. C. Jaeger and T. N. Blalock
08/12/10

20 
1+

 0.15 
10 A  0.1(403) 
Γ = exp
= 0.769 A
 = 403 | I B ≥
11 
20 
 0.025 
 1−

403 

−−−
VT =
1.38x10−23 (273 + 150)
1.60x10−19
 0.05 + 1
= 36.5 mV | VCEMIN = 36.5mV ln
 = 111 mV
 0.05 
−−−
 0.1 
0.25
1
Γ = exp
=
 = 54.6 | α R =
1+ 0.25 5
 0.025 


40
1+

10mA  0.25(54.6) 
10mA
IB ≥
= 1.08 mA | β FOR =
= 9.24


5
40
1.08mA
 1−

54.6 

€
22
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 494


 1mA − I

1mA − I BR
BR
1ns = 6.4ns ln
→ I BR = −5.49 mA
 | 1.169 =
2.5mA
0.0614mA
−
I
BR

− I BR 
 40.7

−−−

V −V
5−0
2.5mA 
iCMAX = CC CE ≅
= 2.5mA | QXS = 6.4ns1mA −
 = 6.01 pC
βF
2500
40.7 

QF = iF τ F = 2.5mA(0.25ns) = 0.625 pC | QXS >> QF
Page 495
€
5 − 0.95
= −1.01 mA | VBE2 = VL + VCESAT1
4000
Using the value of VCESAT in Fig. 9.32, VBE2 = 0.15 + 0.04 = 0.19 V
 2 + 1
A better estimate is VCESAT1 = 25mV ln
 = 10.1 mV
 2 
VBE2 = 0.15 + 0.010 = 0.16 V
For vI = VL = 0.15 V :
iIL = −
Page 496
€
For vI = VH = 5 V :
5 −1.5
= 1.75 mA | VBE2 = 0.8 V
4000
 2
0.875mA + 1− (2.4mA)
 3
= 0.025V ln
= 0.729 V
 1  2 
−15
10 A + 1− 
40  3 
iIH = 2
Using Eq. (5.29), VBESAT
Page 499
€
5 −1.5
= 0.875mA
4000
3.5V
| βR ≤
= 0.2
10(2kΩ)(0.875mA)
5V - N (2kΩ)β R I B ≥ 1.5V | I B =
βR ≤
3.5V
= 0.4
5(2kΩ)(0.875mA)
Page 500
€
I B2 = (2 + 1)
5 −1.5
= 2.63 mA |
4000
2.43mA + N (1.01mA) ≤ 28.3(2.63mA) → N ≤ 71
€
23
©R. C. Jaeger and T. N. Blalock
08/12/10
Page 501
vI = VL and vO = 0 : I B 4 =
5 − VB 4 5 − (0 + 0.7 + 0.7)
=
= 2.25 mA | I L = 41I B 4 = 92.3 mA
1600
1600
IL
− 0.7 − 0.7 ≥ 3 → I L ≤ 15.4 mA
41
5 − (3 + 0.7 + 0.7)
IB4 =
= 0.375 mA | I L = 41I B 4 = 15.4 mA
1600
VCE = 5 −130ΩIC − VO = 5V −130Ω(15.4mA) − 3.7 = −0.702 V
5 −1600
Oops! - the transistor is not in the forward - active region. Assume saturation with VCESAT = 0.15V.
I L = I B + IC =
€
5 − (0.8 + 0.7 + 3.0)
1600
+
5 − (0.15 + 0.7 + 3)
130
= 9.16 mA
Page 513
(a ) BiCMOS NAND gate : Replace the CMOS NOR - gate with a two - input CMOS
NAND - gate, and connect its output to the bases of Q3 and Q 4.
(b)
BiNMOS NAND gate : Replace the input CMOS NOR - gate with a two - input CMOS
NAND - gate, and connect M6 and M 7 in series instead of parallel.
€
24
©R. C. Jaeger and T. N. Blalock
08/12/10
Microelectronic Circuit Design
Fourth Edition - Part III
Solutions to Exercises
Updated - 09/25/10
CHAPTER 10
Page 534
(a) AP = Av Ai = 4x104 2.75x108 = 1.10x1013
(
)
Vo 25.3V
=
= 5.06x103
Vi 0.005V
( b)
Vo = 2Po RL = 2(20W )(16Ω) = 25.3 V | Av =
Io =
Vo 25.3V
Vi
0.005V
I
1.58 A
=
= 1.58 A | I i =
=
= 0.167µA | Ai = o =
= 9.48x106
RL
16Ω
RS + Rin 10kΩ + 20kΩ
I i 0.167µA
AP =
25.3V (1.58 A)
Po
=
= 4.79x1010 | Checking : Ap = 5.06x103 9.48x106 = 4.80x1010
PS 0.005V (0.167µA)
(
)(
)
−−−
(
)
(
)
AvdB = 20 log(5060) = 74.1 dB | AidB = 20 log 9.48x106 = 140 dB | APdB = 10 log 4.80x1010 = 107 dB
−−−
(
)
(
)
(
)
AvdB = 20 log 4x10 4 = 92.0 dB | AidB = 20 log 2.75x108 = 169 dB | APdB = 10 log 1.10x1013 = 130 dB
Page 541
€
Gin = g11 =
1
= 0.262 µS | A = g21 = 0.262µS (76)(50kΩ) = 0.995
20kΩ + 76(50kΩ)
−1
Rout
 1
1
75 
g22
262Ω
= g22 = 
+
+
=−
= −0.0131
 = 262 Ω | g12 = −
 50kΩ 20kΩ 20kΩ
(20kΩ) (20kΩ)
Rin =
€
1
1
= 3.82 MΩ | A = g21 = 0.995 | Rout =
= 262 Ω
g11
g22
1
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 547
(a ) The constant slope region spanning a maximum input range is between - 0.5 V ≤ vID ≤ 1.5 V ,
and the bias voltage VID should be centered in this range : VID =
vID = VID + vid
1.5 + (−0.5)
V = +0.5 V.
2
| - 0.5V ≤ 0.5V + vid → v id ≥ −1 V and 0.5V + vid ≤ 1.5 → v id ≤ +1 V
∴−1 V ≤ vid ≤ +1 V or vid ≤ 1 V and v o ≤ 10 V
(b) For VID = −1 V , the slope of the voltage transfer characteristics is zero, so A = 0.
−−−
vO = 10(vID − 0.5V ) = 10(−0.5 + 0.25 + 0.75sin1000πt ) = (−2.5 + 7.5sin1000πt ) V | VO = −2.5 V
€
€
€
Page 552
10V
10V
vid =
= 0.100V = 100 mV | v id = 4 = 0.001 V = 1.00 mV
100
10
10V
vid = 6 = 1.00x10−5 V = 10.0 µV
10
Page 554
360kΩ
Av = −
= −5.29 | vo = −5.29(0.5V ) = −2.65 V
68kΩ
0.5V
ii =
= 7.35 µA | io = −i2 = −ii = −7.35 µA
68kΩ
Page 556
2V
II =
= 426 µA | I2 = I I = 426 µA | IO = −I2 = −426 µA
4.7kΩ
24kΩ
Av = −
= −5.11 | VO = −5.11(2V ) = −10.2 V
4.7kΩ
Page 558
€
Atr = −R2 = −
5V
= −0.2 MΩ | R2 = 200 kΩ
25µA
(
)
vo = −R2ii = −2x105 5x10−5 sin 2000πt = −10sin 2000πt V
€
2
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 560
36kΩ
−3.80V
Av = 1+
= +19.0 | vo = 19.0(−0.2V ) = − 3.80 V | io =
= −100 µA
2kΩ
36kΩ + 2kΩ
−−−
39kΩ
Av = 1+
= +40.0 | AvdB = 20 log(40.0) = 32.0 dB | Rin = 100kΩ ∞ = 100kΩ
1kΩ
10.0V
vo = 40.0(0.25V ) = 10.0 V | io =
= 0.250 mA
39kΩ + 1kΩ
−−−
54
20
Av = 10 = 501
R1 + R2 ≥ 100kΩ
1+
R2
= 501
R1
R2
= 500
R1
io =
501R1 ≥ 100kΩ → R1 ≥ 200 Ω
vo
R2 + R1
10
≤ 0.1 mA
R2 + R1
There are many possibilities.
( R1 = 200 Ω, R2 = 100 kΩ), but ( R1 = 220 Ω, R2 = 110 kΩ) is a better solution since
resistor tolerances could cause io to exceed 0.1 mA in the first case.
Page 563
€
30kΩ
= −20.0 | Rin = R1 = 1.5 kΩ
1.5kΩ
v
−3.00V
vo = −20.0(0.15V ) = −3.00 V | io = o =
= −100 µA
R2
30kΩ
Inverting Amplifier : Av = −
Non - Inverting Amplifier : Av = 1+
vo = 21.0(0.15V ) = 3.15 V | io =
30kΩ
v 0.15V
= +21.0 | Rin = i =
=∞
1.5kΩ
ii
0A
vo
3.15V
=
= 100 µA
R2 + R1 30kΩ + 1.5kΩ
−−−
Add resistor R3 in parallel with the op amp input as in the schemeatic on page 560 with R3 = 2 kΩ.
€
Page 564
 3kΩ 
 3kΩ 
Vo1 = 2V  −
 = −6V | Vo2 = 4V  −
 = −6V | vo = (−6sin1000πt − 6sin 2000πt ) V
 1kΩ 
 2kΩ 
v
v
The summing junction is a virtual ground : Rin1 = 1 = R1 = 1 kΩ | Rin2 = 2 = R2 = 2 kΩ
i1
i2
I o1 =
Vo1 −6V
V
−6V
=
= −2mA | I o2 = o2 =
= −2mA | io = (−2sin1000πt − 2sin 2000πt ) mA
R3 3kΩ
R3 3kΩ
€
3
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 567
Since i+ = 0, I2 =
3V
= 27.3 µA
10kΩ + 100kΩ
−−−
100kΩ
V − V V − V+
= −10.0 | VO = −10(3V − 5V ) = +20.0 V | IO = O − = O
10kΩ
100kΩ 100kΩ
R4
100kΩ
V+ = V2
=5
= 4.545V
R3 + R4
10kΩ + 100kΩ
Av = −
IO =
€
20.0 − 4.545
5V
= +155 µA | I2 =
= 45.5 µA
100kΩ
10kΩ + 100kΩ
Page 568
36kΩ
V − V V − V+
Av = −
= −18.0 | VO = −18(8V − 8.25V ) = 4.50 V | IO = O − = O
2kΩ
36kΩ
36kΩ
R2
36kΩ
4.50 − 7.816
V+ = V2
= 8.25
= 7.816 V | IO =
= −92.1 µA
R1 + R2
2kΩ + 36kΩ
36kΩ
Page 570
€
€
2π x 106
Av ( s) = −
=
s + 5000π
−400
5000π
→ Amid = −400 | f H =
= 2.50 kHz
s
2π
1+
5000π
BW = f H − f L = 2.50 kHz − 0 = 2.50 kHz | GBW = (400)(2.50kHz) = 1.00 MHz
Page 572
1
1
fH =
= 804 kHz
2π 1kΩ 100kΩ (200 pF )
(
)
Page 573
250
250π
1+
s
€
Av ( s) =
€
Page 575
1
1
fL =
= 15.8 Hz
2π 1kΩ 100kΩ (0.1µF )
(
| Ao = 250 | f L =
250π
= 125 Hz | f H = ∞ | BW = ∞ −125 = ∞
2π
)
€
4
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 577
−1+ 4
150
=
−1+ 2 + j2 1+ j2
Av ( j1) = 50
Av ( j1) =
|
150
2
(1) + (2)
2
= 67.08
 2
A vdB = 20 log(67.08) = 36.5 dB | ∠Av ( j1) = ∠(50) + ∠(3) − tan −1  = 0 + 0 − 63.4 o = −63.4 o
 1
Av ( j5) = 50
−25 + 4
1050
=
−25 + 2 + j10 23 − j10
Av ( j5) =
|
1050
2
(−23) + (10)
2
= 41.87
 10 
o
o
AvdB = 20 log(41.87) = 32.4 dB | ∠Av ( j5) = ∠(1050) + − tan −1
 = 0 − −23.5 = +23.5
 −23
(
)
−−−
A v ( jω ) =
20
0.1ω
1+ j
1− ω 2
20
Av ( j0.95) =
|
2
12 +
(0.1) (0.95 )
= 14.3
2
(1− 0.95 )
2
2
0.1(0.95)
 = 0 − 44.3o = −44.3o
∠Av ( j0.95) = ∠20 − tan −1
2
 1− 0.95 
 0.1(1) 
20
−1
 = 0 − 90 o = −90.0 o
Av ( j1) =
= 0 | ∠Av ( j1) = ∠20 − tan 
2
2 2
 1−1 
0.1) 1
(
2
1 +
2
1−12
(
)
( )
( )
( )
20
Av ( j1.1) =
2
12 +
( )
(1−1.1 )
 0.1(1.1) 
 = 0 − −27.6 o = +27.6 o
| ∠Av ( j1.1) = ∠20 − tan−1
2
 1−1.1 
(
= 17.7
(0.1) 1.12
2
)
2
−−−
−400
100 
s 
1+
1+

s  50000 

(i ) A (s) = 
v
fL =
|
Ao = 400 or 52 dB
100
50000
= 15.9 Hz | f H =
= 7.96 kHz | BW = 7960 −15.9 = 7.94 kHz
2π
2π
€
5
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 577
2x10 7 s
−400
=
(s + 100)(s + 50000) 1+ 100 1+ s 
s  50000 

 ω 
 ω 


 100 
−1
o
−1 100
−1
∠Av ( jω ) = −180 + 90 o − tan −1
 − tan 
 = −90 − tan 
 − tan 

100 
 50000 
 100 
 50000 
Av ( s) = −
∠Av ( j0) = −90 − 0 − 0 = −90 o
 100 
 100 
−1
o
∠Av ( j100) = −90 o − tan−1
 − tan 
 = −90 − 45 − 0.57 = −136
 100 
 50000 
 50000 


−1 50000
o
∠Av ( j50000) = −90 o − tan−1
 − tan 
 = −90 − 89.9 − 45 = −225
 100 
 50000 
∠Av ( j∞) = −90 − 90 − 90 = −270 o
€
Page 581
26
R2
Av = − = −10 20 = −20.0 | R1 = Rin = 10 kΩ
R1
R2 = 20R1 = 200 kΩ | C =
1
= 265 pF
2π (3kHz)(200kΩ)
Closest values : R1 = 10 kΩ | R2 = 200 kΩ | C = 270 pF
€
Page 582
20
R2
Av = − = −10 20 = −10.0 | R1 = Rin = 18 kΩ
R1
R2 = 10R1 = 180 kΩ | C =
1
1
=
= 1.77 nF = 1770 pF
ω L R1 2π (5kHz)(18kΩ)
Closest values : R1 = 10 kΩ | R2 = 180 kΩ | C = 1800 pF
Page 583
€
Rin = R1 = 10 kΩ | ΔV = −
vO
2
(10V 2)  1  1ms = 0.05 µF
I
ΔT | C =

(
)
C
10kΩ  10V 
4
6
8
t (msec)
€
-10V
6
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 567
vO = −RC
dv I
= −(20kΩ)(0.02µF )(2.50V )(2000π )(cos2000πt ) = −6.28cos2000πt V
dt
€
7
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 11
Page 605
1
AvIdeal = = 100
β
| T = Aβ =
105
= 1000
100
vo = Av v i = 99.9(0.1V ) = 9.99 V | vid =
| Av = AvIdeal
T
1000
= 100
= 99.90
1+ T
1001
v o 9.99V
=
= 99.9 µV
A
105
−−−
AvIdeal = −
R2
R1
Av = AvIdeal
|
1
R
R
= 1+ 2 → 2 = 99
β
R1
R1
| AvIdeal = −99 | T = Aβ =
105
= 1000
100
T
1000
= −99
= −98.90
1+ T
1001
vo = Av v i = −98.9(0.1V ) = −9.89 V | v id =
vo −9.89V
=
= −98.9 µV
A
105
−−−
Values taken from OP - 27 specification sheet
(www.jaegerblalock.com or www.analog.com)
−−−
Values taken from OP - 27 specification sheet
(www.jaegerblalock.com or www.analog.com)
€
Page 606
1
R
39kΩ
AvIdeal = = 1+ 2 = 1+
= +40.0
β
R1
1kΩ
FGE =
1
= 0.00398 or 0.398 %
1+ T
| T = Aβ =
| FGE ≅
10 4
= 250
40
| Av = AvIdeal
T
250
= 40
= 39.8
1+ T
251
1
= 0.40 %
T
−−−
AvIdeal = −
R2
39kΩ
=−
= −39.0
R1
1kΩ
Av = AvIdeal
| β=
1
1
=
R
40
1+ 2
R1
|
T = Aβ =
10 4
= 250
40
T
250
1
= −39
= −38.8 | FGE =
= 0.00398 or 0.398 %
1+ T
251
1+ T
|
| FGE ≅
1
= 0.40 %
T
€
8
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 608
Values taken from OP - 77 specification sheet (www.jaegerblalock.com or www.analog.com)
€
Page 609
1
R
50Ω
1+ T = o → T =
−1 = 499 | A = T   = 499( 40) = 2.00x10 4
Rout
0.1Ω
β 
−−−
Av = AvIdeal
T
1+ T
T = Aβ = 10 4
| AvIdeal = 1+
R2
39kΩ
= 1+
= +40.0
R1
1kΩ
1kΩ
= 249.7
0.05kΩ + 39kΩ + 1kΩ
| Av = AvIdeal
T
249.7
= 40
= 39.8
1+ T
250.7
--Avmax = 1+
Avmin = 1+
€
€
39kΩ(1.05)
1kΩ(0.95)
39kΩ(0.95)
1kΩ(1.05)
= 44.1 | GE = 44.2 − 40.0 = 4.20 | FGE =
= 36.3 | GE = 36.3 − 40.0 = −3.70
FGE =
4.20
= 10.5 %
40
−3.70
= −9.3 %
40
Page 610
1
R
200Ω
1+ T = o → T =
−1 = 1999 | A = T   = 1999(100) = 2.00x10 5 or 106 dB
Rout
0.1Ω
β 
Page 612
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
Page 613
€
9
©R. C. Jaeger and T. N. Blalock
09/25/10


10kΩ 1MΩ
10 4
=
Rin = Rid (1+ T ) | T = Aβ = 10 4 
 40.39 = 248 | Rin = 1MΩ [1+ 248] = 249 MΩ
10kΩ
1MΩ
+
390kΩ


vi
1V
βv
T
248
=−
= −4.02 nA | i1 = o | vo = AvIdeal
v i = 40
vi = 39.8v i
Rin
249 MΩ
R1
1+ T
249
βv
39.8  1V 
i1 = o =

 = 98.5 µA | Yes, i1 >> i−
R1 40.4  10kΩ 

 10 4
10kΩ
If we assume 1MΩ >> 10kΩ, T = Aβ = 10 4 
= 250 | Rin = 1MΩ [1+ 250] = 251 MΩ
=
10kΩ + 390kΩ  40
i− = −
vi
1V
βv
T
250
=−
= −3.98 nA | i1 = o | vo = AvIdeal
v i = 40
vi = 39.8v i
Rin
251MΩ
R1
1+ T
251
βv
39.8  1V 
i1 = o =

 = 99.5 µA | Yes, i1 >> i−
R1
40  10kΩ 
i− = −
€
Page 614
Rin ≅ R1 + Rid
R2
100kΩ
= 1kΩ + 1MΩ
= 1001 Ω | Rinideal = R1 = 1000 Ω | 1 Ω or 0.1 %
1+ A
1+ 10 5
Page 622
€
10
©R. C. Jaeger and T. N. Blalock
09/25/10
AvIdeal = 1+
R2
91kΩ
= 1+
= 10.1
R1
10kΩ
RinD = Rid + R1 ( R2 + Ro ) = 25kΩ + 10kΩ (91kΩ + 1kΩ) = 34.0 kΩ
[
]
[
]
D
Rout
= Ro R2 + R1 Rid = 1kΩ 91kΩ + 10kΩ 25kΩ = 990 Ω
T = Ao
R1 Rid
Ro + R2 + R1 Rid
= 10 4
10kΩ 25kΩ
1kΩ + 91kΩ + 10kΩ 25kΩ
= 720
T
720
= 10.1
= 10.1
1+ T
1+ 720
Rin = RinD (1+ T ) = 34.0kΩ(1+ 720) = 24.5 MΩ
Av = AvIdeal
Rout =
D
Rout
990Ω
=
= 1.37 Ω
1+ T 1+ 720
−−−
AvIdeal = 1+
R2
91kΩ
= 1+
= 10.1
R1
10kΩ
(
)
(
)
[ R + R ( R + R )] = 5kΩ 1kΩ [91kΩ + 10kΩ (25kΩ + 2kΩ)] = 826 Ω
RinD = RI + Rid + R1 R2 + Ro RL = 2kΩ + 25kΩ + 10kΩ 91kΩ + 1kΩ 5kΩ = 36.0 kΩ
D
Rout
= RL Ro
2
1
id
I


RL
R1
5kΩ
10kΩ
vth =  Ao
vid 
= 10 4
v id
= 818v id
1kΩ + 5kΩ 1kΩ 5kΩ + 91kΩ + 10kΩ
 Ro + RL  RL Ro + R2 + R1
(
)
(
)
Rth = R1 R2 + RL Ro = 10kΩ 91kΩ + 1kΩ 5kΩ = 9.02 kΩ
v 
Rid
25kΩ
T = − th 
= −818
= −568
9.02kΩ + 25kΩ + 2kΩ
 v id  Rth + Rid + RI
T
568
= 10.1
= 10.1
1+ T
1+ 568
Rin = RinD (1+ T ) = 36.0kΩ(1+ 568) = 20.5 MΩ
Av = AvIdeal
Rout =
D
Rout
826Ω
=
= 1.45 Ω
1+ T 1+ 568
−−−
Continued on next page.
€
11
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 622 cont.
R
91kΩ
AvIdeal = 1+ 2 = 1+
= 10.1
R1
10kΩ
D
RinD = Rid = 25kΩ | Rout
= Ro = 1 kΩ
T = Ao
R1
10kΩ
= 10 4
= 990
R1 + R2
10kΩ + 91kΩ
T
990
= 10.1
= 10.1
1+ T
1+ 990
Rin = RinD (1+ T ) = 25.0kΩ(1+ 990) = 24.8 MΩ
Av = AvIdeal
Rout =
D
Rout
1kΩ
=
= 1.01 Ω
1+ T 1+ 990
−−−
AvIdeal = 1+
R2
91kΩ
= 1+
= 10.1
R1
10kΩ
RinD = RI + Rid + R1 ( R2 + Ro ) = 5kΩ + 25kΩ + 10kΩ (91kΩ + 1kΩ) = 39.0 kΩ
[
]
[
]
D
Rout
= Ro R2 + R1 ( Rid + RI ) = 1kΩ 91kΩ + 10kΩ (25kΩ + 5kΩ) = 990 Ω
vth = Aov id
R1
10kΩ
= 10 4 v id
= 980v id
Ro + R2 + R1
1kΩ + 91kΩ + 10kΩ
Rth = R1 ( R2 + Ro ) = 10kΩ (91kΩ + 1kΩ) = 9.02 kΩ
T=
vth
Rid
25kΩ
= 980
= 628
vid Rth + Rid + RI
9.02kΩ + 25kΩ + 5kΩ
T
628
= 10.1
= 10.1
1+ T
1+ 628
Rin = RinD (1+ T ) = 39.0kΩ(1+ 628) = 24.5 MΩ
Av = AvIdeal
Rout =
D
Rout
990Ω
=
= 1.57 Ω
1+ T 1+ 628
€
12
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 626
RinD =
vi
= RI Rid ( RF + Ro )
ii
|
D
Rout
=
vo
= Ro RF + Rid RI
io
(
)
Page 628
AtrIdeal = −RF = −91kΩ
€
RinD = RI ( RF + Ro ) = 10kΩ (91kΩ + 1kΩ) = 9.02 kΩ
D
Rout
= Ro ( RF + RI ) = 1kΩ (91kΩ + 10kΩ) = 990 Ω
T = Ao
RI
10kΩ
= 10 4
= 980
Ro + RF + RI
1kΩ + 91kΩ + 10kΩ
T
980
= −91kΩ
= −90.9 kΩ
1+ T
1+ 980
D
RinD
9.02kΩ
Rout
990Ω
Rin =
=
= 9.19 Ω Rout =
=
= 1.01 Ω
1+ T 1+ 980
1+ T 1+ 980
−−−
Atr = AtrIdeal
AtrIdeal = −RF = −91kΩ
RinD = Rid ( RF + Ro ) = 25kΩ (91kΩ + 1kΩ) = 19.7 kΩ
D
Rout
= Ro ( RF + Rid ) = 1kΩ (91kΩ + 25kΩ) = 991 Ω
T = Ao
Rid
25kΩ
= 10 4
= 2137
Ro + RF + Rid
1kΩ + 91kΩ + 25kΩ
T
2137
= −91kΩ
= −91.0 kΩ
1+ T
1+ 2137
RD
19.7kΩ
RD
991Ω
Rin = in =
= 9.21 Ω Rout = out =
= 0.464 Ω
1+ T 1+ 2137
1+ T 1+ 2317
Atr = AtrIdeal
€
13
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 633
1
1
=−
= −10−4 S
R
10kΩ
D
Rin = Rid + R Ro = 25kΩ + 10kΩ 1kΩ = 25.9 kΩ
AtcIdeal = −
D
Rout
= Ro + R Rid = 1kΩ + 10kΩ 25kΩ = 8.14 kΩ
v  R
25kΩ
id
T = − th 
= 9090
= 8770
0.909kΩ + 25kΩ
 v id  Rth + Rid
1 T 
1  8770 
Atc = − 
=−

 = −0.100 mS
R  1+ T 
10kΩ  1+ 8770 
Rin = RinD (1+ T ) = 25.9kΩ(1+ 8770) = 227 MΩ
D
Rout = Rout
(1+ T ) = 8.14kΩ(1+ 8770) = 71.4 MΩ
Page 637
€
AiIdeal = 1+
R2
27kΩ
= 1+
= +10
R1
3kΩ
(
)
(
)
RinD = Rid R2 + R1 Ro = 25kΩ 27kΩ + 3kΩ 1kΩ = 13.2 kΩ
D
Rout
= Ro + R1 ( R2 + Rid ) = 1kΩ + 3kΩ (27kΩ + 25kΩ) = 3.84 kΩ
T = Ao
T = 10 4
[ R ( R + R )]
 R

id


R1 ( R2 + Rid ) + Ro  R2 + Rid 
1
2
id
3kΩ (27kΩ + 25kΩ)


25kΩ

 = 3555
1kΩ + 3kΩ (27kΩ + 25kΩ)  27kΩ + 25kΩ 
 T 
 3555 
Ai = +10
 = +10
 = +10.0
1+ T 
 1+ 3555 
Rin =
RinD
13.2kΩ
=
= 3.71 Ω
(1+ T ) 1+ 3555
D
Rout = Rout
(1+ T ) = 3.84kΩ(1+ 3555) = 13.7 MΩ
€
14
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 638
Rid' = RI Rid = 10kΩ 25kΩ = 7.14 kΩ
AiIdeal = 1+
R2
270kΩ
= 1+
= +10
R1
30kΩ
(
)
(
)
RinD = Rid' R2 + R1 Ro = 7.14kΩ 270kΩ + 30kΩ 1kΩ = 6.96 kΩ
(
)
D
Rout
= Ro + R1 R2 + Rid' = 1kΩ + 30kΩ (270kΩ + 7.14kΩ) = 28.1 kΩ
T = Ao
T = 10 4
[ R ( R + R )]
1
(
2
R1 R2 + Rid'
'
id
)
 R' 
id


'
R
+
+ Ro  2 Rid 
30kΩ (270kΩ + 7.14kΩ)


7.14 kΩ

 = 248.5
30kΩ (270kΩ + 7.14kΩ) + 1kΩ  270kΩ + 7.14kΩ 
 T 
 248.5 
Ai = +10
 = +10
 = +9.96
1+ T 
 1+ 248.5 
Rin =
RinD
6.96kΩ
=
= 27.9 Ω
(1+ T ) 1+ 248.5
D
Rout = Rout
(1+ T ) = 28.1kΩ(1+ 248.5) = 7.01 MΩ
Page 644
€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
−−−
VO ≤ 50(0.002V ) → −0.100 V ≤ VO ≤ +0.100 V
Page 647
€
Values taken from op - amp specification sheets (via www.jaegerblalock.com or www.analog.com)
−−−
R = 39kΩ 1kΩ = 975 Ω
R = 1 kΩ is the closest 5% value, or one could use 39 kΩ and 1 kΩ resistors in parallel.
−−−
V
I
1.5mV
100nA
v O (t ) = VOS + OS t + B 2 t | 1.5mV +
t+
t = 15V → t = 6.00 ms
RC
C
10kΩ(100 pF ) 100 pF
Page 648
€
€
Values taken from op - amp specification sheets (via www.jaegerblalock.com or www.analog.com)
15
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 650
Values taken from op - amp specification sheets (via www.jaegerblalock.com or www.analog.com)
−−−
−1
 1
1 
REQ
R1 + R2 ≥ 
−
 = 20kΩ
 4kΩ 5kΩ 
Including 5% tolerances, R1 + R2 ≥ 21kΩ Av = 10 → R2 = 9R1
20V
= RL ( R2 + R1 ) ≥
= 4kΩ
5mA
A few possibilities : 27 kΩ and 3 kΩ, 270 kΩ and 30 kΩ, 180 kΩ and 20 kΩ, etc.
€
€
Page 653

v ic 
vo = A vid +

CMRR 



v ic 
5.000 
vomin = A vid +
 = 25000.002 −
 = 3.750 V
CMRR 
10 4 




v ic 
5.000 
vomax = A v id +
 = 2500 0.002 +
 = 6.250 V |
CMRR 
10 4 


Page 655


1 
1 
A1+
10 41+

4
 2CMRR 
 2x10 
Av =
=
= 1.000


1 
1 
4
1+ A1−
 1+ 10 1−
4
 2CMRR 
 2x10 
3.750 V ≤ v O ≤ 6.250 V

1 
103 1+
3
 2x10 
Av =
= 1.000

1 
3
1+ 10 1−
3
 2x10 
Page 656
€
GE = FGE ( Av ) ≤ 5x10−5 (1) = 5x10−5
Worst case occurs for negative CMRR : GE ≅
1
1
+
A CMRR
1
= 4 x10 4 or 92 dB
2.5x10−5
−1

1 
−5
A =  5x10 −

CMRR 

If both terms make equal contributions: A = CMRR =
−1

1
−5
For other cases : CMRR =  5x10 − 
A

A = 100 dB
−1

1 
−5
CMRR = 5x10 − 5  = 2.5x10 4 or 88 dB
10 

CMRR = 100 dB
€
€
or
−1

1 
−5
A = 5x10 − 5  = 2.5x10 4 or 88 dB
10 

Page 657
Values taken from op - amp specification sheets (via www.jaegerblalock.com or www.analog.com)
16
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 661
Ao = 10
A( s) =
100
20
= 10
5
(
)
6
ωT 2π 5x10
10 7 π
100π
| ωB =
=
=
= 100π | f B =
= 50 Hz
5
5
Ao
2π
10
10
ωT
10 7 π
=
s + ω B s + 100π
−−−
ωT
2π x 10 6
2π x 10 6
=
=
ω
s + 10π
2π x 10 6
s+ T s+
5
Ao
2x10
Av ( s) =
Page 664
€
Ao = 10
90
20
| fB =
= 31600
(
)
fT 5x10 6
=
= 158 Hz | f H ≅ βfT = 0.01(5 MHz) = 50 kHz
Ao 31600
2π 5x10 6
ωT
10 7 π
A( s) =
=
=
s + ω B s + 2π (158) s + 316π
|
Av ( s) =
(
2π 5x10 6
)
(
s + 2π 5x10 4
)
=
10 7 π
s + 10 5 π
−−−
Aβ =
ωT
βωT 1
β | For ω H >> ω B : Aβ ≅
= = − j1 since ω H = βωT
s + ωB
jω H j
Page 666
90
€
fT 5x10 6
5 MHz
=
= 158 Hz | f H ≅ βfT =
= 15.8 kHz
50
Ao 31600
1+ 10 20
2π 5x10 6
2π 5x10 6
10 7 π
10 7 π
A( s) =
=
| Av ( s) =
=
s + 2π (158) s + 316π
s + 3.16x10 4 π
s + 2π 15.8x10 3
Ao = 10 20 = 31600
(
| fB =
)
(
(
)
)
−−−
f H ≅ βfT = 1(10 MHz) = 10 MHz | f H ≅ βfT =
1
(10 MHz) = 5 MHz
2
Page 667
€
100
20
fT 10x10 6
10 MHz 10 MHz
=
= 100 Hz | f H ≅ βfT =
=
= 10 kHz
5
60
Ao
1000
10
20
10
7
7
2π 10
2π 10
ωT
2πx10 7
2x10 7 π
A( s) =
=
=
| Av ( s) =
=
s + ω B s + 2π (100) s + 200π
s + 2x10 4 π
s + 2π 10 4
Ao = 10
= 10 5
| fB =
( )
€
( )
( )
17
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 669
SR 5x10 5V / s
SR
5x10 5 V / s
VM ≤
=
= 3.98 V | f M =
=
= 7.96 kHz
ω 2π (20kHz)
2πVFS
2π (10V )
€
Page 670
Values taken from op - amp specification sheets (via www.jaegerblalock.com or www.analog.com)
Ao = 1.8x10 6
| f T = 8 MHz | ω B =
ωT 2π (8 MHz )
1
=
= 8.89π | RC =
=
s
6
Ao
ω B 8.89π
1.8x10
Page 677
€
2

 ln100 
πζ 
 | Let κ = 
0.01 = exp−

2 
 π 
 1− ζ 
φ M = tan−1
2ζ
(
4ζ 4 + 1 − 2ζ 2
)
0.5
| ζ=
κ
= 0.826
1+ κ
= 70.9 o
( )
cos 45o = 4ζ 4 + 1 − 2ζ 2 → ζ = 0.420

πζ 
| Overshoot = 100% exp −
= 23.4 %
2 
 1− ζ 
−−−
Settling within the 10% error bars requires ω nt ≥ 13. ∴ ω n ≥
ω n = ω Bω 2 (1+ Aoβ ) ≅ ω Bω 2 Aoβ = βω T ω 2
13
= 1.3x106 rad / s
−5
10 s
(
1.3x106 / 2π
f n2
| f2 =
≥
βfT
0.1 10 6
( )
)
2
= 428 kHz
Page 678
€
∠T ( jω1 ) = −180 o → 3tan −1
T ( jω180 ) =
5
3
=
( ω + 1) (
2
1
ω1
= 180 → ω1 = 3
1
5
5
8
=
| GM = = 1.60 or 4.08 dB
3
8
5
3+1
)
€
18
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 682
From the upper graph, the final value of the first step is 5 mV, and the peak of the response is
 9.5mm 
5.7mV − 5mV
approximately 4mV + 2mV 
= 14 %
 = 5.7 mV. Overshoot = 100%
5mV
 11mm 

 − ln 0.14  2
πζ 
κ
 | Let κ = 
0.14 = exp−
=
0.3917
|
ζ
=
= 0.5305

2 
π
1+
κ


1−
ζ


2ζ
φ M = tan−1
= 54.2 o
0.5
4ζ 4 + 1 − 2ζ 2
(
)
−−−
From the lower graph, the final value of the first step is 5 mV, and the peak of the response is
 10mm 
9.2mV − 5mV
approximately 5mV + 5mV 
= 84 %
 = 9.2 mV. Overshoot = 100%
5mV
 12mm 

 − ln 0.84  2
πζ 
κ
 | Let κ = 
0.84 = exp−
= 0.05541
 = 0.3080 | ζ =
2 
π
1+
κ


1−
ζ


2ζ
φ M = tan−1
= 6.34 o
0.5
4ζ 4 + 1 − 2ζ 2
(
)
−−−
The µA741 curves will be distorted by slew rate limiting.
€
19
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 12
Page 700
AvA = AvB = AvC = −
R2
68kΩ
=−
= −25.2 | RinA = RinB = RinC = R1 = 2.7 kΩ
R1
2.7kΩ
The op - amps are ideal : RoutA = RoutB = RoutC = 0
−−−
3
Av = AvA AvB AvC = (−25.2) = −16,000
| Rin = RinA = 2.7 kΩ | Rout = RoutC = 0
−−−
2
2
 2.7kΩ 
3
Av = (−25.2) 
 ≥ 0.99(25.2)
 Rout + 2.7kΩ 
 2.7kΩ 
| 
 ≥ 0.99
 Rout + 2.7kΩ 
3
2.7kΩ
≥ 0.9950
Rout + 2.7kΩ
→ Rout ≥ 13.6 Ω
Page 705
€
Av (0) = 50(25) = 1250
Av (ω H ) =
|
1250
2



ω H2
ω H2
1+
 1+
 = 2 → ω2
H
2
2




 (10000π )   (20000π ) 
= 884
( )
ω H2 = 6.925x10 8 → ω H = 26.3x10 3 → f H =
2
+ 4.935x10 9ω H2 − 3.896x1018 = 0
26.3x10 3
= 4190 Hz
2π
−−−
Av (0) = −100(66.7)(50) = −3.33x10
5
Av (ω H ) =
−3.34 x10 5
2
= −2.36x10 5




ω H2
ω H2
ω H2
1+
 1+
 1+
=2
2
2
2






 (10000π )   (15000π )   (20000π ) 
ω H6 + 7.156x10 9 ω H4 + 1.486x1010 ω H2 − 8.562x10 27 = 0
Using MATLAB, ω H = 21.7x10 3 → f H =
21.7x10 3
= 3450 Hz
2π
−−−
3
Av (0) = (−30) = −2.70x10
4
|
1
3
f H = ( 33.3kHz) 2 −1 = 17.0 kHz
€
20
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 711
130kΩ
Av1 = 1+
= 6.909
22kΩ
2
| vO1 = 0.001(6.909) = 6.91 mV | vO 2 = 0.001V (6.909) = 47.7 mV
3
4
vO 3 = 0.001(6.909) = 330 mV | vO 4 = 0.001V (6.909) = 2.28 V
5
vO 5 = 0.001V (6.909) = 15.7 V > 15 V. ∴vO5 = VOmax = 15 V
vO 6 = 15V (6.909) = 104 V > 15 V. ∴vO6 = VOmax = 15 V
Page 714
€
VA = V1 + IR2
| VB = V2 − IR2
| I=
VA − VB 5.001V − 4.999V
=
= 1.00 µA
2R1
2kΩ
VA = V1 + IR2 = 5.001V + 1.00µA( 49kΩ) = 5.05 V
VB = V2 − IR2 = 4.999V −1.00µA( 49kΩ) = 4.95 V
 R 
 10kΩ 
VO = − 4 (VA − VB ) =  −
(5.05 − 4.95) = −0.100 V
 10kΩ 
 R3 
Page 717
€
ALP ( s) =
For Q =
ω o2
ω
s 2 + s o + ω o2
Q
1
2
: ALP ( jω ) =
| ALP (0) =
ω o2
= 1 or 0 dB
ω o2
ω o2
−ω 2 + jω 2ω o + ω o2
|
ALP ( jω H ) =
 1 2
= 
2
+ 2ω o2ω H2  2 
ω o4
2
(ω
2
o
− ω H2
)
2ω o4 = ω o4 + ω H4 → ω H = ω o
−−−
To increase the cutoff frequency from 5 kHz to 10 kHz while maintaining the resistances
10kHz
the same, we must decrease the capacitances by a factor of
=2
5kHz
0.02µF
0.01µF
∴ C1 =
= 0.01 µF | C2 =
= 0.005 µF
2
2
−−−
ALP ( jω ) =
ω o2
ω
−ω 2 + jω o + ω o2
Q
| ALP ( jω o ) =
ω o2
= − jQ |
ω o2
2
2
−ω o + j
+ ωo
Q
ALP ( jω o ) = Q
€
21
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 718
To decrease the cutoff frequency from 5 kHz to 2 kHz, we must increase the
5kHz
resistances by a factor of
= 2.50 → R1 = R2 = 2.50(2.26kΩ) = 5.65 kΩ
2kHz
2C R 2
2
1
=
=
C 2R
2
2
Q=
| Q is unchanged.
−−−
C R1 R2
→ R12 + 2R1 R2 + R22 = 2R1 R2 → R12 = −R22 - - can't be done!
C
R
+
R
2
1
2
 R (R + R )

R1 R2
dQ
1
1
1
1
2


Q=
=
−
R
R
1 2 = 0 → R2 = R1 → Qmax =
2
R1 + R2
dR2 ( R + R )  2 R1 R2
2

1
2
1
=
Page 719
€
AHP ( jω o ) = K
−ω o2
K
=
2
2
2
−ω o + j (3 − K )ω o + ω o 3 − K
| AHP ( jω o ) =
K
∠90 o
3− K
−−−
fo =
1
2π 10kΩ(20kΩ)(0.0047µF )(0.001µF )
= 5.19 kHz
−1


20kΩ(1.0nF ) 
10kΩ 4.7nF + 1.0nF

Q=
+ (1− 2)
= 0.829
 20kΩ 4.7nF 1.0nF

10kΩ
4.7nF
(
)
(
)


€
Page 720
K dQ
SKQ =
Q dK
Q=
| Q=
1
3− K
|
dQ
−1
K dQ
=
−1) = Q 2 | SKQ =
= KQ
2 (
dK (3 − K )
Q dK
1
3
→ KQ = 3Q −1 SKQ = 3Q −1 =
−1 = 1.12
3− K
2
Page 721
€
Rth = 2kΩ 2kΩ = 1kΩ | f o =
1
2π 1kΩ(82kΩ)(0.02µF )(0.02µF )
= 879 Hz | Q =
1 82kΩ
= 4.53
2 1kΩ
€
22
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 726
The lower gain results in a larger gain error and center frequency shift.
−−−
R
294kΩ
ABP ( jω o ) = KQ = 2 | 10 =
→ R1 = 29.4 kΩ
R1
R1
−−−
R=
1
1
=
= 39.8 kΩ | R2 = QR = 10(39.8kΩ) = 398 kΩ
ω oC 2π (2000)(2000 pF )
R2
R1 =
ABP ( jω o )
=
398kΩ
= 19.9 kΩ | R3 can remain the same. |
20
The nearest 1% values are R = 40.2 kΩ, R2 = 402 kΩ, R1 = 20.0 kΩ, R3 = 49.9 kΩ
fo =
1
1
1
1
=
= 1.98 kHz | BW =
=
= 198 Hz
2πRC 2π ( 40.2kΩ)(2nF )
2πR2C 2π (402kΩ)(2nF )
ABP ( jω o ) = −
R2
402kΩ
=−
= −20.1
R1
20.0kΩ
−−−
Blindly using the equations at the top of page 580 yields
f omin =
1
1
=
= 1946 Hz
2πRC 2π (1.01)(29.4kΩ)(1.02)(2.7nF )
f omax =
1
1
=
= 2067 Hz
2πRC 2π (0.99)(29.4kΩ)(0.98)(2.7nF )
BW min =
1
1
=
= 195 Hz
2πR2C 2π (1.01)(294kΩ)(1.02)(2.7nF )
BW max =
1
1
=
= 207 Hz
2πR2C 2π (0.99)(294kΩ)(0.98)(2.7nF )
min
ABP
=−
294kΩ(1.01)
R2
=−
= −20.4
R1
14.7kΩ(0.99)
max
| ABP
=−
294kΩ(0.99)
R2
=−
= −19.6
R1
14.7kΩ(1.01)
The W/C results are similar if R and C are not the same for example where ωo =
1
RA RB CACB
.
€
23
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 727
SCQ1 =
R1 R2  C1 Q
C1 dQ C1  1
=
= 
= 0.5
Q dC1 Q 2 C1C2 R1 + R2  Q 2C1
SRQ2 =
R2 dQ
Q dR2
| R1 = R2 → Q =
1 C1
→ SRQ2 = 0
2 C2
−−−
SRω o =
SKQ =
R dω o R  −1 
ω
C dω o C  −1 
ω
=  2  = − o = −1 | SCω o =
= 
= − o = −1
2
ω o dR ω o  R C 
ωo
ω o dR ω o  RC 
ωo
K dQ K (−1)(−1) K 2
K
=
= Q = KQ =
2
Q dK Q (3 − K )
Q
3− K
−−−
SRω1o =
R1 dω o R1  ω o  dRth
R1  Rth2 
1 R3
= −
=
−
 2=−

ω o dR1 ω o  2Rth  dR1
2Rth  R1 
2 R1 + R3
R2 dω o R2  ω o 
1
= −
=−
ω o dR2 ω o  2R2 
2
R dω o R3  ω o  dRth
R3  Rth2 
1 R1
= 3
= −
=
−
 2=−

ω o dR1 ω o  2Rth  dR3
2Rth  R3 
2 R1 + R3
SRω2o =
SRω3o
SCω o =
C dω o C  ω o 
= −  = −1
ω o dC ω o  C 
SRQ1 =
R1 dQ R1  Q  dRth
R1  Rth2 
1 R3
= −
=
−
 2=−

Q dR1 Q  2Rth  dR1
2Rth  R1 
2 R1 + R3
R2 dQ R2  Q 
1
= 
=+
Q dR2 Q  2R2 
2
R dQ R3  Q  dRth
R3  Rth2 
1 R1
SRQ1 = 3
= −
=
−
 2=−

Q dR3 Q  2Rth  dR3
2Rth  R3 
2 R1 + R3
SRQ2 =
SCQ =
C dQ C
C dBW
C  BW 
= (0) = 0 | SCBW =
=
−
 = −1
Q dC Q
BW dC
BW  C 
€
24
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 728
(a )
fo =
R1 = R2 = 5(2.26kΩ) = 11.3 kΩ | C1 =
1
fo =
= 4980 Hz
(11.3kΩ)(11.3kΩ)(0.004µF )(0.002µF )
11.3kΩ (0.004 µF )(0.002µF )
2
=
= 0.471
2π
Q=
(b)
0.02µF
0.01µF
= 0.004 µF | C2 =
= 0.002 µF
5
5
11.3kΩ 0.004µF + 0.002µF
3
R1 = R2 = 0.885(2.26kΩ) = 2.00 kΩ | C1 =
0.02µF
0.01µF
= 0.0226 µF | C2 =
= 0.0113 µF
0.885
0.885
1
= 4980 Hz
(2.00kΩ)(2.00kΩ)(0.0226µF )(0.0113µF )
2.00kΩ (0.0226µF )(0.0113µF )
2
=
= 0.471
2π
Q=
2.00kΩ 0.0226µF + 0.0113µF
3
−−−
fo =
1
2π
(2kΩ 2kΩ)(82kΩ)(0.02µF )(0.02µF )
= 879 Hz | Q =
The values of the resistors are unchanged. C1 = C2 =
fo =
€
1
2π
(1kΩ)(82kΩ)(0.005µF )(0.005µF )
82kΩ (0.02µF )(0.02µF )
= 4.53
1kΩ 0.02µF + 0.02µF
0.02µF
= 0.005 µF
4
= 3520 Hz | Q =
82kΩ (0.005µF )(0.005µF )
= 4.53
1kΩ 0.005µF + 0.005µF
Page 728
C
2 pF
ΔvO = − 1 VI = −
0.1V = −0.4 V
C2
0.5 pF
vO (T ) = 0 + ΔvO = −0.4 V | vO (5T ) = 0 + 5ΔvO = −2.0 V | vO (9T ) = 0 + 9ΔvO = −3.6 V
Page 732
€
fo =
Q=
1
C3C4 200kHz 4 pF (0.25 pF )
fC
=
= 10.6 kHz
2π
C1C2
2π
3 pF (3 pF )
3 pF ( 3 pF )
C3 C1C2
4 pF
f 10.6 kHz
=
= 2 | BW = o =
= 5.30 kHz
C4 C1 + C2
0.25 pF 3 pF + 3 pF
Q
2
ABP ( jω o ) = −
R2
C
4 pF
=− 3 =−
= −8.00
2R1
2C4
0.5 pF
25
€
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 734
0.011000012 = 2−2 + 2−3 + 2−8
(
)
10
(
| 0.10001000 2 = 2−1 + 2−5
= 0.3798062510
)
10
= 0.5312510
−−−
VO =
VLSB
5.12V 11
2 + 2 9 + 2 7 + 2 5 + 2 3 + 21 = 3.41250 V
12
2
5.12V
5.12V
= 12 = 1.25 mV | VMSB =
= 2.56 V
2
2
(
)
Page 737
€
VOS = VO (000) = 0.100 VFS
| VLSB =
0.8VFS − 0.1VFS
= 0.1 VFS
7
−−−
2R = 1 kΩ | 4 R = 2 kΩ | 8R = 4 kΩ | 16R = 8 kΩ | 32R = 16 kΩ | 64 R = 32 kΩ
128R = 64 kΩ | 256R = 128 kΩ | R = 500 Ω
€
Page 738
RTotal = R + 2R + 2R + ( n −1)(2R + R) = ( 3n + 2) R | RTotal = ( 3x8 + 2)(1kΩ) = 26 kΩ
−−−
R = 1 kΩ | 2R = 2 kΩ | 4 R = 4 kΩ | 8R = 8 kΩ | 16R = 16 kΩ |
64 R = 64 kΩ | 128R = 128 kΩ | 256R = 256 kΩ | RTotal = 511 kΩ
(
) (
)
(
32R = 32 kΩ
)
In general : RTotal = R 2 0 + 21 + ... + 2 n−1 + 2 n = 2 n+1 −1 R | RTotal = 2 8+1 −1 1kΩ = 511 kΩ
€
Page 739
The general case requires 2 n resistors, and the number of switches is
(2 + 2
1
2
) (
) (
)
+ ... + 2 n = 2 2 0 + 2 2 + ... + 2 n−1 = 2 2 n −1 = 2 n+1 − 2
210 = 1024 resistors | 210+1 − 2 = 2046 switches.
€
Page 740
(a ) In general : CTotal = R 20 + 21 + ... + 2n = 2n+1 −1 C | CTotal = 28+1 −1 1pF = 511 pF
(
(b) In general : R
Total
€
) (
)
= 2C + 2C + ( n −1)(2C + C ) = 2R + n ( 3R)
(
)
| RTotal = 2R + 8( 3kΩ) = 26 kΩ
Page 741
5V
2 8 LSB
VLSB = 8 = 19.53 mV | 1.2V
= 61.44 LSB | The closest code is 6110 = 001111012
5V
2
€
26
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 743
2 n ≥ 10 6
| n≥
6 log10
= 19.93 → n ≥ 20 bits
log2
−−−
The minimum width is 0 corresponding to the missing code 110.
The maximum code width is 2.5 LSB corresponding to output code 101.
DNL = 2.5 −1 = 1.5 LSB
At code 110, the ADC transfer characteristic is 1 LSB off of the fitted line.
∴ INL = 1 LSB
€
€
€
€
€
Page 744
2n
212
1
1
conversions
max
TT =
=
= 2.048 ms | N max = max =
= 488
6
fC 2x10
2.048ms
second
TT
Page 745
n
12
1
1
conversions
TT =
=
= 6.00 µs | N max =
=
= 167,000
6
fC 2x10
TT 6µs
second
Page 748
1
VFS =
RC
T
VR T
∫ VR (t )dt = RC
0
VR
VR 2 n 2.00V  2 8 
| RC =
T=
=

 = 0.100 ms
VFS
VFS f C 5.12V 1MHz 
Page 749
2 n+1
217
1
1
conversions
TTmax =
= 6
= 0.131 s | N max = max =
= 7.63
f C 10 Hz
0.131s
second
TT
Page 750
In general, 2 n resistors and 2 n −1 comparators :
(
(
)
)
210 = 1024 resistors and 210 −1 = 1023 comparators
Page 758
€
fo =
1
= 15.9kHz
2π (10kΩ)(1nF )
|
vo =
3(0.6V )
= 3.00 V
 10kΩ  24kΩ  24kΩ
2 −
1+
−
 10kΩ  12kΩ  10kΩ
−−−
SPICE Results :15.90 kHz, 3.33 V
€
27
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 760
For vI > 0, the diode will conduct and pull the output up to vO = v I = 1.0 V.
v1 = vO + vD = 1.0 + 0.6 = 1.6 V
For a negative input, there is no path for current through R, so vO = 0 V. The op - amp
sees a -1V input so the output will limit at the negative power supply : vO = −10 V.
(Note that the output voltage will actually be determined by the reverse saturation current
of the diode : vO = −I S R ≅ 0.)
The diode has a 10 - V reverse bias across it, so VZ > 10 V.
€
Page 762
vS = +2 V : Diode D2 conducts, and D1 is off. The negative input is a virtual ground.
v1 = −v D2 = −0.6 V.
The current in R is 0, so vO = 0 V.
vI = −2 V : Diode D1 conducts, and D2 is off. The negative input is a virtual ground.
vO = −
R2
68kΩ
vI = −
(−2V ) = +6.18 V | v1 = vO + vD1 = 6.78 V.
R1
22kΩ
The maximum output voltage is vOmax = 15V − 0.6V = 14.4 V.
68kΩ
14.4V
= −3.09 | vI =
= −4.66 V
22kΩ
−3.09
When vO = 15 V , v D 2 = −15 V , so VZ = 15 V.
Av = −
€
Page 763
20kΩ  10.2kΩ  2V
vO =
= 2.00 V


20kΩ  3.24kΩ  π
Page 765
€
VI − = −
R1
1kΩ
VEE = −
10V = −0.990 V
R1 + R2
1kΩ + 9.1kΩ
VI + = +
R1
1kΩ
VCC =
10V = +0.990 V
R1 + R2
1kΩ + 9.1kΩ
Vn = 0.990V − (−0.990V ) = 1.98 V
Page 766
€
€
R1
6.8kΩ
1
=
=
R1 + R2 6.8kΩ + 6.8kΩ 2
 1+ 0.5 
1
T = 2(10kΩ)(0.001µF ) ln
 = 21.97µs | f = = 45.5 kHz
T
 1− 0.5 
T = 2RC ln
1+ β
1− β
| β=
28
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 770

0.7 
1+


R1
22kΩ
5  = 20.4 µs
β=
=
= 0.550 | T = (11kΩ)(0.002µF ) ln
R1 + R2 22kΩ + 18kΩ
1− 0.550 



 5V  
1+ 0.55  
 5V  
Tr = (11kΩ)(0.002µF ) ln
= 13.0 µs | Tmin = 20.4µs + 13.0µs = 33.4 µs

0.7 
 1− 5



€
29
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 13
Page 789
(a) At the Q - point : β F =
(c) Rin =
IC 1.5mA
=
= 100
I B 15µA
vbe 8mV
=
= 1.6 kΩ
ib
5µA
(b) I S =
IC
1.5mA
=
= 1.04 fA
 VBE 
 0.700V 
exp
 exp

 0.025V 
 VT 
(d ) Yes. With the given applied signal, the smallest value of vCE is
min
v CE
= 5V − 0.5mA( 3.3kΩ) = 3.35 V which exceeds v BE = 0.708 V. (e) AvdB = 20 log −206 = 46.3 dB
€
Page 790
(a) No : v min
DS ≅ 2.7V with vGS − VTN = 4 −1 = 3V , so the transistor has entered the triode region.
(b) Choose two points on the i - v characteristics. For example,
2
2
K
K
1.56mA = n ( 3.5 − VTN ) and 1.0mA = n ( 3.0 − VTN ) .
2
2
µA
Solving for Kn and VTN yields 500 2 and 1 V respectively.
V
(c) AvdB = 20 log −4.13 = 12.3 dB
Page 791
€
10kΩ
12V = 3.00 V | REQ = 10kΩ 30kΩ = 7.5 kΩ
10kΩ + 30kΩ
VEQ − VBE
3.0V − 0.7V
IC = β F I B = β F
= 100
= 1.45 mA
REQ + (β F + 1) R4
7.5kΩ + (101)(1.5kΩ)
VEQ =
 101 
VCE = 12 − 4300IC −1500I E = 12 − 4300(1.45mA) −1500
(1.45mA) = 3.57 V
 100 
VB = VEQ − I B REQ = 3.00 −
1.45mA
(7.5kΩ) = 2.89 V
100
€
30
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 792
 101 
vC (t ) = VC + vC = (5.8 −1.1sin 2000πt ) V | v E (t ) = VE + 0 = 1.45mA
(1.5kΩ) = 2.20 V
 100 
1.1V
= 0.256mA | ∠ic = 180 o | ic (t ) = −0.26sin 2000πt mA | vB (t ) = VB + vb (t )
4.3kΩ
1.45mA
VB = VEQ − I B REQ = 3.00 −
(7.5kΩ) = 2.89 V | vB (t ) = (2.89 + 0.005sin sin 2000π ) V
100
−−−
ic =
XC =
€
€
1
1
=
= 0.318 Ω | X C << Rin
ωC 2000π (500µF )
Page 795
RB = 20kΩ 62kΩ = 15.1 kΩ | RL = 8.2kΩ 100kΩ = 7.58 kΩ
Page 799
VT
rd =
ID + IS
| rd =
0.025V
0.025V
= 25.0 TΩ | rd =
= 500 Ω
1 fA
50µA
0.025V
0.025V
= 12.5 Ω | rd =
= 8.33 mΩ
2mA
3A
−−−
0.025V
kT 
V
0.0322V
rd =
= 16.7 Ω |
= 8.62x10−5 (373K ) = 0.0322 V | rd =
= 21.4 Ω
1.5mA
q 
K
1.5mA
rd =
€
31
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 804
g m = 40IC = 40(50µA) = 2.00 mS | rπ =
ro =
βo
75
=
= 37.5 kΩ
g m 2mS
VA + VCE 60V + 5V
=
= 1.30 MΩ | µ f = g mro = 2mS (1.30 MΩ) = 2600
IC
50µA
−−−
g m = 40IC = 40(250µA) = 10.0 mS | rπ =
ro =
βo
50
=
= 5.00 kΩ
g m 10mS
VA + VCE 75V + 15V
=
= 360 kΩ | µ f = g mro = 10mS ( 360kΩ) = 3600
IC
250µA
--The slope of the output characteristics is zero, so VA = ∞ and ro = ∞.
β FO =
βo =
βF
I
1.5mA
Δi
0.5mA
= βF = C =
= 100 | g m = C =
= 62.5 mS
VCE
I B 15µA
ΔvBE
8mV
1+
VA
ΔiC 500µA
=
= 100
ΔiB
5µA
| rπ =
βo
100
=
= 1.60 kΩ |
g m 62.5mS
rπ =
Δv BE
8mV
=
= 1.60 kΩ
ΔiB
0.5mA 100
€
32
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 814
Avt = −g m RL = −9.80mS (18kΩ) = −176
| Ten percent of the input signal is being lost
by voltage division between source resistance RI and the amplifier input resistance.
−−−
Assume the Q - point remains constant.
(a) RiB = rπ =
max
L
( b) R
 104kΩ 12.8kΩ 
125
 = −162
= 12.8 kΩ | Av = −9.80mS (18kΩ)

9.80mS
1kΩ
+
104kΩ
12.8kΩ


= 1.1(18kΩ) = 19.8 kΩ | RLmin = 0.9(18kΩ) = 16.2 kΩ
 104kΩ 10.2kΩ 
 = −143
Avmin = −9.80mS (16.2kΩ)

1kΩ
+
104kΩ
10.2kΩ


 19.8kΩ 
 19.8kΩ 
Avmax = Avmin 
 = −143
 = −175
 16.2kΩ 
 16.2kΩ 
16.2kΩ 


max
nom 19.8kΩ
Checking : Avmin = Avnom 
 = −159(0.9) = −143 | A v = Av 
 = −159(1.1) = −175
 18kΩ 
 18kΩ 


101
(c) VCE = 12V − 22kΩIC −13kΩI E = 12V − 0.275mA22kΩ + 100 13kΩ = 2.34 V
100
g m = 40(0.275mA) = 11.0 mS | RiB = rπ =
= 9.09 kΩ
11.0mS
 104kΩ 9.09kΩ 
 = −177
Av = −11.0mS (18kΩ)

 1kΩ + 104kΩ 9.09kΩ 
−−−
β
100
AvCE ≅ −10VCC = −10(20) = −200 | g m = 40IC = 40(100µA) = 4.00 mS | RiB = rπ = o =
= 25 kΩ
g m 4mS
VA + VCE 50V + 10V
=
= 600 kΩ | µ f = g mro = 4mS (600kΩ) = 2400
IC
100µA
 150kΩ 25kΩ 
RB RiB
 = −278
Av = −g m RC ro
= −4.00mS 100kΩ 600kΩ 

RI + RB RiB
 5kΩ + 150kΩ 25kΩ 
ro =
(
)
(
)
€
33
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 815
160kΩ
12V = 4.17 V | REQ = 160kΩ 300kΩ = 104 kΩ
160kΩ + 300kΩ
VEQ − VBE
4.17V − 0.7V
IC = β F I B = β F
= 100
= 0.245 mA
REQ + (β F + 1) RE
104kΩ + (101)(13kΩ)
VEQ =
 101 
VCE = 12 − 22000IC −13000I E = 12 − 22000(0.245mA) −13000
(0.245mA) = 3.39 V
 100 
−−−
−23
 V  V 
kT 1.38x10 ( 300)
IC = I S exp BE 1+ CB  | VT =
=
= .025875 V
q
1.6x10−19
 VT  VA 
0.245mA
IS =
= 422 fA
 0.7  3.39 − 0.7 
exp
1+

75
 0.025875 

Page 817
€
(a )
(
)
[
]
g m = 2Kn I D (1+ λVDS ) = 2 1mA /V 2 (0.25mA) 1+ 0.02(5) = 0.742 mS
1
+ VDS
50V + 5V
λ
ro =
=
= 220 kΩ | µ f = g mro = 0.742mS (220kΩ) = 163
ID
250µA
(
)
[
]
g m = 2Kn I D (1+ λVDS ) = 2 1mA/V 2 (5mA) 1+ 0.02(10) = 3.46 mS
1
+ VDS
50V + 10V
λ
ro =
=
= 12 kΩ | µ f = g mro = 3.46mS (12kΩ) = 41.5
ID
5mA
(b)
The slope of the output characteristics is zero, so λ = 0 and ro = ∞.
For the positive change in vgs , g m =
ΔiD
ΔvGS
2.1V
≅ 3.3kΩ = 1.3 mS
0.5V
Page 818
€
v gs ≤ 0.2(VGS − VTN ) = 0.2
2(25mA)
2I D
= 0.2
= 1.00 V |
Kn
2.0mA/V 2
vbe ≤ 0.005 V
Page 819
€
€
η=
γ
2 VSB + 2φ F
=
0.75
2 0 + 0.6
= 0.48
| η=
0.75
2 3 + 0.6
34
= 0.20
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 821
gm = 2
I DSS I D (1+ λVDS )
VP
=2
[
] = 3.32 mS
5mA(2mA) 1+ 0.02(5)
−2
1
+ VDS 50V + 5V
ro = λ
=
= 27.5 kΩ | µ f = g mro = 3.32mS (27.5kΩ) = 91.3
ID
2mA


ID 
2mA 
 = −2V 1−
 = −0.735 V
VGS = VP 1−
I DSS 
5mA 


v gs ≤ 0.2(VGS − VP ) = 0.2(−0.735 + 2) = 0.253 V
Page 829
€
1.5 MΩ
12V = 4.87 V | REQ = 1.5 MΩ 2.2 MΩ = 892 kΩ
1.5 MΩ + 2.2 MΩ
Neglect λ in hand calculations of the Q - point.
 5x10−4 
2
4.87 = VGS + 12000I D | 4.87 = VGS + 12000
(VGS −1)
 2 
VEQ =
3VGS2 − 5VGS −1.87 = 0 → VGS = 1.981 V | I D = 241 µA
VDS = 12 − 22000I D −12000I D = 3.81 V | Q - point : (241 µA, 3.81 V )
−−−
The small - signal model appears in Fig. 13.27(c).
−−−
VGS − VTN ≅
2(241µA)
2x10−3
= 0.491 V | AvCS ≅ −
12V
= −24.4
0.491V
| M=
Kn2 2x10−3
=
=4
Kn1 5x10−4
Page 831
€
rπ =
β oVT 100(0.025V )
=
= 3.45 kΩ | RinCE = RB rπ = 104kΩ 3.45kΩ = 3.34 kΩ
IC
0.725mA
Page 832
€
RinCS = 680kΩ 1.0 MΩ = 405 kΩ | VEQnew =
VEQold =
1.5 MΩ
VDD = 0.405VDD
1.5 MkΩ + 2.2 MΩ
680kΩ
VDD = 0.405VDD
680kΩ + 1MΩ
| No change. The gate voltages are the same.
€
35
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 837
From Ex. 13.6, µ f = 230 and Av = −20.3.
Av << µ f
−−−

VGS = VP 1−

ID
I DSS


0.25mA 
 = −1V 1−
 = −0.500 V
1mA 


v gs ≤ 0.2(VGS − VP ) = 0.2(−0.5 + 1) = 0.100 V |
v o ≤ 20.3(0.1) = 2.03 V
−−−
SPICE Results :
λ = 0 : Q - point = (250 µA, 4.75 V ) | λ = 0.02 V −1 : Q - point = (257 µA, 4.54 V )
Page 840
€
 66 
IC = 245 µA | VCE = 3.39 V | I E = 245µA  = 249 µA
 65 
245µA
PD = ICVCE + I BVBE = 245µA(3.39V ) +
(0.7V ) = 0.833 mW
65
V − VB VCC − (VBE + I E RE ) 12 − 0.7 − 0.249mA(13kΩ)
PS = VCC ( IC + I2 ) | I2 = CC
=
=
= 26.9 µA
R2
R2
300kΩ
PS = 12V (245µA + 26.9µA) = 3.26 mW
−−−
PD = I DVDS = 241µA(3.81V ) = 0.918 mW | PS = VDD ( I D + I2 )
I2 =
€
VDD
12V
=
= 3.24 µA | PS = 12V (241µA + 3.24µA) = 2.93 mW
R1 + R2 1.5MΩ + 2.2 MΩ
Page 842
(a) VM ≤ min IC RC , (VCE − VBE ) = min 245µA(22kΩ), (3.39 − 0.7)V = 2.69 V
[
]
[
]
VM is limited by the value of VCE .
( b) V
M
[
]
[
]
≤ min I D RD , (VDS − VDSSAT ) = min 241µA(22kΩ), ( 3.81− 0.982)V = 2.83 V
Limited by the value of VDS .
€
36
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 14
Page 860
160kΩ
12V = 4.17 V | REQ = 160kΩ 300kΩ = 104 kΩ
160kΩ + 300kΩ
VEQ − VBE
4.17V − 0.7V
IC = β F I B = β F
= 100
= 0.245 mA
REQ + (β F + 1) RE
104kΩ + (101)(13kΩ)
VEQ =
 101 
VCE = 12 − 22000IC −13000I E = 12 − 22000(0.245mA) −13000
(0.245mA) = 3.39 V
 100 
g m = 40IC = 40(0.245mA) = 9.80 mS | rπ =
ro =
βo
100
=
= 10.2 kΩ
g m 9.80mS
VA + VCE
53.4V
=
= 218 kΩ | µ f = g mro = 2140
IC
0.245mA
−−−
1.5 MΩ
12V = 4.87 V | REQ = 1.5 MΩ 2.2 MΩ = 892 kΩ
1.5 MΩ + 2.2 MΩ
Neglect λ in hand calculations of the Q - point.
 5x10−4 
2
4.87 = VGS + 12000I D | 4.87 = VGS + 12000
(VGS −1)
 2 
VEQ =
3VGS2 − 5VGS −1.87 = 0 → VGS = 1.981 V | I D = 241 µA
VDS = 12 − 22000I D −12000I D = 3.81 V | Q - point : (241 µA, 3.81 V )
(
)(
)
g m = 2Kn I D = 2 5x10−4 2.41x10−4 = 0.491 mS | ro =
λ−1 + VCE
53.8V
=
= 223 kΩ
IC
0.241mA
µ f = g mro = 110
€
Page 861
RB = 160kΩ 300kΩ = 104 kΩ
| RE = 3.00 kΩ | RL = 22kΩ 100kΩ = 18.0 kΩ
RG = 1.5 MΩ 2.2 MΩ = 892 kΩ | RS = 2.00 kΩ | RL = 22kΩ 100kΩ = 18.0 kΩ
−−−
RB = 160kΩ 300kΩ = 104 kΩ | RL = 13kΩ 100kΩ = 11.5 kΩ
RG = 1.5 MΩ 2.2 MΩ = 892 kΩ | RL = 12kΩ 100kΩ = 10.7 kΩ
€
37
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 863
RI = 2 kΩ
| R6 = 13 kΩ | RL = 22kΩ 100kΩ = 18.0 kΩ
RI = 2 kΩ
| R6 = 12 kΩ | RL = 22kΩ 100kΩ = 18.0 kΩ
€
38
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 875
V −V
1
(a) IC ≅ REQ + RBE or IC ∝ R + R
E
4
E
4
For large gm RE , AvtCE = −
| IC = 0.245mA
13kΩ
RE + R4
RC R3
g m RL
R
≅− L =−
1+ g m RE
RE
RE
For AvtCE max make RC and R3 large and RE small. RL = 1.1(22kΩ) 1.1(100kΩ) = 19.8 kΩ
RE = 0.9( 3kΩ) = 2.7 kΩ | IC = 0.245mA
13kΩ
= 0.251 mA | g m = 40(0.251 mA) = 10.0 mS
12.7kΩ
βo
100
=
= 10.0 kΩ | RiB = 10.0kΩ + 101(2.7kΩ) = 283 kΩ
g m 10.0mS
10.0mS (19.8kΩ)  104kΩ 283kΩ 

 = −6.98
AvCE max = −
1+ 10.0mS (2.7kΩ)  1kΩ + 104kΩ 283kΩ 
rπ =
For AvtCE min make RC and R3 small and RE large. RL = 0.9(22kΩ) 0.9(100kΩ) = 16.2 kΩ
RE = 1.1( 3kΩ) = 3.3 kΩ | IC = 0.245mA
13kΩ
= 0.239 mA | g m = 40(0.239 mA) = 9.56 mS
13.3kΩ
βo
100
=
= 10.5 kΩ | RiB = 10.5kΩ + 101( 3.3kΩ) = 344 kΩ
g m 9.56mS
9.56mS (16.2kΩ)  104kΩ 344kΩ 

 = −4.70
AvCE min = −
1+ 9.56mS ( 3.3kΩ)  1kΩ + 104kΩ 344kΩ 
rπ =
(b) Assume the collector current does not change.
βo
125
=
= 12.8 kΩ | RiB = 12.8kΩ + 126( 3.0kΩ) = 391 kΩ
g m 9.8mS
9.80mS (18kΩ)  104kΩ 391kΩ 

 = −5.73 The gain is essentially unchanged.
AvCE = −
1+ 9.80mS ( 3kΩ)  1kΩ + 104kΩ 391kΩ 


101
(c) VCE = VCC − IC RC − I E ( RE + R4 ) = 12V − 0.275mA22kΩ + 100 13kΩ = 2.34 V
2.34 V > 0.7 V Therefore the transistor is still in the active region.
β
100
g m = 40(0.275mA) = 11.0 mS | rπ = o =
= 9.09 kΩ | RiB = 9.09kΩ + 101( 3.0kΩ) = 312 kΩ
g m 11mS
11.0mS (18kΩ)  104kΩ 312kΩ 
CE

 = −5.75 The gain is essentially unchanged.
Av = −
1+ 11.0mS ( 3kΩ)  1kΩ + 104kΩ 312kΩ 
rπ =
Continued on the next page
€
39
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 875 cont.


100(2kΩ)
 = 5.17 MΩ | µ R = 3140(2kΩ) = 6.28 MΩ
RiC = 320kΩ 1+
f E


1kΩ
104kΩ
+
10.2kΩ
+
2kΩ


(
RiC < µ f RE
)
| Rout = 5.17 MΩ 22kΩ = 21.9 kΩ | Rout << µ f RE
−−−


 β R 
β o RE
lim RiC = lim ro 1+
= ro1+ o E  = (β o + 1)ro

R E →∞
RE 
R E →∞
 Rth + rπ + RE 

€
Page 877
RiB = 10.2kΩ + 101(1kΩ) = 111 kΩ
9.80mS (18kΩ)  104kΩ 111kΩ 

 = −16.0
Av = −
1+ 9.80mS (1kΩ)  1kΩ + 104kΩ 111kΩ 
| R4 = 13kΩ −1kΩ = 12 kΩ.
−−−
−23
 V  V 
kT 1.38x10 ( 300)
BE
CB
IC = I S exp
=
= .025875 V
1+
 | VT =
q
1.60x10−19
 VT  VA 
IS =
0.245mA
= 425 fA
 0.7  3.39 − 0.7 
exp
1+

100 
 0.025875 
−−−
AvCE ≅ −10VCC = −10(20) = −200 | g m = 40IC = 40(100µA) = 4.00 mS | RiB = rπ =
β o 100
=
= 25 kΩ
g m 4mS
VA + VCE 50V + 10V
=
= 600 kΩ | µ f = g mro = 4mS (600kΩ) = 2400
IC
100µA
 150kΩ 25kΩ 
RB RiB
 = −278
AvCE = −g m RC ro
= −4.00mS 100kΩ 600kΩ 

RI + RB RiB
5kΩ
+
150kΩ
25kΩ


ro =
(
)
(
)
€
40
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 884
1.5 MΩ
12V = 4.87 V | REQ = 1.5 MΩ 2.2 MΩ = 892 kΩ
1.5 MΩ + 2.2 MΩ
Neglect λ in hand calculations of the Q - point.
 5x10−4 
2
4.87 = VGS + 12000I D | 4.87 = VGS + 12000
(VGS −1)
 2 
VEQ =
3VGS2 − 5VGS −1.87 = 0 → VGS = 1.981 V | I D = 241 µA
VDS = 12 − 22000I D −12000I D = 3.81 V | Q - point : (241 µA, 3.81 V )
−−−
CS
AvdB
= 20 log −4.50 = −13.1 dB
−−−
RiB = 10.2kΩ + 101(1kΩ) = 111kΩ
CE
v
A
9.80mS (18kΩ)  104kΩ 111kΩ 

 = −16.0 | R4 = 13kΩ −1kΩ = 12 kΩ
=−
1+ 9.80mS (1kΩ)  1kΩ + 104kΩ 111kΩ 
AvCS = −
(iii )
€
0.503mS (18kΩ)  892kΩ 

 = −6.02 | R4 = 12kΩ −1kΩ = 11 kΩ
1+ 0.503mS (1kΩ)  1kΩ + 892kΩ 
RiB = 10.2kΩ + 101(13kΩ) = 1.32 MΩ
AvCE = −
9.80mS (18kΩ)  104kΩ 1.32 MΩ 

 = −1.36 | AvCE ≅ − RL = − 18kΩ = −1.38
RE + R4
13kΩ
1+ 9.80mS (13kΩ)  1kΩ + 104kΩ 1.32 MΩ 
AvCS = −
0.503mS (18kΩ)  892kΩ 
RL
18kΩ
CS
=−
= −1.50

 = −1.29 | Av ≅ −
RS + R4
12kΩ
1+ 0.503mS (12kΩ)  1kΩ + 892kΩ 
Page 885
 1.381x10−23 V 
IC
245µA
VT = 
=
= 0.430 fA
(273K + 27K ) = 25.861 mV | I S =
−19
 VBE 
 0.700V 
1.602x10 K 
exp
 exp

 0.025861V 
 VT 
−−−
18kΩ
g m RL = -9.80mS(18kΩ) = −176 | AvCE ≅ −
= −6.00 | 5.72 < 6.00
3kΩ
18kΩ
g m RL = -0.503mS(18kΩ) = −9.05 | AvCS ≅ −
= −9.00 | 4.50 < 9.00
2kΩ
€
41
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 889
2.5V
1+ 10mS (11.5kΩ) = 1.16 MΩ
0.25mA
[
RB = 160kΩ 300kΩ = 104kΩ | RiB ≅ rπ (1+ g m RL ) =
vi ≤ 0.005V (1+ g m RL )
RI + RB RiB
RB RiB
vi ≤ 0.2(VGS − VTN )(1+ g m RL )
€
[
= 0.005V 1+ 10mS (11.5kΩ)
]
+ 95.4kΩ
= 0.592 V
] 2kΩ95.4kΩ
RI + RG
2kΩ + 892kΩ
= 0.2(1V ) 1+ 0.5mS (10.7kΩ)
= 1.27 V
RG
892kΩ
[
Page 894
2kΩ + 892kΩ
Avt =
0.971 = 0.973 |
892kΩ
]
(0.491ms) R
1+ (0.491ms ) R
L
= 0.973 → RL = 73.4kΩ
L
R6 100kΩ = 73.4kΩ → R6 = 276 kΩ | Note, however, that the 12 kΩ resistor
can't simply be replaced with a 276 kΩ resistor because of Q - point problems.
−−−
RiB = 10.2kΩ + 101(13kΩ) = 1.32 MΩ | RinCC = 104kΩ 1.32 MΩ = 96.4kΩ
AvCE = −
9.80mS (13kΩ)  96.4kΩ 

 = +0.972
1+ 9.80mS (13kΩ)  2kΩ + 96.4kΩ 
AvCS = −
0.491mS (12kΩ)  892kΩ 

 = +0.853
1+ 0.491mS (12kΩ)  2kΩ + 892kΩ 
−−−
BJT : g m RL = 9.80mS (11.5kΩ) = 113 | FET : g m RL = 0.491mS (10.7kΩ) = 5.25
Page 896
€
BJT : v i ≤ 0.005V (1+ g m RI )
 2kΩ + 13kΩ 
RI + R6
= 0.005V 1+ 9.8mS (2kΩ) 
 = 119 mV
R6
 13kΩ 
[
]
[
]
Neglecting R6 , v i ≤ 0.005V (1+ g m RI ) = 0.005V 1+ 9.8mS (2kΩ) = 103 mV
FET : v i ≤ 0.2(VGS − VTN )(1+ g m RI )
RI + R6
2kΩ + 12kΩ
= 0.2(0.982) 1+ 0.491mS (2kΩ)
= 454 mV
R6
12kΩ
[
]
[
]
Neglecting R6 , v i ≤ 0.2(VGS − VTN )(1+ g m RI ) = 0.2(0.982) 1+ 0.491mS (2kΩ) = 389 mV
€
42
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 898


100(1.73kΩ) 
β R 
 = 3.40 MΩ
RiC = ro 1+ o th  = 219kΩ 1+
 Rth + rπ 
 1.73kΩ + 10.2kΩ
[
]
Or more approximately, RiC = ro [1+ g m Rth ] = 219kΩ 1+ 9.8mS (1.73kΩ) = 3.93 MΩ
[
]
RiD = ro [1+ g m Rth ] = 223kΩ 1+ 0.491(1.71kΩ) = 410 kΩ
Page 902
€
AvCB
AvCB
 1
R6 
R6
 gm 
1
1
R6 +
R6 +
R6
gm
gm
= g m RL
= g m RL
= g m RL
 1
R6
R6 (1+ g m RI ) + RI
g m RI +
R6  
1
 gm 
R6 +
RI +
gm
1
R6 +
gm
R6
1
g m RL  R6 
= g m RL
=


g R R 1+ g m Rth  R6 + RI 
R6 + RI
1+ m I 6
R6 + RI
−−−
The voltage gains are proportional to the load resistance
 22kΩ 
 22kΩ 
CG
AvCE = +8.48 
 = +10.4 | Av = +4.12 
 = +5.02
 18kΩ 
 18kΩ 
−−−
R
RL
18kΩ
CB : AvCB ≤ g m RL = 176 | AvCB ≅ L =
=
= 10.4 |
Rth RI R6 1.73kΩ
CG : AvCG ≤ g m RL = 8.84
| AvCB ≅
RL
RL
18kΩ
=
=
= 10.5
Rth RI R6 1.71kΩ
8.48 < 10.4 << 176
|
4.11 < 8.84 < 10.5
Page 909
€
AvCS =
1
1+ η
(W L)
(W L)
1
2
26
| 10 20 =
1
1+ 0.2
(W L)
1
4
=
2290
1
€
43
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 911
η = 0 | I D 2 = I D1 | Both transistors are in the active region since VDS = VGS .
2
2
10−4  2 
10−4  8 
Neglecting λ :
 (5 − VO −1) =
 (VO −1) → VO = 2.00 V
2  1
2  1
2
2
10−4  2 
10−4  8 
Keeping λ :
 (5 − VO −1) 1+ 0.02(5 − VO ) =
 (VO −1) (1+ 0.02VO ) →
2  1
2 1
[
]
VO = 2.0064 V , I D = 421.39 µA → Q − point : (2.01 V , 421 µA)
−−−
I D2 = I D1 | Both transistors are in the active region since VDS = VGS .
 20 
 50 
A
A
Kn = 10−4   = 2x10−3 2 | K p = 4 x10−5   = 2x10−3 2 | The transistors are symmetrical.
V
V
1
1
2
V
3.3V
10−4  20 
∴ VO = DD =
= 1.65 V | I D =
 (1.65 − 0.7) 1+ 0.02(1.65) = 932 µA
2
2
2 1
[
]
Q − point : (1.65 V , 932 µA)
€
Page 914
Since we need high gain, the emitter should be bypassed, and RinCE = RB rπ = 250kΩ.
If we choose RB ≅ rπ , IC =
βo
100
≅
= 5 µA
40rπ 40(500kΩ)
−−−
RinCG ≅
1
gm
| IC ≅
1
= 12.5 µA
40(2kΩ)
€
44
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 918
Common − Emitter :
C1 >>
1
= 8.07nF | Choose C1 = 82 nF = 0.082 µF
2π (250Hz)(1kΩ + 77.9kΩ)
C2 >>
1
= 6.13nF | Choose C2 = 68 nF = 0.068 µF
2π (250Hz)(21.9kΩ + 82kΩ)
C3 >>
1
= 0.269µF | Choose C3 = 2.7 µF


1 
2π (250Hz)10kΩ  3kΩ +

9.80mS 


Common − Source :
C1 >>
1
= 713 pF | Choose C1 = 8200 pF
2π (250Hz)(1kΩ + 892kΩ)
C2 >>
1
= 6.15nF | Choose C2 = 68 nF = 0.068 µF
2π (250Hz)(21.5kΩ + 82kΩ)
C3 >>
1
= 0.221µF | Choose C3 = 2.2 µF



1
2π (250Hz)10kΩ  2kΩ +

0.491mS 


Page 921
€
Common − Collector :
1
C1 >>
= 6.60nF | Choose C1 = 68 nF = 0.068 µF
2π (250Hz)(1kΩ + 95.5kΩ)
C2 >>
1
= 7.75nF | Choose C2 = 82 nF = 0.082 µF
2π (250Hz)(120Ω + 82kΩ)
Common − Drain :
C1 >>
1
= 713 pF | Choose C1 = 8200 pF
2π (250Hz)(1kΩ + 892kΩ)
C2 >>
1
= 7.60nF | Choose C2 = 82 nF = 0.082 µF
2π (250Hz)(1.74kΩ + 82kΩ)
€
45
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 924
Common − Base :
C1 >>
1
= 0.579µF | Choose C1 = 6.8 µF
2π (250Hz)(1kΩ + 0.1kΩ)
C2 >>
1
= 6.13nF | Choose C2 = 0.068 µF
2π (250Hz)(21.9kΩ + 82kΩ)
C3 >>
1


2π (250Hz)160kΩ 300kΩ 10.2kΩ + 101 13kΩ 1kΩ 


[
(
)]
= 12.2nF | Choose C3 = 0.12 µF
Common − Gate :
C1 >>
1
= 0.232µF | Choose C1 = 2.2 µF
2π (250Hz)(1kΩ + 1.74kΩ)
C2 >>
1
= 6.19nF | Choose C2 = 0.068 µF
2π (250Hz)(20.9kΩ + 82kΩ)
C3 >>
1
(
)
2π (250Hz) 1.5 MΩ 2.2 MΩ
= 714 pF | Choose C3 = 8200 pF
Page 925
€
(a ) Common − Source :
C3 =
1
= 55.3nF | Choose C3 = 0.056 µF



1
2π (1000Hz)10kΩ  2kΩ +

0.491mS 


(b) Common − Collector :
C2 >>
1
= 795 pF | Choose C2 = 820 pF
2π (2000Hz)(120Ω + 100kΩ)
(c) Common − Gate :
C1 >>
1
= 42.6nF | Choose C1 = 0.042 µF
2π (1000Hz)(2kΩ + 1.74kΩ)
€
46
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 929
2
0.020
VGS −1.5) → VGS = 2.203 V | I D = 4.94 mA
(
2
= 5 − (−VGS ) = 7.20 V | Q - point : (4.94 mA, 7.20 V) | Rin = RG = 22 MΩ
20V = VGS + 3600I D | 20 = VGS + 3600
VDS
AvCD =
g m RL
1+ g m RL
| gm =
2( 4.94mA)
(2.20 −1.50)V
= 14.2 mS | RL = 3600Ω 3000Ω = 1630 Ω | AvCD = 0.959
−−−
CD
Rout
= 3.6kΩ
1
1
= 3.6kΩ
= 69.1 Ω | v gs ≤ 0.2(2.20 −1.50) 1+ 0.0142(1630) = 3.38 V
gm
0.0142
[
]
−−−
1
1
+ VDS
+ 5 + 2.21
ro = λ
= 0.015
= 14.8 kΩ | RL = 3600Ω 3000Ω 14.8kΩ = 1470 Ω | AvCD = 0.954
ID
0.005
−−−
W Kn 2x10−2 400
=
=
=
L Kn′ 5x10−5
1
€
Page 930
g R
A= m S
1+ g m RS
CD
Rout
= 3.6kΩ
| gm =
2( 4.94mA)
(2.20 −1.50)V
= 14.2 mS | RS = 3600 Ω | AvCD = 0.981 | Rin = RG = 22 MΩ
1
1
3000Ω
= 3.6kΩ
= 69.1 Ω | AvCD = A
= 0.959
gm
0.0142
69.1Ω + 3000Ω
€
47
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 933
Reverse the direction of the arrow on the emitter of the transistor as well
as the values of VEE and VCC .
−−−
RinCG = RE
1
1
75.1Ω
= 13kΩ
= 75.1 Ω | AvCB =
(13.2mS )(7.58kΩ) = 50.1
gm
75Ω + 75.1Ω
40(331µA)
−−−
For vCB ≥ 0, we require v C ≥ 0. VC = 5 − IC RC = 2.29 V ∴ v c ≤ 2.29 V
vo ≤ 5mV ( g m RL ) = 5mV (13.2mS )( 7580Ω) = 0.500 V
−−−
[
]
RE = 75Ω 1+ 40(7.5 − 0.7) = 20.5 kΩ (a standard 1% value) | IC ≅
6.8V
= 332 µA
20.5kΩ
75
→ RL = 7.53kΩ → RC = 8.14 kΩ → 8.06 kΩ (a standard 1% value)
75 + 75
= 0.7 + 7.5 − IC RC = 5.52 V
50 = 40( 332µA) RL
VEC
Page 934
€
5% tolerances ICmax ≅
VEEmax − 0.7V 5(1.05) − 0.7V
=
= 368µA
REmin
13kΩ(0.95)
min
VCmin = VCC
− ICmax RCmax = 5V (0.95) − 368µA(8.2kΩ)(1.05) = 1.58 V | 1.58 ≥ 0, so active region is ok.
10% tolerances ICmax ≅
VEEmax − 0.7V 5(1.1) − 0.7V
=
= 410µA
REmin
13kΩ(0.9)
min
VCmin = VCC
− ICmax RCmax = 5V (0.90) − 410µA(8.2kΩ)(1.1) = 0.802 V | 0.802 ≥ 0, so active region is ok.
−−−
RinCB
75
vth = v i
g R = vi
(13.2mS )(8200Ω) = 54.1vi
CB m C
75Ω + 75
75Ω + Rin
AvCG =
CB
| Rth = Rout
= 8.2 kΩ
v o vth 100kΩ
100kΩ
=
= 54.1
= 50.0
v i vi Rth + 100kΩ
8.2kΩ + 100kΩ
€
48
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 939
ro ≅ ro =
1
1
=
= 333 kΩ
λI D 0.015 2x10−4
(
)
or more exactly VDS = 25 −10 5 I D − 9.1x10 3 I D = 25 −1.09x10 5 (0.2mA) = 3.18 V
1
1
+ VDS
+ 3.18
λ
0.015
ro =
=
= 349 kΩ | RL = 100kΩ 100kΩ 349kΩ = 43.7 kΩ
ID
2x10−4
AvCS = −( g m RL )
 75Ω 
2(0.2mA)
Rin
=−
43.7kΩ)
(
 = −43.7
RI + Rin
0.2V
 75Ω + 75Ω 
−−−
2
0.01
0.25) = 0.3125 mA | VGS − VTN = 0.25 V | VGS = 0.25 − 2 = −1.75 V
(
2
−V
1.75V
A 50(0.25V )
RS = GS =
= 5.60kΩ → 5.6 kΩ | RL = 2 v =
= 40 kΩ | RD 100kΩ = 40 kΩ
ID
0.3125mA
g m 0.3125mA
ID =
RD = 66.7kΩ → 68 kΩ | C1 remains unchanged.
C2 >>
C3 >>
1
= 1.90 pF → Choose C2 = 20 pF
10 π (68kΩ + 100kΩ)
6
1

1 
10 π  5.6kΩ

2.5mS 

= 0.853nF → Choose C3 = 8200 pF
6
−−−
vth = v i
RinCG
75Ω
g R = vi
(2mS )(100kΩ) = 100vi
CG m D
75Ω + 75Ω
75Ω + Rin
AvCG =
v o vth 100kΩ
100kΩ
=
= 100
= 50.0
v i vi Rth + 100kΩ
100kΩ + 100kΩ
CG
| Rth = Rout
= 100 kΩ
€
49
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 941
2
2
Kn
0.01
VGS − VTN ) | VGS = −RS1 I D | I D =
−200I D + 2) → I D = 5.00 mA
(
(
2
2
= 15 − 5mA(820Ω) = 10.9 V
M1 : I D =
VDS
g m = 2Kn I D = 2(0.01)(0.005) = 10.0 mS | ro =
1+ λVDS 1+ 0.02(10.9)
=
= 12.2 kΩ
λI D
0.02(5mA)
22kΩ
(15V ) = 3.30 V | REQ = 22kΩ 78kΩ = 17.2 kΩ
22kΩ + 78kΩ


3.30 − 0.7
151
IC = 150
= 1.52 mA | VCE = 15 −1.52mA 4.7kΩ +
1.6kΩ = 5.41 V
150
17.2kΩ + 151(1.6kΩ)


Q2 : VEQ =
g m = 40(1.52mA) = 60.8 mS | rπ =
150
80 + 5.41
= 2.47 kΩ | ro =
= 56.2 kΩ
60.8mS
1.52mA
120kΩ
(15V ) = 8.53 V | REQ = 120kΩ 91kΩ = 51.8 kΩ
120kΩ + 91kΩ
 81

8.53 − 0.7
IC = 80
= 1.96 mA | VCE = 15 −1.96mA 3.3kΩ = 8.45 V
51.8kΩ + 81( 3.3kΩ)
 80

Q3 : VEQ =
g m = 40(1.96mA) = 78.4 mS | rπ =
80
60 + 8.45
= 1.02 kΩ | ro =
= 34.9 kΩ
78.4mS
1.96mA
−−−
A typical op - amp gain is at least 10,000 which exceeds the amplification factor
of a single transistor.
€
Page 943
RL1 = 478Ω 12.2kΩ = 460 Ω | RL2 = 3.53kΩ 54.2kΩ = 3.31 kΩ | RL3 = 232Ω 34.4kΩ = 230 Ω
 79.6mS (230Ω) 

1MΩ

Av = −10mS ( 460Ω)(−62.8mS )( 3.31kΩ)
 = 898
1+ 79.6mS (230Ω) 10kΩ + 1MΩ 
20 log(898) = 59.1 dB
−−−

VDD 
15
Av ≅ −
(−10VCC )(1) = − (−10)(15)(1) = 2250
1
 VGS − VTN 
−−−
Av = −10mS (2.39kΩ)(−62.8mS )(19.8kΩ)(0.95)(0.99) = 28000
€
50
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 948

1
3990 
Rout = 3300 
+
 = 55.9 Ω
 0.0796S 90.1 
−−−
Note that the answers are obtained directly from SPICE.
−−−
(
)
Av1 = −g m RL1 = − 2(0.01)(0.001) 3kΩ 17.2kΩ 2.39kΩ = 5.52
Av = −5.52(−222)( 3.31kΩ)(0.95)(0.99) = 1150
€
51
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 15
Page 972
60  15 − 0.7 
 = 93.8 µA | VCE = 15 − 93.8µA( 75kΩ) − (−0.7V ) = 8.67 V
IC = α F I E = 
61 2 ( 75kΩ) 
−−−
IC = α F I E =


60  15 − VBE 
IC

 and VBE = 0.025V ln 
 → IC = 94.7 µA, VBE = 0.649 V
-15
61 2 ( 75kΩ) 
 0.5x10 A 
Page 974
€
v1 + v 2 1.01+ 0.99
=
= 1.00 V
2
2
v +v
4.995 + 5.005
vid = v1 − v2 = 4.995 − 5.005 = −0.010 V | v ic = 1 2 =
= 5.00 V
2
2
vod = Add v id + Acd v ic | v oc = Adcv id + Accvic
vid = v1 − v2 = 1.01− 0.990 = 0.020 V | v ic =
2.20
 0.02

 = [ Add Acd ] 
 0 
−0.01
1.002
 0.02

 = [ Adc Acc ] 
5.001
−0.01
€
1.00
 → [ Add Acd ] = [100 0.20]
5.00
1.00
 → [ Adc Acc ] = [0.100 1.00]
5.00
Page 978
Differential output : Adm = Add = −20VCC = −300 | Acm = 0
Single - ended output : Adm =
Add
= +10VCC = 150
2
| CMRR = ∞
| CMRR = 20VEE = 300 | Acm = −
150
= −0.5
300
Page 982
€
 100  R  15 − 0.7 
1−

 C 
101  2RC  15 

VIC = 15V
= 5.30 V


100  RC 
1+




101  2RC 


Page 983
€
€
I DC = I SS −
VO
15V
= 100µA −
= 80 µA
RSS
750kΩ
52
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 985
I
I D = SS = 100 µA | VDS = 12 − I D RD + VGS = 12 −100µA(62kΩ) + VGS = 5.8V + VGS
2
2I D
VGS = VTN +
= VTN + 0.2V | VTN = 1+ 0.75 VSB + 0.6 − 0.6 | VSB = −VGS − (−12V )
Kn
(
VSB = 11.8 − VTN
(
)
)
| VTN = 1+ 0.75 12.4 − VTN − 0.6 → VTN = 2.75V | VDS = 8.75 V
Q − point : (100 µA, 8.75 V )
Page 988
€
Rod = 2ro ≅ 2
VA
60V
=2
= 3.20 MΩ | Roc ≅ 2µ f REE = 2( 40)(60)(1MΩ) = 4.80 GΩ
IC
37.5µA
idm = g mvdm = 40( 37.5µA)vdm = 1.5x10−3 vdm | icm ≅
v cm
v
= cm = 5.00x10−7 v cm
2REE 2 MΩ
Page 993
€
IC1 = IC 2 =
100  150µA 
15V
= 750 µA | VCE 3 = 15 − 0 = 15.0 V

 = 74.3 µA | IC 3 =
101  2 
20kΩ
VCE1 = 15 − 74.3µA(10kΩ) − (−0.7) = 15.0 V | VCE 2 = 15 − (74.3µA − 7.5µA)(10kΩ) − (−0.7) = 15.0 V
VEB 3 = ( 74.3µA − 7.5µA)(10kΩ) = 0.668 V | I S 3 =
750µA
= 1.87x10−15 A
 0.668V 
exp

 0.025 
Page 996
€
max
Adm
= 560 (15) = 8400 | IC1 ≤ 50(1µA) = 50 µA | Adm =
IC1 ≤ 50(1µA) = 50 µA | Adm =
8400
= 2210
28  500µA 
1+


100  50µA 
8400
= 290
28  5mA 
1+


100  50µA 
−−−
Rin = 2rπ = 2
50
15V
= 50 kΩ | Rout ≅
= 30 kΩ
0.5mA
40(50µA)
Rin = 2rπ = 2
50
15V
= 50 kΩ | Rout ≅
= 3.0 kΩ
5mA
40(50µA)
−−−
max
Adm
= 560 (1.5) = 840
53
€
©R. C. Jaeger and T. N. Blalock
09/25/10
54
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 997
CMRR ≅ g m2 R1 = 40 (50µA)( 750kΩ) = 1500 | CMRRdB = 20 log(1500) = 63.5 dB
€
Page 998
 0.7r 
g  Rr 
40  I R r 
560VA 3
π3
Adm = m2  C π 3 (g m3ro3 ) =  C 2 C π 3 ( 40IC 3ro3 ) ≅ 800
(VA3 ) =
R
2  RC + rπ 3 
2  RC + rπ 3 
 RC + rπ 3 
1+ C
rπ 3
Adm =
560VA 3
560VA 3
560VA 3
560VA 3
=
=
=


40IC 3 RC
40(0.7)  IC 3 
40IC 2 RC IC 3
28  IC 3 
1+
1+
1+
  1+
 
 
β o3
β o3  IC 2 
β o3  IC 2 
β o3  IC 2 
−−−
max
Adm
= 560 ( 75) = 42000 | IC1 ≤ 50(1µA) = 50 µA
Adm =
42000
= 11000
28  500µA 
1+


100  50µA 
| Adm =
42000
= 1450
28  5mA 
1+


100  50µA 
−−−
Rin = 2rπ = 2
50
75V + 15V
= 50 kΩ | Rout ≅ ro3 =
= 180 kΩ
0.5mA
40(50µA)
Rin = 2rπ = 2
50
90V
= 50 kΩ | Rout ≅
= 18.0 kΩ
5mA
40(50µA)
€
55
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1002
Avt1 = −3.50 | Avt 2 = −22mS 150kΩ 162kΩ 203kΩ = −1238
(
Avt 3 =
(
)
0.198S 2kΩ 18kΩ
(
)
1+ 0.198S 2kΩ 18kΩ
Rin = 2rπ = 2
)
= 0.9971 | Adm = −3.50(−1238)(0.9971) = 4320
ro3 R2
162kΩ 150kΩ
100
1
1
= 101 kΩ | Rout ≅
+
=
+
= 776 Ω
g m4 β o4 + 1 40( 4.95mA)
101
40( 49.5µA)
P ≅ ( I1 + I 2 + I 3 )(VCC + VEE ) = (100 + 500 + 5000)µA(30V ) = 168 mW
−−−
150 
5mA
0.7V
IC = 50µA
= 533 µA | RC =
= 15.2 kΩ
 = 49.7 µA | IC 3 = 500µA +

151
533
 151 
 49.7 −
µA
150 

rπ 3 =
150
= 7.04 kΩ | Avt 2 = −20( 49.7µA) 15.2kΩ 7.04kΩ = −4.68
40(533µA)
IC 4 =
150
150
75 + 14.3
5mA = 4.97 mA | rπ 4 =
= 755 Ω | ro3 =
= 168 kΩ
151
533µA
40( 4.97mA)
(
[
)
]
Avt 2 = −40(533µA) 168kΩ 755 + 151(2kΩ) = −2304
g m4 = 40(4.97mA) = 0.199 S | Avt 3 =
0.199S (2kΩ)
1+ 0.199S (2kΩ)
= 0.998
Adm = −4.78(−2304 )(0.998) = 11000
Rid = 2rπ 1 = 2
ro3 R2
150
1
1
168kΩ
= 151 kΩ | Rout ≅
+
=
+
= 1.12 kΩ
g m4 β o4 + 1 40(4.95mA)
151
40( 49.7µA)
CMRR is set by the input stage and doesn't change since the bias current is the same.
−−−
ro3 =
2
50 + 14.3
= 117 kΩ | Avt 2 = −22mS 117kΩ 203kΩ = −1630 | Adm = −3.50(−1630)(0.998) = 5700
550µA
(
)
ro3 R2
100
1
1
117kΩ
= 101 kΩ | Rout ≅
+
=
+
= 1.16 kΩ
g m4 β o4 + 1 40(4.95mA)
101
40( 49.5µA)
CMRR and input resistance are set by the input stage and don't change.
−−−
T
6920
R
1.62kΩ
=
= 0.99986 | TOC = T | TSC = 0 | Rout = o =
= 0.234 Ω
1+ T 6921
1+ T 1+ 6920
Rin ≅ Rid (1+ TSC ) = 101kΩ(6921) = Rid (1+ T ) = 699 MΩ (Assuming TOC << 1)
Av =
€
56
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1004
2(500µA)
VGS 3 = 1+
Avt1 = −
= 1.63 V | RD =
2.5mA
1.63V
= 16.3 kΩ
100µA


1
1
 = −316
2(0.005)(100µA) (16.3kΩ) = −8.16 | Avt 2 = −g m3ro3 = − 2(0.0025)(0.0005)

2
 0.01(0.5mA) 
g m4 = 2(0.005mA)(0.005mA) = 7.07 mS | Avt 3 =
7.07mS (2kΩ)
1+ 7.07mS (2kΩ)
Adm = −8.16(−316)(0.934 ) = 2410 | Rid = ∞ | Ro =
= 0.934
1
1
=
= 141 Ω
g m4 7.07mS
CMRR = g m1 R1 = 1.00mS ( 375kΩ) = 375 or 51.5 dB
−−−
P ≅ ( I1 + I 2 + I 3 )(VDD + VSS ) = (5.7mA)(24V ) = 137 mW
Page 1005
€
€
Add1 = −
Kn′
KP′
(W L)
(W L)
| 10 = 2.5
2
L2
(W L)
4
2
→ (W L) 2 =
160
1
Page 1011
VGS1 + VSG 2 = 0.5mA( 4.4kΩ) = 2.2 V | Since the device parameters are the same,
VGS1 = VSG 2 = 1.1 V | I D =
2
0.025
1.1−1) = 125 µA
(
2
−−−
Since the device parameters are the same, VBE1 = VEB 2 =
0.5mA(2.4kΩ)
2
= 0.6 V
 0.6 
IC = 10−14 A exp
 = 265 µA
 0.025 
(
€
Page 1014
2
v
Av1 = d = −g m n 2 RL = − 50mA/V 2 (2V −1V )(10) (8Ω) = −40.0
vg
(
Avo = Av1
vo ≤
€
)
1
40.0
=−
= −4.00 |
n
10
)
v g ≤ 0.2(2 −1)V = 0.200 V |
v d ≤ 0.2V ( 40) = 8.00 V
8V
= 0.800 V
10
57
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1022

150(18.4kΩ) 
 = 32.5 MΩ
RB → 0 | Rout = 432kΩ 1+
 18.8kΩ + 18.4kΩ
−−−
270kΩ
VEQ = −15V
= −10.66 V | REQ = 110kΩ 270kΩ = 78.2 kΩ
110kΩ + 270kΩ
−10.66 − 0.7 − (−15)
195 µA
IC = 150
= 195 µA | VB = VEQ − I B REQ = −10.66 −
(78.2 kΩ) = −10.8 V
150
78.2kΩ + 151(18kΩ)
2
PR1
(−10.8 + 15)
=
PR E
(−10.8 − 0.7 + 15)
=
Rout
110kΩ
= 0.160 mW | PR 2
18kΩ
(−10.8)
=
2
270kΩ
2
= 1.33 mW | ro =
= 0.432 mW
(75 + 11.5)V = 446 kΩ
195µA
| rπ =
150
= 19.3 kΩ
40(195µA)


150(18kΩ)

 = 10.9 MΩ
= 446kΩ 1+
 78.2kΩ + 19.3kΩ + 18kΩ
−−−
15V
= 750kΩ | Using a spreadsheet with I o = 200 µA yields VBB = 9V.
20µA
 9V 
150  9 − 0.7 −1.33µA(180kΩ) 

 = 40.0 kΩ
R1 = 750kΩ
 = 450 kΩ | R1 = 300 kΩ | RE =
151 
200µA
 15V 


 75 + 15 − 8.3 
150(40.0kΩ)

 = 10.7 MΩ
Rout = 
 1+
 2x10−4  180kΩ + 18.75kΩ + 40.0kΩ
R1 + R2 ≅
Page 1026
€
VDS ≥ VGS − VTN = 1+
2(0.2mA)
2.49mA/V 2
= 1.40 V | VD = VS + 1.40 = −15 + 0.2mA(18.2kΩ) + 1.40 = −9.96 V
−−−
W Kn 2.49mA /V 2 99.6
=
=
=
L Kn'
1
25µA/V 2
€
58
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1027
2
PR S = (0.2mA) 18.2kΩ = 0.728 mW | I BIAS =
15V
= 20.1 µA
499kΩ + 249kΩ
2
2
PR 4 = (20.1µA) 499kΩ = 0.202 mW | PR 3 = (20.1µA) 249kΩ = 0.101 mW
510kΩ
= −10.2 V | −10.2 − VGS −18000I D = −15 V
510kΩ + 240kΩ
2
2.49mA
4.8 − VGS −18000
VGS −1) = 0 | VGS = 1.390 V | I D = 189 µA
(
2
VGG = −15V
Rout
(
)
2 2.49x10−3
1
≅ µ f RS ≅
1+ 0.01(11.6) (18kΩ) = 10.3 MΩ
0.01
189x10−6
[
]
€
59
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 16
Page 1049
Ravg = 10kΩ(1+ 0.2) = 12 kΩ | 12kΩ(1− 0.01) ≤ R ≤ 12kΩ(1+ 0.01)
| 11.88 kΩ ≤ R ≤ 12.12 kΩ
Page 1051
€
VDS1 = VTN +
IO = 150µA
2I REF
Kn (1+ λVDS1 )
1+ 0.0133(10)
1+ 0.0133(2.08)
| VDS1 = 1+
2(150µA)
250µA/V 2 [1+ 0.0133VDS1]
→ VDS1 = 2.08 V
= 165 µA
−−−
VDS ≥ VGS − VTN
| VD − (−10V ) ≥
2I D
Kn
| VD ≥ −10V +
2(150µA)
250µA /V 2
= −8.91 V
Page 1052
€
MR =
2(50µA)
1+ 0.02(15)
25 /1
= 8.33 | VDS1 = 1V +
=
2.16
V
|
MR
=
8.33
= 10.4
3/1
1+ 0.02(2.16)
3 25µA/V 2
MR =
2 /1
= 0.400
5 /1
(
| VDS1 = 1V +
)
2(50µA)
(
5 25µA/V
2
)
= 1.89 V | MR = 8.33
1+ 0.02(10)
1+ 0.02(1.89)
= 0.463
Page 1054
€
 V  V
 VBE1
2 
2 
I REF = I S exp BE1 1+ BE1 +
+
 | 100µA = (0.1 fA) exp( 40VBE1 )1+
 → VBE1 = 0.690
 50V 100 
 VT  VA1 β FO 
VCE ≥ VBE → VC ≥ −VEE + 0.690 V
€
€
Page 1055
0.5 A
5A
(a) MR = A = 0.5 | MR = 2 A = 2.50
0.5
2.50
(b) MR = 1.5 = 0.490 | MR = 3.5 = 2.39
1+
1+
75
75
15
15
1+
1+
(c) MR = 0.5 0.7601.5 = 0.606 | MR = 2.5 0.7 60 3.5 = 2.95
1+
+
1+
+
60 75
60 75
60
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1056
 10 /1
 20 /1
IO2 = 100µA
 = 200 µA | IO3 = 100µA
 = 400 µA
 5 /1 
 5 /1 
 40 /1
 2.5 /1
IO4 = 100µA
 = 800 µA | IO5 = 100µA
 = 50 µA
 5 /1 
 5 /1 
−−−
IO2 = 200µA
IO4 = 800µA
1+ 0.02(10)
1+ 0.02(2)
1+ 0.02(12)
1+ 0.02(2)
= 231 µA | IO3 = 400µA
= 954 µA | IO5 = 50µA
1+ 0.02(5)
1+ 0.02(2)
1+ 0.02(8)
1+ 0.02(2)
= 423 µA
= 55.8 µA
−−−
1
= 7.46 µA | IO 3 = 5( 7.46µA) = 37.3 µA | IO 4 = 10( 7.46µA) = 74.6 µA
17
1+
50
10
10
1+
1+
50
50
IO2 = 10µA
= 8.86 µA | IO 3 = 50µA
= 44.3 µA
0.7 17
0.7 17
1+
+
1+
+
50 50
50 50
10
1+
50
IO4 = 100µA
= 88.6 µA
0.7 17
1+
+
50 50
IO2 = 10µA
€
Page 1057
10
10 − 9.957
MR =
= 9.957 | FE =
= 4.3x10−3
11
10
1+
50(51)
| VCE 2 = VBE1 + VBE 3 = 1.4 V
€
61
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1058
MOS
IO2 = 200µA
IO3 = 400µA
1+ 0.02(10)
1+ 0.02(2)
1+ 0.02(5)
1+ 0.02(2)
= 231 µA | Rout 2 =
= 423 µA | Rout 3 =
50V + 10V
= 260 kΩ
231µA
50V + 5V
= 130 kΩ
423µA
BJT
1+
IO2 = 10µA
10
50
= 10.1 µA | Rout 2 =
50V + 10V
= 5.94 MΩ
10.1µA
0.7 17
+
50 100
10
1+
50V + 10V
50
IO3 = 50µA
= 50.7 µA | Rout 3 =
= 1.19 MΩ
0.7 17
50.7µA
1+
+
50 100
1+
Page 1059
€
0.7V
10V
1+
50V
50V
IC1 = 100µA
= 89.4 µA | IC 2 = 500µA
529µA
0.7V
6
0.7V
6
1+
+
1+
+
50V 50V
50V 50V
1
1
529µA
50V + 10V
Rin ≅
=
= 280 Ω | β =
= 5.92 | Rout =
= 113 kΩ
g m1 40(89.4 µA)
89.4 µA
529µA
1+
Page 1060
€
10V
VDS1 = VGS1 = 0.75V +
= 1.20 V | I D 2 = 100µA 50V = 117 µA
1.2
1mA /V 2
1+
50V
1
1
117µA
50V + 10V
Rin ≅
=
= 2.24 kΩ | β =
= 1.17 | Rout =
= 513 kΩ
g m1
100µA
117µA
2 10−3 10−4
1+
2(100µA)
( )( )
€
62
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1061
V I
A  0.025V 100µA 
R = T ln REF E 2  =
ln
5 = 3000 Ω
IO  IO AE1  25µA  25µA 
100µA 
 75V 
K = 1+ ln
5 = 4.00 | Rout = 4
 = 12.0 MΩ
 25µA 
 25µA 
€
Page 1062
V I
A 
0.025V  1000µA 
IO = T ln REF E 2  | IO =
ln
 → IO = 300.54 µA
R  IO AE1 
100Ω  IO 
 100µA

 75V 
K = 1+ ln
10 = 2.202 | Rout = 2.202
 = 550 kΩ
 300.54 µA 
 300.54µA 
Page 1063
€


2(200µA) 
1 2I REF 
IO (W/L)1 
1
IO 1 


IO =
1−
| IO =
1−
R Kn1 
I REF (W/L) 
2kΩ 25µA /V 2 
200µA 10 
2 


IO 
 → IO = 764 µA
IO = 2.00mA1−
2.00mA 

Rout =
€
50V + 10V 

1+ 2000 2 2.5x10−4 7.64 x10−4  = 176 kΩ

764µA 
(
)(
)
Page 1066
β r 150  50V + 15V 
50V + 15V
Rout ≅ o o =
= 1.30 MΩ

 = 97.5 MΩ | Rout = ro =
2
2  50µA 
50µA
Page 1068
€
VDS 2 = VGS 2 = 0.8V +
(
(
2 5x10−5
) = 1.43 V
2.5x10−4
)(
Rout ≅ µ f 4 ro2 = 2 2.5x10−4 5x10−5
Rout = ro =
| VDS 4 = 15 −1.43 = 13.6 V
 1
 1

V + 13.6V 
V + 1.43V 

1+ 0.015(13.6)  0.015
 0.015
 = 379 MΩ
50
µ
A
50µA






)[
]
66.7V + 15V
= 1.63 MΩ
50µA
−−−
Rout ≅
β oro 100  67V + 14.3V 
67V + 15V
=
= 1.64 MΩ

 = 81.3 MΩ | Rout = ro =
2
2 
50µA 
50µA
63
€
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1072
IO = 25.014µA +
10V − 20V
= 25.008 µA
1.66GΩ
−−−
VDS 4 ≥ VGS 4 − VTN = 0.2 V | VD 4 ≥ VS 4 + 0.2V | VD 4 ≥ 0.95 + 0.2 = 1.15 V
−−−
20V
= 400 MΩ | Choose Rout = 1 GΩ.
50nA
2
 0.01
 50µA
50V
1 2Kn
mA
ro ≅
= 1 MΩ → µ f = 1000 | µ f ≅
| Kn = 
1000)
= 2.5 2
(
50µA
λ ID
V
 V
 2
2.5x10−3 50
1  50  25
(W / L)2 = (W / L)4 = 5x10−5 = 1 | (W / L)3 = (W / L)1 = 2  1  = 1
IO = 50µA ± 0.1%
€
| ΔIO ≤ 50 nA | Rout ≥
Page 1073
5V − 0.7V
7.5V − 0.7V
I REF =
= 100 µA | I REF =
= 158 µA
43kΩ
43kΩ
−−−
Since the transistors have the same parameters, VGS1 =
VDD − (−VSS )
3
2
2
4 x10
4 x10−4
I D2 = I D1 =
1.667
−1
=
89.0
µ
A
|
I
=
I
=
2.5 −1) = 450 µA
(
)
(
D2
D1
2
2
−−−
0.025V
5 −1.4
0.025V
7.5 −1.4
IO ≅
ln −16
= 101 µA | IO ≅
ln −16
= 103 µA
6.8kΩ 10 (39kΩ)
6.8kΩ 10 (39kΩ)
−4
€
Page 1074
V I A 
0.025V
IO = T ln C1 E 2  | IO =
ln 10(10) = 115 µA
R  IC 2 AE1 
1000Ω
−−−
VCC + VEE ≥ VBE1 + VBE 4 ≅ 1.4 V
[
]
Page 1076
€
R=
2
(
)( )
5 25x10−6 10−4

5
1−
 = 8.65 kΩ
50


€
64
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1077
V  I A  0.025875V  25 
R = T ln C1 E 2  =
ln  = 925 Ω
IO  IC 2 AE1 
45µA
5
AE1 = A | AE 2 = 25 AE1 = 25 A | AE 3 = A | AE 4 = 5.58 AE 3 = 5.58 A
Page 1081
€
A 
V
82.59mV
VPTAT = VT ln E 2  = (27.57mV ) ln (20) = 82.59 mV | R1 = PTAT =
= 3.30 kΩ
IE
25µA
 AE1 
I 
 25µA 
VBE1 = VT ln C1  = (27.57mV ) ln
 = 0.6792 V
 0.5 fA 
 I S1 
R2 VGO + 3VT − VBE1 1.12 + 3(0.02757) − 0.6792
=
=
= 3.169
R1
2VPTAT
2(0.08259)
VBG = VBE1 + 2
| R2 = 3.169R1 = 10.5 kΩ
R2
VPTAT = 0.6792 + 2(3.169)(0.08259) = 1.203 V
R1
The other resistors remain the same.
Page 1084
€
I D3 = I D 4 = I D1 = I D 2 =
VGS 3 = −0.75V −
2(125µA)
250µA
= 125 µA | VGS1 = 0.75V +
= 1.75 V
2
250µA /V 2
2(125µA)
200µA/V 2
= −1.87 V
VDS1 = VD1 − VS1 = (5 −1.87) − (−1.75) = 4.88 V | VSD 3 = VSG 3 = 1.87 V
M1 and M 2 : (125 µA, 4.88 V )
(
| M 3 and M 4 : (125 µA, 1.87 V )
)(
)
Gm = g m1 = 2 2.5x10−4 1.25x10−4 = 250 µS
Ro = ro2 ro4 =
75.2V + 4.88V 75.2V + 1.87V
= 314 kΩ | Av = Gm Ro = 78.5
125µA
125µA
Page 1085
€
 1 2K 
 1 2K 
n3
n3
 2Kn2 I D 2 RSS = 
 2Kn2 I D 2 RSS
CMRR ≅ µ f 3 g m2 RSS = 
λ
I
λ
I
D3 
D3 


1
Kn3 = Kn2 | I D 2 = I D 3 | CMRR =
2(0.005)10 7 = 5.99x10 6 or 136 dB
0.0167
(
)
(
)
€
65
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1089
For the buffered current mirror, VEC 4 = VEB 3 + VEB11 +
IC11 ≅
 I

| ΔVEB = 0.025 ln C 4  = 80.5 mV
 IC 4 / 25 
2IC 4 2IC 4 IC 4
=
=
β FO 4
50
25
VEC 4 = 0.7 + (0.7 − 0.081) +
€
2VA
β FO 4 (β FO11 + 1)
2(60)
50(51)
= 1.37 V | ΔVEC =
2(60V )
50(51)
= 47.1mV | VOS =
47mV
= 0.47 mV
100
Page 1090

I  150
Av1 ≅  β o5 C 2  =
= 50
IC 5 
3

−−−
For the whole amplifier : Adm ≅ Av1 Av 2 Av3
Adm ≅ 50( 3000)(1) = 150000
| Av 2 ≅ µ f 5 ≅ 40(75) = 3000 | Av 3 ≅ 1
| Note that this assumes R L = ∞.
Page 1091
−1
€
−1



 2  1

1
2 
1
1

 = 5.45x10 6 →135 dB
CMRR =  
−
 =
−
−4
7
100  100( 40)(75) 2( 40) 10 10 
β o3  β o2µ f 2 2g m2 REE 



( )( )
Page 1096
€
Adm ≅ Av1 Av 2 Av 3 | Av1 ≅
40(60 + 14.7)
µ
IC 2
I
β
50
β o5 = REF o5 =
= 5 | Av 2 ≅ f 5 ≅
≅ 1500 | Av 3 ≅ 1
IC 5
2 5I REF 10
2
2
Adm ≅ 5(1500)(1) = 7500 assuming the input resistance of the emitter followers is much greater than
ro5 and VA 8 = VA 5. Checking : ro5 ≅
IC 5 = 10IC 4 = 10IC 3 → AE 5 = 10 A
60V + 14.7V
= 149 kΩ | RiB 6 ≅ β o6 RL = 300 MΩ
500µA
150
| Rid = 2rπ 1 = 2
= 150 kΩ
40(50µA)
€
66
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1098
22 + 22 −1.4
0.025V  1.09mA 
I REF =
= 1.09 mA | I1 =
ln
 → I1 = 20.0 µA
39kΩ
5kΩ
 I1 
23.4V
21.3V
1+
1+
60V
60V
I 2 = 0.75(1.09mA)
= 1.08 mA | I 2 = 0.25(1.09mA)
= 351 µA
0.7V 2
0.7V 2
1+
+
1+
+
60V 50
60V 50
−−−

 733µA 
I
A  60V + 13.5V 
Ro = ro211+ ln C 20 E 20  =
1 = 18.7 MΩ
1+ ln
IC 21 AE 21 
18.4 µA 
 18.4 µA 

Page 1102
€
VCE 6 = VCE 5 +
2(60V )
2VA 6
= 0.7 +
= 1.90 V
β FO6
100
Page 1105
€
€
 60 + 13   60 + 1.3
Rth = Rout 4 Rout 6 = 2ro4 1.3ro6 = 2
 1.3
 = 20.1MΩ 11.1MΩ = 7.15 MΩ
 7.25µA   7.16µA 
Page 1107
Av1 = −1.46x10−4 6.54 MΩ Rin11 = −1.46x10−4 6.54 MΩ 20.7kΩ = −3.01
(
)
(
)
Page 1109
€
Req 2 = rπ 15 + (β o15 + 1) RL =
Req1 = rd14 + (rd13 + R3 )
50(0.025)
+ 51(2kΩ) = 103 kΩ
2mA

0.025V  0.025V
Req 2 =
+
+ 344kΩ 103kΩ = 79.4 kΩ
0.216mA  0.216mA

50(0.025V )
+ 51( 79.4kΩ) = 4.06 MΩ
0.216mA
−1  

  0.025V
r + y22
0.025V
5.79kΩ + 89.1kΩ 
Req 3 = (rd13 + R3 )  rd14 + π 12
+ 344kΩ 
+
 =
 = 1.97 kΩ
β o12 + 1   0.216mA
51
  0.216mA


Rin12 =
rπ 16 = 50
0.025
625 + 1970
= 625 Ω | Rout =
+ 27 = 78 Ω
2mA
51
−−−
I SC + ≅
€
0.7V
0.7V
= 25.9 mA | I SC − ≅ −
= −31.8 mA
27Ω
22Ω
67
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1112
 R 
vo
vo
5
1
vo = 
= 2 = 0.2 | K M =
= 2 =1
v1v 2 = K M v1v 2 | K M =
v1v 2 5
v1v2 1
 I EE R1 R3 
€
68
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 17
Page 1131
2
2
2
2
1
1
fL ≅
10 2 + 1000 2 − 2(50) − 2(0) = 159 Hz | f L ≅
100 2 + 1000 2 − 2(500) − 2(0) = 114 Hz
2π
2π
Page 1132
€
€
€
€
200s( s + 100)
200s
≥ 0.9
(s + 1000)
(s + 10)(s + 1000)
| 1 ≥ 0.9
ω 2 + 100 2
ω 2 + 10 2
→ 0.81 ≤
ω 2 + 10 2
→ ω ≥ 205 rad/s
ω 2 + 100 2
Page 1133
10 6
fH ≅
= 159 kHz
2π
Page 1134
1
1
fH ≅
= 21.7 kHz
2
2
2
2
2π 







1
1
1
1
− 2
− 2 
 5 +
5
5
10   5x10 
 2x10 
∞
Page 1139
The value of C3 does not change A mid , ω P1, ω P2 , ω z1 , or ω Z2 .
ωP3 = −
fL =
1
2π
1
1
= −1000 rad/s | ω Z 3 = −
= −385 rad/s


2µF (1.3kΩ)
1
2µF 1.3kΩ

1.23mS 

(
)
41.0 2 + 95.9 2 + 1000 2 − 2 0 2 + 0 2 + 385 2 = 135 Hz
−−−
13.5
4.732
→ ro = 57.5 kΩ
1.23mS
Note that the SPICE value of gm probably differs from 1.23 mS as well.
Amid = 10 20 = 4.732
|
4.3kΩ 100kΩ ro =
−−−
ωP3 = −
fL =
€
1
2π
1
= −202 rad/s


1
10µF 1.3kΩ
57.5kΩ
1.23mS


(
)
41.0 2 + 95.9 2 + 202 2 − 2 0 2 + 0 2 + 76.9 2 = 31.8 Hz
69
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1142
140(0.025V )
rπ =
= 20.0 kΩ | R1S C1 = 1kΩ + 75kΩ 20.0kΩ 2µF = 33.6 ms | Rth = 75kΩ 1kΩ = 987 Ω
175µA

20.0kΩ + 987Ω 
R2S C2 = ( 43kΩ + 100kΩ)0.1µF = 14.3 ms | R3S C3 = 13kΩ
10µF = 1.47 ms
141


1  1
1
1 
fL ≅
+
+

 = 124 Hz
2π  33.6ms 1.47ms 14.3ms 
(
€
)
Page 1144
Rin  β o   1260  100 
Av = −
 RL  ≅ −

 4.3kΩ 100kΩ = −157
RI + Rin  rπ   2260 1.51kΩ 
Rin  β o   1260  100 
Av = −
 RL  ≅ −

 4.3kΩ 100kΩ 46.8kΩ = −140
RI + Rin  rπ   2260 1.51kΩ 
(
)
(
)
ro is responsible for most of the discrepancy. rπ and βo will also be differ from our hand calculations.
Note that 45% of the gain is lost because of the amplifier's low input resistance.
−−−
gm =
2(1.5mA)
0.5V
= 6.00 mS | R1S C1 = (1kΩ + 243kΩ)0.1µF = 24.4 ms

1 
R2S C2 = ( 4.3kΩ + 100kΩ)0.1µF = 10.4 ms | R3S C3 = 1.3kΩ
10µF = 1.48 ms
6.00mS 

1  1
1
1 
fL ≅
+
+

 = 129 Hz
2π  24.4ms 1.48ms 10.4ms 
Page 1146
€
€

1 
g m = 40(0.1mA) = 4.00 mS | R1S C1 = 100Ω + 43kΩ
 4.7µF = 1.64 ms
4.00mS 

1  1
1 
R2S C2 = (22kΩ + 75kΩ)1µF = 97.0 ms | f L ≅
+

 = 98.7 Hz
2π  1.64ms 97.0ms 
Page 1147

2(1.5mA)
1 
gm =
= 6.00 mS | R1S C1 = 100Ω + 1.3kΩ
1µF = 0.248 ms
0.5V
6.00mS 

R2S C2 = ( 4.3kΩ + 75kΩ)0.1µF = 7.93 ms |
fL ≅
1 
1
1 
+

 = 662 Hz
2π  0.248ms 7.93ms 
€
70
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1148
100
= 2.50 kΩ
.04S


R1S C1 = 1kΩ + 100kΩ 2.5kΩ + 101 3kΩ 47kΩ 0.1µF = 7.52 ms



2.5kΩ + 100kΩ 1kΩ 

100µF = 4.70 s | f ≅ 1  1 + 1  = 21.2 Hz
R2S C2 = 47kΩ + 3kΩ


L


101
2π  7.52ms 4.7s 


g m = 40(1mA) = 40.0 mS | rπ =
[
(
Amid ≅
(β
)]
(
)
(
+ 1) RL
)
101 3kΩ 47kΩ
 RI 

=
Rth + rπ + (β o + 1) RL  RI + RB  990Ω + 2.5kΩ + 101 3kΩ 47kΩ
o
(
)
 100kΩ 

 = +0.978
 1kΩ + 100kΩ 
−−−

1 
R1S C1 = (1kΩ + 243kΩ)0.1µF = 24.4 ms | R2S C2 = 24kΩ + 1.3kΩ
 47µF = 1.15 s
1mS 

1  1
1 
fL ≅
+

 = 6.66 Hz
2π  24.4ms 1.15s 
 RG  g m RL
 243kΩ   1mS 1.3kΩ 24kΩ 
 = +0.550
Amid = +
= +



R
+
R
1+
g
R
244kΩ

 1+ 1mS 1.3kΩ 24kΩ 
 I
G
m L
(
)
(
€
Page 1152
g
Cπ = m − Cµ
ωT
Cµo
Cµ =
VCB
ϕ jc
1+
(100 µA, 8 V ) :
2 pF
Cµ =
1+
(2 mA, 5 V ) :
Cµ =
Cµ =
7.3V
0.6V
2 pF
1+
(50 mA, 8 V ) :
)
4.3V
0.6V
2 pF
1+
7.3V
0.6V
= 0.551 pF | Cπ =
= 0.700 pF | Cπ =
( )
40 10−4
2π (500 MHz)
(
40 2x10−3
)
2π (500 MHz)
= 0.551 pF | Cπ =
(
40 5x10−2
− 0.551x10−12 = 0.722 pF
− 0.700x10−12 = 24.8 pF
)
2π (500 MHz)
− 0.551x10−12 = 636 pF
€
71
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1155
1
CGS = CGD = CISS = 0.5 pF
2
−−−
CGS + CGD =
gm
ωT
| 5CGD + CGD =
2(0.01)(0.01)
2π (200 MHz)
= 11.3 pF | CGD = 1.88 pF | CGS = 9.38 pF
−−−
Cµ =
Cµo
1+
€
VCB
ϕ jc
=
2 pF
1+
7.3V
0.6V
= 0.551 pF | Cπ =
40(20µA)
gm
− Cµ =
− 0.551pF = −0.296 pF
ωT
2π (500 MHz)
Page 1158

Rin  β o
Av = −
RL  | Rin = 7.5kΩ (1.51kΩ + 250Ω) = 1.43kΩ

RI + Rin  rx + rπ 
 1430  100 
≅ −

 4.3kΩ 100kΩ = −139
 2430 1.76kΩ 
Rin  β o   1260  100 
Av = −
 RL  ≅ −

 4.3kΩ 100kΩ = −157
RI + Rin  rπ   2260 1.51kΩ 
(
)
(
)
Page 1166
€
RL
is added to the value of CT .
rπo
 4120 
R
1
CL L = 3 pF 
= 1.39 MHz
 = 18.8 pF | f P1 =
rπo
2π (656Ω)(156 + 18.8) pF
 656 
The term CL
fP 2 =
gm
0.064S
=
= 445 MHz
2π (Cπ + CL ) 2π (19.9 + 3) pF
−−−
Cπ =
0.064
−10−12 = 19.4 pF
2π (500 MHz)

4120 
1
CT = 19.4 + 11+ 0.064 ( 4120) +
= 837 kHz
 = 290 pF | f P1 =
656 
2π (656Ω)290 pF

fP 2 =
gm
0.064S
g
0.064S
=
= 525 MHz | f Z = m =
= 10.2 GHz
2πCπ 2π (19.4 pF )
2πCµ 2π (1pF )
Amid = −135 is not affected by the value of f T .
€
72
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1167

4120 
1
CT = 10 + 2 1+ 1.23mS ( 4.12kΩ) +
= 5.26 MHz
 = 30.4 pF | f P1 =
996 
2π (996Ω) 30.4 pF

fP 2 =
fT =
€
gm
1.23mS
gm
1.23mS
=
= 19.6 MHz | f Z =
=
= 97.9 MHz
2πCGS 2π (10 pF )
2πCGD 2π (2 pF )
gm
1.23mS
=
= 16.3 MHz
2π (CGS + CD ) 2π (12 pF )
Page 1174
1+ g m RE = 1+ 0.064 (100) = 7.40
| RiB = 250 + 1560 + 101(100) = 11.9 kΩ
rπ 0 = 11.9kΩ (882 + 250) = 1030 Ω | Ai =
 264 
Amid = −0.821
 = −29.3 | f H =
 7.4 
10kΩ 30kΩ 11.9kΩ
1kΩ + 10kΩ 30kΩ 11.9kΩ
= 0.821
1
= 6.70 MHz
19.9 pF
 264 4120 
2π (1.03kΩ)
+ 0.5 pF 1+
+

7.4 1030 

 7.4
GBW = 29.3(6.70 MHz) = 196 MHz
€
Page 1176
g m′ RL
Amid ≅
1+ g m′ RE
fH ≅
| g m′ =
βo
=
rx + rπ
3.96mS (17.0kΩ)
100
= 3.96 mS | Amid ≅
= +48.2
100
1+ 3.96mS (100Ω)
250 +
40(0.1mA)
1
= 18.7 MHz | GBW = 903 MHz
2π (17.0kΩ)(0.5 pF )
−−−
RiS = R4
1
1
= 1.3kΩ
= 265 Ω
gm
3mS
Amid = 0.726( g m RL ) = 0.726( 3mS )(4.12kΩ) = 8.98
GBW = 86.7 MHz | f T ≅
|
fH ≅
1
= 9.66 MHz
2π ( 4.12kΩ)( 4 pF )
3mS
= 43.4 MHz
2π (11pF )
€
73
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1179
RL = 1.3kΩ 24kΩ = 1.23kΩ | Amid = 0.998
1
2π
fH ≅
1

10 pF 
1kΩ 430kΩ 1pF +

1+ 3.69 

(
3mS (1.23kΩ)
1+ 3mS (1.23kΩ)
= 0.785
= 50.9 MHz
)
Page 1182
€
€
fZ =
1
1
= 6.37 kHz | f P =
= 6.33 MHz
2π (25 MΩ)1pF
2π (50.25kΩ)0.5 pF
Page 1184
Differential Pair : Adm = −g m RC = −40(99.0µA)(50kΩ) = −198
Cπ =
40(99.0µA)
2π (500 MHz)
− 0.5 pF = 0.761 pF | rπ =
100
= 25.3 kΩ
40(99.0µA)
1
fH =
= 3.18 MHz


50kΩ 
2π (250Ω)0.761pf + 0.5 pF 1+ 198 +

250Ω 


 1 
g m1

198
 g m2 
CC − CB Cascade : Av =
g m RC ) = +
= +99.0
(
 1 
2
1+ g m1

 g m2 
fH ≅
1
= 6.37 MHz
2π (50kΩ)(0.5 pF )
Page 1185
€
g m = 40(1.6mA) = 64.0 mS | rπ =
Amid =
100
64.0mS
= 1.56 kΩ | Cπ =
− 0.5 pF = 19.9 pF
64mS
2π (500 MHz)
rπ
1.56 kΩ
−g m RL ) =
(
(−64.0mS )(4.12kΩ) = −153
RI + rx + rπ
882Ω + 250Ω + 1.56 kΩ
f P1 ≅
1
1
=
= 11.6 MHz
2π rπ 0 (Cπ + 2Cµ ) 2π (656Ω)(19.9 + 1) pF
fP 2 ≅
1
1
=
= 7.02 MHz
2π RL (Cµ + CL ) 2π (4120Ω)(0.5 + 5) pF
€
74
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1186
f P1 ≅
0.02(100µA)
0.02(25µA)
1
1
=
= 159 kHz | f P1 ≅
=
= 39.8 kHz
4 π CGGDro2
4 π CGGDro2
4 π (1pF)
4 π (1pF)
Page 1193
€
XC =
1
= 7.69 MΩ >> 2.39 kΩ
2π (530Hz)39 pF
XC =
1
= 300 MΩ | 51.8kΩ 19.8kΩ = 14.3 kΩ | 300 MΩ >> 14.3 kΩ
2π (530Hz)1pF
−−−
X1 =
1
1
= 23.9 Ω << 1.01 MΩ | X 2 =
= 5.08 mΩ << 66.7 Ω
2π (667kHz)0.01µF
2π (667kHz)47µF
X3 =
1
= 239 mΩ << 2.69 kΩ
2π (667kHz)1µF
Page 1198
€
ZC =
1
= − j3.18 Ω
2π j (5MHz)0.01µF
Page 1199
€
(i )
Q=
fo =
1
2π
(10µH )(100 pF + 20 pF )
100kΩ 100kΩ ro
2π (4.59 MHz)(10µH )
(
= 49.2
= 4.59 MHz | ro =
| BW =
50V + 15V −1.6V
= 19.8 kΩ
3.2mA
4.59 MHz
= 93.3 kHz
49.2
)
(
)
Amid = −g m 100kΩ 100kΩ ro = − 2(0.005)(0.0032) 100kΩ 100kΩ 19.8kΩ = −80.2
−−−
f o is unchanged | ro =
€
€
100kΩ 100kΩ ro
50V + 10V −1.6V
= 18.3 kΩ | Q =
= 46.4
3.2mA
2π (4.59 MHz)(10µH )
Page 1203
 C 
 C 
L
gm
REQ = g m S =
LS 1+ GD  = ωT LS 1+ GD 
CGS CGS + CGD  CGS 
 CGS 
REQ ≅ ωT LS for CGS >> CGD
75
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1204
LS ≅
€
€
€
REQ
ωT
=
L2 REQ
µn (VGS − VTN )
Page 1207
1 A 1
ACG =
=
Aπ π
=
(5x10
(400cm
2
−5
)
2
cm 75Ω
)
V − s (0.25V )
= 1.88x10-9 H
1
| 20log  = −9.94 dB
π 
Page 1209
1 2A 2
ACG =
=
A π
π
 2
| 20log  = −3.92 dB
π 
Page 1210
From Fig. 17.81(a), the amplitude of the output signal is approximately 70 mV, so the conversion
gain is approximately:
 200mV  1.4
1.4 
1
0.7)
| 20log  = −7.02 dB
=
(
100mV
 π  π
π 
From Fig. 17.81(b), the spectral components at 46 and 54 kHz have amplitudes of approximately
45 mV, so the conversion gain is:
ACG =
€
ACG =
45mV
1.4
= 0.45 | 20log(0.45) = −6.93 dB | Note :
= 0.446
100mV
π
Page 1212
€
fC − f m = 20 − 0.01 = 19.99 MHz | f C + f m = 20 + 0.01 = 10.01 MHz
3 f C − f m = 60 − 0.01 = 59.99 MHz | 3 f C + f m = 60 + 0.01 = 60.01 MHz
5 f C − f m = 100 − 0.01 = 99.99 MHz | 5 f C + f m = 100 + 0.01 = 100.01 MHz
−−−
AfC + f m = AfC − f m = 3 V
A3 f C + f m = A3 f C − f m =
A5 f C + f m = A5 f C − f m =
Af C − f m
3
Af C − f m
5
=1 V
= 0.6 V
€
76
©R. C. Jaeger and T. N. Blalock
09/25/10
CHAPTER 18
Page 1231
I1
= 1 mA | VCE 2 = 0.7 V | VEC3 = 0.7 V | VO = 0 V
2
VEC 4 = 5V − VO = 5 − 0 = 5.0 V | VEC1 = 5 − VC1 − VE1 = 5 − 0.7 − (−0.7) = 5.0 V
IC 4 = IC 3 = IC 2 = IC1 =
(1 mA, 5.0 V ), (1 mA, 0.7 V ), (1 mA, 0.7 V ), (1 mA, 5.0 V )
€
Page 1234
With the output shorted, current cannot make it around the loop, so TSC = 0.
−−−
TOC = 0 is zero only if ro1 is neglected since the votlage across rπ must be zero for ib = 0.
If we include ro1, and start at the base of Q2 assuming ro1 >>
vc4 ≅
T≅
1
and a current mirror gain of 1:
g m3

v e1
g r  1 
1) ro4 RiB 2 RL ro4 (1+ g m2ro1 ) =  vb2 m2 o1   ro4 RiB 2 RL
(
ro1
 1+ g m2ro1  ro1 
[
ro4 RiB 2 RL
ro1
]
=
55kΩ 5.1kΩ 10kΩ
55kΩ
[
]
= 0.0579
−−−
(
)
(
)
Evaluating Eq. 18.7 without RL , TOC = g m2 ro1 ro1 RiB 2 = 0.04 55kΩ 55kΩ 5.1kΩ = 172
Page 1239
€
[ (
Tnew = g m1 R3 rπ 2
Tnew = Told
R3 rπ 2
R3 RiB 2
)][−g ([ R + R
m2
(1+ g
m2
2
1
RiE1
]
 R R

1
iE1


R5 RL 

R
R
+
R
2
 1 iE1
)]
1kΩ 2.5kΩ
R4 ) = −2.01
1+ 0.04S (300Ω) = −19.2
970Ω
[
]
19.2
D
= 9.50 | Now RiC2 = ro2 = ∞ | RinD ≅ 31.7 kΩ | Rout
≅ 1.74 kΩ
1+ 19.2
1.86 
1+ 17.8
Scaling using the previous result : TSC = −19.2
= 596kΩ
 = 17.8 | Rin = 31.7kΩ
1+ 0
 2.01
Av = 10
Rout
−1
 1
1+ 0
1 
′ =
= 1.74kΩ
= 86.1 Ω | Removing RL : Rout
−
 = 86.8 Ω
1+ 19.2
 86.1 10 4 
€
77
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1241
1+ TSC
Rx = RxD
1+ TOC
| RxD = R3 RiD1 RiG3 = R3 2ro1 RiG3 = 3kΩ ∞ ∞ = 3kΩ
 1+ 0 
TSC = 0 | TOC = T = 61.1 | Rx = 3kΩ
 = 48.3 Ω
 1+ 61.1
−−−
I D2 = I D1 =
1.63V − (−5V )
I1
= 0.500 mA | I D4 = I2 = 2.00 mA | I D3 =
= 0.510 mA
2
13kΩ
2(0.5mA)
VDS1 = 3.5 + VGS1 | VGS1 = 1+
= 1.32 V | VDS1 = 3.5 + 1.32 = 4.82 V
10mA
VDS2 = 5 + 1.32 = 6.32 V | VDS 3 = 5 −1.63 = 3.37 V | VDS 4 = 5 − VO = 5.00 V
(0.5 mA, 4.82 V ), (0.5 mA, 6.32 V ), (0.51 mA, 3.37 V ), (2 mA, 5.0 V )
€
Page 1245
Ri appears in parallel with rπ : rπ′ = rπ Ri = 4.69kΩ 10kΩ = 3.19 kΩ
 r′ 


3.19kΩ
T = −g m RC ( RF + rπ′ ) ro  π  = −0.032S 5kΩ (50kΩ + 3.19kΩ) 62.4kΩ 
 = −8.17
 3.19kΩ + 50kΩ 
 rπ′ + RF 
T
8.17
Atr = AtrIdeal
= −50kΩ
= −44.5 kΩ
1+ T
1+ 8.17
RinD = rπ′ RF + RC ro = 3.19kΩ 50kΩ + 5kΩ 62.4kΩ = 3.01 kΩ | TSC = 0 |
TOC = T
(
(
)
)
(
)
(
)
1+ 0
= 328 Ω
1+ 8.17
= RC ro ( RF + rπ′ ) = 5kΩ 62.4kΩ (50kΩ + 3.19kΩ) = 4.26 kΩ
Rin = 3.01kΩ
D
Rout
Rout = 4.26kΩ
|
TSC = 0
|
TOC = T
1+ 0
= 463 Ω
1+ 8.17
€
78
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1246
 R  R
  45.1kΩ 

10kΩ
F
L
Av = −

 = −

 = −18.6
 Ri + Rin  Rout + RL   2kΩ + 340Ω  336Ω + 10kΩ 
 R  R
  45.1kΩ 

2kΩ
F
L
Av = −

 = −

 = −3.73
 Ri + Rin  Rout + RL  10kΩ + 340Ω  336Ω + 2kΩ 
−−−
− Atr
48.5kΩ
1  1 
1  1 
=
= 32.3 | g m3 =

=

 = 2.50 mS
RF + Atr 50kΩ − 48.5kΩ
Rout  1+ T  12Ω 1+ 32.3 

1  1  
1  1 
Rin =  RF +

 =  50kΩ +

 = 1.51 kΩ
g m3  1+ T  
2.5mS  1+ 32.3 

T=
Page 1250
€
T is the same : T = 306.
D
Rout
= RF
|
AvIdeal = 1 | Av = AtrIdeal
1
1
= 10kΩ
= 307 Ω
g m5
3.16mS
Rout = 307Ω
|
 306 
T
= 1
 = 0.997
1+ T 1+ 306 
TSC = 0
|
TOC = T
1+ 0
= 1.00 Ω
1+ 306
Page 1251
€
T is the same : T = 306.
|
AvIdeal = 1 | Av = AtrIdeal

2  1
1
RinD = ro11+ g m1
= 600kΩ
= 500 Ω

g m2  g m3
2.00mS

 306 
T
= 1
 = 0.997
1+ T 1+ 306 
|
TOC = 306
Note : The impedance looking in the source of a transistor with a high resistance load of ro is
2/g m rather than 1/gm .
TSC = −
3.16mS (10kΩ)
g m2
g m5 RF
3.16mS
2ro2 ro4
=−
400kΩ 200kΩ
= −204
2
1+ g m5 RF
2
1+ 3.16mS (10kΩ)
(
Rout = 500Ω
)
(
)
1+ 204
= 334 Ω
1+ 306
€
79
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1254
T is the same : T = 152.
D
Rout
= ( R2 + RI ) R1
Rout = 304Ω
|
AvIdeal = −
 152 
R2
T
= −1 | Av = AtrIdeal
= −1
 = −0.993
RI
1+ T
 1+ 152 
1
1
= 40kΩ 10kΩ
= 304 Ω
g m5
3.16mS
|
TSC = 0
|
TOC = T = 152
1+ 0
= 1.99 Ω
1+ 152
Page 1255
€
There are approximately 15 cycles in 0.8 µsec : f ≅
15cycles
= 18.8 MHz
0.8us
Page 1256
€
fT =
1
gm
2π CGS + CGD
fT 3 =
1
2π
| g m = 2Kn I D
2(0.004)(0.0005)
5 pf + 1pF
| f T 2 = fT1 =
= 53.1 MHz |
fT 4 =
1
2π
1
2π
2(0.01)(0.0005)
5 pf + 1pF
2(0.01)(0.002)
5 pf + 1pF
= 83.9 MHz
= 168 MHz
Page 1265
€
( )(
)
Gm = g m2 = 2 10−3 5x10−5 = 0.316 mS | Ro = ro4 ro2 ≅
fT =
1
1
=
= 500 kΩ
2λI D 2(0.02)5x10−5
1  Gm  1  0.316mS 
 =

 = 2.51 MHz
2π  CC  2π  20pF 






 1 
 1
1 
1
1
1

 = 158 Hz

=

≅
fB =


2π  RoCC (1+ Av2 )  2π  RoCC (1+ µ f 2 )  2π 
2(0.001)

500kΩ(20pF)1+ 1
 0.02 5x10−4 




€
80
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1266
1 g m5
1
fz ≅
=
2π CC 2π
2(0.001)5x10−4
20x10
−12
= 7.96 MHz | R =
1
1
=
= 1.00 kΩ
g m5
2(0.001)5x10−4
−−−
(
)
Gm = g m2 = 40 5x10−5 = 2.00 mS | Ro = rπ 5 ≅
100V
(
)
40 5.5x10−4 A
= 4.54 kΩ
−4

1  Gm  1  2.00mS 
1  g m5  1  40 5x10
fT =
 =

=

 = 10.6 MHz | f Z =
2π  CC  2π  30pF 
2π  CC  2π  3x10−11

(
 1 
1 
1
1


≅
fB =
2π  rπ 5CC (1+ µ f 2 )  2π 4.54kΩ(30pF) 1+ 40(50)

[
€
]
) = 106 MHz



 = 584 Hz


Page 1269
100µA
V
V
SR =
= 5.00x106 = 5.00
20 pF
s
µs
−−−
100µA
V
V
SR =
= 5.00x106 = 5.00
20 pF
s
µs
Page 1273
€
(
)
Gm = g m2 = 40 2.5x10−4 = 10.0 mS | f T =
1  Gm / 2  1  10.0mS 

=

 = 15.7 MHz
2π  CC + Cµ3  4π  50.8pF 
 15.7 MHz 






−1 15.7 MHz
−1 15.7 MHz
−1 15.7 MHz
o
φ M = 90 − tan−1
 + tan 
 + tan 
 + tan 
 = 69.5
 142 MHz 
 173MHz 
 192 MHz 
 206 MHz 
€
Page 1277
I1
1mA
V
V
SR ≅
=
= 15.4x106 = 15.4
CC + CGD5 65 pF
s
µs
−−−





fT
fT
fT 
−1
−1
30 o = tan−1
 + tan 
 + tan 
 → f T = 16.6 MHz
 49.2 MHz 
 82.1MHz 
100 MHz 
 8.5MHz 
CC = 65 pF 
 − 2 pF = 31.3 pF
 16.6 MHz 
€
81
©R. C. Jaeger and T. N. Blalock
09/25/10
Page 1283
When the circuit is drawn symmetrically, capacitor 2CGD is replaced with 2 capacitors of value
4CGD in series. The circuit can then be cut vertically down the middle to form a differential mode
half-circuit. The total capacitance at the drain end of inductor L is C+CGS+4CGD.
Page 1285
fP =
1
1
= 5.016 MHz | f P =
= 5.008 MHz
 31.8 fF (7 pF ) 
31.8 fF (25 pF ) 


2π 31.8mH 
2π 31.8mH 
 7.0318 pF 
 25.0318 pF 
€
82
©R. C. Jaeger and T. N. Blalock
09/25/10
MICROELECTRONIC CIRCUIT DESIGN
Fourth Edition
Richard C. Jaeger and Travis N. Blalock
Answers to Selected Problems – Updated 2/27/11
Chapter 1
1.4
1.52 years, 5.06 years
1.5
1.95 years, 6.46 years
1.8
402 MW, 1.83 MA
1.10
2.50 mV, 5.12 V, 5.885 V
1.12
19.53 mV/bit, 100110002
1.14
14 bits, 20 bits
1.16
0.003 A, 0.003 cos (1000t) A
1.19
vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V
1.20
15.7 V, 2.32 V, 75.4 µA, 206 µA
1.22
150 µA, 150 µA, 12.3 V
1.24
39.8 Ω, 0.0251 vs
1.27
56 kΩ, 1.60 x 10-3 vs
1.29
1.50 MΩ, 7.50 x 108 ii
1.36
50/−12°, 10/−45°
1.38
-82.4 sin 750πt mV, 11.0 sin 750πt µA
1.40
1 + R2/R1
1.42
-1.875 V, -2.500 V
1.43
Band-pass amplifier
1.45
50.0 sin (2000πt) + 30.0 cos (8000 πt) V
1.48
0V
1.47
[4653 Ω, 4747 Ω], [4465 Ω, 4935 Ω], [4230 Ω, 5170 Ω]
1.55
6200 Ω, 4.96 Ω/oC
1.61
3.29, 0.995, −6.16; 3.295, 0.9952, −6.155
2
Chapter 2
!
!
2.3
500 mA
2.4
160 Ω, 319 Ω
2.6
For Ge : 2.63 x 10-4 / cm 3 , 2.27 x 1013 / cm 3 , 8.04 x 1015 / cm 3 ,
2.9
305.2 K
2.10
"1.75x10 6 cm s , + 6.25x10 5 cm s , 2.80x10 4 A cm 2 , 1.00x10"10 A cm 2
2.11
1.60 x 106 A/cm2, 1.60 x 10-10 A/cm2
2.13
4 ΜΑ/cm2
2.16
1.60 x 107 A/cm2, 4.00 A
2.17
316.6 K
2.22
Donor, acceptor
2.23
200 V/cm
2.25
1.25 x 104 atoms
2.27
p-type, 6 x 1018/cm3, 16.7/cm3, 5.28 x 109/cm3, 8.80 x 10-10/cm3
2.29
3 x 1017/cm3, 333/cm3
2.30
4 x 1016/cm3, 2.50 x 105/cm3
2.34
40/cm3, 2.5x1018/cm3, 170 cm2/s, 80 cm2/s, p-type, 31.2 mΩ-cm
2.36
1016/cm3, 104/cm3, 800 cm2/s, 220 cm2/s, n-type, 2.84 Ω-cm
2.38
1.16 x 1020/cm3
2.40
1.24 x 1019/cm3
2.41
Yes—add equal amounts of donor and acceptor impurities. Then n = ni = p, but the
mobilities are reduced. See Prob. 2.44.
2.43
6.27 x 1021/cm3, 6.22 x 1021/cm3
2.46
2.00/Ω-cm, 2.50 x 1019/cm3
2.48
75K: 6.64 mV, 150K: 12.9 mV, 300K: 25.8 mV, 400K: 34.5 mV
2.49
-28.0 kA/cm2
2.50
1.20x105 exp (-5000 x/cm) A/cm2, 12.0 mA
2.54
1.108 µm
2.57
8 atoms, 1.60x10-22 cm3, 5.00x1022 atoms/cm3, 3.73x10-23 g, 1.66x10-24 g/proton
3
Chapter 3
3.1
0.0373 µm, 0.0339 µm, 3.39 x 10-3 µm, 0.979 V, 5.24 x 105 V/cm
3.4
1018/cm3, 102/cm3, 1018/cm3, 102/cm3, 0.921 V, 0.0488 µm
3.6
6.80 V, 1.22 µm
3.10
640 kA/cm2
3.13
1.00 x 1021/cm4
3.17
290 K
3.20
312 K
3.21
1.39, 3.17 pA
3.22
0.791 V; 0.721 V; 0 A; 9.39 aA, -10.0 aA
3.25
1.34 V; 1.38 V
3.28
0.518 V; 0.633 V
3.29
335.80 K, 296.35 K
3.33
0.757 V; 0.717 V
3.35
−1.96 mV/K
3.39
0.633 V, 0.949 µm, 3.89 µm, 12.0 µm
3.40
374 V
3.42
4 V, 0 Ω
3.44
10.5 nF/cm2; 232 pF
3.46
800 fF, 20 fC; 20 pF, 0.5 pC
3.50
9.87 MHz; 15.5 MHz
3.51
0.495 V, 0.668 V
3.53
0.708 V, 0.718 V; 0.808 V
3.58
(a) Load line: (450 µA, 0.500 V); SPICE: (443 µA, 0.575 V)
(b) Load line: (-667 µA, -4 V);
(c) Load line: (0 µA, -3 V);
3.61
(0.600 mA, -4 V) , (0.950 mA, 0.5 V) , (-2.00 mA, -4 V)
3.68
Load line: (50 µA, 0.5 V); Mathematical model: (49.9 µA, 0.501 V); Ideal diode
model: (100 µA, 0 V); CVD model: (40.0µA, 0.6 V)
3.72
(a) 0.625 mA, -5 V; 0.625 mA, +3 V; 0 A, 7 V; 0 A, -5 V
4
3.74
(c) (270 µA, 0 V), (409 µA, 0 V); (c) (0 A -3.92 V), (230 µA, 0 V)
3.76
(b) (0.990 mA, 0 V) (0 mA, -1.73 V) (1.09 mA, 0)
(c) (0 A, -0.667 V) (0 A, -1.33 V) (1.21 mA, 0 V)
3.79
(1.50 mA, 0 V) (0 A, -5.00 V) (1.00 mA, 0)
3.82
(IZ, VZ) = (887 µA, 4.00 V)
3.84
12.6 mW
3.86
1.25 W, 3.50 W
3.91
17.6 V
3.95
-7.91 V, 1.05 F, 17.8 V, 3530 A, 840 A (ΔT = 0.628 ms)
3.97
(b) -7.91V, 904 µF, 17.8 V, 3540 A, 839 A
3.99
6.06 F, 8.6 V, 3.04 V, 962 A, 9280 A
3.104 -24.5 V, 1.63 F, 50.1 V, 15600 A, 2200 A
3.107 3.03 F, 8.6 V, 3.04 V, 962 A, 3770 A
3.112 2380 µF, 2800 V, 1980 V, 126 A, 2510 A
3.119 5 mA, 4.4 mA, 3.6 mA, 5.59 ns
3.123 (0.969 A, 0.777 V); 0.753 W; 1 A, 0.864 V
3.125 1.11 µm, 0.875 µm; far infrared, near infrared
5
Chapter 4
4.3
10.5 x 10-9 F/cm2
4.4
43.2 µA/V2, 86.4 µA/V2, 173 µA/V2, 346 µA/V2
4.8
(a) 4.00 mA/V2 (b) 4.00 mA/V2, 8.00 mA/V2
4.11
+840 µA; −880 µA
4.15
23.0 Ω; 50.0 Ω
4.18
125 µA/V2; 1.5 V; enhancement mode; 1.25/1
4.20
0 A, 0 A, 1.88 mA, 7.50 mA, 3.75 mA/V2
4.22
1.56 mA, saturation region; 460 µA, triode region; 0 A, cutoff
4.23
saturation; cutoff; saturation; triode; triode; triode
4.27
6.50 mS, 13.0 mS
4.30
2.48 mA; 2.25 mA
4.34
9.03 mA, 18.1 mA, 10.8 mA
4.37
1.13 mA; 1.29 mA
4.39
Triode region
4.40
99.5 µA; 199 µA; 99.5 µA; 99.5 µA
4.44
202 µA; 184 µA
4.46
5.17 V
4.51
40.0 µA; 72.0 µA; 4.41 µA; 32.8 µA
4.53
5810/1; 2330/1
4.56
235 Ω; 235 Ω
4.57
0.629 A/V2
4.58
400 µA
4.61
The transistor must be a depletion-mode device and the symbol is not correct.
4.65
(a) 6.09 x 10-8 F/cm2; 1.73 fF
4.67
17.3 pF/cm
4.69
20.7 nF
4.71
(a) 1.35 fF, 0.20 fF, 0.20 fF
4.74
50U, 0.5U, 2.5U, 1V, 0
4.77
10U, 0.5U, 25U, 1V, 0
6
4.79
432 µA/V2, 1.94 mA; 864 µA/V2, 0.972 mA
4.82
6.37 GHz, 2.55 Ghz; 637 GHz, 255 GHz
4.85
22λ x 12λ; 15.2%
4.88
12λ x 12λ; 15.2%
4.89
(350 µA, 1.7 V); triode region
4.92
(390 µA, 4.1 V); saturation region
4.95
(572 µA, 7.94 V); (688 µA, 7.52 V)
4.97
(50.3 µA, 8.43 V) ; (54.1 µA, 8.16 V)
4.102 (a) (116 µA, 4.15 V)
4.105 510 kΩ, 470 kΩ, 12 kΩ, 12 kΩ, 5/1
4.107 (124 µA, 2.36 V)
4.109 620 kΩ, 910 kΩ, 2.4 kΩ, 2.7 kΩ
4.111 (109 µA, 1.08 V); (33.5 µA, 0.933 V)
4.113 (42.6 µA, 0.957 V); (42.6 µA, 0.935 V)
4.115 8.8043x10-5 A; 8.323310-5 A
4.118 (73.1 µA, 9.37 V)
4.119 (69.7 µA, 9.49 V); (73.1 µA, 8.49 V)
4.122 (8.17 µA, 7.06 V), (6.74 µA, 7.57 V); (8.36 µA, 6.99 V), (6.89 µA, 7.52 V)
4.124 (91.5 µA, 8.70 V), (70.7 µA, 9.23 V); (97.8 µA, 8.48 V), (81.9 µA, 9.05 V)
4.125 2.25 mA; 16.0 mA; 1.61 mA
4.127 (322 µA, 3.18 V),
4.129 18.1 mA; 45.2 mA; 13.0 mA
4.131 1/3.84
4.132 (153 µA, -3.53 V) ; (195 µA, -0.347 V)
4.134 4.04 V, 10.8 mA; 43.2 mA; 24.5 mA; 98.0 mA
4.135 14.4 mA; 27.1 mA; 10.4 mA
4.137 (1.13 mA, 1.75 V)
4.138 (63.5 µA, -5.48 V) , R ≤ 130 kΩ
4.140 (55.3 µA, -7.09 V) , R ≤ 164 kΩ
4.143 22.3 kΩ  (127 µA, -5.50 V)
7
4.146 35.2 µA , R ≤ 318 kΩ
4.148 One possible design: 220 kΩ, 200 kΩ, 5.1 kΩ, 4.7 kΩ
4.149 (281 µA, -12.2 V)
4.151 (32.1 µA, -1.41 V)
4.153 (36.1 µA, 80.6 mV); (32.4 µA, -1.32 V); (28.8 µA, -2.49 V)
4.155 (431 µA, 6.47 V)
4.156 2.5 kΩ, 10 kΩ
4.157 ID = 1.38 mA, IG = 0.62 mA, VS = -0.7 V
4.159 (76.4 µA, 7.69 V), (76.4 µA, 6.55 V), 5.18 V
4.160 (a) (69.5 µA, 3.52 V)
4.162 (69.5 µA, 5.05 V); (456 µA, 6.20 V),
4.167 10.0 V; 10.0 mA, 501 mA; 13.8 V
4.169 15.0 V; 15.0 mA, 1.00 A; 12.2 V
8
Chapter 5
5.4
0.167, 0.667, 3.00, 0.909, 49.0, 0.995, 0.999, 5000
5.5
0.2 fA; 0.101 fA, −0.115 V
5.6
0.374 µA, -149.6 µA, +150 µA, 0.626 V
5.9
0.404 fA
5.11
1.45 mA; −1.45 mA
5.14
-25 µA, -100 µA, +75 µA, 65.7, 1/3, 0, 0.599 V
5.17
1.77 µA, -33.2 µA, +35 µA, 0.623 V
5.20
(a) 723 µA
5.24
0.990, 0.333, 2.02 fA, 6.00 fA
5.26
83.3, 87.5, 100
5.33
39.6 mV/dec, 49.5 mV/dec, 59.4 mV/dec, 69.3 mV/dec
5.34
5 V, 60 V, 5 V
5.35
2.31 mA; 388 µA; 0
5.36
60.7 V
5.40
Cutoff
5.42
saturation, forward-active region, reverse-active region, cutoff
5.46
25.0 aA, 1.33 aA, 26.3 aA
5.47
IC = 81.4 pA, IE = 81.4 pA, IB = 4.28 pA, forward-active region; although IC, IE, IB are
all very small, the Transport model still yields IC ≅ βFIB
5.48
79.0, 6.83 fA
5.49
83.3, 1.73 fA
5.50
55.3 µA, 0.683 µA, 54.6 µA
5.51
6.67 MHz; 500 MHz
5.53
1.5, 31.1 aA
5.55
-19.9 µA, 26.5 µA, -46.4 µA
5.58
17.3 mV, 0.251 mV
5.60
1.81 A, 10.1 A
5.62
0.768 V, 0.680 V, 27.5 mV
5.65
24.2 µA
5.66
4.0 fF; 0.4 pF; 40 pF
9
5.68
750 MHz, 3.75 MHz
5.71
0.149 µm
5.72
71.7, 43.1 V
5.74
74.1, 40.0 V
5.75
100 µA, 4.52 µA, 95.5 µA, 0.651 V, 0.724
5.77
26.3 µA
5.78
(c) 33.1 mS
5.79
0.388 pF at 1 mA
5.81
(b) 38% reduction
5.83
(80.9 µA, 3.80 V) ; (405 µA, 3.80 V); (16.2 µA, 3.80 V) ; (81.9 µA, 3.72 V);
5.88
(38.8 µA, 5.24 V)
5.90
36 kΩ, 75 kΩ, 3.9 kΩ, 3 kΩ; (0.975 mA, 5.24 V)
5.93
12 kΩ, 20 kΩ, 2.4 kΩ, 1.2 kΩ; (0.870 mA, 1.85 V)
5.95
(7.5 mA, 4.3 V)
5.97
(5.0 mA, 1.3 V)
5.99
30 kΩ, 620 kΩ; 24.2 µA, 0.770 V
5.101 5.28 V
5.103 3.21 Ω
5.104 10 V, 100 mA, 98.5 mA, 10.7 V
5.105
10 V, 109 mA, 109 mA, 14.3 V
10
Chapter 6
6.1
10 µW/gate, 8 µA/gate
6.3
2.5 V, 0 V, 0 W, 62.5 µW; 3.3 V, 0 V, 0 V, 109 µW
6.5
VOL = 0 V, VOH = 2.5 V, VREF = 0.8 V; Z = A
6.7
3 V, 0 V, 2 V, 1 V, −3
6.9
2 V, 0 V, 2 V, 5 V, 3 V, 2 V
6.11
3.3 V, 0 V, 3.0 V, 0.25 V, 1.8 V, 1.5 V, 1.2 V, 1.25 V
6.13
−0.80 V, −1.35 V
6.15
1 ns
6.17
0.152 µW/gate, 22.7 aJ
6.19
2.5 µW/gate, 1.39 µA/gate, 2.5 fJ
6.20
2.20 RC; 2.20 RC
6.22
−0.78 V, −1.36 V; 9.5 ns, 9.5 ns; 4 ns, 4 ns; 4 ns
6.25
Z=01010101
6.27
Z=00010011
6.30
2;1
6.32
84.5 A
6.33
0.583 pF
6.37
3 µW/gate, 1.67 µA/gate
6.38
72 kΩ, 1/1.04
6.39
(b) 2.5 V, 5.48 mV, 15.6 µW
6.43
(a) 0.450 V, 1.57 V
6.46
(a) 0.521 V, 1.81 V
6.49
NML: 0.242 V, 0.134 V, 0.351 V; NMH: 0.941 V, 0.508 V, 1.25 V
6.51
34.1 kΩ; 1.82/1; 1.49 V, 0.266 V
6.53
81.8 kΩ, 1/1.15
6.54
250 Ω; 625 Ω; a resistive channel exists connecting the source and drain; 20/1
6.55
1.44 V
6.57
2.17 V
6.58
1.55 V, 0.20 V, 0.140 mW, 0.260 mW
11
6.62
2.5 V, 0.206 V, 0.434 mW
6.65
1.40/1, 6.67/1
6.67
(b) 2.43/1, 1/3.97
6.69
0.106 V
6.71
1.55 V, 0.20 V, 0.150 mW
6.74
-2.40 at vO = 0.883 V
6.75
-2.44 at vO = 1.08 V
6.77
3.79 V
6.79
3.3 V, 0.296 V, 1.25 mW
6.82
1.75/1, 1/8.79
6.84
1.014
6.85
1.46/1, 1.72/1
6.86
2.5 V, 0.2 V, 0.16 mW
6.89
-5.98 at vO = 1.24 V
6.90
1.80/1, 0.610 V, 0.475 V
6.91
(a) 0.165 V, 80 µA (b) 0.860 V, 0.445 V
6.93
(a) 0.224 V, 88.8 µA (b) 0.700 V, 0.449 V
6.95
1.65/1, 1/2.32, 0.300 V, 0.426 V
6.97
0.103 V, 84.5 µA
6.98
0.196 V
6.102 2.22/1, 1.81/1
6.103 2.22/1, 1.11/1, 0.0643 V
6.104 6.66/1, 1.11/1, 0.203 V, 6.43/1, 6.74/1, 7.09/1
6.107 Y = ( A + B)(C + D) E , 6.66/1, 1.11/1
6.111 Y = ACE + ACDF + BF + BDE , 3.33/1, 26.6/1, 17.8/1
!
!
6.115 1/1.80, 3.33/1
6.117 Y = (C + E ) [A(B + D ) + G ] + F ; 3.62/1, 13.3/1, 4.44/1, 6.67/1
6.120 3.45/1, 6.43/1, 7.09/1, 6.74/1
6.122 1.11/1, 7.09/1, 6.43/1, 6.74/1
6.124 64.9 mV
12
6.126 (a) 5.43/1, 9.99/1, 6.66./1, 20.0/1
6.128 (a) 7.24/1, 26.6/1, 13.3/1
6.132 I D* = 2I D
| PD* = 2PD
6.133 80 mW, 139 mW
!
6.134 1 ns
6.137 60.2 ns, a potentially stable state exists with no oscillation
6.139 31.7 ns, 4.39 ns, 5.86 ns
6.141 5.50 kΩ, 11.6/1
6.144 68,4 ns, 3.55 ns, 9.18 ns
6.146 47.1 ns, 6.14 ns, 5.39 ns
6.148 2.11/1, 16.7/1, 12.8 ns, 0.923 ns
6.149 2.68/1, 3.29/1, 884 µW
6.150 (a) 1/1.68 (d) 1/5.89 (f) 1/1.60
6.152 −1.90 V, −0.156 V
6.153 1/3.30, 1.75/1
6.154 2.30 V, 1.07 V
6.156 Y = A + B
!
13
Chapter 7
7.1
173 µA/V2 ; 69.1 µA/V2
7.3
250 pA; 450 pA; 450 pA
7.6
2.5 V, 0 V
7.8
cutoff, triode, triode, cutoff, saturation, saturation
7.11
1.25 V; 42.3 µA; 1.104 V; 25.4 µA
7.12
0.90 V; 16.0 µA; 0.810 V; 96.2 µA
7.14
1.104 V
7.15
(b) 2.5 V, 92.8 mV
7.17
1.16 V, 0.728 V
7.18
0.810 V
7.22
0.9836 V, 2.77 mA
7.23
6.10/1, 1/5.37
7.24
(a) 1.89 ns, 1.89 ns, 0.630 ns
7.27
9.47 ns, 3.97 ns, 2.21 ns
7.31
2.11/1, 5.26/1
7.33
63.2/1, 158/1
7.35
6.00/1, 15.0/1
7.38
2.76/1
7.41
1.7 ns, 2.3 ns, 1.1 ns, 0.9 ns, C = 138 fF
7.43
0.200 µW/gate; 55.6 A
7.44
1.0 µW/gate; 18.4 fF; 32.0 fF; 61.7 fF
7.46
5.00 W; 8.71 W
7.49
90.3 µA; 25.0 µA
7.52
436 fJ; 425 MHz; 926 µW
7.55
αΔT, α2P, α3PDP
7.58
SPICE: 22.3 ns, 24.2 ns, 16.3 ns, 18.3 ns; Propagation delay formulas: 7.6 ns, 6.9 ns
7.59
2/1, 20/1; 6/1, 60/1
7.61
1/3.75
7.63
1.25/1
14
7.70
3.95 ns, 3.95 ns, 11.8 ns
7.71
4.67/1; 7.5/1
7.72
5 transistors; The CMOS design requires 47% less area.
7.74
Y = ( A + B)(C + D) E = ACE + ADE + BDE + BCE ; 12/1, 20/1, 10/1; 2.4/1; 30/1
7.76
Y = A + B C + D E + F = AB + CD + EF ; 4/1, 15/1; 6/1; 10/1
! 7.78
!
(
)(
)(
)
2/1, 4/1, 6/1, 20/1
7.81
(a) Path through NMOS A-D-E (d) Paths through PMOS A-C and B-E
7.83
24/1, 20/1, 40/1
7.84
6/1, 4/1, 10/1
7.91
5.37 ns, 1.26 ns
7.93
1.26 ns, 0.737 ns, 4.74 ns, 4.16 ns
7.95
4.74 ns, 2.37 ns
7.103 VDD !
2
1
2VIH
2VIH
V ! VDD ; R "
=
, C1 ≥ 83.1C2
3 DD
2
VDD # VIH
N MH
7.109 8; 2.90; 23.2 Ao
7.112 Ao
" N #1
" #1
7.113 263 Ω; 658 Ω
!
7.116 240/1, 96.2/1
7.117 1.41 V, 2.50 V
7.119 Latchup does not occur.
7.122 +0.25; +0.31
7.124 211,000/1; 0.0106 cm2
15
Chapter 8
8.1
268,435,456 bits, 1,073,741,824 bits; 2048 blocks
8.2
3.73 pA/cell , 233 fA/cell
8.5
3 V, 0.667 µV
8.9
1.55 V, 0 V, 3.59 V
8.11
“1” level is discharged by junction leakage current
8.13
1.47 V, 1.43 V
8.14
−19.8 mV; 2.48 V
8.16
0 V, 1.90; Junction leakage will destroy the “1” level
8.18
5.00 V, 1.60 V; −1.84 V
8.22
135 µA, 346 mW
8.24
0.266 V
8.25
0.945 V (The sense amplifier provides a gain of 10.5.)
8.31
0 V, 1.43 V, 3.00 V
8.32
0.8 V, 1.2 V; 0.95 V, 0.95 V
8.34
53,296
8.37
VDD !
8.41
W1 = 010001102 , W3 = 001010112
8.45
1.16/1
2
1
2VIH
2VIH
V ! VDD ; R "
=
; C1 ≥ 2.88C2
3 DD
2
VDD # VIH
N MH
16
Chapter 9
!
9.1
0 V, -0.700 V
9.2
(a) -1.38 V, -1.12
9.3
−1.50 V, 0 V
9.6
0 V, −0.40 V; 3.39 kΩ; Saturation, cutoff; Cutoff, saturation
9.8
−0.70 V, −1.30 V, −1.00 V, 0.60 V
9.11
−0.70 V, −1.50 V, −1.10 V, 2.67 kΩ; 0.289 V; −0.10 V, +0.30 V
9.13
-1.70 V, -2.30 V, 0.60 V, Yes
9.15
11
9.16
11.1 kΩ, 12.0 kΩ, 70.2 kΩ, 252 kΩ
9.18
-1.10 V, -1.50 V, -1.30 V, 0.400 V, 0.107 V, 1.10 mW
9.19
0.383 V
9.21
-0.70 V, -1.50 V, -1.10 V, 11.3 kΩ, 2.67 kΩ, 2.38 kΩ; 0.289 V
9.23
0.413 V
9.25
50.0 µA, -2.30 V
9.26
Standard values: 11 kΩ, 150 kΩ, 136 kΩ
9.30
+0.300 V, −0.535 V, 334 Ω
9.33
5.15 mA
9.36
0.135 mA
9.38
10.7 mA
9.40
400 Ω, 75.0 mA
9.42
(c) 0 V, -0.7 V, 3.93 mA (d) -3.7 V, 0.982 mA (e) 2920 Ω
9.45
Y = A+ B
9.47
−0.850 V; 3.59 pJ
9.49
359 ns
9.50
0 V, −0.600 V, 5.67 mW, 505 Ω, 600 Ω; Y = A + B + C, 5 vs. 6
9.53
5.00 kΩ, 5.40 kΩ, 31.6 kΩ, 113 kΩ
9.54
1 kΩ, 1 kΩ, 1.30 mW
9.56
2.23 kΩ, 4.84 kΩ, 60.1 kΩ
9.59
1.446 mA, 1.476 mA, 29.66 µA; 1.446 mA, 1.476 mA, 29.52 µA
17
!
9.61
-0.9 V, -1.1 V, -1.8 V, -2.0 V, -2.7 V, -2.9 V, -4.2 V
9.64
Y = AB + AC
9.68
0, -0.8. 0, -0.8, 3.8 V
9.69
2.98 pA, 70.5 fA
9.71
160; 0.976; 0.976; 0.773 V
9.72
0.691 V, 0.710 V
9.76
63.3 µA, 265 µA
9.78
40.2 mV, 0.617 mV
9.80
20.6 mW, 4.22 mW
9.82
68.2 mV
9.83
2.5 V, 0.15 V, 0.66 V, 0.80 V
9.85
44.8 kΩ, 22.4 kΩ
9.88
5 V, 0.15 V, 0, −1.06 mA; 31; −1.06 mA vs. −1.01 mA, 0 mA vs. 0.2 mA
9.95
8
9.97
RB ≥ 5 kΩ
9.99
7
9.100 234 mA, 34.9 mA
9.104 (IB, IC): (a) (135 µA, −169µA); (515µA, 0); (169 µA, 506 µA); (0, 0) (b) all 0 except
IB1 = IE1 = 203 µA
9.107 180
9.108 22
9.111 1.85 V, 0.15 V; 62.5 µA, −650 µA; 13
9.113 Y = ABC ; 1.9 V; 0.15 V; 0, −408 µA
9.115 1.5 V, 0.25 V; 0, −1.00 mA; 16
9.116 0.7 V, 191 µA, 59 µA, 1.18 mA
9.117 -1.13 mA, 0, 4.50 mA, 0, 0, 1.80 mA; 0, 0, 0, 0, 1.23 mA, 0
9.119 Y = A + B + C; 0 V, −0.8 V; −0.40 V
9.121 1.05 mA, 26.9 µA
9.122 2 fJ; 10 fJ
9.124 1.67 ns; 0.5 mW
9.126 2.8 ns; 140 mW
18
Chapter 10
10.2
(a) 41.6 dB, 35.6 dB, 94.0 dB, 100 dB, -0.915 dB
10.4
29.35
10.5
Using MATLAB:
t = linspace(0,.004);
vs = sin(1000*pi*t)+0.333*sin(3000*pi*t)+0.200*sin(5000*pi*t);
vo= 2*sin(1000*pi*t+pi/6)+sin(3000*pi*t+pi/6)+sin(5000*pi*t+pi/6); plot(t,vs,t,vo)par
500 Hz: 1 0°, 1500 Hz: 0.333 0°, 2500 Hz: 0.200 0°; 2 30°, 1 30°, 1 30° 2 30°, 3 30°, 5
30° yes
10.7
29.0 dB, 105 dB, 67.0 dB
10.9
19.0 dB, 87.0 dB, 53.0 dB; Vo = 8.94 V, recommend ±10-V or ±12-V supplies
10.11 3.61 x10-8 S, -7.93 x10-3, 1.00, 79.3 Ω
10.13 0.333 mS, -0.333, -1600, 1.78 MΩ
10.15 1.00 mS, −1.00, 3001, 30.0 kΩ
10.16 53.7 dB, 150 dB, 102 dB; 11.7 mV; 31.3 mW
10.17 45.3 mV, 1.00 W
10.21 −5440
10.23 0, ∞, 80 mW, ∞
10.24 182
10.29 −10 (20 dB), 0.1 V; 0, 0 V
10.31 vO = [8 – 4 sin (1000t)] volts; there are only two components; dc: 8 V, 159 Hz: −4 V
10.33 24.1 dB, 2nd and 3rd, 22.4%
10.35 2.4588
0.0012
0.0038 5.3105
0.1863 0.0023
0.0066
1.3341
10.37 59.7 dB, 119 dB, 88.9 dB; 5.66 mV
10.41 Rid ≥ 4.95 MΩ
10.43 50 µV, 140 dB
10.44 (a) −46.8, 4.7 kΩ, 0, 33.4 dB
10.47 (d) (-1.10 + 0.75 sin 2500πt) V
10.49 (a) vO = (4.00 " 20Vi sin 2000#t ) V
(b) 0.3 V
10.53 30.1 kΩ, 604 kΩ Av = -20.1, Rin = 30.1 kΩ
!
10.56 -70.0, 10 kΩ, 0
19
0.0026
0.4427
0.0028
0.0883
10.59 2 MΩ
10.60 83.9, ∞, 0, 38.5 dB
10.63 (d) (5.28 – 2.88 sin 3250πt) V
10.67 2 kΩ, 86.6 kΩ, Av = 44.3
10.69 (-0.47 sin 3770t −0.94 sin 10000t) V, 0 V
10.70 −0.3750 sin 4000πt V; −0.6875 sin 4000πt V; 0 to −0.9375 V in - 62.5-mV steps
10.71 455/1, 50/1
10.72 −10, 110 kΩ, 10 kΩ, (-30 + 15cos 8300πt) V, (-30 + 30cos 8300πt) V
10.73 3.2 V, 3.1 V, 2.82 V, 2.82 V, -1.00 V; 3.82 µA; 3.80 µA, 2.80 µA
10.76 60 dB, 10 kHz, 10 Hz, 9.99 kHz, band-pass amplifier
10.77 80 dB, ∞, 100 Hz, ∞, high-pass amplifier
10.81 60 dB, 100 kHz, 28.3 Hz, 100 kHz
10.83 Using MATLAB: n=[1e4 0]; d=[1 200*pi]; bode(n,d)
10.86 Using MATLAB: n=[-20 0 -2e13]; d=[1 1e4 1e12];
bode(n,d)
10.89 0.030 sin (2πt + 89.4°) V, 1.34 sin (100πt + 63.4°) V, 3.00 sin (104πt + 1.15°) V
10.92 0.956 sin (3.18x105πt + 101°) V, 5.00 sin (105πt + 180°) V, 5.00 sin (4x105πt − 179°) V
10.94 Av ( s) =
2x108 "
s + 10 7 "
|
Av ( s) = -
2x108 "
s + 10 7 "
10.96 66 dB, 12.8 kHz, -60 dB/decade
10.97 3.16 sin (1000πt + 10°) + 1.05 sin (3000πt + 30°)+ 0.632 sin (5000πt + 50°) V
! Using MATLAB:
t = linspace(0,.004);
A=10^(10/20);
vs = sin(1000*pi*t)+0.333*sin(3000*pi*t)+0.200*sin(5000*pi*t);
vo = A*sin(1000*pi*t+pi/18)+3.33*sin(3000*pi*t+3*pi/18)+2.00*sin(5000*pi*t+5*pi/18);
plot(t, A*vs, t, vo)
10.98 -4.44 dB, 26.5 kHz
10.100
10 kΩ, 0.015 µF
10.103
80 dB, 100 Hz
10.105
-1.05 dB, 181 Hz
10.107
(b) -20.7, 105 kHz
10.108
10.5 kΩ, 105 kΩ, 0.015 µF
20
10.113
T(s) = -sRC
10.116
−6.00, 20.0 kΩ, 0; +9.00, 91.0 kΩ, 0; 0, 160 kΩ, 0
10.117
1 A, 2.83 V, > 10 W (choose 15 W)
10.118
0.484 A; 0.730 V; 0.730 V; ≥ 7.03 W (choose 10 W), 7.27 W
21
Chapter 11
11.1
(c) 4, 5.00, 4.00, 20 %
11.3
120 dB
11.4
1/(1+Aβ); 9.99! 10"3percent
11.5
(a) 13.49, 9.11x10-3, 0.0675%
11.7
120 dB
11.9
(a) -9.997, 2.76x10-3, 0.0276%
11.13 103 dB
11.16 100 µA, 100 µA, -48.0 pA, +48.0 pA
11.17 (a) 13.5, 296 MΩ, 135 mΩ
11.19 (a) -17.4, 2.70 kΩ, 36.8 mΩ
11.22 If the gain specification is met with a non-inverting amplifier, the input and output
specifications cannot be met.
11.24 145Vs, 3.56 Ω
11.25 ≤ 0.75 %
11.27 (b) shunt-series feedback (d) series-shunt feedback
11.29 (a) Series-shunt (a) and series-series (c) feedback
11.32 122 dB, 31.5 S
11.33 9.96, 6.55 MΩ, 3.19 Ω
11.35 9130, 3.00, 3.00, 369 MΩ, 0.398 Ω
11.37 (c) -T/(1+T)
11.39 -9.998 kΩ, 1.200 Ω, 0.1500 Ω
11.41 -35.99 kΩ, 4.418 Ω, 0.2799 Ω
11.44 -99.91 µS, 50.04 MΩ, 17.79 MΩ
11.49 10000s ( s + 5000) , 10000 (2s + 1)
11.54 1.100, 1.899 Ω, 24.65 MΩ
!
11.56 10.96, 35.12 Ω, 3.461 MΩ
11.57 680.4, 0.334
11.59 330, 0.0260
11.61 6.25 %, 16.7 %
22
11.62 0.00372 %, 0.0183 %
11.63 0 V, -26 mV, 90.9 kΩ
11.65 +7500, -0.667 mV
11.68 The nearest 5% values are 1 MΩ and 10 kΩ
11.70 -6.2 V, 0 V; 10 V, -1.19 V
11.72 -5.00 V, 0 V; -10.0 V, 0.182 V
11.74 10 V, 0 V; 15 V, 0.125 V
11.77 110 Ω and 22 kΩ represent the smallest acceptable resistor pair.
11.79 39.2 Ω
11.81 0 V, 3 V; 0.105 V; 0 V; 49.0 dB
11.83 (d) [-0.313 sin 120πt - 4.91 sin 5000πt] V
11.85 (b) 124 dB
11.86 60 dB
11.87 20.0 kΩ, 56.0 kΩ
11.89 20 Hz
11.91 50 Hz; 5 MHz; 2.5 MHz
11.93 200; 199
11.95 80 dB, 1 kHz, 1 MHz; 101 MHz, 9.90 Hz; 251 MHz, 3.98 Hz
11.97 100 dB, 1 kHz, 1 MHz; 8.4 Hz, 119 MHz; 5.3 Hz, 188 MHz
[
]
11.99 (a) Ro ( s + " B ) s + " B (1+ Ao# )
[
] (s + " )
11.102
(a) Rid s + " B (1+ Ao# )
!
11.105
Av ( s) = "
!
11.107
3.285x1012
; (2 poles: 2.08 kHz and 2.04 MHz)
s2 + 1.284x10 7 s + 1.675x1011
Av ( s) = "
6.283x1010
; (2 poles: 3.18 mHz and 5.00 MHz)
s2 + 3.142x10 7 s + 6.283x105
!
B
11.109
!
11.111
6.91, 7.53, 6.35; 145 kHz, 157 kHz, 133 kHz
11.113
10 V/µs
11.117
1010 Ω, 7.96 pF, 4x106, Ro not specified
2.51 V/µs; 2.51 V/ s
23
11.119
90.6 o; 90.2o
11.120
8.1 o; 5.1o
11.122
110 kHz; A ≤ 2048; larger
11.123
Yes, but almost no phase margin; 0.4°
11.125
75 o versus 90o; 65 o versus 90o
11.128
5 MHz, 90.0o; 2.5 MHz, 90.0o
11.130
Av ( s) = "
11.132
Yes, but almost no phase margin; 1.83°
11.134
!
11.136
90.0o
11.141
Yes, 24.4o, 50 %
11.143
1.8o
11.144
38.4o, 31 %
11.147
133 pF
11.149
90.4o
11.152
(a) 72.2o
11.153
(a) 11.9 MHz, 5.73o, 85.4%
11.155
(a) 70.2, 23.4%
11.157
(a) 18.8 MHz
5.712x106 s
; 143o
2
5
10
s + 5.741x10 s + 1.763x10
12o; Yes, 50o
24
Chapter 12
12.1
A and B taken together, B and C taken together
12.3
-8000, 2 kΩ, 0
12.5
48.0, 968 MΩ, 46.5 mΩ
12.7
78.1 dB, 2.00 kΩ, 0.105 Ω
12.8
(c) 2.00 mV, -40.0 mV, 4.00 µV, 0.800 V, 80.0 µV, 0 V, -12.0 V, 0.190 V, 0V (ground
node)
12.11 -1080, 3.9 kΩ, 0
12.12 8.62 kΩ, 8.62 kΩ
12.16 2744, 2434, 3094, 1 MΩ, 1.02 MΩ, 980 kΩ, 0
12.17 -1320, 75 kΩ, 0; 5.00 mV, 5.00 mV, 55.0 mV, 0 V, -1.10 V, -1.10 V, -6.60 V, 0V (ground node)
12.19 50.0, 298 kHz; 48.0, 349 kHz; -42.0, 368 kHz
12.21 14.0, 286 kHz, 68.8 dB, 146 kHz
12.23 -2380, 613 MΩ, 98.0 mΩ, 29.6 kHz; 0 V, 10.0 mV, 49.0 mV, 389 µV, -3.89 V,
-3.06 V, -15.0 V, +15.0 V, -15.0 V, 0 V
12.25 3
12.27 20 kΩ, 62 kΩ, 394 kHz
12.30 103 dB, 98.5 dB, 65 kHz, 38 kHz
12.33 (a) In a simulation of 5000 cases, 33.5% of the amplifiers failed to meet one of the
specifications. (b) 1.5% tolerance.
12.36 -12, (-6.00 + 1.20 sin 4000πt) V
12.38 4.500 V, 4.99 V, 5.01 V, 5.500 V, 2.7473 V, 2.7473 V, 0.991 V, -75.4 µA, -375 µA, +175 µA,
0.002, -50.0, 88 dB
12.40 (b) 0.005 µF, 0.0025 µF, 1.13 kΩ
12.44
VO
K
= 2
VS s R1 R2C1C2 + s R1C1 (1" K ) + C2 ( R1 + R2 ) + 1
[
]
12.46 -1
!
12.48 270 pF, 270 pF, 23.2 kΩ
12.49 (a) 51.2 kHz, 7.07, 7.24 kHz
25
|
SKQ =
K
3" K
#
&2
% "6s (
12.52 (a) 1 rad/s, 4.65, 0.215 rad/s; ABP ( s) = %
(
% s2 + s + 1(
$
3 '
12.54 5.48 kHz, 4.09, 1.34 kHz
12.56 10 kΩ, 100 kΩ, 20 kΩ, 0.0133 µF
!
s
R2C2
12.58 T = + K
"
%
1
1
1
2
$
'+
s +s
+
$ R2C2
R1 R2 C1 '& R1 R2C1C2
#
(
)
12.64 -5.5 V, -5.5, 10 %; -5.0 V, -5.0, 0
!
12.66 12.6 kHz, 1.58, 7.97 kHz
12.69 (a) -1-125 V (b) -1.688 V
12.70 10.6 mV, 5 %
12.73 455/1, 50/1
12.74 0.31 LSB, 0.16 LSB
12.76 1.43%, 2.5%, 5%, 10%
12.77 11 resistors, 1024:1
12.79 (a) 1.0742 kΩ, 0.188 LSB, 0.094 LSB
12.81 (a) (2n+1-1)C
12.84 (b) 2.02 inches
12.85 3.415469 V ≤ VX ≤ 3.415781 V
12.86 1.90735 µV, 110100001010001111012, 011111111001110111012
12.89 00010111112, 95 µs
12.91 800 kHz, 125 ns
12.93
#
&
"t
vO (t ) = 2.5x105 %1" exp
( for t ) 0 | RC ) 0.0447 s
5x10 4 RC '
$
12.94 19.1 ns
!
12.97 1/RC, 2R
12.98 0.5774/RC, 1.83
12.100
60 kHz, 6.8 V
26
12.102
17.5 kHz, 11.5 V
12.106
0.759 V
12.107
2.4 Hz
12.112
VO = -V1V2/104IS
12.113
3.11 V, 2.83 V, 0.28 V
12.115
0.445 V, -0.445 V, 0.89 V
12.117
9.86 kHz
12.118
VO = 0 is a stable state, so the circuit does not oscillate. f = 0.
12.120
0, 0.298 V, 69.0 mV
12.122
13 kΩ, 30 kΩ, 51 kΩ, 150 pF
27
Chapter 13
13.1
(0.700 + 0.005 sin 2000πt) V, -1.03 sin 2000πt V, (5.00 −1.03 sin 2000πt) V. 2.82 mA
13.3
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier.
C2 is a coupling capacitor that couples the ac component of the signal at the collector
to the output vO. C3 is a bypass capacitor. (b) The signal voltage at the top of resistor
R4 will be zero.
13.5
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier.
C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of
the signal at the drain to output vO. (b) The signal voltage at the source of M1 will be
vs = 0.
13.7
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier.
C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of
the signal at the collector to output vO. (b) The signal voltage at the emitter terminal
will be ve = 0.
13.9
(a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier.
C2 is a coupling capacitor that couples the ac component of the signal at the drain to
output vO.
13.13 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier.
C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of
the signal at the drain to the output vO. (b) The signal voltage at the top of R4 will be
zero.
13.16 (1.46 mA, 5.36 V)
13.18 (19.2 µA, 6.37 V)
13.20 (62.2 µA, 43.65 V)
13.24 (91.3 µA, 6.05 V)
13.28 (245 µA, 3.73 V)
13.32 (423 µA, -6.76 V)
13.34 (1.25 mA, 8.63 V)
13.46 Thévenin equivalent source resistance, gate-bias voltage divider, gate-bias voltage
divider, source-bias resistor—sets source current, drain-bias resistor—sets drainsource voltage, load resistor
13.51 118 Ω, 3.13 TΩ, ≤ -28.5 mV
13.52 (c) 8.65 Ω
13.53 Errors: +10.7%, -9.37%; +23.0%, - 17.5%
28
13.54 (c) 1.25 µA
13.55 (213 µA, ≥ 0.7 V), 8.52 mS, 469 kΩ
13.60 (b) +16.7%, -13.6%
13.61 90, 120; 95, 75
13.66 [−76.7, −75.8]
13.68 -40.0
13.72 −90
13.74 Yes, using ICRC = (VCC +VEE)/2
13.76 3
13.77 20 mA; 24.7 V
13.78 0.500 V
13.79 No, there will be significant distortion
13.80 −263
13.85 32/1, 0.500 V
13.86 0.800 A
13.87 10%, 20%
13.90 (78 µA, 9 V)
13.91 Virtually any desired Q-point
13.92 400 = 133,000iP + vPK; (1.4 mA, 215 V); 1.6 mS, 55.6 kΩ, 89.0; -62.7
13.93 FET
13.94 BJT
13.95 37.5 µA, 2400
13.96 2000, 200, 8.00 mS, 0.800 mS
13.99 23.5 dB
13.101
(142 µA, 7.5 V)
13.102
0.300 V
13.103
1.0 V, 56 V
13.105
3
13.107
−11.0
13.110
−7.28
29
13.115
29.4 kΩ, 93.4 kΩ
13.118
833 kΩ, 1.46 MΩ
13.120
243 kΩ, 40.1 kΩ
13.122
6.8 MΩ, 45.8 kΩ, independent of Kn
13.124
1 MΩ, 3.53 kΩ
13.125
-281vi, 3.74 kΩ
13.127
-23.6vi, 508 kΩ
13.129
38.9 dB, 6.29 kΩ, 9.57 Ω
13.131
36.4 dB, 62.9 kΩ, 95.7 kΩ
13.135
138 µW, 182 µW, 1.28 mW, 0.780 mW, 0.820 mW, 3.19 mW
13.139
0.552 mW, 0.684 mW, 0.225 mW, 18.7 µW, 50.4 µW, 1.53 mW
13.142
VCC/15
13.143
3.38 V, 13.6 V
13.144
(VCC)2/8RL, (VCC)2/2RL, 25%
13.145
0.992 V
13.147
2.24 V
13.148
1.65 V
13.149
2.93 V
13.153
833 µA
13.154
-4.60, 1 MΩ, 6.82 kΩ
30
Chapter 14
14.1
(a) C-C or emitter-follower (c) C-E (e) not useful, signal is being injected into the
drain (h) C-B (k) C-G (o) C-D or source-follower
14.14 −32.6, 9.58 kΩ, 596 kΩ, -27.1; −17.2, 11.6 kΩ, 1060 kΩ, -17.1
14.15 −3.77, 2 MΩ, 26.5 kΩ, -3770; −8.03, 2 MΩ, 10.0 kΩ, -10000
14.16 (a) −6.91 (e) −240
14.17 3.3 kΩ, 33 kΩ
14.20 −178, - 58, 21.4 kΩ, 39 kΩ, 5.13 mV
14.21 121, - 62, 2.84 kΩ, 7.58 kΩ, 6.76 mV, -120
14.22 −12.3, −9.62, 368 kΩ, 82 kΩ, 141 mV, -7.50
14.24 -2.74, -762, 10 MΩ, 1.80 kΩ, 0.800 V
14.25 -3790, -5.44, 1.30 kΩ, 68.5 kΩ, 5.96 mV
14.27 0.747, 29.8 kΩ, 104 Ω, 29.7
14.28 0.896, 2 MΩ, 125 Ω, 16,000
14.29 0.987, 44.8 kΩ, 15.2 Ω, 1.54 V
14.30 0.960, 1 MΩ, 507 Ω, 6.19 V
14.31 0.992, 12.6 MΩ, 1.32 kΩ, 0.601 V
14.32 
, 7.94 MΩ, 247 Ω, ∞
14.33 vi " (0.005 + 0.2VR E ) V
14.35 0.9992, 30.1 V
!
14.37 190, 980 Ω, 2.52 MΩ, 0.990; 62.1, 980 Ω, 7.57 MΩ, 0.969
14.38 39.4, 1.20 kΩ, ∞, 0.600; 7.94, 1.43 kΩ, ∞, 0.714
14.41 40.7, 185 Ω, 39.0 kΩ, 18.5 mV
14.42 4.11, 1.32 kΩ, 20 kΩ, 354 mV
14.43 5.01, 3.02 kΩ, 24 kΩ, 352 mV
14.46 32.1 Ω, 260 Ω
14.47 633 Ω, 353 Ω
14.49
("
o
+ 1)ro = 244 M#
!
31
14.51 Low Rin, high gain: Either a common-base amplifier operating at a current of 71.4
µA or a common-emitter amplifier operating at a current of approximately 7.14 mA
can meet the specifications with VCC ≈ 14 V.
14.53 Large Rin, moderate gain: Common-source amplifier.
14.55 Common-drain amplifier.
14.56 Low Rin, high gain: Common-emitter amplifier with 5-Ω input "swamping" resistor.
14.57 Cannot be achieved with what we know at this stage in the text.
14.59 1.66 Ω
14.62
vi
5 mV
10 mV
15 mV
1 kHz
621 mV
1.23 V
1.81 V
2 kHz
3 kHz
26.4 mV (4.2%) 0.71 mV (0.11%)
0.104 V (8.5%) 5.5 mV (0.45%)
0.228 V (12.6%) 18.2 mV (1.0%)
THD
4.2%
8.5%
12.7%
14.64 (b) 479vi, 384 kΩ
14.65 vi, 297 Ω
14.68 g m , 0; 400 µS, 0
!
#
1
14.70 "g m %%1+
$ µf
14.71 "
!
gm
1+ g m RE
&
(( | " g o
'
| "
| µ f + 1 ; " 502 µS, " 2.00 µS
$ RE '
$
go
Gm
r '
= µ f &1+ # ) >> 1
&
) |
(1+ g m RE ) % RE + r# ( Gr % RE (
14.74 -0.984, 0.993, 0.766 V
!
14.76 SPICE: (116 µA, 7.53 V), −150, 19.6 kΩ, 37.0 kΩ
14.78 SPICE: (115 µA, 6.30 V), -20.5, 368 kΩ, 65.1 kΩ
14.80 SPICE: (66.7 µA, 4.47 V), −16.8, 1.10 MΩ, 81.0 kΩ
14.83 SPICE: (5.59 mA, -5.93 V), -3.27, 10.0 MΩ, 1.52 kΩ
14.85 SPICE: (6.20 mA, 12.0 V), 0.953, 2.00 MΩ, 388 Ω
14.86 SPICE: (175 µA, 4.29 V), -4.49, 500 kΩ, 17.0 kΩ
14.87 (430 µA, 1.97 V), (430 µA, 3.03 V), -2.89, 257 kΩ, 3.22 kΩ, (Note Atr = 743 kΩ)
14.88 (4.50 mΑ, 2.50 V), (4.50 mA, 2.50 V), -83.5, 6.63 kΩ, 10.3 kΩ,
14.89 0.485, 182 kΩ, 435 Ω
14.92 1.80 µF, 0.068 µF, 120 µF; 2.7 µF
32
14.94 0.20 µF, 270 F; 100 µF, 0.15 µF
14.96 1.5 µF, 0.027 µF
14.98 8200 pF, 820 pF
14.102
33.3 mA
14.103
R1 = 120 kΩ, R2 = 110 kΩ
14.106
45.1 ≤ Av ≤ 55.3 - Only slightly beyond the limits in the Monte Carlo results.
14.108
The second MOSFET
14.111
The supply voltage is not sufficient - transistor will be saturated.
14.113
4.08, 1.00 MΩ, 64.3 Ω
14.116
2.17, 1.00 MΩ, 64.3 Ω
14.121
468, 73.6 kΩ, 18.8 kΩ
14.122
0.670, 107 kΩ, 20.0 kΩ
14.124
7920, 10.0 kΩ, 18.8 kΩ
14.125
140, 94.7 Ω, 113 Ω
14.127
19.2 Hz; 18.0 Hz
14.129
1.56 Hz; 1.22 Hz
14.131
6.40 Hz; 5.72 Hz
14.133
0.497 Hz, 0.427 Hz
14.134
1.70 kHz; 1.68 kHz
33
Chapter 15
15.1
(26.2 µA, 7.08 V); −346, 191 kΩ, 660 kΩ; −0.604, 49.2 dB, 27.3 MΩ
15.2
(5.25 µA, 1.68 V); −21.0, -0.636, 24.4 dB, 572 kΩ, 4.72 MΩ, 200 kΩ, 50.0 kΩ
15.4
(85.6 µA, 10.1 V); −342, -0.494, 50.8 dB, 58.4 kΩ, 10.1 MΩ, 200 kΩ, 50.0 kΩ
15.7
REE = 1.1 MΩ, RC = 1.0 MΩ
15.8
(a) (198 µA, 3.39 V); differential output: −372, 0, ∞ (b) single-ended output: −186,
−0.0862, 66.7 dB; 25.2 kΩ, 27.3 MΩ, 94.0 kΩ, 13.5 kΩ
15.9
2.478 V, 6.258 V, -2.78 V, 4.64 V
15.11 VO = 5.99 V, vo = 0; VO = 5.99 V; vo = 1.80 V; 33.3 mV
15.15 (37.4 µA, 5.22 V); Differential output: −300, 0, ∞; single-ended output: −150,
−0.661, 47.2 dB; 200 kΩ, 22.7 MΩ
15.16 -5.850 V, -3.450 V, -2.40 V
15.18 (4.94 µA, 1.77 V); differential output: −77.2, 0, ∞; single-ended output: −38.6,
−0.0385, 60.0 dB; 810 kΩ, 405 MΩ, [-1.07 V, 1.60 V]
15.21 -283, -.00494, 95.2 dB
15.22 -273.6, -.004429, 94.9 dB
15.24 (107 µA, 10.1 V); differential output: −18.2, 0, ∞; single-ended output: −9.10,
−0.487, 25.4 dB; ∞, ∞
15.27 2.4 kΩ, 5.6 kΩ
15.31 (150 µA, 5.62 V); differential output: −26.0, 0, ∞; single-ended output: −13.0,
−0.232, 35.0 dB; ∞, ∞
15.34 (20.0 µA, 9.05 V); differential output: −38.0, 0, ∞; single-ended output: −19.0,
−0.120, 44.0 dB; ∞, ∞
15.35 312 µA, 27 kΩ
15.38 -20.26, -0.7812, 22.3 dB , ∞, ∞
15.40 −3.80 V, −2.64 V, 48.3 mV
15.44 -79.9, -0.494, 751 kΩ
15.45 (99.0 µA, 8.80 V), −30.4, −0.167, 550 kΩ
15.47 (49.5 µA, 3.29 V), (49.5 µA, 8.70 V); −149, -0.0625, 101 kΩ
15.48 (100 µA, 1.20 V), (100 µA, 3.18 V); −13.4, 0, ∞
15.51 (24.8 µA, 15.0 V), (625 µA, 15.0 V); 896, 202 kΩ; 20.0 kΩ; 153 MΩ; v2
15.52 [-13.6 V, 14.3 V]
34
15.57 (24.8 µA, 14.3 V), (4.95 µA, 14.3 V), (495 µA, 15.0 V); 5010, 202 kΩ; 19.4 kΩ;
150 MΩ; v2
15.58 (98.8 µA, 17.1 V), (360 µA, 17.1 V); 620, 40.5 kΩ; 48.8 kΩ; 37.5 MΩ; v2
15.59 [-16.6 V, 16.4 V]
15.63 (98.8 µA, 15.3 V), (300 µA, 15.3 V); 27700, 40.5 kΩ; 2.59 MΩ
15.67 (250 µA, 18.6 V), (500 µA, 18.0 V); 4490, ∞; 170 kΩ
15.69 5770
15.71 (250 µA, 7.42 V), (6.10 µA, 4.30 V), (494 µA, 5.00 V); 4230, ∞; 97.5 kΩ
15.75 (49.5 µA, 18.0 V), (360 µA, 17.3 V), (990 µA, 18.0 V); 12700, 101 kΩ; 1.88 kΩ;
69.2 MΩ; v2
15.77 (300 µA, 5.10 V), (500 µA, 2.89 V), (2.00 mA, 5.00 V); 528, ∞, 341 Ω
15.79 (300 µA, 5.55 V), (500 µA, 2.89 V), (2.00 mA, 5.00 V), 2930, ∞, 341 Ω
15.81 (250 µA, 10.9 V), (2.00 mA, 9.84 V), (5.00 mA, 12.0 V); 868, ∞; 127 Ω
15.82 (99.0 µA, 4.96 V), (99.0 µA, 5.00 V), (500 µA, 3.41V), (2.00 mA, 5.00 V);
11400, 50.5 kΩ, 224 Ω
15.84 (49.5 µA, 11.0 V), (98.0 µA, 10.3 V), (735 µA, 16.0 V); 2680, 101 kΩ, 3.05
kΩ 
d for an ideal current source, 10.3 V]; 1.44 mV
15.86 No, Rid must be reduced or Rout must be increased.
15.88 (24.8 µA, 17.3 V), (24.8 µA, 17.3 V), (9.62 µA, 15.9 V), (490 µA, 16.6 V),
(49.0 µA, 17.3 V), (4.95 mA, 18.0 V); 88.5 dB, 202 kΩ, 22.0 Ω
15.90 250 µA/V2; 280; 868
15.92 196; 896
15.94 0.9996, 239 MΩ, 21.2 Ω
15.96 9.992, 67.6 MΩ, 1.48 Ω
15.98 36.8 µA
15.101
391 µA
15.104
22.8 µA
15.106
5 mA, 0 mA, 10 mA, 12.5 percent
15.107
66.7 percent
15.110
46.7 mA, 13.5 V
15.112
23.5 µA
35
15.113
6.98 mA, 0 mA
15.114
25.0 mΩ
15.116
(a) 18.7 µA, 61.5 MΩ
15.118
(a) 134 µA, 8.19 MΩ
15.120
Two of many: 75 kΩ, 6.2 kΩ, 150 Ω; 68 kΩ, 12 kΩ, 1 kΩ
15.122
59.4 µA, 15.7 MΩ
15.123
0, ∞
15.125
88.6 µA, 18.6 MΩ
15.127
17.0 µA, 131 MΩ
15.130
390 kΩ, 210 kΩ, 33 kΩ
15.132
157.4 µA, 16.61 MΩ, 31.89 µA, 112.2 MΩ
15.133
44.1 µA, 22.1 MΩ, 10.1 µA, 209 MΩ
15.136
400 µA, 2.89 x 1011 Ω
15.137
(4.64 µA, 7.13 V), (9.38 µA, 9.02 V); 40.9 dB, 96.5 dB
15.140
! o1µ f1 /2 , For typical numbers: 20(100)(70) = 140,000 or 103 dB
15.141
3σ limits: IO = 199 µA ± 32.5 µA, ROUT = 11.8 MΩ ± 2.6 MΩ
3σ limits: IO = 201 µA ± 34.7 µA, ROUT = 21.7 MΩ ± 3.6 MΩ
36
Chapter 16
16.1
[4.28 kΩ, 4.50 kΩ]
16.2
2.50 mV; 5.02 mV; 1%
16.4
7.7%, 0.813 µA, 0.855 µA, (IOS = -42.0 nA)
16.7
25.0 mV; 1.2%; 0.4%
16.8
(a) 94.8 µA, 186 µA, 386 µA, 1.16 MΩ, 580 kΩ, 290 kΩ
16.11 87.5 µA, 175 µA, 350 µA; 0.0834 LSB, 0.126 LSB, 0.411 LSB
16.12 316 µA, 332 kΩ, 662 µA, 166 kΩ
16.16 (a) 693 µA, 93.8 kΩ, 1.11 mA, 56.8 kΩ
16.18 422 kΩ, 112 µA; 498 kΩ, 112 µA
16.21 202 µA, 327 µA
16.22 472 µA, 759 µA; 479 µA, 759 µA; 430 µA, 692 µA
16.24 63.8 kΩ, 11.8 µA, 123 µA
16.26 10
16.28 15 kΩ, 2/3
16.30 138 µA, 514 MΩ
16.32 4.90 kΩ
16.34 172 kΩ, 15.0 kΩ, 0.445
16.36 21.8 µA, 18.4 MΩ; 43.7 µA, 9.17 MΩ
16.38 29.8 µA, 92.9 MΩ; 87.9 µA, 31.5 MΩ; 2770; 1.40 V
16.42 17.0 µA, 80/1; 89.9 MΩ
16.44 2/gm2
16.46 5.49/1
16.49 11.7 MΩ, 0, 6.687, 90.4 MΩ
16.51 1.33 MΩ, 1, 126, 84.4 MΩ
16.52 20.0 µA, 953 MΩ; 19.1 kV; 2.21 V
16.54 21.0 µA, 4.40 nA
16.57 (b) 50 µA, 240 MΩ; 12.0 kV; 3.05 V
16.60 16.9 µA, 163 MΩ, 2750 V; 2VBE = 1.40 V
16.62 2.86 kΩ
37
16.64 (a) 102 GΩ
16.66 (a) 51.0 GΩ
16.68 (a) 64.0 µA, 3.10 MΩ
16.70 5.71 kΩ
16.72 317 µA; 295 µA; 66.5 µA
16.74 " #oro4 2
16.76 14.5 kΩ, 226 kΩ
!
16.78 11.5 kΩ, 313 kΩ
I
I
"2
"2
16.80 IC1 = 137 µA, IC1 = 44.8 µA, SVCCC 1 = 4.40x10 , SVCCC 2 = 1.54 x10
16.82 n > 1/3
!
16.84 6.24 mA
I D1
D2
= 7.75x10"2 SVIDD
= 6.31x10"2
16.86 (b) ID1 = 8.19 µA ID2 = 7.24 µA SVDD
The currents differ considerably from the hand calculations. The currents are quite
sensitive to the value of λ. The hand calculations used λ = 0. If the simulations are
run with λ = 0, then the results are identical to the hand calculations.
!
16.88 14.4 µA, 36.0 µA, 6.56 µA, 72.0 µA, 9.48 µA
16.90 IC2 = 16.9 µA IC1 = 31.5 µA - Similar to hand calculations.
SVICCC 1 = 9.36x10"3 SVICCC 2 = 2.64 x10"3
16.92 (a) 388 µA, 259 µA
!
16.94 110 µA
16.96 4.90 V, 327.4 K
16.98 1.20 V, 304.9 K
16.100
5.07 V, +44.0 µV/K
16.102
-472 µV/K, -199 µV/K
16.104
2.248 kΩ, 10.28 kΩ, 80 kΩ, 23.9 kΩ, 126 kΩ
16.105
79.1, 6.28 x10-5, 122 dB
16.107
47.2, 6.97 x10-5, 117 dB
16.109
1200, 4 x10-3, 110 dB, ±2.9 V
16.113
(100 µA, 8.70 V), (100 µA, 7.45 V), (100 µA, -2.50 V), (100 µA, -1.25 V), 324, 152
16.115
(125 µA, 1.54 V), (125 µA, -2.79 V), (125 µA, 2.50 V), (125 µA, 1.25 V); 19600
38
16.118
171 µA
16.119
(b) 100 µA
16.120
(250 µA, 5.00 V), (250 µA, 5.00 V), (250 µA, -1.75 V), (250 µA, -1.75 V), (500
µA, -3.21 V), (135 µA, 5.00 V), (135 µA, -5.00 V), (250 µA, 2.16 V), (500 µA,
3.25 V), (500 µA, 3.21 V), (500 µA, 3.58 V); 4130; 2065
16.122
12,600
16.124
(250 µA, 7.50 V), (250 µA, 7.50 V), (250 µA, -1.75 V), (250 µA, -1.75 V), (1000
µA, -5.13 V), (330 µA, 7.50 V), (330 µA, -7.50 V), (1000 µA, 4.75 V), (250 µA,
2.16 V), (500 µA, 5.75 V), (1000 µA, 5.13 V), 3160
16.126
(b) 42.9/1 (c) 23000
16.130
7.78, 574 Ω, 3.03 x 105, 60.0 kΩ
16.132
±1.4 V, ±2.4 V
16.133
(a) 9.72 µA, 138 µA, 46.0 µA
16.134
271 kΩ, 255 Ω
16.136
VEE ≥ 2.8 V, VCC ≥ 1.4 V; 3.8 V, 2.4 V
16.138
2.84 MΩ, 356 kΩ. 6.11 x 105
16.141
100 µA, 15.7 V), (100 µA, 15.7 V), ( (50 µA, -12.9 V), (50 µA, -0.700 V), (50 µA,
-0.700 V), (50 µA, -12.9 V), (50 µA, 1.40 V), (50 µA, 1.40 V), (2.00 µA, 29.3 V),
(100 µA, 0.700 V), (100 µA, 13.6 V); 1.00 mS, 752 kΩ
16.142
(50 µA, 15.7 V), (50 µA, 15.7 V), (50 µA, 12.9 V), (50 µA, 12.9 V), (50 µA, 1.40
V), (50 µA, 1.40 V), (1.00 µA, 29.3 V), (100 µA, 1.40 V), (1 µA, 0.700 V), (1 µA,
13.6 V); 1.00 mS, 864 kΩ
16.143
(50 µA, 2.50 V), (25 µA, 3.20 V)
16.144
(a) 125 µA, 75 µA, 62.5 µA, 37.5 µA,
16.146
(500 "195sin 5000#t )µA, (500 + 195sin 5000#t )µA; 0.488 mS
!
39
Chapter 17
17.1
Amid = 50, FL ( s) =
17.4
200,
17.7
200,
!
!
17.9
s2
s
, yes, Av ( s) " 50
, 4.77 Hz, 4.82 Hz
(s + 4)(s + 30)
(s + 30)
1
, yes, 1.59 kHz, 1.58 kHz
" s
%" s
%
$ 4 + 1'$ 5 + 1'
#10
&# 10
&
s2
1
,
, 0.356 Hz, 142 Hz; 0.380 Hz, 133 Hz
(s + 1) s( + 2) !
s $!
s $
#1 + 500& #1+ 1000
&
"
%"
%
(b) -14.3 (23.1 dB), 11.3 Hz
17.10 (b) -15.6, 8.15 Hz
17.11 1.52 µF; 1.50 µF, 49.4 Hz
17.13 0.213 µF; 0.22 µF; 1940 Hz
17.14
Av ( s) = Amid
s2
| " =
(s + "1)(s + "2 ) 1
1
#
1&
C1% RS + RE
(
gm '
$
| "2 =
1
C2 ( RC + R3 )
35.5 dB, 13.2 Hz; -5.0 V, 7.9 V
!
17.16 123 Hz; 91 Hz; (145 µA, 3.57 V)
17.18 -131, 49.9 Hz, 12.0 V
17.19 45.5 Hz
17.21 7.23 dB, 19.8 Hz
17.23 0.739, 11.9 Hz, 10.0 V
17.24 0.15 µF
17.25 1.8 µF
17.27 Cannot reach 2 Hz; fL = 13.1 Hz for C1 = ∞, limited by C3
17.29 0.15 µF
17.30 308 ps
17.33 (a) 22.5 GHz
17.34 −96.7; −110
17.35 0.978; 0.979
40
| 2 zeros at " = 0
17.37 (a) −5100, −98.0, −5000, −100, 2% error (b) −350, −42.9, −300, −50, 18% error
17.39 Real roots: -100, -20, -15, -5
17.41 (3.31 mA, 2.71 V); -89.6, 1.45 MHz; 130 MHz
17.43 -14.3, 540 kHz
17.45 (0.834 mA, 2.41 V); -8.27, 3.29 MHz; 27.2 MHz
17.49 61.0 pF, 303 MHz
17.52 1/105 RC; 1/106 RC; -1/sRC
17.54 39.2 dB, 5.51 MHz
17.56 -114, 1.11 MHz; 127 MHz, 531 MHz
17.59 680 Ω, -18.0, 92.7 MHz
17.60 -29.3, 7.41 MHz, 227 MHz
17.62 300 Ω, 1 kΩ
17.63 −1300; −92.3; −100, −1200
17.64 9.55, 64.4 MHz
17.66 59.7, 1.72 MHz
17.69 2.30, 14.1 MHz
17.70 3.17, 14.1 MHz, 15.4 Hz
17.71 0.960, 114 MHz
17.74 -1.56 dB, 73.3 MHz
17.75 CGD + CGS/(1 + gm RL) for ω << ωT
17.77 Using a factor of 2 margin: 8 GHz, 19.9 ps
17.81 672 mA - not a realistic design. A different FET is needed.
17.83 393 kHz; 640 kHz
17.87 294 kHz
17.89 48.2 kHz
17.90 52.9 kHz
17.92 (a) 568 kHz
17.94 (a) 2.01 MHz
17.96 53.4 dB, 833 Hz, 526 kHz
17.97 1.52 MHz; 323 µH, 3.65 MHz
17.99 -45o; -118o; -105o
41
17.101
22.5 MHz, -41.1, 2.91
17.102
20.1 pF; 12.6; n = 2.81; 21.8 pF
17.103
15.2 MHz; 27.5 MHz
17.104
13.4 MHz, 7.98, 112/−90°; 4.74 MHz, 5.21, 46.1/−90°
17.105
10.1 MHz, 3.96, −35.3; 10.94 MHz, 16.8, −75.1
17.108
65 pF; 240, -4.41x104, 25.1 kHz
17.110
62.3 pF; 152 kHz, 40
17.112
(b) 497 Ω, 108 pF
17.113
100 Ω, 104 fF; 52.2 Ω, 144 fF
17.117
(a) 100 MHz, 1900 MHz
17.119
58.9 dB
17.121
-19.5 dB; -23.9 dB
17.123
0.6 A
17.125
-13.5 dB; -17.9 dB
17.127
0.2 A
17.129
1, 0.5, 0.5
17.131
0.567 V
17.133
0.2I1RC
42
Chapter 18
18.1
(b) 200, 4.975, 0.498%
18.3
1/40, 790.6, -38.95
18.5
104 dB
18.7
1/(1+T); 0.0498 %
18.9
(a) Series-shunt feedback (b) Shunt-series feedback
18.13 857 Ω, 33.3, 57.1, 506 Ω
18.15 245 Ω, 0, 0.952, 126 Ω
18.17 8.33x105, 16.7 S
18.19 12.9, 8.88 MΩ, 1.52 Ω
18.21 31.1 Ω, 2.01, 17.9, 195 Ω
18.23 10.1, 252 kΩ, 358 Ω
18.25 2.66 Ω; (32.2 Ω, 0, 11.1)
18.27 141 Ω; (13.0 kΩ, 0, 91.2)
18.29 0.9987, 110.4 MΩ, 2.845 Ω vs. 0.999, 108 MΩ, 2.70 Ω
18.31 −39.0 kΩ, 8.31 Ω, 0.295 Ω
18.33 33.2 Ω, 45.6 Ω, −38.9 kΩ
18.35 29.7 kΩ, 1.48 kΩ, -672 kΩ
18.37 0.133 mS, 60.4 MΩ, 26.8 MΩ
18.39
SPICE Results :
Atc = 9.92x10"5 S
Rin = 144.1 M# Rout = 11.91 M#
Hand Calculations: Atc = 9.92x10 S
Rin = 148 M#
"5
18.41 0.467 mS, 95.0 MΩ
!
18.42 8.48, 15.1 Ω, 3.51 MΩ
18.43 400 Ω, 12.5 MΩ, 0.992
18.45 29.8 MΩ, 799 MΩ, 1.08 GΩ
18.47 2.97, 14.5 Ω, 24.3 MΩ; 2.99, 14.6 Ω, 18.1 MΩ
18.49 30.39 GΩ; 33.3 GΩ
18.51 47.32 MΩ; 37.5 MΩ
18.53 Tv = 105, Ti = 18.1, T = 15.2, R2 R1 = 5.55
!
43
Rout = 11.1 M#
18.55 Tv = 988, Ti = 109, T = 98.0, R2 R1 = 8.99
18.57 110 kHz, 2048, ≤ 2048
!
18.59 219 pF
18.61 76.6o
18.63 107o
18.67 9.38 MHz, 41.7 V/µS
18.69 (b) 95.5 MHz, 30 V/µS
18.71 ±8.57 V/µS
18.73 71.5 MHz, 11.4 kHz, 236 MHz, 326 MHz, 300 MHz; 84.4 dB; < 0; 16.8 pF
18.75 12 kΩ; 9.05 MHz, 101 MHz, 74.8 MHz, 320 MHz; 2.60 MHz, 44.6 pF
18.77 (a) 81.9o
18.79 6.32 pF; 315 MHz, 91.5 MHz; 89.4o
18.81 17.5 MHz; [20.1 MHz, 36.3 MHz]; 0.211 mS, 5.28 µA
18.82 5.17 MHz, 4.53 MHz
18.84 9.00 MHz, 1.20
18.86 7.96 MHz, 8.11 MHz, 1.05
18.88 7.5 MHz, 80 Vp-p
18.89 7.96 MHz
18.90 11.1 MHz, 18.1 MHz, 1.00
!
!
18.91 LEQ = L1 + L2
| REQ = "# 2 g m L1L2
18.92 " o2 =
1
L(C + CGS + 4CGD )
|
µf # 1+
ro
RP
18.94 5.13 pF; 1 GHz can't be achieved.
18.95 6.33 pF; 2.81 mA; 3.08 mA; 1.32 V
18.97 15.915 mH, 15.915 fF; 10.008 MHz, 10.003 MHz
18.99 9.28 MHz; 9.19 MHz
Congratulations! you have just conquered most of electrical
engineering! except for Ch 18, now where did that go?
44