CHAPTER 13
13.1 Assuming linear operation:
v BE  0.700  0.005 sin 2000 t V
5mV 
v ce  
1.65 sin 2000t = 1.03 sin 2000t V
8mV 
v CE  5. 00 1. 03 sin 2000t V; 10 - 3300I C  0.700  I C  2.82 mA
13.2 Assuming linear operation:
v GS  3.50  0. 25 sin 2000t V
0. 25V
2sin 2000t  = 1.00 sin 2000t V
0. 50V
 4.80  1.00 sin 2000 t V
 v GS  V TN  10  3300I DS  1.00 sin 2000t  3.50  0.25 sin 2000t 1
v ds 
v DS
v DS
For sin 2000t  1, I DS 
10  1  3.5  0. 25 1 V
 1.89 mA
3300

13.3 (a) C1 is a bypass capacitor. C2 is a coupling capacitor that couples the ac component
of vS into the amplifier. C3 is a coupling capacitor which couples the ac component of
the signal at the collector to the output v O. (b) The signal voltage at the base will be v b =
0.
13.4 (a) C1 is a coupling capacitor that couples the ac component of v S into the amplifier. C2
is a bypass capacitor. C3 is a coupling capacitor which couples the ac component of the
signal at the collector to the output vO. (b) The signal voltage at the emitter will be ve =
0.
13.5 (a) C1 is a coupling capacitor that couples the ac component of v S into the amplifier. C2
is a bypass capacitor. C3 is a coupling capacitor which couples the ac component of the
signal at the drain to the output vO. (b) The signal voltage at the source will be v s = 0.
13.6 (a) C1 is a coupling capacitor that couples the ac component of v S into the amplifier. C2
is a coupling capacitor which couples the ac component of the signal at the drain to the
output vO. (b) C1 is a coupling capaictor that couples the ac component of v S into the
amplifier. C2 is a coupling capacitor which couples the ac component of the signal at
the drain to the output vO.
13.7 (a) C1 is a coupling capacitor that couples the ac component of vS into the amplifier. C2
is a bypass capacitor. C3 is a coupling capacitor which couples the ac component of the
signal at the drain to the output vO. (b) The signal voltage at the source will be v s = 0.
1
13.8 (a) C1 is a coupling capacitor that couples the ac component of v S into the amplifier. C2
is a bypass capacitor. C3 is a coupling capacitor which couples the ac component of the
signal at the collector to the output vO. (b) The signal voltage at the emitter will be v e =
0.
13.9
+12 V
V EQ  12V 
6 k
R EQ  5k 10k  3. 33k
R EQ
4  3330I B  0. 7  76 4000 I B 12
Q1
3.33 k
I B  23.8 A | I C  1.78 mA | I E  1.81 mA
V EQ
-4 V
13.10
5k
24 V  4V
5k  10k
4 k
V CE  12  6000I C  4000I E  12   6.08 V
-12 V
Q  po int: 1.78 mA ,6.08 V 
*Problem 13.10 - Common-Emitter Amplifier - Figure P13.1
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
13.11
+9 V
9  86000I B 10182000 IB  0.7
Q1
I B  0.992 A | I C  99.2 A
86 k
13.12
2
82 k
V CE  18  82000 IE  18  82000 100A   9. 80 V
-9 V
Q  po int: 99.2 A, 9.80 V 
*Problem 13.12 - Common-Collector Amplifier - Figure P13.2
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
7.12E-01
13.13
-7.5 V
33 k
7.5  3000I B  66 68000 I B  0.7
Q1
I B  1.51 A | I C  98.4 A
3 k
68 k
+7.5 V
13.14
V EC  15  33000 IC  68000I E  4. 96 V
Q  po int: 98. 4 A,4. 96 V 
*Problem 13.14 - Common-Base Amplifier - Figure P13.3
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
13.15
3
+9 V
62k
 6.80V
62k  20k
 62k 20k  15.1k
V EQ  9V
3.9 k
R EQ
9  3900 136 IB  0.7  15100I B  6.80
Q1
V EQ
R EQ
15.1 k
I B  2.75 A | I C  371 A | I E  374 A
+6.80 V
13 k 
V EC  9  3900I E  13000I C  2. 72 V
Q  po int: 371 A , 2.72 V 
13.16
*Problem 13.16 - Common-Emitter Amplifier - Figure P13.4
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
13.17
+15 V
82 k
1M
 4.05V
1M  2.7M
 1M 2.7M  730k
V EQ  15V
R EQ
R EQ
M1
V EQ
730 k
4.05 V
I DS 
0.25mA
V GS 12
2
V GS  4.05  27000I DS
I DS  0.125mA 3. 05  27000I DS   I DS  82.2 A
2
27 k 
V DS  15  82000 IDS  27000 IDS  6. 04V
Q  po int: 82.2 A, 6.04 V 
13.18
4
*Problem 13.18 - Common-Source Amplifier - Figure P13.5
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
13.19
I DS 
+15 V
VDS
M1
1.81E+00
5x10 4
V GS  22 | V GS  3900I DS
2
V GS  3900
4.3 k
5.96E+00 VGS
I DS = 
5x10 4
V GS  22  VGS  0.990 V
2
VGS
= 254 A
3900
V DS  15  4300IDS  3900 IDS  12.9V
3.9 k
Q  po int: 254 A , 12.9 V 
13.20
*Problem 13.20 - Common-Gate Amplifier - Figure P13.6
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
13.21
+18 V
1.29E+01 VGS
-9.92E-01
3. 3M
 9.00V
3. 3M  3.3M
 3.3M 3.3M  1.65M
V EQ  18V
R EQ
22 k
18  22000I SD  V SG  9
REQ
M1
V EQ
1.65 M
9.00 V
24 k 
4x10 4
V SG  12  V SG
2
 2.24V | I SD  307 A
 18  22000 ISD  24000I SD  3.88 V
9  22000
V SG
V SD
Q  po int: 307 A , 3.88 V 
13.22
*Problem 13.22 - Common-Source Amplifier - Figure P13.7
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
5
.MODEL PFET PMOS KP=400U VTO=-1
.END
Results:
ID
-3.07E-04
VDS
-3.86E+00 VGS
13.23
I SD 
-2.24E+00
12  V SG 200x10 6

V SG 12
33000
2
V SG  0.84V | I SD  338 A
V SD  12  33000I SD  22000I SD  12 
M1
+12 V
33 k
22 k
V SD  5. 41 V
-12 V
Q  po int: 338 A , 5.41 V 
13.24
*Problem 13.24 - Common-Gate Amplifier - Figure P13.8
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
13.25
+18 V
3.9 k
2
J1
I DS
2
 V 
 2000I DS 
 I DSS 1  GS  = 0.005 1
  I DS  1. 25mA
V P 

5


10 M
2 k
V DS  18  3900 IDS  2000I DS  10.6V
Q  po int: 1.25 mA , 10.6 V 
13.26
For the JFET : V TO = V P
and  =
I DSS
V 2P

0. 005
A
 200 2
25
V
*Problem 13.26 - Common-Source Amplifier - Figure P13.9
VDD 4 0 DC 18
RD 4 3 3.9K
RG 2 0 10MEG
6
R1 1 0 2K
J1 3 2 1 NFET
.OP
.MODEL NFET NJF BETA=200U VTO=-5
.END
Results:
ID
1.25E-03
VDS
1.06E+01 VGS
-2.50E+00
13.27
-15 V
7.5 k
J1
2.2 M
Since V SG = 0, ISD  I DSS  1 mA
V SD  15  7500I SD  7.5 V
Q  po int: 1.00 mA,7.50 V 
7
13.28
*Problem 13.28 - Common-Source Amplifier - Figure P13.10
VDD 4 0 DC -15
RD 4 3 7.5K
RG 2 0 2.2MEG
J1 3 2 0 PFET
.OP
*Note that JFET model requires negative sign on VP
.MODEL PFET PJF BETA=111U VTO=-3
.END
Results:
ID
-9.99E-04
VDS
-7.51E+00 VGS
-1.65E-05
13.29
RS
+
Q
R
v
R1
s
R
C
vo
3
R2
-
(a)
c
+
RS
v
s
+
v
R1
R2
r
-
(b)
R
ro
g mv
C
R
vo
3
C1  Coupling
C 2  Bypass
-
C 3  Coupling
13.30 (a)
Q1
RS
v
R1
s
+
R2
(a)
RE
s
vo
3
-
+
RS
v
R
R1
r
c
v
-
ro
gm v
+
(b)
R2
RE
R
vo
3
C1  Coupling
C 2  Coupling
-
C 3  Coupling
8
13.30 (b)
Q1
+
RS
v
R
RE
s
R
C
vo
3
-
(a)
c
ro
v
RE
s
g mv
v
r
(b)
C1  Bypass
+
+
RS
RC
R3
vo
-
C 2  Coupling
-
C 3  Coupling
13.31 (a)
RS
+
Q1
v
R1
s
R
R
C
v
3
o
R2
-
(a)
c
+
RS
v
s
+
v
R1
R2
r
(b)
g mv
-
R
ro
R
C
vo
3
C1  Coupling
C 2  Bypass
-
C 3  Coupling
13.31 (b)
RS
+
M1
RD
v
R1
s
R
3
R2
v
o
-
(a)
c
+
RS
v
s
(b)
R1
+
v
R2
-
g mv
ro
RD
R
vo
3
C1  Coupling
C 2  Bypass
-
C 3  Coupling
9
13.32 (a)
M1
+
RS
v
R
RD
R1
s
vo
3
-
(a)
ro
RS
-
v
s
v
R1
g mv
(b)
vo
R3
RD
+
c
+
C1  Coupling
-
C 2  Coupling
13.32 (b)
RS
+
M1
RD
v
R1
s
R
v
3
o
R2
-
(a)
c
+
RS
v
s
+
v
R1
R2
-
(b)
ro
g mv
RD
R
vo
3
C1  Coupling
C 2  Bypass
-
C 3  Coupling
13.33 (a)
M1
+
RS
v
RD
R1
s
R
vo
3
-
(a)
ro
RS
v
s
(b)
R1
v
+
+
g mv
RD
R3
vo
c
C1  Coupling
-
C 2  Coupling
10
13.33 (b)
RS
+
J1
RD
R
v
3
o
v
-
s
RG
(a)
+
RS
v
s
R1
v
RG
+
ro
g mv
-
vo
R
RD
c
-
3
C1  Coupling
R1
(b)
C 2  Coupling
13.34 The JFET symbol is misdrawn in the first printing. It should be a common source
circuit.
RS
+
J1
RD
R
3
v
v
o
-
s
RG
(a)
+
RS
v
s
(b)
+
v
RG
-
g mv
ro
RD
R
vo
3
c
C1  Coupling
-
C 2  Coupling
13.35 RS: Thévinen equivalent source resistance; R1: base bias voltage divider; R2: base bias
voltage divider; RE: emitter bias resistor - sets emitter current; RC: collector bias
resistor - sets collector-emitter voltage; R3: load resistor
13.36 RS: 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
11
13.37 RS: 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
13.38
a r d 
VT

0. 6  
0. 025
| I D  10 14 exp

 1  264. 9A | r d 
 94. 4 
0.025  
I D  IS

264. 9A  10fA
b  I D  0 r d 
V D  V T ln
0.025
 2.50 T
10fA
0. 025
15
17
 10  I D  I S  2.5x10
A
ID  IS
ID  IS
2.5x10 17
 0.025 ln
 0.150 V
IS
10 14
13.39
VT 
T
VT
rd
c
kT 1. 38x10 23
V
5

T | r d  T  1000V T
19 T  8.63 x10
q
ID
1. 60x10
75K
6.47 mV
6.47 
100K
8.63 mV
8.63 
200K
17.3 mV
17.3 
300K
25.9 mV
25.9 
400K
34.5 mV
34.5 
13.40
rD
1
1

 r D  2.22k |40I D 
 I D  11.3 A | v s = 105mV   50 mV
20k  r D 10
2.22k
13.41
0. 005 
0.005
  1  0. 221 |
 0. 200  10.7% error
0. 025 
0.025
a exp

 0. 005 
0. 005
exp
 1  0.181 | 
 0. 200  9.37% error
 0. 025 
0. 025
0. 010
0.010 
 1  0.492 |
 0.400  23.0% error
b  exp
0.025 
0. 025
0. 010 
0. 010
exp
 0. 400  17.5% error

 1  0.330 | 
 0. 025 
0. 025
IC 
13.42
13.43
IC 
gm
0.03

 0.750 mA  750 A
40
40
o V T 75 0.025 V 

 187.5 A | Q - point: 188 A, V CE > 0.7 V 
4
r
10 


g m  40I C  40 1.875x10 4  7.50 mS | r o 
13.44
12
V A  V CE
V
100V
 A 
 533 k
IC
IC
187.5 A
ro 
V A  V CE
IC
; solving for V A : V A  I C r o  V CE
Using the values from row 1: V A  0.002 40000   10  70 V
Using the values from the second row:  o  g m r   0.12 500   60 and F   o  60.
Row 1: g m  40I C  400.002   0. 08 S | r  
o
60

 750 
g m 0. 08
 F  g m r o  0.08 40000   3200
V  V CE
g
0.12
80
Row 2: I C  m 
 3 mA | r o  A

 26.7 k
40
40
IC
0.003
 F  g m r o  0.1226700   3200
o
60
g m 1.25 x 10 -4
-4
=
=
1.25
x
10
S
|
I


 3.13 A
C
r
4.8 x 10 5
40
40
V  V CE
80
ro  A

 25. 6 M | F  g m r o  1.25 x 10-4 25. 6 x 10 6  3200
IC
3.13 x 10 -6
Row 3: g m =

13.45

0. 005 
0.005
 0. 200  10.7% error
a exp

  1  0. 221 |
0. 025 
0.025
0. 005
 0. 005 
exp
 1  0.181 | 
 0. 200  9.37% error
 0. 025 
0. 025
0.0075 
0.0075
b  exp

 1  0.350 |
 0.300  16.7% error
 0.025 
0.025
0.
0075
0. 0075

exp

  1  0.259 | 
 0.300  13.6% error
 0. 025 
0.025
0.0025 
0.0025
 0.100  5.17% error
c exp

  1  0.105 |
 0.025 
0.025
0.0025
 0. 0025 
exp
  1  0.0952 | 
 0.100  4.84% error
 0. 025 
0.025
13.46
IC 
o V T 75 0.025 V 

 1. 875 A | Q - point: 1.88 A, V CE > 0.7 V 
6
r
10 


g m  40I C  40 1.875x10 6  75. 0 S | r o 
13.47
V A  V CE V A
100 V


 53. 3 M
IC
IC
1.875 A
IC 350A
I
600A  125A

 90 |  o  C 
 120
IB
4A
I B
6A  2A
750A
900A  600A
(b) F 
 95 |  o 
 75
8A
4A
(a )  F 
13.48
(a) t=linspace(0,.004,1024);
ic=.001*exp(40*.005*sin(2000*pi*t));
IC=fft(ic);
z=abs(IC(1:26)/1024);
13
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
(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)
8 x10
-3
6
4
2
0
0
z([5 9 13])
13.49 (a)
NAME
MODEL
IB
IC
VBE
VBC
VCE
BETADC
GM
RPI
RX
RO
BETAAC
14
ans = 0.0016
Q1
NBJT
2.21E-05
1.78E-03
7.28E-01
-5.41E+00
6.14E+00
8.04E+01
6.87E-02
1.17E+03
0.00E+00
4.52E+04
8.04E+01
1
0.0007
2
3
4
x10-3
0.0002
T  27C | V T  8.625 x10 5 300   25. 9mV
gm 
IC
1.78mA

 68.7 mS
VT
25.9mV

V 
 6.14 
 o   FO 1  CE   75 1 
  81.1
V

75 

A 

81.1
r  o 
 1180 
g m 0.0687
ro 
V A  V CE 75  6.14

 45.6 k
IC
1.78mA
13.49 (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  27C | V T  8.625 x10 5 300   25. 9mV
gm 
IC
0.373mA

 14. 4 mS
VT
25.9mV

V 
 2.68 
 o   FO 1  CE   135 1 
  140
V

75 

A 

140
r  o 
 9.72 k
g m 0.0144
ro 
V A  V CE 75  2.68

 208 k
IC
0.373mA
Note: The SPICE model actually is using VCB instead of VCE
5. 41
 1. 99 
a o  75 
1 
  80.4 b   o  135 1
  139

75 

75 
13.50
ix
+
y 11
r
v
-
g mv
vx
ro
re
ie
o ie
1
r
v
v
1  o
 x  o x 
vx
re
re
re
For the hybrid pi model: y 11 
For the T - model: i x
y11 
re 
13.51
13 - 15
ix
1  o


vx
re
1
o
 o 1
1

 r    o  1r e
re


 o 1r e
r
o

 V
V

 o  o T  T
IC
IE
 o 1 g m o  1 g m
A v  10 V CC  10 12   120
©R. C. Jaeger - February 6, 2016
A v  10 V CC  V EE   10 15  15  300
13.52
13.53
A v  10 V CC  V EE   10 1.5  1.5  30; This estimate says no.
However, if we look a bit deeper,
A V  40I CR C   40V R C ,and we let V R C 
then we can achieve AV  401.5   60.
V CC  V EE   1.5V,
2
So with careful design we can probably achieve a gain of50.
13.54
Using our rule-of-thumb estimate:
A v  10 V CC  10 1. 5  15 ; A v  10 1  10
Note that this result really assumes that I C varies with VCC.
13.55
5V
 0.5 mA , but ic  0. 2I C for small - signal operation, or I C  5i c  2.5 mA.
10k
V CC  V BE  i cR L  I CR L  0.7  5  25  30.7 V
ic 
A v  40dB  100 | v o  100 v be  100(0.005 V)  0.500 V.
13.56
13.57
20k
18  4.61V | R EQ  20k 62k  15.1k
62k  20k
4.61  0.7  9
IB 
 6.76A | I C  135I B  913A
15.1k  136 3.9k 
a V EQ  9 
V CE  9  13000I C  3900I E  9  2.54V
g m  40I C  0. 0365S | r  
135
 3.70k | r o  
gm
2.97k

A V   

0.0365 11.5k  314
1k  2. 97k 
b  For V CC  18V, the answers are the same: I C  913A | V EC  2.54 V | A V  314
13.58
13.59
16
For common - emitter stage: A V  50dB  A V  316
15 V
v be 
 47.5mV which is far too big for small - signal operation.
316
The will be significant distortion of the sine wave
.
g m  40 50A  2.00mS | r  
100
75V  10V
 50k | r o 
 1.70M
2.00mS
50A
R BB  100k 50k  33.3k
33.3k


A V   
2mS  1.70M 100k 100k  95. 0
33. 3k  0.75k 

13.60

For  o  100, see Prob. 13.59.
60
r 
 30k | R BB  100k 30k  23.1k
2.00mS
23.1k

A V   

2mS  1.70M 100k 100k  94.1
23.1k  0.75k 
95.0  A V  94.1  only a small variation


17
13.61
75
50  7.5
 750 | r o 
 23.0k
0.1S
2.5mA
4700

v  0. 990v s
4700  50 s
g m  40 2.5mA   0.100S | r  
R th  4.7k 50  49.5 | v th
750
 0.1S 23.0k 4.3k 10k  247
A V  0.990 



750  49.5

13.62

40
50  1.5
 1M | r o 
 51.5M
40S
1A
5M
R th  10k 5M  9.98k | v th 
v  0. 998v s
5M  10k s
1M
 40S 51.5M 1.5M 3. 3M  40.0
A V  0.998 



1M  9.98k 
g m  40 1A  40S | r  

13.63

*Problem 13.63 - Common-Emitter Amplifier - Figure 13.23
VCC 7 0 DC 12
VS 1 0 AC 1
RS 1 2 1K
C1 2 3 100U
R1 3 0 10K
R2 7 3 30K
RE 4 0 1.3K
C3 4 0 100U
RC 7 5 4.3K
C2 5 6 100U
R3 6 0 100K
Q1 5 3 4 NBJT
.OP
.AC DEC 20 1HZ 10KHZ
.MODEL NBJT NPN IS=1E-16 BF=100 VA=75
.PRINT AC VM(6) VP(6)
.END
Results: IC = 1.60 mA, VCE = 3.03 V, AV = -135 -- IC differs by 10% - AV is off by 4%
13.64
13.65
18
20000 
 2.15  N  3 (or possibly 2)
10 V CC N  10 10 N  20000  N  log
log 100 
*Problem 13.65 - Common-Emitter Amplifier - Figure P13.4
VCC 7 0 DC 9
VS 1 0 AC 1
RS 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 A, VEC = 2.68V, AV = -134
Hand calculations in Prob. 13.15 yielded (371 A, 2.72V)
135
 9.12k | r o  
14.8mS
15.1k

v  0. 938v s
15.1  1k s
g m  40 371A   14.8mS | r  
R th  15.1k 1k  938 | v th
9.12k
 14. 8mS  13k 100k  145
A V  0.938 



9.12k  938 


SPICE AV result is lower because ro is included.
13.66
Using the information from Row 1:
1
 I DSr o  V DS  8x10 4 4x10 4  6  26 V ;  = 0.0385 V -1


From Row 2: K n


g 2m


2IDS 1  V DS 

2x10   3. 25x10
6 

2 5x10 1 

 26 
4 2
4
5

A
V2

6 
4
Row 1: g m  2K n I DS 1  V DS   2 3.25x10 4 8x10 4 
1 
  8x10 S
 26 


f  g m r o  8x10 4 4x10 4  32
0. 2V GS  V TN   0.2
2I DS
 0. 2
K n 1  V DS 

2 8x10 4

6 

3.25x10 4 1 

 26 
 0. 40V
1
 V DS
26  6
Row 2: r o  

 640k | f  g m r o  2x10 4 6.4x10 5  128
I DS
5x10 5

0. 2V GS  V TN   0.2
2I DS
 0. 2
K n 1  V DS 
Row 3: g m  2K n I DS 1  V DS  



2 5x10
5


6 

3.25x10 4 1 

 26 
 0.10V

6 
2 3. 25x10 4 10 2 
1 
  2.83x10 3 S
 26 
1
 V DS
26  6
ro  

 3.2k | f  g m r o  2.83x10 3 3.2x10 3  9.06
I DS
0. 01

0. 2V GS
2I DS
 V TN   0.2
 0. 2
K n 1  V DS 

2 10 2

6 

3.25x10 4 1 

 26 

 1. 41V
19
MOSFET Small-Signal Parameters
gm
IDS
( S)
ro
()
f
Small-Signal
Limit vgs (V)
0.8 mA
0.0008
40,000
32
0.40
50 A
0.0002
640,000
128
0.10
10 mA
0.00283
3200
9.06
1.41
13.67


4
1
 2K n 1 V DS  1  2K n
 1  2 2.5x10
 f    V DS
  
| 

 1  IDS  1.25 A
 

IDS
  IDS
.02 
I DS
13.68
1  0. 2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
13.69 From the results of Problem 13.18:
ID = 8.29E-05, VGS = 1.81E+00, VDS=5.96E+00, GM = 2.04E-04, GDS = 0.00E+00
gm 
282.9A 
2I DS

 205 S |  = 0  r o  
V GS  V TN
1. 81 1
13.70 From the results of Problem 13.22:
ID = 3.07E-04, VGS = -2.24E+00, VDS=-3.86E+00, GM = 4.96E-04, GDS = 0.00E+00
gm 
2 307A 
2ISD

 495 S |  = 0  r o  
V SG  V TP
2.24 1
13.71 At virtually any Q-point. 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.)
13.72
R Dr o
V
9V
50V  9V
7. 8V
| Using V DS = DD | R D 
| ro 
 R OUT 
R D  ro
2
I DS
I DS
IDS
7.8V

 156A  R D  57.6k | r o  378k | Q - point: 156 A ,9 V 
50k
R OUT 
I DS
20
21
13.73
2
 v 

1 2I
1 2IDSS 
1
2
2
2
i DS  I DSS 1  GS    DSS
 2 v GS  V P   K n v GS  V TN 
2 V P  v GS  
V
2
2
2
V
V

 P 
 P 
P 
13.74
(a) t=linspace(0,.004,1024);
id=.005*(1-0.5*.4*sin(2000*pi*t)).^2;
ID=fft(id);
y=abs(ID(1:26)/1024);
y(1)
ans = 0.0051
plot(t,id)
8 x10
-3
7
6
5
4
3
0
0.5
y([5 9 13])
ans = 1.0e-03 * 0.9995
1
1.5
0.0500
2
2.5
3
4 x10-3
3.5
0.0007
(b) t=linspace(0,.004,1024);
id=.005*(1-0.5*.4*sin(2000*pi*t)).^2;
ID=fft(id);
y=abs(ID(1:26)/1024);
y(1)
ans = 0.0056
plot(t,id)
0.012
0.01
0.008
0.006
0.004
0.002
0
0
0.5
y([5 9 13 17 21])
ans = 0.0025 0.0003
13.75
22
1
0.0000
1.5
2
2.5
3
3.5
4 x10-3
gm
2

VP
2
I DSSI DS  I DSS
5x10 4 2. 5 
1


  7.81 mA
5x10 5 
2

1
 I DSr o  V DS  10mA 15k   6 V  144V    6.94x10 3 V 1

2
150V
Row 1: g m 
7.81mA 1mA   2.24 mS | r o 
 150 k | f  g m r o  335
2.5
1mA
ID
1mA
 0.2
 71.6 mV
I DSS
7. 81mA
150 V
 0.5 mS | r o 
 3.00 M | f  g m r o  15000
50A
0.2 V GS  V P   0.2
Row 2: g m
0. 2
Row 3: g m 
0. 2
ID
50A
 0. 2
 16. 0 mV
I DSS
7.81mA
2
7.81mA 10mA   7.07 mS | r o  15 k |  f  g m r o  106
2.5
10mA
 226 mV
7.81mA
Note that the gate -channel diode is forward-biased by 0.33 V in the last case.
JFET Small-Signal Parameters
IDS
gm (S)
1 mA
0.00224
50 A
10 mA
ro
()
f
Small-Signal
Limit vgs (V)
150,000
335
0.0716
0.0005
3,000,000
1500
0.0160
0.00707
15,000
106
0.226
13.76 From the results of Problem 13.26:
ID = 1.25E-03, VGS = -2.50E+00, VDS=1.06E+01, GM = 1.00E-03, GDS = 0.00E+00
gm 
2I DS
21. 25mA 

 1. 00 mS |  = 0  r o   | QED
V GS  V P
2.50  5 
13.77 From the results of Problem 13.26:
ID = -9.99E-04, VGS = -1.65E-05, VDS=-7.51E+00, GM = 6.66E-04, GDS = 0.00E+00
gm 
2I SD
20.999mA 

 666 S |  = 0  r o   | QED
V SG  V P 16. 5V  3V
23
13.78
Note that iG  0 for this device.
Load line: 400  133000 iP  v PK
and v GK  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
 55.6k | g m =
 1.6mS | f  89.0
2.15mA  1.25mA
1V  (2V)


A V   g m R P r o  1.6mS 133k 55.6k  62.7
24
13.79
BJT: I C  g m V T  0.5S0.025 V   12.5mA | MOSFET: I DS 
2
g 2m
0.5S

5 A!
2K n
2 25mA / V 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
Note, however, that r is small for the BJT: r  
 120 versus ∞ for the FET.
0.5
13.80 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 r is low:
r 
13.81
 o V T 100. 025V 

 250 
IC
10mA
1
 2K n 1 V DS 
40 V A  V CE     V DS 
 

IDS
40 35   60
2 0.025 1.2 
 I DS  111 A |  f  40(35)  1400
IDS
13.82 Either transistor could be used. For a BJT operating in the common-emitter
configuration or a FET operating in the common-source configuration:
For the BJT: R IN  r  | BJT: IC 
 o V T 1000.025 V 

 33. 3 mA - A fairly high current
r
75
For the FET : R IN = R G and setting RG  75  is satisfactory, particularly if a depletion mode FET is available .
(Note that common-base and common-gate amplifiers from Chapter 14 could also be used.)
13.83
AV  
V DD
12

 12 or 21.6 dB
V GS  V TN
1
13.84
15
7.5V
vd 
 7.5V peak | 15dB  A V  5.62 | v gs 
 1.34 V | VGS  V TN  5 1.34   6.70V
2
5.62
Yes, it is possible although the required value of VGS-VTN is getting rather large.
13.85
For V DS 
I DS 
V DD
V DD
15
, AV  
| 30 
| V GS  V TN  0.5 V
2
V GS  V TN
V GS  V TN
1mA
V GS  V TN 2  125A | Q - point: 125 A,7.5 V 

2
25
13.86
AV 
V DD
9
|
 30  V GS  V TN  0.300 V
V GS  V TN
V GS  V TN
13.87
0.5
 2. 5V
0.2
 10. Using the rule - of - thumb estimate to select VDD :
v gs  0.2V GS  V TN  requires V GS  V TN  
A V  20dB  A V
AV  
V DD
VGS  V TN
13.88
and V DD  102.5  25 V
0.1
 0.5V
0.2
 56.2. Using the rule - of - thumb estimate to select VDD :
v gs  0.2V GS  V TN  requires V GS  V TN  
A V  35dB  A V
AV  
V DD
VGS  V TN
and V DD  56.20.5V   28 V
13.89
N
N
 V DD



10
We desire 
 1000; 

  1000
 V GS  V TN 
 V GS  V TN 
For V GS  V TN  1V, N = 3 meets the requirements, but with no safety margin.
For V GS  V TN  0.75V, N = 3 easily meets the requirements.
13.90
For the bias network: V TH  10 V
430k
 4.343 V | R TH  430k 560k  243k
430k  560k
5x10 4
V GS  12 | V GS  4.343  2x10 4 I DS  V GS  1.72 V | I DS  131 A
2
 10  63k 131A   1.75V  V GS  V TN so saturation region is ok.
I DS 
V DS
gm
AV
AV
1


 1.75



 2 5x10 4 131A   362S | r o  0. 0133
 586k
131A
243k

362S 586k 43k 100k  10.3 | Original A V  4.69
243k  1k
g
increases because m increases as current decreases.
I DS




13.91
A 
50  5V

g m  2500 2 100A

1  0.02(5)  332S | r o 
 550k

100A
V 

6. 8M

A V   
332S 550k 50k 120k  10.9
6.8M  0.1M 

26

13.92


g max
 2 700A / V 2 100A   374S | g min
m
m 


2 300A / V 2 100A   245S
6. 8M


A V   
g  550k 50k 120k  32.7kg m 
6.8M  0.1M  m


A max
 12.2 | A min
 8. 01
V
V
13.93


50  5V
 5.50M
10A
g m  2 100A / V 2 10A 1  0. 02(5)  46.9S | r o 
10M

A V   

46.9S 5.50M 560k 2.2M  19. 2
10M  0.1M 

13.94
g m  2K n I DS 1  V DS  

20.0010.002 1  0. 0157.5  2.11x10 3 S
1
1
 V DS
 7.5

0.015
ro 

 37.1k
I DS
0. 002
v th 
106 
106  10 4 

v s  0. 990v s | R th  10k 1M  9.90 k


A vth  g m r o R D R 3  2.11x10 3 37.1k 3.9k 270k
A v  2.11x10
13.95
3

3.48k0.990   7.27
*Problem 13.95 - Common-Source Amplifier - Figure P13.5
VDD 7 0 DC 15
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 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(VS) IP(VS)
*.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(VS) = 1.369E-06
IP(VS) = -1.800E+02
VM(3) = 9.986E-01
IP(C3) = -1.800E+02
VP(3) = 1.248E-04 IM(C3) = 1.220E-05
27
A V  14.6 | R IN 
13.96
VM 3 
0.9986
1
1

 729 k | R OUT 

 82. 0 k
IM VS 1. 369A
IMC3 12.20A
*Problem 13.96 - Common-Source Amplifier - Figure P13.7
VDD 7 0 DC 18
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 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=500U VTO=-1
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: ID = -3.13E-04
VGS = -2.12E+00
VDS = -3.61E+00
VM(6) = 12.76E+01 VP(6) = -1.799E+02 IM(VS) = 6.057E-07
VM(3) = 9.994E-01 VP(3) = 5.523E-05 IM(C3) = 4.167E-05
A V  12.8 | R IN 
13.97
28
IP(VS) = -1.800E+02
IP(C3) = -1.800E+02
VM3
0. 9994 V
1
1

 1.65 M | R OUT 

 24. 0 k
IM VS 0.6057A
IM C3 41. 67A
*Problem 13.97 - Common-Source Amplifier - Figure P13.9
VDD 7 0 DC 18
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 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 NJF BETA=200U VTO=-5
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VS) IP(VS) 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(VS) = 9.991E-08
VM(3) = 9.990E-01 VP(3) = 9.106E-06
A V  3.52 | R IN 
13.98
IM(C3) = 2.564E-04
IP(C3) = -1.800E+02
VM3 0.9990 V
1
1

 10.0 M | R OUT 

 3.90 k
IM VS 99. 91nA
IMC3 256.4A
*Problem 13.98 - Common-Source Amplifier - Figure P13.10
VDD 7 0 DC -15
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 1 2 10K
C1 2 3 100U
RG 3 0 2.2MEG
RD 7 5 7.5K
C2 5 6 100U
R3 6 0 220K
J1 5 3 0 PFET
.OP
*Note that JFET model requires negative sign on VP
.MODEL PFET PJF BETA=111U VTO=-3
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C2) IP(C2)
.END
Results: ID = -9.99E-04
VGS = -1.65E+05
VDS = -7.51E+00
VM(6) = 4.809E+01 VP(6) = -1.800E+02 IM(VS) = 4.525E-07
VM(3) = 9.955E-01 VP(3) = 4.126E-05 IM(C3) = 1.333E-04
A V  4.81 | R IN 
13.99
IP(VS) = -1.800E+02
gm
AV
IP(C3) = -1.800E+02
VM 3 
0.9955 V
1
1

 2.20 M | R OUT 

 7.50 k
IM VS 0. 4525A
IM C3  133. 3A
i ds  0. 4I DS and i dsR D  5 V or i ds 
13.100
IP(VS) = -1.800E+02
5V
 333A  IDS  833 A
15k
1
 9V
2

1mA 1mA 1  0.015 9  710S | r o  0.015
 75.7k
3
1mA
1M


  
710S 75.7k 7.5k 160k  4.60
1M  10k 


29
13.101
r 
100
75V  10V
 50k | r o 
 1. 70M
2.00mS
50A
R IN  R B r   100k 50k  33.3k | R OUT  1.7M 100k  94.4k
13.102
R IN  R B r  | r min


60
100
 30k | r max

 50k

4050A 
40 50A 
max
R min
 R B r   100k 50k  33.3k
IN  R B r   100k 30k  23.1k | R IN
R OUT  R C r o  100k
13.103
r 
75  10
 100k 1.7M  94. 4k independent of o
50A
75 0.025 V 
50  7.5 V
 750 | r o 
 23.0k
2.5mA
2.5
mA
R IN  R B r   4.7k 0.75k  647  | R OUT  R C r o  4.3k 23k  3.62k
13.104
r 
40 0.025V 
50  1.5 V
 1.00M | r o 
 51.5M
1A
1
A
R IN  R B r   5M 1M  833 k | R OUT  R C r o  1.5M 51.5M  1.46 M
13.105
r 
40 0.025V 
50  1.5 V
 1.00M | r o 
 51.5M
1A
1
A
z11 
v1
i1
z 21 
v2
i1
z 22 
 R B r   5M 1M  833 k | z12 
i 2 0





i2  0
v2
i2
30

 R C r o  1.5M 51.5M  1.46 M
i1 0
20.0010.002 1  0. 0157.5  2.11x10 3 S
1
1
 V DS
 7.5
v

0.015
ro 

 37.1k | z11  1
I DS
0. 002
i1
z 21 
0
i1 0
 R B r   g m  r o R C  833 k 40x10 6 51.5M 1. 5M  48.7 M
13.106
g m  2K n I DS 1  V DS  
z 22 
v1
i2
v2
i2
v2
i1
 R G  1 M | z12 
i2  0
 R D r o  3.9k 37.1k = 3.53 k
i1  0
 1M 2.11mS 3.53k  7. 45 M
i 2 0
v1
i2
0
i1  0
13.107
From Prob. 13.90: Q - Point = 131A,1.75V 
R IN  R1 R 2  430k 560k  243 k | R OUT  43k r o
 1


1.75 V
0.0133

ro 
 587k | R OUT  43k 587k  40.1k
0.131mA
At the input there is no change in voltage division
. However, at the output
the voltage division between ROUT and R 3 is much worse due to the higher
output resistance (40.1k versus 3. 98k). This is more than compensated
for by the increase in gm / I DS .
13.108
R IN  R G  6. 8M | R OUT  50k r o
ro 
50  5 V  550k | R
OUT  50k 550k  45. 8k
0.1mA
13.109
R IN  R G  6. 8M which is independent of Kn | R OUT  R D r o
 1


 5V
0.02

ro 
 550k | R OUT  50k 550k  45.8 k, also independent of Kn
0.1mA
13.110
R IN  R G  10M | R OUT  R D r o
 1


 5V
0.02

| ro 
 5.50 M
10A
R OUT  560k 5.50M  508 k
13.111
R IN  R G  1M | R OUT  R D r o
 1  7.5V


0.015

| ro 
 37.1 k
2mA
R OUT  3.9k 37.1k  3.53 k
13.112
R IN  R G  1M | R OUT  R D r o
 1  9V


0.015

| ro 
 75.7 k
1mA
R OUT  7.5k 75. 7k  6.82 k
13.113
g m  40 50A  2.00mS | r  
100
75V  10V
 50k | r o 
 1.70M
2.00mS
50A
R BB  R B r   100 k 50k  33.3k
33.3k


vth   vs 
2mS 1.70M 100k  185 vs
33. 3k  0.75k 
R th  1.70M 100k  94. 4 k
31
13.114
g m  40 2.5mA   100 mS | r  
75
50V  7.5V
 750 | r o 
 23. 0k
100mS
2.5mA
R BB  R B r   4.7k 0.75k  647
647


vth   vs 
100mS

23.0k 4.3k  336 v s
647  50 
R th  23.0k 4.3k  3.62 k
13.115


g m  2 500A / V 2 100A 1  0.02(5)  332S | r o 
50  5 V
100A
 550k
6. 8M


vth   vs 
332S 550k 50k  15.0v s
6. 8M  0.1M
R th  550 k 50k  45.8 k
13.116


g m  2 100A / V 2 10A 1  0. 02(5)  46.9S | r o 
50  5V
10A
 5.50M
10M


vth   vs 
46.9S 5.50M 560k  23.6v s
10M  0.1M 
R th  5.50M 560k  508 k
13.117
gm
 1


 9V
2
0. 015


1mA 1mA 1  0. 015(9)  710S | r o 
 75.7k
3
1mA
1M


vth   vs 
710S 75. 7k 7.5k  4.79vs
1M  0. 01M 
R th  75.7k 7.5k  6.82 k
13.118
Note: f = 1000 Hz is more nearly midband
*Problem 13.118 - Common-Emitter Amplifier - Figure 13.42
VCC 7 0 DC 5
VEE 8 0 DC -5
VS 1 0 AC 1
*VO 6 0 AC 1
*VS 1 0 AC 0
RS 1 2 330
C1 2 3 100U
RB 3 0 100K
RE 4 8 16K
C2 4 0 100U
RC 7 5 10K
C3 5 6 100U
R4 6 0 220K
Q1 5 3 4 NBJT
.OP
.AC LIN 1 1000HZ 1000HZ
32
.MODEL NBJT NPN IS=1E-16 BF=65 VA=50
.PRINT AC VM(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: IC = 2.41E-04
VCE = 3.68E+00
VM(6) = 8.148E+01 VP(6) = -1.792E+02 IM(VS) = 1.388E-04
IP(VS) = -1.793E+02
VM(3) = 9.542E-01 VP(3) = -2.259E-02 IM(C3) = 1.046E-04
A V  81.5 | R IN 
IP(C3) = -1.800E+02
VM3  0.9542 V
1
1

 6.88 k | R OUT 

 9.40 k
IM  VS 138. 8A
IM C3 106. 4A
The results are quite close. The variations are caused by small differences in the values of
the small-signal parameters due to T = 27 C and o.
13.119
IB 
5  0.7
 37.2 A | I C  65I B = 2.42 mA | I E  66I B = 2. 46 mA
10000  66 1600 
V CE  10  1000I C  1600I E  5  8.65 V
65
50  8.75
g m  40 0. 00242   0. 0968S | r  
 672 | r o 
 24.2k
0. 0968S
0.00242
R IN  R B r   10k 672  630 | R OUT  1k 24.2k  960 
 630

A V   
0.0968  1k 24.2k 220k  60.7
330  630 


Note that the gain has been reduced by 25% by the lower value of R IN. Also note that RIN and
ROUT have changed directly with the current.
13.120
Note: f = 10 kHz is more nearly midband because of the smaller resistor values.
*Problem 13.120 - Common-Emitter Amplifier - Figure 13.119
VCC 7 0 DC 5
VEE 8 0 DC -5
VS 1 0 AC 1
*VO 6 0 AC 1
*VS 1 0 AC 0
RS 1 2 330
C1 2 3 100U
RB 3 0 10K
RE 4 8 1.6K
C2 4 0 100U
RC 7 5 1K
C3 5 6 100U
R4 6 0 220K
Q1 5 3 4 NBJT
.OP
.AC LIN 1 10KHZ 10KHZ
.MODEL NBJT NPN IS=1E-16 BF=65 VA=50
.PRINT AC VM(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: IC = 2.38E-03
VCE = 3.76E+00
VM(6) = 5.942E+01 VP(6) = -1.794E+02 IM(VS) = 9.738E-04
IP(VS) = -1.795E+02
33
VM(3) = 6.787E-01 VP(3) = -2.345E-01 IM(C3) = 1.045E-03
A V  59.4 | R IN 
IP(C3) = -1.800E+02
VM 3  0. 6787V
1
1

 697  | R OUT 

 957 
IM VS 973. 8A
IM C3 1.045mA
The results are quite close. The variations are caused by small differences in the values of
the small-signal parameters due to T = 27 C and o.
13.121
IB 
5  0.7
10  66 160k 
6
 0.372 A | I C  65I B = 24.2 A | I E  66I B = 24.6 A
V CE  10  10 5 IC  1.6x10 5 I E  5  8.65 V
65
50  8.75
g m  40 24.2 A  968S | r  
 67.1k | r o 
 2.42M
968S
24.2 A
R IN  R B r   1M 67.1k  62.9k | R OUT  100k 2. 42M  96.0 k
62. 9k


A V   
968S 100k 2.42M 220k  64. 4
0. 330k  62.9k 


Note that the gain has been increased due to the higher value of RIN relative to that of RS,
but gain has been lost ecause ROUT is now larger relative to the 220k load resistor. Also
note that RIN and ROUT have changed directly with the current.
13.122
*Problem 13.122 - Common-Emitter Amplifier - Figure 13.121
VCC 7 0 DC 5
VEE 8 0 DC -5
VS 1 0 AC 1
*VO 6 0 AC 1
*VS 1 0 AC 0
RS 1 2 330
C1 2 3 100U
RB 3 0 1MEG
RE 4 8 160K
C2 4 0 100U
RC 7 5 100K
C3 5 6 100U
R4 6 0 220K
Q1 5 3 4 NBJT
.OP
.AC LIN 1 1000HZ 1000HZ
.MODEL NBJT NPN IS=1E-16 BF=65 VA=50
.PRINT AC VM(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: IC = 2.44E-05
VCE = 3.59E+00
VM(6) = 6.264E+01 VP(6) = -1.799E+02 IM(VS) = 1.467E-05
VM(3) = 9.952E-01 VP(3) = 9.552E-04 IM(C3) = 1.046E-05
A V  62.6 | R IN 
34
IP(VS) = -1.799E+02
IP(C3) = -1.800E+02
VM3  0.9952 V
1
1

 67.8 k | R OUT 

 95.6 k
IM  VS 14. 67A
IM C3 10.46A
The results are quite close. The variations are caused by small differences in the values of
the small-signal parameters due to T = 27 C and o.
13.123 AV = - gmR; We need to increase gm by a factor or 10. Since gm is proportional to the
square root of K , K must increase by a factor of 100: K = 50 mA/V2.
n
V GS  V TN 
n
2I DS
 1
K n 1  V DS 
n
2 250A 
 1. 096 V | V GS  V TN  0. 096 V
0.05A 1  0.0167 5
(Not really strong inversion) | V GS 
R G1
10  5V  R G1  842 k
R G1  3M
13.124 (a)

V 
5  0.7
5  0.7
 F   FO 1  CE  | I C   F
| I E   F  1
V A 
100000   F  116000
100000   F  116000

V CE  10  10000I C  16000I E
Iterative procedure: Pick VCE, then calculate F, IC, IE and find the updated value of VCE.
Results: IC = 243 A VCE = 3.62 V compared to IC = 241 A VCE = 3.67 V
(b)
*Problem 13.124 - Common-Emitter Amplifier - Figure P13.42
*Monte Carlo Analysis for the Q-Point
VEE 8 0 DC -5
VCC 7 0 DC 5
RC 7 5 RRAN 10K
RB 3 0 RRAN 100K
RE 4 8 RRAN 16K
Q1 5 3 4 NBJT
.OP
.DC VCC 5 5 0.01
.MODEL NBJT NPN IS=1E-16 BF=65 VA=50
.MODEL RRAN RES R=1 DEV 10%
.MC 500 DC IC(Q1) YMAX OUTPUT(ALL)
*.MC 500 DC VCE(Q1) YMAX OUTPUT(ALL)
.END
Results: Nominal IC = 241 A Mean IC = 240 A
Nominal VCE = 3.68 V Mean VCE = 3.68 V
 = 12.5 A
 = 0.185 V
3limits: 240 ± 37.5 A
3limits: 3.68 ± 0.555 V 
Adding VA to the hand calcualtions makes only a small change in our estimate of the Q-point,
and the change is far smaller than that caused by the circuit tolerances.
The added
complexity in the hand calculations is not worth the extra effort.
13.125
*Problem 13.125 - Common-Emitter Amplifier - Figure P13.42
*Monte Carlo Analysis
*Generate VCC and VEE with 5% Tolerances
ICC 0 9 DC 5
RCC 9 0 RRA 1
ECC 7 0 9 0 1
*
IEE 10 0 DC 5
REE 10 0 RRA 1
EEE 8 0 10 0 1
35
*
VS 1 0 AC 1
RS 1 2 RRB 330
C1 2 3 100U
RC 7 5 RRB 10K
RB 3 0 RRB 100K
RE 4 8 RRB 16K
C2 4 0 100U
C3 5 6 100U
R4 6 0 RRB 220K
Q1 5 3 4 NBJT
.OP
.AC LIN 1 1000 1000
.PRINT AC VM(6) VP(6)
.MODEL NBJT NPN IS=1E-16 BF=70 DEV 42.86% VA=65 DEV 23.08%
.MODEL RRA RES R=1 DEV 5%
.MODEL RRB RES R=1 DEV 10%
.MC 250 AC VM(6) YMAX OUTPUT(ALL)
.END
Results: Nominal Gain = 82.9 Mean Gain = 82.3
13.126
 = 7.22
3limits: 82.3 ± 21.7
*Problem 13.126 - Common-Source Amplifier - Figure 13.52
VDD 7 0 DC 10
VS 1 0 AC 1
*VO 6 0 AC 1
*VS 1 0 AC 0
RS 1 2 10K
C1 2 3 100U
RG1 3 0 2MEG
RG2 3 4 2MEG
C2 4 0 100U
RG3 4 5 1MEG
RD 7 5 20K
C3 5 6 100U
R3 6 0 100K
M1 5 3 0 0 NFET
.OP
.MODEL NFET NMOS KP=500U VTO=1 LAMBDA=0.0167
.AC LIN 1 1000HZ 1000HZ
.PRINT AC VM(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: IDS = 2.53E-04
VDS = 4.92E+00
VM(6) = 7.983E+00 VP(6) = -1.800E+02 IM(VS) = 9.901E-07
VM(3) = 9.901E-01 VP(3) = 9.370E-05 IM(C3) = 5.491E-05
A V  79.8 | R IN 
IP(VS) = -1.800E+02
IP(C3) = -1.800E+02
VM3
0.9901V
1
1

 1. 00 M | R OUT 

 18.2 k
IM VS 0.9901A
IM C3  54. 91A
The results are almost identical.
13.127
36
*Problem 13.127 - Common-Source Amplifier - Figure P13.90
VDD 7 0 DC 10
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 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(VS) IP(VS)
*.PRINT AC IM(C2) IP(C2)
.END
Results: IDS = 1.31E-04
VDS = 1.73E+00
VM(6) = 1.044E+01 VP(6) = -1.800E+02 IM(VS) = 4.094E-06
VM(3) = 9.959E-01 VP(3) = 3.734E-04 IM(C3) = 2.496E-05
A V  10.4 | R IN 
IP(VS) = -1.800E+02
IP(C3) = -1.800E+02
VM 3  0.9959V
1
1

 243 k | R OUT 

 40.1 k
IM VS 4.094 A
IM C3  24.96A
13.128 For the JFET model,  = IDSS/VP2.
*Problem 13.128 - Common-Source Amplifier - Figure P13.59
VDD 7 0 DC 12
*FOR OUTPUT RESISTANCE
*VO 6 0 AC 1
*VS 1 0 AC 0
*
VS 1 0 AC 1
RS 1 2 1K
C1 2 3 100U
RG 3 0 1MEG
R1 4 0 2K
C2 4 0 100U
RD 7 5 27K
C3 5 6 100U
R3 6 0 100K
J1 5 3 4 NFET
.OP
.MODEL NFET NJF BETA=1M VTO=-1 LAMBDA=0.02
.AC LIN 1 1000 1000
.PRINT AC VM(6) VDB(6) VP(6) IM(VS) IP(VS) VM(3) VP(3)
*.PRINT AC IM(C3) IP(C3)
.END
Results: IDS = 2.57E-04
VDS = 4.54E+00
VM(6) = 2.045E+01 VP(6) = -1.799E+02 IM(VS) = 9.990E-07
IP(VS) = -1.800E+02
37
VM(3) = 9.990E-01 VP(3) = 9.110E-05 IM(C3) = 4.175E-05
A V  20.5 | R IN 
VM3
0.9990 V
1
1

 1.00 M | R OUT 

 24. 0 k
IM VS 0.9990A
IM C3 41. 75A
The results are almost identical.
38
IP(C3) = -1.800E+02
13.129
I B  3.71A
IC  241A I E  245A V CE  3.67V
2
2
P R B  I 2BR B  3.71A  100k  1. 38 W | P R C  I 2CR C  241A 10k  0. 581 mW
P R E  I 2ER E  245A  16k  0. 960 mW
2
P BJT  I C V CE  I B V BE  241A 3.67V   3.71A 0.7V   0. 887 mW
P S  5V 241A  5V 245A  2.43 mW | P S  P R B  P R C  P R E  P Q
13.130
I DS  250A V DS  4.75V
P JFET  I DS V DS  250A 4.75V   1.19 mW | P R D  I2DSR D  250A 2 27k  1.69 mW
P R 4  I 2DS R 4  250A  2k  0.125 mW | P R G  0
2
P S  12V 250A   3.00 mW | P JFET  P R D  P R 4  P R G  3. 00 mW
13.131 Use values from Prob. 13.9
I B  23.8A
PR1 
V12
R1
IC  1.78mA
I E  1.81mA
4.09  (12)
2

5000
V
2
V CE  6.08 V V B  4.09V
 12.5 mW | P R 2 
2
V 22 12  (4. 09) V 2

 25.9 mW
R2
10000
P R C  I2C R C  1. 78mA 2 6k   19.0 mW | P R E  I 2ER E  1.81mA 2 4k  13.0 mW
P BJT  I C V CE  I B V BE  1.78mA 6. 08V   23.8A0.7V   10.8 mW

12  (4. 09)V 

4. 09  (12)V 
P S  12V 1.78mA 
12 V 1.81mA 
 81.3 mW






10000
5000
P S  P R1  PR 2  P R C  P R E  PBJT  81. 2 mW
13.132 Use values from Prob. 13.13
I B  1.51A
PRB 
I 2BR B
IC  98. 4A
IE  99. 9A V CE  6.96 V
2
2
 1.51A  3k  6. 84 nW | P R C  I2C R C  98.4A 33k  0.320 mW
P R E  I 2ER E  99.9A  68k  0.679 mW
2
P BJT  I C V CE  I B V BE  98.4A 6.96 V   1.51A0.7V   0.489 mW
P S  7.5 98. 4A   7. 5V99.9A  1.49 mW | P S  P R B  P R C  P R E  PBJT  1. 49 mW
13.133 Use values from Prob. 13.17
39
I DS  82.2A V DS  6.04V
P FET  I DS V DS  82.2A 6.04 V   0. 497 mW | P R D  I 2DSR D  82. 2A 2 82k  0.554 mW
P R 4  I 2DS R 4  82.2A  27k  0.182 mW | I 2 
2
15V
 4.05A
3.7M
P R  I 22 R1  4.05A 2 1M  16.4 W | P R  I22 R 2  4.05A 2 2.7M  44.3 W
1
2
P S  15 V82.2A  4.05A   1.29 mW | P FET  P R D  P R 4  P R1 + P R 2  1. 29 mW
13.134 Use values from Prob. 13.21
I SD  307A V SD  3. 88V
P FET  I SD V SD  307A 3.88V   1.19 mW | P R D  I2SD R D  307A 2 24k  2.26 mW
P R 4  I 2SD R 4  307A  22k  2.07 mW | I 2 
2
18V
 2.73A
6.6M
P R  I 22 R1  2.73A 2 3.3M  24. 6 W | P R  I 22 R 2  2.73A 2 3.3M  24.6 W
1
2
P S  18 V307A  2. 73A   5.58 mW | P FET  P R D  P R 4  P R 1 + P R 2  5. 57 mW
13.135 Use values from Prob. 13.25
I DS  1.25mA V DS  10.6V
2
P JFET  I DS V DS  1.25mA 10. 6V   13.3 mW | P R D  I DSR D  1. 25mA  3.9k  6.09 mW
2
2
P R 4  I 2DS R 4  1.25mA  2k  3.13 mW | P R G  0
P S  18 V1.25mA   22.5 mW | P JFET  P R D  P R 4  P R G  22.5 mW
13.136
IC 
V CC
V
V
| i c  0.2I C | v c  icR C  0.2 CC R C  CC
3R C
3R C
15
13.137
ids  0. 4I DS | I DS 
500x10 6
2
0  1.5  563 A
2
v gs  0.2V GS  V TN   0.2 0  1.5   0.3 V
v ds  ids R D  0.4 563A 15k  3.38 V
To insure saturation: v DS  v GS  V TN  v gs  V TN  0. 3  (1.5)  1.8 V
V DD  1.8  3.38  563A 15k  13.6 V
13.138
vo 
13.139
40
V 2CC
8R L
2
 V  V 2
V CC
1  V  1
V2
| P dc  V CC  CC   CC | P ac   CC 
 CC |  =100% 2  25 %
2
2  2  R L 8R L
V CC
2R L  2R L
2R L
The Q- point from problem 13.9 is (1.78mA ,6.08 V).
ic  0.2IC  0.356 mA | v c  0. 2I CR L  0.356mA 100k 6k  2. 02V
13.140
The Q- point from problem 13.9 is (371A, 2.72V).




v o  g m v be r o R C R 3  40371A 5mV   13k 100k  0.854 V
13.141
The Q- point from Problem 13.17 is (82. 2A,6. 04V).
ids  0. 4I DS  32.9A | v ds  0.4I DSR L  32. 9A  82k 420k  2. 30V
13.142
The Q- point from Problem 13.21 is (307A ,3.88V).
isd  0. 4I SD  307A | v o  0.4I SDR L  307A 24k 470k  2.80V
Checking the bias point: V R D  307A24k  7. 37V | 2. 80 < 7. 37 & 2. 80 < 3.88 -1
13.143
The Q- point from Problem 13.25 is (1. 25mA ,10. 6V)
ids  0. 4I DS  0.500mA | v ds  0.4I DS R L  0.500mA  3. 9k 36k  1.76 V
Checking the bias point: I DS R D  1.25mA 3.9k  4.88V & V DS  10.6 V
13.144
The Q- point from Problem 13.27 is (1. 00mA,7.50 V).
isd  0. 4I SD  0.400A | v o  0. 4I SDR L  307A  7.5k 2200 k  2. 90V
Checking the bias point: V R D  1.25mA 7.5k  9.38V | 2.90 < 9.38 & 2.90 < 7.50 - 3
13.145
V CE  20  20000I C : Two points on the load line (0mA ,20 V) , (1mA ,0V)
At IB = 2A, the maximum swing is approximately 2.5 V limited by V R C .
For I B  5A ,the maximum swing is approximately - 8.5 V limited by V CE .
41
IC
I B= 10 A
1000 A
I = 8 A
B
750 A
I = 6 A
B
500 A
I B = 5 A
I = 4 A
B
250 A
8.5 V
9V
I B= 2 A
V
0
CE
0
42
10 V
2.5 V
20 V