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Chapter 1
Exercise Problems
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
EX1.1
GaAs: ni = ( 2.1× 1014 ) ( 300 )
Ge: ni = (1.66 × 1013 ) ( 300 )
3/ 2
3/ 2
⎛
⎞
−1.4
⎟ or ni = 1.8 × 106 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠
⎛
⎞
−0.66
⎟ or ni = 2.40 × 1013 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠
EX1.2
(a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons,
n 2 (1.5 × 10
no = i =
1017
po
)
10 2
= 2.25 × 103 cm −3
(b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes,
n 2 (1.5 × 10 )
po = i =
= 4.5 × 104 cm −3
no
5 × 1015
10 2
EX1.3
For n-type, drift current density J ≅ eμ n nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields
E = 17.9 V / cm
EX1.4
Diffusion current density due to holes:
dp
J p = −eD p
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟
⎜L ⎟
⎜L ⎟
⎝ p⎠
⎝ p⎠
(a) At x = 0
(1.6 ×10 ) (10 ) (10 ) = 16 A / cm
=
−19
16
10−3
−3
(b) At x = 10 cm
Jp
2
⎛ −10−3 ⎞
J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2
⎝ 10 ⎠
EX1.5
⎡N N
Vbi = VT ln ⎢ a 2 d
⎣ ni
EX1.6
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
and
⎡ (1016 )(1017 ) ⎤
⎤
⎢
⎥ or Vbi = 1.23 V
=
0.026
ln
(
)
⎥
6 2 ⎥
⎢
⎦
⎣ (1.8 × 10 ) ⎦
−1/ 2
⎡N N ⎤
Vbi = VT ln ⎢ a 2 d ⎥
⎣ ni ⎦
⎡ (1017 )(1016 ) ⎤
⎥ = 0.757 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
5 ⎞
⎛
Then 0.8 = C jo ⎜ 1 +
⎟
⎝ 0.757 ⎠
or
C jo = 2.21 pF
−1/ 2
= C jo ( 7.61)
−1/ 2
EX1.7
⎡
⎛v ⎞ ⎤
iD = I S ⎢exp ⎜ D ⎟ − 1⎥
⎝ VT ⎠ ⎥⎦
⎣⎢
⎡
⎛ v ⎞ ⎤
so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥
⎝ 0.026 ⎠ ⎦
⎣
⎡ 10−3
⎤
Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥
⎣10
⎦
or
vD ≅ ( 0.026 ) ln (1010 )
which yields
vD = 0.599 V
EX1.8
⎛V ⎞
VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
( 4 − VD )
so 4 = I D ( 4 ×103 ) + VD ⇒ I D =
4 ×103
and
⎛ V ⎞
I D = (10 −12 ) exp ⎜ D ⎟
⎝ 0.026 ⎠
By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V
EX1.9
(a)
ID =
(b)
ID =
Then R =
(c)
VPS − Vγ
R
VPS − Vγ
R
5 − 0.7
⇒ I D = 1.08 mA
4
VPS − Vγ
⇒R=
ID
=
8 − 0.7
= 6.79 kΩ
1.075
ID(mA)
Diode curve
1.25
1.08
Load lines
(b)
(a)
0
0.7
2
4
VD(v)
6
8
EX1.10
PSpice analysis
EX1.11
Quiescent diode current I DQ =
VPS − Vγ
=
10 − 0.7
= 0.465 mA
20
R
Time-varying diode current:
V
0.026
We find that rd = T =
= 0.0559 kΩ
I DQ 0.465
Then id =
vI
0.2sin ω t (V )
=
⋅
or id = 9.97sin ω t ( μ A)
rd + R 0.0559 + 20 ( kΩ )
EX1.12
⎛I ⎞
⎛ 1.2 × 10−3 ⎞
or VD = 0.6871 V
For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 4 × 10 ⎠
⎝ IS ⎠
The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V
⎛V ⎞
Now I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
or
1.2 × 10−3
IS =
⇒ I S = 1.07 × 10−10 A
0.4221
⎛
⎞
exp ⎜
⎟
⎝ 0.026 ⎠
EX1.13
P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA
Also I =
10 − 5.6
= 1.79 ⇒ R = 2.46 kΩ
R
Test Your Understanding Exercises
TYU1.1
(a) T = 400K
⎛ − Eg ⎞
Si: ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
ni = ( 5.23 × 1015 ) ( 400 )
or
ni = 4.76 × 1012 cm −3
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
Ge: ni = (1.66 × 1015 ) ( 400 )
or
ni = 9.06 × 1014 cm −3
GaAs:
ni = ( 2.1× 1014 ) ( 400 )
3/ 2
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
⎡
⎤
−1.4
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
or
ni = 2.44 × 109 cm −3
(b) T = 250 K
Si: ni = ( 5.23 × 1015 ) ( 250 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 1.61× 108 cm −3
Ge: ni = (1.66 × 1015 ) ( 250 )
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
or
ni = 1.42 × 1012 cm −3
GaAs: ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 6.02 × 103 cm −3
TYU1.2
(a)
n = 5 × 1016 cm −3 , p <<< n, so σ ≅ eμ n n = (1.6 × 10 −19 ) (1350 ) ( 5 × 1016 )
σ = 10.8 ( Ω − cm )
or
(b)
−1
p = 5 × 1016 cm −3 , n <<< p, so σ ≅ eμ p p = (1.6 × 10−19 ) ( 480 ) ( 5 × 1016 )
σ = 3.84 ( Ω − cm )
or
−1
TYU1.3
J = σ E = (10 )(15 ) or J = 150 A / cm2
TYU1.4
(a)
J n = eDn
or
J n = 202 A / cm 2
(b)
J p = −eD p
or
J p = −24.5 A / cm2
TYU1.5
(
)
⎛ 1015 − 1016 ⎞
Δn
dn
so J n = 1.6 × 10−19 ( 35 ) ⎜
= eDn
−4 ⎟
Δx
dx
⎝ 0 − 2.5 × 10 ⎠
(
)
⎛ 1014 − 5 × 1015 ⎞
Δp
dp
so J p = − 1.6 × 10−19 (12.5 ) ⎜
= −eD p
−4 ⎟
Δx
dx
⎝ 0 − 4 × 10 ⎠
no = N d = 8 × 1015 cm −3
10
n 2 (1.5 × 10 )
po = i =
= 2.81× 10 4 cm −3
no
8 × 1015
(a)
2
(b) n = no + δ n = 8 × 1015 + 0.1× 1015
or
n = 8.1×1015 cm−3
p = po + δ p = 2.81 × 10 4 + 1014
or
p ≅ 1014 cm −3
TYU1.6
(a)
⎡ (1015 )(1017 ) ⎤
⎡N N ⎤
⎥ = 0.697 V
Vbi = VT ln ⎢ a 2 d ⎥ so Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣ ni ⎦
⎣
⎦
⎡ (1017 )(1017 ) ⎤
⎥ = 0.817 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
(b)
TYU1.7
(a)
⎡
⎛V
I D = I S ⎢exp ⎜ D
⎝ VT
⎣⎢
⎞ ⎤
⎟ − 1⎥
⎠ ⎦⎥
⎛ 0.5 ⎞
I D ≅ 10−14 exp ⎜
⎟
⎝ 0.026 ⎠
Then, for
VD = 0.5 V, I D = 2.25 μ A
VD = 0.6 V, I D = 0.105 mA
VD = 0.7 V, I D = 4.93 mA
(b)
I D ≅ − I S = −10 −14 A
for both cases.
TYU1.8
ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV
Then VD = 0.650 − 0.20 = 0.450 V
TYU1.9
ID(mA)
Diode
1.0
艐0.87
Load line
0
1
艐0.54v
2
VD(v)
VD
ID
0.45
0.50
0.55
0.033
0.225
1.54
3
TYU1.10
P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA
4
Now R =
VPS − Vγ
ID
=
10 − 0.7
⇒ R = 6.2 kΩ
1.5
TYU1.11
I
0.8
gd = D =
= 30.8 mS
VT 0.026
TYU1.12
V
0.026
0.026
rd = T ⇒ 50 =
⇒ ID =
ID
ID
50
or
I D = 0.52 mA
TYU1.13
For the pn junction diode,
4 − 0.7
ID =
= 0.825 mA
4
For the Schottky diode, I D =
4 − 0.3
= 0.925 mA
4
TYU1.14
Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10 −3 ) ( 20 ) = 5.18 V
Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V
Chapter 1
Problem Solutions
1.1
− E / 2 kT
ni = BT 3 / 2 e g
(a)
Silicon
(i)
ni = ( 5.23 × 1015 ) ( 250 )
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
= 2.067 × 1019 exp [ −25.58]
3/ 2
ni = 1.61× 108 cm −3
(ii)
ni = ( 5.23 × 1015 ) ( 350 )
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= 3.425 × 1019 exp [ −18.27 ]
3/ 2
ni = 3.97 ×1011 cm −3
(b)
(i)
ni = ( 2.10 × 1014 ) ( 250 )
GaAs
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
= ( 8.301× 1017 ) exp [ −32.56]
ni = 6.02 × 103 cm −3
(ii)
ni = ( 2.10 × 1014 ) ( 350 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= (1.375 × 1018 ) exp [ −23.26]
ni = 1.09 × 108 cm −3
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
⎛
⎞
−1.1
1012 = 5.23 × 1015 T 3 / 2 exp ⎜
⎟
−6
×
2(86
10
)(
T
)
⎝
⎠
1.2
a.
⎛ 6.40 × 103 ⎞
1.91×10−4 = T 3 / 2 exp ⎜ −
⎟
T
⎝
⎠
By trial and error, T ≈ 368 K
b.
ni = 109 cm −3
⎛
⎞
−1.1
⎟
109 = 5.23 × 1015 T 3 / 2 exp ⎜
⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝
⎠
⎛ 6.40 × 103 ⎞
1.91× 10−7 = T 3 / 2 exp ⎜ −
⎟
T
⎝
⎠
By trial and error, T ≈ 268° K
1.3
Silicon
ni = ( 5.23 × 1015 ) (100 )
(a)
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦
= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3
ni = ( 5.23 × 1015 ) ( 300 )
(b)
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 300 ) ⎥⎦
= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3
ni = ( 5.23 × 1015 ) ( 500 )
(c)
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦
= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3
ni = (1.66 × 1015 ) (100 )
Germanium.
(a)
3/ 2
ni = 35.9 cm −3
ni = (1.66 × 1015 ) ( 300 )
(b)
3/ 2
⎡
⎤
−0.66
⎥ = ( 8.626 × 1018 ) exp [ −12.79]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 300 ) ⎥⎦
3/ 2
⎡
⎤
−0.66
⎥ = (1.856 × 1019 ) exp [ −7.674]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦
ni = 2.40 × 1013 cm −3
ni = (1.66 × 1015 ) ( 500 )
(c)
ni = 8.62 ×1015 cm −3
1.4
a.
⎡
⎤
−0.66
⎥
= (1.66 × 1018 ) exp [ −38.37 ]
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦
N d = 5 × 1015 cm −3 ⇒ n − type
n0 = N d = 5 × 1015 cm −3
n 2 (1.5 × 10 )
p0 = i =
⇒ p0 = 4.5 × 10 4 cm −3
5 × 1015
n0
10 2
N d = 5 × 1015 cm −3 ⇒ n − type
b.
no = N d = 5 × 1015 cm −3
ni = ( 2.10 × 1014 ) ( 300 )
= ( 2.10 × 1014 ) ( 300 )
= 1.80 × 106 cm −3
3/ 2
3/ 2
⎛
⎞
−1.4
exp ⎜
⎟
−6
⎝ 2(86 × 10 )(300) ⎠
(1.65 ×10 )
−12
6
ni2 (1.8 × 10 )
=
⇒ p0 = 6.48 × 10 −4 cm −3
p0 =
15
n0
5 × 10
2
1.5
(a)
(b)
n-type
no = N d = 5 × 1016 cm −3
10
n 2 (1.5 × 10 )
po = i =
= 4.5 × 103 cm −3
no
5 × 1016
2
no = N d = 5 × 1016 cm −3
(c)
From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3
( 3.97 × 10 )
=
= 3.15 × 106 cm −3
11 2
5 × 1016
po
1.6
a.
N a = 1016 cm −3 ⇒ p − type
p0 = N a = 1016 cm −3
n 2 (1.5 × 10 )
⇒ n0 = 2.25 × 10 4 cm −3
n0 = i =
p0
1016
b.
Germanium
N a = 1016 cm −3 ⇒ p − type
10 2
p0 = N a = 1016 cm −3
ni = (1.66 × 1015 ) ( 300 )
= (1.66 × 1015 ) ( 300 )
3/ 2
3/ 2
= 2.4 × 1013 cm −3
⎛
⎞
−0.66
⎟
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠
( 2.79 ×10 )
−6
13
n 2 ( 2.4 × 10 )
n0 = i =
⇒ n0 = 5.76 × 1010 cm −3
p0
1016
2
1.7
(a)
(b)
p-type
po = N a = 2 × 1017 cm −3
10
ni2 (1.5 × 10 )
no =
=
= 1.125 × 103 cm −3
po
2 × 1017
2
(c)
po = 2 × 1017 cm −3
From Problem 1.1(a)(i) ni = 1.61 × 108 cm −3
no
1.8
(a)
(1.61×10 )
=
2 × 10
8 2
17
= 0.130 cm −3
no = 5 × 1015 cm −3
10
ni2 (1.5 × 10 )
po =
=
⇒ po = 4.5 × 104 cm −3
no
5 × 1015
2
(b)
(c)
no po ⇒ n-type
no ≅ N d = 5 × 1015 cm −3
1.9
Add Donors
a.
N d = 7 × 1015 cm −3
Want po = 106 cm −3 = ni2 / N d
So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21
b.
⎛ − Eg ⎞
= B 2T 3 exp ⎜
⎟
⎝ kT ⎠
⎛
⎞
2
−1.1
⎟
7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜
6
−
⎜ ( 86 × 10 ) (T ) ⎟
⎝
⎠
By trial and error, T ≈ 324° K
1.10
I = J ⋅ A = σ EA
I = ( 2.2 )(15 ) (10−4 ) ⇒ I = 3.3 mA
1.11
J =σE ⇒σ =
J 85
=
E 12
σ = 7.08 (ohm − cm) −1
1.12
g≈
1
1
1
⇒ Na =
=
eμ p N a
eμ p g (1.6 × 10 −19 ) ( 480 )( 0.80 )
N a = 1.63 × 10 16 cm −3
1.13
σ = eμ n N d
Nd =
σ
eμ n
=
( 0.5)
(1.6 ×10 ) (1350 )
−19
N d = 2.31× 1015 cm −3
1.14
(a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d
For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm )
(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 × 103 A / cm2
1.15
J n = eDn
Δn
dn
= eDn
Δx
dx
⎡10 15 −10 2 ⎤
= (1.6 × 10 −19 ) (180 ) ⎢
−4 ⎥
⎣ 0.5 × 10 ⎦
J n = 576 A/cm 2
1.16
−1
J p = −eD p
dp
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟
⎜ Lp ⎟
⎜ Lp ⎟
⎝ ⎠
⎝ ⎠
Jp =
(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞
−19
J p = 2.4 e
10 × 10 −4
− x / Lp
⎜⎜ ⎟⎟
⎝ Lp ⎠
15
J p = 2.4 A/cm2
(a)
x=0
(b)
(c)
x = 30 μ m
1.17
a.
N a = 1017 cm −3 ⇒ po = 1017 cm −3
x = 10 μ m
J p = 2.4 e−1 = 0.883 A/cm 2
J p = 2.4 e−3 = 0.119 A/cm 2
6
ni2 (1.8 × 10 )
no =
=
⇒ no = 3.24 × 10−5 cm −3
po
1017
2
n = no + δ n = 3.24 × 10 −5 + 1015 ⇒ n = 1015 cm −3
b.
p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3
1.18
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
(a)
(b)
(c)
⎡ (10 16 )(10 16 ) ⎤
⎥ = 0.697 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦
⎡ (10 18 )(10 16 ) ⎤
⎥ = 0.817 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦
⎡ (10 18 )(10 18 ) ⎤
⎥ = 0.937 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 10 10 )2 ⎥
⎣
⎦
1.19
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
a.
b.
c.
⎡ (1016 )(1016 ) ⎤
⎥ ⇒ Vbi = 1.17 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦
⎡ (1018 )(1016 ) ⎤
⎥ ⇒ Vbi = 1.29 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦
⎡ (1018 )(1018 ) ⎤
⎥ ⇒ Vbi = 1.41 V
Vbi = ( 0.026 ) ln ⎢
6 2
⎢⎣ (1.8 × 10 ) ⎥⎦
⎡ N a (1016 ) ⎤
⎛ Na Nd ⎞
⎥
Vbi = VT ln ⎜ 2 ⎟ = ( 0.026 ) ln ⎢
10 2
⎢⎣ (1.5 × 10 ) ⎥⎦
⎝ ni ⎠
1.20
For N a = 1015 cm −3 , Vbi = 0.637 V
For N a = 1018 cm −3 , Vbi = 0.817 V
Vbi (V)
0.817
0.637
1015
1016
1017
1018 Na(cm⫺3)
1.21
⎛ T ⎞
kT = (0.026) ⎜
⎟
⎝ 300 ⎠
T
kT
(T)3/2
200
0.01733
2828.4
250
0.02167
3952.8
300
0.026
5196.2
350
0.03033
6547.9
400
0.03467
8000.0
450
0.0390
9545.9
500
0.04333
11,180.3
⎛
⎞
−1.4
⎟
ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜
−
6
⎜ 2 ( 86 × 10 ) (T ) ⎟
⎝
⎠
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
T
200
250
300
350
400
450
500
ni
1.256
6.02 × 103
1.80 × 106
1.09 × 108
2.44 × 109
2.80 × 1010
2.00 × 1011
Vbi
1.405
1.389
1.370
1.349
1.327
1.302
1.277
Vbi (V)
1.45
1.35
1.25
200
1.22
250
300
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
−1/ 2
350
400
450
500 T(⬚C)
⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤
⎥ = 0.684 V
Vbi = ( 0.026 ) ln ⎢
⎢⎣
(1.5 ×10 10 ) 2 ⎥⎦
(a)
(b)
(c)
1.23
(a)
1 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠
−1/ 2
5 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
0.684
⎝
⎠
−1/ 2
3 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
−1/ 2
= 0.255 pF
= 0.172 pF
= 0.139 pF
−1 / 2
5 ⎞
⎛
For VR = 5 V, C j = (0.02) ⎜ 1 +
⎟
⎝ 0. 8 ⎠
−1 / 2
= 0.00743 pF
⎛ 1. 5 ⎞
For VR = 1.5 V, C j = (0.02) ⎜1 +
= 0.0118 pF
⎟
⎝ 0. 8 ⎠
0.00743 + 0.0118
C j (avg ) =
= 0.00962 pF
2
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ
−1 / 2
where
τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 )
τ = 4.52 ×10−10 s
or
Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e − ti / τ
5
+ r /τ
⎛ 5 ⎞
= e 1 ⇒ t1 = τ ln ⎜ ⎟
1.5
⎝ 1.5 ⎠
−10
t1 = 5.44 × 10 s
(b)
For VR = 0 V, Cj = Cjo = 0.02 pF
−1/ 2
⎛ 3.5 ⎞
For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 +
= 0.00863 pF
⎟
⎝ 0.8 ⎠
0.02 + 0.00863
C j (avg ) =
= 0.0143 pF
2
τ = RC j ( avg ) = 6.72 ×10−10 s
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ
(
3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ
so that t2 = 8.09 × 10
−10
)
s
⎡ (1018 )(1015 ) ⎤
⎥ = 0.757 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
a.
VR = 1 V
1.24
1 ⎞
⎛
C j = (0.25) ⎜ 1 +
⎟
⎝ 0.757 ⎠
−1/ 2
= 0.164 pF
f0 =
1
2π LC
=
f 0 = 8.38 MHz
( 2.2 ×10 )( 0.164 ×10 )
2π
−3
1
−12
b.
VR = 10 V
−1/ 2
10 ⎞
⎛
= 0.0663 pF
C j = (0.25) ⎜ 1 +
⎟
⎝ 0.757 ⎠
1
f0 =
−3
2π ( 2.2 × 10 )( 0.0663 × 10−12 )
f 0 = 13.2 MHz
1.25
⎡
⎛V
I = I S ⎢ exp ⎜ D
⎝ VT
⎣
a.
⎞ ⎤
⎛ VD
⎟ − 1⎥ − 0.90 = exp ⎜
⎠ ⎦
⎝ VT
⎛V ⎞
exp ⎜ D ⎟ = 1 − 0.90 = 0.10
⎝ VT ⎠
VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V
b.
IF
IR
⎡
⎛ VF
⎢ exp ⎜
I
⎝ VT
= S ⋅⎣
IS ⎡
⎛ VR
⎢exp ⎜
⎝ VT
⎣
=
⎞ ⎤
⎛ 0.2 ⎞
⎟ − 1⎥ exp ⎜
⎟ −1
⎠ ⎦
⎝ 0.026 ⎠
=
⎛ −0.2 ⎞
⎞ ⎤
⎟ − 1⎥ exp ⎝⎜ 0.026 ⎠⎟ − 1
⎠ ⎦
2190
−1
IF
= 2190
IR
1.26
a.
⎛ 0.5 ⎞
I ≅ (10−11 ) exp ⎜
⎟ ⇒ I = 2.25 mA
⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−11 ) exp ⎜
⎟ ⇒ I = 0.105 A
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−11 ) exp ⎜
⎟ ⇒ I = 4.93 A
⎝ 0.026 ⎠
b.
⎛ 0.5 ⎞
I ≅ (10−13 ) exp ⎜
⎟ ⇒ I = 22.5 μ A
⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−13 ) exp ⎜
⎟ ⇒ I = 1.05 mA
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−13 ) exp ⎜
⎟ ⇒ I = 49.3 mA
⎝ 0.026 ⎠
1.27
(a)
150 × 10
(
)
I = I S eVD / VT − 1
−6
= 10
−11
(e
VD / VT
)
− 1 ≅ 10−11 eVD / VT
⎞
⎟ −1
⎠
⎛ 150 × 10−6
Then VD = VT ln ⎜
−11
⎝ 10
Or VD = 0.430 V
⎞
⎛ 150 × 10−6 ⎞
⎟ = (0.026) ln ⎜
⎟
−11
⎠
⎝ 10
⎠
(b)
⎛ 150 × 10−6 ⎞
VD = VT ln ⎜
⎟
−13
⎝ 10
⎠
Or VD = 0.549 V
1.28
⎛ 0.7 ⎞
10−3 = I S exp ⎜
⎟
⎝ 0.026 ⎠
I S = 2.03 × 10 −15 A
(b)
VD
I D ( A ) ( n = 1)
(a)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1.29
(a)
I S = 10 −12 A
VD(v)
0.10
0.20
0.30
0.40
0.50
0.60
0.70
(b)
I S = 10 −14 A
VD(v)
0.10
0.20
0.30
0.40
0.50
0.60
0.70
I D ( A )( n = 2 )
9.50 ×10 −14
4.45 ×10 −12
2.08 ×10 −10
9.75 ×10 −9
4.56 ×10 −7
2.14 ×10 −5
10 −3
1.39 ×10 −14
9.50 ×10 −14
6.50 ×10 −13
4.45 ×10 −12
3.04 ×10 −11
2.08 ×10 −10
1.42 ×10 −9
ID(A)
4.68 ×10−11
2.19 ×10−9
1.03 ×10−7
4.80 ×10−6
2.25 ×10−4
1.05 ×10−2
4.93 ×10−1
log10ID
−10.3
−8.66
−6.99
−5.32
−3.65
−1.98
−0.307
ID(A)
4.68 ×10−13
2.19 ×10−11
1.03 ×10−9
4.80 ×10−8
2.25 ×10−6
1.05 ×10−4
4.93 ×10−3
log10ID
−12.3
−10.66
−8.99
−7.32
−5.65
−3.98
−2.31
1.30
a.
⎛ V − VD1 ⎞
ID2
= 10 = exp ⎜ D 2
⎟
I D1
⎝ VT
⎠
ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV
b.
1.31
(a) (i)
(ii)
(b) (i)
(ii)
(iii)
(iv)
1.32
ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV
⎛I
VD = Vt ln ⎜ D
⎝ IS
VD = 0.669 V
⎞
⎛ 150 × 10−6 ⎞
⎟ = ( 0.026 ) ln ⎜
⎟
−15
⎝ 10
⎠
⎠
⎛ 25 × 10−6 ⎞
VD = ( 0.026)ln ⎜
⎟
−15
⎝ 10
⎠
VD = 0.622 V
⎛ 0.2 ⎞
−12
I D = (10−15 )exp ⎜
⎟ = 2.19 × 10 A
⎝ 0.026 ⎠
ID = 0
I D = −10 −15 A
I D = −10 −15 A
⎛I
VD = Vt ln ⎜ D
⎝ IS
⎞
⎛ 2 × 10−3 ⎞
=
= 0.6347 V
.
(0
026)
ln
⎟
⎜
−14 ⎟
⎝ 5 × 10 ⎠
⎠
⎛ 2 × 10−3 ⎞
= 0.5150 V
VD = (0.026) ln ⎜
−12 ⎟
⎝ 5 × 10 ⎠
0.5150 ≤ VD ≤ 0.6347 V
1.33
⎛V ⎞
I D = I S exp ⎜ D ⎟
⎝ Vt ⎠
⎛ 1.10 ⎞
−21
12 ×10−3 = I S exp ⎜
⎟ ⇒ I S = 5.07 × 10 A
⎝ 0.026 ⎠
(a)
(b)
1.34
(a)
(b)
(c)
⎛ 1.0 ⎞
I D = ( 5.07 × 10−21 ) exp ⎜
⎟
⎝ 0.026 ⎠
I D = 2.56 × 10−4 A = 0.256 mA
⎛ 1.0 ⎞
−7
I D = 10−23 exp ⎜
⎟ = 5.05 × 10 A
⎝ 0.026 ⎠
⎛ 1.1 ⎞
−5
I D = 10−23 exp ⎜
⎟ = 2.37 × 10 A
0
.
026
⎝
⎠
1
.
2
⎛
⎞
−3
I D = 10−23 exp ⎜
⎟ = 1.11× 10 A
⎝ 0.026 ⎠
1.35
IS doubles for every 5C increase in temperature.
I S = 10 −12 A at T = 300K
For I S = 0.5 × 10 −12 A ⇒ T = 295 K
For I S = 50 × 10 −12 A, (2) n = 50 ⇒ n = 5.64
Where n equals number of 5C increases.
Then ΔT = ( 5.64 )( 5 ) = 28.2 K
So 295 ≤ T ≤ 328.2 K
1.36
I S (T )
= 2ΔT / 5 , ΔT = 155° C
I S (−55)
I S (100)
= 2155 / 5 = 2.147 × 109
I S (−55)
VT @100°C ⇒ 373°K ⇒ VT = 0.03220
VT @ − 55°C ⇒ 216°K ⇒ VT = 0.01865
I D (100)
= (2.147 × 109 ) ×
I D (−55)
=
⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.0322 ⎠
⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.01865 ⎠
( 2.147 ×10 )(1.237 ×10 )
( 9.374 ×10 )
9
8
13
I D (100)
= 2.83 × 103
I D (−55)
1.37
3.5 = ID (105) + VD
⎛ V ⎞
⎛ ID ⎞
I D = 5 ×10−9 exp ⎜ D ⎟ ⇒ VD = 0.026 ln ⎜
−9 ⎟
⎝ 0.026 ⎠
⎝ 5 × 10 ⎠
Trial and error.
VD
ID
VD
−5
0.50
0.226
3 ×10
0.40
0.227
3.1×10−5
−
5
0.250
0.228
3.25 ×10
−5
0.229
0.2284
3.271×10
5
−
0.2285
0.2284
3.2715 ×10
(a)
So VD ≅ 0.2285 V
I D ≅ 3.272 × 10−5 A
(b)
I D = I S = 5 × 10−9 A
VR = ( 5 × 10−9 )(105 ) = 5 × 10−4 V
VD = 3.4995 V
⎛ I ⎞
10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D−12 ⎟
⎝ 10 ⎠
Trial and error.
VD(v)
ID(A)
VD(v)
0.50
0.5194
4.75 ×10−4
−4
0.517
0.5194
4.7415 ×10
0.5194
0.5194
4.740 ×10−4
1.38
VD = 0.5194 V
I D = 0.4740 mA
1.39
I s = 5 × 10 −13 A
R1 ⫽ 50 K
⫹
1.2 V
⫺
⫹
VD
⫺
R2 ⫽ 30 K
ID
RTH ⫽ R1 兩兩 R2 ⫽ 18.75 K
⫹
VTH
⫺
⫹
VD
⫺
ID
⎛ R2
VTH = ⎜
⎝ R1 + R2
⎞
⎛ 30 ⎞
⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V
80 ⎠
⎝
⎠
⎛I ⎞
0.45 = I D RTH + VD , VD = VT ln ⎜ D ⎟
⎝ IS ⎠
By trial and error:
I D = 2.56 μ A, VD = 0.402 V
1.40
⫹ VD⫺ ⫹ VD⫺
⫹
VI
⫺
I1
1K
IR
I2
⫹
V0
⫺
I S = 2 × 10 −13 A
V0 = 0.60 V
⎛V ⎞
⎛ 0.60 ⎞
I 2 = I S exp ⎜ 0 ⎟ = ( 2 × 10−13 ) exp ⎜
⎟
⎝ 0.026 ⎠
⎝ VT ⎠
= 2.105 mA
IR =
0.6
= 0.60 mA
1K
I1 = I 2 + I R = 2.705 mA
⎛I ⎞
⎛ 2.705 × 10−3 ⎞
VD = VT ln ⎜ 1 ⎟ = (0.026) ln ⎜
⎟
−13
⎝ 2 × 10
⎠
⎝ IS ⎠
= 0.6065
VI = 2VD + V0 ⇒ VI = 1.81 V
1.41
(a) Assume diode is conducting.
Then, VD = Vγ = 0.7 V
So that I R 2 =
0. 7
⇒ 23.3 μ A
30
1.2 − 0.7
⇒ 50 μ A
10
Then I D = I R1 − I R 2 = 50 − 23.3
I R1 =
Or I D = 26.7 μ A
(b) Let R1 = 50 k Ω Diode is cutoff.
VD =
30
⋅ (1.2) = 0.45 V
30 + 50
Since VD < Vγ , I D = 0
1.42
⫹5 V
3 k⍀
2 k⍀
ID
VB ⫽VA⫺Vr
VA
2 k⍀
2 k⍀
V
5 − VA
= ID + A
2
2
A& VA − Vr
A&VA:
(1)
(2)
5 − (VA − Vr )
+ ID =
(VA − Vr )
2
2
5 − (VA − Vr ) ⎡ 5 − VA VA ⎤ VA − Vr
+⎢
− ⎥=
So
3
2⎦
2
⎣ 2
Multiply by 6:
10 − 2 (VA − Vr ) + 15 − 6VA = 3 (VA − Vr )
25 + 2Vr + 3Vr = 11VA
(a)
Vr = 0.6 V
11VA = 25 + 5 ( 0.6 ) = 28 ⇒ VA = 2.545 V
5 − VA VA
−
= 2.5 − VA ⇒ I D Neg. ⇒ I D = 0
2
2
Both (a), (b) I D = 0
From (1) I D =
VA = 2.5, VB =
2
⋅ 5 = 2 V ⇒ VD = 0.50 V
5
1.43
Minimum diode current for VPS (min)
I D (min) = 2 mA, VD = 0.7 V
I2 =
0.7
5 − 0.7 4.3
=
, I1 =
R2
R1
R1
We have I1 = I 2 + I D
4.3 0.7
=
+2
R1
R2
Maximum diode current for VPS (max)
P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA
so (1)
I1 = I 2 + I D
or
9.3 0.7
=
+ 14.3
(2)
R1
R2
Using Eq. (1),
9. 3 4 . 3
=
− 2 + 14.3 ⇒
R1
R1
Then R2 = 82.5Ω 82.5Ω
R1 = 0.41 kΩ
1.44
(a)
Vo = 0.7 V
5 − 0.7
I=
⇒ I = 0.215 mA
20
10 − 0.7
(b)
I=
⇒ I = 0.2325 mA
20 + 20
Vo = I (20 K) − 5 ⇒ Vo = −0.35 V
(c)
(d)
1.45
(a)
VD
0.6
0.482
(b)
Vo
0.5
0.484
(c)
Vo
0.480
0.496
10 − 0.7
⇒ I = 0.372 mA
5 + 20
Vo = 0.7 + I (20) − 8 ⇒ Vo = +0.14 V
I =0
Vo = I (20) − 5 ⇒ Vo = −5 V
I=
5 = I ( 2 × 109 ) + VD
→
ID
→
2.2 ×10−4
2.259 ×10−4
10 = I ( 4 × 10 4 ) + VD
→
I
→
2.375 ×10−4
2.379 ×10−4
I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 ×10 ⎠
VD
Vo = VD = 0.482 V
0.481
0.482
I = 0.226 mA
I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 ×10 ⎠
VD
VD = 0.483 V
10 = I ( 2.5 × 10 4 ) + VD
→
0.4834
0.4834
I = 0.238 mA
Vo = −0.24 V
I
⎛
⎞
VD = ( 0.026 ) ln ⎜
−12 ⎟
×
2
10
⎝
⎠
VD
VD = 0.496 V
I
→
0.496
I = 0.380 mA
3.808 ×10−4
−4
0.496
Vo = −0.10 V
3.802 ×10
(d)
I = − I S ⇒ I = 2 × 10−12 A
1.46
(a)
Diode forward biased VD = 0.7 V
Vo ≅ −5 V
5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V
P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω
(b)
1.47
(a)
I R 2 = I D1 =
ID2
0.65
= 0.65 mA = I D1
1
= 2(0.65) = 1.30 mA
ID2 =
IR2 =
VI − 2Vr − V0 5 − 3(0.65)
=
= 1.30 ⇒ R1 = 2.35 K
R1
R1
0.65
= 0.65 mA
1
8 − 3(0.65)
ID2 =
⇒ I D 2 = 3.025 mA
2
I D1 = I D 2 − I R 2 = 3.025 − 0.65
I D1 = 2.375 mA
(b)
1.48
a.
τd =
VT
(0.026)
=
= 0.026 kΩ = 26Ω
1
I DQ
id = 0.05 I DQ = 50 μ A peak-to-peak
vd = idτ d = (26)(50) μ A ⇒ vd = 1.30 mV peak-to-peak
For I DQ = 0.1 mA ⇒ τ d =
b.
id = 0.05 I DQ = 5 μ A peak-to-peak
(0.026)
= 260Ω
0. 1
vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak
1.49
RS
␯S
⫹
␯d
⫺
⫹
⫺
a.
diode resistance rd = VT /I
b.
RS = 260Ω
⎛
⎜ VT /I
⎛ rd ⎞
vd = ⎜
⎟ vS = ⎜ V
⎝ rd + RS ⎠
⎜⎜ T + RS
⎝ I
⎛ VT
⎞
vd = ⎜
⎟ vs = vo
⎝ VT + IRS ⎠
⎞
⎟
⎟ vS
⎟⎟
⎠
⎞
v0 ⎛ VT
v
0.026
=⎜
⇒ 0 = 0.0909
⎟=
vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26)
vS
v
v
0.026
⇒ 0 = 0.50
I = 0.1 mA, 0 =
vs 0.026 + ( 0.1)( 0.26 )
vS
I = 1 mA,
I = 0.01 mA.
1.50
⎛V
I ≅ I S exp ⎜ a
⎝ VT
v0
v
0.026
=
⇒ 0 = 0.909
vS 0.026 + (0.01)(0.26)
vS
⎛ I ⎞
⎞
⎟ , Va = VT ln ⎜ ⎟
⎠
⎝ IS ⎠
⎛ 100 × 10−6 ⎞
pn junction, Va = (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
Va = 0.599 V
⎛ 100 × 10−6 ⎞
Schottky diode, Va = (0.026) ln ⎜
⎟
−9
⎝ 10
⎠
Va = 0.299 V
1.51
Schottky
pn junction
I
⎛V ⎞
Schottky: I ≅ I S exp ⎜ a ⎟
⎝ VT ⎠
⎛ I ⎞
⎛ 0.5 ×10−3 ⎞
Va = VT ln ⎜ ⎟ = (0.026) ln ⎜
−7 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
= 0.1796 V
Then
Va of pn junction = 0.1796 + 0.30
= 0.4796
0.5 × 10−3
I
=
⎛V ⎞
⎛ 0.4796 ⎞
exp ⎜ a ⎟ exp ⎜
⎟
⎝ 0.026 ⎠
⎝ VT ⎠
I S = 4.87 × 10 −12 A
IS =
1.52
(a)
⫹ VD ⫺
I1
0.5 mA
I2
I1 + I 2 = 0.5 × 10 −3
⎛V
5 × 10−8 exp ⎜ D
⎝ VT
⎞
⎛ VD ⎞
−12
−3
⎟ + 10 exp ⎜ ⎟ = 0.5 × 10
V
⎠
⎝ T ⎠
⎛V ⎞
5.0001× 10 −8 exp ⎜ D ⎟ = 0.5 × 10 −3
⎝ VT ⎠
⎛ 0.5 × 10−3 ⎞
⇒ VD = 0.2395
VD = (0.026) ln ⎜
−8 ⎟
⎝ 5.0001× 10 ⎠
Schottky diode, I 2 = 0.49999 mA
pn junction, I1 = 0.00001 mA
(b)
⫹ VD1 ⫺
⫹ VD2 ⫺
I
⫹
⫺
0.90 V
⎛V ⎞
⎛V ⎞
I = 10 −12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ D 2 ⎟
⎝ VT ⎠
⎝ VT ⎠
VD1 + VD 2 = 0.9
⎛V ⎞
⎛ 0.9 − VD1 ⎞
10−12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜
⎟
⎝ VT ⎠
⎝ VT
⎠
⎛ 0.9 ⎞
⎛ −VD1 ⎞
= 5 ×10−8 exp ⎜
⎟
⎟ exp ⎜
⎝ VT ⎠
⎝ VT ⎠
⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞
⎛ 0.9 ⎞
exp ⎜ D1 ⎟ = ⎜
exp ⎜
⎟
−12 ⎟
⎝ 0.026 ⎠
⎠
⎝ VT ⎠ ⎝ 10
⎛ 5 × 10−8 ⎞
+ 0.9 = 1.1813
2VD1 = VT ln ⎜
−12 ⎟
⎝ 10
⎠
VD1 = 0.5907 pn junction
VD 2 = 0.3093 Schottky diode
⎛ 0.5907 ⎞
I = 10−12 exp ⎜
⎟ ⇒ I = 7.35 mA
⎝ 0.026 ⎠
1.53
R ⫽ 0.5 K
V0
⫹
VPS ⫽ 10 V
⫺
I
⫹
VZ
⫺
IZ
VZ = VZ 0 = 5.6 V at I Z = 0.1 mA
rZ = 10Ω
I Z rZ = ( 0.1)(10 ) = 1 mV
VZ0 = 5.599
a.
RL → ∞ ⇒
IZ =
10 − 5.599
4.401
=
= 8.63 mA
0.50 + 0.01
R + rZ
VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 )
VZ = V0 = 5.685 V
RL
IL
11 − 5.599
= 10.59 mA
0.51
VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V
b.
VPS = 11 V ⇒ I Z =
9 − 5.599
= 6.669 mA
0.51
VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V
VPS = 9 V ⇒ I Z =
ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V
I = IZ + IL
V0
V − V0
V − VZ 0
IL =
, I = PS
, IZ = 0
RL
R
rZ
c.
10 − V0 V0 − 5.599 V0
=
+
0.50
0.010
2
10 5.599
1
1⎤
⎡ 1
+
= V0 ⎢
+
+ ⎥
0.50 0.010
⎣ 0.50 0.010 2 ⎦
20.0 + 559.9 = V0 (102.5)
V0 = 5.658 V
1.54
a.
9 − 6.8
⇒ I Z = 11 mA
0.2
PZ = (11)( 6.8 ) ⇒ PZ = 74.8 mW
IZ =
12 − 6.8
⇒ I Z = 26 mA
0.2
b.
26 − 11
%=
× 100 ⇒ 136%
11
PZ = ( 26 )( 6.8 ) = 176.8 mW
IZ =
%=
176.8 − 74.8
× 100 ⇒ 136%
74.8
1.55
I Z rZ = ( 0.1)( 20 ) = 2 mV
VZ 0 = 6.8 − 0.002 = 6.798 V
a.
RL = ∞
10 − 6.798
⇒ I Z = 6.158 mA
0.5 + 0.02
V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 )
IZ =
V0 = 6.921 V
b.
I = IZ + IL
10 − V0 V0 − 6.798 V0
=
+
0.50
0.020
1
10 6.798
1
1⎤
⎡ 1
+
= V0 ⎢
+
+ ⎥
.
.
0.30 0.020
0
50
0
020
1⎦
⎣
359.9 = V0 (53)
V0 = 6.791 V
ΔV0 = 6.791 − 6.921
ΔV0 = −0.13 V
1.56
For VD = 0, I SC = 0.1 A
⎛ 0.2
⎞
+ 1⎟
For ID = 0 VD = VT ln ⎜
−14
⎝ 5 × 10
⎠
VD = VDC = 0.754 V
Chapter 2
Exercise Problems
30 − 12 − 0.6
= 87 mA
0.2
vR ( max ) = 30 + 12 = 42 V
iD ( peak ) =
EX2.1
⎛ 12.6 ⎞
⎟ ⇒ 24.8°
⎝ 30 ⎠
By symmetry
ω t2 = 180 − 24.8 = 155.2°
ω t1 = sin −1 ⎜
% time =
155.2 − 24.8
× 100% = 36.2%
360
EX2.2
(a) vO = 12sin θ1 − 1.4 = 0
or sin θ1 =
1.4
= 0.1166
12
which yields
θ1 = 6.7°
By symmetry, θ 2 = 180 − 6.7 = 173.3°
173.3 − 6.7
× 100% = 46.3%
360
1.4
(b)
sinθ1 =
= 0.35
4
which yields
θ1 = 20.5°
Then % time =
By symmetry, θ 2 = 180 − 20.5 = 159.5°
Then % time =
159.5 − 20.5
× 100% = 38.6%
360
EX2.3
VM
24
C=
=
⇒ or C = 500 μ F
2 fRVr 2 ( 60 ) (103 ) ( 0.4 )
EX2.4
Vr =
VM
VM
75
⇒R=
or R =
f RC
f CVr
60
50
( ) ( ×10−6 ) ( 4 )
Then R = 6.25 k Ω
EX2.5
10 ≤ VPS ≤ 14 V , VZ = 5.6 V , 20 ≤ RL ≤ 100 Ω
I L ( max ) =
5.6
= 0.28 A,
20
5.6
I L ( min ) =
= 0.056 A
100
⎡VPS ( max ) − VZ ⎤⎦ ⋅ I L ( max )
⎡⎣VPS ( min ) − VZ ⎤⎦ ⋅ I L ( min )
−
I Z (max) = ⎣
VPS ( min ) − 0.9VZ − 0.1VPS ( max ) VPS ( min ) − 0.9VZ − 0.1VPS ( max )
or I L ( max ) =
(14 − 5.6 )( 280 ) − (10 − 5.6 )( 56 )
10 − ( 0.9 )( 5.6 ) − ( 0.1)(14 )
or I L ( max ) = 591.5 mA
Power(min) = I Z ( max ) ⋅ VZ = ( 0.5915)( 5.6 )
So Power(min) = 3.31 W
VPS ( max ) − VZ
14 − 5.6
=
Now Ri =
or Ri ≅ 13 Ω
I Z ( max ) + I L ( min ) 0.5915 + 0.056
EX2.6
13.6 − 9
= 0.2383 A
15.3 + 4
= 9 + ( 4 )( 0.2383) = 9.9532 V
For vPS = 13.6 V , I Z =
vL ,max
11 − 9
= 0.1036 A
15.3 + 4
= 9 + ( 4 )( 0.1036 ) = 9.4144 V
For vPS = 11 V , I Z =
vL ,min
ΔvL
9.9532 − 9.4144
× 100% =
× 100%
ΔvPS
13.6 − 11
or Source Reg = 20.7%
Source Reg =
13.6 − 9
= 0.2383 A
15.3 + 4
= 9 + ( 4 )( 0.2383) = 9.9532 V
For I L = 0, I Z =
vL , noload
For I L = 100 mA, I Z =
13.6 − ⎡⎣9 + I Z ( 4 ) ⎤⎦
15.3
which yields
I Z = 0.1591 A
vL , full load = 9 + ( 4 )( 0.1591A ) = 9.6363 V
vL , noload − vL , full load
× 100%
Load Reg =
vL , full load
9.9532 − 9.6363
× 100%
9.6363
or Load Reg = 3.29%
=
EX2.7
R1
VO
␯I
⫹
⫺
D1
⫹
V1
⫺
For vI < 5 V , D2 on ⇒ VO = −5 V
Then, V2 = 4.3 V.
D1 turns on when v1 = 2.5 V,
Then, V1 = 1.8 V.
D2
⫺
V2
⫹
− 0.10
For vI > 2.5 V ,
ΔvO 1
R2
1
= ⇒
=
ΔvI 3
R1 + R2 3
So that R1 = 2R2
EX2.8
For Vγ = 0, vO ( max ) = −2 V
Now, ΔvO = 8 V , so that vO ( min ) = −10 V
10 − 4.4
= 0.5895 mA
9.5
Set I = ID1, then vI = 4.4 − 0.6 − ( 0.5895 )( 0.5 ) = 3.505 V
EX2.9
vO = 4.4 V , I =
Summary: For 0 ≤ vI ≤ 3.5 V , vO = 4.4 V
For vI > 3.5 V , D2 turns on and when vI ≥ 9.4 V , vO = 10 V
␯O(V)
10
4.4
0
3.505
EX2.10
VO = −0.6 V , I D1 = 0
ID2 = I =
−0.6 − ( −10 )
2.2
9.4 ␯I (V)
⇒ I D 2 = I = 4.27 mA
EX2.11
At the VA node:
15 − VA
V
(1)
= ID2 + A
5
15
At the VB node:
15 − (VB + 0.7 )
V
+ I D 2 = B (2)
5
10
We see that
VB = VA − 0.7, so Equation (2) becomes
15 − VA
V − 0.7
(2’)
+ ID2 = A
5
10
Solving for ID2 from Equation (1) and substituting into Equation (2’), we find
⎛ 15 − VA ⎞ VA VA − 0.7
=
2⎜
⎟−
10
⎝ 5 ⎠ 15
Then VA = 10.71 V and VB = 10.01 V
15 − 10.71
Solving for the diode currents, we obtain I D1 =
⇒ I D1 = 0.858 mA
5
Also I D 2 =
15 − 10.71 10.71
−
⇒ I D 2 = 0.144 mA
5
15
EX2.12
Assuming all diodes are conducting, we have VB = −0.7 V and VA = 0
Summing currents, we have
5 − VA
(1)
= I D1 + I D 2
5
V − ( −5 )
I D 2 + I D3 = B
(2)
5
V − ( −10 ) 10 − 0.7
=
I D1 = B
(3)
10
10
From ( 3) , we find
I D1 = 0.93 mA
From (1) , we obtain
I D 2 = 0.07 mA
From ( 2 ) , we have
I D 3 = 0.79 mA
EX2.13
(a)
I ph = η eΦ
⎡ 6.4 × 10 −2 ⎤
⎥ ( 0.5 )
so I = ( 0.8 ) (1.6 × 10−19 ) ⎢
⎢⎣ ( 2 ) (1.6 × 10−19 ) ⎥⎦
or I ph = 12.8 mA
(b)
We have vO = (12.8)(1) = 12.8 V .
The diode must be reverse biased so that VPS > 12.8 V
EX2.14
The equivalent circuit is
⫹5 V
⫹
Vr ⫽ 1.7 V
⫺
rf ⫽ 15 ⍀
I
R
0.2 V
So I =
5 − 1.7 − 0.2
= 15 mA
rf + R
15 − 1.7 − 0.2 3.1
=
= 0.207 kΩ
15
15
Then R = 207 − 15 ⇒ R = 192 Ω
2.15
40 − 12
IZ =
= 0.233 A
(a)
120
P = ( 0.233)(12 ) = 2.8 W
Or rf + R =
I R = 0.233 A, I L = ( 0.9 )( 0.233) = 0.21 A
(b)
So 0.21 =
12
⇒ RL = 57.1 Ω
RL
P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W
(c)
Test Your Understanding Exercises
TYU2.1
vI = 120sin ( 2π 60t ) , Vγ = 0.7 V , and R = 2.5 kΩ
Full-wave rectifier: Turns ratio 1:2 so that
vS = v I
VM = 120 − 0.7 = 119.3 V
Vr = 119.3 − 100 = 19.3 V
So C =
VM
119.3
=
or C = 20.6 μ F
2 f RVr 2 ( 60 ) ( 2.5 x103 ) (19.3)
TYU2.2
vI = 50sin ( 2π 60t ) , Vγ = 0.7 V , and R = 10 kΩ. Full-wave rectifier
C=
( 50 − 1.4 )
VM
=
2 f RVr 2 ( 60 ) (10 × 103 ) ( 2 )
or C = 20.3 μ F
TYU2.3
Using Equation (2.10)
(a)
ω Δt =
2Vr
=
VM
2 ( 4)
75
= 0.327
⎛ 0.327 ⎞
Percent time = ⎜
⎟ × 100% = 5.2%
⎝ 2π ⎠
(b)
ω Δt =
2Vr
=
VM
2 (19.3)
119.3
= 0.569
⎛ 0.569 ⎞
Percent time = ⎜
⎟ × 100% = 18.1%
⎝ π ⎠
(c)
ω Δt =
2Vr
=
VM
2 ( 2)
48.6
= 0.287
⎛ 0.287 ⎞
Percent time = ⎜
⎟ × 100% = 9.14%
⎝ π ⎠
TYU2.4
V − VZ
I Z = PS
− IL
Ri
11 − 9
− 0.1 = 0 (Minimum Zener current is zero.)
20
13.6 − 9
For VPS (max) and IL (min), then I Z ( max ) =
− 0 ⇒ I Z ( max ) = 230 mA
20
The characteristic of the minimum Zener current being one-tenth of the maximum value is violated.
The proper circuit operation is questionable.
For VPS (min) and IL (max), then I Z ( min ) =
TYU2.5
I Z ( min ) =
so 30 =
VPS ( min ) − VZ
Ri
− I L ( max )
10 − 9
− I L ( max ) which yields I L ( max ) = 35.4 mA
0.0153
TYU2.6
␯O(V)
4.35
2.7
⫺4.7
2.7
6
␯I (V)
⫺4.7
TYU2.7
As vS goes negative, D turns on and vO = +5 V . As vS goes positive, D turns off. Output is a square
wave oscillating between +5 and +35 volts.
TYU2.8
␯O(V)
4.4
0
0.6
5
␯I (V)
5
␯I (V)
(a)
␯O(V)
4.4
2.4
0
(b)
3
TYU2.9
(a)
VO = 0.6 V for all V1.
VO (V)
3.6
0.6
0
(b)
PSpice Exercises
3
VI (V)
Chapter 2
Problem Solutions
2.1
␯I
␯0
R⫽1K
␯I
10
0
⫺10
Vγ = 0.6 V, rf = 20 Ω
⎛ R ⎞
For vI = 10 V, v0 = ⎜
10 − 0.6 )
⎜ R + r ⎟⎟ (
f ⎠
⎝
⎛ 1 ⎞
=⎜
⎟ ( 9.4 )
⎝ 1 + 0.02 ⎠
v0 = 9.22
10
0.6
␯0
9.22
2.2
⫹ ␯D ⫺
␯I
␯0
D
R
v0 = vI − vD
⎛i ⎞
v
vD = VT ln ⎜ D ⎟ and iD = 0
R
⎝ IS ⎠
⎛ v ⎞
v0 = vI − VT ln ⎜ 0 ⎟
⎝ IS R ⎠
2.3
80sin ω t
= 13.33sin ω t
6
13.33
Peak diode current id (max) =
R
(a)
vs =
(b)
PIV = vs (max) = 13.3 V
(c)
vo ( avg ) =
1
To
∫ v (t )dt = 2π ∫ 13.33sin × dt
To
1
π
o
o
o
13.33
13.33
13.33
π
=
⎡⎣ − ( −1 − 1) ⎤⎦ =
[ − cos x ]o =
2π
2π
π
vo ( avg ) = 4.24 V
(d)
50%
2.4
v0 = vS − 2Vγ ⇒ vS ( max ) = v0 ( max ) + 2Vγ
a.
For v0 ( max ) = 25 V ⇒ vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V
N1 160
N
=
⇒ 1 = 6.06
N 2 26.4
N2
b.
For v0 ( max ) = 100 V ⇒ vS ( max ) = 101.4 V
N1
N
160
=
⇒ 1 = 1.58
N 2 101.4
N2
From part (a) PIV = 2vS ( max ) − Vγ = 2 ( 26.4 ) − 0.7
or PIV = 52.1 V or, from part (b) PIV = 2 (101.4 ) − 0.7 or PIV = 202.1 V
2.5
(a)
vs (max) = 12 + 2(0.7) = 13.4 V
13.4
vs ( rms ) =
⇒ vs (rms) = 9.48 V
2
(b)
VM
VM
Vr =
⇒C =
2 f RC
2 f Vr R
C=
12
⇒ C = 2222 µ F
2 ( 60 )( 0.3)(150 )
⎡
2VM ⎤
⎢1 + π
⎥
Vr ⎥⎦
⎣⎢
2 (12 ) ⎤
12 ⎡
⎢1 + π
⎥
=
150 ⎢
0.3 ⎥
⎣
⎦
id , peak = 2.33 A
(c)
id , peak =
VM
R
2.6
(a)
vS ( max ) = 12 + 0.7 = 12.7 V
vS ( rms ) =
(b)
vS ( max )
Vr =
2
⇒ vS ( rms ) = 8.98 V
VM
V
12
⇒C = M =
or C = 4444 µ F
fRC
fRVr ( 60 )(150 )( 0.3)
(c)
For the half-wave rectifier iD , max =
VM
R
⎛
VM
⎜⎜ 1 + 4π
2
Vr
⎝
iD , max = 4.58 A
2.8
(a)
vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V
16.4
vs ( rms ) =
= 11.6 V
2
VM
15
(b)
=
= 2857 µ F
C=
2 f RVr 2 ( 60 )(125 )( 0.35 )
2.9
␯S
26
0.6
⫺0.6
⫺26
25.4
⫹
␯O
0
0
⫺
␯O
⫺25.4
2.10
R
iD
⫹
VB ⫽ 12 V
⫺
␯S ⫽ ⫹
⫺
␯S
iD
␻t
x2
x1
vS ( t ) = 24sin ω t
1
iD ( t ) dt
T ∫0
T
Now iD ( avg ) =
24sin x − 12.7
R
To find x1 and x2, 24sin x1 = 12.7
We have for x1 ≤ ω t ≤ x2 iD =
x1 = 0.558 rad
x2 = π − 0.558 = 2.584 rad
Then
⎞ 12 ⎛
12 ⎞
⎜ 1 + 4π
⎟ or
⎟⎟ =
⎜
2 ( 0.3) ⎟⎠
⎠ 150 ⎝
1
2π
iD ( avg ) = 2 =
∫ ⎢⎣
⎡ 24sin x − 12.7 ⎤
⎥dx
R
⎦
x1
x2
1 ⎛ 24 ⎞
1 ⎛ 12.7 ⎞ x2
6.482 4.095
x2
−
⇒ R = 1.19Ω
⎜ ⎟ ( − cos x ) x1 −
⎜
⎟ ⋅ xx or 2 =
2π ⎝ R ⎠
2π ⎝ R ⎠ 1
R
R
x −x
2.584 − 0.558
Fraction of time diode is conducting = 2 1 × 100% =
× 100% or Fraction = 32.2%
2π
2π
Power rating
=
R 2
R 2 ⎡ 24sin x − 12.7 ⎤
iD dt =
∫
⎥ dx
T 0
R
2π x∫1 ⎢⎣
⎦
=
⎡( 24 )
2π R ∫ ⎣
x2
1
x1
2
x
T
2
=
Pavg = R ⋅ irms
2
2
sin 2 x − 2 (12.7 )( 24 ) sin x + (12.7 ) ⎤dx
⎦
1 ⎡
sin 2 x ⎤ 2
2 ⎡x
2
x2
− 2 (12.7 )( 24 )( − cos x ) x + (12.7 ) xxx12 ⎦⎤
24 ) ⎢ −
(
⎥
1
2π R ⎣
2
4
⎣
⎦x
x
=
1
For R = 1.19 Ω, then Pavg = 17.9 W
2.11
(a)
15
= 150Ω
0.1
vS ( max ) = vo ( max ) + Vγ = 15 + 0.7 or vS ( max ) = 15.7 V
R=
Then vS ( rms ) =
Now
15.7
2
= 11.1 V
N1 120
N
=
⇒ 1 = 10.8
N 2 11.1
N2
VM
VM
15
⇒C =
=
or C = 2083 µ F
2 fRC
2 fRVr 2 ( 60 )(150 )( 0.4 )
(b)
Vr =
(c)
PIV = 2vS ( max ) − Vγ = 2 (15.7 ) − 0.7 or PIV = 30.7 V
2.12
␯0
␯i
⫹
⫺
D2
R2
RL
R1
␯0
␯i
⫹
⫺
RL
R2
R1
For vi > 0
Vγ = 0
Voltage across RL + R1 = vi
⎛ RL ⎞
1
Voltage Divider ⇒ v0 = ⎜
⎟ vi = vi
2
⎝ RL + R1 ⎠
␯0
20
2.13
For vi > 0, (Vγ = 0 )
R1
␯i
⫹
⫺
␯0
R2
RL
a.
⎛ R2 || RL ⎞
v0 = ⎜
⎟ vi
⎝ R2 || RL + R1 ⎠
R2 || RL = 2.2 || 6.8 = 1.66 kΩ
⎛ 1.66 ⎞
v0 = ⎜
⎟ vi = 0.43 vi
⎝ 1.66 + 2.2 ⎠
␯0
4.3
v0 ( rms ) =
b.
v0 ( max )
2
⇒ v0 ( rms ) = 3.04 V
2.14
3.9
⇒ I 2 = 0.975 mA
4
20 − 3.9
IR =
= 1.3417 mA
12
I Z = 1.3417 − 0.975 ⇒ I Z = 0.367 mA
IL =
PT = I Z ⋅ VZ = ( 0.367 )( 3.9 ) ⇒ PT = 1.43 mW
2.15
(a)
40 − 12
= 0.233 A
120
P = ( 0.233)(12 ) = 2.8 W
IZ =
(b)
IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A
12
So 0.21 =
⇒ RL = 57.1Ω
RL
P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W
(c)
2.16
Ri
VI
V0
II
IZ
⫹
VZ
⫺
RL
IL
VI = 6.3 V, Ri = 12Ω, VZ = 4.8
a.
6.3 − 4.8
⇒ 125 mA
12
I L = I I − I Z = 125 − I Z
II =
25 ≤ I L ≤ 120 mA ⇒ 40 ≤ RL ≤ 192Ω
b.
PZ = I Z VZ = (100 )( 4.8 ) ⇒ PZ = 480 mW
PL = I LV0 = (120 )( 4.8 ) ⇒ PL = 576 mW
2.17
a.
20 − 10
⇒ I I = 45.0 mA
222
10
IL =
⇒ I L = 26.3 mA
380
I Z = I I − I L ⇒ I Z = 18.7 mA
II =
b.
PZ ( max ) = 400 mW ⇒ I Z ( max ) =
⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40
⇒ I L ( min ) = 5 mA =
⇒ RL = 2 kΩ
(c)
400
= 40 mA
10
10
RL
For Ri = 175Ω I I = 57.1 mA
I L = 26.3 mA
I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA
RL =
I Z = 30.8 mA
10
⇒ RL = 585Ω
17.1
2.18
a.
From Eq. (2-31)
500 [ 20 − 10] − 50 [15 − 10]
I Z ( max ) =
15 − ( 0.9 )(10 ) − ( 0.1)( 20 )
5000 − 250
4
I Z ( max ) = 1.1875 A
I Z ( min ) = 0.11875 A
=
From Eq. (2-29(b)) Ri =
20 − 10
⇒ Ri = 8.08Ω
1187.5 + 50
b.
PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W
PL = I L ( max ) V0 = ( 0.5 )(10 ) ⇒ PL = 5 W
2.19
(a)
As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18.
V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ
= 10 + (1.1875)(2) = 12.375 V
V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ
= 10 + (0.11875)(2) = 10.2375 V
b.
% Reg =
12.375 − 10.2375
× 100% ⇒ % Reg = 21.4%
10
2.20
% Reg =
=
VL ( max ) − VL ( min )
VL ( nom )
× 100%
VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz )
VL ( nom )
⎡ I Z ( max ) − I Z ( min ) ⎦⎤ ( 3)
=⎣
= 0.05
6
So I Z ( max ) − I Z ( min ) = 0.1 A
6
6
= 0.012 A, I L ( min ) =
= 0.006 A
500
1000
VPS ( min ) − VZ
Now I L ( max ) =
Now Ri =
or 280 =
I Z ( min ) + I L ( max )
15 − 6
⇒ I Z ( min ) = 0.020 A
I Z ( min ) + 0.012
Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri =
or 280 =
VPS ( max ) − 6
0.12 + 0.006
⇒ VPS ( max ) = 41.3 V
VPS ( max ) − VZ
I Z ( max ) + I L ( min )
2.21
Using Figure 2.21
a.
VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V
For VPS ( min ) :
I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA
Ri =
b.
VPS ( min ) − VZ
II
=
15 − 10
⇒ Ri = 200Ω
25
For VPS ( max ) ⇒ I I ( max ) =
25 − 10
⇒ I I ( max ) = 75 mA
Ri
For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA
VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V
V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25
V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90
ΔV0 = 0.35 V
c.
% Reg =
ΔV0
× 100% ⇒ % Reg = 3.5%
V0 ( nom )
2.22
From Equation (2.29(a))
VPS ( min ) − VZ
24 − 16
or Ri = 18.2Ω
Ri =
=
I Z ( min ) + I L ( max ) 40 + 400
Also Vr =
VM
VM
⇒C =
2 fRC
2 fRVr
R ≅ Ri + rz = 18.2 + 2 = 20.2Ω
Then
C=
24
⇒ C = 9901 µ F
2 ( 60 )(1)( 20.2 )
2.23
VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V
8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V
Ii =
VS ( max ) − VZ ( nom )
Ri
=
12 − 8
= 1.333 A
3
For I L = 0.2 A ⇒ I Z = 1.133 A
For I L = 1 A ⇒ I Z = 0.333 A
VL ( max ) = VZ 0 + I Z ( max ) rZ
= 7.95 + (1.133)( 0.5 ) = 8.5165
VL ( min ) = VZ 0 + I Z ( min ) rZ
= 7.95 + ( 0.333)( 0.5 ) = 8.1165
ΔVL = 0.4 V
ΔVL
0.4
=
⇒ % Reg = 5.0%
% Reg =
V0 ( nom )
8
Vr =
VM
VM
⇒C =
2 fRC
2 fRVr
R = Ri + rz = 3 + 0.5 = 3.5Ω
Then C =
12
⇒ C = 0.0357 F
2 ( 60 )( 3.5 )( 0.8 )
2.24
(a)
For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0
For 0 ≤ vI ≤ 3, Zener not in breakdown, so i1 = 0, vO = 0
vI − 3
mA
20
1
⎛ v −3⎞
vo = ⎜ I
⎟ (10 ) = vI − 1.5
20
2
⎝
⎠
At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA
For vI > 3 i1 =
␯O(V)
4
3.5
⫺10
(b)
3.0
For vI < 0, both diodes forward biased
0 − vI
. At vI = −10 V , i1 = −1 mA
10
v −3
For vI > 3, i1 = I
. At vI = 10 V , i1 = 0.35 mA
20
−i1 =
10
␯I(V)
i1(mA)
0.35
10 ␯I(V)
3
⫺10
⫺1
2.25
(a)
1K
␯I
␯0
1K
2K
⫹15
V1
1
V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI
3
For vI > 5.7 V
vI − (V1 + 0.7 ) 15 − V1 V1
+
= , v0 = V1 + 0.7
1
2
1
−
−
15
0.7
v
(0
) v0 − 0.7
vI − v0
+
=
1
2
1
vI 15.7 0.7
⎛ 1 1 1⎞
+
+
= v0 ⎜ + + ⎟ = v0 ( 2.5 )
1
2
1
⎝ 1 2 1⎠
1
vI + 8.55 = v0 ( 2.5 ) ⇒ v0 =
vI + 3.42
2.5
vI = 5.7 ⇒ v0 = 5.7
vI = 15 ⇒ v0 = 9.42
␯0(V)
9.42
5.7
0
5.7
(b)
iD = 0 for 0 ≤ vI ≤ 5.7
Then for vI > 5.7 V
15
␯I (V)
⎛ v
⎞
vI − ⎜ I + 3.42 ⎟
vI − vO
2.5
⎝
⎠ or i = 0.6vI − 3.42 For v = 15, i = 5.58 mA
=
iD =
I
D
D
1
1
1
iD(mA)
5.58
5.7
15
␯I (V)
2.26
⎛ 20 ⎞
For D off, vo = ⎜ ⎟ (20) − 10 = 3.33 V
⎝ 30 ⎠
Then for vI ≤ 3.33 + 0.7 = 4.03 V ⇒ vo = 3.33 V
For vI > 4.03, vo = vl − 0.7;
For vI = 10, vo = 9.3
(a)
␯O(V)
9.3
3.33
0
(b)
4.03
10 ␯I (V)
For vI ≤ 4.03 V , iD = 0
10 − vo vo − ( −10 )
=
10
20
3
Which yields iD =
vI − 0.605
20
For vI = 10, iD = 0.895 mA
For vI > 4.03, iD +
iD(mA)
0.895
0
4.03
2.27
␯O
⫺30
12.5
10.7
10.7
30
␯I
⫺30
30 − 10.7
= 0.175 A
100 + 10
v0 = i(10) + 10.7 = 12.5 V
For vI = 30 V, i =
10 ␯I(V)
b.
␯O
12.5
10.7
0
⫺30
2.28
⫹ 5⫺
␯I
␯O
R ⫽ 6.8 K
Vγ = 0.6 V
vI = 15sin ω t
␯O
␯O
⫺4.4
⫺19.4
2.29
a.
Vγ = 0
0
⫺3 V
Vγ = 0.6
0
⫺2.4
b.
Vγ = 0
20
5
Vγ = 0.6
19.4
5
2.30
10
6.7
0
⫺4.7
⫺10
2.31
One possible example is shown.
Ri
L
D
Ii
⫹
⫺
DZ
Vign
⫹
VZ ⫽ 14 V
⫺
⫹
⫺
RADIO
VRADIO
L will tend to block the transient signals
Dz will limit the voltage to +14 V and −0.7 V.
Power ratings depends on number of pulses per second and duration of pulse.
2.32
␯O(V)
40
(a)
0
␯O(V)
35
(b)
0
⫺5
2.33
C
␯I
␯O
⫹
Vx
⫺
For Vγ = 0 ⇒ Vx = 2.7 V
a.
For Vγ = 0.7 V ⇒ Vx = 2.0 V
b.
2.34
C
⫹
␯I
⫺
2.35
⫹
⫺
10 V
⫹
␯O
⫺
20
␯O
10
VB ⫽ 0
0
␯I
⫺10
20
␯O
13
10
3
0
VB ⫽ ⫹3 V
␯I ⫹ VB
⫺7
20
␯O
10
7
VB ⫽ ⫺3 V
0
⫺3
␯I ⫹ V B
⫺13
2.36
For Figure P2.32(a)
10
␯I
0
⫺10
␯O
⫺20
2.37
a.
10 − 0.6
⇒ I D1 = 0.94 mA
9.5 + 0.5
V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V
I D1 =
ID2 = 0
b.
5 − 0.6
⇒ I D1 = 0.44 mA
9.5 + 0.5
V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V
I D1 =
c.
d.
10 =
ID2 = 0
Same as (a)
(I )
( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA
V0 = I ( 9.5 ) ⇒ V0 = 9.16 V
2
I D1 = I D 2 =
2.38
a.
b.
I
⇒ I D1 = I D 2 = 0.482 mA
2
I = I D1 = I D 2 = 0 V0 = 10
10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) ⇒ I = I D 2 = 0.94 mA
V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V
I D1 = 0
c.
10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 ⇒ I = I D 2 = 0.44 mA
V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V
d.
10 = I ( 9.5 ) + 0.6 +
I D1 = 0
I
( 0.5) ⇒ I = 0.964 mA
2
I
⇒ I D1 = I D 2 = 0.482 mA
2
V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V
I D1 = I D 2 =
2.39
a.
V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V
10 − 4.4
⇒ I = 0.589 mA
9.5
4.4 − 0.6
= ID2 =
⇒ I D1 = I D 2 = 7.6 mA
0.5
= I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA
I=
I D1
I D3
b.
V1 = V2 = 5 V D1 and D2 on, D3 off
I
10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA
2
I
I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA
2
I D3 = 0
V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V
c.
V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V
10 − 4.4
⇒
9.5
4.4 − 0.6
=
⇒
0.5
I = 0.589 mA
I=
I D2
I D 2 = 7.6 mA
I D1 = 0
I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA
d.
V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V
10 − 4.4
⇒
9.5
4.4 − 0.6 − 2
=
⇒
0.5
I = 0.589 mA
I=
I D2
I D 2 = 3.6 mA
I D1 = 0
I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA
2.40
(a)
D1 on, D2 off, D3 on
So I D 2 = 0
Now V2 = −0.6V , I D1 =
10 − 0.6 − ( −0.6 )
R1 + R2
=
10
⇒ I D1 = 1.25 mA
2+6
V1 = 10 − 0.6 − (1.25 )( 2 ) ⇒ V1 = 6.9 V
I D3
−0.6 − ( −5 )
= 2.2 mA
2
= I R 3 − I D1 = 2.2 − 1.25 ⇒ I D 3 = 0.95 mA
I R3 =
(b)
D1 on, D2 on, D3 off
So I D 3 = 0
V1 = 4.4 V , I D1 =
10 − 0.6 − 4.4 5
=
R1
6
or I D1 = 0.833 mA
I R2 =
4.4 − ( −5 )
R2 + R3
=
9.4
= 0.94 mA
10
I D 2 = I R 2 − I D1 = 0.94 − 0.833 ⇒ I D 2 = 0.107 mA
V2 = I R 2 R3 − 5 = ( 0.94 )( 5 ) − 5 ⇒ V2 = −0.3 V
(c)
All diodes are on V1 = 4.4V , V2 = −0.6 V
I D1 = 0.5 mA =
10 − 0.6 − 4.4
⇒ R1 = 10 k Ω
R1
I R 2 = 0.5 + 0.5 = 1 mA =
I R 3 = 1.5 mA =
4.4 − ( −0.6 )
−0.6 − ( −5 )
R3
R2
⇒ R2 = 5 k Ω
⇒ R3 = 2.93 k Ω
2.41
⎛ 0.5 ⎞
For vI small, both diodes off vO = ⎜
⎟ vI = 0.0909vI
⎝ 0.5 + 5 ⎠
When vI − vO = 0.6, D1 turns on. So we have vI − 0.0909vI = 0.6 ⇒ vI = 0.66, vO = 0.06
vI − 0.6 − vO vI − vO vO
2v − 0.6
+
=
which yields vO = I
5
5
0.5
12
2vI − 0.6
When vO = 0.6, D2 turns on. Then 0.6 =
⇒ vI = 3.9 V
12
v − 0.6 − vO vI − vO vO vO − 0.6
+
=
+
Now for vI > 3.9 I
5
5
0.5
0.5
2vI + 5.4
; For vI = 10 ⇒ vO = 1.15 V
Which yields vO =
22
For D1 on
2.42
⫹10 V
10 K
D2
D1
␯I
␯0
D3
D4
10 K
⫺10 V
10 K
For vI > 0. when D1 and D4 turn off
10 − 0.7
= 0.465 mA
20
v0 = I (10 kΩ ) = 4.65 V
I=
␯0
4.65
⫺10
⫺4.65
4.65
10
␯I
⫺4.65
v0 = vI for − 4.65 ≤ vI ≤ 4.65
2.43
a.
R1
D2
⫹10 V
V0
D1
ID1
R2
⫺10 V
R1 = 5 kΩ, R2 = 10 kΩ
D1 and D2 on ⇒ V0 = 0
10 − 0.7 0 − ( −10 )
−
= 1.86 − 1.0
5
10
= 0.86 mA
I D1 =
I D1
b.
R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0
10 − 0.7 − ( −10 )
= 1.287
15
V0 = IR2 − 10 ⇒ V0 = −3.57 V
I=
2.44
If both diodes on
(a)
VA = −0.7 V, VO = −1.4 V
IR2
I R1 + I D1
10 − ( −0.7 )
= 1.07 mA
10
−1.4 − ( −15 )
=
= 2.72 mA
5
= I R 2 ⇒ I D1 = 2.72 − 1.07
I R1 =
I D1 = 1.65 mA
D1 off, D2 on
10 − 0.7 − ( −15 )
= 1.62 mA
I R1 = I R 2 =
5 + 10
VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V
(b)
VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off ,
I D1 = 0
2.45
(a)
D1 on, D2 off
10 − 0.7
= 0.93 mA
I D1 =
10
VO = −15 V
(b)
D1 on, D2 off
10 − 0.7
= 1.86 mA
I D1 =
5
VO = −15 V
2.46
15 − (V0 + 0.7 )
V0 + 0.7 V0
+
10
20
20
15 0.7 0.7
1
1
1 ⎞
⎛
⎛ 4.0 ⎞
−
−
= V0 ⎜ +
+ ⎟ = V0 ⎜
⎟
10 10 20
⎝ 10 20 20 ⎠
⎝ 20 ⎠
V0 = 6.975 V
ID =
=
V0
⇒ I D = 0.349 mA
20
2.47
10 K
Va
⫺ VD ⫹
10 K
Vb
V1
V2
ID
10 K
10 K
a.
V1 = 15 V, V2 = 10 V Diode off
Va = 7.5 V, Vb = 5 V ⇒ VD = −2.5 V
ID = 0
b.
V1 = 10 V, V2 = 15 V Diode on
2.48
vI = 0, D1 off, D2 on
V2 − Vb Vb Va Va − V1
= + +
⇒ Va = Vb − 0.6
10
10 10
10
15 10
⎛1 1⎞
⎛1 1⎞
⎛1 1⎞
+ = Vb ⎜ + ⎟ + Vb ⎜ + ⎟ − 0.6 ⎜ + ⎟
10 10
10
10
10
10
⎝
⎠
⎝
⎠
⎝ 10 10 ⎠
⎛ 4⎞
2.62 = Vb ⎜ ⎟ ⇒ Vb = 6.55 V
⎝ 10 ⎠
15 − 6.55 6.55
ID =
−
⇒ I D = 0.19 mA
10
10
VD = 0.6 V
10 − 2.5
= 0.5 mA
15
vo = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V
I=
For vI > 7.5 V , Both D1 and D2 on
vI − vo vo − 2.5 vo − 10
=
+
or vI = vo ( 5.5 ) − 33.75
15
10
5
When vo = 10 V, D2 turns off
vI = (10 )( 5.5 ) − 33.75 = 21.25 V
For vI > 21.25 V, vo = 10 V
2.49
a.
V01 = V02 = 0
b.
V01 = 4.4 V, V02 = 3.8 V
c.
V01 = 4.4 V, V02 = 3.8 V
Logic “1” level degrades as it goes through additional logic gates.
2.50
a.
V01 = V02 = 5 V
b.
V01 = 0.6 V, V02 = 1.2 V
c.
V01 = 0.6 V, V02 = 1.2 V
Logic “0” signal degrades as it goes through additional logic gates.
2.51
(V1 AND V2 ) OR (V3 AND V4 )
2.52
10 − 1.5 − 0.2
= 12 mA = 0.012
I=
R + 10
8.3
= 691.7Ω
R + 10 =
0.012
R = 681.7Ω
2.53
10 − 1.7 − VI
=8
0.75
VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V
I=
2.54
⫹ VR ⫺
⫹
VPS
⫺
R⫽ 2 K
VR = 1 V, I = 0.8 mA
VPS = 1 + ( 0.8 )( 2 )
VPS = 2.6 V
2.55
I Ph = η eΦA
0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A
A = 3.75 × 10−2 cm 2
Chapter 3
Exercise Solutions
EX3.1
VTN = 1 V , VGS = 3 V , VDS = 4.5 V
VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V
Transistor biased in the saturation region
I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2
2
2
(a) VGS = 2 V, VDS = 4.5 V
Saturation region:
I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA
2
(b) VGS = 3 V, VDS = 1 V
Nonsaturation region:
2
I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA
⎣
⎦
EX3.2
VTP = −2 V , VSG = 3 V
VSD ( sat ) = VSG + VTP = 3 − 2 = 1 V
(a)
(b)
(c)
VSD = 0.5 V ⇒ Nonsaturation
VSD = 2 V ⇒ Saturation
VSD = 5 V ⇒ Saturation
EX3.3
⎛ R2 ⎞
⎛ 160 ⎞
VG = ⎜
⎟ (VDD ) = ⎜
⎟ (10 ) = 3.636 V = VGS
⎝ 160 + 280 ⎠
⎝ R1 + R2 ⎠
2
I D = 0.25 ( 3.636 − 2 ) = 0.669 mA
VDS = 10 − ( 0.669 )(10 ) = 3.31 V
P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW
EX3.4
I DQ = K P (VSG + VTP )
2
1.2 = 0.4 (VSG − 1.2 ) ⇒ VSG = 2.932 V
2
⎛ R1 ⎞
1
VSG = ⎜
⋅ VTTN − VDD
⎟ VDD =
R2
⎝ R1 + R2 ⎠
Note K = kΩ
1
2.932 =
( 200 )(10 ) ⇒ R2 = 682 K
R2
682 R1
= 200 ⇒ R1 = 283 K
682 + R1
RD =
10 − 4
=5K
1.2
EX3.5
⎛ R2 ⎞
⎛ 40 ⎞
VG = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5 = −1 V
R
R
+
⎝ 40 + 60 ⎠
⎝ 1
2 ⎠
V − ( −5 )
2
= K n (VGS − VTN )
ID = S
RS
(a)
VS = VG − VGS
( 5 − 1) − VGS = ( 0.5 )(1) (VGS2 − 2VGS + 1)
0.5VGS2 − 3.5 = 0 VGS2 = 7 VGS = 2.646 V
I D = ( 0.5 )( 2.646 − 1) ⇒ I D = 1.354 mA
VDS = 10 − (1.354 )( 3) = 5.937 V
2
4 − VGS = K n (1)(VGS − VTN )
(b)
2
(1) K n = (1.05 )( 0.5 ) = 0.525
(2) K n = ( 0.95)( 0.5 ) = 0.475
(3) VTN = (1.05 )(1) = 1.05 V
(4) VTN = ( 0.95 )(1) = 0.95 V
(1)-(3)
4 − VGS = 0.525 (VGS2 − 2.1VGS + 1.1025 )
0.525VGS2 − 0.1025VGS − 3.421 = 0
VGS =
0.1025 ± 0.010506 + 7.1841
= 2.652 V
2 ( 0.525 )
2
I D = 0.525 ( 2.652 − 1.05 ) = 1.348 mA
VDS = 10 − (1.348 )( 3) = 5.957 V
(2)-(4)
4 − VGS = 0.475 (VGS2 − 1.9VGS + 0.9025 )
0.475VGS2 + 0.0975VGS − 3.5713 = 0
VGS =
−0.0975 ± 0.00950625 + 6.78547
2 ( 0.475 )
VGS = 2.641 V
2
I D = 0.475 ( 2.641 − 0.95 ) = 1.359 mA
VDS = 10 − (1.359 )( 3) = 5.924 V
(1)-(4)
4 −VGS = ( 0.525) (VGS2 −1.9VGS + 0.9025)
0.525 VGS2 + 0.0025VGS − 3.5262 = 0
VGS =
−0.0025 ± 0.00000625 + 7.40502
2 ( 0.525)
= 2.5893 V
2
I D = ( 0.525)( 2.5893 − 0.95) = 1.411
VDS = 10 − I D ( 3) = 5.7678 V
(2)-(3)
4 − VGS = 0.475 (VGS2 − 2.1VGS +1.1025 )
0.475VGS2 + 0.0025VGS − 3.4763 = 0
−0.0025 ± 0.00000625 + 6.60499
2(0.475)
= 2.7027
VGS =
VGS
I D = (0.475)(2.7027 − 1.05) 2 = 1.2973 mA
VDS = 10 − I D (3) = 6.108 V
1.297 ≤ I DQ ≤ 1.411 mA
5.768 ≤ VDS ≤ 6.108 V
EX3.6
⎛ R2 ⎞
VG = ⎜
⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛ 200 ⎞
=⎜
⎟ (10 ) − 5 = 0.714 V
⎝ 350 ⎠
VS = 5 − I D RS = 5 − (1.2 ) I D
So
VSG = VS − VG = 5 − (1.2 ) I D − 0.714
= 4.286 − (1.2 ) I D
ID =
4.286 − VSG
1.2
I D = K p (VSG + VTP )
2
(
2
4.286 − VSG = (1.2 )( 0.25 ) × VSG
− 2VSG ( −1) + ( −1)
2
)
2
SG
4.286 − VSG = ( 0.3) V − 0.6VSG + 0.3
2
0.3VSG
+ 0.4VSG − 3.986 = 0
VSG =
( 0.4 )
−0.4 ±
2
+ 4 ( 0.3)( 3.986 )
2 ( 0.3)
Must use + sign ⇒ VSG = 3.04 V
I D = ( 0.25 )( 3.04 − 1) ⇒ I D = 1.04 mA
2
VSD = 10 − I D ( RS + RD ) = 10 − (1.04 )(1.2 + 4 ) ⇒ VSD = 4.59 V
VSD > VSD ( sat ) , Yes
EX3.7
VSD = 10 − I DQ ( RS + RP )
2
VSD = 10 − K P (VSG + VTP ) ( RS + RP )
Set VSD = VSG + VTP
2
VSG + VTP = 10 − ( 0.25 )(VSG + VTP ) ( 5.2 )
2
1.3 (VSG + VTP ) + (VSG + VTP ) − 10 = 0
(VSG + VTP ) =
−1 ± 1 + 4 (1.3)(10 )
2 (1.3)
= 2.415 V
( 3.42 V )
VSD = 2.415 V ( 2.42 V )
2
I D = ( 0.25 )( 2.415 ) = 1.46 mA
VSG = 3.415 V
EX3.8
⎛ R2 ⎞
⎛ 240 ⎞
VG = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5
R
R
+
⎝ 240 + 270 ⎠
⎝ 1
2 ⎠
VG = −0.294 V
ID =
VS − ( −5 )
RS
=
VG − VGS + 5
2
= K n (VGS − VTN )
RS
⎛ 0.08 ⎞
2
4.706 − VGS = ⎜
⎟ ( 4 )( 3.9 ) (VGS − 2.4VGS + 1.44 )
⎝ 2 ⎠
0.624VGS2 − 0.4976VGS − 3.80744 = 0
VGS =
0.4976 ± 0.2476 + 9.50337
2 ( 0.624 )
VGS = 2.90 V
2
⎛ 0.08 ⎞
ID = ⎜
⎟ ( 4 )( 2.90 − 1.2 ) ⇒ I D = 0.463 mA
2
⎝
⎠
VDS = 10 − I D ( 3.9 + 10 ) ⇒ VDS = 3.57 V
EX3.9
10 − VSG
2
ID =
and I D = K p (VSG + VTP )
RS
0.12 = ( 0.050 )(VSG − 0.8 )
VSG = 2.35 V
2
10 − 2.35
⇒ RS = 63.75 kΩ
0.12
VSD = 8 = 20 − I D ( RS + RD )
8 = 20 − ( 0.12 )( 63.75 ) − ( 0.12 ) RD
RS =
RD =
(1)
(2)
(3)
(4)
⇒ RD = 36.25 kΩ
0.12
= ( 0.05 )(1.05 ) = 0.0525
= ( 0.05 )( 0.95 ) = 0.0475
= −0.8 (1.05 ) = −0.84 V
= −0.8 ( 0.95 ) = −0.76 V
10 − VSG
2
=
= K P (VSG + VTP )
RS
20 − ( 0.12 )( 63.75 ) − 8
KP
KP
VTP
VTP
ID
(1)-(3)
2
− 1.68VSG + 0.7056 ⎤⎦
10 − VSG = ( 63.75 )( 0.0525 ) ⎡⎣VSG
2
− 4.623VSG − 7.6384 = 0
3.347VSG
VSG =
4.623 ± 21.372 + 102.263
2 ( 3.347 )
VSG = 2.352 V ⇒ I D ≅ 0.120 mA
VSD ≈ 8.0 V
(2)-(4)
2
− 1.52VSG + 0.5776 ⎤⎦
10 − VSG = ( 63.75 )( 0.0475 ) ⎡⎣VSG
2
− 3.603VSG − 8.251 = 0
3.028VSG
VSG =
3.603 ± 12.9816 + 99.936
2 ( 3.028 )
VSG = 2.35 V
I D ≈ 0.120
VSD ≈ 8.0
(1)-(4)
2
− 1.52VSG + 0.5776 ⎤⎦
10 − VSG = ( 63.75 )( 0.0525 ) ⎡⎣VSG
2
− 4.087VSG − 8.06685 = 0
3.347VSG
VSG =
4.087 ± 16.7036 + 107.999
2 ( 3.347 )
VSG = 2.279 V
I D = 0.121 mA
VSD = 7.89 V
(2)-(3)
2
− 1.68VSG + 0.7056 ⎤⎦
10 − VSG = ( 63.75 )( 0.0475 ) ⎡⎣VSG
2
− 4.0873VSG − 7.8634 = 0
3.028VSG
VSG =
4.0873 ± 16.706 + 95.242
2 ( 3.028 )
VSG = 2.422 V
I D = 0.119 mA
VSD = 8.11 V
Summary 0.119 ≤ I D ≤ 0.121 mA
7.89 ≤ VSD ≤ 8.11 V
EX3.10
V − VGS
I D = DD
,
RS
I D = K n (VGS − VTN )
2
10 − VGS = (10 )( 0.2 ) (VGS2 − 2VGSVTN + VTN2 )
10 − VGS = 2VGS2 − 8VGS + 8
2VGS2 − 7VGS − 2 = 0
VGS =
7±
2
(7) + 4 ( 2) 2
2 ( 2)
Use + sign: VGS = VDS = 3.77 V
10 − 3.77
⇒ I D = 0.623 mA
10
Power = I DVDS = ( 0.623)( 3.77 ) ⇒ Power = 2.35 mW
ID =
EX3.11
(a) VI = 4 V, Driver in Non ⋅ Sat.
K nD ⎣⎡ 2 (VI − VTND ) VO − VO2 ⎦⎤ = K nL [VDD − VO − VTNL ]
2
5 ⎡⎣ 2 ( 4 − 1) VD − VD2 ⎤⎦ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2
2
2
6VD2 − 38VO + 16 = 0
VD =
38 ± 1444 − 384
2 ( 6)
VD = 0.454 V
(b) VI = 2 V Driver: Sat
2
2
K nD [VI − VTND ] = K nL [VDD − VO − VTNL ]
2
2
5 [ 2 − 1] = [5 − VO − 1]
5 = 4 − VO ⇒ VO = 1.76 V
EX3.12
If the transistor is biased in the saturation region
2
I D = K n (VGS − VTN ) = K n ( −VTN )
2
I D = ( 0.25 )( 2.5 ) ⇒ I D = 1.56 mA
2
VDS = VDD − I D RS = 10 − (1.56 )( 4 ) ⇒ VDS = 3.75
VDS > VGS − VTN = −VTN
3.75 > − ( −2.5 )
Yes — biased in the saturation region
Power = I DVDS = (1.56 )( 3.75 ) ⇒ Power = 5.85 mW
EX3.13
(a) For VI = 5 V, Load in saturation and driver in nonsaturation.
I DD = I DL
K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL ( −VTNL )
2
I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡⎣ − ( −2 ) ⎤⎦
2
K nD ⎡
K
2
2 ( 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06
⎣
⎦
K nL
K nL
(b)
2
K nL = 50 µ A / V 2 and K nD = 103 µ A / V 2
EX3.14
For M N
I DN = I DP
2
K n (VGSN − VTN ) = K p (Vscop + VTP )
2
VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI
Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V
For M P : VI = 1.75 V
VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V
So Vot = 5 − 0.75 ⇒ Vot = 4.25 V
EX3.15
For RD = 10 k Ω, VDD = 5 V, and Vo = 1 V
5 −1
= 0.4 mA
10
2
⎤⎦
I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS
ID =
2
I D = 0.4 = K n ⎡ 2 ( 5 − 1)(1) − (1) ⎤ ⇒ K n = 0.057 mA / V 2
⎣
⎦
P = I D ⋅ VDS = ( 0.4 )(1) ⇒ P = 0.4 mW
EX3.16
a.
V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0
I D = K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ =
5 − VO
RD
( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦ = 5 − V0
1.5V02 − 13V0 + 5 = 0
V0 =
13 ±
(13) − 4 (1.5 )( 5)
⇒ V0 = 0.40 V
2 (1.5 )
2
5 − 0.40
⇒ I R = I D1 = 0.153 mA
30
V1 = V2 = 5 V
I R = I D1 =
b.
5 − VO
= 2 K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦
RD
{
}
5 − V0 = 2 ( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦
3V02 − 25V0 + 5 = 0
V0 =
25 ±
( 25) − 4 ( 3)( 5 )
⇒ V0 = 0.205 V
2 ( 3)
2
5 − 0.205
⇒ I R = 0.160 mA
30
= I D 2 = 0.080 mA
IR =
I D1
EX3.17
M 2 & M 3 watched ⇒ I Q1 = I REF 1 = 0.4 mA
0.4 = 0.3 (VGS 3 − 1) ⇒ VGS 3 = VGS 2 = 2.15 V
2
0.4 = 0.6 (VGS 1 − 1) ⇒ VGS1 = 1.82 V
2
EX3.18
2
⎛ 0.04 ⎞
0.1 = ⎜
⎟ (15 )(VSGC − 0.6 )
⎝ 2 ⎠
VSGC = 1.177 V = VSGB
2
⎛ 0.04 ⎞ ⎛ W ⎞
0.2 = ⎜
⎟ ⎜ ⎟ (1.177 − 0.6 )
⎝ 2 ⎠ ⎝ L ⎠B
⎛W ⎞
⎜ ⎟ = 30
⎝ L ⎠B
2
⎛ 0.04 ⎞
0.2 = ⎜
⎟ ( 25 )(VSGA − 0.6 )
⎝ 2 ⎠
VSGA = 1.23 V
EX3.19
2
(a)
I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN )
VGS 3 = 2 V ⇒ VGS 4 = 3 V
K
K
1
2
2
( 2 − 1) = n 4 ( 3 − 1) ⇒ n 4 =
K n3
K n3 4
(b)
I Q = K n 2 (VGS 2 − VTN )
But VGS 2 = VGS 3 = 2 V
2
2
0.1 = K n 2 ( 2 − 1) ⇒ K n 2 = 0.1 mA / V 2
2
(c)
0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2
2
0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2
2
EX3.20
VS 2 = 5 − 5 = 0
RS 2 =
I D 2 = K n 2 (VGS 2 − VTN 2 )
5
= 16.7 K
0.3
2
0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V
2
5 − 2.425
= 25.8 K
0.1
VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V
RD1 =
RS 1 =
−2.575 − ( −5 )
0.1
⇒ RS 1 = 24.3 K
I D1 = K n1 (VGS 1 − VTN 1 )
2
0.1 = 0.5 (VGS1 − 1.2 ) ⇒ VGS 1 = 1.647 V
VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V
2
⎛ R2 ⎞
1
VG1 = ⎜
⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5
R1
⎝ R1 + R2 ⎠
1
−0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K
R1
491 R2
= 200 ⇒ R2 = 337 K
491 + R2
EX3.21
VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V
I DQ = K n (VGS1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V
VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V
⎛
⎞
R3
R3
(5) ⇒ R3 = 50.7 K
VG1 = ⎜
⎟ (5) ⇒ 0.507 =
500
⎝ R1 + R2 + R3 ⎠
VS 2 = VS 1 + VDS1 = −1 + 2.5 = 1.5 V
VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V
⎛ R2 + R3 ⎞
⎛ R2 + R3 ⎞
VG 2 = ⎜
(5)
⎟ (5) ⇒ 3.007 = ⎜
R
R
R
500 ⎟⎠
+
+
⎝
2
3 ⎠
⎝ 1
R2 + R3 = 300.7
R2 = 300.7 − 50.7 ⇒ R2 = 250 K
R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K
VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V
5−4
⇒ RD = 4 K
RD =
0.25
EX3.22
VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V
⎛ ( −1.2 ) ⎞
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟ = 12 ⎜1 −
⎟⎟ ⇒ I D = 6.45 mA
⎜
VP ⎠
⎝
⎝ ( −4.5 ) ⎠
2
2
EX3.23
Assume the transistor is biased in the saturation region.
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
VP ⎠
⎝
2
⎛
⎞
V
8 = 18 ⎜1 − GS ⎟ ⇒ VGS = −1.17 V ⇒ VS = −VGS = 1.17
⎜ ( −3.5 ) ⎟
⎝
⎠
VD = 15 − ( 8 )( 0.8 ) = 8.6
VDS = 8.6 − (1.17 ) = 7.43 V
VDS = 7.43 > VGS − VP = −1.17 − ( −3.5 ) = 2.33
2
Yes, the transistor is biased in the saturation region.
EX3.24
I D = 2.5 mA
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
VP ⎠
⎝
2
⎛
V ⎞
2.5 = 6 ⎜1 − GS ⎟ ⇒ VGS = −1.42 V
⎜ ( −4 ) ⎟
⎝
⎠
VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5
VS = −4.375
VDS = 6 ⇒ VD = 6 − 4.375 = 1.625
5 − 1625
RD =
⇒ RD = 1.35 kΩ
2.5
2
( 20 )
2
R1 + R2
= 2 ⇒ R1 + R2 = 200 kΩ
VG = VGS + VS = −1.42 − 4.375 = −5.795
⎛ R2 ⎞
VG = ⎜
⎟ ( 20 ) − 10
⎝ R1 + R2 ⎠
⎛ R ⎞
−5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ
⎝ 200 ⎠
R1 = 157.95 kΩ → 158 kΩ
EX3.25
VS = −VGS . I D =
0 − VS VGS
=
RS
RS
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛ V
VGS
V2 ⎞
⎛ V ⎞
= 6 ⎜ 1 − GS ⎟ = 6 ⎜ 1 − GS + GS ⎟
1
4 ⎠
2
16 ⎠
⎝
⎝
2
0.375VGS − 4VGS + 6 = 0
2
VGS =
4 ± 16 − 4 ( 0.375 )( 6 )
2 ( 0.375 )
VGS = 8.86 or VGS = 1.806 V
impossible
ID =
VGS
= 1.806 mA
RS
VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278
VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V
VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19
So VSD > VSD ( sat )
EX3.26
Rin = R1 & R2 =
R1 R2
= 100 kΩ
R1 + R2
I DQ = 5 mA, VS = − I DQ RS = − ( 5 )(1.2 ) = −6 V
VSDQ = 12 V ⇒ VD = VS − VSDQ
= −6 − 12 = −18 V
RD =
−18 − ( −20 )
5
⇒ RD = 0.4 kΩ
⎛ V ⎞
⎛ V ⎞
I DQ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜1 − GS ⎟
VP ⎠
4 ⎠
⎝
⎝
VGS = 0.838 V
2
2
VG = VGS + VS = 0.838 − 6 = −5.162
⎛ R2 ⎞
VG = ⎜
⎟ ( −20 )
⎝ R1 + R2 ⎠
1
−5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ
R1
R1 R2
= 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2
R1 + R2
( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ
TYU3.1
VTN = 1.2 V , VGS = 2 V
(a)
V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V
(i) VDS = 0.4 ⇒ Nonsaturation
(ii) VDS = 1 ⇒ Saturation
(iii) VDS = 5 ⇒ Saturation
VTN = −1.2 V , VGS = 2 V
(b)
V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V
(i) VDS = 0.4 ⇒ Nonsaturation
(ii) VDS = 1 ⇒ Nonsaturation
(iii) VDS = 5 ⇒ Saturation
TYU3.2
(a)
W µ n Cox
2L
−14
∈ox ( 3.9 ) ( 8.85 × 10 )
Cox =
=
= 7.67 × 10−8 F / cm
−8
tox
450 × 10
Kn =
Kn =
(100 )( 500 ) ( 7.67 ×10−8 )
⇒ K n = 0.274 mA / V 2
2 (7)
(b) VTN = 1.2 V, VGS = 2 V
(i)
(ii)
VDS = 0.4 V ⇒ Nonsaturation
2
I D = ( 0.274 ) ⎡ 2 ( 2 − 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.132 mA
⎣
⎦
VDS = 1 V ⇒ Saturation
I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA
2
(iii) VDS = 5 V ⇒ Saturation
I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA
2
VTN = −1.2 V , VGS = 2 V
(i)
(ii)
VDS = 0.4 V ⇒ Nonsaturation
2
I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.658 mA
⎣
⎦
VDS = 1 V ⇒ Nonsaturation
2
I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⇒ I D = 1.48 mA
⎣
⎦
(iii) VDS = 5 V ⇒ Saturation
I D = ( 0.274 )( 2 + 1.2 ) ⇒ I D = 2.81 mA
2
TYU3.3
(a) VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V
(i) Non Sat (ii) Sat
(iii) Sat
(b) VSD (sat) = 2 + 1.2 = 3.2 V
(i) Non Sat (ii) Non Sat
(iii) Sat
TYU3.4
(a)
(3.9)(8.85 × 10−14 )
⎛ W ⎞ ⎛ µ p Cox ⎞
KP = ⎜ ⎟ ⎜
⎟ Cox =
350 × 10−8
⎝ L ⎠⎝ Z ⎠
= 9.861× 10−8
KP =
(40) ⎛ ( 300 ) ( 9.861× 10
⎜
(2) ⎜
2
⎝
K P = 0.296 mA / V
(b)
(i)
−8
) ⎞⎟
⎟
⎠
2
I D = (0.296) ⎡⎣ 2(2 − 1.2)(0.4) − (0.4) 2 ⎤⎦
= 0.142 mA
(ii)
I D = (0.296) [ 2 − 1.2] ⇒ I D = 0.189 mA
2
(iii) ID = 0.189 mA
2
(i) I D = (0.296) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤
⎣
⎦
= 0.710 mA
(ii)
2
I D = (0.296) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤
⎣
⎦
=1.60 mA
(iii) I D = ( 0.296 )( 2 + 1.2 )
= 3.03 mA
TYU3.5
2
(a)
λ = 0, VDS ( sat ) = 2.5 − 0.8 = 1.7 V
For VDS = 2 V , VDS = 10 V ⇒ Saturation Region
I D = ( 0.1)( 2.5 − 0.8 ) ⇒ I D = 0.289 mA
2
(b)
λ = 0.02 V −1
2
I D = K n (VGS − VTN ) (1 + λVDS )
For VDS = 2 V
I D = ( 0.1)( 2.5 − 0.8 ) ⎡⎣1 + ( 0.02 )( 2 ) ⎦⎤ ⇒ I D = 0.300 mA
VDS = 10 V
2
(c)
2
I D = ( 0.1) ⎡( 2.5 − 0.8 ) (1 + ( 0.02 )(10 ) ) ⎤ ⇒ I D = 0.347 mA
⎣
⎦
For part (a), λ = 0 ⇒ ro = ∞
For part (b), λ = 0.02 V −1 ,
2
2
ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡( 0.02 )( 0.1)( 2.5 − 0.8 ) ⎤
⎣
⎦
⎣
⎦
−1
−1
or ro = 173 k Ω
TYU3.6
VTN = VTNO + γ ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦
2φ f = 0.70 V , VTNO = 1 V
(a)
(b)
(c)
VSB = 0 ⇒, VTN = 1 V
VSB = 1 V , VTN = 1 + ( 0.35 ) ⎣⎡ 0.7 + 1 − 0.7 ⎦⎤ ⇒ VTN = 1.16 V
VSB = 4 V , VTN = 1 + ( 0.35 ) ⎣⎡ 0.7 + 4 − 0.7 ⎦⎤ ⇒ VTN = 1.47 V
TYU3.7
I D = K n (VGS − VTN )
2
0.4 = 0.25 (VGS − 0.8 ) ⇒ VGS = 2.06 V
2
⎛ R2 ⎞
VGS = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
2.06 = ⎜ 2 ⎟ ( 7.5 ) ⇒ R2 = 68.8 kΩ
⎝ 250 ⎠
R1 = 181.2 kΩ
VDS = 4 = VDD − I D RD
7.5 − 4
RD =
⇒ RD = 8.75 kΩ
0.4
VDS > VDS ( sat ) , Yes
TYU3.8
VS − ( −5 )
ID =
and VS = −VGS
RS
So RS =
5 − VGS
0.1
I D = K n (VGS − VTN )
2
0.1 = ( 0.080 )(VGS − 1.2 ) ⇒ VGS = 2.32 V
2
5 − 2.32
⇒ RS = 26.8 kΩ
0.1
= VD − VS ⇒ VD = VDS + VS = 4.5 − 2.32
So RS =
VDS
VD = 2.18
5 − VD 5 − 2.18
RD =
=
⇒ RD = 28.2 kΩ
ID
0.1
VDS > VDS ( sat ) , Yes
TYU3.9
For VDS = 2.2 V
5 − 2.2
⇒ I D = 0.56 mA
ID =
5
I D = K n (VGS − VTN )
0.56 = K n ( 2.2 − 1)
2
2
K n = 0.389 mA / V =
W µ n Cox
⋅
2
L
W ( 389 )( 2 )
W
=
⇒
= 19.4
L
L
( 40 )
TYU3.10
(a) The transition point is
(
VDD − VTNL + VTND 1 + K nD / K nL
VIt =
=
)
1 + K nD /K nL
(
5 − 1 + 1 1 + 0.05/ 0.01
)
1 + 0.05/ 0.01
7.236
=
⇒ VIt = 2.236 V
3.236
VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V
(b) We may write
I D = K n D (VGSD − VTND ) = ( 0.05 )( 2.236 − 1) ⇒ I D = 76.4 µ A
2
2
TYU3.11
(
VDD − VTNL + VTND 1 + K nD /K nL
VIt =
)
1 + K nD /K nL
2.5 =
(
5 − 1 + 1 1 + K nD /K nL
)
1 + K nD /K nL
2.5 + 2.5 K nD /K nL = 5 + K nD /K nL ⇒ K nD /K nL =
b.
For VI = 5, driver in nonsaturated region.
5 − 2.5
= 1.67 ⇒ K nD /K nL = 2.78
1.5
I DD = I DL
K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL (VGSL − VTNL )
2
K nD
2
⎡ 2 (VI − VTND ) VO − VO2 ⎤⎦ = [VDD − VO − VTNL ]
K nL ⎣
2.78 ⎡⎣ 2 ( 5 − 1) V0 − V02 ⎤⎦ = [5 − V0 − 1]
2
2
22.24V0 − 2.78V02 = ( 4 − V0 )
= 16 − 8V0 + V02
3.78V02 − 30.24V0 + 16 = 0
V0 =
30.24 ±
( 30.24 ) − 4 ( 3.78 )(16 )
⇒ V0 = 0.57 V
2 ( 3.78 )
2
TYU3.12
We have VDS = 1.2 V < VGS − VTN = −VTN = 1.8 V
Transistor is biased in the nonsaturation region.
VDD − VDS 5 − 1.2
2
=
⇒ I D = 0.475 mA
I D = K n ⎣⎡ 2 (VGS − VTN ) VDS − VDS
⎦⎤ and I D =
8
RS
0.475 = K n ⎡ 2 ( 0 − ( −1.8 ) ) (1.2 ) − (1.2 ) ⎤
⎣
⎦
0.475 = K n ( 2.88 ) ⇒ K n = 0.165 mA/V 2
2
W µ n Cox
⋅
2
L
165
)( 2 ) W
W (
=
⇒
= 9.43
35
L
L
Kn =
TYU3.13
(a) Transition point for the load transistor – Driver is in the saturation region.
I DD = I DL
2
K nD (VGSD − VTND ) = K nL (VGSL − VTNL )
2
VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V
Then VOt = 5 − 2 = 3 V , VOt = 3 V
K nD
(VIt − 1) = ( −VTNL )
K nL
0.08
(VIt − 1) = 2 ⇒ VIt = 1.89 V
0.01
(b) For the driver:
VOt = VIt − VTND
VIt = 1.89 V , VOt = 0.89 V
TYU3.14
2
⎤⎦
I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS
2
= ( 0.050 ) ⎡ 2 (10 − 0.7 )( 0.35 ) − ( 0.35 ) ⎤
⎣
⎦
I D = 0.319 mA
RD =
VDD − Vo 10 − 0.35
=
⇒ RD = 30.3 kΩ
ID
0.319
TYU3.15
(a) Transistor biased in the nonsaturation region
5 − 1.5 − VDS
= 12
R
2
⎤⎦
I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS
ID =
2
⎤⎦
12 = 4 ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS
2
− 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V
4VDS
Then R =
TYU3.16
a.
ID =
5 − 1.5 − 0.374
⇒ R = 261 Ω
12
5 − VO
= K n ⎡⎣ 2 (V2 − VTN ) VO − VO2 ⎤⎦
RD
5 − ( 0.10 )
b.
2
= K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2
⎣
⎦
25
5 − V0
= 2 ( 0.248 ) ⎣⎡ 2 ( 5 − 1) V0 − V02 ⎦⎤
25
5 − V0 = 12.4 ⎡⎣8V0 − V02 ⎤⎦
12.4V02 − 100.2V0 + 5 = 0
V0 =
100.2 ±
(100.2 ) − 4 (12.4 )( 5 )
⇒ V0 = 0.0502 V
2 (12.4 )
2
TYU3.17
2
2
I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V
VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V
VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V
VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V
TYU3.18
2
⎤⎦
I D = K ⎡⎣ 2 (VGS − VTN ) VDS − VDS
2
= 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤
⎣
⎦
ID = 9 µA
RD =
2.5 − 0.1
⇒ RD = 267 kΩ
0.009
Chapter 3
Problem Solutions
3.1
⎛ W ⎞ ⎛ k ′ ⎞ ⎛ 10 ⎞ ⎛ 0.08 ⎞
2
Kn = ⎜ ⎟ ⎜ n ⎟ = ⎜
⎟⎜
⎟ = 0.333 mA/V
L
2
1.2
2
⎝ ⎠⎝ ⎠ ⎝
⎠⎝
⎠
For VDS = 0.1 V ⇒ Non Sat Bias Region
(a)
VGS = 0 ⇒ I D = 0
(b)
2
VGS = 1 V I D = 0.333 ⎡ 2 (1 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.01 mA
⎣
⎦
(c)
VGS = 2 V
(d)
VGS = 3 V
2
I D = 0.333 ⎡ 2 ( 2 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.0767 mA
⎣
⎦
2
I D = 0.333 ⎡ 2 ( 3 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.143 mA
⎣
⎦
3.2
All in Sat region
⎛ 10 ⎞⎛ 0.08 ⎞
2
Kn = ⎜
⎟⎜
⎟ = 0.333 mA/V
⎝ 1.2 ⎠⎝ 2 ⎠
(a)
ID = 0
2
(b)
I D = 0.333[1 − 0.8] = 0.0133 mA
(c)
I D = 0.333[ 2 − 0.8] = 0.480 mA
(d)
I D = 0.333[3 − 0.8] = 1.61 mA
2
2
3.3
(a)
Enhancement-mode
(b)
From Graph VT = 1.5 V
Now
2
0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12
2
K n = 0.0666
2
K n = 0.0624
2
K n = 0.0629
0.15 = K n ( 3 − 1.5 ) = 2.25 K n
0.39 = K n ( 4 − 1.5 ) = 6.25 K n
0.77 = K n ( 5 − 1.5 ) = 12.25 K n
From last three, K n (Avg) = 0.0640 mA/V 2
(c)
3.4
a.
iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V
iD (sat) = 0.0640(4.5 − 1.5) 2 ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V
VGS = 0
VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V
i.
ii.
VDS = 0.5 V ⇒ Biased in nonsaturation
2
I D = (1.1) ⎡ 2 ( 0 − (−2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA
⎣
⎦
VDS = 2.5 V ⇒ Biased in saturation
I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA
2
iii.
VDS = 5 V Same as (ii) ⇒ I D = 6.88 mA
b.
VGS = 2 V
VDS ( sat ) = 2 − ( −2.5 ) = 4.5 V
i.
VDS = 0.5 V ⇒ Nonsaturation
I D = (1.1) ⎣⎡ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎦⎤ ⇒ I D = 4.68 mA
ii.
VDS = 2.5 V ⇒ Nonsaturation
iii.
VDS = 5 V ⇒ Saturation
I D = (1.1) ⎡⎣ 2(2 − (−2.5))(2.5) − (2.5) 2 ⎤⎦ ⇒ I D = 17.9 mA
I D = (1.1) ( 2 − ( −2.5 ) ) ⇒ I D = 22.3 mA
2
3.5
VDS > VGS − VTN = 0 − ( −2 ) = 2 V
Biased in the saturation region
k′ W
2
I D = n ⋅ (VGS − VTN )
2 L
2
W
⎛ 0.080 ⎞ ⎛ W ⎞
= 9.375
1.5 = ⎜
⎟ ⎜ ⎟ ⎡⎣0 − ( −2 ) ⎤⎦ ⇒
L
⎝ 2 ⎠⎝ L ⎠
3.6
kn′ = µ n Cox =
µ n ∈ox
tox
=
( 600 )( 3.9 ) (8.85 ×10−14 )
tox
(a)
500 A
kn′ = 41.4 µ A/V 2
(b)
250
kn′ = 82.8 µ A/V 2
(c)
100
kn′ = 207 µ A/V 2
(d)
50
kn′ = 414 µ A/V 2
(e)
25
kn′ = 828 µ A/V 2
=
2.071× 10−10
tox
3.7
a.
Cox =
Kn =
∈ox ( 3.9 ) ( 8.85 × 10
=
t0 x
450 × 10−8
−14
)⇒∈
ox
t0 x
= 7.67 ×10−8 F/cm 2
µ n Cox W
2
⋅
L
1
64
( 650 ) ( 7.67 ×10−8 ) ⎛⎜ ⎞⎟
2
⎝ 4 ⎠
K n = 0.399 mA / V 2
=
b.
VGS = VDS = 3 V ⇒ Saturation
I D = K n (VGS − VTN ) = ( 0.399 )( 3 − 0.8 ) ⇒ I D = 1.93 mA
2
3.8
2
⎛ ω ⎞⎛ k′ ⎞
I D = ⎜ ⎟ ⎜ n ⎟ (VGS − VTN )
2
2
⎝ ⎠⎝ ⎠
2
⎛ ω ⎞ ⎛ 0.08 ⎞
1.25 ⎜
⎟⎜
⎟ ( 2.5 − 1.2 ) ⇒ ω = 23.1 µ m
⎝ 1.25 ⎠ ⎝ 2 ⎠
3.9
∈
Cox = ox
t0 x
=
( 3.9 ) (8.85 ×10−14 )
400 × 10−8
= 8.63 × 10−8 F/cm 2
2
Kn =
µ n Cox W
2
⋅
L
1
⎛W ⎞
= ( 600 ) ( 8.63 × 10−8 ) ⎜
⎟
2
⎝ 2.5 ⎠
K n = (1.036 × 10−5 ) W
I D = K n (VGS − VTN )
2
1.2 × 10 −3 = (1.036 × 10 −5 ) W ( 5 − 1) ⇒ W = 7.24 µ m
2
3.10
Biased in the saturation region in both cases.
k p′ W
2
I D = ⋅ (VSG + VTP )
2 L
2
⎛ 0.040 ⎞⎛ W ⎞
0.225 = ⎜
(1)
⎟⎜ ⎟ ( 3 + VTP )
2
L
⎝
⎠⎝ ⎠
2
⎛ 0.040 ⎞ ⎛ W ⎞
1.40 = ⎜
(2)
⎟ ⎜ ⎟ ( 4 + VTP )
⎝ 2 ⎠⎝ L ⎠
Take ratio of (2) to (1):
(4 + VTP ) 2
1.40
= 6.222 =
0.225
(3 + VTP ) 2
6.222 = 2.49 =
4 + VTP
⇒ VTP = −2.33 V
3 + VTP
W
2
⎛ 0.040 ⎞ ⎛ W ⎞
Then 0.225 = ⎜
= 25.1
⎟ ⎜ ⎟ ( 3 − 2.33) ⇒
L
⎝ 2 ⎠⎝ L ⎠
3.11
VS = 5 V, VG = 0 ⇒ VSG = 5 V
VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V
a.
VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation
I D = 2 ( 5 − 0.5 ) ⇒ I D = 40.5 mA
2
b.
c.
VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation
2
I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA
⎣
⎦
VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation
d.
2
I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA
⎣
⎦
VD = 5 V ⇒ VSD = 0 ⇒ I D = 0
3.12
(a)
(b)
Enhancement-mode
From Graph VTP = + 0.5 V
0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p =
2
0.20
2
0.20
1.25 = k p ( 3 − 0.5 ) = 6.25 K p
2
2.45 = k p ( 4 − 0.5 ) = 12.25 K p
2
4.10 = k p ( 5 − 0.5 ) = 20.25 K p
(c)
0.20
0.202
Avg K p = 0.20 mA/V 2
iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA
iD (sat) = 0.20 (4.5 − 0.5) 2 = 3.2 mA
3.13
VSD ( sat ) = VSG + VTP
VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V
(a)
VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V
(b)
(c)
ID =
(a)
(b)
(c)
VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V
k p′ W
k ′p W
2
2
⋅ (VSG + VTP ) = ⋅ ⋅ ⎡⎣VSD ( sat ) ⎤⎦
2 L
2 L
2
⎛ 0.040 ⎞
ID = ⎜
⎟ ( 6 )(1) ⇒ I D = 0.12 mA
⎝ 2 ⎠
2
⎛ 0.040 ⎞
ID = ⎜
⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA
⎝ 2 ⎠
2
⎛ 0.040 ⎞
ID = ⎜
⎟ ( 6 )( 3) ⇒ I D = 1.08 mA
⎝ 2 ⎠
3.14
VSD (sat) = VSG + VTP = 3 − 0.8 = 2.2 V
⎛ 15 ⎞⎛ 0.04 ⎞
2
KP = ⎜
⎟⎜
⎟ = 0.25 mA/V
⎝ 1.2 ⎠⎝ 2 ⎠
a)
b)
c)
d)
e)
2
VSD = 0.2 Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = 0.21 mA
⎣
⎦
2
VSD = 1.2 V Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )(1.2 ) − (1.2 ) ⎤ = 0.96 mA
⎣
⎦
2
VSD = 2.2 V Sat I D = 0.25(3 − 0.8) = 1.21 mA
VSD = 3.2 V Sat ID = 1.21 mA
VSD = 4.2 V Sat ID = 1.21 mA
3.15
k ′p = µ p Cox =
(a)
(b)
(c)
(d)
(e)
µ p ∈ox
t0 x
=
( 250 )( 3.9 ) (8.85 ×10−14 )
t0 x
=
tox = 500Å ⇒ k ′p = 17.3 µ A/V 2
250Å ⇒ k ′p = 34.5 µ A/V 2
100Å ⇒ k ′p = 86.3 µ A/V 2
50Å ⇒ k p′ = 173 µ A/V 2
25Å ⇒ k ′p = 345 µ A/V 2
3.16
−14
∈ox ( 3.9 ) ( 8.85 × 10 )
=
= 6.90 × 10−8 F/cm 2
Cox =
−8
t0 x
500 × 10
kn′ = ( µ n Cox ) = ( 675 ) ( 6.90 × 10−8 ) ⇒ 46.6 µ A/V 2
k ′p = ( µ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 µ A/V 2
PMOS:
8.629 × 10−11
t0 x
ID =
k ′p ⎛ W ⎞
2
⎜ ⎟ (VSG + VTP )
2 ⎝ L ⎠p
2
⎛ 0.0259 ⎞⎛ W ⎞
⎛W ⎞
0.8 = ⎜
⎟⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19
⎝ 2 ⎠⎝ L ⎠ p
⎝ L ⎠p
L = 4 µ m ⇒ W p = 12.8 µ m
⎛ 0.0259 ⎞
2
Kp = ⎜
⎟ ( 3.19 ) ⇒ K p = 41.3 µ A/V = K n
⎝ 2 ⎠
Want Kn = Kp
k ′p ⎛ W ⎞
kn′ ⎛ W ⎞
⎜ ⎟ = ⎜ ⎟ = 41.3
2 ⎝ L ⎠N
2 ⎝ L ⎠p
⎛ 46.6 ⎞ ⎛ W ⎞
⎛W ⎞
⎜
⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77
2
L
⎝
⎠ ⎝ ⎠N
⎝ L ⎠N
L = 4 µ m ⇒ WN = 7.09 µ m
3.17
2
VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA
1
1
r0 =
=
⇒ r0 = 781 kΩ
λ I D ( 0.01)( 0.128 )
2
VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA
1
r0 =
⇒ r = 63.7 kΩ
( 0.01)(1.57 ) 0
VA =
1
λ
=
1
⇒ VA = 100 V
( 0.01)
3.18
2
2
⎛ 0.080 ⎞
ID = ⎜
⎟ ( 4 )( 3 − 0.8 ) = ( 0.16 )( 3 − 0.8 ) ⇒ I D = 0.774 mA
2
⎝
⎠
1
1
1
⇒λ =
=
⇒ λ (max) = 0.00646 V −1
r0 =
λ ID
r0 I D ( 200 )( 0.774 )
VA ( min ) =
1
λ ( max )
=
1
⇒ VA ( min ) = 155 V
0.00646
3.19
VTN = VTNO + γ ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦
ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤
⎣
⎦
2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V
3.20
VTN = VTNo + r ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦
= 0.75 + 0.6 ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤
⎣
⎦
= 0.75 + 0.6 [1.934 − 0.860]
VTN = 1.39 V
VDS (sat) = 2.5 − 1.39 = 1.11 V
2
⎛ 0.08 ⎞
Sat Region I D = (15 ) ⎜
⎟ ( 2.5 − 1.39 )
⎝ 2 ⎠
I D = 0.739 mA
(a)
2
⎛ 0.08 ⎞ ⎡
Non-Sat I D = (15 ) ⎜
⎟ ⎣ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎤⎦
2
⎝
⎠
I D = 0.296 mA
(b)
3.21
a.
VG = %ox t0 x = ( 6 × 106 )( 275 × 10−8 )
VG = 16.5 V
b.
VG =
16.5
⇒ VG = 5.5 V
3
3.22
Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 × 106 ) t0 x
t0 x = 1.2 ×10−5 cm = 1200 Angstroms
3.23
⎛ R2 ⎞
⎛ 18 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) = 3.6 V
⎝ 18 + 32 ⎠
⎝ R1 + R2 ⎠
Assume transistor biased in saturation region
V
V − VGS
2
= K n (VGS − VTN )
ID = S = G
RS
RS
3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 )
= VGS2 − 1.6VGS + 0.64
2
VGS2 − 0.6VGS − 2.96 = 0
VGS =
ID =
0.6 ±
VG − VGS
RS
⇒ VGS = 2.046 V
2
3.6 − 2.046
=
⇒ I D = 0.777 mA
2
( 0.6 )
2
+ 4 ( 2.96 )
VDS = VDD − I D ( RD + RS )
= 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V
VDS > VDS ( sat )
3.24
ID(mA)
4
(a)
Q-pt
Q-pt
1.67
(b)
4
(a)
5 V (V)
DS
VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V
2
If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56
VDS = 1.44 ×
Non-Sat
2
⎤⎦ + VDS
4 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT ) VDS − VDS
2
⎤⎦ + VDS
4 = ( 0.25 )(1) ⎡⎣ 2 ( 4 − 0.8 ) VDS − VDS
2
4 = 2.6VDS − 0.25VDS
2
0.25VDS
− 2.6VDS + 4 = 0
2.6 ± 6.76 − 4
= 1.88 V
2 ( 0.25 )
VDS =
4 − 1.88
= 2.12 mA
1
(b)
Non-Sat region
2
⎤⎦ + VDS
5 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT )VDS − VDS
ID =
5 = ( 0.25 )( 3) ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS2 ⎤⎦ + VDS
2
5 = 7.3VDS − 0.75VDS
2
− 7.3VDS + 5 = 0
0.75 VDS
VDS =
7.3 ± 53.29 − 15
2 ( 0.75 )
VDS = 0.741 V
5 − 0.741
ID =
= 1.42 mA
3
3.25
ID(mA)
2.92
(a)
Q-pt
1.25
(b)
3.5
5 V (V)
SD
(a)
VSG = VDD = 3.5
VSD ( sat ) = 3.5 − 0.8 = 2.7 V
If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 )
= 1.46 mA
VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V
2
×
Biased in Non-Sat Region.
2
⎤⎦
3.5 = VSD + I D RD = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD
2
⎤⎦
3.5 = VSD + ( 0.2 )(1.2 ) ⎡⎣ 2 ( 3.5 − 0.8 ) VSD − VSD
2
3.5 = VSD + 1.296 VSD − 0.24 VSD
2
− 2.296 VSD + 3.5 = 0
0.24 VSD
VSD =
+2.296 ± 5.272 − 3.36
use − sign VSD = 1.90 V
2 ( 0.24 )
2
I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61]
⎣
⎦
3.5 − 1.90
= 1.33 mA
ID =
1.2
I D = 1.33 mA
VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V
(b)
2
If Sat Region I D = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0
Non-Sat Region.
2
⎤⎦
5 = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD
2
⎤⎦
5 = VSD + ( 0.2 )( 4 ) ⎡⎣ 2 ( 5 − 0.8 ) VSD − VSD
2
5 = VSD + 6.72 VSD − 0.8 VSD
2
0.8 VSD
− 7.72 VSD + 5 = 0
VSD =
ID =
7.72 ± 59.598 − 16
use − sign VSD = 0.698 V
2 ( 0.8 )
5 − 0.698
⇒ I D = 1.08 mA
4
3.26
10 − VS
2
= K p (VSG + VTP )
RS
Assume transistor biased in saturation region
⎛ R2 ⎞
VG = ⎜
⎟ ( 20 ) − 10
⎝ R1 + R2 ⎠
ID =
⎛ 22 ⎞
=⎜
⎟ ( 20 ) − 10 ⇒ VG = 4.67 V
⎝ 8 + 22 ⎠
VS = VG + VSG
10 − ( 4.67 + VSG ) = (1)( 0.5 )(VSG − 2 )
2
2
− 4VSG + 4 )
5.33 − VSG = 0.5 (VSG
2
− VSG − 3.33 = 0
0.5VSG
VSG =
1±
(1)
2
+ 4 ( 0.5 )( 3.33)
2 ( 0.5 )
⇒ VSG = 3.77 V
⇒ I D = 3.12 mA
0.5
= 20 − I D ( RS + RD )
= 20 − ( 3.12 )( 0.5 + 2 ) ⇒ VSD = 12.2 V
ID =
VSD
10 − ( 4.67 + 3.77 )
VSD > VSD ( sat )
3.27
VG = 0, VSG = VS
Assume saturation region
2
I D = 0.4 = K p (VSG + VTP )
0.4 = ( 0.2 )(VS − 0.8 )
2
0.4
+ 0.8 ⇒ VS = 2.21 V
0.2
VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V
VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V
VS =
VSD > VSD ( sat )
3.28
VDD = I DQ RD + VDSQ + I DQ RS
2
⎛ k ′ ⎞⎛ W ⎞
(1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
⎝ 2 ⎠⎝ L ⎠
2
⎛ 0.060 ⎞ ⎛ W ⎞
or (2) I DQ = ⎜
⎟ ⎜ ⎟ (VGS − 1.2 )
⎝ 2 ⎠⎝ L ⎠
Let VGS = 2.5 V
Then from (1), 10 = I DQ ( 5 ) + 5 + 2.5 ⇒ I D = 0.5 mA
W
2
⎛ 0.060 ⎞⎛ W ⎞
= 9.86
Then from (2), 0.5 = ⎜
⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒
2
L
L
⎝
⎠⎝ ⎠
V
2.5
I DQ RS = VGS ⇒ RS = GS =
⇒ RS = 5 k Ω
I DQ 0.5
IR =
10
= ( 0.5 )( 0.05 ) = 0.025 mA
R1 + R2
Then R1 + R2 =
10
= 400 k Ω
0.025
⎛ R2 ⎞
⎛ R2 ⎞
⎜
⎟ (VDD ) = 2VGS ⇒ ⎜
⎟ (10 ) = 2 ( 2.5 ) ⇒ R1 = R2 = 200 k Ω
R
R
+
⎝ 400 ⎠
2 ⎠
⎝ 1
3.29
⎛ 75 ⎞
K n = ( 25 ) ⎜ ⎟ ⇒ 0.9375 mA/V 2
⎝ 2⎠
⎛ 6 ⎞
VG = ⎜
⎟ (10 ) − 5 = −2 V
⎝ 6 + 14 ⎠
(VG − VGS ) − ( −5)
2
= I D = K n (VGS − VTN )
RS
−2 − VGS + 5 = ( 0.9375 )( 0.5 )(VGS − 1)
3 − VGS = 0.469 (VGS2 − 2VGS + 1)
2
0.469 VGS2 + 0.0625 VGS − 2.53 = 0
VGS =
−0.0625 ± 0.003906 + 4.746
⇒ VGS = 2.26 V
2 ( 0.469 )
I D = 0.9375 ( 2.26 − 1) ⇒ I D = 1.49 mA
2
VDS = 10 − (1.49 )(1.7 ) ⇒ VDS = 7.47 V
3.30
20 = I DQ RS + VSDQ + I DQ RD
(1) 20 = VSG + 10 + I DQ RD
⎛ k ′p ⎞ ⎛ W ⎞
2
I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
L
2
⎝ ⎠⎝ ⎠
2
⎛ 0.040 ⎞ ⎛ W ⎞
(2) I DQ = ⎜
⎟ ⎜ ⎟ (VSG − 2 )
⎝ 2 ⎠⎝ L ⎠
For example, let I DQ = 0.8 mA and VSG = 4 V
⎛ 0.040 ⎞ ⎛ W
Then 0.8 = ⎜
⎟⎜
⎝ 2 ⎠⎝ L
I DQ RS = VSG ⇒ ( 0.8 ) RS
W
2
⎞
= 10
⎟ ( 4 − 2) ⇒
L
⎠
= 4 ⇒ RS = 5 k Ω
From (1) 20 = 4 + 10 + ( 0.8 ) RD ⇒ RD = 7.5 k Ω
IR =
20
= ( 0.8 )( 0.1) ⇒ R1 + R2 = 250 k Ω
R1 + R2
⎛ R1 ⎞
⎜
⎟ ( 20 ) = 2VSG = ( 2 )( 4 )
⎝ R1 + R2 ⎠
R1
( 20 ) = 8 ⇒ R1 = 100 k Ω, R2 = 150 k Ω
250
3.31
I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V
2
(a)
(i)
(ii)
I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V
VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V
2
VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V
(b)
(i) Same as (a) VGS = VDS = 1.516 V
(ii)
VGS = VDS = 2.61 V
3.32
I D = K n (VGS − VTN )
2
0.25 = ( 0.2 )(VGS − 0.6 )
2
0.25
+ 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V
0.2
VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V
VGS =
3.33
(a)
ID(mA)
1.0
0.808
Q-pt
0.5
3.81
RD =
10 V (V)
DS
5 −1
⇒ RD = 8 K
0.5
I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V
2
RS =
⇒ RS = 4.38 K
0.5
Let RD = 8.2 K, RS = 4.3 K
−2.81 − ( −5 )
(b)
Now
−VGS − ( −5 )
5 − VGS
= I D = 0.25 (VGS − 1.4 )
4.3
= 1.075 (VGS2 − 2.8 VGS + 1.96 )
2
1.075 VGS2 − 2.01 VGS − 2.89 = 0
VGS =
2.01 ± 4.04 + 12.427
⇒ VGS = 2.82 V
2 (1.075 )
I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA
2
VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V
(c)
If RS = 4.3 + 10% = 4.73 K
5 − VGS = 1.18 (VGS2 − 2.8VGS + 1.96 )
1.18 VGS2 − 2.31 VGS − 2.68 = 0
VGS =
2.31 ± 5.336 + 12.65
= 2.78 V
2 (1.18 )
I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA
2
If Rs = 4.3 − 10% = 3.87 K
5 − VGS = ( 0.9675 ) (VGS2 − 2.8VGS + 1.96 )
0.9675VGS2 − 1.71VGS − 3.10 = 0
VGS =
1.71 ± 2.924 + 12.0
= 2.88 V
2 ( 0.9675 )
2
I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA
3.34
VDD = VSD + I DQ R
9 = 2.5 + ( 0.1) R ⇒ R = 65 k Ω
⎛ k ′p ⎞ ⎛ W ⎞
2
I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
⎝ 2 ⎠⎝ L ⎠
W
0.025 ⎞ ⎛ W ⎞
2
=8
( 0.1) = ⎜⎛
⎟ ⎜ ⎟ ( 2.5 − 1.5 ) ⇒
L
⎝ 2 ⎠⎝ L ⎠
Then for L = 4 µ m, W = 32 µ m
3.35
5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5
I DQ = 1.25 mA
IR =
10
= (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω
R1 + R2
I DQ = K p (VSG + VTP )
2
1.25 = 0.5 (VSG + 1.5 ) ⇒
2
1.25
− 1.5 = VSG
0.5
VSG = 0.0811 V
VG = VS − VSG = 2.5 − 0.0811 = 2.42 V
⎛ R2 ⎞
VG = ⎜
⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛R ⎞
2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω
⎝ 80 ⎠
3.36
(a)
ID(mA)
0.429
Q-pt
RD =
VD − ( −5 )
I DQ
=
5−2
⇒ RD = 12 K
0.25
2
⎞ ⎛ k ′p ⎞
⎟ ⎜ ⎟ (VSG + VTP )
⎠⎝ 2 ⎠
2
⎛ 0.035 ⎞
0.25 = (15 ) ⎜
⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V
⎝ 2 ⎠
5 − 2.18
RS =
⇒ RS = 11.3 K
0.25
VSD = 2.18 − ( −2 ) = 4.18 V
(b)
k p′ = 35 + 5% = 36.75 µ A/V 2
⎛W
ID = ⎜
⎝L
5 − VSG
2
⎛ 0.03675 ⎞
I D = (15 ) ⎜
⎟ (VSG − 1.2 ) =
2
11.3
⎝
⎠
2
− 2.4VSG + 1.44 ) = 5 − VSG
3.11(VSG
2
− 6.46VSG − 0.522 = 0
3.11VSG
5 V (V)
SD
VSG =
6.46 ± 41.73 + 6.49
= 2.155 V
2 ( 3.11)
5 − 2.155
= 0.252 mA
11.3
= 10 − ( 0.252 )(12 + 11.3) = 4.13 V
ID =
VSD
k p′ = 35 − 5% = 33.25 µ A/V 2
5 − VSG
2
⎛ 0.03325 ⎞
I D = (15 ) ⎜
⎟ (VSG − 1.2 ) =
2
11.3
⎝
⎠
2
2.82 (VSG
− 2.4VSG + 1.44 ) = 5 − VSG
2
2.82VSG
− 5.77VSG − 0.939 = 0
VSG =
5.77 ± 33.29 + 10.59
= 2.198 V
2 ( 2.82 )
5 − 2.198
= 0.248 mA
11.3
= 10 − ( 0.248 )(12 + 11.3) = 4.22 V
ID =
VSD
3.37
ID =
−VSD − ( −10 )
RD
⇒5=
−6 + 10
⇒ RD = 0.8 kΩ
RD
I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 )
2
VSG =
2
5
+ 1.75 = 3.04 V ⇒ VG = −3.04
3
⎛ R2 ⎞
VG = ⎜
⎟ (10 ) − 5 = −3.04
⎝ R1 + R2 ⎠
Rin = R1 || R2 = 80 kΩ
1
⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ
R1
408 R2
= 80 ⇒ R2 = 99.5 kΩ
408 + R2
3.38
⎛ 60 ⎞
K n1 = ⎜ ⎟ ( 4 ) = 120 µ A/V 2
⎝ 2 ⎠
⎛ 60 ⎞
K n 2 = ⎜ ⎟ (1) = 30 µ A/V 2
⎝ 2 ⎠
For vI = 1 V , M1 Sat. region, M2 Non-sat region.
(a)
I D 2 = I D1
30 ⎡ 2 ( −VTNL )( 5 − vO ) − ( 5 − vO ) ⎤ = 120 (1 − 0.8 )
⎣
⎦
We find vO2 − 6.4vO + 7.16 = 0 ⇒ vO = 4.955 V
2
(b)
2
For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1
30 ⎣⎡ − ( −1.8 ) ⎦⎤ = 120 ⎡⎣ 2 ( 3 − 0.8 ) vO − vO2 ⎤⎦
2
We find 4vO2 − 17.6vO + 3.24 = 0 ⇒ vO = 0.193 V
(c)
For vI = 5 V , biasing same as (b)
30 ⎡⎣ − ( −1.8 ) ⎤⎦ = 120 ⎡⎣ 2 ( 5 − 0.8 ) vO − vO2 ⎤⎦
2
We find 4vO2 − 33.6vO + 3.24 = 0 ⇒ vO = 0.0976 V
3.39
For vI = 5 V , M1 Non-sat region, M2 Sat. region.
I D1 = I D 2
2
⎛ kn′ ⎞ ⎛ W ⎞
⎛ kn′ ⎞ ⎛ W ⎞
2
⎜ 2 ⎟ ⎜ L ⎟ ⎡⎣ 2 (VGS1 − VTN 1 ) VDS 1 − VDS 1 ⎤⎦ = ⎜ 2 ⎟ ⎜ L ⎟ (VGS 2 − VTN 2 )
⎝ ⎠ ⎝ ⎠1
⎝ ⎠ ⎝ ⎠2
2
W
2
⎛ ⎞ ⎡
⎜ ⎟ ⎣ 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎦⎤ = (1) ⎡⎣0 − ( −2 ) ⎤⎦
⎝ L ⎠1
⎛W ⎞
which yields ⎜ ⎟ = 3.23
⎝ L ⎠1
3.40
a.
M1 and M2 in saturation
2
2
K n1 (VGS 1 − VTN 1 ) = K n 2 (VGS 2 − VTN 2 )
K n1 = K n 2 , VTN 1 = VTN 2 ⇒ VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V
I D = (15 )( 40 )( 2.5 − 0.8 ) ⇒ I D = 1.73 mA
2
b.
⎛W ⎞ ⎛W ⎞
⎜ ⎟ > ⎜ ⎟ ⇒ VGS1 < VGS 2
⎝ L ⎠1 ⎝ L ⎠ 2
2
40 (VGS1 − 0.8 ) = (15 )(VGS 2 − 0.8 )
2
VGS 2 = 5 − VGS 1
1.633 (VGS 1 − 0.8 ) = ( 5 − VGS1 − 0.8 )
2.633VGS 1 = 5.506 ⇒ VGS 1 = 2.09 V
VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V
I D = (15 )(15 )( 2.91 − 0.8 ) ⇒ I D = 1.0 mA
2
3.41
(a)
V1 = VGS 3 = 2.5 V
2
⎛ W ⎞ ⎛ 0.06 ⎞
I D = 0.5 = ⎜ ⎟ ⎜
⎟ ( 2.5 − 1.2 )
⎝ L ⎠3 ⎝ 2 ⎠
⎛W ⎞
⎜ ⎟ = 9.86
⎝ L ⎠3
V2 = 6 V ⇒ VGS 2 = V2 − V1 = 6 − 2.5 = 3.5 V
2
⎛ W ⎞ ⎛ 0.06 ⎞
⎛W ⎞
0.5 = ⎜ ⎟ ⎜
⎟ ( 3.5 − 1.2 ) ⇒ ⎜ ⎟ = 3.15
⎝ L ⎠2 ⎝ 2 ⎠
⎝ L ⎠2
VGS1 = 10 − V2 = 10 − 6 = 4 V
2
⎛ W ⎞ ⎛ 0.06 ⎞
⎛W ⎞
0.5 = ⎜ ⎟ ⎜
⎟ ( 4 − 1.2 ) ⇒ ⎜ ⎟ = 2.13
⎝ L ⎠1 ⎝ 2 ⎠
⎝ L ⎠1
(b)
kn′1 = 0.06 + 5% = 0.063 mA/V 2
kn′2 = k n′3 = 0.6 − 5% = 0.057 mA/V 2
2
⎛ 0.057 ⎞
For M3: I D = ( 9.86 ) ⎜
⎟ (V1 − 1.2 )
⎝ 2 ⎠
2
⎛ 0.057 ⎞
For M2: I D = ( 3.15 ) ⎜
⎟ (V2 − V1 − 1.2 )
2
⎝
⎠
2
⎛ 0.063 ⎞
For M1: I D = ( 2.13) ⎜
⎟ (10 − V2 − 1.2 )
⎝ 2 ⎠
2
2
0.281(V1 − 1.2 ) = 0.0898 (V2 − V1 − 1.2 ) = 0.0671( 8.8 − V2 )
Take square root.
0.530 (V1 − 1.2 ) = 0.300 (V2 − V1 − 1.2 ) = 0.259 ( 8.8 − V2 )
(1) 0.830V1 = 0.300V2 + 0.276
(2) 0.559V2 = 0.300V1 + 2.64
From (2) ⇒ V2 = 0.537V1 + 4.72
Substitute into (1)
0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69
V1 = 2.53 V
Then
V2 = 0.537 ( 2.53) + 4.72
V2 = 6.08 V
3.42
ML in saturation
MD in nonsaturation
2
⎛W ⎞
⎛W ⎞
2
⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣⎡ 2 (VGSD − VTND )VDSD − VDSD ⎦⎤
⎝ L ⎠L
⎝ L ⎠D
W
2
2
(1)( 5 − 0.1 − 0.8) = ⎛⎜ ⎞⎟ ⎡⎣ 2 ( 5 − 0.8)( 0.1) − ( 0.1) ⎤⎦
L
⎝ ⎠D
⎛W ⎞
16.81 = ⎜ ⎟ [ 0.83]
⎝ L ⎠D
⎛W ⎞
⎜ ⎟ = 20.3
⎝ L ⎠D
3.43
ML in saturation
MD in nonsaturation
2
⎛W ⎞
⎛W ⎞
2
⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣⎡ 2 (VGSD − VTND )VDSD − VDSD ⎦⎤
⎝ L ⎠L
⎝ L ⎠D
W
2
2
(1)(1.8 ) = ⎜⎛ ⎟⎞ ⎣⎡ 2 ( 5 − 0.8 )( 0.05) − ( 0.05) ⎦⎤
⎝ L ⎠D
⎛W ⎞
3.24 = ⎜ ⎟ [ 0.4175]
⎝ L ⎠D
⎛W ⎞
⎜ ⎟ = 7.76
⎝ L ⎠D
3.44
VDD − V0 5 − 0.1
=
= 0.49 mA
10
RD
Transistor biased in nonsaturation
I D = 0.49
ID =
2
⎛W ⎞
= ( 0.015 ) ⎜ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤
⎣
⎦
⎝L⎠
W
⎛W ⎞
0.49 = ⎜ ⎟ 0.01005 ⇒
= 48.8
L
⎝L⎠
3.45
2
5 = I D RD + Vγ + VDS
5 = (12 ) RD + 1.6 + 0.2 ⇒ RD = 267 Ω
2
⎛ k′ ⎞⎛ W ⎞
I D = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
⎝ 2 ⎠⎝ L ⎠
W
2
⎛ 0.040 ⎞ ⎛ W ⎞
12 = ⎜
= 34
⎟ ⎜ ⎟ ( 5 − 0.8 ) ⇒
2
L
L
⎝
⎠⎝ ⎠
3.46
5 = VSD + I D RD + Vγ
5 = 0.15 + (15 ) RD + 1.6 ⇒ RD = 217 Ω
⎛ k ′p ⎞ ⎛ W ⎞
2
I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
⎝ 2 ⎠⎝ L ⎠
W
2
⎛ 0.020 ⎞⎛ W ⎞
= 85
15 = ⎜
⎟⎜ ⎟ ( 5 − 0.8 ) ⇒
L
⎝ 2 ⎠⎝ L ⎠
3.47
(a)
VDD − VO
⎛W
= 2⎜
RD
⎝L
5 − 0.2
⎛W
= 2⎜
20
⎝L
⎞⎛ 0.060 ⎞
2
⎟⎜
⎟ ⎡⎣( 2 )(VGS − VTN ) VO − VO ⎤⎦
⎠⎝ 2 ⎠
2
⎞
⎟ ( 0.030 ) ⎡⎣ 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤⎦
⎠
⎛W ⎞ ⎛W ⎞ ⎛W ⎞
0.24 = 0.0984 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.44
⎝ L ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2
(b)
5 − VO
⎛ 0.06 ⎞
2
= ( 2.44 ) ⎜
⎟ ⎡⎣ 2 ( 5 − 0.8 ) VO − VO ⎤⎦
20
2
⎝
⎠
5 − VO = 12.30VO − 1.464VO2
1.464VO2 − 13.30VO + 5 = 0
VO =
13.30 ± 176.89 − 29.28
2 (1.464 )
VO = 0.393 V
3.48
2
⎞ ⎛ k n′ ⎞
⎟ ⎜ ⎟ (VGS 1 − VTN )
2
⎠1 ⎝ ⎠
2
⎞ ⎛ kn′ ⎞
⎟ ⎜ ⎟ (VDS 2 ( sat ) )
⎠2 ⎝ 2 ⎠
2
⎛ W ⎞ ⎛ 0.08 ⎞
⎛W ⎞
⎛W ⎞
0.1 = ⎜ ⎟ ⎜
⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 10 = ⎜ ⎟
L
L
2
⎝ ⎠2 ⎝
⎠
⎝ ⎠2
⎝ L ⎠1
⎛ W ⎞ ⎛ 200 ⎞ ⎛ W ⎞
⎜ ⎟ =⎜
⎟ ⎜ ⎟ = 20
⎝ L ⎠3 ⎝ 100 ⎠ ⎝ L ⎠ 2
M1 & M2 matched.
2
⎛ 0.08 ⎞
Then 0.1 = (10 ) ⎜
⎟ (VGS 1 − 0.25 )
2
⎝
⎠
VGS1 = 0.75 V
⎛W
I Q1 = ⎜
⎝L
⎛W
I Q1 = ⎜
⎝L
VD1 = −0.75 + 2 = 1.25 V
RD =
2.5 − 1.25
⇒ RD = 12.5 K
0.1
3.49
(a)
2
⎛ W ⎞ ⎛ k ′p ⎞
I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSDB ( sat ) )
⎝ L ⎠B ⎝ 2 ⎠
2
⎛ W ⎞ ⎛ 0.04 ⎞
0.25 = ⎜ ⎟ ⎜
⎟ ( 0.8 ) ⇒
⎝ L ⎠B ⎝ 2 ⎠
⎛W ⎞
⎛W ⎞
⎜ ⎟ = 19.5 = ⎜ ⎟
⎝ L ⎠B
⎝ L ⎠A
I
⎛W ⎞
KQ 2 ⎛ W ⎞
⎜ ⎟ =
⎜ ⎟
L
I
⎝ ⎠C
Q 2 ⎝ L ⎠B
2
⎛ W ⎞ ⎛ k ′p ⎞
I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSGA + VTP )
⎝ L ⎠A ⎝ 2 ⎠
2
⎛ 0.04 ⎞
0.25 = (19.5 ) ⎜
⎟ (VSGA − 0.5 )
⎝ 2 ⎠
VSGA = 1.30 V
(b)
VDA = 1.3 − 4 = −2.7 V
RD =
−2.7 − ( −5 )
0.25
⎛ 100 ⎞
=⎜
⎟ (19.5 ) = 7.81
⎝ 250 ⎠
⇒ RD = 9.2 K
3.50
2
⎞ ⎛ kn′ ⎞
⎟ ⎜ ⎟ (VDS 2 ( sat ) )
2
⎠2 ⎝ ⎠
2
⎞ ⎛ 0.06 ⎞
⎛W ⎞
⎛W ⎞
⎟ ⎜
⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 53.3 = ⎜ ⎟
⎠2 ⎝ 2 ⎠
⎝ L ⎠2
⎝ L ⎠1
2
⎛ W ⎞ ⎛ k′ ⎞
I Q = ⎜ ⎟ ⎜ n ⎟ (VGS 1 − VTN )
2
L
⎝ ⎠1 ⎝ ⎠
2
⎛ 0.06 ⎞
0.4 = ( 53.3) ⎜
⎟ (VGS 1 − 0.75 )
⎝ 2 ⎠
VGS1 = 1.25 V VD1 = −1.25 + 4 = 2.75 V
5 − 2.75
⇒ RD = 5.625 K
RD =
0.4
⎛W
IQ = ⎜
⎝L
⎛W
0.4 = ⎜
⎝L
3.51
VDS ( sat ) = VGS − VP
So VDS > VDS ( sat ) = −VP , I D = I DSS
3.52
VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat )
a.
VGS = 0 ⇒ I D = I DSS = 6 mA
⎛ V ⎞
⎛ −1 ⎞
I D = I DSS ⎜1 − GS ⎟ = 6 ⎜1 − ⎟ ⇒ I D = 2.67 mA
V
⎝ −3 ⎠
P ⎠
⎝
2
b.
⎛ −2 ⎞
I D = 6 ⎜1 − ⎟ ⇒ I D = 0.667 mA
⎝ −3 ⎠
ID = 0
2
c.
d.
2
3.53
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
⎛
1 ⎞
2.8 = I DSS ⎜1 − ⎟
⎝ VP ⎠
2
2
⎛
3 ⎞
0.30 = I DSS ⎜1 − ⎟
⎝ VP ⎠
⎛
1
⎜1 −
2.8 ⎝ VP
=
0.30 ⎛
3
⎜1 −
V
P
⎝
⎛
1
⎜1 −
V
P
⎝
⎛
3
⎜1 −
⎝ VP
2
⎞
⎟
⎠ = 9.33
2
⎞
⎟
⎠
2
⎞
⎟
⎠ = 3.055
⎞
⎟
⎠
1
9.165
= 3.055 −
1−
VP
VP
8.165
= 2.055 ⇒ VP = 3.97 V
VP
1 ⎞
⎛
2.8 = I DSS ⎜1 −
⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA
⎝ 3.97 ⎠
2
3.54
VS = −VGS , VSD = VS − VDD
Want VSD ≥ VSD ( sat ) = VP − VGS
VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP
So VDD ≤ −2.5 V
⎛ V ⎞
I D = 2 = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
2 = 6 ⎜1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V
⎝ 2.5 ⎠
2
3.55
I D = K n (VGS − VTN )
2
18.5 = K n ( 0.35 − VTN )
86.2 = K n ( 0.5 − VTN )
2
2
Then
( 0.35 − VTN )
18.5
= 0.2146 =
⇒ VTN = 0.221 V
2
86.2
( 0.50 − VTN )
2
18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2
2
3.56
I D = K (VGS − VTN )
2
250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2
2
3.57
⎛ V ⎞
V
V
I D = I DSS ⎜1 − GS ⎟ = S = − GS
V
R
RS
P ⎠
S
⎝
2
V
⎛ V ⎞
10 ⎜1 − GS ⎟ = − GS
0.2
−5 ⎠
⎝
2
⎛ 2V
V ⎞
2 ⎜1 + GS + GS ⎟ = −VGS
5
25 ⎠
⎝
2
2 2 9
VGS + VGS + 2 = 0
25
5
2VGS2 + 45VGS + 50 = 0
VGS =
−45 ±
ID = −
( 45 ) − 4 ( 2 )( 50 )
⇒ VGS
2 ( 2)
2
= −1.17 V
VGS 1.17
=
⇒ I D = 5.85 mA
0.2
RS
VD = 20 − ( 5.85 )( 2 ) = 8.3 V
VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V
3.58
VDS = VDD − VS
8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ −1 ⎞
5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA
⎝ VP ⎠
2
⎛ −1 ⎞
5 = 10 ⎜1 − ⎟ ⇒ VP = −3.41 V
⎝ VP ⎠
VG = VGS + VS = −1 + 2 = 1 V
2
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R1
⎝ R1 + R2 ⎠
1
1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ
R1
5R2
= 0.5 ⇒ R2 = 0.556 MΩ
5 + R2
3.59
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
5 = 8 ⎜1 − GS ⎟ ⇒ VGS = 0.838 V
4 ⎠
⎝
VSD = VDD − I D ( RS + RD )
= 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V
2
VS = 20 − ( 5 )( 0.5 ) = 17.5 V
VG = VS + VGS = 17.5 + 0.838 = 18.3 V
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R1
⎝ R1 + R2 ⎠
1
18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ
R1
109 R2
= 100 ⇒ R2 = 1.21 MΩ
109 + R2
3.60
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛ V ⎞
5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V
3 ⎠
⎝
VSD = VDD − I D ( RS + RD )
2
6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ
VS = 12 − ( 5 )( 0.3) = 10.5 V
VG = VS + VGS = 10.5 + 0.465 = 10.965 V
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ
⎝ 100 ⎠
3.61
⎛ R2 ⎞
⎛ 60 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ ( 20 ) ⇒ VG = 6 V
⎝ 140 + 60 ⎠
⎝ R1 + R2 ⎠
⎛ V ⎞
V
V − VGS
I D = I DSS ⎜1 − GS ⎟ = S = G
V
R
RS
P ⎠
S
⎝
2
⎛
(8 )( 2 ) ⎜⎜1 −
⎝
VGS ⎞
⎟ = 6 − VGS
( −4 ) ⎟⎠
2
⎛ V
V2 ⎞
16 ⎜ 1 + GS + GS ⎟ = 6 − VGS
2
16 ⎠
⎝
2
VGS + 9VGS + 10 = 0
VGS =
−9 ±
(9)
2
− 4 (10 )
2
⇒ VGS = −1.30
⎛ ( −1.30 ) ⎞
I D = 8 ⎜⎜ 1 −
⎟ ⇒ I D = 3.65 mA
( −4 ) ⎟⎠
⎝
VDS = VDD − I D ( RS + RD )
2
= 20 − ( 3.65 )( 2 + 2.7 )
VDS = 2.85 V
VDS > VDS ( sat ) = VGS − VP
= −1.30 − ( −4 )
= 2.7 V (Yes)
3.62
VDS = VDD − I D ( RS + RD )
5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA
VS = I D RS = ( 4.67 ) ( 0.5 ) ⇒ VS = 2.33 V
⎛ R2 ⎞
⎛ 20 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (12 ) ⇒ VG = 0.511 V
+
R
R
⎝ 450 + 20 ⎠
⎝ 1
2 ⎠
VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ ( −1.82 ) ⎞
4.67 = 10 ⎜⎜ 1 −
⎟ ⇒ VP = −5.75 V
VP ⎟⎠
⎝
2
3.63
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟ , VGS = 0
⎝ VP ⎠
I D = I DSS = 4 mA
2
RD =
VDD − VDS 10 − 3
=
⇒ RD = 1.75 kΩ
4
ID
3.64
VSD = VDD − I D RS
10 = 20 − (1) RS ⇒ RS = 10 kΩ
R1 + R2 =
VDD 20
=
= 200 kΩ
I
0.1
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V
2 ⎠
⎝
VG = VS + VGS = 10 + 0.586 = 10.586
2
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ
⎝ 200 ⎠
R1 = 94 kΩ
3.65
VDS = VDD − I D ( RS + RD )
2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ
I D = K (VGS − VTN )
2
40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V
2
VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ
⎝ 150 ⎠
R1 = 100 kΩ
3.66
For VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN
0.75 − 0.15 = 0.6
Biased in the saturation region
V − VDS 3 − 0.7
I D = DD
=
⇒ I D = 46 µ A
50
RD
I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 µ A / V 2
2
2
Chapter 4
Exercise Solutions
EX4.1
g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN )
2
0.75 = 0.5 (VGS − 0.8 ) ⇒ VGS = 2.025 V
g m = 2 ( 0.5 )( 2.025 − 0.8 ) ⇒ g m = 1.22 mA / V
2
ro =
1
λ I DQ
1
(0.01)(0.75)
= 133 k Ω
ro = 133 k Ω
=
EX4.2
Av = − g m RD
g m = 2 K n I DQ = 2
( 0.5)( 0.4 ) = 0.8944 mA/V
Av = − ( 0.8944 )(10 ) = −8.94
EX4.3
(a)
⎛ R2 ⎞
⎛ 320 ⎞
VGS = ⎜
⎟ VDD = ⎜
⎟ ( 5 ) = 1.905 V
R
R
520 + 320 ⎠
+
⎝
⎝ 1
2 ⎠
2
I DQ = 0.20 (1.905 − 0.8 ) = 0.244 mA
g m = 2 K n I DQ = 2
( 0.2 )( 0.244 ) = 0.442 mA/V
ro = ∞
(b)
Av = − g m RD = − ( 0.422 )(10 ) = −4.22
(c)
(d)
RO = RD = 10 K
Ri = R1 & R2 = 520 & 320 = 198 K
EX4.4
At transition point, I D = 1 mA
I D = K n (VGSt − VTN ) = K n (VDS ( sat ) )
2
2
1 = 0.2 (VDS ( sat ) ) ⇒ VDS ( sat ) = 2.236 V
2
5 − 2.236
+ 2.236 = 3.62 V
2
5 − 3.62
= 2.76 K
RD =
0.5
Want VDSQ =
0.5 = 0.2 (VGSQ − 0.8 ) ⇒ VGSQ = 2.38 V
2
⎛ R2 ⎞
1
VGSQ = ⎜
⎟ VDD = ( R1 & R2 ) VDD
R1
⎝ R1 + R2 ⎠
1
So 2.38 = ( 200 )( 5 ) ⇒ R1 = 420 K and R2 = 382 K
R1
Av = − g m RD
( 0.2 )( 0.5 ) = 0.6325 mA/V
Av = − ( 0.6325 )( 2.76 )
g m = 2 K n I DQ = 2
= −1.75
EX4.5
(a)
⎛ R2 ⎞
250
⎛
⎞
VG = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5 = −3 V
+
+
250
1000
R
R
⎝
⎠
⎝ 1
2 ⎠
(V − VGS ) − ( −5 )
2
= K n (VGS − VTN )
ID = G
2
−3 − VGS + 5 = 2 ( 0.5 )(VGS − 0.6 )
2 − VGS = VGS2 − 1.2VGS + 0.36
VGS2 − 0.2VGS − 1.64 = 0
VGS =
0.2 ±
( 0.04 ) + 4 (1.64 )
2
2
= 1.385 V
I DQ = ( 0.5 )(1.385 − 0.6 ) ⇒ I DQ = 0.308 mA
VDSQ = 10 − ( 0.308 )(10 + 2 ) ⇒ VDSQ = 6.30 V
2
(b)
Av =
Av =
− g m RD
1 + g m RS
g m = 2 K n I DQ = 2
− ( 0.7849 )(10 )
1 + ( 0.7849 )( 2 )
⇒
g m = 0.7849 mA/V
Av = −3.05
EX4.6
VSDQ = 3 V and I DQ = 0.5 mA ⇒ RD =
I DQ = K P (VSG − VTP
)
( 0.5 )( 0.308 )
5−3
⇒ RD = 4 kΩ
0.5
2
0.5 = 1(VSG − 1) ⇒ VSG = 1.71 V ⇒ VGG = 5 − 1.71 ⇒ VGG = 3.29 V
2
Av = − g m RD
g m = 2 K P I DQ = 2 (1)( 0.5 )
g m = 1.414 mA/V
Av = − (1.414 )( 4 ) ⇒ Av = −5.66
Av =
v0 −vsd
0.46sin ω t
=
=−
= −5.66 ⇒ vi = 0.0813sin ω t
vi
vi
vi
EX4.7
a.
VSG = 9 − I DQ RS , I DQ = K P (VSG − VTP
VSG = 9 − ( 2 )(1.2 )(VSG − 2 )
2
2
= 9 − 2.4 (VSG
− 4VSG + 4 )
2
2.4VSG
− 8.6VSG + 0.6 = 0
)
2
2
(8.6 ) − 4 ( 2.4 )( 0.6 )
2 ( 2.4 )
2
VSG = 3.51 V, I DQ = 2 ( 3.51 − 2 ) ⇒ I DQ = 4.57 mA
VSDQ = 9 + 9 − I DQ (1.2 + 1) = 18 − ( 4.57 )( 2.2 ) ⇒ VSDQ = 7.95 V
VSG =
8.6 ±
b.
V0
⫺
VSG
Vi
gmVSG
⫹
⫹
⫺
RD
RS
CS
( 2 )( 4.57 ) = 6.046 mA/V
g m = 2 K P I DQ = 2
V0 = g mVSG RD
Av = − g m RD = − ( 6.046 )(1) ⇒ Av = −6.05
EX4.8
VDSQ = VDD − I DQ RS
5 = 10 − (1.5 ) RS ⇒ RS = 3.33 kΩ
I DQ = K n (VGS − VTN ) ⇒ 1.5 = (1)(VGS − 0.8 )
2
2
⎛ R2
VGS = 2.025 V = VG − VS = VG − 5 ⇒ VG = 7.025 V = ⎜
⎝ R1 + R2
So R2 = 281 kΩ, R1 = 119 kΩ
Neglecting RSi , Av =
g m ( RS & r0 )
1 + g m ( RS & r0 )
r0 = ⎡⎣λ I DQ ⎤⎦ = ⎣⎡( 0.015 )(1.5 ) ⎦⎤ = 44.4 kΩ
RS & r0 = 3.33 & 44.4 = 3.1 kΩ
−1
−1
g m = 2 K n I DQ = 2 (1)(1.5 ) = 2.45 mA / V
Av =
( 2.45)( 3.1)
⇒ Av = 0.884
1 + ( 2.45 )( 3.1)
EX4.9
I DQ = K P (VSG − VTP
)
2
3 = 2 (VSG − 2 ) ⇒ VSG = 3.22 V
5 − VSG
5 − 3.22
⇒3=
⇒ RS = 0.593 kΩ
I DQ =
RS
RS
2
r0 = ⎡⎣λ I DQ ⎤⎦ = ⎡⎣( 0.02 )( 3) ⎤⎦ = 16.7 kΩ
−1
−1
( 2 )( 3) = 4.9 mA / V
g m ( r0 & RS )
RL = ∞, Av =
1 + g m ( r0 & RS )
g m = 2 K P I DQ = 2
For
r0 & RS = 16.7 & 0.593 = 0.573 kΩ
⎞
R2
⋅10
⎟ VDD =
400
⎠
Av =
( 4.9 )( 0.573)
⇒ Av = 0.737
1 + ( 4.9 )( 0.573)
If Av is reduced by 10% ⇒ Av = 0.737 − 0.0737 = 0.663
g m ( r0 RS RL )
Av =
1 + g m ( r0 RS RL )
Let r0 & RS & RL = x
0.663 =
( 4.9 ) x
⇒ 0.663 = 4.9 x (1 − 0.663)
1 + ( 4.9 ) x
x = 0.402 = 0.573 RL
0.573RL
= 0.402 ⇒ ( 0.573 − 0.402 ) RL = ( 0.402 )( 0.573) ⇒ RL = 1.35 kΩ
RL + 0.573
EX4.10
⎛ R2 ⎞
9.3 ⎞
⎛
VG = ⎜
⎟ VDD = ⎜
⎟ (5)
⎝ 70.7 + 9.3 ⎠
⎝ R1 + R2 ⎠
= 0.581 V
I DQ = K p (VSG − VTP
)
2
= K P (VS − VG − VTP
=
)
2
5 − VS
RS
2
Then ( 0.4 )( 5 )(VS − 0.581 − 0.8 ) = 5 − VS
2
2 (VS − 1.381) = 5 − VS
2 (VS2 − 2.762VS + 1.907 ) = 5 − VS
2VS2 − 4.52VS − 1.19 = 0
VS =
4.52 ±
2
( 4.52 ) + 4 ( 2 )(1.19 )
2 ( 2)
VS = 2.50 V ⇒ I DQ =
g m = 2 K P I DQ = 2
Av =
=
5 − 2.5
= 0.5 mA
5
( 0.4 )( 0.5 ) = 0.894 mA / V
g m RS
R1 & R2
⋅
1 + g m RS R1 & R2 + RSi
( 0.894 )( 5)
70.7 & 9.3
⋅
⇒ Av = 0.770
1 + ( 0.894 )( 5 ) 70.7 & 9.3 + 0.5
Neglecting RSi , Av = 0.817
R0 = RS
1
1
=5
= 5 & 1.12 ⇒ R0 = 0.915 kΩ
0.894
gm
EX4.11
gmVsg
V0
⫹
Vi
⫹
⫺
RS
Vsg
⫺
RD
RL
V0 = g mVsg ( RD & RL ) and Vsg = Vi
Av = g m ( RD & RL )
I DQ =
5 − VSG
= K p (VSG − VTP
RS
5 − VSG = (1)( 4 )(VSG − 0.8 )
)
2
2
2
5 − VSG = 4 (VSG
− 1.6VSG + 0.64 )
2
4VSG
− 5.4VSG − 2.44 = 0
VSG =
5.4 ± (5.4) 2 + ( 4 )( 4 )( 2.44 )
2 ( 4)
VSG = 1.71 V
5 − 1.71
I DQ =
= 0.822 mA
4
g m = 2 K p I DQ = 2 (1)( 0.822 ) = 1.81 mA / V
Av = (1.81)( 2 & 4 ) = (1.81)(1.33) ⇒ Av = 2.41
Rin = RS
1
1
=4
= 4 & 0.552 ⇒ Rin = 0.485 kΩ
gm
1.81
µ n Cox ⎛ W ⎞
EX4.12
Kn2 =
Av = −
2
⋅ ⎜ ⎟ = ( 0.015 )( 2 ) = 0.030 mA / V 2
⎝ L ⎠2
K n1
K
= −6 ⇒ n1 = 36
Kn2
Kn2
K n1 = ( 36 )( 0.030 ) = 1.08 mA / V 2
⎛W ⎞
⎛W ⎞
1.08 = ( 0.015 ) ⎜ ⎟ ⇒ ⎜ ⎟ = 72
⎝ L ⎠1 ⎝ L ⎠1
The transition point is found from vGSt − 1 = (10 − 1) − ( 6 )( vGSt − 1)
10 − 1 + 6 + 1
= 2.29 V
1+ 6
2.29 − 1
+ 1 ⇒ VGS = 1.645 V
For Q-point in middle of saturation region VGS =
2
vGSt =
EX4.13
(a)
Transition points:
For M 2 : vOtB = VDD − VTNL = 5 − 1.2 = 3.8 V
2
2
For M 1 : K n1 ⎡( vOtA ) (1 + λ vOtA ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vOtA ]) ⎤
⎣
⎦
⎣
⎦
2
2
3
⎤⎦ = 25 ⎡(1.2 ) (1 + ( 0.01)( 5 ) − ( 0.01) vOtA ) ⎤
+ ( 0.01) vOtA
250 ⎡⎣ vOtA
⎣
⎦
2
3
3
2
⎤⎦ = 1.512 − 0.0144vOtA ( 0.01) vOtA
+ ( 0.01) vOtA
+ vOtA
+ 0.00144vOtA − 0.512 = 0
10 ⎡⎣vOtA
which yields vOtA ≅ 0.388 V
3.8 − 0.388
+ 0.388 ⇒ VDSQ1 = 2.094 V
2
2
2
K n1 ⎡(VGS 1 − VTND ) (1 + λ1vO ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vO ]) ⎤
⎣
⎦
⎣
⎦
Then middle of saturation region V0Q =
250 ⎡(VGS 1 − 0.8 ) (1 + [ 0.01][ 2.094]) ⎤ = 25 ⎡(1.2 ) (1 + [ 0.01][5 − 2.094]) ⎤
⎣
⎦
⎣
⎦
2
2
2
10 ⎡(VGS1 − 0.8 ) = (1.0209 ) ⎤ = 1.482
⎣
⎦
(VGS1 − 0.8 )
2
= 0.145 ⇒ VGS1 = 1.18 V
I DQ = K n1 ⎡(VGS1 − 0.8 ) (1 + ( 0.01)( 2.094 ) ) ⎤
⎣
⎦
2
b.
2
I DQ = ( 0.25 ) ⎡( 0.145 ) (1.02094 ) ⎤ ⇒ I DQ = 37.0 µ A
⎣
⎦
− g m1
Av =
= − g m1 ( r01 & r02 )
I DQ ( λ1 + λ2 )
c.
g m1 = 2 K n1 (VGS 1 − VTND ) = 2 ( 0.25 )(1.18 − 0.8 ) = 0.19 mA/V
Av =
−0.19
⇒ Av = −257
( 0.037 )( 0.01 + 0.01)
EX4.14
Av = − g m ( ron & rop )
ron = rop =
1
= 666.7 K
( 0.015)( 0.1)
−250 = − g m ( 666.7 & 666.7 )
g m = 0.75 mA/V = 2 K n I DQ = 2 K n ( 0.1)
K n = 1.406 mA/V 2 =
⎛W ⎞
⎜ ⎟ = 35.2
⎝ L ⎠1
k n′ ⎛ W
⎜
2⎝L
⎞ ⎛ 0.080 ⎞ ⎛ W ⎞
⎟=⎜
⎟⎜ ⎟
⎠ ⎝ 2 ⎠⎝ L ⎠
EX4.15
(a)
RO =
So g m1 =
1
1
ro 2 ro1 ≈
g m1
g m1
1 1
= = 0.5 mA/V
R0 2
g m1 = 2 K n I D
( 0.2 ) I D
0.5 = 2
⇒ I D = 0.3125 mA
(b)
Av =
g m1 ( ro1 & ro 2 )
1 + g m1 ( ro1 & ro 2 )
ro1 = ro 2 =
Av =
1
= 320 K
( 0.01)( 0.3125 )
0.5 ( 320 & 320 )
1 + ( 0.5 )( 320 & 320 )
Av = 0.988
EX4.16
(a)
1
ro1 2 K n I D + λ1 I D
=
Av =
1
1
λ2 I D + λ1 I D
+
ro 2 ro1
g m1 +
120 =
2 0.2 I D + 0.01I D
0.01I D + 0.01I D
2.4 I D − 0.01I D = 2 0.2 I D
2.39 I D = 2 0.2 ⇒ I D = 0.140 mA
g m1 = 2
( 0.2 )( 0.14 ) ⇒ g m1 = 0.335 mA/V
(b)
Ro = ro1 & ro 2
ro1 = ro 2 =
1
= 714 K
0.01
( )( 0.14 )
Ro = 357 K
EX4.17
R0 = RS 2
1
gm2
g m 2 = 0.632 mA/V, RS 2 = 8 kΩ
R0 = 8
1
= 8 &1.58 ⇒ R0 = 1.32 kΩ
0.632
EX4.18
a.
I DQ1 = K n1 (VGS 1 − VTN 1 )
2
1 = 1.2 (VGS 1 − 2 ) ⇒ VGS 1 = VGS 2 = 2.91 V
2
RS = 10 kΩ ⇒ VS1 = I DQ RS − 10 = (1)(10 ) − 10 = 0
⎛
⎞
R3
VG1 = 2.91 = ⎜
⎟ (10 )
⎝ R1 + R2 + R3 ⎠
⎛ R ⎞
= ⎜ 3 ⎟ (10 ) ⇒ R3 = 145.5 kΩ
⎝ 500 ⎠
VDSQ1 = 3.5 ⇒ VS 2 = 3.5 V ⇒ VG 2 = 3.5 + 2.91 ⇒ VG 2 = 6.41
⎛ R2 + R3 ⎞
VG 2 = ⎜
⎟ (10 ) = 6.41
⎝ R1 + R2 + R3 ⎠
⎛ R + R3 ⎞
=⎜ 2
⎟ (10 )
⎝ 500 ⎠
R2 + R3 = 320.5 = R2 + 145.5 ⇒ R2 = 175 kΩ
Then R1 + R2 + R3 = 500 = R1 + 175 + 145.5 ⇒ R1 = 179.5 kΩ
Now VS 2 = 3.5 ⇒ VD 2 = VS 2 + VSDQ 2 = 3.5 + 3.5 = 7 V
So RD =
b.
10 − 7
⇒ RD = 3 kΩ
1
From Example 6-18:
Av = − g m1 RD
g m1 = 2 K n1 I DQ = 2 (1.2 )(1) = 2.19 mA / V
Av = − ( 2.19 )( 3) ⇒ Av = −6.57
EX4.19
VS = I DQ RS = (1.2 )( 2.7 ) = 3.24 V
VD = VS + VDSQ = 3.24 + 12 = 15.24
20 − 15.24
RD =
⇒ RD = 3.97 kΩ
1.2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛ V ⎞
V
1.2 = 4 ⎜1 − GS ⎟ ⇒ GS = 0.4523
V
VP
P ⎠
⎝
VGS = ( 0.4523)( −3) = −1.357
2
VG = VS + VGS = 3.24 − 1.357 = 1.883 V
⎛ R2 ⎞
⎛ R2 ⎞
VG = ⎜
⎟ ( 20 ) = ⎜
⎟ ( 20 ) = 1.88 ⇒ R2 = 47 kΩ. R1 = 453 kΩ
⎝ 500 ⎠
⎝ R1 + R2 ⎠
1
1
=
= 167 kΩ
r0 =
λ I DQ ( 0.005 )(1.2 )
2 I DSS ⎛ VGS ⎞ 2 ( 4 ) ⎛ 1.357 ⎞
=
1−
1−
⎟ = 1.46 mA/V
3 ⎠
( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝
Av = − g m ( r0 & RD & RL ) = − (1.46 )(167 & 3.97 & 4 ) ⇒ Av = −2.87
gm =
EX4.20
a.
⎛ V ⎞
⎛ V ⎞
V
I DQ = I DSS ⎜1 − GS ⎟ 2 = 8 ⎜1 − GS ⎟ ⇒ GS = 0.5
V
V
VP
P ⎠
P ⎠
⎝
⎝
VGS = ( 0.5 )( −3.5 ) ⇒ VGS = −1.75
2
Also I DQ =
2
−VGS − ( −10 )
RS
2=
1.75 + 10
⇒ RS = 5.88 kΩ
RS
b.
⫹
gmVgs
Vgs
Vi
⫹
⫺
V0
⫺
RS
gm =
r0 =
2 I DSS
VP
⎛ VGS
⎜1 −
⎝ VP
⎞ 2 ( 8 ) ⎛ 1.75 ⎞
⎟=
⎜1 −
⎟ = 2.29 mA/V
3.5 ⎠
⎠ 3.5 ⎝
1
= 50 kΩ
( 0.01)( 2 )
Vi = Vgs + g m RS Vgs ⇒ Vgs =
Av =
Vi
1 + g m RS
( 2.29 ) [5.88 & 50]
V0
g m RS & r0
=
=
⇒ Av = 0.9234
V1 1 + g m RS & r0 1 + ( 2.29 ) [5.88 & 50]
c.
Av =
g m ( RS & RL ) & ro
1 + g m ( RS & RL & ro )
= ( 0.80 )( 0.9234 ) = 0.7387
( 2.29 )( RS & RL & ro )
= 0.7387 ⇒ RS & RL & ro = 1.235 kΩ
1 + ( 2.29 )( RS & RL & ro )
RS & ro = 5.261 kΩ
5.261RL
= 1.235 kΩ ⇒ RL = 1.61 kΩ
5.261 + RL
TYU4.1
g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN ) ⇒ VGS − VTN =
2
and g m = 2 K n
I DQ
Kn
I DQ
Kn
= 2 K n I DQ
2
( 3.4 )
g2
= 1.445 mA / V
Kn = m =
4 I DQ
4 ( 2)
K
n
=
µ n Cox W
2
⋅
L
⎛W
1.445 = ( 0.018 ) ⎜
⎝L
⎞ W
= 80.3
⎟⇒
L
⎠
2
ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡⎣λ I DQ ⎤⎦
⎣
⎦
−1
−1
ro = ⎡⎣( 0.015 )( 2 ) ⎤⎦ ⇒ ro = 33.3 k Ω
−1
TYU4.2
a.
I DQ = K n (VGS − VTN )
2
0.4 = 0.5 (VGS − 2 ) ⇒ VGS = 2.894 V
2
VDSQ = VDD − I DQ RD = 10 − ( 0.4 )(10 ) ⇒ VDSQ = 6 V
b.
g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2.894 − 2 ) ⇒ g m = 0.894 mA/V
r0 = ⎡⎣ λ I DQ ⎤⎦ , λ = 0 ⇒ r0 = ∞
v
Av = 0 = − g m RD = − ( 0.894 )(10 ) ⇒ Av = −8.94
vi
−1
c.
vi = 0.4sin ω t ⇒ vds = − ( 8.94 )( 0.4 ) sin ω t
vds = −3.58sin ω t
At VDS 1 = 6 − 3.58 = 2.42
VGS1 = 2.89 + 0.4 = 3.29
VGS1 − VTN = 3.29 − 2 = 1.29 = VDS ( sat )
So VDS 1 > VGS 1 − VTN ⇒ Biased in saturation region
TYU4.3
I DQ = K n (VGS − VTN ) = ( 0.25 )( 2 − 0.8 ) ⇒ I DQ = 0.36 mA
2
a.
2
VDSQ = VDD − I DQ RD = 5 − ( 0.36 )( 5 ) ⇒ VDSQ = 3.2 V
b.
g m = 2 K n (VGS − VTN ) = 2 ( 0.25 )( 2 − 0.8 ) ⇒ g m = 0.6 mA/V,
c.
Av =
v0
= − g m RD = − ( 0.6 )( 5 ) ⇒ Av = −3.0
vi
r0 = ∞
TYU4.4
vi = vgs = 0.1sin ω t
id = g m vgs = ( 0.6 )( 0.1) sin ω t
id = 0.06sin ω t mA
vds = ( −3)( 0.1) sin ω t = −0.3sin ω t ( V )
Then iD = I DQ + id = 0.36 + 0.06sin ω t
= iD mA
vDS = VDSQ + vds = 3.2 − 0.3sin ω t = vDS
TYU4.5
a.
VSDQ = VDD − I DQ RD
7 = 12 − I DQ ( 6 ) ⇒ I DQ = 0.833 mA
I DQ = K P (VSG − VTP
)
2
0.833 = 2 (VSG − 1) ⇒ VSG = 1.65 V
2
g m = 2 K P (VSG − VTP ) = 2 ( 2 )(1.65 − 1) ⇒ g m = 2.58 mA/V, r0 = ∞
b.
Av =
v0
= − g m RD = − ( 2.58 )( 6 ) ⇒ Av = −15.5
vi
V0
⫺
Vi
⫹
⫺
gmVsg
VSG
RD
⫹
TYU4.6
I DQ = K n (VGS − VTN ) ⇒ VGS − VTN =
2
g m = 2 K n (VGS − VTN ) = 2 K n
So g m = 2 K n I DQ
TYU4.7
η=
γ
2 2φ f + vSB
I DQ
Kn
I DQ
Kn
η=
(a)
0.40
2 2 ( 0.35 ) + 1
η=
(b)
⇒ η = 0.153
0.40
2 2 ( 0.35 ) + 3
⇒ η = 0.104
( 0.5 )( 0.75) = 1.22 mA / V
g mb = g m η = (1.22 )( 0.153) ⇒ g mb = 0.187 mA / V
g mb = (1.22 )( 0.104 ) ⇒ g mb = 0.127 mA / V
g m = 2 K n I DQ = 2
For (a),
For (b),
TYU4.8
I DQ = I Q = 0.5 mA
W
Let
= 25
L
K n = ( 20 )( 25 ) = 500 µ A / V 2
0.5
+ 1.5 = 2.5 V ⇒ VS = −2.5 V
0.5
Av = − g m RD
VGS =
g m = 2 ( 0.5 )( 2.5 − 1.5 ) = 1 mA/V
For Av = −4.0 ⇒ RD = 4 kΩ
VD = 5 − ( 0.5 )( 4 ) = 3 V ⇒ VDSQ = 3 − ( −2.5 ) = 5.5 V
TYU4.9
a.
With RG ⇒ VGS = VDS ⇒ transistor biased in sat. region
2
I D = K n (VGS − VTN ) = K n (VDS − VTN )
2
VDS = VDD − I D RD = VDD − K n RD (VDS − VTN )
VDS = 15 − ( 0.15 )(10 )(VDS − 1.8 )
2
2
2
= 15 − 1.5 (VDS
− 3.6VDS + 3.24 )
2
1.5VDS
− 4.4VDS − 10.14 = 0
VDS =
4.4 ±
( 4.4 )
2
+ ( 4 )(1.5 )(10.14 )
2 (1.5 )
⇒ VDSQ = 4.45 V
I DQ = ( 0.15 )( 4.45 − 1.8 ) ⇒ I DQ = 1.05 mA
2
b.
Neglecting effect of RG:
Av = − g m ( RD & RL )
g m = 2 K n (VGS − VTN ) = 2 ( 0.15 )( 4.45 − 1.8 ) ⇒ g m = 0.795 mA/V
Av = − ( 0.795 )(10 & 5 ) ⇒ Av = −2.65
c.
RG ⇒ establishes VGS = VDS ⇒ essentially no effect on small-signal voltage gain.
TYU4.10
a.
2
I DQ = K n (VGS − VTN )
2
I DQ = 0.8 ( 2 − VSG ) =
VSG VSG
=
4
RS
2
3.2 ( 4 − 4VSG + VSG
) = VSG
2
− 13.8VSG + 12.8 = 0
3.2VSG
VSG
2
(13.8) − 4 ( 3.2 )(12.8 )
2 ( 3.2 )
2
= 1.35 V ⇒ I DQ = 0.8 ( 2 − 1.35 ) ⇒ I DQ = 0.338 mA
VSG =
13.8 ±
b.
VDSQ = VDD − I DQ ( RD + RS )
6 = 10 − ( 0.338 )( RD + 4 )
RD =
10 − ( 0.338 )( 4 ) − 6
0.338
⇒ RD = 7.83 kΩ
c.
V0
⫹
Vgs
Vi
⫹
⫺
gmVgs
RD
⫺
RS
Vi = Vgs + g mVgs RS ⇒ Vgs =
Vi
1 + g m RS
V0 = − g mVgs RD
g m = 2 K n (VGS − VTN ) = 2 ( 0.8 )( −1.35 + 2 ) = 1.04 mA/V
Av =
− (1.04 )( 7.83)
V0
− g m RD
=
=
⇒ Av = −1.58
1 + (1.04 )( 4 )
Vi 1 + g m RS
TYU4.11
a.
5 = I DQ RS + VSG and I DQ = K p (VSG + VTP ) 2
0.8 = 0.5(VSG + 0.8) 2 ⇒ VSG = 0.465 V
5 = ( 0.8 ) RS + 0.465 ⇒ RS = 5.67 kΩ
VSDQ = 10 − I DQ ( RS + RD )
3 = 10 − ( 0.8 )( 5.67 + RD )
RD =
10 − ( 0.8 )( 5.67 ) − 3
0.8
⇒ RD = 3.08 kΩ
b.
V0
⫺
Vi
⫹
⫺
VSG
gmVsg
r0
RD
⫹
V0 = g mVsg ( RD & r0 ) = − g mVi ( RD & r0 )
V
Av = 0 = − g m ( RD & r0 )
Vi
g m = 2 K p (VSG + VTP ) = 2(0.5)(0.465 + 0.8) = 1.265 mA/V
1
1
=
= 62.5 kΩ
r0 =
λ I 0 ( 0.02 )( 0.8 )
Av = − (1.265 )( 3.08 & 62.5 ) ⇒ Av = −3.71
TYU4.12
(a)
V0 = g mVgs r0
Vi = Vgs + V0 ⇒ Vgs = Vi − V0
So V0 = g m r0 (Vi − V0 )
Av =
( 4 )( 50 )
V0
g m r0
=
=
⇒ Av = 0.995
Vi 1 + g m r0 1 + ( 4 )( 50 )
⫹ Vgs ⫺
Ix
I x + g mVgs =
I x = g mVx +
⫹
⫺
r0
gmVgs
Vx
and Vgs = −Vx
r0
Vx
1
1
⇒ R0 = r0
= 50 ⇒ R0 ≅ 0.25 kΩ
4
r0
gm
With RS = 4 kΩ ⇒ Av =
(b)
Vx
r0 || Rs = 50 || 4 = 3.7 kΩ ⇒ Av =
g m ( r0 & RS )
1 + g m ( r0 & RS )
( 4 )( 3.7 )
1 + ( 4 )( 3.7 )
⇒ Av = 0.937
TYU4.13
(a)
g m = 2 K n I DQ ⇒ 2 = 2 K n ( 0.8 ) ⇒ K n = 1.25 mA / V 2
Kn =
So
⎛W ⎞
⇒ 1.25 = ( 0.020 ) ⎜ ⎟
L
⎝L⎠
µ n Cox W
2
⋅
W
= 62.5
L
I DQ = K n (VGS − VTN ) ⇒ 0.8 = 1.25 (VGS − 2 ) ⇒ VGS = 2.8 V
2
2
b.
r0 = ⎣⎡ λ I DQ ⎦⎤ = ⎡⎣( 0.01)( 0.8 ) ⎤⎦ = 125 kΩ
g m ( r0 & RL )
Av =
1 + g m ( r0 & RL )
−1
−1
r0 & RL = 125 & 4 = 3.88
Av =
R0 =
( 2 )( 3.88 )
⇒ Av = 0.886
1 + ( 2 )( 3.88 )
1
1
r0 = 125 ⇒ R0 ≈ 0.5 kΩ
gm
2
TYU4.14
Rin =
1
= 0.35 kΩ ⇒ g m = 2.86 mA/V
gm
V0
4 RD
= RD & RL = 2.4 = RD & 4 =
4 + RD
Ii
( 4 − 2.4 ) RD = ( 2.4 )( 4 ) ⇒ RD = 6 kΩ
g m = 2 K n I DQ
2.86 = 2 K n ( 0.5 ) ⇒ K n = 4.09 mA / V 2
I DQ = K n (VGS − VTN )
2
0.5 = 4.09 (VGS − 1) ⇒ VGS = 1.35 V ⇒ VS = −1.35 V, VD = 5 − ( 0.5 )( 6 ) = 2 V
2
VDS = VD − VS = 2 − ( −1.35 ) = 3.35 V
We have VDS = 3.35 > VGS − VTN = 1.35 − 1 = 0.35 V ⇒ Biased in the saturation region
⎛W
⋅⎜
⎝L
µ C ⎛W
= n ox ⋅ ⎜
2 ⎝L
TYU4.15
µC
K n1 = n ox
2
Kn2
Av = −
⎞
2
⎟ = ( 0.020 )( 80 ) = 1.6 mA / V
⎠1
⎞
2
⎟ = ( 0.020 )(1) = 0.020 mA / V
⎠2
K n1
1.6
=−
⇒ Av = −8.94
Kn2
0.020
The transition point is determined from vGSt − VTND = VDD − VTNL −
K n1
( vGSt − VTND )
Kn2
vGSt − 0.8 = ( 5 − 0.8 ) − ( 8.94 )( vGSt − 0.8 )
vGSt =
( 5 − 0.8) + ( 8.94 )( 0.8) + 0.8
1 + 8.94
vGSt = 1.22 V
For Q-point in middle of saturation region VGS =
1.22 − 0.8
+ 0.8 ⇒ VGS = 1.01 V
2
TYU4.16
a.
I DQ 2 = 2mA, VDSQ 2 = 10 V
I DQ 2 ⋅ RS 2 = 10 = 2 RS 2 ⇒ RS 2 = 5 kΩ
I DQ 2 = K n 2 (VGS 2 − VTN 2 )
2
2 = 1(VGS 2 − 2 ) ⇒ VSG 2 = 3.41 V ⇒ VG 2 = 3.41 V
2
10 − 3.41
⇒ RD1 = 3.29 kΩ
2
= 10 V ⇒ VS 1 = 3.41 − 10 = −6.59 V
Then RD1 =
For VDSQ1
Then RS 1 =
−6.59 − ( −10 )
2
I D1 = K n1 (VGS 1 − VTN 1 )
⇒ RS1 = 1.71 kΩ
2
2 = 1(VGS 1 − 2 ) ⇒ VGS 1 = 3.41 V
2
⎛ R2 ⎞
R2
1
VGS1 = ⎜
= ⋅ Rin
⎟ ( 20 ) − I DQ1 RS 1
R
R
R
R
R
+
+
⎝ 1
2 ⎠
1
2
1
1
3.41 = ( 200 )( 20 ) − ( 2 )(1.71) ⇒ R1 = 586 kΩ
R1
586 R2
= 200 ⇒ ( 586 − 200 ) R2 = ( 200 )( 586 ) ⇒ R2 = 304 kΩ
586 + R2
g m1 = 2 K n1 I DQ1 = 2 (1)( 2 ) ⇒ g m1 = g m 2 = 2.828 mA/V
b.
From Example 6.17
− g g R ( R & RL )
Av = m1 m 2 D1 S 2
1 + g m 2 ( RS 2 & RL )
RS 2 & RL = 5 & 4 = 2.222 kΩ
− ( 2.828 )( 2.828 )( 3.29 )( 2.222 )
Av =
1 + ( 2.828 )( 2.222 )
⇒ Av = −8.03
1
1
RS 2 =
5 = 0.3536 & 5 ⇒ R0 = 0.330 kΩ
gm2
2.828
R0 =
TYU4.17
From Example 6.19
g m = 3.0 mA/V, r0 = 41.7 kΩ
R1 & R2 = 420 & 180 = 126 kΩ
⎛ R1 & R2 ⎞
Vgs = ⎜
⎟ Vi
⎝ R1 & R2 + Ri ⎠
⎛ 126 ⎞
=⎜
⎟ Vi = 0.863Vi
⎝ 126 + 20 ⎠
− g mVgs ( r0 & RD & RL )
Av =
Vi
= − ( 3.0 )( 0.863)( 41.7 & 2.7 & 4 )
= − ( 2.589 )( 41.7 & 1.61) = − ( 2.589 )(1.55 ) ⇒ Av = −4.01
TYU4.18
a.
⎛ R2 ⎞
VG1 = ⎜
⎟ (VDD )
⎝ R1 + R2 ⎠
⎛ 430 ⎞
VG1 = ⎜
⎟ ( 20 ) = 17.2 V
⎝ 430 + 70 ⎠
⎛ V ⎞
⎛ V −V ⎞
I DQ1 = I DSS ⎜1 − GS ⎟ = 6 ⎜1 − G1 S1 ⎟
2
V
⎝
⎠
P ⎠
⎝
2
2
20 − VS1
⎛ 17.2 VS 1 ⎞
⎛ VS 1
⎞
= 6 ⎜1 −
+
⎟ = 6 ⎜ 2 − 7.6 ⎟ and I DQ1 =
2
2
4
⎝
⎠
⎝
⎠
2
2
20 − VS 1
⎛V
⎞
= 6 ⎜ S1 − 7.6 ⎟
4
2
⎝
⎠
2
⎛V
⎞
20 − VS1 = 24 ⎜ S 1 − 7.6VS 1 + 57.76 ⎟
4
⎝
⎠
2
Then
= 6VS21 − 182.4VS 1 + 1386.24
6VS21 − 181.4VS 1 + 1366.24 = 0
VS 1 =
181.4 ±
2
(181.4 ) − 4 ( 6 )(1366.24 )
2 ( 6)
VS 1 = 14.2 V ⇒ VGS 1 = 17.2 − 14.2 = 3 V > VP
So VS1 = 16.0 ⇒ VGS1 = 17.2 − 16 = 1.2 < VP − Q
20 − 16
⇒ I DQ1 = 1 mA
4
= 20 − I DQ1 ( RS 1 + RD1 )
= 20 − (1)( 8 ) ⇒ VSDQ1 = 12 V
on I DQ1 =
VSDQ1
VD1 = I DQ1 RD1 = (1)( 4 ) = 4 V = VG 2
⎛ V − VS 2 ⎞
⎛ V ⎞
I DQ 2 = I DSS ⎜ 1 − GS ⎟ = 6 ⎜⎜ 1 − G 2
⎟
VP ⎠
( −2 ) ⎟⎠
⎝
⎝
2
2
V
V
⎛ 4 V ⎞
⎛ V ⎞
= 6 ⎜ 1 + − S 2 ⎟ = 6 ⎜ 3 − S 2 ⎟ and I DQ 2 = S 2 = S 2
RS 2
2 ⎠
4
⎝ 2 2 ⎠
⎝
2
Then
2
VS 2
⎛ V ⎞
= 6⎜3 − S2 ⎟
4
2 ⎠
⎝
2
⎛
V2 ⎞
VS 2 = 24 ⎜ 9 − 3VS 2 + S 2 ⎟
4 ⎠
⎝
2
6VS 2 − 73VS 2 + 216 = 0
VS 2 =
( 73)
73 ±
2
− 4 ( 6 )( 216 )
2 ( 6)
⇒ VS 2 = 7.09 V or = 5.08 V
For VS 2 = 5.08 V ⇒ VGS 2 = 4 − 5.08 = −1.08 transistor biased ON
5.08
⇒ I DQ 2 = 1.27 mA
4
= 20 − VS 2 = 20 − 5.08 ⇒ VDS 2 = 14.9 V
I DQ 2 =
VDS 2
b.
Vg2
⫺
gm2Vgs2
⫹
Vgs2
Vi
⫹
⫺
RD1
V0
Vsg1
gm1Vsg1
RS2
⫹
Vg 2 = g m1Vsg1 RD1 = − g m1V1 RD1
V0 = g m 2Vgs 2 ( RS 2 & RL )
Vg 2 = Vgs 2 + V0 ⇒ Vgs 2 =
Vg 2
1 + g m 2 ( RS 2 & RL )
Av =
− g m1 RD1
V0
=
Vi 1 + g m 2 ( RS 2 & RL )
g m1 =
2 I DSS
VP
⎛ VGS ⎞
⎜1 −
⎟
⎝ VP ⎠
2 ( 6 ) ⎛ 1.2 ⎞
⎜1 −
⎟ = 2.4 mA/V
2 ⎝
2 ⎠
2 ( 6 ) ⎛ 1.08 ⎞
gm2 =
⎜1 −
⎟ = 2.76 mA/V
2 ⎝
2 ⎠
− ( 2.4 )( 4 )
= −2.05 = Av
Then Av =
1 + ( 2.76 )( 4 )( 2 )
=
⫺
RL
Chapter 4
Problem Solutions
4.1
(a)
⎛ µ C ⎞⎛W ⎞
g m = 2 K n I D = 2 ⎜ n ox ⎟ ⎜ ⎟ I D
⎝ 2 ⎠⎝ L ⎠
0.5 = 2
( 0.040 ) ⎜⎛
ω⎞
W
= 3.125
⎟ ( 0.5 ) ⇒
L
⎝L⎠
(b)
2
⎛ µ C ⎞⎛W ⎞
I D = ⎜ n ox ⎟ ⎜ ⎟ (VGS − VTN )
L
2
⎝
⎠⎝ ⎠
0.5 = ( 0.04 )( 3.125 )(VGS − 0.8 ) ⇒ VGS = 2.80 V
2
4.2
⎛ µ p Cox ⎞ ⎛ W ⎞
gm = 2 K p I D = 2 ⎜
⎟ ⎜ ⎟ ID
⎝ 2 ⎠⎝ L ⎠
(a)
W
⎛ 50 ⎞
⎛W ⎞
= 0.3125
⎜ ⎟ = ( 20 ) ⎜ ⎟ (100 ) ⇒
L
⎝ 2⎠
⎝L⎠
⎛ µ p Cox ⎞ ⎛ W ⎞
2
ID = ⎜
⎟ ⎜ ⎟ (VSG + VTP )
L
2
⎝
⎠
⎝
⎠
2
(b)
100 = ( 20 )( 0.3125 )(VSG − 1.2 ) ⇒ VSG = 5.2 V
2
4.3
2
I D = K n (VGS − VTN ) (1 + λVDS )
I D1 1 + λVDS1
3.4 1 + λ (10 )
=
⇒
=
I D 2 1 + λVDS 2
3.0 1 + λ ( 5 )
3.4 [1 + 5λ ] = 3.0 [1 + 10λ ]
3.4 − 3.0 = λ ( 3 ⋅10 − ( 3.4 ) ⋅ 5 ) ⇒ λ = 0.0308
ro =
ΔVDS
5
=
= 12.5 kΩ
0.4
ΔI D
4.4
r0 =
ID =
1
λ ID
1
λ r0
=
1
( 0.012 )(100 )
⇒ I D = 0.833 mA
4.5
2
I D = K n (VGS − VTN ) (1 + λVDS )
I D1 1 + λVDS 1
=
I D 2 1 + λVDS 2
0.20 1 + λ ( 2 )
=
0.22 1 + λ ( 4 )
0.20 (1 + 4λ ) = 0.22 (1 + 2λ )
( 0.8 − 0.44 ) λ = 0.22 − 0.20
λ = 0.0556 V −1
ro =
ΔVDS
2
=
⇒ ro = 100 K
ΔI D
0.02
4.6
(a)
(i)
ro =
1
1
=
= 1000 K
λ I D ( 0.02 )( 0.05 )
(ii)
ro =
1
= 100 K
( 0.02 )( 0.5)
(b)
(i)
ΔI D =
ΔVDS
1
=
= 0.001 mA = 1.0 µ A
1000
ro
ΔI D 1.0
=
⇒ 2%
50
ID
(ii)
ΔI D =
ΔVDS
1
=
= 0.01 ⇒ 10 µ A
ro
100
ΔI D
10
=
⇒ 2%
ID
500
4.7
I D = 1.0 mA
1
1
=
= 100 K
ro =
λ I D ( 0.01)(1)
4.8
Av = − g m ( r0 || RD ) = − (1)( 50 ||10 ) ⇒ Av = −8.33
4.9
a.
RD =
VDD − VD SQ
I DQ
=
10 − 6
⇒ RD = 8 kΩ
0.5
For VGSQ = 2 V, for example,
2
⎛ µ C ⎞⎛W ⎞
I DQ = ⎜ n ox ⎟ ⎜ ⎟ (VGSQ − VTN )
2
L
⎝
⎠⎝ ⎠
W
2
⎛W ⎞
0.5 = ( 0.030 ) ⎜ ⎟ ( 2 − 0.8 ) ⇒
= 11.6
L
⎝L⎠
b.
g m = 2 K n I DQ = 2 K n (VGSQ − VTN )
g m = 2 ( 0.030 )(11.6 )( 2 − 0.8 ) ⇒ g m = 0.835 mA/V
ro =
1
c.
=
⇒ r = 133 kΩ
1
( 0.015)( 0.50 ) o
Av = − g m ( r0 & RD ) = − ( 0.835 )(133 & 8 ) ⇒ Av = −6.30
λ I DQ
4.10
K n vgs2 = K n ⎡⎣Vgs sinω t ⎤⎦ = K nVgs2 sin 2ω t
2
1
[1 − cos 2ω t ]
2
K nVgs2
So K n vgs2 =
[1 − cos 2ω t ]
2
sin 2 ω t =
K nVgs2
Ratio of signal at 2ω to that at ω :
2
⋅ cos 2ω t
2 K n (VGSQ − VTN ) Vgs ⋅ sin ω t
The coefficient of this expression is then:
Vgs
4 (VGSQ − VTN )
4.11
0.01 =
Vgs
4 (VGSQ − VTN )
So Vgs = ( 0.01)( 4 )( 3 − 1) ⇒ Vgs = 0.08 V
4.12
Vo = − g mVgs ( ro RD )
⎛ 50 ⎞
⋅ Vi = ⎜
⎟ ⋅ Vi = ( 0.9615 ) Vi
R1 R2 + RSi
⎝ 50 + 2 ⎠
Av = − g m ( 0.9615 ) ( ro RD ) = − (1)( 0.9615 ) ( 50 10 ) ⇒ Av = −8.01
R1 R2
Vgs =
4.13
Av = − g m ( r0 || RD )
−10 = − g m (100 || 5 ) ⇒ g m = 2.1 mA/V
4.14
a.
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ 200 ⎞
VG = ⎜
⎟ (12 ) = 4.8 V
⎝ 200 + 300 ⎠
V − VGS
2
ID = G
= K n (VGS − VTN )
RS
4.8 − VGS = (1)( 2 ) (VGS2 − 4VGS + 4 )
2VGS2 − 7VGS + 3.2 = 0
VGS =
7±
(7)
2
− 4 ( 2 )( 3.2 )
2 ( 2)
= 2.96 V
I D = (1)( 2.96 − 2 ) ⇒ I D = 0.920 mA
2
VDS = VDD − I D ( RD + RS )
= 12 − ( 0.92 )( 3 + 2 ) ⇒ VDS = 7.4 V
(b)
Vo =
− g mVg ( RD RL )
1 + g m RS
where Vg =
Then
R1 R2
R1 R2 + RSi
⋅ Vi =
300 200
300 200 + 2
⋅ Vi =
120
⋅ Vi = ( 0.9836 ) Vi
120 + 2
Av =
− g m ( 0.9836 )( RD || RL )
1 + g m RS
g m = 2 (1)( 2.96 − 2 ) = 1.92 mA / V
So Av =
− (1.92 )( 0.9836 )( 3 || 3)
1 + (1.92 )( 2 )
⇒ Av = −0.585
c.
AC load line
⫺1
⫺1
⫽
Slope ⫽
3.5 K
3兩兩3 ⫹ 2
0.92
7.4
12
−1
Δi D =
⋅ Δvds
3.5 kΩ
Δi D = 0.92 mA ⇒ Δvds = ( 0.92 )( 3.5 ) = 3.22 ⇒ 6.44 V peak-to-peak
4.15
a.
I D Q = 3 mA ⇒ VS = I DQ RS = ( 3)( 0.5 ) = 1.5 V
I DQ = K n (VGS − VTN )
2
3 = ( 2 )(VGS − 2 ) ⇒ VGS = 3.225 V
2
VG = VGS + VS = 3.225 + 1.5 = 4.725 V
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R1
⎝ R1 + R2 ⎠
1
4.725 = ( 200 )(15 ) ⇒ R1 = 635 kΩ
R1
635R2
= 200 ⇒ R2 = 292 kΩ
635 + R2
b.
− g m ( RD || RL )
Av =
1 + g m RS
g m = 2 ( 2 )( 3.225 − 2 ) = 4.90 mA / V
Av =
− ( 4.90 )( 2 || 5 )
1 + ( 4.90 )( 0.5 )
⇒ Av = −2.03
4.16
From Problem 4.14: I D = 0.920 mA
VDS = 7.4 V
g m = 1.92 mA/V
⎛ R1 || R2 ⎞
Av = − g m ( RD || RL ) ⎜
⎟
⎝ R1 || R2 + RSi ⎠
⎛ 200 || 300 ⎞
= − (1.92 )( 3 || 3) ⎜
⎟ = − ( 2.88 )( 0.9836 )
⎝ 200 || 300 + 2 ⎠
Av = −2.83
AC load line
⫺1
⫺1
Slope ⫽
⫽
3兩兩3 1.5 K
0.92
7.4
12
−1
⋅ ΔvDS , ΔiD = 0.92 mA ⇒ ΔvDS = ( 0.92 ) (1.5 ) = 1.38
1.5 kΩ
⇒ 2.76 V peak-to-peak output voltage swing
ΔiD =
4.17
(a)
Av = − g m RD
−15 = −2 RD ⇒ RD = 7.5 K
(b)
Av =
−5 =
− g m RD
1 + g m RS
− ( 2 )( 7.5 )
1 + ( 2 ) RS
⇒ RS = 1 K
4.18
(a)
Av =
− g m RD
1 + g m RS
(1)
−8 =
− g m RD
1 + g m (1)
(2)
−16 = − g m RD
Then 8 =
16
⇒
1 + g m (1)
g m = 1 mA/V
RD = 16 K
(b)
Av = −10 =
− (1)(16 )
1 + (1) RS
RS = 0.6 K
4.19
a.
AC load line
⫺1
Slope ⫽
5K
IDQ
VDS(sat)
VDSQ
VDSQ = V + − I DQ RD − (−VGSQ )
Output Voltage Swing = VDSQ − VDS ( sat )
= ⎡⎣V + − I DQ RD + VGSQ ⎤⎦ − (VGSQ − VTN )
= V + − I DQ RD + VTN
So ΔI D = I DQ =
1
1
⎡V + − I DQ RD + VTN ⎤⎦
⋅ ΔVDS =
5 kΩ
5 kΩ ⎣
1
⎡5 − I DQ (10) + 1⎤⎦
5 kΩ ⎣
= 1.2 − 2 I DQ = I DQ ⇒ I DQ = 0.4 mA = I Q
I DQ =
b.
( 0.5 )( 0.4 ) = 0.8944 mA / V
g m = 2 K n I DQ = 2
r0 =
1
1
=
= 250 kΩ
λ I DQ ( 0.01)( 0.4 )
Av = − g m ( RD || RL || r0 )
= − ( 0.8944 )(10 ||10 || 250 ) ⇒ Av = −4.38
4.20
(a)
I DQ = K n (VGSQ − VTN )
2
0.1 = 0.85 (VGSQ − 0.8 )
VGSQ = 1.143 V
RS =
2
⇒ RS = 38.6 K
−1.143 − ( −5 )
0.1
ΔV = Δ I ( RD RL ) ro
ro =
1
= 500 K
( 0.02 )( 0.1)
1 = 0.1( RD RL ) ro
RD RL ro = 10 K =
40 500 RD
40 500 + RD
⇒
37.04 RD
= 10
37.04 + RD
RD = 13.7 K
(b)
gm = 2 Kn I D = 2
( 0.85)( 0.1)
g m = 0.583 mA/V
ro = 500 K
(c)
Av = − g m ( RD RL ro )
= − ( 0.583) (13.7 40 500 ) = − ( 0.583)(10 )
Av = −5.83
4.21
a.
2
I D = K n (VGS − VTN )
2 = 4 (VGS − ( −1) )
2
VGS = −0.293 V ⇒ VS = 0.293 V = I D RS = (2) RS ⇒ RS = 0.146 kΩ
VD = VDS + VS = 6 + 0.293 = 6.293
RD =
b.
Av =
10 − 6.293
⇒ RD = 1.85 kΩ
2
− g m ( RD RL )
1 + g m RS
g m = 2 K n (VGS − VTN )
g m = 2 ( 4 )( −0.293 + 1) = 5.66 mA/V
Av =
− ( 5.66 ) (1.85 2 )
1 + ( 5.66 )( 0.146 )
⇒ Av = −2.98
c.
AC load line
⫺1
1.85兩兩2 ⫹ 0.146
⫺1
⫽
1.107 K
Slope ⫽
2 mA
10
6
Δv0 = ( ΔiD )(1.85 || 2 ) = ( 2 )(1.85 || 2 ) = 1.92 V
Δvi =
1.92
= 0.645 ⇒ Vi = 0.645 V
2.98
4.22
a.
VDSQ = VDD − I DQ ( RD + RS )
2.5 = 5 − I DQ ( 2 + RS )
2.5
I DQ =
2 + RS
2
I DQ = K n (VGS − VTN ) =
−VGS
−2.5 RS
⇒ VGS = − I DQ RS =
2 + RS
RS
⎡ −2.5RS
⎤
2.5
Kn ⎢
− VTN ⎥ =
2
2
R
+
+ RS
S
⎣
⎦
2
⎡ 2.5 RS ⎤
2.5
4 ⎢1 −
⎥ =
+
+ RS
2
2
R
S ⎦
⎣
2
⎡ 2 + RS − 2.5 RS ⎤
2.5
4⎢
⎥ =
2 + RS
2 + RS
⎣
⎦
2
4
( 2 − 1.5RS )
2 + RS
2
= 2.5
4 ( 4 − 6 RS + 2.25 RS2 ) = 2.5 ( 2 + RS )
9 RS2 − 26.5RS + 11 = 0
RS =
26.5 ±
2
( 26.5) − 4 ( 9 )(11)
2 (9)
RS = 0.5 kΩ or 2.44 kΩ
But RS = 2.44 kΩ ⇒ VGS = −1.37 Cut off. ⇒ RS = 0.5 kΩ,
I DQ = 1.0 mA
b.
Av =
− g m ( RD || RL )
1 + g m RS
( 4 )(1) = 4 mA / V
g m = 2 K n I DQ = 2
Av =
− ( 4 )( 2 || 2 )
1 + ( 4 )( 0.5 )
⇒ Av = −1.33
4.23
a.
5 = I DQ RS + VSDQ + I DQ RD − 5
5 = I DQ RS + 6 + I DQ (10 ) − 5
1.
I DQ =
4
RS + 10
VS = VSDQ + I DQ RD − 5 = VSGQ
2.
1 + I DQ (10 ) = VSGQ
3.
I DQ = K p (VSGQ − 2 )
2
Choose RS = 10 kΩ ⇒ I DQ =
4
= 0.20 mA
20
VSGQ = 1 + (0.2)(10) = 3 V
0.20 = K P (3 − 2) 2 ⇒ K P = 0.20 mA / V 2
b.
2
I DQ = ( 0.20 )( 3 − 2 ) = 0.20 mA
( 0.2 )( 0.2 ) = 0.4 mA / V
Av = − g m ( RD || RL ) = − ( 0.4 )(10 ||10 ) ⇒ Av = −2.0
g m = 2 K P I DQ = 2
c.
Choose RS = 20 kΩ ⇒ I DQ =
VSGQ
4
= 0.133 mA
30
= 1 + (0.133)(10) = 2.33 V
0.133 = K p (2.33 − 2) 2 ⇒ K p = 1.22 mA / V 2
g m = 2 (1.22)(0.133) = 0.806 mA/V
Av = −(0.806)(10 10) ⇒ Av = −4.03
A larger gain can be achieved.
4.24
(a)
I DQ = K p (VSGQ + VTP )
2
0.25 = 0.8 (VSGQ − 0.5 )
2
VSGQ = 1.059 V
3 − 1.059
⇒ RS = 7.76 K
RS =
0.25
VD = VS − VSDQ = 1.059 − 1.5 = −0.441 V
RD =
(b)
−0.441 − ( −3)
0.25
⇒ RD = 10.2 K
Av = − g m ( RD RL )
( 0.8)( 0.25 ) = 0.8944 mA/V
Av = − ( 0.8944 )(10.2 || 2 )
g m = 2 K p I DQ = 2
Av = −1.50
(c)
ΔVO = ΔI ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418
So ΔVO = 0.836 peak-to-peak
4.25
I DQ = K n (VGSQ − VTN )
2
g m = 2 K n I DQ
2.2 = 2 K n ( 6 ) ⇒ K n = 0.202 mA / V 2
6 = 0.202 ( 2.8 − VTN ) ⇒ VTN = −2.65 V
2
VDSQ = 18 − I DQ ( RS + RD )
18 − 10
= 1.33 kΩ ⇒ RS = 1.33 − RD
RS + RD =
6
g m ( RD RL )
Av = −
1 + g m RS
⎛ R ⋅1 ⎞
−2.2 ⎜ D ⎟
⎝ RD + 1 ⎠
−1 =
1 + ( 2.2 )(1.33 − RD )
1 + 2.93 − 2.2 RD =
2.2 RD
1 + RD
( 3.93 − 2.2 RD )(1 + RD ) = 2.2 RD
3.93 + 1.73RD − 2.2 RD2 = 2.2 RD
2.2 RD2 + 0.47 RD − 3.93 = 0
RD =
−0.47 +
( 0.47 ) + 4 ( 2.2 )( 3.93)
⇒ RD = 1.23 kΩ,
2 ( 2.2 )
= 2.8 + ( 6 )( 0.1) = 3.4 V
2
VG = VGS + VS
1
1
VG = ⋅ Rin ⋅ VDD = (100 )(18 ) = 3.4 ⇒ R1 = 529 kΩ
R1
R1
529 R2
= 100 ⇒ R2 = 123 kΩ
529 + R2
4.26
a.
VSD(sat)
VSDQ
V1
RS = 0.10 kΩ
V1 = 9 + VSG , VSD ( sat ) = VSG + VTP
VSDQ =
=
V1 − VSD ( sat )
2
+ VSD ( sat )
( 9 + VSG ) − (VSG + VTP )
2
+ (VSG + VTP )
9 + 1.5
+ VSG − 1.5
2
= 3.75 + VSG = 9 + VSG − I DQ RD
=
VSDQ
2
I DQ = K p (VSG + VTP )
Set RD = 0.1RL = 2 kΩ
3.75 = 9 − I DQ ( 2 ) ⇒ I DQ = 2.625 mA
b.
g m = 2 K p I DQ = 2
r0 =
( 2 )( 2.625) = 4.58 mA / V
1
1
=
= 38.1 kΩ
λ I DQ ( 0.01)( 2.625 )
Open circuit.
Av = − g m ( RD || r0 )
Av = −4.58 ( 2 || 38.1) ⇒ Av = −8.70
c.
With RL
Av = −4.58 ( 2 || 20 || 38.1) ⇒ Av = −7.947 ⇒ Change = 8.66%
4.27
(a)
I DQ = K p (VSGQ + VTP )
2
0.5 = 0.25 (VSGQ + 0.8 )
2
VSGQ = 0.614 V = VS
10 − 0.614
⇒ RS = 18.8 K
RS =
0.5
VD = VS − VSDQ = 0.614 − 3 = −2.386 V
RD =
−2.386 − ( −10 )
0.5
⇒ RD = 15.2 K
(b)
Av = − g m RD
( 0.25)( 0.5 ) = 0.7071 mA/V
Av = − ( 0.7071)(15.2 )
g m = 2 K p I DQ = 2
Av = −10.7
4.28
Av = − g m ( RD RL )
VDSQ = VDD − I DQ ( RS + RD )
10 = 20 − (1)( RS + RD ) ⇒ RS + RD = 10 kΩ
Let RD = 8 kΩ, RS = 2 kΩ
Av = −10 = − g m ( 8 & 20 )
g m = 1.75 mA / V = 2 K n I DQ = 2 K n (1) ⇒ K n = 0.766 mA / V 2
VS = I DQ RS = (1)( 2 ) = 2 V
I DQ = K n (VGS − VTN ) ⇒ 1 = 0.766 (VGS − 2 ) ⇒ VGS = 3.14 V
2
2
VG = VGS + VS = 3.14 + 2 = 5.14
VG =
1
1
⋅ Rin ⋅ VDD ⇒ ( 200 )( 20 ) = 5.14 ⇒ R1 = 778 kΩ
R1
R1
778R2
= 200 ⇒ R2 = 269 kΩ
778 + R2
4.29
Av =
Ro =
Av =
=
R0 =
( 4 )( 50 )
g m r0
=
⇒ Av = 0.995
1 + g m r0 1 + ( 4 )( 50 )
1
1
r0 = 50 ⇒ R0 = 0.249 kΩ
4
gm
g m ( r0 & RS )
1 + g m ( r0 & RS )
4 ( 2.38 )
1 + 4 ( 2.38 )
=
4 ( 50 & 2.5 )
1 + 4 ( 50 & 2.5 )
⇒ Av = 0.905
1
1
r0 RS = 50 2.5 ⇒ R0 = 0.226 kΩ
4
gm
4.30
Av =
0.98 =
g m ( RL ro )
1 + g m ( RL ro )
g m ro
⇒ g m ro = 49
1 + g m ro
Also 0.49 =
0.49 =
0.49 =
g m ( RL ro )
1 + g m ( RL ro )
⎛ Rr ⎞
gm ⎜ L o ⎟
⎝ RL + ro ⎠
=
⎛ Rr ⎞
1 + gm ⎜ L o ⎟
⎝ RL + ro ⎠
g m ( RL ro )
RL + ro + g m ( RL ro )
( 49 )(1)
49
=
1 + ro + ( 49 )(1) 50 + ro
ro = 50 K
g m = 0.98 mA/V
4.31
(a)
Av =
( 2 )( 25)
g m ro
=
1 + g m ro 1 + ( 2 )( 25 )
Av = 0.98
Ro =
1
1
ro = 25 = 0.5 || 25
gm
2
Ro = 0.49 K
(b)
Av =
g m ( ro || RL )
1 + g m ( ro || RL )
=
2 ( 25 || 2 )
1 + 2 ( 25 || 2 )
=
2 (1.852 )
1 + 2 (1.852 )
Av = 0.787
4.32
Av =
g m ( RL || ro )
1 + g m ( RL || ro )
=
5 (10 || 100 )
1 + ( 5 )(10 || 100 )
=
5 ( 9.09 )
1 + 5 ( 9.09 )
Av = 0.978
Ro =
1
1
RL ro = 10 || 100 = 0.2 || 9.0909
gm
5
Ro = 0.196 K
4.33
a.
⎛ R2 ⎞
396
⎛
⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) ⇒ VG = 2.42 V
⎝ 396 + 1240 ⎠
⎝ R1 + R2 ⎠
10 − (VSG + 2.42 )
2
I DQ =
= K p (VSG + VTP )
RS
2
− 4VSG + 4 )
7.58 − VSG = ( 2 )( 4 ) (VSG
2
− 31VSG + 24.4 = 0
8VSG
VSG =
31 ±
( 31)
2
− 4 ( 8 )( 24.4 )
2 (8)
⇒ VSG = 2.78 V
I DQ = ( 2 )( 2.78 − 2 ) ⇒ I DQ = 1.21 mA
2
VSDQ = 10 − I DQ RS = 10 − (1.21)( 4 ) ⇒ VSDQ = 5.16 V
b.
g m = 2 K p I DQ = 2
r0 =
Av =
=
( 2 )(1.21) = 3.11 mA / V
1
1
=
= 41.3 kΩ
λ I DQ ( 0.02 )(1.21)
g m ( RS RL r0 )
1 + g m ( RS RL r0 )
3.11( 4 4 41.3)
1 + ( 3.11) ( 4 4 41.3)
⇒ Av = 0.856
⫺
Ii
Vsg
Vi
⫹
⫺
R1兩兩R2
r0
gmVsg
⫹
RS
RL
I0
⎛ RS r0
I 0 = − ( g mVsg ) ⎜
⎜R r +R
L
⎝ S 0
−Vi
Vsg =
1 + g m ( RS RL r0 )
⎞
⎟⎟
⎠
Vi = I i ( R1 R2 )
Ai =
g m ( R1 R2 )
RS r0
I0
=
⋅
I i 1 + g m ( RS RL r0 ) RS r0 + RL
( 3.11) ( 396 1240 ) 4 41.3
⋅
1 + ( 3.11) ( 4 4 41.3) 4 41.3 + 4
( 3.11)( 300 )
3.647
=
⋅
⇒ Av = 64.2
1 + ( 3.11)(1.908 ) 3.647 + 4
=
R0 =
1
1
4 41.3 ⇒ R0 = 0.295 kΩ
RS r0 =
3.11
gm
4.34
g m = 2 K n I Q = 2 (1)(1) = 2 mA / V
1
1
=
= 100 kΩ
λ I Q ( 0.01)(1)
r0 =
Av =
g m ( r0 RL )
1 + g m ( r0 RL )
R0 =
=
2 (100 4 )
1 + 2 (100 4 )
⇒ Av = 0.885
1
1
r0 = 100 ⇒ R0 = 0.498 kΩ
2
gm
4.35
a.
Av =
gm ( 4)
g m RL
⇒ 0.95 =
1 + g m RL
1 + gm ( 4)
0.95 = 4 (1 − 0.95 ) g m ⇒ g m = 4.75 mA/V
⎛1
⎞⎛W
g m = 2 ⎜ µ n Cox ⎟ ⎜
⎝2
⎠⎝ L
( 0.030 ) ⎜⎛
⎞
⎟ IQ
⎠
W⎞
W
= 47.0
⎟ ( 4) ⇒
L
⎝L⎠
4.75 = 2
⎛1
⎞⎛ W
g m = 2 ⎜ µ n Cox ⎟⎜
2
⎝
⎠⎝ L
b.
4.75 = 2
⎞
⎟ IQ
⎠
( 0.030 )( 60 ) I Q
⇒ I Q = 3.13 mA
4.36
2
I DQ = K n (VGS − VTN )
5 = 5 (VGS + 2 ) ⇒ VGS = −1 V ⇒ VS = −VGS = 1 V
2
I DQ =
VS − ( −5 )
RS
⇒ RS =
g m = 2 K n I DQ = 2
r0 =
1
λ I DQ
=
1+ 5
⇒ RS = 1.2 kΩ
5
( 5 )( 5) = 10 mA / V
1
= 20 kΩ
0.01
( )( 5 )
Av =
=
R0 =
g m ( r0 RS RL )
1 + g m ( r0 RS RL )
(10 ) ( 20 1.2 2 )
⇒ Av = 0.878
1 + (10 ) ( 20 1.2 2 )
1
1
|| r0 || RS = || 20 ||1.2 ⇒ Ro = 91.9 Ω
10
gm
4.37
Av =
g m (10 )
g m RS
⇒ 0.90 =
1 + g m RS
1 + g m (10 )
0.90 = 10 (1 − 0.90 ) g m ⇒ g m = 0.90 mA/V
1
1
RS =
10 ⇒ R0 = 1 kΩ
gm
0.90
With RL :
R0 =
Av =
( 0.90 )(10 || 2 )
⇒ Av = 0.60
1 + g m ( RS || RL ) 1 + ( 0.90 )(10 || 2 )
g m ( RS || RL )
=
4.38
R0 =
1
RS
gm
Output resistance determined primarily by gm
1
Set
= 0.2 kΩ ⇒ g m = 5 mA/V
gm
g m = 2 K n I DQ ⇒ 5 = 2
I DQ = K n (VGS − VTN )
( 4 ) I DQ
⇒ I DQ = 1.56 mA
2
2
1.56 = 4 (VGS + 2 )
VGS = −1.38 V, VS = −VGS = 1.38 V
⇒ RS = 4.09 kΩ
1.56
5 ( 4.09 2 )
g m ( RS RL )
Av =
=
⇒ Av = 0.870
1 + g m ( RS RL ) 1 + 5 ( 4.09 2 )
RS =
1.38 − ( −5 )
4.39
a.
g m = 2 K p I DQ = 2
( 5)( 5 ) = 10 mA / V
1
1
=
⇒ Ro = 100 Ω
g m 10
b.
Open-circuit gain
g m r0
But r0 = ∞ so Av = 1.0
Av =
1 + g m r0
With RL:
g m RL
Av =
1 + g m RL
Ro =
0.50 =
4.40
10 RL
⇒ 0.50 = 10 (1 − 0.5 ) RL ⇒ RL = 0.1 kΩ
1 + 10 RL
−1
⋅ Δv0
RS RL
ΔiD = I DQ =
Δv0 = − I DQ ⋅ RS RL = −
v0 ( min ) = −
Av =
vi =
I DQ ⋅ RS RL
RS + RL
I DQ RS
R
1+ S
RL
g m ( RS RL )
1 + g m ( RS RL )
=
v0
vi
− I DQ ( RS RL ) ⎡⎣1 + g m ( RS RL ) ⎤⎦
g m ( RS RL )
vi ( min ) = −
⎡1 + g m ( RS RL ) ⎤⎦
gm ⎣
I DQ
4.41
(a)
VDD = VDSQ + I DQ RS
3 = 1.5 + ( 0.25 ) RS ⇒ RS = 6 K
VS = 1.5 V
I DQ = K n (VGSQ − VTN )
0.25 = 0.5 (VGSQ − 0.4 )
2
2
VGSQ = 1.107 V
VG = VGSQ + VS + 1.107 + 1.5 = 2.607 V
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = − RL − VDD
R
R
R
+
2 ⎠
1
⎝ 1
1
2.607 = ( 300 )( 3) ⇒ R1 = 345.2 K ⇒ R2 = 2291 K
R1
(b)
g m RS
Av =
g m = 2 K n I DQ = 2 ( 0.5 )( 0.25 ) = 0.7071 mA/V
1 + g m RS
Av =
( 0.7071)( 6 )
⇒ Av = 0.809
1 + ( 0.7071)( 6 )
Ro =
1
1
RS =
6 = 1.414 || 6
gm
( 0.7071)
Ro = 1.14 K
4.42
Av = + g m RD = ( 5 )( 4 )
Av = 20
Ri =
1
1
10 = 10 = 0.2 ||10
5
gm
Ri = 0.196 K
4.43
a.
VGS + I DQ RS = 5
5 − VGS
2
= K n (VGS − VTN )
I DQ =
RS
5 − VGS = (10 )( 3) (VGS2 − 2VGS + 1)
30VGS2 − 59VGS + 25 = 0
VGS =
59 ±
( 59 )
2
− 4 ( 30 )( 25 )
2 ( 30 )
⇒ VGS = 1.35 V
I DQ = ( 3)(1.35 − 1) ⇒ I DQ = 0.365 mA
2
VDSQ = 10 − ( 0.365 )( 5 + 10 ) ⇒ VDSQ = 4.53 V
b.
g m = 2 K n I DQ = 2
r0 =
c.
1
( 3)( 0.365 ) ⇒ g m = 2.093 mA / V
⇒r =∞
1
=
( 0 )( 0.365 ) 0
Av = g m ( RD RL ) = ( 2.093) ( 5 4 ) ⇒ Av = 4.65
λ I DQ
4.44
a.
I DQ = K p (VSG + VTP )
2
0.75 = ( 0.5 )(VSG − 1) ⇒ VSG = 2.225 V
5 − 2.225
5 = I DQ RS + VSG ⇒ RS =
⇒ RS = 3.70 kΩ
0.75
VSDQ = 10 − I DQ ( RS + RD )
6 = 10 − ( 0.75 )( 3.70 + RD ) ⇒ RD = 1.63 kΩ
2
b.
Ri =
1
gm
g m = 2 K p I DQ = 2
( 0.5 )( 0.75) = 1.225 mA / V
1
⇒ Ri = 0.816 kΩ
1.225
Ro = RD ⇒ Ro = 1.63 kΩ
Ri =
c.
⎞
⎞⎛
RS
⎟⎟ ⋅ ii
⎟ ⎜⎜
⎠ ⎝ RS + [1/ g m ] ⎠
3.70
⎛ 1.63 ⎞ ⎛
⎞
i0 = ⎜
⎟⎜
⎟ ii
⎝ 1.63 + 2 ⎠ ⎝ 3.70 + 0.816 ⎠
i0 = 0.368ii = i0 = 1.84sin ω t ( µ A )
⎛ RD
i0 = ⎜
⎝ RD + RL
v0 = i0 RL = (1.84 )( 2 ) sin ω t ⇒ v0 = 3.68sin ω t ( mV )
4.45
a.
2
I DQ = K n (VGS − VTN )
5 = 4 (VGS − 2 ) ⇒ VGS = 3.12 V
V + = I DQ RD + VDSQ − VGS
2
10 = ( 5 ) RD + 12 − 3.12 ⇒ RD = 0.224 kΩ
b.
g m = 2 K n I DQ = 2
Ri =
( 4 )( 5 ) ⇒
g m = 8.944 mA / V
1
1
=
⇒ Ri = 0.112 kΩ
g m 8.94
Av = g m ( RD RL ) = ( 8.944) ( 0.224 2 ) ⇒ Av = 1.80
c.
4.46
a.
2
I DQ = K p (VSG + VTP )
3 = 2 (VSG − 2 ) ⇒ VSG = 3.225 V
V + = I DQ RS + VSG
10 − 3.225
RS =
⇒ RS = 2.26 kΩ
3
VSDQ = 20 − I DQ ( RS + RD )
10 = 20 − ( 3)( 2.26 + RD ) ⇒ RD = 1.07 kΩ
2
b.
Av = g m ( RD RL )
( 2 )( 3) = 4.90 mA / V
Av = ( 4.90 )(1.07 ||10 ) ⇒ Av = 4.74
g m = 2 K p I DQ = 2
4.47
a.
Av =
(W / L )D
(W / L )L
= 5 ⇒ (W / L ) D = 25 ( 0.5 ) ⇒ (W / L ) D = 12.5
1
⎛W ⎞
µ n Cox ⎜ ⎟ = ( 30 )(12.5 ) = 375 µ A/V 2
2
⎝ L ⎠D
K L = ( 30 )( 0.5 ) = 15 µ A / V 2
KD =
Transition point:
vGSD − VTND = (VDD − VTNL ) −
KD
( vGSD − VTND )
KL
375
( vGSD − 2 )
15
vGSD (1 + 5 ) = (10 − 2 ) + 2 + 5 ( 2 )
vGSD = 3.33 V and vDSD = 1.33 V
3.33 − 2
+ 2 ⇒ VGSQ = 2.67 V
VGSQ =
2
vGSD − 2 = (10 − 2 ) −
␯O
8
Q-point
VDSQ
1.33
3.33
2
VGSQ
b.
␯GSD
2
I DQ = K D (VGSQ − VTND )
I DQ = 0.375 ( 2.67 − 2 ) ⇒ I DQ = 0.166 mA
2
VDSQ =
4.48
(a)
8 − 1.33
+ 1.33 ⇒ VDSQ = 4.67 V
2
Transition point: Load: VOtB = VDD − VTNL = 10 − 2 = 8 V
Driver:
2
2
K D ⎢⎡( vOt A ) (1 + λD vOt A ) ⎤⎥ = K L ⎡( −VTNL ) 1 + λL (VDD − vOt A ) ⎤
⎣
⎦
⎣
⎦
2
3
0.5 ⎡⎣v0t A + ( 0.02 ) v0t A ⎤⎦ = 0.1 ⎡⎣( 4 )(1 + 0.02 ) (10 − v0t A ) ⎤⎦
We have 0.01v03t A + 0.5v02t A + 0.008v0t A − 0.48 = 0
(
)
Therefore v0t A = 0.963 V
8 − 0.963
+ 0.963 = 4.48 V = VDSQ
2
2
2
Then K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤ = K L ⎡( −VTNL ) 1 + λL (VDD − vOQ ) ⎤
⎣
⎦
⎣
⎦
2
⎡
⎤
0.5 (VGSD − 1.2 ) (1 + ( 0.02 )( 4.48 ) ) = 0.1 ⎡⎣( 4 ) (1 + 0.02 (10 − 4.48 ) ) ⎤⎦ which yields VGSD = 2.103 V
⎣
⎦
b.
2
I DQ = K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤
⎣
⎦
2
I DQ = 0.5 ⎡( 2.103 − 1.2 ) (1 + ( 0.02 )( 4.48 ) ) ⎤
⎣
⎦
So I DQ = 0.444 mA
Now v0 Q =
(
)
c.
r0 D = r0 L =
1
1
=
= 112.6 kΩ
λ I DQ ( 0.02 )( 0.444 )
g mD = 2 K D (VGSD − VTND ) = 2 ( 0.5 )( 2.103 − 1.2 ) ⇒ g mD = 0.903 mA / V
Then Av = − g mD ( r0 D r0 L ) = − ( 0.903) (112.6 112.6 ) or Av = −50.8
4.49
2
I D = K n (VGS − VTN ) , VDS = VGS
I D = 0 when VDS = 1.5 V ⇒ VTN = 1.5 V
0.8 = K n ( 3 − 1.5 ) ⇒ K n = 0.356 mA / V 2
dI
I
go = D =
= 2 K n (VDS − VTN ) = 2 ( 0.356 )( 3 − 1.5 ) ⇒ Ro = 0.936 k Ω
dVDS Ro
2
4.50
a.
2
I DQ = K nD (VGS − VTND ) = ( 0.5 ) ( 0 − ( −1) )
2
I DQ = 0.5 mA
2
I DQ = K nL (VGSL − VTNL ) = K nL (VDD − VO − VTNL )
0.5 = 0.030 (10 − V0 − 1)
2
0.5
= 9 − V0 ⇒ V0 = 4.92 V
0.030
b.
2
I DD = I DL
2
K nD (Vi − VTND ) = K nL (VDD − Vo − VTNL )
2
K nD
(Vi − VTND ) = VDD − Vo − VTNL
K nL
K nD
(Vi − VTND )
K nL
Vo = VDD − VTNL −
Av =
(W / L )D
(W / L )L
dVo
K nD
=−
=−
dVi
K nL
Av = −
500
⇒ Av = −4.08
30
4.51
(a)
2
I DQ = K L (VGSL − VTNL ) = K L (VDSL − VTNL )
2
2
I D = ( 0.1)( 4 − 1) = 0.9 mA
I DQ = K D (VGSD − VTND )
2
0.9 = (1)(VGSD − 1) ⇒ VGSD = 1.95 V
VGG = VGSD + VDSL = 1.95 + 4 ⇒ VGG = 5.95 V
2
b.
I DD = I DL
2
K D (VGSD − VTND ) = K L (VGSL − VTNL )
2
KD
(VGG + Vi − Vo − VTND ) = Vo − VTNL
KL
⎛
KD ⎞
Vo ⎜ 1 +
⎟=
⎜
K L ⎟⎠
⎝
Av =
(c)
RLD
KD
(VGG + Vi − VTND ) + VTNL
KL
KD / KL
dVo
=
⇒
dVi 1 + K D / K L
Av =
1
1+ KL / KD
From Problem 4.49.
1
=
2 K L (VDSL − VTNL )
=
1
= 1.67 k Ω
2 ( 0.1)( 4 − 1)
"
g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V
Av =
g m ( RLD || RL )
1 + g m ( RLD || RL )
4.52
a.
=
(1.90 )(1.67 || 4 )
1 + (1.90 )(1.67 || 4 )
From Problem 4.51.
g m ( RLD || RL )
(1.90 )(1.67 ||10 )
Av =
=
1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 ||10 )
Av = 0.731
b.
⇒ Av = 0.691
R0 =
1
1
RLD =
1.67 = 0.526 ||1.67
gm
1.90
R0 = 0.40 kΩ
4.53
⎛ 85 ⎞
K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2
⎝ 2⎠
g m1 = 2 K n1 I D1 = 2
ro1 =
ro 2 =
1
λ1 I D1
=
( 2.125 )( 0.1) = 0.9220
1
= 200 K
0.05
( )( 0.1)
1
1
=
= 133.3 K
λ2 I D 2 ( 0.075 )( 0.1)
Av = − g m1 ( ro1 || ro 2 ) = − ( 0.922 )( 200 ||133.3)
Av = −73.7
4.54
K p1 =
k ′p ⎛ w ⎞ ⎛ 40 ⎞
2
⎜ ⎟ = ⎜ ⎟ ( 50 ) ⇒ 1.0 mA/V
2 ⎝L⎠ ⎝ 2 ⎠
g m1 = 2 K p1 I D1 = 2 (1)( 0.1) = 0.6325 mA/V
ro1 =
ro 2 =
1
λ1 I D1
=
1
= 133.3 K
0.075
(
)( 0.1)
1
1
=
= 200 K
λ2 I D 2 ( 0.05 )( 0.1)
Av = − g m1 ( ro1 ro 2 ) = − ( 0.6325 )(133.3 || 200 )
Av = −50.6
4.55
(a)
⎛ 85 ⎞
K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2
⎝ 2⎠
g m1 = 2 K n1 I D1 = 2
ro1 =
ro 2 =
1
λ1 I D1
=
( 2.125 )( 0.1) = 0.922 mA/V
1
( 0.05 )( 0.1)
= 200 K
1
1
=
= 133.3 K
λ2 I D 2 ( 0.075 )( 0.1)
(b)
Ri1 =
1
1
=
= 1.085 K
g m1 0.922
⎛
⎞
Ri1
1.085
⎛
⎞
Vgs1 = − ⎜
⎟ Vi = − ⎜
⎟ Vi = −0.956Vi
+
+
0.050
1.085
0.050
R
⎝
⎠
i
1
⎝
⎠
Vgs1
= + ( 0.956 )( 0.922 )( 200 )(133.3)
Av = − g m1 ( ro1 ro 2 ) ⋅
Vi
Av = 70.5
(c)
(d)
Ri = 0.05 +
1
1
= 0.05 +
⇒ Ri = 1.135 K
g m1
0.922
Ro ≈ ro1 ro 2 = 200 133.7 ⇒ Ro ≈ 80 K
4.56
(a)
g m1 = 2 K n I D1 = 2
( 2 )( 0.1) = 0.8944 mA/V
gm2 = 2 K p I D 2 = 2
( 2 )( 0.1) = 0.8944 mA/V
ro1 = ro 2 =
1
=
1
= 100 K
λ I D ( 0.1)( 0.1)
(b) The small-signal equivalent circuit
gm2Vsg2
Vo
⫹
⫹
rO2
gm1Vi
Vi
Vsg2
rO1
⫺
⫺
(1) g m1Vi +
Vsg 2
ro1
+ g m 2Vsg 2 +
Vsg 2 − Vo
ro 2
=0
(2)
Vo Vo − Vsg 2
+
= g m 2Vsg 2
ro
ro 2
⎛1 1 ⎞
⎛ 1
⎞
+ gm2 ⎟
Vo ⎜ + ⎟ = Vsg 2 ⎜
⎝ ro ro 2 ⎠
⎝ ro 2
⎠
1
1
1
⎛
⎞
⎛
⎞
+ 0.8944 ⎟ ⇒ Vsg 2 = Vo ( 0.03317 )
Vo ⎜ +
⎟ = Vsg 2 ⎜
⎝ 50 100 ⎠
⎝ 100
⎠
(1)
⎛ 1
1 ⎞ V
g m1Vi + Vsg 2 ⎜ + g m 2 + ⎟ = o
r
r
ro 2
o2 ⎠
⎝ o1
1 ⎞ Vo
⎛ 1
0.8944 Vi + Vo ( 0.03317 ) ⎜
+ 0.8944 +
⎟=
100 ⎠ 100
⎝ 100
0.8944 Vi = Vo ( 0.01 − 0.03033)
Vo
= −44
Vi
(c) For output resistance, set Vi = 0.
gm2Vsg2
RO
Ix
rO2
⫹
Vsg2
⫺
rO1
rO
⫹
⫺
Vx
(1)
(2)
g m 2Vsg 2 + I x =
Vsg 2
ro1
Vx Vx − Vsg 2
+
ro
ro 2
+ g m 2Vsg 2 +
Vsg 2 − Vx
ro 2
=0
(2)
⎛ 1
1 ⎞ V
Vsg 2 ⎜ + g m 2 + ⎟ = x
ro 2 ⎠ ro 2
⎝ ro1
1 ⎞ Vx
⎛ 1
+ 0.8944 +
Vsg 2 ⎜
⎟=
100
100
⎝
⎠ 100
Vsg 2 = Vx ( 0.010936 )
(1)
⎛1 1 ⎞
⎛ 1
⎞
I x = Vx ⎜ + ⎟ − Vsg 2 ⎜
+ gm2 ⎟
⎝ ro ro 2 ⎠
⎝ ro 2
⎠
1
1
⎛
⎞
⎛ 1
⎞
+ 0.8944 ⎟
I x = Vx ⎜ +
⎟ − Vx ( 0.010936 ) ⎜
⎝ 50 100 ⎠
⎝ 100
⎠
I x = Vx ( 0.03 − 0.0098905 )
V
Ro = x = 49.7 K
Ix
4.57
a.
2
I D1 = K1 (VGS 1 − VTN 1 )
0.4 = 0.1(VGS 1 − 2 ) ⇒ VGS1 = 4 V
VS 1 = I D1 RS 1 = ( 0.4 )( 4 ) = 1.6 V
VG1 = VS 1 + VGS 1 = 1.6 + 4 = 5.6 V
2
⎛ R2 ⎞
1
VG1 = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R1
⎝ R1 + R2 ⎠
1
5.6 = ( 200 )(10 ) ⇒ R1 = 357 kΩ
R1
357 R2
= 200 ⇒ R2 = 455 kΩ
357 + R2
VDS1 = VDD − I D1 ( RS1 + RD1 )
4 = 10 − ( 0.4 )( 4 + RD1 ) ⇒ RD1 = 11 kΩ
VD1 = 10 − ( 0.4 )(11) = 5.6 V
I D 2 = K 2 (VSG 2 + VTP 2 )
2
2 = 1(VSG 2 − 2 ) ⇒ VSG 2 = 3.41 V
VS 2 = VD1 + VSG 2 = 5.6 + 3.41 = 9.01
10 − 9.01
RS 2 =
⇒ RS 2 = 0.495 kΩ
2
VSD 2 = VDD − I D 2 ( RS 2 + RD 2 )
5 = 10 − ( 2 )( 0.495 + RD 2 ) ⇒ RD 2 = 2.01 kΩ
2
b.
Vi
⫹
⫺
V0
⫺
⫹
Vgs1
RD1
R1兩兩R2
gm1Vgs1
Vgs2
RD2
gm2Vsg2
⫹
⫺
V0 = g m 2Vsg 2 RD 2 = ( g m 2 RD 2 ) ( g m1Vgs1 RD1 )
Vgs1 = Vi
V
Av = 0 = g m1 g m 2 RD1 RD 2
Vi
g m1 = 2 K1 I D1 = 2
( 0.1)( 0.4 ) = 0.4 mA / V
g m 2 = 2 K 2 I D 2 = 2 (1)( 2 ) = 2.828 mA / V
Av = ( 0.4 )( 2.828 )(11)( 2.01) ⇒ Av = 25.0
c.
4
2
VSD(sat)
5
10
VSD ( sat ) = VSG + VTh
= VDD − I D 2 ( RD 2 + RS 2 )
2
= VDD − K p 2 ( RD 2 + RS 2 )VSD
( sat )
So
(1)( 2.01 + 0.495 )VSD2 ( sat ) + VSD ( sat ) − VDD = 0
2
2.50VSD
( sat ) + VSD ( sat ) − 10 = 0
VSD ( sat ) =
−1 ± 1 + 4 ( 2.501)(10 )
2 ( 2.501)
VSD ( sat ) = 1.81 V
5 − 1.81 = 3.19 ⇒ Max. output swing = 6.38 V peak-to-peak
4.58
a.
10
⫽ 4 mA
2.5
VSD2(sat)
10
2
VSD 2 ( sat ) = VDD − I D 2 ( RD 2 + RS 2 ) = VDD − K p 2 ( RD 2 + RS 2 ) VSD
2 ( sat )
2
(1)( 2 + 0.5)VSD 2 ( sat ) + VSD 2 ( sat ) − 10 = 0
2
2.5VSD
2 ( sat ) + VSD 2 ( sat ) − 10 = 0
VSD 2 ( sat ) =
−1 ± 1 + 4 ( 2.5 )(10 )
2 ( 2.5 )
= 1.81 V
10 − 1.81
+ 1.81 ⇒ VSDQ 2 = 5.91 V
2
VSDQ 2 = VDD − I DQ 2 ( RD 2 + RS 2 )
5.91 = 10 − I DQ 2 ( 2 + 0.5 ) ⇒ I DQ 2 = 1.64 mA
VS 2 = 10 − (1.64 )( 0.5 ) = 9.18 V
VSDQ 2 =
I DQ 2 = K p 2 (VSG 2 + VTP 2 )
2
1.64 = (1)(VSG 2 − 2 ) ⇒ VSG 2 = 3.28 V
VD1 = VS 2 − VSG 2 = 9.18 − 3.28 = 5.90 V
10 − 5.90
RD1 =
⇒ RD1 = 10.25 kΩ
0.4
2
I DQ1 = K n1 (VGS1 − VTN 1 )
2
0.4 = ( 0.1)(VGS 1 − 2 ) ⇒ VGS1 = 4 V
VS 1 = ( 0.4 )(1) = 0.4 V
VG1 = VS 1 + VGS 1 = 0.4 + 4 = 4.4 V
2
⎛ R2 ⎞
1
VG1 = ⎜
⎟ VDD = ⋅ Rm ⋅ VDD
R
R
R
+
2 ⎠
1
⎝ 1
1
4.4 = ⋅ ( 200 )(10 ) ⇒ R1 = 455 kΩ
R1
455 R2
= 200 ⇒ R2 = 357 kΩ
455 + R2
b.
I DQ 2 = 1.64 mA
VSDQ 2 = 10 − (1.64 )( 2 + 0.5 ) ⇒ VSDQ 2 = 5.90 V
VDSQ1 = 10 − ( 0.4 )(10.25 + 1) ⇒ VDSQ1 = 5.50 V
(c)
g m1 = 2 K n1 I DQ1 = 2
( 0.1)( 0.4 ) = 0.4 mA / V
g m 2 = 2 K p 2 I DQ 2 = 2 (1)(1.64 ) = 2.56 mA / V
Av = g m1 g m 2 RD1 RD 2 = ( 0.4 )( 2.56 )(10.25 )( 2 ) ⇒ Av = 21.0
4.59
a.
VDSQ 2 = 7 = VDD − I DQ 2 RS 2 = 10 − I DQ 2 ( 6 )
I DQ 2 = 0.5 mA
I DQ 2 = K 2 (VGS 2 − VTN 2 )
2
0.5 = ( 0.2 )(VGS 2 − 0.8 ) ⇒ VGS 2 = 2.38 V
VS 1 = VS 2 + VGS 2 = 3 + 2.38 = 5.38
2
I DQ1 =
VS 1 5.38
=
= 0.269 mA
RS1
20
I DQ1 = K1 (VGS 1 − VTN 1 )
2
0.269 = ( 0.2 )(VGS 1 − 0.8 ) ⇒ VGS 1 = 1.96 V
VG1 = VS 1 + VGS 1 = 5.38 + 1.96 = 7.34 V
2
⎛ R2 ⎞
1
VG1 = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R
R
R
+
⎝ 1
2 ⎠
1
1
7.34 = ( 400 )(10 ) ⇒ R1 = 545 kΩ
R1
545 R2
= 400 ⇒ R2 = 1.50 MΩ
545 + R2
b.
I DQ1 = 0.269 mA, I DQ 2 = 0.5 mA
VDSQ1 = 10 − ( 0.269 )( 20 ) ⇒ VDSQ1 = 4.62 V
c.
Av =
g m1 RS 1
g R
⋅ m2 S 2
1 + g m1 RS1 1 + g m 2 RS 2
g m1 = 2 K1 I DQ1 = 2
( 0.2 )( 0.269 ) = 0.464 mA / V
( 0.2 )( 0.5) = 0.6325 mA / V
( 0.464 )( 20 ) ( 0.6325)( 6 )
⋅
Av =
1 + ( 0.464 )( 20 ) 1 + ( 0.6325 )( 6 )
= ( 0.9027 )( 0.7915 )
g m 2 = 2 K 2 I DQ 2 = 2
= Av = 0.714
gm1Vgs1
⫹
Vgs1
⫹ Vgs2 ⫺
⫺
Ix
RS2
RS1
R0 =
gm2Vgs2
⫹
⫺
Vx
1
1
RS 2 =
6 = 1.581 6 ⇒ Ro = 1.25 kΩ
gm2
0.6325
4.60
(a)
I DQ1 =
10 − VGS 1
2
= K n1 (VGS 1 − VTN 1 )
RS 2
10 − VGS 1 = ( 4 )(10 ) (VGS2 1 − 4VGS 1 + 4 )
40VGS2 1 − 159VGS 1 + 150 = 0
VGS1 =
159 ±
(159 ) − 4 ( 40 )(150 )
⇒ VGS 1 = 2.435 V
2 ( 40 )
2
I DQ1 = ( 4 )( 2.435 − 2 ) ⇒ I DQ1 = 0.757 mA
2
VDSQ1 = 20 − ( 0.757 )(10 ) ⇒ VDSQ1 = 12.4 V
Also I DQ 2 = 0.757 mA
VDSQ 2 = 20 − ( 0.757 )(10 + 5 ) ⇒ VDSQ 2 = 8.65 V
(b)
g m1 = g m 2 = 2 KI DQ = 2
( 4 )( 0.757 ) ⇒ g m1 = g m 2 = 3.48 mA / V
c.
⫹
Vgs1
Vi
⫹
⫺
gm1Vgs1
gm2Vgs2
⫺
RG
V0
⫺
RS1
Vgs2
RS2
RD
⫹
V0 = − ( g m 2Vgs 2 ) ( RD RL )
Vgs 2 = ( − g m1Vgs1 − g m 2Vgs 2 ) ( RS1 RS 2 )
Vi = Vgs1 − Vgs 2 ⇒ Vgs1 = Vi + Vgs 2
Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) = − g m1 (Vi + Vgs 2 ) ( RS 1 RS 2 )
Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) + g m1Vgs 2 ( RS 1 Rs 2 ) = − g m1Vi ( RS1 RS 2 )
Vgs 2 =
Av =
− g m1Vi ( RS 1 RS 2 )
1 + g m 2 ( RS 1 RS 2 ) + g m1 ( RS 1 RS 2 )
V0 g m1 g m 2 ( RS 1 RS 2 )( RD RL )
=
Vi
1 + ( g m1 + g m 2 ) ( RS1 RS 2 )
( 3.48 ) (10 10 )( 5 2 )
Av =
⇒ Av = 2.42
1 + ( 3.48 + 3.48 ) (10 10 )
2
4.61
a.
I DQ = 3 mA
VS 1 = I DQ RS − 5 = ( 3)(1.2 ) − 5 = −1.4 V
I DQ = K1 (VGS − VTN )
2
3 = 2 (VGS − 1) ⇒ VGS = 2.225 V
2
VG1 = VGS + VS 1 = 2.225 − 1.4 = 0.825 V
⎛
⎞
R3
⎛ R3 ⎞
VG1 = ⎜
⎟ ( 5 ) ⇒ 0.825 = ⎜
⎟ ( 5 ) ⇒ R3 = 82.5 kΩ
⎝ 500 ⎠
⎝ R1 + R2 + R3 ⎠
VD1 = VS1 + VDSQ1 = −1.4 + 2.5 = 1.1 V
VG 2 = VD1 + VGS = 1.1 + 2.225 = 3.325 V
⎛ R2 + R3 ⎞
⎛ R2 + R3 ⎞
VG 2 = ⎜
⎟ ( 5 ) ⇒ 3.325 = ⎜
⎟ ( 5)
⎝ 500 ⎠
⎝ R1 + R2 + R3 ⎠
R2 + R3 = 332.5 ⇒ R2 = 250 kΩ
RL
R1 = 500 − 250 − 82.5 ⇒ R1 = 167.5 kΩ
VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V
RD =
5 − 3.6
⇒ RD = 0.467 kΩ
3
b.
Av = − g m1 RD
( 2 )( 3) = 4.90 mA / V
g m1 = 2 K n I DQ = 2
Av = − ( 4.90 )( 0.467 ) ⇒ Av = −2.29
4.62
a.
VS 1 = I DQ RS − 10 = ( 5 )( 2 ) − 10 ⇒ VS 1 = 0
I DQ = K1 (VGS 1 − VTN )
2
5 = 4 (VGS1 − 1.5 ) ⇒ VGS 1 = 2.618 V
VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 ⇒ R3 = 26.2 kΩ
2
VD1 = VS1 + VDSQ1 = 0 + 3.5 = 3.5 V
VG 2 = VD1 + VGS = 3.5 + 2.62 = 6.12 V
= ( 0.1)( R2 + R3 )
R2 + R3 = 61.2 kΩ ⇒ R2 = 35 kΩ
VD 2 = VD1 + VDSQ 2 = 3.5 + 3.5 = 7.0 V
10 − 7
RD =
⇒ RD = 0.6 kΩ
5
10 − 6.12
R1 =
⇒ R1 = 38.8 kΩ
0.1
b.
Av = − g m1 RD
( 4 )( 5 ) = 8.944 mA / V
g m1 = 2 K n I DQ = 2
Av = − ( 8.944 )( 0.6 ) ⇒ Av = −5.37
4.63
a.
⎛ V ⎞
I DQ = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
⎛
V ⎞
4 = 6 ⎜⎜ 1 − GS ⎟⎟
⎝ ( −3) ⎠
2
2
⎡
4⎤
VGS = ( −3) ⎢1 −
⎥ ⇒ VGS = −0.551 V
6⎦
⎣
VDSQ = VDD − I DQ RD
6 = 10 − ( 4 ) RD ⇒ RD = 1 kΩ
b.
2 I DSS ⎛ VGS ⎞ 2 ( 6 ) ⎛ −0.551 ⎞
=
1−
1−
⎟ ⇒ g m = 3.265 mA/V
−3 ⎠
( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝
1
1
=
⇒ r0 = 25 kΩ
r0 =
λ I DQ ( 0.01)( 4 )
gm =
c.
Av = − g m ( r0 & RD ) = − ( 3.265 )( 25 & 1) ⇒ Av = −3.14
4.64
VGS + I DQ ( RS1 + RS 2 ) = 0
⎛ V ⎞
I DQ = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛ V ⎞
+ I DSS ( RS1 + RS 2 ) ⎜ 1 − GS ⎟ = 0
VP ⎠
⎝
2
VGS
⎛ V ⎞
VGS + ( 2 )( 0.1 + 0.25 ) ⎜1 − GS ⎟ = 0
VP ⎠
⎝
2
⎛ 2
V2 ⎞
VGS + 0.7 ⎜ 1 − VGS + GS 2 ⎟ = 0
⎜
⎟
⎝ ( −2 ) ( −2 ) ⎠
0.175VGS2 + 1.7VGS + 0.7 = 0
VGS =
−1.7 ±
(1.7 ) − 4 ( 0.175)( 0.7 )
⇒ VGS
2 ( 0.175 )
VGS ⎞ 2 ( 2 ) ⎛ −0.431 ⎞
2
= −0.4314 V
2 I DSS ⎛
⎜1 −
⎟=
⎜1 −
⎟ ⇒ g m = 1.569 mA/V
−VP ⎝
−2 ⎠
VP ⎠
2 ⎝
− g m ( RD RL ) − (1.569 ) ( 8 4 )
=
⇒ Av = −3.62
Av =
1 + g m RS1
1 + (1.569 )( 0.1)
gm =
Ai =
i0 ( v0 / RL ) v0 RG
⎛ 50 ⎞
=
= ⋅
= ( −3.62 ) ⎜ ⎟ ⇒ Ai = −45.2
ii ( vi / RG ) vi RL
⎝ 4⎠
4.65
I DSS
= 4 mA
2
V
= DD = 10 V
2
= VDD − I DQ ( RS + RD )
I DQ =
VDSQ
VDSQ
10 = 20 − ( 4 )( RS + RD ) ⇒ RS + RD = 2.5 kΩ
VS = 2 V = I DQ RS = 4 RS ⇒ RS = 0.5 kΩ, RD = 2.0 kΩ
I DQ
⎛ V ⎞
= I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛
⎞
⎛
V
4⎞
4 = 8 ⎜⎜1 − GS ⎟⎟ ⇒ VGS = ( −4.2 ) ⎜⎜1 −
⎟ ⇒ VGS = −1.23 V
8 ⎠⎟
⎝
⎝ ( −4.2 ) ⎠
VG = VS + VGS = 2 − 1.23
2
⎛ R2 ⎞
⎛ R2 ⎞
VG = 0.77 V = ⎜
⎟ ( 20 ) = ⎜
⎟ ( 20 ) ⇒ R2 = 3.85 kΩ, R1 = 96.2 KΩ
⎝ 100 ⎠
⎝ R1 + R2 ⎠
4.66
a.
I DSS
= 5 mA
2
V
12
VDSQ = DD =
=6V
2
2
12 − 6
RS =
⇒ RS = 1.2 kΩ
5
I DQ =
I DQ
⎛ V ⎞
= I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
⎛
⎛
V ⎞
5 ⎞
5 = 10 ⎜⎜ 1 − GS ⎟⎟ ⇒ VGS = ( −5 ) ⎜⎜ 1 −
⎟ ⇒ VGS = −1.464 V
10 ⎠⎟
⎝
⎝ ( −5 ) ⎠
VG = VS + VGS = 6 − 1.464 = 4.536 V
2
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R1
⎝ R1 + R2 ⎠
1
4.536 = (100 )(12 ) ⇒ R1 = 265 kΩ
R1
265 R2
= 100 ⇒ R2 = 161 kΩ
265 + R2
b.
⎛ V ⎞ 2 (10 ) ⎛ −1.46 ⎞
2I
g m = DSS ⎜ 1 − GS ⎟ =
⎜1 −
⎟ ⇒ g m = 2.83 mA/V
−
−5 ⎠
V
5 ⎝
( P ) ⎝ VP ⎠
r0 =
1
1
=
= 20 kΩ
λ I DQ ( 0.01)( 5 )
Av =
Av =
(
g m r0 Rs RL
)
(
1 + g m r0 RS RL
)
( 2.83) ( 20 1.2 0.5 )
⇒ Av = 0.495
1 + ( 2.83) ( 20 1.2 0.5 )
R0 =
1
1
RS =
1.2 = 0.353 1.2 ⇒ R0 = 0.273 kΩ
gm
2.83
4.67
a.
⎛ R2 ⎞
⎛ 110 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) = 5.5 V
⎝ 110 + 90 ⎠
⎝ R1 + R2 ⎠
I DQ =
10 − (VG − VGS )
RS
⎛ V ⎞
= I DSS ⎜1 − GS ⎟
VP ⎠
⎝
⎛ V ⎞
10 − 5.5 + VGS = ( 2 )( 5 ) ⎜1 − GS ⎟
⎝ 1.75 ⎠
2
2
4.5 + VGS = 10 (1 − 1.143VGS + 0.3265VGS2 )
3.265VGs2 − 12.43VGS + 5.5 = 0
VGS =
12.43 ±
(12.43) − 4 ( 3.265 )( 5.5)
⇒ VGS
2 ( 3.265 )
2
⎛ 0.511 ⎞
I DQ = ( 2 ) ⎜1 −
⎟ ⇒ I DQ = 1.00 mA
1.75 ⎠
⎝
VSDQ = 10 − (1.00 )( 5 ) ⇒ VSDQ = 5.0 V
2
= 0.511 V
b.
2 I DSS ⎛ VGS ⎞ 2 ( 2 ) ⎛ 0.511 ⎞
⎜1 −
⎟=
⎜1 −
⎟ ⇒ g m = 1.618 mA/V
VP ⎝ VP ⎠ 1.75 ⎝
1.75 ⎠
g m ( RS RL )
(1.618 ) ( 5 10 )
Av =
=
⇒ Av = 0.844
1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 )
gm =
⎛R ⎞
i0 ( v0 / RL )
=
= Av ⋅ ⎜ i ⎟
ii
( vi / Ri )
⎝ RL ⎠
Ri = R1 R2 = 90 110 = 49.5 kΩ
Ai =
⎛ 49.5 ⎞
Ai = ( 0.844 ) ⎜
⎟ ⇒ Ai = 4.18
⎝ 10 ⎠
c.
AC load line
⫺1
Slope ⫽
5兩兩10
⫺1
⫽
3.33 K
1.0
5.0
10
Δid = 1.0 mA
vsd = ( 3.33)(1.0 ) = 3.33 V
Maximum swing in output voltage = 6.66 V peak-to-peak
4.68
⎛ V ⎞
I DQ = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
2
⎛
4⎞
⎛ V ⎞
4 = 8 ⎜1 − GS ⎟ ⇒ VGS = 4 ⎜⎜ 1 −
⎟ ⇒ VGS = 1.17 V
4 ⎠
8 ⎟⎠
⎝
⎝
VSDQ = VDD − I DQ ( RS + RD )
7.5 = 20 − 4 ( RS + RD ) ⇒ RS + RD = 3.125 kΩ
⎛ VGS
⎜1 −
⎝ VP
RS = 3.125 − RD
− g m RD
Av =
1 + g m RS
gm =
2 I DSS
VP
⎞ 2 ( 8 ) ⎛ 1.17 ⎞
⎟=
⎜1 −
⎟ ⇒ g m = 2.83 mA/V
4 ⎝
4 ⎠
⎠
−3 (1 + g m RS ) = − g m RD
3 ⎡⎣1 + ( 2.83)( 3.125 − RD ) ⎤⎦ = ( 2.83) RD
9.844 − 2.83RD = 0.9433RD ⇒ RD = 2.61 kΩ RS = 0.516 kΩ
VS = 20 − ( 4 )( 0.516 ) ⇒ VS = 17.94 V
VG = VS − VGS = 17.94 − 1.17 = 16.77 V
⎛ R2 ⎞
⎛ R2 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ ( 20 ) ⇒ R2 = 335 kΩ. R1 = 65 kΩ
R
R
+
⎝ 400 ⎠
2 ⎠
⎝ 1
Chapter 5
Exercise Problems
EX5.1
β=
α
1−α
0.980
= 49
1 − 0.980
0.995
= 199
For α = 0.995, β =
1 − 0.995
So 49 ≤ β ≤ 199
For α = 0.980, β =
EX5.2
BVCEO =
BVCBO
n
=
β
200
3
120
or BVCEO = 40.5 V
EX5.3
V − VBE ( on ) 2 − 0.7
=
⇒ 6.5 µ A
I B = BB
RB
200
I C = β I = (120 )( 6.5 µ A ) ⇒ I C = 0.78 mA
VCE = VCC − I C RC = 5 − ( 0.78 )( 4 ) or VCE = 1.88 V
P = I BVBE ( on ) + I CVCE = ( 0.0065 )( 0.7 ) + ( 0.78 )(1.88 ) ≅ 1.47 mW
EX5.4
V + − VEB ( on ) − VBB 5 − 0.7 − 2.8
IB =
=
or I B = 4.62 µ A
RB
325
I C = β I B = ( 80 )( 4.615 ) ⇒ I C = 0.369 mA
VEC = 2 = 5 − ( 0.369 ) RC which yields RC = 8.13 k Ω
EX5.5
(a)
IB =
VBB − VBE ( on )
RB
=
2 − 0.7
⇒ 5.91 µ A
220
I C = β I B = (100 )( 5.91 µ A ) ⇒ 0.591 mA
I E = (1 + β ) I B = 0.597 mA
VCE = VCC − I C RC = 10 − ( 0.591)( 4 ) or VCE = 7.64 V
6.5 − 0.7
⇒ 26.4 µ A
220
Transistor is biased in saturation mode, so
(b)
IB =
VCE = VCE ( sat ) = 0.2 V
IC =
VCC − VCE ( sat )
RC
=
10 − 0.2
or I C = 2.45 mA
4
I E = I C + I B = 2.45 + 0.0264 ≅ 2.48 mA
EX5.6
For 0 ≤ VI < 0.7 V , Qn is cutoff, VO = 9 V
When Qn is biased in saturation, we have 0.2 = 9 −
So, for VI ≥ 5.1 V , VO = 0.2 V
EX5.7
(100 )(VI − 0.7 )( 4 )
200
⇒ VI = 5.1 V
IB =
VBB − VBE ( on )
RB + (1 + β ) RE
=
8 − 0.7
30 + ( 76 )(1.2 )
or I B = 60.2 µ A
I C = β I B = ( 75 )( 60.23 µ A ) ⇒ 4.52 mA
I E = (1 + β ) I B = 4.58 mA
VCE = VCC − I C RC − I E RE
= 12 − ( 4.517 )( 0.4 ) − ( 4.577 )(1.2 )
or VCE = 4.70 V
EX5.8
For VC = 4 V and ICQ = 1.5 mA,
V + − VC 10 − 4
RC =
=
⇒ RC = 4 k Ω
1.5
I CQ
⎛ 1+ β ⎞
⎛ 101 ⎞
IE = ⎜
⎟ IC = ⎜
⎟ (1.5 ) = 1.515 mA
β
⎝ 100 ⎠
⎝
⎠
−V ( on ) − V −
also I E = BE
RE
Then RE =
−0.7 − ( −10 )
1.515
⇒ RE = 6.14 k Ω
EX5.9
3 − 0.7
= 0.25
I EQ =
RE
RE = 9.2 kΩ
⎛ 75 ⎞
I CQ = ⎜ ⎟ ( 0.25 ) = 0.2467 mA
⎝ 76 ⎠
−0.7 + VCEQ + I CQ RC − 3 = 0
RC =
3 + 0.7 − 2
⇒ RC = 6.89 kΩ
0.2467
EX5.10
5 = I E RE + VEB ( on ) + I B RB − 2
(a)
(b)
(c)
(d)
EX5.11
180 ⎞
⎛
5 + 2 − 0.7 = I E ⎜ 2 +
⎟ I E = 0.9859 mA
41 ⎠
⎝
I C = 0.962 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.2725 mA
61 ⎠
⎝
I C = 1.25 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.6657 mA
101 ⎠
⎝
I C = 1.64 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.97365 mA
151 ⎠
⎝
I C = 1.94 mA
IE =
VBB − VEB ( on )
RE
⇒ RE =
4 − 0.7
1.0
or RE = 3.3 k Ω
I C = α I E = ( 0.992 )(1) = 0.992 mA
I B = I E − I C = 1.0 − 0.992 or I B = 8 µ A
VCB = I C RC − VCC = ( 0.992 )(1) − 5
or VCB = −4.01 V
EX5.12
V + − (Vγ + VCE ( sat ) ) 5 − (1.5 + 0.2 )
=
R=
IC
15
or R = 220 Ω
15
= 1 mA
IB =
15
v − VBE ( on ) 5 − 0.7
=
= 4.3 k Ω
RB = I
IB
1
P = I BVBE (on) + I CVCE
= (1)( 0.7 ) + (15 )( 0.2 ) = 3.7 mW
EX5.13
(a)
For V1 = V2 = 0, All currents are zero and VO = 5 V.
(b)
For V1 = 5 V, V2 = 0; IB2 = IC2 = 0,
5 − 0.7
I B1 =
= 4.53 mA
0.95
5 − 0.2
I C1 =
⇒ I C1 = I R = 8 mA
0.6
VO = 0.2 V
(c)
For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V
EX5.14
vO = 5 − iC RC = 5 − β iB RC
ΔvO = − βΔiB RC
iB =
VBB + ΔvI − VBE ( on )
RB
ΔiB =
ΔvI
RB
Then
ΔvO − β RC
=
RB
ΔvI
Let β = 100, RC = 5 k Ω, RB = 100 k Ω
So
Δvo − (100 )( 5 )
=
= −5
100
ΔvI
Want Q-point to be vo ( Q − pt ) = 2.5 = 5 − (100 ) I BQ ( 5 )
Then I BQ = 0.005 mA, I BQ = 0.005 =
VBB − 0.7
100
so VBB = 1.2 V
Also, I CQ = β I BQ = (100 )( 0.005 ) = 0.5 mA
EX5.15
VCEQ = 2.5 = 5 − I CQ RC
5 − 2.5
= 10 k Ω
0.25
I CQ 0.25
=
=
= 0.002083 mA
β
120
or RC =
I BQ
Then RB =
5 − 0.7
⇒ RB = 2.06 M Ω
0.002083
EX5.16
(a)
RTH = R1 & R2 = 9 & 2.25 = 1.8 k Ω
or VTH
(b)
⎛ R2 ⎞
⎛ 2.25 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ ( 5)
R
R
+
⎝ 9 + 2.25 ⎠
2 ⎠
⎝ 1
= 1.0 V
I BQ =
VTH − VBE ( on )
RTH + (1 + β ) RE
=
1 − 0.7
1.8 + (151)( 0.2 )
or I BQ = 9.375 µ A
I CQ = β I BQ = (150 )( 9.375 µ A )
or I CQ = 1.41 mA
I EQ = (1 + β ) I BQ ⇒ I EQ = 1.42 mA
VCEQ = 5 − I CQ RC − I EQ RE
= 5 − (1.41)(1) − (1.42 )( 0.2 )
or VCEQ = 3.31 V
(c)
For β = 75
I BQ =
1 − 0.7
= 17.6 µ A
1.8 + ( 76 )( 0.2 )
I CQ = β I BQ = ( 75 )(17.6 µ A )
or I CQ = 1.32 mA
I EQ = (1 + β ) I BQ = ( 76 )(17.6 µ A )
or I EQ = 1.34 mA
VCEQ = 5 − (1.32 )(1) − (1.34 )( 0.2 )
or VCEQ = 3.41 V
EX5.17
VCEQ ≅ VCC − I CQ ( RC + RE )
or 2.5 ≅ 5 − I CQ (1 + 0.2 )
which yields
I CQ = 2.08 mA,
I CQ 2.08
=
= 0.0139 mA
I BQ =
150
β
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(151)( 0.2 )
or RTH = 3.02 k Ω
⎛ R2 ⎞
1
Now VTH = ⎜
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
2 ⎠
1
⎝ 1
1
so VTH = ( 3.02 )( 5 )
R1
We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
or
1
( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 )
R1
We obtain R1 = 13 k Ω and then R2 = 3.93 k Ω
EX5.18
β = 150, RTH = R1 || R2
5−0
I CQ =
= 5 mA
1
I CQ
5
=
= 0.0333 mA
I BQ =
β 150
I BQ =
VTH − VBE ( on ) − ( −5 )
RTH + (1 + β ) RE
Set RTH = ( 0.1)(1 + β ) RE
⎛ R2 ⎞
We have VTH = ⎜
⎟ (10 ) − 5
⎝ R1 + R2 ⎠
Then I BQ
⎛ R2 ⎞
⎜
⎟ (10 ) − 0.7
R + R2 ⎠
= 0.0333 = ⎝ 1
(1.1)(151)( 0.2 )
⎛ R2 ⎞
which yields ⎜
⎟ = 0.1806
⎝ R1 + R2 ⎠
⎛ RR ⎞
Now RTH = ⎜ 1 2 ⎟ = ( 0.1)(151)( 0.2 ) = 3.02 k Ω
⎝ R1 + R2 ⎠
so R1 ( 0.1806 ) = 3.02 k Ω
We obtain R1 = 16.7 k Ω and R2 = 3.69 k Ω
EX5.19
V + − V − − VECQ 5 − ( −5 ) − 5
I CQ ≅
=
4.5 + 0.5
RC + RE
so I CQ = 1 mA, and
I BQ =
I CQ
β
=
1
= 0.00833 mA
120
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.5 )
or RTH = 6.05 k Ω
We can write V + = I EQ RE + VEB ( on ) + I BQ RTH + VTH
We have VTH =
1
⋅ RTH (10 ) − 5 and if we let I EQ ≅ I CQ = 1 mA,
R1
then we have 5 = (1)( 0.5 ) + 0.7 + ( 0.00833)( 6.05 ) +
which yields R1 = 6.91 k Ω and R2 = 48.6 k Ω
EX5.20
⎛
2⎞
2 ⎞
⎛
I1 = I Q ⎜1 + ⎟ = ( 0.25 ) ⎜1 + ⎟ = 0.2625 mA
⎝ 40 ⎠
⎝ β⎠
−
0 − VBE ( on ) − V
0 − 0.7 − ( −5 )
=
R1 =
0.2625
I1
or R1 = 16.38 k Ω
1
( 6.05 )(10 ) − 5
R1
For VCEO = 3 V , then VCO = 2.3 V
⎛ β ⎞
⎛ 40 ⎞
I CO = ⎜
⎟ ⋅ I Q = ⎜ ⎟ ( 0.25 ) = 0.2439 mA
⎝ 41 ⎠
⎝ 1+ β ⎠
+
V − VCO 5 − 2.3
RC =
=
= 11.07 k Ω
I CO
0.2439
EX5.21
RTH 50 ||100 = 33.3 k Ω
⎛ 50 ⎞
VTH = VTH = ⎜
⎟ (10 ) − 5 = −1.67 V
⎝ 50 + 100 ⎠
−1.67 − 0.7 − ( −5 )
⇒ 11.2 µ A
I B1 =
33.3 + (101)( 2 )
I C1 = 1.12 mA, I E1 = 1.13 mA
VE1 = I E1 RE1 − 5 = (1.13)( 2 ) − 5 = −2.74 V
VCE1 = 3.25 V ⇒ VC1 = 0.51 V
Now VE 2 = 0.51 + 0.7 = 1.21 V
5 − 1.21
= 1.90 mA ⇒ I B 2 = 18.8 µ A
2
= 1.88 mA
IE2 =
IC 2
I R1 = I C1 − I B 2 = 1.12 − 0.0188 = 1.10 mA
5 − 0.51
= 4.08 k Ω
1.10
= 2.5 ⇒ VC 2 = VE 2 − VEC 2
RC1 =
VEC 2
= 1.21 − 2.5 = −1.29 V
RC 2 =
−1.29 − ( −5 )
1.88
= 1.97 k Ω
EX5.22
12
= 240 k Ω = R1 + R2 + R3
0.05
Then VB1 = ( 0.5)( 2 ) + 0.7 = 1.7 V
We find
1.7
= 34 k Ω
0.05
= ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 V
R3 =
Also VB 2
ΔVR 2 = 5.7 − 1.7 = 4 V
4
= 80 k Ω
0.05
and R1 = 240 − 80 − 34 = 126 k Ω
so R2 =
VC 2 = 1 + 4 + 4 = 9 V
Then RC =
V + − VC 2 12 − 9
=
= 6 kΩ
I CQ
0.5
Test Your Understanding Exercises
TYU5.1
α=
β
1+ β
For β = 75, α =
75
= 0.9868
76
For β = 125, α =
125
= 0.9921
126
TYU5.2
I E = (1 + β ) I B
IE
0.780
=
= 81.25 then β = 80.3
I B 0.00960
so 1 + β =
Now
α=
β
=
1+ β
80.25
= 0.9877
81.25
I C = α I E = ( 0.9877 )( 0.78 ) = 0.770 mA
TYU5.3
β=
α
0.990
=
= 99
1 − α 1 − 0.990
Now I B =
IE
2.15
=
⇒ 21.5 µ A and I C = α I E = ( 0.990 )( 2.15 ) = 2.13 mA
1 + β 100
TYU5.4
V
150
ro = A =
IC
IC
For I C = 0.1 mA ⇒ ro = 1.5 M Ω
For I C = 1.0 mA ⇒ ro = 150 k Ω
For I C = 10 mA ⇒ ro = 15 k Ω
TYU5.5
⎛ V ⎞
I C = I O ⎜1 + CE ⎟
VA ⎠
⎝
At VCE = 1 V , I C = 1 mA
1 ⎞
⎛
For VA = 75 V , I C = 1 = I O ⎜1 + ⎟ ⇒ I O = 0.9868 mA
75
⎝
⎠
Then, at VCE = 10 V
(a)
⎛ 10 ⎞
I C = ( 0.9868 ) ⎜ 1 + ⎟ = 1.12 mA
⎝ 75 ⎠
(b)
At VCE
1 ⎞
⎛
For VA = 150 V , I C = 1 = I O ⎜ 1 +
⎟ ⇒ I O = 0.9934 mA
⎝ 150 ⎠
10 ⎞
⎛
= 10 V , I C = ( 0.9934 ) ⎜ 1 +
⎟ = 1.06 mA
⎝ 150 ⎠
TYU5.6
BVCEO =
BVCBO
n β
so BVCBO = 3 100 ( 30 ) = 139 V
TYU5.7
(a)
For VI = 0.2 V < VBE ( on ) ⇒ I B = I C = 0, VO = 5 V and P = 0
(b)
IB =
For VI = 3.6 V , transistor is driven into saturation, so
VI − VBE ( on )
RB
=
V + − VCE ( sat ) 5 − 0.2
3.6 − 0.7
= 4.53 mA and I C =
=
= 10.9 mA
RC
0.440
0.64
I C 10.9
=
= 2.41 < β which shows that the transistor is indeed driven into saturation. Now,
I B 4.53
Note that
P = I BVBE ( on ) + I CVCE ( sat )
= ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mW
TYU5.8
For VBC = 0 ⇒ VO = 0.7 V
Then I C =
I
5 − 0.7
9.77
= 9.77 mA and I B = C =
= 0.195 mA
0.44
β
50
Now VI = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7 or VI = 0.825 V
Also P = I BVBE ( on ) + I CVCE
= ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mW
TYU5.9
V + − VC 10 − 6.34
IC =
=
= 0.915 mA
RC
4
−VBE ( on ) − V −
And I E =
Now α =
RE
=
−0.7 − ( −10 )
10
or IE = 0.930
I C 0.915
α
0.9839
=
= 0.9839 and β =
=
= 61
I E 0.930
1 − α 1 − 0.9839
Also I B = I E − I C = 0.930 − 0.915 ⇒ 15 µ A and VCE = VC − VE = 6.34 − ( −0.7 ) = 7.04 V
TYU5.10
10 − 0.7
IE =
= 1.16 mA
8
1.1625
IB =
= 22.8 µ A
51
I C = ( 50 )( 22.8 µ A ) = 1.14 mA
VE = V + − I E RE = 10 − (1.1625 )( 8 ) = 0.7 V
VC = I C RC − 10 = (1.1397 )( 4 ) − 10 = −5.44
VEC = 0.7 − ( −5.44 )
VEC = 6.14 V
TYU5.11
VBB = I B RB + VBE ( on ) + I E RE
or VBB = I B RB + VBE ( on ) + (1 + β ) I B RE
Then I B =
VBB − VBE ( on )
RB + (1 + β ) RE
=
2 − 0.7
10 + ( 76 )(1)
or I B = 15.1 µ A
Also I C = ( 75 )(15.1 µ A ) = 1.13 mA and I E = ( 76 )(15.1 µ A ) = 1.15 mA
Now VCE = VCC + VBB − I C RC − I E RE
= 8 + 2 − (1.13)( 2.5 ) − (1.15 )(1) = 6.03 V
TYU5.12
VCE = 2.5 V ⇒ VE = 2.5 V = I E RE
We have VBB = I B RB + VBE ( on ) + VE
so I B =
VBB − VBE ( on ) − VE
RB
=
5 − 0.7 − 2.5
10
or I B = 0.18 mA
Then I E = (101)( 0.18 ) = 18.2 mA
And RE =
2.5
= 0.138 k Ω ⇒ 138 Ω
18.18
TYU5.13
VBB = I E RE + VEB ( on ) + I B RB
I E = 2.2 mA ⇒ I B =
2.2
= 0.0431 mA
51
⎛ β ⎞
⎛ 50 ⎞
and I C = ⎜
⎟ ⋅ I E = ⎜ ⎟ ( 2.2 ) = 2.16 mA
⎝ 51 ⎠
⎝ 1+ β ⎠
Then VBB = ( 2.2 )(1) + 0.7 + ( 0.0431)( 50 )
or VBB = 5.06 V
Now VEC = 5 − I E RE = 5 − ( 2.2 )(1) = 2.8 V
TYU5.14
(a)
For vI = 0, iB = iC = 0, vO = 12 V , P = 0
(b)
iC =
For vI = 12 V , iB =
VCC − VCE ( sat )
RC
=
vI − VBE ( on )
RB
=
12 − 0.7
= 47.1 mA
0.24
12 − 0.1
= 2.38 A
5
vO = 0.1 V
and
P = iBVBE ( on ) + iCVCE ( sat )
= ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 W
TYU5.15
(a)
For VCEQ = 2.5 V , I CQ =
I BQ =
I CQ
β
=
5 − 2.5
= 1.25 mA
2
1.25
⇒ I BQ = 12.5 µ A
100
5 − 0.7
= 344 k Ω
0.0125
IBQ is independent of β .
Then RB =
(b)
For VCEQ = 1 V , I C =
β=
5 −1
= 2 mA
2
IC
2
=
⇒ β = 160
I B 0.0125
For VCEQ = 4 V , I C =
5−4
= 0.5 mA
2
IC
0.5
=
⇒ β = 40
I B 0.0125
So 40 ≤ β ≤ 160
β=
TYU5.16
5 − 0.7
I BQ =
= 0.005375 mA
800
For β = 75, I CQ = β I BQ = ( 75 )( 0.005375 )
Or I CQ = 0.403 mA
For β = 150, I CQ = (150 )( 0.005375 )
Or I CQ = 0.806 mA
Largest I CQ ⇒ Smallest VCEQ
5 −1
= 4.96 k Ω
0.806
5−4
For β = 75, RC =
= 2.48 k Ω
0.403
For β = 150, RC =
5 − 2.5
= 4.14 k Ω
0.604
= 5 − ( 0.403)( 4.14 ) = 3.33 V
For a nominal I CQ = 0.604 mA and VCEQ = 2.5 V , RC =
Now for I CQ = 0.403 mA, VCEQ
For I CQ = 0.806 mA, VCEQ = 5 − ( 0.806 )( 4.14 ) = 1.66 V
So, for RC = 4.14 k Ω, 1.66 ≤ VCEQ ≤ 3.33 V
TYU5.17
(a)
⎛ β ⎞
⎛ 100 ⎞
I CQ = ⎜
⎟ ⋅ I EQ = ⎜
⎟ (1) = 0.99 mA
⎝ 101 ⎠
⎝1+ β ⎠
I EQ
1
=
⇒ 9.90 µ A
I BQ =
1 + β 101
VB = − I BQ RB = − ( 0.0099 )( 50 )
or VB = −0.495 V
⎛ I CQ ⎞
⎛ 0.99 × 10−3 ⎞
VBE = VT ln ⎜
⎟ = ( 0.026 ) ln ⎜
−14 ⎟
⎝ 3 × 10
⎠
⎝ IS ⎠
or VBE = 0.630 V
Then VE = VB − VBE = −0.495 − 0.630 = −1.13 V
VC = 10 − ( 0.99 )( 5 ) = 5.05 V
Then VCEQ = VC − VE = 5.05 − ( −1.13) = 6.18 V
(b)
1
= 0.0196 mA
1 + β 51
VB = − ( 0.0196 )( 50 ) = −0.98 V
I EQ = 1 mA, I BQ =
⎛ β
I CQ = ⎜
⎝1+ β
I EQ
=
⎞
⎛ 50 ⎞
⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.98 mA
⎝ 51 ⎠
⎠
⎛ 0.98 × 10−3 ⎞
VBE = ( 0.026 ) ln ⎜
= 0.629 V
−14 ⎟
⎝ 3 × 10
⎠
VE = −0.98 − 0.629 = −1.61 V
VC = 10 − ( 0.98 )( 5 ) = 5.1 V
VCEQ = VC − VE = 5.1 − ( −1.61) = 6.71 V
TYU5.18
IQ
IQ
IB =
=
1 + β 121
and
⎛ IQ ⎞
VB = ⎜
⎟ ( 20 ) = I Q ( 0.165 )
⎝ 121 ⎠
VE = I Q ( 0.165 ) + 0.7
⎛ β ⎞
⎛ 120 ⎞
I CQ = ⎜
⎟ ⋅ I EQ = ⎜
⎟ ⋅ I Q = ( 0.992 ) I Q
⎝ 121 ⎠
⎝ 1+ β ⎠
VC = I CQ RC − 5 = ( 0.992 ) I Q ( 4 ) − 5
= 3.97 I Q − 5
VECQ = VE − VC
= ⎣⎡ I Q ( 0.165 ) + 0.7 ⎦⎤ − ⎣⎡ I Q ( 3.97 ) − 5⎦⎤
= −3.805 I Q + 5.7
Then 3 = 5.7 − 3.805I Q
which yields I Q = 0.710 mA
Chapter 5
Problem Solutions
5.1
(a)
β=
iC 510
=
⇒ β = 85
6
iB
85
⇒ α = 0.9884
1 + β 86
iE = (1 + β ) iB = ( 86 )( 6 ) ⇒ iE = 516 µ A
α=
(b)
β
=
2.65
⇒ β = 53
0.050
53
α = ⇒ α = 0.9815
54
iE = (1 + β ) iB = ( 54 )( 0.050 ) ⇒ iE = 2.70 mA
β=
5.2
(a)
For β = 110: α =
β
1+ β
=
110
= 0.99099
111
180
= 0.99448
181
0.99099 ≤ α ≤ 0.99448
For β = 180: α =
(b)
I C = β I B = 110 ( 50 µ A ) ⇒ I C = 5.50 mA
or I C = 180 ( 50 µ A ) ⇒ I C = 9.00 mA
so 5.50 ≤ I C ≤ 9.0 mA
5.3
(a)
iE
α
(b)
1.12
⇒ 9.33 µ A
120
⎛ 121 ⎞
= (1.12 ) ⎜
⎟ = 1.13 mA
⎝ 120 ⎠
120
=
= 0.9917
121
50
=
= 2.5 mA
20
⎛ 21 ⎞
= ⎜ ⎟ ( 50 ) = 52.5 mA
⎝ 20 ⎠
20
=
= 0.9524
21
iB =
iB
iE
α
5.4
(a)
α
β=
0.9
0.95
0.98
0.99
0.995
0.999
9
19
49
99
199
999
(b)
α
1−α
α=
β
β
1+ β
0.9524
0.9804
0.9901
0.9934
0.9955
0.9975
20
50
100
150
220
400
5.5
1.2
⇒ 14.8 µ A
81
⎛ 80 ⎞
I C = (1.2 ) ⎜ ⎟ = 1.185 mA
⎝ 81 ⎠
80
α=
= 0.9877
81
VC = 5 − (1.185 )( 2 ) = 2.63 V
IB =
(a)
0.80
⇒ 9.88 µ A
81
⎛ 80 ⎞
I C = ( 0.80 ) ⎜ ⎟ = 0.790 mA
⎝ 81 ⎠
80
α=
= 0.9877
81
VC = 5 − ( 0.790 )( 2 ) = 3.42 V
IB =
(b)
Yes, VC > VB so B-C junction is reverse biased in both areas.
(c)
5.6
For VC = 0, I C =
IE =
IC
α
=
2.5
⇒ I E = 2.546 mA
0.982
5.7
(a)
α
VC
VC
(c)
0.75
⇒ 12.3 µ A
61
⎛ 60 ⎞
= ( 0.75 ) ⎜ ⎟ = 0.738 mA
⎝ 61 ⎠
60
=
= 0.9836
61
= I C RC − 10 = ( 0.738 )( 5 ) − 10
= −6.31 V
IB =
IC
(b)
5
= 2.5 mA
2
1.5
⇒ 24.6 µ A
61
⎛ 60 ⎞
I C = (1.5 ) ⎜ ⎟ = 1.475 mA
⎝ 61 ⎠
⎛ 60 ⎞
α = ⎜ ⎟ = 0.9836
⎝ 61 ⎠
VC = (1.475 )( 5 ) − 10 ⇒ VC = −2.625 V
IB =
Yes, VC < 0 in both cases so that B-C junction is reverse biased.
5.8
IC =
IE =
VC − ( −10 )
RC
IC
α
=
=
10 − 1.2
= 1.76 mA
5
1.76
⇒ I E = 1.774 mA
0.992
5.9
I C = I S eVRE / VT
⎛ 0.685 ⎞
= 10−13 exp ⎜
⎟ ⇒ I C = 27.67 mA
⎝ 0.026 ⎠
⎛1+ β ⎞
⎛ 91 ⎞
IE = ⎜
⎟ I C = ⎜ ⎟ ( 27.67 )
⎝ 90 ⎠
⎝ β ⎠
I D = 27.98 mA
I
27.67
⇒ I B = 0.307 mA
IB = C =
β
90
5.10
Device 1: iE = I Eo1evEB / VT ⇒ 0.5 × 10−3 = I Eo1e0.650 / 0.026
So that
I EO1 = 6.94 × 10−15 A
Device 2: 12.2 × 10−3 = I Eo 2 e0.650 / 0.026
Or
I Eo 2 = 1.69 × 10−13 A
Ratio of areas =
5.11
(a)
ro =
(b)
ro =
I Eo 2 1.69 × 10−13
=
⇒ Ratio = 24.4
I Eo1 6.94 × 10−15
VA 250
=
⇒ ro = 250 k Ω
IC
1
VA 250
=
⇒ ro = 2.50 M Ω
IC
0.1
5.12
BVC E 0 =
BVC B 0
3
β
=
60
3
100
BVC E 0 = 12.9 V
5.13
BVC E 0 =
56 =
220
3
β
β = 60.6
5.14
BVC B 0
3
β
⇒3β =
220
= 3.93
56
BVC E 0 =
BVC B 0
3
β
BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50
BVC B 0 = 184 V
5.15
(a)
IE =
−0.7 − ( −10 )
= 1.86 mA
5
⎛ 75 ⎞
I C = (1.86 ) ⎜ ⎟ = 1.836 mA
⎝ 76 ⎠
VC = −0.7 + 4 = 3.3 V
10 − 3.3
⇒ RC = 3.65 K
1.836
0.5
= 0.00658 mA
IB =
76
VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V
RC =
(b)
(c)
⎛ 75 ⎞
I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA
⎝ 76 ⎠
−1 − ( −5 )
⇒ RC = 8.11 K
RC =
0.493
I
O = E (10 ) + 0.7 + I E ( 4 ) − 8
76
7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA
⎛ 75 ⎞
I C = (1.767 ) ⎜ ⎟ = 1.744 mA
⎝ 76 ⎠
VCE = 8 − (1.744 )( 4 ) − ⎡⎣(1.767 )( 4 ) − 8⎤⎦
= 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V
(d)
⎛I ⎞
5 = I E (10 ) + ⎜ E ⎟ ( 20 ) + 0.7 + I E ( 2 )
= I E (10 + 0.263 + 2 ) + 0.7
⎝ 76 ⎠
I E = 0.3506 mA ⇒ I B = 4.61 µ A
VC = 5 − ( 0.3506 )(10 )
VC = 1.49 V
5.16
For Fig. 5.15 (a) RE = 5 + 5% = 5.25 K
IE =
−0.7 − ( −10 )
= 1.77 mA
5.25
I C = 1.75 mA
10 − 3.3
= 3.83 K
RC =
1.75
RE = 5 − 5% = 4.75 K
IE =
−0.7 − ( −10 )
= 1.96 mA
4.75
I C = 1.93 mA
10 − 3.3
= 3.47 K
RC =
1.93
So 1.75 ≤ I C ≤ 1.93 mA 3.47 ≤ RC ≤ 3.83 K
For Fig. 5.15(c) RE = 4 + 5% = 4.2 K
IB =
8 − 0.7
= 0.0222 mA
10 + ( 76 )( 4.2 )
I C = 1.66 mA
I E = 1.69 mA
VCE = 16 − (1.66 )( 4 ) − (1.69 )( 4.2 )
= 16 − 6.64 − 7.098 ⇒ VCE = 2.26 V
RE = 4 − 5% = 3.8 K
8 − 0.7
IB =
= 0.0244 I C = 1.83 mA
10 + ( 76 )( 3.8 )
I E = 1.86 mA
VCE = 16 − (1.83)( 4 ) − (1.86 )( 3.8 )
= 16 − 7.32 − 7.068
VCE = 1.61 V
So 1.66 ≤ I C ≤ 1.83 mA 1.61 ≤ VCE ≤ 2.26 V
5.17
RB =
VBB − VEB 2.5 − 0.7
=
⇒ RB = 120 k Ω
IB
0.015
I CQ = ( 70 )(15µ A ) ⇒ 1.05 mA
RC =
VCC − VECQ
I CQ
=
5 − 2.5
⇒
1.05
RC = 2.38 k Ω
5.18
(a)
VB = − I B RB ⇒ I B =
−VB − ( −1)
=
RB
500
I B = 2.0 µ A
VE = −1 − 0.7 = −1.7 V
IE =
VE − ( −3)
RE
=
−1.7 + 3
= 0.2708 mA
4.8
IE
0.2708
= (1 + β ) =
= 135.4 ⇒ β = 134.4
IB
0.002
⇒ α = 0.9926
1+ β
I C = β I B ⇒ I C = 0.269 mA
α=
β
VCE = 3 − VE = 3 − ( −1.7 ) ⇒ VCE = 4.7 V
(b)
5−4
⇒ I E = 0.5 mA
2
4 = 0.7 + I B RB + ( I B + I C ) RC − 5
IE =
I B + IC = I E
I B + IC = I E
4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) − 5
I B = 0.043 ⇒
IE
0.5
= (1 + β ) =
= 11.63
IB
0.043
β = 10.63, α =
5.19
β
1+ β
⇒ α = 0.9140
VB − 0.7 − ( −5 )
VB + 4.3
3
3
4.3
V
+
⎛
⎞ ⎛ 50 ⎞
⎛ 50 ⎞
IC = ⎜ ⎟ I E = ⎜ B
⎟ ⎜ 51 ⎟
3
⎝ 51 ⎠
⎝
⎠⎝ ⎠
⎛ V + 4.3 ⎞ ⎛ 50 ⎞
VC = 5 − I C RC = 5 − ⎜ B
⎜ ⎟ (10 )
3 ⎟⎠ ⎝ 51 ⎠
⎝
IE =
=
Now VB = VC , so VB [1 + 3.27 ] = 5 − 14.1 = −9.05
VB = −2.12 V
IE =
−2.12 + 4.3
⇒ I E = 0.727 mA
3
5.20
10 − VE 10 − 2
=
⇒ I E = 0.80 mA
10
10
VB = VE − 0.7 = 2 − 0.7 = 1.3 V
IE =
IB =
VB 1.3
=
⇒ I B = 0.026 mA
RB 50
β=
I C 0.774
=
⇒ β = 29.77
I B 0.026
I C = I E − I B = 0.80 − 0.026 ⇒ I C = 0.774 mA
α=
β
1+ β
=
29.77
⇒ α = 0.9675
30.77
VEC = VE − VC = VE − ( I C RC − 10 )
= 2 − ⎡⎣( 0.774 )(10 ) − 10 ⎤⎦
VEC = 4.26 V
Load line developed assuming the VB voltage can change and the RB resistor is removed.
1 mA
Q-point
0.774
IC
4.26
5.21
5 − 0.7
⇒ 17.2 µ A
250
I C = (120 )( 0.0172 ) = 2.064 mA
IB =
VC = ( 2.064 )(1.5 ) − 5 = −1.90 V
VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V
20
VEC
IC (mA)
6.67
Q-point
2.06
6.9
10 V
EC (V)
5.22
⎛ 50 ⎞
I C = ⎜ ⎟ (1) = 0.98 mA
⎝ 51 ⎠
VC = I C RC − 9 = ( 0.98 )( 4.7 ) − 9 or VC = −4.39 V
1
= 0.0196 mA
51
VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V
IB =
5.23
0.5
⎛ 50 ⎞
= 0.0098 mA
I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B =
51
⎝ 51 ⎠
VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V
VC = I C RC − 9 = ( 0.49 )( 4.7 ) − 9 = −6.70 V
Then VEC = VE − VC = 1.19 − ( −6.7 ) == 7.89 V
PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW
Power Dissipated = PS = I Q ( 9 − VE ) = ( 0.5 )( 9 − 1.19 )
Or PS = 3.91 mW
5.24
I
⇒ I E1 = I E 2 = 0.5 mA
2
≈ 0.5 mA
= 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V
I E1 = I E 2 =
I C1 = I C 2
VC1 = VC 2
5.25
(a)
RE = 0 I B =
2 − 0.7 1.3
=
RB
RB
⎛ 1.3 ⎞ 5 − 2
= 0.8 ⇒ RC = 3.75 K
I C = ( 80 ) ⎜ ⎟ =
⎝ RB ⎠ RC
RB = 130 K
(b)
0.8
⎛ 81 ⎞
= 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA
80
⎝ 80 ⎠
2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K
RE = 1 K
IB =
5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K
(c)
For part (a)
IB =
2 − 0.7
= 0.01 mA
130
I C = (120 )( 0.01) ⇒ I C = 1.20 mA
VCE = 5 − (1.2 )( 3.75 ) ⇒ VCE = 0.5 V
2 = I B ( 49 ) + 0.7 + (121) I B (1)
For part (b)
I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA
VCE = 5 − ( 0.918 )( 2.74 ) − ( 0.925 )(1) ⇒ VCE = 1.56 V
Including RE result in smaller changes in Q-point values.
5.26
I BQ =
a.
VCC − VBE ( on )
RB
I CQ
2
= 0.0333 mA
60
β
24 − 0.7
RB =
⇒ RB = 699 kΩ
0.0333
VCC − VCEQ
24 − 12
I CQ =
⇒ RC =
⇒ RC = 6 kΩ
RC
2
I BQ =
I BQ =
b.
=
VCC − VBE ( on )
RB
24 − 0.7
699
=
= 0.0333 mA ( Unchanged )
I CQ = β I BQ = (100 )( 0.0333) ⇒ I CQ = 3.33 mA
VCEQ = VCC − I CQ RC = 24 − ( 3.33)( 6 ) ⇒ VCEQ = 4.02 V
VCE = VCC − I C RC = 24 − I C ( 6 )
(c)
IC (mA)
4
3.33
Q-pt (␤ ⫽ 100)
Q-pt (␤ ⫽ 60)
2
4.02
12
5.27
a.
VB = 0 ⇒ Cutoff ⇒ I E = 0, VC = 6 V
b.
VB = 1 V, I E =
c.
VB = 2 V. Assume active-mode
1 − 0.7
⇒ I E = 0.3 mA
1
I C ≈ I E ⇒ VC = 6 − ( 0.3)(10 ) ⇒ VC = 3 V
2 − 0.7
= I E = 1.3 mA ≈ I C
1
VC = 6 − (1.3)(10 ) = −7 V!
IE =
24
VCE
Transistor in saturation
2 − 0.7
⇒ I E = 1.3 mA
1
VE = 1.3 V, VCE ( sat ) = 0.2 V
IE =
VC = VE + VCE ( sat ) = 1.3 + 0.2 ⇒ VC = 1.5 V
5.28
a.
VBB = 0.
b.
VBB = 1 V
⎛ RL
Cutoff V0 = ⎜
⎝ RC + RL
V0 = 3.33 V
⎞
⎛ 10 ⎞
⎟ VCC = ⎜
⎟ ( 5)
⎝ 10 + 5 ⎠
⎠
1 − 0.7
⇒ 6 μA
50
I C = β I B = ( 75 )( 6 ) ⇒ I C = 0.45 mA
IB =
5 − V0
V
= IC + 0
5
10
⎛1 1 ⎞
1 − 0.45 = V0 ⎜ + ⎟ ⇒ V0 = 1.83 V
⎝ 5 10 ⎠
Transistor in saturation V0 = VCE ( sat ) = 0.2 V
c.
5.29
(a) β = 100
⎛ 100 ⎞
(i) I Q = 0.1 mA I C = ⎜
⎟ ( 0.1) = 0.0990 mA
⎝ 101 ⎠
VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V
⎛ 100 ⎞
(ii) I Q = 0.5 mA I C = ⎜
⎟ ( 0.5 ) = 0.495 mA
⎝ 101 ⎠
VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V
(iii) I Q = 2 mA Transistor is in saturation
VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V
(b) β = 150
⎛ 150 ⎞
(i) I Q = 0.1 mA I C = ⎜
⎟ ( 0.1) = 0.09934 mA
⎝ 151 ⎠
VO = 5 − ( 0.09934 )( 5 ) ⇒ VO = 4.503 V
4.503 − 4.505
× 100% = −0.044%
4.503
⎛ 150 ⎞
(ii) I Q = 0.5 mA I C = ⎜
⎟ ( 0.5 ) = 0.4967 mA
⎝ 151 ⎠
VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V
% change =
2.517 − 2.525
× 100% = −0.32%
2.525
(iii) I Q = 2 mA Transistor in saturation
No change
Vo = −8.5 V
% change =
5.30
VCB = 0.5 V ⇒ VO = 0.5 V , I C =
5 − 0.5
= 0.90 mA
5
⎛ 101 ⎞
IQ = ⎜
⎟ ( 0.90 ) ⇒ I Q = 0.909 mA
⎝ 100 ⎠
5.31
For I Q = 0, then PQ = 0
⎛ 50 ⎞
For I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA
⎝ 51 ⎠
0.5
= 0.0098 mA, VB = 0.490 V , VE = 1.19 V
IB =
51
VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V
P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒ P = 3.87 mW
For I Q = 1.0 mA, Using the same calculations as above, we find P = 5.95 mW
For I Q = 1.5 mA, P = 6.26 mW
For I Q = 2 mA, P = 4.80 mW
For I Q = 2.5 mA, P = 1.57 mW
For I Q = 3 mA, Transistor is in saturation.
0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) − 9
I E = IQ = I B + IC ⇒ I B = 3 − IC
Then, 0.7 + ( 3 − I C ) ( 50 ) = 0.2 + I C ( 4.7 ) − 9
Which yields I C = 2.916 mA and I B = 0.084 mA
P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 ) or P = 0.642 mW
5.32
IE =
VEE − VEB ( on )
RE
=
9 − 0.7
⇒ I E = 2.075 mA
4
I C = α I E = ( 0.9920 ) ( 2.075 ) ⇒ I C = 2.06 mA
VBC + I C RC = VCC
VBC = 9 − ( 2.06 ) ( 2.2 ) ⇒ VBC = 4.47 V
5.33
I CQ =
I BQ =
IR2 =
VCC − VCEQ
RC
I CQ
β
=
=
12 − 6
= 2.73 mA
2.2
2.73
⇒ I BQ = 0.091 mA
30
0.7 − ( −12 )
= 0.127 mA
100
I R1 = I R 2 + I BQ = 0.127 + 0.091 = 0.218 mA
V1 = I R1 R1 + 0.7 = ( 0.218 )(15 ) + 0.7 ⇒ V1 = 3.97 V
5.34
For VCE = 4.5
5 − 4.5
= 0.5 mA
I CQ =
1
0.5
= 0.02 mA
I BQ =
25
0.7 − ( −5 )
= 0.057 mA
I R2 =
100
I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA
V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V
For VCE = 1.0
5 −1
= 4 mA
1
4
=
= 0.16 mA
25
= 0.057 mA
I CQ =
I BQ
I R2
I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mA
V1 = ( 0.217 )(15 ) + 0.7 ⇒ 3.96 V
So 1.86 ≤ V1 ≤ 3.96 V
IC
5
4
Range of
Q-pt values
0.5
0
1
4.5
5
5.35
5 − 2.5
=5K
0.5
0.5
IB =
= 0.00417 mA
120
5 − 0.7
RB =
= 1032 K
0.00417
RC =
(a)
IC (mA)
1.0
Q-point
0.5
2.5
(b)
Choose RC = 5.1 K
RB = 1 MΩ
5 V (V)
CE
For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K
5 − 0.7
I BQ =
= 3.91 µ A ⇒ I CQ = 0.469 mA
1.1
VCEQ = 2.37 V
RB = 1 MΩ + 10% = 1.1M, RC = 5.1 K − 10% = 4.59 K
I BQ = 3.91 µ A ⇒ I CQ = 0.469 mA
VCEQ = 2.85 V
RB = 1 MΩ − 10% = 0.90 MΩ
RC = 5.1 k + 10% = 5.61 K
5 − 0.7
= 4.78 µ A ⇒ I C = 0.573 mA
I BQ =
0.90
VCEQ = 1.78 V
RB = 1 MΩ − 10% = 0.90 MΩ
RC = 5.1 k − 10% = 4.59 K
I BQ = 4.78 µ A ⇒ I C = 0.573 mA
VCEQ = 2.37 V
IC (mA)
1.09
0.891
0.573
0.469
1.78 2.37 2.85
5 V (V)
CE
5.36
VE 2 = 5 − VBE 2
VE1 = 5 − VBE1
VO = VE 2 − VE1 = ( 5 − VBE 2 ) − ( 5 − VBE1 )
VO = VBE1 − VBE 2
⎛I ⎞
We have VBE1 = VE ln ⎜ E1 ⎟
⎝ I EO ⎠
⎛I ⎞
VBE 2 = VT ln ⎜ E 2 ⎟
⎝ I EO ⎠
⎡ ⎛I ⎞
⎛I
VO = VT ⎢ln ⎜ E1 ⎟ − ln ⎜ E 2
⎝ I EO
⎣ ⎝ I EO ⎠
⎞⎤
⎟⎥
⎠⎦
⎛I ⎞
⎛ 10 ⎞
VO = VT ln ⎜ E1 ⎟ = VT ln ⎜ I ⎟
⎝ I ⎠
⎝ IE2 ⎠
VO =
5.37
(a)
kT
ln (10 )
e
RE = 0
VI − 0.7
(120 )( 4 )
VO = 5 − I C ( 4 ) = 5 −
IC = β I B
(VI − 0.7 )
200
200
When VO = 0.2, 0.2 = 5 − 2.4 VI + 1.68 ⇒ VI = 2.7
IB =
VO(V)
5
0.2
0.7
2.7
5
VI (V)
RE = 1 K
VI − 0.7
V − 0.7
IB =
I C = β IB
= I
200 + (121)(1)
321
(b)
(120 )( 4 )
(VI − 0.7 )
321
When VO = 0.2 = 5 − 1.495VI + 1.047
VO = 5 −
VI = 3.91 V
VO(V)
5
0.2
0.7
3.91
5.38
For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0
VO = 0
If Q reaches saturation, VO = 4.8
4.8
IC =
= 1.2 mA
4
5 − 0.7 − VI
1.2
IB =
= 0.015 =
⇒ VI = 1.6
80
180
So VI ≤ 1.6, VO = 4.8
5
VI (V)
VO(V)
4.8
1.6
5.39
(a)
4.3
5
VI (V)
For VI ≥ 4.3, Q is off and VO = 0
⎛ 101 ⎞
When transistor enters saturation, 5 = ⎜
⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA
⎝ 100 ⎠
VO = 3.832 V
I B = 0.00958 mA
⎛ 101 ⎞
5=⎜
⎟ ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI
⎝ 100 ⎠
VI = 5 − 0.7 − 0.9676 − 1.7244 ⇒ VI = 1.61 V
For VI = 0, transistor in saturation
5 = I E (1) + 0.2 + I C ( 4 ) ⇒ 5 = I C (1) + I B (1) + 0.2 + I C ( 4 )
5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 )
I E = IC + I B
4.8 = 5 I C + I B (1)
4.3 = 1I C + 181I B
I B = 4.8 − 5 I C
4.3 = I C + (181)( 4.8 − 5 I C )
904 I C = 864.5
I C = 0.956 mA
VO = 3.825 V
VO(V)
3.832
3.825
1.61
4.3
5
VI (V)
5.40
RTH = R1 & R2 = 33 & 10 = 7.67 kΩ
⎛ R2 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (18 ) = 4.186 V
R
R
+
⎝ 10 + 33 ⎠
⎝ 1
2 ⎠
V − VBE ( on )
4.186 − 0.7
I BQ = TH
=
RTH + (1 + β ) RE 7.67 + ( 51)(1)
I BQ = 0.0594 mA
I CQ = β I BQ ⇒ I CQ = 2.97 mA
I EQ = 3.03 mA
VCEQ = VCC − I CQ RC − I EQ RE
= 18 − ( 2.97 )( 2.2 ) − ( 3.03)(1) ⇒ VCEQ = 8.44 V
5.41
I CQ = 1.2 mA, VCEQ = 9 V , RTH = 50 k Ω
1.2
= 0.015 mA
Also I B =
80
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
⎛ R2 ⎞
1
1
VTH = ⎜
⎟ (VCC ) = ⋅ RTH ⋅ VCC = ( 50 )(18 )
R1
R1
⎝ R1 + R2 ⎠
1
Then
( 50 )(18) = ( 0.015)( 50 ) + 0.7 + ( 81)( 0.015)(1) or R1 = 338 k Ω.
R1
Then
338R2
= 50 ⇒ R2 = 58.7 k Ω
338 + R2
⎛ 81 ⎞
I EQ = ⎜ ⎟ (1.2 ) = 1.215 mA
⎝ 80 ⎠
18 = I CQ RC + VCEQ + I EQ RE
18 = (1.2 ) RC + 9 + (1.215 )(1) ⇒ RC = 6.49 k Ω
5.42
RTH = R1 & R2 = 20 & 15 = 8.57 k Ω
⎛ R2 ⎞
⎛ 15 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (10 ) = 4.29 V
⎝ 15 + 20 ⎠
⎝ R1 + R2 ⎠
I EQ
VCC = I EQ RE + VEB ( on ) +
⋅ RTH + VTH
1+ β
⎛ 8.57 ⎞
10 = I EQ (1) + 0.7 + I EQ ⎜
⎟ + 4.29
⎝ 101 ⎠
10 − 0.7 − 4.29 5.01
Then I EQ =
=
⇒ I EQ = 4.62 mA
8.57
1.085
1+
101
I EQ
⎛ 4.62 ⎞
VB =
⋅ RTH + VTH = ⎜
⎟ ( 8.57 ) + 4.29 or VB = 4.68 V
1+ β
⎝ 101 ⎠
5.43
(a)
RTH = 42 & 58 = 24.36 K
⎛ 42 ⎞
VTH = ⎜
⎟ ( 24 ) = 10.08 V
⎝ 100 ⎠
10.08 − 0.7
9.38
I BQ =
=
⇒ 7.30 µ A
24.36 + (126 )(10 ) 1284.36
I CQ = 0.913 mA
VCEQ = 14.8 V
I EQ = 0.9202
IC (mA)
2.38
Q-point
0.913
14.8
24 V
(b)
R1 + 5% = 60.9, R2 + 5% = 44.1
RTH = 25.58 K
10.08 − 0.7
9.38
=
⇒ 7.30 µ A
I BQ =
25.58 + 126 (10 ) 1285.58
I CQ = 0.912 mA
VCEQ = 14.81
I EQ = 0.919
CE (V)
VTH = 10.08
R1 + 5% = 60.9, R2 − 5% = 39.90
RTH = 24.11 K
9.50 − 0.7
8.8
I BQ =
=
= 6.85 µ A
24.11 + (126 )(10 ) 1284.11
I CQ = 0.857 mA
VCEQ = 15.37 V
I EQ = 0.8635 mA
R1 − 5% = 55.1 K
R2 + 5% = 44.1 K
10.67 − 0.7
9.97
=
= 7.76 µ A
I BQ =
24.50 + 1260 1284.5
I CQ = 0.970 mA
I EQ = 0.978 mA
VCEQ = 14.22 V
R1 − 5% = 55.1 K
R2 − 5% = 39.90
10.08 − 0.7
9.38
=
= 7.31 µ A
I BQ =
23.14 + 1260 1283.14
I CQ = 0.914 mA
I EQ = 0.9211 mA
VCEQ = 14.79 V
So we have 0.857 ≤ I CQ ≤ 0.970 mA
14.22 ≤ VCEQ ≤ 15.37 V
5.44
a.
RTH = R1 & R2 = 25 & 8 = 6.06 kΩ
⎛ R2 ⎞
⎛ 8 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ ( 24 )
⎝ 8 + 25 ⎠
⎝ R1 + R2 ⎠
= 5.82 V
V − VBE (on)
5.82 − 0.7
=
I BQ = TH
RTH + (1 + β ) RE 6.06 + ( 76 )(1)
I BQ = 0.0624 mA, I CQ = 4.68 mA
I EQ = 4.74
VCEQ = VCC − I CQ RC − I EQ RE
= 24 − ( 4.68 )( 3) − ( 4.74 )(1)
VCEQ = 5.22 V
b.
I BQ =
5.82 − 0.7
⇒ I BQ = 0.0326 mA
6.06 + (151)(1)
I CQ = 4.89 mA
I EQ = 4.92
VCEQ = 24 − ( 4.89 )( 3) − ( 4.92 )(1)
VCEQ = 4.41 V
5.45
(a)
VTH = 9.50
RTH = 24.50 K
RTH = 23.14 K
VTH = 10.67 V
VTH = 10.08
I CQ ≅ I EQ = 0.4 mA
3
3
⇒ RC = 7.5 k Ω; RE =
⇒ RE = 7.5 k Ω
0.4
0.4
9
= 112.5 k Ω
R1 + R2 ≅
( 0.2 )( 0.4 )
RC =
⎛ R2 ⎞
VTH = ⎜
⎟ (VCC ) = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
⎝ R1 + R2 ⎠
(112.5 − R2 ) R2
RR
0.4
, I BQ =
= 0.004 mA
RTH = 1 2 =
112.5
100
R1 + R2
⎡ (112.5 − R2 ) R2 ⎤
⎛ 9 ⎞
R2 ⎜
⎥ + 0.7 + (101)( 0.004 )( 7.5 )
⎟ = ( 0.004 ) ⎢
112.5
⎝ 112.5 ⎠
⎦
⎣
We obtain R2 ( 0.08 ) = 0.004 R2 − 3.56 × 10−5 R22 + 3.73
From this quadratic, we find R2 = 48 k Ω ⇒ R1 = 64.5 k Ω
(b)
Standard resistor values:
Set RE = RC = 7.5 k Ω and R1 = 62 k Ω, R2 = 47 k Ω
Now RTH = R1 & R2 = 62 & 47 = 26.7 k Ω
⎛ R2 ⎞
⎛ 47 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ ( 9 ) = 3.88 V
⎝ 47 + 62 ⎠
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
So I BQ =
3.88 − 0.7
= 0.00406 mA
26.7 + (101)( 7.5 )
Then I CQ = 0.406 mA
VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V
5.46
(a)
RTH = R1 & R2 = 12 & 2 = 1.714 K
⎛ R2 ⎞
⎛ 2⎞
VTH = ⎜
⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = −3.571 V
R
R
+
⎝ 14 ⎠
2 ⎠
⎝ 1
(b)
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
−3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) − 5
I BQ =
5 − 0.7 − 3.571
0.729
=
⇒ 13.96 µ A
1.714 + (101)( 0.5 ) 52.21
I CQ = 1.396 mA, I EQ = 1.410 mA
VCEQ = 10 − (1.396 )( 5 ) − (1.41)( 0.5 ) ⇒ VCEQ = 2.32 V
(d)
RE = 0.5 + 5% = 0.525 K
I BQ =
RC = 5 + 5% = 5.25 K
0.729
⇒ 13.32 µ A
1.714 + (101)( 0.525 )
I CQ = 1.332 mA
I EQ = 1.345 mA
VCEQ = 10 − (1.332 )( 5.25 ) − (1.345 )( 0.525 )
= 10 − 6.993 − 0.7061 ⇒ VCEQ = 2.30 V
RE = 0.5 + 5% = 0.525 K
I CQ = 1.332 mA
RC = 5 − 5% = 4.75 K
I EQ = 1.345 mA
VCEQ = 10 − (1.332 )( 4.75 ) − (1.345 )( 0.525 )
= 10 − 6.327 − 0.7061 ⇒ VCEQ = 2.97 V
RE = 0.5 − 5% = 0.475 K
I BQ =
RC = 5 + 5% = 5.25 K
0.729
⇒ 14.67 µ A
1.714 + (101)( 0.475 )
I CQ = 1.467 mA
I EQ = 1.482 mA
VCEQ = 10 − (1.467 )( 5.25 ) − (1.482 )( 0.475 )
= 10 − 7.70175 − 0.70395 ⇒ VCEQ = 1.59 V
RE = 0.5 − 5% = 0.475 K
I CQ = 1.467 mA
RC = 5 − 5% = 4.75 K
I EQ = 1.482 mA
VCEQ = 10 − (1.467 )( 4.75 ) − (1.482 )( 0.475 )
= 10 − 6.96825 − 0.70395 ⇒ VCEQ = 2.33 V
5.47
RTH = R1 & R2 = 9 & 1 = 0.91 kΩ
⎛ R2 ⎞
⎛ 1 ⎞
VTH = ⎜
⎟ ( −12 ) = ⎜
⎟ ( −12 ) = −1.2 V
+
R
R
⎝ 1+ 9 ⎠
⎝ 1
2 ⎠
I EQ RE + VEB ( on ) + I BQ RTH + VTH = 0
I BQ =
−VTH − VEB (on)
1.2 − 0.7
=
RTH + (1 + β ) RE 0.90 + ( 76 )( 0.1)
I BQ = 0.0588,
I CQ = 4.41 mA
I EQ = 4.47 mA
Center of load line ⇒ VECQ = 6 V
I EQ RE + VECQ + I CQ RC − 12 = 0
( 4.47 ) ( 0.1) + 6 + ( 4.41) RC
= 12 ⇒ RC = 1.26 kΩ
5.48
(a)
RTH = 36 & 68 = 23.5 K
⎛ 36 ⎞
VTH = ⎜
⎟ (10 ) = 3.46 V
⎝ 36 + 68 ⎠
3.46 − 0.7
I BQ =
= 0.00178 mA
23.5 + ( 51)( 30 )
I CQ = 0.0888 mA
I EQ = 0.0906 mA
VCE = 10 − ( 0.0888 )( 42 ) − ( 0.0906 )( 30 )
= 10 − 3.73 − 2.72 ⇒ VCE = 3.55 V
(b)
R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K
RTH = 7.85 k
VTH = 3.46
3.46 − 0.7
I BQ =
= 0.00533 mA
7.85 + ( 51)(10 )
I CQ = 0.266 mA
I EQ = 0.272 mA
VCE = 10 − ( 0.266 )(14 ) − ( 0.272 )(10 )
VCE = 3.56 V
5.49
(a)
⎛ R2 ⎞
⎛ 68 ⎞
RTH = 36 & 68 = 23.5 K VTH = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5 = 1.54 V
R
R
+
⎝ 36 + 68 ⎠
⎝ 1
2 ⎠
5 = ( 51) I BQ ( 30 ) + 0.7 + I B ( 23.5 ) + 1.54
I BQ =
2.76
= 1.78 μA ⇒ I CQ = 0.0888 mA
1553.5
I EQ = 0.0906 mA
VECQ = 10 − ( 0.0906 )( 30 ) − ( 0.0888 )( 42 )
= 10 − 2.718 − 3.7296 ⇒ VECQ = 3.55 V
(b)
RTH = 12 & 22.7 = 7.85 K
VTH = 1.54
RE = 10 K
RC = 14 K
5 = ( 51) I BQ (10 ) + 0.7 + I B ( 7.85 ) + 1.54
I BQ =
2.76
= 5.33 µ A I CQ = 0.266 mA
517.85
I EQ = 0.272 mA
VECQ = 10 − ( 0.272 )(10 ) − ( 0.266 )(14 )
= 10 − 2.72 − 3.724 ⇒ VECQ = 3.56 V
5.50
(a)
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
R1
I BQ =
I CQ
β
=
0.8
= 0.008 mA
100
1
Then
( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008)( 0.5)
R1
or R1 = 44.1 k Ω,
44.1R2
= 5.05 ⇒ R2 = 5.70 k Ω
44.1 + R2
⎛ 101 ⎞
Now I EQ = ⎜
⎟ ( 0.8 ) = 0.808 mA
⎝ 100 ⎠
VCC = I CQ RC + VCEQ + I EQ RE
10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 )
RC = 5.75 k Ω
(b)
For 75 ≤ β ≤ 150
⎛ R2 ⎞
⎛ 5.7 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (10 ) = 1.145 V
5.7 + 44.1 ⎠
R
R
+
⎝
⎝ 1
2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
For β = 75, I BQ =
1.145 − 0.7
= 0.0103 mA
5.05 + ( 76 )( 0.5 )
Then I CQ = ( 75 )( 0.0103) = 0.775 mA
For β = 150, I BQ =
1.145 − 0.7
= 0.00552 mA
5.05 + (151)( 0.5 )
Then I CQ = 0.829 mA
% Change =
ΔI CQ
=
I CQ
0.829 − 0.775
× 100% ⇒ % Change = 6.75%
0.80
For RE = 1 k Ω
(c)
RTH = ( 0.1)(101)(1) = 10.1 k Ω
VTH =
1
1
⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1)
R1
R1
which yields R1 = 63.6 k Ω
And
63.6 R2
= 10.1 ⇒ R2 = 12.0 k Ω
63.6 + R2
⎛ R2 ⎞
⎛ 12 ⎞
Now VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (10 ) = 1.587 V
R
R
12 + 63.6 ⎠
+
⎝
⎝ 1
2 ⎠
1.587 − 0.7
= 0.0103 mA
For β = 75, I BQ =
10.1 + ( 76 )(1)
So I CQ = 0.773 mA
For β = 150, I BQ =
1.587 − 0.7
= 0.00551 mA
10.1 + (151)(1)
Then I CQ = 0.826 mA
% Change =
ΔI CQ
I CQ
=
0.826 − 0.773
× 100% ⇒ % Change = 6.63%
0.8
5.51
VCC ≅ I CQ ( RC + RE ) + VCEQ
10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω
Let RE = 0.875 k Ω
Then, for bias stable RTH = ( 0.1)(121)( 0.875 ) = 10.6 k Ω
I BQ =
0.8
= 0.00667 mA
120
1
(10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 )
R1
So R1 = 71.8 k Ω and
71.8R2
= 10.6 ⇒ R2 = 12.4 k Ω
71.8 + R2
10
= 0.119 mA
71.8 + 12.4
This is close to the design specification.
Then I R ≅
5.52
I CQ ≈ I EQ ⇒ VCEQ
= VCC − I CQ ( RC + RE )
6 = 12 − I CQ ( 2 + 0.2 )
I CQ = 2.73 mA,
I BQ = 0.0218 mA
VCEQ = 6 V
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
⎛ R2 ⎞
VTH = ⎜
⎟ (12 ) − 6,
⎝ R1 + R2 ⎠
RTH = R 1& R2
Bias stable ⇒ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 kΩ
⎛ 1⎞
VTH = ⎜ ⎟ ( RTH )(12 ) − 6
⎝ R1 ⎠
1
( 2.52 )(12 ) − 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) − 6
R1
1
( 30.24 ) = 0.7549 + 0.5494
R1
R1 = 23.2 kΩ,
23.2R 2
= 2.52
23.2 + R 2
R2 = 2.83 kΩ
5.53
a.
⎛ 81 ⎞
I CQ = 1 mA. I EQ = ⎜ ⎟ (1) = 1.01 mA
⎝ 80 ⎠
VCEQ = 12 − (1)( 2 ) − (1.01)( 0.2 ) ⇒ VCEQ = 9.80 V
1
= 0.0125 mA
80
= + ( 0.1) (1 + β ) RE = ( 0.1)( 81)( 0.2 ) = 1.62 kΩ
I BQ =
RTH
⎛ R2 ⎞
1
1
VTH = ⎜
( RTH )(12 ) − 6 = (19.44 ) − 6
⎟ (12 ) − 6 =
R
R
R
R
+
⎝ 1
2 ⎠
1
1
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
1
(19.44 ) − 6 = ( 0.0125 )(1.62 ) + 0.7 + (81)( 0.0125)( 0.2 ) − 6
R1
1
(19.44 ) = 0.923
R1
R1 = 21.1 kΩ,
21.1R2
= 1.62
21.1 + R2
R2 = 1.75 kΩ
b.
R1 = 22.2 kΩ or R1 = 20.0 kΩ
R2 = 1.84 kΩ or R2 = 1.66 kΩ
R2 ( max ) , R1 ( min )
RTH = (1.84 ) & ( 20.0 ) = 1.685 kΩ
⎛ 1.84
⎞
VTH = ⎜
⎟ (12 ) − 6 = −4.99 V
⎝ 1.84 + 20.0 ⎠
−4.99 − 0.7 − ( −6 ) 0.31
=
= 0.0173 mA
I BQ =
1.685 + ( 81)( 0.2 ) 17.89
I CQ = 1.39 mA, I EQ = 1.40 mA
For max, RC ⇒ VCE = 12 − (1.39 )( 2 ) − (1.40 )( 0.2 )
VCE = 8.94 V
R2 ( min ) , R1 ( max )
RTH = (1.66 ) & ( 22.2 ) = 1.545 kΩ
⎛ 1.66 ⎞
VTH = ⎜
⎟ (12 ) − 6 = −5.165 V
⎝ 1.66 + 22.2 ⎠
−5.165 − 0.7 + 6
0.135
=
= 0.00761 mA ⇒ I CQ = 0.609 mA, I E = 0.616
I BQ =
1.545 + ( 81)( 0.20 ) 17.745
VCEQ = 12 − ( 0.609 )( 2 ) − ( 0.616 )( 0.2 )
VCEQ = V 10.7 V
So 0.609 ≤ I C ≤ 1.39 mA
8.94 ≤ VCEQ ≤ 10.7 V
5.54
VCEQ ≅ VCC − I CQ ( RC + RE )
5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω
Let RE = 0.333 k Ω and RC = 2 k Ω
Nominal value of β = 100
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 k Ω
3
= 0.03 mA
100
1
1
= ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6
R1
R1
I BQ =
VTH
Then VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
1
( 3.36 )(12 ) − 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) − 6
R1
which yields R1 = 22.3 k Ω and R2 = 3.96 k Ω
⎛ R2 ⎞
⎛ 3.96 ⎞
Now VTH = ⎜
⎟ (12 ) − 6 = ⎜
⎟ (12 ) − 6 or VTH = −4.19 V
⎝ 3.96 + 22.3 ⎠
⎝ R1 + R2 ⎠
For β = 75, VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
I BQ =
VTH + 6 − 0.7
−4.19 + 6 − 0.7
=
= 0.0387 mA ⇒ I C = 2.90 mA
RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333)
For β = 150, I BQ =
−4.19 + 6 − 0.7
= 0.0207 mA
3.36 + (151)( 0.333)
Then I C = 3.10 mA
Specifications are met.
5.55
RTH = R1 & R2 = 3 & 12 = 2.4 k Ω
⎛ R2 ⎞
⎛ 12 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ ( 20 ) = 16 V
⎝ 12 + 3 ⎠
⎝ R1 + R2 ⎠
(a)
For β = 75
20 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
20 − 0.7 − 16 = I BQ ⎡⎣( 76 )( 2 ) + 2.4 ⎤⎦
So I BQ = 0.0214 mA, I CQ = 1.60 mA, I EQ = 1.62 E
VECQ = 20 − (1.6 )(1) − (1.62 )( 2 ) or VECQ = 15.16 V
(b) For β = 100, we find I BQ = 0.0161 mA, I CQ = 1.61 mA, VECQ = 15.13 V , I EQ = 1.63 mA
5.56
I CQ = 4.8 mA → I EQ = 4.84 mA
VCEQ = VCC − I CQ RC − I EQ RE
6 = 18 − ( 4.8 )( 2 ) − ( 4.84 ) RE ⇒ RE = 0.496 kΩ
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kΩ
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
I BQ = 0.040 mA
VTH =
1
1
⋅ RTH ⋅VCC = ( 6.0 )(18 )
R1
R1
1
( 6.0 )(18) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 )
R1
1
(108 ) = 3.34
R1
R1 = 32.3 kΩ,
32.3 R2
= 6.0
32.3 + R2
R2 = 7.37 kΩ
5.57
For nominal β = 70
2
= 0.0286 mA → I EQ = 2.03 mA
70
= VCC − I CQ RC − I EQ RE
I BQ =
VCEQ
10 = 20 − ( 2 )( 4 ) − ( 2.03) RE ⇒ RE = 0.985 K
RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 71)( 0.985 ) = 6.99 K
VTH = I BQ RTH + VBE ( on ) + I EQ RE
1
⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE
R1
1
( 6.99 )( 20 ) = ( 0.0286 )( 6.99 ) + 0.70 + ( 2.03)( 0.985 )
R1
1
(139.8 ) = 2.90
R1
R1 = 48.2 K,
48.2 R2
= 6.99
48.2 + R2
R2 = 8.18 K
Check: For β = 50
⎛ 8.18 ⎞
VTH = ⎜
⎟ ( 20 ) = 2.90
⎝ 8.18 + 48.2 ⎠
V − VBE ( on )
2.90 − 0.7
I BQ = TH
=
= 0.0384 mA
RTH + (1 + β ) RE 6.99 + ( 51)( 0.985 )
I CQ = 1.92 mA
For β = 90
I BQ =
2.90 − 0.7
= 0.0228 mA
6.99 + ( 91)( 0.985 )
I CQ = 2.05 mA
Design criterion is satisfied.
5.58
I CQ = 1 mA → I EQ = 1.017 mA
VCEQ = VCC − I CQ RC − I EQ RE
5 = 15 − (1)( 5 ) − (1.017 ) RE ⇒ RE = 4.92 kΩ
Bias stable: RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 kΩ
1
= 0.0167 mA
I BQ =
60
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE
R1
1
( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 )
R1
1
( 448.5) = 6.197
R1
R1 = 72.5 kΩ,
72.5 R2
= 30.0
72.5 + R2
R2 = 51.2 kΩ
Check: For β = 45
⎛ 51.2 ⎞
VTH = ⎜
⎟ (15 ) = 6.21V
⎝ 51.2 + 72.5 ⎠
V − VBE ( on )
6.21 − 0.7
=
= 0.0215 mA
I BQ = TH
RTH + (1 + β ) RE 30 + ( 46 )( 4.92 )
I CQ = 0.967 mA,
ΔI C
= 3.27%
IC
Check: For β = 75
6.21 − 0.7
I BQ =
= 0.0136 mA
30.0 + ( 76 )( 4.92 )
ΔI C
= 2.31%
IC
Design criterion is satisfied.
I CQ = 1.023 mA,
5.59
(a)
VCC ≅ I CQ ( RC + RE ) + VCEQ
3 = ( 0.1)( 5 RE + RE ) + 1.4 ⇒ RE = 2.67 k Ω
100
= 0.833 µ A
120
= ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω
RC = 13.3 k Ω, I BQ =
RTH
VTH =
1
1
⋅ RTH ⋅ VCC = ( 32.3)( 3)
R1
R1
= I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
= ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833)( 2.67 )
which gives R1 = 97.3 k Ω, and R2 = 48.4 k Ω
(b)
IR ≅
3
3
=
⇒ 20.6 µ A
R1 + R2 97.3 + 48.4
I CQ = 100 µ A
P = ( I CQ + I R )VCC = (100 + 20.6 )( 3)
or P = 362 µW
5.60
IE =
5 − VE 5
= = 1.67 mA
RE
3
RTH = R1 || R2 = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 3) = 30.3 kΩ
⎛ R2 ⎞
1
VTH = ⎜
⎟ ( 4 ) − 2 = ⋅ RTH ⋅ ( 4 ) − 2
R
R
R
+
⎝ 1
2 ⎠
1
I EQ
I BQ =
= 0.0165 mA
1+ β
5 = I EQ RE + VEB ( on ) + I B RTH + VTH
1
5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + ( 30.3)( 4 ) − 2
R1
0.80 =
1
( 30.3)( 4 ) ⇒ R1 = 152 kΩ
R1
152 R2
= 30.3 ⇒ R2 = 37.8 kΩ
152 + R2
5.61
a.
RTH = R1 & R2 = 10 & 20 ⇒ RTH = 6.67 kΩ
⎛ R2 ⎞
⎛ 20 ⎞
VTH = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5 ⇒ VTH = 1.67 V
⎝ 20 + 10 ⎠
⎝ R1 + R2 ⎠
b.
10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
10 − 0.7 − 1.67 7.63
I BQ =
=
⇒ I BQ = 0.0593 mA
6.67 + ( 61)( 2 ) 128.7
I CQ = 3.56 mA, I EQ =3.62 mA
VE = 10 − I EQ RE = 10 − ( 3.62 )( 2 )
VE = 2.76 V
VC = I CQ RC − 10 = ( 3.56 )( 2.2 ) − 10
VC = −2.17 V
5.62
V + − V − ≅ I CQ ( RC + RE ) + VECQ
20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω
Let RE = 10 k Ω then RC = 14 k Ω
Let β = 60 from previous problem.
RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 )
Or RTH = 61 k Ω
0.5
= 0.00833 mA
60
⎛ R2 ⎞
1
=⎜
⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5
R1
⎝ R1 + R2 ⎠
I BQ =
VTH
Now 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
10 = ( 61)( 0.00833)(10 ) + 0.7 + ( 0.00833) ( 61) +
1
( 61) (10 ) − 5
R1
Then R1 = 70.0 k Ω and R2 = 474 k Ω
10
10
=
⇒ 18.4 µ A
R1 + R2 70 + 474
So the 40 µ A current limit is met.
IR ≅
RTH = R1 & R2 = 35 & 20 ⇒ RTH = 12.7 kΩ
5.63
a.
⎛ R2 ⎞
⎛ 20 ⎞
VTH = ⎜
⎟ (7) − 5 = ⎜
⎟ ( 7 ) − 5 ⇒ VTH = −2.45 V
+
R
R
⎝ 20 + 35 ⎠
⎝ 1
2 ⎠
b.
I BQ =
=
VTH − VBE ( on ) − ( −10 )
RTH + (1 + β ) RE
−2.45 − 0.7 + 10
⇒ I BQ = 0.136 mA
12.7 + ( 76 )( 0.5 )
I CQ = 10.2 mA, I EQ = 10.4 mA
VCEQ = 20 − I CQ RC − I EQ RE
= 20 − (10.2 )( 0.8 ) − (10.4 )( 0.5 )
VCEQ = 6.64 V
c.
R2 = 20 + 5% = 21 kΩ
R1 = 35 − 5% = 33.25 kΩ
RE = 0.5 − 5% = 0.475 kΩ
RTH = R1 & R2 = 21 & 33.25 = 12.9 kΩ
⎛ R2 ⎞
VTH = ⎜
⎟ (7) − 5
⎝ R1 + R2 ⎠
21
⎛
⎞
=⎜
⎟ ( 7 ) − 5 = −2.29 V
⎝ 21 + 33.25 ⎠
I BQ =
−2.29 − 0.7 − ( −10 )
12.9 + ( 76 )( 0.475 )
= 0.143 mA
I CQ = 10.7 mA, I EQ = 10.9 mA
For RC = 0.8 + 5% = 0.84 kΩ
VCEQ = 20 − (10.7 )( 0.84 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 5.83 V
For RC = 0.8 − 5% = 0.76 kΩ
VCEQ = 20 − (10.7 )( 0.76 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 6.69 V
R2 = 20 − 5% = 19 kΩ
R1 = 35 + 5% = 36.75 kΩ
RE = 0.5 + 5% = 0.525 kΩ
RTH = R1 & R2 = 19 & 36.75 = 12.5 kΩ
19
⎛
⎞
VTH = ⎜
⎟ ( 7 ) − 5 = −2.61 V
⎝ 19 + 36.75 ⎠
−2.61 − 0.7 − ( −10 )
= 0.128 mA
I BQ =
12.5 + ( 76 )( 0.525 )
I CQ = 9.58 mA, I EQ = 9.70 mA
For RC = 0.84 kΩ
VCEQ = 20 − ( 9.58 )( 0.84 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 6.86 V
For RC = 0.76 kΩ
VCEQ = 20 − ( 9.58 )( 0.76 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 7.63 V
So 9.58 ≤ I CQ ≤ 10.7 mA and 5.83 ≤ VCEQ ≤ 7.63 V
5.64
a.
RTH = 500 kΩ & 500 kΩ & 70 kΩ = 250 kΩ & 70 kΩ ⇒ RTH = 54.7 kΩ
5 − VTH 3 − VTH VTH − ( −5 )
+
=
500
500
70
5
3
5
1
1 ⎞
⎛ 1
+
−
= VTH ⎜
+
+ ⎟ − 0.0554 = VTH ( 0.0183)
500 500 70
⎝ 500 500 70 ⎠
VTH = −3.03 V
b.
I BQ =
=
VTH − VBE ( on ) − ( −5 )
RTH + (1 + β ) RE
−3.03 − 0.7 + 5
54.7 + (101)( 5 )
I BQ = 0.00227 mA
I CQ = 0.227 mA, I EQ = 0.229
VCEQ = 20 − ( 0.227 )( 50 ) − ( 0.229 )( 5 )
VCEQ = 7.51 V
5.65
RTH = 30 || 60 || 20 ⇒ RTH = 10 kΩ
5 − VTH 5 − VTH VTH
+
=
30
60
20
5 ⎞
1
1 ⎞
⎛ 5
⎛ 1
⎜ + ⎟ = VTH ⎜ + + ⎟
⎝ 30 60 ⎠
⎝ 30 60 20 ⎠
VTH = 2.5 V
For β = 100
I BQ =
=
VTH − VBE ( on ) − ( −5 )
RTH + (1 + β ) RE
2.5 − 0.7 + 5
10 + (101)( 0.2 )
I BQ = 0.225 mA
I CQ = 22.5 mA, I EQ = 22.7 mA
VCEQ = 15 − ( 22.5 )( 0.5 ) − ( 22.7 ) ( 0.2 )
VCEQ = −0.79! In saturation ⇒ VCEQ = 0.2 V
VTH = I BQ RTH + VBE + I EQ RE − 5
2.5 + 5 − 0.7 = I BQ (10 ) + I EQ ( 0.2 )
6.8 = I BQ (10 ) + I EQ ( 0.2 )
14.8 = I CQ ( 0.5 ) + I EQ ( 0.2 )
Transistor in saturation,
I EQ = I BQ + I CQ
6.8 = I BQ (10 ) + I BQ ( 0.2 ) + I CQ ( 0.2 )
6.8 = I BQ (10.2 ) + I CQ ( 0.2 )
51× 14.8 = I BQ ( 0.2 ) + I CQ ( 0.7 )
754.8 = I BQ (10.2 ) + I CQ ( 35.7 )
748 = I CQ ( 35.5 )
I CQ = 21.1 mA
VCEQ = 0.2 V
5.66
I CQ = 50 µ A, I BQ = 0.625 µ A, I EQ = 50.6 µ A
(a)
1
= 19.8 K
0.0506
5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) − 5
RE =
RC = 80 K
RTH = R1 & R2 Design bias stable circuit.
RTH = ( 0.1)( 51)(19.8 ) = 101 K
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10 ) − 5 = ⋅ RTH ⋅ (10 ) − 5
R1
⎝ R1 + R2 ⎠
1
So
(101)(10 ) − 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8 ) − 5
R1
1
(1010 ) = 0.0631 + 0.7 + 1
R1
R1 = 573 K
R2 = 123 K
(b)
573 R2
= 101
573 + R2
RTH = 101 K, VTH = −3.23 V
VTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ − 5
1.07 = I BQ (101 + 2395.8 ) ⇒ I BQ = 0.429 µ A
I CQ = 0.0514 mA, I EQ = 0.0519 mA
VCEQ = 10 − ( 0.0514 )( 80 ) − ( 0.0519 )(19.8 )
= 10 − 4.11 − 1.03 ⇒ VCEQ = 4.86 V
5.67
(a)
2
= 2.5 K
0.8
+ I EQ RE
I EQ = 0.80 mA, RE =
12 = I CQ RC + VCEQ
12 = ( 0.8 ) RC + 7 + 2 ⇒ RC = 3.75 K
VTH = I BQ RTH + VBE + I EQ RE
For a bias stable circuit
RTH = ( 0.1)(12.1)( 2.5 ) = 30.25 K
1
1
⋅ RTH ⋅ VCC = ( 30.25 )(12 ) = ( 0.00667 )( 30.25 ) + 0.7 + 2
R1
R1
1
( 363) = 2.90 ⇒ R1 = 125 K
R1
125 R2
= 30.25 ⇒ R2 = 39.9 K
125 + R2
(b) Let RE = 2.4 K, RC = 3.9 K
R1 = 120 K
R2 = 39 K
Then
RTH = R1 & R2 = 120 & 39 = 29.4 K
⎛ 39 ⎞
VTH = ⎜
⎟ (12 ) = 2.94 V
⎝ 120 + 39 ⎠
2.94 − 0.7
2.24
=
⇒ 7.00 µ A
I BQ =
29.4 + (121)( 2.4 ) 319.8
I CQ = 0.841 mΑ I EQ = 0.848 mA
VCEQ = 12 − ( 0.841)( 3.9 ) − ( 0.848 )( 2.4 ) = 12 − 3.28 − 2.04
VCEQ = 6.68 V
5.68
(a)
I CQ = 100 µ A, I BQ = 1.18 µ A, I EQ = 101 µ A
2
⇒ RE = 19.8 K
0.101
9 = ( 0.101)(19.8 ) + 6 + ( 0.1) RC − 9
RE =
RC = 100 K
Design a bias stable circuit.
BTH = ( 0.1)( 86 )(19.8 ) = 170 K
⎛ R2 ⎞
1
VTH = ⎜
⎟ (18 ) − 9 = (170 )(18 ) − 9
+
R
R
R
⎝ 1
2 ⎠
1
9 = ( 0.101)(19.8 ) + 0.7 + ( 0.00118 )(170 ) +
1
(170 )(18 ) − 9
R1
203R2
= 170
203 + R2
R1 = 203 K
R2 = 1046 K
(b)
β = 125
9 = (126 ) I BQ (19.8 ) + 0.7 + I BQ (170 ) + (15.07 − 9 )
2.23
= 0.837 µ A
2664.8
= 0.1054 mA
I BQ =
I EQ
I CQ = 0.1046 mA
VECQ = 18 − ( 0.1046 )(100 ) − ( 0.1054 )(19.8 )
VECQ = 5.45 V
5.69
(a)
3
⎛ 51 ⎞
I EQ ≈ ⎜ ⎟ ( 20 ) = 20.4 mA ⇒ RE =
= 0.147 K
20.4
⎝ 50 ⎠
6
VRC = 18 − 9 − 3 = 6 V
RC =
= 0.3 K
20
For a bias stable circuit.
RTH = ( 0.1)( 51)( 0.147 ) = 0.750 K
V + = I EQ RE + VEB + I BQ RTH + VTH
18 = 3 + 0.7 + ( 0.4 )( 0.75 ) +
14 =
1
( 0.75 )(18)
R1
1
(13.5) ⇒ R1 = 0.964 K
R1
( 0.964 ) R2
0.964 + R2
= 0.75 ⇒ R2 = 3.38 K
(b) Let RE = 0.15 K, RC = 0.3 K
R1 = 1.0 K, R2 = 3.3 K
⎛ 3.38 ⎞
RTH = 1/13.3 = 0.767 K, VTH = ⎜
⎟ (18 ) = 13.8 V
⎝ 1 + 3.38 ⎠
18 − 13.8 − 0.7
3.5
I BQ =
=
= 0.416 mA
0.767 + ( 51)( 0.15 ) 8.417
I CQ = 20.8 mA, I EQ = 21.2 mA
VECQ = 18 − ( 20.8 )( 0.3) − ( 21.2 )( 0.15 )
= 18 − 6.24 − 3.18
VECQ = 8.58 V
5.70
RTH = R1 & R2 = 100 & 40 = 28.6 kΩ
⎛ R2 ⎞
⎛ 40 ⎞
VTH = ⎜
⎟ (10 ) = ⎜
⎟ (10 ) = 2.86 V
⎝ 40 + 100 ⎠
⎝ R1 + R2 ⎠
VTH − VBE ( on )
2.86 − 0.7
I B1 =
=
RTH + (1 + β ) RE1 28.6 + (121) (1)
I B1 = 0.0144 mA
I C1 = 1.73 mA,
I E1 = 1.75 mA
10 − VB 2
= I C1 + I B 2
3
VB 2 − VBE ( on ) − ( −10 )
IE2 =
5
10 − VB 2
VB 2 − 0.7 + 10
= I C1 +
3
(121) ( 5 )
⎛1
⎞
10
9.3
1
− 1.73 −
= VB 2 ⎜⎜ +
⎟⎟
3
605
⎝ 3 (121) ( 5 ) ⎠
1.588 = VB 2 ( 0.335 ) ⇒ VB 2 = 4.74 V
IE2 =
I B2
4.74 − 0.7 − ( −10 )
5
= 0.0232 mA
⇒ I E 2 = 2.808 mA
I C 2 = 2.785 mA
VCEQ1 = 4.74 − (1.75 ) (1) ⇒ VCEQ1 = 2.99 V
VCEQ 2 = 10 − ( 4.74 − 0.7 ) ⇒ VCEQ 2 = 5.96 V
5.71
VE1 = −0.7
I R1 =
VE 2
−0.7 − ( −5 )
= 0.215 mA
20
= −0.7 − 0.7 = −1.4
IE2 =
−1.4 − ( −5 )
1
⇒ I E 2 = 3.6 mA
I B 2 = 0.0444 mA
I C 2 = 3.56 mA
I E1 = I R1 + I B 2 = 0.215 + 0.0444
I E1 = 0.259 mA
I B1 = 0.00320 mA
I C1 = 0.256 mA
5.72
Current through V − source = I E1 + I E 2 and I E1 = I E 2 = (1 + β ) I B1 = ( 51)( 8.26 ) µ A
So total current = 2 ( 51)( 8.26 ) µ A=843 µ A
P − = I ⋅ V − = ( 0.843)( 5 ) ⇒ P − = 4.22 mW
(From V − source)
From Example 5.19, I Q = 0.413 mA
⎛ 50 ⎞
So I C 0 = ⎜ ⎟ ( 0.413) = 0.405 mA
⎝ 51 ⎠
P + = I ⋅ V + = ( 0.405 )( 5 ) ⇒ P + = 2.03 mW
(From V + source)
5.73
RTH = R1 & R2 = 50 & 100 = 33.3 kΩ
⎛ R2 ⎞
VTH = ⎜
⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛ 100 ⎞
=⎜
⎟ (10 ) − 5 = 1.67 V
⎝ 100 + 50 ⎠
5 = I E1 RE1 + VEB ( on ) + I B1 RTH + VTH
⎛ 101 ⎞
I E1 = ⎜
⎟ ( 0.8 ) = 0.808 mA
⎝ 100 ⎠
I B1 = 0.008 mA
5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67
RE1 = 2.93 kΩ
VE1 = 5 − ( 0.808 ) ( 2.93) = 2.63 V
VC1 = VE1 − VECQ1 = 2.63 − 3.5 = −0.87 V
VE 2 = −0.87 − 0.70 = −1.57 V
IE2 =
−1.57 − ( −5 )
RE 2
= 0.808 ⇒ RE 2 = 4.25 kΩ
VCEQ 2 = 4 ⇒ VC 2 = −1.57 + 4 = 2.43 V
5 − 2.43
RC 2 =
⇒ RC 2 = 3.21 kΩ
0.8
I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA
RC1 =
−0.87 − ( −5 )
0.792
⇒ RC1 = 5.21 kΩ
Chapter 6
Exercise Problems
EX6.1
(a)
I BQ =
VBB − VBE ( on )
RB
=
2 − 0.7
⇒ 2 µA
650
Then
I CQ = β I BQ = (100 )( 2 µ A ) = 0.20 mA
VCEQ = VCC − I CQ RC = 5 − ( 0.2 )(15 ) = 2 V
(b)
Now
I CQ
0.20
gm =
=
= 7.69 mA / V
VT
0.026
and
β VT (100 )( 0.026 )
rπ =
=
= 13 k Ω
0.20
I CQ
(c)
We find
⎛ r
⎞
Av = − g m RC ⎜ π ⎟
⎝ rπ + RB ⎠
⎛ 13 ⎞
= − ( 7.69 )(15 ) ⎜
⎟ = −2.26
⎝ 13 + 650 ⎠
EX6.2
V − VBE ( on ) 0.92 − 0.7
I BQ = BB
=
100
RB
or I BQ = 0.0022 mA
and I CQ = β I BQ = (150 )( 0.0022 ) = 0.33 mA
(a)
gm =
rπ =
ro =
I CQ
VT
β VT
I CQ
=
0.33
= 12.7 mA / V
0.026
=
(150 )( 0.026 )
0.33
= 11.8 k Ω
VA
200
=
= 606 k Ω
I CQ 0.33
(b)
⎛ r
⎞
vo = − g m vπ ( ro & RC ) and vπ = ⎜ π ⎟ ⋅ vs
⎝ rπ + RB ⎠
so
⎛ rπ ⎞
v
Av = o = − g m ⎜
⎟ ( ro & RC )
vπ
⎝ rπ + RB ⎠
⎛ 11.8 ⎞
= − (12.7 ) ⎜
⎟ ( 606 & 15 )
⎝ 11.8 + 100 ⎠
which yields Av = −19.6
EX6.3
(a)
V − VEB ( on ) 1.145 − 0.7
I BQ = BB
=
50
RB
or I BQ = 0.0089 mA
Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA
Now
gm =
rπ =
ro =
I CQ
VT
β VT
I CQ
=
0.801
= 30.8 mA / V
0.026
=
( 90 )( 0.026 )
0.801
= 2.92 k Ω
VA
120
=
= 150 k Ω
I CQ 0.801
We have Vo = g mVπ ( ro & RC )
(b)
⎛ r
⎞
and Vπ = − ⎜ π ⎟ Vs
⎝ rπ + RB ⎠
so
⎛ rπ ⎞
V
Av = o = − g m ⎜
⎟ ( ro & RC )
Vs
⎝ rπ + RB ⎠
⎛ 2.92 ⎞
= − ( 30.8 ) ⎜
⎟ (150 & 2.5 )
⎝ 2.92 + 50 ⎠
which yields Av = −4.18
EX6.4
Using Figure 6.23
(a)
For I CQ = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10−4 < hre < 50 × 10−4 ,
5 < hoe < 13 µ mhos
For I CQ = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10−4 < hre < 1.6 ×10−4 ,
(b)
22 < hoe < 35 µ mhos
EX6.5
RTH = R1 & R2 = 35.2 || 5.83 = 5 k Ω
⎛ R2 ⎞
⎛ 5.83 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (5)
R
R
+
⎝ 5.83 + 35.2 ⎠
2 ⎠
⎝ 1
or
VTH = 0.7105 V
Then
V − VBE ( on ) 0.7105 − 0.7
I BQ = TH
=
RTH
5
or
I BQ = 2.1 µ A
and
I CQ = β I BQ = (100 )( 2.1 µ A ) = 0.21 mA
VCEQ = VCC − I CQ RC = 5 − ( 0.21)(10 )
and
VCEQ = 2.9 V
Now
gm =
ro =
I CQ
VT
=
0.21
= 8.08 mA
0.026
VA
100
=
= 476 k Ω
I CQ 0.21
And
Av = − g m ( ro & RC ) = − ( 8.08 ) ( 476 & 10 )
so
Av = −79.1
EX6.6
RTH = R1 & R2 = 250 & 75 = 57.7 k Ω
⎛ R2 ⎞
⎛ 75 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (5)
⎝ 75 + 250 ⎠
⎝ R1 + R2 ⎠
or
VTH = 1.154 V
I BQ =
VTH − VBE ( on )
RTH + (1 + β ) RE
=
1.154 − 0.7
57.7 + (121)( 0.6 )
or
I BQ = 3.48 µ A
I CQ = β I BQ = (120 )( 3.38 µ A ) = 0.418 mA
(a)
Now
gm =
rπ =
I CQ
VT
=
0.418
= 16.08 mA / V
0.026
βVT (120 )( 0.026 )
=
= 7.46 k Ω
0.418
I CQ
We have
Vo = − g mVπ RC
We find
Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 )
or
Rib = 80.1 k Ω
Also
R1 & R2 = 250 & 75 = 57.7 k Ω
R1 & R2 & Rib = 57.7 & 80.1 = 33.54 k Ω
We find
⎛ R1 & R2 & Rib ⎞
⎛ 33.54 ⎞
Vs′ = ⎜
⎟ ⋅ Vs = ⎜
⎟ ⋅ Vs
⎝ 33.54 + 0.5 ⎠
⎝ R1 & R2 & Rib + RS ⎠
or
Vs′ = ( 0.985 ) Vs
Now
⎡ ⎛1+ β ⎞ ⎤
⎡ ⎛ 121 ⎞
⎤
Vs′ = Vπ ⎢1 + ⎜
⎟ RE ⎥ = Vπ ⎢1 + ⎜
⎟ ( 0.6 ) ⎥
r
7.46
⎝
⎠
⎢⎣ ⎝ π ⎠ ⎥⎦
⎣
⎦
or
Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 ) Vs
So
Av =
Vo
= − (16.08 )( 0.0932 )( 0.985 )( 5.6 )
Vs
or
Av = −8.27
EX6.7
RTH = R1 & R2 = 15 & 85 = 12.75 k Ω
⎛ R2 ⎞
⎛ 85 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (12 )
R
R
+
⎝ 15 + 85 ⎠
2 ⎠
⎝ 1
or
VTH = 10.2 V
Now
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
so
12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2
which yields
I BQ = 0.0174 mA
and
I CQ = β I BQ = (100 )( 0.0174 ) = 1.74 mA
I EQ = (101)( 0.0174 ) = 1.76 mA
VECQ = VCC − I EQ RE − I CQ RC
= 12 − (1.76 )( 0.5 ) − (1.74 )( 4 )
or
VECQ = 4.16 V
Now
rπ =
β VT
I CQ
=
(100 )( 0.026 )
1.74
= 1.49 k Ω
We have
Vo = g mVπ ( RC || RL )
⎛
V ⎞
Vs = −Vπ − ⎜ g mVπ + π ⎟ RE
rπ ⎠
⎝
Solving for Vπ and noting that β = g m rπ , we find
Av =
Vo
Vs
=
=
− β ( RC & RL )
rπ + (1 + β ) RE
− (100 )( 4 & 2 )
1.49 + (101)( 0.5 )
or Av = −2.56
EX6.8
Dc analysis
10 − 0.7
I BQ =
= 0.00439 mA
100 + (101)( 20 )
I CQ = 0.439 mA, I EQ = 0.443 mA
Now
(100 )( 0.026 )
= 5.92 k Ω
0.439
0.439
gm =
= 16.88 mA / V
0.026
100
ro =
= 228 k Ω
0.439
(a)
Now
Vo = − g mVπ ( ro & RC )
rπ =
⎛ RB & rπ ⎞
Vπ = ⎜
⎟ ⋅ Vs
⎝ RB & rπ + RS ⎠
RB & rπ = 100 & 5.92 = 5.59 k Ω
⎛ 5.59 ⎞
Then Vπ = ⎜
⎟ ⋅ Vs = ( 0.918 ) Vs
⎝ 5.59 + 0.5 ⎠
V
Then Av = o = − (16.88 ) ( 0.918 ) ( 228 & 10 )
Vs
or Av = −148
(b)
Rin = RS + RB & rπ = 0.5 + 100 & 5.92 = 6.09 k Ω
and
Ro = RC & ro = 10 & 228 = 9.58 k Ω
EX6.9
(a)
I CQ
0.25
gm =
=
= 9.615 mA / V
0.026
VT
ro =
VA
100
=
= 400 k Ω
I CQ 0.25
Av = − g m ( ro & rc ) = − ( 9.615 )( 400 & 100 ) = −769
(b)
Av = − g m ( ro & rc & rL ) = − ( 9.615 )( 400 & 100 & 100 )
Av = −427
EX6.10
5 − 0.7
I BQ =
= 0.00672 mA
10 + (126 )( 5 )
I CQ = 0.84 mA, I EQ = 0.847 mA
VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 )
or
VCEQ = 3.83 V
dc load line
VCE ≅ (V + − V − ) − I C ( RC + RE )
or
VCE = 10 − I C ( 7.3)
ac load line (neglecting ro)
vce = −ic ( RC & RL ) = −ic ( 2.3 & 5 ) = −ic (1.58 )
IC (mA)
1.37
0.84
AC load line
Q-point
DC load line
3.83
10 V
CE (V)
EX6.11
(a)
dc load line
VEC ≅ VCC − I C ( RC + RE ) = 12 − I C ( 4.5 )
ac load line
vec ≅ −ic ( RE + RC & RL ) or vec = −ic ( 0.5 + 4 & 2 ) = −ic (1.83)
Q-point values
12 − 0.7 − 10.2
= 0.0150 mA
I BQ =
12.75 + (121)( 0.5 )
I CQ = 1.80 mA, I EQ = 1.82 mA
VECQ = 3.89 V
IC (mA)
AC load line
2.67
Q-point
1.80
DC load line
3.89
(b)
12 V (V)
EC
ΔiC = I CQ = 1.80 mA
ΔvEC = (1.8 )(1.83) = 3.29 V
vEC ( min ) = 3.89 − 3.29 = 0.6 V
So maximum symmetrical swing = 2 × 3.29 = 6.58 V peak-to-peak
EX6.12
⎛ R2 ⎞
1
VTH = ⎜
(a)
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
R1
⎝ R1 + R2 ⎠
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1)
so
RTH = 12.1 k Ω, VTH =
1
(12.1)(12 )
R1
We can write
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
We have
I CQ = 1.6 mA, I BQ =
1.6
= 0.0133 mA
120
Then
1
(12.1)(12 )
R1
12.1 + (121)(1)
12 − 0.7 −
I BQ = 0.0133 =
which yields
R1 = 15.24 k Ω
Since RTH = R1 & R2 = 12.1 k Ω, we find R2 = 58.7 k Ω
Also VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V
(b)
ac load line vec = −ic ( RC & RL )
Want Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA
Also Δvec = 3.99 − 0.5 = 3.49 V
Now
Δvec 3.49
=
= 2.327 k Ω = RC & RL
Δic
1.5
So 4 & RL = 2.327 k Ω which yields RL = 5.56 k Ω
EX6.13
RTH = R1 & R2 = 25 & 50 = 16.7 k Ω
⎛ R2 ⎞
⎛ 50 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (5)
⎝ 25 + 50 ⎠
⎝ R1 + R2 ⎠
or
VTH = 3.33 V
I BQ =
VTH − VBE ( on )
RTH + (1 + β ) RE
=
3.33 − 0.7
16.7 + (121)(1)
or
I BQ = 0.0191 mA
Also
I CQ = (120 )( 0.0191) = 2.29 mA
Now
gm =
rπ =
ro =
I CQ
VT
=
β VT
I CQ
2.29
= 88.1 mA / V
0.026
=
(120 )( 0.026 )
2.29
= 1.36 k Ω
VA
100
=
= 43.7 k Ω
I CQ 2.29
(a)
⎛ R1 & R2 & Rib ⎞
Vs′ = ⎜
⎟ ⋅ Vs
⎝ R1 & R2 & Rib + RS ⎠
Rib = rπ + (1 + β ) ( RE & ro ) = 1.36 + (121)(1 & 43.7 )
or
Rib = 120 k Ω and R1 & R2 = 16.7 k Ω
Then
R1 & R2 & Rib = 16.7 & 120 = 14.7 k Ω
Now
⎛ 14.7 ⎞
Vs′ = ⎜
⎟ ⋅ Vs = ( 0.967 ) Vs
⎝ 14.7 + 0.5 ⎠
and
⎛V
⎞
⎛ 1+ β
Vo = ⎜ π + g mVπ ⎟ ( RE & ro ) = Vπ ⎜
⎝ rπ
⎠
⎝ rπ
We have
Vs′ = Vπ + Vo
then
⎞
⎟ RE & ro
⎠
Vs′
( 0.967 )Vs
=
⎛ 1+ β ⎞
⎛1+ β ⎞
1+ ⎜
⎟ RE & ro 1 + ⎜
⎟ RE & ro
r
⎝ π ⎠
⎝ rπ ⎠
We then obtain
Vπ =
Av =
=
Vo
=
Vs
⎛1+ β
⎝ rπ
( 0.967 ) ⎜
⎞
⎟ RE & ro
⎠
⎛1+ β ⎞
1+ ⎜
⎟ RE & ro
⎝ rπ ⎠
( 0.967 )(1 + β ) RE & ro
rπ + (1 + β ) RE & ro
Now
RE & ro = 1 & 43.7 = 0.978 k Ω
Then
Av =
( 0.967 )(121)( 0.978 )
= 0.956
1.36 + (121)( 0.978 )
(b)
Rib = rπ + (1 + β )( RE & ro )
or
Rib = 1.36 + (121)( 0.978 ) = 120 k Ω
EX6.14
For RS = 0, then
r
Ro = RE & ro π
1+ β
Using the parameters from Exercise 4.13, we obtain
1.36
Ro = 1 43.7
121
or
Ro = 11.1 Ω
EX6.15
(a)
For I CQ = 1.25 mA and β = 100, we find
I EQ = 1.26 mA and I BQ = 0.0125 mA
Now VCEQ = 10 − I EQ RE
Or
4 = 10 − (1.26 ) RE
which yields
RE = 4.76 k Ω
Then
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 )
or
RTH = 48.1 k Ω
We have
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5
R1
⎝ R1 + R2 ⎠
or
VTH =
1
( 481) − 5
R1
We can write I BQ =
VTH − 0.7 − ( −5 )
RTH + (1 + β ) RE
Or
1
( 481) − 5 − 0.7 + 5
R1
0.0125 =
48.1 + (101)( 4.76 )
which yields
R1 = 65.8 k Ω
Since R1 & R2 = 48.1 k Ω, we obtain
R2 = 178.8 k Ω
(b)
rπ =
ro =
β VT
I CQ
=
(100 )( 0.026 )
1.25
= 2.08 k Ω
VA
125
=
= 100 k Ω
I CQ 1.25
We may note that
g mVπ = g m ( I b rπ ) = β I b
Also
Rib = rπ + (1 + β )( RE & RL & ro )
= 2.08 + (101) ( 4.76 1 100 )
or
Rib = 84.9 k Ω
Now
⎛ RE ro ⎞
Io = ⎜
1 + β ) Ib
⎜ R r + R ⎟⎟ (
L ⎠
⎝ E o
where
⎛ R1 R2 ⎞
Ib = ⎜
⋅I
⎜ R R + R ⎟⎟ s
ib ⎠
⎝ 1 2
We can then write
⎛ RE ro ⎞
⎛ R1 R2
I
AI = o = ⎜
1+ β )⎜
(
⎟
⎜R R +R
I s ⎜⎝ RE ro + RL ⎟⎠
ib
⎝ 1 2
We have
RE ro = 4.76 100 = 4.54 k Ω
so
⎞
⎟⎟
⎠
48.1 ⎞
⎛ 4.54 ⎞
⎛
AI = ⎜
⎟ (101) ⎜
⎟
4.54
1
48.1
+
+ 84.9 ⎠
⎝
⎠
⎝
or
AI = 29.9
r
2.08
(c)
Ro = RE & ro π = 4.76 100
1+ β
101
or Ro = 20.5 Ω
EX6.16
(a)
RTH = R1 R2 = 70 6 = 5.53 k Ω
⎛ R2 ⎞
⎛ 6 ⎞
VTH = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5
⎝ 70 + 6 ⎠
⎝ R1 + R2 ⎠
or
VTH = −4.2105 V
We find
−4.2105 − 0.7 − ( −5 )
I BQ1 =
⇒ 2.91 µ A
5.53 + (126 )( 0.2 )
and
I CQ1 = β I BQ1 = (125 )( 2.91 µ A ) = 0.364 mA
I EQ1 = (1 + β ) I BQ1 = 0.368 mA
At the collector of Q1,
V − 0.7 − ( −5 )
5 − VC1
= I CQ1 + C1
RC1
(1 + β )( RE 2 )
or
V − 0.7 − ( −5 )
5 − VC1
= 0.364 + C1
5
(126 )(1.5)
which yields
VC1 = 2.99 V
also
VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5
or
VE1 = −4.93 V
Then
VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V
We find
V − 0.7 − ( −5 )
I EQ 2 = C1
= 4.86 mA
1.5
and
⎛ β ⎞
⎛ 125 ⎞
I CQ 2 = ⎜
⎟ ⋅ I EQ1 = ⎜
⎟ ( 4.86 ) = 4.82 mA
+
β
1
⎝ 126 ⎠
⎝
⎠
We find
VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V
and
VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V
(b)
The small-signal transistor parameters are:
rπ 1 =
β VT
I CQ1
g m1 =
rπ 2 =
gm2 =
=
I CQ1
VT
β VT
I CQ 2
I CQ 2
VT
(125 )( 0.026 )
0.364
0.364
= 14.0 mA / V
0.026
=
=
= 8.93 k Ω
(125 )( 0.026 )
=
4.82
= 0.674 k Ω
4.82
= 185 mA / V
0.026
We find Rib1 = rπ 1 + (1 + β ) RE1 = 8.93 + (126 )( 0.2 )
Or
Rib1 = 34.1 k Ω
and
Rib 2 = rπ 2 + (1 + β )( RE 2 & RL )
= 0.674 + (126 )(1.5 & 10 ) = 165 k Ω
The small-signal equivalent circuit is:
⫹
⫹
V␲1
r␲1
gm1V␲1
V␲2
⫹
⫺
R1兩兩R2
RC1
RE1
We can write
Vo = (1 + β ) I b 2 ( RE 2 & RL )
where
⎛ RC1 ⎞
Ib2 = ⎜
⎟ ( − g m1Vπ 1 )
⎝ RC1 + Rib 2 ⎠
V
Vπ 1 = s ⋅ rπ 1
Rib1
Then
V
Av = o
Vs
so
⎛ RC1 ⎞ ⎛ − g m1rπ 1 ⎞
= (1 + β )( RE 2 & RL ) ⎜
⎟⎜
⎟
⎝ RC1 + Rib 2 ⎠ ⎝ Rib1 ⎠
⎛ 5 ⎞ ⎛ 125 ⎞
Av = − (126 )(1.5 & 10 ) ⎜
⎟⎜
⎟
⎝ 5 + 165 ⎠ ⎝ 34.1 ⎠
or
Av = −17.7
(c)
Ri = R1 & R2 & Rib1 = 70 & 6 & 34.1 = 4.76 k Ω
⎛ r + RC1 ⎞
⎛ 0.676 + 5 ⎞
and Ro = RE 2 ⎜ π 2
⎟ = 1.5 ⎜
⎟
⎝ 126 ⎠
⎝ 1+ β ⎠
or
Ro = 43.7 Ω
Test Your Understanding Exercises
TYU6.1
I CQ
0.25
=
= 9.62 mA / V
gm =
0.026
VT
rπ =
ro =
β VT
I CQ
=
(120 )( 0.026 )
0.25
= 12.5 k Ω
VA
150
=
= 600 k Ω
I CQ 0.25
TYU6.2
V
V
75
ro = A ⇒ I CQ = A =
I CQ
ro 200 k Ω
or
ICQ = 0.375 mA
gm2V␲2
⫺
⫺
Vs
r␲2
Vo
RE2
RL
TYU6.3
RC
RE
Resulting gain is always smaller than this value. The effect of RS is very small.
Set
RC
= 10
RE
Now
5 ≅ I C ( RC + RE ) + VCEQ
As a first approximation, Av ≅ −
or
5 = ( 0.5 )( RC + RE ) + 2.5
which yields
RC + RE = 5 k Ω
So
10 RE + RE = 5
or
RE = 0.454 k Ω and RC = 4.54 k Ω
We have
I CQ 0.5
I BQ =
=
= 0.005 mA
β 100
and
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 )
or
RTH = 4.59 k Ω
also
⎛ R2 ⎞
1
VTH = ⎜
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
⎝ 1
2 ⎠
1
or
1
23
( 4.59 )( 5 ) =
R1
R1
We can write
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
VTH =
or
23
= ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 )
R1
which yields
R1 = 24.1 k Ω
and since R1 & R2 = 4.59 k Ω
we find R2 = 5.67 k Ω
TYU6.4
dc analysis
RTH = R1 & R2 = 15 & 85 = 12.75 k Ω
⎛ R2 ⎞
⎛ 85 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (12 )
⎝ 15 + 85 ⎠
⎝ R1 + R2 ⎠
or
VTH = 10.2 V
We can write
12 − 0.7 − VTH
12 − 0.7 − 10.2
=
I BQ =
RTH + (1 + β ) RE 12.75 + (101)( 0.5 )
or
IBQ = 0.0174 mA
and
I CQ = β I BQ = 1.74 mA
ac analysis
Vo = h fe I b ( RC & RL )
and
Ib =
−Vs
hie + (1 + h fe ) RE
so
Av =
−h fe ( RC & RL )
Vo
=
Vs hie + (1 + h fe ) RE
For ICQ = 1.74 mA, we find
h fe ( max ) = 110, h fe ( min ) = 70
hie ( max ) = 2 k Ω, hie ( min ) = 1.1 k Ω
We obtain
−110 ( 4 & 2 )
= −2.59
Av ( max ) =
1.1 + (111)( 0.5 )
and
Av ( min ) =
−70 ( 4 & 2 )
2 + ( 71)( 0.5 )
= −2.49
TYU6.5
First approximation, Av ≅ −
RC
RE
This predicts a low value, so set
RC
=9
RE
Now
VCC ≅ I CQ ( RC + RE ) + VECQ
or
7.5 = ( 0.6 )( 9 RE + RE ) + 3.75
which yields
RE = 0.625 k Ω and RC = 5.62 k Ω
We have
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 )
or
RTH = 6.31 k Ω
Also
1
1
VTH = ⋅ RTH ⋅ VCC = ( 6.31)( 7.5 )
R1
R1
We have
I CQ 0.6
I BQ =
=
= 0.006 mA
β 100
The KVL equation around the E-B loop gives
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
or
7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) +
1
( 6.31)( 7.5)
R1
which yields
R1 = 7.41 k Ω
Since RTH = R1 & R2 = 6.31 k Ω, then R2 = 42.5 k Ω
TYU6.6
⎛R ⎞
− β RC
= − ( 0.95 ) ⎜ C ⎟
rπ + (1 + β ) RE
⎝ RE ⎠
⎛ 2 ⎞
or Av = − ( 0.95 ) ⎜
⎟ = −4.75
⎝ 0.4 ⎠
Assume rπ = 1.2 k Ω from Example 4.6.
Then
−β ( 2)
= −4.75
1.2 + (1 + β )( 0.4 )
We have Av =
or
β = 76
TYU6.7
dc analysis: By symmetry, VTH = 0
RTH = R1 & R2 = 20 & 20 = 10 k Ω
We can write
0 − 0.7 − ( −5 )
I BQ =
= 0.00672 mA
10 + (126 )( 5 )
I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA
Small-signal transistor parameters:
β VT (125 )( 0.026 )
=
= 3.87 k Ω
rπ =
0.84
I CQ
gm =
ro =
I CQ
VT
=
0.84
= 32.3 mA / V
0.026
VA
200
=
= 238 k Ω
I CQ 0.84
(a)
We have
Vo = − g mVπ ( ro & RC & RL ) and Vπ = Vs
so
Av =
Vo
Vs
= − g m ( ro & RC & RL )
= − ( 32.3)( 238 & 2.3 & 5 )
or
Av = −50.5
(b)
Ro = ro & RC = 238 & 2.3 = 2.28 k Ω
TYU6.8
We find I CQ = 0.418 mA, then
⎛ 121 ⎞
VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜
⎟ ( 0.418 )( 0.6 )
⎝ 120 ⎠
or
VCEQ = 2.41 V
So
ΔvCE = ( 2.41 − 0.5 ) × 2
or
ΔvCE = 3.82 V peak-to-peak
TYU6.9
dc load line
VCE ≅ (10 + 10 ) − I CQ ( RC + RE )
or
VCE = 20 − I C (10 + RE )
ac load line
vce = −ic RC = −ic (10 )
Now
ΔvCE = VCEQ − 0.7 = ΔiC (10 ) = I CQ (10 )
so
VCEQ − 0.7 = I CQ (10 )
We have that
VCEQ = 20 − I CQ (10 + RE )
then
I CQ (10 ) + 0.7 = 20 − I CQ (10 + RE )
or
I CQ ( 20 + RE ) = 19.3
(1)
The base current is found from
10 − 0.7
I BQ =
100 + (101) RE
so
I CQ =
(100 )( 9.3)
100 + (101) RE
Substituting into Equation (1),
⎡ (100 )( 9.3) ⎤
⎢
⎥ ( 20 + RE ) = 19.3
⎢⎣100 + (101) RE ⎥⎦
which yields
RE = 16.35 k Ω
Then
(100 )( 9.3)
I CQ =
= 0.531 mA
100 + (101)(16.35 )
and
VCEQ ≅ 20 − ( 0.531)(10 + 16.35 ) = 6.0 V
Now
ΔvCE = VCEQ − 0.7 = 6 − 0.7 = 5.3 V
or, maximum symmetrical swing = 2 × 5.3 = 10.6 V peak-to-peak
TYU6.10
We can write
0 − 0.7 − ( −10 )
⇒ 6.60 µ A
I BQ =
100 + (131)(10 )
and
I CQ = (130 )( 6.60 µ A ) = 0.857 mA
Assume nominal small-signal parameters of:
hie = 4 k Ω, h fe = 134
hre = 0, hoe = 12 µ S ⇒
1
= 83.3 k Ω
hoe
We find
⎛
1 ⎞
Rib = hie + (1 + h fe ) ⎜ RE RL
⎟
hoe ⎠
⎝
= 4 + (135 )(10 ||10 || 83.3) = 641 k Ω
To find the voltage gain:
RB & Rib
100 & 641
⋅ Vs =
⋅ Vs
RB & Rib + RS
100 & 641 + 10
Vs′ =
or
Vs′ = ( 0.896 ) Vs
Also
(1 + h fe ) R′
Vo
=
Vs′ hie + (1 + h fe ) R′
where
R ′ = RE & RL
1
= 10 & 10 & 83.3 = 4.72 k Ω
hoe
Then
Av =
Vo ( 0.896 )(135 )( 4.72 )
=
= 0.891
4 + (135 )( 4.72 )
Vs
To find the current gain:
⎛
⎞
1
⎜ RE
⎟
hoe ⎟
⎛ RB ⎞
Io ⎜
Ai = =
1 + h fe ) ⎜
(
⎟
⎟
Ii ⎜
1
⎝ RB + Rib ⎠
+ RL ⎟⎟
⎜⎜ RE
hoe
⎝
⎠
⎛ 10 & 83.3 ⎞
⎛ 100 ⎞
=⎜
⎟ (135 ) ⎜ 100 + 641 ⎟
&
+
10
83.3
10
⎝
⎠
⎝
⎠
or
Ai = 8.59
To find the output resistance:
1 hie + RS RB
Ro = RE
1 + h fe
hoe
= 10 83.3
4 + 10 & 100
⇒ 96.0 Ω
135
TYU6.11
We find
RTH = R1 & R2 = 50 & 50 = 25 k Ω
⎛ R2 ⎞
⎛1⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜ ⎟ ( 5 ) = 2.5 V
⎝ 2⎠
⎝ R1 + R2 ⎠
Now
V − VEB ( on ) − VTH
I BQ = CC
RTH + (1 + β ) RE
=
5 − 0.7 − 2.5
= 0.00793 mA
25 + (101)( 2 )
and
I CQ = (100 )( 0.00793) = 0.793 mA
The small-signal transistor parameters:
I CQ 0.793
=
= 30.5 mA / V
gm =
0.026
VT
rπ =
ro =
(a)
β VT
I CQ
=
(100 )( 0.026 )
0.793
= 3.28 k Ω
VA
125
=
= 158 k Ω
I CQ 0.793
Define R ′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω
Now
Av =
(1 + β ) R′
(101)( 0.4 )
=
rπ + (1 + β ) R ′ 3.28 + (101)( 0.4 )
or
Av = 0.925
(b)
Rib = rπ + (1 + β ) R ′ = 3.28 + (101)( 0.4 )
or
Rib = 43.7 k Ω
Also Ro = RE & ro
rπ
3.28
= 2 & 158
1+ β
101
or Ro = 32.0 Ω
TYU6.12
For VECQ = 2.5, I EQ =
VCC − VECQ
RE
=
5 − 2.5
= 5 mA
0.5
then
⎛ 75 ⎞
I CQ = ⎜ ⎟ ( 5 ) = 4.93 mA
⎝ 76 ⎠
5
I BQ =
= 0.0658 mA
76
Small-signal transistor parmaters:
β VT ( 75 )( 0.026 )
=
= 0.396 k Ω
rπ =
I CQ
4.93
ro =
VA
75
=
= 15.2 k Ω
I CQ 4.93
Define the small-signal base current into the base, then
g mVπ = − β I b
Now,
⎛ RE & ro ⎞
Io = ⎜
⎟ (1 + β ) I b
⎝ RE & ro + RL ⎠
and
⎛ R1 & R2 ⎞
Ib = ⎜
⎟ ⋅ Ii
⎝ R1 & R2 + Rib ⎠
The current gain is
⎛ RE & ro ⎞
⎛ R1 & R2 ⎞
I
AI = o = ⎜
⎟ (1 + β ) ⎜
⎟
I i ⎝ RE & ro + RL ⎠
⎝ R1 & R2 + Rib ⎠
We have
RE = RL = 0.5 k Ω
Rib = rπ + (1 + β )( RE & RL & ro )
= 0.396 + ( 76 )( 0.5 & 0.5 & 15.2 ) = 19.1 k Ω
and
RE & ro = 0.5 & 15.2 = 0.484 k Ω
Then
⎛ R1 & R2
⎞
⎛ 0.484 ⎞
AI = 10 = ⎜
⎟
⎟ ( 76 ) ⎜
⎝ 0.484 + 0.5 ⎠
⎝ R1 & R2 + 19.1 ⎠
which yields
R1 & R2 = 6.975 k Ω
Now
⎛ R2 ⎞
1
VTH = ⎜
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
⎝ 1
2 ⎠
1
or
1
VTH = ( 6.975 )( 5 )
R1
We can write
V − VEB ( on ) − VTH
I BQ = CC
RTH + (1 + β ) RE
or
0.0658 =
5 − 0.7 − VTH
6.975 + ( 76 )( 0.5 )
which yields
VTH = 1.34 =
1
( 6.975)( 5 )
R1
or
R1 = 26.0 k Ω and R2 = 9.53 k Ω
TYU6.13
(a)
dc analysis:
V − VEB ( on ) 10 − 0.7
I EQ = EE
=
= 0.93 mA
10
RE
⎛ β ⎞
⎛ 100 ⎞
I CQ = ⎜
⎟ I EQ = ⎜
⎟ ( 0.93) = 0.921 mA
⎝ 101 ⎠
⎝ 1+ β ⎠
VECQ = VEE − I EQ RE − I CQ RC − ( −VCC )
= 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 )
or
VECQ = 6.1 V
(b)
Small-signal transistor parameters:
β VT (100 )( 0.026 )
=
= 2.82 k Ω
rπ =
I CQ
0.921
0.921
= 35.42 mA / V
0.026
VT
Small-signal current gain:
I o = g mVπ and Vπ = Vs
also
⎛ 1
⎞
Vs
+ g mVπ = Vs ⎜⎜
+ g m ⎟⎟
Ii =
RE rπ
⎝ RE rπ
⎠
Then
g m ( RE rπ )
I
g mVπ
AI = o =
=
Ii
⎛ 1
⎞ 1 + g m ( RE rπ )
Vπ ⎜
+ gm ⎟
⎝ RE & rπ
⎠
gm =
=
I CQ
=
( 35.42 ) (10 2.82 )
1 + ( 35.42 ) (10 2.82 )
or
AI = 0.987
(c)
Small-signal voltage gain:
Vo = g mVπ RC = g mVs RC
or
V
Av = o = g m RC = ( 35.42 )( 5 )
Vs
or
Av = 177
TYU6.14
(a)
V − VBE ( on )
10 − 0.7
=
I BQ = EE
RB + (1 + β ) RE 100 + (101)(10 )
or
I BQ = 8.38 µ A and I CQ = 0.838 mA
rπ =
gm =
ro =
β VT
=
I CQ
I CQ
VT
=
(100 )( 0.026 )
0.838
= 3.10 k Ω
0.838
= 32.23 mA / V
0.026
VA
∞
=
=∞
I CQ 0.838
(b)
Summing currents, we have
⎛ −V ⎞ ⎛ −V − Vs ⎞
V
g mVπ + π = ⎜ π ⎟ + ⎜ π
⎟
rπ ⎝ RE ⎠ ⎝ RS ⎠
or
⎡⎛ 1 + β ⎞ 1
V
1 ⎤
Vπ ⎢⎜
+ ⎥=− s
⎟+
r
R
R
RS
⎥
E
S ⎦
⎣⎢⎝ π ⎠
We can then write
⎤
V ⎡⎛ r ⎞
Vπ = − s ⎢⎜ π ⎟ RE RS ⎥
RS ⎣⎝ 1 + β ⎠
⎦
Now
Vo = − g mVπ ( RC & RL )
So
( RC RL ) ⎡⎛ rπ ⎞ R R ⎤
Vo
= gm
⎢⎜
⎟ E S⎥
Vs
RS
⎣⎝ 1 + β ⎠
⎦
32.23
10
1
(
) ( ) ⎡ 3.10
⎤
=
10 1⎥
⎢
1
101
()
⎣
⎦
Av =
or
Av = 0.870
Now
⎛ RE
Ie = ⎜
⎝ RE + rπ
⎛ β
Ic = ⎜
⎝ 1+ β
⎞
⎛ 10 ⎞
⎟ ⋅ Ii = ⎜
⎟ ⋅ I i = ( 0.763) I i
⎝ 10 + 3.10 ⎠
⎠
⎞
⎛ 100 ⎞
⎟ ⋅ Ie = ⎜
⎟ ⋅ Ie
⎝ 101 ⎠
⎠
⎛ RC ⎞
⎛ 10 ⎞
Io = ⎜
⎟ ⋅ Ic = ⎜
⎟ ⋅ I c = ( 0.909 ) I c
+
R
R
⎝ 10 + 1 ⎠
L ⎠
⎝ C
So we have
I
⎛ 100 ⎞
AI = o = ( 0.909 ) ⎜
⎟ ( 0.763)
Ii
⎝ 101 ⎠
or
AI = 0.687
(c)
We have
r
3.10
Ri = RE π = 10
1+ β
101
or
Ri = 30.6 Ω
Also
Ro = RC = 10 k Ω
TYU6.15
dc analysis:
We can write 5 = I BQ RB + VBE ( on ) + I EQ RE
or
I BQ =
5 − 0.7
4.3
=
RB + (101) RE RB + (101) RE
I CQ =
(100 )( 4.3)
RB + (101) RE
Also
5 = I CQ RC + VCEQ + I EQ RE − 5
or
⎡
⎛ 101 ⎞ ⎤
VCEQ = 10 − I CQ ⎢ RC + ⎜
⎟ RE ⎥
⎝ 100 ⎠ ⎦
⎣
ac analysis:
Vo = − g mVπ ( RC RL )
and
Vs = −Vπ −
Vπ
⋅ RB = −Vπ
rπ
or
⎛ RB ⎞
⎜1 +
⎟
rπ ⎠
⎝
⎛ r
⎞
Vπ = − ⎜ π ⎟ ⋅ Vs
+
r
R
B ⎠
⎝ π
Then
V
β
Av = o =
( RC RL )
Vs rπ + RB
where β = g m rπ
For I CQ = 1 mA, rπ =
β VT
I CQ
Now
Av = 20 =
(100 ) ( 2 2 )
2.6 + RB
which yields
RB = 2.4 k Ω
Also
(100 )( 4.3)
I CQ = 1 =
2.4 + (101) RE
which yields
RE = 4.23 k Ω
TYU6.16
(a)
=
(100 )( 0.026 )
1
= 2.6 k Ω
dc analysis:
For I EQ 2 = 1 mA,
⎛ 100 ⎞
I CQ 2 = ⎜
⎟ (1) = 0.990 mA
⎝ 101 ⎠
I EQ 2
1
I EQ1 =
=
= 0.0099 mA
1 + β 101
and
I EQ1 0.0099
I BQ1 =
=
= 0.000098 mA
1+ β
101
and
I CQ1 = (100 )( 0.000098 ) = 0.0098 mA
and
VB1 = − I BQ1 RB = − ( 0.000098 )(10 )
or
VB1 = −0.00098 ≅ 0
so
VE1 = −0.7 V
and
VE 2 = −1.4 V
Now
I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA
then
VO = 5 − (1)( 4 ) = 1 V
and
VCEQ 2 = 1 − ( −1.4 ) = 2.4 V
VCEQ1 = 1 − ( −0.7 ) = 1.7 V
(b)
Small-signal transistor parameters:
β VT (100 )( 0.026 )
rπ 1 =
=
= 265 k Ω
I CQ1
0.0098
g m1 =
rπ 2 =
gm2 =
I CQ1
VT
β VT
I CQ 2
I CQ 2
VT
0.0098
= 0.377 mA
0.026
=
=
(100 )( 0.026 )
=
0.990
= 2.63 k Ω
0.990
= 38.1 mA / V
0.026
ro1 = ro 2 = ∞
(c)
Small-signal voltage gain: From Figure 4.73 (b)
Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC
Vs = Vπ 1 + Vπ 2
and
⎛V
⎞
⎛1+ β ⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜
⎟ ⋅ Vπ 1rπ 2
⎝ rπ 1
⎠
⎝ rπ 1 ⎠
Then
⎡
⎛1+ β
Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜
⎝ rπ 1
⎣
Also
⎤
⎞
⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC
⎠
⎦
⎛1+ β
Vs = Vπ 1 + ⎜
⎝ rπ 1
or
⎞
⎟ ⋅ rπ 2Vπ 1
⎠
⎡
⎛ r ⎞⎤
= Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥
⎝ rπ 1 ⎠ ⎦
⎣
Vπ 1 =
Now
Vs
⎛r ⎞
1 + (1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
⎡
⎛ rπ 2 ⎞ ⎤
⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC
V
⎝ rπ 1 ⎠ ⎦
Av = o = − ⎣
Vs
⎛r ⎞
1 + (1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
so
⎡
⎛ 2.63 ⎞ ⎤
⎢0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 )
⎝
⎠⎦
Av = − ⎣
⎛ 2.63 ⎞
1 + (101) ⎜
⎟
⎝ 265 ⎠
or
Av = −77.0
(d)
Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63)
or
Ri = 531 k Ω
TYU6.17
(a)
dc analysis:
For R1 + R2 + R3 = 100 k Ω
I1 =
VCC
12
=
= 0.12 mA
R1 + R2 + R3 100
Now
VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V
VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V
and
VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V
So RC =
VCC − VC 2 12 − 8.25
=
= 7.5 k Ω
0.5
I CQ
Also
VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V
And
⎛
⎞
R3
R3
VB1 = ⎜
(12 )
⎟ ⋅ VCC ⇒ 0.95 =
100
⎝ R1 + R2 + R3 ⎠
which yields
R3 = 7.92 k Ω
We have
VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 V
and
⎛ R2 + R3 ⎞
VB 2 = ⎜
⎟ ⋅ VCC
⎝ R1 + R2 + R3 ⎠
or
⎛ R + 7.92 ⎞
4.95 = ⎜ 2
⎟ (12 )
⎝ 100 ⎠
which yields
R2 = 33.3 k Ω
Then
R1 = 100 − 33.3 − 7.92 = 58.8 k Ω
(b)
Small-signal transistor parameters:
(100 )( 0.026 )
βV
rπ 1 = rπ 2 = T =
0.5
I CQ
or
rπ 1 = rπ 2 = 5.2 k Ω
and
I CQ
0.5
g m1 = g m 2 =
=
VT
0.026
or
g m1 = g m 2 = 19.23 mA/V
and
ro1 = ro 2 = ∞
(c)
Small-signal voltage gain:
We have
Vo = − g m 2Vπ 2 ( RC & RL )
Also
Vπ 2
+ g m 2Vπ 2 = g m1Vπ 1
rπ 2
or
⎛ r ⎞
Vπ 2 = g m1Vπ 1 ⎜ π 2 ⎟ and Vπ 1 = Vs
⎝ 1+ β ⎠
We find
⎛ r ⎞
V
Av = o = − g m 2 ( RC RL ) g m1 ⎜ π 2 ⎟
Vs
⎝1+ β ⎠
⎛ β ⎞
= − g m 2 ( RC RL ) ⎜
⎟
⎝ 1+ β ⎠
We obtain
⎛ 100 ⎞
Av = − (19.23) ( 7.5 2 ) ⎜
⎟
⎝ 101 ⎠
or
Av = −30.1
TYU6.18
(a)
dc analysis: with vs = 0
RTH = R1 R2 = 125 30 = 24.2 k Ω
⎛ R2 ⎞
⎛ 30 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (12 )
⎝ 125 + 30 ⎠
⎝ R1 + R2 ⎠
or
VTH = 2.32 V
Now
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
or
I BQ =
2.32 − 0.7
= 0.0250 mA
24.2 + ( 81)( 0.5 )
and
I CQ = ( 80 )( 0.025 ) = 2.00 mA
Also
⎡
⎛1+ β ⎞ ⎤
VCEQ = VCC − I CQ ⎢ RC + ⎜
⎟ RE ⎥
⎝ β ⎠ ⎦
⎣
⎡ ⎛ 81 ⎞
⎤
= 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥
⎣ ⎝ 80 ⎠
⎦
or
VCEQ = 6.99 V
Power dissipated in RC:
2
2
PRC = I CQ
RC = ( 2.0 ) ( 2 ) = 8.0 mW
Power dissipated in RL:
I LQ = 0 ⇒ PRL = 0
Power dissipated in transistor:
PQ = I BQVBEQ + I CQVCEQ
= ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW
(b)
With vs = 18cos ω t ( mV )
β VT
rπ =
I CQ
=
(80 )( 0.026 )
2.0
= 1.04 k Ω
We can write
β
( RC RL )VP cos ω t
rπ
Power dissipated in RL:
vce =
pRL =
=
vce ( rms )
RL
1
1
⋅
2 2 × 103
2
⎤
1 1 ⎡β
= ⋅ ⋅ ⎢ ( RC RL ) VP ⎥
2 RL ⎣ rπ
⎦
⎡ 80
⋅⎢
( 2 2 ) ( 0.018)⎤⎥
⎣1.04
⎦
2
2
or
pRL = 0.479 mW
Power dissipated in RC:
Since RC = RL = 2 k Ω, we have pRC = 8.0 + 0.479 = 8.48 mW
Power dissipated in transistor: From the text, we have
⎛β ⎞ ⎛V ⎞
pQ ≅ I CQVCEQ − ⎜ ⎟ ⎜ P ⎟ ( RC RL )
⎝ rπ ⎠ ⎝ 2 ⎠
2
2
80
⎛
⎞ ⎛ 0.018 ⎞
3
= ( 2 × 10−3 ) ( 6.99 ) − ⎜
⎟ ( 2 × 10 2 ×
3 ⎟ ⎜
⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠
2
or
pQ = 13.0 mW
TYU6.19
(a)
dc analysis:
2
3
)
RTH = R1 R2 = 53.8 10 = 8.43 k Ω
⎛ R2 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (5)
53.8 + 10 ⎠
R
R
+
⎝
⎝ 1
2 ⎠
or
VTH = 0.7837 V
Now
0.7837 − 0.7
I BQ =
= 0.00993 mA
8.43
and
I CQ = (100 )( 0.00993) = 0.993 mA
We have
VCEQ = VCC − I CQ RC
or
2.5 = 5 − ( 0.993) RC
which yields
RC = 2.52 k Ω
(b)
Power dissipated in RC:
2
2
PRC = I CQ
RC = ( 0.993) ( 2.52 )
or
PRC = 2.48 mW
Power dissipated in transistor:
PQ ≅ I CQVCEQ = ( 0.993)( 2.5 )
or
PQ = 2.48 mW
(c)
ac analysis:
Maximum ac collector current:
ic = ( 0.993) cos ω t ( mA )
average ac power dissipated in RC:
1
1
2
2
pRC = ( 0.993) RC = ( 0.993) ( 2.52 )
2
2
or
pRC = 1.24 mW
Now
pRC
1.24
Fraction =
=
= 0.25
PRC + PQ 2.48 + 2.48
Chapter 6
Problem Solutions
6.1
a.
gm =
rπ =
r0 =
I CQ
VT
β VT
I CQ
=
2
⇒ g m = 76.9 mA/V
0.026
=
(180 )( 0.026 )
2
⇒ rπ = 2.34 kΩ
VA 150
=
⇒ r0 = 75 kΩ
I CQ
2
b.
0.5
⇒ g m = 19.2 mA/V
0.026
(180 )( 0.026 )
⇒ rπ = 9.36 kΩ
rπ =
0.5
150
⇒ r0 = 300 kΩ
r0 =
0.5
gm =
6.2
(a)
gm =
rπ =
ro =
I CQ
VT
β VT
I CQ
=
0.8
= 30.8 mA/V
0.026
=
(120 )( 0.026 )
0.8
= 3.9 K
VA 120
=
= 150 K
I CQ 0.8
(b)
0.08
= 3.08 mA/V
0.026
(120 )( 0.026 )
rπ =
= 39 K
0.08
120
ro =
= 1500 K
0.08
gm =
6.3
gm =
rπ =
r0 =
6.4
I CQ
VT
β VT
I CQ
⇒ 200 =
=
I CQ
0.026
⇒ I CQ = 5.2 mA
(125)( 0.026 )
5.2
⇒ rπ = 0.625 kΩ
VA 200
=
⇒ r0 = 38.5 kΩ
I CQ 5.2
gm =
rπ =
⇒ 80 =
I CQ
VT
I CQ
0.026
⇒ 1.20 =
β VT
I CQ
⇒ I CQ = 2.08 mA
β ( 0.026 )
2.08
⇒ β = 96
6.5
(a)
2 − 0.7
= 0.0052 mA
250
I C = (120 )( 0.0052 ) = 0.624 mA
I BQ =
0.624
⇒ g m = 24 mA / V
0.026
(120 )( 0.026 )
⇒ rπ = 5 k Ω
rπ =
0.624
ro = ∞
gm =
⎛ r
⎞
⎛ 5 ⎞
Av = − g m RC ⎜ π ⎟ = − ( 24 )( 4 ) ⎜
⎟ ⇒ Av = −1.88
r
R
+
⎝ 5 + 250 ⎠
B ⎠
⎝ π
v
v
vS = O = O ⇒ vS = −0.426sin100t V
Av −1.88
(b)
(c)
6.6
gm =
I CQ
VT
, 1.08 ≤ I CQ ≤ 1.32 mA
1.08
1.32
≤ gm ≤
⇒ 41.5 ≤ g m ≤ 50.8 mA/V
0.026
0.026
(120 )( 0.026 )
β VT
rπ =
; rπ ( max ) =
= 2.89 kΩ
I CQ
1.08
rπ ( min ) =
(80 )( 0.026 )
1.32
= 1.58 kΩ
1.58 ≤ rπ ≤ 2.89 kΩ
6.7
a.
rπ = 5.4 =
β VT
I CQ
=
(120 )( 0.026 )
I CQ
⇒ I CQ = 0.578 mA
1
1
VCEQ = VCC = ( 5 ) = 2.5 V
2
2
VCEQ = VCC − I CQ RC ⇒ 2.5 = 5.0 − ( 0.578 ) RC ⇒ RC = 4.33 kΩ
I CQ
0.578
= 0.00482 mA
β
120
VBB = I BQ RB + VBE ( on )
= ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V
I BQ =
b.
=
rπ =
gm =
r0 =
β VT
I CQ
I CQ
=
0.578
= 5.40 kΩ
0.578
= 22.2 mA/V
0.026
=
VT
(120 )( 0.026 )
VA
100
=
= 173 kΩ
I CQ 0.578
⎛ r
⎞
V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS
⎝ rπ + RB ⎠
β ( r0 RC )
⎛ r
⎞
Av = − g m ⎜ π ⎟ ( r0 RC ) = −
rπ + RB
⎝ rπ + RB ⎠
Av = −
(120 ) ⎡⎣173
4.33⎤⎦
5.40 + 25
=−
(120 )( 4.22 )
30.4
⇒ Av = −16.7
6.8
a.
1
VECQ = VCC = 5 V
2
VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ
I CQ
0.5
= 0.005
100
VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V
I BQ =
=
β
b.
gm =
rπ =
r0 =
c.
I CQ
=
VT
β VT
I CQ
0.5
⇒ g m = 19.2 mA/V
0.026
=
(100 )( 0.026 )
0.5
⇒ rπ = 5.2 kΩ
VA
∞
=
⇒ r0 = ∞
I CQ 0.5
Av = −
(100 )(10 ) ⇒ A = −18.1
β RC
=−
v
5.2 + 50
rπ + RB
6.9
10 − 4
= 1.5 mA
4
1.5
I BQ =
= 0.015 mA
100
(100 )( 0.026 )
rπ =
= 1.73 K
1.5
5sin ω t ( mV )
v
ib = be =
= 2.89sin ω t ( µ A )
rπ
1.73 kΩ
So
I CQ =
iB ( t ) = I BQ + iEb = 15 + 2.89sin ω t ( µ A )
iC1 ( t ) = β iB ⇒ iC1 ( t ) = 1.5 + 0.289sin ω t ( mA )
vC ( t ) = 10 − iC1 ( t ) RC = 10 − [1.5 + 0.289sin ω t ] (γ )
vC1 ( t ) = 4 − 1.156sin ω t ( v )
Av =
vC ( t )
vbe ( t )
=
−1.156
⇒ Av = −231
0.005
6.10
vo = 1.2sin ω t ( V )
iC ( t ) RC + vo = 0 ⇒ iC ( t ) =
−1.2sin ω t
2
iC ( t ) = −0.60sin ω t ( mA )
ib ( t ) =
iC ( t )
β
= −6sin ω t ( µ A )
vbe ( t ) = ib ( t ) ⋅ rπ
g m rπ = β
100
=2K
50
vbe ( t ) = −12sin ω t ( mV )
rπ =
6.11
a.
I CQ ≈ I EQ
VCEQ = 5 = 10 − I CQ ( RC + RE )
= 10 − I CQ (1.2 + 0.2)
I CQ = 3.57 mA
3.57
= 0.0238 mA
150
R1 & R2 = RTH = ( 0.1)(1 + β ) RE
I BQ =
= ( 0.1)(151)( 0.2 ) = 3.02 kΩ
VTH =
1
⋅ RTH ⋅ (10) − 5
R1
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
(3.02)(10) − 5 = ( 0.0238)(3.02) + 0.7 + (151)( 0.0238)( 0.2) − 5
R1
1
( 30.2 ) = 1.50 ⇒ R1 = 20.1 k Ω
R1
20.1R2
= 3.02 ⇒ R2 = 3.55 kΩ
20.1 + R2
b.
(150 )( 0.026 )
rp =
= 1.09 kΩ
3.57
3.57
gm =
= 137 mA/V
0.026
V0
⫹
V␲
r␲
gmV␲
⫺
⫹
⫺
VS
R1兩兩R2
RC
RE
Av =
(150 )(1.2 )
2 β RC
=2
⇒ Av = 2 5.75
rp + (1 + β ) RE
1.09 + (151)( 0.2 )
6.12
a.
⎛ R2 ⎞
⎛ 50 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (12 ) = 10 V
+
R
R
⎝ 50 + 10 ⎠
⎝ 1
2 ⎠
RTH = R1 R2 = 50 10 = 8.33 kΩ
I BQ =
12 − 0.7 − 10
= 0.0119 mA
8.33 + (101)(1)
I CQ = 1.19 mA, I EQ = 1.20 mA
VECQ = 12 − (1.20 )(1) − (1.19 )( 2 )
VECQ = 8.42 V
iC
4
1.19
8.42
12
␯EC
b.
V0
⫺
V␲
r␲
gmV␲
⫹
VS
⫹
⫺
R1兩兩R2
RC
RE
rp =
(100 )( 0.026 )
1.19
V0 = g mVp RC
= 2.18 kΩ
⎛V
⎞
VS = 2 Vp − ⎜ p + g mVp ⎟ RE
⎝ rp
⎠
⎡ rπ + (1 + β ) RE ⎤
= −Vp ⎢
⎥
rp
⎣
⎦
2 (100 )( 2 )
2 β RC
=
⇒ Av = 2 1.94
Av =
rp + (1 + β ) RE 2.18 + (101)(1)
c.
Approximation: Assume rp does not vary significantly.
RC = 2 kΩ ± 5% = 2.1 kΩ or 1.9 kΩ
RE = 1 kΩ ± 5% = 1.05 kΩ or 0.95 kΩ
For RC ( max ) = 2.1 kΩ and RE ( min )
Av =
− (100 )( 2.1)
2.18 + (101)( 0.95 )
= −2.14
For RC ( min ) = 1.9 kΩ and RE ( max ) = 1.05 kΩ
Av =
− (100 )(1.9 )
2.18 + (101)(1.05 )
= −1.76
So 1.76 ≤ Av ≤ 2.14
6.13
(a)
⎛
⎞
VCC = ⎜⎜ 1+ β ⎟⎟ I CQ RE + VECQ + I CQ RC
⎝
β
⎠
⎛ 101 ⎞
12 = ⎜
⎟ I CQ (1) + 6 + I CQ ( 2 )
⎝ 100 ⎠
so that I CQ = 1.99 mA
1.99
= 0.0199 mA
100
= ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω
I BQ =
RTH
⎛ R2 ⎞
1
1
VTH = ⎜
⎟ VCC = ⋅ RTH ⋅ VCC = (10.1)(12 )
R
R
R
R
+
2 ⎠
1
1
⎝ 1
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
121.2
12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) +
R1
which yields R1 = 13.3 k Ω and R2 = 41.6 k Ω
(b)
6.14
Av =
2 (100 )( 2 )
2 β RC
=
⇒ Av = 2 1.95
rp + (1 + β ) RE 1.31 + (101)(1)
I CQ = 0.25 mA, I EQ = 0.2525 mA
I BQ = 0.0025 mA
I BQ RB + VBE ( on ) + I EQ ( RS + RE ) − 5 = 0
( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5
RE = 16.4 kΩ
VE = − ( 0.0025 )( 50 ) − 0.7 = −0.825 V
VC = VCEQ + VE = 3 − 0.825 = 2.175 V
5 − 2.175
⇒ RC = 11.3 kΩ
RC =
0.25
− β RC
Av =
rπ + (1 + β ) RS
rπ =
(100 )( 0.026 )
= 10.4 kΩ
0.25
− (100 )(11.3)
Av =
⇒ Av = −55.1
10.4 + (101)( 0.1)
Ri = RB ⎡⎣ rπ + (1 + β ) RS ⎤⎦
= 50 ⎡⎣10.4 + (101)( 0.1) ⎤⎦
Ri = 50 20.5 ⇒ Ri = 14.5 kΩ
6.15
(a)
VCC > I CQ ( RC + RE ) + VCEQ
9 = I CQ ( 2.2 + 2 ) + 3.75 So that
I CQ = 1.25 mA
Assume circuit is to be designed to be bias stable.
RTH = R1 & R2 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 Ω
1.25
I BQ =
= 0.01042 mA
120
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I BQ (121)( RE )
R1
1
( 24.2 )( 9 ) = ( 0.01042 )( 24.2 ) + 0.7 + ( 0.01042 )(121)( 2 )
R1
= 0.2522 + 0.7 + 2.5216
= 3.474
R1 = 62.7 K
R2 = 39.4 K
(b)
62.7 R2
= 24.2
62.7 + R2
1.25
= 48.08 mA / V
0.026
(120 )( 0.026 )
rp =
= 2.50 k Ω
1.25
100
ro =
= 80 k Ω
1.25
gm =
Vo
⫹
IS
V␲
r␲
R1兩兩R2⫺
ro
RC
RL
gmV␲
Vo = 2 g mVp ( ro RC RL )
Vp = I S ( R1 R2 rp )
Then
Rm =
Vo
= 2 g m ( R1 R2 rp )( ro RC RL )
Is
Rm = 2 48.08 ( 24.2 2.5 )( 80 2.2 1) = 2 48.08 ( 2.266 )( 0.6816 )
or
Rm =
Vo
= −74.3 k Ω = −74.3 V / mA
Is
6.16
a.
I EQ = 0.80 mA, I BQ =
0.80
= 0.0121 mA
66
I CQ = 0.788 mA
VB = I BQ RB ⇒ RB =
0.3
⇒ RB = 24.8 kΩ
0.0121
V − ( −5 ) 5 − 3
=
⇒ RC = 2.54 kΩ
RC = C
I CQ
0.788
b.
0.788
= 30.3 mA / V
0.026
( 65)( 0.026)
= 2.14 kΩ
rπ =
0.788
75
= 95.2 kΩ
r0 =
0.788
⎛ RC r0 ⎞
i0 = ⎜
g V , V = − vS
⎜ R r + R ⎟⎟ m π π
L ⎠
⎝ C 0
gm =
Gf =
⎛ RC r0
i0
= − gm ⎜
⎜R r +R
vS
L
⎝ C 0
⎞
⎟⎟
⎠
⎛ 2.54 95.2 ⎞
= − ( 30.3) ⎜
⎜ 2.54 95.2 + 4 ⎟⎟
⎝
⎠
G f = −11.6 mA/V
6.17
(a)
I CQ = 0.8 mA ⇒ I BQ = 0.00667 mA
I BQ RS + 0.7 + (121) I BQ RE − 15 = 0
( 0.01667 )( 2.5) + 0.7 + (121)( 0.00667 ) RE = 15
RE = 17.7 K
VE = − ( 0.00667 )( 2.5 ) − 0.7 = −0.717 V
VC = −0.717 + 7 = 6.283 V
15 − 6.283
RC =
= 10.9 K
0.8
(120 )( 0.026 )
0.8
gm =
= 30.77 mA/V rπ =
= 3.9 K
0.026
0.8
⎛ r
⎞
vo = − g m ( RC RL ) ⋅ vπ
vπ = ⎜ π ⎟ vS
⎝ rπ + RS ⎠
Av =
− β ( RC RL )
rπ + RS
=
− (120 ) (10.9 5 )
3.9 + 2.5
Av = −64.3
(b) For RS = 0
0.7 + (121)( 0.00667 ) RE = 15
RE = 17.7 K
VE = −0.7 ⇒ VC = −0.7 + 7 = 6.3
15 − 6.3
⇒ RC = 10.9 K
0.8
− β ( RC RL )
= −30.77 (10.9 5 )
Av =
rπ
RC =
Av = −105
6.18
(a)
15 = ( 81) I BQ (10 ) + 0.7 + I BQ ( 2.5 )
15 − 0.7
I BQ =
= 0.0176 mA
2.5 + ( 81)(10 )
I CQ = 1.408 mA
(80 )( 0.026 )
1.408
= 54.15 mA/V rπ =
0.026
1.408
rπ = 1.48 K
gm =
RS
V0
IS
⫹
VS
⫹
⫺
V␲
⫺
r␲
gmV␲
RC
Io
RL
Vo = − g mVσ ( RC RL ) ⇒ Av =
i
AI = o =
iS
− β ( RC RL )
rπ + RS
=
⎛ RC ⎞
− g mVπ ⎜
⎟
⎝ RC + RL ⎠ = − β ⎛ RC ⎞
⎜
⎟
V
⎝ RC + RL ⎠
π
r
− ( 80 ) ( 5 5 )
1.48 + 2.5
⇒ Av = −50.3
π
AI = −40
vo ( t ) = ( −50.3)( 4sin ω t )
vo ( t ) = −0.201sin ω t ( V )
4 sin ω t ( mV )
= 1.005sin ω t ( μA )
2.5 + 1.48
io = −40.2 sin ω t ( μA )
is =
(b)
15 − 0.7
= 1.43 mA
10
⎛ 80 ⎞
= ⎜ ⎟ (1.43) = 1.412 mA
⎝ 81 ⎠
I EQ =
I CQ
gm =
(80 )( 0.026 )
1.412
= 54.3 mA/V rπ =
= 1.47 K
0.026
1.412
Av = − g m ( RC RL ) = − ( 54.3) ( 5 5 ) ⇒ Av = −136
⎛ RC ⎞
⎛ 5 ⎞
AI = − β ⎜
⎟ = −80 ⎜
⎟ ⇒ AI = −40
⎝5+5⎠
⎝ RC + RL ⎠
vo ( t ) = ( −136 )( 4sin ω t ) ⇒ vo ( t ) = −544sin ω t ⇒ vo ( t ) = −0.544sin ω t ( V )
4sin ω t ( mV )
= 2.72sin ω t ( μA )
1.47 k
io ( t ) = ( −40 )( 2.72sin ω t )
is ( t ) =
io ( t ) = −109sin ω t ( μA )
6.19
RTH = R1 R2 = 27 15 = 9.64 K
⎛ R2 ⎞
⎛ 15 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ ( 9 ) = 3.214 V
15 + 27 ⎠
R
R
+
⎝
2 ⎠
⎝ 1
V − VBE ( on )
3.214 − 0.7
2.514
I BQ = TH
=
=
RTH + (1 + β ) RE 9.64 + (101)(1.2 ) 130.84
I BQ = 0.0192 mA I CQ = 1.9214 mA
gm =
(100 )( 0.026 )
1.92
= 73.9 mA/V rπ =
= 1.35 K
0.026
1.92
RS
V0
IS
⫹
⫹
⫺
VS
V␲
RTH
r␲
gmV␲
⫺
ro =
100
= 52.1 K
1.92
⎛ r R
Vπ = ⎜ π TH
⎜r R +R
S
⎝ π TH
= 1.35 9.64 = 1.184 K
(
Vo = − g mVπ r0 RC RL
rπ RTH
)
⎞
⎟⎟ VS
⎠
⎛ 1.184 ⎞
Vπ = ⎜
⎟ VS
⎝ 1.184 + 10 ⎠
= 0.1059VS
(
Av = − ( 73.9 ) ( 0.1059 ) 52.1 2.2 2
)
= − ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 )
= − ( 73.9 ) ( 0.1059 ) (1.027 )
Av = −8.04
⎛ ro RC
− g mVπ ⎜⎜
I
⎝ ro RC + RL
AI = o =
Vπ
IS
RTH rπ
⎞
⎟⎟
⎠
⎛ ro RC ⎞
AI = − g m ( RTH rπ ) ⎜
⎜ r R + R ⎟⎟
L ⎠
⎝ o C
ro RC = 52.1 2.2 = 2.11 K
RTH rπ = 9.64 1.35 = 1.184 K
⎛ 2.11 ⎞
AI = − ( 73.9 ) (1.184 ) ⎜
⎟
⎝ 2.11 + 2 ⎠
AI = −44.9
Ri = RTH rπ = 9.64 1.35
Ri = 1.184 K
6.20
a.
0.35
= 0.00347 mA
101
VB = 2 I B RB = 2 ( 0.00347 )(10 ) ⇒ VB = 2 0.0347 V
I E = 0.35 mA, I B =
VE = VB − VBE ( on ) ⇒ VE = 2 0.735 V
b.
r0
RC
I0
RL
VC = VCEQ + VE = 3.5 − 0.735 = 2.77 V
⎛ b ⎞
⎛ 100 ⎞
IC = ⎜
⎟ IE = ⎜
⎟ ( 0.35 ) = 0.347 mA
⎝ 101 ⎠
⎝ 1+ b ⎠
V 1 − VC 5 − 2.77
RC =
=
⇒ RC = 6.43 kΩ
IC
0.347
(c)
⎛ RB rp ⎞
Av = 2 g m ⎜
R r
⎜ R r + R ⎟⎟ ( C o )
S ⎠
⎝ B π
0.347
100
gm =
= 13.3 mA/V , ro =
= 288 k Ω
0.026
0.347
(100 )( 0.026 )
= 7.49 k Ω
rp =
0.347
RB rp = 10 7.49 = 4.28 k Ω
⎛ 4.28 ⎞
Av = 2 (13.3) ⎜
⎟ ( 6.43 288 ) ⇒ Av = 2 81.7
⎝ 4.28 + 0.1 ⎠
d.
⎛ RB rp ⎞
Av = 2 g m ⎜
R r
⎜ R r + R ⎟⎟ ( C 0 )
S ⎠
⎝ B p
RB rp = 10 7.49 = 4.28 kΩ
⎛ 4.28 ⎞
Av = 2 (13.3) ⎜
⎟ ( 6.43 288 ) ⇒ Av = 2 74.9
⎝ 4.28 + 0.5 ⎠
6.21
a.
RTH = R1 R2 = 6 1.5 = 1.2 kΩ
⎛ R2 ⎞ + ⎛ 1.5 ⎞
VTH = ⎜
⎟V = ⎜
⎟ ( 5 ) = 1.0 V
⎝ 1.5 + 6 ⎠
⎝ R1 + R2 ⎠
V − VBE ( on )
1.0 − 0.7
I BQ = TH
=
= 0.0155 mA
RTH + (1 + β ) RE 1.2 + (181)( 0.1)
I CQ = 2.80 mA, I EQ = 2.81
VCEQ = V + − I CQ RC − I EQ RE
= 5 − ( 2.8 )(1) − ( 2.81)( 0.1) ⇒ VCEQ = 1.92 V
b.
⇒ rp = 1.67 kΩ
2.80
2.80
⇒ g m = 108 mA/V, r0 =`
gm =
0.026
(c)
R1 R2 rp
⎛
⎞
Av = 2 g m ⎜
⎟ ( RC RL )
⎝ R1 & R2 & rp + RS ⎠
R1 R2 rp = 6 1.5 1.67 = 0.698 k V
rp =
(180 )( 0.026 )
⎛ 0.698 ⎞
Av = 2 (108 ) ⎜
⎟ (1 1.2 ) ⇒ Av = 2 45.8
⎝ 0.698 + 0.2 ⎠
6.22
a.
9 = I EQ RE + VEB ( on ) + I BQ RS
0.75
= 0.00926 mA
I EQ = 0.75 mA, I BQ =
81
I CQ = 0.741 mA
9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) ⇒ RE = 11.0 kΩ
b.
VE = 9 − ( 0.75 )(11) = 0.75 V
VC = VE − VECQ = 0.75 − 7 = −6.25 V
RC =
VC − ( −9 )
I CQ
=
9 − 6.25
⇒ RC = 3.71 kΩ
0.741
c.
⎛ rp ⎞
Av = 2 g m ⎜
⎟ ( RC || RL || r0 )
⎝ rp + RS ⎠
(80 )( 0.026 )
rp =
= 2.81 kΩ
0.741
80
r0 =
= 108 kV
0.741
2 80
Av =
( 3.71||10 ||108 )
2.81 + 2
Av = 2 43.9
d.
Ri = RS + rp = 2 + 2.81 ⇒ Ri = 4.81 kΩ
6.23
I BQ =
4 − 0.7
= 0.00647
5 + (101)( 5 )
I CQ = 0.647 mA
80 ≤ h fe ≤ 120, 10 ≤ h0e ≤ 20 mS
a.
2
.45 kΩ ≤ hie ≤ 3
.7 k Ω
low gain
high gain
RS
V0
IS
⫹
VS
⫹
⫺
V␲
⫺
r␲
gmV␲
RC
Io
RL
⎛ 1
⎞
V0 = 2 h fe I b ⎜
RC RL ⎟
⎝ hoe
⎠
RB
?VS
R + RS
Ib = B
RTH + hie
RTH = RB RS = 5 1 = 0.833 kΩ
High-gain
⎛ 5 ⎞
⎜
⎟ VS
5 +1⎠
Ib = ⎝
= 0.1838VS
0.833 + 3.7
Low-gain
⎛ 5 ⎞
⎜
⎟ VS
5 +1⎠
⎝
Ib =
= 0.2538VS
0.833 + 2.45
1
1
For hoe = 10 ⇒
|| Rc || RL =
|| 4 || 4
0.010
hoe
= 100 || 2 = 1.96 kΩ
For hoe = 20 ⇒
Av
Av
max
min
1
|| 4 || 4 = 50 || 2 = 1.92 kΩ
0.020
= (120 )( 0.1838 )(1.96 ) = 43.2
= ( 80 )( 0.2538 )(1.92 ) = 39.0
39.0 ≤ Av ≤ 43.2
b.
Ri = RB hie = 5 3.7 = 2.13 kV or Ri = 5 2.45 = 1.64 kΩ
1.64 ≤ Ri ≤ 2.13 kΩ
R0 =
1
1
4 = 100 || 4 = 3.85 kΩ
RC =
0.010
hoe
1
|| 4 = 50 || 4 = 3.70 kΩ
0.020
3.70 ≤ R0 ≤ 3.85 kΩ
or R0 =
6.24
VCC ⫽ 10 V
RC
R1
␯o
RS ⫽ 1 k⍀
CC
␯s
⫹
⫺
R2
RE
CE
Assume an npn transistor with b = 100 and VA = ∞. Let VCC = 10 V .
0.5
= 50
0.01
Bias at I CQ = 1 mA and let RE = 1 k Ω
Av =
For a bias stable circuit
RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)(1) = 10.1 k Ω
1
1
101
VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) =
R1
R1
R1
1
I BQ =
= 0.01 mA
100
VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
101
= ( 0.01)(10.1) + 0.7 + (101)( 0.01)(1)
R1
which yields R1 = 55.8 k Ω and R2 = 12.3 k Ω
Now
rp =
(100 )( 0.026 )
1
= 2.6 k Ω
1
= 38.46 mA/V
0.026
Vo = − g mVp RC
gm =
⎛ R1} R2 } rp ⎞
⎛ 10.1} 2.6 ⎞
where Vp = ⎜
⎟ ⋅ Vs = ⎜
⎟ .Vs
⎝ 10.1} 2.6 + 1 ⎠
⎝ R1} R2 } rp + RS ⎠
or Vp = 0.674 Vs
V
Then Av = o = − ( 0.674 ) g m RC = − ( 0.674 )( 38.46 ) RC = −50
Vs
which yields RC = 1.93 k Ω
With this RC, the dc bias is OK.
Finish Design, Set RC = 2 K
RE = 1 K
R1 = 56 K
R2 = 12 K
RTH = R1 R2 = 9.88 K
⎛ R2 ⎞
⎛ 12 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 1.765 V
⎝ 12 + 56 ⎠
⎝ R1 + R2 ⎠
1.765 − 0.7
I BQ =
= 9.60 µ A
9.88 + (101)(1)
I CQ = 0.9605 mA
rπ =
RTH
(100 )( 0.026 )
0.9605
rπ = 2.125 K
= 2.707 K
gm =
0.9605
= 36.94
0.026
⎛ RTH & rπ ⎞
⎛ 2.125 ⎞
Vπ = ⎜
⎟ Vi = ⎜
⎟ Vi = ( 0.680 ) Vi
⎝ 2.125 + 1 ⎠
⎝ RTH & rπ + RS ⎠
Av = − ( 0.680 ) g m RC = − ( 0.680 )( 36.94 )( 2 ) = −50.2
Design specification met.
6.25
a.
I BQ =
6 − 0.7
= 0.0169 mA
10 + (101)( 3)
I CQ = 1.69 mA, I EQ = 1.71 mA
VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3)
VCEQ = 5.38 V
b.
1.69
⇒ g m = 65 mA/V
0.026
(100 )( 0.026 )
rp =
⇒ rp = 1.54 kV ,
1.69
(c)
− β ( RC RL )
RB Rib
Av =
⋅
rπ + (1 + β ) RE RB Rib + RS
gm =
r0 = ∞
Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω
RB Rib = 10 304.5 = 9.68 k Ω
Then
Av =
− (100 )( 6.8 & 6.8 ) ⎛ 9.68 ⎞
⋅⎜
⎟ ⇒ Av = −1.06
1.54 + (101)( 3) ⎝ 9.68 + 0.5 ⎠
⎛ RC ⎞
i0 = ⎜
⎟ ( − β ib )
⎝ RC + RL ⎠
⎛
RB
ib = ⎜
R
r
+
+
⎝ B π (1 + β ) RE
⎞
⎟ iS
⎠
⎞
⎞⎛
RB
⎟⎟
⎟ ⎜⎜
⎠ ⎝ RB + rπ + (1 + β ) RE ⎠
⎞
10
⎛ 6.8 ⎞ ⎛
= − (100 ) ⎜
⎟ ⇒ Ai = −1.59
⎟ ⎜⎜
⎝ 6.8 + 6.8 ⎠ ⎝ 10 + 1.54 + (101)( 3) ⎠⎟
(d)
Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω
⎛ RC
Ai = − ( β ) ⎜
⎝ RC + RL
(e)
Av =
Av =
2 b ( RC RL )
rp + (1 + b ) RE
2 (100 ) ( 6.8} 6.8 )
1.54 + (101)( 3)
⇒ Av = 2 1.12
Ai = same as ( c ) ⇒ Ai = 2 1.59
6.26
ie
⫹
⫹
vCE
⫹
vbe ⫺
⫺
vCE
gmv␲
ris
␲o
⫺
⫹
vCE
gmv␲
⫺
r=
vCe
1
=
g m vCe g m
⎛ 1 ⎞
So re = rp ⎜
⎟ r0
⎝ gm ⎠
6.27
Let b = 100, VA = ∞
VCC
RC
R1
␯o
RS ⫽ 100 ⍀
CC
␯s
⫹
⫺
R2
RE
Let VCC = 2.5 V
P = ( I R + I C ) VCC ⇒ 0.12 = ( I R + I C )( 2.5 ) ⇒ I R + I C = 48 mA, Let I R = 8mA, I C = 40 mA
R1 + R2 >
I BQ =
VCC 2.5
=
⇒ 312.5 k Ω
IR
8
40
= 0.4 mA
100
Let RE = 2 k Ω. For a bias stable circuit
RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)( 2 ) = 20.2 k Ω
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
R1
1
( 20.2 )( 2.5 ) = ( 0.0004 )( 20.2 ) + 0.7 + (101)( 0.0004 )( 2 )
R1
which yields R1 = 64 k V and R2 = 29.5 k Ω
(100 )( 0.026 )
rπ =
0.04
Av =
= 65 k Ω
Neglect RS
2 b RC
Vo
>
Vs
rπ + (1 + b ) RE
−10 =
2 100 RC
⇒ RC = 26.7 k Ω
65 + (101)( 2 )
With this RC , dc biasing is OK.
6.28
100
= 20.
5
Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31.
Let RS = 0. Need an input resistance of
Need a voltage gain of
5 × 102 3
= 25 × 103 = 25 k Ω
0.2 × 102 6
Ri = RTH Rib . Let RTH = 50 k Ω, Rib = 50 k Ω
Ri =
Rib = rp + (1 + b ) RE > (1 + b ) RE
Rib
50
=
= 0.495 k Ω
1 + b 101
Let RE = 0.5 k V , VCC = 10 V , I CQ = 0.2 mA
For b = 100, RE =
0.2
= 0.002 mA
100
= I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
Then I BQ =
VTH
1
1
⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002 )( 0.5)
R
R1
1
which yields R1 = 555 k Ω and R2 = 55 k Ω
Now Av =
(100)( 0.026)
− β RC
= 13 k Ω
, rπ =
rπ + (1 + β ) RE
0.2
So
−20 =
− (100 ) RC
13 + (101)( 0.5)
⇒ RC = 12.7 k Ω
[Note: I CQ RC = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.]
6.29
VCC ⫽ 10 V
R1
RE
CC
␯s
␯o
⫹
⫺
R2
RC
b = 80, Av =
2 b RC
rp + (1 + b ) RE
First approximation:
R
( Av ) ≈ C = 10 ⇒ RC = 10 RE
RE
Set RC = 12 RE
VEC ≈ VCC − I C ( RC + RE ) = 10 − I C (13RE )
1
For VEC = VCC = 5
2
5 = 10 − I C (13RE )
For I C = 0.7 mA
I E = 0.709, I B = 0.00875 mA ⇒ RE = 0.55 kΩ − RC = 6.6 kΩ
Bias stable ⇒
R1 R2 = RTH = ( 0.1)(1 + β ) RE
= ( 0.1)(81)( 0.55 ) = 4.46 kΩ
1
10 = ( 0.709 )( 0.55 ) + 0.7 + ( 0.00875 )( 4.46 ) + ( 4.46 )(10 )
R1
8.87 =
1
( 4.46 ) ⇒ R1 = 5.03 kΩ
R1
5.03R2
= 4.46 ⇒ R2 = 39.4 kΩ
5.03 + R2
10
10
=
= 0.225 mA
R1 + R2 5.03 + 39.4
0.7 + 0.225 ≅ 0.925 mA from VCC source.
Now rπ =
Av =
6.30
(80 ) ( 0.026 )
0.7
(80 )( 6.6 )
= 2.97 kΩ
2.97 + ( 81)( 0.55 )
= 11.1
⫹5V
RC
R1
CC2
␯o
CC1
RL ⫽ 10 K
␯s
⫹
⫺
R2
CE
RE
⫺5V
β = 120
Let I CQ = 0.35 mA, I EQ = 0.353 mA
I BQ = 0.00292 mA
Let RE = 2 kΩ. For VCEQ = 4 V ⇒ 10 = 4 + ( 0.35) RC + ( 0.353)( 2)
RC = 15.1 kΩ, rπ =
Av =
− β ( RC RL )
rπ
(120 )( 0.026 )
=−
= 8.91 kΩ
0.35
(120 ) (15.1 10 )
8.91
Av = −81.0
For bias stable circuit:
R1 R2 = RTH = ( 0.1)(1 + β ) RE
= ( 0.1)(121)( 2 ) = 24.2 kΩ
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10) − 5 = ⋅ RTH ⋅ (10 ) − 5
R1
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
( 24.2 )(10 ) − 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) − 5
R1
1
( 242 ) = 1.477, R1 = 164 kΩ
R1
164 R2
= 24.2 ⇒ R2 = 28.4 kΩ
164 + R2
10
= 0.052, 0.35 + 0.052 = 0.402 mA
164 + 28.4
So bias current specification is met.
6.31
From Prob. 6.12,
RTH = R1} R2 = 10}50 = 8.33 kΩ
⎛ R2 ⎞
⎛ 50 ⎞
VTH = ⎜
⎟ (12 ) = ⎜
⎟ (12 ) = 10 V
⎝ 50 + 10 ⎠
⎝ R1 + R2 ⎠
12 − 0.7 − 10
= 0.0119 mA
I BQ =
8.33 + (101)(1)
I CQ = 1.19 mA, I EQ = 1.20 mA
VECQ = 12 − (1.19 )( 2 ) − (120 )(1) = 8.42 V
1.19
1
8.42
11 12
For 1 ≤ vEC ≤ 11
DvEC = 11 − 8.42 = 2.58
⇒ Output voltage swing = 5.16 V (peak-to-peak)
6.32
I BQ =
5 − 0.7
= 0.00315 mA
50 + (101)( 0.1 + 12.9)
I CQ = 0.315 mA, I EQ = 0.319 mA
VCEQ = ( 5 + 5 ) − ( 0.315 )( 6 ) − ( 0.319 )(13)
VCEQ = 3.96 V
AC load line
⫺1
Slope ⫽
6.1 K
0.315
3.96
10
1
Δv
6.1 eC
For ΔiC = 0.315 − 0.05 = 0.265 ⇒ ΔvEC = 1.62
ΔiC = −
vEC ( min ) = 3.96 − 1.62 = 2.34
Output signal swing determined by current:
Max. output swing = 3.24 V peak-to-peak
6.33
From Problem 4.18, I CQ = 1.408 mA, I EQ = 1.426 mA
(a) VECQ = 30 − (1.408 )( 5 ) − (1.426 )(10 ) = 8.7 V
IC (mA)
AC load line
⫺1
Slope ⫽
RC 兩兩RL
⫺1
⫽
2.5 kΩ
1.408
8.7
␯EC (V)
vEC ( max ) = 8.7 + ΔI C ⋅ ( 2.5 ) = 8.7 + (1.408 )( 2.5 ) = 12.22
Set vEC ( max ) = 12 = 8.7 + ΔI C ( 2.5 ) ⇒ ΔI C = 1.32 mA
So ΔvEC (peak-to-peak) = 2(12 − 8.7) = 6.6 V
(b)
ΔiC (peak-to-peak) = 2(1.32) = 2.64 mA
6.34
I EQ
I BQ
VE
VC
VECQ
= 0.80 mA, I CQ = 0.792 mA
= 0.00792 mA
= 0.7 + ( 0.00792 )(10 ) = 0.779 V
= I CQ RC − 5 = ( 0.792 )( 4 ) − 5 = 2 1.83 V
= 0.779 − ( −1.83) = 2.61 V
Load line: Assume VE remains constant at ≈ 0.78 V
IC (mA)
AC load line
⫺1
Slope ⫽
RC 兩兩RL
⫺1
⫽
2.5 kΩ
1.408
8.7
␯EC (V)
21
?vec
2 kV
Collector current swing = 0.792 − 0.08
= 0.712 mA
Dvec = ( 0.712 )( 2 ) = 1.424 V
DiC =
Output swing determined by current.
Max. output swing = 2.85 V peak-to-peak
2.85
4
= 0.712 mA peak-to-peak
Swing in i0 current =
6.35
I BQ =
6 − 0.7
= 0.0169 mA
10 + (101)( 3)
I CQ = 1.69 mA, I EQ = 1.71 mA
VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3)
VCEQ = 5.38 V
AC load line
⫺1
Slope ⫽
3.4 ⫹ 3
⫺1
⫽
6.4 K
1.69
5.38
22
1
DiC = 2
Dvce
6.4
For vce ( min ) = 1 V, Dvce = 5.38 − 1 = 4.38 V ⇒ DiC =
4.38
= 0.684 mA
6.4
Output swing limited by voltage:
Δvce = Max. swing in output voltage
= 8.76 V peak-to-peak
1
ΔiC ⇒ Δi0 = 0.342 mA
2
or Δi0 = 0.684 mA (peak-to-peak)
Δi0 =
6.36
AC load line
⫺1
Slope ⫽
1.05 K
2.65
Q-point
ICQ
VCEQ
ro =
9
100
I CQ
Neglect ro as (E) approx. dc load line VCE = 9 − I C ( 3.4 )
ΔI C = I CQ − 0.1
ΔVCE = VCEQ − 1
Also ΔVCE = ΔI C ( RC RL ) = ΔI C (1.05 )
Or VCEQ − 1 = ( I CQ − 0.1) (1.05 )
Substituting the expression for the dc load line.
⎣⎡9 − I CQ ( 3.4 ) − 1⎦⎤ = ( I CQ − 0.1) (1.05 )
8.105 = I CQ ( 4.45 ) ⇒ I CQ = 1.821 mA
VCEQ = 2.81 V
1.821
I BQ =
= 0.01821
100
RTH = ( 0.1)(101)(1.2 ) = 12.12 K
1
1
VTH = ⋅ RTH ⋅ VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 )
R1
R1
= 0.2207 + 0.7 + 2.20705
R1 = 34.9 K
R2 = 18.6 K
34.9 R2
= 12.12
34.9 + R2
6.37
dc load line
5
⫽ 4.55 mA
1 ⫹ 0.1
AC load line
⫺1
Slope ⫽
1兩兩1.2
⫺1
⫽
0.545 K
ICQ
5
VCEQ
For maximum symmetrical swing
ΔiC = I CQ − 0.25
ΔvCE = VCEQ − 0.5 and ΔiC =
I CQ − 0.25 =
1
⋅ | ΔvCE |
0.545 kΩ
VCEQ − 0.5
0.545
VCEQ = 5 − I CQ (1.1)
0.545 ( I CQ − 0.25 ) = ⎡⎣5 − I CQ (1.1) ⎤⎦ − 0.5
( 0.545 + 1.1) I CQ = 5 − 0.5 + 0.136
I CQ = 2.82 mA,
I BQ = 0.0157 mA
RTH = R1 & R2 = ( 0.1)(1 + β ) RE
= ( 0.1)(181)( 0.1) = 1.81 kΩ
VTH =
1
⋅ RTH ⋅ V + = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
R1
1
(1.81)( 5 ) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1)
R1
1
( 9.05 ) = 1.013 ⇒ R1 = 8.93 kΩ
R1
8.93R2
= 1.81 ⇒ R2 =2.27 kΩ
8.93 + R2
6.38
I CQ = 0.647 mA , VCEQ > 10 − ( 0.647 )( 9 ) = 4.18 V
DiC = I CQ = 0.647 mA
So DvCE = DiC ( 4} 4 ) = ( 0.647 )( 2 ) = 1.294 V
Voltage swing is well within the voltage specification. Then DvCE = 2 (1.294 ) = 2.59 V peak-to-peak
6.39
a.
RTH = R1} R2 = 10}10 = 5 kΩ
⎛ R2 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ (18 ) − 9 = ⎜
⎟ (18 ) − 9 = 0
⎝ 10 + 10 ⎠
⎝ R1 + R2 ⎠
0 − 0.7 − ( −9 )
= 0.0869 mA
I BQ =
5 + (181)( 0.5 )
I CQ = 15.6 mA, I EQ = 15.7 mA
VCEQ = 18 − (15.7 )( 0.5 ) ⇒ VCEQ = 10.1 V
b.
AC load line
⫺1
Slope ⫽
0.5兩兩0.3
⫺1
⫽
0.188 K
15.6
10.1
18
c.
rπ =
(180 )( 0.026 )
= 0.30 kΩ
15.6
(1 + β )( RE & RL ) ⎛ R1 & R2 & Rib ⎞
⋅⎜
Av =
⎟
rπ + (1 + β ) ( RE RL ) ⎝ R1 & R2 & Rib + RS ⎠
Rib = rπ + (1 + β )( RE & RL ) = 0.30 + (181)( 0.5 & 0.3) or Rib = 34.2 k Ω
R1 & R2 & Rib = 5 & 34.2 = 4.36 k Ω
Av =
d.
(181)( 0.5 & 0.3) ⎛ 4.36 ⎞
⋅⎜
⎟ ⇒ Av
0.3 + (181)( 0.5 & 0.3) ⎝ 4.36 + 1 ⎠
= 0.806
Rib = rp + (1 + b ) ( RE } RL )
Rib = 0.30 + (181)( 0.188 ) ⇒ Rib = 34.3 kΩ
Ro = RE
rp + R1} R2 } RS
0.3 + 5}1
= 0.5
⇒ Ro = 6.18 Ω
1+ b
181
6.40
a.
RTH = R1} R2 = 10}10 = 5 kΩ
⎛ R2 ⎞
VTH = ⎜
⎟ ( −10 ) = 2 5 V
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE − 10
I BQ =
2 5 − 0.7 − ( −10 )
5 + (121)( 2 )
= 0.0174 mA
I CQ = 2.09 mA, I EQ = 2.11 mA
VCEQ = 10 − ( 2.09 )(1) − ( 2.11)( 2 ) ⇒ VCEQ = 3.69 V
b.
AC load line
⫺1
Slope ⫽
2兩兩2
⫺1
⫽
1K
2.09
3.69
10
c.
rπ =
(120 )( 0.026 )
= 1.49 kΩ
2.09
(1 + β ) ( RE RL ) ⎛ R1 R2 Rib ⎞
Av =
⋅⎜
⎟
rπ + (1 + β ) ( RE RL ) ⎝⎜ R1 R2 Rib + RS ⎠⎟
Rib = rπ + (1 + β ) ( RE RL ) = 1.49 + (121) ( 2 2)
Rib = 122.5 k Ω,
Av =
R1 R2 Rib = 5 122.5 = 4.80 k Ω
(121) ( 2 2) ⎛ 4.80 ⎞
⋅⎜
⎟ ⇒ Av
1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠
= 0.484
d.
Rib = rπ + (1 + b ) ( RE } RL )
Rib = 1.49 + (121) ( 2} 2 ) 1 Rib = 122 kΩ
Ro = RE
6.41
a.
rπ + R1} R2 } RS
1.49 + 5}5
1 Ro = 32.4 Ω
=2
1+ b
121
RTH = R1 R2 = 60 40 = 24 kΩ
⎛ R2 ⎞
⎛ 40 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ ( 5) = 2 V
⎝ 40 + 60 ⎠
⎝ R1 + R2 ⎠
5 − 0.7 − 2
I BQ =
= 0.0130 mA
24 + ( 51)( 3)
I CQ = 0.650 mA, I EQ = 0.663 mA
VECQ = 5 − I EQ RE = 5 − ( 0.663)(3) ⇒ VECQ = 3.01 V
b.
1.63
AC load line
⫺1
Slope ⫽ 51
冸 50 冹 冸3兩兩4冹
⫺1
⫽
1.75 K
0.65
5
3.01
c.
( 50 )( 0.026 )
80
= 2 kΩ, r0 =
= 123 kΩ
0.650
0.65
Define RL′ = RE RL r0 = 3 4 123 = 1.69 kΩ
rπ =
Av =
(1 + β ) RL′
rπ (1 + β ) RL′
=
( 51)(1.69 )
2 + ( 51)(1.69 )
⎛ RE r0
Ai = (1 + β ) I b ⎜
⎜R r +R
L
⎝ E 0
⇒ Av = 0.977
⎞
⎟⎟
⎠
⎛ RTH ⎞
Ib = I S ⎜
⎟
⎝ RTH + Rib ⎠
Rib = rπ + (1 + β ) RL′ = 2 + ( 51)(1.69 ) = 88.2
RE r0 = 3 r0 = 3 123 = 2.93
24 ⎞
⎛ 2.93 ⎞ ⎛
Ai = ( 51) ⎜
⎟⎜
⎟ ⇒ Ai = 4.61
2.93
4
24
88.2 ⎠
+
+
⎝
⎠⎝
d.
Rib = rπ + (1 + β ) RE RL r0 = 2 + ( 51)(1.69 ) ⇒ Rib = 88.2 kΩ
rπ
⎛ 2⎞
RE = ⎜ ⎟ 3 = 0.0392 3
1+ β
⎝ 51 ⎠
R0 = 38.7 Ω
e.
Assume variations in rπ and r0 have negligible effects
R1 = 60 ± 5% R1 = 63 kΩ,
R1 = 57 kΩ
R0 =
R2 = 40 ± 5% R2 = 42 kΩ,
R2 = 38 kΩ
RE = 3 ± 5%
RE = 3.15 kΩ, RE = 2.85 kΩ
RL = 4 ± 5%
RL = 4.2 kΩ,
RL = 3.8 kΩ
⎛ RE r0 ⎞ ⎛ RTH ⎞
Ai = (1 + β ) ⎜
⎜ R r + R ⎟⎟ ⎜ R + R ⎟
L ⎠ ⎝ TH
ib ⎠
⎝ E 0
Rib = rπ + (1 + β ) ( RE RL r0 )
RTH ( max ) = 25.2 kΩ, RTH ( min ) = 22.8 kΩ
Rib ( max ) = 92.5 kΩ, Rib ( min ) = 84.0 kΩ
RE ( max ) , RL ( min ) , Rib = 88.6 kΩ
RE ( min ) , RL ( max ) , Rib = 87.4 kΩ
RE ( max ) & r0 = 3.07 kΩ
RE ( min ) & r0 = 2.79 kΩ
For RE ( min ) , RL ( max ) , RTH ( min )
22.8 ⎞
⎛ 2.79 ⎞ ⎛
Ai = ( 51) ⎜
⎟⎜
⎟ ⇒ Ai = 4.21
⎝ 2.79 + 4.2 ⎠ ⎝ 22.8 + 87.4 ⎠
For RE ( max ) , RL ( min ) , RTH ( max )
25.2 ⎞
⎛ 3.07 ⎞⎛
Ai = ( 51) ⎜
⎟⎜
⎟ ⇒ Ai = 5.05
3.07
+
3.8
25.2
+ 88.6 ⎠
⎝
⎠⎝
6.42
(a)
0.5
= 0.00617 mA
81
VB = I BQ RB = ( 0.00617 )(10 ) ⇒ VB = 0.0617 V
I BQ =
VE = VB + 0.7 ⇒ VE = 0.7617 V
(b)
⎛ 80 ⎞
I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA
⎝ 81 ⎠
I CQ 0.494
gm =
=
⇒ g m = 19 mA / V
VT
0.026
rπ =
ro =
β VT
I CQ
=
(80 )( 0.026 )
0.494
⇒ rπ = 4.21 k Ω
VA
150
=
⇒ ro = 304 k Ω
I CQ 0.494
(c)
RS
Vs
⫹
⫺
IS
V⬘S
⫺
RB V␲
r␲
ro
gmV␲
⫹
Vo
RL
Io
For RS = 0
⎛V
⎞
Vo = − ⎜ π + g mVπ ⎟ ( RL & ro )
⎝ rπ
⎠
−Vo
⎛1+ β ⎞
⎜
⎟ ( RL & ro )
⎝ rπ ⎠
Now Vs + Vπ = Vo
so that Vπ =
or Vs = Vo − Vπ = Vo +
Vo
⎛ 1+ β ⎞
⎜
⎟ ( RL & ro )
⎝ rπ ⎠
We find
(1 + β )( RL & ro )
(81)( 0.5 & 304 )
=
rπ + (1 + β )( RL & ro ) 4.21 + ( 81)( 0.5 & 304 )
(81)( 0.5 )
≅
⇒ Av = 0.906
4.21 + ( 81)( 0.5 )
Rib = rπ + (1 + β )( RL & ro ) ≅ 4.21 + ( 81)( 0.5 ) = 44.7 k Ω
Av =
Vo
Vs
=
⎛ RB ⎞
⎛ ro ⎞
Ib = ⎜
⎟ ⋅ I s and I o = ⎜
⎟ (1 + β ) I b
⎝ RB + Rib ⎠
⎝ ro + RL ⎠
Then
⎛ RB ⎞⎛ ro ⎞
Io
= (1 + β ) ⎜
⎟⎜
⎟
Is
⎝ RB + Rib ⎠⎝ ro + RL ⎠
⎛ 10 ⎞
Ai ≅ ( 81) ⎜
⎟ (1) ⇒ Ai = 14.8
⎝ 10 + 44.7 ⎠
(d)
⎛ RB + Rib ⎞
⎛ 10 44.7 ⎞
Vs′ = ⎜
⋅V =
⋅ V = 0.803) Vs
⎜ R R + R ⎟⎟ s ⎜⎜ 10 44.7 + 2 ⎟⎟ s (
s ⎠
⎝ B ib
⎝
⎠
Then Av = ( 0.803)( 0.906 ) ⇒ Av = 0.728
Ai =
Ai = 14.8 (Unchanged)
6.43
(a)
I CQ = 1.98 mA
ro =
rπ =
(100 )( 0.026 )
1.98
= 1.313 K
VA
100
=
I CQ 1.98
= 50.5 K
rπ + RS
1.31 + 10
ro =
50.5 ⇒ Ro = 112 Ω
1+ β
101
0.112 & 50.5 ⇒ Ro ≅ 112 Ω
(b)
From Equation 4.68
(1 + β ) ( ro RL )
100
Av =
ro =
= 50.5 K
1.98
rπ + (1 + β ) ( ro RL )
Ro =
(i)
RL = 0.5 K
(101) ( 50.5 0.5)
1.31 + (101) ( 50.5 0.5 )
(101)( 0.4951)
⇒ Av = 0.974
Av =
1.31 + (101)( 0.4951)
Av =
(ii)
RL = 5 K
Av =
ro RL = 50.5 5 = 4.5495
(101)( 4.55)
⇒ Av = 0.997
1.31 + (101)( 4.55 )
6.44
5 − 0.7
I CQ = 1.293 mA
= 1.303
3.3
(125 )( 0.026 )
rπ =
= 2.51 K
1.293
1.293
gm =
= 49.73 mA/V
0.026
(a)
Rib = rπ + (1 + β ) ( RE RL ) = 2.51 + (126 ) ( 3.3 1)
I EQ =
Rib = 99.2 K
Ro = RE
rπ
2.51
= 3.3
= 3.3 0.01992
1+ β
126
Ro = 19.8 Ω
(b)
v
2sin ω t
is = s =
⇒ is ( t ) = 20.2sin ω t ( µ A )
Rib
99.2
veb ( t ) = −is ( t ) rπ = ( −20.2 )( 2.51) sin ω t
veb ( t ) = −50.6sin ω t ( mV )
(126 ) ( 3.3 1)
(1 + β ) ( RE RL )
(126 )( 0.7674 )
=
=
rπ + (1 + β ) ( RE RL ) 2.51 + (126 ) ( 3.3 1) 2.51 + (126 )( 0.7674 )
Av = 0.9747 ⇒ vo ( t ) = 1.95sin ω t ( V )
v (t )
⇒ io ( t ) = 1.95sin ω t ( mA )
io ( t ) = o
Av =
RL
6.45
a.
I EQ = 1 mA , VCEQ = VCC − I EQ RE
5 = 10 − (1)( RE ) ⇒ RE = 5 kΩ
1
= 0.0099 mA
101
10 = I BQ RB + VBE ( on ) + I EQ RE
10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ
I BQ =
b.
␯b
␯0
RB
rπ =
RE
(100 )( 0.026 )
0.99
(1 + β ) RE
= 2.63 kΩ
(101)( 5 )
v0
=
=
= 0.995
vb rπ + (1 + β ) RE 2.63 + (101)( 5 )
⇒ vb =
v0
4
=
⇒ vb = 4.02 V peak-to-peak at base
0.995 0.995
RS
␯b
␯S
⫹
⫺
RB兩兩Rib
Rib = rπ + (1 + β ) RE = 508 kΩ
RB Rib = 434 508 = 234 kΩ
vb =
RB Rib
RB Rib + RS
⋅ vS =
vb = 0.997vS ⇒ vS =
234vS
234
=
vS
234 + 0.7 234.7
4.02
⇒ vS = 4.03 V peak-to-peak
0.997
c.
Rib = rπ + (1 + β ) ( RE RL )
Rib = 2.63 + (101) ( 5 1) = 86.8 kΩ
RB Rib = 434 86.8 = 72.3 kΩ
⎛ 72.3 ⎞
vb = ⎜
⎟ vS = 0.99vS = ( 0.99 )( 4.03)
⎝ 72.3 + 0.7 ⎠
vb = 3.99 V peak-to-peak
(1 + β )( RE & RL )
⋅ vb
rπ + (1 + β )( RE & RL )
(101)( 0.833)
=
( 3.99 )
2.63 + (101)( 0.833)
v0 =
v0 = 3.87 V peak-to-peak
6.46
RTH = R1 & R2 = 40 & 60 = 24 kΩ
⎛ 60 ⎞
VTH = ⎜
⎟ (10 ) = 6 V
⎝ 60 + 40 ⎠
6 − 0.7
β = 75 I BQ =
= 0.0131 mA
24 + ( 76 )( 5 )
I CQ = 0.984 mA
β = 150 I BQ =
6 − 0.7
= 0.00680 mA
24 + (151)( 5 )
I CQ = 1.02 mA
β = 75 rπ =
( 75 )( 0.026 )
0.984
β = 150 rπ = 3.82 kΩ
= 1.98 kΩ
β = 75 Rib = rπ + (1 + β )( RE & RL ) = 65.3 kΩ
β = 150 Rib = 130 kΩ
(1 + β )( RE & RL )
R1 & R2 & Rib
⋅
Av =
rπ + (1 + β )( RE & RL ) R1 & R2 & Rib + RS
For β = 75, R1 & R2 & Rib = 40 & 60 & 65.3 = 17.5 k Ω
Av =
( 76 )( 0.833)
17.5
⋅
⇒ Av = 0.789
1.98 + ( 76 )( 0.833) 17.5 + 4
For β = 150, R1 & R2 & Rib = 40 & 60 & 130 = 20.3 k Ω
Av =
(151)( 0.833)
20.3
⋅
⇒ Av = 0.811
3.82 + (151)( 0.833) 20.3 + 4
So 0.789 ≤ Av ≤ 0.811
⎛ RE ⎞ ⎛ RTH ⎞
Ai = (1 + β ) ⎜
⎟
⎟⎜
⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠
β = 75
24 ⎞
⎛ 5 ⎞⎛
Ai = ( 76 ) ⎜
⎟⎜
⎟ ⇒ Ai = 17.0
⎝ 5 + 1 ⎠ ⎝ 24 + 65.3 ⎠
β = 150
⎛ 5 ⎞ ⎛ 24 ⎞
Ai = (151) ⎜ ⎟ ⎜
⎟ ⇒ Ai = 19.6
⎝ 6 ⎠ ⎝ 24 + 130 ⎠
17.0 ≤ Ai ≤ 19.6
6.47
(a)
⎛ I ⎞
9 = ⎜ E ⎟ (100 ) + VBE ( on ) + I E RE
⎝ 1+ β ⎠
9 − 0.7
IE =
⎛ 100 ⎞
⎜
⎟ + RE
⎝ 1+ β ⎠
8.3
= 2.803 mA
⎛ 100 ⎞
1
+
⎜
⎟
⎝ 51 ⎠
8.3
β = 200 I E =
= 5.543 mA
⎛ 100 ⎞
⎜
⎟ +1
⎝ 201 ⎠
2.80 ≤ I E ≤ 5.54 mA
β = 50 I E =
VE = I E RE , β = 50, VE = 2.80 V
β = 200, VE = 5.54 V
(b)
β = 50, I CQ = 2.748 mA, rπ = 0.473 K
β = 200, I CQ = 5.515 mA, rπ = 0.943 K
Ri = RB & ⎡⎣ rπ + (1 + β ) RE & RL ⎤⎦
β = 50 ⇒ Ri = 100 ⎡⎣0.473 + ( 51)(1 & 1) ⎤⎦ = 100 25.97 = 20.6 K
β = 200 ⇒ Ri = 100 ⎣⎡0.943 + ( 201)(1 & 1) ⎦⎤ = 100 101.4 = 50.3 K
From Fig. (4.68)
(1 + β ) ( RE RL ) ⎛ Ri ⎞
Av =
⋅⎜
⎟
rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠
=
( 51) (1 1)
⎛ 20.6 ⎞
⋅⎜
⎟
0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠
β = 50 ⇒ Av = 0.661
β = 200 ⇒ Av =
( 201) (1 1) ⎛ 50.3 ⎞
⎜
⎟
0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠
Av = 0.826
6.48
Vo = (1 + β ) I b RL
Vs
Ib =
rπ + (1 + β ) RL
so Av =
(1 + β ) RL
rπ + (1 + β ) RL
For β = 100, RL = 0.5 k Ω
rπ =
(100 )( 0.026 )
0.5
= 5.2 k Ω
(101)( 0.5 )
= 0.9066
5.2 + (101)( 0.5 )
Then Av ( min ) =
Then β = 180, RL = 500 k Ω
rπ =
(180 )( 0.026 )
0.5
Then Av ( max ) =
= 9.36 k Ω
(181)( 500 )
= 0.9999
9.36 + (181)( 500 )
6.49
Rib
IS
⫹
Ib
V␲
gmV␲ ⫽ ␤Ib
r␲
⫺
␯S
⫹
⫺
R1兩兩R2
RE
RL
I0
⎛ RE ⎞
I 0 = (1+ β ) I b ⎜
⎟
⎝ RE + RL ⎠
⎛ R1 & R2 ⎞
Ib = I S ⎜
⎟
⎝ R1 & R2 + Rib ⎠
Rib = rπ + (1 + β )( RE & RL )
VCC = 10 V, For VCEQ = 5 V
⎛1+ β ⎞
5 = 10 − ⎜
⎟ I CQ RE
⎝ β ⎠
β = 80, For RE = 0.5 kΩ
I CQ = 9.88 mA, I EQ = 10 mA, I BQ = 0.123 mA
rπ =
(80 )( 0.026 )
= 0.211 kΩ
9.88
Rib = 0.211 + ( 81)( 0.5 & 0.5 ) ⇒ Rib = 20.46 kΩ
Ai =
⎛ RE ⎞ ⎛ R1 & R2 ⎞
I0
= (1 + β ) ⎜
⎟
⎟⎜
IS
⎝ RE + RL ⎠⎝ R1 & R2 + Rib ⎠
⎞
R1 & R2
⎛ 1 ⎞⎛
8 = ( 81) ⎜ ⎟ ⎜
⎟
⎝ 2 ⎠ ⎝ R1 & R2 + 20.46 ⎠
0.1975 ⎡⎣ R1} R2 + 20.46 ⎤⎦ = R1} R2
R1} R2 ⇒ 5.04 kΩ
VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
1
( 5.04 )(10 ) = ( 0.123)( 5.04 ) + 0.7 + (10 )( 0.5 ) ⇒ R1 = 7.97 kΩ
R1
7.97 R2
= 5.04 ⇒ R2 = 13.7 kΩ
7.97 + R2
Now Ro = RE
rπ
0.211
= 0.5
or Ro = 2.59 Ω
1+ b
81
(b)
Rib = 0.211 + (81) ( 0.5&2) = 32.6 k Ω
⎛ 0.5 ⎞ ⎛ 5.04 ⎞
Ai = ( 81) ⎜
⎟⎜
⎟ = ( 81)( 0.2 )( 0.134 )
⎝ 0.5 + 2 ⎠ ⎝ 5.04 + 32.6 ⎠
Ai = 2.17
6.50
Ri = RTH Rib where Rib = rπ + (1 + β ) RE
5 − 3.5
VCEQ = 3.5, I CQ
= 0.75 mA
2
(120 )( 0.026 )
rπ =
= 4.16 k Ω
0.75
Rib = 4.16 + (121) ( 2 ) = 246 k Ω
Then Ri = 120 = RTH 246 ⇒ RTH = 234 k Ω
0.75
= 0.00625 mA
I BQ =
120
VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE
1
1
⋅ RTH ⋅ VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 )
R1
R1
which yields R1 = 318 k Ω and R2 = 886 k Ω
6.51
a.
Let RE = 24 Ω and VCEQ = 12 VCC = 12 V ⇒ I EQ =
12
= 0.5 A
24
I CQ = 0.493 A, I BQ = 6.58 mA
rπ =
( 75)( 0.026 )
0.493
= 3.96 Ω
Reb
Is
Ib
⫹
V␲
gmV␲ ⫽ ␤Ib
r␲
⫺
VS
⫹
⫺
R1 兩兩 R2
⫽ Rrn
RE
Io
RL
⎛ RE ⎞
I 0 = (1 + β ) I b ⎜
⎟
⎝ RE + RL ⎠
⎛ RTH ⎞
Ib = I S ⎜
⎟
⎝ RTH + Rib ⎠
Rib = rπ + (1 + β ) ( RE & RL )
= 3.96 + ( 76 )( 24 & 8 ) ⇒ Rib = 460 Ω
Ai =
⎛ RE ⎞ ⎛ RTH ⎞
I0
= (1 + β ) ⎜
⎟
⎟⎜
IS
⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠
⎞
⎛ 24 ⎞ ⎛ RTH
8 = ( 76) ⎜
⎟⎜
⎟
⎝ 24 + 8 ⎠ ⎝ RTH + 460 ⎠
RTH
0.140 =
⇒ RTH = 74.9 Ω (Minimum value)
RTH + 460
dc analysis:
1
VTH = ⋅ RTH ⋅ VCC
R1
= I BQ RTH + VBE ( on ) + I EQ RE
1
( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5 )( 24 )
R1
= 13.19
R1 = 136 Ω,
136 R2
= 74.9 ⇒ R2 = 167 Ω
136 + R2
b.
0.493
AC load line
⫺1
Slope ⫽
24兩兩8
⫺1
⫽
6⍀
12
24
1
ΔiC = − Δvce
6
For ΔiC = 0.493 ⇒ Δvce = ( 0.493)( 6 ) ⇒ Max. swing in output voltage for this design
= 5.92 V peak-to-peak
c.
R0 =
rπ
3.96
RE =
24 = 0.0521 24 ⇒ R0 = 52 mΩ
1+ β
76
6.52
The output of the emitter follower is
⎛ RL ⎞
vo = ⎜
⎟ ⋅ vTH
⎝ RL + Ro ⎠
Ro
⫹
⫹
⫺
␯TH
␯O
RL
⫺
For vO to be within 5% for a range of RL , we have
RL ( min )
RL ( min ) + Ro
= ( 0.95 )
RL ( max )
RL ( max ) + Ro
4
10
= ( 0.95 )
which yields Ro = 0.364 k Ω
4 + Ro
10 + Ro
⎛ r + R1 & R2 & RS ⎞
We have Ro = ⎜ π
⎟ RE ro
1+ β
⎝
⎠
The first term dominates
Let R1 & R2 & RS ≅ RS , then
rπ + RS
r +4
⇒ 0.364 = π
1+ β
1+ β
Ro ≅
or 0.364 =
0.364 ≅
rπ
β VT
4
4
+
=
+
1 + β 1 + β I CQ (1 + β ) 1 + β
VT
4
+
I CQ 1 + β
The factor
V
4
4
4
= 0.044 to
= 0.0305. We can set Ro ≅ 0.32 = T
is in the range of
91
131
I CQ
1+ β
Or I CQ = 0.08125 mA. To take into account other factors, set I CQ = 0.15 mA,
I BQ =
0.15
= 0.00136 mA
110
5
= 33.3 k Ω
0.15
Design a bias stable circuit.
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10) − 5 = ( RTH )(10) − 5
R1
⎝ R1 + R2 ⎠
For VCEQ ≅ 5 V , set RE =
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(111)(33.3) = 370 k Ω
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
( 370 )(10 ) − 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) − 5
R
1
which yields R1 = 594 k Ω and R2 = 981 k Ω
So
Now Av =
(1 + β ) ( RE RL ) ⎛ RTH Rib ⎞
⋅⎜
⎟
rπ + (1 + β ) ( RE RL ) ⎝ RTH Rib + RS ⎠
Rib = rπ + (1 + β ) ( RE RL )
and rπ
=
β VT
I CQ
For β = 90, RL = 4 k Ω,
rπ = 15.6 k Ω, Rib = 340.6 k Ω
Av =
( 91)( 33.3 & 4 )
370 & 340.6
⋅
⇒ Av = 0.9332
15.6 + ( 91)( 33.3 & 4 ) 370 & 340.6 + 4
For β = 90, RL = 10 k Ω
Rib = 715.4 k Ω
Av =
( 91)( 33.3 & 10 )
370 & 715.4
⋅
⇒ Av = 0.9625
15.6 + ( 91)( 33.3 & 10 ) 370 & 715.4 + 4
For β = 130, RL = 4 k Ω
rπ = 22.5 k Ω, Rib = 490 k Ω
Av =
(131)( 33.3 & 4 )
370 & 490
⋅
⇒ Av = 0.9360
22.5 + (131)( 33.3 & 4 ) 370 & 490 + 4
For β = 130, RL = 10 k Ω
Rib = 1030 k Ω
(131)( 33.3 & 10 )
370 & 1030
⋅
⇒ Av = 0.9645
22.5 + (131)( 33.3 & 10 ) 370 & 1030 + 4
Now vO ( min ) = Av ( min ) .vS = 3.73sin ω t
vO ( max ) = Av ( max ) .vS = 3.86sin ω t
Av =
ΔvO
= 3.5%
vO
6.53
PAVG = iL2 ( rms ) RL ⇒ 1 = iL2 ( rms )(12 )
so iL ( rms ) = 0.289 A ⇒ iL ( peak ) = 2 ( 0.289 )
iL ( peak ) = 0.409 A
vL ( peak ) = iL ( peak ) ⋅ RL = ( 0.409 )(12 ) = 4.91 V
4.91
= 0.982
5
With RS = 10 k Ω, we will not be able to meet this voltage gain requirement. Need to insert a buffer or an
Need a gain of
op-amp voltage follower (see Chapter 9) between RS and CC1 .
1
Set I EQ = 0.5 A, VCEQ = (12 − ( −12 ) ) = 8 V
3
24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω
50
( 0.5 ) = 0.49 A
51
β VT ( 50 )( 0.026 )
=
= 2.65 Ω
rπ =
I CQ
0.49
Let β = 50, I CQ =
Rib = rπ + (1 + β ) ( RE RL ) = 2.65 + ( 51) ( 32 12 )
Rib = 448 Ω
Av =
(1 + β ) ( RE RL )
( 51) ( 32 12 )
=
= 0.994
rπ + (1 + β ) ( RE RL ) 2.65 + ( 51) ( 32 12 )
So gain requirement has been met.
0.49
= 0.0098 A = 9.8 mA
50
24
≅ 10 I B = 98 mA
Let I R ≅
R1 + R2
I BQ =
So that R1 + R2 = 245 Ω
VTH =
R2
( 24 ) − 12 = I BQ RTH + VBE ( on ) + I EQ RE − 12
R1 + R2
( 0.0098) R1 R2
⎛ R2 ⎞
+ 0.7 + ( 0.5 )( 32 )
⎜ 245 ⎟ ( 24 ) =
245
⎝
⎠
Now R1 = 245 − R2
So we obtain
4 × 10−5 R22 + 0.0882 R2 − 16.7 = 0 which yields R2 = 175 Ω and R1 = 70 Ω
6.54
(a)
RTH = R1 R2 = 25.6 10.4 = 7.40 k Ω
⎛ R2 ⎞
⎛ 10.4 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (18 ) = 5.2 V
⎝ 10.4 + 25.6 ⎠
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
I BQ =
5.2 − 0.7
= 0.0117 mA
7.40 + (126 )( 3)
Then I CQ = 1.46 mA and I EQ = 1.47 mA
VCEQ = VCC − I CQ RC − I EQ RE
VCEQ = 18 − (1.46 )( 4 ) − (1.47 )( 3) ⇒ VCEQ = 7.75 V
(b)
rπ =
(125) ( 0.026 )
= 2.23 k Ω
1.46
1.46
gm =
= 56.2 mA / V
0.026
Re
Vo
Ie
Is
RS
RE
r␲
Ib
RTH
␤I b
RC
RL
rπ + RTH 2.23 + 7.40
=
= 0.0764 k Ω
1+ β
126
Re =
Ie =
− ( RS RE )
(R
S
RE ) + Re
⋅ Is =
− (100 3)
(100 & 3) + 0.0764
or I e = − ( 0.974 ) I s
⎛ β
Vo = − I c ( RC RL ) = − ⎜
⎝ 1+ β
⋅ Is
⎞
⎟ I e ( RC RL )
⎠
⎛ β ⎞
Vo
⎛ 125 ⎞
= −⎜
⎟ ( −0.974 )( RC & RL ) = ⎜
⎟ ( 0.974 )( 4 & 4 )
Is
1
126 ⎠
+
β
⎝
⎝
⎠
V
Then Rm = o = 1.93 k Ω = 1.93 V / mA
Is
Then
(c)
(
) (
)
Vs = I s RS R E Re = I s 100 3 0.0764 = I s ( 0.0744 )
Vs
0.0744
V V
which yields o = o ( 0.0744 ) = 1.93
I s Vs
or I s =
or Av =
Vo
= 25.9
Vs
6.55
(a)
β ( RC RL )
Av =
rπ + R1 R2
, RL = 12 k Ω, β = 100
Let R1 & R2 = 50 k Ω, I CQ = 0.5 mA
VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE
(100 )( 0.026 )
0.5
= 0.005 mA, rπ =
= 5.2 k Ω
100
0.5
1
1
⋅ RTH ⋅ VCC = ( 50 )(12 ) = ( 0.005 )( 50 ) + 0.7 + (101)( 0.005 )( 0.5 )
R1
R1
I BQ =
which yields R1 = 500 k Ω
and R2 = 55.6 k Ω
Av =
(100 )(12 & 12 )
5.2 + 50
= 10.9, Design criterion is met.
(b)
I CQ
= 0.5 mA, I EQ = 0.505 mA
VCEQ = 12 − ( 0.5)(12) − ( 0.505)( 0.5) ⇒ VCEQ = 5.75 V
0.5
= 19.23 mA / V
0.026
Av = (19.23) (12 12 ) ⇒ Av = 115
Av = g m ( RC RL ) , g m =
6.56
a. Emitter current
I EQ = I CC = 0.5 mA
0.5
= 0.00495 mA
101
VE = I EQ RE = ( 0.5 )(1) ⇒ VE = 0.5 V
I BQ =
VB = VE + VBE ( on ) = 0.5 + 0.7 ⇒ VB = 1.20 V
VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) ⇒ VC = 1.7 V
b.
rπ =
(100 )( 0.026 )
= 5.25 kΩ
(100 )( 0.00495 )
(100 )( 0.00495 )
gm =
= 19.0 mA/V
0.026
Ri
gmV␲
VS
RS
⫹
⫺
⫺
RE V␲
r␲
RB
⫹
gmV␲
冢
冣
RE
V
RE ⫹ RS S
⫹
⫺
RS 兩兩RE ⫺
V␲
r␲
⫹
Vo = − g mVπ ( RB & RL )
Vπ = −
RE & Rie
⋅ VS = − ( 0.4971) VS
RE & Rie + RS
Vo = (19 )( 0.4971) VS (100 & 1)
Av = 9.37
c.
VX
⫹
⫺
IX
gmV␲
⫺
RE V␲
r␲
⫹
IX =
VX VX
+
− g mVπ , Vπ = −VX
RE rπ
IX
1
1
1
=
=
+ + gm
VX Ri RE rπ
or Ri = RE rπ
1
1
= 1 5.253
19
gm
Ri = 0.84 & 0.05252 ⇒ Ri = 49.4 Ω
6.57
(a)
V0
I EQ = 1 mA, I CQ = 0.9917 mA
RL
VC = 5 − ( 0.9917 )( 2 ) = 3.017 V
VE = −0.7 V
VCEQ = 3.72 V
(b)
Av = g m ( RC RL )
0.9917
= 38.14 mA/V
0.026
Av = ( 38.14 ) ( 2 10 ) ⇒ Av = 63.6
gm =
6.58
(a)
10 − 0.7
= 0.93 mA
10
= 0.921 mA
I EQ =
I CQ
VECQ = 20 − ( 0.93)(10 ) − ( 0.921)( 5 )
VECQ = 6.10 V
(b)
0.921
= 35.42 mA/V
0.026
Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 )
gm =
Av = 161
6.59
(a)
I EQ = 0.93 mA, I CQ = 0.921 mA
VECQ = 6.10 V
(b)
gm =
0.921
= 35.42 mA/V rπ = 2.82 K
0.026
From Eq. 6.90
( RC RL ) ⎡ rπ R R ⎤
Av = g m
⎢1 + β E S ⎥
RS
⎣
⎦
( 35.42 ) ( 50 5 ) ⎡ 2.82
⎤
=
10 0.1⎥
⎢
0.1
⎣ 101
⎦
( 35.42 )( 4.545 )
Av =
[0.0218]
0.1
Av = 35.1
6.60
(a)
⎛ 60 ⎞
I CQ = ⎜ ⎟ (1) ⇒ I CQ = 0.984 mA
⎝ 61 ⎠
⎛ 1⎞
VCEQ = I BQ RB + VBE ( on ) = ⎜ ⎟ (100 ) + 0.7
⎝ 61 ⎠
VCEQ = 2.34 V
(b)
Av = g m
(R
RL ) ⎡ rπ
⎤
RS ⎥
⎢
RS
⎣1 + β
⎦
B
0.984
= 37.85 mA/V
0.026
rπ = 1.59 K
gm =
⎤
⎢ 61 0.05⎥
0.05
⎣
⎦
= 1484 ⎡⎣ 0.0261 0.05⎤⎦
Av = 25.4
Av =
( 37.85) (100 2 ) ⎡1.59
6.61
is ( peak ) = 2.5 mA, Vo ( peak ) = 5 mV
vo
5 × 102 3
=
= 2 × 103 = 2 k Ω
is 2.5 × 102 6
From Problem 4.54
⎛ RS RE ⎞
Vo ⎛ β ⎞
=⎜
⎟⎟
⎟ ( RC & RL ) ⎜⎜
Is ⎝ 1+ β ⎠
⎝ RS RE + Rie ⎠
Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω
So we need Rm =
Now β = 120, so we have
⎛ RS RE
⎛ 120 ⎞
2=⎜
⎟ ( 4 5 ) ⎜⎜
⎝ 121 ⎠
⎝ RS RE + Rie
RS RE
= 0.9075
Then
RS RE + Re
⎞
⎛ RS RE ⎞
= 2.204 ⎜
⎟⎟
⎜ R R + R ⎟⎟
ie ⎠
⎠
⎝ S E
RS RE = 50 2 = 1.923 k Ω, so that Rie = 0.196 k Ω
Assume VCEQ = 3 V
VCC ≅ I CQ ( RC + RE ) + VCEQ
5 = I CQ ( 4 + 2 ) + 3 ⇒ I CQ = 0.333 mA
rπ =
(120 )( 0.026 )
= 9.37 k Ω
0.333
r + RTH
9.37 + RTH
Rie = π
⇒ 0.196 =
1+ β
121
which yields RTH = 14.35 k Ω
Now VTH = I BQ RTH + VBE ( on ) + I EQ RE
1
⎛ 121 ⎞
= 0.00833 mA, I EQ = ⎜
⎟ (1) = 1.008 mA
120
⎝ 120 ⎠
1
1
= ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 )
R1
R1
I BQ =
VTH
which yields R1 = 25.3 k Ω
and R2 = 33.2 k Ω
6.62
a.
20 − 0.7
= 1.93 mA
10
= 1.91 mA
I EQ =
I CQ
VECQ = VCC + VEB ( on ) − I C RC
= 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V
b.
RS
VS
⫹
⫺
IS
Rie
h fe Ib
V0
Ie
RE
hie
RC
RL
Ib
Neglect effect hoe
From Problem 6-16, assume
2.45 ≤ hie ≤ 3.7 kΩ
80 ≤ h fe ≤ 120
Vo = ( h fe I b ) ( RC & RL )
Rie =
⎛ RE ⎞
hie
, Ie = ⎜
⎟ IS
1 + h fe
⎝ RE + Rie ⎠
⎛ I ⎞
VS
Ib = ⎜ e ⎟ , I S =
⎜ 1+ h ⎟
RS + RE & Rie
fe ⎠
⎝
⎛ h fe ⎞
⎛ RE ⎞ ⎛
⎞
1
Av = ⎜
R & RL ) ⎜
⎟×⎜
⎟
⎜ 1 + h ⎟⎟ ( C
fe ⎠
⎝ RE + Rie ⎠ ⎝ RS + RE & Rie ⎠
⎝
High gain device: hie = 3.7 kΩ, h fe = 120
3.7
= 0.0306 kΩ
121
RE & Rie = 10 & 0.0306 = 0.0305
Rie =
10
1
⎛ 120 ⎞
⎛
⎞⎛
⎞
Av = ⎜
⎟ ( 6.5 & 5 ) ⎜
⎟⎜
⎟ ⇒ Av = 2.711
.
.
121
10
0
0306
1
0
0305
+
+
⎝
⎠
⎝
⎠⎝
⎠
Low gain device: hie = 2.45 kΩ, h fe = 80
2.45
= 0.03025 kΩ
81
RE & Rie = 10 & 0.03025 = 0.0302
Rie =
10
1
⎛ 80 ⎞
⎛
⎞⎛
⎞
Av = ⎜ ⎟ ( 6.5 & 5 ) ⎜
⎟⎜
⎟ ⇒ Av = 2.70 So Av ≈ constant
⎝ 81 ⎠
⎝ 10 + 0.03025 ⎠ ⎝ 1 + 0.0302 ⎠
2.70 ≤ Av ≤ 2.71
c.
Ri = RE & Rie
We found 0.0302 ≤ Ri ≤ 0.0305 kΩ
Neglecting hoe , Ro = RC = 6.5 kΩ
6.63
a.
Small-signal voltage gain
Av = g m ( RC & RL ) ⇒ 25 = g m ( RC & 1)
For VECQ = 3 V ⇒ VC = −VECQ + VEB ( on ) = −3 + 0.7 ⇒ VC = −2.3
VCC − I CQ RC + VC = 0 ⇒ I CQ =
5 − 2.3 2.7
=
= I CQ
RC
RC
For I CQ = 1 mA, RC = 2.7 kΩ
1
= 38.5 mA/V
0.026
Av = ( 38.5 )( 2.7 & 1) = 28.1
gm =
Design criterion satisfied and VECQ satisfied.
⎛ 101 ⎞
IE = ⎜
⎟ (1) = 1.01 mA
⎝ 100 ⎠
VEE = I E RE + VEB ( on ) ⇒ RE =
b.
rπ =
β VT
I CQ
=
(100)( 0.026)
1
5 − 0.7
⇒ RE = 4.26 kΩ
1.01
⇒ rπ = 2.6 kΩ, g m = 38.5 mA/V, ro = ∞
6.64
a.
⎛ R2 ⎞
⎛ 20 ⎞
VTH 1 = ⎜
⎟ VCC = ⎜
⎟ (10 ) ⇒ VTH 1 = 2.0 V
⎝ 20 + 80 ⎠
⎝ R1 + R2 ⎠
RTH 1 = R1 & R2 = 20 & 80 = 16 kΩ
I B1 =
2 − 0.7
= 0.0111 mA
16 + (101)(1)
I C1 = 1.11 mA ⇒ g m1 =
rπ 1 =
(100)( 0.026)
1.11
1.11
⇒ g m1 = 42.74 mA/V
0.026
⇒ rπ 1 = 2.34 kΩ
∞
⇒ r01 = ∞
1.11
⎛ R4 ⎞
⎛ 15 ⎞
VTH 2 = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 1.50 V
⎝ 15 + 85 ⎠
⎝ R3 + R4 ⎠
RTH 2 = R3 & R4 = 15 & 85 = 12.75 kΩ
r01 =
IB2 =
1.50 − 0.70
= 0.01265 mA
12.75 + (101)( 0.5 )
I C 2 = 1.265 mA ⇒ g m 2 =
rπ 2 =
(100 )( 0.026 )
1.26
1.265
⇒ g m2 = 48.65 mA/V
0.026
⇒ rπ 2 = 2.06 kΩ
r02 = ∞
b.
Av1 = − g m1 RC1 = − ( 42.7 )( 2 ) ⇒ Av1 = −85.48
Av 2 = − g m 2 ( RC 2 RL ) = − ( 48.5) ( 4 4) ⇒ Av 2 = −97.3
c.
Input resistance of 2nd stage
Ri 2 = R3 & R4 & rπ 2 = 15 & 85 & 2.06
= 12.75 & 2.06 ⇒ Ri 2 = 1.773 kΩ
′
Av1 = − g m1 ( RC1 Ri 2 ) = − ( 42.7 ) ( 2 1.77B)
Av′1 = −40.17
Overall gain: Av = ( −40.17 )( −97.3) ⇒ Av = 3909
If we had Av1 ⋅ Av 2 = ( −85.48)( −97.3) = 8317
Loading effect reduces overall gain
6.65
a.
⎛ R2 ⎞
⎛ 12.7 ⎞
VTH 1 = ⎜
⎟ VCC = ⎜
⎟ (12) ⇒ VTH 1 = 1.905 V
⎝ 12.7 + 67.3 ⎠
⎝ R1 + R2 ⎠
RTH 1 = R1 & R2 = 12.7 & 67.3 = 10.68 kΩ
1.905 − 0.70
= 0.00477 mA
I B1 =
10.68 + (121)( 2 )
I C1 = 0.572 mA
0.572
⇒ g m1 = 22 mA/V
g m1 =
0.026
(120 )( 0.026 )
⇒ rπ 1 = 5.45 kΩ
rπ 1 =
0.572
∞
⇒ r01 = ∞
r01 =
0.572
⎛ R4 ⎞
⎛ 45 ⎞
VTH 2 = ⎜
⎟ VCC = ⎜
⎟ (12) ⇒ VTH 2 = 9.0 V
⎝ 45 + 15 ⎠
⎝ R3 + R4 ⎠
RTH 2 = R3 & R4 = 15 & 45 = 11.25 kΩ
9.0 − 0.70
I B2 =
= 0.0405 mA
11.25 + (121)(1.6)
I C2 = 4.86 mA
4.86
gm2 =
⇒ g m 2 = 187 mA/V
0.026
(120 )( 0.026 )
rπ 2 =
⇒ rπ 2 = 0.642 kΩ
4.86
r02 = ∞
b.
I E1 = 0.577 mA
VCEQ1 = 12 − ( 0.572 ) (10 ) − ( 0.577 ) ( 2 ) ⇒ VCEQ1 = 5.13 V
I E 2 = 4.90
VCEQ 2 = 12 − ( 4.90 )(1.6 ) ⇒ VCEQ 2 = 4.16 V
Q1
AC load line
⫺1
10兩兩7.92
⫺1
⫽
4.42 K
Slope ⫽
0.572
5.13
12
Q2
4.86
AC load line
⫺1
Slope ⫽
1.6兩兩0.25
⫺1
⫽
0.216 K
4.16
12
Ri 2 = R3 R4 Rib
Rib = rπ 2 + (1 + β ) ( RE 2 & RL )
= 0.642 + (121) (1.6 & 0.25 )
Rib = 26.8
Ri 2 = 15 45 26.8
Ri 2 = 7.92 kΩ
c.
Av1 = − g m1 ( RC1 & Ri 2 ) = − ( 22 )(10 & 7.92 ) ⇒ Av 2 = −97.2
(1 + β )( RE 2 & RL )
rπ 2 + (1 + β )( RE 2 & RL )
(121)( 0.216 )
=
= 0.976
0.642 + (121)( 0.216 )
Overall gain = ( −97.2 )( 0.976 ) = −94.9
Av 2 =
d.
RiS = R1 R2 rπ 1 = 67.3 12.7 5.45 ⇒ RiS = 3.61 kΩ
Ro =
rπ 2 + RS
RE 2 where
1+ β
RS = R3 R4 RC1
= 15 45 10 ⇒ RS = 5.29 kΩ
Ro =
0.642 + 5.29
1.6 ⇒ 0.049 & 1.6 ⇒ Ro = 47.6 Ω
121
e.
−1
⋅ Δvce , ΔiC = 4.86
0.216 kΩ
Δvce = ( 4.86 )( 0.216 ) = 1.05 V
Max. output voltage swing = 2.10 V peak-to-peak
ΔiC =
6.66
(a)
I R1 =
IR2
IC 2
5 − 2 ( 0.7 )
= 72 mA
0.050
0.7
=
= 1.4 mA
0.5
⎛ β ⎞
=⎜
⎟ ( 72 − 1.4 ) ⇒ I C 2 = 69.9 mA
⎝ 1+ β ⎠
69.9
= 0.699 mA
100
⎛ β ⎞
=⎜
⎟ (1.4 + 0.699 ) ⇒ I C1 = 2.08 mA
⎝ 1+ β ⎠
IB2 =
I C1
(b)
⫹
Vs
⫹
⫺
V␲1
r␲1
gm1V␲1
gm2V␲2
⫺
r␲2
0.5 k⍀
⫹
V␲2
⫺
Vo
50 Ω
Vs = Vπ 1 + Vπ 2 + Vo
(1)
⎛V
⎞
V
Vo = ⎜ π 2 + π 2 + g m 2Vπ 2 ⎟ ( 0.05 )
⎝ 0.5 rπ 2
⎠
rπ 2 =
(100 )( 0.026 )
= 0.0372 k Ω
69.9
69.9
gm2 =
= 2688 mA / V
0.026
V
1
⎛ 1
⎞
Vo = Vπ 2 ⎜
+
+ 2688 ⎟ ( 0.05 ) so that (1) Vπ 2 = o
135.8
⎝ 0.5 0.0372
⎠
(2)
Vπ 1
V
V
+ g m1Vπ 1 = π 2 + π 2
rπ 1
0.5 rπ 2
rπ 1 =
(100 )( 0.026 )
= 1.25 k Ω
2.08
2.08
g m1 =
= 80 mA / V
0.026
1 ⎞
⎛ 1
⎞
⎛ 1
Vπ 1 ⎜
+ 80 ⎟ = Vπ 2 ⎜
+
⎟
1.25
0.5
0.0372
⎝
⎠
⎝
⎠
⎛ V ⎞
Vπ 1 ( 80.8 ) = Vπ 2 ( 28.88 ) = ⎜ o ⎟ ( 28.88 ) or (2) Vπ 1 = Vo ( 0.00261)
⎝ 136.7 ⎠
V
V
Then Vs = Vo ( 0.00261) + o + Vo = Vo (1.00993) or Av = o = 0.990
Vs
136.7
(c)
Rib = rπ 1 (1 + β ) [ Rx ]
Ix
⫹
Vx
⫹
⫺
0.5 k⍀ V␲2
r␲2
⫺
gm2V␲2
Vo
50 ⍀
⎛ 1
Vπ 2 Vπ 2
1 ⎞
+
= Vπ 2 ⎜
+
⎟
r
0.5 rπ 2
0.5
π2 ⎠
⎝
Vo
V − Vπ 2
= x
= I x + g m 2Vπ 2
0.05
0.05
Ix =
⎛ 1
⎞
Ix ⎜
+ gm2 ⎟
Vx
⎛ 1
⎞
⎝ 0.05
⎠
− I x = Vπ 2 ⎜
+ gm2 ⎟ =
0.05
0.05
⎛ 1
⎝
⎠
1 ⎞
+
⎜
⎟
r
0.5
π2 ⎠
⎝
We find
Vx
= Rx = 4.74 k Ω
Ix
Then Rib = 1.25 + (101) ( 2.89 ) ⇒ Rib = 480 k Ω
⫹
V␲1
r␲1
gm1V␲1
⫺
r␲2
0.5 k⍀
gm2V␲2
⫹
V␲2
⫺
Ix
50 ⍀
⫹
⫺
Vx
To find Ro:
Vx
V
− g m 2Vπ 2 − π 2
0.05
0.5 rπ 2
(1)
Ix =
(2)
⎛V
⎞
⎛ 1
⎞
+ 80 ⎟ ( 0.5 0.0372 ) or Vπ 2 = ( 2.77 ) Vπ 1
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ( 0.5 rπ 2 ) = Vπ 1 ⎜
⎝ 1.25
⎠
⎝ rπ 1
⎠
Vπ 1 + Vπ 2 + Vx = 0 ⇒ Vπ 1 + ( 2.77 ) Vπ 1 + Vx = 0
(3)
so that Vπ 1 = − ( 0.2653) Vx
and Vπ 2 = ( 2.77 ) ⎡⎣ − ( 0.2653) Vx ⎤⎦ = − ( 0.735 ) Vx
Now I x =
⎛
Vx
1 ⎞
− Vπ 2 ⎜ g m 2 +
⎟
⎜
0.05
0.5 rπ 2 ⎟⎠
⎝
So that I x =
⎡
⎤
Vx
Vx
1
+ ( 0.735 ) Vx ⎢ 2688 +
= 0.496 Ω
⎥ which yields Ro =
Ix
0.05
0.5 0.0372 ⎥⎦
⎢⎣
6.67
a.
RTH = R1 R2 = 335 125 = 91.0 kΩ
⎛ R2 ⎞
VTH = ⎜
⎟ VCC
⎝ R1 + R2 ⎠
⎛ 125 ⎞
=⎜
⎟ (10 ) = 2.717 V
⎝ 125 + 335 ⎠
VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2
2
I E 2 = (1 + β ) I E1 = (1 + β ) I B1
I B1 =
2.717 − 1.40
2
91.0 + (101) (1)
⇒ I B1 = 0.128 µΑ
I C1 = 12.8 µΑ
I C 2 = β I E1 = β (1 + β ) I B1 = (100 )(101)( 0.128 µΑ )
I C 2 = 1.29 mΑ, I E 2 = 1.31 mΑ
I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mΑ
VC = 10 − I RC RC = 10 − (1.30 )( 2.2 ) = 7.14 V
VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V
VCE 2 = 7.14 − 1.30 = 5.84 V
VCE1 = VCE 2 − VBE 2 = 5.84 − 0.7
VCE1 = 5.14 V
Summary:
I C1 = 12.8 µΑ
I C 2 = 1.29 mΑ
VCE1 = 5.14 V
VCE 2 = 5.84 V
b.
0.0128
g m1 =
= 0.492 mΑ / V
0.026
1.292
gm2 =
= 49.7 mΑ / V
0.026
Rib
Ib
VS
V0
⫹
V␲1
r␲1
gm1V␲1
⫺
⫹
⫺
RC
⫹
R1兩兩 R2
V␲2
r␲2
gm2V␲2
⫺
V0 = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC
VS = Vπ 1 + Vπ 2 , Vπ 1 = VS − Vπ 2
⎛V
⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2
r
⎝ π1
⎠
⎛1+ β ⎞
Vπ 2 = Vπ 1 ⎜
⎟ rπ 2
⎝ rπ 1 ⎠
V0 = − ⎡⎣ g m1 (VS − Vπ 2 ) + g m 2Vπ 2 ⎤⎦ RC
V0 = − ⎡⎣ g m1VS + ( g m 2 − g m1 ) Vπ 2 ⎤⎦ RC
⎛r ⎞
Vπ 2 = (VS − Vπ 2 )(1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
⎡
⎛ r ⎞⎤
⎛r ⎞
Vπ 2 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ = VS (1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠ ⎦
⎝ rπ 1 ⎠
⎣
⎧
⎛r ⎞⎫
VS (1 + β ) ⎜ π 2 ⎟ ⎪
⎪
⎪
⎝ rπ 1 ⎠ ⎪ R
V0 = − ⎨ g m1VS + ( g m 2 − g m1 ) ⋅
⎬ C
⎛r ⎞
⎪
1 + (1 + β ) ⎜ π 2 ⎟ ⎪
⎪
⎝ rπ 1 ⎠ ⎪⎭
⎩
V
Av = 0
VS
2.01 ⎞ ⎫
⎧
( 49.7 − 0.492 )(101) ⎛⎜
⎟⎪
⎪⎪
⎝ 203 ⎠ ⎪ 2.2
= − ⎨( 0.492 ) +
⎬
⎛ 2.01 ⎞
⎪
⎪
1 + (101) ⎜
⎟
⎪⎩
⎪⎭
⎝ 203 ⎠
Av = −55.2
c.
Ris = R1 R2 Rib
Rib = rπ 1 + (1 + β ) rπ 2
= 203 + (101)( 2.01) = 406 kΩ
Ris = 91 406 = 74.3 kΩ = Ris
R0 = RC = 2.2 kΩ
6.68
R0
Ix
⫹
V␲1
r␲1
⫹
⫺
ro1
gm1V␲1
⫺
Vx
VA
⫹
V␲2
r␲2
⫺
gm2V␲2
ro2
Vx Vx − VA
+
+ g m1Vπ 1
ro 2
ro1
(1)
I x = g m 2Vπ 2 +
(2)
Vx − VA
VA
+ g m1Vπ 1 =
ro1
rπ 1 rπ 2
Vπ 2 = VA = −Vπ 1
(3)
Then from (2)
⎛ 1
Vx
1 ⎞
= VA ⎜ + g m1 +
⎟
⎜
ro1
rπ 1 rπ 2 ⎟⎠
⎝ ro1
⎛ 1
⎛
⎞
Vx Vx VA
1 ⎞
1
+ − − g m1VA or I x = Vx ⎜ + ⎟ + VA ⎜ g m 2 − − g m1 ⎟
ro 2 ro1 ro1
ro1
⎝ ro1 ro 2 ⎠
⎝
⎠
Solving for VA from Equation (2) and substituting into Equation (1), we find
(1) I x = g m 2VA +
1
1
+ g m1 +
V
ro1
rπ 1 & rπ 2
Ro = x =
Ix
⎞
1 ⎛ 1
1 ⎞ 1 ⎛ 1
+ gm2 ⎟
⎜ + g m1 +
⎟+ ⎜
ro 2 ⎝ ro1
rπ 1 & rπ 2 ⎠ ro1 ⎝ rπ 1 & rπ 2
⎠
For β = 100, VA = 100 V , I C1 = I Bias = 1 mA
ro1 = ro 2 =
rπ 1 = rπ 2 =
100
= 100 k Ω
1
(100 )( 0.026 )
1
g m1 = g m 2 =
Then Ro =
= 2.6 k Ω
1
= 38.46 mA/V
0.026
1
1
+ 38.46 +
100
2.6 2.6
⎞
1 ⎛ 1
1 ⎞ 1 ⎛ 1
+ 38.46 +
+ 38.46 ⎟⎟
⎜
⎟+
⎜
100 ⎜⎝ 100
2.6 2.6 ⎟⎠ 100 ⎜⎝ 2.6 2.6
⎠
or Ro = 50.0 k Ω
Now I C 2 = 1 mA, I Bias = 0
Replace I Bias by
I
β
⋅
= C 2 , I C1 ≅ 0.01 mA
β 1+ β 1+ β
IC 2
100
100
= 100 k Ω, ro1 =
= 10, 000 k Ω
1
0.01
1
gm2 =
= 38.46 mA/V , g m1 = 0.3846 mA/V
0.026
(100 )( 0.026 )
rπ 2 =
= 2.6 k Ω, rπ 1 = 260 k Ω
1
Then Ro = 66.4 k Ω
ro 2 =
6.69
a.
RTH = R1 R2 = 93.7 6.3 = 5.90 k Ω
⎛ R2 ⎞
VTH = ⎜
⎟ VCC
⎝ R1 + R2 ⎠
6.3 ⎞
⎛
=⎜
⎟ (12 ) = 0.756 V
⎝ 6.3 + 93.7 ⎠
0.756 − 0.70
I BQ =
= 0.00949 mA
5.90
I CQ = 0.949 mA
VCEQ = 12 − ( 0.949 )( 6 ) ⇒ VCEQ = 6.305 V
Transistor:
PQ ≈ I CQVCEQ = ( 0.949 )( 6.305 ) ⇒ PQ = 5.98 mW
2
RC : PR = I CQ
RC = ( 0.949 ) ( 6 ) ⇒ PR = 5.40 mW
2
b.
2
AC load line
⫺1
Slope ⫽
6兩兩105
⫺1
⫽
5.68 K
0.949
6.31
12
100
r0 =
= 105 kΩ
0.949
Peak signal current = 0.949 mA
V0 ( max ) = ( 5.68 )( 0.949 ) = 5.39 V
PRC =
2
2
1 V0 ( max ) 1 ⎡ ( 5.39 ) ⎤
⋅
= ⎢
⎥ ⇒ PRC = 2.42 mW
RC
2
2 ⎢⎣ 6 ⎥⎦
6.70
(a)
10 = I BQ RB + VBE ( on ) + (1 + β ) I BQ RE
10 − 0.7
I BQ =
= 0.00369 mA
100 + (121)( 20 )
I CQ = 0.443 mA, I EQ = 0.447 mA
For RC : PRC = ( 0.443) (10 ) ⇒ PRC = 1.96 mW
2
For RE : PRE = ( 0.447 ) ( 20 ) ⇒ PRE = 4.0 mW
2
(b)
ΔiC = 0.667 − 0.443 = 0.224 mA
1
1
2
2
( ΔiC ) RC = ( 0.224 ) (10 )
2
2
= 0.251 mW
Then P RC =
P RC
6.71
a.
I BQ =
10 − 0.7
= 0.00596 mA
50 + (151)(10 )
I CQ = 0.894 mA, I EQ = 0.90 mA
VECQ = 20 − ( 0.894 )( 5 ) − ( 0.90 )(10 ) ⇒ VECQ = 6.53 V
PQ ≅ I CQVECQ = ( 0.894 )( 6.53) ⇒ PQ = 5.84 mW
2
PRC ≅ I CQ
RC = ( 0.894 ) ( 5 ) ⇒ PRC = 4.0 mW
2
2
PRE ≅ I EQ
RE = ( 0.90 ) (10 ) ⇒ PRE = 8.1 mW
2
b.
AC load line
⫺1
5兩兩2
⫺1
⫽
1.43 K
Slope ⫽
0.894
6.53
20
−1
ΔiC =
⋅ Δvec
1.43 kΩ
ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V
⎛ 5 ⎞
Δi0 = ⎜
⎟ ΔiC = 0.639 mA
⎝5+2⎠
1
2
PRL = ( 0.639 ) ( 2 ) ⇒ PRL = 0.408 mW
2
1
2
PRC = ⋅ ( 0.894 − 0.639 ) ( 5 ) ⇒ PRC = 0.163 mW
2
PRE = 0
PQ = 5.84 − 0.408 − 0.163 ⇒ PQ = 5.27 mW
6.72
I BQ =
10 − 0.70
= 0.00838 mA
100 + (101)(10 )
I CQ = 0.838 mA, I EQ = 0.846 mA
VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V
AC load line
⫺1
RE 兩兩RL兩兩r0
Slope ⫽
0.838
3.16
20
100
r0 =
= 119 kΩ
0.838
Neglecting base currents:
a.
RL = 1 kΩ
slope =
−1
−1
=
10 1 119 0.902 kΩ
−1
⋅ ΔVce
0.902 kΩ
ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V
ΔiC =
1 ( 0.756 )
⇒ PRL = 0.286 mW
2
1
2
PRL =
b.
RL = 10 kΩ
slope =
−1
−1
=
10 10 119 4.80
For ΔiC = 0.838 ⇒ Δvce = ( 0.838 )( 4.80 ) = 4.02
1 ( 3.16 )
⇒ PRL = 0.499 mW
2 10
2
Max. swing determined by voltage PRL =
6.73
a.
I BQ =
10 − 0.7
= 0.00838 mA
100 + (101)(10 )
I CQ = 0.838 mA, I EQ = 0.846 mA
VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V
PQ ≅ I CQVCEQ = ( 0.838 )( 3.16 ) ⇒ PQ = 2.65 mW
2
PRC ≅ I CQ
RC = ( 0.838 ) (10 ) ⇒ PRC = 7.02 mW
2
b.
AC load line
⫺1
RC 兩兩RL
⫺1
⫺1
⫽
⫽
10兩兩1 0.909 K
Slope ⫽
0.838
3.16
20
−1
ΔiC =
⋅ Δvce
0.909 kΩ
For ΔiC = 0.838 ⇒ Δvce = ( 0.909 )( 0.838 ) = 0.762 V
⎛ RC ⎞
⎛ 10 ⎞
Δi0 = ⎜
⎟ ΔiC = ⎜
⎟ ΔiC = 0.762 mA
⎝ 10 + 1 ⎠
⎝ RC + RL ⎠
1
2
PRL = ( 0.762 ) (1) ⇒ PRL = 0.290 mW
2
1
2
PRC = ⋅ ( 0.838 − 0.762 ) (10 ) ⇒ PRC = 0.0289 mW
2
PQ = 2.65 − 0.290 − 0.0289 ⇒ PQ = 2.33 mW
Chapter 7
Exercise Solutions
EX7.1
(a)
(i)
RS = RP = 4 kΩ
1
1
ω= =
rS ( RS + RP ) CS
CS =
1
2π f ( RS + RP )
=
1
2π ( 20 )( 4 + 4 ) × 10 3
CS = 0.995 µ F
(ii)
⎛ RP ⎞ ω rS
T ( jω ) = ⎜
⎟
⎝ RS + RP ⎠ 1 + ω 2 rS2
rS = ( RS + RP ) CS = 7.96 × 10 −3
RP
4
=
= 0.5
RS + RP 4 + 4
f = 40 Hz
T ( jω ) =
( 0.5)( 2π )( 40 ) ( 7.96 × 10 −3 )
1 + ⎡⎣ 2π ( 40 ) 7.96 × 10 −3 ⎤⎦
T ( jω ) = 0.447
(
)
2
f = 80 Hz
T ( jω ) =
( 0.5)( 2π )(80 ) ( 7.96 × 10 −3 )
1 + ⎡⎣ 2π ( 80 ) 7.96 × 10 −3 ⎤⎦
T ( jω ) = 0.485
(
)
2
f = 200 Hz
T ( jω ) =
( 0.5 )( 2π )( 200 ) ( 7.96 × 10 −3 )
1 + ⎡⎣ 2π ( 200 ) 7.96 × 10 −3 ⎤⎦
T ( jω ) = 0.498
(
)
(b)
ω=
1
1
=
rP ( RS & RP ) CP
CP =
=
1
2π f ( RS & RP )
1
(
2π 500 × 10
CP = 63.7 pF
EX7.2
a.
3
) (10 & 10 ) × 10
3
2
⎛ RP ⎞
RP
RP
= 0.891 =
⇒ (1 − 0.891) RP = 0.891 ⇒ RP = 8.20 kΩ
20 log10 ⎜
⎟ = −1 ⇒
+
+
R
R
R
R
R
S ⎠
P
S
P +1
⎝ P
1
1
⇒ CS =
fL =
2π ( RS + RP ) CS
2π (100 )(1 + 8.20 ) × 10 3
CS = 0.173 µ F
fH =
1
1
⇒ CP =
6
2π ( RS & RP ) CP
2π 10 (1 || 8.20 ) × 10 3
( )
CP = 179 pF
b.
rS = ( RS + RP ) CS
(
)(
rS = 1 × 10 3 + 8.20 × 10 3 0.173 × 10 −6
)
rS = 1.59 ms open-circuit time-constant
rP = ( RS & RP ) CP
(
rP = (1 & 8.20 ) × 10 3 179 × 10 −12
)
rP = 0.160 µ s short-circuit time-constant
EX7.3
a.
rS = ( Ri + RS i ) CC
b.
f =
1
2π rS
RTH = R1 & R2 = 2.2 & 20 = 1.98 kΩ
⎛ R2 ⎞
⎛ 2.2 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 0.991 V
⎝ 2.2 + 20 ⎠
⎝ R1 + R2 ⎠
VTH − VBE ( on )
0.991 − 0.7
=
= 0.0132 mA
I BQ =
RTH + (1 + β ) RE 1.98 + ( 201)( 0.1)
I CQ = 2.636 mA
rπ =
gm =
β VT
I CQ
ICQ
VT
=
=
( 200 )( 0.026 )
2.636
= 1.97 kΩ
2.636
= 101.4 mA/V
0.026
Rib = τ π + (1 + β ) RE = 1.97 + ( 201)( 0.1) = 22.1 kΩ
RB = R1 & R2 = 1.98 kΩ
Ri = RB & Rib = 1.98 & 22.1 = 1.817 kΩ
rS = ( Ri + RSi ) CC
(
= (1.817 + 0.1) ×10 3
)( 47 × 10 )
−6
= 90.1 ms
1
f =
⇒ f = 1.77 Hz
2π 90.1 × 10 −3
(
Midband Gain
)
Aν =
=
− β RC
Ri
⋅
rπ + (1 + β ) RE Ri + RSi
− ( 200 )( 2 )
1.82
1.97 + ( 201)( 0.1) 1.82 + 0.1
⋅
Aν = −17.2
EX7.4
I DQ = K n (VGS − VTN )
2
a.
0.8
+ 2 = VGS ⇒ VGS = 3.265 V
0.5
VS − ( −5 )
VS = −3.265 ⇒ I DQ =
RS
−3.265 + 5
⇒ RS = 2.17 kΩ
0.8
5
⇒ RD = 6.25 kΩ
VD = 0 ⇒ RD =
0.8
b.
rS = ( RD + RL ) CC = (10 + 6.25 ) × 10 3 × CC
1
1
⇒ CC =
f =
2π rS
2π f 16.25 × 10 3
RS =
(
CC =
)
1
(
2π ( 20 ) 16.25 × 10 3
)
⇒ CC = 0.49 µ F
EX7.5
rS = ( RL + Ro ) CC 2
1
1
⇒ CC 2 =
f =
2π rS
2π f ( RL + Ro )
⎧⎪ rπ + ( RS & RB ) ⎫⎪
R0 = RE & r0 ⎨
⎬
1+ β
⎪⎩
⎪⎭
From Example 7-5, R0 = 35.5 Ω
CC 2 =
1
2π (10 ) ⎡⎣10 × 10 3 + 35.5⎤⎦
CC 2 = 1.59 µ F
EX7.6
a.
I DQ = K P (VSG + VTP )
2
1
− ( −2 ) = VSG ⇒ VSG = 3.41 V
0.5
VS = 3.41
5 − 3.41
⇒ RS = 1.59 kΩ
RS =
1
5
For VSDG = VSGQ ⇒ VD = 0 ⇒ RD = ⇒ RD = 5 kΩ
1
rP = ( RD & RL ) CL
b.
f =
1
1
⇒ CL =
2π rP
2π f ( RD & RL )
CL =
1
( )
2π 10 6
( 5 & 10 ) × 10 3
⇒ CL = 47.7 pF
EX7.7
(a)
RTH = 5 K
VTH = −3.7527
I BQ =
−3.7527 − 0.7 − ( −5) 0.54726
=
5 + (101)( 0.5)
55.5
= 0.00986
I CQ = 0.986 mA
gm = 37.925 rπ = 2.637 K
5 − Vo Vo
− = 1 − Vo ( 0.4 )
5
5
Vo = 0.035
(b)
0.986 =
Rib
RS ⫽ 0.1 k⍀
V0
⫹
Vi
⫹
⫺
RTH ⫽
5 k⍀
V␲
r␲
gmV␲
⫺
RE ⫽
0.5 k⍀
Vo = − gmVπ ( RC & RL )
Rib = rπ + (1 + β ) RE = 2.64 + (101)( 0.5) = 53.14 K
Vπ =
Vb
Vb
Vb
=
=
101
14.885
⎛1+ β ⎞
⎛
⎞
1+ ⎜
⎟ RE 1 + ⎜ 2.637 ⎟ ( 0.5)
⎝
⎠
r
⎝ π ⎠
⎛ RTH & Rib ⎞
⎛ 5 & 53.14 ⎞
Vb = ⎜
⎟ Vi = ⎜
⎟ Vi
&
+
R
R
R
⎝ 5 & 53.14 + 0.1 ⎠
⎝ TH
ib
S ⎠
⎛ 4.57 ⎞
=⎜
⎟ Vi = 0.9786
⎝ 4.57 + 0.1 ⎠
(37.925)( 2.5)
Av = −
( 0.9786 ) = Av = −6.23
14.885
(c)
EX7.8
a.
RC 兩兩RL ⫽
2.5 k⍀
I BQ =
0 − 0.7 − ( −10 )
= 0.0230 mA
0.5 + (101)( 4 )
I CQ = 2.30 mA
β VT
rπ =
I CQ
I CQ
gm =
rB =
VT
=
=
(100 )( 0.026 )
2.30
= 1.13 kΩ
2.30
= 88.46 mA / V
0.026
RE ( RS + rπ ) CE
RS + rπ + (1 + β ) RE
( 4 × 10 ) ( 0.5 + 1.13) C
=
3
E
0.5 + 1.13 + (101)( 4 )
rB =
1
1
=
= 0.7958 ms
2π f B 2π ( 200 )
rB = 16.07CE ⇒ CE =
b.
rA = RE CE = 4 × 10 3
(
fA =
0.796 × 10 −3
⇒ CE = 49.5 µ F
16.07
)( 49.5 × 10 ) ⇒ r
−6
A
= 0.198 s
1
1
=
⇒ f A = 0.804 H Z
2π rA 2π ( 0.198 )
EX7.9
rπ =
fβ =
=
β 0VT
I CQ
=
=
fβ =
=
0.5
= 7.8 kΩ
1
2π rπ ( Cπ + Cµ )
1
(
2π 7.8 × 10
EX7.10
rπ
(150 )( 0.026 )
β 0VT
I CQ
=
3
) ( 2 + 0.3) × 10
−12
(150)(0.026)
⇒ rπ = 3.9 kΩ
1
1
2π rπ ( Cπ + Cµ )
1
(
⇒ f β = 8.87 MH Z
2π 3.9 × 10
3
) ( 4 + 0.5) (10 )
−12
⇒ f β = 9.07 MHz
fT = β 0 f β = (150 )( 9.07 ) ⇒ fT = 1.36 GHz
EX7.11
RTH = R1 & R2 = 200 & 220 = 104.8 kΩ
⎛ R2 ⎞
⎛ 220 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (5) = 2.619 V
⎝ 200 + 220 ⎠
⎝ R1 + R2 ⎠
2.62 − 0.7
I BQ =
= 0.009316 mA
105 + (101)(1)
I CQ = 0.9316 mA
gm =
rπ =
I CQ
=
VT
β VT
I CQ
0.9316
⇒ gm = 35.83 mA/V
0.026
=
(100)(0.026)
⇒ rπ = 2.79 kΩ
0.932
a.
C M = Cµ ⎡⎣1 + gm ( RC & RL ) ⎤⎦
C M = ( 2 ) ⎡⎣1 + ( 35.83 )( 2.2 & 4.7 ) ⎤⎦ ⇒ C M = 109 pF
b.
RB = rS & R1 & R2 = 100 & 200 & 220 = 51.17 kΩ
1
f3 dB =
2π ( RB & rπ ) ( Cπ + Cµ )
=
1
⇒ f3dB = 0.506 MHz
2π [ 51.17 & 2.79] × 10 3 × (10 + 109 ) × 10 −12
EX7.12
fT =
gm
2π ( Cgs + Cgd )
gm = 2 K n I DQ
=2
( 0.2 )( 0.4 )
= 0.5657 mA/V
fT =
0.5657 × 10 −3
⇒ fT = 333 MHz
2π ( 0.25 + 0.02 ) × 10 −12
EX7.13
dc analysis
⎛ R2 ⎞
⎛ 166 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) = 4.15 V
+
R
R
⎝ 166 + 234 ⎠
⎝ 1
2 ⎠
V
I D = S and VS = VG − VGS
RS
2
K n (VGS − VTN ) =
VG − VGS
RS
( 0.5)( 0.5 ) (VGS2 − 4VGS + 4 ) = 4.15 − VGS
0.25VGS2 − 3.15 = 0 ⇒ VGS = 3.55 V
gm = 2K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 )
= 1.55 mA/V
Small-signal equivalent circuit.
Ri ⫽ 10 k⍀
V0
Vi
⫹
⫺
RG ⫽
R1兩兩R2
Cgs
⫹ Cgd
Vgs
RD
gmVgs
⫺
a.
b.
C M = Cgd (1 + gm ( RD RL ) )
C M = ( 0.1) ⎡⎣1 + (1.55) ( 4 20 )⎤⎦ ⇒ C M = 0.617 pF
RL
1
fH =
2πτ P
τ P = ( RG & Ri ) ( Cgs + C M )
RG = R1 & R2 = 234 & 166 = 97.1 kΩ
RG & Ri = 97.1 & 10 = 9.07 kΩ
(
)
rP = 9.07 × 10 3 (1 + 0.617 ) × 10 −12 = 14.7 ns
fH =
1
(
2π 14.7 × 10 −9
)
⇒ f H = 10.9 MHz
EX7.14
dc analysis
VTH = 0, RTH = 10 kΩ
I BQ =
0 − 0.7 − ( −5 )
10 + (126 )( 5 )
= 0.00672 mA
I CQ = 0.840 mA
rπ =
gm =
r0 =
β VT
I CQ
I CQ
VT
=
=
(125)( 0.026 )
0.840
= 3.87 kΩ
0.840
= 32.3 mA/V
0.026
VA
200
=
= 238 kΩ
I CQ 0.84
High-frequency equivalent circuit
C␮
RS ⫽ 1 k⍀
V0
⫹
Vi
RB ⫽
V
R1兩兩R2 ␲
⫺
⫹
⫺
r␲
r0
C␲
a.
Miller Capacitance
C M = Cµ (1 + gm RL′ )
RL′ = r0 & RC & RL
RL′ = 238 & 2.3 & 5 = 1.565 kΩ
C M = ( 3 ) ⎡⎣1 + ( 32.3 )(1.57 ) ⎤⎦ ⇒ Cµ = 155 pF
b.
Req = RS & RB & rπ = RS & R1 & R2 & rπ
Req = 1 & 20 & 20 & 3.87 = 0.736 kΩ
rP = Re q ( Cπ + C M )
(
)
= 0.736 × 10 3 ( 24 + 155 ) × 10 −12
= 1.314 × 10
fH =
c.
−7
1
(
2π 1.314 × 10 −7
)
⇒ f H = 1.21 MHz
gmV␲
RC
RL
( Av )M
( Av )M
⎡ RB & τ π ⎤
= − gm RL′ ⎢
⎥
⎣⎢ RB τ π + RS ⎦⎥
⎡ 10 & 3.87 ⎤
= − ( 32.3 )(1.565 ) ⎢
⎥ ⇒ ( Av ) M = −37.2
⎣10 & 3.87 + 1 ⎦
EX7.15
The dc analysis
10 − 0.7
I BQ =
= 0.00838 mA
100 + (101)(10 )
I CQ = 0.838 mA
rπ =
β VT
I CQ
=
(100 )( 0.026 )
0.838
= 3.10 kΩ
I CQ
= 32.22 mA/V
VT
For the input
⎡⎛ r ⎞
⎤
rPπ = ⎢⎜ π ⎟ RE RS ⎥ Cπ
⎣⎝ 1 + β ⎠
⎦
gm =
⎡ 3.10
⎤
=⎢
10 1⎥ × 10 3 × 24 × 10 −12
⎣ 101
⎦
= 7.13 × 10 −10 s
f Hπ =
1
1
=
⇒ f Hπ = 223 MHz
2π rPπ 2π 7.13 × 10 −10
(
)
For the output
rP µ = [ RC & RL ] Cµ = (10 & 1) × 10 3 × 3 × 10 −12
= 2.73 × 10 −9
1
1
fHµ =
=
⇒ f H µ = 58.4 MHz
2π rP µ 2π 2.73 × 10 −9
(
( Aν )M
)
⎡
⎛ rπ ⎞ ⎤
⎢ RE & ⎜
⎟ ⎥
⎝1+ β ⎠ ⎥
= gm ( RC & RL ) ⎢
⎢
⎥
⎛ rπ ⎞
⎢ RE & ⎜
⎟ + RS ⎥
⎝1+ β ⎠
⎣⎢
⎦⎥
⎡
⎛ 3.1 ⎞ ⎤
⎢ 10 & ⎜ 101 ⎟ ⎥
⎝
⎠ ⎥⇒ A
= ( 32.22 )(10 & 1) ⎢
( ν )M = 0.870
⎛ 3.1 ⎞ ⎥
⎢
⎢ 10 & ⎜⎝ 101 ⎟⎠ + 1 ⎥
⎣
⎦
EX7.16
⎛
⎞
R3
7.92
⎛
⎞
VB1 = ⎜
⎟ (12 ) = ⎜
⎟ (12 ) = 0.9502 V
+
+
+
+
R
R
R
58.8
33.3
7.92
⎝
⎠
2
3 ⎠
⎝ 1
Neglecting lose currents
0.9502 − 0.7
IC =
= 0.50 mA
0.5
β VT (100 )( 0.026 )
rπ =
=
= 5.2 K
IC
0.5
IC
0.5
=
= 19.23 mA/V
VT 0.026
From Eq (7.127(a)),
gm =
τ Pπ
= [ RS & RB1 & rπ ][Cπ 1 + C M 1 ]
RB1 = R2 & R3 = 33.3 & 7.92 = 6.398 K
C M 1 = 2Cµ 1 = 6 pF
Then
3
−12
τ Pπ = [1 & 6.398 & 5.2] × 10 × [24 + 6] × 10
⇒ 22.24 ns
1
1
=
⇒ 7.15 MHz
f Hπ =
2πτ Pπ 2π ( 22.24 × 10 −9 )
From Eq (7.128(a))
3
−12
τ P µ = [ RC & RL ] C µ 2 = ( 7.5 & 2 ) × 10 × 3 × 10
⇒ 4.737 ns
1
1
=
= 33.6 MHz
fHµ =
2πτ Pµ 2π ( 4.737 × 10 −9 )
From Eq. 7.133
Av
Av
M
M
⎡ RB1 & rπ 1 ⎤
= gm 2 ( RC & RC ) ⎢
⎥
⎣ RB1 & rπ 1 + RS ⎦
⎡ 6.40 & 5.2 ⎤
= (19.23)( 7.5 & 2 ) ⎢
⎥
⎣ 6.40 & 5.2 + 1 ⎦
⎡ 2.869 ⎤
= (19.23)(1.579 ) ⎢
⎣ 2.869 + 1 ⎥⎦
= 22.5
TYU7.1
a.
V0 = − ( gmVπ ) RL
rπ
Vπ =
× Vi
1
rπ +
+ RS
sCC
T (s) =
=
V0 ( s )
Vi ( s )
=
− gm rπ RL ( sCC )
1 + s ( rπ + RS ) CC
gm rπ = β
T (s) =
− gm rπ RL
rπ + RS + (1 / sCC )
− β RL
rπ + RS
⎛ s ( rπ + RS ) CC
×⎜
⎜ 1 + s (r + R ) C
π
S
C
⎝
⎞
⎟⎟
⎠
Then τ = ( rπ + RS ) CC
b.
1
f3− dB =
2π ( rπ + RS ) CC
1
2π ⎡⎣2 × 10 + 1 × 10 3 ⎤⎦ ⎡⎣10 −6 ⎤⎦
( 2 )( 50 )( 4 )
rg R
T ( jω ) max = π m L =
2 +1
rπ + RS
f3− dB =
T ( jω ) max = 133
c.
3
⇒ f3 dB = 53.1 Hz
兩T( j␻)兩
133
f
53.1 Hz
TYU7.2
a.
⎛
1 ⎞
V0 = − g mVπ ⎜ RL &
⎟
sCL ⎠
⎝
⎛ r
⎞
Vπ = ⎜ π ⎟ × Vi
⎝ rπ + RS ⎠
1
⎛
RL ×
⎜
V0 ( s )
rπ
sCL
⎜
T (s) =
= − gm
1
Vi ( s )
rπ + RS ⎜
⎜ RL + sC
L
⎝
T (s) =
⎞
− β RL ⎛
1
×⎜
⎟
rπ + RS ⎝ 1 + sRL CL ⎠
⎞
⎟
⎟
⎟
⎟
⎠
Then τ = RL CL
b.
1
1
=
⇒ f3dB = 3.18 MH Z
f3− dB =
3
2π RL CL 2π 5 × 10 10 × 10 −12
(
T ( jω ) =
)(
)
gm rπ RL ( 75)(1.5)( 5)
=
1.5 + 0.5
rπ + RS
T ( jω ) max = 281
c.
281
兩T( j␻)兩
3.18 MHz
f
TYU7.3
a.
Open-circuit time constant ( CL → open )
rS = ( RS + rπ ) CC
(
)
= ( 0.25 + 2 ) × 10 3 2 × 10 −6 = 4.5 ms
Short-circuit time constant ( CC → short )
(
)(
rP = RL CL = 4 × 10 3 50 × 10 −12
rP = 0.2 µ s
b.
Midband gain
)
⎛ rπ ⎞
V0 = − gmVπ RL , Vπ = ⎜
⎟ Vi
⎝ τ π + RS ⎠
V
−g r R
Av = 0 = m π L
Vi
rπ + RS
=
− ( 65 )( 2 )( 4 )
2 + 0.25
Av = −231
c.
fL =
1
1
=
⇒ f L = 35.4 Hz
2π rS 2π 4.5 × 10 −3
fH =
1
1
⇒ f H = 0.796 MHz
=
2π rP 2π 0.2 × 10 −6
(
)
(
)
TYU7.4 Computer Analysis
TYU7.5 Computer Analysis
TYU7.6
(100 )( 0.026 )
βV
rπ = 0 T =
= 10.4 kΩ
I CQ
0.25
fβ =
1
2π rπ ( Cπ + C µ )
Cπ + Cµ =
1
1
=
2π f β rπ 2π 11.5 × 10 6 10.4 × 10 3
(
)(
Cπ + Cµ = 1.33 pF
Cµ = 0.1 pF ⇒ Cπ = 1.23 pF
TYU7.7
h fe =
β0
1 + j ( f / fβ )
f β = 5 MH Z , β 0 = 100
At f = 50 MH Z
h fe =
100
⎛ 50 ⎞
1+ ⎜ ⎟
⎝ 5 ⎠
2
⇒ h fe = 9.95
⎛ 50 ⎞
Phase = − tan −1 ⎜ ⎟ ⇒ Phase = −84.3°
⎝ 5 ⎠
TYU7.8
f
500
fβ = T =
⇒ f β = 4.17 MHz
β 0 120
fβ =
1
2π rπ (Cπ + C µ )
Cπ + Cµ =
1
1
=
2π f β rπ 2π (4.167 × 10 6 )(5 × 10 3 )
Cπ + Cµ = 7.639 pF
Cµ = 0.2 pF ⇒ Cπ = 7.44 pF
)
TYU7.9
(a)
gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 3 − 1) ⇒ gm = 1.6 mA/V
gm′ = 80% of gm = 1.28 mA/V
gm
gm′ =
1 + gm rS
1 + gm rS =
gm
gm′
⎛ gm
⎞ 1 ⎛ 1.6
⎞
− 1⎟ =
⎜
⎜ 1.28 − 1 ⎟
′
g
1.6
⎝
⎠
⎝ m
⎠
rS = 0.156 kΩ ⇒ rS = 156 ohms
1
gm
rS =
(b)
gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 5 − 1) ⇒ gm = 3.2 mA/V
gm
3.2
gm′ =
=
= 2.134
1 + gm rS 1 + ( 3.2 )( 0.156 )
Δgm 3.2 − 2.134
=
⇒ A 33.3% reduction
3.2
gm
TYU7.10
fT =
=
Cgs =
=
gm
2π ( CgsT + C gdT )
gm
2π ( Cgs + Cgsp + Cgdp )
gm
− Cgsp − C gdp
2π fT
0.5 × 10 −3
(
2π 500 × 10
6
)
− ( 0.01 + 0.01) × 10 −12 ⇒ Cgs = 0.139 pF
TYU7.11
fT =
gm
2π (Cgs + Cgsp + Cgdp )
Cgsp = Cgdp
2Cgsp =
gm
1× 10−3
− C gs =
− 0.4 × 10−12
2π fT
2π (350 × 106 )
2Cgsp = 0.0547 pF ⇒ Cgsp = Cgdp ≅ 0.0274 pF
TYU7.12
dc analysis
⎛ 50 ⎞
VG = ⎜
⎟ (10 ) − 5 = −2.5
⎝ 50 + 150 ⎠
VS = VG − VGS . I D =
2
K n (VGS − VTN ) =
VS − ( −5 )
RS
VG − VGS + 5
RS
(1)( 2 ) ⎣⎡VGS2 − 1.6VGS + 0.64 ⎦⎤ = −2.5 − VGS + 5
2
− 2.2VGS − 1.22 = 0
2VGS
2.2 ±
VGS =
( 2.2 )
2
+ 4 ( 2 )(1.22 )
2 (2)
⇒ VGS = 1.505 V
gm = 2 K n (VGS − VTN ) = 2 (1)(1.505 − 0.8 )
= 1.41 mA/V
Equivalent circuit
Cgd
Ri
V0
⫹
Vi
RG ⫽
Vgs
R1兩兩R2
⫹
⫺
Cgs
gmVgs
RD
⫺
C M = Cgd (1 + gm RD ) = ( 0.2 ) ⎣⎡1 + (1.42 )( 5 ) ⎦⎤ ⇒ C M = 1.61 pF
(a)
(b)
τ P = ( Ri & RG ) ( Cgs + CM )
rP = [ 20 & 50 & 150 ] × 10 3 × ( 2 + 1.61) × 10 −12
(
)(
)
= 13 × 10 3 3.62 × 10 −12 = 4.71 × 10 −8 s
fH =
c.
1
1
=
⇒ f H = 3.38 MHz
2π rP 2π 4.71 × 10 −8
(
( Av )M
( Av )M
)
⎛ RG ⎞
= − gm RD ⎜
⎟
⎝ RG + RS ⎠
⎛ 37.5 ⎞
= − (1.41)( 5 ) ⎜
⎟ ⇒ ( Av ) M = −4.60
⎝ 37.5 + 20 ⎠
TYU7.13 Computer Analysis
Chapter 7
Problem Solutions
7.1
a.
T (s) =
T (s) =
V0 ( s )
Vi ( s )
=
1/ ( sC1 )
⎣⎡1/ ( sC1 ) ⎦⎤ + R1
1
1 + sR1C1
b.
1
兩T 兩
fH ⫽ 159 Hz
f
fH =
1
1
=
⇒ f H = 159 Hz
2π R1C1 2π (103 )(10−6 )
c.
V0 ( s ) = Vi ( s ) ⋅
1
1 + sR1C1
For a step function Vi ( s ) =
1
s
K
K2
1
1
= 1+
V0 ( s ) = ⋅
s 1 + sR1C1
s 1 + sR1C1
=
=
K1 (1 + sR1C1 ) + K 2 s
s (1 + sR1C1 )
K1 + s ( K1 R1C1 + K 2 )
s (1 + sR1C1 )
K 2 = − K1 R1C1 and K1 = 1
V0 ( s ) =
=
− R1C1
1
+
s 1 + sR1C1
1
1
−
1
s
+s
R1C1
v0 ( t ) = 1 − e− t / R1C1
7.2
a.
T (s) =
T (s) =
b.
V0 ( s )
Vi ( s )
=
R2
R2 + ⎡⎣1/ ( sC2 ) ⎤⎦
sR2 C2
1 + sR2 C2
1
兩T 兩
fL ⫽ 1.59 Hz
1
1
=
⇒ f L = 1.59 Hz
fL =
4
2π R2 C2 2π (10 )(10 × 10−6 )
c.
V0 ( s ) = Vi ( s ) ⋅
Vi ( s ) =
V0 ( s ) =
sR2 C2
1 + sR2 C2
1
s
R2 C2
1
=
1 + sR2 C2 s + 1
R2 C2
v0 ( t ) = e− t / R2C2
7.3
a.
1
sCP
T (s) =
=
Vi ( s )
⎛
1
1 ⎞
RP &
+ ⎜ RS +
⎟
sCP ⎝
sCS ⎠
1
RP ⋅
sCP
RP
1
RP &
=
=
1 + sRP CP
sCP R + 1
P
sCP
Then
RP
T (s) =
⎛
1 ⎞
RP + ⎜ RS +
⎟ (1 + sRP CP )
sCS ⎠
⎝
V0 ( s )
=
RP &
RP
RP CP
1
+
+ sRS RP CP
RP + RS +
CS
sCS
⎛ RP
T (s) = ⎜
⎝ RP + RS
b.
⎤⎞
⎞ ⎛ ⎡
sRP RS
RP
C
1
⋅ P+
+
⋅ CP ⎥ ⎟
⎟ × ⎜⎜ 1/ ⎢1 +
⎟
⎠ ⎝ ⎣⎢ RP + RS CS s ( RS + RP ) CS RS + RP
⎦⎥ ⎠
−11
⎛
⎤⎞
1
⎛ 10 ⎞ ⎜ ⎡ 10 10
3
−11 ⎟
⎢
⎥
T (s) = ⎜
s
1/
1
5
10
10
×
+
⋅
+
+
×
⋅
(
)
⎟
−6
s ( 2 × 104 ) ⋅10−6
⎝ 10 + 10 ⎠ ⎜ ⎢⎣ 20 10
⎥⎦ ⎟
⎝
⎠
1
1
≅ ⋅
1
2 1+
+ s ( 5 × 10−8 )
s ( 0.02 )
s = jω
1
⋅
2
1
⎡
⎤
1
1 + j ⎢ω ( 5 × 10−8 ) −
⎥
ω ( 0.02 ) ⎦
⎣
1
1
For ω L =
=
= 50
4
( RS + RR ) CS ( 2 ×10 )(10−6 )
T ( jω ) =
1
⋅
2
1
⎡
⎤
1
1 + j ⎢( 50 ) ( 5 × 10−8 ) −
⎥
( 50 )( 0.02 ) ⎦
⎣
1 1
1 1
≈ ⋅
⇒ T ( jω ) = ⋅
2 1− j
2 2
T ( jω ) =
For
ωH =
1
( RS & RP ) CP
=
1
( 5 ×103 )(10−11 )
= 2 × 107
1
⎡
⎤
1
⎥
1 + j ⎢( 2 × 107 )( 5 × 10−8 ) −
⎢⎣
( 2 ×107 ) ( 0.02 ) ⎥⎦
1 1
1 1
⇒ T ( jω ) = ⋅
T ( jω ) ≈ ⋅
2 1+ j
2 2
T ( jω ) =
1
⋅
2
In each case, T ( jω ) =
1
RP
2 RP + RS
⋅
c.
RS = RP = 10 kΩ, CS = CP = 0.1 µ F
T (s) =
⎛ ⎡
⎤⎞
1 ⎜ ⎢ 1 1
1
3
−7 ⎥ ⎟
⋅ 1/ 1 + ⋅ +
+
×
s
5
10
10
(
)( )⎥ ⎟
2 ⎜⎜ ⎢ 2 1 s 2 × 104 (10−7 )
⎟
⎦⎠
⎝ ⎣
(
)
s = jω
1
⎡
⎤
1
1
⎥
1 + + j ⎢ω ( 5 × 10−4 ) −
2
ω ( 2 × 10−3 ) ⎥⎦
⎢⎣
1
= 500
For ω =
4
( 2 ×10 )(10−7 )
T ( jω ) =
1
⋅
2
1
⋅
2
1
⎡
⎤
1
⎥
1.5 + j ⎢( 500 ) ( 5 × 10−4 ) −
⎢⎣
( 500 ) ( 2 ×10−3 ) ⎥⎦
1
1
= ⋅
⇒ T ( jω ) = 0.298
2 1.5 − j ( 0.75 )
T ( jω ) =
For ω =
1
( 5 ×10 )(10 )
3
−7
= 2 × 103
⎧
⎡
⎤ ⎞ ⎪⎫
1 ⎪ ⎜⎛
1
⎟⎬
⎥
⋅ ⎨1/ 1.5 + j ⎢( 2 × 103 )( 5 × 10−4 ) −
2 ⎪ ⎜
2 × 103 )( 2 × 10−3 ) ⎥⎦ ⎟ ⎪
⎢⎣
(
⎠⎭
⎩ ⎝
1
1
= ⋅
⇒ T ( jω ) = 0.298
2 1.5 + j ( 0.75 )
T ( jω ) =
In each case, T ( jω ) <
7.4
Circuit (a):
V
R2
T= 0 =
Vi
1
R2 +
sC1
=
1
RP
2 RP + RS
⋅
R2
R1 (1/ sC1 )
R2 +
R1 + (1/ sC1 )
R2
=
R2 +
R1
1 + sR1C1
=
R2 (1 + sR1C1 )
R2 + sR1 R2 C1 + R1
or
(1 + sR1C1 )
Vo
R2
=
⋅
Vi R1 + R2 1 + sR1 R2 C1
Low frequency:
Vo
R2
20
2
=
=
=
Vi
R1 + R2 10 + 20 3
High frequency:
Vo
=1
Vi
τ 1 = R1C1 = (104 )(10 × 10−6 ) = 0.10 ⇒ f1 =
1
= 1.59 Hz
2πτ 1
τ 2 = ( R1 R 2 ) C1 = (10 20 ) × 103 × (10 × 10−6 ) ⇒ τ 2 = 0.0667 ⇒ f 2 =
兩T 兩
1.0
0.67
1.59
2.39
f
Circuit (b):
1
sC2
R2
V
1 + sR2 C2
T= o =
=
R2
Vi
1
+ R1
R2
+ R1
sR
+
1
sC2
2 C2
R2
⎞
⎛ R2 ⎞ ⎛
1
=⎜
⎟
⎟ ⎜⎜
⎝ R1 + R2 ⎠ ⎝ 1 + s ( R1 R2 ) C2 ⎠⎟
Low frequency:
Vo
R2
20
2
=
=
=
Vi
R1 + R2 20 + 10 3
τ = ( R1 R2 ) C2 = (10 20 ) × 103 × 10 × 10−6 = 0.0667
f =
1
2πτ
= 2.39 Hz
1
= 2.39 Hz
2πτ 2
0.67
兩T 兩
2.39
f
7.5
a.
rS = ( Ri + RP ) CS = [30 + 10] × 103 × 10 × 10−6 ⇒ rS = 0.40 s
rP = ( Ri RP ) CP = ⎡⎣30 10 ⎤⎦ × 103 × 50 × 10−12 ⇒ rP = 0.375 µ s
b.
1
1
fL =
=
⇒ f L = 0.398 Hz
2π rS 2π ( 0.4 )
fH =
1
1
=
⇒ f H = 424 kHz
2π rP 2π ( 0.375 × 10−6 )
At midband. CS → short, CP → open
Vo = I i ( Ri RP )
T ( s ) = Ri R P = 30 10 ⇒ T ( s ) = 7.5 k Ω
c.
7.5 k⍀
兩T 兩
fH
fL
7.6
(a)
T=
T
1
(1 + j 2π f τ )
max
2
⇒T =
1
(
1 + ( 2π f τ )
2
)
2
=
1
1 + ( 2π f τ )
2
=1
At f =
1
1
1
⇒T =
=
2
2πτ
2
1 + (1)
⎛1⎞
= 20 log10 ⎜ ⎟ ⇒ T dB ≅ −6 dB
⎝ 2⎠
−1
Phase = 2 tan ( 2π f τ ) = −2 tan −1 (1) = −2 ( 45° ) ⇒ Phase = −90°
(b)
Slope = −2 ( 6dB / oct ) = −12dB / oct = −40dB / decade
T
dB
Phase = −2 ( 90° ) ⇒ Phase = −180°
7.7
(a)
−10 ( jω )
jω ⎞
jω ⎞
⎛
⎛
20 ⎜1 +
⎟ ( 2000 ) ⎜1 +
⎟
20
2000
⎝
⎠
⎝
⎠
ω
j
⎛
⎞
−5 × 10−3 ⎜
−4
⎟
⎝ ω ⎠ = 2.5 × 10 ( jω )
=
jω ⎞ ⎛
jω ⎞ ⎛
jω ⎞⎛
jω ⎞
⎛
⎜1 +
⎟ ⎜1 +
⎟ ⎜1 +
⎟⎜ 1 +
⎟
ω
2000
20
2000
⎝
⎠⎝
⎠ ⎝
⎠⎝
⎠
2.5 × 10−4 (ω )
T ( jω ) =
T =
⎛ω ⎞
⎛ ω ⎞
1+ ⎜ ⎟ ⋅ 1+ ⎜
⎟
⎝ 20 ⎠
⎝ 2000 ⎠
2
2
兩T 兩
5 ⫻ 10⫺3
2.5 ⫻ 10⫺4
(b)
1
20
2000
␻
(10 )(10 ) ⎛⎜1 +
jω ⎞
⎟
10
⎝
⎠
T ( jω ) =
jω ⎞
⎛
1000 ⎜1 +
⎟
⎝ 1000 ⎠
( 0.10 )
⎛ω ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
⎛ ω ⎞
1+ ⎜
⎟
⎝ 1000 ⎠
T =
2
2
兩T 兩
10
0.10
10
␻
1000
7.8
(a)
T (s) =
105
=
10
( 5 + 10 )( 5 + 500 ) (10 )( 500 )
⎛ jω ⎞
⎜
⎟
10
⎝ 10 ⎠
=
⋅
500 ⎛ jω ⎞⎛ jω
⎞
+ 1⎟⎜
+ 1⎟
⎜
⎝ 10
⎠⎝ 500 ⎠
⋅
5
⎛ 5
⎞⎛ 5
⎞
+ 1⎟
⎜ + 1⎟⎜
⎝ 10 ⎠⎝ 500 ⎠
兩T 兩
0.02
10
(b)
(c)
(d)
500
␻(rod/s)
Midband gain = 0.02
ω = 500 rad/s
ω = 10 rad/s
7.9
(a)
T ( s) =
=
T ( jω ) =
2 × 104
( S + 10 )( S + 10 )
3
5
2 × 104
(10 )(10 )
3
5
⋅
1
⎛ S
⎞⎛ S
⎞
⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟
10
10
⎝
⎠⎝
⎠
2 × 10−4
⎛ jω
⎞ ⎛ jω
⎞
⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟
10
10
⎝
⎠⎝
⎠
兩T 兩
2 ⫻ 10⫺4
10 3
(b)
(c)
(d)
2 × 10
10 5
␻(rod/s)
−4
ω = 103 rad/s
no low freq −3dB freq.
7.10
a.
⎛ r
⎞
V0 = − g mVπ RL Vπ = ⎜ π ⎟ Vi
+
r
R
S ⎠
⎝ π
⎛ r
⎞
⎛ 5.2 ⎞
T = g m RL ⎜ π ⎟ = ( 29 )( 6 ) ⎜
⎟
⎝ 5.2 + 0.5 ⎠
⎝ rπ + RS ⎠
Tmidband = 159
b.
rS = ( RS + rπ ) CC
1
1
1
fL =
⇒ rS =
=
⇒ rS = 5.31 ms Open-circuit
2π rS
2π f L 2π ( 30 )
rP =
1
1
=
⇒ τ P = 0.332 µ s Short-circuit
2π f H 2π ( 480 × 103 )
c.
rS
5.31× 10−3
=
⇒ CC = 0.932 µ F
( RS + τ π ) ( 0.5 + 5.2 ) ×103
CC =
rP = RL CL
rP 0.332 × 10−6
=
⇒ CL = 55.3 pF
RL
6 × 103
CL =
7.11 Computer Analysis
7.12 Computer Analysis
7.13
a.
RTH = R1 R2 = 10 1.5 = 1.304 kΩ
⎛ R2 ⎞
⎛ 1.5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (12 ) = 1.565 V
⎝ 1.5 + 10 ⎠
⎝ R1 + R2 ⎠
1.565 − 0.7
I BQ =
= 0.0759 mA
1.30 + (101)( 0.1)
I CQ = 7.585 mA
(100 )( 0.026 )
= 0.343 kΩ
7.59
7.59
= 292 mA/V
gm =
0.026
rπ =
Ri = R1 R2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦
= 10 1.5 ⎡⎣0.343 + (101)( 0.1) ⎤⎦
= 1.30 10.44 ⇒ Ri = 1.159 kΩ
r = ( RS + Ri ) CC = [ 0.5 + 1.16] × 103 × 0.1× 10−6
r = 1.659 × 10−4 s
fL =
1
1
=
⇒ f L = 959 Hz
2π r 2π (1.66 × 10−4 )
b.
RS
Rib
V␲
Vi
⫹
⫺
V0
⫹
Ib
r␲
gmV␲
⫺
R1兩兩R2
RC
RE
V0 = − ( β I b ) RC
R1b = rπ + (1 + β ) RE
= 0.343 + (101)( 0.1) = 10.44 kΩ
⎛ R1 R2 ⎞
Ib = ⎜
I
⎜ R R + R ⎟⎟ i
ib ⎠
⎝ 1 2
⎛ 1.30 ⎞
=⎜
⎟ I i = ( 0.111) I i
⎝ 1.30 + 10.4 ⎠
Vi
Ii =
RS + R1 R2 Rib
=
Ii =
V0
=
Vi
Vi
0.5 + (1.3) (10.44 )
Vi
1.659
β RC ( 0.111)
1.659
⇒
V0
Vi
=
(100 )(1)( 0.111)
midband
1.659
⇒
V0
Vi
= 6.69
midband
c.
6.69
兩 VV 兩
0
i
fL ⫽ 959 Hz
f
7.14
I DQ = 0.5 mA ⇒ VS = ( 0.5 )( 0.5 ) = 0.25 V
I DQ = K n (VGS − VTN ) ⇒ VGS =
0.5
+ 1.5 = 3.08 V
0.2
+ VS = 3.08 + 0.25 ⇒ VG = 3.33 V
2
VG = VGS
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD ⇒ 3.33 = ⋅ Rin ⋅ VDD
+
R
R
R
⎝ 1
2 ⎠
1
1
3.33 = ( 200 )( 9 ) ⇒ R1 = 541 kΩ
R1
541R2
= 200 ⇒ R2 = 317 kΩ
541 + R2
VD = VDSQ + VS = 4.5 + 0.25 = 4.75
9 − 4.75
⇒ RD = 8.5 kΩ
0.5
1
1
1
⇒ rL =
=
= 7.96 ms
fL =
2π rL
2π f L 2π ( 20 )
RD =
rL = Rin ⋅ CC ⇒ CC =
rL
7.96 × 10−3
=
⇒ CC = 0.0398 µ F
200 × 103
Rin
g m = 2 ( 0.2 )( 3.08 − 1.5 ) = 0.632 mA/V
Av
midband
=
( 0.632 )(8.5)
g m RD
=
⇒ Av = 4.08
1 + g m RS 1 + ( 0.632 )( 0.5 )
4.08
兩A␯兩
fL ⫽ 20 Hz
Phase
f
fL
⫺90
⫺135
⫺180
7.15
I DQ = K n (VGS − VTN ) ⇒ VGS =
I DQ
2
Kn
+ VTN =
1
+ 1 = 2.414 V
0.5
VS = −2.414 V
⇒ RS = 2.59 kΩ
1
+ VS = 3 − 2.414 = 0.586 V
−2.414 − ( −5 )
RS =
VD = VDSQ
5 − 0.59
⇒ RD = 4.41 kΩ
1
RD =
b.
CC
⫹
Vi
⫹
⫺
RG
Vgs
gmVgs
I0
RD
⫺
RS
⎛
⎜
RD
I 0 = − ( g mVgs ) ⎜
⎜R +R + 1
L
⎜ D
sCC
⎝
Vi
Vgs =
1 + g m RS
I0 ( s )
Vi ( s )
=
T (s) =
=
c.
⎞
⎟
⎟
⎟
⎟
⎠
⎡
⎤
− gm
sCC
⋅ RD ⎢
⎥
1 + g m RS
⎣⎢1 + s ( RD + RL ) CC ⎦⎥
I0 ( s )
Vi ( s )
s ( RD + RL ) CC
− g m RD
1
⋅
⋅
1 + g m RS RD + RL 1 + s ( RD + RL ) CC
RL
fL =
1
1
1
→ rL =
=
= 15.92 ms
2π rL
2π f L 2π (10 )
rL = ( RD + RL ) CC ⇒ CC =
rL
15.9 × 10−3
=
⇒ CC = 1.89 µ F
RD + RL ( 4.41 + 4 ) × 103
7.16
a.
9 − VSG
2
= I D = K P (VSG + VTP )
RS
2
− 4VSG + 4 )
9 − VSG = ( 0.5 )(12 ) (VSG
2
− 23VSG + 15 = 0
6VSG
VSG =
( 23)
23 ±
2
− 4 ( 6 )(15 )
2 ( 6)
⇒ VSG = 3 V
g m = 2 K P (VSG + VTP ) = 2 ( 0.5 )( 3 − 2 ) ⇒ g m = 1 mA/V
Ro =
1
RS = 1 12 ⇒ Ro = 0.923 kΩ
gm
b.
r = ( R0 + RL ) CC
c.
fL =
CC =
1
2πτ
⇒r=
1
2π f L
=
1
= 7.96 ms
2π ( 20 )
7.96 × 10−3
r
=
⇒ CC = 0.729 µ F
Ro + RL ( 0.923 + 10 ) × 103
7.17
a.
1
= 0.00833 mA
120
R1 || R2 = ( 0.1)(1 + β )( RE )
= ( 0.1)(121)( 4 ) = 48.4 kΩ
I CQ = 1 mA, I BQ =
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
1
⋅ RTH ⋅ VCC = ( 0.00833)( 48.4 ) + 0.7 + (121)( 0.00833)( 4 )
R1
1
( 48.4 )(12 ) = 5.135
R1
R1 = 113 kΩ
113R2
= 48.4 ⇒ R2 = 84.7 k Ω
113 + R2
b.
r
R0 = π RE r0
1+ β
rπ =
(120 )( 0.026 )
1
= 3.12 kΩ
80
= 80 kΩ
1
3.12
R0 =
4 80 = 0.02579 4 80 ⇒ R0 = 25.6 Ω
121
c.
r0 =
r = ( R0 + RL ) CC 2
r = ( 0.0256 + 4 ) × 103 × 2 × 10−6 = 8.05 × 10−3 s
f =
1
1
=
⇒ f = 19.8 Hz
2π r 2π ( 8.05 × 10−3 )
7.18
(a)
5 − 0.7
= 1.075 mA I CQ = 1.064 mA
4
= 10 − (1.064 )( 2 ) − (1.075 )( 4 )
= 3.57 V
I EQ =
VCEQ
VCEQ
gm =
rπ =
I CQ
=
VT
β VT
1.064
= 40.92 mA/V
0.026
=
I CQ
(100 )( 0.026 )
1.064
= 2.44 K
(b)
For CC1 , Req1 = RS + RE
rπ
2440
= 200 + 4000
1+ β
101
Req1 = 224.0r , τ 1 = Req1CC1 = 1.053 ms
For CC 2 , Req 2 = RC + RL = 2 + 47 = 49 K
τ 2 = Req 2 ⋅ Cc 2 = 49 ms
(c)
f1 =
1
2πτ 1
=
1
2π (1.053 × 10−3 )
⇒ f1 = 151 Hz
7.19
(a)
τ H = ( RC RL ) CL = ( 2 47 ) × 103 × 10 × 10−12
= 1.918 × 10−8 s
fH =
1
2πτ H
=
1
2π (1.918 × 10−8 )
(b)
1
1 + ( f .2πτ H )
2
= 0.1
2
⎛ 1 ⎞
⎜
⎟ = 100 = 1 + ( f .2πτ H )
⎝ 0.1 ⎠
2
f =
99
2πτ H
=
99
2π (1.918 × 10−8 )
f = 82.6 MHz
7.20
(a)
⇒ f H = 8.30 MHz
5 − VSG
2
= K P (VSG + VTP )
R1
2
2
5 − VSG = (1)(1.2 )(VSG − 1.5 ) = (1.2 ) (VSG
− 3VSG + 2.25 )
2
1.2VSG
− 2.6VSG − 2.3 = 0⇒ VSG = 2.84 V
I DQ = 1.8 mA
VSDQ = 10 − (1.8 )(1.2 + 1.2 ) ⇒ VSDQ = 5.68 V
g m = 2 K P I DQ = 2 (1)(1.8 ) = 2.683 mA / V
ro = ∞
(b)
Ris =
1
1
=
= 0.3727 k Ω
g m 2.68
Ri = 1.2 0.373 = 0.284 k Ω
For CC1 , τ s1 = ( 284 + 200 ) ( 4.7 × 10−6 ) = 2.27 ms
For CC 2 , τ s 2 = (1.2 x103 + 50 × 103 )(10−6 ) = 51.2 ms
(c)
CC2 dominates,
1
1
f 3− dB =
=
= 3.1 Hz
2πτ s 2 2π ( 51.2 × 10−3 )
7.21
Assume VTN = 1V , kn′ = 80µ A / V 2 , λ = 0
Neglecting RSi = 200Ω , Midband gain is:
Av = g m RD
Let I DQ = 0.2 mA, VDSQ = 5V
Then RD =
9−5
⇒ RD = 20 k Ω
0.2
We need g m =
Av
RD
=
10
⎛ k′ ⎞⎛ W ⎞
= 0.5 mA / V 2 and g m = 2 K n I DQ = 2 ⎜ n ⎟ ⎜ ⎟ I DQ
20
⎝ 2 ⎠⎝ L ⎠
W
⎛ 0.080 ⎞ ⎛ W ⎞
or 0.5 = 2 ⎜
= 7.81
⎟ ⎜ ⎟ ( 0.2 ) ⇒
L
⎝ 2 ⎠⎝ L ⎠
Let
9
9
=
= 225 k Ω
R1 + R2 =
( 0.2 ) I DQ ( 0.2 )( 0.2 )
⎛ R2 ⎞
2
⎛ R2 ⎞
⎛ 0.080 ⎞
I DQ = 0.2 = ⎜
⎟ (9) = ⎜
⎟ ( 7.81)(VGS − 1) ⇒ VGS = 1.80 = ⎜
⎟ (9) ⇒
⎝ 2 ⎠
⎝ 225 ⎠
⎝ R1 + R2 ⎠
R2 = 45 k Ω, R1 = 180 k Ω
RTH = R1 R2 = 180 45 = 36 k Ω
τ1 =
1
2π f1
=
1
7.96 × 10−4
= 7.958 × 10−4 s = ( RSi + RTH ) CC or CC =
⇒
2π ( 200 )
( 200 + 36 ×103 )
CC = 0.022 µ F
τ2 =
7.22
a.
1
2π f 2
=
1
2π ( 3 x103 )
= 5.305 × 10−5 s = RD CL or CL =
5.31× 10−5
⇒ CL = 2.65 nF
20 × 103
T (s) =
T (s) =
R2 + (1/ sC )
R2 + (1/ sC ) + R1
1 + sR2 C
1 + s ( R1 + R2 ) C
rA = R2 C , rB = ( R1 + R2 ) C
b.
兩T 兩
1
0.0909
fB
fA
f
c.
rA = R2 C = (103 )(100 × 10−12 ) = 10−7 s = rA
rB = ( R1 + R2 ) C = [10 + 1] × 103 × 100 × 10−12
= 1.1× 10−6 s = rB
1
1
=
⇒ f A = 1.59 MHz
2π rA 2π (10−7 )
fA ≈
1
1
=
⇒ f B = 0.145 MHz
2π rB 2π (1.1× 10−6 )
fB ≈
7.23
I BQ =
10 − 0.7
= 0.00997 mA
430 + ( 201)( 2.5 )
I CQ = ( 200 ) I BQ = 1.995 mA
( 200 )( 0.026 )
= 2.61 k Ω
1.99
Rib = 2.61 + ( 201)( 2.5 ) = 505 k Ω
rπ =
τs =
1
=
2π f L
1
= 0.0106 s
2π (15 )
= Req CC = ( 0.5 + 505 430 ) × 103 CC = 232.7 × 103 CC
Or CC = 4.55 × 10−8 F ⇒ 45.5 nF
7.24
10 = I BQ (300) + 0.7 + ( 201) I BQ (1) ⇒ I BQ = 0.0186 mA ⇒ I CQ = 3.7126 mA
rπ =
β VT
I CQ
=
( 200 )( 0.026 )
3.71
= 1.40 K
Ri = rπ + (1 + β ) RE = 1.40 + ( 201)(1) = 202.4 K
Req = RS + RB Ri = 0.1 + 300 202.4 = 121 K
τ L = Req ⋅ CC
fL =
CC =
7.25
1
2πτ L
τL
Req
⇒τL =
=
1
2π f
=
1
= 7.958 × 10−3 s
2π ( 20 )
7.958 × 10−3
⇒ CC = 0.0658 µ F
121× 103
RTH = R1 R2 = 1.2 1.2 = 0.6 k Ω
⎛ R2 ⎞
⎛ 1.2 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ ( 5 ) = 2.5 V
⎝ 1.2 + 1.2 ⎠
⎝ R1 + R2 ⎠
2.5 − 0.7
= 0.319 mA
I BQ =
0.6 + (101)( 0.05 )
I CQ = 31.9 mA
rπ =
(100 )( 0.026 )
31.9
= 0.0815 k Ω
1
so that f 3− dB ( CC1 ) << f3− dB ( CC 2 )
2πτ
Then, for f 3− dB ( CC1 ) ⇒ CC 2 acts as an open and for f 3− dB ( CC 2 ) ⇒ CC1 acts as a short circuit.
τ C >> τ C and f =
C1
C2
f 3− dB ( CC 2 ) = 25 Hz =
1
2πτ 2
, so that τ 2 =
1
= 0.006366 s = Req CC 2
2π ( 25 )
where
⎛ r + R1 R2 RS ⎞
Req = RL + RE ⎜ π
⎟
1+ β
⎝
⎠
⎛ 81.5 + 600 300 ⎞
= 10 + 50 ⎜
⎟ = 10 + 50 2.787 ⇒
101
⎝
⎠
0.00637
Req = 12.64 Ω ⇒ CC 2 =
⇒ CC 2 = 504 µ F
12.6
Rib = rπ + (1 + β ) RE Assume CC 2 an open
Rib = 81.5 + (101)( 50 ) = 5132 Ω
τ 1 = (100 )τ 2 = (100 )( 0.006366 ) = 0.6366 s = Req1CC1
Req1 = RS + RTH Rib = 300 + 600 5132 = 837.2 Ω
So CC1 =
0.6366
⇒ CC1 = 760 µ F
837.2
7.26
From Problem 7.25 RTH = 0.6 K, I CQ = 31.9 mA, rπ = 81.5 Ω
1
so f 3− dB ( CC 2 ) f 3− dB ( CC1 )
2πτ
Then f 3− dB ( CC 2 ) ⇒ CC1 acts as an open circuit and for f 3− dB ( CC1 ) ⇒ CC 2 acts as a short circuit.
τ C 2 τ C1 and f =
f 3− dB ( CC1 ) = 20 Hz =
1
2πτ C1
⇒ τ C1 = 0.007958 s
Rib = rπ + (1 + β ) ( RE RL ) = 81.5 + (101) ( 50 10 ) = 923.2 Ω
τ C1 ⇒ Req1 = RS + RTH Rib = 300 + 600 923.2 = 663.7 Ω
0.007958
⇒ CC1 = 12 µ F
663.7
= 100τ C1 = 0.7958 s
CC1 =
τC2
⎛ r + RTH ⎞
Req 2 = RL + RE ⎜ π
⎟ = 10 + 50
⎝ 1+ β ⎠
Req 2 = 10 + 50 6.748 = 15.95 Ω
0.7958
CC 2 =
⇒ CC 2 = 0.050 F
15.95
7.27
a.
⎛ 81.5 + 600 ⎞
⎜
⎟
⎝ 101 ⎠
I D = K n (VGS − VTN )
VGS =
2
ID
0.5
+ VTN =
+ 0.8 = 1.8 V
Kn
0.5
5 − 1.8
⇒ RS = 6.4 kΩ
0.5
0.5
VD = VDSQ + VS = 4 − 1.8 = 2.2 V
5 − 2.2
RD =
⇒ RD = 5.6 kΩ
0.5
(b)
RS =
−VGS − ( −5 )
gm = 2 Kn I D = 2
=
( 0.5 )( 0.5 ) = 1 mA / V
rA = RS CS = ( 6.4 × 103 )( 5 × 10−6 )
= 3.2 × 10−2 s
1
1
=
⇒ f A = 4.97 Hz
fA =
2π rA 2π ( 3.2 × 10−2 )
⎡ 6.4 × 103 ⎤
⎞
−6
⎥ ( 5 × 10 )
⎟ CS = ⎢
+
1
1
6.4
(
)(
)
⎢⎣
⎠
⎦⎥
= 4.32 × 10−3 s
1
1
=
⇒ f B = 36.8 Hz
fB =
2π rB 2π ( 4.32 × 10−3 )
⎛ RS
rB = ⎜
⎝ 1 + g m RS
c.
Av =
g m RD (1 + sRS CS )
⎡
⎛
(1 + g m RS ) ⎢1 + s ⎜
RS
1
+
⎝ g m RS
⎞ ⎤
⎟ CS ⎥
⎠ ⎦
⎣
As RS becomes large
g m RD ( sRS CS )
Av →
⎡
⎛ R ⎞ ⎤
( g m RS ) ⎢1 + s ⎜ S ⎟ CS ⎥
⎝ g m RS ⎠ ⎦
⎣
⎡ ⎛ 1 ⎞ ⎤
⎟ CS ⎥
⎣ ⎝ gm ⎠ ⎦
( g m RD ) ⎢ s ⎜
Av =
⎛ 1
1+ s ⎜
⎝ gm
⎞
⎟ CS
⎠
The corner frequency f B =
gm = 2 Kn I D = 2
fB =
1
and the corresponding f A → 0
2π (1/ g m ) CS
( 0.5 )( 0.5 ) = 1 mA / V
1
⇒ f B = 31.8 Hz
⎛ 1 ⎞
−6
2π ⎜ −3 ⎟ ( 5 × 10 )
⎝ 10 ⎠
7.28
a.
fB =
RE ( RS + rπ ) CE
1
and rB =
2π rB
RS + rπ + (1 + β ) RE
For RS = 0 rB =
I EQ =
RE rπ CE
rπ + (1 + β ) RE
−0.7 − ( −10 )
5
= 1.86 mA
β = 75 ⇒ I CQ = 1.84 mA
β = 125 ⇒ I CQ = 1.85 mA
For f B ≤ 200 Hz ⇒ rB ≥
1
= 0.796 ms
2π ( 200 )
rπ αβ so smallest rB will occur for smallest β .
β = 75; rπ =
( 75)( 0.026 )
= 1.06 kΩ
1.84
( 5 ×103 ) (1.06 ) CE ⇒ C = 57.2 µ F
0.796 × 10−3 =
E
1.06 + ( 76 )( 5 )
b.
(125 )( 0.026 )
For β = 125; rπ =
= 1.76 kΩ
1.85
5 × 103 ) (1.76 ) ( 57.2 × 10−6 )
(
= 0.797 ms
rB =
1.76 + (126 )( 5 )
fB =
1
1
=
⇒ f B = 199.7 Hz Essentially independent of β .
2π rB 2π ( 0.797 × 10−3 )
rA = RE CE = ( 5 × 103 )( 57.2 × 10−6 ) = 0.286 sec
fA =
1
1
=
⇒ f A = 0.556 Hz Independent of β .
2π rA 2π ( 0.286 )
7.29
a.
Expression for the voltage gain is the same as Equation (7.66) with RS = 0.
b.
rA = RE CE
RE rπ CE
rB =
rπ + (1 + β ) RE
7.30
5 − 0.7
I C = 0.99 mA
= 1 mA = I E
4.3
β VT (100 )( 0.026 )
rπ =
=
= 2.626 K
0.99
I CQ
rA = RE CE = ( 4.3 × 103 )( 5 × 10−6 ) = 2.15 ×10 −2 s
rB =
( 4.3 ×103 )( 2.626 ×103 )( 5 ×10−6 ) = 1.292 ×10−4 s
RE rπ CE
=
rπ + (1 + β ) RE
2.626 × 103 + (101) ( 4.3 × 103 )
fA =
1
1
=
= 7.40 Hz
2π rA 2π ( 2.15 × 10−2 )
fB =
1
1
=
= 1.23 × 103 = 1.23 kHz
2π rB 2π (1.292 × 10−4 )
Av
w→0
Av
w →∞
=
=
(100 )( 2 )
β RC
=
= 0.458
rπ + (1 + β ) RE 2.626 + (101)( 4.3)
β RC
rπ
=
(100 )( 2 )
2.626
= 76.2
兩A␯兩
76.2
0.458
7.40 Hz
1.23 kHz
f
7.31
rH = ( RL RC ) CL = (10 5 ) × 103 × 15 × 10−12
rH = 5 × 10−8 s
1
1
=
⇒ f H = 3.18 MHz
2π rH 2π ( 5 × 10−8 )
fH =
10 − 0.7
= 0.93 mA, I CQ = 0.921 mA
10
0.921
gm =
= 35.4 mA/V
0.026
Av = g m ( RC RL ) = 35.4 ( 5 10 ) ⇒ Av = 118
I EQ =
7.32
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
ID =
⎛ 166 ⎞
=⎜
⎟ (10 )
⎝ 166 + 234 ⎠
= 4.15 V
VG − VGS
2
= K n (VGS − VTN )
RS
4.15 − VGS = ( 0.5 )( 0.5 ) (VGS2 − 4VGS + 4 )
0.25VGS2 − 3.15 = 0 ⇒ VGS = 3.55 V
g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 )
g m = 1.55 mA / V
1
1
= 0.5
= 0.5 0.645
1.55
gm
R0 = RS
R0 = 0.282 kΩ
r = ( R0 RL ) CL and f H =
βω ≈ f H = 5 MHz ⇒ r =
CL =
7.33
r
R0 RL
=
1
2π r
1
2π ( 5 × 106 )
= 3.18 × 10−8 s
3.18 × 10−8
⇒ CL = 121 pF
( 0.282 4 ) ×103
(a) Low-frequency
RS
CC
Vo
⫹
⫹
⫺
Vs
RB
V␲
RC
r␲
⫺
RL
gmV␲
Mid-Band
RS
Vo
⫹
⫹
⫺
Vs
RB
V␲
r␲
⫺
RC
RL
gmV␲
High-frequency
RS
Vo
⫹
⫹
⫺
Vs
RB
V␲
⫺
r␲
RC
gmV␲
(b)
兩Am兩
fL
fH
f
(c)
12 − 0.7
= 11.3 µ A
1 MΩ
= 1.13 mA
I BQ =
I CQ
(100 )( 0.026 )
= 2.3 k Ω
1.13
1.13
gm =
= 43.46 mA / V
0.026
rπ =
Am =
⎛ R R ⎞
V0
( midband ) = − g m ( RC RL ) ⎜⎜ B π ⎟⎟
Vs
⎝ RB rπ + RS ⎠
⎛ 1000 2.3 ⎞
= − ( 43.46 ) ( 5.1 500 ) ⎜
⎜ 1000 2.3 + 1 ⎟⎟
⎝
⎠
⎛ 2.29 ⎞
= ( 43.46 )( 5.05 ) ⎜
⎟ ⇒ Am = 153
⎝ 2.29 + 1 ⎠
Am dB = 43.7 dB
RL
CL
fL =
1
2πτ L
, τ L = ( RS + RB rπ ) CC or τ L = (1 + 1000 2.3) × 103 (10 × 10−6 )
⇒ τ L = 3.29 × 10−2 s ⇒ f L = 4.83 Hz
fH =
1
, τ H = ( RC RL ) CL ⇒ τ L = ( 5.1 500 ) × 103 (10 × 10−12 )
2πτ H
= 5.05 × 10−8 s ⇒ f H = 3.15 MHz
7.34
a.
V0
⫺
Vi
⫹
⫺
Vsg
RD
gmVsg
RL
CL
⫹
⎛
1 ⎞
V0 = ( g mVsg ) ⎜ R0 RL
⎟
sCL ⎠
⎝
Vsg = −Vi
Av ( s ) =
⎛
1 ⎞
= − g m ⎜ RD RL
⎟
Vi ( s )
sCL ⎠
⎝
V0 ( s )
⎡
1
⎢ RD RL ⋅
sC
L
= − gm ⎢
⎢
1
⎢ RD RL +
sCL
⎣⎢
Av ( s ) = − g m ( RD RL ) ⋅
b.
c.
⎤
⎥
⎥
⎥
⎥
⎦⎥
1
1 + s ( RD RL ) CL
r = ( RD RL ) CL
r = (10 20 ) × 103 × 10 × 10−12 ⇒ r = 6.67 × 10 −8 s
fH =
1
1
=
⇒ f H = 2.39 MHz
2π r 2π ( 6.67 × 10−8 )
From Example 7.6, gm = 0.705 mA/V
Av = g m ( RD RL ) = ( 0.705 ) (10 20 ) ⇒ Av = 4.7
7.35 Computer Analysis
7.36 Computer Analysis
7.37 Computer Analysis
7.38
fT =
gm
2π ( Cπ + Cµ )
I CQ
gm =
fT =
1
= 38.46 mA/V
0.026
=
VT
38.46 × 10−3
2π (10 + 2 ) × 10−12
fT = 510 MHz
fT
fβ =
=
β
7.39
fT
fβ =
fT =
=
β
510
⇒ f β = 4.25 MHz
120
5000 MHz
⇒ f β = 33.3 MHz
150
gm
2π ( Cπ + Cµ )
0.5
= 19.23 mA/V
0.026
19.2 × 10 −3
5 × 109 =
2π ( Cπ + 0.15 ) × 10−12
gm =
Cπ + 0.15 =
19.2 × 10−3
2π (10−12 )( 5 × 109 )
= 0.612 pF
Cπ = 0.462 pF
7.40
a.
fβ =
fT
β
=
2000 MHz
= 13.3 MHz = f β
150
b.
h fe =
150
1 + j ( f / fβ )
h fe =
150
1 + ( f / fβ )
2
= 10
⎛ f ⎞ ⎛ 150 ⎞ 2
1+ ⎜ ⎟ = ⎜
= 225
⎜ f ⎟ ⎝ 10 ⎟⎠
⎝ β⎠
2
f = f β ⋅ 224 = (13.33) 224 ⇒ f = 199.6 MHz
7.41
(a)
V0 = − g mVπ RL where
1
sC1
rπ
1 + srπ C1
Vπ =
⋅ Vi =
⋅ Vi
rπ
1
+ rb
rπ
+ rb
1 + srπ C1
sC1
rπ
=
⎞
⎛ r ⎞⎛
rπ
1
⋅ Vi = ⎜ π ⎟ ⎜
⎟ ⋅ Vi
rπ + rb + srb rπ C1
⎝ rπ + rb ⎠ ⎜⎝ 1 + s ( rb rπ ) C1 ⎟⎠
So Av ( s ) =
⎞
⎛ r ⎞⎛
1
= − g m RL ⎜ π ⎟ ⎜
⎟
⎜
⎟
Vi ( s )
⎝ rπ + rb ⎠ ⎝ 1 + s ( rb rπ ) C1 ⎠
V0 ( s )
(100 )( 0.026 )
(b) Midband gain: rπ =
(i)
(ii)
1
= 2.6 k Ω, g m =
1
= 38.46 mA / V
0.026
For rb = 100 Ω
⎛ 2.6 ⎞
Av1 = − ( 38.46 )( 4 ) ⎜
⎟ ⇒ Av1 = −148.1
⎝ 2.6 + 0.1 ⎠
For rb = 500 Ω
(i)
⎛ 2.6 ⎞
Av 2 = − ( 38.46 )( 4 ) ⎜
⎟ ⇒ Av 2 = −129.0
⎝ 2.6 + 0.5 ⎠
1
f 3− dB =
, τ = ( rb rπ ) C1
2πτ
For rb = 100 Ω
(ii)
For rb = 500 Ω
(c)
τ 1 = ( 0.1 2.6 ) × 103 ( 2.2 ×10−12 ) = 2.12 × 10−10 s ⇒ f 3− db = 751 MHz
τ 2 = ( 0.5 2.6 ) × 103 ( 2.2 ×10−12 ) = 9.23 ×10−10 s f 3− dB = 173 MHz
7.42
(b)
f = 10 kHz = 104
Z i = 200 +
(
2500 1 − j (104 )(1.333 × 10−6 )
1 + (10
4 2
)
−6 2
) (1.333 ×10 )
= 200 + 2500 − j 33.3 = 2700 − j 33.3
f = 100 kHz = 105
(c)
Z i = 200 +
(
2500 1 − j (105 )(1.333 × 10−6 )
1 + (10
5 2
)
−6 2
) (1.333 ×10 )
Z i = 200 + 2456 − j 327 = 2656 − j 327
f = 1 MHz = 106
(d)
Z i = 200 +
(
2500 1 − j (106 )(1.333 × 10−6 )
1 + (10
6 2
)
−6 2
) (1.333 ×10 )
Z i = 200 + 900 − j1200 = 1100 − j1200
7.43
a.
CM = Cµ (1 + g m RL )
b.
RB兩兩RS
rb
V0
⫹
RB
• Vi
RB ⫹ RS
⫹
⫺
V␲
⫺
r␲
C␲
CM
gmV␲
RL
V0 = − g mVπ RL
1
sCi
⎛ RB ⎞
⋅⎜
⎟ Vi
R + RS ⎠
1
+ RB RS + rb ⎝ B
rπ
sC1
rπ
Vπ =
Let Cπ + CM = Ci
Av ( s ) =
Vo ( s )
Vi ( s )
1
⎡
⎤
rπ ⋅
⎢
⎥
sCi
⎢
⎥
1
⎢
⎥
rπ +
⎥
⎛ RB ⎞ ⎢
sCi
= − g m RL ⎜
⎥
⎟⎢
⎝ RB + RS ⎠ ⎢ rπ ⋅ 1
⎥
⎢
⎥
sCi
+ RB & RS + rb ⎥
⎢
1
⎢ rπ +
⎥
sCi
⎢⎣
⎥⎦
⎤
⎛ RB ⎞ ⎡
rπ
= − g m RL ⎜
⎥
⎟× ⎢
⎝ RB + RS ⎠ ⎢⎣ rπ + (1 + srπ Ci ) ( RB RS + rb ) ⎥⎦
Let Req = ( RB RS + rb )
⎛ RB
Av ( s ) = − β RL ⎜
⎝ RB + RS
Av ( s ) =
c.
⎡
⎤
⎞ ⎢
1
⎥
×
⎟ ⎢
⎥
⎡
⎤
⎠ ⎢ ( rπ + Req ) 1 + s rπ Req Ci ⎥
⎣
⎦⎦
⎣
(
− β RL ⎛ RB ⎞
1
⋅⎜
⎟⋅
rπ + Req ⎝ RB + RS ⎠ 1 + s rπ Req Ci
(
fH =
)
)
1
(
)
2π rπ Req Ci
7.44
High Freq. ⇒ CC1 , CC 2 , CE → short circuits
C␮
V0
⫹
IS
R1兩兩R2 V␲
⫺
r␲
C␲
gmV␲
RC
RL
I0
gm =
fT =
I CQ
VT
=
5
= 192.3 mA/V
0.026
gm
2π ( Cπ + Cµ )
⇒ 250 × 106 =
192 × 10−3
2π ( Cπ + Cµ )
Cπ + Cµ = 122.4 pF ⇒ Cµ = 5 pF, Cπ = 117.4 pF
(
CM = Cµ 1 + g m ( RC RL )
)
= 5 ⎣⎡1 + (192.3) (1 1) ⎦⎤ ⇒ CM = 485.8 pF
Ci = Cπ + CM = 117 + 485 = 603 pF
rπ =
( 200 )( 0.026 )
= 1.04 kΩ
5
Req = R1 R2 rπ = 5 1.04 = 0.861 kΩ
r = Req ⋅ Ci
= ( 0.861× 103 )( 603 × 10−12 )
= 5.19 × 10 −7 s
1
f =
2π r
=
1
2π ( 5.19 × 10−7 )
⇒ f = 307 kHz
7.45
RTH = R1 R2 = 60 5.5 = 5.04 kΩ
⎛ R2 ⎞
⎛ 5.5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (15 ) = 1.26 V
⎝ 5.5 + 60 ⎠
⎝ R1 + R2 ⎠
1.26 − 0.7
I BQ =
= 0.0222 mA
5.04 + (101)( 0.2 )
I CQ = 2.22 mA
(100 )( 0.026 )
= 1.17 kΩ
2.22
2.22
gm =
= 85.4 mA/V
0.026
Lower 3 – dB frequency:
rL = Req ⋅ CC1
rπ =
Req = RS + R1 R2 rπ
= 2 + 60 5.5 1.17 = 2.95 k Ω
rL = ( 2.95 ×103 )( 0.1× 10−6 ) = 2.95 × 10−4 s
fL =
1
1
=
⇒ f L = 540 Hz
2π rL 2π ( 2.95 × 10−4 )
Upper 3 – dB frequency:
gm
fT =
2π ( Cπ + Cµ )
⇒ 400 × 106 =
85.4 × 10−3
2π ( Cπ + Cµ )
Cπ + Cµ = 34 pF; Cµ = 2 pF ⇒ Cπ = 32 pF
CM = Cµ (1 + g m RC ) = 2 (1 + ( 85.4 )( 4 ) ) ⇒ CM = 685 pF
Ci = Cπ + CM = 32 + 685 = 717 pF
Req = RS R1 R2 rπ
= 2 60 5.5 1.17
= 0.644 kΩ
= ( 0.644 × 103 )( 717 × 10−12 )
r = Req ⋅ Ci
= 4.62 × 10−7 s
fH =
1
2π r
⇒ f H = 344 kHz
7.46
RTH = R1 R2 = 600 55 = 50.38 K
⎛ R2 ⎞
⎛ 55 ⎞
VTH = ⎜
⎟ (15 ) = ⎜
⎟ (15 ) = 1.2595 V
⎝ 600 + 55 ⎠
⎝ R1 + R2 ⎠
1.26 − 0.7
I BQ =
= 0.00222 mA
50.4 + (101)( 2 )
I CQ = 0.2217 mA
(100 )( 0.026 )
= 11.73 K
0.222
0.2217
gm =
= 8.527 mA/V
0.026
Lower – 3dB Fig.
τ L = R e q1 Cc1 ; Req1 = RS + RTH r π
rπ =
= 0.50 + 50.38 11.73 = 10.0 K
τ L = (10 × 10
3
)( 0.1×10 ) = 10
−6
−3
s
τ L = R e q1 Cc1 ; Req1 = RS + RTH r π
= 0.50 + 50.38 11.73 = 10.0 K
τ L = (10 × 103 )( 0.1×10−6 ) = 10−3 s
1
fL =
2πτ 2
=
1
2π (10−3 )
⇒ f L = 159 Hz
Upper – 3dB Fig.
gm
8.527 × 10−3
=
= 400 × 106
fT =
−12
2π ( Cπ + Cµ ) 2π ( Cπ + 2 ) × 10
Cπ + Cµ = 3.393 pF ⇒ Cπ = 1.393 pF
CM = Cµ (1 + g m RC ) = 2 ⎡⎣1 + ( 8.527 )( 40 ) ⎤⎦ = 684 pF
CT = Cπ + CM = 1.393 + 684 = 685.4 pF
Req 2 = RS RTH rπ = 0.5 50.38 11.73
= 50.38 0.480 = 0.4750 K
τ H = R eq 2 .CT
fH =
= ( 0.4750 × 103 ) ( 685.4 × 10−12 )
= 3.256 × 10−7 s
1
1
=
⇒ f H = 489 KHz
2πτ H 2π ( 3.256 × 10−7 )
7.47
fT =
gm
2π ( Cgs + Cgd )
⎛ 40 ⎞
g m = 2 K n I D , K n = (15 ) ⎜ ⎟ = 60 µ A / V 2
⎝ 10 ⎠
gm = 2
fT =
( 60 )(100 ) = 154.9 µ A / V
154.9 × 10−6
⇒ fT = 44.8 MHz
2π ( 0.5 + 0.05 ) × 10−12
7.48
fT =
gm
2π ( Cgs + Cgd )
g m = 2 K n (VGs − VT )
ID
then g m = 2 K n I D
Kn
2
I D = K n (VGs − VTN ) or VGs − VTN =
So fT =
2 Kn I D
2π ( Cgs + Cgd )
=
Kn I D
π ( Cgs + Cgd )
( 0.2 ×10 )( 20 ×10 )
−3
(a)
fT =
(b)
fT =
(c)
9
−6
π ( 0.5 + 0.1) × 10−12
( 0.2 ×10 )( 250 ×10 )
−3
−6
π ( 0.5 + 0.1) × 10−12
( 0.2 ×10 ) I
−3
10 =
⇒ fT = 33.6 MHz
D
π ( 0.5 + 0.1) × 10−12
⇒ fT = 118.6 MHz
⇒ I D = 17.8 mA
7.49
fT =
gm
2π ( Cgs + Cgd )
Cgs + Cgd = WLCox
⎛W ⎞⎛ µ C ⎞
g m = 2 K n (VGS − VTN ) = 2 ⎜ ⎟ ⎜ n ox ⎟ (VGS − VTN )
⎝ L ⎠⎝ 2 ⎠
⎛W ⎞
⎜ ⎟ ( µ n Cox )(VGS − VTN )
L
Then fT = ⎝ ⎠
2π WLCox
fT =
µ n (VGS − VTN )
2π L2
(a)
fT =
(b)
fT =
7.50
a.
450 ( 0.5 )
2π (1.2 × 10−4 )
2
450 ( 0.5 )
2π ( 0.18 × 10−4 )
⇒ fT = 2.49 GHz
⇒ fT = 111 GHz
(
CM = Cgd ′ 1 + g m ( ro RD )
)
CM = 5 ⎡⎣1 + ( 3) (15 10 ) ⎤⎦ ⇒ CM = 95 pF
b.
r = ( ri ) ( Cgs + CM )
r = (10 × 103 ) ( 50 + 95 ) × 10−12 = 1.45 × 10−6 s
f =
1
1
=
⇒ f = 110 kHz
2π r 2π (1.45 × 10−6 )
7.51
fT =
gm
2π ( CgsT + CgdT )
( Eq. 7.104 )
⎛ 2⎞
Let CgdT = 0 and CgsT = ⎜ ⎟ (WLCox )
⎝ 3⎠
⎛µ C
g m = 2 K n I D = 2 ⎜ n ox
⎝ 2
⎞ ⎡W ⎤
⎟ ⎢ L ⎥ ID
⎠⎣ ⎦
⎛1
⎞⎛ W ⎞
2 ⎜ µ n Cox ⎟⎜ ⎟ I D
⎝2
⎠⎝ L ⎠
So fT =
⎛ 2⎞
2π ⎜ ⎟ (W LCox )
⎝ 3⎠
⎛1
⎞⎛ W ⎞
⎜ µ n Cox ⎟⎜ ⎟ I D
2
3
⎝
⎠⎝ L ⎠
=
⋅
W Cox
2π L
fT =
3
2π L
⋅
µn I D
2W Cox L
7.52
(a)
g m′ =
gm
1 + g m rS
g m = 2 K n (VGS − VTN )
−8
⎛ µ C ⎞ ⎛ W ⎞ ⎛ ( 400 ) ( 7.25 × 10 ) ⎞⎟
K n = ⎜ n ox ⎟ ⎜ ⎟ = ⎜
(10 )
⎟
2
⎝ 2 ⎠ ⎝ L ⎠ ⎜⎝
⎠
K n = 1.45 × 10−4 A / V 2
For VGS = 5 V
g m = 2 (1.45 × 10 −4 ) ( 5 − 0.65 ) = 1.26 × 10−3 A/V
g m′ = ( 0.80 ) g m = 1.01× 10−3 A/V
1.01× 10−3 =
1.26 × 10−3
1 + (1.26 × 10−3 ) rS
1 + (1.26 × 10−3 ) rS = 1.25 ⇒ rS = 196 Ω
b.
For VGS = 3 V
g m = 2 (1.45 × 10−4 ) ( 3 − 0.65 ) = 0.6815 × 10−3 A/V
g m′ =
0.6815 × 10−3
1 + ( 0.6815 × 10−3 ) (196 )
g m′ = 0.60 × 10−3 A/V
Re duced by ≈ 12%
7.53
a.
⫹
Vgs
Ii
Ri
gmVgs
⫺
RL
I0
rS
I 0 = g mVgs and Vgs = I i Ri − g mVgs rS so Vgs =
Then Ai =
b.
I i Ri
1 + g m rS
I0
g R
= m i
I i 1 + g m rS
As an approximation, consider
⫹
Ii
Vgs
Ri
CgsT
I0
CM
RL
g⬘mVgs
⫺
In this case
I
gm
1
Ai = 0 = g m′ Ri ⋅
where CM = C gdT (1 + g m′ RL ) and g m′ =
Ii
1 + g m rs
1 + sRi ( C gsT + CM )
c.
As rS increases, CM decreases, so the bandwidth increases, but the current gain magnitude
decreases.
7.54
⎛ R2 ⎞
⎛ 225 ⎞
VGS = ⎜
⎟ VDD = ⎜
⎟ (10 )
⎝ 225 + 500 ⎠
⎝ R1 + R2 ⎠
VGS = 3.10 V
g m = 2 K n (VGS − VTN ) = 2 (1)( 3.10 − 2 )
g m = 2.207 mA / V
a.
CgdT
Ri
V0
⫹
Vi
⫹
⫺
R1兩兩R2
Vgs
⫺
b.
CgsT
gmVgs
RD
CM = C gdT (1 + g m RD ) = (1) ⎡⎣1 + ( 2.207 )( 5 ) ⎤⎦
CM = 12 pF
c.
r = ( Ri R1 R2 ) ( CgsT + CM )
Ri R1 R2 = 1 500 225 = 1 155 = 0.9936 kΩ
r = ( 0.9936 × 103 ) ( 5 + 12 ) × 10−12 = 1.69 × 10 −8 s
fH =
Av =
1
2π r
=
1
2π (1.69 × 10−8 )
− g mVgs RD
Vi
and Vgs =
⇒ f H = 9.42 MHz
R1 R2
R1 R2 + Ri
Av = − ( 2.2 )( 5 )( 0.994 ) ⇒ Av = −10.9
⋅ Vi =
155
⋅ Vi = 0.994 Vi
155 + 1
7.55
RTH = R1 R2 = 33 22 = 13.2 kΩ
⎛ R2 ⎞
⎛ 22 ⎞
VTH = ⎜
⎟ ( 5) = ⎜
⎟ ( 5) = 2 V
⎝ 22 + 33 ⎠
⎝ R1 + R2 ⎠
2 − 0.7
I BQ =
= 0.00261 mA
13.2 + (121)( 4 )
I CQ = 0.3138
(120 )( 0.026 )
= 9.94 kΩ
0.3138
0.3138
gm =
= 12.07 mA/V
0.026
100
r0 =
= 318 kΩ
0.3138
a.
gm
fT =
2π ( Cπ + Cµ )
rπ =
Cπ + Cµ =
gm
12.07 × 10−3
=
2π fT 2π ( 600 × 106 )
Cπ + Cµ = 3.20 pF; Cµ = 1 pF ⇒ Cπ = 2.20 pF
CM = Cµ ⎡1 + g m ro RC RL ⎤
⎣
⎦
= (1) ⎡1 + (12.07 ) 318 4 5 ⎤
⎣
⎦
CM = 27.6 pF
b.
(
)
(
)
r = Req ( Cπ + CM )
Req = R1 R2 Rs rπ = 33 22 2 rπ
= 1.74 9.94 kΩ ⇒ Req = 1.48 kΩ
r = (1.48 × 103 ) ( 2.20 + 27.6 ) × 10−12
r = 4.41× 10 −8 s
fH =
1
2π r
1
=
2π ( 4.41× 10−8 )
(
V0 = − g mVπ ro RC RL
⇒ f H = 3.61 MHz
)
⎛ R1 R2 rπ ⎞
⎟V
Vπ = ⎜
⎜ R1 R2 rπ + RS ⎟ i
⎝
⎠
R1 R2 rπ = 33 22 9.94 = 5.67 kΩ
⎛ 5.67 ⎞
vπ = ⎜
⎟ Vi = 0.739Vi
⎝ 5.67 + 2 ⎠
r0 RC RL = 318 4 5 = 2.18 kΩ
Av = − (12.07 )( 0.739 )( 2.18 )
Av = −19.7
7.56
RTH = R1 R2 = 40 5 = 4.44 kΩ
⎛ R2 ⎞
⎛ 5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 1.111 V
⎝ 5 + 40 ⎠
⎝ R1 + R2 ⎠
1.111 − 0.7
= 0.00633 mA
I BQ =
4.44 + (121)( 0.5 )
I CQ = 0.760 mA
(120 )( 0.026 )
= 4.11 kΩ
0.760
0.760
= 29.23 mA/V
gm =
0.026
r0 = ∞
rπ =
fT =
gm
2π ( Cπ + Cµ )
Cπ + Cµ =
gm
29.23 × 10−3
=
2π fT 2π ( 250 × 106 )
Cπ + Cµ = 18.6 pF; Cµ = 3 pF ⇒ Cπ = 15.6 pF
a.
CM = Cµ ⎡⎣1 + g m ( RC RL ) ⎤⎦
CM = 3 ⎡⎣1 + ( 29.2 ) ( 5 2.5 ) ⎤⎦ ⇒ CM = 149 pF
For upper frequency:
x
rH = Req ( Cπ + CM )
Req = rπ R1 R2 RS = 4.11 40 5 0.5
Req = 0.405 kΩ
rH = ( 0.405 × 103 ) (15.6 + 149 ) × 10−12
= 6.67 ×10 −8 s
1
⇒ f H = 2.39 MHz
2π rH
For lower frequency:
rL = Req CC1
fH =
Req = RS + R1 R2 rπ = 0.5 + 40 5 4.11
Req = 2.64 kΩ
rL = ( 2.64 × 103 )( 4.7 × 10−6 ) = 1.24 × 10−2 s
fL =
1
⇒ f L = 12.8 Hz
2π rL
b.
兩A␯兩
39.5
fL
fH
V0 = − g mVπ ( RC RL )
⎛ R1 R2 rπ ⎞
Vπ = ⎜
⎟V
⎜ R1 R2 rπ + RS ⎟ i
⎝
⎠
⎛ 2.135 ⎞
Vπ = ⎜
⎟ Vi = 0.8102Vi
⎝ 2.135 + 0.5 ⎠
AV = ( 29.23)( 0.8102 ) ( 5 2.5 )
AV = 39.5
7.57
2
I D = K P (VSG + VTP ) =
9 − VSG
RS
( 2 )(1.2 ) (VSG2 − 4VSG + 4 ) = 9 − VSG
2
2.4VSG
− 8.6VSG + 0.6 = 0
VSG =
8.6 ±
2
(8.6 ) − 4 ( 2.4 )( 0.6 )
2 ( 2.4 )
VSG = 3.512 V
g m = 2 K P (VSG + VTP ) = 2 ( 2 )( 3.512 − 2 )
g m = 6.049 mA / V
2
I D = ( 2 )( 3.512 − 2 ) = 4.572 mA
1
1
⇒ r0 = 21.9 kΩ
=
r0 =
λ I o ( 0.01)( 4.56 )
f
(
CM = CgdT 1 + g m ( ro RD )
a.
)
CM = (1) ⎣⎡1 + ( 6.04 ) ( 21.9 1) ⎦⎤ ⇒ CM = 6.785 pF
b.
rH = ( Ri RG ) ( CgsT + CM )
rH = ( 2 100 ) × 103 (10 + 6.78 ) × 10−12
rH = 3.29 × 10−8 s
1
fH =
→ f H = 4.84 MHz
2πτ H
V0 = − g m ( ro RD ) ⋅ Vgs
⎛ RG ⎞
⎛ 100 ⎞
Vgs = ⎜
⎟ Vi = ⎜
⎟ Vi
⎝ 102 ⎠
⎝ RG + Ri ⎠
⎛ 100 ⎞
Av = − ( 6.04 ) ⎜
⎟ ( 21.9 1)
⎝ 102 ⎠
Av = −5.67
7.58
⎛ R2 ⎞
⎛ 22 ⎞
VG = ⎜
⎟ ( 20 ) − 10 = ⎜
⎟ ( 20 ) − 10
+
R
R
⎝ 22 + 8 ⎠
2 ⎠
⎝ 1
VG = 4.67 V
ID =
10 − VSG − 4.67
2
= K P (VSG + VTP )
RS
2
5.33 − VSG = (1)( 0.5 ) (VSG
− 4VSG + 4 )
2
0.5VSG
− VSG − 3.33 = 0
VSG =
1 ± 1 + 4 ( 0.5 )( 3.33)
2 ( 0.5 )
⇒ VSG = 3.77 V
g m = 2 K p (VSG + VTP ) = 2 (1)( 3.77 − 2 )
g m = 3.54 mA / V
b.
(
CM = CgdT 1 + g m ( RD RL )
)
CM = ( 3) ⎡⎣1 + ( 3.54 ) ( 2 5 ) ⎤⎦ ⇒ CM = 18.2 pF
a.
r = Req ( CgsT + CM )
Req = Ri R1 R2 = 0.5 8 22 = 0.461 kΩ
r = ( 0.461× 103 ) (15 + 18.2 ) × 10−12
= 1.53 × 10−8 s
fH =
1
2π r
⇒ f H = 10.4 MHz
c.
V0 = − g mVgs ( RD RL )
⎛ R R ⎞
⎛ 5.87 ⎞
Vgs = ⎜ 1 2 ⎟ Vi = ⎜
⎟ Vi ⇒ Vgs = ( 0.9215 ) Vi
⎜
⎟
⎝ 5.87 + 0.5 ⎠
⎝ R1 R2 Ri ⎠
Av = − ( 3.54 )( 0.9215 ) ( 2 5 ) ⇒ Av = −4.66
7.59
⎛ 100 ⎞
I E = 0.5 mA ⇒ I CQ = ⎜
⎟ ( 0.5 ) = 0.495 mA
⎝ 101 ⎠
0.495
= 19.0 mA/V
gm =
0.026
(100 )( 0.026 )
= 5.25 kΩ
rπ =
0.495
a.
Input: From Eq. 7.107b
⎡ rπ
⎤
rPπ = ⎢
RE RS ⎥ Cπ
⎣1 + β
⎦
⎡ 5.25
⎤
=⎢
0.5 0.05⎥ × 103 × 10 × 10−12
⎣ 101
⎦
= 2.43 × 10−10 s
1
⇒ f H π = 656 MHz
2π rpπ
Output: From Eq. 7.108b
rPµ = ( RB RL ) Cµ = (100 1) × 103 × 10−12
f Hπ =
= 9.90 × 10−10 s
fH µ =
1
⇒ f H µ = 161 MHz
2π rPµ
b.
RS
gmV␲
V0
⫺
Vi
⫹
⫺
RE
V␲
r␲
⫹
V0 = − g m Vπ ( RB & RL )
g mVπ +
Vπ Vπ Vi − ( −Vπ )
+
+
=0
rπ RE
RS
⎡
1
1
1 ⎤ −V
+ ⎥= i
Vπ ⎢ g m + +
rπ RE RS ⎦ RS
⎣
1
1
1 ⎤ −Vi
⎡
+
+
=
Vπ ⎢19 +
5.25 0.5 0.05 ⎥⎦ 0.05
⎣
Vπ ( 41.19 ) = −Vi ( 20 )
Vπ = − ( 0.4856 ) Vi
V0
= − (19 )( −0.4856 )(100 & 1)
Vi
Av = 9.14
c.
r = CL ( RL & RB ) = (15 × 10−12 ) (1 & 100 ) × 103
r = 1.485 × 10−8 s
1
→ f = 10.7 MHz
2π r
Since f < f H µ ⇒ 3d B freq. dominated by CL .
f =
7.60
RB
RL
20 − 0.7
= 1.93 mA
10
⎛ 100 ⎞
I CQ = ⎜
⎟ (1.93) = 1.91 mA
⎝ 101 ⎠
1.91
gm =
= 73.5 mA/V
0.026
(100 )( 0.026 )
rπ =
= 1.36 kΩ
1.91
a.
Input:
⎡ rπ
⎤
rPπ = ⎢
RE RS ⎥ ⋅ Cπ
⎣1 + β
⎦
I EQ =
rPπ
⎡1.36
⎤
10 1⎥ × 103 × 10 × 10−12
=⎢
⎣ 101
⎦
= 1.327 × 10−10 s
1
⇒ f Pπ = 1.20 GHz
2π rPπ
Output:
rPπ = ( RC RL ) Cµ = ( 6.5 & 5 ) × 103 × 10−12
f Pπ =
rPπ = 2.826 × 10−9 s
1
f Pµ =
→ f Pµ = 56.3 MHz
2π rPπ
b.
V0 = − g m Vπ ( RC & RL )
g mVπ +
Vπ Vπ Vi − ( −Vπ )
+
+
=0
rπ RE
RS
⎛
Vi
1
1
1 ⎞
+
Vπ ⎜ g m + +
⎟=−
r
R
R
R
π
E
S ⎠
S
⎝
1
1 1 ⎤ −V
⎡
+ + ⎥= i
Vπ ⎢ 73.5 +
1.36
10
1 ⎦ (1)
⎣
Vπ ( 75.34 ) = −Vi ⇒ Vπ = − ( 0.01327 ) Vi
V0 = − ( 73.5 )( −0.01327 ) ( 6.5 5 ) Vi
Av = 2.76
c.
r = CL ( RL Rc ) = (15 × 10−12 ) ( 6.5 5 ) × 103
r = 4.24 × 10−8 s
1
→ f = 3.75 MHz
2π r
Since f < fp µ , 3dB frequency is dominated by CL.
f =
7.61
VGS + I D RS = 5
5 − VGS
2
ID =
= K n (VGS − VTN )
RS
5 − VGS = ( 3)(10 ) (V 2GS − 2VGS + 1)
30V 2GS − 59VGS + 25 = 0
VGS =
59 ±
( 59 )
2
− 4 ( 30 )( 25 )
2 ( 30 )
⇒ VGS = 1.349 V
g m = 2 K n (VGS − VTN ) = 2 ( 3)(1.35 − 1)
g m = 2.093 mA / V
On the output:
rPµ = ( RD RL ) Cgd T = ( 5 4 ) × 103 × 4 × 10−12
rPµ = 8.89 × 10−9 s
f Pµ =
1
→ f Pµ = 17.9 MHz
2π rPµ
Ri
gmVgs
V0
⫺
Vi
⫹
⫺
RS
Vgs
RD
⫹
V0 = − g mVgs ( RD RL )
g mVgs +
Vgs
RS
+
Vi − ( −Vgs )
RS
=0
⎛
V
1
1⎞
+ ⎟=− i
Vgs ⎜ g m +
RS Ri ⎠
Ri
⎝
V
1 1⎞
⎛
Vgs ⎜ 2.093 + + ⎟ = − i
10
2
2
⎝
⎠
Vgs = ( 0.1857 )Vi
Av =
V0
= ( 2.093)( 0.1857 ) ( 5 4 )
Vi
Av = 0.864
7.62
dc analysis
ID =
V + − VSG
2
= K P (VSG + VTP )
RS
5 − VSG = (1)( 4 )(VSG − 0.8 )
2
2
= 4 (VSG
− 1.6VSG + 0.64 )
2
4VSG
− 5.4VSG − 2.44 = 0
VSG =
5.4 ±
( 5.4 )
2
+ 4 ( 4 )( 2.44 )
2 ( 4)
= 1.707
g m = 2 K P (VSG + VTP ) = 2 (1)(1.707 − 0.8 )
g m = 1.81 mA / V
RL
Ri
gmVgs
⫺
RS
Vgs
CgsT
CgdT
RD
⫹
3 ⋅ dB frequency due to CgsT : Req =
1
RS Ri
gm
1
2π Req ⋅ CgsT
fA =
1
4 0.5 = 0.246 kΩ
1.81
1
= 162 MHz
fA =
2π ( 246 ) ( 4 × 10−12 )
Req =
3 − dB frequency due to CgdT
1
2π ( RD RL ) CgdT
fB =
=
1
2π ( 2 4 ) × 103 × 10−12
f = 119 MHz
Midband gain
Ri
gmVgs
V0
⫺
Vi
⫹
⫺
Vgs
RS
RD
⫹
1
RS
gm
1
4
1.81
Vgs =
⋅ Vi =
⋅ Vi
1
1
RS + R i
4 + 0.5
gm
1.81
−
−
= −0.492Vi
V0 = − g mVgs ( RD RL )
Av = ( 0.492 )(1.81) ( 4 2 ) ⇒ Av = 1.19
7.63
rπ =
(120 )( 0.026 )
1.02
g m = 39.23 mA/V
a.
= 3.059 kΩ
RL
RL
Input: f H π =
1
2π rπ
rπ = ⎡ Rs R2 R3 rπ ⎤ ( Cπ + 2Cµ )
⎣
⎦
Req = 0.1 20.5 28.3 3.06 = 0.096 kΩ
rπ = ( 96 ) (12 + 2 ( 2 ) ) × 10−12 = 1.537 × 10−9 s
f Hπ =
1
2π (1.536 × 10−9 )
Output: f H µ =
= 103.6 MHz
1
2π rµ
rµ = ( RC RL ) Cµ
= (15 10 ) × 103 × 2 × 10−12
= 6.67 × 10−9
fH µ =
1
2π ( 6.67 × 10−9 )
= 23.9 MHz
b.
⎡ R2 & R3 & rπ ⎤
A = g m ( RC & RL ) ⎢
⎥
⎣ R2 & R3 & rπ + RS ⎦
R2 R3 rπ = 20.5 28.3 3.059 = 2.433 kΩ
⎡ 2.433 ⎤
A = ( 39.23) ( 5 10 ) ⎢
⎥ ⇒ A = 125.6
⎣ 2.433 + 0.1 ⎦
c.
CL = 15 pF > Cµ ⇒ CL dominates frequency response.
Chapter 8
Exercise Solutions
EX8.1
VCC = 30 V, VCE = 30 − I C RC , I CVCE = 10
1
Maximum power at VCE = VCC = 15
2
10 10 2
IC =
=
=
VCE 15 3
a.
2
So 15 = 30 − RL ⇒ RL = 22.5 Ω ⇒ Maximum Power = 10 W
3
b.
VCC = 15 V, I C ,max = 2 A
VCE = 15 − I C RL
0 = 15 − 2 RL ⇒ RL = 7.5 Ω ⇒ Maximum Power = (1)( 7.5 ) = 7.5 W
EX8.2
(a)
(b)
ΔT = Pθ = ( 8 )( 2.4 )
ΔT = 19.2° C
ΔT = Pθ
ΔT 85
=
P=
θ
3.7
P = 23.0 W
EX8.3
Power = iD ⋅ vDS = (1)(12 ) = 12 watts
c.
Tsink = Tamb + P ⋅θ snk − amb
b.
Tcase = Tsink + P ⋅θ case − snk
Tcase = 73 + (12 )(1) ⇒ Tcase = 85°C
a.
Tdev = Tcase + P ⋅θ dev − case
EX8.4
Tsink = 25 + (12 )( 4 ) ⇒ Tsink = 73°C
Tdev = 85 + (12 )( 3) ⇒ Tdev = 121°C
θ dev − case =
PD ,max =
TJ ,max − Tamb
PD,rated
=
200 − 25
= 3.5°C/W
50
TJ ,max − Tamb
θ dev − case + θ case −snk + θ snk − amb
200 − 25
⇒ PD ,max = 29.2 W
3.5 + 0.5 + 2
= Tamb + PD ,max (θ case −snk + θ snk − amb )
= 25 + ( 29.2 )( 0.5 + 2 ) ⇒ Tcase = 98°C
=
Tcase
EX8.5
I DQ =
a.
10 − 4
⇒ I DQ = 60 mA
0.1
b.
⎛9⎞
vds = − ⎜ ⎟ ( 60 )( 0.050 ) = −2.7 V ⇒ vDS ( min ) = 4 − 2.7 = 1.3 V
⎝ 10 ⎠
So maximum swing is determined by drain-to-source voltage.
VPP = 2 × ( 2.5 ) = 5.0 V
c.
1 VP2 1 ( 2.5 )
⋅
= ⋅
⇒ PL = 31.25 mW
2 RL 2 0.1
2
PL =
PS = VDD ⋅ I DQ = (10 )( 60 ) = 600 mW
η=
PL
PS
=
31.25
⇒ η = 5.2%
600
EX8.6 Computer Analysis
EX8.7 No Exercise Problem
EX8.8
a.
vI = v0 + vGSn −
VBB
2
dv
dvI
= 1 + GSn
dv0
dv0
iDn = K n ( vGSn − VTN )
vGSn =
iDn
+ VTN
Kn
dvGSn dvGSn diDn
=
⋅
dvo
diDn dvo
2
So
dvGSn
=
diDn
1
1 1
⋅ ⋅
K n 2 iDn
At v0 = 0, iDn = 0.050 A
So
iDn
dvGSn
1 1
1
=
⋅ ⋅
=5
2
diDn
0.2
0.050
= iL + iDp
For a small change in v0 → Δ iL = Δ iDn − ( −Δ iDp )
1
Δ iL
2
di
1 di
1 1 1 1
= ⋅
= 0.025
or Dn = ⋅ L = ⋅
dv0 2 dv0 2 RL 2 20
So Δ iDn =
Then
Then
dvGSn
= ( 5 )( 0.025 ) = 0.125
dv0
dv
dvI
1
= 1 + 0.125 = 1.125 and Av = 0 =
⇒ Av = 0.889
dv0
dvI 1.125
For v0 = 5 V, iL = 0.25 A = iDn , and iDp = 0
b.
dvGSn dvGSn diDn
=
⋅
dv0
diDn dv0
dvGSn
=
diDn
1
1 1
1 1
1
⋅ ⋅
=
⋅ ⋅
= 2.24
0.2 2 0.25
K n 2 iDn
diDn diL
1
=
=
= 0.05
dv0 dv0 20
dvGSn
= ( 2.24 )( 0.05 ) = 0.112
dv0
dvI
= 1 + 0.112 = 1.112
dv0
Av =
dv0
1
=
⇒ Av = 0.899
dvI 1.112
EX8.9
a.
2
Rb = rπ + (1 + β ) RE′ and RE′ = a 2 RL = (10 ) ( 8 ) = 800 Ω
Ri = 1.5 kΩ = RTH Rb
V
18
= 22.5 mA
I Q = 2CC =
a RL (10 )2 ( 8 )
rπ =
(100 )( 0.026 )
= 0.116 kΩ
22.5
Rb = 0.116 + (101)( 0.8 ) = 80.9 kΩ
1.5 = RTH 80.9 =
RTH ( 80.9 )
RTH + ( 80.9 )
⇒ ( 80.9 − 1.5 ) RTH = (1.5 )( 80.9 ) ⇒ RTH = 1.53 kΩ
⎛ R2 ⎞
1
VTH = ⎜
⎟ VCC = ⋅ RTH ⋅ VCC
R
R
R
+
2 ⎠
1
⎝ 1
I BQ
IQ
22.5
= 0.225 mA
β 100
V − 0.7
1
= TH
⇒ (1.53)(18 ) = ( 0.225 )(1.53) + 0.7 ⇒ R1 = 26.4 kΩ
RTH
R1
I BQ =
=
26.4 R2
= 1.53
26.4 + R2
( 26.4 − 1.53) R2 = (1.53)( 26.4 ) ⇒ R2 = 1.62 kΩ
b.
vE = 0.9VCC = ( 0.9 )(18 ) = 16.2 V
iE = 0.9 I CQ = ( 0.9 )( 22.5 ) = 20.25 mA
v
16.2
v0 = E =
⇒ VP = 1.62 V
10
a
i0 = aiE = (10 )( 20.25 ) ⇒ I P = 203 mA
PL =
1
(1.62 )( 0.203) ⇒ PL = 0.164 W
2
EX8.10
a.
⎛ IC ⎞
⎞
⎟⎟
⎟ ⇒ VBE = VT ln ⎜⎜
⎠
⎝ I SQ ⎠
⎛ 5 × 10−3 ⎞
= ( 0.026 ) ln ⎜
= 0.6225 V ⇒ VD1 = VD 2 = 0.6225
−13 ⎟
⎝ 2 × 10 ⎠
⎛V
I C = I SQ exp ⎜ BE
⎝ VT
VBE
⎛ 0.6225 ⎞
I Bias = I D = I SD exp ⎜
⎟
⎝ 0.026 ⎠
⎛ 0.6225 ⎞
= 5 × 10−13 exp ⎜
⎟
⎝ 0.026 ⎠
I Bias = 12.5 mA
b.
V0 = 2 V, iL =
2
= 26.7 mA
0.075
1st approximation:
iCn ≅ 26.7 mA, iBn = 0.444 mA
⎛ 26.7 × 10−3 ⎞
VBE = ( 0.026 ) ln ⎜
= 0.6661
−13 ⎟
⎝ 2 × 10
⎠
I D = 12.5 − 0.444 = 12.056 mA
⎛ 12.056 × 10−3 ⎞
VD = ( 0.026 ) ln ⎜
⎟ = 0.6216
−13
⎝ 5 × 10
⎠
2VD = 1.243 V
VEB = 2VD − VBE = 0.5769
⎛ 0.5769 ⎞
icP = 2 × 10−13 exp ⎜
⎟ = 0.866 mA
⎝ 0.026 ⎠
2nd approximation:
iEn = iL + iCP = 26.7 + 0.866 ≅ 27.6 mA = iEn
⎛ 60 ⎞
iCn = ⎜ ⎟ ( 27.6 ) ⇒ iCn = 27.1 mA
⎝ 61 ⎠
iBn = 0.452 mA
I D = 12.5 − 0.452 ⇒ I D = 12.05 mA
⎛ 27.1× 10−3 ⎞
⇒ VBEn = 0.6664 V
VBEn = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 2 × 10
⎠
⎛ 12.05 × 10−3 ⎞
VD = ( 0.026 ) ln ⎜
⎟ = 0.6215 V
−13
⎝ 5 × 10
⎠
2VDD = 1.243 V
VEB = 1.243 − 0.6664 ⇒ VEBp = 0.5766 V
⎛ 0.5766 ⎞
iCP = 2 × 10−13 exp ⎜
⎟ ⇒ iCP = 0.856 mA
⎝ 0.026 ⎠
c.
10
= 133 mA
0.075
≅ iL = 133 mA ⇒ iCN = 131 mA
V0 = 10 V, iL =
iEn
iBn = 2.18 mA ⇒ I D = 12.5 − 2.18 ⇒ I D = 10.3 mA
⎛ 10.3 × 10−3 ⎞
= 0.6175
VD = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10
⎠
2VDD = 1.235 V
⎛ 131× 10−3 ⎞
⇒ VBEn = 0.7074 V
VBEn = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 2 × 10 ⎠
VEBp = 1.235 − 0.7074 ⇒ VEBp = 0.5276 V
⎛ 0.5276 ⎞
iCP = 2 × 10−13 exp ⎜
⎟ ⇒ iCP = 0.130 mA
⎝ 0.026 ⎠
EX8.11 No Exercise Problem
EX8.12
a.
vI = 0 = v0 , vB 3 = 0.7 V
12 − 0.7 11.3
=
⇒ I R1 = 45.2 mA
I R1 =
0.25
R1
If transistors are matched, then
iE1 = iE 3
i
iR1 = iE1 + iB 3 = iE1 + E 3
1+ β
⎛
1 ⎞
1⎞
⎛
iR1 = iE1 ⎜ 1 +
⎟ = iE1 ⎜ 1 + ⎟
1
41
+
β
⎝
⎠
⎝
⎠
45.2
iE1 =
⇒ iE1 = iE 2 = 44.1 mA
1.024
i
44.1
iB1 = iB 2 = E1 =
⇒ iB1 = iB 2 = 1.08 mA
1+ β
41
b.
For vI = 5 V ⇒ v0 = 5 V
5
⇒ i0 = 0.625 A
8
0.625
iE 3 ≅ 0.625 A, iB 3 =
⇒ iB 3 = 15.2 mA
41
12 − 5.7
= 25.2 mA
vB 3 = 5.7 V ⇒ iR1 =
0.25
10
= 0.244 mA
iE1 = 25.2 − 15.2 ⇒ iE1 = 10.0 mA ⇒ iB1 =
41
vB 4 = 5 − 0.7 = 4.3 V
i0 =
I R2 =
4.3 − ( −12 )
0.25
= 65.2 mA ≅ iE 2
65.2
= 1.59 mA
41
iI = iB 2 − iB1 = 1.59 − 0.244 ⇒ iI = 1.35 mA
iB 2 =
i0 625
=
⇒ AI = 463
iI 1.35
From Equation (8.54)
(1 + β ) R ( 41)( 250 )
AI =
=
= 641
2 RL
2 (8)
AI =
c.
TYU8.1
24
= 1.2 A = I D ( max )
20
For I D = 0 ⇒ VDS ( max ) = 24 V
For VDS = 0, I D ( max ) =
Maximum power when
VDS =
ID =
VDS ( max )
2
I D ( max )
2
= 12 V and
= 0.6 A ⇒ PD ( max ) = (12 )( 0.6 ) = 7.2 Watts
TYU8.2
Maximum power at center of load line
Pmax = ( 0.05 )(10 ) ⇒ Pmax = 0.5 W
TYU8.3
a.
PQ = VCEQ ⋅ I CQ = ( 7.5 )( 7.5 )
PQ = 56.3 mW
1 VP2 1 ( 6.5 )
⋅
= ⋅
⇒ PL = 21.1 mW
2 RL 2
1
2
b.
PL =
PS = (15 )( 7.5 ) ⇒ PS = 113 mW
η=
PL
PS
=
21.1
⇒ η = 18.7%
113
PQ = 56.3 − 21.1 = 35.2 mW
TYU8.4
1 VP2
20
⋅
⇒ VP = 2 RL PL = 2 ( 8 )( 25 ) ⇒ VP = 20 V ⇒ VCC =
⇒ VCC = 25 V
2 RL
0.8
a.
PL =
b.
IP =
VP 20
=
⇒ I P = 2.5 A
8
RL
c.
PQ =
VCCVP VP2
−
π RL 4 RL
2
( 25 )( 20 ) ( 20 )
−
PQ =
4 (8)
π (8)
d.
η=
= 19.9 − 12.5 ⇒ PQ = 7.4 W
π VP π 20
= ⋅ ⇒ η = 62.8%
4VCC 4 25
TYU8.5
( 4)
1 VP2
⋅
=
⇒ PL = 80 mW
2 RL 2 ( 0.1)
2
a.
PL =
b.
IP =
VP
4
=
⇒ I P = 40 mA
RL 0.1
c.
PQ =
VCCVP VP2
−
π RL 4 RL
PQ =
( 5 )( 4 ) ( 4 )
−
= 63.7 − 40 ⇒ PQ = 23.7 mW
π ( 0.1) 4 ( 0.1)
2
d.
η=
π VP
4VCC
TYU8.6
a.
1 ⎛ 2V
I CQ ≅ ⋅ ⎜ CC
2 ⎝ RL
RTH = R1 R2
=
⋅ ⇒ η = 62.8%
4 5
π 4
⎞ VCC 12
=
= 8 mA
⎟=
⎠ RL 1.5
⎛ R2 ⎞
1
VTH = ⎜
⎟ VCC = ⋅ RTH ⋅ VCC
R1
⎝ R1 + R2 ⎠
I CQ
VTH − VBE
8
=
= 0.107 mA =
I BQ =
β
75
RTH + (1 + β ) RE
Let RTH = (1 + β ) RE = ( 76 )( 0.1) = 7.6 kΩ
1
⋅ ( 7.6 )(12 ) − 0.7
R1
0.107 =
7.6 + 7.6
1
⋅ ( 91.2 ) = 2.33 ⇒ R1 = 39.1 kΩ
R1
39.1R2
= 7.6 ⇒ ( 39.1 − 7.6 ) R2 = ( 7.6 )( 39.1) ⇒ R2 = 9.43 kΩ
39.1 + R2
b.
2
2
1
1
PL = ⋅ ( 0.9 I CQ ) RL = ⎡⎣( 0.9 )( 8 ) ⎤⎦ (1.5 ) ⇒ PL = 38.9 mW
2
2
PS = VCC I CQ = (12 )( 8 ) = 96 mW
PQ = PS − PL = 96 − 38.9 ⇒ PQ = 57.1 mW
η=
PL
PS
=
38.9
⇒ η = 40.5%
96
TYU8.7
I E = I E 3 + IC 4 + IC 5
= I E 3 + IC 4 + β5 I B5
= I E 3 + β 4 I B 4 + β 5 (1 + β 4 ) I B 4
I E = (1 + β 3 ) I B 3 + β 4 β 3 I B 3 + β 5 (1 + β 4 ) β 3 I B 3
If β 4 and β 5 are large, then I E ≅ β 3 β 4 β 5 I B 3
So that composite current gain is β ≅ β 3 β 4 β 5
Chapter 8
Problem Solutions
8.1
a.
b.
i.
VD D = 80 V
Maximum power at VDS =
ID =
RD =
ii.
RD =
8.2
a.
2
= 40 V
PT
25
=
= 0.625 A
VD S 40
80 − 40
⇒ RD = 64 Ω
0.625
VD D = 50 V
Maximum power at VD S =
ID =
VD D
PT
25
=
=1 A
VDS 25
VD D
50 − 25
⇒ RD = 25 Ω
1
2
= 25 V
PQ (max) = I C Q ⋅
So I C Q =
RL =
VC C
2
2 PQ (max)
VCC
=
2(20)
= 1.67 A
24
VCC − (VCC / 2) 24 − 12
=
⇒ RL = 7.2 Ω
1.67
I CQ
1.67
⇒ 20.8 mA
80
24 − 0.7
RB =
⇒ RB = 1.12 kΩ
20.8
b.
I CQ ⋅ RL (1.67 )( 7.2 )
=
= 462
Av = g m RL =
0.026
VT
IB =
I CQ
β
=
V0 (max) = 12 V ⇒ VP =
V0 (max) 12
=
⇒ VP ≅ 26 mV
Av
462
8.3
a.
For maximum power delivered to the load, set VC EQ =
VC C
2
Set VC C = 25 V = VCE ( sus )
Then I Cm =
VCC 25
=
RL 0.1
I Cm = 250 mA < I C ,max
25 − 12.5
= 125 mA
0.1
V
PQ ( max ) = I CQ ⋅ CC = ( 0.125 )(12.5 )
2
= 1.56 W < PD,max
I CQ =
125
= 1.25 mA
100
25 − 0.7
RB =
⇒ RB = 19.4 kΩ
1.25
1 2
1
2
b.
PL ( max ) = ⋅ I CQ
⋅ RL = ( 0.125 ) (100 ) ⇒ PL ( max ) = 0.781 W ( rms )
2
2
I BQ =
8.4
Point (b): Maximum power delivered to load.
Point (a): Will obtain maximum signal current output.
Point (c): Will obtain maximum signal voltage output.
8.5
a.
b.
2
VGG = 5 V, I D = 0.25 ( 5 − 4 ) = 0.25 A, VD S = 37.5 V, P = 9.375 W
2
VGG = 6 V, I D = 0.25 ( 6 − 4 ) = 1.0 A, VD S = 30 V, P = 30 W
2
VGG = 7 V, I D = 0.25 ( 7 − 4 ) = 2.25 A, VD S = 17.5 V, P = 39.375 W
VGG = 8 V, I D
= 0.25 ⎡⎣ 2 ( 8 − 4 )VD S − VD2 S ⎤⎦
=
40 − VD S
10
I D = 3.71 A, P = 10.8 W
VGG = 9 V, I D
⇒ VD S = 2.92
= 0.25 ⎡⎣ 2 ( 9 − 4 ) VD S − VD2S ⎤⎦
⇒ VD S = 1.88 V
10
I D = 3.81 A, P = 7.16 W
c.
Yes, at VGG = 7 V, P = 39.375 W > PD ,max = 35 W
=
40 − VD S
8.6
a.
VDD
= 25 V
2
50 − 25
=
= 1.25 A
20
Set VDSQ =
I DQ
I DQ = K n (VGS − VTN )
2
1.25
+ 4 = VGS = 6.5 V
0.2
⎛ R2 ⎞
VGS = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
Let R1 + R2 = 100 kΩ
⎛ R ⎞
6.5 = ⎜ 2 ⎟ ( 50 ) ⇒ R2 = 13 kΩ
⎝ 100 ⎠
R1 = 87 kΩ
b.
PD = I DQVDSQ = (1.25 )( 25 ) ⇒ PD = 31.25 W
c.
I D ,max = 2 I DQ ⇒ I D ,max = 2.5 A
VDS ,max = VDD ⇒ VDS ,max = 50 V
PD ,max = 31.25 W
d.
V0
= g m RL
Vi
g m = 2 K n I DQ = 2
( 0.2 )(1.25) = 1 A / V
V0 = (1)( 20 )( 0.5 ) = 10 V
1 V02 1 (10 )
⋅
= ⋅
⇒ PL = 2.5 W
2 RL 2 20
2
PL =
PQ = 31.25 − 2.5 ⇒ PQ = 28.75 W
8.7
(a)
PD = PD ,max − ( Slope ) (T j − 25 )
(b)
60
+ 25 ⇒ T j ,max = 145°C
0.5
T j ,max − Tcase
145 − 25
⇒ θ dev − amb = 2°C/W
=
or θ dev − amb =
60
θ dev − amb
At PD = 0, T j ,max =
(c)
PD ,max
8.8
PD ,rated =
T j ,max − Tamb
θ dev − case
or θ dev − case =
T j ,max − Tamb
PD ,rated
150 − 25
= 2.5°C/W
50
− Tamb = PD (θ dev − case + θ case − amb )
=
Then Tdev
150 − 25 = PD ( 2.5 + θ case − amb ) ⇒ 125 = PD ( 2.5 + θ case − amb )
8.9
PD = I D ⋅ VDS = ( 4 )( 5 ) = 20 W
Tdev − Tamb = PD (θ dev − case + θ case − snk + θ snk − amb )
Tdev − 25 = 20 (1.75 + 0.8 + 3) = 111 ⇒ Tdev = 136°C
Tdev − Tcase = PD ⋅θ dev − case = ( 20 )(1.75 ) = 35
Tcase = Tdev − 35 = 136 − 35 ⇒ Tcase = 101°C
Tcase − Tsink = PD ⋅θ case − snk = ( 20 )( 0.8 ) = 16°C
Tsink = Tcase − 16 = 101 − 16 ⇒ Tsink = 85°C
8.10
Tdev − Tamb = PD (θ dev − case + θ case − amb )
200 − 25 = 25 ( 3 + θ case − amb ) ⇒ θ case − amb = 4°C/W
8.11
θ dev − case =
PD =
=
T j ,max − Tamb
PD ,rated
=
175 − 25
= 10°C/W
15
T j ,max − Tamb
θ dev − case + θ case −snk + θ snk − amb
175 − 25
⇒ PD = 10 W
10 + 1 + 4
8.12
η=
PL
PS
PS = VCC ⋅ I Q
⎛V ⎞
PL = VP ⋅ I P = ⎜ CC ⎟ ( I Q )
⎝ 2 ⎠
1
⋅ VCC ⋅ I Q
⇒ η = 50%
η= 2
VCC ⋅ I Q
8.13
vo ( max ) = 4.8 V
iC 3 = iC 2 =
vI = vo + 0.7
−0.7 − ( −5 )
1
iL ( max ) = −4.3 mA =
so − 3.6 ≤ vI ≤ 5.5 V vo ( min ) = −4.3 V
8.14
= 4.3 mA
vS ( min )
1
2
I D 3 = K (VGS 3 − VTN ) =
0 − VGS 3 − ( −5 )
R
2
12 (VGS 3 − 0.5 ) = 5 − VGS 3
2VGS2 3 − 11VGS − 2 = 0
VGS 3 =
11 ±
(11)
2
+ 4 (12 )( 2 )
2 (12 )
VGS 3 = VGS 2 = 1.072 V
2
I D 3 = I D 2 = 12 (1.072 − 0.5 ) = 3.93 mA
VDS 2 ( sat ) = VGS 2 − VTN = 1.072 − 0.5 = 0.572 V
vo ( min ) : i2 ( max ) = −3.93 =
vI ( min ) = vo ( min ) + VTN
vI ( min ) = −3.43 V
V0 ( min )
1
= −3.93 + 0.5
⇒ V0 ( min ) = −3.93 V
vo ( max ) = 5 − VDS ( sat ) = 5 − 0.572
vo ( max ) = 4.43 V
4.43
= 8.36 mA
I D1 ( max ) = 3.93 +
1
I D1 = 8.36 = 12 (VGS 1 − 0.5 ) ⇒ VGS 1 = 1.33 V
vI ( max ) = vo + VGS1 = 4.43 + 1.33 ⇒ vI ( max ) = 5.76 V
2
8.15
a.
Neglect base currents.
v0 ( max ) = V + − VCE (sat) = 10 − 0.2 = 9.8 V
9.8 9.8
=
⇒ I Q = 9.8 mA
iL (max) = I Q =
RL
1
⇒ R = 949 Ω
9.8
iE1 ( max ) = 2 I Q ⇒ iE1 ( max ) = 19.6 mA
R=
0 − 0.7 − ( −10 )
iE1 ( min ) = 0
iL ( max ) = I Q = 9.8 mA
iL ( min ) = − I Q = −9.8 mA
b.
2
1
1
( iL ( max ) ) RL = 2 ( 9.8)2 (1) ⇒ PL = 48.02 mW
2
PS = I Q (V + − V − ) + I Q ( 0 − V − )
PL =
= 9.8 ( 20 ) + 9.8 (10 ) ⇒ PS = 294 mW
η=
PL
PS
=
48.02
⇒ η = 16.3%
294
8.16
a.
I Q ( min ) =
R=
b.
v0 ( max )
RL
0 − 0.7 − ( −12 )
100
=
10
⇒ I Q ( min ) = 100 mA
0.1
⇒ R = 113 Ω
PQ1 = I Q ⋅ VCE1 = (100 )(12 ) ⇒ PQ1 = 1.2 W
P (source) = 2 I Q (12 ) = 2.4 W
c.
2
PL =
(10 )
1 VP2
⋅
=
= 0.5 W
2 RL 2 (100 )
PS = 1.2 + 2.4 = 3.6 W
η=
PL
PS
=
0.5
⇒ η = 13.9%
3.6
8.17
I D1 = K n (VGS − VTN ) = 12 ( 0 − ( −1.8 ) )
2
2
I D1 = 38.9 mA
(a)
For RL = ∞
vo ( max ) = 4.8 V
VDS ( sat ) = VGS − VTN = 1.8 V
vo ( min ) = −5 + 1.8 = −3.2 V
vI = vo + 0.7 ⇒ −2.5 ≤ vI ≤ 5.5 V
For RL = 500 Ω vo ( max ) = 4.8 V
(b)
For vo < 0, vo ( min ) = −3.2 V
I 2′ =
vo −3.2
=
= −6.4 mA
0.5
RL
−2.5 ≤ vI ≤ 5.5 V
(c)
For vo = −2V , I 2′ ( max ) = −38.9 mA
−2
R2 ( min ) =
⇒ RL ( min ) = 51.4 Ω
−38.9
1 v2 1 ( 2)
PL = ⋅ o = ⋅
⇒ PL = 38.9 mW
2 RL 2 51.4
2
PL = 10 ( 38.9 ) = 389 mW % =
38.9
= 10%
389
8.18
+
V 2 (V )
PL = P =
RL
RL
2
+ 2
− 2
1 (V ) 1 (V )
+ ⋅
, V − = −V +
PS = ⋅
2 RL
2 RL
+ 2
So PS
η=
8.19
(a)
PL
PS
(V )
=
RL
⇒ η = 100%
As maximum conversion efficiency
π V
η = , P = 0.785
4 VCC
⎛4⎞
So V p ( max ) = ( 0.785 )( 5 ) ⎜ ⎟
⎝π ⎠
V p ( max ) = 5 V
(b)
Maximum power dissipation occurs when V p =
(c)
Pθ ( max ) =
2=
2VCC
π
2
VCC
2
π RL
( 5)
2
π 2 RL
⇒ RL = 1.27 Ω
8.20
P=
(a)
2
1 Vp
⋅
2 RL
2
1 Vp
⋅
⇒ V p = 49 V ⇒ V + = 52 V, V − = −52 V
2 24
V
49
IP = P =
= 2.04 A
RL 24
50 =
(b)
η=
(c)
π VP
⋅
=
4 VCC
π ⎛ 49 ⎞
⎜ ⎟
4 ⎝ 52 ⎠
η = 74.0%
8.21
(a)
VDS ≥ VDS ( sat ) = VGS − VTN = VGS
VDS = 10 − Vo ( max ) and I D = I L = K n (VGS )
Vo ( max )
RL
VGS =
= K n (VGS )
2
2
Vo ( max )
RL ⋅ K n
So 10 − Vo ( max ) =
⎡⎣10 − V0 ( max ) ⎤⎦ =
2
Vo ( max )
RL ⋅ K n
=
Vo ( max )
( 5 )( 0.4 )
V0 ( max )
2
100 − 20V0 ( max ) + V02 ( max ) =
V0 ( max )
2
V02 ( max ) − 20.5V0 ( max ) + 100 = 0
V0 ( max ) =
iL =
( 20.5 )
8
⇒ iL = 1.6 mA
5
VGS =
b.
20.5 ±
2
2
− 4 (100 )
⇒ V0 ( max ) = 8 V
iL
1.6
=
= 2 V ⇒ VI = 10 V
Kn
0.4
=
2 ( 5)
π
= 3.183 V
2
1 (8)
= 6.4 mW
PL = ⋅
2 5
20 (1.6 )
= 10.2 mW
PS =
π
η=
PL
PS
=
6.4
⇒ η = 62.7%
10.2
8.22
2
2
vO = iL RL and iL = iD = K n ( vGS − VTN ) or iL = K n ( vGS ) and vGS = vI − vO
Then
2
vO = K n RL ( vI − vO ) or vO = 2 ( vI − vO )
2
⎛ dv ⎞
dv0
= ( 2 )( 2 )( vI − v0 ) ⎜1 − 0 ⎟
dvI
⎝ dvI ⎠
dv0
⎡1 + 4 ( vI − v0 ) ⎤⎦ = 4 ( vI − v0 )
dvI ⎣
or
4 ( vI − v0 )
dv0
=
dvI 1 + 4 ( vI − v0 )
For vI = 10 V, v0 = 8 V ⇒
At vI = 0, v0 = 0 ⇒
dv0
=0
dvI
At vI = 1, v0 = 0.5 ⇒
8.23
a.
4 (10 − 8 )
dv0
dv
=
⇒ 0 = 0.889
dvI 1 + 4 (10 − 8 ) dvI
dv0
= 0.667
dvI
⎛i ⎞
⎛ 5 × 10−3 ⎞
VBE = VT ln ⎜ C ⎟ = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
V
VBE = BB = 0.5987 V ⇒ VBB = 1.1973 V
2
PQ = iC ⋅ vCE = ( 5 )(10 ) ⇒ PQ = 50 mW
b.
v0 = −8 V
iL =
−8
⇒ iL = −80 mA
0.1
iCp ≈ 80 mA
⎛ iCp ⎞
⎛ 80 × 10−3 ⎞
vEB = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
vEB = 0.6708 V
VBB
− vEB + v0 = 0.5987 − 0.6708 − 8⇒ vI = −8.072 V
2
= VBB − vEB = 1.1973 − 0.6708 = 0.5265 V
vI =
VBE
⎛v ⎞
⎛ 0.5265 ⎞
iCn = I S exp ⎜ BE ⎟ = 5 × 10−13 exp ⎜
⎟ ⇒ iCn = 0.311 mA
⎝ 0.026 ⎠
⎝ VT ⎠
PL = iL2 RL = ( 80 ) ( 0.1) ⇒ PL = 640 mW
2
PQn = iCn ⋅ vCE = ( 0.311) (10 − ( −8 ) ) ⇒ PQn = 5.60 mW
PQp = iCp ⋅ vEC = ( 80 )( 2 ) ⇒ PQp = 160 mW
8.24
(a)
iDn = K n ( vGSn − VTN )
2
V
0.5
+ 2 = vGSn = 2.5 V = BB ⇒ VBB = 5.0 V
2
2
Pn = ( 0.5 )(10 ) ⇒ Pn = Pp = 5 mW
(b)
VDS = VGS − VTN ⇒ VDS = VGS − 2 VDS = 10 − vo ( max )
and
v ( max )
v ( max )
iL
+ VTN = O
+2 = O
+2
Kn
RL K n
( 2 )(1)
VGS =
so
10 − v0 ( max ) =
v0 ( max )
2
+2−2 =
v0 ( max )
2
so v0 ( max ) = 8 V
iDn = iL =
8
⇒ iDn = iL = 8 mA
1
8
+ 2 ⇒ VGS = 4 V
2
V
Then vI = vo + VGS − BB = 8 + 4 − 2.5 ⇒ vI = 9.5 V
2
VBB ⎞
⎛
vSGp = vo − ⎜ vI −
⎟ = 8 − ( 9.5 − 2.5 )
2 ⎠
⎝
vSGp = 1 V ⇒ M p cutoff ⇒ iDp = 0
VGS =
PL = iL2 RL = ( 8 ) (1) ⇒ PL = 64 mW
2
PMn = iDn ⋅ vDS = ( 8 )(10 − 8 ) ⇒ PMn = 16 mW
PMp = iDp ⋅ vSD ⇒ PMp = 0
8.25
a.
v0 = 24 V ⇒ iL =
24
⇒ iL ≈ iN = 3 A
8
3
⇒ iBn = 73.2 mA
41
For iD = 25 mA ⇒ iR1 = 25 + 73.2 = 98.2 mA
iBn =
⎛i
VBE = VT ln ⎜ N
⎝ IS
Then 98.2 =
⎞
3
⎛
⎞
⎟ = ( 0.026 ) ln ⎜
−12 ⎟
6
10
×
⎝
⎠
⎠
= 0.7004 V
30 − ( 24 + 0.7 )
R1
⇒ R1 =
5.3
⇒ R1 = 53.97 Ω
98.2
⎛ 25 × 10−3 ⎞
= 0.5759 V
VD = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 6 × 10 ⎠
VEB = 2VD − VBE = 2 ( 0.5759 ) − 0.7004
= 0.4514 V
⎛V ⎞
⎛ 0.4514 ⎞
iP = I S exp ⎜ EB ⎟ = ( 6 × 10−12 ) exp ⎜
⎟ ⇒ iP = 0.208 mA
V
⎝ 0.026 ⎠
⎝ T ⎠
b.
Neglecting base current
30 − 0.6 30 − 0.6
iD ≈
=
⇒ iD ≈ 545 mA
R1
53.97
⎛ 0.545 ⎞
VD = ( 0.026 ) ln ⎜
= 0.656 V
−12 ⎟
⎝ 6 × 10 ⎠
Approximation for iD is okay.
Diodes and transistors matched ⇒ iN = iP = 545 mA
8.26
(a)
I D1 = K1 (VGS 1 − VTN )
VGS1 =
2
5
+1 = 2 V
5
I D 3 = K 3 (VGS 3 − VTN )
2
200 = K 3 ( 2 − 1) ⇒ K n3 = K p 4 = 200 µ A / V 2
2
(b)
vI + VSG 4 + VGS 3 − VGS1 = vO
For vo large, iL = i1 = K n1 (VGS1 − VTN )
VGS1 =
iL
+ VTN =
K n1
⎛
So vI + 2 + 2 − ⎜
⎜
⎝
vo
+ VTN
RL K n1
⎞
vo
+ 1⎟ = v0
( 0.5)( 5) ⎟⎠
v0
−3
2.5
dv 1
dv
dvI
1
=1= 0 + ⋅
⋅ 0
dvI
dvI 2 2.5v0 dvI
vI = v0 +
⎡
⎤
1
⎢1 +
⎥
⎢⎣ 2 2.5v0 ⎥⎦
For vO = 5 V :
1=
dv0
dvI
2
1=
⎤ dv
dv0 ⎡
dv
1
⎢1 +
⎥ = 0 (1.1414 ) ⇒ 0 = 0.876
dvI ⎢ 2 2.5 ( 5 ) ⎥ dvI
dvI
⎣
⎦
8.27
vO = vI +
I Dn
+ VTN
Kn
VBB
− VGS and VGS =
2
For vO ≈ 0, I Dn = I DQ + iL = I DQ +
Then vO = vI +
I DQ + ( vO / RL )
I DQ
v
VBB
V
⋅ 1+ O
− VTN −
or vO = vI + BB − VTN −
2
2
Kn
I DQ RL
Kn
For vO small, vO ≅ vI +
⎡ 1 I DQ
vO ⎢1 + ⋅
⎢⎣ 2 K n
Now
dvO
=
dvI ⎡ 1
⎢1 + ⋅
⎣⎢ 2
So
⋅
I DQ
VBB
1 v
− VTN −
⋅ 1+ ⋅ O
Kn
2
2 I DQ RL
I DQ
V
1 ⎤
⎥ = vI + BB − VTN −
2
I DQ RL ⎥⎦
Kn
1
1 ⎤
⋅
⎥
K n I DQ RL ⎦⎥
= 0.95
I DQ
1 I DQ
1
1
⋅
⋅
=
− 1 = 0.0526
2 K n I DQ RL 0.95
For RL = 0.1 k Ω, then
Or
vO
RL
1
K n I DQ
= 0.01052
K n I DQ = 95.1
We can write g m = 2 K n I DQ = 190 mA/V
This is the required transconductance for the output transistor. This implies a very large transistor.
8.28
Av = − g m RL
So −12 = − g m ( 2 ) ⇒ g m = 6 mA/V=
I CQ = ( 6 )( 0.026 ) ⇒ I CQ = 0.156 mA
I CQ
VT
But for maximum symmetrical swing, set I CQ =
Maximum power to the load:
2
(10 )
1 VCC
⋅
=
⇒ PL ( max ) = 25 mW
2 RL
2 ( 2)
2
PL ( max ) =
PS = VCC ⋅ I CQ = (10 )( 5 ) = 50 mW
So η = 50%
VCC 10
=
= 5 mA ⇒ Av > 12
2
RL
I CQ
5
= 0.0278 mA
β 180
R1 = RTH = 6 kΩ
I BQ =
=
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
Set RE = 20 Ω
VTH = ( 0.0278 )( 6 ) + 0.7 + (181)( 0.0278 )( 0.020 )
VTH = 0.967 V
1
⋅ RTH ⋅ VCC
R1
VTH =
0.967 =
1
( 6 )(10 ) ⇒ R1 = 62.0 kΩ
R1
R2 = 6.64 kΩ
8.29
I CQ =
VCC 15
= = 15 mA
1
RL
I BQ =
15
= 0.15 mA
100
(15)
1 V2
⇒ PL ( max ) = 112.5 mW
PL ( max ) = ⋅ CC =
2 RL
2 (1)
2
Let RTH = 10 kΩ
VTH = I BQ RTH + VBE + (1 + β ) I BQ RE
= ( 0.15 )(10 ) + 0.7 + (101)( 0.15 )( 0.1)
VTH = 3.715 =
1
1
⋅ RTH ⋅ VCC = ⋅ (10 )(15 )
R1
R1
R1 = 40.4 kΩ
R2 = 13.3 kΩ
8.30
⎛ R2 ⎞
⎛ 1.55 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 )
+
R
R
⎝ 1.55 + 0.73 ⎠
⎝ 1
2 ⎠
= 6.80 V
RTH = R1 R2 = 0.73 1.55 = 0.496 kΩ
I BQ =
VTH − VBE
6.80 − 0.70
=
RTH + (1 + β ) RE 0.496 + ( 26 )( 0.02 )
I BQ = 6.0 mA, I CQ = 150 mA
2
Av = − g m RL′ and RL′ = a 2 RL = ( 3) ( 8 ) = 72 Ω
gm =
I CQ
VT
=
150
⇒ 5.77 A/V
0.026
Av = − ( 5.77 )( 72 ) = −415
Vo ′ = Av ⋅ Vi = ( 415 )( 0.017 ) = 7.06 V
7.06
= 2.35 V
3
7.06
Vo =
= 2.35 V
3
7.06
Vo =
= 2.35 V
3
PS = I CQ ⋅ VCC = ( 0.15 )(10 ) = 1.5 W
Vo =
η=
PL
PS
=
0.345
⇒ η = 23%
1.5
8.31
a.
Assuming the maximum power is being delivered, then
36
9
Vo′ ( peak ) = 36 V ⇒ Vo =
= 9 V ⇒ Vrms =
⇒ Vrms = 6.36 V
4
2
36
Vo =
⇒ Vo = 25.5 V
b.
2
c.
Secondary I rms =
Primary I P =
PL
2
=
⇒ I rms = 0.314 A
Vrms 6.36
0.314
⇒ I P = 78.6 mA
4
d.
PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W
2
⇒ η = 37%
η=
5.4
8.32
a.
⎛V
ve = ⎜ π + g mVπ
⎝ rπ
⎞
⎟ RE′ = Vπ
⎠
⎛1
⎞
⎜ + g m ⎟ RE′
r
⎝ π
⎠
⎛1+ β
= Vπ ⎜
⎝ rπ
vi = Vπ + ve ⇒ Vπ = vi − ve
⎞
⎟ RE′
⎠
⎛ 1+ β ⎞
ve = (vi − ve ) ⎜
⎟ RE′
⎝ rπ ⎠
1+ β
⋅ RE′
2
⎛n ⎞
(1 + β ) RE′
ve
rπ
v
=
=
= e where RE′ = ⎜ 1 ⎟ RL
vi 1 + 1 + β ⋅ R ′ rπ + (1 + β ) RE′ vi
⎝ n2 ⎠
E
rπ
⎛n ⎞
ve
so ve − v0 ⎜ 1 ⎟
⎛ n1 ⎞
⎝ n2 ⎠
⎜ ⎟
⎝ n2 ⎠
(1 + β ) RE′
v
1
⋅
so 0 =
vi ⎛ n1 ⎞ rπ + (1 + β ) RE′
⎜ ⎟
⎝ n2 ⎠
v0 =
b.
I
n1
1 2
1
2
RL
⋅ I P RL , a = , I CQ = P so PL = .a 2 I CQ
n2
a
2
2
PS = I CQ .VCC
For η = 50% :
PL =
1 2 2
⋅ a I CQ RL a 2 I R
VCC
VCC
V
PL
CQ L
so a 2 =
= 0.5 = 2
=
=
⇒ a 2 = CC
I CQ ⋅ VCC
I CQ ⋅ RL ( 0.1)( 50 )
2VCC
5
PS
c.
49 ( 0.026 )
r
β VT
R0 = π =
=
⇒ R0 = 0.255 Ω
1 + β (1 + β ) I CQ ( 50 )( 0.1)
8.33
a.
With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.
⎛ R1 R2 ⎞
⎛ R1 R2
i
ie = (1 + β ) ib and ib = ⎜
ii so we need e = 8 = (1 + β ) ⎜
⎟
⎜R R +R ⎟
⎜R R +R
ii
ib
ib ⎠
⎝ 1 2
⎝ 1 2
Rib = rπ + (1 + β ) RL′ ≈ (1 + β ) RL′ = (101)( 0.8 ) = 80.8
⎞
⎟⎟ where
⎠
⎛
⎞
R1 R2
Then 8 = (101) ⎜⎜
⎟⎟
⎝ R1 R2 + 80.8 ⎠
R1 R2
= 0.0792 or R1 R2 = 6.95 kΩ
R1 R2 + 80.8
2VCC
V
12
= RL′ ⇒ I CQ = CC =
= 15 mA
2 I CQ
RL′ 0.8
Set
15
= 0.15 mA
100
= I BQ RTH + VBE
I BQ =
VTH
1
⋅ RTH ⋅ VCC = I BQ RTH + VBE
R1
1
( 6.95)(12 ) = ( 0.15)( 6.95) + 0.7 ⇒ R1 = 47.9 kΩ then R2 = 8.13 kΩ
R1
b.
I
I e = 0.9 I CQ = 13.5 mA = L ⇒ I L = 135 mA
a
1
2
PL = ( 0.135 ) ( 8 ) ⇒ PL = 72.9 mW
2
PS = VCC I CQ = (12 )(15 ) ⇒ PS = 180 mW
η=
PL
PS
⇒ η = 40.5%
8.34
a.
VP = 2 RL PL
VP = 2 ( 8 )( 2 ) = 5.66 V = peak output voltage
IP =
VP 5.66
=
= 0.708 A = peak output current
RL
8
Set Ve = 0.9VCC = aVP to minimize distortion
Then a =
( 0.9 )(18)
5.66
⇒ a = 2.86
b.
1 ⎛ I P ⎞ 1 ⎛ 0.708 ⎞
=
⎜
⎟ ⇒ I CQ = 0.275 A
0.9 ⎜⎝ a ⎟⎠ 0.9 ⎝ 2.86 ⎠
Then PQ = VCC I Q = (18 )( 0.275 ) ⇒ PQ = 4.95 W Power rating of transistor
Now I CQ =
8.35
a.
Need a current gain of 8 through the transistor.
⎛ R1 R2
ib
= 8 = (1 + β ) ⎜
⎜R R +R
ii
ib
⎝ 1 2
⎞
⎟⎟ where Rib ≈ (1 + β )( 0.9 ) = 90.9 kΩ
⎠
⎛
⎞
R1 R2
8
=⎜
⎟ = 0.0792 or R1 R2 = 7.82 kΩ
⎜
101 ⎝ R1 R2 + 90.9 ⎟⎠
2VCC
12
= 0.9 kΩ ⇒ I CQ =
= 13.3 mA
2 I CQ
0.9
Set
13.3
= 0.133 mA
100
1
Then
( 7.82 )(12 ) = ( 0.133)( 7.82 ) + 0.7 ⇒ R1 = 53.9 kΩ and R 2 = 9.15 kΩ
R1
b.
I
I e = ( 0.9 ) I CQ = 12 mA = L ⇒ I L = 120 mA
a
1
2
PL = ( 0.12 ) ( 8 ) ⇒ PL = 57.6 mW
2
PS = VCC I CQ = (12 )(13.3) ⇒ PS = 159.6 mW
I BQ =
η=
PL 57.6
=
⇒ η = 36.1%
PS 159.6
8.36
a.
All transistors are matched.
iC
⎛1+ β ⎞
3 mA = iE1 + iB 3 = ⎜
⎟ iC +
β
⎝ β ⎠
⎛ 61 1 ⎞
3 = ⎜ + ⎟ iC ⇒ iC = 2.90 mA
⎝ 60 60 ⎠
b.
For vo = 6 V , let RL = 200 Ω.
6
= 0.03 A = 30 mA ≅ iE 3
200
30
iB 3 =
= 0.492 mA
61
iE1 = 3 − 0.492 = 2.508 mA
io =
2.508
⇒ iB1 = 41.11 µ A
61
3
iE 2 ≅ 3 mA ⇒ iB 2 =
⇒ 49.18 µ A
61
iI = iB 2 − iB1 = 49.18 − 41.11 ⇒ iI = 8.07 µ A
iB1 =
Current gain
Ai =
VBE 3
30 × 10−3
⇒ Ai = 3.72 × 103
8.07 × 10−6
⎛i ⎞
⎛ 30 × 10−3 ⎞
= VT ln ⎜ E 3 ⎟ = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
VBE 3 = 0.6453 V
⎛i ⎞
⎛ 2.508 × 10−3 ⎞
VEB1 = VT ln ⎜ E1 ⎟ = ( 0.026 ) ln ⎜
⎟
−13
⎝ 5 × 10
⎠
⎝ IS ⎠
VEB1 = 0.5807 V
vI = v0 + VBE 3 − VEB1 = 6 + 0.6453 − 0.5807
vI = 6.0646 V
Voltage gain
Av =
v0
6
=
⇒ Av = 0.989
vI 6.0646
8.37
a.
For i0 = 1 A, I B 3 ≅
1
⇒ 20 mA
50
⎡10 − ( v0,max + VBE 3 )
⎤
10 − VEB1
= 2⎢
− 20 ⎥
R1
R1
⎢⎣
⎦⎥
10 − VBE 2vo,max
If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then
=
+ 40
R1
R1
We can then write
If we assume v0,max = 4 V, then
b.
9.3 2 ( 4 )
=
+ 40 which yields R1 = R2 = 32.5 Ω
R1
R1
9.3
⇒ I E1 = 0.286 A = I E 2
32.5
= 10 I S1,2 , then I E 3 = I E 4 = 2.86 A
For vI = 0, I E1 =
Since I S 3,4
c.
We can write
⎧
rπ 1 ⎫
⎪ rπ 3 + R1
⎪
1 + β1 ⎪
1⎪
R0 = ⎨
⎬
2⎪
1 + β3
⎪
⎪
⎪
⎩
⎭
( 50 )( 0.026 )
βV
= 0.4545 Ω
Now rπ 3 = 3 T =
2.86
IC 3
rπ 1 =
β1VT
I C1
=
(120 )( 0.026 )
0.286
= 10.91 Ω
So
⎧
10.91 ⎫
0.4545 + 32.5
1 ⎪⎪
121 ⎪⎪
R0 = ⎨
⎬
2⎪
51
⎪
⎪⎭
⎩⎪
10.91
= 32.5 0.0902 = 0.0900
32.5
121
Then R0 =
8.38
1 ⎧ 0.4545 + 0.0900 ⎫
⎨
⎬ or R 0 = 0.00534 Ω
2⎩
51
⎭
1
rπ 1 + (1 + β ) ⎡ R1 ( rπ 3 + (1 + β ) 2 RL ) ⎤
⎣
⎦
2
iC1 ≈ 7.2 mA and iC 3 ≈ 7.2 mA
Ri =
{
}
Then rπ =
( 60 )( 0.026 )
7.2
= 0.217 kΩ
1
0.217 + ( 61) ⎡ 2 ( 0.217 + ( 61)( 0.2 ) ) ⎤
⎣
⎦
2
1
= 0.217 + 61 ⎡⎣ 2 12.4 ⎤⎦ or Ri = 52.6 kΩ
2
So Ri =
{
}
{
}
8.39
a.
b.
2
I1 = K1 (VSG + VTP ) =
V + − VSG
R1
5 = 10 (VSG − 2 ) ⇒ VSG = 2.707 V
2
10 − 2.707
⇒ R1 = R2 = 1.46 kΩ
R1
c.
RL = 100 Ω For a sinusoidal output signal:
5=
1 (v )
1 ( 5)
⇒ PL = 125 mW
PL = ⋅ o = ⋅
2 RL
2 0.1
2
iD 3 ≈
2
( vo ) ( 5 )
RL
=
0.1
⇒ iD 3 = 50 mA
50
+ 2 = 4.236 V
10
10 − ( 4.236 + 5 )
I1 =
⇒ I D1 = 0.523 mA
1.46
0.523
VSG1 =
+ 2 = 2.229 V
10
vI = 5 + 4.236 − 2.229 ⇒ vI = 7.007 V
VGS 3 =
I D2 =
(VI − VGS ) − ( −10 )
= 10 (VGS − 2 )
2
1.46
17.007 − VGS
= 10 (VGS2 − 4VGS + 4 )
1.46
14.6VGS2 − 57.4VGS + 41.4 = 0
VGS =
57.4 ±
2
( 57.4 ) − 4 (14.6 )( 41.4 )
2 (14.6 )
VGS 2 = 2.98 V
I D 2 = 10 ( 2.98 − 2 ) ⇒ I D 2 = 9.60 mA
2
VG 4 = vI − VGS 2 = 7 − 2.98 = 4.02 V
VSG 4 = 5 − 4.02 = 0.98 V ⇒ I D 4 = 0
8.40
For v0 = 0
I Q = I C 3 + I C 2 + I E1
⎛ 1+ βn ⎞
IC 3
I B3 = I E 2 = ⎜
⎟ IC 2 =
β
βn
⎝ n ⎠
I C 3 = (1 + β n ) I C 2
⎛ β ⎞
I
I B 2 = I C 1 = ⎜ P ⎟ I E1 = C 2
βn
⎝ 1+ βP ⎠
⎛ β ⎞
I C 2 = β n ⎜ P ⎟ I E1
⎝ 1+ βP ⎠
⎛ β ⎞
I C 3 = (1 + β n ) β n ⎜ P ⎟ I E1
⎜1+ β ⎟
p ⎠
⎝
⎛ β ⎞
⎛ β ⎞
I Q = (1 + β n ) β n ⎜ P ⎟ I E1 + β n ⎜ P ⎟ I E1 + I E1
⎝1+ βP ⎠
⎝ 1+ βP ⎠
⎛ 10 ⎞
⎛ 10 ⎞
= ( 51)( 50 ) ⎜ ⎟ I E1 + ( 50 ) ⎜ ⎟ I E1 + I E1
⎝ 11 ⎠
⎝ 11 ⎠
I Q = 2318.18I E1 + 45.45 I E1 + I E1
I E1 = 1.692 µ A ⇒ I C1 = 1.534 µ A
⎛ 10 ⎞
I C 2 = ( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 2 = 76.9 µ A
⎝ 11 ⎠
⎛ 10 ⎞
I C 3 = ( 51)( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 3 = 3.92 mA
⎝ 11 ⎠
Because of rπ 1 and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find
I X = g m 3Vπ 3 + g m 2Vπ 2 + g m1Vπ 1 +
VX
rπ 1 + Z
Now
⎛ r
⎞
Vπ 1 = ⎜ π 1 ⎟ VX , Vπ 2 ≅ g m1Vπ 1rπ 2
⎝ rπ 1 + Z ⎠
and
Vπ 3 = ( g m1Vπ 1 + g m 2Vπ 2 ) rπ 3
= ⎡⎣ g m1Vπ 1 + g m 2 ( g m1Vπ 1rπ 2 ) ⎤⎦ rπ 3
⎛ r
⎞
Vπ 3 = ⎜ π 1 ⎟ [ g m1 + g m1 g m 2 rπ 2 ] rπ 3 ⋅ VX
⎝ rπ 1 + Z ⎠
( β + β1 β 2 ) rπ 3
⋅ VX
Vπ 3 = 1
rπ 1 + Z
⎛ r
⎞
⎛ βr ⎞
and Vπ 2 = g m1 ⎜ π 1 ⎟ rπ 2VX = ⎜ 1 π 2 ⎟ VX
⎝ rπ 1 + Z ⎠
⎝ rπ 1 + Z ⎠
( β + β1 β 2 ) β 3
ββ
β1
VX
Then I X = 1
⋅ V X + 1 2 ⋅ VX +
⋅ VX +
rπ 1 + Z
rπ 1 + Z
rπ 1 + Z
rπ 1 + Z
Then
R0 =
rπ 1 =
rπ 1 + Z
VX
=
I X 1 + β1 + β1 β 2 + ( β1 + β1 β 2 ) β 3
(10 )( 0.026 )
1.534
Z = 25 kΩ
Then
= 0.169 MΩ
R0 =
169 + 25
1 + (10 ) + (10 )( 50 ) + ⎡⎣10 + (10 )( 50 ) ⎤⎦ ( 50 )
R0 =
194
= 0.00746 kΩ or Ro = 7.46 Ω
26, 011
8.41
a
VBB
Neglect base currents.
⎛I ⎞
= 2VD = 2VT ln ⎜ Bias ⎟
⎝ IS ⎠
⎛ 5 × 10−3 ⎞
= 2 ( 0.026 ) ln ⎜
⇒ VBB = 1.281 V
−13 ⎟
⎝ 10
⎠
VBE1 + VEB 3 = VBB
I E1 = I E 3 + I C 2
⎛ β ⎞
I B2 = IC 3 = ⎜ P ⎟ I E 3
⎝ 1+ βP ⎠
⎛ β ⎞
IC 2 = β n I B 2 = β n ⎜ P ⎟ I E 3
⎝ 1+ βP ⎠
⎛ β
I E1 = I E 3 + β n ⎜ P
⎝ 1+ βP
⎞
⎟ IE3
⎠
⎡
⎛ β
I E1 = I E 3 ⎢1 + β n ⎜ P
⎝ 1+ βP
⎣
⎞⎤
⎟⎥
⎠⎦
⎡
⎛ 1+ βn ⎞
⎛ 1+ βP ⎞
⎛ βP
⎜
⎟ I C1 = ⎜
⎟ I C 3 ⎢1 + β n ⎜
⎝ βP ⎠
⎝ 1+ βP
⎝ βn ⎠
⎣
⎡I ⎤
⎡I ⎤
VBE1 = VT ln ⎢ C1 ⎥ , VEB 3 = VT ln ⎢ C 3 ⎥
I
⎣ S ⎦
⎣ IS ⎦
(1.01) I C1 = ⎛⎜
21 ⎞
⎟ IC 3
⎝ 20 ⎠
I C1
⎞⎤
⎟⎥
⎠⎦
⎡
⎛ 20 ⎞ ⎤
⎢1 + (100 ) ⎜ 21 ⎟ ⎥
⎝ ⎠⎦
⎣
⎡ 21
⎤
= I C 3 ⎢ + 100 ⎥ = 101.05 I C 3
⎣ 20
⎦
= 100.05 I C 3
⎛ 100.05 I C 3 ⎞
⎛ IC 3 ⎞
VT ln ⎜
⎟ + VT ln ⎜
⎟ = VBB
I
S
⎝
⎠
⎝ IS ⎠
⎛ 100.05 I C2 3 ⎞
VT ln ⎜
⎟ = VBB
I S2
⎝
⎠
100.05 I C2 3
I
2
S
⎛V ⎞
= exp ⎜ BB ⎟
⎝ VT ⎠
⎛V ⎞
exp ⎜ BB ⎟ = 0.4995 mA = I C3
100.05
⎝ VT ⎠
Then I E 3 = 0.5245 mA
Now I C1 = 100.05 I C 3 = 49.97 mA = I C1
IC 3 =
IS
⎛ 20 ⎞
I C 2 = (100 ) ⎜ ⎟ ( 0.5245 ) = 49.95 mA = I C 2
⎝ 21 ⎠
⎛I ⎞
⎛ 49.97 × 10−3 ⎞
VBE1 = VT ln ⎜ C1 ⎟ = 0.026 ln ⎜
⎟
−13
⎝ 10
⎠
⎝ IS ⎠
= 0.70037
⎛I ⎞
⎛ 0.4995 × 10−3 ⎞
VEB 3 = VT ln ⎜ C 3 ⎟ = 0.026 ln ⎜
⎟
10−13
⎝
⎠
⎝ IS ⎠
= 0.58062
Note: VBE1 + VEB 3 = 0.70037 + 0.58062 = 1.28099
= VBB
b.
v0 = 10 V ⇒ iE1 ≈
10
= 0.10 A = iC1
100
100
= 1 mA
100
⎛ 4 × 10−3 ⎞
= 2 ( 0.026 ) ln ⎜
= 1.2694 V
−13 ⎟
⎝ 10
⎠
iB1 =
VBB
⎛ 0.1 ⎞
VBE1 = ( 0.026 ) ln ⎜ −13 ⎟ = 0.7184
⎝ 10 ⎠
VEB 3 = 1.2694 − 0.7184 = 0.55099 V
⎛ 0.55099 ⎞
I C 3 = 10−13 exp ⎜
⎟ = 0.1598 mA
⎝ 0.026 ⎠
V 2 (10 )
PL = 0 =
⇒ PL = 1 W
RL
100
2
PQ1 = iC1 ⋅ vCE1 = ( 0.1)(12 − 10 ) ⇒ PQ1 = 0.2 W
PQ 3 = iC 3 ⋅ vEC 3 = ( 0.1598 ) (10 − [ 0.7 − 12]) ⇒ PQ 3 = 3.40 mW
iC 2 = (100 )( iC 3 ) = (100 )( 0.1598 ) = 15.98 mA
PQ 2 = iC 2 ⋅ vCE 2 = (15.98 ) (10 − [ −12]) ⇒ PQ 2 = 0.352 W
8.42
a.
⎛ 10 × 10−3 ⎞
⇒ VBB = 1.74195 V
VBB = 3 ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 × 10 ⎠
VBE1 + VBE 2 + VEB 3 = VBB
I C1 ≈
IC 2
βn
, IC 3 ≈
IC 2
β n2
⎛I ⎞
⎛I ⎞
⎛I ⎞
VT ln ⎜ C1 ⎟ + VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 3 ⎟ = VBB
⎝ IS ⎠
⎝ IS ⎠
⎝ IS ⎠
⎡ I3 ⎤
VT ln ⎢ 3C 2 3 ⎥ = VBB
⎣ βn IS ⎦
⎛V ⎞
I C 2 = β n I S 3 exp ⎜ BB ⎟
⎝ VT ⎠
⎛ 1.74195 ⎞
= ( 20 ) ( 20 × 10−12 ) 3 exp ⎜
⎟
⎝ 0.026 ⎠
I C 2 = 0.20 A, I C1 ≈ 10 mA, I C 3 ≈ 0.5 mA
⎛ 10 × 10−3 ⎞
⇒ VBE1 = 0.58065 V
VBE1 = ( 0.026 ) ln ⎜
−12 ⎟
⎝ 2 × 10 ⎠
⎛ 0.2 ⎞
VBE 2 = ( 0.026 ) ln ⎜
⇒ VBE 2 = 0.6585 V
−12 ⎟
⎝ 2 × 10 ⎠
⎛ 0.5 × 10−3 ⎞
VEB 3 = ( 0.026 ) ln ⎜
⇒ VEB 3 = 0.50276 V
−12 ⎟
⎝ 2 × 10 ⎠
b.
1 V2 1 V2
PL = 10 W= ⋅ 0 = ⋅ 0 ⇒ V0 ( max ) = 20 V
2 RL 2 20
For v0 ( max ) :
v02 ( 20 )
=
⇒ PL = 20 W
20
RL
2
PL =
20
= −1 A
20
iC 5 + iC 4 + iE 3 = −io ( max ) = 1 A
i0 ( max ) = −
⎞
⎟⎟ = 1
⎠
i ⎛ β ⎞ i ⎛ β ⎞ ⎡ 1 ⎛ 1+ β p
iC 5 + C 5 ⎜ n ⎟ + C 5 ⎜ n ⎟ ⎢ ⎜
β n ⎝ 1 + β n ⎠ β n ⎝ 1 + β n ⎠ ⎢⎣ β n ⎜⎝ β p
iC 5 +
iC 5 ⎛ β n ⎞ iC 4 ⎛ 1 + β p
⋅⎜
⎜
⎟+
β n ⎝ 1 + β n ⎠ β n ⎜⎝ β p
1 ⎛ 20 ⎞ ⎤ ⎛ 1 ⎞ ⎡ 1 ⎛ 6 ⎞ ⎤
⎡
iC 5 ⎢1 + ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ = 1
⎣ 20 ⎝ 21 ⎠ ⎦ ⎝ 21 ⎠ ⎣ 20 ⎝ 5 ⎠ ⎦
iC 5 (1.05048 ) = 1 iC 5 = 0.952 A
⎞⎤
⎟⎟ ⎥ = 1
⎠ ⎥⎦
iC 4 = 0.0453 A
iE 3 = 0.00272 A
⎛5⎞
iC 3 = 0.00272 ⎜ ⎟
⎝6⎠
= 0.002267 A
⎛ 2.267 × 10−3 ⎞
VEB 3 = ( 0.026 ) ln ⎜
⎟ = 0.54206 V
−12
⎝ 2 × 10
⎠
VBE1 + VBE 2 = 1.74195 − 0.54206 = 1.19989
⎛ I ⎞
⎛I ⎞
VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 2 ⎟ = 1.19989
⎝ βn IS ⎠
⎝ IS ⎠
⎛ 1.19989 ⎞
iC 2 = β n ⋅ I S exp ⎜
⎟
⎝ 0.026 ⎠
= 20 (18.83) mA
iC 2 = 93.9 mA
iC 2 ⎛ β n ⎞ 93.9
= 4.47 mA
⎜
⎟=
β n ⎝ 1 + β n ⎠ 21
PQ 2 = I C 2 ( 24 − ( −20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 W
iC1 =
PQ 5 = ( 0.952 ) ( −10 − ( −24 ) ) = 13.3 W
Chapter 9
Exercise Solutions
EX9.1
Av = −15 = −
R2
R1
R1 = Ri = 20 K ⇒ R2 = 15 R1 = 15 ( 20 ) = 300 K
EX9.2
We can write ACL = −
R2 ⎛ R3 ⎞ R3
⎜1 + ⎟ −
R1 ⎝ R4 ⎠ R1
R1 = R1 = 10 kΩ
Want ACL = −50 Set R2 = R3 = 50 kΩ
⎛ R ⎞
ACL = −50 = −5 ⎜1 + 3 ⎟ − 5
⎝ R4 ⎠
R
R
50
1+ 3 = 9 ⇒ 3 = 8 =
R4
R4
R4
R4 = 6.25 kΩ
EX9.3
R2
1
⋅
R1 ⎡
⎛
R2 ⎞ ⎤
1
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Ad ⎝
R
Ri = R1 = 25 kΩ Let 2 = x
R1
We have ACL = −
x
1
1+
(1 + x )
5 × 103
−x
=
x
1.0002 +
5 × 103
x ⎞
⎛
12 ⎜ 1.0002 +
⎟=x
×
5
103 ⎠
⎝
−12 = −
12.0024 = x − ( 2.4 × 10 −3 ) x
R2
12.0024
= 12.0313 =
0.9976
25 kΩ
R2 = 300.78 kΩ
x=
EX9.4
⎛R
⎞
R
R
R
v0 = − ⎜ F vI 1 + F vI 2 + F vI 3 + F VI 4 ⎟
R2
R3
R4
⎝ R1
⎠
We need
RF
R
R
R
= 7, F = 14, F = 3.5, F = 10
R1
R2
R3
R4
Set RF = 280 kΩ
280
7
280
R2 =
14
280
R3 =
3.5
280
R4 =
10
Then R1 =
= 40 kΩ
= 20 kΩ
= 80 kΩ
= 28 kΩ
EX9.5
We may note that
R
R3
R
R
3
20
= 2 so that 3 = F
=
= 2 and F =
R2 R1
R1 10
R2 1.5
Then
iL =
− ( −3)
− vI
=
⇒ iL = 2 mA
R2 1.5 kΩ
vL = iL Z L = ( 2 × 10−3 ) ( 200 ) = 0.4 V
i4 =
vL
0.4
=
= 0.267 mA
R2 1.5 kΩ
i3 = i4 + iL = 0.267 + 2 = 2.267 mA
v0 = i3 R3 + vL = ( 2.267 × 10−3 )( 3 × 103 ) − 0.4 ⇒ v0 = 7.2 V
EX9.6
Refer to Fig. 9.24
R1 = 2 R1 = 5 kΩ
Let R1 = R3 = 2.5 kΩ
Set R2 = R4
Differential Gain =
EX9.7
v0 R2
R2
=
= 100 =
⇒ R2 = R4 = 250 kΩ
2.5 kΩ
v1 R1
We have the general relation that
⎛ R ⎞ ⎛ [ R4 / R3 ] ⎞
R2
v0 = ⎜ 1 + 2 ⎟ ⎜
⎟⎟ vI 2 − vI 1
⎜
R1 ⎠ ⎝ 1 + [ R4 / R3 ] ⎠
R1
⎝
R1 = R3 = 10 kΩ, R2 = 20 kΩ, R4 = 21 kΩ
⎛ 20 ⎞ ⎛ [ 21/10] ⎞
⎛ 20 ⎞
v0 = ⎜1 + ⎟ ⎜
⎟⎟ vI 2 − ⎜ ⎟ vI 1
⎜
+
10
1
21/10
[
]
⎝
⎠⎝
⎝ 10 ⎠
⎠
v0 = 2.0323vI 2 − 2.0vI
a.
vI 1 = 1, vI 2 = −1
v0 = −2.0323 − 2.0 ⇒ v0 = −4.032 V
b.
vI 1 = vI 2 = 1 V
v0 = 2.0323 − 2.0 ⇒ v0 = 0.0323 V
c.
vcm = vI 1 = vI 2 so common-mode gain
v0
= 0.0323
vcm
Acm =
d.
⎛ A ⎞
C M R RdB = 20 log10 ⎜ d ⎟
⎝ Acm ⎠
2.0323
⎛ 1⎞
Ad =
− ( 2.0 ) ⎜ − ⎟ = 2.016
2
⎝ 2⎠
⎛ 2.016 ⎞
C M R RdB = 20 log10 ⎜
⎟ = 35.9 d B
⎝ 0.0323 ⎠
EX9.8
v0 = −
R4 ⎛ 2 R2 ⎞
⎜1 +
⎟ ( vI 1 − vI 2 )
R3 ⎝
R1 ⎠
Differential gain (magnitude) =
R4 ⎛ 2 R2 ⎞
⎜1 +
⎟
R3 ⎝
R1 ⎠
Minimum Gain ⇒ Maximum R1 = 1 + 50 = 51 kΩ
So Ad =
20 ⎛ 2 (100 ) ⎞
⎜1 +
⎟ ⇒ Ad = 4.92
20 ⎝
51 ⎠
Maximum Gain ⇒ Minimum R1 = 1 kΩ
Ad =
20 ⎛ 2 (100 ) ⎞
⎜1 +
⎟ ⇒ Ad = 201
20 ⎝
1 ⎠
Range of Differential Gain = 4.92 − 201
EX9.9
Time constant = r = R1C2 = (104 )( 0.1×10−6 )
−1
×t
0 ≤ t ≤ 1 ⇒ v0 =
R1 C2
= 1 m sec
At t = 1 m sec ⇒ v0 = −1 V
0 ≤ t ≤ 2 ⇒ v0 = −1 +
1
× ( t − 1)
R1 C2
At t = 2 m sec ⇒ v0 = −1 +
( 2 − 1)
1
=0
EX9.10
v0 = vI 1 + 10vI 2 − 25vI 3 − 80vI 4
From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB.
From Equation (9.94)
RF
R
= 25 and F = 80
R1
R2
Set RF = 500 kΩ, then R1 = 20 kΩ, and R2 = 6.25 kΩ.
⎛ R ⎞⎛ R ⎞
⎛ R ⎞⎛ R ⎞
Also ⎜1 + F ⎟⎜ P ⎟ = 1 and ⎜1 + F ⎟ ⎜ P ⎟ = 10
⎝ RN ⎠ ⎝ RA ⎠
⎝ RN ⎠ ⎝ RB ⎠
where RN = R1 & R2 = 20 & 6.25 = 4.76 kΩ and RP = RA & RB & RC
We find that
RA
= 10
RB
Let RA = 200 kΩ, RB = 20 kΩ
500 ⎞ ⎛ RP ⎞
⎛ RP ⎞
⎛
Now ⎜1 +
⎟ = 1 = (106 ) ⎜
⎟⎜
⎟
⎝ 4.76 ⎠ ⎝ RA ⎠
⎝ 200 ⎠
Then RP = 1.89 kΩ
RA & RB = 200 & 20 = 18.2 kΩ
So RP = 1.89 =
18.2 RC
⇒ RC = 2.11 kΩ
18.2 + RC
EX9.11 Computer Analysis
TYU9.1
ACL = −
R2 −100 kΩ
=
⇒ ACL = −10
R1
10 kΩ
vI = 0.25 V ⇒ vo = −2.5 V
i1 =
vI
0.25
=
= 0.025 mA ⇒ i1 = 25 µ A
R1 10 kΩ
i2 = i1 = 25 µ A
Ri = R1 = 10 kΩ
TYU9.2
(a)
− R2
Av =
R1 + RS
−100
= −4.926
19 + 1.3
−100
= −5.076
Av ( max ) =
19 + 0.7
so 4.926 ≤ Av ≤ 5.076
Av ( min ) =
(b)
0.1
= 5.076 µ A
19 + 0.7
0.1
= 4.926 µ A
i1 ( min ) =
19 + 1.3
so 4.926 ≤ i1 ≤ 5.076 µ A
i1 ( max ) =
(c)
Maximum current specification is violated.
TYU9.3
v0 = Ad ( v2 − v1 )
Ad = 103
a.
v2 = 0, v0 = 5
v
5
v1 = − 0 = − 3 ⇒ v1 = −5 mV
Ad
10
b.
v1 = 5, v0 = −10
v0
= v2 − v1
Ad
−10
= v2 − 5 ⇒ v2 = 4.99 V
103
c.
v1 = 0.001, v2 = −0.001
v0 = 103 ( −0.001 − 0.001)
v0 = −2 V
d.
v2 = 3, v0 = 3
v0 = Ad ( v2 − v1 )
v0
= v2 − v1
Ad
3
= 3 − v1 ⇒ v1 = 2.997 V
103
TYU9.4
⎛R
⎞
R
R
v0 = − ⎜ 4 vI 1 + 4 vI 2 + 4 vI 3 ⎟
R2
R3
⎝ R1
⎠
⎡⎛ 40 ⎞
⎤
⎛ 40 ⎞
⎛ 40 ⎞
v0 = − ⎢⎜ ⎟ ( 250 ) + ⎜ ⎟ ( 200 ) + ⎜ ⎟ ( 75 ) ⎥
⎝ 20 ⎠
⎝ 30 ⎠
⎣⎝ 10 ⎠
⎦
v0 = − [1000 + 400 + 100]
v0 = −1500 µ V = −1.5 mV
TYU9.5
vI 1 + vI 2 + vI 3 RF
=
( vI 1 + vI 2 + vI 3 )
R
3
RF 1
= ⇒ R1 = R2 = R3 ≡ R = 1 M Ω
R 3
1
Then RF = M Ω = 333 k Ω
3
vO =
TYU9.6
v ⎛ R ⎞
R
Av = 0 = ⎜1 + 2 ⎟ = 5 so that 2 = 4
vI ⎝
R1 ⎠
R1
For v0 = 10 V, vI = 2 V
Then i1 =
2
= 50 µ A ⇒ R1 = 40 kΩ
R1
Then R2 = 160 kΩ we find i2 =
v0 − vI 10 − 2
=
= 50 µ A
R2
160
TYU9.7
v0 = Aod ( v2 − v1 ) = Aod ( vI − v1 )
v0
v
− vI = −v1 or v1 = vI − 0
Aod
Aod
i1 =
v −v
v1
= i2 and i2 = 0 1
R1
R2
⎛ 1 ⎞ v −v
Then v1 ⎜ ⎟ = 0 1
R2
⎝ R1 ⎠
⎛ 1
1 ⎞ v
v1 ⎜ + ⎟ = 0
⎝ R1 R2 ⎠ R2
⎛ R ⎞
⎛ R ⎞⎛
v ⎞
v0 ⎜ 1 + 2 ⎟ v1 = ⎜ 1 + 2 ⎟ ⎜ vI − 0 ⎟
R1 ⎠
R1 ⎠⎝
Aod ⎠
⎝
⎝
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
v0
⎝
So Av = =
vI
1 ⎛ R2 ⎞
1+
⎜1 + ⎟
Aod ⎝
R1 ⎠
TYU9.8
⎛ Rb
For vI 2 = 0, v2 = ⎜
⎝ Rb + Ra
⎞
⎟ vI 1 and
⎠
⎛ R ⎞ ⎛ Rb ⎞
v0 ( vI 1 ) = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 1
R1 ⎠⎝ Rb + Ra ⎠
⎝
⎛ 70 ⎞⎛ 50 ⎞
= ⎜ 1 + ⎟⎜
⎟ vI 1
5 ⎠⎝ 50 + 25 ⎠
⎝
= 10vI 1
⎛ Ra ⎞
For vI 1 = 0, v2 = ⎜
⎟ vI 2
⎝ Rb + Ra ⎠
⎛ R ⎞ ⎛ Ra ⎞
v0 ( vI 2 ) = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 2
R1 ⎠⎝ Rb + Ra ⎠
⎝
⎛ 70 ⎞⎛ 25 ⎞
= ⎜ 1 + ⎟⎜
⎟ vI 2
5 ⎠⎝ 25 + 50 ⎠
⎝
= 5vI 2
Then
v0 = v0 ( vI 1 ) + v0 ( vI 2 )
v0 = 10vI 1 + 5vI 2
TYU9.9
RS Ri so i1 = i2 = iS = 100 µ A
v0 = −iS RF
We want −10 = − (100 × 10−6 ) RF ⇒ RF = 100 kΩ
TYU9.10
We want iL = 1 mA when vI = −5 V
iL =
− ( −5 )
−VI
−v
⇒ R2 = I =
⇒ R2 = 5 kΩ
R2
i2
10−3
vL = iL Z L = (10−3 ) ( 500 ) ⇒ vL = 0.5 V
i4 =
vL
0.5
=
⇒ i4 = 0.1 mA
R2 5 kΩ
i3 = i4 + iL = 0.1 + 1 = 1.1 mA
If op-amp is biased at ±10 V, output must be limited to ≈ 8 V.
So v0 = i3 R3 + vL
8 = (1.1× 10−3 ) R3 + 0.5 ⇒ R3 = 6.82 kΩ
Let R3 = 7.0 kΩ
Then we want
R3 RF 7
=
= = 1.4
R2 R1 5
Can choose R1 = 10 kΩ and RF = 14 kΩ
TYU9.11
a.
v −v
i1 = I 1 I 2
R1
v01 = vI 1 + i1 R2′ , v02 = vI 2 − i1 R2 and v0 =
v0 =
v0 =
R4
( v02 − v01 )
R3
R4
[vI 2 − i1 R2 − vI 1 − i1 R2′ ]
R3
R4
⎡( vI 2 − vI 1 ) − i1 ( R2 + R2′ ) ⎤⎦
R3 ⎣
⎤
⎛ vI 2 − vI 1 ⎞
R4 ⎡
⎢ ( vI 2 − vI 1 ) − ⎜
⎟ ( R2 + R2′ ) ⎥
R3 ⎣
⎝ R1 ⎠
⎦
For common-mode input vI 2 = vI 1 ⇒ v0 = 0 ⇒ Common Gain = 0, C M R R = ∞
v0 =
b.
Ad ( min ) ⇒ R2′ min, R1 max
⎛ 20 ⎞ ⎡ 100 + 95 ⎤
Ad = ⎜ ⎟ ⎢1 +
= 4.82
51 ⎦⎥
⎝ 20 ⎠ ⎣
⎛ 20 ⎞ ⎡ 100 + 105 ⎤
Ad ( max ) = ⎜ ⎟ ⎢1 +
⎥ = 206
1
⎝ 20 ⎠ ⎣
⎦
c.
A
CM RR = d
Acm
Acm = 0 ⇒ C M R R = ∞
TYU9.12
Differential Gain =
R4 ⎛ 2 R2 ⎞
⎜1 +
⎟
R3 ⎝
R1 ⎠
Let R3 = R4 so the difference amplifier gain is unity.
Minimum Gain ⇒ Maximum R1
⎛
2 R2 ⎞
So ⎜⎜1 +
⎟⎟ = 2
⎝ R1 ( max ) ⎠
We want 2 R2 = R1 ( max )
Maximum Gain ⇒ Minimum R1
⎛
2 R2 ⎞
So ⎜⎜1 +
⎟⎟ = 1000 or 2 R2 = 999 R1 ( min )
R
1 ( min ) ⎠
⎝
If R2 = 50 kΩ, let R1 ( min ) = 100 Ω fixed resistor and let R1 ( max ) = 100
kΩ + 100 Ω = 100.1
pot
Then actual differential gain is in the range of 1.999 − 1001
TYU9.13
−1 10 µ sec −10 × 10−6
×t 0
=
r
r
− (10 ) (10 × 10−6 )
After 10 pulses: v0 = −5 =
r
End of 1st pulse: v0 =
100 × 10−6
= 20 µ sec = r
5
r = R1C2 = 20 µ sec = 20 × 10−6
So r =
For example, C2 = 0.01× 10−6 = 0.01 µ F ⇒ R1 = 2 kΩ
TYU9.14
⎡
⎤ +
R − ΔR
R + ΔR
−
v01 = ⎢
⎥V
⎣⎢ ( R − ΔR ) + ( R + ΔR ) ( R + ΔR ) + ( R − ΔR ) ⎦⎥
⎡ R − ΔR R + ΔR ⎤ +
V
=⎢
−
2 R ⎥⎦
⎣ 2R
⎛ R − ΔR − R − ΔR ⎞ +
=⎜
⎟V
2R
⎝
⎠
⎛ ΔR ⎞ +
v01 = − ⎜
⎟V
⎝ R ⎠
For V + = 3.5 V, ΔR = 50, R = 10 × 103
⎛ 50 ⎞
v01 = − ⎜ 4 ⎟ ( 3.5 ) = −1.75 × 10−2
⎝ 10 ⎠
Need an amplifier with a gain of Ad =
v0
5
=
⇒ Ad = −285.7
vi −1.75 × 10−2
Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( −v01 ) to vI2.
⎛ R ⎞ ⎛ 2R ⎞
Ad = ⎜ 4 ⎟ ⎜ 1 + 2 ⎟ = 285.7
R1 ⎠
⎝ R3 ⎠ ⎝
Let R4 = 150 kΩ, R3 = 10 kΩ Then
R2
= 9.02
R1
Let R2 = 100 kΩ, R1 = 11.1 kΩ
TYU9.15
⎡1
⎤ +
R
v01 = ⎢ −
⎥V
⎣⎢ 2 R + R (1 + δ ) ⎦⎥
⎡ R + R (1 + δ ) − 2 R ⎤ +
=⎢
⎥V
⎣⎢ 2 ( R + R (1 + δ ) ) ⎦⎥
=
Rδ
×V +
2R ( 2 + δ )
⎛δ ⎞
v01 ≈ ⎜ ⎟ V +
⎝4⎠
+
V = 5 For δ = 0.01
⎛ 0.01 ⎞
v01 = ⎜
⎟ ( 5 ) = 0.0125
⎝ 4 ⎠
v
5
= 400
Need a gain Ad = 0 =
v01 0.0125
⎛ R ⎞ ⎛ 2R ⎞
Use an instrumentation amplifier Ad = 400 = ⎜ 4 ⎟ ⎜1 + 2 ⎟
R1 ⎠
⎝ R3 ⎠ ⎝
R
Let R4 = 150 kΩ, R3 = 10 kΩ then 2 = 12.8
R1
Let R2 = 150 kΩ, R1 = 11.7 kΩ
Chapter 9
Problem Solutions
9.1
(a)
vO = Ad ( v2 − v1 )
1 = Ad 10−3 − ( −10−3 ) ⇒ Ad = 500
(
)
(b)
1 = 500 ( v2 − 10−3 ) = 1 + 0.5 = 500v2
v2 = 3 mV
(c)
5 = 500 (1 − v1 ) ⇒ 500v1 = 495
v1 = 0.990 V
(d)
(e)
vO = 0
− 3 = 500 ( v2 − ( −0.5 ) )
−250 − 3 = 500v2
v2 = −0.506 V
9.2
(a)
1
⎛
⎞
−3
v2 = ⎜
⎟ vI = ( 0.49975 × 10 ) ( 3)
+
1
2000
⎝
⎠
v2 = 1.49925 × 10−3
vO = Aod ( v2 − v1 ) = ( 5 × 103 )(1.49925 × 10−3 − 0 )
vO = 7.49625 V
(b)
vO = Aod ( v2 − v1 )
3 = Aod (1.49925 × 10−3 − 0 )
Aod = 2 × 103
9.3
Av = −
R2
= −12 ⇒ R2 = 12 R1
R1
Ri = R1 = 25 kΩ ⇒ R2 = (12 )( 25 ) = 300 kΩ
9.3
(a)
(b)
v2 = 3.00 V
vO = Aod ( v2 − v1 )
2.500 = Aod ( 3.010 − 3.00 )
Aod = 250
9.4
⎛ Ri ⎞
vid = ⎜
⎟ vI
⎝ Ri + 25 ⎠
⎛ Ri ⎞
0.790 = ⎜
⎟ ( 0.80 )
⎝ Ri + 25 ⎠
0.9875 ( Ri + 25 ) = Ri
24.6875 = 0.0125 Ri
Ri = 1975 K
9.5
200
⎫
= −10 ⎪
20
⎪
and
⎬ for each case
⎪
Ri = 20 kΩ
⎪
⎭
Av = −
9.6
a.
100
= −10
10
Ri = R1 = 10 kΩ
b.
100 & 100
Av = −
= −5
10
Ri = R1 = 10 kΩ
c.
100
Av = −
= −5
10 + 10
Ri = 10 + 10 = 20 K
Av = −
9.7
I1 =
vI
0.5
⇒ R1 =
⇒ R1 = 5 K
R1
0.1
R2
= 15 ⇒ R2 = 75 K
R1
9.8
Av = −
(a)
(b)
(c)
(d)
(e)
(f)
R2
R1
Av = −10
Av = −1
Av = −0.20
Av = −10
Av = −2
Av = −1
9.9
Av = −
(a)
(b)
(c)
(d)
R2
R1
R1 = 20 K, R2 = 40 K
R1 = 20 K, R2 = 200 K
R1 = 20 K, R2 = 1000 K
R1 = 80 K, R2 = 20 K
9.10
Av = −
R2
= −8 ⇒ R2 = 8 R1
R1
For vI = −1, i1 =
9.11
Av = −
1
= 15 µ A ⇒ R1 = 66.7 kΩ ⇒ R2 = 533.3 kΩ
R1
R2
= −30 ⇒ R2 = 30 R1
R1
Set R2 = 1 MΩ ⇒ R1 = 33.3 kΩ
9.12
a.
Av =
⎛R ⎞
R2
1.05R2
⇒
= 1.105 ⎜ 2 ⎟
R1
0.95 R1
⎝ R1 ⎠
⎛R ⎞
0.95R2
= 0.905 ⎜ 2 ⎟
1.05R1
⎝ R1 ⎠
Deviation in gain is +10.5% and − 9.5%
b.
⎛R ⎞
⎛R ⎞
1.01R2
0.99 R2
= 1.02 ⎜ 2 ⎟ ⇒
= 0.98 ⎜ 2 ⎟
Av ⇒
0.99 R1
1.01R1
⎝ R1 ⎠
⎝ R1 ⎠
Deviation in gain = ±2%
9.13
(a)
Av =
vO −15
=
= −15
vl
1
vO = −15vl ⇒ vO = −150sin ω t ( mV )
(b)
i2 = i1 =
vI
= 10sin ω t ( µ A )
R1
vO
⇒ iL = −37.5sin ω t ( µ A )
RL
iL =
iO = iL − i2
iO = −47.5sin ω t ( µ A )
9.14
Av = −
R2
R1 + R5
Av = −30 ± 2.5% ⇒ 29.25 ≤ Av ≤ 30.75
So
R2
R2
= 29.25 and
= 30.75
R1 + 2
R1 + 1
We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1)
Which yields R1 = 18.5 k Ω and R2 = 599.6 k Ω
For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V
9.15
vO1 = −
R2
120
, vI = −
( 0.2 ) ⇒ vO1 = −1.2 V
20
R1
R4
⎛ −75 ⎞
, vO1 = ⎜
⎟ ( −1.2 ) ⇒ vO = +6 V
R3
⎝ 15 ⎠
0.2
⇒ i1 = i2 = 10 µ A
i1 = i2 =
20
v
−1.2
i3 = i4 = O1 =
⇒ i3 = i4 = −80 µ A
15
R3
1st op-amp: 90 µ A into output terminal
2nd op-amp: 80 µ A out of output terminal.
vO = −
9.16
(a)
R2
22
=−
⇒ Av = −22
1
R1
(b)
From Eq. (9.23)
R2
1
1
= −22 ⋅
Av = − ⋅
1
R1 ⎡
⎡
⎤
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎢⎣1 + 104 ( 23) ⎥⎦
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
Av = −
Av = −21.95
(c)
Want Av = −22 ( 0.98 ) = −21.56
So − 21.56 =
1+
−22
1
1+
( 23)
Aod
1
22
( 23) =
Aod
21.56
1
( 23) = 0.020408 ⇒ Aod = 1127
Aod
9.17
(a)
R2
1
⋅
R1 ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
100
1
=−
⋅
1
25 ⎡
⎤
⎢⎣1 + 5 × 103 ( 5 ) ⎥⎦
Av = −3.9960
Av = −
(b)
vO = −3.9960 (1.00 ) ⇒ vO = −3.9960 V
(c)
4 − 3.9960
× 100% = 0.10%
4
(d)
vO = Aod ( v2 − v1 ) = − Aod v1
− ( −3.9960 )
vO
=
Aod
5 × 10+3
v1 = −
v1 = 0.7992 mV
9.18
vO = Aod ( v2 − v1 ) = − Aod v1
v
−5
v1 = − O =
Aod 5 × 10+3
v1 = −1 mV
9.19
Av = −
a.
R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
R1 ⎝ R4 R2 ⎠
−10 = −
10 =
b.
9.20
a.
R2 ⎛ 100 100 ⎞
+
⎜1 +
⎟
100 ⎝ 100 R2 ⎠
2 R2
+ 1 ⇒ R2 = 450 kΩ
100
2R
100 = 2 + 1 ⇒ R2 = 4.95 MΩ
100
R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
R1 ⎝ R4 R2 ⎠
R1 = 500 kΩ
Av = −
R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
500 ⎝ R4 R2 ⎠
Set R2 = R3 = 500 kΩ
80 =
⎛ 500 ⎞
500
80 = 1⎜1 +
+ 1⎟ = 2 +
⇒ R4 = 6.41 kΩ
R
R4
⎝
⎠
4
b.
For vI = −0.05 V
−0.05
⇒ i1 = i2 = −0.1 µ A
i1 = i2 =
500 kΩ
v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05
i4 = −
vX
0.05
=−
⇒ i4 = −7.80 µ A
6.41
R4
i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 µ A
9.21
(a)
Av = −1000 =
− R2 −500
=
R1
R1
R1 = 0.5 K
(b)
− R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
R1 ⎝ R4 R2 ⎠
−250 ⎛ 500 500 ⎞ −1250
−1000 =
+
⎜1 +
⎟=
R1 ⎝ 250 250 ⎠
R1
R1 = 1.25 K
Av =
9.22
i1 =
vI
= i2
R
⎛v ⎞
v A = −i2 R = − ⎜ I ⎟ R = −vI
⎝R⎠
v
v
i3 = − A = I
R R
2v
2v
vA vA
−
=− A = I
R R
R
R
⎛ 2vI ⎞
vB = v A − i4 R = −vI − ⎜
⎟ ( R ) = −3vI
⎝ R ⎠
i4 = i2 + i3 = −
( −3vI ) 3vI
vB
=−
=
R
R
R
2vI 3vI 5vI
i6 = i4 + i5 =
+
=
R
R
R
v
5
v
⎛
⎞
v0 = vB − i6 R = −3vI − ⎜ I ⎟ R ⇒ 0 = −8
vI
⎝ R ⎠
i5 = −
From Figure 9.12 ⇒ Av = −3
9.23
(a)
R2
1
⋅
R1 ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
50
1
=− ⋅
⇒ Av = −4.99985
10 ⎡
1 ⎛ 50 ⎞ ⎤
1
1
+
+
⎟⎥
5 ⎜
⎢
⎣ 2 × 10 ⎝ 10 ⎠ ⎦
Av = −
(b)
vO = − ( 4.99985 ) (100 × 10−3 ) ⇒ vO = −499.985 mV
(c)
Error =
0.5 − 0.499985
× 100% ⇒ 0.003%
0.5
9.24
a.
From Equation (9.23)
R2
1
Av = − ⋅
R1 ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
100
1
=−
⋅
= −0.9980
100 ⎡
1 ⎛ 100 ⎞ ⎤
+
+
1
1
⎢ 103 ⎜ 100 ⎟ ⎥
⎝
⎠⎦
⎣
Then v0 = Av ⋅ vI = ( −0.9980 )( 2 ) ⇒ v0 = −1.9960 V
b.
v0 = Aod ( v A − vB )
⎛ 1
vB v0 − vB
1 ⎞ v
=
⇒ vB ⎜ + ⎟ = 0
R1
R2
R
R
R2
2 ⎠
⎝ 1
v0
vB =
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
⎝
Then v0 = Aod v A −
Aod v0
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
⎝
⎡
⎤
⎢
⎥
Aod ⎥
v0 ⎢1 +
= Aod v A
⎢ ⎛ R ⎞⎥
2
⎢ ⎜1 + ⎟ ⎥
R1 ⎠ ⎥⎦
⎢⎣ ⎝
⎡ ⎛ R2 ⎞
⎤
⎢ ⎜1 + ⎟ + Aod ⎥
R
1 ⎠
⎥=A v
v0 ⎢ ⎝
od A
⎢ ⎛ R ⎞ ⎥
⎢ ⎜1 + 2 ⎟ ⎥
R1 ⎠ ⎥⎦
⎢⎣ ⎝
⎛ R ⎞
Aod ⎜1 + 2 ⎟ v A
R1 ⎠
⎝
v0 =
⎛ R ⎞
Aod + ⎜ 1 + 2 ⎟
R1 ⎠
⎝
⎛ R2 ⎞
⎜1 + ⎟ vA
R1 ⎠
v0 = ⎝
1 ⎛ R2 ⎞
1+
⎜1 + ⎟
Aod ⎝
R1 ⎠
⎛ 10 ⎞ ⎛ vI ⎞
⎜1 + ⎟ ⎜ ⎟
10 ⎠ ⎝ 2 ⎠
= 0.9980vI
So v0 = ⎝
1 ⎛ 10 ⎞
1 + 3 ⎜1 + ⎟
10 ⎝ 10 ⎠
For vI = 2 V
v0 = 1.9960 V
9.25
(a)
ii =
vl
v
v
R
= i2 = − O ⇒ O = − 2
R1
R2
vl
R1
(b)
⎞
vl
v
1 ⎛ R2
= i3 + O = i3 +
⎜ − ⋅ vl ⎟
R1
RL
RL ⎝ R1
⎠
v ⎛ R ⎞
Then i3 = l ⎜ 1 + 2 ⎟
R1 ⎝ RL ⎠
i2 = i1 =
9.26
⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞
VX .max = ⎜
⋅V = ⎜
10 ⇒ VX .max = 0.09008 V
⎜ R R + R ⎟⎟
⎜ 0.1 1 + 10 ⎟⎟ ( )
4 ⎠
⎝ 3 1
⎝
⎠
R
vO = 2 ⋅ VX .max
R1
10 =
R2
R
( 0.09008 ) ⇒ 2 = 111
R1
R1
So R2 = 111 k Ω
9.27
(a)
⎛R
⎞
R
R
vO = − ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟
R2
R3
⎝ R1
⎠
⎡⎛ 100 ⎞
⎤
⎛ 100 ⎞
⎛ 100 ⎞
= − ⎢⎜
⎟ ( 0.5 ) + ⎜
⎟ ( 0.75 ) + ⎜
⎟ ( 2.5 ) ⎥
50
20
100
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
= − [1 + 3.75 + 2.5]
vO = −7.25 V
(b)
⎡⎛ 100 ⎞
⎛ 100 ⎞
⎛ 100 ⎞ ⎤
−2 = − ⎢⎜
⎟ (1) + ⎜
⎟ ( 0.8 ) + ⎜
⎟ vI 3 ⎥
⎝ 20 ⎠
⎝ 100 ⎠ ⎦
⎣⎝ 50 ⎠
2 = 2 + 4 + vI 3
vI 3 = −4 V
9.28
vo =
R
R
− RF
⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1
R2
R3
= −4vI 1 − 8vI 2 − 2vI 3
RF
=4
R1
RF
=8
R2
RF
=2
R3
Largest resistance = RF = 250 K ⇒ R1 = 62.5 K
R2 = 31.25 K
9.29
v0 = −4vI 1 − 0.5vI 2 = −
RF
=4
R1
RF
R
vI 1 − F vI 2
R1
R2
RF
= 0.5 ⇒ R1 is the smallest resistor
R2
i = 100 µ A =
vI
2
=
R1 R1
⇒ R1 = 20 kΩ
⇒ RF = 80 kΩ
⇒ R2 = 160 kΩ
9.30
vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft )
1
1
⇒ 10 ms
f = 1 kHz ⇒ T = 3 ⇒ 1 ms vI 2 ⇒ T2 =
100
10
R
R
10
10
vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2
R1
R2
1
5
vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V )
vO = −0.707 sin ( 2π ft ) − ( ±2 V )
R3 = 125 K
9.31
vO = −
RF
R
R
⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1
R2
R3
20
20
20
⋅ vI 1 − ⋅ vI 2 − ⋅ vI 3
10
5
2
K sin ω t = −2vI 1 − 4 [ 2 + 100sin ω t ] − 0
vO = −
Set vI 1 = −4 mV
9.32
Only two inputs.
⎡R
⎤
R
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
R2
⎣ R1
⎦
1
⎡
⎤
= − ⎢3vI 1 + ⋅ vI 2 ⎥
4
⎣
⎦
RF
RF 1
=3
=
R1
R2 4
Smallest resistor = 10 K = R1
RF = 30 K
R2 = 120 K
9.33
⎡R
⎤
R
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
R
R
2
⎣ 1
⎦
R
− RF
−R
−5 − 5sin ω t =
( 2.5sin ω t ) F ⋅ ( 2 ) ⇒ F = 2
R1
R2
R1
RF = largest resistor ⇒ RF = 200 K
R1 = 100 K
RF
= 2.5
R2
R2 = 80 K
9.34
a.
v0 = −
RF
R
R
R
⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 )
R3
R2
R1
R0
So v0 =
b.
RF ⎡ a3 a2 a1 a0 ⎤
+ + +
( 5)
10 ⎢⎣ 2 4 8 16 ⎥⎦
R 1
v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ
10 2
c.
i.
v0 =
10 1
⋅ ⋅ 5 ⇒ v0 = 0.3125 V
10 16
ii.
v0 =
10 ⎡ 1 1 1 1 ⎤
+ + +
( 5 ) ⇒ v0 = 4.6875 V
10 ⎢⎣ 2 4 8 16 ⎥⎦
9.35
(a)
10
⋅ vI 1
1
20
20
vO = − ⋅ vO1 − ⋅ vI 2 = − ( 20 )( −10 ) vI 1 − ( 20 ) vI 2
1
1
vO = 200vI 1 − 20vI 2
vO1 = −
(b)
vI 1 = 1 + 2sin ω t ( mV )
vI 2 = −10 mV
Then vO = 200 (1 + 2sin ω t ) − 20 ( −10 )
So vO = 0.4 + 0.4sin ω t (V )
9.36
For one-input
v1 = −
v0
Aod
v −v
vI 1 − v1
v1
=
+ 1 0
R1
R2 & R3
RF
⎡1
VI 1
1
1 ⎤ v0
= v1 ⎢ +
+
⎥−
R1
R
R
R
R
RF
&
2
3
F ⎦
⎣ 1
=−
v0
Aod
⎡1
1
1 ⎤ v0
+
⎢ +
⎥−
&
R
R
R
R
RF
2
3
F ⎦
⎣ 1
1
1 ⎛ 1
1 ⎞ ⎪⎫
⎪⎧ 1
= −v0 ⎨
+
+
⎜ +
⎟⎬
⎪⎩ Aod RF RF Aod ⎝ R1 R2 & R3 ⎠ ⎪⎭
=−
⎫
v0 ⎧ 1
RF
1
+1+
⋅
⎨
⎬
RF ⎩ Aod
Aod R1 & R2 & R3 ⎭
⎧
⎫
⎪
⎪
R
1
⎪
⎪
v0 = − F ⋅ vI 1 ⋅ ⎨
⎬ where RP = R1 & R2 & R3
R1
⎛
⎞
R
1
F ⎪
⎪1 +
1+
⎪ Aod ⎜⎝ RP ⎟⎠ ⎪
⎩
⎭
Therefore, for three-inputs v0 =
9.37
⎛R
⎞
R
R
−1
× ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟
R2
R3
1 ⎛ RF ⎞ ⎝ R1
⎠
1+
⎜1 +
⎟
Aod ⎝ RP ⎠
⎛ R ⎞
R
Av = 12 = ⎜ 1 + 2 ⎟ ⇒ 2 = 11
R
R1
1 ⎠
⎝
v
v
0.5
i1 = I ⇒ R1 = I =
R1
i1 0.15
R1 = 3.33 K
R2 = 36.7 K
9.38
⎛ 1 ⎞
vB = ⎜
⎟ vI
⎝ 1 + 500 ⎠
a.
b.
⎛ 1 ⎞
v0 = Aod ⎜
⎟ vi
⎝ 501 ⎠
⎛ 1 ⎞
2.5 = Aod ⎜
⎟ ( 5 ) ⇒ Aod = 250.5
⎝ 501 ⎠
⎛ 1 ⎞
v0 = 5000 ⎜
⎟ ( 5 ) ⇒ v0 = 49.9 V
⎝ 501 ⎠
9.39
⎛ R ⎞
Av = ⎜ 1 + 2 ⎟
R1 ⎠
⎝
(a)
Av = 11
(b)
(c)
(d)
(e)
(f)
9.40
(a)
(b)
Av = 2
Av = 1.2
Av = 11
Av = 3
Av = 2
R2
= 1 ⇒ R1 = R2 = 20 K
R1
R2
= 9 ⇒ R1 = 20 K, R2 = 180 K
R1
(c)
R2
= 49 ⇒ R1 = 20 K, R2 = 980 K
R1
(d)
R2
= 0 can set R2 = 20 K, R1 = ∞ (open circuit)
R1
9.41
⎛ 50 ⎞ ⎡⎛ 20 ⎞
⎛ 40 ⎞ ⎤
v0 = ⎜1 + ⎟ ⎢⎜
⎟ vI 2 + ⎜
⎟ vI 1 ⎥
⎝ 50 ⎠ ⎣⎝ 20 + 40 ⎠
⎝ 20 + 40 ⎠ ⎦
v0 = 1.33vI 1 + 0.667vI 2
9.42
(a)
vI 1 − v2 vI 2 − v2 v2
+
=
20
40
10
⎛ 100 ⎞
vO = ⎜ 1 +
⎟ v2 = 3v2
50 ⎠
⎝
Now 2vI 1 − 2v2 + vI 2 − v2 = 4v2
⎛v ⎞
2vI 1 + vI 2 = 7v2 = 7 ⎜ o ⎟
⎝3⎠
6
3
So vO = ⋅ vI 1 + ⋅ vI 2
7
7
(b)
(c)
6
3
( 0.2 ) + ⎜⎛ ⎟⎞ ( 0.3) ⇒ vO = 0.3 V
7
7
⎝ ⎠
⎛6⎞
⎛ 3⎞
vO = ⎜ ⎟ ( 0.25 ) + ⎜ ⎟ ( −0.4 ) ⇒ vO = 42.86 mV
⎝7⎠
⎝7⎠
vO =
9.43
⎛ R4 ⎞
v2 = ⎜
⎟ vI
⎝ R3 + R4 ⎠
⎛ R ⎞
⎛ R ⎞ ⎛ R4 ⎞
vO = ⎜1 + 2 ⎟ v2 = ⎜ 1 + 2 ⎟ ⎜
⎟ vI
R1 ⎠
R1 ⎠⎝ R3 + R4 ⎠
⎝
⎝
Av =
vO ⎛ R2 ⎞ ⎛ R4 ⎞
= ⎜1 + ⎟ ⎜
⎟
vI ⎝
R1 ⎠ ⎝ R3 + R4 ⎠
9.44
(a)
vO ⎛
50 x ⎞
= ⎜⎜ 1 +
⎟
vI ⎝ (1 − x ) 50 ⎟⎠
vO ⎛
x ⎞ 1− x + x
= ⎜1 +
⎟=
vI ⎝ 1 − x ⎠
1− x
v
1
Av = O =
vI 1 − x
(b)
1 ≤ Av ≤ ∞
(c)
If x = 1, gain goes to infinity.
9.45
Change resister values as shown.
i1 =
vI
= i2
R
⎛v ⎞
vx = i2 2 R + vI = ⎜ I ⎟ 2 R + vI = 3vI
⎝R⎠
v x 3I
i3 = =
R R
4v
v 3v
i4 = i2 + i3 = I + I = I
R
R
R
⎛ 4vI ⎞
v0 = i4 2 R + vx = ⎜
⎟ 2 R + 3vI
⎝ R ⎠
v0
= 11
vI
9.46
(a)
(b)
vO
=1
vI
From Exercise TYU9.7
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
⎝
vO
=
vI ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
But R2 = 0, R1 = ∞
vO
v
1
1
=
=
⇒ O = 0.999993
1
1
vI 1 +
vI
1+
Aod
1.5 × 105
(b)
9.47
Want
vO
1
= 0.990 =
⇒ Aod = 99
1
vI
1+
Aod
v0 = Aod ( vI − v0 )
⎛ 1
⎞
+ 1⎟ v0 = vI
⎜
⎝ Aod
⎠
v0
1
=
vI ⎛
1 ⎞
⎜1 +
⎟
A
od ⎠
⎝
v
Aod = 104 ; 0 = 0.99990
vI
Aod = 103 ;
v0
= 0.9990
vI
Aod = 102 ;
v0
= 0.990
vI
Aod = 10;
v0
= 0.909
vI
9.48
⎛ R ⎞
v0 A = ⎜ 1 + 2 ⎟ vI
R1 ⎠
⎝
⎛ R ⎞
⎛ R ⎞
v01 = ⎜1 + 2 ⎟ vI , v02 = − ⎜ 1 + 2 ⎟ vI
R1 ⎠
R1 ⎠
⎝
⎝
So v01 = −v02
9.49
(a)
iL =
vI
R1
(b)
vO1 = iL RL + vI = iL RL + iL R1
vOI ( max ) ≅ 10 V = iL (1 + 9 ) = 10iL
So iL ( max ) ≅ 1 mA
Then vI ( max ) ≅ iL R1 = (1)( 9 ) ⇒ vI ( max ) ≅ 9 V
9.50
(a)
⎛ 20 ⎞
⎛ 20 ⎞
vX = ⎜
⎟ ⋅ vI = ⎜ ⎟ ( 6 ) = 2
⎝ 20 + 40 ⎠
⎝ 60 ⎠
vO = 2 V
(b)
(c)
Same as (a)
⎛ 6 ⎞
vX = ⎜
⎟ ( 6 ) = 0.666 V
⎝ 6 + 48 ⎠
⎛ 10 ⎞
vO = ⎜ 1 + ⎟ ⋅ v X ⇒ vO = 1.33 V
⎝ 10 ⎠
9.51
a.
Rin =
v −v
v1
and 1 0 = i1 and v0 = − Aod v1
i1
RF
So i1 =
v1 − ( − Aod v1 )
Then Rin =
RF
=
v1 (1 + Aod )
RF
v1
RF
=
i1 1 + Aod
b.
⎛ RS ⎞
RF
⋅ i1
i1 = ⎜
⎟ iS and v0 = − Aod ⋅
1 + Aod
⎝ RS + Rin ⎠
⎛ A ⎞⎛ RS ⎞
So v0 = − RF ⎜ od ⎟⎜
⎟ iS
⎝ 1 + Aod ⎠⎝ RS + Rin ⎠
Rin =
RF
10
=
= 0.009990
1 + Aod 1001
⎞
RS
⎛ 1000 ⎞ ⎛
v0 = − RF ⎜
⎟ iS
⎟⎜
⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠
⎞
RS
⎛ 1000 ⎞ ⎛
Want ⎜
⎟ ≤ 0.990
⎟⎜
⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠
which yields RS ≥ 1.099 kΩ
9.52
vO = iC RF , 0 ≤ iC ≤ 8 mA
For vO ( max ) = 8 V, Then RF = 1 k Ω
9.53
v
10
i = I so 1 =
⇒ R = 10 kΩ
R
R
In the ideal op-amp, R1 has no influence.
⎛ R ⎞
Output voltage: v0 = ⎜1 + 2 ⎟ vI
R⎠
⎝
v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input
voltage vI in which the output is valid.
9.54
(a)
iL =
− vI
R2
10mA =
− ( −10V )
R2
R2 = 1 K
R
1
= F
R2 R1 R3
Also
vL = (10mA )( 0.05k ) = 0.5 V
0.5
= 0.5 mA
i2 =
1
iR 3 = 10 + 0.5 = 10.5 mA
v −v
13 − 0.5
R3 = 1.19 K
Limit vo to 13V ⇒ R3 = O L =
iR 3
10.5
Then
RF R3 1.19
R
=
=
= 1.19 = F
1
R1 R2
R1
For example, RF = 119 K, R1 = 100 K
(b)
From part (a), vO = 13 V when vI = −10 V
9.55
(a)
i1 = i2 and i2 =
vx
+ iD , vx = −i2 RF
R2
⎛R
Then i1 = −i1 ⎜ F
⎝ R2
⎛
R
Or iD = i1 ⎜1 + F
R2
⎝
(b)
R1 =
⎞
⎟ + iD
⎠
⎞
⎟
⎠
vI 5
= ⇒ R1 = 5 k Ω
i1 1
⎛
R ⎞
R
12 = (1) ⎜ 1 + F ⎟ ⇒ F = 11
R2 ⎠
R2
⎝
For example, R2 = 5 k Ω, RF = 55 k Ω
9.56
VX VX − vO
+
R2
R3
(1)
IX =
(2)
VX VX − vO
+
=0
R1
RF
⎛ R ⎞
From (2) vO = VX ⎜1 + F ⎟
R1 ⎠
⎝
⎛ 1
⎛ R ⎞
1 ⎞ 1
Then (1) I X = VX ⎜ + ⎟ − ⋅ VX ⎜1 + F ⎟
R1 ⎠
⎝
⎝ R2 R3 ⎠ R3
IX
R
R
1
1
1
1
1
=
=
+
− − F =
− F
VX R0 R2 R3 R3 R1 R3 R2 R1 R3
=
R1 R3 − R2 RF
R1 R2 R3
or Ro =
Note: If
R1 R2 R3
R1 R3 − R2 RF
RF
1
=
⇒ R2 RF = R1 R3 then Ro = ∞, which corresponds to an ideal current source.
R1 R3 R2
9.57
R2 R4
=
=5
R1 R3
Minimum resistance seen by vI1 is R1.
Set R1 = R3 = 25 kΩ Then R2 = R4 = 125 kΩ
Ad =
iL =
v0
⇒ v0 = iL RL = ( 0.5 )( 5 ) = 2.5 V
RL
v0 = 5 ( vI 2 − vI 1 )
2.5 = 5 ( vI 2 − 2 ) ⇒ vI 2 = 2.5 V
9.58
vO =
R2
( vI 2 − vI 1 )
R1
R2
R
R
and 2 = 4 with R2 = R4 and R1 = R3
R1
R1 R3
Differential input resistance
R 20
Ri = 2 R1 ⇒ R1 = i =
= 10 K
2
2
R2
(a)
= 50 ⇒ R2 = R4 = 500 K
R1
Ad =
R1 = R3 = 10 K
(b)
R2
= 20 ⇒ R2 = R4 = 200 K
R1
R1 = R3 = 10 K
(c)
R2
= 2 ⇒ R2 = R4 = 20 K
R1
R1 = R3 = 10 K
(d)
R2
= 0.5 ⇒ R2 = R4 = 5 K
R1
R1 = R3 = 10 K
9.59
We have
⎞
⎛ R ⎞⎛ R / R ⎞
⎛R ⎞
⎛ R ⎞⎛
⎛ R2 ⎞
1
vO = ⎜ 1 + 2 ⎟ ⎜ 4 3 ⎟ vI 2 − ⎜ 2 ⎟ vI 1 or vO = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 2 − ⎜ ⎟ vI 1
R1 ⎠ ⎝ 1 + R4 / R3 ⎠
R1 ⎠ ⎝ 1 + R3 / R4 ⎠
⎝
⎝ R1 ⎠
⎝
⎝ R1 ⎠
Set R2 = 50 (1 + x ) , R1 = 50 (1 − x )
R3 = 50 (1 − x ) , R4 = 50 (1 + x )
⎡
⎤
⎢
⎥
⎡ ⎛ 1 + x ⎞⎤
1+ x ⎞
1
⎥ vI 2 − ⎛⎜
vO = ⎢1 + ⎜
⎟⎥ ⎢
⎟ vI 1
−
x
1
−
1− x ⎠
x
1
⎛
⎞
⎢
⎥
⎝
⎠
⎝
⎣
⎦ 1+
⎢ ⎜⎝ 1 + x ⎟⎠ ⎥
⎣
⎦
⎤
⎡1 − x + (1 + x ) ⎤ ⎡
1+ x
⎛ 1+ x ⎞
vO = ⎢
⎥ vI 2 − ⎜
⎥⋅⎢
⎟ vI 1
1− x
⎝ 1− x ⎠
⎣
⎦ ⎢⎣1 + x + (1 − x ) ⎥⎦
⎛ 1+ x ⎞
⎛1+ x ⎞
=⎜
⎟ vI 2 − ⎜
⎟ vI 1
⎝1− x ⎠
⎝ 1− x ⎠
For vI 1 = vI 2 ⇒ vO = 0
Set
R2 = 50 (1 + x )
R1 = 50 (1 − x )
R3 = 50 (1 + x ) R4 = 50 (1 − x )
⎛
⎞
1 ⎟
⎛ 1+ x ⎞⎜
⎛ 1+ x ⎞
vO = ⎜1 +
⎟ ⎜ 1 + x ⎟ vI 2 − ⎜
⎟ vI 1
⎝ 1− x ⎠⎜1+
⎝ 1− x ⎠
⎜
⎟⎟
⎝ 1− x ⎠
⎛ 1+ x ⎞
= vI 2 − ⎜
⎟ vI 1
⎝ 1− x ⎠
vI 1 = vI 2 = vcm
vO
1 + x 1 − x − (1 + x ) −2 x
= 1−
=
=
1− x
1− x
1− x
vcm
Set
R2 = 50 (1 − x )
R1 = 50 (1 + x )
R3 = 50 (1 − x ) R4 = 50 (1 + x )
⎛
⎞
1 ⎟
⎛ 1− x ⎞⎜
⎛ 1− x ⎞
vO = ⎜ 1 +
⎟ ⎜ 1 − x ⎟ vI 2 − ⎜
⎟ vI 1
⎝ 1+ x ⎠⎜ 1+
⎝ 1+ x ⎠
⎜
⎟⎟
⎝ 1+ x ⎠
⎛ 1− x ⎞
= ⎜1 −
⎟ vcm
⎝ 1+ x ⎠
1 + x − (1 − x )
2x
1+ x
1+ x
Worst common-mode gain
Acm =
Acm =
(b)
−2 x
1− x
=
−2 x −2 ( 0.01)
=
= −0.0202
1 − x 1 − 0.01
−2 ( 0.02 )
For x = 0.02, Acm =
= −0.04082
1 − 0.02
−2 ( 0.05 )
For x = 0.05, Acm =
= −0.1053
1 − 0.05
1
1
For this condition, set vI 2 = + , vI 1 = − ⇒ vd = 1 V
2
2
1
1 ⎡ ⎛ 1 + x ⎞ ⎤ 1 ⎡1 − x + (1 + x ) ⎤ 1 2
=
Ad = ⎢1 + ⎜
⎥= ⋅
⎟⎥ = ⎢
2 ⎣ ⎝ 1 − x ⎠⎦ 2 ⎣
1− x
⎦ 2 1− x 1− x
For x = 0.01,
Acm =
For x = 0.01 Ad = 1.010
C M R RdB = 20 log10
1.010
= 33.98 dB
0.0202
C M R RdB = 20 log10
1.020
= 27.96 dB
0.04082
For x = 0.02, Ad =
1
= 1.020
0.98
For x = 0.05 Ad =
1
1.0526
= 1.0526 C M R RdB = 20 log10
≅ 20 dB
0.95
0.1053
9.60
⎛ 10R ⎞
⎛ 10 ⎞
vy = ⎜
⎟ v2 = ⎜ ⎟ ( 2.65 ) ⇒ v y = vx = 2.40909 V
10R+R
⎝
⎠
⎝ 11 ⎠
v2 − v y 2.65 − 2.40909
=
= 0.0120 mA
i3 = i4 =
R
20
v −v
2.50 − 2.40909
= 0.0045455 mA
i1 = i2 = 1 x =
R
20
vO = vx − i2 (10R ) = ( 2.40909 ) − ( 0.0045455 )( 200 )
vO = 1.50 V
9.61
iE = (1 + β )( iB ) = ( 81)( 2 ) = 162 mA =
10
R
R = 61.73 Ω
9.62
a.
From superposition:
R2
v01 = − ⋅ vI 1
R1
⎛ R ⎞ ⎛ R1 ⎞
v02 = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 2
R1 ⎠ ⎝ R3 + R4 ⎠
⎝
Setting vI 1 = vI 2 = vcm
⎡
⎛
⎢⎛ R ⎞ ⎜ 1
v0 = v01 + v02 = ⎢⎜1 + 2 ⎟ ⎜
R3
R1 ⎠ ⎜
⎢⎝
⎜ 1+ R
⎢
4
⎝
⎣
⎤
⎞
⎟ R ⎥
⎟ − 2 ⎥ vcm
⎟ R1 ⎥
⎟
⎥
⎠
⎦
⎛
v
R ⎛ R ⎞⎜ 1
Acm = 0 = 4 ⋅ ⎜1 + 2 ⎟ ⎜
R4
vcm R3 ⎝
R1 ⎠ ⎜
⎜ 1+ R
3
⎝
R4 ⎛ R2 ⎞ R2 ⎛ R4 ⎞
⎜1 + ⎟ − ⎜1 + ⎟
R
R1 ⎠ R1 ⎝ R3 ⎠
= 3⎝
⎛ R4 ⎞
⎜1 + ⎟
⎝ R3 ⎠
⎞
⎟ R
⎟− 2
⎟ R1
⎟
⎠
R4 R2
−
R
R1
Acm = 3
⎛ R4 ⎞
⎜1 + ⎟
⎝ R3 ⎠
b.
Max. Acm
Max. Acm ⇒ Min.
R4
R
and Max. 2
R3
R1
47.5 52.5
−
10.5
9.5 = 4.5238 − 5.5263 ⇒ A
=
cm
47.5
1 + 4.5238
1+
10.5
9.63
vI 1 − v A v A − vB vA − v0
=
+
R1 + R2
Rv
R2
(1)
vI 2 − vB vB − v A vB
=
+
R1 + R2
Rv
R2
(2)
⎛ R1 ⎞
⎛ R2 ⎞
v− = ⎜
⎟ vA + ⎜
⎟ vI 1
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
⎛ R1 ⎞
⎛ R2 ⎞
v+ = ⎜
⎟ vB + ⎜
⎟ vI 2
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
(3)
(4)
max
= 0.1815
Now v− = v+ ⇒ R1vA + R2 vI 1 = R1vB + R2 vI 2
R
So that v A = vB + 2 ( vI 2 − vI 1 )
R1
⎛ 1
v
vI 1
1
1 ⎞ v
= vA ⎜
+
+ ⎟− B − 0
R1 + R2
R
R
R
R
R
R
+
V
V
2
2 ⎠
2
⎝ 1
⎛ 1
vI 2
1
1 ⎞ v
= vB ⎜
+
+ ⎟− A
R1 + R2
R
R
R
R
RV
+
V
2
2 ⎠
⎝ 1
(1)
( 2)
Then
⎛ 1
v ⎛ R ⎞⎛ 1
vI 1
1
1 ⎞ v
1
1 ⎞
= vB ⎜
+
+ ⎟ − B − 0 + ⎜ 2 ⎟⎜
+
+ ⎟ ( vI 2 − vI 1 )
R1 + R2
R
R
R
R
R
R
R
R
R
R
+
+
R
V
V
⎝ 1 ⎠⎝ 1
2
2 ⎠
2
2
2 ⎠
V
⎝ 1
⎛ 1
⎤
vI 2
R2
1
1 ⎞ 1 ⎡
= vB ⎜
+
+ ⎟−
⎢ vB + ( vI 2 − vI 1 ) ⎥
R1 + R2
R
R
R
R
R
R
+
2
2 ⎠
1
V
V ⎣
⎦
⎝ 1
Subtract (2) from (1)
⎛ R ⎞⎛ 1
v
1
1
1 ⎞
1 R2
+
+ ⎟ ( vI 2 − vI 1 ) − 0 +
⋅ ( vI 2 − vI 1 )
( vI 1 − vI 2 ) = ⎜ 2 ⎟ ⎜
R1 + R2
R
R
R
R
R
R
R
R1
+
2
2 ⎠
2
V
V
⎝ 1 ⎠⎝ 1
⎧⎪⎛ R ⎞ ⎛ 1
v0
1
1 ⎞
1
1 R2 ⎫⎪
= ( vI 2 − vI 1 ) ⎨⎜ 2 ⎟ ⎜
+
+ ⎟+
+
⋅ ⎬
R2
⎩⎪⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R1 + R2 RV R1 ⎭⎪
⎛ R ⎞ ⎧ R2
R
R1
R ⎫
v0 = ( vI 2 − vI 1 ) ⎜ 2 ⎟ ⎨
+ 2 +1+
+ 2⎬
R1 + R2 RV ⎭
⎝ R1 ⎠ ⎩ R1 + R2 RV
v0 =
9.64
R2 ⎞
2 R2 ⎛
⎜1 +
⎟ ( vI 2 − vI 1 )
R1 ⎝ RV ⎠
(1)
(2)
i1 =
vI 1 − vI 2 ( 0.50 − 0.030sin ω t ) − ( 0.50 + 0.030sin ω t )
=
20
R1
−0.060sin ω t
20
i1 = −3sin ω t ( µ A )
vO1 = i1 R2 + vI 1 = ( −0.0030sin ω t )(115 ) + 0.50 − 0.030sin ω t
vO1 = 0.50 − 0.375sin ω t
=
vO 2 = vI 2 − i1 R2 = 0.50 + 0.030sin ω t − ( −0.003sin ω t )(115 )
vO 2 = 0.50 + 0.375sin ω t
R4
200
⎡ 0.50 + 0.375sin ω t − ( 0.50 − 0.375sin ω t ) ⎤⎦
( vO 2 − vO1 ) =
R3
50 ⎣
vO =
vO = 3sin ω t ( V )
i3 =
vO 2
0.50 + 0.375sin ω t
=
50 + 200
R3 + R4
i3 = 2 + 1.5sin ω t ( µ A )
i2 =
vO1 − vO ( 0.5 − 0.375sin ω t ) − ( 3sin ω t )
=
250
R3 + R4
i2 = 2 − 13.5sin ω t ( µ A )
9.65
⎛ 40 ⎞
vOB = ⎜1 + ⎟ vI = 2.1667 sin ω t
⎝ 12 ⎠
30
vOC = − vI = −1.25sin ω t
12
(a)
(b)
(c)
vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t )
vO = 3.417 sin ω t
vO 3.417
=
= 6.83
0.5
vI
(d)
9.66
iO =
vI
R
9.67
vO
R ⎛ 2R ⎞
= 4 ⎜1 + 2 ⎟
vI 2 − vI 1 R3 ⎝
R1 ⎠
200 ⎛ 2 (115 ) ⎞
vO =
⎜1 +
⎟ ( 0.06sin ω t )
R1 ⎠
50 ⎝
230
For vO = 0.5
= 1.0833 ⇒ R1 = 212.3 K
R1
Ad =
vO = 8 V
9.68
230
= 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K
R1
⎛ 2 R2 ⎞
⎜1 +
⎟ ( vI 2 − vI 1 )
R1 ⎠
⎝
Set R2 = 15 K, Set R1 = 2 K + 100 k ( Rot )
R4
R3
vO =
Want
R4
≈8
R3
Set R3 = 10 K
R 4 = 75 K
Now
Gain (min) =
Gain ( max ) =
75 ⎛ 2 (15 ) ⎞
⎜1 +
⎟ = 9.71
10 ⎝
102 ⎠
75 ⎛ 2 (15 ) ⎞
⎜1 +
⎟ = 120
10 ⎝
2 ⎠
9.69
For a common-mode gain, vcm = vI 1 = vI 2
Then
⎛ R ⎞
R
v01 = ⎜1 + 2 ⎟ vcm − 2 vcm = vcm
R1 ⎠
R1
⎝
⎛ R ⎞
R
v02 = ⎜1 + 2 ⎟ vcm − 2 vcm = vcm
R
R1
1 ⎠
⎝
From Problem 9.62 we can write
R4 R4
−
R3 R3′
Acm =
⎛
R4 ⎞
⎜1 + ⎟
R3 ⎠
⎝
R3 = R4 = 20 kΩ, R3′ = 20 kΩ ± 5%
20
1−
R3′ 1 ⎛ 20 ⎞
Acm =
= 1−
1
+
( 1) 2 ⎝⎜ R3′ ⎠⎟
For R3′ = 20 kΩ − 5% = 19 kΩ
1 ⎛ 20 ⎞
⎜ 1 − ⎟ = −0.0263
2 ⎝ 19 ⎠
For R3′ = 20 kΩ + 5% = 21 kΩ
Acm =
1 ⎛ 20 ⎞
⎜ 1 − ⎟ = 0.0238
2⎝
21 ⎠
So Acm max = 0.0263
Acm =
9.70
a.
v0 =
1
⋅ vI ( t ′ ) dt ′
R1C2 ∫
∫ 0.5sin ω t dt = − ω
0.5
v0 = 0.5 =
f =
cos ω t
1 ( 0.5 )
0.5
⋅
=
R1C2 ω
2π R1C2 f
1
1
=
⇒ f = 31.8 Hz
2π R1C2 2π ( 50 × 103 )( 0.1× 10−6 )
Output signal lags input signal by 90°
b.
i.
f =
ii.
f =
0.5
2π ( 50 × 103 )( 0.1× 10−6 )
⇒ f = 15.9 Hz
0.5
( 0.1)( 2π ) ( 50 ×103 )( 0.1×10−6 )
9.71
−v ⋅ t
1
vI ( t ) dt = I
∫
RC
RC
vI = −0.2
vO = −
Now
8=
− ( −0.2 )( 2 )
(a)
RC
RC = 0.05 s
(b)
14 =
( 0.2 ) t
0.05
⇒ t = 3.5 s
9.72
a.
v0
=
vI
− R2
1
jω C2
R1
R2 ⋅
=−
1
jω C2
⎛
1 ⎞
R1 ⎜ R2 +
⎟
jω C2 ⎠
⎝
v0
R
1
=− 2⋅
vI
R1 1 + jω R2 C2
b.
v0
R
=− 2
vI
R1
c.
f =
1
2π R2 C2
9.73
a.
R ( jω C1 )
v0
− R2
=
=− 2
vI R + 1
1 + jω R1C1
1
jω C1
v0
R
jω R1C1
=− 2⋅
vI
R1 1 + jω R1C1
b.
v0
R
=− 2
vI
R1
c.
f =
1
2π R1C1
9.74
Assuming the Zener diode is in breakdown,
⇒ f = 159 Hz
vO = −
i2 =
R2
1
⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V
R1
1
0 − vO 0 − ( −6.8 )
=
⇒ i2 = 6.8 mA
R2
1
10 − Vz
10 − 6.8
− i2 =
− 6.8 ⇒ iz = −6.2 mA!!!
Rs
5.6
Circuit is not in breakdown. Now
10 − 0
10
= i2 =
⇒ i2 = 1.52 mA
5.6 + 1
Rs + R1
iz =
vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V
iz = 0
9.75
⎡
⎤
⎛ v ⎞
v
⎛ v ⎞
vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −I 10 ⎟
I
R
10 ⎠
10
10
⎢
⎥
⎝
)( ) ⎦
⎝ s 1⎠
⎣(
For vI = 20 mV , vO = 0.497 V
For vI = 2 V , vO = 0.617 V
9.76
⎛ 333 ⎞
v0 = ⎜
⎟ ( v01 − v02 ) = 16.65 ( v01 − v02 )
⎝ 20 ⎠
⎛i ⎞
v01 = −vBE1 = −VT ln ⎜ C1 ⎟
⎝ IS ⎠
⎛i ⎞
v02 = −vBE 2 = −VT ln ⎜ C 2 ⎟
⎝ IS ⎠
⎛i ⎞
⎛i ⎞
v01 − v02 = −VT ln ⎜ C1 ⎟ = VT ln ⎜ C 2 ⎟
⎝ iC 2 ⎠
⎝ iC1 ⎠
v
v
iC 2 = 2 , iC1 = 1
R2
R1
⎛v R ⎞
So v01 − v02 = VT ln ⎜ 2 ⋅ 1 ⎟
⎝ R2 v1 ⎠
Then
⎛v R ⎞
v0 = (16.65 )( 0.026 ) ln ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
⎛v R ⎞
v0 = 0.4329 ln ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
ln ( x ) = log e ( x ) = ⎣⎡log10 ( x ) ⎦⎤ ⋅ ⎣⎡log e (10 ) ⎦⎤
= 2.3026 log10 ( x )
⎛v R ⎞
Then v0 ≅ (1.0 ) log10 ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
9.77
vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT
(
vO = (10
−10
)
)e
vI / 0.026
For vI = 0.30 V ,
vo = 1.03 × 10−5 V
For vI = 0.60 V ,
vo = 1.05 V
Chapter 10
Exercise Solutions
EX10.1
V + − VBE ( on ) 10 − 0.7
=
I REF =
15
R1
I REF = 0.62 mA
I REF
0.62
=
2
2
1+
1+
β
75
I 0 = 0.604 mA
I0 =
EX10.2
V + − VBE ( on ) − V − 5 − 0.7 − ( −5 )
I REF =
=
12
R1
I REF = 0.775 mA
I
0.775
I 0 = REF =
= 0.7549 mA
2
2
1+
1+
75
β
ΔI 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and ΔI 0 =
r0 =
ΔV
1
ΔVCE 2 ⇒ r0 = CE 2
ΔI 0
r0
V
4
= 265 kΩ = A ⇒ VA = ( 265 )( 0.7549 ) ⇒ VA ≅ 200 V
0.0151
I0
EX10.3
V + − 2VBE ( on ) 9 − 2 ( 0.7 )
I REF =
=
R1
12
I REF = 0.6333 mA
I0 =
I REF
0.6333
=
= 0.6331 mA
2
2
1+
1+
β (1 + β )
75 ( 76 )
I 0 = 0.6331 mA = I C1
I B1 = I B 2 =
I0
β
⇒ I B1 = I B 2 = 8.44 µ A
I E 3 = I B1 + I B 2 ⇒ I E 3 = 16.88 µ A
I B3 =
I E3
⇒ I B 3 = 0.222 µ A
1+ β
EX10.4
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⎛ I ⎞ 0.026 ⎛ 0.75 ⎞
V
ln ⎜
RE = T ln ⎜ REF ⎟ =
⎟ ⇒ RE = 3.54 kΩ
I 0 ⎝ I 0 ⎠ 0.025 ⎝ 0.025 ⎠
5 − 0.7
R1 =
⇒ R1 = 5.73 kΩ
0.75
VBE1 − VBE 2 = I 0 RE = ( 0.025 )( 3.54 ) ⇒ VBE1 − VBE 2 = 88.5 mV
EX10.5
⇒ I REF = 0.775 mA
12
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⎛ 0.775 ⎞
I 0 ( 6 ) = ( 0.026 ) ln ⎜
⎟ ⇒ I 0 ≅ 16.6 µ A
⎝ I0 ⎠
I REF =
5 − 0.7 − ( −5 )
EX10.6
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
0.026 ⎛ 0.70 ⎞
RE =
ln ⎜
⎟ ⇒ RE = 3.465 kΩ
0.025 ⎝ 0.025 ⎠
I
0.025
gm2 = 0 =
⇒ g m 2 = 0.9615 mA/V
VT 0.026
β VT
rπ 2 =
r02 =
I0
=
(150 )( 0.026 )
0.025
= 156 kΩ
VA
100
=
= 4000 kΩ
I 0 0.025
RE′ = RE || rπ 2 = 3.47 ||156 = 3.39 kΩ
R0 = r02 (1 + g m 2 RE′ ) = 4000 ⎡⎣1 + ( 0.962 )( 3.39 ) ⎤⎦
R0 = 17.04 MΩ
1
3
dI 0 =
⋅ dVC 2 =
⇒ dI 0 = 0.176 µ A
R0
17, 040
EX10.7
I REF = I R + I BR + I B1 + ... + I BN
I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN =
⎛I ⎞
⎛ N +1⎞
I REF = I 01 + ( N + 1) ⎜ 01 ⎟ = I 01 ⎜ 1 +
⎟
β
β ⎠
⎝ ⎠
⎝
I REF
So I 01 = I 02 = ... = I 0 N =
N +1
1+
β
I 01
1
= 0.90 =
N +1
I REF
1+
50
N +1 1
1+
=
50
0.9
1
⎛
⎞
N +1 = ⎜
− 1⎟ ( 50 )
⎝ 0.9 ⎠
⎛ 1
⎞
N =⎜
− 1⎟ ( 50 ) − 1
0.9
⎝
⎠
N = 4.55 ⇒ N = 4
EX10.8
I 01
β
VDS ( sat ) = 1 V = VGS 2 − VTN = VGS 2 − 2 ⇒ VGS 2 = 3 V
2
2
⎛ µ C ⎞⎛W ⎞
I O = K n 2 (VGS 2 − VTN ) = ⎜ n ox ⎟ ⎜ ⎟ (VGS 2 − VTN )
⎝ 2 ⎠ ⎝ L ⎠2
2
⎛W ⎞
⎛W ⎞
0.20 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 10
L
⎝ ⎠2
⎝ L ⎠2
2
⎛ µ C ⎞⎛W ⎞
I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 1 − VTN )
2
L
⎝
⎠ ⎝ ⎠1
VGS 1 = VGS 2
2
⎛W ⎞
⎛W ⎞
0.5 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 25
⎝ L ⎠1
⎝ L ⎠1
VGS 3 = V + − VGS 1 = 10 − 3 = 7 V
⎛ µ C ⎞⎛W ⎞
2
I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 3 − VTN )
⎝ 2 ⎠ ⎝ L ⎠3
2
⎛W ⎞
⎛W ⎞
0.5 = ( 0.020 ) ⎜ ⎟ ( 7 − 2 ) ⇒ ⎜ ⎟ = 1
L
⎝ ⎠3
⎝ L ⎠3
EX10.9
a.
I REF = K n (VGS − VTN )
2
0.020 = 0.080 (VGS − 1)
2
VGS = 1.5 V all transistors
b.
VG 4 = VGS 3 + VGS1 + V − = 1.5 + 1.5 − 5 = −2 V
VS 4 = VG 4 − VGS 4 = −2 − 1.5 = −3.5 V
VD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 − VTN = 1.5 − 1 = 0.5 V
So VD 4 ( min ) = −3.5 + 0.5 ⇒ VD 4 ( min ) = −3.0 V
c.
R0 = r04 + r02 (1 + g m r04 )
1
1
r02 = r04 =
=
= 2500 kΩ
λ I 0 ( 0.02 )( 0.020 )
g m = 2 K n (VGS − VTN ) = 2 ( 0.080 )(1.5 − 1) ⇒ g m = 0.080 mA / V
R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) ⇒ R0 = 505 MΩ
EX10.10
For Q2 : vDS ( min ) = VP = 2 V ⇒ VS ( min ) = vDS ( min ) − 5 = 2 − 5 ⇒ VS ( min ) = −3 V
I 0 = I DSS 2 (1 + λ vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) ⇒ I 0 = 0.65 mA
⎛ v ⎞
I 0 = I DSS1 ⎜1 − GS 1 ⎟
⎝ VP1 ⎠
2
⎛ v ⎞
0.65 = 0.80 ⎜1 − GS1 ⎟
−2 ⎠
⎝
vGS 1
= 0.0986 ⇒ vGS 1 = −0.197 V
−2
vGS 1 = VI − VS − 0.197 = VI − ( −3) ⇒ VI ( min ) = −3.2 V
2
Vgs 2 = 0, Vgs1 = −VX
IX =
VX VX − V1
+
+ g m1VX
r02
r01
V1 V1 − VX
+
= g m1VX
RD
r01
(1)
(2)
⎛ 1
⎞
VX ⎜ + g m1 ⎟
r
⎝ 01
⎠
V1 =
1
1
+
RD r01
⎞
1 ⎛ 1
⎜ + g m1 ⎟
r
r
IX
1
1
1
⎠
=
=
+ + g m1 − 01 ⎝ 01
1
1
VX R0 r02 r01
+
RD r01
1
⎡
⎤
⎢
⎞
r01 ⎥
1 ⎛ 1
⎥
=
+ ⎜ + g m1 ⎟ ⎢1 −
1
1 ⎥
r02 ⎝ r01
⎠⎢
+
⎢⎣ RD r01 ⎥⎦
1
⎛
⎞
⎞ ⎜ RD ⎟
1 ⎛ 1
⎟
=
+ ⎜ + g m1 ⎟ ⎜
r02 ⎝ r01
⎠⎜ 1 + 1 ⎟
⎜R
⎟
⎝ D r01 ⎠
For RD r01 ⇒
⎞
1
1 ⎛ 1
≅
+ ⎜ + g m1 ⎟
R0 r02 ⎝ r01
⎠
For Q1:
2I
g m1 = DSS 1
VP
⎛ VGS 1 ⎞ 2 ( 0.8 ) ⎛ −0.197 ⎞
⎜1 −
⎟=
⎜1 −
⎟
−2 ⎠
2 ⎝
VP ⎠
⎝
g m1 = 0.721 mA/V
1
1
=
= 10.3 kΩ
r0 =
λ I 0 ( 0.15 )( 0.65 )
1
1
1
=
+
+ 0.721 = 0.915 ⇒ R0 = 1.09 kΩ
R0 10.3 10.3
EX10.11
a.
⎛V ⎞
I REF = I S exp ⎜ EB 2 ⎟
⎝ VT ⎠
⎛I ⎞
⎛ 0.5 × 10−3 ⎞
VEB 2 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
⎟ ⇒ VEB 2 = 0.521 V
−12
⎝ 10
⎠
⎝ IS ⎠
5 − 0.521
R1 =
⇒ R1 = 8.96 kΩ
b.
0.5
c.
Combining Equations (10.79), (10.80), and (10.81), we find
⎛ VEC 2 ⎞
⎜1 +
⎟
⎡
VAP ⎠
⎛ V ⎞⎤ ⎛ V ⎞
I S 0 ⎢exp ⎜ I ⎟ ⎥ ⎜1 + CEo ⎟ = I REF × ⎝
⎛ VEB 2 ⎞
⎝ VT ⎠ ⎦ ⎝ VAN ⎠
⎣
⎜1 +
⎟
⎝ VAP ⎠
2.5 ⎞
⎛
1+
⎜
⎟
⎡
⎤
⎛V ⎞ ⎛
2.5 ⎞
⎝ 100 ⎠
−3
=
×
10−12 ⎢exp ⎜ I ⎟ ⎥ ⎜ 1 +
0.5
10
) ⎛ 0.521 ⎞
⎟ (
⎝ VT ⎠ ⎦ ⎝ 100 ⎠
⎣
⎜1 +
⎟
100 ⎠
⎝
⎛V
1.025 ×10 −12 exp ⎜ I
⎝ VT
d.
⎞
⎛ VI
−4
⎟ = 5.098 × 10 exp ⎜
⎠
⎝ VT
⎞
8
⎟ = 4.974 × 10 ⇒ VI = 0.521 V
⎠
⎛ 1 ⎞
1
−⎜ ⎟
−
VT ⎠
−38.46
⎝
0.026
=
=
⇒ Av = −1923
Av =
1
1
1
1
+ 0.01
0.01
+
+
VAN VAP 100 100
EX10.12
I CQ
0.8
gm =
=
= 30.77 mA/V
0.026
VT
r0 = r02 =
VA
80
=
= 100 kΩ
I CQ 0.8
a.
V0 = − g mVπ 1 ( r0 || r02 ) , Vπ 1 = Vi
Av = − g m ( r0 || r02 ) = − ( 30.77 ) [100 ||100] ⇒ Av = −1538
b.
Av = − g m ( r0 || r02 || RL )
Av = −
1540
= −770 − 770 = − ( 30.77 )( 50 || RL ) ⇒ ( 50 || RL ) = 25 ⇒ RL = 50 kΩ
2
EX10.13
(a) Neglecting effect of λ and RL
I O = I REF = K n (VIQ − VTN )
0.40 = 0.25 (VIQ − 1)
2
2
Then VIQ = 2.265 V
r0 = r02 =
b.
1
1
=
= 125 kΩ
λ I 0 ( 0.02 )( 0.4 )
g m = 2 K n (VIQ − VTN ) = 2 ( 0.25 )( 2.26 − 1) = 0.632 mA/V
Av = − g m ( r0 || r02 ) = − ( 0.632 )(125 ||125 ) ⇒ Av = −39.5
Av = − g m ( r0 || r02 || RL ) −
c.
39.4
= − ( 0.632 )( 62.5 || RL ) ⇒ 62.5 || RL = 31.25 ⇒ RL = 62.5 kΩ
2
TYU10.1
For I 0 = 0.75 mA
⎛
2⎞
2 ⎞
⎛
I REF = I 0 ⎜1 + ⎟ = ( 0.75 ) ⎜1 +
⎟
⎝ 100 ⎠
⎝ β⎠
I REF = 0.765 mA
I REF =
R1 =
V + − VBE ( on ) − V −
R1
5 − 0.7 − ( −5 )
0.765
R1 = 12.2 kΩ
TYU10.2
10 − ( 0.7 )( 2 )
I REF =
= 0.717 mA
12
I 0 ≈ I REF = 0.717
V
100
r0 = A =
⇒ r0 = 139.5 kΩ
I 0 0.717
ΔI 0 =
1
4
ΔVCE 2 =
⇒ ΔI 0 = 0.0287 mA
r0
139.5
TYU10.3
I 0 = I REF ⋅
I B3 =
I0
β
1
0.50
=
⇒ I 0 = 0.4996 mA
⎛
⎞
⎛
2
2 ⎞
⎜⎜ 1 + β (1 + β ) ⎟⎟ ⎜⎜ 1 + 50 ( 51) ⎟⎟
⎝
⎠ ⎝
⎠
⇒ I B 3 = 9.99 µ A
⎛1+ β ⎞
I E3 = ⎜
⎟ I C 3 = I E 3 = 0.5096 mA
⎝ β ⎠
IE3
0.5096
IC 2 =
=
⇒ I C 2 = 0.490 mA = I C1
2⎞
⎛
2⎞ ⎛
⎜ 1 + ⎟ ⎜ 1 + 50 ⎟
⎠
⎝ β⎠ ⎝
I B1 = I B 2 =
TYU10.4
IC 2
β
⇒ I B1 = I B 2 = 9.80 µ A
I REF =
k n′1 ⎛ W ⎞
kn′3 ⎛ W ⎞
2
2
⎜ ⎟ (VGS 1 − VTN 1 ) =
⎜ ⎟ ( 5 − VGS 1 − VTN 3 )
2 ⎝ L ⎠1
2 ⎝ L ⎠3
⎡⎛ 40 ⎞ ⎛ 17.3 ⎞ ⎤
⎢⎜ 38 ⎟ ⎜ 2.70 ⎟ ⎥ (VGS1 − 0.98 ) = ( 3.98 − VGS 1 ) ⇒ VGS 1 = 1.814 V
⎠⎦
⎣⎝ ⎠ ⎝
2
⎛ 0.040 ⎞
I REF = ⎜
⎟ (17.3)(1.814 − 0.98 )
⎝ 2 ⎠
I REF = 0.241 mA
1
2
kn′2 ⎛ W ⎞
2
⎜ ⎟ (VGS 1 − VTN 2 )
2 ⎝ L ⎠2
2
⎛ 0.042 ⎞
=⎜
⎟ ( 6.92 )(1.814 − 1.0 )
⎝ 2 ⎠
I O = 0.0963 mA
IO =
TYU10.5
a.
From Equation (10.52),
⎛
3
3 ⎞
⎜ 1−
⎟
12 × 10 + ⎜
12 ⎟ × 1.8
VGS 1 =
( )
⎜
3
3 ⎟
1+
⎜1+
⎟
12
12 ⎠
⎝
⎛ 0.5 ⎞
⎛ 1 − 0.5 ⎞
VGS 1 = ⎜
⎟ (10 ) + ⎜
⎟ × (1.8 )
⎝ 1 + 0.5 ⎠
⎝ 1 + 0.5 ⎠
VGS 1 = 3.93 V also VDS1 = 3.93 V
I REF = (12 )( 0.020 ) [3.93 − 1.8] ⎡⎣1 + ( 0.01)( 3.93) ⎤⎦ ⇒ I REF = 1.13 mA
(W / L )2 (1 + λVDS 2 )
b.
×
I 0 = I REF ×
(W / L )1 (1 + λVDS1 )
2
⎛ 6 ⎞ ⎡1 + ( 0.01)( 2 ) ⎤⎦
⇒ I 0 = 0.555 mA
I 0 = (1.13) × ⎜ ⎟ × ⎣
⎝ 12 ⎠ ⎡⎣1 + ( 0.01)( 3.93) ⎤⎦
c.
For VDS 2 = 6 V ⇒ I o = 0.576 mA
TYU10.6
2
K n1 (VGS 1 − VTN ) = K n3 (VGS 3 − VTN )
2
⎛ 0.10 ⎞
VGS 1 − 2 = ⎜⎜
⎟⎟ (VGS 3 − 2 )
⎝ 0.25 ⎠
VGS 1 − 2 = ( 0.6325 )(VGS 3 − 2 )
VGS 3 = 10 − VGS 1
VGS 1 − 2 = ( 0.6325 )(10 − VGS 1 ) − ( 0.6325 )( 2 )
1.6325VGS1 = 7.06 ⇒ VGS 1 = 4.325 V
I REF = K n1 (VGS 1 − VTN ) = ( 0.25 ) ( 4.325 − 2 ) ⇒ I REF = 1.35 mA
2
2
2
I O = 3K n 2 (VGS 1 − VTN ) = 3 ( 0.25 )( 4.325 − 2 )
I O = 3I REF ⇒ I O = 4.05 mA
TYU10.7
2
I REF = 0.20 = K n1 (VGS 1 − VTN ) = 0.15 (VGS 1 − 1) ⇒ VGS 1 = VGS 2 = 2.15 V
2
2
I O = K n 2 (VGS 2 − VTN ) =
I O = K n 3 (VGS 3 − VTN )
2
0.15
2
( 2.15 − 1) ⇒ I O = 0.10 mA
2
2
0.10 = 0.15 (VGS 3 − 1) ⇒ VGS 3 = 1.82 V
2
TYU10.8
All transistors are identical
⇒ I 0 = I REF = 250 µ A
I REF = K n (VGS − VTN )
2
0.25 = 0.20 (VGS − 1) ⇒ VGS = 2.12 V
2
TYU10.9
For Q1 : iD = I DSS 1 (1 + λ vDS1 )
⎛ v ⎞
For Q2 : iD = I DSS 2 ⎜ 1 − GS 2 ⎟ (1 + λ vDS 2 )
VP ⎠
⎝
vGS 2 = −vDS1 and vDS 2 = VDS − vDS 1
So
2
⎡ −v ⎤
I DSS 1 (1 + λ vDS1 ) = I DSS 2 ⎢1 − DS 1 ⎥ ⎡⎣1 + λ (VDS − vDS1 ) ⎤⎦
VP ⎦
⎣
I DSS 1 = I DSS 2
2
⎡ v ⎤
⎡⎣1 + ( 0.1) vDS1 ⎤⎦ = ⎢1 − DS1 ⎥ ⎡⎣1 + ( 0.1) ( 3) − ( 0.1) vDS 1 ⎤⎦
2 ⎦
⎣
2
2
1 + 0.1vDS 1 = (1 − vDS 1 + 0.25vDS
1 ) (1.3 − 0.1vDS 1 )
3
2
This becomes 0.025vDS
1 − 0.425vDS 1 + 1.5vDS 1 − 0.3 = 0
We find vDS 1 = 0.2127 V, vDS 2 = 2.787 V, vGS 2 = −0.2127 V
iD = I DSS 1 (1 + λ vDS 1 ) = 2 ⎡⎣1 + ( 0.1)( 0.2127 ) ⎤⎦
iD = 2.04 mA
R0 = r02 + r01 (1 + g m 2 r02 )
2 I DSS ⎛ vGS 2 ⎞ 2 ( 2 ) ⎛ −0.2127 ⎞
⎜1 −
⎟=
⎜1 −
⎟
VP ⎠
−VP ⎝
−2 ⎠
2 ⎝
g m = 1.787 mA/V
1
1
r02 = r04 =
=
= 5 kΩ
λ I DSS ( 0.1)( 2 )
gm =
R0 = 5 + 5 ⎡⎣1 + (1.787 )( 5 ) ⎤⎦ ⇒ R0 = 54.7 kΩ
TYU10.10
a.
b.
c.
⎛ 0.1× 10−3 ⎞
⇒ VEB 2 = 0.557 V
VEB 2 = ( 0.026 ) ln ⎜
−14 ⎟
⎝ 5 × 10 ⎠
5 − 0.557
R1 =
⇒ R1 = 44.4 kΩ
0.1
⎛ VEC 2 ⎞
⎜ 1+ V ⎟
⎤
⎞ ⎛ VCE 0 ⎞
AP ⎟
⎜
I
+
=
×
1
⎟ REF
⎟⎥ ⎜
V
V
⎜
EB 2 ⎟
AN ⎠
⎠⎦ ⎝
⎜ 1+ V ⎟
AP ⎠
⎝
2.5 ⎞
⎛
⎜ 1 + 100 ⎟
⎡
⎛ VI ⎞ ⎤ ⎛
2.5 ⎞
−14
−3
5 × 10 ⎢ exp ⎜ ⎟ ⎥ ⎜1 +
⎟ = ( 0.1× 10 ) ⎜ 0.557 ⎟
⎝ VT ⎠ ⎦ ⎝ 100 ⎠
⎣
⎜⎜ 1 +
⎟⎟
100 ⎠
⎝
⎛ ⎞
( 5.125 ×10−14 ) exp ⎜ VVI ⎟ = 1.019 ×10−4
⎝ T⎠
⎡
⎛V
I S 0 ⎢ exp ⎜ I
⎝ VT
⎣
⎛V
exp ⎜ I
⎝ VT
1
−
0.026
Av =
⇒
1
1
+
100 100
d.
⎞
9
⎟ = 1.988 × 10 ⇒ VI = 0.557 V
⎠
Av = −1923
TYU10.11
I REF = K p1 (VSG + VTP )
a.
2
0.25 = 0.20 (VSG − 1) ⇒ VSG = 2.12 V
2
b.
VDSO
From Equation (10.89)
⎡1 + λP (V + − VSG ) ⎤ K (V − V )2
TN
⎣
⎦− n I
= Vo =
I REF ( λn + λP )
λn + λP
5=
1 + ( 0.015 )(10 − 2.12 )
0.030
2
( 0.2 )(VI − 1)
0.25 ( 0.030 )
−
0.15 = 1.12 − 0.8 (VI − 1) ⇒ VI = 2.10 V
2
c.
Av =
−2 K n (VI − VTN )
I REF ( λn + λP )
Av = −
2 ( 0.2 )( 2.10 − 1.0 )
0.25 ( 0.030 )
⇒ Av = −58.7
TYU10.12
(a)
I REF = K p1 (VSG + VTP )
2
80 = 50 (VSG − 1) ⇒ VSG = 2.26 V
2
(b)
VDSo
⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2
TN
⎦− n I
= Vo = ⎣
λn + λ p
I REF ( λn + λ p )
⎡1 + ( 0.015 )(10 − 2.26 ) ⎤⎦ ( 50 )(VI − 1)
−
5= ⎣
0.030
(80 )( 0.030 )
2
20.83 (VI − 1) = 32.2 ⇒ VI = 2.243 V
2
(c)
Av =
TYU10.13
a.
−2 K n (VI − VTN )
I REF ( λn + λ p )
=
−2 ( 50 )( 2.243 − 1)
(80 )( 0.030 )
⇒ Av = −51.8
gm =
r0 =
r02 =
IC 0
0.5
=
⇒ g m = 19.2 mA/V
0.026
VT
VAN 120
=
⇒ r0 = 240 kΩ
I CQ 0.5
VAP 80
=
⇒ r02 = 160 kΩ
I CQ 0.5
Av = − g m ( r0 || r02 || RL ) = − (19.2 ) [ 240 ||160 || 50] ⇒ Av = −631
b.
TYU10.14
1
= 38.46 mA/V
0.026
(100 )( 0.026 )
rπ 1 = rπ 2 =
= 2.6 K
1
80
rO1 = rO 2 =
= 80 K
1
120
rO =
= 120 K
1
1
80 = 0.0257 K
RO1 = 2.6
38.46
For R1 = 9.3 K
I C = 1mA, g m =
( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K
RE′′ = 1 [ 2.6 + 0.924] = 0.779 K
RO 2 = 80 ⎡⎣1 + ( 38.46 )( 0.779 ) ⎤⎦ = 2476.7 K
Av = − g m ( rO || RO 2 ) = − ( 38.46 )(120 || 2476.7 ) = − ( 38.46 )(114.5 )
R ′ = R1
Av = −4404
For RL = 100 K
Av = −38.46 ⎡⎣114.5 100 ⎤⎦ = −2053
For RL = 10 K
Av = −38.46 [114.5 ||10] = −354
TYU10.15
M 1 and M 2 identical ⇒ I o = I REF
a.
I O = K n (VI − VYN )
2
2
0.25 = 0.2 (VI − 1)
VI = 2.12 V
g m = 2 K n (VI − VTN ) = 2 ( 0.2 )( 2.12 − 1) ⇒ g m = 0.447 mA/V
r0 n =
r0 p =
b.
1
λn I 0
=
1
( 0.01)( 0.25 )
⇒ r0 n = 400 kΩ
1
1
=
⇒ r0 p = 200 kΩ
λ p I 0 ( 0.02 )( 0.25 )
Av = − g m ( r0 || r02 || RL )
Av = − ( 0.447 ) [ 400 || 200 ||100] ⇒ Av = −25.5
Chapter 10
Problem Solutions
10.1
a.
I1 = I 2 =
0 − 2Vγ − V −
R1 + R2
2Vγ + I 2 R2 = VBE + I C R3
R2
2Vγ +
( −2Vγ − V − ) = VBE + IC R3
R1 + R2
b.
⎧⎪
⎫⎪
R2 ⎞
− ⎛
⎨2Vγ − ( 2Vγ + V ) ⎜
⎟ − VBE ⎬
⎪⎩
⎪⎭
⎝ R1 + R2 ⎠
Vγ = VBE and R1 = R2
IC =
1
R3
IC =
1 ⎧
1
⎫
−
⎨2Vγ − ( 2Vγ + V ) − VBE ⎬
2
R3 ⎩
⎭
or I C =
c.
−V −
2 R3
I C = 2 mA =
− ( −10 )
2 R3
I1 = I 2 = 2 mA =
10.2
⇒ R3 = 2.5 kΩ
−2 ( 0.7 ) − ( −10 )
R1 + R2
⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ
(a)
⎛I ⎞
VBE1 = VT ln ⎜ C1 ⎟
⎝ IS ⎠
(i)
I REF = I C1 = 10 µ A,
(ii)
I REF = I C1 = 100 µ A,
(iii)
⎛ 10−3 ⎞
I REF = I C1 = 1 mA, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.6585 V
⎝ 10 ⎠
I O = 1 mA
(b)
IO =
⎛ 10 × 10−6 ⎞
VBE1 = ( 0.026 ) ln ⎜
⎟ = 0.5388 V
−14
⎝ 10
⎠
I O = 10 µ A
⎛ 100 × 10−6 ⎞
VBE1 = ( 0.026 ) ln ⎜
⎟ = 0.5987 V
−14
⎝ 10
⎠
I O = 100 µ A
I REF
2
1+
β
(i)
(ii)
IO =
IO =
10
⇒ I O = 9.615 µ A VBE1 = VBE 2
2
1+
50
⎛I ⎞
= VT ln ⎜ O ⎟
⎝ IS ⎠
⎛ 9.615 × 10−6 ⎞
= ( 0.026 ) ln ⎜
⎟
−14
⎝ 10
⎠
= 0.5378 V
⎛ 96.15 × 10−6 ⎞
100
⇒ I O = 96.15 µ A VBE1 = ( 0.026 ) ln ⎜
⎟
−14
2
⎝ 10
⎠
1+
50
= 0.5977 V
IO =
(iii)
⎛ 0.9615 × 10−3 ⎞
⇒ I O = 0.9615 mA VBE1 = ( 0.026 ) ln ⎜
⎟
2
10−14
⎝
⎠
1+
50
= 0.6575 V
1
10.3
I REF =
V + − VBE ( on ) − V −
R1
⇒ 0.250 =
3 − 0.7 − ( −3)
R1
R1 = 21.2 K
I REF
0.250
=
⇒ I C1 = I C 2 = 0.2419 mA
2
2
1+
1+
β
60
= 4.03 µ A
I C1 = I C 2 =
I B1 = I B 2
10.4
I REF =
V + − VBE ( on ) − V −
R1
=
5 − 0.7 − ( −5 )
18.3
I REF = 0.5082 mA
I REF
0.5082
=
⇒ I C1 = I C 2 = 0.4958 mA
2
2
1+
1+
80
β
= ( 6.198 µ A )
I C1 = I C 2 =
I B1 = I B 2
10.5
(a)
I REF =
(b)
R1 =
V + − VBE ( on ) − V −
R1
V + − VBE ( on ) − V −
=
or R1 =
15 − 0.7 − ( −15 )
0.5
0 − 0.7 − ( −15 )
0.5
I REF
Advantage: Requires smaller resistance.
(c)
For part (a):
29.3
= 0.526 mA
I O ( max ) =
( 58.6 )( 0.95 )
I O ( min ) =
29.3
= 0.476 mA
( 58.6 )(1.05 )
I O ( min ) =
14.3
= 0.476 mA
( 28.6 )(1.05 )
⇒ R1 = 28.6 k Ω
ΔI O = 0.526 − 0.476 = 0.05 mA ⇒ ±5%
For part (b):
14.3
I O ( max ) =
= 0.526 mA
( 28.6 )( 0.95)
ΔI O = 0.05 mA ⇒ ±5%
10.6
a.
⇒ R1 = 58.6 k Ω
⎛
2⎞
2 ⎞
⎛
I REF = I 0 ⎜1 + ⎟ = 2 ⎜1 +
⎟ or I REF = 2.04 mA
⎝ 100 ⎠
⎝ β⎠
15 − 0.7
⇒ R1 = 7.01 kΩ
R1 =
2.04
r0 =
b.
VA 80
=
= 40 kΩ
2
I0
ΔI 0
1
⎛ 1 ⎞
= ⇒ ΔI 0 = ⎜ ⎟ ( 9.3) = 0.2325 mA
ΔVCE r0
⎝ 40 ⎠
ΔI 0 0.2325
ΔI
=
⇒ 0 = 11.6%
2
I0
I0
10.7
I 0 = nI C1
I REF = I C1 + I B1 + I B 2 = I C1 +
I C1
β
+
I0
β
⎛
⎛ 1+ n ⎞
1 n⎞
I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 +
⎟
β ⎠
⎝ β β⎠
⎝
=
10.8
IO =
I0 ⎛ 1 + n ⎞
nI REF
⎜1 +
⎟ or I 0 =
n⎝
β ⎠
⎛ 1+ n ⎞
⎜1 +
β ⎟⎠
⎝
I REF
2 ⎞
⎛
⇒ I REF = ( 0.20 ) ⎜ 1 + ⎟ = 0.210 mA
2
40
⎝
⎠
1+
β
5 − 0.7 4.3
=
⇒ R1 = 20.5 K
R1 =
I REF
0.21
10.9
a.
b.
5 − 0.7
= 0.239 mA
18
0.239
⇒ I 0 = 0.230 mA
I0 =
2
1+
50
VA
50
r0 =
=
= 218 kΩ
I 0 0.230
I REF =
1
⎛ 1 ⎞
⋅ ΔVEC = ⎜
⎟ (1.3) = 0.00597 mA ⇒ I 0 = 0.236 mA
r0
⎝ 217 ⎠
⎛ 1 ⎞
ΔI 0 = ⎜
⎟ ( 3.3) = 0.01516 mA ⇒ I 0 = 0.245 mA
⎝ 217 ⎠
ΔI 0 =
c.
10.10
a.
b.
c.
10.11
I REF = 1 =
5 − 0.7 − ( −5 )
R1
I 0 = 2 I REF ⇒ I 0 = 2 mA
⇒ R1 = 9.3 kΩ
For VEC 2 ( min ) = 0.7 ⇒ RC 2 =
5 − 0.7
⇒ RC 2 = 2.15 kΩ
2
I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA
⎛
3⎞
3 ⎞
⎛
I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟
⎝ 60 ⎠
⎝ β⎠
I REF = 0.2625 mA
R1 =
2.5 − 0.7
⇒ R1 = 6.86 K
0.2625
10.12
R1 =
10 − 0.7
= 37.2 K
0.25
10.13
I 2 = 2 I1 and I3 = 3I1
(a) I 2 = 1.0 mA, I 3 = 1.5 mA
(b) I1 = 0.25 mA, I 3 = 0.75 mA
(c) I1 = 0.167 mA, I 2 = 0.333 mA
10.14
a.
I 0 = I C1 and I REF = I C1 + I B 3 = I C1 +
VBE 2 I C1 VBE
=
+
R2
R2
β
I E 3 = I B1 + I B 2 +
I REF = I C1 +
I REF −
2 I C1
β (1 + β )
+
VBE
+
1
( β ) R2
⎛
⎞
VBE
2
= I 0 ⎜⎜1 +
⎟⎟
(1 + β ) R2
⎝ β (1 + β ) ⎠
I REF −
I0 =
IE3
1+ β
VBE
+
1
( β ) R2
⎛
⎞
2
⎜⎜ 1 + β (1 + β ) ⎟⎟
⎝
⎠
⎛
⎞
2
0.7
I REF = ( 0.70 ) ⎜⎜ 1 +
⎟⎟ + ( 81)(10 )
80
81
(
)(
)
⎝
⎠
I REF = 0.700216 + 0.000864
b.
I REF = 0.7011 mA =
10 − 2 ( 0.7 )
R1
⇒ R1 = 12.27 kΩ
10.15
a.
I ES
1+ β
= (1 + N ) I BR
I 0i = I CR and I REF = I CR + I BS = I CR +
I ES = I BR + I B1 + I B 2 + ... + I BN
=
(1 + N ) I CR
β
Then I REF = I CR +
or I 0i =
b.
I REF
⎛
(1 + N ) ⎞
⎜⎜1 +
⎟⎟
⎝ β (1 + β ) ⎠
⎡
⎤
6
I REF = ( 0.5 ) ⎢1 +
⎥ = 0.5012 mA
⎣⎢ ( 50 )( 51) ⎦⎥
R1 =
10.16
(1 + N ) I CR
β (1 + β )
5 − 2 ( 0.7 ) − ( −5 )
0.5012
⇒ R1 = 17.16 kΩ
⎛
⎞
⎡
⎤
2
2
I REF = I 0 ⎜⎜ 1 +
⎟⎟ = ( 0.5 ) ⎢1 +
⎥ ⇒ I REF = 0.5004 mA
⎝ β (1 + β ) ⎠
⎣⎢ ( 50 )( 51) ⎦⎥
5 − 2 ( 0.7 ) − ( −5 )
R1 =
⇒ R1 = 17.19 kΩ
0.5004
10.17
1
⎛
⎞
2
⎜⎜ 1 +
⎟⎟
⎝ β (2 + β ) ⎠
For I 0 = 0.8 mA
I 0 = I REF ⋅
⎛
2 ⎞
I REF = ( 0.8 ) ⎜⎜ 1 +
⎟⎟ ⇒ I REF = 0.8024 mA
⎝ 25 ( 27 ) ⎠
18 − 2 ( 0.7 )
⇒ R1 = 20.69 kΩ
R1 =
0.8024
10.18
The analysis is exactly the same as in the text. We have
1
I 0 = I REF ⋅
⎛
⎞
2
⎜⎜ 1 +
⎟⎟
⎝ β (2 + β ) ⎠
10.19
2
= 0.0267 mA
75
1
= 0.0133 mA
I C1 = 1 mA, I B1 =
75
I E 3 = I B1 + I B 2 = 0.0133 + 0.0267 = 0.04 mA
I 0 = 2 mA, I B 2 =
I E3
0.04
=
= 0.000526 mA
1+ β
76
= I C1 + I B 3 ⇒ I REF = 1.000526 ≈ 1 mA
I B3 =
I REF
R1 =
10 − 2 ( 0.7 )
I REF
=
8.6
⇒ R1 = 8.6 kΩ
1
10.20
(a)
Assuming RO ≈
rO 3 =
RO =
(b)
RO =
VA
V
= A
I O I REF
β ro3
2
100
=
= 400 K
0.25
(100 )( 400 )
2
⇒ RO = 20 MΩ
ΔV
ΔV
5
⇒ ΔI O =
=
20 MΩ 20 MΩ
ΔI O
ΔI O = 0.25 µ A
10.21
I REF =
V + − VBE1 − V − 5 − 0.7
=
9.3
R1
I REF = 0.4624 mA
VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞
ln ⎜
ln ⎜
⎟=
⎟
1.5
RE ⎝ I O ⎠
⎝ IO ⎠
⎛ 0.4624 ⎞
I O = 0.01733ln ⎜
⎟
⎝ IO ⎠
IO =
By trial and error
I O 41.7 µ A
VBE 2 = 0.7 − I O RE
VBE 2 = 0.7 − ( 0.0417 )(1.5 )
VBE 2 = 0.6375 V
10.22
(a)
I REF =
V + − VBE1 − V − 5 − 0.7 − ( −5 )
=
⇒ I REF = 93 µ A
100
R1
⎛I ⎞
⎛ 93 × 10−3 mA ⎞
I O RE = VT ln ⎜ REF ⎟ ⇒ I O (10 ) = 0.026 ln ⎜
⎟
IO
⎝ IO ⎠
⎝
⎠
By trial and error, I O ≅ 6.8 µ A
Ro = ro 2 (1 + g m 2 RE′ )
Now
30
= 4.41 M Ω
6.8
0.0068
= 0.2615 mA / V
gm2 =
0.026
(100 )( 0.026 )
= 382.4 k Ω
rπ 2 =
0.0068
So
RE′ = rπ 2 || RE = 382 || 10 = 9.74 k Ω
Then
Ro = 4.41 ⎡⎣1 + ( 0.262 )( 9.74 ) ⎤⎦ ⇒ Ro = 15.6 M Ω
ro 2 =
VBE1 − VBE 2 = I o RE = ( 0.0068 )(10 ) ⇒ VBE1 − VBE 2 = 0.068 V
(b)
10.23
I
⋅ ΔVC
R0
ΔI 0 =
R0 = r02 (1 + g m 2 RE′ )
V
80
= 11.76 MΩ
r02 = A =
I 0 6.8
gm2 =
rπ 2 =
I 0 0.0068
=
= 0.2615 mA/V
0.026
VT
(80 )( 0.026 )
= 306 kΩ
0.0068
RE′ = RE & rπ 2 = 10 & 306 = 9.68 K
R0 = (11.76 ) ⎡⎣1 + ( 0.2615 )( 9.68 ) ⎤⎦ = 41.54 MΩ
Now
⎛ 1 ⎞
ΔI 0 = ⎜
⎟ ( 5 ) ⇒ ΔI 0 = 0.120 µ A
⎝ 41.54 ⎠
10.24
(a)
I REF =
5 − 0.7 − ( −5 )
R1 = 18.6 K
R1
= 0.50
⎛I ⎞
I O RE = VT ln ⎜ REF ⎟
⎝ IO ⎠
0.026 ⎛ 0.50 ⎞
RE =
ln ⎜
⎟
0.050 ⎝ 0.050 ⎠
RE = 1.20 K
(b)
RO = rc 2 [1 + RE′ g m 2 ]
RE′ = RE rπ 2
rπ 2 =
( 75 )( 0.026 )
= 39 K
0.050
VA 100
ro 2 =
=
⇒ 2 MΩ
I O 0.05
(c)
gm2 =
0.050
= 1.923 mA/V
0.026
RE′ = 1.20 39 = 1.164 K
RO = 2 ⎡⎣1 + (1.164 )(1.923) ⎤⎦ ⇒ RO = ( 6.477 ) MΩ
ΔV
5
ΔI O =
=
= 0.772 µ A
RO 6.477
ΔI O
0.772
× 100% =
× 100 = 1.54%
50
IO
10.25
Let R1 = 5 k Ω, Then
I REF =
12 − 0.7 − ( −12 )
5
⇒ I REF = 4.66 mA
Now
⎛I ⎞
0.026 ⎛ 4.66 ⎞
ln ⎜
I O RE = VT ln ⎜ REF ⎟ ⇒ RE =
⎟ ⇒ RE ≅ 1 k Ω
0.10 ⎝ 0.10 ⎠
⎝ IO ⎠
10.26
⎛I ⎞
VBE = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
−15
0.7 = ( 0.026 ) ln ⎜
⎟ ⇒ I S = 2.03 × 10 A
I
⎝ S ⎠
⎛ 2 × 10−3 ⎞
At 2 mA, VBE = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 2.03 × 10 ⎠
= 0.718 V
15 − 0.718
⇒ R1 = 7.14 kΩ
2
⎛ I ⎞ 0.026 ⎛ 2 ⎞
V
RE = T ln ⎜ REF ⎟ =
⋅ ln ⎜
⎟ ⇒ RE = 1.92 kΩ
I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠
R1 =
10.27
a.
10 − 0.7
= 0.465 mA
20
Let V − = 0
I REF ≈
⎛I ⎞
VBE ≅ VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
−15
0.7 = ( 0.026 ) ln ⎜
⎟ ⇒ I S = 2.03 ×10 A
I
⎝ S ⎠
Then
⎛ 0.465 × 10−3 ⎞
= 0.680 V
VBE ≅ ( 0.026 ) ln ⎜
−15 ⎟
⎝ 2.03 × 10 ⎠
Then
I REF ≅
b.
10 − 0.680
⇒ I REF = 0.466 mA
20
RE =
VT ⎛ I REF
ln ⎜
I0 ⎝ I0
⎞ 0.026
⎛ 0.466 ⎞
⋅ ln ⎜
⎟=
⎟ ⇒ RE = 400Ω
0.10
⎝ 0.10 ⎠
⎠
10.28
I REF ≈
10 − 0.7 − ( −10 )
40
= 0.4825 mA
⎛I ⎞
VBE ≅ VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
−15
0.7 = ( 0.026 ) ln ⎜
⎟ ⇒ I S = 2.03 × 10 A
⎝ IS ⎠
Now
⎛ 0.4825 × 10−3 ⎞
VBE = ( 0.026 ) ln ⎜
= 0.681 V
−15 ⎟
⎝ 2.03 × 10
⎠
VBE1 = 0.681 V
So
⇒ I REF = 0.483 mA
40
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⎛ 0.483 ⎞
I 0 (12 ) = ( 0.026 ) ln ⎜
⎟
⎝ I0 ⎠
I REF ≅
10 − 0.681 − ( −10 )
By trial and error.
⇒ I 0 ≅ 8.7 µ A
VBE 2 = VBE1 − I 0 RE = 0.681 − ( 0.0087 )(12 ) ⇒ VBE 2 = 0.5766 V
10.29
VBE1 + I REF RE1 = VBE 2 + I 0 RE 2
VBE1 − VBE 2 = I 0 RE 2 − I REF RE1
For matched transistors
⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛I ⎞
VBE 2 = VT ln ⎜ 0 ⎟
⎝ IS ⎠
⎛I ⎞
Then VT ln ⎜ REF ⎟ = I 0 RE 2 − I REF RE1
⎝ I0 ⎠
Output resistance looking into the collector of Q2 is increased.
10.30
V + − VBE1 − V − 5 − 0.7 − ( −5 )
=
= 0.3174 mA
R1 + RE1
27.3 + 2
(a)
I REF =
(b)
I O = I REF = 0.3174 mA
Using the same relation as for the widlar current source.
RO = ro 2 ⎡⎣1 + g m 2 ( RE rπ 2 ) ⎤⎦
ro 2 =
rπ 2 =
VA
80
=
= 252 K
I O 0.3174
gm2 =
0.3174
= 12.21 mA/V
0.026
(100 )( 0.026 )
= 8.192 K RE || rπ 2 = 2 || 8.192 = 1.608 K
0.3174
RO = 252 ⎡⎣1 + (12.21)(1.608 ) ⎤⎦ ⇒ RO = 5.2 MΩ
(c)
I O = I REF =
5 − 0.7 − ( −5 )
= 0.3407 mA
27.3
V
80
RO = ro 2 = A =
⇒ RO = 235 K
I O 0.3407
10.31
Assume all transistors are matched.
a.
2VBE1 = VBE 3 + I 0 RE
⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛I ⎞
VBE 3 = VT ln ⎜ 0 ⎟
⎝ IS ⎠
⎛I ⎞
⎛I ⎞
2VT ln ⎜ REF ⎟ − VT ln ⎜ 0 ⎟ = I 0 RE
I
⎝ S ⎠
⎝ IS ⎠
⎡ ⎛I ⎞
⎛I
VT ⎢ ln ⎜ REF ⎟ − ln ⎜ 0
⎢⎣ ⎝ I S ⎠
⎝ IS
2
⎛ I 2 REF ⎞
VT ln ⎜
⎟ = I 0 RE
⎝ I0 I S ⎠
b.
⎞⎤
⎟ ⎥ = I 0 RE
⎠ ⎥⎦
⎛ 0.7 ⎞
−15
VBE = 0.7 V at 1 mA ⇒ 10−3 = I S exp ⎜
⎟ or I S = 2.03 × 10 A
⎝ 0.026 ⎠
⎛ 0.1× 10−3 ⎞
VBE at 0.1 mA ⇒ VBE = ( 0.026 ) ln ⎜
= 0.640 V
−15 ⎟
⎝ 2.03 × 10 ⎠
0.640
Since I 0 = I REF , then VBE = I 0 RE ⇒ RE =
or RE = 6.4 kΩ
0.1
10.32
(a)
I REF =
5 − 0.7 − ( −5 )
R1
= 0.80 mA
R1 = 11.6 K
RE 2 =
RE 3 =
0.026 ⎛ 0.80 ⎞
ln ⎜
⎟ ⇒ RE 2 = 1.44 K
0.050 ⎝ 0.050 ⎠
0.026 ⎛ 0.80 ⎞
ln ⎜
⎟ ⇒ RE 2 = 4.80 K
0.020 ⎝ 0.020 ⎠
(b)
VBE 2 = 0.7 − ( 0.05 )(1.44 ) ⇒ VBE 2 = 0.628 V
VBE 3 = 0.7 − ( 0.02 )( 4.80 ) ⇒ VBE 3 = 0.604 V
10.33
(a)
VBE1 = VBE 2
I REF =
V + − 2VBE1 − V −
R1 + R2
Now
2VBE1 + I REF R2 = VBE 3 + I O RE
or
I O RE = 2VBE1 − VBE 3 + I REF R2
We have
⎛I ⎞
⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟ and VBE 3 = VT ln ⎜ O ⎟
⎝ IS ⎠
⎝ IS ⎠
(b)
Let R1 = R2 and I O = I REF ⇒ VBE1 = VBE 3 ≡ VBE
Then
VBE = I O RE − I REF R2 = I O ( RE − R2 )
so
I REF = I O =
V + − V − − 2 I O ( RE − R2 )
=
2 R2
⎛R ⎞
V −V
− IO ⎜ E ⎟ + IO
2 R2
⎝ R2 ⎠
+
−
Then
IO =
V + −V −
2 Rε
(c)
Want I O = 0.5 mA
So RE =
2 R2 =
5 − ( −5 )
2 ( 0.5 )
⇒ RE = 10 k Ω
5 − 2 ( 0.7 ) − ( −5 )
0.5
Then R1 = R2 = 8.6 k Ω
= 17.2 k Ω
10.34
a.
20 − 0.7 − 0.7
= 1.55 mA
12
I 01 = 2 I REF = 3.1 mA
I 02 = I REF = 1.55 mA
I 03 = 3I REF = 4.65 mA
b.
VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V
I REF =
VEC 2 = 10 − I 02 RC 2 = 10 − (1.55 )( 3) ⇒ VEC 2 = 5.35 V
VEC 3 = 10 − I 03 RC 3 = 10 − ( 4.65 )(1) ⇒ VEC 3 = 5.35 V
10.35
a.
Ist approximation
20 − 1.4
= 2.325 mA
8
⎛ 2.32 ⎞
Now VBE − 0.7 = ( 0.026 ) ln ⎜
⎟ ⇒ VBE = VEB = 0.722 V
⎝ 1 ⎠
Then 2nd approximation
20 − 2 ( 0.722 )
= 2.32 mA
I REF ≅
8
I 01 = 2 I REF = 4.64 mA
I 02 = I REF = 2.32 mA
I 03 = 3I REF = 6.96 mA
b.
At the edge of saturation,
VCE = VBE = 0.722 V
I REF ≅
⇒ RC1 = 2.0 kΩ
4.64
10 − 0.722
⇒ RC 2 = 4.0 kΩ
=
2.32
10 − 0.722
⇒ RC 3 = 1.33 kΩ
=
6.96
RC1 =
RC 2
RC 3
0 − 0.722 − ( −10 )
10.36
I C1 = I C 2 =
10 − 0.7 − 0.7 − ( −10 )
10
= 1.86 mA
I C 3 = I C 4 = 1.86 mA
⎛ 1.86 ⎞
I C 5 ( 0.5 ) = 0.026 ln ⎜
⎟
⎝ IC 5 ⎠
By Trial and error.
⇒ I C 5 = 0.136 mA = I C 6 = I C 7
2 I C 3 ( 0.8 ) + VCE 3 = 10 ⇒ VCE 3 = 10 − 2 (1.86 )( 0.8 )
VCE 3 = 7.02 V
5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) − 10
VCE 5 = 5 + 10 − 0.7 − ( 0.136 )( 0.5 )
VCE 5 = 14.2 V
5 = VEC 7 + I C 7 ( 0.8 )
VEC 7 = 5 − ( 0.136 )( 0.8 )
VEC 7 = 4.89 V
10.37
I C1 = I C 2 =
10 − 0.7 − 0.7 − ( −10 )
10
⇒ I C1 = I C 2 = 1.86 mA
I C 4 = I C 5 = 1.86 mA
⎛I ⎞
⎛ 1.86 ⎞
I C 3 RE1 = VT ln ⎜ C1 ⎟ ⇒ I C 3 ( 0.3) = 0.026 ln ⎜
⎟
⎝ IC 3 ⎠
⎝ IC 3 ⎠
By trial and error I C 3 = 0.195 mA
⎛I ⎞
⎛ 1.86 ⎞
I C 6 RE 2 = VT ln ⎜ C 5 ⎟ ⇒ I C 6 ( 0.5 ) = 0.026 ln ⎜
⎟
⎝ IC 6 ⎠
⎝ IC 6 ⎠
By trial and error I C 6 = 0.136 mA
10.38
10 − 0.7
= 1 mA
6.3 + 3
VBE ( QR ) = 0.7 V as assumed
VRER = I REF ⋅ RER = (1)( 3) = 3 V
I REF =
VRE1 = 3 V ⇒ RE1 =
VRE 2 = 3 V ⇒ RE 2 =
VRE 3 = 3 V ⇒ RE 3 =
VRE1 3
= ⇒ RE1 = 3 kΩ
I 01 1
VRE 2 3
= ⇒ RE 2 = 1.5 kΩ
I 02
2
VRE 3 3
= ⇒ RE 3 = 0.75 kΩ
I 03
4
I 01 = 1 mA
I 02 = 2 mA
I 03 = 4 mA
10.38
VDS 2 ( sat ) = 2 V = VGS 2 − VTN 2 = VGS 2 − 1.5 ⇒ VGS 2 = 3.5 V
2
⎛1
⎞⎛W ⎞
I O = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 2 )
L
2
⎝
⎠ ⎝ ⎠2
2
⎛W ⎞
⎛W ⎞
250 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 3.125
⎝ L ⎠2
⎝ L ⎠2
⎛1
⎞⎛W ⎞
I REF = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 1 )
⎝2
⎠ ⎝ L ⎠1
2
⎛W ⎞
⎛W ⎞
100 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 1.25
L
⎝ ⎠2
⎝ L ⎠1
Now VGS 3 = 10 − VGS 2 = 10 − 3.5 = 6.5 V
2
⎛W ⎞
⎛W ⎞
So 100 = ( 20 ) ⎜ ⎟ ( 6.5 − 1.5 ) ⇒ ⎜ ⎟ = 0.2
⎝ L ⎠3
⎝ L ⎠3
10.39
I REF =
2.5 − VGS ⎛ 0.08 ⎞
2
=⎜
⎟ ( 6 )(VGS − 0.5 )
15
⎝ 2 ⎠
2.5 − VGS = 3.6 (VGS2 − VGS + 0.25 )
3.6VGS2 − 2.6VGS − 1.6 = 0
VGS =
2.6 ± 6.76 + 23.04
2 ( 3.6 )
VGS = 1.12 V (1.1193)
2.5 − 1.1193
⇒ I REF = 92.0 µ A ( 92.05 )
I REF =
15
I o = 92.0 µ A
VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5
VDS 2 ( sat ) = 0.619 V
10.39
a.
From Equation (10.50),
VGS1 = VGS 2
VGS 1 = VGS 2
⎛
⎛
5 ⎞
5 ⎞
⎜
⎟
⎜ 1−
⎟
25 ⎟ 5 + ⎜
25 ⎟ 0.5
=⎜
( )
( )
⎜
⎜
5 ⎟
5 ⎟
1
1
+
+
⎜
⎟
⎜
⎟
25 ⎠
25 ⎠
⎝
⎝
⎛ 0.447 ⎞
⎛ 1 − 0.447 ⎞
=⎜
⎟ (5) + ⎜
⎟ ( 0.5 )
1
0.447
+
⎝
⎠
⎝ 1 + 0.447 ⎠
= 1.74 V
I REF ≅ K n1 (VGS 1 − VTN ) = (18 )( 25 )(1.74 − 0.5 ) ⇒ I REF = 0.692 mA
2
2
2
⎛1
⎞⎛W ⎞
I O = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN ) (1 + λVDS 2 )
2
L
⎝
⎠ ⎝ ⎠2
b.
I 0 = (18 )(15 )(1.74 − 0.5 ) ⎡⎣1 + ( 0.02 )( 2 ) ⎤⎦
= ( 415 )(104 ) ⇒ I 0 = 0.432 mA
2
I 0 = ( 415 ) ⎡⎣1 + ( 0.02 )( 4 ) ⎤⎦ ⇒ I 0 = 0.448 mA
c.
10.40
(a)
2
⎛ 80 ⎞ ⎛ W ⎞
I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VGS − 0.5 )
⎝ 2 ⎠ ⎝ L ⎠1
2.0 − VGS
I REF = 0.050 =
R
Design such that VDS 2 ( sat ) = 0.25 = VGS − 0.5
VGS = 0.75 V
2 − 0.75
⇒ R = 25 K
R
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20
⎝ 2 ⎠ ⎝ L ⎠1
⎝ L ⎠1
So 0.050 =
⎛W ⎞
⎜ ⎟
20
50
⎛W ⎞
⎝ L ⎠1 I REF
=
⇒
=
⇒ ⎜ ⎟ = 40
100 ⎝ L ⎠ 2
IO
⎛W ⎞
⎛W ⎞
⎜ ⎟
⎜ ⎟
⎝ L ⎠2
⎝ L ⎠2
1
1
=
⇒ RO = 667 K
(b)
RO =
λ I O ( 0.015 )( 0.1)
(c)
ΔI O =
ΔV
1
=
⇒ 1.5 µ A
RO 666
ΔI O
⎛ 1.5 ⎞
× 100% = ⎜
⎟ × 100% ⇒ 1.5%
IO
⎝ 100 ⎠
10.41
(a)
2
⎛ 80 ⎞
I REF = 250 = ⎜ ⎟ ( 3)(VGS − 1)
2
⎝ ⎠
VGS = 2.44 V
I O = 250 µ A at VDS 2 = VGS = 2.44 V
1
1
=
= 200 K
RO =
λ I O ( 0.02 )( 0.25 )
ΔI O =
(i)
ΔV 3 − 2.44
=
⇒ 2.8 µ A
RO
200
I O = 252.8 µ A
ΔI O =
(ii)
ΔV 4.5 − 2.44
=
⇒ 10.3 µ A
200
RO
I O = 260.3 µ A
ΔI O =
(iii)
ΔV 6 − 2.44
=
⇒ 17.8 µ A
RO
200
I O = 267.8 µ A
4.5
( 250 ) = 375 µ A at VDS = 2.44 V
3
1
1
=
= 133.3 K
RO =
λ I O ( 0.02 )( 0.375 )
IO =
(b)
ΔI O =
(i)
ΔV 3 − 2.44
=
⇒ 4.20 µ A
RO
133.3
I O = 379.2 µ A
ΔI O =
(ii)
ΔV 4.5 − 2.44
=
⇒ 15.5 µ A
RO
133.3
I O = 390.5 µ A
ΔI O =
(iii)
ΔV 6 − 2.44
=
⇒ 26.7 µ A
133.3
RO
I O = 401.7 µ A
10.41
VSD 2 ( sat ) = 0.25 = VSG + VTP = VSG − 0.4 ⇒ VSG 2 = 0.65 V
k ′p ⎛ W ⎞
2
I O = ⎜ ⎟ (VSG 2 + VTP )
2 ⎝ L ⎠2
40 ⎛ W ⎞
2
⎛W ⎞
25 =
⎜ ⎟ ( 0.65 − 0.4 ) ⇒ ⎜ ⎟ = 20
2 ⎝ L ⎠2
⎝ L ⎠2
I REF = 75 µ A =
(W /L )1
⎛W ⎞
⋅ I O ⇒ ⎜ ⎟ = 60
(W /L )2
⎝ L ⎠1
k ′p ⎛ W ⎞
2
⎜ ⎟ (VSG 3 + VTP )
2 ⎝ L ⎠3
= 3 − 0.65 = 2.35 V
I REF =
VSG 3
Then 75 =
40 ⎛ W ⎞
2
⎛W ⎞
⎜ ⎟ ( 2.35 − 0.4 ) ⇒ ⎜ ⎟ = 0.986
2 ⎝ L ⎠3
⎝ L ⎠3
10.42
(a)
VGS = VTN 1 +
I REF
0.5
= 1+
=2V
0.5
K n1
⎛ I
I O = K n 2 ⎜ REF
⎜ K
n1
⎝
⎞
⎛ I REF ⎞
⎟⎟ = K n 2 ⎜
⎟
⎝ K n1 ⎠
⎠
2
⎛ 0.5 ⎞
I 0 ( max ) = ( 0.5 )(1.05 ) ⎜
⎟ ⇒ I 0 ( max ) = 0.525 mA
⎝ 0.5 ⎠
⎛ 0.5 ⎞
I 0 ( min ) = ( 0.5 )( 0.95 ) ⎜
⎟ ⇒ I 0 ( min ) = 0.475 mA
⎝ 0.5 ⎠
So 0.475 ≤ I 0 ≤ 0.525 mA
(b)
⎡ I
⎤
I O = K n 2 ⎢ REF + VTN 1 − VTN 2 ⎥
⎣⎢ K n1
⎦⎥
2
⎡ 0.5
⎤
I 0 ( min ) = ( 0.5 ) ⎢
+ 1 − 1.05⎥ ⇒ I 0 (min) = 0.451 mA
⎣ 0.5
⎦
2
⎡ 0.5
⎤
I 0 ( max ) = ( 0.5 ) ⎢
+ 1 − 0.95⎥ ⇒ I 0 (max) = 0.551 mA
⎣ 0.5
⎦
So 0.451 ≤ I 0 ≤ 0.551 mA
2
10.43
(1)
(2)
Ix =
Vx − VA
+ g mVgs 2
ro
Ix =
VA
+ g mVgs1
ro
Vgs1 = Vx , Vgs 2 = −VA
So
(1)
Ix =
(2)
Ix =
⎛1
⎞
Vx
− VA ⎜ + g m ⎟
ro
⎝ ro
⎠
VA
+ g mVx ⇒ VA = ro [ I x − g mVx ]
ro
Then
Ix =
⎛1
⎞
Vx
− ro ( I x − g mVx ) ⎜ + g m ⎟
ro
⎝ ro
⎠
Ix =
⎡I
⎤
Vx
g
− ro ⎢ x + g m I x − m ⋅ Vx − g m2 Vx ⎥
ro
ro
⎣ ro
⎦
Ix =
Vx
− I x − g m ro I x + g mVx + g m2 roVx
ro
⎡1
⎤
I x [ 2 + g m ro ] = Vx ⎢ + g m + g m2 ro ⎥
r
⎣ o
⎦
1
Since g m >>
ro
I x [ 2 + g m ro ] ≅ Vx ( g m )(1 + g m ro )
Then
Vx
2 + g m ro
= Ro =
Ix
g m (1 + g m ro )
Usually, g m ro >> 2, so that Ro ≅
1
gm
10.44
VDS 2 (sat) = 2 = VGS 2 − 0.8 ⇒ VGS 2 = 2.8 V
I O = 200 =
60 ⎛ W ⎞
2
⎛W ⎞
⎜ ⎟ ( 2.8 − 0.8 ) ⇒ ⎜ ⎟ = 1.67
2 ⎝ L ⎠2
⎝ L ⎠2
⎛W ⎞
⎛W ⎞
⎜ ⎟
I REF ⎜⎝ L ⎟⎠1
0.4 ⎝ L ⎠1
⎛W ⎞
=
⇒
=
⇒ ⎜ ⎟ = 3.33
0.2 (1.67 ) ⎝ L ⎠1
IO
⎛W ⎞
⎜ ⎟
L
⎝ ⎠2
VGS 3 = 6 − 2.8 = 3.2 V
2
⎛ 60 ⎞ ⎛ W ⎞
⎛W ⎞
I REF = 400 = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 0.8 ) ⇒ ⎜ ⎟ = 2.31
⎝ 2 ⎠ ⎝ L ⎠3
⎝ L ⎠3
10.45
(a)
2
2
⎛ 60 ⎞
⎛ 60 ⎞
I REF = ⎜ ⎟ ( 20 )(VGS 1 − 0.7 ) = ⎜ ⎟ ( 3)(VGS 3 − 0.7 )
⎝ 2 ⎠
⎝ 2 ⎠
VGS1 + VGS 3 = 5
20
(VGS1 − 0.7 ) = 5 − VGS1 − 0.7
3
3.582VGS 1 = 6.107 ⇒ VGS1 = VGS 2 = 1.705 V
2
⎛ 60 ⎞
I O = ⎜ ⎟ (12 )(1.705 − 0.7 ) = 363.6 µ A at VDS 2 = 1.705 V
⎝ 2 ⎠
2
⎛ 60 ⎞
I REF = ⎜ ⎟ ( 20 )(1.705 − 0.7 ) = 606 µ A
⎝ 2 ⎠
(b)
RO =
ΔI O =
1
λ IO
=
1
= 183.4 K
0.015
(
)( 0.3636 )
ΔV 1.5 − 1.705
=
⇒ −1.12 µ A
RO
183.4
I O = 362.5 µ A
(c)
ΔI O =
ΔV 3 − 1.705
=
⇒ 7.06 µ A
RO
183.4
I O = 370.7 µ A
10.46
2
2
⎛ 50 ⎞
⎛ 50 ⎞
I REF = ⎜ ⎟ (15 )(VSG1 − 0.5 ) = ⎜ ⎟ ( 3)(VSG 3 − 0.5 )
⎝ 2⎠
⎝ 2⎠
VSG1 + VSG 3 = 10 ⇒ VSG 3 = 10 − VSG1
15
(VSG1 − 0.5) = 10 − VSG1 − 0.5
3
3.236VSG1 = 10.618 ⇒ VSG1 = 3.28 V
2
⎛ 50 ⎞
I REF = ⎜ ⎟ (15 )( 3.28 − 0.5 ) ⇒ I REF = 2.90 mA
⎝ 2⎠
I O = I REF = 2.90 mA
VSD 2 (sat) = VSG 2 + VTP = 3.28 − 0.5 ⇒ VSD 2 (sat) = 2.78 V
10.47
VSD 2 (sat) = 1.2 = VSG 2 − 0.35 ⇒ VSG 2 = 1.55 V
2
⎛ 50 ⎞⎛ W ⎞
⎛W ⎞
I O = 100 = ⎜ ⎟⎜ ⎟ (1.55 − 0.35 ) ⇒ ⎜ ⎟ = 2.78
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
W
W
I REF
200
⎛W ⎞
L1
L1
=
⇒
=
⇒ ⎜ ⎟ = 5.56
W
IO
100
2.78
⎝ L ⎠1
L 2
VSG1 + VSG 3 = 4 ⇒ VSG 3 = 2.45 V
( )
( )
( )
2
⎛ 50 ⎞⎛ W ⎞
⎛W ⎞
I REF = 200 = ⎜ ⎟⎜ ⎟ ( 2.45 − 0.35 ) ⇒ ⎜ ⎟ = 1.81
⎝ 2 ⎠⎝ L ⎠3
⎝ L ⎠3
10.48
2
2
⎛ 80 ⎞
⎛ 80 ⎞
I REF = ⎜ ⎟ ( 25 )(VSG1 − 1.2 ) = ⎜ ⎟ ( 4 )(VSG 3 − 1.2 )
2
2
⎝ ⎠
⎝ ⎠
10 − VSG1
VSG1 + 2VSG 3 = 10 ⇒ VSG 3 =
2
10 − VSG1
25
− 1.2
Then
(VSG1 − 1.2 ) =
4
2
3VSG1 = 6.8 ⇒ VSG1 = 2.27 V
2
⎛ 80 ⎞
I REF = ⎜ ⎟ ( 25 )( 2.267 − 1.2 ) ⇒ I REF = I O = 1.14 mA
⎝ 2⎠
VSD 2 (sat) = VSG 2 + VTP = 2.27 − 1.2 ⇒ VSD 2 ( sat ) = 1.07 V
10.49
VSD 2 (sat) = 1.8 = VSG 2 − 1.4 ⇒ VSG 2 = 3.2 V
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
I O = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 1.4 ) = 100 ⇒ ⎜ ⎟ = 0.772
⎝ 2 ⎠ ⎝ L ⎠2
⎝ L ⎠2
W
I REF
L1
=
W
IO
L 2
W
200
⎛W ⎞
L1
=
⇒ ⎜ ⎟ = 1.54
100 0.772
⎝ L ⎠1
( )
( )
( )
Assume M3 and M4 are matched.
10 − 3.2
= 3.4 V
2VSG 3 + VSG1 = 10 ⇒ VSG 3 =
2
2
⎛ 80 ⎞ ⎛ W ⎞
I REF = 200 = ⎜ ⎟ ⎜ ⎟ ( 3.4 − 1.4 )
L
2
⎝ ⎠ ⎝ ⎠3,4
⎛W ⎞
⎜ ⎟ = 1.25
⎝ L ⎠3,4
10.50
(a)
⎛ k ′p ⎞ ⎛ W ⎞
2
I REF = ⎜ ⎟ ⎜ ⎟ (VSG1 + VTP )
⎝ 2 ⎠ ⎝ L ⎠1
⎛ k ′p ⎞ ⎛ W ⎞
2
= ⎜ ⎟ ⎜ ⎟ (VSG 3 + VTP )
L
2
⎝
⎠
3
⎝ ⎠
But VSG 3 = 3 − VSG1
2
So 25 (VSG1 − 0.4 ) = 5 ( 3 − VSG1 − 0.4 )
2
which yields VSG1 = 1.08 V
and VSG 3 = 1.92 V
I REF = 20 ( 25 )(1.08 − 0.4 ) ⇒ I REF = 231 µ A
2
(W / L )2 15
=
= 0.6
I REF (W / L )1 25
Then I O = ( 0.6 )( 231) = 139 µ A
IO
=
(b)
VDS 2 ( sat ) = 1.08 − 0.4 = 0.68 V
VR = 3 − 0.68 = 2.32 = I O R
then
R=
2.32
⇒ R = 16.7 k Ω
0.139
10.51
VSD 2 (sat) = 0.35 = VSG 2 − 0.4 ⇒ VSG 2 = 0.75 V
2
⎛W ⎞
⎛W ⎞
I O = 80 = ( 20 ) ⎜ ⎟ ( 0.75 − 0.4 ) ⇒ ⎜ ⎟ = 32.7
L
⎝ ⎠2
⎝ L ⎠2
( )
( )
( )
W
W
I REF
50
L1
L1
=
⇒
=
⇒ W
= 20.4
L1
W
80
32.7
IO
L 2
VSG 3 = 3 − 0.75 = 2.25
( )
2
⎛W ⎞
⎛W ⎞
I REF = 50 = ( 20 ) ⎜ ⎟ ( 2.25 − 0.4 ) ⇒ ⎜ ⎟ = 0.730
⎝ L ⎠3
⎝ L ⎠3
10.52
a.
I REF = K n (VGS − VTN )
2
100 = 100 (VGS − 2 ) ⇒ VGS = 3 V
2
For VD 4 = −3 V, I 0 = 100 µ A
R0 = r04 + r02 (1 + g m r04 )
b.
r02 = r04 =
1
1
=
= 500 kΩ
λ I 0 ( 0.02 )( 0.1)
g m = 2 K n (VGS − VTN ) = 2 ( 0.1)( 3 − 2 ) = 0.2 mA / V
R0 = 500 + 500 ⎡⎣1 + ( 0.2 )( 500 ) ⎤⎦
R0 = 51 MΩ
ΔI 0 =
1
6
⋅ ΔVD 4 =
⇒ ΔI 0 = 0.118 µ A
51
R0
10.53
Vgs 4 = − I X r02
VS 6 = ( I X − g mVgs 4 ) r04 + I X r02
= ( I X + g m I X r02 ) r04 + I X r02
VS 6 = I X ⎣⎡ r02 + (1 + g m r02 ) r04 ⎦⎤ = −Vgs 6
⎛
V − VS 6 VX
1 ⎞
I X = g mVgs 6 + X
=
− VS 6 ⎜ g m + ⎟
r06
r06
r06 ⎠
⎝
⎛
1 ⎞
⎜ g m + ⎟ ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦
r06 ⎠
⎝
⎧⎪ ⎛
⎫⎪ V
1 ⎞
I X ⎨1 + ⎜ g m + ⎟ ⎣⎡ r02 + (1 + g m r02 ) r04 ⎦⎤ ⎬ = X
r06 ⎠
⎩⎪ ⎝
⎭⎪ r06
VX
= R0 = r06 + (1 + g m r06 ) ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦
IX
IX =
VX
− IX
r06
I 0 ≈ I REF = 0.2 mA = 0.2 (VGS − 1)
2
VGS = 2 V
g m = 2 K n (VGS − VTN ) = 2 ( 0.2 )( 2 − 1) = 0.4 mA / V
r02 = r04 = r06 =
1
=
1
= 250 kΩ
( 0.02 )( 0.2 )
R0 = 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ × {250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ ( 250 )}
λ I0
R0 = 2575750 kΩ ⇒ R0 = 2.58 × 109 Ω
10.54
kn′ ⎛ W ⎞
kn′ ⎛ W ⎞
2
2
⎜ ⎟ (VGS1 − VTN ) = ⎜ ⎟ (VGS 3 − VTN )
2 ⎝ L ⎠1
2 ⎝ L ⎠3
k ′p ⎛ W ⎞
2
= ⎜ ⎟ (VGS 4 + VTP )
2 ⎝ L ⎠4
2
(1) 50 ( 20 )(VGS 1 − 0.5 ) = 50 ( 5 )(VGS 3 − 0.5 )
2
2
(2) 50 ( 20 )(VGS 1 − 0.5 ) = 20 (10 )(VGS 4 − 0.5 )
2
(3) VSG 4 + VGS 3 + VGS1 = 6
From (1) 4 (VGS 1 − 0.5 ) = (VGS 3 − 0.5 ) ⇒ VGS 3 = 2 (VGS 1 − 0.5 ) + 0.5
2
2
From (2) 5 (VGS1 − 0.5 ) = (VGS 4 − 0.5 ) ⇒ VSG 4 = 5 (VGS1 − 0.5 ) + 0.5
2
2
Then (3) becomes
5 (VGS1 − 0.5 ) + 0.5 + 2 (VGS 1 − 0.5 ) + 0.5 + VGS1 = 6
which yields VGS 1 = 1.36 V and VGS 3 = 2.22 V , VSG 4 = 2.42 V
k′ ⎛ W ⎞
2
2
Then I REF = n ⎜ ⎟ (VGS1 − VTN ) = 50 ( 20 )(1.36 − 0.5 ) or I REF = I O = 0.740 mA
2 ⎝ L ⎠1
VGS 1 = VGS 2 = 1.36 V
VDS 2 ( sat ) = VGS 2 − VTN = 1.36 − 0.5 ⇒ VDS 2 ( sat ) = 0.86 V
10.55
VDS 2 ( sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.5 ⇒ VGS 2 = 1 V
kn′ ⎛ W ⎞
2
⎜ ⎟ (VGS 2 − VTN )
2 ⎝ L ⎠2
2
⎛W ⎞
⎛W ⎞
= 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 4
⎝ L ⎠2
⎝ L ⎠2
I O = 50 µ A =
VGS1 = VGS 2 = 1 V ⇒ I REF = 150 =
kn′ ⎛ W ⎞
2
2
⎛W ⎞
⎛W ⎞
⎜ ⎟ (VGS 1 − VTN ) = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 12
2 ⎝ L ⎠1
⎝ L ⎠1
⎝ L ⎠1
VGS 3 + VSG 4 + VGS 1 = 6
2VGS 3 = 6 − 1 = 5 V ⇒ VGS 3 = 2.5 V
2
⎛W ⎞
⎛W ⎞
I REF = 150 = 50 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ = 0.75
⎝ L ⎠3
⎝ L ⎠3
k ′p ⎛ W ⎞
2
⎜ ⎟ (VSG 4 + VTP )
2 ⎝ L ⎠4
2
⎛W ⎞
⎛W ⎞
150 = 20 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ 1.88
L
⎝ ⎠4
⎝ L ⎠4
I REF =
10.56
a.
As a first approximation
2
I REF = 80 = 80 (VGS 1 − 1) ⇒ VGS 1 = 2 V
Then VDS 1 ≅ 2 ( 2 ) = 4 V
The second approximation
80 = 80 (VGS1 − 1) ⎡⎣1 + ( 0.02 )( 4 ) ⎤⎦
80
2
= (VGS 1 − 1) ⇒ VGS 1 = 1.962
Or
86.4
2
Then
2
I O = K n (VGS1 − VTN ) (1 + λnVGS 1 )
= 80 (1.962 − 1) ⎡⎣1 + ( 0.02 )(1.962 ) ⎤⎦
Or I 0 = 76.94 µ A
2
b.
From a PSpice analysis, I 0 = 77.09 µ A for VD 3 = −1 V and I 0 = 77.14 µ A for VD 3 = 3 V. The
change is ΔI 0 ≈ 0.05 µ A or 0.065%.
10.57
a.
For a first approximation,
I REF = 80 = 80 (VGS 4 − 1) ⇒ VGS 4 = 2 V
2
As a second approximation
I REF = 80 = 80 (VGS 4 − 1) ⎡⎣1 + ( 0.02 )( 2 ) ⎤⎦
Or VGS 4 = 1.98 V = VGS 1
2
2
I O = K n (VGS 2 − VTN ) (1 + λVGS 2 )
To a very good approximation I 0 = 80 µ A
b.
From a PSpice analysis, I 0 = 80.00 µ A for VD 3 = −1 V and the output resistance is
R0 = 76.9 MΩ.
Then
For VD = +3 V
1
4
ΔI 0 =
⋅ VD 3 =
= 0.052 µ A
76.9
R0
I 0 = 80.05 µ A
10.58
(a)
(b)
(c)
10.59
(a)
K n1 =
R=
=
VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0
k′ ⎛ W ⎞
2
I D = n ⎜ ⎟ (VGS 3 − VTN )
2⎝L⎠
2
⎛W ⎞
⎛W ⎞
50 = 48 ⎜ ⎟ ( 0.2 ) ⇒ ⎜ ⎟ = 26
⎝L⎠
⎝ L ⎠3
VGS 5 − VTN = 2 (VGS 3 − VTN )
VGS 5 = 0.8 + 2 ( 0.2 ) ⇒ VGS 5 = 1.2 V
VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) ⇒ VD1 ( min ) = 0.4 V
kn′ ⎛ W ⎞
2
⎜ ⎟ = 50 ( 5 ) = 250 µ A / V
2 ⎝ L ⎠1
1
K n1 I D1
⎛
⎜1 −
⎜
⎝
(W / L )1 ⎞⎟
(W / L )2 ⎟⎠
⎛
5 ⎞
⎜⎜1 −
⎟ = ( 8.944 )( 0.6838 )
50 ⎟⎠
( 0.25 )( 0.05 ) ⎝
1
R = 6.12 k Ω
(b)
V + − V − = VSD 3 ( sat ) + VGS 1
VSD 3 ( sat ) = VSG 3 + VTP
I D1 = 50 = 20 ( 5 )(VSG 3 − 0.5 ) ⇒ VSG 3 = 1.207 V
Then VSD 3 ( sat ) = 1.21 − 0.5 = 0.707 V
2
Also I D1 = 50 = 50 ( 5 )(VGS1 − 0.5 ) ⇒ VGS 1 = 0.9472 V
2
Then (V + − V − )
min
= 0.71 + 0.947 = 1.66 V
(c)
2
⎛W ⎞
⎛W ⎞
I O1 = 25 = 50 ⎜ ⎟ ( 0.947 − 0.5 ) ⇒ ⎜ ⎟ = 2.5
⎝ L ⎠5
⎝ L ⎠5
2
⎛W ⎞
⎛W ⎞
I O 2 = 75 = 20 ⎜ ⎟ (1.207 − 0.5 ) ⇒ ⎜ ⎟ = 7.5
⎝ L ⎠6
⎝ L ⎠6
10.60
1
( 5 ) = 1.667 V
3
2
⎛1
⎞⎛W ⎞
I REF = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN )
⎝2
⎠ ⎝ L ⎠3
2
⎛W ⎞
⎛W ⎞ ⎛W ⎞ ⎛W ⎞
100 = ( 20 ) ⎜ ⎟ (1.667 − 1) ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 11.25
⎝ L ⎠3
⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ L ⎠5
VGS 3 =
2
⎛1
⎞⎛W ⎞
I O1 = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN )
⎝2
⎠ ⎝ L ⎠1
⎛W ⎞
⎜ ⎟
I REF ⎝ L ⎠3
Or
=
I 01
⎛W ⎞
⎜ ⎟
⎝ L ⎠1
⎛ W ⎞ ⎛ I 01
⎜ ⎟ =⎜
⎝ L ⎠1 ⎝ I REF
⎞ ⎛ W ⎞ ⎛ 0.2 ⎞
⎛W ⎞
⎟⎜ ⎟ = ⎜
⎟ (11.25 ) ⇒ ⎜ ⎟ = 22.5
⎝ L ⎠1
⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠
⎛ W ⎞ ⎛ I ⎞ ⎛ W ⎞ ⎛ 0.3 ⎞
⎛W ⎞
And ⎜ ⎟ = ⎜ 02 ⎟ ⎜ ⎟ = ⎜
⎟ (11.25 ) ⇒ ⎜ ⎟ = 33.75
0.1
L
I
L
⎝ ⎠ 2 ⎝ REF ⎠ ⎝ ⎠3 ⎝
⎠
⎝ L ⎠2
10.61
I REF =
24 − VSGP − VGSN
R
Also I REF = 40 (1)(VGSN − 1.2 )
I REF = 18 (1)(VSGP − 1.2 )
2
2
Then 40 (VGSN − 1.2 ) = 18 (VSGP − 1.2 )
which yields VSGP =
6.325
(VGSN − 1.2 ) + 1.2
4.243
2
Then ⎡ 0.040 (VGSN − 1.2 ) ⎤ ⋅ R = 24 − VGSN − 1.49 (VGSN − 1.2 ) − 1.2
⎣
⎦
which yields VGSN = 2.69 V
and VSGP = 3.43 V
24 − 3.42 − 2.69
⇒ I REF = 89.4 µ A
200
Now I REF =
89.4
= 17.9 µ A
5
I 2 = (1.25 )( 89.4 ) = 112 µ A
I 3 = ( 0.8 )( 89.4 ) = 71.5 µ A
I 4 = 4 ( 89.4 ) = 358 µ A
I1 =
10.61
a.
g m ( M 0 ) = 2 K n I REF
gm ( M 0 ) = 2
( 0.25)( 0.2 ) ⇒ g m ( M 0 ) = 0.447 mA/V
1
1
=
⇒ r0 n = 250 kΩ
λn I REF ( 0.02 )( 0.2 )
r0 n =
r0 p =
1
1
=
⇒ r0 p = 167 kΩ
λ p I REF ( 0.03)( 0.2 )
b.
Av = − g m ( r0 n || r0 p ) = − ( 0.447 )( 250 ||167 ) ⇒ Av = −44.8
c.
RL = 205 ||167 = r0 n || r0 p or RL = 100 kΩ
10.62
We have VGSN = 2.69 V and VSGP = 3.43 V
10 − 2.69 − 3.43 3.88
So I REF =
=
⇒ I REF = 19.4 µ A
200
R
Then I1 = ( 0.2 )(19.4 ) = 3.88 µ A
I 2 = (1.25 )(19.4 ) = 24.3 µ A
I 3 = ( 0.8 )(19.4 ) = 15.5 µ A
I 4 = 4 (19.4 ) = 77.6 µ A
10.63
I D2
W )
(
L
=
W
( L)
(W L )
=
(W L )
2
1
IO
4
⋅ I REF =
9
( 200 ) ⇒ I D 2 = 120 µ A
15
⎛ 20 ⎞
⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 µ A
⎝ 9 ⎠
2
⎛ 40 ⎞
I O = 266.7 = ⎜ ⎟ ( 20 )(VSG 4 − 0.6 )
2
⎝ ⎠
VSG 4 = 1.416 V
VSD 4 (sat) = 1.416 − 0.6 ⇒ VSD 4 ( sat ) = 0.816 V
3
10.64
2
⎛ 40 ⎞ ⎛ W ⎞
I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 )
⎝ 2 ⎠ ⎝ L ⎠1
1.75 − VSG1
= 50
I REF =
R
VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V
1.75 − 0.95
⇒ R = 16 K
R=
0.05
2
⎛ 40 ⎞⎛ W ⎞
⎛W ⎞
50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4
L
2
⎝ ⎠⎝ ⎠1
⎝ L ⎠1
W
I O1 120
⎛W ⎞
L 2
=
=
⇒
= 49
50
I REF
( 20.4 ) ⎜⎝ L ⎟⎠2
( )
W
I D 3 25
⎛W ⎞
L 3
=
=
⇒ ⎜ ⎟ = 10.2
I REF 50 ( 20.4 )
⎝ L ⎠3
( )
VDS 5 (sat) = 0.35 = VGS 5 − 0.4 ⇒ VGS 5 = 0.75 V
2
⎛ 100 ⎞⎛ W ⎞
⎛W ⎞
IO 2 = ⎜
⎟⎜ ⎟ ( 0.75 − 0.4 ) = 150 ⇒ ⎜ ⎟ = 24.5
L
2
⎝
⎠⎝ ⎠5
⎝ L ⎠5
W
I D4 I D3
25
⎛W ⎞
L 4
=
=
=
⇒ ⎜ ⎟ = 4.08
I O 2 I O 2 150
24.5
⎝ L ⎠4
( )
10.65
For vGS = 0, iD = I DSS (1 + λ vDS )
a.
b.
c.
VD = −5 V, vDS = 5
iD = ( 2 ) ⎡⎣1 + ( 0.05 )( 5 ) ⎤⎦ ⇒ iD = 2.5 mA
VD = 0, vDS = 10
iD = ( 2 ) ⎡⎣1 + ( 0.05 )(10 ) ⎤⎦ ⇒ iD = 3 mA
VD = 5 V, vDS = 15 V
iD = ( 2 ) ⎡⎣1 + ( 0.05 )(15 ) ⎤⎦ ⇒ iD = 3.5 mA
10.66
⎛ V ⎞
I 0 = I DSS ⎜1 − GS ⎟
VP ⎠
⎝
⎛ V ⎞
2 = 4 ⎜1 − GS ⎟
VP ⎠
⎝
2
2
VGS
2
= 1−
= 0.293
4
VP
So VGS = ( 0.293)( −4 ) = −1.17 V
VS
and VS = −VGS
R
( −1.17 )
−V
⇒ R = 0.586 kΩ
R = GS = −
2
I0
Finish solution:
See solution
Then I 0 =
10.66
Completion of solution
Need vDS ≥ vDS ( sat ) = vGS − VP
= −1.17 − ( −4 )
vDS ≥ 2.83 V
So VD ≥ vDS ( sat ) + VS = 2.83 + 1.17 ⇒ VD ≥ 4 V
10.67
a.
⎛I ⎞
⎛ 1× 10−3 ⎞
or VEB1 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
⇒ VEB1 = 0.5568
−13 ⎟
⎝ 5 × 10 ⎠
⎝ I S1 ⎠
5 − 0.5568
R1 =
⇒ R1 = 4.44 kΩ
1
From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V
b.
c.
10
⎛V ⎞
I REF = I S 1 exp ⎜ EB1 ⎟
⎝ VT ⎠
−12
2.5 ⎞
⎛
⎜ 1 + 80 ⎟
⎛ VI ⎞ ⎡ 2.5 ⎤
−3
exp ⎜ ⎟ ⎢1 +
⎥ = 10 ⎜ 0.5568 ⎟
⎝ VT ⎠ ⎣ 120 ⎦
⎜⎜ 1 +
⎟⎟
80 ⎠
⎝
⎛V ⎞
⎛ 1.03125 ⎞
1.0208333 × 10−12 exp ⎜ I ⎟ = (10−3 ) ⎜
⎟
⎝ 1.00696 ⎠
⎝ VT ⎠
Then VI = 0.026 ln (1.003222 × 109 )
So VI = 0.5389 V
d.
Av =
− (1/ VT )
(1/ VAN ) + (1/ VAP )
1
−38.46
Av = 0.026 =
1
1
0.00833 + 0.0125
+
120 80
Av = −1846
−
10.68
a.
b.
c.
other.
⎛I ⎞
⎛ 0.5 × 10−3 ⎞
VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
⎟ ⇒ VBE = 0.5208
−12
⎝ 10
⎠
⎝ I S1 ⎠
5 − 0.5208
R1 =
⇒ R1 = 8.96 kΩ
0.5
Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each
⎛ V ⎞⎛ V
I CO = I SO exp ⎜ EBO ⎟⎜ 1 + ECO
VAP
⎝ VT ⎠⎝
⎞
⎛ VBE ⎞ ⎛ VCE 2 ⎞
⎟
⎟ = I C 2 = I S 2 exp ⎜
⎟ ⎜1 +
⎠
⎝ VT ⎠ ⎝ VAN ⎠
⎛ V ⎞ ⎛ 2.5 ⎞
⎛ VBE ⎞ ⎛
2.5 ⎞
−12
5 × 10−13 exp ⎜ EBO ⎟ ⎜ 1 +
⎟ ⎜1 +
⎟ = 10 exp ⎜
⎟
80
120
V
V
⎠
⎠
⎝ T ⎠⎝
⎝ T ⎠⎝
⎛V ⎞
⎛V ⎞
5.15625 × 10−13 exp ⎜ EBO ⎟ = 1.020833 × 10−12 exp ⎜ BE ⎟
V
⎝ T ⎠
⎝ VT ⎠
⎛V ⎞
exp ⎜ EBO ⎟
⎝ VT ⎠ = 1.9798 = exp ⎛ VEBO − VBE ⎞
⎜
⎟
VT
⎛V ⎞
⎝
⎠
exp ⎜ BE ⎟
V
⎝ T ⎠
VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 )
VEBO = 0.5386 ⇒ VI = 5 − 0.5386 ⇒ VI = 4.461 V
Av =
d.
− (1/ VT )
(1/ VAN ) + (1/ VAP )
1
−38.46
0.026
Av =
=
1
1
0.00833
+ 0.0125
+
120 80
Av = −1846
−
10.69
a.
M1 and M2 matched.
For I REF = I 0 , we have VSD 2 = VSG = VSG 3 = VDS 0 = 2.5 V
For M1 and M3:
2
⎛1
⎞⎛W ⎞
I REF = ⎜ µ p Cox ⎟ ⎜ ⎟ (VSG + VTP ) (1 + λPVSD )
⎝2
⎠ ⎝ L ⎠1
2
⎛W ⎞
⎛W ⎞
⎛W ⎞ ⎛W ⎞
100 = 10 ⎜ ⎟ ( 2.5 − 1) ⎡⎣1 + ( 0.02 )( 2.5 ) ⎤⎦ ⇒ ⎜ ⎟ = 4.23 = ⎜ ⎟ = ⎜ ⎟
⎝ L ⎠1
⎝ L ⎠1
⎝ L ⎠3 ⎝ L ⎠ 2
For M0:
2
⎛1
⎞⎛W ⎞
I O = ⎜ µ n Cox ⎟ ⎜ ⎟ (VGS − VTN ) (1 + λnVDS )
⎝2
⎠ ⎝ L ⎠0
2
⎛W ⎞
⎛W ⎞
100 = 20 ⎜ ⎟ ( 2 − 1) ⎡⎣1 + ( 0.02 )( 2.5 ) ⎤⎦ ⇒ ⎜ ⎟ = 4.76
L
⎝ ⎠0
⎝ L ⎠0
b.
r0 n = r0 p =
1
λ I0
=
1
( 0.02 )( 0.1)
= 500 kΩ
⎛1
⎞⎛W ⎞
g m = 2 K n I O = 2 ⎜ µ n Cox ⎟ ⎜ ⎟ I O
2
⎝
⎠ ⎝ L ⎠o
( 0.02 )( 4.76 )( 0.1)
=2
g m = 0.195 mA/V
Av = − g m ( r0 n || r0 p ) = − ( 0.195 )( 500 || 500 ) ⇒ Av = −48.8
10.70
I REF = K p1 (VSG + VTP )
a.
2
100 = 100 (VSG − 1) ⇒ VSG = 2 V
2
b.
From Eq. 10.89
⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2
TN
⎦− n I
VO = ⎣
λn + λ p
I REF ( λn + λ p )
⎡1 + ( 0.02 )(10 − 2 ) ⎤⎦
100 (VI − 1)
−
5= ⎣
0.02 + 0.02
100 ( 0.02 + 0.02 )
2
5 = 29 −
(VI − 1)
2
(VI − 1)
0.04
= 0.96
VI = 1.98 V
2
Av = − g m ( r0 n || r0 p )
c.
r0 n = r0 p =
1
λ I REF
=
1
= 500 kΩ
0.02
( )( 0.1)
( 0.1)( 0.1) = 0.2 mA / V
Av = − ( 0.2 )( 500 || 500 ) ⇒ Av = −50
g m = 2 K n I REF = 2
10.71
5 − 0.6 5 − 0.6
=
= 0.22 mA
R1
20
From Eq. 10.96
⎛I ⎞
0.22
−⎜ C ⎟
−
VT ⎠
⎝
0.026
Av =
=
⎛ IC
I C ⎞ 0.22 + 1 + 0.22
1
+
+
⎜
⎟
90
⎝ VAN RL VAP ⎠ 140 RL
I REF =
=
−8.4615
−8.4615
=
1
1
0.0015714 +
+ 0.002444 0.004016 +
RL
RL
RL = ∞, Av = −2107
(a)
(b)
(c)
RL = 250 K, Av = −1056
RL = 100 K, Av = −604
10.72
I REF =
5 − 0.6
= 0.1257 mA
35
Then
I CO = 2 I REF = 0.2514 mA
From Eq. 10.96
−0.2514
−9.6692
0.026
Av =
=
0.2514 0.2514 1
1
0.002095 + 0.0031425 +
+
+
RL
RL
120
80
Av =
−9.6692
1
RL
RL = ∞ Av = −1846
0.0052375 +
(a)
(b)
RL = 250 K, Av = −1047
10.73
(a)
To a good approximation, output resistance is the same as the widlar current source.
R0 = r02 ⎡⎣1 + g m 2 ( rπ 2 || RE ) ⎤⎦
(b)
Av = − g m 0 ( r0 || RL || R0 )
10.74
Output resistance of Wilson source
β r03
R0 ≅
2
Then
Av = − g m ( r0 || R0 )
V
80
r03 = AP =
= 400 kΩ
I REF 0.2
r0 =
VAN 120
=
= 600 kΩ
I REF 0.2
gm =
I REF
0.2
=
= 7.692 mA/V
0.026
VT
⎡
(80 )( 400 ) ⎤
Av = −7.69 ⎢600
⎥ = −7.69 [ 600 ||16, 000] ⇒ Av = −4448
2
⎢⎣
⎥⎦
10.75
(a)
I D 2 = I D 0 = I REF = 200 µ A
For M 2 ; ro 2 =
1
λP I D 2
=
1
= 250 K
0.02
( )( 0.2 )
⎛ 0.04 ⎞
gm2 = 2 K P I D2 = 2 ⎜
⎟ ( 35 )( 0.2 )
⎝ 2 ⎠
g m 2 = 0.748 mA/V
1
1
For M 0 ; r∞ =
=
= 333 K
λn I Do ( 0.015 )( 0.2 )
⎛ 0.08 ⎞
g mo = 2 ⎜
⎟ ( 20 )( 0.2 ) ⇒ g mo = 0.80 mA/V
⎝ 2 ⎠
Av = − g mo ( ro 2 || roo ) = − ( 0.80 )( 250 || 333)
(b)
Av = −114.3
Want Av = −57.15 = −0.80 (142.8 || RL )
142.8 RL
142.8 || RL = 71.375 =
⇒ RL = 143 K
142.8 + RL
(c)
10.76
Assume M1, M2 matched
I REF = I D 2 = I Do = 200 µ A
1
1
=
= 250 K
ro 2 =
λ p I D 2 ( 0.02 )( 0.2 )
roo =
1
1
=
= 333 K
λn I D 0 ( 0.015 )( 0.2 )
Av = − g mo ( ro 2 roo )
−100 = − g mo ( 250 333) ⇒ g mo = 0.70 mA/V
⎛ 0.08 ⎞⎛ W ⎞
g mo = 2 ⎜
⎟⎜ ⎟ ( 0.2 ) = 0.70
⎝ 2 ⎠⎝ L ⎠0
⎛W ⎞
⎜ ⎟ = 15.3
⎝ L ⎠0
⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′p ⎞ ⎛ W ⎞
Now ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ L ⎠0 ⎝ 2 ⎠ ⎝ L ⎠ 2
⎛ 80 ⎞
⎛ 40 ⎞⎛ W ⎞
⎜ ⎟ (15.3) = ⎜ ⎟⎜ ⎟
⎝ 2⎠
⎝ 2 ⎠⎝ L ⎠ 2
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 30.6
⎝ L ⎠ 2 ⎝ L ⎠1
10.77
Since Vsg 3 = 0, the circuit becomes
I x = − g m′ Vsg 2 +
Vx − Vsg 2
r02
and Vsg 2 = I x ro 3
Then
⎛
r ⎞ V
I x ⎜ 1 + g m′ ro3 + o3 ⎟ = x
ro 2 ⎠ ro 2
⎝
so that
⎛
Vx
r ⎞
= Ro = ro 2 ⎜1 + g m′ ro 3 + o 3 ⎟
Ix
r
o2 ⎠
⎝
or
Ro = ro 2 + ro 3 (1 + g m′ ro 2 )
vo
= − g m1 ( ro1 Ro )
vi
Av =
Now
g m1 = 2
ro1 =
( 0.050 )( 20 )( 0.10 ) = 0.632 mA / V
1
1
=
= 500 k Ω
λn I DQ ( 0.02 )( 0.10 )
g m′ = 2 K p I DQ = 2
ro 2 = ro 3 =
( 0.020 )(80 )( 0.1) = 0.80 mA / V
1
1
=
= 500 k Ω
λ p I DQ ( 0.020 )( 0.1)
Then
Ro = 500 + 500 ⎡⎣1 + ( 0.8 )( 500 ) ⎤⎦ ⇒ 201 M Ω
Av = − ( 0.632 ) ( 500 201000 ) ⇒ Av = −315
10.78
From Eq. 10.105
− g m2
Av =
1
1
+
ro3 ro 4 ro1 ro 2
⎛ k′ ⎞⎛W ⎞
g m = 2 ⎜ n ⎟ ⎜ ⎟ I D1
⎝ 2 ⎠⎝ L ⎠
=2
( 0.050 ) ( 20 ) ( 0.08 )
g m = 0.5657 mA / V
ro =
1
1
=
= 625 K
λ I D ( 0.02 )( 0.08 )
2
− ( 0.5657 )
−0.3200
Av =
=
1
1
2 ( 0.00000256 )
+
2
2
( 625) ( 625)
Av = −62,500
10.79
V − ( −Vπ 2 )
Vπ 2 Vπ 2
+
+ g m 2Vπ 2 + O
rπ 2 ro1
ro 2
(1)
g m1Vi =
(2)
VO VO − ( −Vπ 2 )
+
+ g m 2Vπ 2 = 0
RO 3
ro 2
⎛ 1
1
1 ⎞ V
+ + gm2 + ⎟ + O
g m1Vi = Vπ 2 ⎜
ro 2 ⎠ ro 2
⎝ rπ 2 ro1
(1)
⎛ 1
⎛ 1
⎞
1 ⎞
VO ⎜
+ ⎟ + Vπ 2 ⎜
+ gm2 ⎟ = 0
⎝ RO 3 ro 2 ⎠
⎝ ro 2
⎠
(2)
g m >>
(1)
1
ro
⎛ 1 + β ⎞ VO
g m1Vi = Vπ 2 ⎜
⎟+
⎝ rπ 2 ⎠ ro 2
(2)
⎛ 1
1 ⎞
VO ⎜
+ ⎟ + Vπ 2 ⋅ g m 2 = 0
⎝ RO 3 ro 2 ⎠
(3)
Vπ 2 = −
Then
VO
gm2
⎛ 1
1 ⎞
+ ⎟
⎜
⎝ RO 3 ro 2 ⎠
⎛ 1
1 ⎞⎛ 1 + β ⎞ VO
+ ⎟⎜
⎜
⎟+
⎝ RO 3 ro 2 ⎠⎝ rπ 2 ⎠ ro 2
⎛ 1
1 ⎞ ⎛ 1 + β ⎞ VO
= −VO ⎜
+ ⎟⎜
⎟+
R
r
⎝ O 3 o 2 ⎠ ⎝ β ⎠ ro 2
(1) g m1Vi = −
VO
gm2
VO ⎛ 1 + β ⎞
⎜
⎟
RO 3 ⎝ β ⎠
⎛ β ⎞
VO
= − g m1 RO 3 ⎜
⎟
Vi
⎝1+ β ⎠
From Equation (10.20) RO 3 ≈ β rO 3
So
≈−
Av =
Av =
VO − g m1ro3 β 2
=
1+ β
Vi
gm =
0.25
= 9.615 mA/V
0.026
ro3 =
80
= 320 K
0.25
− ( 9.615 )( 320 )(120 )
121
2
= −366,165
Chapter 11
Exercise Solutions
EX11.1
vE = −VBE ( on ) ⇒ vE = −0.7 V
I C1 = I C 2 = 0.5 mA
vC1 = vC 2 = 10 − ( 0.5 )(10 ) ⇒ vC1 = vC 2 = 5 V
EX11.2
iC 2
1
=
= 0.99
IQ
⎛ vd ⎞
1 + exp ⎜ ⎟
⎝ VT ⎠
⎛v ⎞
1
1 + exp ⎜ d ⎟ =
0.99
V
⎝ T⎠
⎛v
exp ⎜ d
⎝ VT
⎞
1
−1
⎟=
0.99
⎠
⎡ 1
⎤
− 1⎥ ⇒ vd = −119.5 mV
vd = VT ln ⎢
⎣ 0.99 ⎦
EX11.3
a.
v1 = v2 = 0 ⇒ vE = 0.7 V
ΔVRC = ( 0.25 )( 8 ) = 2 V ⇒ vC1 = vC 2 = −3 V ⇒ vEC1 = 3.7 V
v1 = v2 = 2.5 V ⇒ vE = 3.2 V ⇒ vEC1 = 6.2 V
b.
v1 = v2 = −2.5 V ⇒ vE = −1.8 V ⇒ vEC1 = 1.2 V
c.
EX11.4
Let I Q = 1 mA, then I CQ1 = I CQ 2 = 0.5 mA
0.5
= 19.23 mA / V
0.026
v
1
At vC 2 , Ad = c 2 = g m RC 2
vd
2
g m1 = g m 2 =
1
(19.23) RC 2 ⇒ RC 2 = 15.6 k Ω
2
v
1
At vC1 , Ad = c1 = − g m RC1
vd
2
So, 150 =
1
(19.23) RC1 ⇒ RC1 = 10.4 k Ω
2
If V + = +10 V and V − = −10 V , dc biasing is OK.
So, −100 = −
EX11.5
Acm =
1+
⎛ I Q RC ⎞
−⎜
⎟
⎝ 2VT ⎠
=
(1 + β ) IQ RO
VT β
⎡ ( 0.8 )(12 ) ⎤
−⎢
⎥
⎣ 2 ( 0.026 ) ⎦
= −0.0594
101)( 0.8 )(100 )
(
1+
( 0.026 )(100 )
vo = Acm vcm = ( −0.0594 )(1sin ω t )( mV ) = −59.4sin ω t ( µV )
EX11.6
vd = 40 µV , vcm = 0
vo = ( 50 )( 40 ) ⇒ 2.0 mV
(a)
(b)
vd = 40 µV vcm = 200 µV
Assuming Acm > 0
50
Acm
60 = 20 log10
Acm = 0.05
vo = ( 50 )( 40 ) + ( 0.05 )( 200 )
vo = 2.010 mV
EX11.7
Diff. Gain Ad =
a.
I Q RC
4VT
For v1 = v2 = 5 V ⇒ Minimum collector voltage
vC 2 = 5 V ⇒
IQ
⋅ RC = 15 − 5 = 10 V or I Q RC = 20 V for max. Ad
2
20
⇒ Ad ( max ) = 192
Then Ad =
2 ( 0.026 )
If I Q = 0.5 mA, RC = 40 kΩ
b.
⎛ I Q RC ⎞
−⎜
⎟
⎝ 2VT ⎠
Acm =
⎡ (1 + β ) I Q R0 ⎤
⎢1 +
⎥
VT β
⎣
⎦
Then
⎛
20 ⎞
− ⎜⎜
⎟
2 ( 0.026 ) ⎟⎠
⎝
Acm =
⇒ Acm = −0.199
⎡ ( 201)( 0.5 )(100 ) ⎤
⎢1 +
( 0.026 )( 200 ) ⎥⎦
⎣
⎛ 192 ⎞
and C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 59.7 dB
⎝ 0.199 ⎠
EX11.8
Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦
rπ =
β VT
I CQ
=
(100 )( 0.026 )
0.25
= 10.4 K
Rid = 2 ⎡⎣10.4 + (101)( 0.5 ) ⎤⎦ = 122 K
EX11.9
Ad =
g m RC
2 (1 + g m RE )
( 9.62 )(10 )
2 ⎣⎡1 + ( 9.62 ) RE ⎦⎤
1 + ( 9.62 ) RE = 4.81 ⇒ RE = 0.396 K
Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦ = 2 ⎡⎣10.4 + (101)( 0.396 ) ⎤⎦
10 =
Rid = 100.8 K
EX11.10
10 − VGS 4
2
= K n 3 (VGS 4 − VTN )
I1 =
R1
10 − VGS 4 = ( 0.1)( 80 )(VGS 4 − 0.8 )
2
10 − VGS 4 = 8 (VGS2 4 − 1.6VGS 4 + 0.64 )
8VGS2 4 − 11.8VGS 4 − 4.88 = 0
VGS 4 =
11.8 ±
2
(11.8) + 4 ( 8)( 4.88)
2 (8)
= 1.81 V
10 − 1.81
= 0.102 mA
80
0.102
= ID2 =
= 0.0512 mA
2
I1 = I Q =
I D1
= K n1 (VGS 1 − VTN )
2
0.0512 = 0.050 (VGS 1 − 0.8 ) ⇒ VGS 1 = 1.81 V
2
v01 = v02 = 5 − ( 0.0512 )( 40 ) = 2.95 V
Max vcm : VDS 1 ( sat ) = VGS1 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( max ) = v01 − VDS 1 ( sat ) + VGS 1
= 2.95 − 1.01 + 1.81
vcm ( max )
= 3.75 V
Min vcm : VDS 4 ( sat ) = VGS 4 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( min )
= VGS 1 + VDS 4 ( sat ) − 5
= 1.81 + 1.01 − 5
vcm ( min )
= −2.18 V
−2.18 ≤ vcm ≤ 3.75 V
EX11.11
(1)( 2 )
⇒ g f ( max ) = 1 mA / V
2
( 2)
= g f RD = (1)( 5 ) ⇒ Ad = 5
g f ( max ) =
Ad
EX11.12
K n IQ
=
⎛ K
Kn
iD1 1
= +
⋅ vd 1 − ⎜ n
⎜ 2 IQ
IQ 2
2IQ
⎝
⎞ 2
⎟⎟ vd
⎠
Using the parameters in Example 11.11, K n = 0.5 mA / V 2 , I Q = 1 mA, then
⎛ 0.5 ⎞ 2
iD1
1
0.5
= 0.90 = +
⋅ vd 1 − ⎜⎜
⎟⎟ vd
2
2 (1)
IQ
⎝ 2 (1) ⎠
By trial and error, vd = 0.894 V
EX11.13
a.
gm =
IQ
2VT
=
0.5
= 9.615 mA/V
2 ( 0.026 )
r02 =
VA 2 125
=
= 500 kΩ
I C 2 0.25
r04 =
VA 4
85
=
= 340 kΩ
I C 4 0.25
Ad = g m ( r02 r04 ) = ( 9.615 ) ( 500 340 ) ⇒ Ad = 1946
b.
c.
(
Ad = g m r02 r04 RL
)
Ad = ( 9.615 ) ⎡⎣500 340 100 ⎤⎦ ⇒ Ad = 644
β VT (150 )( 0.026 )
rπ =
=
= 15.6 kΩ
I CQ
0.25
Rid = 2rπ ⇒ Rid = 31.2 kΩ
d.
R0 = r02 r04 = 500 340 ⇒ R0 = 202 kΩ
EX11.14
80
80
2
2
I REF = (10 )(VGS 4 − 0.5 ) = ( 0.33)(VGS 5 − 0.5 )
2
2
VGS 4 + 2VGS 5 = 6
1
VGS 5 = ( 6 − VGS 4 )
2
10
1
(VGS 4 − 0.5) = ( 6 − VGS 4 ) − 0.5
0.33
2
6.0VGS 4 = 5.252 ⇒ VGS 4 = 0.8754 V
80
2
I REF = (10 )( 0.8754 − 0.5 ) = 53.37 µ A
2
Also I Q = I REF = 53.37 µ A
⎛ 80 ⎞
g m = 2kn I Q = 2 ⎜ ⎟ (10 )( 53.37 ) ⇒ 0.2066 mA/V
⎝ 2⎠
1
1
=
= 1873.7 K
ro1 = ro8 =
λ ID
⎛ 0.05337 ⎞
( 0.02 ) ⎜
⎟
2
⎝
⎠
Ad = g m ( ro1 & ro8 ) = ( 0.2066 )(1873.7 & 1873.7 )
Ad = 194
EX11.15
Ad = g m ( ro 2 & RO )
400 = g m ( 500 & 101000 ) ⇒ g m = 0.8 mA / V
g m = 2 K n I DQ
⎛ 0.08 ⎞ ⎛ W ⎞
⎛W ⎞ ⎛W ⎞
0.8 = 2 ⎜
⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 40
⎝ 2 ⎠ ⎝ L ⎠1
⎝ L ⎠1 ⎝ L ⎠ 2
EX11.16
I e 6 = (1 + β ) I b 6 = I b 7
I c 7 = β I b 7 = β (1 + β ) I b 6
Ic7
= β (1 + β ) = (100 )(101) = 1.01× 104
Ib6
EX11.17
r + 300
ROA = π A
1+ β
⎛ β
I CA = ⎜
⎝1+ β
⎛ β
=⎜
⎝1+ β
⎞
⎟ I EA
⎠
⎞⎛ 1 ⎞
⎟⎜
⎟ (1) ⇒ 9.803 µ A
⎠⎝ 1 + β ⎠
(100 )( 0.026 )
rπ A =
= 265.2 K
0.009803
265.2 + 300
ROA =
= 5.596 K
101
r + ROA
RO = π B
10
1+ β
rπ B =
(100 )( 0.026 )
= 2.626 K
0.99
2.626 + 5.596
RO =
10
101
RO = 81.4 & 104 = 80.7 Ω
EX11.18
10 − 0.7 − ( −10 )
I1 =
= 0.6 ⇒ R1 = 32.2 K
R1
⎛I
I Q R2 = VT ln ⎜ 1
⎜I
⎝ Q
⎞
⎟⎟
⎠
.
( 0.2 ) R2 = ( 0.026 ) ln ⎜⎛
I R 6 = I1 ⇒ R3 = 0
0.6 ⎞
⎟ ⇒ R2 = 143 Ω
⎝ 0.2 ⎠
10 = I C1 RC + VCE1 − 0.7
10.7 = ( 0.1) RC + 4 ⇒ RC = 67 K
vo 2 = −0.7 + 4 = 3.3 V
vE 4 = 3.3 − 1.4 = 1.9 V
I R4 =
vE 4
1.9
⇒ R4 =
⇒ R4 = 3.17 K
0.6
R4
vC 3 = vO 2 − 1.4 + vCE 4
= 3.3 − 1.4 + 3 = 4.9 V
I R 5 = I R 4 = 0.6 =
10 − 4.9
⇒ R5 = 8.5 K
R5
vE 5 = 4.9 − 0.7 = 4.2 V ⇒ R6 =
R7 =
0 − ( −10 )
5
⇒ R7 = 2 K
4.2 − 0.7
= 5.83 K
0.6
EX11.19
Ri 2 = rπ 3 + (1 + β ) rπ 4
(100 )( 0.026 )
rπ 4 =
0.6
β 2VT
rπ 3 ≈
I R4
= 4.333 K
2
=
(100 ) ( 0.026 )
0.6
= 433.3 K
Ri 2 = 433.3 + (101)( 4.333) ⇒ Ri 2 = 871 K
Ri 3 = rπ 5 + (1 + β ⎣⎡ R6 + rπ 6 + (1 + β ) R7 ⎦⎤
(100 )( 0.026 )
rπ 5 =
0.6
(100 )( 0.026 )
rπ 6 =
= 4.333 K
= 0.52 K
5
Ri 3 = 4.333 + (101) ⎣⎡5.83 + 0.52 + (101)( 2 ) ⎦⎤
Ri 3 = 21.0 MΩ
gm
0.1
RC Ri 2 ) g m =
= 3.846 mA/V
(
2
0.026
3.846
Ad 1 =
( 67 871) = 119.64
2
⎛I ⎞
0.6
A2 = ⎜ R 4 ⎟ ( R5 Ri 3 ) =
(8.5 21000 ) = 98.037
2 ( 0.026 )
⎝ 2VT ⎠
Ad 1 =
A = Ad 1 ⋅ A2 = (119.64 )( 98.037 ) = 11, 729
EX11.20
fz =
1
1
=
2π RO CO 2π (10 × 106 )( 0.2 × 10−12 )
f z = 79.6 kHz
fp =
1
2π Req CO
Req = 51.98 Ω from Example 11.20
1
fp =
2π ( 51.98 ) ( 0.2 × 10−12 )
f p = 15.3 GHz
TYU11.1
Vd = V1 − V2
= 2 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) ⇒ Vd = 1.5 + 0.010sin ω t ( V )
V1 + V2
2
2 + 0.005sin ω t + 0.5 − 0.005sin ω t
=
⇒ Vcm = 1.25 V
2
Vcm =
TYU11.2
For v1 = v2 = +4 V ⇒ Minimum vC1 = vC 2 = 4 V
IQ
= 1 mA
I C1 = I C 2 =
2
10 − 4
⇒ RC = 6 kΩ
RC =
1
TYU11.3
From Equation (11.41)
g R
CMRR = m O
⎛ ΔRC ⎞
⎜
⎟
⎝ RC ⎠
For CMRR |dB = 75 dB ⇒ CMRR = 5623.4
Then 5623.4 =
( 3.86 )(100 )
ΔRC
⋅ (10 )
Or ΔRC = 0.686 K
TYU11.4
From Equation (11.49)
1 + 2 RO g m
CMRR =
⎛ Δg ⎞
2⎜ m ⎟
⎝ gm ⎠
For CMRR |dB = 90 dB ⇒ CMRR = 31622.8
⎛ 1 + 2 (100 )( 3.86 ) ⎞
Then 31622.8 = ⎜
⎟ ( 3.86 )
2Δg m
⎝
⎠
Δg m 0.0472
Or Δg m = 0.0472 mA/V or
=
= 0.0122 ⇒ 1.22%
3.86
gm
TYU11.5
For v1 = v2 = 5 V ⇒ min vC1 = vC 2 = 5 V
So I C1 RC = 10 − 5 = 0.25 RC ⇒ RC = 20 kΩ
Ad =
I Q RC
=
4VT
( 0.5 )( 20 )
⇒ Ad
4 ( 0.026 )
= 96.2 Let I Q = 0.5 mA
C M RRdB = 95 db ⇒ C M RR = 5.62 × 104 ⇒ Acm =
Acm
⎛ I Q RC ⎞
⎜
⎟
⎝ 2VT ⎠
=
= 1.71× 10−3
⎡ (1 + β ) I Q R0 ⎤
⎢1 +
⎥
VT β
⎣
⎦
⎡ ( 0.5 )( 20 ) ⎤
⎢
⎥
⎣ 2 ( 0.026 ) ⎦
= 1.71× 10−3
⎡ ( 201)( 0.5 ) R0 ⎤
⎢1 +
⎥
⎣ ( 0.026 )( 200 ) ⎦
1 + 19.33R0 = 1.125 × 105 ⇒ R0
= 5.82 × 103 kΩ
= 5.82 MΩ
We have R0 = r04 ⎡⎣1 + g m 4 ( R2 rπ 4 ) ⎤⎦
r04 =
VA 125
=
= 250 kΩ
I Q 0.5
gm4 =
rπ 2 =
IQ
VT
β VT
IQ
=
0.5
= 19.23 mA/V
0.026
=
( 200 )( 0.026 )
0.5
= 10.4 kΩ
96.2
⇒ Acm = 1.71× 10−3
5.62 × 104
5.820 = 250 ⎡⎣1 + g m 4 ( R2 rπ 4 ) ⎤⎦
19.23 ( R2 rπ 2 ) = 22.28
(R
2
rπ 2 ) = 1.159 kΩ
R2 (10.4 )
R2 + 10.4
= 1.159
R2 (10.4 − 1.16 ) = (1.16 )(10.4 ) ⇒ R2 = 1.30 kΩ
Let I1 = 1 mA
⎛I
I Q R2 − I1 R3 = VT ln ⎜ 1
⎜I
⎝ Q
⎞
⎟⎟
⎠
( 0.5)(1.304 ) − (1) R3 = ( 0.026 ) ln ⎜⎛
If VBE ( Q3 ) ≅ 0.7 V
10 − 0.7 − ( −10 )
R1 + R3 =
1
1 ⎞
⎟ ⇒ R3 = 0.634 kΩ
⎝ 0.5 ⎠
= 19.3 ⇒ R1 ≅ 18.7 kΩ
TYU11.6
a.
v0 = Ad vd + Acm vcm
vd = v1 − v2 = 0.505sin ω t − 0.495sin ω t
= 0.01sin ω t
v +v
0.505sin ω t + 0.495sin ω t
vcm = 1 2 =
2
2
= 0.50sin ω t
v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5sin ω t ) ⇒ v0 = 0.85sin ω t ( V )
b.
vd = v1 − v2
= 0.5 + 0.005sin ω t − ( 0.5 − 0.005sin ω t )
= 0.01sin ω t
v +v
vcm = 1 2
2
0.5 + 0.005sin ω t + 0.5 − 0.005sin ω t
=
2
= 0.5
v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5 ) ⇒ v0 = 0.25 + 0.6sin ω t ( V )
TYU11.7
a.
b.
I B1 = I B 2 =
rπ =
β VT
I CQ
IQ / 2
=
1
⇒ I B1 = I B 2 = 6.62 µ A
151
(1 + β )
(150 )( 0.026 )
=
1
= 3.9 kΩ
Rid = 2rπ = 2 ( 3.9 ) = 7.8 kΩ
Ib =
c.
Vd 10sin ω t ( mV )
=
⇒ I b = 1.28sin ω t ( µ A )
Rid
7.8 kΩ
Ricm ≅ 2 (1 + β ) R0 = 2 (151)( 50 ) ⇒ 15.1 MΩ
Ib =
Vcm
3sin ω t
=
⇒ I b = 0.199sin ω t ( µ A )
Ricm 15.1 MΩ
TYU11.8
Ad = g f RD
8 = g f ( 4 ) ⇒ g f ( max ) = 2 mA / V
g f ( max ) =
( 2)
2
Kn IQ
2
⎛ 4⎞
= K n ⎜ ⎟ ⇒ K n = 2 mA / V 2
⎝ 2⎠
TYU11.9
From Example 11-10, I Q = 0.587 mA
⋅ (16) ⇒ Ad = 2.74
2
1
1
=
⇒ R0 = 85.2 kΩ
For M 4 , R0 =
λ4 I Q ( 0.02)( 0.587 )
Ad =
Kn2 IQ
2
⋅ RD =
( 0.1)( 0.587 )
g m = 2 K n (VGS 2 − VTN ) = 2 ( 0.1)( 2.71 − 1)
= 0.342 mA/V
Acm =
− ( 0.342)(16)
− g m RD
=
⇒ Acm = −0.0923
1 + 2 g m R0 1 + 2 ( 0.342)(85.2)
⎛ 2.74 ⎞
C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 29.4 dB
⎝ 0.0923 ⎠
TYU11.10
1
CMRR = ⎡⎣1 + 2 2 K n I Q ⋅ Ro ⎤⎦
2
CMRRdB = 60 dB ⇒ C M RR = 1000
1
1000 = ⎡1 + 2 2 ( 0.1)( 0.2 ) ⋅ R0 ⎤
⎦
2⎣
2000 = 1 + 0.4 R0 ⇒ R0 ≅ 5 MΩ
TYU11.11
Ro = ro 4 + ro 2 (1 + g m 4 ro 4 )
Assume I REF = I O = 100 µ A and λ = 0.01 V −1
ro 2 = ro 4 =
1
1
=
⇒1 MΩ
λ I D ( 0.01)( 0.1)
Let K n ( all devices ) = 0.1 mA / V 2
( 0.1)( 0.1) = 0.2 mA / V
Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) ⇒ 202 M Ω
Then g m 4 = 2 K n I D = 2
Now VGS1 = VGS 2 =
ID
0.05
+ VTN =
+ 1 = 1.707 V
Kn
0.1
VDS1 ( sat ) = VGS 1 − VTN = 1.707 − 1 = 0.707 V
So vo1 ( min ) = +4 − VGS 1 + VDS 1 ( sat ) = 4 − 1.707 + 0.707
vo1 ( min ) = 3 V = 10 − I D RD = 10 − ( 0.05 ) RD ⇒ RD = 140 kΩ
For a one-sided output, the differential gain is:
Ad =
1
g m1 RD where g m1 = 2 K n I D
2
=2
( 0.1)( 0.05) = 0.1414 mA / V
1
( 0.1414 )(140 ) ⇒ Ad = 9.90
2
The common-mode gain is:
2 K n I Q ⋅ RD
2 ( 0.1)( 0.1) ⋅ (140 )
Acm =
=
⇒ Acm = 0.0003465
1 + 2 2 K n I Q ⋅ Ro 1 + 2 2 ( 0.1)( 0.1) ⋅ ( 202000 )
Ad =
Ad
⇒ CMRRdB = 89.1 dB
Acm
Then CMRRdB = 20 log10
TYU11.12
I B5 =
a.
IQ
β (1 + β )
=
0.5
⇒ 15.3 nA
(180 )(181)
So I 0 = 15.3 nA
b.
For a balanced condition
VEC 4 = VEC 3 = VEB 3 + VEB 5 ⇒ VEC 4 = 1.4 V
VCE 2 = VC 2 − VE 2 = (10 − 1.4 ) − ( −0.7 ) ⇒ VCE 2 = 9.3 V
TYU11.13
Ad = 2 g f ( r02 r04 )
gf =
IQ
4VT
=
0.2
= 1.923 mA/V
4 ( 0.026 )
r02 =
VA 2 120
=
= 1200 kΩ
I C 2 0.1
r04 =
VA 4 80
=
= 800 kΩ
I C 4 0.1
Ad = 2 (1.923) (1200 800 ) ⇒ Ad = 1846
TYU11.14
P = ( I Q + I REF ) ( 5 − ( −5 ) )
10 = ( 0.1 + I REF )(10 ) ⇒ I REF = 0.9 mA
R1 =
5 − 0.7 − ( −5 )
I REF
=
9.3
⇒ R1 = 10.3 k Ω
0.9
⎛I ⎞
I Q RE = VT ln ⎜ REF ⎟
⎜ I ⎟
⎝ Q ⎠
0.026 ⎛ 0.9 ⎞
ln ⎜
RE =
⎟ ⇒ RE = 0.571 kΩ
0.1
⎝ 0.1 ⎠
ro 2 =
VA 2 120
=
⇒ 2.4 M Ω
I C 2 0.05
ro 4 =
VA 4
80
=
⇒ 1.6 M Ω
I C 4 0.05
gm =
0.05
= 1.923 mA / V
0.026
Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 ⇒ Ad = 158
(
)
(
)
TYU11.15
a.
R0 = r02 r04
120
r02 =
= 1.2 MΩ
0.1
80
r04 =
= 0.8 MΩ
0.1
R0 = 1.2 0.8 ⇒ R0 = 0.48 MΩ
b.
Ad ( open circuit ) = 2 g f ( r02 r04 )
(
Ad ( with load ) = 2 g f r02 r04 RL
For Ad ( with load ) =
)
1
Ad ( open circuit ) ⇒ RL = ( r02 r04 ) ⇒ RL = 0.48 MΩ
2
TYU11.16
2Kn
1
Ad = 2
⋅
I Q ( λ2 + λ4 )
=2
2 ( 0.1)
0.1
⋅
1
( 0.01 + 0.015)
⇒ Ad = 113
TYU11.17
For the MOSFET, I D = 25 −
g m1 = 2 K n I D = 2
75
= 24.26 µ A
101
( 20 )( 24.26 ) = 44.05 µ A/V
For the Bipolar, I E = 100 − 25 = 75 µ A
⎛ 100 ⎞
IC = ⎜
⎟ ( 75 ) = 74.26 µ A
⎝ 101 ⎠
(100 )( 0.026 )
rπ 2 =
= 35.0 K
0.07426
0.07426
= 2.856 mA/V
gm2 =
0.026
g (1 + g m 2 rπ ) ( 44.05 ) ⎡⎣1 + ( 2.856 )( 35 ) ⎤⎦
g mC = m1
=
1 + g m1rπ
1 + ( 0.04405 )( 35 )
g mC = 1.75 mA/V
TYU11.18
From Figure 11.43
80
r04 =
= 160 kΩ
0.5
R0 ≅ β r04 = (150 )(160 ) kΩ ⇒ R0 = 24 MΩ
From Figure 11.44
1
1
r06 =
=
⇒ r06 = 160 kΩ
λ I D ( 0.0125 )( 0.5 )
0.5 = 0.5 (VGS − 1) ⇒ VGS = 2 V
2
g m 6 = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2 − 1) = 1 mA / V
r04 = 160 kΩ
R0 = ( g m 6 )( r06 )( β r04 ) = (1)(160 )(150 )(160 ) ⇒ R0 = 3.840 MΩ
TYU11.19
From Equation (11.126)
2 (1 + β ) β VT 2 (121)(120 )( 0.026 )
=
⇒ Ri = 1.51 MΩ
Ri =
0.5
IQ
rπ 11 =
β VT
=
IQ
(120 )( 0.026 )
0.5
= 6.24 kΩ
RE′ = rπ 11 & R3 = 6.24 & 0.1 = 0.0984 kΩ
g m11 =
r011 =
IQ
VT
=
0.5
= 19.23 mA/V
0.026
VA 120
=
= 240 kΩ
I Q 0.5
Then RC11 = r011 (1 + g m11 RE′ )
= 240 ⎡⎣1 + (19.23)( 0.0984 ) ⎤⎦
= 694 kΩ
rπ 8 =
β VT
IC 8
=
(120 )( 0.026 )
2
= 1.56 kΩ
Rb8 = rπ 8 + (1 + β ) R4 = 1.56 + (121)( 5 )
= 607 kΩ
Then RL 7 = RC11 & Rb8 = 694 & 607 = 324 kΩ
⎡ 0.5 ⎤
⎛ IQ ⎞
Then Av = ⎜
⎥ ( 324 ) ⇒ Av = 3115
⎟ RL 7 = ⎢
⎢⎣ 2 ( 0.026 ) ⎦⎥
⎝ 2VT ⎠
⎛r +Z ⎞
R0 = R4 || ⎜ π 8
⎟
⎝ 1+ β ⎠
Z = RC11 & RC 7
V
120
= 240 kΩ
RC 7 = A =
I Q 0.5
Z = 694 & 240 = 178 kΩ
⎛ 1.56 + 178 ⎞
R0 = 5 || ⎜
⎟ = 5 & 1.48 ⇒ R0 = 1.14 kΩ
⎝ 121 ⎠
TYU11.20
⎛ IQ
Av = ⎜
⎝ 2VT
⎞
⎟ RL 7
⎠
⎛ 0.5 ⎞
103 = ⎜⎜
⎟⎟ RL 7 ⇒ RL 7 = 104 kΩ
⎝ 2 ( 0.026 ) ⎠
Chapter 11
Problem Solutions
11.1
(a)
−0.7 − ( −3)
RE
= 0.1 ⇒ RE = 23 K
3 − 1.5
= 0.05 ⇒ RC = 30 K
RC
(b)
vCE 2 = 6 − iC 2 ( RC + 2 RE ) = 6 − iC 2 ( 76 )
(c)
vcm ( max ) ⇒ vCB 2 = 0 ⇒ vCE 2 = 0.7 V
So 0.7 = 6 − iC 2 ( 76 ) ⇒ iC 2 = 69.74 µ A
( 0.06974 ) ⇒ vCM ( max ) = 0.908 V
23
( min ) ⇒ VS = −3 V ⇒ vCM ( min ) = −2.3 V
( v ( max ) − 0.7 ) − ( −3) = 2
CM
vCM
11.2
Ad = 180, C M RRdB = 85 dB
Ad
180
=
⇒ Acm = 0.01012
Acm
Acm
Assume the common-mode gain is negative.
v0 = Ad vd + Acm vcm
C M RR = 17, 783 =
= 180vd − 0.01012vcm
v0 = 180 ( 2sin ω t ) mV − ( 0.01012 )( 2sin ω t ) V
v0 = 0.36sin ω t − 0.02024sin ω t
Ideal Output:
v0 = 0.360sin ω t ( V )
Actual Output:
v0 = 0.340sin ω t ( V )
11.3
a.
I1 =
⇒ I1 = 1.01 mA
10 − 2 ( 0.7 )
IC 2 =
8.5
I1
2
1+
β (1 + β )
=
1.01
⇒ I C 2 ≅ 1.01 mA
2
1+
(100 )(101)
⎛ 100 ⎞ ⎛ 1.01 ⎞
IC 4 = ⎜
⎟⎜
⎟ ⇒ I C 4 ≅ 0.50 mA
⎝ 101 ⎠ ⎝ 2 ⎠
VCE 2 = ( 0 − 0.7 ) − ( −5 ) ⇒ VCE 2 = 4.3 V
VCE 4 = ⎣⎡5 − ( 0.5 )( 2 ) ⎦⎤ − ( −0.7 ) ⇒ VCE 4 = 4.7 V
b.
For VCE 4 = 2.5 V ⇒ VC 4 = −0.7 + 2.5 = 1.8 V
5 − 1.8
⇒ I C 4 = 1.6 mA
IC 4 =
2
⎛ 1+ β ⎞
⎛ 101 ⎞
IC 2 + ⎜
⎟ ( 2IC 4 ) = ⎜
⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA
100 ⎠
β
⎝
⎝
⎠
I1 ≈ I C 2 = 3.23 mA
R1 =
10 − 2 ( 0.7 )
3.23
⇒ R1 = 2.66 kΩ
11.4
a.
Neglecting base currents
30 − 0.7
I1 = I 3 = 400 µ A ⇒ R1 =
⇒ R1 = 73.25 kΩ
0.4
VCE1 = 10 V ⇒ VC1 = 9.3 V
15 − 9.3
RC =
⇒ RC = 28.5 kΩ
0.2
b.
(100 )( 0.026 )
= 13 kΩ
rπ =
0.2
50
= 125 kΩ
r0 ( Q3 ) =
0.4
We have
Ad =
(100 )( 28.5)
⇒ Ad
2 ( rπ + RB )
2 (13 + 10 )
β RC
=
= 62
⎧
⎫
⎪
⎪
β RC ⎪
1
⎪
Acm = −
⎨
2
1
β
r
+
rπ + RB ⎪
(
) ⎬⎪
1+ 0
rπ + RB ⎭⎪
⎩⎪
⎧
⎫
⎪⎪
(100 )( 28.5 ) ⎪⎪
1
=−
⎨
⎬ ⇒ Acm = −0.113
13 + 10 ⎪ 2 (125 )(101) ⎪
1+
13 + 10 ⎭⎪
⎩⎪
⎛ 62 ⎞
C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 54.8 dB
⎝ 0.113 ⎠
c.
Rid = 2 ( rπ + RB ) = 2 (13 + 10 ) ⇒ Rid = 46 kΩ
1
⎡ rπ + RB + 2 (1 + β ) r0 ⎤⎦
2⎣
1
= ⎡⎣13 + 10 + 2 (101)(125 ) ⎤⎦ ⇒ Ricm = 12.6 MΩ
2
Ricm =
11.5
vCM ( max ) ⇒ VCB = 0 so that vCM ( max ) = 5 −
(a)
vCM ( max ) = 3 V
(b)
IQ
2
( RC ) = 5 −
Vd ⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞
=⎜
=⎜
⎟⋅
⎟⎜
⎟ = 0.08654 mA
2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠
= ΔI ⋅ RC = ( 0.08654 ) ( 8 ) = 0.692 V
ΔI = g m ⋅
ΔVC 2
(c)
⎛ 0.25 ⎞ ⎛ 0.010 ⎞
ΔI = ⎜
⎟⎜
⎟ = 0.04808 mA
⎝ 0.026 ⎠ ⎝ 2 ⎠
ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V
11.6
P = ( I1 + I C 4 ) (V + − V − )
I1 ≅ I C 4 so 1.2 = 2 I1 ( 6 ) ⇒ I1 = I C 4 = 0.1 mA
R1 =
3 − 0.7 − ( −3)
0.1
⇒ R1 = 53 k Ω
For vCM = +1V ⇒ VC1 = VC 2 = 1 V ⇒ RC =
3 −1
⇒ RC = 40 k Ω
0.05
One-sided output
1
0.05
Ad = g m RC where g m =
= 1.923 mA / V
2
0.026
Then
1
Ad = (1.923)( 40 ) ⇒ Ad = 38.5
2
11.7
a.
IE
( 2 ) + I E (85) − 5
2
5 − 0.7
⇒ I E = 0.050 mA
IE =
85 + 1
⎛ β ⎞ ⎛ I E ⎞ ⎛ 100 ⎞⎛ 0.050 ⎞
I C1 = I C 2 = ⎜
⎟⎜ ⎟ = ⎜
⎟⎜
⎟
⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 101 ⎠⎝ 2 ⎠
0 = 0.7 +
Or I C1 = I C 2 = 0.0248 mA
VCE1 = VCE 2 = ⎡⎣5 − I C1 (100 ) ⎤⎦ − ( −0.7 )
So VCE1 = VCE 2 = 3.22 V
b.
vcm ( max ) for VCB = 0 and VC = 5 − I C1 (100 ) = 2.52 V
So vcm ( max ) = 2.52 V
vcm ( min ) for Q1 and Q2 at the edge of cutoff ⇒ vcm ( min ) = −4.3 V
(c) Differential-mode half circuits
( 0.5)
2
(8)
⎛V
⎞
vd
= Vπ + ⎜ π + g mVπ ⎟ .RE′
2
⎝ rπ
⎠
⎡ (1 + β ) ⎤
RE′ ⎥
= Vπ ⎢1 +
rπ
⎣
⎦
Then
−
Vπ =
− ( vd / 2 )
⎡ (1 + β ) ⎤
RE′ ⎥
⎢1 +
rπ
⎣
⎦
vo = − g mVπ RC ⇒ Ad =
rπ =
β VT
=
I CQ
β RC
1
⋅
2 rπ + (1 + β ) RE′
(100 )( 0.026 )
= 105 k Ω RE′ = 2 k Ω
0.0248
Then
Ad =
11.8
a.
RE =
1 (100 )(100 )
⋅
⇒ Ad = 16.3
2 105 + (101)( 2 )
For v1 = v2 = 0 and neglecting base currents
−0.7 − ( −10 )
0.15
⇒ RE = 62 kΩ
b.
Ad =
rπ =
Ad =
β RC
v02
=
vd 2 ( rπ + RB )
β VT
I CQ
=
(100 )( 0.026 )
0.075
(100 )( 50 )
⇒ Ad
2 ( 34.7 + 0.5 )
= 34.7 kΩ
= 71.0
⎡
⎤
⎢
⎥
β RC ⎢
1
⎥
Acm = −
rπ + RB ⎢ 2 RE (1 + β ) ⎥
⎢1 +
⎥
rπ + RB ⎥⎦
⎢⎣
⎡
⎤
⎥
1
=−
⎢
⎥ ⇒ Acm = −0.398
34.7 + 0.5 ⎢ 2 ( 62 )(101) ⎥
⎢⎣1 + 34.7 + 0.5 ⎥⎦
71.0
⇒ C M RRdB = 45.0 dB
C M RRdB = 20 log10
0.398
c.
Rid = 2 ( rπ + RB )
(100 )( 50 ) ⎢
Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩ
Common-mode input resistance
1
Ricm = ⎡⎣ rπ + RB + 2 (1 + β ) RE ⎤⎦
2
1
= ⎣⎡34.7 + 0.5 + 2 (101)( 62 ) ⎦⎤ ⇒ Ricm = 6.28 MΩ
2
11.9
(a)
v1 = v2 = 1 V ⇒ VE = 1.6
IE =
9 − 1.6
⇒ 18.97 µ A
390
IE
= 9.49 µ A I C1 = I C 2 = 9.39 µ A
2
vC1 = vC 2 = ( 9.39 )( 0.51) − 9 = −4.21 V
(b)
9.39
gm =
⇒ 0.361 mA/V
0.026
V
ΔI = g m d = ( 0.361× 10−3 ) ( 0.005 ) = 1.805 µ A
2
ΔvC = (1.805 × 10−6 )( 510 × 103 ) = 0.921 V ⇒ vC 2 = −4.21 + 0.921 ⇒ −3.29 V
vC1 = −4.21 − 0.921 ⇒ −5.13 V
11.10
(a)
v1 = v2 = 0
I E1 = I E 2 ≅ 6 µ A
β = 60
I C1 = I C 2 = 5.90 µ A
vC1 = vC 2 = ( 5.90 )( 0.360 ) − 3
= −0.875 V
VEC1 = VEC 2 = +0.6 − ( −0.875 )
= 1.475 V
(b)
(i)
5.90
⇒ 0.227 mA/V
gm =
0.026
Ad = g m RC = ( 0.227 )( 360 ) = 81.7
Acm = 0
(ii)
( 60 )( 0.026 )
g R
Ad = m C = 40.8 rπ =
2
0.0059
= 264 K
Acm =
11.11
− ( 0.227 )( 360 )
= −0.0442
2 ( 61)( 4000 )
1+
264
For v1 = v2 = 0.20 V
I C1 = I C 2 = 0.1 mA
vC1 = vC 2 = ( 0.1)( 30 ) − 10
= −7 V
0.1
= 3.846 mA/V
gm =
0.026
v
ΔI = g m d = ( 3.846 )( 0.008 ) ⇒ 30.77 µ A
2
ΔvC = ΔI ⋅ RC = ( 30.77 × 10−6 )( 30 × 103 ) = 0.923 V
v2 ↑⇒ I C 2 ↓⇒ vC 2 ↓⇒ vC1
= −7 + 0.923
= −6.077 V
vC 2 = −7 − 0.923
= −7.923 V
11.12
RC = 50 K
For v1 = v2 = 0
IE =
−0.7 − ( −10 )
75
= 0.124 mA
I C1 = I C 2 = 0.0615 mA
0.0615
= 2.365 mA/V
gm =
0.026
(120 )( 0.026 )
= 50.7 K
rπ =
0.0615
Differential Input
v
V
v1 = d v2 = − d
2
2
Half-circuit.
V
ΔR ⎞
⎛
ΔI = + g m d ⇒ ΔvC1 = −ΔI ⎜ RC +
⎟
2
2 ⎠
⎝
vo = ΔvC1 − ΔvC 2
= −2ΔIRC
ΔR ⎞
⎛
ΔvC 2 = +ΔI ⎜ RC −
⎟
2 ⎠
⎝
ΔR ⎞
ΔR ⎞
⎛
⎛
= −ΔI ⎜ RC +
⎟ − ΔI ⎜ RC −
⎟
2 ⎠
2 ⎠
⎝
⎝
⎛ V ⎞
= −2 ⎜ g m d ⎟ RC
2⎠
⎝
Ad = − g m RC = − ( 2.365 )( 50 ) = −118.25
Common-mode input.
⎛V
⎞
vcm = Vπ + ⎜ π + g mVπ ⎟ ( 2 RE )
⎝ rπ
⎠
vcm
Vπ =
⎛ β⎞
1 + ⎜ 1 + ⎟ ( 2 RE )
⎝ rπ ⎠
ΔI = g mVπ =
β vcm
g m vcm
=
rπ + (1 + β )( 2 RE )
⎛1+ β ⎞
1+ ⎜
⎟ ( 2 RE )
⎝ rπ ⎠
ΔR ⎞
⎛
− β ⎜ RC +
⎟ ⋅ vcm
2 ⎠
⎝
ΔvC1 = −ΔIR1 =
rπ + (1 + β )( 2 RE )
ΔvC 2
ΔR ⎞
⎛
− β ⎜ RC −
⎟ vcm
2 ⎠
⎝
= −ΔIR2 =
rπ + (1 + β )( 2 RE )
vo = ΔvC1 − ΔvC 2
ΔR ⎞
ΔR ⎞
⎛
⎛
− β ⎜ RC +
⎟ vcm
⎟ vcm + β ⎜ RC −
2 ⎠
2 ⎠
⎝
⎝
=
⎛ ΔR ⎞
−2 β ⎜
⎟ vcm
⎝ 2 ⎠
=
rπ + (1 + β )( 2 RE )
Acm =
[ ]
− (120 )( 0.5 )
− βΔR
=
rπ + (1 + β )( 2 RE ) 50.7 + (121)( 2 )( 75 )
= −0.0032966
118.25
C M RR =
= 35,870.5
0.0032966
C M R R ∫ = 91.1 dB
dB
11.13
v1 = v2 = 0
IE =
−0.7 − ( −10 )
75
= 0.124 mA
I C1 = I C 2 = 0.0615 mA
0.0615
gm =
= 2.365 mA/V
0.026
Δg m
= 0.01
gm
g m1 = 2.377 mA/V
g m 2 = 2.353 mA/V
rπ =
[ ]
(120 )( 0.026 )
0.0615
= 50.7 K
Vd
2
V
= − g m1 d Rc
2
Vd
= + gm2
Rc
2
ΔI = g m
ΔvC1
ΔvC 2
vo = ΔvC1 − ΔvC 2 = − g m1
Vd
V
RC − g m 2 d RC
2
2
Vd
RC ( g m1 + g m 2 )
2
R
−50
Ad = − C ( g m1 + g m 2 ) =
( 2.377 + 2.353) ⇒ Ad = −118.25
2
2
Common-Mode
− g m1 RC vcm
− g m 2 RC vcm
ΔvC1 =
ΔvC 2 =
⎛ 1+ β ⎞
⎛ 1+ β ⎞
1+ ⎜
1+ ⎜
⎟ ( 2 RE )
⎟ ( 2 RE )
⎝ rπ ⎠
⎝ rπ ⎠
=−
− ( g m1 − g m 2 ) RC
− ( 2.377 − 2.353) ( 50 )
vo
= Acm =
=
vcm
⎛ 1+ β ⎞
⎛ 121 ⎞
1+ ⎜
1+ ⎜
⎟ ( 2 )( 75 )
⎟ ( 2 RE )
⎝ 50.7 ⎠
⎝ rπ ⎠
−1.2
=
⇒ Acm = −0.003343
358.99
C M R R ∫ = 91 dB
dB
11.14
(a)
v1 = v2 = 0
vE = +0.7 V
5 − 0.7
= 4.3 mA
IE =
1
I C1 = I C 2 = 2.132 mA
vC1 = vC 2 = ( 2.132 )(1) − 5
= −2.87 V
v1 = 0.5, v2 = 0 Q2 on
(b)
Q1 off
⎛ 120 ⎞
I C1 = 0, I C 2 = 4.3 ⎜
⎟ mA = 4.264 mA
⎝ 121 ⎠
vC1 = −5 V vC 2 = ( 4.264 ) (1) − 5
vC 2 = −0.736 V
2.132
= 82.0 mA/V
0.026
(82.0 )
v
V
ΔI = g m d ΔvC = ΔI ⋅ RC = g m d ⋅ RC =
⋅ Vd (1) = 41.0Vd
2
2
2
Vd = 0.015 ⇒ Δvc = 0.615 V
(c)
vE ≈ 0.7 V
gm =
vC 2 ↓ vC1 ↑
vC1 = −2.87 + 0.615 = −2.255 V
vC 2 = −2.87 − 0.615 = −3.485 V
11.15
(a)
gm =
IC
1
=
= 38.46 mA/V
VT 0.026
Ad =
vo
1
=
= 100
vd 0.01
Ad = g m RC
100 = 38.46 RC
Rc = 2.6 K
(b)
With v1 = v2 = 0
vC1 = vC 2 = 10 − (1)( 2.6 ) = 7.4 V ⇒ vcm ( max ) = 7.4 V
11.16
a.
i.
( v01 − v02 ) = 0
ii.
I C1 = I C 2 = 1 mA
v01 − v02 = ⎡⎣V + − I C1 RC1 ⎤⎦ − ⎡⎣V + − I C 2 RC 2 ⎤⎦
= I C ( RC 2 − RC1 ) = (1)( 7.9 − 8 ) ⇒ v01 − v02 = −0.1 V
b.
⎛v ⎞
I 0 = ( I S 1 + I S 2 ) exp ⎜ BE ⎟
⎝ VT ⎠
⎛v ⎞
2 × 10−3
So exp ⎜ BE ⎟ = −13
−13
⎝ VT ⎠ 10 + 1.1× 10
= 9.524 × 109
⎛v
I C1 = I S 1 exp ⎜ BE
⎝ VT
⎞
−13
9
⎟ = (10 )( 9.524 × 10 ) ⇒ I C1 = 0.952 mA
⎠
I C 2 = (1.1× 10−13 )( 9.524 × 109 ) ⇒ I C 2 = 1.048 mA
i.
v01 − v02 = I C 2 RC 2 − I C1 RC1 ⇒ v01 − v02 = (1.048 − 0.952 )( 8 ) ⇒ v01 − v02 = 0.768 V
ii.
v01 − v02 = (1.048 )( 7.9 ) − ( 0.952 )( 8 )
v01 − v02 = 8.279 − 7.616 ⇒ v01 − v02 = 0.663 V
11.17
From Equation (11.12(b))
IQ
iC 2 =
1 + evd / VT
1
0.90 =
1 + evd / VT
1
So evd / VT =
− 1 = 0.111
0.90
vd = VT ln ( 0.111) = ( 0.026 ) ln ( 0.111) ⇒ vd = −0.0571 V
11.18
From Example 11.2, we have
vd ( max )
1
−
− vd ( max ) / 0.026
4 ( 0.026 ) 1 + e
= 0.02
v ( max )
0.5 + d
4 ( 0.026 )
0.5 +
⎡
v ( max ) ⎤
1
0.98 ⎢0.5 + d
⎥=
− vd ( max ) / 0.026
4 ( 0.026 ) ⎦⎥ 1 + e
⎣⎢
0.490 + 9.423vd ( max ) =
1
1+ e
− vd ( max ) / 0.026
By trial and error
vd ( max ) = 23.7 mV
11.19
a.
For I1 = 1 mA, VBE3 = 0.7 V
20 − 0.7
R1 =
⇒ R1 = 19.3 kΩ
1
⎛ I ⎞ 0.026 ⎛ 1 ⎞
V
R2 = T ⋅ ln ⎜ 1 ⎟ =
⋅ ln ⎜
⎟ ⇒ R2 = 0.599 kΩ
⎜
⎟
0.1
IQ
⎝ 0.1 ⎠
⎝ IQ ⎠
b.
(180 )( 0.026 )
rπ 4 =
= 46.8 kΩ
0.1
0.1
gm =
= 3.846 mA/V
0.026
100
r04 =
⇒ 1 MΩ
0.1
From Chapter 10
R0 = r04 ⎡⎣1 + g m ( RE & rπ 4 ) ⎤⎦
RE & rπ 4 = 0.599 & 46.8 = 0.591
R0 = (1) ⎣⎡1 + ( 3.846 )( 0.591) ⎦⎤ = 3.27 MΩ
r01 =
100
⇒ 2 MΩ
0.05
⎡
⎛ r ⎞⎤
Ricm ≅ ⎡⎣(1 + β ) R0 ⎤⎦ & ⎢(1 + β ) ⎜ 01 ⎟⎥
⎝ 2 ⎠⎦
⎣
= ⎣⎡(181)( 3.27 ) ⎦⎤ & ⎣⎡(181)(1) ⎦⎤
= 592 & 181 ⇒ Ricm = 139 MΩ
(c)
From Eq. (11.32(b))
− g m RC
Acm =
2 (1 + β ) Ro
1+
rπ + RB
0.05
= 1.923 mA / V
0.026
(180 )( 0.026 )
= 93.6 k Ω
rπ =
0.05
RB = 0
gm =
Then
Acm =
− (1.923)( 50 )
⇒ Acm = −0.00760
2 (181)( 3270 )
1+
93.6
11.19
For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need the
quiescent collector voltage to be
VC = 3.5 + 1 = 4.5 V
Assume the bias is ±10 V , and I Q = 0.5 mA.
Then I C = 0.25 mA
10 − 4.5
⇒ RC = 22 k Ω
0.25
(100 )( 0.026 )
In this case, rπ =
= 10.4 k Ω
0.25
Then
(100 )( 22 )
Ad =
= 101 So gain specification is met.
2 (10.4 + 0.5 )
Now RC =
For CMRRdB = 80 dB ⇒
1 ⎡ (1 + β ) I Q Ro ⎤ 1 ⎡ (101)( 0.5 ) Ro ⎤
⎥ ⇒ Ro = 1.03 M Ω
⎢1 +
⎥ = ⎢1 +
2⎣
VT β
⎦ 2 ⎣⎢ ( 0.026 )(100 ) ⎦⎥
Need to use a Modified Widlar current source.
Ro = ro ⎡⎣1 + g m ( RE1 & rπ ) ⎤⎦
CMRR = 104 =
If VA = 100V , then ro =
rπ =
100
= 200 k Ω
0.5
(100 )( 0.026 )
= 5.2 k Ω
0.5
0.5
= 19.23 mA / V
gm =
0.026
Then 1030 = 200 ⎡⎣1 + (19.23)( RE1 & rπ ) ⎤⎦ ⇒ RE1 & rπ = 0.216 k Ω = RE1 & 5.2 ⇒ RE1 = 225 Ω
Also let RE 2 = 225 Ω and I REF ≅ 0.5 mA
11.20
(a)
(b)
RE =
−0.7 − ( −10 )
0.25
⇒ RE = 37.2 k Ω
⎛1+ β ⎞
Vπ 1
V
V
Ve
+ g mVπ 1 + π 2 + g mVπ 2 = e or (1) ⎜
⎟ (Vπ 1 + Vπ 2 ) =
rπ
rπ
RE
r
R
E
⎝ π ⎠
⎛ r
⎞
Vπ 1 V1 − Ve
=
⇒ Vπ 1 = ⎜ π ⎟ (V1 − Ve )
+
rπ
RB + rπ
r
R
B ⎠
⎝ π
Vπ 2 = V2 − Ve
Then
⎛1+ β
⎝ rπ
⎞ ⎡ rπ
⎤ V
(V1 − Ve ) + (V2 − Ve )⎥ = e
⎟⎢
⎠ ⎣ rπ + RB
⎦ RE
From this, we find
(1) ⎜
V1 +
rπ + RB
⋅ V2
rπ
⎡ rπ + RB
r + RB ⎤
+1+ π
⎢
⎥
rπ ⎦
⎣ RE (1 + β )
Ve =
Now
Vo = − g mVπ 2 RC = − g m RC (V2 − Ve )
We have
rπ =
(120 )( 0.026 )
0.125
≅ 25 k Ω,
gm =
0.125
= 4.81 mA / V
0.026
(i)
Set V1 =
Vd
V
and V2 = − d
2
2
Then
⎛ ⎛ 25 + 0.5 ⎞ ⎞
Vd
⎜ 1 − ⎜ 25 ⎟ ⎟
( −0.02 )
⎝
⎠
⎝
⎠
Ve =
= 2
2.026
⎡ 25 + 0.5
25 + 0.5 ⎤
+1+
⎢
⎥
25 ⎦
⎣ ( 37.2 )(121)
Vd
2
So
Ve = −0.00494Vd
Now
V
⎛ V
⎞
Vo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 )Vd ⎟ ⇒ Ad = o = 119
Vd
⎝ 2
⎠
(ii)
Set V1 = V2 = Vcm
Then
⎛ 25 + 0.5 ⎞
Vcm ⎜ 1 +
⎟
V ( −2.02 )
25 ⎠
⎝
= cm
Ve =
2.02567
⎡ 25 + 0.5
25 + 0.5 ⎤
+1+
⎢
⎥
25 ⎦
⎣ ( 37.2 )(121)
Ve = Vcm ( 0.9972 )
Then
Vo = − ( 4.81)( 50 ) ⎡⎣Vcm − Vcm ( 0.9972 ) ⎤⎦
or Acm =
Vo
= −0.673
Vcm
11.21
From Equation (11.18)
v0 = vC 2 − vC1 = g m RC vd
gm =
I CQ
VT
For I Q = 2 mA, I CQ = 1 mA
1
= 38.46 mA/V
0.026
Now 2 = ( 38.46 ) RC ( 0.015 )
Then g m =
So RC = 3.47 kΩ
Now VC = V + − I C RC
= 10 − (1)( 3.47 )
= 6.53 V
For VCB = 0 ⇒ vcm ( max ) = 6.53 V
11.22
The small-signal equivalent circuit is
A KVL equation: v1 = Vπ 1 − Vπ 2 + v2
v1 − v2 = Vπ 1 − Vπ 2
A KCL equation
Vπ 1
V
+ g mVπ 1 + π 2 + g mVπ 2 = 0
rπ
rπ
⎛1
⎞
+ g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2
⎝ rπ
⎠
(Vπ 1 + Vπ 2 ) ⎜
Then v1 − v2 = 2Vπ 1 ⇒ Vπ 1 =
1
1
( v1 − v2 ) and Vπ 2 = − ( v1 − v2 )
2
2
At the v01 node:
v01 v01 − v02
+
+ g mVπ 1 = 0
RC
RL
⎛ 1
⎛ 1
1 ⎞
+
v01 ⎜
⎟ − v02 ⎜
⎝ RL
⎝ RC RL ⎠
At the v02 node:
⎞ 1
⎟ = g m ( v2 − v1 )
⎠ 2
(1)
v02 v02 − v01
+
+ g mVπ 2 = 0
RC
RL
⎛ 1
⎛ 1 ⎞ 1
1 ⎞
+
v02 ⎜
⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 )
⎝ RL ⎠ 2
⎝ RC RL ⎠
From (1):
(2)
⎛ R ⎞ 1
v02 = v01 ⎜ 1 + L ⎟ − g m RL ( v2 − v1 )
⎝ RC ⎠ 2
Substituting into (2)
⎛ R ⎞⎛ 1
⎛ 1
⎛ 1
1 ⎞ 1
1 ⎞
+
+
v01 ⎜1 + L ⎟ ⎜
⎟ − g m RL ( v2 − v1/ ) ⎜
⎟ − v01 ⎜
R
R
R
R
R
2
⎝ RL
C ⎠⎝ C
L ⎠
L ⎠
⎝
⎝ C
⎡ ⎛ RL
⎛ 1 RL
⎞⎤
1 ⎞ 1
+ 2 +
+ 1⎟ ⎥
v01 ⎜
⎟ = g m ( v1 − v2 ) ⎢1 − ⎜
⎝ RC RC RC ⎠ 2
⎠⎦
⎣ ⎝ RC
v01 ⎛
RL ⎞
1 ⎛ RL ⎞
⎜2+
⎟ = − gm ⎜
⎟ ( v1 − v2 )
RC ⎝
RC ⎠
2 ⎝ RC ⎠
For v1 − v2 = vd
1
− g m RL
v01
Av1 =
= 2
vd ⎛
RL ⎞
⎜2+
⎟
RC ⎠
⎝
1
g m RL
v02
From symmetry: Av 2 =
= 2
vd ⎛
RL ⎞
⎜2+
⎟
RC ⎠
⎝
Then Av =
v02 − v01
g m RL
=
vd
⎛
RL ⎞
⎜2+
⎟
RC ⎠
⎝
11.23
The small-signal equivalent circuit is
KVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2
KCL equation:
⎞ 1
⎟ = g m ( v1 − v2 )
⎠ 2
Vπ 1
V
+ g mVπ 1 + g mVπ 2 + π 2 = 0
rπ
rπ
⎛1
⎞
+ g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2
⎝ rπ
⎠
(Vπ 1 + Vπ 2 ) ⎜
Then v1 − v2 = −2Vπ 2 or Vπ 2 = −
Now
1
( v1 − v2 )
2
v0 = − g mVπ 2 ( RC & RL )
1
g m ( RC & RL )( v1 − v2 )
2
v
1
For v1 − v2 ≡ vd ⇒ Ad = 0 = g m ( RC & RL )
vd 2
=
11.23
a.
10 − 7
⇒ RD = 6 kΩ
0.5
I Q = I D1 + I D 2 ⇒ I Q = 1 mA
RD =
b.
10 = I D ( 6 ) + VDS − VGS
and VGS =
ID
+ VTN
Kn
For I D = 0.5 mA, VGS =
0.5
+ 2 = 3.12 V
0.4
and VDS = 10.12
Load line is actually nonlinear.
c.
Maximum common-mode voltage when M 1 and M 2 reach the transition point, or
VDS ( sat ) = VGS − VTN = 3.12 = 2 = 1.12V
Then
vcm = v02 − vDS ( sat ) + VGS = 7 − 1.12 + 3.12
Or vcm ( max ) = 9 V
Minimum common-mode voltage, voltage across I Q becomes zero.
So vcm ( min ) = −10 + 3.12
⇒ vcm ( min ) = −6.88 V
11.24
We have VC 2 = − g mVπ 2 RC = − g m (Vb 2 − Ve ) RC
and
VC1 = − g mVπ 1 RC = − g m (Vb1 − Ve ) RC
Then
V0 = VC 2 − VC1
= − g m (Vb 2 − Ve ) RC − ⎡⎣ − g m (Vb1 − Ve ) RC ⎤⎦
= g m RC (Vb1 − Vb 2 )
Differential gain Ad =
V0
= g m RC
Vb1 − Vb 2
Common-mode gain Acm = 0
11.25
(a)
vcm = 3 V ⇒ VC1 = VC 2 = 3 V
10 − 3
Then RC =
⇒ RC = 70 k Ω
0.1
(b)
CMRRdB = 75 dB ⇒ CMRR = 5623
Now
CMRR =
5623 =
1 ⎡ (1 + β ) I Q Ro ⎤
⎢1 +
⎥
β VT
2⎣
⎦
1 ⎡ (151)( 0.2 ) Ro ⎤
⎢1 +
⎥ ⇒ Ro = 1.45 M Ω
2 ⎣⎢ (150 )( 0.026 ) ⎦⎥
Use a Widlar current source.
Ro = ro [1 + g m RE′ ]
Let VA of current source transistor be 100 V.
100
0.2
= 500 k Ω, g m =
= 7.69 mA / V
Then ro =
0.2
0.026
(150 )( 0.026 )
rπ =
= 19.5 k Ω
0.2
So 1450 = 500 ⎡⎣1 + ( 7.69 ) RE′ ⎤⎦ ⇒ RE′ = 0.247 k Ω
Now RE′ = RE & rπ ⇒ 0.247 = RE & 19.5 ⇒ RE = 250Ω
⎛I
Then I Q RE = VT ln ⎜ REF
⎜ I
⎝ Q
⎞
⎟⎟
⎠
⎛ I REF ⎞
⎟⎟ ⇒ I REF = 1.37 mA
⎝ ( 0.2 ) ⎠
( 0.2 )( 0.250 ) = ( 0.026 ) ln ⎜⎜
Then R1 =
10 − 0.7 − ( −10 )
11.26
At terminal A.
RTHA = RA & R =
1.37
⇒ R1 = 14.1 k Ω
R (1 + δ ) ⋅ R
R (1 + δ ) + R
=
R (1 + δ )
2+δ
Variation in RTH is not significant
≅
R
= 5 kΩ
2
⎛ RA ⎞ + R (1 + δ )( 5 ) 5 (1 + δ )
=
VTHA = ⎜
⎟V =
R (1 + δ ) + R
2+δ
⎝ RA + R ⎠
At terminal B.
R
= 5 kΩ
2
⎛ R ⎞ +
VTHB = ⎜
⎟ V = 2.5 V
⎝R+R⎠
From Eq. (11.27)
− β RC (V2 − V1 )
where V2 = VTHB and V1 = VTHA
VO =
2 ( rπ + RB )
RTHB = R & R =
RB = 5 k Ω, rπ =
So VO =
(120 )( 0.026 )
0.25
− (120 )( 3)(V2 − V1 )
2 (12.5 + 5 )
= 12.5 k Ω
= −10.3 (V2 − V1 )
We can find V2 − V1 = VTHB − VTHA
⎡ 5 (1 + δ ) ⎤
VTHB − VTHA = 2.5 − ⎢
⎥
⎣ 2+δ ⎦
2.5 ( 2 + δ ) − 5 (1 + δ ) 2.5δ − 5δ
=
=
2+δ
2+δ
−2.5δ
≅
= −1.25δ
2
Then VO = − (10.3)( −1.25 ) δ = 12.9δ
So for −0.01 ≤ δ ≤ 0.01
We have −0.129 ≤ VO 2 ≤ 0.129 V
11.27
a.
Rid = 2rπ
rπ =
(180 )( 0.026 )
0.2
So Rid = 46.8 kΩ
= 23.4 kΩ
Assuming rµ → ∞, then
b.
Ricm ≅ ⎡⎣(1 + β ) R0 ⎤⎦
Ricm = ⎡⎣(181)(1) ⎤⎦
= 181 ⇒ Ricm = 181 MΩ
11.28
(a)
10 − 0.7 − ( −10 )
I1 =
= 0.5 ⇒ R1 = 38.6 K
R1
R2 =
(b)
0.026 ⎛ 0.5 ⎞
ln ⎜
⎟ ⇒ R2 = 236 Ω
0.14 ⎝ 0.14 ⎠
Ricm ≈ (1 + β ) Ro
0.14
= 5.385 mA/V
0.026
(180 )( 0.026 )
rπ 4 =
= 33.4 K
0.14
RE′ = 33.4 0.236 = 0.234 K
Ro = ro 4 (1 + g m 4 RE′ ) g m 4 =
100
= 714 K
0.14
Ro = 714 ⎡⎣1 + ( 5.385 )( 0.234 ) ⎤⎦
ro 4 =
= 1614 K
Ricm = (181)(1614 ) ≈ 292 MΩ
(c)
Acm =
− g m1 RC
2 (1 + β ) Ro
1+
rπ 1
g m1 =
rπ 1 =
0.07
= 2.692 mA/V
0.026
(180 )( 0.026 )
0.07
= 66.86 K
− ( 2.692 )( 40 )
2 (181)(1614 )
1+
66.86
Acm = −0.0123
Acm =
11.29
Ad 1 = g m1 ( R1 rπ 3 )
g m1 =
rπ 3 =
Ad 2 =
I Q1 / 2
VT
β VT
IQ2 / 2
= 19.23I Q1
=
2 (100 )( 0.026 )
IQ 2
=
5.2
IQ 2
IQ 2 / 2
g m 3 R2
, g m3 =
= 19.23I Q 2
2
VT
⋅ R2 ⇒ I Q 2 R2 = 3.12 V
2
Maximum vo 2 − vo1 = ±18 mV for linearity
Then 30 =
(19.23) I Q 2
vo3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 V
so I Q 2 R2 = 3.12 V is OK.
From Ad 1 :
⎛
⎛ 5.2 ⎞ ⎞
⎜ R1 ⎜⎜
⎟ ⎟
I Q 2 ⎟⎠ ⎟
⎜
⎝
20 = 19.23I Q1 ( R1 & rπ 3 ) = 19.23I Q1 ⎜
⎟
⎜ R + ⎛⎜ 5.2 ⎞⎟ ⎟
⎜ 1 ⎜ IQ 2 ⎟ ⎟
⎝
⎠⎠
⎝
19.23I Q1 R1 ( 5.2 )
20 =
I Q 2 R1 + 5.2
I Q1
Let
2
⋅ R1 = 5V ⇒ I Q1 R1 = 10 V
Then 20 =
19.23 (10 )( 5.2 )
I Q 2 R1 + 5.2
Now I Q1 R1 = 10 ⇒ R1 =
⇒ I Q 2 R1 = 44.8 V
10
I Q1
⎛ IQ2 ⎞
⎛ 10 ⎞
= 44.8 ⇒ ⎜
= 4.48
So I Q 2 ⎜
⎟
⎜ I ⎟⎟
⎜I ⎟
⎝ Q1 ⎠
⎝ Q1 ⎠
Let I Q1 = 100 µ A, I Q 2 = 448 µ A
Then
I Q 2 R2 = 3.12 ⇒ R2 = 6.96 k Ω
I Q1 R1 = 10 ⇒ R1 = 100 k Ω
11.30
a.
20 − VGS 3
2
I1 =
= 0.25 (VGS 3 − 2 )
50
20 − VGS 3 = 12.5 (VGS2 3 − 4VGS 3 + 4 )
12.5VGS2 3 − 49VGS 3 + 30 = 0
VGS 3 =
49 ±
( 49 )
2
− 4 (12.5 )( 30 )
2 (12.5 )
⇒ VGS 3 = 3.16 V
20 − 3.16
⇒ I1 = I Q = 0.337 mA
50
IQ
=
⇒ I D1 = 0.168 mA
2
I1 =
I D1
0.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS1 = 2.82 V
VDS 4 = −2.82 − ( −10 ) ⇒ VDS 4 = 7.18 V
2
VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V
VDS1 = 5.97 − ( −2.82 ) ⇒ VDS 1 = 8.79 V
(b)
(c)
Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS1 − VTN
2.82 − 2 = 0.82 V
Now VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V
VS ( max ) = 5.97 − VDS1 ( sat ) = 5.97 − 0.82
VS ( max ) = 5.15 V
vCM ( max ) = VS ( max ) + VGS1 = 5.15 + 2.82
vCM ( max ) = 7.97 V
vCM ( min ) = V − + VDS 4 ( sat ) + VGS 1
VDS 4 ( sat ) = VGS 4 − VTN = 3.16 − 2 = 1.16 V
Then vCM ( min ) = −10 + 1.16 + 2.82 ⇒ vCM ( min ) = −6.02 V
11.31
a.
I D1 = I D 2 = 120 µ A = 100 ( VGS1 − 1.2 ) ⇒ VGS 1 = VGS 2 = 2.30 V
2
For v1 = v2 = −5.4 V and VDS1 = VDS 2 = 12 V ⇒ −5.4 − 2.30 + 12 = 4.3 V = VD
10 − 4.3
⇒ RD = 47.5 kΩ
0.12
I Q = I D1 + I D 2 ⇒ I Q = I1 = 240 µ A
RD =
I1 = 240 = 200 (VGS 3 − 1.2 ) ⇒ VGS 3 = 2.30 V
2
R1 =
20 − 2.3
⇒ R1 = 73.75 kΩ
0.24
b.
r04 =
1
λ IQ
ΔI Q =
=
1
( 0.01)( 0.24 )
= 416.7 kΩ
1
5.4
⋅ ΔVDS =
⇒ ΔI Q ≅ 13 µ A
r04
416.7
11.32
(a)
I Q = 160 µ A
k′ ⎛ W ⎞
2
I D = n ⎜ ⎟ (VGS − VTN )
2⎝L⎠
80
2
80 = ( 4 )(VOS − 0.5 )
2
80 = 160 (Vo5 − 0.5 )
2
80
+ 0.5 = 1.207 V
160
5−2
= 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V
RD =
0.08
(c)
VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 V
VGS =
Then VS = VO 2 − VDS ( sat ) = 2 − 0.707 = +1.29 V
And v1 = v2 = vcm = VGS + VS = 1.207 + 1.29
vcm = 2.50 V
(b)
11.33
vD = 5 − ( 0.2 )( 8 ) = 3.4 V
VGS =
ID
+ VTN
Kn
0.2
+ 0.8 = 1.694 V
0.25
VDS ( sat ) = VGS − VTN = 1.694 − 0.8
= 0.894 V
VS = VD − VDS ( sat ) = 3.4 − 0.894
= 2.506
=
vCM = VS + VGS = 2.506 + 1.694 ⇒ vCM = 4.2 V
(b)
ΔvD = ΔI D ⋅ RD
ΔI D = g m ⋅
Vd
2
gm = 2 Kn I D
ΔI D = ( 0.4472 )( 0.05 ) ⇒ 22.36 µ A
=2
( 0.25)( 0.2 ) = 0.4472 mA/V
ΔvD = ( 22.36 × 10−6 )( 8 × 103 ) = 0.179 V
vD 2 = 3.4 + ΔvD
vD 2 = 3.4 + 0.179 ⇒ vD 2 = 3.58 V
(c)
vd = −50 mV
ΔI D = − ( 0.4472 )( 0.025 ) ⇒ −11.18 µ A
ΔvD = − (11.18 × 10−6 )( 8 × 103 ) = −0.0894 V
vD 2 = 3.4 − 0.0894 ⇒ vD 2 = 3.31 V
11.34
a.
I D1 = I D 2 = 0.5 mA
v01 − v02 = ⎡⎣V + − I D1 RD1 ⎤⎦ − ⎡⎣V + − I D 2 RD 2 ⎤⎦
v01 − v02 = I D 2 RD 2 − I D1 RD1 = I D ( RD 2 − RD1 )
i.
RD1 − RD 2 = 6 kΩ, v01 − v02 = 0
ii.
RD1 = 6 kΩ, RD 2 = 5.9 kΩ
v01 − v02 = ( 0.5 )( 5.9 − 6 ) ⇒ v01 − v02 = −0.05 V
b.
K n1 = 0.4 mA / V 2 , K n 2 = 0.44 mA / V 2
VGS1 = VGS 2
I Q = ( K n1 + K n 2 )(VGS − VTN )
2
1 = ( 0.4 + 0.44 )(VGS − VTN ) ⇒ (VGS − VTN ) = 1.19
2
2
I D1 = ( 0.4 )(1.19 ) = 0.476 mA
I D 2 = ( 0.44 )(1.19 ) = 0.524 mA
i.
RD1 = RD 2 = 6 kΩ
v01 − v02 = ( 0.524 − 0.476 )( 6 ) ⇒ v01 − v02 = 0.288 V
ii.
RD1 = 6 kΩ,
RD 2 = 5.9 kΩ
v01 − v02 = ( 0.524 )( 5.9 ) − ( 0.476 )( 6 )
= 3.0916 − 2.856 ⇒ v01 − v02 = 0.236 V
11.35
(a)
From Equation (11.69)
⎛ K
Kn
iD 2 1
= −
⋅ vd 1 − ⎜ n
⎜ 2IQ
IQ 2
2IQ
⎝
0.90 = 0.50 −
⎞ 2
⎟⎟ vd
⎠
⎡ 0.1 ⎤ 2
0.1
⋅ vd 1 − ⎢
⎥ vd
2 ( 0.25 )
⎣⎢ 2 ( 0.25 ) ⎦⎥
+0.40 = − ( 0.4472 ) vd 1 − ( 0.2 ) vd2
0.8945 = −vd 1 − ( 0.2 ) vd2
Square both sides
0.80 = vd2 (1 − [ 0.2] vd2 )
( 0.2 ) ( vd2 )
vd2 =
2
− vd2 + 0.80 = 0
1 ± 1 − 4 ( 0.2 )( 0.80 )
2 ( 0.2 )
= 4V 2 or 1V 2
Then vd = ± 2 V or ± 1 V
But vd
max
=
IQ
kn
=
0.25
= 1.58
0.1
So vd = ±1V, ⇒ vd = −1V
b.
11.36
From part (a), vd ,max = 1.58 V
⎛i ⎞
d ⎜ D1 ⎟
⎜I ⎟
⎝ Q⎠=
dvd
⎛ K
Kn
⋅ 1− ⎜ n
⎜ 2I
2IQ
⎝ Q
) vd
vd =0
Kn
2IQ
=
So linear
⎞ 2
⎟⎟ vd + (
⎠
Kn
iD1 1
= +
⋅ vd
IQ 2
2 IQ
⎡1
⎤
⎛K ⎞
Kn
Kn
1
+
⋅ vd ( max ) − ⎢ +
⋅ vd ( max ) ⋅ 1 − ⎜ n ⎟vd2( max ) ⎥
2
2IQ
2 IQ
⎢⎣ 2
⎥⎦
⎝ 2I n ⎠
Then
= 0.02
Kn
1
+
⋅v
2
2 I Q d ( max )
⎡1
⎤ ⎡1
⎛K
Kn
Kn
0.98 ⎢ +
⋅ vd ( max ) ⎥ = ⎢ +
⋅ vd ( max ) ⋅ 1 − ⎜ n
⎜ 2I
2IQ
2IQ
⎝ Q
⎣⎢ 2
⎦⎥ ⎢⎣ 2
0.49 + 0.98
⎤
⎞ 2
⎟⎟ vd ( max ) ⎥
⎥
⎠
⎦
⎡1
⎤
⎛ 0.15 ⎞ 2
0.15
0.15
⋅ vd ( max ) = ⎢ +
⋅ vd ( max ) ⋅ 1 − ⎜⎜
⎟⎟ vd ( max ) ⎥
2 ( 0.2 )
2 ( 0.2 )
⎢2
⎥
⎝ 2 ( 0.2 ) ⎠
⎣
⎦
0.49 + 0.600 vd ( max ) = 0.50 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd2( max )
0.600 vd ( max ) = 0.010 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd2( max )
By trial and error vd ( max ) ≈ 0.429 V
11.37
(b)
gm = 2 K p I D = 2
( 0.05 )( 0.008696 )
= 0.0417 mA/V
Vd
= ( 0.0417 )( 0.05 ) = 0.002085 mA
2
ΔvD = ( 0.002085 )( 510 ) = 1.063
ΔI = g m
vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 V
vD1 = −1.063 − 4.565 = −5.628 V
9 = I S RS + VSG + 1
I S = 2I D
8 = 2 K P RS (VSG + VTP ) + VSG
2
8 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG
2
2
− 1.6VSG + 0.64 ) + VSG
8 = 39 (VSG
2
− 61.4VSG + 16.96 = 0
39VSG
VSG =
61.4 ± 3769.96 − 4 ( 39 )(16.96 )
2 ( 39 )
= 1.217 V VS = 2.217
I S = 0.01739 mA
I D1 = I D 2 ⇒ 8.696 µ A
vD1 = vD 2 = ( 8.696 )( 0.510 ) − 9 = −4.565 V
(b)
g m = 2 K P I DQ = 2
( 0.05 )( 0.008696 ) = 0.0417 mA/V
Vd
= ( 0.0417 )( 0.05 ) = 0.002085 mA
2
ΔvD = ( 0.002085 )( 510 ) = 1.063 V
ΔvD = ΔI D ⋅ RD
ΔI D = g m ⋅
v1 ↑, I D1 ↓, vD1 ↓
vD1 = −4.565 − 1.063 = −5.628 V
vD 2 = −4.565 + 1.063 = −3.502 V
11.38
(a)
v1 = v2 = 0
I D = K n (VSG + VTP )
ID = 6 µA
2
6
+ 0.4 = VSG
30
VSG = 0.847 V
VS = +0.847 V
vD = I D RD − 3
= ( 6 )( 0.36 ) − 3 = −0.84 V
VSD = VS − vD = 0.847 − ( −0.84 )
vSD = 1.69 V
(b)
(i)
Ad = g m RD g m = 2 K n I D
=2
( 30 )( 6 ) = 26.83 µ A/V
Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66
Acm = 0
(ii)
( 26.83)( 0.36 )
g R
Ad = m D =
⇒ Ad = 4.83
2
2
− ( 26.83)( 0.36 )
− g m RD
Acm =
=
= −0.0448
1 + 2 g m RO 1 + 2 ( 26.83)( 4 )
11.39
For v1 = v2 = −0.30 V
I D1 = I D 2 = 0.1 mA
VSG =
ID
− VTP
KP
0.1
+1 = 2 V
0.1
= vD 2 = ( 0.1)( 30 ) − 10
= −7 V
=
vD1
gm = 2 K p I D = 2
( 0.1)( 0.1) = 0.2 mA/V
⎛V ⎞
ΔI D = g m ⎜ d ⎟ = ( 0.2 )( 0.1) = 0.02 mA
⎝ 2⎠
ΔvD = ( ΔI D ) RD = ( 0.02 )( 30 ) = 0.6 V
vD 2 ↑⇒ vD 2 = −7 + 0.6 ⇒ vD 2 = −6.4 V
vD1 = −7 − 0.6 ⇒ vD1 = −7.6 V
11.40
For v1 = v2 = 0
0 = VGS + 2 I D RS − 10
10 = VGS + 2 K n RS (VGS − VTN )
2
= VGS + 2 ( 0.15 )( 75 )(VGS − 1)
2
22.5VGS2 − 44VGS + 12.5 = 0
So VGS = 1.61 V and I D = ( 0.15 )(1.61 − 1) ⇒ 55.9 µ A
2
gm = 2 Kn I D = 2
( 0.15 )( 0.0559 )
g m = 0.1831 mA/V
Use Half-circuits – Differential gain
ΔR ⎞
⎛V ⎞⎛
vD1 = − g m ⎜ d ⎟ ⎜ RD +
⎟
2 ⎠
⎝ 2 ⎠⎝
ΔR ⎞
⎛V ⎞⎛
vo 2 = g m ⎜ d ⎟ ⎜ RD −
⎟
2 ⎠
⎝ 2 ⎠⎝
vo = vD1 − vD 2 = − g mVd RD
v
Ad = o = − g m RD
Vd
Now – Common-Mode Gain
Vi = Vgs + g mVgs ( 2 RS ) = Vcm
Vcm
Vgs =
1 + g m ( 2 RS )
vD1
vD 2
ΔR ⎞
⎛
− g m ⎜ RD + D ⎟ Vcm
2 ⎠
⎝
=
1 + g m ( 2 RS )
ΔR ⎞
⎛
− gm ⎜ RD − D ⎟ Vcm
2 ⎠
⎝
=
1 + g m ( 2 RS )
vO = vD1 − vD 2
So vo =
Acm =
− g m ( ΔRD ) Vcm
1 + g m ( 2 RD )
− g m ( ΔRD )
vo
=
Vcm 1 + g m ( 2 RS )
Then
Ad = − ( 0.1831)( 50 ) = −9.16
Acm =
− ( 0.1831)( 0.5 )
1 + ( 0.1831)( 2 )( 75 )
C M R R ∫ = 69.1 dB
= −0.003216
bB
11.41
a.
Ad = g m ( r02 r04 )
r02 =
VA 2 150
=
= 375 kΩ
I C 2 0.4
r04 =
VA 4 100
=
= 250 kΩ
I C 4 0.4
gm =
IC 2
0.4
=
= 15.38 mA/V
VT
0.026
Ad = (15.38 ) ( 375 250 ) ⇒ Ad = 2307
b.
RL = r02 r04 = 375 250 ⇒ RL = 150 kΩ
11.41
From 11.40
I D1 = I D 2 = 55.9 µ A
g m = 0.183 mA/V
⎛ +V ⎞
ΔvD 2 = + g m 2 ⎜ d ⎟ RD
⎝ 2 ⎠
V
V
vO = ΔvD1 − ΔvD 2 = − g m1 d RD − g m 2 d RD
2
2
−V
−V
⎛
Δg
Δg ⎞ ⎞
⎛
vO = d ⋅ RD ( g m 2 + g m1 ) = d ⋅ RD ⎜ g m − m + ⎜ g m − m ⎟ ⎟
2
2
2
2 ⎠⎠
⎝
⎝
Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15
Ad : ΔvD1 = − g m1
Vd
⋅ RD
2
Δg ⎞
Δg ⎞
⎛
⎛
− ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM
2 ⎠
2 ⎠
⎝
⎝
=
+
1 + g m ( 2 RS )
1 + g m ( 2 RS )
ACM : vO = ΔvD1 − ΔvD 2
−Δg m RD
vO
=
vcm 1 + g m ( 2 RS )
Acm =
Acm =
Δg m = ( 0.01) ( 0.183) = 0.00183
− ( 0.00183) ( 50 )
1 + ( 0.183)( 2 ) ( 75 )
C M R R ∫ = 69.1 dB
= −0.003216
dB
11.42
(a)
v1 = v2 = 0
5 = 2 I D RS + VSG
5 = 2 K p RS (VSG + VTP ) + VSG
2
2
5 = 2 ( 0.5 )( 2 ) (VSG
− 1.6VSG + 0.64 ) + VSG
2
5 = 2VSG
− 2.2VSG + 1.28
2
2VSG − 2.2VSG − 3.72 = 0
VSG =
2.2 ± 4.84 + 4 ( 2 )( 3.72 )
2 ( 2)
VSG = 2.02 V
vS = 2.02 V,
5 − 2.02
= 1.49 mA
2
= I D 2 = 0.745 mA
IS =
I D1
vD1 = vD 2 = ( 0.745 (1) − 5 ) ⇒ vD1 = vD 2 = −4.26 V
(b)
5 = I S RS + VSG 2
5 = ( I D1 + I D 2 ) RS + VSG 2
2
2
5 = ⎡ K p (VSG1 + VTP ) + K p (VSG 2 + VTP ) ⎤ RS + VSG 2
⎣
⎦
VSG1 = VSG 2 − 1
5 = ( 0.5 )( 2 ) ⎡(VSG 2 − 1.8 ) + (VSG 2 − 0.8 ) ⎤ + VSG 2
⎣
⎦
2
2
5 = ⎡⎣VSG 2 − 3.6VSG 2 + 3.24 + VSG 2 − 1.6VSG 2 + 0.64 ⎤⎦ + VSG 2
2
2
5 = 2VSG
2 − 4.2VSG 2 + 3.88
2
2VSG 2 − 4.2VSG 2 − 1.12 = 0
VSG 2 =
4.2 ± 17.64 + 4 ( 2 ) (1.12 )
2 ( 2)
2
VSG 2 = 2.339 V VSG1 = 1.339 V
vS = 2.339 V
I D1
I D1
vD1
vD1
= 0.5 (1.339 − 0.8 )
= 0.1453 mA
= ( 0.1453)(1) − 5
= −4.855 V
2
I D2
I D2
vD 2
vD 2
= 0.5 ( 2.339 − 0.8 )
= 1.184 mA
= (1.184 ) (1) − 5
= −3.816 V
2
(c)
ΔI = g m
Vd
2
gm = 2 K p I D
vS ≈ 2.02 V = 2
( 0.5 )( 0.745 )
g m = 1.22 mA/V
ΔI = (1.22 )( 0.1) = 0.122 mA
ΔvD = ( ΔI ) RD = ( 0.122 )(1) = 0.122 V
vD 2 ↓ vD1 ↑
vD1 = −4.26 + 0.122 vD 2 = −4.26 − 0.122
vD1 = −4.138 V
vD 2 = −4.382 V
11.43
a.
gf =
IQ
4VT
⇒ I Q = 0.832 mA
⇒ I Q = g f ( 4VT ) = ( 8 )( 4 )( 0.026 )
Neglecting base currents.
30 − 0.7
R1 =
⇒ R1 = 35.2 kΩ
0.832
V
100
b.
r04 = r02 = A =
= 240 kΩ
I CQ 0.416
gm =
I CQ
VT
=
0.416
= 16 mA / V
0.026
Ad = g m ( r02 || r04 ) = 16 ( 240 || 240 )
⇒ Ad = 1920
Rid = 2rπ , rπ =
(180 )( 0.026 )
⇒ Rid = 22.5 kΩ
0.416
= 11.25 kΩ
R0 = r02 || r04 ⇒ R0 = 120 kΩ
c.
Max. common-mode voltage when
VCB = 0 for Q1 and Q2 .
Therefore
vcm ( max ) = V + − VEB ( Q3 ) = 15 − 0.7
vcm ( max ) = 14.3 V
Min. common-mode voltage when
VCB = 0 for Q5 .
Therefore
vcm ( min ) = 0.7 + 0.7 + ( −15 ) = −13.6 V
So −13.6 ≤ vcm ≤ 14.3 V
1
(1 + β )( 2 R0 )
2
V
100
R0 = A =
= 120 kΩ
I Q 0.832
Ricm ≅
Ricm = (181)(120 ) ⇒ Ricm = 21.7 MΩ
11.43
(a)
gm = 2 Kn I D
=2
( 0.4 )(1)
g m = 1.265 mA/V
v
1
Ad = o =
= 10
vd 0.1
Ad = g m RD
10 = (1.265 ) RD
RD = 7.91 K
(b)
Quiescent v1 = v2 = 0
vD1 = vD 2 = 10 − (1)( 7.91) = 2.09 V
VGS =
ID
1
+ VTN =
+ 0.8 = 2.38 V
Kn
0.4
VDS ( sat ) = 2.38 − 0.8 = 1.58
So vcm = vD − VDS ( sat ) + VGS
= 2.09 − 1.58 + 2.38
vcm = 2.89 V
11.44
g m RD
2
For vCM = 2.5 V
IQ
= 0.25 mA
I D1 = I D 2 =
2
Ad =
Let VD1 = VD 2 = 3 V , then RD =
Then 100 =
g m ( 28 )
2
k′ ⎛ W
And g m = 2 n ⎜
2⎝L
10 − 3
⇒ RD = 28 k Ω
0.25
⇒ g m = 7.14 mA / V
⎞
⎟ ID
⎠
⎛ 0.080 ⎞ ⎛ W ⎞
7.14 = 2 ⎜
⎟ ⎜ ⎟ ( 0.25 ) ⇒
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 1274 (Extremely large transistors to meet the gain requirement.)
⎝ L ⎠1 ⎝ L ⎠ 2
Need ACM = 0.10
From Eq. (11.64(b))
g m RD
ACM =
1 + 2 g m Ro
( 7.14 )( 28)
⇒ Ro = 140 k Ω
1 + 2 ( 7.14 ) Ro
For the basic 2-transistor current source
1
1
Ro = ro =
=
= 200 k Ω
λ I Q ( 0.01)( 0.5 )
So 0.10 =
This current source is adequate to meet common-mode gain requirement.
11.45
Not in detail, Approximation looks good.
a.
−V − ( −5 )
2
and I S = 2 I D = 2 K n (VGS 1 − VTN )
I S = GS 1
RS
5 − VGS 1
2
= 2 ( 0.050 )(VGS 1 − 1)
20
5 − VGS 1 = 2 (VGS2 1 − 2VGS1 + 1)
2VGS2 1 − 3VGS 1 − 3 = 0
VGS1 =
3±
( 3)
2
+ 4 ( 2 )( 3)
2 ( 2)
⇒ VGS1 = 2.186 V
5 − 2.186
⇒ I S = 0.141 mA
20
I
I D1 = I D 2 = S ⇒ I D1 = I D 2 = 0.0704 mA
2
v02 = 5 − ( 0.0704 )( 25 ) ⇒ v02 = 3.24 V
IS =
b.
g m = 2 K n (VGS − VTN ) = 2 ( 0.05 )( 2.186 − 1)
g m = 0.119 mA/V
1
1
r0 =
=
= 710 kΩ
λ I DQ ( 0.02 )( 0.0704 )
Vgs1 = v1 − VS , Vgs 2 = v2 − VS
v01
v −V
+ g mVgs1 + 01 S = 0
RD
r0
⎛ 1
V
1⎞
v01 ⎜
+ ⎟ + g m ( v1 − VS ) − S = 0
r0
⎝ RD r0 ⎠
v02
v − VS
+ g mVgs 2 + 02
=0
RD
r0
⎛ 1
V
1⎞
v02 ⎜
+ ⎟ + g m ( v2 − VS ) − S = 0
R
r
r0
⎝ D 0⎠
v − V v − VS
V
+ g mVgs 2 = S
g mVgs1 + 01 S + 02
r0
r0
RS
(1)
(2)
g m ( v1 − VS ) +
v01 v02 2VS
V
+
−
+ g m ( v2 − VS ) = S
r0
r0
r0
RS
g m ( v1 + v2 ) +
v01 v02
+
= VS
r0
r0
⎧
2 1 ⎫
⎨2 g m + + ⎬
r
RS ⎭
0
⎩
(3)
From (1)
⎛
1⎞
VS ⎜ g m + ⎟ − g m v1
r
0 ⎠
v01 = ⎝
⎛ 1
1⎞
+ ⎟
⎜
⎝ RD r0 ⎠
Then
⎛
1⎞
VS ⎜ g m + ⎟ − g m v1
r0 ⎠
⎧
v
2 1 ⎫
g m ( v1 + v2 ) + ⎝
+ 02 = VS ⎨2 g m + + ⎬ (3)
r0
r0 RS ⎭
⎛ 1
1⎞
⎩
r0 ⎜
+ ⎟
R
r
⎝ D 0⎠
⎛ 1
⎛
⎛ 1
⎧
1⎞
1⎞
1⎞
2 1 ⎫ ⎛ 1
1⎞
g m ( v1 + v2 ) r0 ⎜
+ ⎟ + VS ⎜ g m + ⎟ − g m v1 + v02 ⎜
+ ⎟ = VS ⎨2 g m + + ⎬ ⋅ r0 ⎜
+ ⎟
r0 ⎠
r0 RS ⎭ ⎝ RD r0 ⎠
⎝ RD r0 ⎠
⎝
⎝ RD r0 ⎠
⎩
⎛ 1
⎛
r ⎞
r0 ⎞ ⎛
1⎞
2 1 ⎞⎛
1 ⎞ ⎪⎫
⎪⎧⎛
g m ( v1 + v2 ) ⎜1 + 0 ⎟ − g m v1 + v02 ⎜
+ ⎟ = VS ⎨⎜ 2 g m + +
⎟ ⎜1 +
⎟ − ⎜ gm + ⎟⎬
R
R
r
r
R
R
r
⎪⎩⎝
0
0 ⎠⎪
D ⎠
S ⎠⎝
D ⎠ ⎝
⎝
⎝ D 0⎠
⎭
⎛ 1
⎧
⎛
r
r ⎞
r
r
1⎞
2 1
2
1⎫
g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜
+ ⎟ = VS ⎨2 g m + +
+ 2gm ⋅ 0 +
+ 0 − gm − ⎬
RD ⎠
r0 RS
RD RD RS RD
r0 ⎭
⎝ RD
⎝ RD r0 ⎠
⎩
⎧⎪
⎫
⎛ 1
⎛
r
r ⎞
r0 ⎞ 2
1⎞
1 1 ⎛
+ ⎟ = VS ⎨2 g m + +
g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜
(1 + g m r0 )⎪⎬ (4)
⎜1 +
⎟+
RD ⎠
r0 RS ⎝ RD ⎠ RD
⎪⎩
⎝ RD
⎝ RD r0 ⎠
⎭⎪
⎛ 1
⎛
1⎞
1⎞
+ ⎟ + g m v2 = VS ⎜ g m + ⎟
Then substituting into (2), v02 ⎜
r0 ⎠
⎝ RD r0 ⎠
⎝
710 ⎤
1 ⎤
⎡ 710
⎡1
(4)
Substitute numbers: ( 0.119 ) ⎢v1
+ v2 + v2
+ v02 ⎢ +
⎥
⎥
25 ⎦
⎣ 25
⎣ 25 710 ⎦
1
1 ⎛ 710 ⎞ 2
⎧
⎫
= VS ⎨0.119 +
+ ⎜1 +
⎟ + ⎣⎡1 + ( 0.119 )( 710 ) ⎦⎤ ⎬
710 20 ⎝
25 ⎠ 25
⎩
⎭
( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392}
= VS ( 8.4296 )
or VS = 0.4010v1 + 0.4150v2 + 0.00491v02
1 ⎞
1 ⎞
⎛ 1
⎛
(2)
Then v02 ⎜ +
⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 +
⎟
710 ⎠
⎝ 25 710 ⎠
⎝
v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ]
v02 ( 0.0408 ) = ( 0.04828 ) v1 − ( 0.0690 ) v2
v02 = (1.183) v1 − (1.691) v2
vd
2
vd
v2 = vcm −
2
v ⎞
v ⎞
⎛
⎛
So v02 = (1.183) ⎜ vcm + d ⎟ − (1.691) ⎜ vcm − d ⎟
2
2⎠
⎝
⎠
⎝
Or v02 = 1.437vd − 0.508vcm ⇒ Ad = 1.437, Acm = −0.508
Now
v1 = vcm +
⎛ 1.437 ⎞
C M R RdB = 20 log10 ⎜
⎟ ⇒ C M R RdB = 9.03 dB
⎝ 0.508 ⎠
11.46
KVL:
v1 = Vgs1 − Vgs 2 + v2
So v1 − v2 = Vgs1 − Vgs 2
KCL:
g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2
1
1
So Vgs1 = ( v1 − v2 ) , Vgs 2 = − ( v1 − v2 )
2
2
Now
v02 v02 − v01
+
= − g mVgs 2
RD
RL
⎛ 1
1 ⎞ v01
= v02 ⎜
+
⎟−
⎝ RD RL ⎠ RL
v01 v01 − v02
+
= − g mVgs1
RD
RL
⎛ 1
1 ⎞ v02
= v01 ⎜
+
⎟−
R
R
RL
L ⎠
⎝ D
⎛
R ⎞
From (1): v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2
R
⎝
D ⎠
(1)
(2)
Substitute into (2):
⎛
⎛ 1
v02
R ⎞⎛ 1
1 ⎞
1 ⎞
− g mVgs1 = v02 ⎜1 + L ⎟ ⎜
+
+
⎟ + g m RL ⎜
⎟ Vgs 2 −
R
R
R
R
R
RL
D ⎠⎝ D
L ⎠
L ⎠
⎝
⎝ D
⎛
⎛ 1
R ⎞⎛ 1 ⎞
R
1 ⎞
− g m ⋅ ( v1 − v2 ) + g m ⎜ 1 + L ⎟ ⎜ ⎟ ( v1 − v2 ) = v02 ⎜
+ L2 +
⎟
⎝ RD ⎠ ⎝ 2 ⎠
⎝ RD RD RD ⎠
1
⋅ g m RL
v02 ⎛
v02
RL ⎞
1 ⎛ RL ⎞
2
2
gm ⎜
v
v
A
−
=
+
⇒
=
=
(
)
⎟ 1 2
⎜
⎟
d2
2 ⎝ RD ⎠
RD ⎝
RD ⎠
v1 − v2 ⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
1
− ⋅ g m RL
v01
= 2
From symmetry Ad 1 =
v1 − v2 ⎛
RL ⎞
⎜2+
⎟
R
D ⎠
⎝
Then Av =
11.47
v02 − v01
g m RL
=
v1 − v2
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
v1 − v2 = Vgs1 − Vgs 2 and g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2
Then v1 − v2 = −2Vgs 2
Or Vgs 2 = −
1
( v1 − v2 )
2
v0 = − g mVgs 2 ( RD RL ) =
Or Ad =
gm
( RD RL ) ( v1 − v2 )
2
gm
( RD RL )
2
11.48
From Equation (11.64(a)), Ad =
We need Ad =
Then 10 =
Kn IQ
2
⋅ RD
2
= 10
0.2
K n ( 0.5 )
⋅ RD or K n ⋅ RD = 20
2
If we set RD = 20 k Ω, then K n = 1 mA / V 2
For this case VD = 10 − ( 0.25 )( 20 ) = 5 V
0.25
+ 1 = 1.5 V
1
VDS ( sat ) = VGS − VTN = 1.5 − 1 = 0.5 V
VGS =
Then vcm ( max ) = VD − VDS ( sat ) + VGS
= 5 − 0.5 + 1.5
Or vcm ( max ) = 6 V
11.49
Vd 1 = − g mVgs1 RD = − g m RD (V1 − Vs )
Vd 2 = − g mVgs 2 RD = − g m RD (V2 − Vs )
Now Vo = Vd 2 − Vd 1 = − g m RD (V2 − Vs ) − ( − g m RD (V1 − Vs ) )
Vo = g m RD (V1 − V2 )
Define V1 − V2 ≡ Vd
V
Then Ad = o = g m RD and Acm = 0
Vd
11.49
Ad = g m ( r02 r04 )
g m = 2 kn I DQ
=2
( 0.12 )( 0.075 )
= 0.1897 mA/V
1
1
=
= 889 kΩ
r02 =
λn I DQ ( 0.015 )( 0.075 )
r04 =
1
λ p I DQ
=
1
= 667 kΩ
0.02
( )( 0.075)
Ad = ( 0.1897 ) ( 889 667 ) ⇒ Ad = 72.3
11.50
(a)
⎛ K′ ⎞⎛W
K n1 = K n 2 = ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
VGS1 = VGS 2 =
⎞ ⎛ 0.080 ⎞
2
⎟=⎜
⎟ (10 ) = 0.40 mA / V
⎠ ⎝ 2 ⎠
ID
0.1
+ VTN =
+ 1 = 1.5 V
Kn
0.4
VDS1 ( sat ) = 1.5 − 1 = 0.5 V
For vCM = +3 V ⇒ VD1 = VD 2 = vCM − VGS 1 + VDS 1 ( sat )
= 3 − 1.5 + 0.5 ⇒ VD1 = VD 2 = 2 V
RD =
10 − 2
⇒ RD = 80 k Ω
0.1
(b)
1
g m RD and g m = 2 ( 0.4 )( 0.1) = 0.4 mA / V
2
1
Then Ad = ( 0.4 )( 80 ) = 16
2
16
C M R RdB = 45 ⇒ C M R R = 177.8 =
Acm
Ad =
So Acm = 0.090
Acm =
g m RD
1 + 2 g m Ro
0.090 =
( 0.4 )(80 )
1 + 2 ( 0.4 ) Ro
⇒ Ro = 443 k Ω
If we assume λ = 0.01 V −1 for the current source transistor, then
1
1
ro =
=
= 500 k Ω
λ I Q ( 0.01)( 0.2 )
So the CMRR specification can be met by a 2-transistor current source.
⎛W ⎞ ⎛W ⎞
Let ⎜ ⎟ = ⎜ ⎟ = 1
⎝ L ⎠3 ⎝ L ⎠ 4
IQ
0.2
⎛ 0.080 ⎞
2
Then K n 3 = K n 4 = ⎜
+ VTN =
+ 1 = 3.24 V
⎟ (1) = 0.040 mA / V and VGS 3 =
2
0.04
K
⎝
⎠
n3
For vCM = −3 V , VD 3 = −3 − VGS1 = −3 − 1.5 = −4.5 V ⇒ VDS 3 ( min ) = −4.5 − ( −10 ) = 5.5 V > VDS 3 ( sat )
So design is OK.
⎛W ⎞
On reference side: For ⎜ ⎟ ≥ 1, VGS ( max ) = 3.24 V
⎝L⎠
20 − VGS 3 = 20 − 3.24 = 16.76 V
Then
16.67
= 5.17 ⇒ We need six transistors in series.
3.24
20 − 3.24
= 2.793 V
6
2
⎛ K′ ⎞⎛W ⎞
= ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
⎝ 2 ⎠⎝ L ⎠
VGS =
I REF
2
⎛ 0.080 ⎞⎛ W ⎞
⎛W ⎞
0.2 = ⎜
⎟⎜ ⎟ ( 2.793 − 1) ⇒ ⎜ ⎟ = 1.56 for each of the 6 transistors.
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
11.51
Ad =
1
g m RD
2
gm = 2 Kn I D = 2
( 0.25 )( 0.25) = 0.50 mA / V
1
( 0.50 )( 3) = 0.75
2
From Problem 11.26
Ad =
V1 = VA =
5 (1 + δ )
2+δ
, V2 = VB = 2.5 V and V1 − V2 = 1.25δ
Then
Vo 2 = Ad ⋅ (V1 − V2 ) = ( 0.75 )(1.25δ ) = 0.9375δ
So for −0.01 ≤ δ ≤ 0.01
−9.375 ≤ Vo 2 ≤ 9.375 mV
11.52
From previous results
v −v
Ad 1 = o 2 o1 = g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20
v1 − v2
and Ad 2 =
Set
I Q1 R1
2
vo3
1
1
2 K n3 I Q 2 ⋅ R2 = 30
= g m 3 R2 =
2
vo 2 − vo1 2
I Q 2 R2
= 5 V and
2
= 2.5 V
Let I Q1 = I Q 2 = 0.1 mA
Then R1 = 100 k Ω, R2 = 50 k Ω
⎛ 0.06 ⎞ ⎛ W ⎞
⎛ 20 ⎞
⎛W ⎞ ⎛W ⎞
Then 2 ⎜
⎟ ⎜ ⎟ ( 0.1) = ⎜
⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67
⎝ 2 ⎠ ⎝ L ⎠1
⎝ 100 ⎠
⎝ L ⎠1 ⎝ L ⎠ 2
2
⎛ 2 ( 30 ) ⎞
⎛ 0.060 ⎞ ⎛ W ⎞
⎛W ⎞ ⎛W ⎞
and 2 ⎜
⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240
⎟ ⎜ ⎟ ( 0.1) = ⎜
⎝ 2 ⎠ ⎝ L ⎠3
⎝ L ⎠3 ⎝ L ⎠ 4
⎝ 50 ⎠
2
11.53
a.
iD1
⎛ v ⎞
= I DSS ⎜ 1 − GS 1 ⎟
VP ⎠
⎝
⎛ v ⎞
iD 2 = I DSS ⎜ 1 − GS 2 ⎟
VP ⎠
⎝
iD1 − iD 2
2
2
⎛ v ⎞
⎛ v ⎞
= I DSS ⎜1 − GS 1 ⎟ − I DSS ⎜ 1 − GS 2 ⎟
VP ⎠
VP ⎠
⎝
⎝
I DSS
=
( vGS 2 − vGS1 )
VP
I DSS
=−
VP
⋅ vd =
iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1
(
iD1 − I Q − iD1
)
2
=
I DSS
( −VP )
2
I DSS
( −VP )
⋅ vd2
iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) =
Then iD1 ( I Q − iD1 ) =
Square both sides
⋅ vd
I DSS
( −VP )
2
⋅ vd2
⎤
I
1⎡
⎢ I Q − DSS 2 ⋅ vd2 ⎥
2 ⎢⎣
( −VP ) ⎦⎥
⎤
I
1⎡
i − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd2 ⎥ = 0
4 ⎣⎢
( −VP ) ⎦⎥
2
2
D1
⎤
I
⎛ 1⎞⎡
I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd2 ⎥
⎝ 4 ⎠ ⎢⎣
( −VP ) ⎥⎦
2
2
Q
iD1 =
2
2
2
⎡
⎤
1 2 ⎢ 2 2 I Q I DSS vd ⎛ I DSS vd2 ⎞ ⎥
⎜
⎟
iD1 =
IQ − IQ −
±
+
2
2
⎢
2 2
( −VP ) ⎜⎝ ( −VP ) ⎟⎠ ⎥⎦
⎣
Use + sign
IQ
iD1
1 2 I Q I DSS 2
=
+
⋅ vd
2 2 ( −VP )2
iD1 =
IQ
IQ
2
+
⎛ I
⎞
− ⎜ DSS 2 ⋅ vd2 ⎟
⎜ ( −V )
⎟
P
⎝
⎠
2
2 I DSS ⎛ I DSS
1 IQ
vd
−⎜
⎜
2 ( −VP )
IQ
⎝ IQ
⎞ ⎛ vd ⎞
⎟⎟ ⎜ ⎟
⎠ ⎝ VP ⎠
⎞
2 I DSS ⎛ I DSS
−⎜
⎟ ⋅ vd
⎜ I
IQ
⎠
⎝ Q
⎞ ⎛ vd ⎞
⎟⎟ ⎜ ⎟
⎠ ⎝ VP ⎠
⎞
2 I DSS ⎛ I DSS
−⎜
⎟ ⋅ vd
⎜ I
IQ
⎠
⎝ Q
⎞ ⎛ vd ⎞
⎟⎟ ⎜ ⎟
⎠ ⎝ VP ⎠
Or
iD1 1 ⎛ 1
= +⎜
I Q 2 ⎝ −2VP
2
2
2
2
We had
iD 2 = I Q − iD1
Then
iD 2 1 ⎛ 1
= −⎜
I Q 2 ⎝ −2VP
b.
If iD1 = I Q , then
1 ⎛ 1
1= +⎜
2 ⎝ −2VP
⎞
2 I DSS ⎛ I DSS
−⎜
⎟ ⋅ vd
⎜ I
IQ
⎠
⎝ Q
2 I DSS ⎛ I DSS
VP = vd
−⎜
⎜ I
IQ
⎝ Q
Square both sides
⎞ ⎛v ⎞
⎟⎟ ⎜ d ⎟
⎠ ⎝ VP ⎠
2
2
2
⎞ ⎛ vd ⎞
⎟⎟ ⎜ ⎟
⎠ ⎝ VP ⎠
2
2
2
VP
2
⎛ I DSS
⎜⎜
⎝ IQ
vd2 =
⎡ 2I
⎛I
= v ⎢ DSS − ⎜ DSS
⎜ I
⎢ IQ
⎝ Q
⎣
2
d
2
2
⎞ ⎛ vd ⎞ ⎤
⎟⎟ ⎜ ⎟ ⎥
⎠ ⎝ VP ⎠ ⎥⎦
⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2
⋅ vd + VP
⎟⎟ ⎜ ⎟ ( vd ) −
IQ
⎠ ⎝ VP ⎠
2
2
⎛ 2I
2 I DSS
± ⎜ DSS
⎜ I
IQ
⎝ Q
2 ⎛ IQ ⎞
vd2 = (VP ) ⎜
⎟
⎝ I DSS ⎠
⎞
⎛I
⎟⎟ − 4 ⎜⎜ DSS
⎠
⎝ IQ
2
⎛ 2I
2 ⎜ DSS
⎜ IQ
⎝
2
=0
⎞ ⎛ 1 ⎞
2
⎟⎟ ⎜ ⎟ (VP )
⎠ ⎝ VP ⎠
⎞ ⎛ 1 ⎞
⎟⎟ ⎜ ⎟
⎠ ⎝ VP ⎠
2
2
2
2
⎛ IQ ⎞
Or vd = VP ⎜
⎟
⎝ I DSS ⎠
c.
For vd small,
IQ 1 IQ
2 I DSS
iD1 ≈
+ ⋅
⋅ vd
IQ
2 2 ( −VP )
1/ 2
gf =
diD1
d vd
vd → 0
=
2 I DSS
1 IQ
⋅
⋅
IQ
2 ( −VP )
⎛ 1 ⎞ I Q I DSS
Or ⇒ g f ( max ) = ⎜
⎟
2
⎝ −VP ⎠
11.53
Ad = g m ( ro 2 Ro )
Want Ad = 400
From Example 11.15, ro 2 = 1 M Ω
Assuming that g m = 0.283 mA / V for the PMOS from Example 11.15, then Ro = 285 M Ω.
⎛ k ′ ⎞⎛ W ⎞
So 400 = g m (1000 285000 ) ⇒ g m = 0.4014 mA / V = 2 ⎜ n ⎟ ⎜ ⎟ I DQ
⎝ 2 ⎠ ⎝ L ⎠1
⎛ 0.080 ⎞ ⎛ W ⎞
⎛W ⎞ ⎛W ⎞
0.04028 = ⎜
⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 10.1
⎝ 2 ⎠ ⎝ L ⎠1
⎝ L ⎠1 ⎝ L ⎠ 2
11.54
a.
I Q = I D1 + I D 2 ⇒ I Q = 1 mA
v0 = 7 = 10 − ( 0.5 ) RD ⇒ RD = 6 kΩ
b.
⎛ 1 ⎞ I Q ⋅ I DSS
g f ( max ) = ⎜
⎟
2
⎝ −VP ⎠
⎛ 1 ⎞ (1)( 2 )
g f ( max ) = ⎜ ⎟
⇒ g f ( max ) = 0.25 mA/V
2
⎝ 4⎠
c.
g R
Ad = m D = g f ( max ) ⋅ RD
2
Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5
11.55
a.
IS =
−VGS − ( −5 )
RS
⎛ V ⎞
= ( 2 ) I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
⎛
V ⎞
5 − VGS = ( 2 )( 0.8 )( 20 ) ⎜⎜ 1 − GS ⎟⎟
⎝ ( −2 ) ⎠
1
⎛
⎞
5 − VGS = ( 2 )16 ⎜1 + VGS + VGS2 ⎟
4
⎝
⎠
2
8VGS + 33VGS + 27 = 0
VGS =
2
2
−33 ± 1089 − 4 ( 8 )( 27 )
2 (8)
= −1.125 V
IS =
5 − ( −1.125 )
20
= 0.306 mA
I D1 = I D 2 = 0.153 mA
vo 2 = 1.17 V
(b)
11.56
Equivalent circuit and analysis is identical to that in problem 11.36.
1
⋅ g m RL
Ad 2 = 2
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
1
− ⋅ g m RL
Ad 1 = 2
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
Av =
v02 − v01
g m RL
=
vd
⎛
RL ⎞
⎜2+
⎟
R
⎝
D ⎠
11.57
(a)
Ad = g m ( ro 2 ro 4 )
0.1
= 3.846 mA/V
0.026
120
ro 2 =
= 1200 K
0.1
80
ro 4 =
= 800 K
0.1
Ad = ( 3.846 ) (1200 800 )
gm =
Ad = 1846
(b)
For Ad = 923 = ( 3.846 ) (1200 800 RL )
240 = 480 & RL =
480 RL
⇒ RL = 480 K
480 + RL
11.58
(a)
⎛
2⎞
I Q = 250 µ A I REF = I Q ⎜1 + ⎟
⎝ β⎠
2 ⎞
⎛
= 250 ⎜1 +
⎟ = 252.8 µ A
⎝ 180 ⎠
5 − ( 0.7 ) − ( −5 )
R1 =
⇒ R1 = 36.8 K
0.2528
(b)
0.125
= 4.808 mA/V
Ad = g m ( ro 2 ro 4 )
gm =
0.026
150
= 1200 K
ro 2 =
0.125
100
= 800 K
Ad = ( 4.808 ) (1200 800 ) ro 4 =
0.125
Ad = 2308
(c)
Rid = 2rπ =
2 (180 )( 0.026 )
0.125
⇒ Rid = 74.9 K
Ro = ro 2 ro 4 = 1200 800 = 480 K = Ro
(d)
vcm ( max ) = 5 − 0.7 = 4.3 V
vcm ( min ) = 0.7 + 0.7 − 5 = −3.6 V
11.59
a.
⎛ IQ ⎞ ⎛ 1 ⎞
I 0 = I B3 + I B 4 ≈ 2 ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠⎝ β ⎠
I Q 0.2
I0 =
=
⇒ I0 = 2 µ A
β 100
b.
V
100
= 1000 kΩ
r02 = r04 = A =
I CQ 0.1
gm =
I CQ
VT
=
0.1
= 3.846 mA/V
0.026
Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) ⇒ Ad = 1923
c.
Ad = g m r02 r04 RL
(
)
Ad = ( 3.846 ) (1000 1000 250 ) ⇒ Ad = 641
11.60
a.
Ad = g m ( r02 r04 RL )
gm =
I CQ
VT
=
IQ
2VT
V
125
r02 = A 2 =
I CQ I CQ
r04 =
VA 4 80
=
I CQ I CQ
If I Q = 2 mA, then g m = 38.46 mA/V
r02 = 125 kΩ, r04 = 80 kΩ
So Ad = 38.46 ⎡⎣125 80 200 ⎤⎦
Or Ad = 1508
For each gain of 1000. lower the current level
For I Q = 0.60 mA, I CQ = 0.30 mA
0.3
gm =
= 11.54 mA/V
0.026
125
r02 =
= 417 kΩ
0.3
80
r04 =
= 267 kΩ
0.3
Ad = 11.54 ⎡⎣ 417 267 200 ⎤⎦ = 1036
So I Q = 0.60 mA is adequate
b.
For V + = 10 V, VBE = VEB = 0.6 V
For VCB = 0, vcm ( max ) = V + − 2VEB = 10 − 2 ( 0.6 )
Or vcm ( max ) = 8.8 V
11.61
a.
From symmetry.
VGS 3 = VGS 4 = VDS 3 = VDS 4 =
0.1
+1
0.1
Or VDS 3 = VDS 4 = 2 V
0.1
+1 = 2 V
0.1
= VSD 2 = VSG1 − (VDS 3 − 10 )
= 2 − ( 2 − 10 )
VSG1 = VSG 2 =
VSD1
Or VSD1 = VSD 2 = 10 V
b.
r0 n =
r0 p =
1
λn I DQ
1
=
=
1
⇒ 1 MΩ
0.01
( )( 0.1)
1
⇒ 0.667 MΩ
( 0.015 )( 0.1)
g m = 2 K p (VSG + VTP )
= 2 ( 0.1)( 2 − 1) = 0.2 mA / V
Ad = g m ( ron rop ) = ( 0.2 ) (1000 667 ) ⇒ Ad
(c)
λP I DQ
= 80
IQ
I D 2 = I D1 =
2
= 0.1 mA
1
1
=
= 1000 k Ω
λn I D 4 ( 0.01)( 0.1)
ro 4 =
1
ro 2 =
λP I D 2
=
1
( 0.015)( 0.1)
= 667 k Ω
Ro = ro 2 ro 4 = 667 1000 = 400 k Ω
11.62
Ad = g m ( ro 4 ro 2 )
⎛ 0.08 ⎞
gm = 2 ⎜
⎟ ( 2.5 )( 0.05 )
⎝ 2 ⎠
= 0.1414 mA/V
1
= 1000 K
ro 4 =
( 0.02 )( 0.05 )
ro 2 =
1
= 1333 K
( 0.015)( 0.05 )
Ad = ( 0.1414 ) (1000 1333)
Ad = 80.8
11.63
R04 = r04 ⎡⎣1 + g m 4 ( R rπ 4 ) ⎤⎦
80
= 800 K
0.1
0.1
gm4 =
= 3.846
0.026
(100 )( 0.026 )
rπ 4 =
0.1
= 26 K
r04 =
R rπ 4 = 1 26 = 0.963 K
Assume β = 100
rπ 3 =
(100 )( 0.026 )
0.1
= 26 kΩ
0.1
= 3.846 mA/V
0.026
R04 = 800 ⎡⎣1 + ( 3.846 )( 0.963) ⎤⎦ ⇒ 3.763 MΩ
g m3 =
⇒ R0 = 3.763MΩ
Then
Av = − g m ( r02 R0 )
120
= 1200 kΩ
0.1
0.1
gm =
= 3.846 mA/V
0.026
Av = − ( 3.846 ) ⎡⎣1200 3763⎤⎦ ⇒ Av = −3499
r02 =
b.
For
R = 0, r04 =
Av = − g m ( r02
80
= 800 kΩ
0.1
r04 )
= − ( 3.846 ) ⎡⎣1200 800 ⎤⎦ ⇒ Av = −1846
(c)
For part (a), Ro = ( 3.763 1.2 ) = 0.910 M Ω
For part (b), Ro = (1.2 0.8 ) = 0.48 M Ω
11.64
I B5 =
IE5
I +I
I +I
= B3 B4 = C 3 C 4
1+ β
1+ β
β (1 + β )
Now I C 3 + I C 4 ≈ I Q
IQ
So I B 5 ≈
β (1 + β )
I B6 =
I Q1
IE6
=
1 + β β (1 + β )
For balance, we want I B 6 = I B 5
So that I Q1 = I Q
11.65
Resistance looking into drain of M4.
Vsg 4 ≅ I X R1
I X ± g m 4Vsg 4 =
VX − Vsg 4
r04
⎡
R ⎤ V
I X ⎢1 + g m 4 R1 + 1 ⎥ = X
r04 ⎦ r04
⎣
⎡
R ⎤
Or R0 = r04 ⎢1 + g m 4 R1 + 1 ⎥
r04 ⎦
⎣
a.
Ad = g m 2 ( ro 2 Ro )
g m 2 = 2 K n I DQ = 2
( 0.080 )( 0.1)
= 0.179 mA / V
1
1
ro 2 =
=
= 667 k Ω
λn I DQ ( 0.015 )( 0.1)
g m 4 = 2 K P I DQ = 2
ro 4 =
1
λ p I DQ
( 0.080 )( 0.1)
= 0.179 mA / V
1
=
= 500 k Ω
0.02
( )( 0.1)
1 ⎤
⎡
R0 = 500 ⎢1 + ( 0.179 )(1) +
⎥ = 590.5 kΩ
500
⎣
⎦
Ad = ( 0.179 ) ⎡⎣667 590.5⎤⎦ ⇒ Ad = 56.06
b.
When R1 = 0, R0 = r04 = 500 kΩ
Ad = ( 0.179 ) ⎡⎣667 500 ⎤⎦ ⇒ Ad = 51.15
(c)
For part (a), Ro = ro 2 Ro = 667 590.5 ⇒ Ro = 313 k Ω
For part (b), Ro = ro 2 ro 4 = 667 500 ⇒ Ro = 286 kΩ
11.66
Let β = 100, VA = 100 V
ro 2 =
VA 100
=
= 1000 k Ω
I CQ 0.1
Ro 4 = ro 4 [1 + g m RE′ ] where RE′ = rπ RE
Now
rπ =
(100 )( 0.026 )
= 26 k Ω
0.1
0.1
gm =
= 3.846 mA / V
0.026
RE′ = 26 1 = 0.963 k Ω
Then Ro 4 = 1000 ⎡⎣1 + ( 3.846 )( 0.963) ⎤⎦ = 4704 k Ω
Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) ⇒ Ad = 3172
11.67
(a) For Q2, Q4
Vx − Vπ 4
V
+ g m 2Vπ 2 + g m 4Vπ 4 + x
ro 2
ro 4
(1)
Ix =
(2)
g m 2Vπ 2 +
(3)
Vπ 4 = −Vπ 2
From (2)
Vx − Vπ 4
V
= π4
ro 2
rπ 4 rπ 2
⎡ 1
⎤
Vx
1
= Vπ 4 ⎢
+
+ gm2 ⎥
ro 2
r
r
r
⎣⎢ π 4 π 2 o 2
⎦⎥
Now
⎞ ⎛ 120 ⎞
⎟=⎜
⎟ ( 0.5 ) = 0.496 mA
⎠ ⎝ 121 ⎠
⎛ 120 ⎞
⎛ IQ ⎞ ⎛ 1 ⎞ ⎛ β ⎞
⎟ ⇒ I C 2 = 0.0041 mA
= ⎜ ⎟⎜
⎟⎜
⎟ = ( 0.5 ) ⎜⎜
2
⎟
⎝ 2 ⎠⎝ 1+ β ⎠⎝ 1+ β ⎠
⎝ (121) ⎠
⎛ β
IC 4 = ⎜
⎝ 1+ β
IC 2
⎞ ⎛ IQ
⎟⎜
⎠⎝ 2
So
rπ 2 =
(120 )( 0.026 )
= 761 k Ω
0.0041
0.0041
gm2 =
= 0.158 mA/V
0.026
100
ro 2 =
⇒ 24.4 M Ω
0.0041
(120 ) ( 0.026 )
= 6.29 k Ω
rπ 4 =
0.496
0.496
= 19.08 mA / V
gm4 =
0.026
100
= 202 k Ω
ro 4 =
0.496
Now
⎡
⎤
Vx
Vx
1
1
= Vπ 4 ⎢
+
+ 0.158⎥ ⇒ which yields Vπ 4 =
ro 2
( 0.318) ro 2
⎣⎢ 6.29 761 24400
⎦⎥
From (1),
⎛
V V
1 ⎞
I x = x + x + Vπ 4 ⎜ g m 4 − g m 2 − ⎟
ro 2 ro 4
ro 2 ⎠
⎝
1 ⎞⎤
⎡
⎛
⎜19.08 − 0.158 −
⎟
Ix ⎢ 1
V
1
24400 ⎠ ⎥
⎥ which yields Ro 2 = x = 135 k Ω
=⎢
+
+⎝
Vx ⎢ 24400 202
Ix
( 0.318)( 24400 ) ⎥
⎢
⎥
⎣
⎦
80
Now ro 6 =
= 160 k Ω
0.5
Then Ro = Ro 2 ro 6 = 135 160 ⇒ Ro = 73.2 k Ω
(b)
Ad = g mc Ro where g mc =
Δi
vd / 2
Δi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 =
⎛V
⎞
Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3
⎝ rπ 1
⎠
⎛1+ β ⎞
So Vπ 1 ⎜
⎟ rπ 3 = Vπ 3
⎝ rπ 1 ⎠
vd
2
⎛ 121 ⎞
Or Vπ 1 ⎜
⎟ ( 6.29 ) = Vπ 3 ≅ Vπ 1
⎝ 761 ⎠
v
v
Then 2Vπ 1 = d ⇒ Vπ 1 = d
2
4
⎛v ⎞
⎛v ⎞
So Δi = ( g m1 + g m 3 ) Vπ 1 = ( 0.158 + 19.08 ) ⎜ d ⎟ = 9.62 ⎜ d ⎟
⎝ 4⎠
⎝ 2⎠
Δi
= 9.62 ⇒ Ad = ( 9.62 )( 73.2 ) ⇒ Ad = 704
So g mc =
vd / 2
Now Rid = 2 Ri where Ri = rπ 1 + (1 + β ) rπ 3
Ri = 761 + (121)( 6.29 ) = 1522 k Ω
Then Rid = 3.044 M Ω
11.69
(a)
Ad = 100 = g m ( ro 2 ro 4 )
Let I Q = 0.5 mA
1
1
=
= 200 k Ω
ro 2 =
λn I D ( 0.02 )( 0.25 )
ro 4 =
1
1
=
= 160 k Ω
λP I D ( 0.025 )( 0.25 )
Then 100 = g m ( 200 160 ) ⇒ g m = 1.125 mA / V
⎛ K′ ⎞⎛W
gm = 2 ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
⎞
⎟ ID
⎠
⎛ 0.080 ⎞ ⎛ W ⎞
⎛W ⎞
1.125 = 2 ⎜
⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎜ ⎟ = 31.6
⎝ 2 ⎠⎝ L ⎠
⎝ L ⎠n
⎛W ⎞
⎛W ⎞
Now ⎜ ⎟ somewhat arbitrary. Let ⎜ ⎟ = 31.6
L
⎝ ⎠P
⎝ L ⎠P
11.70
Ad = g m ( ro 2 ro 4 )
P = ( I Q + I REF ) (V + − V − )
Let I Q = I REF
Then 0.5 = 2 I Q ( 3 − ( −3) ) ⇒ I Q = I REF = 0.0417 mA
ro 2 =
1
1
=
= 3205 k Ω
λn I D ( 0.015 )( 0.0208 )
ro 4 =
1
1
=
= 2404 k Ω
λP I D ( 0.02 )( 0.0208 )
Then
Ad = 80 = g m ( 3205 2404 ) ⇒ g m = 0.0582 mA/V
⎛ k ′ ⎞⎛ W ⎞
gm = 2 ⎜ n ⎟ ⎜ ⎟ I D
⎝ 2 ⎠ ⎝ L ⎠n
⎛ 0.080 ⎞⎛ W ⎞
⎛W ⎞
0.0582 = 2 ⎜
⎟⎜ ⎟ ( 0.0208 ) ⇒ ⎜ ⎟ = 1.02
⎝ 2 ⎠⎝ L ⎠ n
⎝ L ⎠n
11.71
Ad = g m ( ro 2 Ro )
≈ g m ro 2
1
ro 2 =
λn I D
=
1
= 666.7 K
( 0.015)( 0.1)
Ad = 400 = g m ( 666.7 )
g m = 0.60 mA/V
⎛ k′ ⎞⎛ W
= 2 ⎜ n ⎟⎜
⎝ 2 ⎠⎝ L
⎞
⎟ ID
⎠
⎛ 0.08 ⎞ ⎛ W ⎞
0.60 = 2 ⎜
⎟ ⎜ ⎟ ( 0.1)
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
0.090 = 0.004 ⎜ ⎟
⎝L⎠
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 22.5
⎝ L ⎠1 ⎝ L ⎠ 2
11.72
Ad = g m ( Ro 4 Ro 6 )
where
Ro 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ]
Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ]
We have
1
= 1667 k Ω
ro 2 = ro 4 =
( 0.015 )( 0.040 )
ro 6 = ro8 =
1
= 1250 k Ω
0.02
( )( 0.040 )
⎛ 0.060 ⎞
gm4 = 2 ⎜
⎟ (15 )( 0.040 ) = 0.268 mA/V
⎝ 2 ⎠
⎛ 0.025 ⎞
gm6 = 2 ⎜
⎟ (10 )( 0.040 ) = 0.141 mA/V
⎝ 2 ⎠
Then
Ro 4 = 1667 + 1667 ⎡⎣1 + ( 0.268 )(1667 ) ⎤⎦ ⇒ 748 M Ω
Ro 6 = 1250 + 1250 ⎡⎣1 + ( 0.141)(1250 ) ⎤⎦ ⇒ 222.8 M Ω
(a)
Ro = Ro 4 Ro 6 = 748 222.8 ⇒ Ro = 172 M Ω
(b)
Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) ⇒ Ad = 46096
11.73
Ad = g m ( ro 2 ro 4 )
ro 2 = ro 4 =
=
1
λ ID
1
( 0.02 )( 0.1)
gm = 2 Kn I D = 2
= 500 K
( 0.5)( 0.1)
= 0.4472 mA/V
Ad = ( 0.4472 ) ( 500 500 ) ⇒ Ad = 112
Ro = ro 2 ro 4 = 500 500 ⇒ Ro = 250 K
11.74
(a)
I DP = K p (VSG + VTP )
2
0.4
+ 1 = VSG 3 = 1.894 V
0.5
I DN = K n (VGS − VTN )
2
0.4
+ 1 = VGS 1 = 1.894 V
0.5
VDS1 ( sat ) = VGS1 − VTN = 1.894 − 1 = 0.894 V
V + = VSG 3 + VDS1 ( sat ) − VGS 1 + vCM
V + = 1.894 + 0.894 − 1.894 + 4 ⇒ V + = 4.89 V = −V −
(b)
Ad = g m ( ro 2 ro 4 )
ro 2 = ro 4 =
1
λ ID
gm = 2 Kn I D
1
=
= 166.7 K
( 0.015 )( 0.4 )
= 2 ( 0.5 )( 0.4 ) = 0.8944 mA/V
Ad = ( 0.8944 ) (166.7 166.7 ) ⇒ Ad = 74.5
11.75
(a)
For vcm = +2V ⇒ V + = 2.7 V
If I Q is a 2-transistor current source,
V − = vcm − 0.7 − 0.7
V − = −3.4 V ⇒ V + = −V − = 3.4 V
(b)
100
= 1000 K
Ad = g m ( ro 2 ro 4 ) ro 2 =
0.1
60
= 600 K
ro 4 =
0.1
0.1
= 3.846 mA/V
gm =
0.026
Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 1442
11.76
(a)
(b)
V + = −V − = 3.4 V
75
= 1250 K
0.06
40
ro 4 =
= 666.7 K
0.06
0.06
gm =
= 2.308 mA/V
0.026
Ad = ( 2.308 ) (1250 666.7 )
ro 2 =
Ad = 1004
11.77
g m1 = 2 K n I Bias1 = 2
gm2 =
rπ 2 =
I CQ
VT
β VT
I CQ
=
=
( 0.2 )( 0.25 ) = 0.447 mA/V
0.75
= 28.85 mA/V
0.026
(120 )( 0.026 )
0.75
= 4.16 kΩ
i0 = g m1Vgs1 + g m 2Vπ 2
Vπ 2 = g m1Vgs1rπ 2 and vi = Vgs1 + Vπ 2
i0 = Vgs1 ( g m1 + g m 2 ⋅ g m1rπ 2 )
vi = Vgs1 + g m1Vgs1rπ 2 and Vgs1 =
i0 = vi ⋅
g mC =
vi
1 + g m1rπ 2
g m1 (1 + β )
1 + g m1rπ 2
( 0.447 )(121)
i0 g m1 (1 + β )
=
=
1 + g m1rπ 2 1 + ( 0.447 )( 4.16 )
vi
⇒ g mC = 18.9 mA/V
11.78
r0 ( M 2 ) =
r0 ( Q2 ) =
1
λn I DQ
=
1
( 0.01)( 0.2 )
= 500 kΩ
VA
80
=
= 400 kΩ
I CQ 0.2
g m ( M 2 ) = 2 K n I DQ = 2
( 0.2 )( 0.2 )
= 0.4 mA/V
Ad = g m ( M 2 ) ⎡⎣ r0 ( M 2 ) r0 ( Q2 ) ⎤⎦
= 0.4 ⎣⎡500 400 ⎤⎦ ⇒ Ad = 88.9
If the IQ current source is ideal, Acm = 0 and C M RRdB = ∞
11.79
a.
b.
Assume RL is capacitively coupled. Then
I CQ + I DQ = I Q
I DQ =
VBE 0.7
=
= 0.0875 mA
R1
8
I CQ = 0.9 − 0.0875 = 0.8125 mA
g m1 = 2 K P I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V
gm2 =
rπ 2 =
I CQ
VT
β VT
I CQ
=
0.8125
⇒ g m 2 = 31.25 mA/V
0.026
=
(100 )( 0.026 )
0.8125
⇒ rπ 2 = 3.2 kΩ
c.
V0 = ( − g m1Vsg − g m 2Vπ 2 ) RL
Vi + Vsg = V0 ⇒ Vsg = V0 − Vi
Vπ 2 = ( g m1Vsg ) ( R1 rπ 2 )
V0 = − ⎡⎣ g m1Vsg + g m 2 g m1Vsg ( R1 rπ 2 ) ⎤⎦ RL
V0 = − (V0 − Vi ) ⎡⎣ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤⎦ RL
⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦⎤ RL
V0
= ⎣
Vi 1 + ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ RL
⎣
⎦
We find
Av =
g m1 + g m 2 g m1 ( R1 rπ 2 ) = 0.592 + ( 31.25 )( 0.592 ) ( 8 3.2 )
= 42.88
Then Av =
( 42.88 )( RL )
1 + ( 42.88 )( RL )
11.80
a.
I DQ
I CQ
Assume RL is capacitively coupled.
0.7
=
= 0.0875 mA
8
= 1.2 − 0.0875 = 1.11 mA
g m1 = 2 K p I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V
gm2 =
rπ 2 =
b.
I CQ
VT
β VT
I CQ
=
=
1.11
⇒ g m 2 = 42.7 mA/V
0.026
(100 )( 0.026 )
1.11
⇒ rπ 2 = 2.34 kΩ
Vsg = VX
I X = g m 2Vπ 2 + g m1Vsg
(g
V
)(R
rπ 2 ) = Vπ 2
I X = VX ⎡⎣ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤⎦
V
1
R0 = X =
IX
g m1 + g m 2 g m1 ( R1 rπ 2 )
m1 sg
=
1
1
0.592 + ( 0.592 )( 42.7 ) ( 8 2.34 )
⇒ R0 = 21.6 Ω
11.81
(a)
(1)
g m 2Vπ +
(2)
g m 2Vπ +
Then Vπ =
From (1)
Vo − ( −Vπ )
ro 2
Vo − ( −Vπ )
g m1Vi
⎛ 1 1⎞
⎜ + ⎟
⎝ ro1 rπ ⎠
ro 2
=0
= g m1Vi +
⎛ 1 1⎞
−Vπ −Vπ
+
or 0 = g m1Vi − Vπ ⎜ + ⎟
ro1
rπ
⎝ ro1 rπ ⎠
⎛
Vo
1 ⎞
=0
⎜ g m 2 + ⎟ Vπ +
ro 2 ⎠
ro 2
⎝
⎛
1 ⎞
⎜ gm2 + ⎟
ro 2 ⎠
⎛
1 ⎞
Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = −ro 2 g m1Vi ⎝
ro 2 ⎠
⎛ 1 1⎞
⎝
⎜ + ⎟
⎝ ro1 rπ ⎠
⎛
1 ⎞
− g m1ro 2 ⎜ g m 2 + ⎟
ro 2 ⎠
V
⎝
Av = o =
Vi
⎛ 1 1⎞
⎜ + ⎟
⎝ ro1 rπ ⎠
Now
( 0.25)( 0.025 ) = 0.158 mA / V
g m1 = 2 K n I Q = 2
gm2 =
ro1 =
VT
1
λ IQ
=
=
0.025
= 0.9615 mA / V
0.026
1
( 0.02 )( 0.025)
= 2000 k Ω
VA
50
=
= 2000 k Ω
I Q 0.025
ro 2 =
rπ =
IQ
β VT
Then
IQ
=
(100 )( 0.026 )
0.025
= 104 k Ω
1 ⎞
⎛
− ( 0.158 )( 2000 ) ⎜ 0.9615 +
⎟
2000
⎝
⎠ ⇒ A = −30039
Av =
v
1 ⎞
⎛ 1
+
⎜
⎟
⎝ 2000 104 ⎠
To find Ro; set Vi = 0 ⇒ g m1Vi = 0
I x = g m 2Vπ +
Vx − ( −Vπ )
Vπ = − I x ( ro1 rπ )
ro 2
Then
⎛
V
1 ⎞
I x = ⎜ g m 2 + ⎟ ( − I x ) ( ro1 rπ ) + x
ro 2 ⎠
ro 2
⎝
Combining terms,
Ro =
⎡
⎛
Vx
1 ⎞⎤
= ro 2 ⎢1 + ( ro1 rπ ) ⎜ g m 2 + ⎟ ⎥
Ix
ro 2 ⎠ ⎦
⎝
⎣
⎡
1 ⎞⎤
⎛
= 2000 ⎢1 + ( 2000 104 ) ⎜ 0.9615 +
⎟ ⎥ ⇒ Ro = 192.2 M Ω
2000
⎝
⎠⎦
⎣
(b)
Vo − ( −Vgs 3 )
(1)
g m 3Vgs 3 +
(2)
g m 3Vgs 3 +
(3)
−Vgs 3 − ( −Vπ 2 )
( −Vπ 2 )
Vπ 2
+ g m 2Vπ 2 +
= g m1Vi +
rπ 2
ro 2
ro1
From (2), Vπ 2 =
Then
(3)
or
ro3
Vo − ( −Vgs 3 )
ro3
=0
= g m 2Vπ 2 +
−Vgs 3 − ( −Vπ 2 )
ro 2
Vgs 3
⎛
1 ⎞
ro 2 ⎜ g m 2 + ⎟
ro 2 ⎠
⎝
Vgs 3
⎛ 1
1
1⎞
+ gm2 +
+ ⎟ = g m1Vi +
Vπ 2 ⎜
ro 2 ro1 ⎠
ro 2
⎝ rπ 2
⎛
1 ⎞ Vgs 3
or 0 = Vπ 2 ⎜ g m 2 + ⎟ −
r
ro 2
o2 ⎠
⎝
Vgs 3
⎡ 1
1
1⎤
+ ⎥ = g m1Vi +
⎢ + gm2 +
ro 2 ro1 ⎦
ro 2
⎛
1 ⎞ r
ro 2 ⎜ g m 2 + ⎟ ⎣ π 2
ro 2 ⎠
⎝
Vgs 3
Vgs 3
1
1 ⎤
⎡ 1
+ 0.9615 +
+
= 0.9615Vi +
⎢
⎥
1 ⎞ ⎣104
2000 2000 ⎦
2000
⎛
2000 ⎜ 0.9615 +
⎟
2000
⎝
⎠
Then Vgs 3 = 1.83 × 105 Vi
Vgs 3
⎛
−V
1 ⎞
1 ⎞
⎛
5
From (1), ⎜ g m 3 + ⎟ Vgs 3 = o or Vo = −2000 ⎜ 0.158 +
⎟ (1.83 ×10 ) Vi
r
r
2000
⎝
⎠
o3 ⎠
o3
⎝
V
Av = o = −5.80 × 107
Vi
To find Ro
Vx − ( −Vgs 3 )
(1)
I x = g m 3Vgs 3 +
(2)
g m 3Vgs 3 +
(3)
Vπ 2 = − I x ( ro1 rπ 2 )
ro3
Vx − ( −Vgs 3 )
ro 3
= g m 2Vπ 2 +
⎛
1 ⎞ V
From (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x
ro 3 ⎠ ro3
⎝
1 ⎞ Vx
⎛
I x = Vgs 3 ⎜ 0.158 +
⎟+
2000 ⎠ 2000
⎝
V
Ix − x
2000
So Vgs 3 =
0.1585
−Vgs 3 − ( −Vπ 2 )
ro 2
From (2),
⎡
⎛
1
1 ⎤ V
1 ⎞
Vgs 3 ⎢ g m 3 +
+ ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟
ro 3 ro 2 ⎦ ro 3
ro 2 ⎠
⎣
⎝
1
1 ⎤ Vx
1 ⎞
⎡
⎛
Vgs 3 ⎢ 0.158 +
+
+
= Vπ 2 ⎜ 0.9615 +
⎟
⎥
2000 2000 ⎦ 2000
2000 ⎠
⎣
⎝
Vx
⎡ I − Vx / 2000 ⎤
Then ⎢ x
⎥ ( 0.159 ) + 2000 = − I x ( 2000 104 ) ( 0.962 )
⎣ 0.1585 ⎦
V
We find Ro = x = 6.09 × 1010 Ω
Ix
11.82
Assume emitter of Q1 is capacitively coupled to signal ground.
⎛ 80 ⎞
I CQ = 0.2 ⎜ ⎟ = 0.1975 mA
⎝ 81 ⎠
0.2
I DQ =
= 0.00247 mA
81
(80 )( 0.026 )
rπ =
= 10.5 k Ω
0.1975
0.1975
= 7.60 mA / V
g m ( Q1 ) =
0.026
gm ( M1 ) = 2 K n I D = 2
( 0.2 )( 0.00247 )
g m ( M 1 ) = 0.0445 mA / V
Vi = Vgs + Vπ and Vπ = g m ( M 1 ) Vgs rπ or Vgs =
⎛
1
Then Vi = Vπ ⎜⎜ 1 +
g
M
(
m
1 ) rπ
⎝
Vπ
g m ( M 1 ) rπ
⎞
Vi
⎟⎟ or Vπ =
⎛
⎞
1
⎠
⎜⎜ 1 + g ( M ) r ⎟⎟
m
1 π ⎠
⎝
− g m ( Q1 ) RC
V
Vo = − g m ( Q1 ) Vπ RC ⇒ Av = o =
Vi ⎛
⎞
1
⎜⎜1 +
⎟⎟
⎝ g m ( M 1 ) rπ ⎠
− ( 7.60 )( 20 )
Then Av =
⇒ Av = −48.4
⎛
⎞
1
⎜⎜ 1 +
⎟⎟
⎝ ( 0.0445 )(10.5 ) ⎠
11.83
Using the results from Chapter 4 for the emitter-follower:
rπ 9 + r07 R011 ⎤
⎡
⎢ rπ 8 +
⎥
1+ β
⎥
R0 = R4 || ⎢
⎢
⎥
1+ β
⎢
⎥
⎣
⎦
β VT (100 )( 0.026 )
rπ 8 =
=
= 2.6 kΩ
IC8
1
IC 8
IC 9 ≈
rπ 9 =
β
=
1
= 0.01 mA
100
(100 )( 0.026 )
= 260 kΩ
0.01
V
100
r07 = A =
= 500 kΩ
I Q 0.2
r011 =
VA 100
=
= 500 kΩ
I Q 0.2
R011 = r011 [1 + g m RE′ ] , g m =
0.2
= 7.69
0.026
(100 ) ( 0.026 )
= 13 kΩ
0.2
RE′ = 0.2 13 = 0.197 kΩ
R011 = 500 ⎡⎣1 + ( 7.69 )( 0.197 ) ⎤⎦ = 1257 kΩ
rπ 11 =
Then
⎡
260 + 500 1257 ⎤
⎢ 2.6 +
⎥
101
⎥
R0 = 5 || ⎢
101
⎢
⎥
⎢⎣
⎥⎦
= 5 0.0863 ⇒ R0 = 0.0848 K ⇒ 84.8 Ω
11.84
Ri = rπ 1 + (1 + β ) rπ 2
rπ 2 =
(100 )( 0.026 )
0.5
= 5.2 kΩ
(100 )( 0.026 ) (100 ) ( 0.026 )
=
= 520 kΩ
0.5
( 0.5 /100 )
Ri = 520 + (101)( 5.2 ) ⇒ Ri ≅ 1.05 MΩ
(100 )( 0.026 )
rπ 3 + 50
2
rπ 1 =
R0 = 5
R0 = 5
101
, rπ 3 =
1
= 2.6 kΩ
2.6 + 50
= 5 0.521 ⇒ R0 = 0.472 kΩ
101
⎛V
⎞
V0 = − ⎜ π 3 + g m 3Vπ 3 ⎟ ( 5 )
⎝ rπ 3
⎠
⎛1+ β ⎞
V0 = −Vπ 3 ⎜
⎟ ( 5)
⎝ rπ 3 ⎠
(1)
(V − V )
Vπ 3
= g m 2Vπ 2 + 0 π 3
50
rπ 3
⎛ 1
1 ⎞ V
+ ⎟− 0
g m 2Vπ 2 = Vπ 3 ⎜
⎝ rπ 3 50 ⎠ 50
⎛V
⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2
⎝ rπ 1
⎠
⎛ 1+ β ⎞
= Vπ 1 ⎜
⎟ rπ 2
⎝ rπ 1 ⎠
and
Vin = Vπ 1 + Vπ 2
gm2 =
(2)
(3)
(4)
0.5
= 19.23 mA/V
0.026
Then
⎛ 101 ⎞
V0 = −Vπ 3 ⎜
⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 )
⎝ 2.6 ⎠
And
1 ⎞ V
⎛ 1
+ ⎟− 0
19.23Vπ 2 = −V0 ( 0.005149 ) ⎜
⎝ 2.6 50 ⎠ 50
= −V0 ( 0.02208 )
(1)
(2)
Or Vπ 2 = −V0 ( 0.001148 )
And
Vπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 )
(4)
So
⎛ 101 ⎞
−V0 ( 0.001148 ) = ⎡⎣Vin + V0 ( 0.001148 ) ⎤⎦ ⎜
⎟ ( 5.2 )
⎝ 520 ⎠
−V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av =
11.85
(3)
V0
= −438
Vin
I2 =
5
= 1 mA
5
1
+ 0.8 = 2.21 V
0.5
2.21 − ( −5 )
= 0.206 mA
I1 =
35
VGS 2 =
V0 = ( g m 2Vgs 2 ) ( R2 r02 )
Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) − V0 and Vsg1 = −Vin
So Vgs 2 = − ( g m1Vin ) ( r01 R1 ) − V0
Then
V0 = g m 2 ( R2 r02 ) ⎡⎣ − ( g m1Vin ) ( r01 R1 ) − V0 ⎤⎦
Av =
V0 − g m 2 ( R2 r02 ) g m1 ( r01 R1 )
=
Vin
1 + g m 2 ( R2 r02 )
gm2 = 2 Kn2 I D 2 = 2
( 0.5 )(1) = 1.414 mA / V
g m1 = 2 K p1 I D1 = 2
( 0.2 )( 0.206 ) = 0.406 mA / V
r01 =
1
λ1 I D1
r02 =
=
1
( 0.01)( 0.206 )
= 485 kΩ
1
1
=
= 100 kΩ
λ2 I D 2 ( 0.01)(1)
R2 r02 = 5 100 = 4.76 kΩ
R1 r01 = 35 485 = 32.6 kΩ
Then Av =
− (1.414 )( 4.76 )( 0.406 )( 32.6 )
1 + (1.414 )( 4.76 )
So ⇒ Av = −11.5
Output Resistance—From the results for a source follower in Chapter 6.
1
1
R0 =
R2 r02 =
5 100
gm2
1.414
= 0.707 4.76
So R0 = 0.616 kΩ
11.86
a.
R2 =
5
⇒ R2 = 10 kΩ
0.5
I D2
0.5
− VTP 2 =
+ 1 = 2.41 V
K p2
0.25
VSG 2 =
R1 =
5 − ( −2.41)
0.1
⇒ R1 = 74.1 kΩ
b.
V0 = − ( g m 2Vsg 2 ) ( r02 R2 )
Vsg 2 = V0 − ⎡⎣ − ( g m1Vgs1 ) ( r01 R1 ) ⎤⎦ and Vgs1 = Vin
Av =
V0 − ( g m 2 ) ( r02 R2 ) ( g m1 ) ( r01 R1 )
=
Vin
1 + ( g m 2 ) ( r02 R2 )
g m1 = 2 K n1 I D1 = 2
gm2 = 2 K p 2 I D 2 = 2
r01 =
r02 =
1
λ1 I D1
=
( 0.1)( 0.1) = 0.2 mA / V
( 0.25)( 0.5) = 0.707 mA / V
1
( 0.01)( 0.1)
= 1000 kΩ
1
1
=
= 200 kΩ
λ2 I D 2 ( 0.01)( 0.5 )
r02 R2 = 200 10 = 9.52 kΩ
r01 R1 = 1000 74.1 = 69.0 kΩ
Then Av =
− ( 0.707 )( 9.52 )( 0.2 )( 69 )
1 + ( 0.707 )( 9.52 )
So ⇒ Av = −12.0
R0 =
1
1
R2 r02 =
10 200
gm2
0.707
= 1.414 9.52
Or R0 = 1.23 kΩ
11.87
a.
I C 2 = 0.25 mA
5−2
R=
⇒ R = 12 kΩ
0.25
v − VBE ( on )
2 − 0.7
I C 3 = 02
⇒ RE1 =
⇒ RE1 = 2.6 kΩ
0.5
RE1
RC =
5 − v03 5 − 3
=
⇒ RC = 4 kΩ
0.5
IC 3
⎡ v03 − VBE ( on ) ⎤⎦ − ( −5 )
IC 4 = ⎣
RE 2
RE 2 =
3 − 0.7 + 5
⇒ RE 2 = 2.43 kΩ
3
b.
Input resistance to base of Q3,
Ri 3 = rπ 3 + (1 + β ) RE1
rπ 3 =
(100 )( 0.026 )
= 5.2 kΩ
0.5
Ri 3 = 5.2 + (101)( 2.6 ) = 267.8 kΩ
v
1
Ad 1 = 02 = g m 2 ( R Ri 3 )
vd 2
0.25
= 9.62 mA/V
0.026
1
Ad 1 = ( 9.62 ) (12 267.8 ) ⇒ Ad 1 = 55.2
2
− β ( RC Ri 4 )
v
Now 03 =
v02 rπ 3 + (1 + β ) RE1
gm2 =
where Ri 4 = rπ 4 + (1 + β ) RE 2
and
(1 + β ) RE 2
v0
=
v03 rπ 4 + (1 + β ) RE 2
(100 )( 0.026 )
rπ 4 =
= 0.867 kΩ
3
(101)( 2.43)
v0
=
= 0.9965
v03 0.867 + (101)( 2.43)
Ri 4 = 0.867 + (101)( 2.43) = 246.3 kΩ
rπ 3 = 5.2 kΩ
So
v03 − (100 ) ( 4 246.3)
=
= −1.47
5.2 + (101)( 2.6 )
v02
v0
= ( 55.2 )( 0.9965 )( −1.47 ) ⇒ Ad = −80.9
vd
Using Equation (11.32b)
− g m 2 ( R Ri 3 )
So Ad =
c.
Acm1 =
1+
rπ 2 =
2 (1 + β ) R0
rπ 2
(100 )( 0.026 )
= 10.4 kΩ
0.25
− ( 9.62 ) (12 267.8 )
= −0.0569 = Acm1
Acm1 =
2 (101)(100 )
1+
10.4
⎛ v0 ⎞⎛ v03 ⎞
Then Acm = ⎜ ⎟⎜ ⎟ ⋅ Acm1
⎝ v03 ⎠⎝ v02 ⎠
= ( 0.9965 )( −1.47 )( −0.0569 ) ⇒ Acm = 0.08335
⎛ 80.9 ⎞
C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 59.7 dB
⎝ 0.08335 ⎠
11.88
a.
RC1 =
RC 2 =
10 − v01 10 − 2
=
⇒ RC1 = 80 kΩ
I C1
0.1
10 − v04 10 − 6
=
⇒ RC 2 = 20 kΩ
IC 4
0.2
b.
Ad 1 =
v01 − v02
= − g m1 ( RC1 rπ 3 )
vd
0.1
= 3.846 mA/V
0.026
(180 )( 0.026 )
rπ 3 =
= 23.4 kΩ
0.2
Ad 1 = − ( 3.846 ) ( 80 23.4 ) ⇒ Ad 1 = −69.6
g m1 =
Ad 2 =
v04
1
= g m 4 RC 2
v01 − v02 2
0.2
= 7.692 mA/V
0.026
1
Ad 2 = ( 7.692 )( 20 ) = 76.9
2
Then Ad = ( 76.9 )( −69.6 ) ⇒ Ad = −5352
gm4 =
11.89
a.
Neglect the effect of r0 in determining the differential-mode gain.
v02 1
Ad 1 =
= g m 2 ( RC Ri 3 ) where Ri 3 = rπ 3 + (1 + β ) RE
vd 2
A2 =
I1 =
− β RC 2
rπ 3 + (1 + β ) RE
12 − 0.7 − ( −12 )
R1
=
23.3
= 1.94 mA ≈ I C 5
12
1
⋅ (1.94 )
= 37.3 mA/V
gm2 = 2
0.026
( 200 )( 0.026 )
rπ 3 =
IC 3
1
(1.94 )(8) = 4.24 V
2
4.24 − 0.7
= 1.07 mA
IC 3 =
3.3
( 200 )( 0.026 )
= 4.86 kΩ
rπ 3 =
1.07
Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ
v02 = 12 −
Ad 1 =
1
( 37.3) ⎡⎣8 668⎤⎦ = 147.4
2
Then
Ad = Ad 1 ⋅ A2 = (147.4 )( −1.197 ) ⇒ Ad = −176
R0 = r05 =
Acm1 =
− g m 2 ( RC Ri 3 )
1+
rπ 2 =
VA
80
=
= 41.2 kΩ
I C 5 1.94
2 (1 + β ) R0
rπ 2
( 200 )( 0.026 )
1
⋅ (1.94 )
2
= 5.36 kΩ
Acm1 =
− ( 37.3) ( 8 668 )
= −0.09539
2 ( 201)( 41.2 )
1+
5.36
A2 = −1.197
Acm = ( −0.09539 )( −1.197 ) ⇒ Acm = 0.114
b.
vd = v1 − v2 = 2.015sin ω t − 1.985sin ω t
vd = 0.03sin ω t ( V )
v +v
vcm = 1 2 = 2.0sin ω t
2
v03 = Ad vd + Acm vcm
= ( −176 )( 0.03) + ( 0.114 )( 2 )
Or v03 = −5.052sin ω t
Ideal, Acm = 0
So
v03 = Ad vd = ( −176 )( 0.03)
v03 = −5.28sin ω t
c.
Rid = 2rπ 2 = 2 ( 5.36 ) ⇒ Rid = 10.72 kΩ
2 Ricm ≅ 2 (1 + β ) R0 (1 + β ) r0
VA
80
=
= 82.5 kΩ
1
IC 2
⋅ (1.94 )
2
2 Ricm = ⎡⎣ 2 ( 201)( 41.2 ) ⎤⎦ ⎣⎡( 201)( 82.5 ) ⎦⎤
r0 =
= 16.6 MΩ 16.6 MΩ
So ⇒ Ricm = 4.15 MΩ
11.90
a.
24 − VGS 4
2
= kn (VGS 4 − VTh )
I1 =
R1
24 − VGS 4 = ( 55 )( 0.2 )(VGS 4 − 2 )
2
24 − VGS 4 = 11 (VGS2 4 − 4VGS 4 + 4 )
11VGS2 4 − 43VGS 4 + 20 = 0
VGS 4 =
43 ±
( 43)
2
− 4 (11)( 20 )
2 (11)
= 3.37 V
24 − 3.37
= 0.375 mA = I Q
55
⎛ 0.375 ⎞
v02 = 12 − ⎜
⎟ ( 40 ) = 4.5 V
⎝ 2 ⎠
v02 − VGS 3
2
= I D 3 = kn (VGS 3 − VTh )
R5
I1 =
4.5 − VGS 3 = ( 0.2 ) ( 6 ) (VGS2 3 − 4VGS 3 + 4 )
1.2VGS2 3 − 3.8VGS 3 + 0.3 = 0
VGS 3 =
I D3 =
3.8 ±
( 3.8)
2
− 4 (1.2 ) ( 0.3)
2 (1.2 )
4.5 − 3.09
= 0.235 mA
6
gm2 = 2 Kn I D2 = 2
( 0.2 ) ⎛⎜
⎝
= 3.09 V
0.375 ⎞
⎟
2 ⎠
= 0.387 mA/V
1
1
g m 2 RD = ( 0.387 )( 40 ) ⇒ Ad 1 = 7.74
2
2
− g m 3 RD 2
A2 =
1 + g m 3 R5
Ad 1 =
g m3 = 2 K n I D3 = 2
( 0.2 )( 0.235 )
= 0.434 mA/V
A2 =
− ( 0.434 ) ( 4 )
1 + ( 0.434 )( 6 )
= −0.482
So Ad = Ad 1 ⋅ A2 = ( 7.74 ) ( −0.482 ) ⇒ Ad = −3.73
R0 = r05 =
Acm1 =
1
1
=
= 133 kΩ
λ I Q ( 0.02 )( 0.375 )
− ( 0.387 ) ( 40 )
− g m 2 RD
=
1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133)
= −0.149
Acm = ( −0.149 )( −0.482 ) ⇒ Acm = 0.0718
b.
vd = v1 − v2 = 0.3sin ω t
v +v
vcm = 1 2 = 2sin ω t
2
v03 = Ad vd + Acm vcm
= ( −3.73)( 0.3) + ( 0.0718 )( 2 ) ⇒ v03 = −0.975sin ω t ( V )
Ideal, Acm = 0
v03 = Ad vd = ( −3.73)( 0.3)
Or
⇒ v03 = −1.12sin ω t ( V )
11.91
The low-frequency, one-sided differential gain is
⎛ r
⎞
v02 1
= g m RC ⎜ π ⎟
vd 2
⎝ rπ + RB ⎠
1
⋅ β RC
= 2
rπ + RB
Av 2 =
rπ =
(100 )( 0.026 )
0.5
= 5.2 kΩ
1
⋅ (100 )(10 )
⇒ Av 2 = 87.7
Av 2 = 2
5.2 + 0.5
CM = Cµ (1 + g m RC )
0.5
= 19.23 mA/V
0.026
CM = 2 ⎡⎣1 + (19.23)(10 ) ⎤⎦ ⇒ CM = 387 pF
gm =
1
2π ⎡⎣ rπ RB ⎤⎦ ( Cπ + CM )
1
=
So ⇒ f H = 883 kHz
3
2π ⎣⎡5.2 0.5⎦⎤ × 10 × ( 8 + 387 ) × 10−12
fH =
11.92
From Equation (11.117), f Z =
a.
1
1
=
2π R0 C0 2π ( 5 × 106 )( 0.8 × 10−12 )
Or f Z = 39.8 kHz
b.
From Problem 11.69, f H = 883 kHz. From Equation (11.116(b)), the low-frequency
common- mode gain is
− g m RC
Acm =
⎡⎛ RB ⎞ 2 (1 + β ) R0 ⎤
⎢⎜ 1 +
⎥
⎟+
rπ ⎠
rπ
⎣⎝
⎦
rπ = 5.2 kΩ, g m = 19.23 mA/V
So
− (19.23)(10 )
⎡⎛ 0.5 ⎞ 2 (101) ( 5 × 106 ) ⎤
⎢⎜ 1 +
⎥
⎟+
5.2 × 103
⎢⎣⎝ 5.2 ⎠
⎥⎦
−4
= −9.9 × 10
Acm =
⎛ 87.7 ⎞
C M RRdB = 20 log10 ⎜
= 98.9 dB
−4 ⎟
⎝ 9.9 × 10 ⎠
11.93
a.
From Equation (7.72), fT =
gm
2π ( Cπ + Cµ )
1
= 38.46 mA/V
0.026
38.46 × 10−3
Then 800 × 106 =
2π ( Cπ + Cµ )
gm =
Or Cπ + Cµ = 7.65 × 10−12 F = 7.65 pF
And Cπ = 6.65 pF
CM = Cµ (1 + g m RC ) = 1 ⎡⎣1 + ( 38.46 )(10 ) ⎤⎦
= 386 pF
1
fH =
2π ⎡⎣ rπ RB ⎤⎦ ( Cπ + CM )
rπ =
(120 )( 0.026 )
1
= 3.12 kΩ
1
2π ⎡⎣3.12 1⎤⎦ × 10 × ( 6.65 + 386 ) × 10−12
Or f H = 535 kHz
fH =
b.
3
From Equation (11.140), f Z =
1
1
=
2π R0 C0 2π (10 × 106 )(10−12 )
Or f Z = 15.9 kHz
11.94
The differential-mode half circuit is:
⎛v ⎞
g m ⎜ d ⎟ RC
⎝ 2⎠
or Av =
v02 =
rπ
⎛1+ β ⎞
1+ ⎜
R
⎟ E
r
⎝ π ⎠
rπ =
(100 )( 0.026 )
0.5
= 5.2 kΩ
⎛1⎞
⎜ ⎟ β RC
⎝ 2⎠
+ (1 + β ) RE
⎛1⎞
⎜ ⎟ (100 )(10 )
500
2
=
Av = ⎝ ⎠
5.2 + (101) RE 5.2 + (101) RE
a.
For RE = 0.1 kΩ : Av = 32.7
b.
For RE = 0.25 kΩ : Av = 16.4
Chapter 12
Exercise Solutions
EX12.1
A
1 + Aβ
A
A
1 + Aβ =
⇒ Aβ =
−1
Af
Af
a.
Af =
β=
1 1
1
1
− =
−
= 0.05 − 0.0001 ⇒ β = 0.0499
Af A 20 104
Af
b.
=
20
(1/ β ) (1/ 0.0499 )
=
20
= 0.998
20.040
EX12.2
A
1 + Aβ
1 1
1
1
− =
− 6 = 0.01 − 10−6
β=
Af A 100 10
Af =
β = 0.009999
dA ⎛ Af ⎞ dA
=
⋅
Af
(1 + β A) A ⎜⎝ A ⎟⎠ A
dAf ⎛ 100 ⎞
dAf
= ⎜ 6 ⎟ ( 20 ) % ⇒
= 0.002%
Af
Af
⎝ 10 ⎠
dAf
=
1
⋅
EX12.3
Bandwidth = ω H (1 + β A0 )
⎛A ⎞
⎛ 105 ⎞
= ω H ⎜ 0 ⎟ = ( 2π )(10 ) ⎜
⎟
⎝ 100 ⎠
⎝ At ⎠
ω = ( 2π ) (104 ) rad / sec ⇒ f = 10 kHz
EX12.4
(a)
vOA = A1 A2 vi + A2 vn
= (100 )(10 ) vi + (10 ) vn
So 1000vi
S
=
= 100 i
10vn
No
Ni
(b)
A1 A2
A2
vOC =
vi +
vn
1 + β Α1 Α2
1 + β Α1 Α2
=
105
10
v +
vn
5 i
1 + ( 0.001)10
1 + ( 0.001) (105 )
So 103 vi
S
=
= 104 i
N o 0.1vn
Ni
EX12.5
a.
Vε = VS − V fb = 100 − 99 = 1 m V
V0 = AvVε ⇒ Av =
5
⇒ Av = 5000 V/V
0.001
V fb 0.099
V fb = β V0 ⇒ β =
=
⇒ β = 0.0198 V/V
V0
5
Avf =
Av
1 + β Av
=
5000
⇒ Avf = 50 V/V
1 + ( 0.0198 )( 5000 )
Rif = Ri (1 + β Av ) = ( 5 ) ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ⇒ Rif = 500 kΩ
b.
R0 f =
R0
4
=
⇒ R0 f ⇒ 40 Ω
1 + β Av 1 + ( 0.0198 )( 5000 )
EX12.6
a.
I ε = I S − I fb = 100 − 99 = 1 µ A
Ai =
β=
Aif =
I0
5
=
⇒ Ai = 5000 A/A
I ε 0.001
I fb
I0
=
0.099
⇒ β = 0.0198 A/A
5
Ai
5000
=
⇒ Aif = 50 A/A
1 + Ai β 1 + ( 5000 )( 0.0198 )
b.
Rif =
Ri
5
=
⇒ Rif ⇒ 50 Ω
1 + β Ai 1 + ( 0.0198 )( 5000 )
R0 f = (1 + β Ai ) R0 = ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ( 4 ) ⇒ R0 f = 400 kΩ
EX12.7
Avf =
Avf =
Av
104
=
⇒ Avf = 3.9984
Av
104
1+
1+
1 + ( R2 / R1 )
1 + ( 30 /10 )
105
= 3.99984
105
1+
1 + ( 30 /10 )
3.99984 − 3.9984
× 100% ⇒ 0.0360%
3.9984
EX12.8
Use a non inverting op-amp.
R
R
1 + 2 = 15 ⇒ 2 = 14
R1
R1
Let
β=
R2 = 140 K
R1 = 10 K
1
= 0.066667
R2
1+
R1
Input resistance.
Rif = 5 ( 0.06667 ) ( 5 × 103 ) ≅ 1.67 MΩ
Rof =
50
( 0.066667 ) ( 5 × 103 )
≡ 0.15Ω
EX12.9
⎛
⎞
⎟
⎞⎜
RE
⎜
⎟ ⋅ ii
⎟
⎠ ⎜ R + rk ⎟
⎜ E 1+ h ⎟
⎝
FE ⎠
80
0.026
( )(
)
rk =
= 4.16 k Ω
0.5
Then
rk
4.16
=
= 0.0514 k Ω
1 + hFE
81
Then we want
⎞
io
RE
⎛ 80 ⎞ ⎛
= 0.95 = ⎜ ⎟ ⎜
⎟
ii
⎝ 81 ⎠ ⎝ RE + 0.0514 ⎠
⎛ h
io = ⎜ FE
⎝ 1 + hFE
or
⎛
⎞
RE
⎜
⎟ = 0.9619
⎝ RE + 0.0514 ⎠
which yields
RE ( min ) = 1.30 k Ω
and
⎛ 81 ⎞
V + = I E RE + 0.7 = ⎜ ⎟ ( 0.5 )(1.3) + 0.7 ⇒ V + ( min ) = 1.36 V
⎝ 80 ⎠
EX12.10
Use the configuration shown in figure 12.20.
RS = 500 Ω, RL = 200 Ω
Let 1 +
RF
= 15
R1
For example, let R1 = 2 K
RF = 28 K
EX12.11
a.
⎛ R2 ⎞
⎛ 20 ⎞
VG = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5
⎝ 20 + 30 ⎠
⎝ R1 + R2 ⎠
VG = −1 V
VS = −1 − VGS
ID =
VS − ( −5 )
RS
= K n (VGS − VTN )
2
( −1) − VGS + 5 = (1.5)( 0.4 )(VGS − 2 )
2
4 − VGS = 0.6 (V 2GS − 4VGS + 4 )
0.6V 2GS − 1.4VGS − 1.6 = 0
VGS =
1.4 ±
(1.4 )
2
+ 4 ( 0.6 )(1.6 )
2 ( 0.6 )
VGS = 3.174 V
g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 3.17 − 2 )
g m = 3.52 mA / V
⎛ RD ⎞
⎛ 2 ⎞
I0 = − ⎜
⎟ ( g mVgs ) = − ⎜
⎟ ( 3.52 ) Vgs
⎝ 2+2⎠
⎝ RD + RL ⎠
I 0 = −1.76Vgs
Vi = Vgs + g mVgs RS ⇒ Vgs =
Vgs =
Vi
1 + g m RS
Vi
= ( 0.4153) Vi
1 + ( 3.52 )( 0.4 )
I 0 = − (1.76 )( 0.4153)Vi ⇒ Agf =
I0
= −0.731 mA / V
Vi
b.
For K n = 1 mA / V 2
From dc analysis:
4 − VGS = (1)( 0.4 )(VGS − 2 )
2
4 − VGS = 0.4 (V 2GS − 4VGS + 4 )
0.4V 2GS − 0.6VGS − 2.4 = 0
VGS =
0.6 ±
2
( 0.6 ) + 4 ( 0.4 )( 2.4 )
2 ( 0.4 )
VGS = 3.31V
g m = 2 (1)( 3.31 − 2 ) = 2.623 mA / V
⎛ 2 ⎞
I0 = − ⎜
⎟ ( 2.623) Vgs = −1.311Vgs
⎝ 2+2⎠
Vi
Vgs =
= Vi ( 0.488 )
1 + ( 2.623)( 0.4 )
I 0 = − (1.31)( 0.488 )Vi
Agf =
I0
= −0.6398 mA / V
Vi
% change =
0.731 − 0.6398
⇒ 12.5%
0.731
EX12.12
Use the circuit with the configuration shown in Figure 12.27.
The LED replaces RL .
1
⇒ RE = 100 Ω
RE
Agf = 10 mS = 10 × 10−3 =
EX12.13
dc analysis:
V0
10 − V0
= ID +
4.7
47 + 20
I D = K n (VGS − VTN )
(1)
2
(2)
⎛ 20 ⎞
VGS = ⎜
(3)
⎟ V0 = 0.2985V0
⎝ 20 + 47 ⎠
2.13 − V0 ( 0.213) = I D + ( 0.0149 ) V0
(1)
I D = 2.13 − V0 ( 0.2279 )
From (2):
2.13 − V0 ( 0.2279 ) = 1 ⎡⎣( 0.2985V0 ) − 1.5⎤⎦
2
2
2.13 − V0 ( 0.2279 ) = 0.0891V0 − 0.8955V0 + 2.25
2
0.0891V0 − 0.6676V0 + 0.12 = 0
V0 =
0.6676 ±
2
( 0.6676 ) − 4 ( 0.0891)( 0.12 )
2 ( 0.0891)
V0 = 7.31 V
10 − 7.31 7.31
−
= 0.572 − 0.109
4.7
67
I D = 0.463 mA
ID =
VGS =
a.
0.463
+ 1.5 = 2.18
1
g m = 2 K n (VGS − VTN ) = 2 (1)( 2.18 − 1.5 ) ⇒ g m = 1.36 mA/V
V0 − Vgs
V0
+ g mVgs +
= 0 (1)
RD
RF
Vgs − Vi
RS
+
Vgs − V0
RF
(2)
=0
⎛ 1
1 ⎞ V0 Vi
+
+
Vgs ⎜
⎟=
⎝ RS RF ⎠ RF RS
V
1 ⎞ V
⎛ 1
Vgs ⎜ + ⎟ = 0 + i
⎝ 20 47 ⎠ 47 20
Vgs ( 0.0713) = V0 ( 0.0213) + Vi ( 0.050 )
Vgs = V0 ( 0.299 ) + Vi ( 0.701)
From (1):
V0
V Vgs
+ (1.36 ) Vgs + 0 −
=0
4.7
47 47
V0 ( 0.213) + (1.36 ) Vgs + V0 ( 0.213) − Vgs ( 0.0213) = 0
V0 ( 0.234 ) + Vgs (1.34 ) = 0
V0 ( 0.234 ) + (1.34 ) ⎡⎣(V0 )( 0.299 ) + (Vi )( 0.701) ⎤⎦ = 0
V0 ( 0.635 ) + Vi ( 0.939 ) = 0 ⇒ Avf =
V0
= −1.48
Vi
b.
For K n = 1.5 mA / V 2
From dc analysis:
2.13 − V0 ( 0.2279 ) = 1.5 ⎡⎣( 0.2985V0 ) − 1.5⎤⎦
2
= 1.5 ⎡⎣0.0891V0 − 0.8955V0 + 2.25⎤⎦
2
2
= 0.1337V0 − 1.343V0 + 3.375
2
0.1337V0 − 1.115V0 + 1.245 = 0
V0 =
V0 =
1.115 ±
2
(1.115) − 4 ( 0.1337 )(1.245 )
2 ( 0.1337 )
1.115 ± 0.7597
⇒ V0 = 7.01 V
2 ( 0.1337 )
10 − 7.01 7.01
−
= 0.636 − 0.105
4.7
67
I D = 0.531 mA
ID =
0.531
+ 1.5 = 2.09
1.5
g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 2.09 − 1.5 )
VGS =
= 1.77 mA/V
From ac analysis:
V0
V Vgs
+ (1.77 )Vgs + 0 −
=0
4.7
47 47
V0 ( 0.213) + (1.77 ) Vgs + V0 ( 0.0213) − Vgs ( 0.0213) = 0
V0 ( 0.234 ) + Vgs (1.75 ) = 0
V0 ( 0.234 ) + (1.75 ) ⎡⎣V0 ( 0.299 ) + Vi ( 0.701) ⎤⎦ = 0
V0 ( 0.757 ) + Vi (1.23) = 0 ⇒ Avf =
% change =
1.62 − 1.48
⇒ 9.46%
1.48
V0
= −1.62
Vi
EX12.14
a.
Input resistance.
V0 − Vgs
V0
+ g mVgs +
= 0 (1)
RD
RF
Ii =
Vgs − V0
RF
(2)
So V0 = Vgs − I i RF
⎛ 1
⎛
1 ⎞
1 ⎞
V0 ⎜
+
⎟ + Vgs ⎜ g m −
⎟ = 0 (1)
RF ⎠
⎝ RD RF ⎠
⎝
1 ⎞
1 ⎞
⎛ 1
⎛
⎡⎣Vgs − I i ( 47 ) ⎤⎦ ⎜
+ ⎟ + Vgs ⎜1.36 − ⎟ = 0
47 ⎠
⎝ 4.7 47 ⎠
⎝
⎣⎡Vgs − I i ( 47 ) ⎦⎤ ( 0.234 ) + Vgs (1.34 ) = 0
Vgs
Vgs (1.57 ) = I i (11.0 ) ⇒ Rif =
= 7.0 kΩ
Ii
Output Resistance.
IX =
VX
VX
+ g mVgs +
RD
RS + RF
⎛ RS ⎞
⎛ 20 ⎞
Vgs = ⎜
⎟ VX = ⎜
⎟ VX = 0.2985VX
⎝ 20 + 47 ⎠
⎝ RS + RF ⎠
V
VX
I X = X + (1.36)(0.2985)VX +
4.7
20 + 47
I X = VX [ 0.213 + 0.406 + 0.0149]
R0 f =
b.
VX
= 1.58 kΩ
IX
From part (a)
1 ⎞
⎛
⎡⎣Vgs − I i ( 47 ) ⎤⎦ ( 0.234 ) + Vgs ⎜ 1.77 − ⎟ = 0
47 ⎠
⎝
Vgs
Vgs (1.98 ) = I i (11) ⇒ Rif =
= 5.56 kΩ
Ii
IX =
VX
VX
+ (1.77 )( 0.2985 ) VX +
4.7
20 + 47
I X = VX [ 0.213 + 0.528 + 0.0149] ⇒ R0 f =
VX
= 1.32 kΩ
IX
EX12.15
Use the circuit with the configuration shown in Figure 12.41
Let RF = 10 K
EX12.16
⎛ 5.5 ⎞
VTH = ⎜
⎟ (10 ) = 0.9735 V
⎝ 5.5 + 51 ⎠
RTH = 5.5 || 51 = 4.965 kΩ
I BQ =
0.973 − 0.7
= 0.00217 mA
4.96 + (121)(1)
I CQ = 0.2605 mA
rπ = 11.98 kΩ, g m = 10.02 mA / V
Req = RS || R1 || R2 || rπ
= (10 ) || 51|| 5.5 || 12
= 2.598 kΩ
From Equation (12.99(b)):
⎛
⎞
Req
T = ( g m RC ) ⎜
⎜ R + R + R ⎟⎟
F
eq ⎠
⎝ C
2.598
⎛
⎞
= (10 )(10 ) ⎜
⎟ ⇒ T = 2.75
10
82
2.598
+
+
⎝
⎠
EX12.17 Computer Analysis
EX12.18
Vε = −Vt , V0 = AvVε = − AvVt
⎛ R1 || Ri ⎞
⎛ R1 || Ri ⎞
Vr = ⎜
⎟ V0 = − ⎜
⎟ ( AvVt )
+
R
||
R
R
2 ⎠
⎝ 1 i
⎝ R1 || Ri + R2 ⎠
T =−
⎛ R1 || Ri ⎞
Vr
= Av ⎜
⎟
Vt
⎝ R1 || Ri + R2 ⎠
or
T=
Av
R
1+ 2
R1 Ri
⎛ f′ ⎞
Phase = −180° = −3 tan −1 ⎜ 5 ⎟
⎝ 10 ⎠
⎛ f′ ⎞
or tan −1 ⎜ 5 ⎟ = 60° ⇒ f ′ = 1.732 × 105 Hz
⎝ 10 ⎠
β (100 )
T ( f ′) = 1 =
3
2
⎡
f′ ⎞ ⎤
⎛
⎢ 1+ ⎜ 5 ⎟ ⎥
⎢
⎝ 10 ⎠ ⎥⎦
⎣
β (100 )
=
⇒ β = 0.08
3
⎡ 1 + (1.732 )2 ⎤
⎣⎢
⎦⎥
EX12.19
EX12.20
Afo =
1000
= 7.092
1 + ( 0.140 )(1000 )
( 0.140 )(1000 )
T =1=
⎛ f ⎞
⎛ f ⎞
1 + ⎜ 3 ⎟. 1 + ⎜
4 ⎟
10
⎝
⎠
⎝ 5 × 10 ⎠
2
⎛ f ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
By trial and error, at f = 7.65 × 104 Hz
( 0.140 )(1000 )
2
2
1 + ( 76.5 ) 1 + (1.53) 1 + ( 0.0765 )
( 0.140 )(1000 )
=
= 1.00
76.5
( )(1.828 )(1.0 )
T =
2
Then
φ = − ⎡⎣ tan −1 (76.5) + tan −1 (1.53) + tan −1 (0.0765) ⎤⎦
= − [89.25 + 56.83 + 4.37 ] = −150.45
Phase Marge = 180 − 150.45 = 29.5°
EX12.21
The new loop gain function is
105
1
T ′( f ) =
×
⎛
f ⎞⎛
f ⎞ ⎛1 + j ⋅ f ⎞ ⎛ 1 + j ⋅ f ⎞
⎟⎜
⎟
⎜1 + j ⋅
⎟⎜1 + j ⋅
⎟ ⎜
107 ⎠ ⎝
5 × 108 ⎠
f PD ⎠ ⎝
5 × 105 ⎠ ⎝
⎝
⎛ f ⎞
f ⎞
⎪⎧
⎛ f ⎞
⎛ f ⎞ ⎪⎫
−1 ⎛
Phase = − ⎨ tan −1 ⎜
+ tan −1 ⎜ 7 ⎟ + tan −1 ⎜
⎟ + tan ⎜
5 ⎟
8 ⎟⎬
⎝ 5 × 10 ⎠
⎝ 10 ⎠
⎝ 5 × 10 ⎠ ⎪⎭
⎪⎩
⎝ f PD ⎠
For a phase margin 45° ⇒ Phase = −135°, the poles are far apart so this will occur at approximately
f ′ = 5 × 105 Hz. Then
105
T ′( f ) = 1 =
⎛ 5 × 105 ⎞
1+ ⎜
⎟ × 1+1× 1× 1
⎝ f PD ⎠
2
105
1=
(1.414 )
⎛ 5 × 105 ⎞
9
1+ ⎜
⎟ = 5 × 10
⎝ f PD ⎠
⎛ 5 × 105 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
2
2
f PD ≅
5 × 105
5 × 109
⇒ f PD = 7.07 Hz
EX12.22
Af ( 0 ) =
A0
1 + β A0
=
2 × 105
1 + ( 0.05 ) ( 2 × 105 )
⇒ Af (0) ≅ 20
f C = f PD (1 + β A0 ) = 100 ⎡⎣1 + ( 0.05 ) ( 2 × 105 ) ⎤⎦ ⇒ fC ≅ 1 MHz
EX12.23
Phase Margin = 45° ⇒ Phase = −135°
This will occur at approximately f ′ = 107 Hz
105
T ′( f ) = 1 =
⎛ 107 ⎞
1+ ⎜
⎟ × 2× 1
⎝ f PD ⎠
2
⎛ 107 ⎞
9
1+ ⎜
⎟ = 5 × 10
f
⎝ PD ⎠
107
⇒ f PD = 141 Hz
f PD ≅
5 × 109
2
TYU12.1
A
Af =
1 + Aβ
A f + A f Aβ = A
Af = A (1 − Af β )
A=
Af
1 − Af β
=
80
1 − ( 80 )( 0.0120 )
A = 2000
TYU12.2
dAf
dA ⎛ Af ⎞ dA
1
=
⋅
=⎜
⎟.
Af
(1 + β A) A ⎝ A ⎠ A
dA dAf
=
A
Af
⎛ A
⋅⎜
⎜A
⎝ f
⎞
⎛ 5 × 105 ⎞
dA
= ±5%
⎟⎟ = ( 0.001) ⎜
⎟⇒
A
⎝ 100 ⎠
⎠
TYU12.3
Af ⋅ f H = A0 ⋅ f1
Af =
6
A0 ⋅ f1 (10 ) ( 8 )
=
⇒ Af ( 0 ) = 32
fH
250 × 103
TYU12.4
Vε = VS − V fb = 100 − 99 = 1 mV
Ag =
β=
Agf =
I 0 5 mA
=
⇒ Ag = 5 A/V
Vε 1 mV
V fb
=
I0
99 mV
⇒ β = 19.8 V/A
5 mA
Ag
1 + β Ag
TYU12.5
=
5
⇒ Agf = 0.05 A/V = 50 mA/V
1 + (19.8 )( 5 )
I ε = I S − I fb = 100 − 99 = 1 µ A
Az =
β=
Azf =
V0
5V
=
⇒ Az = 5 × 106 V/A
Iε 1 µ A
I fb
=
V0
99 µ A
⇒ β = 1.98 × 10−5 A/V
5V
Az
1 + β Az
=
5 × 106
1 + (1.98 × 10−5 )( 5 × 106 )
⇒ Azf = 5 × 104 V/A = 50 V/mA
TYU12.6
a.
rπ =
gm =
I CQ
=
VT
hFEVT (100 )( 0.026 )
=
= 5.2 kΩ
I CQ
0.5
0.5
= 19.23 mA / V
0.026
⎛1
⎞
⎛ 1
⎞
+ 19.23 ⎟ ( 2 )
⎜ + g m ⎟ RE
⎜
r
5.2
⎠
⎠
= ⎝
Avf = ⎝ π
⎛1
⎞
⎛ 1
⎞
+ 19.23 ⎟ ( 2 )
1 + ⎜ + g m ⎟ RE 1 + ⎜
⎝ 5.2
⎠
⎝ rπ
⎠
(19.42 )( 2 )
⇒ Avf = 0.97490
1 + (19.42 )( 2 )
= rπ + (1 + hFE ) RE = 5.2 + (101)( 2 ) ⇒ Rif
=
Rif
R0 f = RE
b.
= 207.2 kΩ
rπ
5.2
=2
⇒ R0 f = 0.0502 kΩ ⇒ 50.2 Ω
1 + hFE
101
hFE = 150 ⇒ rπ = 7.8 kΩ, g m = 19.23mA/V
⎛ 1
⎞
+ 19.23 ⎟ ( 2 )
⎜
(19.36 )( 2 )
7.8
⎠
Avf = ⎝
=
1 + (19.36 )( 2 )
⎛ 1
⎞
1+ ⎜
+ 19.23 ⎟ ( 2 )
7.8
⎝
⎠
Avf = 0.97482 ⇒ 0.0082% change in Avf
Rif = 7.8 + (101)( 2 ) = 209.8 kΩ ⇒ 1.25% change in Rif
R0 f = RE
rπ
7.8
=2
= 2 0.0517
1 + hFE
151
R0 f = 50.36 Ω ⇒ 0.319% change in R0 f
TYU12.7
V0 = ( g mVgs ) RS
Vi = Vgs + g m RSVgs
Vi
Vgs =
1 + g m RS
g m = 2 K n I DQ = 2
Avf =
Avf =
( 0.2 )( 0.25 ) = 0.447 mA / V
V0
g m RS
=
Vi 1 + g m RS
( 0.447 )( 5 )
⇒ Avf
1 + ( 0.447 )( 5 )
= 0.691
Rif = ∞
I X = g mVgs =
VX
RS
Vgs = −VX
⎛
1 ⎞
I X = VX ⎜ g m +
⎟
RS ⎠
⎝
1
1
R0 f =
RS =
5
gm
0.447
R0 f = 1.55 kΩ
TYU12.8 Computer Analysis
TYU12.9 Computer Analysis
TYU12.10
a.
Agf =
b.
Agf =
hFE ⋅ Ag
1 + ( hFE Ag ) RE
=
( 200 ) (103 )
⇒ Agf
1 + ( 200 ) (103 )(103 )
( 200 ) (104 )
⇒ Agf
1 + ( 200 ) (104 )(103 )
≅ 1 mA/V
= 10−3 A/V=1 mA/V
Percent change is negligible. ( 4.5 × 10−7 % )
( 200 ) (104 )
( 200 ) (103 )
−
1 + ( 200 ) (104 )(103 ) 1 + ( 200 ) (103 )(103 )
10−3
=
−
=
( 200 ) (104 ) ⎡⎣1 + ( 200 ) (104 )(103 )⎤⎦
(10 )( 2 ×10 )( 2 ×10 )
−3
9
8
(200)(103 )[1 + (200)(104 )(103 )]
(10−3 )(2 × 109 )(2 × 108 )
200 × 104 − 200 × 103
(10−3 )( 2 ×109 )( 2 ×108 )
= 4.5 × 10−9
TYU12.11
T ( f1 )
Ai 0 β
(10 ) ( 0.01)
5
⎛
⎛ f ⎞
f ⎞
⎜ 1 + j ⋅ ⎟ 1 + j ⋅ ⎜ 10 ⎟
⎝ ⎠
f1 ⎠
⎝
3
10
=1=
2
⎛ f′⎞
1+ ⎜ E ⎟
⎝ 10 ⎠
T = Ai β =
=
⎛ f′⎞
1 + ⎜ E ⎟ = 106
⎝ 10 ⎠
2
f E′ = 10 106 − 1 ⇒ f E′ ≈ 104 Hz
⎛ 104 ⎞
⎛ f′⎞
Phase = φ = − tan −1 ⎜ E ⎟ = − tan −1 ⎜
⎟
⎝ 10 ⎠
⎝ 10 ⎠
= − tan −1 (103 )
φ ≈ 90°
Phase Margin = 180 − 90 ⇒ Phase Margin = 90°
TYU12.12
Αι 0 β
⎛
f ⎞⎛
f ⎞
⎜1+ j ⋅ ⎟ ⎜1 + j ⋅ ⎟
f1 ⎠ ⎝
f2 ⎠
⎝
⎡
⎛ f ⎞
⎛ f ⎞⎤
Phase = − ⎢ tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ ⎥
⎝ f1 ⎠
⎝ f2 ⎠⎦
⎣
T = Ai β =
Phase Margin = 60° ⇒ Phase = −120°
⎡
⎛ f ⎞
⎛ f ⎞⎤
−120° = − ⎢ tan −1 ⎜ 4 ⎟ + tan −1 ⎜ 5 ⎟ ⎥
10
10 ⎠ ⎦
⎝
⎠
⎝
⎣
At f ′ = 7.66 × 104 Hz,
Phase = − ⎡⎣ tan −1 ( 7.66 ) + tan −1 ( 0.766 ) ⎤⎦
= − [82.56 + 37.45]
= −120°
(10 ) β
5
T ( f ′) = 1 =
2
1+ ( 7.66 ) × 1 + ( 0.766 )
(10 ) β
5
1=
( 7.725 )(1.26 )
2
⇒ β = 9.73× 10−5
TYU12.13
Phase Margin = 60° ⇒ Phase = −120°
⎛ f′ ⎞
Phase = −120° = −3 tan −1 ⎜ 5 ⎟
⎝ 10 ⎠
⎛ f′ ⎞
tan −1 ⎜ 5 ⎟ = 40° ⇒ f ′ = 0.839 × 105 Hz
⎝ 10 ⎠
β (100 )
T ( f ′) = 1 =
3
2
⎡
f′ ⎞ ⎤
⎛
⎢ 1+ ⎜ 5 ⎟ ⎥
⎢
⎝ 10 ⎠ ⎥⎦
⎣
β (100 )
=
⇒ β = 0.0222
3
⎡ 1 + ( 0.839 )2 ⎤
⎣⎢
⎦⎥
Chapter 12
Problem Solutions
12.1
(a)
Af =
A
1 + Aβ
120 =
5 × 105
1 + (5 × 105 ) β
120 + (120)(5 × 105 ) β = 5 × 105
β = 0.008331
(b)
5 × 103
120 =
1 + (5 × 103 ) β
120 + (120)(5 × 103 ) β = 5 × 103
β = 0.008133
12.2
Af =
(a)
A
1 + Aβ
β = 0.15
T = Aβ
(i)
T =∞
A = 80 dB ⇒ A = 104 ⇒ T = 1.5 × 103
(ii)
(iii)
T = 15
1
(i)
Af = = 6.667
β
(ii)
Af = 6.662
(iii)
Af = 6.25
(b)
(i)
(ii)
(iii)
(i)
T =∞
T = 2.5 × 103
T = 25
1
Af = = 4.00
(ii)
Af = 3.9984
(iii)
Af = 3.846
β
12.3
(a)
A
1
≅ = 125
1 + Aβ β
β = 0.0080
(b)
Af =
Af = (125)(0.9975) = 124.6875
124.6875 =
A
1 + (0.008) A
124.6875 [1 + (0.008) A] = A
124.6875 = A [1 − 0.9975]
A = 49,875
12.4
(a)
100 =
104
1 + (104 ) β
β = 9.9 × 10−3
(b)
⎛ Af ⎞ dA
=⎜
⎟
⎝ A⎠ A
Af
dAf
dAf
⎛ 100 ⎞
= ⎜ 4 ⎟ (−0.10) = −0.001 ⇒
= −0.10%
Af
⎝ 10 ⎠
12.5
⎛ Af ⎞ dA
=⎜
⎟
⎝ A⎠ A
Af
dAf
−0.001 =
Af
5 × 104
(−0.10)
Af = 500
500 =
5 × 104
1 + (5 × 104 ) β
β = 1.98 ×10−3
12.6
(a)
For Fig. P12.6(a)
vo1
vo
200
10
=
=
vi 1 + 200 β1 vo1 1 + 10 β1
⎛ 200 ⎞ ⎛ 10 ⎞
Af = ⎜
⎟⎜
⎟ = 40
⎝ 1 + 200 β1 ⎠ ⎝ 1 + 10 β1 ⎠
50 = (1 + 200 β1 )(1 + 10 β1 ) = 1 + 210 β1 + 2000 β12
2000 β12 + 210 β1 − 49 = 0
−210 ± 44100 + 4(2000)(49)
2(2000)
β1 = 0.1126
For Fig P12.6(b)
(200)(10)
40 =
1 + (200)(10) β 2
β1 =
β 2 = 0.0245
Fig. P12.6(a)
(b)
A1 = 180
180
10
⎛
⎞⎛
⎞
= (8.4634)(4.704)
Af = ⎜
⎟
⎜
⎟
⎝ 1 + (180)(0.1126) ⎠ ⎝ 1 + (10)(0.1126) ⎠
Af = 39.81
dAf
Af
=
39.81 − 40
⇒ −0.475%
40
Fig. P12.6(b)
(180)(10)
= 39.91
Af =
1 + (180)(10)(0.0245)
dAf 39.91 − 40
=
⇒ −0.225%
40
Af
(c)
Fig. P12.6(b) is a better feedback circuit.
12.7
(a)
VO = (−10)(−15)(−20)Vε = −3000Vε
Vε = β VO + VS
So VO = −3000( β VO + VS )
We find
V
−3000
Avf = O =
VS 1 + 3000 β
For Avf = −120 =
(b)
Then
Avf =
−3000
⇒ β = 0.008
1 + 3000 β
Now VO = (−9)(−13.5)(−18)Vε = −2187Vε
−2187
−2187
=
= −118.24
1 + 2187 β 1 + 2187(0.008)
% change =
120 − 118.24
× 100 ⇒ 1.47% change
120
12.8
(105 )(4) = (50) f B ⇒ f B = 8 kHz
12.9
(a)
(b)
(50) f3− dB = (105 )(4) ⇒ f 3− dB = 8 kHz
(10) f3− dB = (105 )(4) ⇒ f3− dB = 40 kHz
12.10
(50)(20 × 103 ) = 5A0 so A0 = 2 × 105
12.11
Low freq. Af =
A0
1 + A0 β
5000
⇒ β = 0.0098
1 + (5000) β
Freq. response
100 =
Af =
A
1 + Aβ
5000
f
f ⎞
⎛
⎞⎛
⎜1 + j f ⎟⎜ 1 + j f ⎟
1 ⎠⎝
2 ⎠
= ⎝
(5000)(0.0098)
1+
f ⎞⎛
f ⎞
⎛
⎜1 + j f ⎟⎜ 1 + j f ⎟
1 ⎠⎝
2 ⎠
⎝
5000
f ⎞⎛
f ⎞
⎛
⎜1 + j f ⎟ ⎜ 1 + j f ⎟ + 49
1 ⎠⎝
2 ⎠
⎝
5000
=
f
f ⎛ jf ⎞ ⎛ jf
1+ j + j + ⎜ ⎟⎜
f1
f 2 ⎝ f1 ⎠ ⎝ f 2
=
=
⎞
⎟ + 49
⎠
5000
f
f ⎛ jf ⎞ ⎛ jf ⎞
50 + j + j + ⎜ ⎟ ⎜ ⎟
f1
f 2 ⎝ f1 ⎠ ⎝ f 2 ⎠
Also
Af =
Af 0
f ⎞⎛
f ⎞
⎛
⎜1 + j f ⎟ ⎜ 1 + j f ⎟
A ⎠⎝
B ⎠
⎝
=
100
f
f ⎛ f ⎞⎛ f ⎞
1+ j
j
+j
+ j
fA
f B ⎜⎝ f A ⎟⎠ ⎜⎝ f B ⎟⎠
So
100
100
=
1 ⎛ jf ⎞ ⎛ jf ⎞
f
f ⎛ f ⎞⎛ f ⎞
f
f
+j
+ ⎜ j ⎟⎜ j ⎟ 1 + j
+j
+ ⎜ ⎟⎜ ⎟
1+ j
fA
f B ⎝ f A ⎠⎝ f B ⎠
50 f1
50 f 2 50 ⎝ f1 ⎠ ⎝ f 2 ⎠
Then
1
1
1
1
+
=
+
f A f B 50 f1 50 f 2
and
1
1
=
f A f B 50 f1 f 2
f1 = 10 and f 2 = 2000
1
1
1
1
+
=
+
= 0.002 + 0.000010 = 0.002010
f A f B 50(10) 50(2000)
and
f
1
1
1
=
⇒
= B
f A f B (50)(10)(2000)
f A 106
Then
fB
1
+
= 0.002010
6
fB
10
10−6 f B2 + 1 = 2.01 + 10−3 f B
10−6 f B2 − 2.01× 10−3 f B + 1 = 0
fB =
2.01× 10−3 ± 4.0401× 10 −6 − 4(10−6 )(1)
2(10−6 )
fB =
2.01× 10−3 ± 2.0025 × 10−4
2(10−6 )
+ sign
f B = 1.105 × 103 Hz
+ sign
f A = 9.05 × 102 Hz
12.12
(a)
Fig. P12.6(a)
⎡
⎤
⎛ 200 ⎞
⎢
⎥
⎜
f ⎟
⎢
⎥
⎜⎜ 1 + j ⎟⎟
f1 ⎠
10
⎤
⎢
⎥⎡
⎝
Af = ⎢
⎢
⎥
⎥
200
+
1
(10)(0.1126)
⎛
⎞
⎣
⎦
⎢1 + ⎜
⎟ (0.1126) ⎥
⎢ ⎜ 1+ j f ⎟
⎥
⎢⎣ ⎜⎝
⎥⎦
f1 ⎟⎠
200
⎡
⎤
=⎢
⎥ (4.704)
f ⎞
⎛
⎢ ⎜1 + j ⎟ + 22.52 ⎥
f1 ⎠
⎣⎢ ⎝
⎦⎥
=
940.73
23.52 + j
f
f1
=
940.73
1
⋅
f
23.52 1 + j
(23.52) f1
40
f −3dB = (23.52)(100) ⇒ 2.352 kHz
jf
1+
(23.52) f1
Fig P12.6(b)
(200)(10)
f
1+ j
f1
2000
=
Af =
f
(0.0245)(200)(10)
1+
1 + j + 49
f
f1
1+ j
f1
=
=
(b)
2000
1
⋅
f −3dB = (50)(100) ⇒ 5 KHz
50 1 + j f
(50) f1
Overall feedback ⇒ wider bandwidth.
12.13
v0 = A1 A2 vi + A1vn
v0 = (100)vi + (1)vn = (100)(10) + (1)(1) ⇒
12.14
(a)
S0 1000
=
= 1000
N0
1
(b)
Circuit (b) – less distortion
12.15
(a)
Low input R ⇒ Shunt input
Low output R ⇒ Shunt output
Or a Shunt-Shunt circuit
(b)
High input R ⇒ Series input
High output R ⇒ Series output
Or a series-Series circuit
(c)
Shunt-Series circuit
(d)
Series-Shunt circuit
12.16
(a)
Ri (max) = Ri (1 + T ) = 10(1 + 104 ) ⇒ Ri (max) ≅ 105 k Ω
Ri
10
=
≅ 10−3 k Ω
1 + T 1 + 104
Or Ri (min) = 1Ω
Ri (min) =
Ro (max) = Ro (1 + T ) = 1(1 + 104 ) ⇒ Ro (max) ≅ 104 k Ω
(b)
Ro
1
=
≅ 10−4 k Ω
1 + T 1 + 104
Or Ro (min) = 0.1Ω
Ro (min) =
12.17
io
Series output = current signal and Shunt input = current
vi
signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are
shown.
Overall Transconductance Amplifier, Ag =
12.18
⎛ R1 || Ri ⎞
Vε = − ⎜
⎟ ⋅ Vx
⎝ R1 || Ri + R2 ⎠
12.19
Vε = Vi − V fb = 50 − 48 = 2mV
Av =
β=
Vo
5
=
= 2.5 × 103 V/V
Vε 0.002
V fb
Vo
=
0.048
= 0.0096 V/V
5
V
5
Arf = o =
= 100 V/V
Vi 0.05
12.20
⎛ R ⎞
R
Avf ≈ ⎜1 + 2 ⎟ = 20 ⇒ 2 = 19
R
R1
⎝
1 ⎠
vd = iS Ri
vS − vd (vs − vd ) − v0
+
R1
R2
(1)
v0 − A0 L vd v0 − (vs − vd )
+
=0
R0
R2
(2)
is =
⎛ 1
(v − v )
1 ⎞ A v
v0 ⎜ + ⎟ = 0 L d + S d
R0
R2
⎝ R0 R2 ⎠
A0 L vd (vS − vd )
+
R0
R2
v0 =
⎛ 1
1 ⎞
⎜ + ⎟
⎝ R0 R2 ⎠
From (1):
1 ⎡ A0 L vd (vS − vd ) ⎤
⋅⎢
+
R2 ⎦⎥
vS − vd vS − vd R2 ⎣ R0
+
−
iS =
R1
R2
⎛ 1
1 ⎞
⎜ + ⎟
⎝ R0 R2 ⎠
A0 L 1
1 ⎞
⎛
⎛
−
⎜ 1
⎜ 1
R0 R2
R2 ⎟
1
1
⎟ − vd ⎜ +
iS = vS ⎜ +
−
+
R
⎜ R1 R2
⎜ R1 R2 1 + R2 ⎟
1+ 2
⎜
⎟
⎜
R
R0
0 ⎠
⎝
⎝
vd = iS Ri
⎞
⎟
⎟
⎟
⎟
⎠
⎧
⎡⎛ 1
⎡⎛ 1
1 ⎞ ⎛ R ⎞ A0 L 1 ⎤ ⎫
1 ⎞ ⎛ R2 ⎞ 1
− ⎥⎪
⎪ Ri ⎢⎜ + ⎟ ⎜ 1 + 2 ⎟ +
⎢ ⎜ + ⎟ ⎜1 + ⎟ −
R
R
R
R
R
⎪
⎝ 1
2 ⎠⎝
0 ⎠
0
2 ⎦⎪
⎢ ⎝ R1 R2 ⎠ ⎝ R0 ⎠ R2
iS ⎨1 + ⎣
⎬ = vS ⎢
R
R
⎪
⎪
1+ 2
1+ 2
⎢
R0
R0
⎪
⎪
⎣⎢
⎩
⎭
⎧⎪ R
⎡1 R 1
⎡ 1 R2 1
1 A0 L ⎤ ⎫⎪
1⎤
iS ⎨1 + 2 + Ri ⎢ + 2 ⋅ +
+
⎥ ⎬ = vS ⎢ + ⋅ + ⎥
⎪⎩ R0
⎣ R1 R1 R0 R0 R0 ⎦ ⎪⎭
⎣ R1 R1 R0 R0 ⎦
⎧⎪
⎡R ⎛ R ⎞
⎤ ⎫⎪
⎡ R ⎛ R ⎞⎤
iS ⎨ R0 + R2 + Ri ⎢ 0 + ⎜1 + 2 ⎟ + A0 L ⎥ ⎬ = vS ⎢ 0 + ⎜ 1 + 2 ⎟ ⎥ (1)
R1 ⎠
R1 ⎠ ⎦
⎣ R1 ⎝
⎦ ⎭⎪
⎣ R1 ⎝
⎩⎪
Let R2 = 190 kΩ , R1 = 10 kΩ
⎧
⎡ 0.1
⎤⎫
⎡ 0.1
⎤
+ 20 + 105 ⎥ ⎬ = vS ⎢
+ 20 ⎥
iS ⎨0.1 + 190 + 100 ⋅ ⎢
⎣ 10
⎦⎭
⎣ 10
⎦
⎩
7
iS (1.000219 × 10 ) = vS (20.01)
vS
≅ 5 × 105 kΩ ⇒ Rif ≅ 500 MΩ
iS
Output Resistance
Rif =
⎤
⎥
⎥
⎥
⎥
⎦⎥
IX =
VX − A0 L vd
VX
+
R0
R2 + R1 || Ri
vd =
− R1 || Ri
⋅ VX
R1 || Ri + R2
A0 L ⋅ R1 || Ri
IX
1
1
1
=
=
+
+
VX R0 f R0 R0 ( R1 || Ri + R2 ) R2 + R1 || Ri
R1 || Ri = 10 ||100 = 9.09
1
1 105 ⎛ 9.09 ⎞
1
=
+
⋅⎜
⎟+
R0 f 0.1 0.1 ⎝ 9.09 + 190 ⎠ 190 + 9.09
= 10 + 4.566 × 104 + 0.00502
R0 f = 2.19 × 10−5 kΩ ⇒ R0 f = 0.0219 Ω
12.21
a.
v
vS − vd v0 − (vS − vd )
=
and vd = 0
R1
R2
A
⎛ 1
vS vS
v
1 ⎞
+
= 0 + vd ⎜ + ⎟
R1 R2 R2
R
R
⎝ 1
2 ⎠
v0 v0 ⎛ 1
1 ⎞
+ ⎜ + ⎟
R2 A ⎝ R1 R2 ⎠
⎛ 1
1 ⎞ v ⎡ 1 ⎛ R ⎞⎤
vS ⎜ + ⎟ = 0 ⎢1 + ⎜ 1 + 2 ⎟ ⎥
R1 ⎠ ⎦
⎝ R1 R2 ⎠ R2 ⎣ A ⎝
=
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
v0
= ⎝
vS
1⎛ R ⎞
1 + ⎜1 + 2 ⎟
A⎝
R1 ⎠
which can be written as
v
A
Avf = 0 =
vS
⎡ ⎛ R ⎞⎤
1 + ⎢ A / ⎜1 + 2 ⎟ ⎥
R1 ⎠ ⎦
⎣ ⎝
1
b.
β=
R
1+ 2
R1
20 =
c.
105
1 + (105 ) β
105
−1
So β = 20 5 ⇒ β = 0.04999
10
R2 1
R
1
Then
= −1 =
− 1 ⇒ 2 = 19.004
R1 β
0.04999
R1
A → 9 × 104
9 × 104
Af =
= 19.99956
1 + (9 × 104 )(0.04999)
d.
ΔAf
Af
=
ΔAf
−4.444 × 10−4
= −2.222 × 10 −3 % ⇒
= −0.005%
20
Af
12.22
Aif =
Rif =
Ai
1000
=
= 90.9 A/A
1 + β i Ai 1 + (1000)(0.01)
Ri
1 + β i Ai
=
1
⇒ Rif = 90.9 Ω
1 + (0.01)(1000)
Rof = Ro (1 + β i Ai ) = 10 [1 + (0.01)(1000) ] ⇒ Rof = 110 kΩ
12.23
I ε = I i − I fb = 50 − 47.5 = 2.5 µ A
Ai =
βi =
Io
5
=
= 2000 A/A
I ε 0.0025
I fb
Io
=
0.0475
= 0.0095 A/A
5
I
5
Aif = o =
= 100 A/A
I i 0.05
12.24
a.
Assume that V1 is at virtual ground.
V0 = − I fb RF
Now
I fb = I 0 +
I fb RF
V0
= I0 −
R3
R3
I fb = I S − I ε
and
I 0 = Ai I ε =
I0
Ai
so
I0
Ai
From above
⎛ R ⎞
I fb ⎜1 + F ⎟ = I 0
R3 ⎠
⎝
I fb = I S −
⎛
I 0 ⎞ ⎛ RF ⎞
⎜ I S − ⎟ ⎜1 +
⎟ = I0
Ai ⎠ ⎝
R3 ⎠
⎝
⎡
⎛ R ⎞
1 ⎛ R ⎞⎤
I S ⎜1 + F ⎟ = I 0 ⎢1 + ⎜ 1 + F ⎟ ⎥
R3 ⎠
R3 ⎠ ⎦
⎝
⎣ Ai ⎝
or
Aif =
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
=
⎡
1 ⎛ RF ⎞ ⎤
⎢1 + ⎜1 +
⎟⎥
R3 ⎠ ⎦
⎣ Ai ⎝
Ai
=
= Aif
Ai
1+
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
I0
IS
b.
βi =
c.
25 =
1
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
105
1 + (105 ) β i
105
−1
so β i = 25 5 ⇒ β i = 0.03999
10
RF
R
1
1
=
−1 =
− 1 ⇒ F = 24.0
so
R3 β i
R3
0.03999
Ai = 105 − (0.15)(105 ) = 8.5 × 104
d.
so Aif =
so
ΔAif
Aif
8.5 × 104
= 24.9989
1 + (8.5 × 104 )(0.03999)
=−
1.10 × 10−3
= −4.41× 10 −5 ⇒ −4.41× 10 −3 %
25
12.24
b.
V0 = ( g m1 Vgs + g m 2 Vπ ) RL (1)
Vi = Vgs + V0
(2)
⎛ 1 + hFE ⎞ V1
Vπ
V
+ g m 2 Vπ + 1 = 0 ⇒ Vπ ⎜
=0
⎟+
rπ
RE 2
⎝ rπ ⎠ RE 2
Vπ
V −V
= g m1 Vgs + 1 π = 0
rπ
RD1
or
⎡V
⎤
V
V1 = RD1 ⎢ π + π − g m1 Vgs ⎥
(4)
⎣ rπ RD1
⎦
Then
⎛ 1 + hFE
Vπ ⎜
⎝ rπ
⎞ RD1
⎟+
⎠ RE 2
⎪⎧⎛ 1 + hFE
Vπ ⎨⎜
⎪⎩⎝ rπ
Let
⎡ Vπ Vπ
⎤
− g m1 Vgs ⎥ = 0 (3)
⎢ +
⎣ rπ RD1
⎦
⎫
⎞ RD1
⎛ RD1 ⎞
1 ⎪
+
⎬ = g m1 ⎜
⎟+
⎟ Vgs
R
r
R
E2 π
E2 ⎪
⎝ RE 2 ⎠
⎠
⎭
(3)
Vπ ⋅
⎛R ⎞
1
= g m1 ⎜ D1 ⎟ Vgs
Req
⎝ RE 2 ⎠
⎛R ⎞
so Vπ = g m1 Req ⎜ D1 ⎟ Vgs
⎝ RE 2 ⎠
Then
⎡
⎛R ⎞ ⎤
V0 = ⎢ g m1 Vgs + g m1 g m 2 Req ⎜ D1 ⎟ Vgs ⎥ RL
⎝ RE 2 ⎠ ⎦
⎣
so
⎡
⎛ R ⎞⎤
V0 = g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ (Vi − V0 )
⎝ RE 2 ⎠ ⎦
⎣
so
⎡
⎛ R ⎞⎤
g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥
V
⎝ RE 2 ⎠ ⎦
⎣
Av = 0 =
Vi
⎡
⎛ R ⎞⎤
1 + g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥
⎝ RE 2 ⎠ ⎦
⎣
c.
Set Vi = 0
I X + g m1 Vgs + g m 2Vπ =
VX
RL
Vgs = −VX
From part (b), we have
⎛R ⎞
⎛R ⎞
Vπ = g m1 Req ⎜ D1 ⎟ Vgs = − g m1 Req ⎜ D1 ⎟ VX
R
⎝ E2 ⎠
⎝ RE 2 ⎠
Then
⎛R ⎞
IX
1
1
=
=
+ g m1 g m 2 Req ⎜ D1 ⎟
VX R0 RL
⎝ RE 2 ⎠
or
R0 = RL
1
g m1
1
⎛R ⎞
g m1 g m 2 Req ⎜ D1 ⎟
⎝ RE 2 ⎠
12.25
I S = I ε + I fb , V1 = I ε Ri
I fb = I 0 +
V0
and V0 = V1 − I fb RF
R3
I 0 = Ai I ε ⇒ I ε =
I0
Ai
(1)
Now
I fb = Ai I ε +
1
(V1 − I fb RF )
R3
⎡ R ⎤
V
I fb ⎢1 + F ⎥ = Ai Iε + 1
R
R3
3 ⎦
⎣
I fb = I S − I e
⎡ R ⎤
V
( I S − I e ) ⎢1 + F ⎥ = Ai I e + 1
R3
⎣ R3 ⎦
⎡⎛ R ⎞
⎤ V
⎡ R ⎤
I S ⎢1 + F ⎥ = I e ⎢⎜1 + F ⎟ + Ai ⎥ + 1
R3 ⎠
⎣ R3 ⎦
⎣⎝
⎦ R3
V
Iε = 1
Ri
⎤ 1 ⎪⎫
⎤
⎪⎧ 1 ⎡⎛ RF ⎞
⎟ + Ai ⎥ + ⎬
⎥ = V1 ⎨ ⋅ ⎢⎜ 1 +
R3 ⎠
⎦
⎦ R3 ⎪⎭
⎩⎪ Ri ⎣⎝
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
V1
⎝
Rif =
=
I S ⎪⎧ 1 ⎡⎛ RF ⎞
⎤ 1 ⎪⎫
⎨ ⋅ ⎢⎜ 1 +
⎟ + Ai ⎥ + ⎬
R3 ⎠
⎦ R3 ⎭⎪
⎩⎪ Ri ⎣⎝
⎡ R
I S ⎢1 + F
⎣ R3
The 1/ R3 term in the denominator will be negligible. Using the results of Problem 12.15:
25
Rif =
⎧1 ⎡
5 ⎫
⎨ ⎣ (25) + 10 ⎤⎦ ⎬
2
⎩
⎭
Rif ≅ 5 × 10−4 kΩ ⇒ Rif = 0.5 Ω
Output Resistance (Let Z L = 0)
IX =
VX
VX
+ Ai I ε +
R3
RF + Ri
Iε =
VX
RF + Ri
so
A + 1 RF
IX
1
1
=
=
+ i
= 24
,
VX R0 f R3 RF + Ri R3
Let RF = 240 kΩ, R3 = 10 k Ω
1
1 105 + 1
= +
Rof 10 240 + 2
so R0 f ≈
RF + Ri 240 + 2
= 5
⇒ R0 f ≈ 2.42 × 10−3 kΩ or R0 f ≈ 2.42 Ω
10 + 1
Ai + 1
12.26
Agf =
Ag
5
=
= 0.2 A/V
1 + β z Ag 1 + (4.8)(5)
Rif = Ri [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ
Rof = Ro [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ
12.27
Vε = Vi − V fb = 40 − 38 = 2.0 mV
I o 8 mA
=
= 4 A/V
Vε 2 mV
Ag =
V fb
βz =
I0
Agf =
=
38 mV
= 4.75 V/A
8 mA
I0
8 mA
=
= 0.2 A/V
Vi 40 mV
12.28
IE =
V − Vε
(1 + hFE )
⋅ I0 = S
hFE
RE
Also I 0 = hFE ( AgVε ) so Vε =
I0
hFE Ag
Then
V
I0
1 + hFE
⋅ I0 = S −
hFE
RE hFE Ag RE
⎡1 + hFE
⎤
VS
1
+
⎢
⎥ I0 =
h
h
A
R
RE
⎥
FE g E ⎦
⎣⎢ FE
⎡ Ag (1 + hFE ) RE + 1 ⎤
VS
⎢
⎥ I0 =
hFE Ag RE
RE
⎥⎦
⎣⎢
I0
1
=
VS RE
⎡
⎤
hFE Ag RE
hFE Ag
I0
⋅⎢
≈
⎥⇒
⎢⎣1 + Ag (1 + hFE ) RE ⎥⎦ VS 1 + (hFE Ag ) RE
b.
Bz = RE
c.
10 =
5 × 105
1 + (5 × 105 ) β z
5 × 105
−1
β z = 10 5 ⇒ β z = RE = 0.099998 kΩ
5 × 10
d.
If Ag → 5.5 × 105 then
Agf =
ΔAgf
Agf
5.5 × 105
= 10.0000182
1 + (5.5 × 105 )(0.099998)
=
1.82 × 10−5
⇒ 1.82 × 10−4 %
10
12.29
I E = (1 + hFE ) AgVε , I E =
Now (1 + hFE ) Ag I S Ri =
Vε
− I S and Vε = I S Ri , Vε = VS − Vε = VS − I S Ri
RE
1
⋅ (VS − I S Ri ) − I S
RE
⎡
⎤
V
Ri
+ 1⎥ I S = S
⎢(1 + hFE ) Ag Ri +
R
RE
E
⎣
⎦
⎡
⎤
VS
R
= RE ⎢(1 + hFE ) Ag Ri + i + 1⎥
IS
RE ⎦
⎣
From Problem 12.16:
(1 + hFE ) Ag ≈ hFE Ag = 5 × 105 mS
Rif =
RE ≈ 0.1 kΩ
20 ⎤
⎡
+ 1⎥
so Rif = (0.1) ⎢(5 × 105 )(20) +
0.1
⎣
⎦
or Rif = 106 kΩ
Vπ
= AgVε
rπ
I X = g mVπ +
VX − (−Vε )
R0
(1)
Vε = −( I X + AgVε )( RE || Ri )
(2)
or Vε ⎡⎣1 + Ag ( Rε || Ri ) ⎤⎦ = − I X ( RE || Ri )
Now:
V
V
I X = g m Ag rπ Vε + X + ε (1)
R0 R0
⎛
1 ⎞ ⎡ − I X ( RE || Ri ) ⎤ VX
I X = ⎜ g m Ag rπ + ⎟ ⎢
⎥+
R0 ⎠ ⎣⎢1 + Ag ( RE || Ri ) ⎦⎥ R0
⎝
V
R0 f = X
IX
⎧⎪ ⎛
1 ⎞ ⎡ ( RE || Ri ) ⎤ ⎫⎪
= R0 ⎨1 + ⎜ g m Ag rπ + ⎟ ⎢
⎥⎬
R0 ⎠ ⎣⎢1 + Ag ( RE || Ri ) ⎦⎥ ⎭⎪
⎪⎩ ⎝
g m rπ Ag = hFE Ag = 5 × 105 mS
Let hFE = 100 so Ag = 5× 103 mS
RE || Ri = 0.1 || 20 ≈ 0.1 kΩ
Then
⎧⎪ ⎛
⎤ ⎫⎪
1 ⎞⎡
0.1
R0 f = 50 ⎨1 + ⎜ 5 × 105 + ⎟ ⎢
⎥⎬
3
50 ⎠ ⎣1 + (5 × 10 )(0.1) ⎦ ⎪⎭
⎪⎩ ⎝
or R0 f = 5.04 MΩ
12.30
Azf =
Rif =
Rof =
12.31
Az
1 + β g Az
=
5
= 0.2 V/μA
1 + (4.8)(5)
Ri
1
=
⇒ Rif = 40 Ω
1 + β g Az 1 + (4.8)(5)
Ro
1
=
⇒ Rof = 40 Ω
1 + β g Az 1 + (4.8)(5)
I ε = I i − I fb = 40 − 38 = 2 μA
Az =
Vo 8
V
= =4
μA
Iε 2
βg =
Azf =
I fb
Vo
=
38
μA
= 4.75
8
V
Vo
8
V
=
= 0.2
μA
I i 40
12.32
a.
Assuming V1 is at virtual ground
(−V0 ) = − I fb RF and (−V0 ) = − Az Iε ⇒ I ε =
I fb = I S − I ε
⎛V ⎞
So V0 = ( I S − I ε ) RF = I S RF − ⎜ 0 ⎟ RF
⎝ Az ⎠
⎡ R ⎤
V0 ⎢1 + F ⎥ = I S RF
⎣ Az ⎦
V
RF
AR
so Azf = 0 =
= z F
I S ⎡ RF ⎤ Az + RF
⎢1 + A ⎥
z ⎦
⎣
Az
Az
or Azf =
=
⎛ 1 ⎞ 1 + Az β g
1 + Az ⎜
⎟
⎝ RF ⎠
1
RF
b.
βg =
c.
5 × 104 =
5 × 106
1 + (5 × 106 ) β g
5 × 106
−1
4
β g = 5 × 10 6 ⇒ β g = 1.98 × 10−5
5 × 10
1
RF =
⇒ RF = 50.5 kΩ
βg
d.
Az = (0.9)(5 × 106 ) = 4.5 × 106
V0
Az
Azf =
ΔAzf
Azf
4.5 × 106
= 4.994 × 104
1 + (4.5 × 106 )(1.98 × 10−5 )
=−
55.4939
= −1.11× 10−3 ⇒ −0.111%
5 × 104
12.33
V1 = I ε Ri , − V0 = − Az I ε ⇒ V0 = Ax I ε
I fb = I S − I ε and −V0 = V1 − I fb RF
− Az I ε = V1 − ( I S − I ε ) RF
⎛V ⎞
⎛V ⎞
− Az ⎜ 1 ⎟ = V1 − I S RF + ⎜ 1 ⎟ RF
⎝ Ri ⎠
⎝ Ri ⎠
⎡ A R ⎤
I S RF = V1 ⎢1 + z + F ⎥
⎣ Ri Ri ⎦
V
RF
Rif = 1 =
I S ⎡ Az RF ⎤
⎢1 + R + R ⎥
i
i ⎦
⎣
Or, using the results from problem 12.18.
50.5 × 103
Rif =
⎡
5 × 106 50.5 × 103 ⎤
⎢1 + 10 × 103 + 10 × 103 ⎥
⎣
⎦
=
50.5 × 103
⇒ Rif = 99.79 Ω
[1 + 500 + 5.05]
12.34
Assume I CQ = 0.2 mA
Then rπ =
(100)(0.026)
= 13 k Ω
0.2
(1)
⎛ 1 1
Vi − VA VA VA − Vo
V V
1 ⎞
=
+
⇒ i + o = VA ⎜ + + ⎟
Ri
R1
R2
Ri R2
⎝ Ri R1 R2 ⎠
Now
Vi Vo
⎛1 1 1⎞
+ = VA ⎜ + + ⎟ ⇒ Vi + Vo = VA (12)
10 10
⎝ 10 1 10 ⎠
1
or VA = (Vi + Vo )
12
⎛ AvVε − Vo ⎞
Vo − VA
(2)
⎜
⎟ (1 + hFE ) =
R2
⎝ Ro + rπ ⎠
where Vε = Vi − VA
Then
⎛ Av (Vi − VA ) − Vo ⎞
Vo − VA
⎜
⎟ (1 + hFE ) =
R0 + rπ
R2
⎝
⎠
we find
AvVi (1 + hFE ) Vo (1 + hFE ) Vo AvVA (1 + hFE ) VA
−
−
=
−
Ro + rπ
Ro + rπ
R2
Ro + rπ
R2
Then
(5 × 103 )(101)Vi Vo (101) Vo ⎛ (5 × 103 )(101) 1 ⎞
−
− =⎜
− ⎟ VA
14
14
10 ⎝
14
10 ⎠
Rearranging terms, we find
V
Avf = o =10.97
Vi
⎛ Vi ⎞
Vi
Vi
=
=⎜
⎟ Ri
I i ⎛ Vi − VA ⎞ ⎝ Vi − VA ⎠
⎜
⎟
⎝ Ri ⎠
1
1
VA = (Vi + Vo ) = (Vi + 10.97Vi ) = 0.9975Vi
12
12
Then
1
⎛
⎞
Rif = ⎜
⎟ (10 k Ω) ⇒ Rif = 4 M Ω
⎝ 1 − 0.9975 ⎠
To find the output resistance:
Rif =
Ix +
( AvVε )(1 + hFE )
Vx
=
Ro + rπ
R2 + R1 || Ri
⎛ R1 || Ri ⎞
Vε = − ⎜
⎟ ⋅ Vx
⎝ R1 || Ri + R2 ⎠
Now
R1 || Ri = 1 || 10 = 0.909
Then
Vε = −0.0833Vx
Now
3
1
⎪⎧ ⎡ (5 × 10 )(0.0833) + 1 ⎤
⎪⎫
I x = Vx ⎨ ⎢
⎬
⎥ (101) +
1 + 13
10 + 0.909 ⎭⎪
⎦
⎩⎪ ⎣
= Vx {3.012 × 103 + 0.0917}
Or
Vx
= Rof = 3.32 × 10−4 k Ω ⇒ Rof = 0.332 Ω
Ix
12.35
a.
Neglecting base currents
I C 2 = 0.5 mA, VC 2 = 12 − (0.5)(22.6) = 0.7 V
I C1 = 0.5 mA ⇒ v0 = 0
Then I C 3 = 2 mA
b.
rπ 1 = rπ 2 =
hFE ⋅ VT (100)(0.026)
=
= 5.2 kΩ
I C1
0.5
0.5
= 19.23 mA / V
0.026
(100)(0.026)
= 1.3 kΩ
rπ 3 =
2
2
= 76.92 mA / V
g m3 =
0.026
g m1 = g m 2 =
Vπ 1
V
+ g m1Vπ 1 + g m1Vπ 2 + π 2 = 0
rπ 1
rπ 1
⎛ 1
⎞
(Vπ 1 + Vπ 2 ) ⎜ + g m1 ⎟ = 0 ⇒ Vπ 1 = −Vπ 2
⎝ rπ 1
⎠
Vπ 1
Vi =
( RS + rπ 1 ) − Vπ 2 + Vb 2
rπ 1
(1)
⎛ R ⎞
or Vi = Vπ 1 ⎜1 + S ⎟ − Vπ 2 + Vb 2
⎝ rπ 1 ⎠
But Vπ 2 = −Vπ 1
so
⎛
R ⎞
Vi = Vπ 1 ⎜ 2 + S ⎟ + Vb 2
(2)
rπ 1 ⎠
⎝
V02
V −V
+ g m1Vπ 2 + 02 0 = 0
RC
rπ 3
(3)
Vπ 3
V V −V
+ g m 3Vπ 3 = 0 + 0 b 2
rπ 3
RL
R2
Vπ 3 = V02 − V0
so
⎛ 1 + hFE
(V02 − V0 ) ⎜
⎝ rπ 3
⎞
⎛ 1
1 ⎞ V
+ ⎟ − b2
⎟ = V0 ⎜
R
R
R2
⎝ L
2 ⎠
⎠
(4)
Vb 2 − V0 Vb 2 Vπ 2
+
+
= 0 (5)
R2
R1 rπ 1
Substitute numbers into (2), (3), (4) and (5):
1 ⎞
⎛
Vi = −Vπ 2 ⎜ 2 +
⎟ + Vb 2
5.2 ⎠
⎝
(2)
Vi = −Vπ 2 (2.192) + Vb 2
1 ⎞
⎛ 1
⎛ 1 ⎞
+
V02 ⎜
⎟ + (19.23)Vπ 2 − V0 ⎜ ⎟ = 0
⎝ 22.6 1.3 ⎠
⎝ 1.3 ⎠
V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3)
⎛ 101 ⎞
⎛ 101 1 1 ⎞
⎛ 1 ⎞
V02 ⎜
+ + ⎟ − Vb 2 ⎜ ⎟
⎟ = V0 ⎜
⎝ 1.3 ⎠
⎝ 1.3 4 50 ⎠
⎝ 50 ⎠
V02 (77.69) = V0 (77.96) − Vb 2 (0.02) (4)
⎛ 1 1⎞
⎛ 1 ⎞
⎛ 1 ⎞
Vb 2 ⎜ + ⎟ − V0 ⎜ ⎟ + Vπ 2 ⎜
⎟=0
⎝ 50 10 ⎠
⎝ 50 ⎠
⎝ 5.2 ⎠
Vb 2 (0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0
(5)
From (2): Vb 2 = Vi + Vπ 2 (2.192). Substitute in (4) and (5) to obtain:
(4′)
V02 (77.69) = V0 (77.96) − [Vi + Vπ 2 (2.192)](0.02)
(5′)
[Vi + Vπ 2 (2.192)](0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0
So we now have the following three equations:
V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3)
V02 (77.69)
= V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384)
(4′)
(5′)
(0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0
From (3): V02 = V0 (0.9455) − Vπ 2 (23.64). Substitute for V02 in (4′) to obtain:
(77.69)[Vo (0.9455) − Vπ 2 (23.64)] = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384)
or
0 = V0 (4.504) − Vi (0.02) + Vπ 2 (1836.5)
Next, solve (5′) for Vπ 2 :
(0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0
Vπ 2 = V0 (0.04393) − Vi (0.2636)
Finally,
0 = V0 (4.504) − Vi (0.02) + (1836.5)[V0 (0.04393) − Vi (0.2636)]
0 = V0 (85.18) − Vi (484.12)
So
V
484.12
Avf = 0 =
⇒ Avf = 5.68
Vi
85.18
12.36
a.
RTH = R1 || R2 = 400 || 75 = 63.2 kΩ
⎛ R2 ⎞
⎛ 75 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10) = 1.579 V
R
R
+
75 + 400 ⎠
⎝
⎝ 1
2 ⎠
1.579 − 0.7
= 0.007106 mA
I BQ1 =
63.2 + (121)(0.5)
I CQ1 = 0.853 mA
VC1 = 10 − (0.853)(8.8) = 2.49 V
2.49 − 0.7
= 0.497 mA
3.6
= 10 − (0.497)(13) = 3.54 V
IC 2 ≈
VC 2
IC 3 ≈
Then
3.54 − 0.7
= 2.03 mA
1.4
(120)(0.026)
= 3.66 kΩ
0.853
0.853
g m1 =
= 32.81 mA / V
0.026
(120)(0.026)
rπ 2 =
= 6.28 kΩ
0.497
0.497
gm2 =
= 19.12 mA / V
0.026
(120)(0.026)
rπ 3 =
= 1.54 kΩ
2.03
2.03
g m3 =
= 78.08 mA / V
0.026
b.
rπ 1 =
Vi = Vπ 1 + Vε 1 ⇒ Vε 1 = Vi − Vπ 1
(1)
Vπ 1
V
V −V
+ g m1Vπ 1 = ε 1 + ε 1 0
rπ 1
RE1
RF
(2)
Vπ 2 = −( g m1Vπ 1 )( RC1 || rπ 2 )
(3)
g m 2Vπ 2 +
Vπ 3 + V0 Vπ 3
+
=0
RC 2
rπ 3
(4)
Vπ 3
V
V −V
+ g m 3Vπ 3 = 0 + 0 ε 1
(5)
rπ 3
RE 3
RF
Substitute numbers in (2), (3), (4) and (5):
1⎞ V
⎛ 1
⎞
⎛ 1
Vπ 1 ⎜
+ 32.81⎟ = (Vi − Vπ 1 ) ⎜
+ ⎟− 0
⎝ 3.66
⎠
⎝ 0.5 10 ⎠ 10
or Vπ 1 (35.18) = Vi (2.10) − V0 (0.10)
(2)
Vπ 2 = −(32.81)Vπ 1 (88 || 6.28)
or Vπ 2 = −Vπ 1 (120.2)
(3)
(19.12)Vπ 2 +
Vπ 3 V0 Vπ 3
+ +
=0
13 13 1.54
or
Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4)
1 ⎞ V −V
⎛ 1
⎞
⎛ 1
Vπ 3 ⎜
+ 78.08 ⎟ = V0 ⎜
+ ⎟ − i π1
10
⎝ 1.54
⎠
⎝ 1.4 10 ⎠
or
(5)
Vπ 3 (78.73) = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10)
Now substituting Vπ 2 = −Vπ 1 (120.2) in (4):
(19.12)[−Vπ 1 (120.2)] + Vπ 3 (0.7263) + V0 (0.07692) = 0
or
−Vπ 1 (2298.2) + Vπ 3 (0.7263) + V0 (0.07692) = 0
Then
Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059)
Substituting Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) in (5):
(78.73)[Vπ 1 (3164.3) − V0 (0.1059)] = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10)
or Vπ 1 (2.49 × 105 ) − V0 (9.152) = −Vi (0.10)
Then
Vπ 1 = V0 (3.674 × 10−5 ) − Vi (4.014 × 10−7 )
Now substituting Vπ 1 = V0 (3.674 × 10−5 )
−Vi (4.014 × 10−7 ) in (2):
(35.18)[V0 (3.674) ×10−5 ) − Vi (4.014 ×10 −7 )]
= Vi (2.10) − V0 (0.10)
or V0 (0.1013) = Vi (2.10)
So
V0
= 20.7
Vi
Rif =
c.
I RB1 =
I b1 =
Vi
and I i = I RB1 + I b1
Ii
Vi
RB1
Vπ 1
rπ 1
Now
Vπ 1 = (20.7Vi )(3.674 × 10−5 ) − Vi (4.014 × 10−7 )
Vπ 1 = Vi (7.60 × 10−4 )
Then
Vi
Rif =
Vi
V (7.60 × 10−4 )
+ i
63.2
3.66
1
=
0.01582 + 2.077 × 10−4
or Rif = 62.4 kΩ
d.
To determine R0 f :
Equation (1) is modified to Vπ 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to:
(5)
Vπ 3 (78.73) + I X = V0 (0.8143) + Vπ 1 (0.10)
Now
Vπ 1 (35.18) = −V0 (0.10)
(2)
(3)
Vπ 2 = −Vπ 1 (120.2)
Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4)
Now
Vπ 1 = −V0 (0.002843)
so
Vπ 2 = −(−V0 )(0.002843)(120.2)
Vπ 2 = V0 (0.3417)
Then
V0 (0.3417)(19.12) + Vπ 3 + (0.7263) + V0 (0.07692) = 0
or Vπ 3 = −V0 (9.101)
(4)
So then
−V0 (9.101)(78.73) + I X
= V0 (0.8143) + (0.10)(−V0 )(0.002843)
or
I X = V0 (717.3) (5)
or
V
R0 f = 0 = 0.00139 kΩ ⇒ R0 f = 1.39 Ω
IX
12.37
(a)
(1)
Vi − VA
V −V
V
+ g m1Vπ 1 = A + A O
rπ 1
RE
RF
(2)
VB
V
+ g m1Vπ 1 + B = 0
RC1
rπ 2
(3)
VC
V − Vo
+ g m 2Vπ 2 + C
=0
RC 2
rπ 3
(4)
g m 3Vπ 3 +
VC − VO VO − VA
=
rπ 3
RF
Vπ 1 = Vi − VA Vπ 2 = VB Vπ 3 = VC − VO
(1)
(2)
(3)
⎛ 1
⎞ V
V −V
+ g m1 ⎟ = A + A O
r
R
RF
E
⎝ π1
⎠
(Vi − VA ) ⎜
⎛ 1
1 ⎞
VB ⎜
+
⎟ + g m1 (Vi − VA ) = 0
R
r
⎝ C1 π 2 ⎠
⎛ 1
VO
1 ⎞
VC ⎜
+
=0
⎟ + g m 2VB −
rπ 3
⎝ RC 2 rπ 3 ⎠
⎛
1 ⎞ VO − VA
⎟=
rπ 3 ⎠
RF
(4)
(VC − VO ) ⎜ g m3 +
(1)
(2)
(3)
(4)
(Vi − VA )(555.5) = VA (20) + (VA − VO )(0.8333)
VB (5.109) + 550(Vi − VA ) = 0
VC (3.257) + 178VB − VO (1.718) = 0
(VC − VO )(173.7) = (VO − VA )(0.8333)
(1)
(2)
(3)
(4)
Vi (555.5) + VO (0.8333) = VA (576.3)
VB (5.109) + 550Vi − VA (550) = 0
VC (3.257) + 178VB − VO (1.718) = 0
VC (173.7) + VA (0.8333) = VO (174.5)
⎝
(100)(0.026)
14.3
rπ 1 =
= 0.182 K g m1 =
= 550 mA/V
14.3
0.026
(100)(0.026)
4.62
= 0.563 K g m 2 =
= 178 mA/V
rπ 2 =
4.62
0.026
(100)(0.026)
4.47
rπ 3 =
= 0.582 K g m 3 =
= 172 mA/V
4.47
0.026
V −V
1
⎞ V
+ 550 ⎟ = A + A O
(1)
(Vi − VA ) ⎛⎜
1.2
⎝ 0.182
⎠ 0.05
1
1
⎛
⎞
(2)
VB ⎜
+
⎟ + (550)(Vi − VA ) = 0
⎝ 0.3 0.563 ⎠
VO
1 ⎞
⎛ 1
VC ⎜
+
=0
(3)
⎟ + 178 VB −
0.65
0.582
0.582
⎝
⎠
1 ⎞ VO − VA
(4)
(VC − VO ) ⎛⎜172 +
⎟=
0.582
1.2
⎝
⎠
From (2)
VB = VA (107.7) − Vi (107.7)
From (4)
VC = VO (1.0046) − VA (0.004797)
Substitute into (3)
(3.257) [VO (1.0046) − VA (0.004797) ]
+(178)[VA (107.7) − Vi (107.7)] − Vo (1.718) = 0
VO (3.272) − VA (0.01562) + VA (19170.6) − Vi (19170.6) − VO (1.718) = 0
VA (19170.6) = Vi (19170.6) − VO (1.554)
VA = Vi (1.00) − VO (0.00008106)
Substitute into (1)
Vi (555.5) + Vo (0.8333) = (576.3) [Vi (1.00) − Vo (0.00008106) ]
= Vi (576.3) − Vo (0.0467)
Vo (0.880) = Vi (20.8)
Vo
= Avf = 23.6
Vi
Ideal
R + RE 1.2 + 0.05
=
= 25.0
Avf = F
RE
0.05
Rif =
(b)
V
V − VA
Vi
and I i = π 1 = i
rπ 1
rπ 1
Ii
We have
VA = Vi (1.00) − Vo (0.00008106)
= Vi (1.00) − (23.6)Vi (0.00008106)
VA = Vi (0.99809)
Vi (1 − 0.99809)
= Vi (0.01051)
0.182
Vi
Rif =
⇒ Rif = 95.1 K
Vi (0.01051)
Then I i =
To find Rof , set Vi = 0
I X + g m 3Vπ 3 +
Vπ 3 Vx − VA
=
rπ 3
RF
Vπ 3 = VC − VX
I X + (VC − VX )( g m 3 +
V − VA
1
)= X
rπ 3
RF
For Vi = 0, we have
VC = VX (1.0046) − VA (0.004797)
VA (576.3) = VX (0.8333)
VA = VX (0.001446)
VC = VX (1.0046) − VX (0.001446)(0.004797)
VC = VX (1.0046)
1 ⎞ VX (1 − 0.004797)
⎛
I X + VX (1.0046 − 1.0) ⎜172 +
⎟=
0.582
1.2
⎝
⎠
I X + VX (0.7991) = VX (0.8293)
I X = VX (0.03024)
Rof =
VX
= 33.1 K
IX
12.38
RD1 = RE1 = 8 K
VGG = VGS + ( I D + I C ) RL
ID =
10 − VD
8
IC =
10 − (VD + 0.7)
8
VGG =
4.5 =
36 =
ID
+ VTN + ( I D + I C ) RL
kn
⎛ 10 − VD ⎞
⎛ 9.3 − VD ⎞
+1+ ⎜
⎟ (1.8) + ⎜
⎟ (1.8)
8(0.4)
8 ⎠
⎝ 8 ⎠
⎝
(10 − VD )
8
3.2
10 − VD + 8 + 18 − 1.8VD + 16.74 − 1.8VD
36 = 4.472 10 − VD + 42.74 − 3.6VD
3.6VD − 6.74 = 4.472 10 − VD
12.96VD2 − 48.528VD + 45.4276 = 19.999(10 − VD )
12.96VD2 − 28.528VD − 154.6 = 0
28.528 ± 813.847 + 4(12.96)(154.6)
2(12.96)
VD = 4.726 V
VD =
10 − 4.726
= 0.6593 mA
8
9.3 − 4.726
IC =
8
= 0.5718 mA
0.6593
VGS =
+ 1 = 2.284 V
0.4
VO = 4.5 − 2.284 = 2.216 V
ID =
VDS = VD − VO = 4.726 − 2.216 = 2.51V
(V ( sat ) = 1.28 V )
DS
Also
VO = ( I D + I C ) RL = (0.6593 + 0.5718)(1.8) = 2.216 V OK
VECQ 2 = (4.726 + 0.7) − 2.216 = 3.21 V
(b)
(1)
Vi = Vgs + Vo
(2)
V
VA
+ g m1Vgs = π
RD1
rπ 2
(3)
VB Vπ
+
+ g m 2Vπ = 0
RE 2 rπ 2
Vπ = VB − VA
Vgs = Vi − Vo
(2)
VA
V − VA
+ g m1 (Vi − Vo ) = B
RD1
rπ 2
(3)
VB VB − VA
+
+ g m 2 (VB − VA ) = 0
RE 2
rπ 2
(4)
g m1Vgs + g m 2Vπ =
VO
RL
g m1 (Vi − Vo ) + g m 2 (VB − VA ) =
VO
RL
g m1 = 2 K n I D1 = 2 (0.4)(0.6593) = 1.027 mA/V
gm2 =
rπ 2 =
I C 2 0.5718
=
= 21.99 mA/V
VT
0.026
(100)(0.026)
= 4.547 K
0.5718
(2)
(3)
VA
V − VA
+ (1.027)(Vi − Vo ) = B
8
4.547
VB VB − VA
+
+ (21.99)(VB − VA ) = 0
8
4.547
0.125VA + 1.027Vi − 1.027Vo = 0.2199VB − 0.2199VA
(2)
(3)
0.125VB + 0.2199VB − 0.2199VA + 21.99VB − 21.99VA = 0
(2)
0.3449VA − 0.2199VB = −1.027Vi + 1.027Vo
(3)
22.21VA = 22.335VB
VA = 1.0056VB
Then (0.3449)(1.0056)VB − 0.2199VB = −1.027Vi + 1.027Vo
0.1269VB = −1.027Vi + 1.027Vo
VB = −8.093Vi + 8.093Vo
VA = −8.138Vi + 8.138Vo
Then
(4)
(1.027)(Vi − Vo ) + (21.99) [ −8.093Vi + 8.093Vo + 8.138Vi − 8.138Vo ] =
Vo
1.8
1.027Vi − 1.027Vo − 177.965Vi + 177.965Vo + 178.955Vi − 178.955Vo = 0.5556Vo
2.017Vi − 2.5726Vo = 0
Vo
= 0.784 = AV
Vi
Set Vi = 0
I X + g m 2Vπ + g m1Vgs =
VX
RL
I X + g m 2 (VB − VA ) + g m1 (−VX ) =
VX
RL
VB = 8.093VX
VA = 8.138VX
I X + (21.99) [8.093VX − 8.138VX ] + 1.027( −VX ) = VX (0.5556)
I X = 2.572VX
VX
= Rof = 0.389 K
IX
12.39
a.
g m = 2 K n I DQ = 2 (0.5)(0.5) = 1 mA / V
V0 = ( g mVgs ) RS
Vi = Vgs + V0 so Vgs = Vi − V0
Then
V0 = g m RS (Vi − V0 )
Av =
g m RS
1(2)
=
⇒ Av = 0.667
1 + g m RS 1 + (1)(2)
To determine R0 f
I X + g mVgs =
VX
and Vgs = −VX
RS
IX
1
1
=
= gm +
VX R0 f
RS
so R0 f =
1
1
RS = 2 ⇒ R0 f = 0.667 kΩ
1
gm
For K n = 0.8 mA / V 2
b.
g m = 2 (0.8)(0.5) = 1.265 mA / V
Av =
(1.265)(2)
= 0.7167
1 + (1.265)(2)
ΔAf
=
Af
0.7167 − 0.667
⇒ 7.45% increase
0.667
1
|| 2 = 0.7905 || 2
1.265
= 0.5666 kΩ
R0 f =
R0 f
ΔR0 f
R0 f
=
0.5666 − 0.667
⇒ 15.05% decrease
0.667
12.40
dc analysis:
RTH 1 = 150 || 47 = 35.8 kΩ ,
⎛ 47 ⎞
VTH 1 = ⎜
⎟ (25) = 5.96 V
⎝ 47 + 150 ⎠
RTH 2 = 33 || 47 = 19.4 kΩ ,
⎛ 33 ⎞
VTH 2 = ⎜
⎟ (25) = 10.3 V
⎝ 33 + 47 ⎠
5.96 − 0.7
= 0.0187 mA
I B1 =
35.8 + (51)(4.8)
I C1 = (50)(0.0187) = 0.935 mA
10.3 − 0.7
= 0.03705 mA
19.4 + (51)(4.7)
= (50)(0.03705) = 1.85 mA
I B2 =
IC 2
(50)(0.026)
= 1.39 kΩ;
0.935
(50)(0.026)
= 0.703 kΩ
rπ 2 =
1.85
rπ 1 =
0.935
= 35.96 mA / V
0.026
1.85
=
= 71.15 mA / V
0.026
g m1 =
gm2
VS = Vπ 1 + Ve
(1)
Vπ 1
V V − V0
+ g m1Vπ 1 = e + e
rπ 1
R1
RF
(2)
g m1Vπ 1 +
Vπ 2 Vπ 2 Vπ 2
+
+
=0
RC1 RB 2 rπ 2
(3)
V0 V0 − Ve
+
=0
(4)
RC 2
RF
Substitute numerical values in (2), (3) and (4):
(1)
Ve = VS − Vπ 1
g m 2Vπ 2 +
Vπ 1
1 ⎞
⎛ 1
⎛ 1 ⎞
+ (35.96)Vπ 1 = (VS − Vπ 1 ) ⎜
+
⎟ − V0 ⎜
⎟
1.39
⎝ 0.1 4.7 ⎠
⎝ 4.7 ⎠
or
(2)
Vπ 1 (46.89) = VS (10.213) − V0 (0.2128)
1
1 ⎞
⎛1
+
(35.96)Vπ 1 + Vπ 2 ⎜ +
⎟=0
⎝ 10 19.4 0.703 ⎠
or
(35.96)Vπ 1 + Vπ 2 (1.574) = 0
(3)
1 ⎞
⎛ 1
⎛ 1 ⎞
+
(71.15)Vπ 2 + V0 ⎜
⎟ − (VS − Vπ 1 ) ⎜
⎟=0
⎝ 4.7 4.7 ⎠
⎝ 4.7 ⎠
or
(4)
(71.15)Vπ 2 + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0
From (3): Vπ 2 = −Vπ 1 (22.85)
Then substitute in (4):
−(71.15)Vπ 1 (22.85) + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0
or
−Vπ 1 (1625.6) + V0 (0.4255) − VS (0.2128) = 0
From (2):
Vπ 1 = VS (0.2178) − V0 (0.004538)
Then
−(1625.6)[VS (0.2178) − V0 (0.004538)] + V0 (0.4255) − VS (0.2128) = 0
or −VS (354.3) + V0 (7.802) = 0
Finally
V
⇒ 0 = 45.4
VS
12.41
For example, use a z-stage amplifier. Each stage shown in Fig. 12.29.
12.42
For M 3 : K n 3 =
kn′
2
⎛W ⎞
⎛W ⎞
⋅ ⎜ ⎟ Let ⎜ ⎟ = 25
L
⎝ L ⎠3
⎝ ⎠3
⎛ 0.080 ⎞
2
Then K n 3 = ⎜
⎟ (25) = 1 mA / V
⎝ 2 ⎠
Want vo = 0 for vi = 0, so that
I D 3 = I Q 2 = 0.4 = 1 ⋅ (VGS 3 − VTN ) 2
0.4
+ 2 = 2.63 V
1
For VGS 3 = 2.63V ⇒ VG 3 = 12 − I D 2 RD
Or 2.63 = 12 − (0.1) RD ⇒ RD = 93.7 k Ω
Then VGS 3 =
g m 3 = 2 K n 3 I D 3 = 2 (1)(0.4) = 1.26 mA / V
⎛ R1 ⎞
⎛ 10 ⎞
VA = ⎜
⎟ (Vo ) = ⎜
⎟ (Vo ) = 0.0769Vo
R
R
+
⎝ 120 + 10 ⎠
2 ⎠
⎝ 1
(Small amount of feedback)
(1)
Vi = Vgs1 − Vgs 2 + VA
g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2
(2)
Then
Vi = −2Vgs 2 + VA ⇒ Vgs 2 =
1
(VA − Vi )
2
Vgs 2 = 0.0384 Vo − 0.5Vi
(3)
VB = − g mVgs 2 RD = − g m RD [0.03846Vo − 0.5Vi ]
(4)
Vgs 3 = VB − Vo and Vo = g m 3Vgs 3 [ RL || ( R1 + R2 )]
So
Vo = g m 3 [ RL || ( R1 + R2 ) ] (VB − Vo )
Then
Vo = g m 3 ⎡⎣ RL || ( R1 + R2 ) ⎤⎦ ⎡⎣ − g m RD ( 0.03846Vo − 0.5Vi ) − Vo ⎤⎦
Or
Vo ⎡⎣1 + g m 3 [ RL || ( R1 + R2 )][ g m RD (0.03846) + 1]⎤⎦ = g m 3 [ RL || ( R1 + R2 ) ][ 0.5 g m RD ] ⋅ Vi
Now
RL || ( R1 + R2 ) = 4 ||130 = 3.88
So
Vo ⎡⎣1 + (1.26)(3.88) [ g m (93.7)(0.03846) + 1]⎤⎦ = (1.26)(3.88)(0.5) g m (93.7)Vi
Rearranging terms, we find
229 g m
Vo
=
= 10 ⇒ g m = 1.11 mA / V
Vi 5.89 + 17.6 g m
We have
⎛ k ′ ⎞⎛ W
gm = 2 ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
⎞
⎛ 0.080 ⎞ ⎛ W
⎟ I D ⇒ 1.11 = 2 ⎜
⎟⎜
⎠
⎝ 2 ⎠⎝ L
⎞
⎛W ⎞ ⎛W ⎞
⎟ (0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 77
⎠
⎝ L ⎠1 ⎝ L ⎠ 2
12.43
Assuming an ideal op-amp, then from Equation (12.58)
I0
R
20
= 1+ 1 =
= 100
IS
R2 0.2
Then
R1
= 99
R2
For example, set R2 = 5 kΩ and R1 = 495 kΩ
12.44
(a)
⎛ h
⎞
⎛ 100 ⎞
I C1 = ⎜ FE ⎟ I E1 = ⎜
⎟ (0.2) = 0.198 mA
⎝ 101 ⎠
⎝ 1 + hFE ⎠
VC1 = 10 − (0.198)(40) = 2.08 V
2.08 − 0.7
= 1.38 mA
1
⎛ 100 ⎞
=⎜
⎟ (1.38) = 1.37 mA
⎝ 101 ⎠
IE2 =
IC 2
For Q1 :
(100)(0.026)
= 13.1 k Ω
0.198
0.198
=
= 7.62 mA / V
0.026
rπ 1 =
g m1
For Q2 :
(100)(0.026)
= 1.90 k Ω
1.37
1.37
=
= 52.7 mA / V
0.026
rπ 2 =
gm2
(b)
IS =
Vπ 1 Vπ 1 Vπ 1 − Ve
+
+
RS rπ 1
RF
g m1Vπ 1 +
Vπ 2 + Ve Vπ 2
+
=0
RC1
rπ 2
(1)
(2)
Vπ 2
V V −V
+ g m 2Vπ 2 = e + e π 1
(3)
rπ 2
RE
RF
Substitute numerical values in (1), (2), and (3):
1
1⎞
⎛1
⎛1⎞
I S = Vπ 1 ⎜ +
+ ⎟ − Ve ⎜ ⎟
10
13.1
10
⎝
⎠
⎝ 10 ⎠
I S = Vπ 1 (0.2763) − Ve (0.10)
(1)
1 ⎞
⎛ 1
⎛ 1 ⎞
(7.62)Vπ 1 + Vπ 2 ⎜ +
⎟ + Ve ⎜ ⎟ = 0
⎝ 40 1.90 ⎠
⎝ 40 ⎠
(7.62)Vπ 1 + Vπ 2 (0.5513) + Ve (0.025) = 0
(2)
⎛ 1
⎞
⎛1 1 ⎞
⎛1⎞
Vπ 2 ⎜
+ 52.7 ⎟ = Ve ⎜ + ⎟ − Vπ 1 ⎜ ⎟
⎝ 1.90
⎠
⎝ 1 10 ⎠
⎝ 10 ⎠
Vπ 2 (53.23) = Ve (1.10) − Vπ 1 (0.10)
(3)
From (3),
Ve = Vπ 2 (48.39) + Vπ 1 (0.0909)
Substituting into (1),
I S = Vπ 1 (0.2763) − (0.10) [Vπ 2 (48.39) + Vπ 1 (0.0909)]
or
(1′)
I S = Vπ 1 (0.2672) − Vπ 2 (4.839)
and substituting into (2),
(7.62)Vπ 1 + Vπ 2 (0.5513) + ( 0.025 ) ⎡⎣Vπ 2 ( 48.39 ) + Vπ 1 ( 0.0909 ) ⎤⎦ = 0
or
(7.622)Vπ 1 + Vπ 2 (1.761) = 0 ⇒ Vπ 1 = −Vπ 2 (0.2310)
Then substituting (2′) into (1′), we obtain
I S = (0.2672)(−Vπ 2 )(0.2310) − Vπ 2 (4.839)
or
I S = −Vπ 2 (4.901)
Now
(2′)
⎛ RC 2 ⎞
I O = − g m 2Vπ 2 ⎜
⎟
⎝ RC 2 + RL ⎠
⎛ 2 ⎞
= −(52.7) ⎜
⎟ Vπ 2 = −(42.16)Vπ 2
⎝ 2 + 0.5 ⎠
Then
⎛ −IS ⎞
I O = −(42.16) ⎜
⎟
⎝ 4.901 ⎠
or
I
Aif = O = 8.60
Is
(c)
Ri =
Vπ 1
and Ri = RS || Rif
IS
We had
Vπ 1 = −Vπ 2 (0.2310) and I S = −Vπ 2 (4.901)
so
⎛ −Vπ 1 ⎞
IS = − ⎜
⎟ (4.901) = Vπ 1 (21.22)
⎝ 0.2310 ⎠
Then
V
1
Ri = π 1 =
= 0.04713
21.22
IS
Finally
10 Rif
0.04713 =
⇒ Rif = 47.4 Ω
10 + Rif
12.45
(a)
Vπ 1
V −V
+ π 1 e2
RF
RS RB1 rπ 1
(1)
Ii =
(2)
g m1Vπ 1 +
(3)
Vπ 2
V
V −V
+ g m 2Vπ 2 = e 2 + e 2 π 1
rπ 2
RE 2
RF
(4)
VC1
V
+ π2 = 0
RC1 RB 2 rπ 2
⎛ RC 2 ⎞
I o = −( g m 2Vπ 2 ) ⎜
⎟
⎝ RC 2 + RL ⎠
Now
(1)′
Ii =
Vπ 1
V
− e2
RS RB1 rπ 1 RF RF
⎛
⎞
RF
So Ve 2 = ⎜
⎟V − I R
⎜ RS RB1 rπ 1 RF ⎟ π 1 i F
⎝
⎠
Now, from (2)
V + Ve 2 Vπ 2
+
=0
g m1Vπ 1 + π 2
Rc1 RB 2 rπ 2
(2)′
⎛
Vπ 2
Ve 2
1 ⎞
+
=0
⎜ g m1 +
⎟ Vπ 1 +
r
R
R
R
r
C1 RB 2
π2 ⎠
⎝
C1
B2 π 2
Also
(3)′
⎛
⎛ 1
Vπ 1
1 ⎞
1 ⎞
= Ve 2 ⎜
+
⎜ gm2 +
⎟ Vπ 2 +
⎟
rπ 2 ⎠
RF
⎝ RE 2 RF ⎠
⎝
And
⎛ I ⎞ ⎛ R + RL ⎞
Vπ 2 = − ⎜ O ⎟ ⎜ C 2
⎟
⎝ g m 2 ⎠ ⎝ RC 2 ⎠
Substitute (1)′ into (2)′ and (3)′
(4)′
⎡
⎤
RF
⎢
− Vπ 1 − I i RF ⎥ = 0
⎢ RS RB1 rπ 1 RF
⎥
⎣
⎦
⎡
⎤
⎛
Vπ 1 ⎛ 1
RF
1 ⎞
1 ⎞
=⎜
+
− Vπ 1 − I i RF ⎥
(3)″
⎜ gm2 +
⎟ Vπ 2 +
⎟⎢
rπ 2 ⎠
RF ⎝ RE 2 RF ⎠ ⎢ RS RB1 rπ 1 RF
⎥
⎝
⎣
⎦
Solve for Vπ 1 from (2)″ and substitute into (3)″. Also use Equation (4)′.
(2)″
⎛
Vπ 2
1 ⎞
1
+
⎜ g m1 +
⎟ Vπ 1 +
r
R
R
R
r
C1 RB 2
π2 ⎠
⎝
C1
B2 π 2
(b)
RB1 = R1 R2 = 20 80 = 16 K
⎛ 20 ⎞
VTH 1 = ⎜
⎟ (10) = 2 V
⎝ 100 ⎠
2 − 0.7
I BQ1 =
= 0.0111 mA
16 + (101)(1)
I CQ1 = 1.11 mA
RTH 2 = 15 85 = 12.75 K
⎛ 15 ⎞
VTH 2 = ⎜
⎟ (10) = 1.5 V
⎝ 100 ⎠
1.5 − 0.7
= 0.01265 mA
I BQ 2 =
12.75 + (101)(0.5)
I CQ 2 = 1.265 mA
1.11
= 42.69 mA/V
0.026
1.265
gm2 =
= 48.65 mA/V
0.026
(100)(0.026)
= 2.34 K
rπ 1 =
1.11
(100)(0.026)
= 2.06 K
rπ 2 =
1.265
Now RC1 RB 2 = 2 12.75 = 1.729 K
g m1 =
RS RB1 rπ 1 RF ≅ RB1 rπ 1 RF = 16 2.34 10 = 1.695 K
RC1 RB 2 rπ 2 = 1.729 2.06 = 0.940 K
Now
(2)″
(3)″
Vπ 2
1 ⎞
1 ⎡ 10
⎛
⎤
+
⋅ Vπ 1 − I i (10) ⎥
⎜ 42.69 +
⎟ Vπ 1 +
2.06 ⎠
0.940 1.729 ⎣⎢1.695
⎝
⎦
46.587Vπ 1 + 1.064Vπ 2 − 5.784 I i = 0
Vπ 1 ⎛ 1
1 ⎞ ⎡ 10
⎛ 101 ⎞
⎤
=⎜
+ ⎟⎢
⋅ Vπ 1 − I i (10) ⎥
⎜
⎟ Vπ 2 +
2.06
10
0.5
10
1.695
⎝
⎠
⎝
⎠⎣
⎦
49.03Vπ 2 = 12.29Vπ 1 − 21I i
From (2)″ Vπ 1 = (0.1242) I i − (0.02284)Vπ 2
Then
49.03Vπ 2 = 12.29 [ (0.1242) I i − (0.02284)Vπ 2 ] − 21I i
(3)″
49.31Vπ 2 = −19.47 I i
From (4)′
⎛ I ⎞⎛ 4 + 4 ⎞
Vπ 2 = − ⎜ o ⎟ ⎜
⎟ = −(0.0411) I o
⎝ 48.65 ⎠ ⎝ 4 ⎠
Then
(49.31) [ −(0.0411) I o ] = −19.47 I i
Io
= Ai = 9.61
Ii
12.46
a.
RTH = 13.5 || 38.3 = 9.98 kΩ
⎛ 13.5 ⎞
VTH = ⎜
⎟ (10) = 2.606 V
⎝ 13.5 + 38.3 ⎠
(120)(2.606 − 0.7)
= 1.75 mA
I C1 =
9.98 + (121)(1)
VC1 = 10 − (1.75 )( 3) = 4.75 V
4.75 − 0.7
= 0.50 mA
8.1
(120)(0.026)
= 1.78 kΩ
rπ 1 =
1.75
1.75
= 67.31 mA / V
g m1 =
0.026
(120)(0.026)
= 6.24 kΩ
rπ 2 =
0.50
0.50
= 19.23 mA / V
gm2 =
0.026
b.
IC 2 ≈
VS − Vπ 1
Vπ 1
V −V
=
+ π 1 e2
RS
RB || rπ 1
RF
g m1Vπ 1 +
Vπ 2 + Ve 2 Vπ 2
+
=0
RC1
rπ 2
Vπ 2
V
V −V
+ g m 2Vπ 2 = ε 2 + ε 2 π 1
rπ 2
RE 2
RF
and
(1)
(2)
(3)
V0 = −( g m 2Vπ 2 ) RC 2
(4)
Substitute numerical values in (1), (2), and (3)
VS
Vπ 1
⎡ 1
1 ⎤ Ve 2
= Vπ 1 ⎢
+
+
⎥−
0.6
⎣ 0.6 9.98 ||1.78 1.2 ⎦ 1.2
VS (1.67) = Vπ 1 (4.011) − Ve 2 (0.8333)
1 ⎞ Ve 2
⎛1
(67.31)Vπ 1 + Vπ 2 ⎜ +
=0
⎟+
⎝ 3 6.24 ⎠ 3
or
Vπ 1 (67.31) + Vπ 2 (0.4936) + Ve 2 (0.3333) = 0
(1)
(2)
V
V
⎛ 1
⎞ V
Vπ 1 ⎜
+ 19.23 ⎟ = e 2 + e 2 − π 2
⎝ 6.24
⎠ 8.1 1.2 1.2
or
Vπ 2 (19.39) = Ve 2 (0.9568) − Vπ 1 (0.8333)
(3)
From (1)
Ve 2 = Vπ 1 (4.813) − VS (2.00)
Then
Vπ 1 (67.31) + Vπ 2 (0.4936) + (0.3333)[Vπ 1 (4.813) − VS (2.00)] = 0
or
Vπ 1 (68.91) + Vπ 2 (0.4936) − VS (0.6666) = 0 (2′)
and
Vπ 2 (19.39) = (0.9568) [Vπ 1 (4.813) − VS (2.00) ] − Vπ 1 (0.8333)
or
Vπ 2 (19.39) = Vπ 1 (3.772) − VS (1.914)
(3′)
We find
Vπ 1 = VS (0.009673) − Vπ 2 (0.007163)
Then
Vπ 2 (19.39) = (3.772)[VS (0.009673) − Vπ 2 (0.007163)] − VS (1.914)
Vπ 2 (19.42) = VS ( −1.878) or Vπ 2 = −VS (0.09670)
so that
V0 = −(19.23)(4)(−VS )(0.09670)
Then
V0
= 7.44
VS
12.47
Using the circuit from Problem 12.32, we have Rif =
VS − Vπ 1
RS
From Problem 12.32
Vπ 1 = VS (0.009673) − Vπ 2 (0.007163)
Where I S =
= VS (0.009673) − (0.007163)(−VS )(0.09670)
= VS (0.01037)
So
Rif =
VS (0.01037) ⋅ (0.6)
= 0.00629 kΩ
VS − VS (0.01037)
Vπ 1
.
IS
or
Rif = 6.29 Ω
12.48
RTH = 1.4 ||17.9 = 1.298 kΩ
⎛ 1.4 ⎞
VTH = ⎜
⎟ (10) = 0.7254 V
⎝ 1.4 + 17.9 ⎠
0.7254 − 0.7
= 0.0196 mA
I B1 =
1.298
I C1 = (50)(0.0196) = 0.98 mA
Neglecting dc base currents,
VB 2 = 10 − (0.98)(7) = 3.14 V
3.14 − 0.7
= 3.25 mA
0.25 + 0.5
⎛ 50 ⎞
I C 2 = ⎜ ⎟ (3.25) = 3.19 mA
⎝ 51 ⎠
(50)(0.026)
rπ 1 =
= 1.33 kΩ
0.98
0.98
g m1 =
= 37.7 mA / V
0.026
(50)(0.026)
= 0.408 kΩ
rπ 2 =
3.19
3.19
= 123 mA / V
gm2 =
0.026
IE2 =
IS =
Vπ 1
V −V
+ π1 1
R1 || R2 || rπ 1
RF
g m1Vπ 1 +
Vπ 2 Vπ 2 + Ve 2
+
=0
rπ 2
RC1
Vπ 2
V −V
+ g m 2Vπ 2 = e 2 1
rπ 2
RE1
(1)
(2)
(3)
Ve 2 − Vπ 1
V −V
V
= 1 + 1 π1
(4)
RE1
RE 2
RF
Enter numerical values in (1), (2), (3) and (4):
Vπ 1
V −V
+ π1 1
IS =
17.9 ||1.4 ||1.33
5
or
I S = Vπ 1 (1.722) − V1 (0.20) (1)
(37.7)Vπ 1 +
Vπ 2
V + Ve 2
+ π2
=0
0.408
7
or
Vπ 1 (37.7) + Vπ 2 (2.594) + Ve 2 (0.1429) = 0
Vπ 2
V −V
+ (123)Vπ 2 = e 2 1
0.408
0.25
or
Vπ 2 (125.5) = Ve 2 (4) − V1 (4)
(2)
(3)
Ve 2 − V1
V −V
V
= 1 + 1 π1
0.25
0.50
5
or
Ve 2 (4) = V1 (6.20) − Vπ 1 (0.20)
(4)
From (4):
Ve 2 = V1 (1.55) − Vπ 1 (0.05)
Then substituting in (3):
Vπ 2 (125.5) = (4)[V1 (1.55) − Vπ 1 (0.05)] − V1 (4)
or
Vπ 2 (125.5) = V1 (2.20) − Vπ 1 (0.20) (3′)
and substituting in (2):
Vπ 1 ( 37.7 ) + Vπ 2 ( 2.594 ) + ( 0.1429 ) ⎡⎣V1 (1.55 ) − Vπ 1 ( 0.05 ) ⎤⎦ = 0
or
Vπ 1 (37.69) + Vπ 2 (2.594) + V1 (0.2215) = 0
Now
V1 = −Vπ 1 (170.16) − Vπ 2 (11.71)
Then substituting in (1):
I S = Vπ 1 (1.722) − (0.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)]
or
I S = Vπ 1 (35.75) + Vπ 2 (2.342)
and substituting in (3′):
Vπ 2 (125.5) = (2.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)] − Vπ 1 (0.20)
or Vπ 2 (151.3) = −Vπ 1 (374.55) so that
Vπ 1 = −Vπ 2 (0.4040)
Then
I S = (35.75)[−Vπ 2 (0.4040)] + Vπ 2 (2.342)
I S = −Vπ 2 (12.10)
⎛ RC 2 ⎞
I 0 = −( g m 2Vπ 2 ) ⎜
⎟
⎝ RC 2 + RL ⎠
⎛ 2.2 ⎞
= −(123) ⎜
⎟ Vπ 2 = −(64.43)Vπ 2
⎝ 2.2 + 2 ⎠
or Vπ 2 = −(0.01552) I 0
Then
I0
I
1
=
⇒ 0 = 5.33
I S (0.01552)(12.10)
IS
12.49
For example, use the circuit shown in Figure P12.30
12.50
rπ 1 = 6.24 kΩ, rπ 2 = 3.12 kΩ, rπ 3 = 1.56 kΩ
g m1 = 19.23mA / V , g m 2 = 38.46 mA / V,
g m 3 = 76.92 mA / V
VS = Vπ 1 + Ve1
(1)
Vπ 1
V
V −V
+ g m1Vπ 1 = e1 + e1 e 3
rπ 1
RE1
RF
Vπ 2 = − g m1Vπ 1 ( RC1 || rπ 2 )
g m 2Vπ 2 +
(2)
(3)
Vπ 3 + Ve 3 Vπ 3
+
=0
RC 2
rπ 3
(4)
Vπ 3
V
V −V
+ g m 3Vπ 3 = e 3 + e 3 e1
(5)
rπ 3
RE 2
RF
Enter numerical values in (2)-(5):
Vπ 1
1 ⎞
⎛ 1
⎛ 1 ⎞
+ (19.23)Vπ 1 = Ve1 ⎜
+
⎟ − Ve 3 ⎜
⎟
6.24
0.1
0.8
⎝
⎠
⎝ 0.8 ⎠
or
Vπ 1 (19.39) = Ve1 (11.25) − Ve 3 (1.25) (2)
(3)
Vπ 2 = −(19.23)Vπ 1 (5 || 3.12) = −(36.94)Vπ 1
1 ⎞
⎛1
⎛1⎞
(38.46)Vπ 2 + Vπ 3 ⎜ +
⎟ + Ve 3 ⎜ ⎟ = 0
⎝ 2 1.56 ⎠
⎝ 2⎠
or
Vπ 2 (38.46) + Vπ 3 (1.141) + Ve 3 (0.5) = 0
(4)
1 ⎞
⎛ 1
⎞
⎛ 1
⎛ 1 ⎞
+ 76.92 ⎟ = Ve 3 ⎜
+
Vπ 3 ⎜
⎟ − Ve 3 ⎜
⎟
⎝ 1.56
⎠
⎝ 0.1 0.8 ⎠
⎝ 0.8 ⎠
or
Vπ 3 (77.56) = Ve 3 (11.25) − Ve1 (1.25) (5)
From (1) Vπ 1 = VS − Ve1
Then
(VS − Ve1 )(19.39) = Ve1 (11.25) − Ve3 (1.25)
or VS (19.39) = Ve1 (30.64) − Ve 3 (1.25)
(2′)
Vπ 2 = −VS (36.94) + Ve1 (36.94)
(3′)
(38.46)[−VS (36.94) + Ve1 (36.94)] + Vπ 3 (1.141) + Ve 3 (0.5) = 0
From (5): Ve 3 = Vπ 3 (6.894) + Ve1 (0.1111)
Then
VS (19.39) = Ve1 (30.64) − (1.25)[Vπ 3 (6.894) + Ve1 (0.1111)]
(4′)
or
VS (19.39) = Ve1 (30.50) − Vπ 3 (8.6175)
(2″)
and
−VS (1420.7) + Ve1 (1420.7) + Vπ 3 (1.141) + (0.5)[Vπ 3 (6.894) + Ve1 (0.1111)] = 0
or
−VS (1420.7) + Ve1 (1420.76) + Vπ 3 (4.588) = 0 (4″)
From (2″):
Ve1 = VS (0.6357) + Vπ 3 (0.2825)
Then substituting in (4″):
−VS (1420.7) + (1420.76)[VS (0.6357) + Vπ 3 (0.2825)] + Vπ 3 (4.588) = 0
−VS (517.5) + Vπ 3 (405.95) = 0
Now
I 0 = g m 3Vπ 3 = 76.92Vπ 3 or Vπ 3 = I 0 (0.0130)
Then −VS (517.5) + I 0 (0.0130)(405.95) = 0
or
I0
= 98.06 mA / V
VS
12.52
(100)(0.026)
= 5.2 kΩ
0.5
0.5
= 19.23 mA / V
g m1 = g m 2 =
0.026
(100)(0.026)
= 1.3 kΩ
rπ 3 =
2
2
= 76.92 mA / V
g m3 =
0.026
rπ 1 = rπ 2 =
Vπ 1
V
+ g m1Vπ 1 + g m 2Vπ 2 + π 2 = 0
rπ 1
rπ 2
(1)
Since rπ 1 = rπ 2 and g m1 = g m 2 , then Vπ 1 = −Vπ 2
VS = Vπ 1 − Vπ 2 + Ve 3 = −2Vπ 2 + Ve 3 (2)
g m 2Vπ 2 +
Vπ 3 Vπ 3 + Ve 3
+
=0
rπ 3
RC 2
Vπ 3
V
V
+ g m 3Vπ 3 = e3 + π 2
rπ 3
RF rπ 2
(4)
(3)
⎛ RC 3 ⎞
I0 = − ⎜
⎟ ( g m 3Vπ 3 )
⎝ RC 3 + RL ⎠
From (2): Ve 3 = VS + 2Vπ 2
(19.23) Vπ 2 +
(5)
Vπ 3 Vπ 3
1
+
+
(VS + 2Vπ 2 ) = 0
1.3 18.6 18.6
or
(19.23)Vπ 2 + (0.8230)Vπ 3 + (0.05376)VS = 0 (3′)
V
⎛ 1
⎞ ⎛1⎞
Vπ 3 ⎜
+ 76.92 ⎟ = ⎜ ⎟ (VS + 2Vπ 2 ) + π 2
1.3
10
5.2
⎝
⎠ ⎝ ⎠
or
(77.69)Vπ 3 = (0.3923)Vπ 2 + (0.1)VS (4′)
⎛ 2 ⎞
I0 = − ⎜
(5′)
⎟ (76.92)Vπ 3 = −(51.28)Vπ 3
⎝ 2 +1⎠
From (3′):
Vπ 2 = −(0.04255)Vπ 3 − (0.002780)VS
Then
(77.69)Vπ 3 = (0.3923)[ −(0.04255)Vπ 3 − (0.002780)VS ] + (0.1)VS
(77.71)Vπ 3 = (0.0989)VS
or
Vπ 3 = (0.001273)VS
so that
I 0 = −(51.28)(0.001273)VS
or
I0
= −(0.0653) mA/V
VS
12.52
5 × 103
f ⎞⎛
f ⎞
⎛
⎜1 + j 4 ⎟ ⎜1 + j 5 ⎟
10
10
⎝
⎠⎝
⎠
f
⎛
⎞
⎛ f ⎞
Phase = φ = − tan −1 ⎜ 4 ⎟ − 2 tan −1 ⎜ 5 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
By trial and error, when f = 1.095 × 105 Hz , φ ≅ 180°
A=
2
For T = 1 at f = 1.095 × 105 Hz,
1=
β ( 5 × 103 )
⎛ f ⎞
1+ ⎜ 4 ⎟
⎝ 10 ⎠
2
⎡ ⎛ f ⎞
⋅ ⎢1 + ⎜ 5 ⎟
⎢⎣ ⎝ 10 ⎠
2
⎤
⎥
⎥⎦
⇒1=
β ( 5 × 103 )
(10.996 )( 2.199 )
12.53
Use the basic circuit shown in Figure 12.27.
⇒ β = 4.84 × 10−3
For the ideal case
I0
1
=
Vi RE
we want
I0
= 10−3 A/V = 1 mA/V
Vi
Set RE = 1 kΩ
Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain
is slightly less than the ideal. RE will need to be slightly decreased to increase the gain.
12.54
dc analysis
I E RE + VEB (on) + I B RB + VCC = 0
5 − 0.7
= 0.0343
100 + (51)(0.5)
I C = (50)(0.0343) = 1.71 mA
IB =
(50)(0.026)
= 0.760 kΩ
1.71
1.71
gm =
= 65.77 mA/V
0.026
a.
Then rπ =
To determine Rif :
IS +
Vπ
V − (−Vπ )
+ 0
=0
RB || rπ
RF
(1)
V0 V0 − (−Vπ )
+
(2)
RC
RF
Now from (2):
V
⎛1 1 ⎞
(65.77)Vπ − π = V0 ⎜ + ⎟
10
⎝ 1 10 ⎠
(65.67)Vπ = V0 (1.10)
or
V0 = (59.7)Vπ
and from (1):
Vπ
V V
IS +
+ π + 0 =0
100 0.760 10 10
g mVπ =
I S + Vπ (0.8543) + (0.1)(59.7)Vπ = 0
I S = −Vπ (6.824)
Now
(−Vπ )
⇒ Rif = 147 Ω
Rif =
IS
To determine R0 f :
IX =
VX
VX
+
− g mVπ
RC RF + RB rπ
(3)
⎛ −( RB rπ ) ⎞
Vπ = ⎜⎜
⎟⎟ (VX ) (4)
⎝ ( RB rπ ) + RF ⎠
Now
⎛ −(100 0.760) ⎞
Vπ = ⎜⎜
⎟⎟ (VX ) = −(0.07014)VX
⎝ (100 0.760) + 10 ⎠
so
1
⎛1
⎞
I X = VX ⎜ +
+ (65.77)(0.07014) ⎟
1
10.754
⎝
⎠
VX
R0 f =
⇒ R0 f = 175 Ω
IX
b.
From part (a), we find
I
Vπ = − S
6.824
then
⎛ −IS ⎞
V0 = (59.7) ⎜
⎟
⎝ 6.824 ⎠
or
V0
= −8.75 kΩ
VS
c.
If capacitance is finite, a phase shift will be introduced.
12.55
dc analysis: VGS = VDS
ID =
VDD − VGS
= K n (VGS − VTN ) 2
RD
10 − VGS = (0.20)(8)(VGS − 2) 2
10 − VGS = 1.6(VGS2 − 4VGS + 4)
1.6VGS2 − 5.4VGS − 3.6 = 0
5.4 ± (5.4) 2 + 4(1.6)(3.6)
= 3.95 V
2(1.6)
10 − 3.95
ID =
= 0.756 mA
8
g m = 2 K n I D = 2 (0.2)(0.756) ⇒ g m = 0.778 mA/V
VGS =
a.
Vgs − VS
RS
+
Vgs − V0
RF
=0
⎛ 1
1 ⎞ VS V0
Vgs ⎜
+
+
⎟=
R
R
RS RF
F ⎠
⎝ S
V0 − Vgs
V0
+ g mVgs +
=0
RD
RF
(1)
⎛ 1
⎛ 1
⎞
1 ⎞
V0 ⎜
+
− g m ⎟ (2)
⎟ = Vgs ⎜
⎝ RD RF ⎠
⎝ RF
⎠
So from (1):
1 ⎞ VS V0
⎛1
Vgs ⎜ +
+
⎟=
⎝ 10 100 ⎠ 10 100
or
Vgs (0.11) = VS (0.10) + V0 (0.010)
Vgs = VS (0.909) + V0 (0.0909)
Then from (2):
⎛1 1 ⎞
⎛ 1
⎞
V0 ⎜ +
− 0.778 ⎟
⎟ = Vgs ⎜
⎝ 8 100 ⎠
⎝ 100
⎠
V0 (0.135) = Vgs (−0.768)
= (−0.768)[VS (0.909) + V0 (0.0909)]
V0 (0.2048) = −VS (0.6981)
so
V
Av = 0 = −3.41
VS
b.
We have
Vgs = VS (0.909) + V0 (0.0909)
= VS (0.909) + (0.0909) − (3.41VS )
= 0.599VS
Now
V
V0
(−3.41VS ) RS
Azf = 0 =
=
I S VS − Vgs VS − 0.599VS
RS
or
(−3.41)(10)
Azf =
⇒ Azf = −85.0 V/ma
0.401
Vgs
Vgs 0.401VS 0.599VS
c.
Rif =
=
=
⋅ (10) ⇒ Rif = 14.9 kΩ
0.401VS
IS
RS
d.
IX =
VX
VX
+ g mVgs +
RD
RS + RF
⎛ RS ⎞
⎛ 10 ⎞
Vgs = ⎜
⎟ VX = ⎜
⎟ VX
R
R
+
⎝ 10 + 100 ⎠
F ⎠
⎝ S
= (0.0909)VX
1 ⎤
⎡1
I X = VX ⎢ + (0.778)(0.0909) +
10 + 100 ⎥⎦
⎣8
IX
1
=
= 0.2048 ⇒ R0 f = 4.88 kΩ
VX R0 f
12.56
As g m → ∞,
V0 − RF −100
=
=
= −10
10
VS
RS
To be within 10% of ideal,
V0
= −10(0.9) = −9
VS
From Problem 12.41, we had
Vgs = VS (0.909) + V0 (0.0909)
= VS (0.909) + ( −9VS )(0.0909)
= 0.0909VS
Also from Problem 12.41, we had
V0 (0.135) = Vgs (0.010 − g m )
or
(−9VS )(0.135) = (0.0909)VS (0.010 − g m )
−1.215 = 0.000909 − 0.0909 g m
or
g m = 13.36 mA/V
12.57
dc analysis
RTH = 24 || 150 = 20.7 kΩ
⎛ 24 ⎞
VTH = ⎜
⎟ (12) = 1.655 V
⎝ 24 + 150 ⎠
1.655 − 0.7
I BQ =
= 0.00556 mA
20.7 + (151)(1)
so I CQ = 0.834 mA
(150)(0.026)
= 4.68 kΩ
0.834
0.834
= 32.08 mA / V
gm =
0.026
rπ =
VS − Vπ
Vπ
V − V0
=
+ π
RS
RB || rπ
RF
(1)
V0 V0 − Vπ
+
= 0 (2)
RC
RF
From (1):
⎡1
VS
1
1 ⎤ V0
= Vπ ⎢ +
+
⎥−
R
RF
5
5
20.7
||
4.68
⎣
F ⎦
g mVπ +
or
⎛
1 ⎞ V0
VS (0.20) = Vπ ⎜ 0.4620 +
⎟−
R
RF
⎝
F ⎠
From (2):
⎛
⎛1 1 ⎞
1 ⎞
⎜ 32.08 −
⎟ Vπ + V0 ⎜ +
⎟=0
R
⎝
⎝ 6 RF ⎠
F ⎠
⎛
1 ⎞
−V0 ⎜ 0.1667 +
⎟
R
⎝
F ⎠
(2)
Vπ =
⎛
1 ⎞
⎜ 32.08 −
⎟
RF ⎠
⎝
Then
⎡
⎛
1
⎢ −V0 ⎜ 0.1667 +
RF
⎛
1 ⎞⎢
⎝
VS (0.20) = ⎜ 0.4620 +
⎟⎢
R
⎛
1 ⎞
⎝
F ⎠
⎢ ⎜ 32.08 −
⎟
RF ⎠
⎢⎣ ⎝
so
⎞⎤
⎟⎥
⎠ ⎥ − V0
⎥ RF
⎥
⎥⎦
Neglect the RF in the denominator term. Now
V0
V
= −5 ⇒ VS = − 0 = −V0 (0.20)
5
VS
⎡ −V ( 0.1667 RF + 1) ⎤
−V0 (0.20)(0.20) RF = (0.4620 RF + 1) ⎢ 0
⎥ − V0
32.08RF
⎣
⎦
2
−1.283RF = −(0.4620 RF + 1)(0.1667 RF + 1) − 32.08 RF
1.206 RF2 − 32.71RF − 1 = 0
32.71 ± (32.71) 2 + 4(1.206)(1)
2(1.206)
so that
RF = 27.2 kΩ
RF =
12.58
dc analysis
RTH = 4 || 15 = 3.16 kΩ = RB
⎛ 4 ⎞
VTH = ⎜
⎟12 = 2.526 V
⎝ 4 + 15 ⎠
2.526 − 0.7
= 0.00251
I BQ =
3.16 + (181)(4)
I CQ = 0.452 mA
(180)(0.026)
= 10.4 kΩ
0.452
0.452
gm =
= 17.4 mA/V
0.026
rπ =
Vi − Vπ 1
Vπ 1
V −V
=
+ π1 0
RS
RB || rπ
RF
(1)
g mVπ 1 +
Vπ 2
=0
RC || RB || rπ
(2)
g mVπ 2 +
Vπ 3
=0
RC || RB || rπ
(3)
g mVπ 3 +
V0 V0 V0 − Vπ 1
+
+
=0
RC RL
RF
(4)
Now
RC || RB || rπ = 8 || 3.16 ||10.4 = 1.86 kΩ
RB || rπ = 3.16 ||10.4 = 2.42 kΩ
Now substituting in (2):
V
(17.4)Vπ 1 + π 2 = 0 or Vπ 2 = −(32.36)Vπ 1
1.86
and substituting in (3):
V
(17.4)Vπ 2 + π 3 = 0
1.86
V
(17.4)[−(32.36)Vπ 1 ] + π 3 = 0
1.86
or Vπ 3 = (1047.3)Vπ 1
Substitute numerical values in (1):
⎛1
Vi
1
1 ⎞ V0
= Vπ 1 ⎜ +
+
⎟−
10
10
2.42
R
RF
⎝
F ⎠
or
⎛
1 ⎞ V0
Vi (0.10) = Vπ 1 ⎜ 0.513 +
⎟−
R
RF
⎝
F ⎠
Substitute numerical values in (4):
⎛ 1 1 1 ⎞ Vπ 1
=0
(17.4)(1047.3)Vπ 1 + V0 ⎜ + +
⎟−
⎝ 8 4 RF ⎠ RF
⎛
⎛
1 ⎞
1 ⎞
Vπ 1 ⎜ 1.822 × 104 −
⎟ + V0 ⎜ 0.375 +
⎟=0
RF ⎠
RF ⎠
⎝
⎝
⎛
1 ⎞
−V0 ⎜ 0.375 +
⎟
RF ⎠
⎝
Vπ 1 =
1
1.822 × 104 −
RF
so that
⎡
⎛
1 ⎞⎤
⎢ −V0 ⎜ 0.375 +
⎟⎥
RF ⎠ ⎥ V0
⎛
1 ⎞⎢
⎝
−
Vi (0.10) = ⎜ 0.513 +
⎟
RF ⎠ ⎢ 1.822 × 104 − 1 ⎥ RF
⎝
⎢
⎥
RF ⎥
⎣⎢
⎦
V
V
We have 0 = −80 or Vi = − 0
80
Vi
⎡ ⎛
1 ⎞⎤
⎢ − ⎜ 0.375 +
⎟⎥
R
(0.10) ⎛
1 ⎞⎢ ⎝
1
F ⎠ ⎥
−
= ⎜ 0.513 +
−
⎟⎢
⎥
1
80
R
R
4
⎝
F ⎠ 1.822 × 10 −
F
⎢
⎥
RF ⎥
⎢⎣
⎦
Neglect that 1/ RF term in the denominator.
(0.513RF + 1)(0.375 RF + 1)
−(0.00125 RF ) = −
−1
1.822 × 104 RF
22.775 RF2 = (0.513RF + 1)(0.375 RF + 1) + 1.822 × 10 4 RF
We find
22.58 RF2 − 1.822 × 104 RF − 1 = 0
RF =
1.822 × 104 ± (1.822 ×10 4 ) 2 + 4(22.58)(1)
2(22.58)
or
RF = 0.807 MΩ
12.59
a.
VS − (−Vd ) −Vd − V1
=
RS
RF
or
⎛ 1
1 ⎞ VS V1
+
+
=0
Vd ⎜
⎟+
⎝ RS RF ⎠ RS RF
V0 − V1 V1 V1 − (−Vd )
=
+
R1
R2
RF
or
⎛ 1
V0
1
1 ⎞ Vd
= V1 ⎜ +
+
⎟+
R1
⎝ R1 R2 RF ⎠ RF
(1)
(2)
V0
A0 L
Substitute numerical values in (1) and (2):
V0 ⎛ 1 1 ⎞ VS V1
⋅⎜ + ⎟ + + = 0
104 ⎝ 5 10 ⎠ 5 10
or
V0 (0.3 × 10−4 ) + VS (0.20) + V1 (0.10) = 0
and V0 = A0 LVd or Vd =
V0
⎛ 1 1 1⎞ V ⎛1⎞
= V1 ⎜ + + ⎟ + 04 ⋅ ⎜ ⎟
50
⎝ 50 10 10 ⎠ 10 ⎝ 10 ⎠
or
V0 (0.02 − 10−5 ) = V1 (0.22) (2)
(1)
⎛ 0.02 − 10−5 ⎞
Then V1 = V0 ⎜
⎟
⎝ 0.22 ⎠
and
⎡ ⎛ 0.02 − 10−5 ⎞ ⎤
V0 (0.3 × 10−4 ) + VS (0.20) + (0.10) ⎢V0 ⎜
⎟⎥ = 0
⎣ ⎝ 0.22 ⎠ ⎦
V0 ⎡⎣0.3 × 10−4 − 0.4545 × 10−5 + 0.00909⎤⎦ + VS (0.20) = 0
Then
b.
V0
V
−0.20
=
⇒ 0 = −21.94
−3
VS 9.115 × 10
VS
Rif =
Now Vd =
−Vd
−V ⋅ R
= d S
VS − (−Vd ) VS + Vd
RS
V0
21.94VS
=−
104
A0 L
(21.94 × 10−4 )(5)
1 − 21.94 × 10−4
or Rif = 1.099 × 10−2 kΩ ⇒ Rif = 10.99 Ω
Then Rif =
c.
Because of the A0LVd source,
R0 f = 0
12.60
For example, use the circuit shown in Figure 12.41
12.61
Break the loop
It = Iε
Now Ai I t +
V0
V0
+
=0
R1 RF + RS Ri
⎛ RS ⎞
V0
Ir = ⎜
⎟⋅
⎝ RS + Ri ⎠ RF + RS Ri
⎛ R + Ri ⎞
or V0 = I r ⎜ S
⎟ ⋅ ( RF + RS Ri )
⎝ RS ⎠
Then
⎛ 1
⎞ ⎡ ⎛ RS + Ri ⎞
⎤
1
Ai I t + ⎜⎜ +
⎟⎟ × ⎢ I r ⎜
⎟ ( RF + RS Ri ) ⎥ = 0
⎦
⎝ R1 RF + RS Ri ⎠ ⎣ ⎝ RS ⎠
Ai
I
T = − r ⇒T =
It
⎡1
⎤
⎛ RS + Ri ⎞
1
⎢ +
⎥⎜
⎟ ( RF + RS Ri )
R
R
R
R
RS ⎠
+
F
S
i ⎦⎝
⎣ 1
12.62
Vπ 1
V
V −V
+ g m1Vπ 1 = ε 1 + ε 1 0
rπ 1
RE1
RF
g m1Vπ 1 +
Vr
RC1 rπ 2
(1)
= 0 ⇒ Vr = −( g m1Vπ 1 )( RC1 rπ 2 )
Vπ 2 = Vt so that
g m 2Vt +
Vπ 3 + V0 Vπ 3
+
= 0 (3)
RC 2
rπ 3
Vπ 3
V
V −V
+ g m 3Vπ 3 = 0 + 0 ε 1
(4)
rπ 3
RE 3
RF
From (4):
⎛ 1
⎛ 1
⎞ V
1 ⎞
V0 ⎜
+
+ g m3 ⎟ + ε 1
⎟ = Vπ 3 ⎜
⎝ RE 3 RF ⎠
⎝ rπ 3
⎠ RF
But Vε 1 = −Vπ 1
⎛ 1
⎞ V
+ g m3 ⎟ − π 1
Vπ 3 ⎜
⎝ rπ 3
⎠ RF
so V0 =
⎛ 1
1 ⎞
+
⎜
⎟
R
R
F ⎠
⎝ E3
Then
(2)
⎡⎛ 1
⎞ ⎛ 1
1 ⎞⎤
+
Vπ 1 ⎢⎜ + g m1 ⎟ − ⎜
⎟⎥ =
⎠ ⎝ RE1 RF ⎠ ⎦
⎣⎝ rπ 1
⎛ 1
⎞ V
−Vπ 3 ⎜
+ g m3 ⎟ + π 1
r
⎝ π3
⎠ RF
⎛ 1
1 ⎞
RF ⋅ ⎜
+
⎟
⎝ RE 3 RF ⎠
(1′)
and
⎛ 1
⎞ V
+ g m3 ⎟ − π 1
Vπ 3 ⎜
⎛ 1
1 ⎞
⎝ rπ 3
⎠ RF = 0
+
g m 2Vt + Vπ 3 ⎜
⎟+
⎛ 1
1 ⎞
⎝ RC 2 rπ 3 ⎠
+
RC 2 ⋅ ⎜
⎟
R
R
F ⎠
⎝ E3
(3′)
From (3′), solve for Vπ 3 and substitute into (1′). Then from (1′), solve for Vπ 1 and substitute into (2). Then
T =−
Vr
.
Vt
12.63
Vr Vr Vr − Ve
+
+
=0
RS rπ 1
RF
g m1Vt +
(1)
Vπ 2 + Ve Vπ 2
+
= 0 (2)
RC1
rπ 2
Vπ 2
V V − Vr
+ g m 2Vπ 2 = e + e
(3)
rπ 2
RE
RF
Using the parameters from Problem 12.29, we obtain
1
1⎞ V
⎛1
Vr ⎜ +
+ ⎟− e = 0
⎝ 10 15.8 10 ⎠ 10
or
Vr (0.2633) = Ve (0.10)
(1)
1 ⎞ Ve
⎛ 1
=0
(7.62)Vt + Vπ 2 ⎜ +
⎟+
⎝ 40 2.28 ⎠ 40
or
Vt (7.62) + Vπ 2 (0.4636) + Ve (0.025) = 0
(2)
⎛ 1
⎞
⎛1 1 ⎞ V
Vπ 2 ⎜
+ 52.7 ⎟ = Ve ⎜ + ⎟ − r
2.28
⎝
⎠
⎝ 1 10 ⎠ 10
or
Vπ 2 (53.14) = Ve (1.10) − Vr (0.10)
Then
Vπ 2 = Ve (0.0207) − Vr (0.001882) (3)
Substituting in (2):
Vt (7.62) + (0.4636)[Ve (0.0207) − Vr (0.001882)] + Ve (0.025) = 0
or
Vt (7.62) + Ve (0.03460) − Vr (0.0008725) = 0
From (1) Ve = Vr (2.633)
Then
Vt (7.62) + Vr (2.633)(0.03460) − Vr (0.0008725) = 0
Vt (7.62) + Vr (0.09023) = 0
Vr
= −84.45
Vt
Now
V
T = − r ⇒ T = 84.45
Vt
or
12.64
Vπ = −Vt
g mVπ =
V0
V0
+
RC RF + RB rπ
(1)
and
⎛ RB rπ ⎞
(2)
Vr = ⎜
V
⎜ R r + R ⎟⎟ 0
F ⎠
⎝ B π
Now
⎛1
⎞
1
(65.77)Vπ = V0 ⎜ +
⎜ 1 10 + 100 0.760 ⎟⎟
⎝
⎠
or (65.77)Vπ = V0 (1.0930)
and
⎛ 0.754 ⎞
Vr = ⎜
⎟ V0 = (0.07011)V0
⎝ 10 + 0.754 ⎠
so V0 = (14.26)Vr
Then (65.77)(−Vt ) = (14.26)Vr (1.0930)
Vr
= −4.22 so that T = 4.22
Vt
12.64
Want f1 = 12 MHz for a phase margin of 45°
TdB ( 0 ) = 80 dB ⇒ T ( 0 ) = 104
Then
⎛
f
⎜1 + j
f PD
⎝
Set f = f1 and T
T(f)=
T ( 0)
⎞⎛
f
⎞
⎟⎜1 + j
6 ⎟
12 × 10 ⎠
⎠⎝
=1
So
104
⎛ 12 × 106 ⎞
1+ ⎜
⎟ ⋅ 2
⎝ f PD ⎠
which yields
12 × 106 104
=
⇒ f PD = 1.70 kHz
f PD
2
T =1=
12.65
a.
2
⎛ f ⎞
⎛ f ⎞
− 2 tan −1 ⎜ 4 ⎟
φ = − tan −1 ⎜
2 ⎟
5
10
×
⎝
⎠
⎝ 10 ⎠
⎛ f
⎞
⎛ f ⎞
−180 = − tan −1 ⎜ 180 2 ⎟ − 2 tan −1 ⎜ 1804 ⎟ ⇒ f180 ≈ 1.05 × 104 Hz
⎝ 5 × 10 ⎠
⎝ 10 ⎠
β (105 )
T ( f180 ) = 1 =
b.
2
2
⎛ 1.05 × 104 ⎞ ⎡ ⎛ 1.05 ×104 ⎞ ⎤
1+ ⎜
1
+
⎢
⎥
⎟
⎜
⎟
2
4
⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10
⎠ ⎥⎦
β (105 )
or
1=
(21.02)(2.105)
or
β = 4.42 × 10−4
12.65
A0 = 80 dB ⇒ A0 = 104
A0
Af ( 0 ) =
1 + β A0
or 5 =
104
1 + β (104 )
⇒ β ≈ 0.2
Then T ( 0 ) = β A0 = 0.2 × 104
Inserting a dominate pole
⎛ f ⎞
−1 ⎛ f ⎞
−1 ⎛ f ⎞
φ = − tan −1 ⎜
⎟ − tan ⎜ 6 ⎟ − tan ⎜ 7 ⎟
10
f
⎝
⎠
⎝ 10 ⎠
⎝ PD ⎠
If we want a phase margin of 45°, then
⎛ f ⎞
⎛ f ⎞
−135° ≈ −90° − tan −1 ⎜ 6 ⎟ − tan −1 ⎜ 7 ⎟
10
⎝
⎠
⎝ 10 ⎠
By trial and error, f ≈ 0.845 MHz
Then
0.2 × 104
1
×
T =1=
2
2
2
⎛ 0.845 × 106 ⎞
⎛ 0.845 ⎞
⎛ 0.845 ⎞
+
1
+
1+ ⎜
1
⎜
⎟
⎟
⎜
⎟
⎝ 10 ⎠
f PD
⎝ 1 ⎠
⎝
⎠
0.845 × 106
0.2 ×104
≈
f PD
(1.309 )(1.0036 )
so f PD = 555 Hz
12.66
β (103 )
f ⎞
⎛
⎜1 + j 4 ⎟
10 ⎠
⎝
f
f
φ = −2 tan −1 4 − tan −1 5
10
10
Set φ = −180°
T = β Av =
(a)
2
f ⎞
⎛
⎜1 + j 5 ⎟
10 ⎠
⎝
By trial and error, f = 4.58 ×10 4 Hz
Set T = 1 at f = 4.58 ×10 4 Hz
(b)
β (103 )
⎛
4 2 ⎞
4 2
⎜ 1 + ⎜⎛ 4.58 ×10 ⎟⎞ ⎟ 1 + ⎜⎛ 4.58 ×10 ⎟⎞
4
5
⎜
⎝ 10
⎠ ⎟⎠
⎝ 10
⎠
⎝
β (103 )
⇒ β = 0.02417
1=
(21.976)(1.10)
1=
2
103
= 39.7
1 + (10 )(0.02417)
(c)
Avfo =
(d)
Wait T < 1 at f = 4.58 × 104 Hz, so system is stable for smaller values of β .
3
12.67
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
− tan −1 ⎜
− tan −1 ⎜ 5 ⎟
4 ⎟
4 ⎟
⎝ 10 ⎠
⎝ 5 × 10 ⎠
⎝ 10 ⎠
4
At f = 8.1× 10 Hz, φ = −180.28°
φ = − tan −1 ⎜
Determine T ( f ) at this frequency.
T = β (103 ) ×
=
1
⎛ 8.1× 104 ⎞
1+ ⎜
⎟
4
⎝ 10
⎠
2
×
1
⎛ 8.1× 104 ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
β (103 )
a.
(8.161)(1.904)(1.287)
For β = 0.005
b.
For β = 0.05
T ( f ) = 0.250 < 1 ⇒ Stable
T ( f ) = 2.50 > 1 ⇒ Unstable
12.68
(b)Phase margin = 80° ⇒ φ = −100°
⎛ f ⎞
⎛ f ⎞
− tan −1 ⎜
3 ⎟
4 ⎟
⎝ 10 ⎠
⎝ 5 × 10 ⎠
3
By trial and error, f = 1.16 × 10 Hz
φ = −100 = −2 tan −1 ⎜
2
×
1
⎛ 8.1× 104 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
2
Then
β (5 × 103 )
⎛
3 2 ⎞
3 2
⎜ 1 + ⎛⎜ 1.16 × 10 ⎞⎟ ⎟ ⋅ 1 + ⎛⎜ 1.16 × 10 ⎞⎟
3
4
⎜
⎝ 10
⎠ ⎟⎠
⎝ 5 × 10 ⎠
⎝
β (5 × 103 )
=
⇒ β = 4.7 × 10−4
(2.35)(1.00)
T =1=
2
12.69
For β = 0.005,
c.
T ( f ) = 1(0 dB) at f ≈ 2.10 × 104 Hz
Then
4
4
⎛ 2.10 × 104 ⎞
−1 ⎛ 2.10 × 10 ⎞
−1 ⎛ 2.10 × 10 ⎞
⎟ − tan ⎜
⎟ − tan ⎜
⎟
3
4
5
⎝ 10
⎠
⎝ 10
⎠
⎝ 10
⎠
φ = − tan −1 ⎜
= −87.27 − 64.54 − 11.86
or
φ = −163.7
System is stable.
Phase margin = 16.3°
For β = 0.05,
T ( f ) = 1 (0 dB) at f ≈ 6.44 × 104 Hz
Then
4
4
⎛ 6.44 × 104 ⎞
−1 ⎛ 6.44 × 10 ⎞
−1 ⎛ 6.44 × 10 ⎞
tan
tan
−
−
⎟
⎜
⎟
⎜
⎟
3
4
5
⎝ 10
⎠
⎝ 10
⎠
⎝ 10
⎠
φ = − tan −1 ⎜
= −89.11 − 81.17 − 32.78
φ = −203.1° ⇒ System is unstable.
or
12.70
β (105 )
f ⎞⎛
f ⎞⎛
f ⎞
⎛
1 + j 5 ⎟ ⎜1 + j
⎜1 + j
⎟
4 ⎟⎜
5 × 10 ⎠ ⎝
10 ⎠ ⎝
5 × 105 ⎠
⎝
Phase Margin = 60° ⇒ φ = −120°
So
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
−120 = − tan −1 ⎜
− tan −1 ⎜ 5 ⎟ − tan −1 ⎜
4 ⎟
5 ⎟
⎝ 5 × 10 ⎠
⎝ 10 ⎠
⎝ 5 × 10 ⎠
By trial and error, at f = 105 Hz , φ ≅ −120°
Then
β (105 )
T = 1=
2
2
2
⎛ 105 ⎞
⎛ 105 ⎞
⎛ 105 ⎞
⋅ 1+ ⎜ 5 ⎟ ⋅ 1+ ⎜
1+ ⎜
4 ⎟
5 ⎟
⎝ 5 × 10 ⎠
⎝ 10 ⎠
⎝ 5 × 10 ⎠
T = Aβ =
1=
β (105 )
( 2.236 )(1.414 )(1.02 )
12.71
(a)
100 =
⇒ β = 3.22 × 10−5
105
⇒ β = 9.99 × 10−3
1 + (105 ) β
T = 1 = β Av =
(9.99 × 10−3 )(105 )
⎛ f ⎞
1+ ⎜ 3 ⎟
⎝ 10 ⎠
999
2
⎛ f ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
⎛ f ⎞
⎛ f ⎞
1+ ⎜ 3 ⎟ 1+ ⎜ 5 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
5
f = 3.08 × 10 Hz
1=
2
2
2
Phase
f
f
− tan −1 5
3
10
10
5
5
3.08 ×10
−1 3.08 × 10
tan
= − tan −1
−
103
105
= −89.81 − 72.01
φ = −161.8
Stable
(b)
Phase Margin = 180 − 161.8 = 18.2°
φ = − tan −1
12.72
φ = − tan −1
(a)
f
f
f
− tan −1 6 − tan −1
4
5 × 10
10
5 × 107
For φ = −180°
By trial and error, f180 = 7.25 × 106 Hz
T =
(b)
=
⎛ 7.25 × 106 ⎞
1+ ⎜
⎟
4
⎝ 5 × 10 ⎠
(0.10)(105 )
2
⎛ 7.25 ×106 ⎞
1+ ⎜
⎟
6
⎝ 10
⎠
2
⎛ 7.25 ×106 ⎞
1+ ⎜
⎟
7
⎝ 5 × 10 ⎠
104
(145)(7.319)(1.01)
T = 9.33
System is unstable
T =1=
⎛ f ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
104
2
⎛ f ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
⎛ f ⎞
1+ ⎜
7 ⎟
⎝ 5 × 10 ⎠
2
f = 2.14 ×107 Hz
7
7
⎛ 2.14 × 107 ⎞
−1 ⎛ 2.14 × 10 ⎞
−1 ⎛ 2.14 × 10 ⎞
⎟ − tan ⎜
⎟ − tan ⎜
⎟
4
6
7
⎝ 5 × 10 ⎠
⎝ 10
⎠
⎝ 5 × 10 ⎠
φ = − tan −1 ⎜
= −89.87 − 87.32 − 23.17
φ = −200.4°
For f = 7.25 × 106 Hz
(c)
(0.0010)(105 )
⎛ 7.25 × 106 ⎞
⎛ 7.25 ×106 ⎞
1+ ⎜
⎟ 1+ ⎜
⎟
4
6
⎝ 5 × 10 ⎠
⎝ 10
⎠
100
=
(145)(7.319)(1.01)
T =
2
2
⎛ 7.25 ×106 ⎞
1+ ⎜
⎟
7
⎝ 5 × 10 ⎠
T = 0.0933
System is stable.
T =1=
⎛ f ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
100
2
⎛ f ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
⎛ f ⎞
1+ ⎜
7 ⎟
⎝ 5 × 10 ⎠
2
f = 2.13 ×106 Hz
6
6
2.13 × 106
−1 2.13 × 10
−1 2.13 × 10
−
−
tan
tan
5 × 104
106
5 × 107
= −88.66 − 64.85 − 2.44
φ = −155.9°
φ = − tan −1
12.73
(a)
f180
f
− 2 tan −1 1805
5 × 103
10
= 1.05 × 105 Hz
φ = −180 = − tan −1
f180
2
2
(0.0045)(2 × 103 )
2
2
⎛ 1.05 × 105 ⎞ ⎡ ⎛ 1.05 × 105 ⎞ ⎤
1+ ⎜
⎢1 + ⎜
⎟ ⎥
3 ⎟
5
⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10
⎠ ⎥⎦
9
=
(21.02)(2.1025)
T =
(b)
0
T = f180
= 0.204
System is stable
9
T =1=
⎛ f ⎞
1+ ⎜
3 ⎟
⎝ 5 × 10 ⎠
2
⎡ ⎛ f ⎞2 ⎤
⎢1 + ⎜ 5 ⎟ ⎥
⎣⎢ ⎝ 10 ⎠ ⎦⎥
f = 3.88 × 104 Hz
3.88 × 104
3.88 ×104
− 2 tan −1
3
5 × 10
105
= −82.66 − 42.41
φ = −125.1°
φ = − tan −1
(c)
T =
=
(0.15)(2 × 103 )
⎛ 1.05 × 105 ⎞
1+ ⎜
3 ⎟
⎝ 5 × 10 ⎠
2
⎡ ⎛ 1.05 × 105 ⎞ 2 ⎤
⎢1 + ⎜
⎟ ⎥
5
⎢⎣ ⎝ 10
⎠ ⎥⎦
300
(21.02)(2.1025)
T = 6.79
System is unstable
T = 1=
300
⎛ f ⎞
1+ ⎜
3 ⎟
⎝ 5 × 10 ⎠
2
⎡ ⎛ f ⎞2 ⎤
⎢1 + ⎜ 5 ⎟ ⎥
⎢⎣ ⎝ 10 ⎠ ⎥⎦
f = 2.33 × 105 Hz
5
2.33 × 105
−1 2.33 × 10
2
tan
−
5 × 103
105
= −88.77 − 133.54
φ = − tan −1
φ = −222.3°
12.74
Phase Margin = 45° ⇒ φ = −135°
φ = −135°
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
= − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
4
At f = 10 Hz, φ = −135.6°
T =1
1
= β (103 ) ×
1
⎛ 104 ⎞
⎛ 104 ⎞
1+ ⎜ 3 ⎟
1+ ⎜ 4 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
1
1
×
×
2
4 2
⎛ 10 ⎞
⎛ 104 ⎞
1+ ⎜ 5 ⎟
1+ ⎜ 6 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
1=
2
×
2
×
β (103 )
(10.05)(1.414)(1.005)(1.00)
or
β = 0.01428
12.75
1
1
×
f
⎛
⎞
f ⎞ ⎛1 + j
3 ⎟
⎜1 + j
⎟ ⎜
300 × 10 ⎠
f PD ⎠ ⎝
⎝
1
1
×
×
f
f
⎛
⎞ ⎛
⎞
⎜1 + j
⎜1 + j
⎟
6 ⎟
2 × 10 ⎠ ⎝
25 × 106 ⎠
⎝
Phase Margin = 45° ⇒ φ = −135° at f = 300 kHz
T = 5000 ×
3
⎛ 300 × 103 ⎞
−1 ⎛ 300 × 10 ⎞
−135° = − tan −1 ⎜
−0−0
⎟ − tan ⎜
3 ⎟
⎝ 300 × 10 ⎠
⎝ f PD ⎠
= −90° − 45°
Now
T = 1@ f = 300 kHz
T =1≈
5000
⎛ 300 × 103 ⎞
1+ ⎜
⎟ ⋅ 2 ⋅1 ⋅1
⎝ f PD ⎠
2
⎛ 300 × 103 ⎞ ⎛ 5000 ⎞
1+ ⎜
⎟ =⎜
⎟
⎝ f PD ⎠ ⎝ 2 ⎠
2
f PD ≈
12.76
(a)
2
300 × 103 2
⇒ f PD = 84.8 Hz
5000
At f = 106 Hz,
6
106
−1 10
2
tan
−
104
106
= −89.4 − 90 ≈ −180°
φ = − tan −1
T =
=
103
⎛ 106 ⎞
1+ ⎜ 4 ⎟
⎝ 10 ⎠
2
⎡ ⎛ 106 ⎞ 2 ⎤
⎢1 + ⎜ 6 ⎟ ⎥
⎢⎣ ⎝ 10 ⎠ ⎥⎦
103
=5
(100)(2)
T > 1 at f = f180 = 106 Hz, System is unstable.
103
⎛
f ⎞⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j 6 ⎟
f PD ⎠ ⎝
10 ⎠ ⎝
10 ⎠
⎝
Phase margin = 45° ⇒ φ = −135
T=
(b)
φ = −135 = − tan −1
f
f PD
− tan −1
2
f
f
− 2 tan −1 6
4
10
10
4
f ≅ 10 Hz
T =1=
⎛ 104 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
103
2
⎛ 104 ⎞
1+ ⎜ 4 ⎟
⎝ 10 ⎠
2
⎡ ⎛ 104 ⎞ 2 ⎤
⎢1 + ⎜ 6 ⎟ ⎥
⎢⎣ ⎝ 10 ⎠ ⎥⎦
103
⎛ 104 ⎞
⎜
⎟ (1.414)(1.0)
⎝ f PD ⎠
(104 )(1.414)
f PD =
103
f PD = 14.14 Hz
1≅
12.77
(a)
⎛ f180 ⎞
⎛ f
⎞
⎛ f ⎞
− tan −1 ⎜ 180 4 ⎟ − tan −1 ⎜ 1805 ⎟
4 ⎟
⎝ 10 ⎠
⎝ 5 × 10 ⎠
⎝ 10 ⎠
φ = −180 = − tan −1 ⎜
f180 ≅ 8.06 × 104 Hz
(b)
500
⎛ 8.06 × 10 ⎞
⎛ 8.06 × 104 ⎞
1+ ⎜
1
+
⎟
⎜
⎟
4
4
⎝ 10
⎠
⎝ 5 × 10 ⎠
500
=
(8.122)(1.897)(1.284)
T =
4
2
2
⎛ 8.06 ×104 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
T = 25.3
500
⎛
⎞
f ⎛
f ⎞⎛
f ⎞⎛
f ⎞
1+ j 5 ⎟
⎜1 + j
⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j
4 ⎟⎜
×
10
5
10
10
f
⎝
⎠
⎝
⎠
⎝
⎠
⎝
PD ⎠
Phase Margin = 60° ⇒ φ = −120°
(c)
T=
−120 = − tan −1
f
f
f
f
− tan −1 4 − tan −1
− tan −1 5
f PD
10
5 × 104
10
Assume tan −1
f
f PD
≅ 90°
2
Then f ≅ 4.2 × 103 Hz
500
⎛ 4.2 × 103 ⎞
⎛ 4.2 × 103 ⎞
1+ ⎜
⎟ 1+ ⎜
⎟
4
⎝ 10
⎠
⎝ f PD ⎠
500
T =1=
2
2
⎛ 4.2 × 103 ⎞
1+ ⎜
⎟ (1.085)(1.004)(1.0)
⎝ f PD ⎠
1=
2
4.2 × 103
500
≅
(1.0846)(1.0035)(1.0)
f PD
f PD = 9.14 Hz
12.78
50 =
(a)
T=
104
⇒ β = 0.0199
1 + (104 ) β
(0.0199)(104 )
⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟⎜ 1 + j 5 ⎟
10 ⎠
f PD ⎠ ⎝
⎝
Phase margin = 45° ⇒ φ = −135°
−135 = − tan −1
f
f PD
− tan −1
f
105
5
f = 10 Hz
T =1=
(0.0199)(104 )
⎛ 105 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
2
⎛ 105 ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
2
105 (0.0199)(104 )
=
f PD
1.414
f PD = 711 Hz
20 =
(b)
T=
104
⇒ β = 0.0499
1 + (104 ) β
(0.0499)(104 )
f ⎞⎛
f ⎞
⎛
⎜1 + j
⎟⎜ 1 + j 5 ⎟
711 ⎠⎝
10 ⎠
⎝
T =1=
(0.0499)(104 )
⎛ f ⎞
1+ ⎜
⎟
⎝ 711 ⎠
2
⎛ f ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
2
f = 1.76 × 105 Hz
5
⎛ 1.76 × 105 ⎞
−1 ⎛ 1.76 × 10 ⎞
⎟ − tan ⎜
⎟
5
⎝ 711 ⎠
⎝ 10
⎠
φ = − tan −1 ⎜
= −89.77 − 60.40
φ = −150.2
Phase Margin = 180 − 150.2 = 29.8°
⎛ 4.2 × 103 ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
2
⎛ 4.2 × 103 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
2
12.79
(a)
20 =
AO = 100 dB ⇒ AO = 105
105
⇒ β = 0.04999
1 + (105 ) β
(0.04999)(105 )
⎛
f ⎞⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟⎜1 + j 6 ⎟⎜ 1 + j 7 ⎟
f
10
10
⎝
⎠⎝
⎠
PD ⎠
⎝
Phase Margin = 45° ⇒ φ = −135°
T=
−135 = − tan −1
f
f
f
− tan −1 6 − tan −1 7
10
10
f PD
f ≈ 106 Hz
T =1=
1=
(0.04999)(105 )
⎛ 106 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
2
⎛ 106 ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
⎛ 106 ⎞
1+ ⎜ 7 ⎟
⎝ 10 ⎠
2
(0.04999)(105 )
⎛ 106 ⎞
1+ ⎜
⎟ (1.414)(1.005)
⎝ f PD ⎠
2
106 (0.04999)(105 )
=
f PD (1.414)(1.005)
f PD = 2.84 Hz
(b)
T =1=
5=
105
⇒ β = 0.19999
1 + (105 ) β
(0.19999)(105 )
⎛ f ⎞
1+ ⎜
⎟
⎝ 284 ⎠
2
⎛ f ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
⎛ f ⎞
1+ ⎜ 7 ⎟
⎝ 10 ⎠
2
f = 2.25 × 106 Hz
6
6
⎛ 2.25 × 106 ⎞
−1 ⎛ 2.25 × 10 ⎞
−1 ⎛ 2.25 × 10 ⎞
⎟ − tan ⎜
⎟ − tan ⎜
⎟
6
7
⎝ 284 ⎠
⎝ 10
⎠
⎝ 10
⎠
φ = − tan −1 ⎜
= −89.99 − 66.04 − 12.68
φ = −168.7
Phase Margin = 180 − 168.7 = 11.3°
12.80
a.
T( f ) =
T (0) = 100 dB ⇒ T (0) = 105
105
f ⎞⎛
f ⎞⎛
f
⎛
⎞
1+ j
⎜1 + j ⎟ ⎜1 + j
⎟
6 ⎟⎜
10 ⎠ ⎝
5 × 10 ⎠ ⎝
10 × 106 ⎠
⎝
T =1=
= 105 ×
1
⎛ f ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
2
×
1
⎛ f ⎞
1+ ⎜
6 ⎟
⎝ 5 × 10 ⎠
2
×
1
f
⎛
⎞
1+ ⎜
6 ⎟
⎝ 10 × 10 ⎠
2
By trial and error
f = 0.976 MHz
⎛ 0.976 × 106 ⎞
−1 ⎛ 0.976 ⎞
−1 ⎛ 0.976 ⎞
⎟ − tan ⎜
⎟ − tan ⎜
⎟
10
5
⎝
⎠
⎝ 10 ⎠
⎝
⎠
φ = − tan −1 ⎜
= −90° − 11.05° − 5.574° = −106.6°
Phase Margin = 180° − 106.6° = 73.4°
b.
f P′1 ∝
1
10 75
so
=
CF
f P′1 20
or
f P′1 = 2.67 Hz
Now
T =1=
= 105 ×
1
⎛ f ⎞
1+ ⎜
⎟
⎝ 2.67 ⎠
2
×
1
⎛ f ⎞
1+ ⎜
6 ⎟
⎝ 5 × 10 ⎠
2
1
f
⎛
⎞
1+ ⎜
6 ⎟
⎝ 10 × 10 ⎠
By trial and error
f ≈ 2.66 ×105 Hz
then
⎛ 2.66 × 105 ⎞
−1 ⎛ 0.266 ⎞
−1 ⎛ 0.266 ⎞
φ = − tan −1 ⎜
⎟ − tan ⎜
⎟ − tan ⎜
⎟
⎝ 5 ⎠
⎝ 10 ⎠
⎝ 2.67 ⎠
×
2
= −90° − 3.045° − 1.524° = −94.57°
Phase Margin = 180° − 94.57° = 85.4°
12.81
1
where τ = ( Ro1 Ri 2 ) Ci
2πτ
= (500 1000) × 103 × 2 × 10−12 ⇒ τ = 6.67 × 10−7 s
(a)
f 3− dB =
Then
1
⇒ f 3− dB = 239 kHz
2π (6.67 × 10−7 )
(b)
For
1
1
f PD = 10 Hz, τ =
=
= 0.0159 s
2π f PD 2π (10)
f 3− dB =
Then τ = ( Ro1 Ri 2 ) ( Ci + CM )
0.0159 = ( 500 1000 ) × 103 × ( Ci + CM )
or
( Ci + CM ) = 4.77 ×10−8 = 2 x10−12 + CM ⇒ CM = 0.0477 µ F
12.82
Assuming a phase margin of 45°.
f
⎛ f ⎞
⎛
⎞
−135° ≈ −90° − tan −1 ⎜
− tan −1 ⎜
6 ⎟
6 ⎟
⎝ 2 × 10 ⎠
⎝ 25 × 10 ⎠
By trial and error, f ≈ 1.74 MHz
Then
T =1
1
⎛ 1.74 × 106 ⎞
1+ ⎜
⎟
f PD ⎠
⎝
1
1
×
×
2
2
⎛ 1.74 ⎞
⎛ 1.74 ⎞
1+ ⎜
1+ ⎜
⎟
⎟
⎝ 2 ⎠
⎝ 25 ⎠
= 5000 ×
2
or
1.74 × 106
5000
≈
f PD
(1.325)(1.0024)
so f PD = 462 Hz
12.83
20 =
105
⇒ β = 0.04999
1 + (105 ) β
Phase Margin = 60° ⇒ φ = −120°
f
−120 = − tan −1
Assume tan −1
f PD
− tan −1
f
f
− tan −1
6
10
5 × 107
f
= 90°
f PD
f
≅ 30° ⇒ f = 5.77 × 105 Hz
106
(0.04999)(105 )
T = 1=
2
2
2
⎛ 5.77 × 105 ⎞
⎛ 5.77 × 105 ⎞
⎛ 5.77 ×105 ⎞
1+ ⎜
1
1
+
+
⎟
⎜
⎟
⎜
⎟
6
7
f PD
⎝ 10
⎠
⎝ 5 × 10 ⎠
⎝
⎠
tan −1
1=
4.999 × 103
⎛ 5.77 × 105 ⎞
1+ ⎜
⎟ (1.155)(1.0)
f PD
⎝
⎠
5.77 × 105 4.999 ×103
(1.155)(1.0)
f PD
f PD = 133.3 Hz
2
Chapter 13
Exercise Solutions
EX13.1
I C1 = I C 2 ≅ 9.5 µ A
9.5 µ A
= 0.0475 µ A ⇒ I B1 = I B 2 = 47.5 nA
I B1 = I B 2 =
200
EX13.2
5 − 0.6 − 0.6 − (−5)
= 0.22 mA
I REF =
40
I C17 = I C13 B = 0.75I REF = (0.75)(0.22) = 0.165 mA
0.165 (0.165)(0.1) + 0.6
+
I C16 =
200
50
= 0.000825 + 0.01233
I C16 = 13.2 µ A
EX13.3
⎛V ⎞
0.18 ×10−3 = 10−14 exp ⎜ D ⎟
⎝ VT ⎠
⎛ 0.18 × 10−3 ⎞
VD = VT ln ⎜
⎟
−14
⎝ 10
⎠
⎛ 0.18 × 10−3 ⎞
= (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
VD = 0.6140
VBB = 2VDD ≅ 1.228 V
⎛ V /2 ⎞
I C14 = I C 20 = I S exp ⎜ BB ⎟
⎝ VT ⎠
⎛ 0.6140 ⎞
= 3 × 10−14 exp ⎜
⎟
⎝ 0.026 ⎠
I C14 = I C 20 = 0.541 mA
EX13.4
100
ro 6 =
⇒ 10.5 MΩ
0.0095
Then, using results from Example 13.4
Ract1 = ro 6 [1 + g m 6 ( R2 || rπ 6 ) ] = 10.5 [1 + (0.365)(1|| 547) ]
= 14.3 MΩ
V
100
⇒ 10.5 MΩ
ro 4 = A =
I C 4 0.0095
⎛ 9.5 ⎞
Ad = − ⎜
⎟ (10.5 ||14.3 || 4.07 ) = −889
⎝ 0.026 ⎠
EX13.5
100
= 556 K
0.18
Ri 3 = rπ 22 + (1 + β P )[ R19 || R20 ]
= 7.22 + (51)(556 ||111) ⇒ 4.73 MΩ
100
= 185 K
Ract 2 =
0.54
100
= 185 K
Ro17 =
0.54
−(200)(201)(50)(185 || 4730 ||185)
Av 2 =
4070[50 + [9.63 + (201)(0.1)]]
−182358786.9
=
32450.1
Av 2 = −562
R19 = ro13 A =
EX13.6
100
100
ro17 =
= 185 K ro13 B =
= 185 K
0.54
0.54
RC17 = 185 [1 + (20.8)(0.1|| 9.63) ]
RC17 = 566 K
7.22 + 566 ||185
= 2.88 K
Re 22 =
51
100
= 556 K
RC19 =
0.18
0.65 + 2.88 || 556
Re 20 =
= 0.0689
51
= 68.9 Ω
RO = 22 + 68.9 = 90.9 Ω
EX13.7
Ci = C1 (1+ | A2 |) = 30(1 + 562) = 16890 pF
Ri 2 = 4.07 MΩ
Ro1 = Ract1 || ro 4 = 14.3 ||10.5 = 6.05 MΩ
Then Req = Ro1 || Ri 2 = 6.05 || 4.07 = 2.43 MΩ
Then
f PD =
1
1
=
6
2π Req Ci 2π (2.43 × 10 )(16890 × 10−12 )
= 3.88 Hz
EX13.8
For M 5 , M 6
⎛ 40 ⎞
K P = (12.5) ⎜ ⎟ = 250 µ A / V 2
⎝ 2 ⎠
5 + 5 − VSG 5
0.25(VSG 5 − 0.5) 2 =
225
2
56.25(VSG
0.25)
V
−
+
= 10 − VSG 5
SG 5
5
2
56.25VSG
5 − 55.25VSG 5 + 4.0625 = 0
55.25 ± 3052.5625 − 914.0625
2(56.25)
= 0.902 V
VSG 5 =
VSG 5
10 − 0.902
⇒ 40.4 µ A
225
Current in M 1 − M 4 = 20.2 µ A
I set = I Q =
EX13.9
⎛ 40 ⎞
K P1 = K P 2 = (12.5) ⎜ ⎟ = 250 µ A/V 2
⎝ 2 ⎠
From Exercise Ex 13.8, I D1 = I D 2 = 20.2 µ A
ro 4 = ro 2 =
1
1
=
= 2.48 MΩ
λ I D (0.02)(0.0202)
Ad = 2 K P1 I Q (ro 2 || ro 4 ) = 2(0.25)(0.0404)(2480 || 2480)
Ad = 176
⎛ 80 ⎞
K n 7 = (12.5) ⎜ ⎟ = 500 µ A / V 2
⎝ 2⎠
g m 7 = 2 K n 7 I Q = 2 (0.50)(0.0404) = 0.284 mA/V
ro 7 = ro8 =
1
λ ID7
=
1
⇒ 1.24 MΩ
(0.02)(0.0404)
Av 2 = g m 7 (ro 7 || ro8 ) = (0.284)(1240 || 1240) = 176
AV = Ad ⋅ AV 2 = (176)(176) = 30,976
EX13.10
(a)
⎛ k ′ ⎞⎛ W ⎞ ⎛ I Q1 ⎞
g m1 = 2 ⎜ n ⎟ ⎜ ⎟ ⎜
⎟
⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 2 ⎠
⎛ 0.08 ⎞
⎛ 0.25 ⎞
=2 ⎜
⎟ (20) ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= 0.6325 mA/V
1
= 800 K
(0.01)(0.25/ 2)
1
= 533.3 K
ro 4 =
(0.015)(0.25/ 2)
ro 2 =
Ad 1 = g m1 (ro 2 || ro 4 ) = (0.6325)(800 || 533.3)
Ad 1 = 202.4
⎛ k′ ⎞⎛ W ⎞
g m5 = 2 ⎜ n ⎟ ⎜ ⎟ IQ 2
⎝ 2 ⎠ ⎝ L ⎠5
⎛ 0.04 ⎞
=2 ⎜
⎟ (80)(0.25)
⎝ 2 ⎠
= 1.265 mA/V
1
= 266.7 K
(0.015)(0.25)
1
ro9 =
= 400 K
(0.01)(0.25)
ro 5 =
A2 = − g m 5 (ro 5 || ro 9 ) = −(1, 265)(266.7 || 400)
A2 = −202.4
Overall gain (assuming A3 = 1)
A = Ad 1 ⋅ A2 = (202.4)(−202.4) = −40,966
EX13.11
I D1 = I D 2 = 25 µ A
g m1 = g m8 = 2
k ′p ⎛ W
⎜
2 ⎝L
⎛ k′ ⎞⎛ W
gm6 = 2 ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
⎞
⎛ 40 ⎞
⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m1 = g m8 = 224 µ A / V
⎠
⎝ 2 ⎠
⎞
⎛ 80 ⎞
⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m 6 = 316 µ A / V
⎠
⎝ 2⎠
1
1
ro1 = ro 6 = ro8 = ro10 =
=
= 2 MΩ
λ I D (0.02)(25)
ro 4 =
1
λ ID4
=
1
=1 MΩ
(0.02)(50)
Ro8 = g m8 (ro8 ro10 ) = (224)(2)(2) = 896 M Ω
Ro 6 = g m 6 (ro 6 ) ( ro 4 || ro1 ) = 316(2) (1|| 2 ) = 421 M Ω
Then
Ad = g m1 ( Ro 6 Ro8 ) = 224 ( 421 896 ) ⇒ Ad = 64,158
EX13.12
V + − V − = VEB1 + VEB 6 + VBE 7 + I1 R1
= 0.6 + 0.6 + 0.6 + (0.236)(8) = 3.69 V
So
V + = −V − = 1.845 V
EX13.13
Ad = 2 K P I Q 5 ( Ri 2 )
= 2(1)(0.2)(26)
Ad = 16.4
EX13.14
rπ 13 =
β VT
I C13
=
(200)(0.026)
0.20
= 26 kΩ
Ri 2 = rπ 13 + (1 + β ) RE13 = 26 + 201(1)
= 227 kΩ
r010 =
r012 =
g m12 =
rπ 12 =
1
λ I D10
=
1
= 500 kΩ
(0.02)(0.1)
VA
50
=
= 500 kΩ
I C12 0.1
I C12
0.1
=
= 3.85 mA / V
VT
0.026
β VT
I C12
=
(200)(0.026)
= 52 kΩ
0.1
Ract1 = r012 [1 + g m12 (rπ 12 || R5 )]
= 500[1 + (3.85)(52 || 0.5)] = 1453 kΩ
Ad = 2 K n I Q 5 ⋅ ( ro10 || Ract1 || R12 )
= 2(0.6)(0.2) ⋅ (500 ||1453 || 227)
= (0.490)(141) ⇒ Ad = 69.1
EX13.15
From Example 13.15, f PD = 265 Hz
Av = Adi ⋅ A2 = (16.4)(1923) = 31,537
fT = f PD ⋅ Av = (265)(31,537) = 8.36 MHz
TYU13.1 Computer Analysis
TYU13.2 Computer Analysis
TYU13.3
Vi N (max) = V + − VEB (on) = 15 − 0.6 = 14.4 V
Vi N (min) ≅ 4VBE (on) + V +
= 4(0.6) − 15 = −12.6 V
−12.6 ≤ Vi N (cm) ≤ 14.4 V
TYU13.4
a.
V0 (max) ≅ V + − 2VBE (on) = 15 − 2(0.6)
V0 (max) = 13.8 V
V0 (min)
= 3VBE (on) + V − = 3(0.6) − 15
V0 (min)
≅ −13.2 V
−13.2 ≤ V0 ≤ 13.8 V
b.
V0 (max) = 5 − 1.2 = 3.8 V
V0 (min) ≅ 3VBE + V − = 3(0.6) − 5 = −3.2 V
−3.2 ≤ V0 ≤ 3.8 V
TYU13.5
15 − 2(0.6) − (−15)
= 0.72 mA
I REF ≅
40
⎛I ⎞
⎛ 0.72 × 10−3 ⎞
VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
⎝ IS ⎠
= 0.650 V
So
I REF =
30 − 2(0.65)
⇒ I REF = 0.718 mA
40
VBE11 = 0.650 V
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
⎛ 0.718 ⎞
I C10 (5) = (0.026) ln ⎜
⎟
⎝ I C10 ⎠
By trial and error: I C10 = 18.9 µ A
VBE10 = VBE11 − I C10 R4 = 0.650 − (0.0189)(5) ⇒ VBE10 ≅ 0.556 V
I C10 18.9
=
= 9.45 µ A
2
2
⎛I ⎞
⎛ 9.45 × 10−6 ⎞
= VT ln ⎜ C 6 ⎟ = (0.026)ln ⎜
⎟ ⇒ VBE 6 = 0.537 V
−14
⎝ 10
⎠
⎝ IS ⎠
IC 6 ≅
VBE 6
TYU13.6
10 − 0.6 − 0.6 − (−10)
⇒ I REF = 0.47 mA
I REF =
40
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
⎛ 0.47 ⎞
I C10 (5) = (0.026) ln ⎜
⎟
⎝ I C10 ⎠
By trial and error:
⇒ I C10 ≅ 17.2 µ A
IC 6 ≅
I C13 B
I C10
⇒ I C 6 = 8.6 µ A
2
= (0.75) I REF ⇒ I C13 B = 0.353 mA
I C13 A = (0.25) I REF ⇒ I C13 A = 0.118 mA
TYU13.7
R0 = R6 + RE14
RE14 =
rπ 14 + R0 d R013 A
1 + βn
The diode resistance can be found as
⎛V ⎞
I D = I S exp ⎜ D ⎟
⎝ VT ⎠
⎛ 1
1 ∂I D
=
= IS ⎜
rd ∂VD
⎝ VT
or
rd =
⎞
⎛ VD
⎟ ⋅ exp ⎜
⎠
⎝ VT
⎞ ID
⎟=
⎠ VT
VT
V
0.026
= T =
⇒ 144 Ω
0.18
I D I C13 A
RE 22 =
rπ 22 + R017 R013 B
1+ βP
R013 B = r013 B = 92.6 kΩ
R017 = r017 ⎡⎣1 + g m17 ( R8 rπ 17 ) ⎦⎤ = 283 kΩ
From previous calculations
RE 22 = 1.51 kΩ
R0 d = 2rd + RE 22 = 2(0.144) + 1.51 = 1.80 kΩ
R013 A = r013 A = 278 kΩ
βV
(200)(0.026)
= 1.04 kΩ
rπ 14 = n T =
5
I C14
1.04 + 1.8 || 278
⇒ 14.1 Ω
201
R0 = R6 + RE14 = 27 + 14.1 ⇒ R0 ≅ 41 Ω
RE14 =
TYU13.8
For M 6 we have VSG 5 = VSG 6 = 1.06 V
So
VSD 6 (sat) = 1.06 − 0.5 = 0.56 V
For M 1 and M 2
ID =
IQ
= K P (VSG1 + VTP )
2
2
0.0397
= 0.125(VSG1 − 0.5) 2 ⇒ VSG1 = 0.898 V
2
So maximum input voltage
= V + − VSD 6 (sat) − VSG1
= 5 − 0.56 − 0.898
⇒ Vi N (max) = 3.54 V
For M 3,
K p = (6.25)(20) = 125 µ A / V 2
I D3 =
IQ
=
39.7
µA
2
2
39.7
= 125(VGS 3 − VTN ) 2
2
VTN = 0.5V ⇒ VGS 3 = 0.898 V
VSD1 (sat) = 0.898 − 0.5 = 0.398 V
Vi N (min) = V − + VGS 3 + VSD1 (sat) − VSG1
= −5 + 0.898 + 0.398 − 0.898
Vi N (min) = −4.60 V
−4.60 ≤ Vi N (cm) ≤ 3.54 V
TYU13.9
V0 (max) = V + − VSD 8 (sat)
VSG 8 = VSG 5 = 1.06 V
VSD 8 (sat) = 1.06 − 0.5 = 0.56 V
V0 (max) = 5 − 0.56 = 4.44 V
V0 (min) = V − + VDS 7 (sat)
VGS 7 = 1.06 ⇒ VDS 7 (sat) = 1.06 − 0.5 = 0.56
V0 (min) = −5 + 0.56 = −4.44
−4.44 ≤ V0 ≤ 4.44 V
TYU13.10
(a)
For M 5, K p 5 = 125 µ A / V 2
V + − V − − VSG 5
Rset
K p 5 (VSG 5 + VTP ) 2 =
5 + 5 − VSG 5
100
2
12.5(VSG 5 − VSG 5 + 0.25) = 10 − VSG 5
0.125(VSG 5 − 0.5) 2 =
2
12.5VSG
5 − 11.5VSG 5 − 6.875 = 0
11.5 ± (11.5) 2 + 4(12.5)(6.875)
2(12.5)
= 1.33 V
VSG 5 =
VSG 5
Then
10 − 1.33
⇒ 86.7 µ A
100
IQ
= ID4 =
= 43.35 µ A
2
I REF = I Q = I D 8 = I D 7 =
I D1 = I D 2 = I D 3
K p1 = K p 2 = 125µ A / V 2
(b)
ro 2 = ro 4 =
I
λ ID
=
1
= 1153 k Ω
(0.02)(0.04335)
Input stage gain
Ad = 2 K p1 I Q ⋅ (ro 2 ro 4 )
= 2(0.125)(0.0867) ⋅ (1153 1153) ⇒ Ad = 84.9
Transconductance of M 7
g m 7 = 2 K n 7 I D 7 = 2 (0.250)(0.0867)
= 0.294 mA / V
ro 7 = ro8 =
1
λ I D7
=
1
= 577 k Ω
(0.02)(0.0867)
Second stage gain
Av 2 = g m 7 (ro 7 ro8 ) = (0.294)(577 577) ⇒ Av 2 = 84.8
Overall gain = Ad ⋅ Av 2 = (84.9)(84.8) ⇒ A = 7, 200
TYU13.11
(a)
Ad = Bg m1 ( ro 6 ro8 )
⎛ k ′ ⎞⎛ W
g m1 = 2 ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
g m1 = 400 µ A / V
ro 6 =
ro8 =
1
λP I D 6
=
⎞
⎛ 80 ⎞
⎟ I D1 = 2 ⎜ ⎟ (20)(50)
⎠
⎝ 2⎠
1
= 0.333 M Ω
(0.02)(150)
1
1
=
= 0.333 M Ω
λn I D 8 (0.02)(150)
Ad = 3(400) ( 0.333 || 0.333) ⇒ Ad = 200
(b)
f PD =
1
2π Ro (CL + CP )
where Ro = ro 6 || ro8 = 0.333 || 0.333 M Ω
f PD =
1
⇒ f PD = 477 kHz
2π ( 0.333 || 0.333) × 106 × 2 × 10−12
f PD ⋅ Ad = (477 × 103 )(200) ⇒ 95.5 MHz
TYU13.12
(a)
From Exercise TYU 13.11, g m1 = 400 µ A / V
ro 6 = ro8 = ro10 = ro12 = 0.333 M Ω
⎛ k′ ⎞⎛ W
g m10 = 2 ⎜ P ⎟ ⎜
⎝ 2 ⎠⎝ L
⎞
⎛ 40 ⎞
⎟ I D10 = 2 ⎜ ⎟ (20)(150) ⇒ g m10 = 490 µ A / V
⎠
⎝ 2 ⎠
⎛ k ′ ⎞⎛ W ⎞
⎛ 80 ⎞
g m12 = 2 ⎜ n ⎟ ⎜ ⎟ I D12 = 2 ⎜ ⎟ (20)(150) ⇒ g m12 = 693 µ A / V
⎝ 2⎠
⎝ 2 ⎠⎝ L ⎠
Ro10 = g m10 (ro10 ro 6 ) = (490)(0.333)(0.333) = 54.4 M Ω
Ro12 = g m12 (ro12 ro8 ) = (693)(0.333)(0.333) = 77.0 M Ω
Ad = Bg m1 ( Ro10 || Ro12 ) = 3(400)(54.4 || 77.0) ⇒ Ad = 38, 254
(b)
Ro = Ro10 || Ro12 = 54.4 || 77.0 = 31.9 M Ω
1
= 2.50 kHz
2π (31.9 × 106 )(2 × 10−12 )
f PD ⋅ Ad = (2.5 × 103 )(38, 254) ⇒ 95.6 MHz
f PD =
TYU13.13
(a)
Ad = g m1 ( Ro 6 || Ro8 )
From Example 13.10,
g m1 = 316 µ A / V , Ro8 = 316 M Ω
Now
Ro 6 = g m 6 (ro 6 )(ro 4 || ro1 )
ro1 = 1 M Ω, ro 4 = 0.5 M Ω
I
50
gm6 = C 6 =
⇒ 1.923 mA / V
VT
0.026
ro 6 =
VA6 80
=
= 1.6 M Ω
I C 6 50
Then
Ro 6 = (1.923)(1600)(0.5 || 1) = 1026 M Ω
Ad = (316)(1026 || 316) ⇒ Ad = 76,343
(b)
1
⇒ f PD = 329 Hz
2π (316 || 1026) × 106 × 2 × 10−12
f PD ⋅ Ad = (329)(76,343) ⇒ 25.1 MHz
f PD =
TYU13.14
For Q7 and R1
VSG = VBE 7 + I1 R1 = 0.6 + I1 (5)
For M 8 :
I 2 = K p (VSG + VTP ) 2
I 2 = 0.3(VSG − 1.4) 2
By trial and error:
VSG = 2.54 V
I1 = I 2 = 0.388 mA
TYU13.15
For J 6 biased in the saturation region
⇒ I C 3 = I DSS = 300 µ A
Q1 , Q2 , Q3 are matched
⇒ I C1 = I C 2 = I C 3 = 300 µ A
Chapter 13
Problem Solutions
13.1
Computer Simulation
13.2
Computer Simulation
13.3
(
(a)
Ad = g m1 ro 2 ro 4 Ri 6
)
g m1 =
I C1
20
=
⇒ 0.769 mA / V
VT 0.026
ro 2 =
VA 2 80
=
= 4 MΩ
I C 2 20
ro 4 =
VA 4 80
=
= 4 MΩ
I C 2 20
rπ 6 =
(120)(0.026)
= 104 k Ω
0.030
Ri 6 = rπ 6 + (1 + β n ) ⎡⎣ R1 rπ 7 ⎤⎦
(120)(0.026)
= 15.6 k Ω
rπ 7 =
0.2
V (on) 0.6
=
= 0.030 mA
I C 6 ≅ BE
20
R1
Then
Ri 6 = 104 + (121) ⎡⎣ 20 15.6 ⎤⎦ ⇒ 1.16 M Ω
Then
Ad = 769 4 4 1.16 ⇒ Ad = 565
(
)
Now
⎛ R1
Vo = − I c 7 ro 7 = −( β n I b 7 )ro 7 = − β n ro 7 ⎜
⎝ R1 + rπ 7
⎛ R1 ⎞
Vo1
= − β n (1 + β n )ro 7 ⎜
⎟ I b 6 and I b 6 =
+
R
r
Ri 6
⎝ 1 π7 ⎠
Then
V
− β n (1 + β n )ro 7 ⎛ R1 ⎞
Av 2 = o =
⎜
⎟
Vo1
Ri 6
⎝ R1 + rπ 7 ⎠
ro 7 =
⎞
⎟ Ic6
⎠
VA
80
=
= 400 k Ω
I C 7 0.2
So
Av 2 =
−(120)(121)(400) ⎛ 20 ⎞
⎜
⎟ ⇒ Av 2 = −2813
1160
⎝ 20 + 15.6 ⎠
Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 ×106
(b)
Rid = 2rπ 1 and rπ 1 =
(80)(0.026)
= 104 k Ω
0.020
Rid = 208 k Ω
(c)
f PD =
1
and CM = (10)(1 + 2813) = 28,140 pF
2π Req CM
Req = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M Ω
f PD =
1
= 7.71 Hz
2π (0.734 × 10 )(28,140 × 10−12 )
6
Gain-Bandwidth Product = (7.71)(1.59 × 106 ) ⇒ 12.3 MHz
13.4
a.
b.
Q3 acts as the protection device.
Same as part (a).
13.5
If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5
So breakdown voltage ≈ 56.4 V.
13.6
(a)
I REF =
15 − 0.6 − 0.6 − (−15)
= 0.50 ⇒ R5 = 57.6 k Ω
R5
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
0.026 ⎛ 0.50 ⎞
R4 =
ln ⎜
⎟ ⇒ R4 = 2.44 k Ω
0.030 ⎝ 0.030 ⎠
5 − 0.6 − 0.6 − (−5)
⇒ I REF = 0.153 mA
57.6
⎛ 0.153 ⎞
I C10 (2.44) = (0.026) ln ⎜
⎟
⎝ I C10 ⎠
(b)
I REF =
By trial and error, I C10 ≅ 21.1 µ A
13.7
(a)
I REF ≅ 0.50 mA
(b)
From Problem 13.6, I REF ≅ 0.15 mA
⎛I ⎞
⎛ 0.50 × 10−3 ⎞
VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜
⎟ ⇒ VBE11 = 0.641V = VEB12
−14
⎝ 10
⎠
⎝ IS ⎠
Then
15 − 0.641 − 0.641 − (−15)
⇒ R5 = 57.4 k Ω
R5 =
0.50
0.026 ⎛ 0.50 ⎞
R4 =
ln ⎜
⎟ ⇒ R4 = 2.44 k Ω
0.030 ⎝ 0.030 ⎠
⎛ 0.030 × 10−3 ⎞
VBE10 = 0.026 ln ⎜
⎟ ⇒ VBE10 = 0.567 V
−14
⎝ 10
⎠
⎛ 0.15 × 10−3 ⎞
VBE11 = VEB12 = 0.026 ln ⎜
⎟ = 0.609 V
−14
⎝ 10
⎠
5 − 0.609 − 0.609 − (−5)
⇒ I REF = 0.153 mA
Then I REF =
57.4
Then I C10 ≅ 21.1 µ A from Problem 13.6
13.8
5 − 0.6 − 0.6 − (−5)
⇒ I REF = 0.22 mA
40
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
a.
I REF =
⎛ 0.22 ⎞
I C10 (5) = (0.026) ln ⎜
⎟
⎝ I C10 ⎠
By trial and error;
I C10 ≅ 14.2 µ A
I C10
⇒ I C 6 = 7.10 µ A
2
= 0.75 I REF ⇒ I C17 = 0.165 mA
IC 6 ≅
I C17
I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA
(b)
Using Example 13.4
rπ 17 = 31.5 kΩ
RE′ = 50 [31.5 + (201)(0.1)] = 50 51.6 = 25.4 kΩ
rπ 16 =
β nVT
I C16
and
0.165 (0.165)(0.1) + 0.6
+
= 0.0132 mA
200
50
rπ 16 = 394 kΩ
Then
Ri 2 = 394 + (201)(25.4) ⇒ 5.5 MΩ
rπ 6 = 732 kΩ
0.00710
= 0.273 mA / V
gm6 =
0.026
50
= 7.04 MΩ
r06 =
0.0071
Then
Ract1 = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ
50
r04 =
= 7.04 MΩ
0.0071
Then
⎛ 7.1 ⎞
Ad = − ⎜
⎟ (7.04 8.96 5.5)
⎝ 0.026 ⎠
or
Ad = −627 Gain of differential amp stage
I C16 =
Using Example 13.5, and neglecting the input resistance to the output stage:
V
50
= 303 kΩ
Ract 2 = A =
I C13 B 0.165
Av 2 =
−(200)(201)(50)(303) (303)
(5500)[50 + 31.5 + (201)(0.1)]
or
Av 2 = −545 Gain of second stage
13.9
I C10 = 19 µ A
From Equation (13.6)
⎡ β 2 + 2β P + 2 ⎤
⎡ (10) 2 + 2(10) + 2 ⎤
=
I C10 = 2 I ⎢ P2
I
2
⎥
⎢
⎥
2
⎣ (10) + 3(10) + 2 ⎦
⎣ β P + 3β P + 2 ⎦
So
⎡122 ⎤
= 2I ⎢
⎥
⎣132 ⎦
⎛ 132 ⎞
2 I = (19) ⎜
⎟ = 20.56 µ A
⎝ 122 ⎠
I C 2 = I = 10.28 µ A
I B9
2I
20.56
⇒ I C 9 = 17.13 µ A
2⎞
⎛
1
+
⎜
⎟
⎝ 10 ⎠
⎛
2 ⎞
⎜1 +
⎟
β
⎝
P ⎠
I
17.13
= C9 =
⇒ I B 9 = 1.713 µ A
10
βP
IC 9 =
IB4 =
=
10.28
I
=
⇒ I B 4 = 0.9345 µ A
(1 + β P )
11
⎛ β
IC 4 = I ⎜ P
⎝1+ βP
⎞
⎛ 10 ⎞
⎟ = (10.28) ⎜ ⎟ ⇒ I C 4 = 9.345 µ A
11 ⎠
⎝
⎠
13.10
VB 5 − V − = VBE (on) + I C 5 (1)
= 0.6 + (0.0095)(1) = 0.6095
0.6095
⇒ I C 7 = 12.2 µ A
IC 7 =
50
I C 8 = I C 9 = 19 µ A
I REF = 0.72 mA
I E13 = I REF = 0.72 mA
I C14 = 138 µ A
Power = (V + − V − ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ]
= 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138]
⇒ Power = 48.8 mW
Current supplied by V + and V − = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14
= 1.63 mA
13.11
(a)
vcm (min) = −15 + 0.6 + 0.6 + 0.6 + 0.6 = −12.6 V
vcm (max) = +15 − .6 = 14.4 V
So − 12.6 ≤ vcm ≤ 14.4 V
(b)
vcm (min) = −5 + 4(0.6) = −2.6 V
vcm (max) = 5 − 0.6 = 4.4 V
So − 2.6 ≤ vcm ≤ 4.4 V
13.12
If v0 = V − = −15 V , the base voltage of Q14 is pulled low, and Q18 and Q19 are effectively cut off. As
a first approximation
0.6
I C14 =
= 22.2 mA
0.027
22.2
I B14 =
= 0.111 mA
200
Then
I C15 = I C13 A − I B14 = 0.18 − 0.111 = 0.069 mA
Now
⎛I ⎞
VBE15 = VT ln ⎜ C15 ⎟
⎝ 15 ⎠
⎛ 0.069 × 10−3 ⎞
= (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
= 0.589 V
As a second approximation
0.589
I C14 =
⇒ I C14 = 21.8 mA
0.027
21.8
I B14 =
= 0.109 mA
200
and
I C15 = 0.18 − 0.109 ⇒ I C15 = 0.071 mA
13.13
a.
Neglecting base currents:
I D = I BIAS
Then
⎛I ⎞
VBB = 2VD = 2VT ln ⎜ D ⎟
⎝ IS ⎠
⎛ 0.25 × 10−3 ⎞
= 2(0.026) ln ⎜
−14 ⎟
⎝ 2 × 10
⎠
or
VBB = 1.2089 V
⎛V / 2⎞
I CN = I CP = I S exp ⎜ BB ⎟
⎝ VT ⎠
⎛ 1.2089 ⎞
= 5 × 10−14 exp ⎜
⎟
⎝ 2(0.026) ⎠
So
I CN = I CP = 0.625 mA
b.
For vI = 5 V, v0 ≅ 5 V
5
= 1.25 mA
4
As a first approximation
I CN ≈ iL = 1.25 mA
iL ≅
⎛ 1.25 × 10−3 ⎞
VBEN = (0.026) ln ⎜
= 0.6225 V
−14 ⎟
⎝ 5 × 10
⎠
Neglecting base currents,
VBB = 1.2089 V
Then VEBP = 1.2089 − 0.6225 = 0.5864 V
⎛ 0.5864 ⎞
I CP = 5 × 10−14 exp ⎜
⎟ ⇒ I CP = 0.312 mA
⎝ 0.026 ⎠
As a second approximation,
I CN = iL + I CP = 1.25 + 0.31 ⇒ I CN ≅ 1.56 mA
13.14
R1 + R2 =
VBB
1.157
=
= 64.28 kΩ
(0.1) I BIAS 0.018
⎛I
VBE = VT ln ⎜ C
⎝ IS
⎞
⎛ (0.9) I BIAS ⎞
⎟ = (0.026) ln ⎜
⎟
IS
⎠
⎝
⎠
⎛ 0.162 × 10−3 ⎞
= (0.026) ln ⎜
⎟
−14
⎝ 10
⎠
VBE = 0.6112 V
⎛ R2 ⎞
VBE = ⎜
⎟ VBB
⎝ R1 + R2 ⎠
⎛ R ⎞
0.6112 = ⎜ 2 ⎟ (1.157)
⎝ 64.28 ⎠
So
R2 = 33.96 kΩ
Then
R1 = 30.32 kΩ
13.15
(
(a)
Ad = − g m ro 4 ro 6 Ri 2
)
From example 13.4
9.5
gm =
= 365 µ A / V , ro 4 = 5.26 M Ω
0.026
Now
ro 6 = ro 4 = 5.26 M Ω
Assuming R8 = 0, we find
Ri 2 = rπ 16 + (1 + β n ) RE′
= 329 + (201) ( 50 9.63) ⇒ 1.95 M Ω
Then
Ad = −(365) 5.26 5.26 1.95 ⇒ Ad = −409
(
(b)
Av 2 =
)
From Equation (13.20),
(
− β n (1 + β n ) R9 Ract 2 Ri 3 R017
Ri 2 R9 + ⎡⎣ rπ 17 + (1 + β n ) Rg ⎤⎦
{
)
}
For Rg = 0, Ri 2 = 1.95 M Ω
Using the results of Example 13.5
Av 2 =
(
−200(201)(50) 92.6 4050 92.6
(1950){50 + 9.63}
)⇒A
v2
= −792
13.16
Let I C10 = 40 µ A, then I C1 = I C 2 = 20 µ A. Using Example 13.5,
Ri 2 = 4.07 MΩ
(200)(0.026)
= 260 kΩ
0.020
0.020
= 0.769 mA/V
gm6 =
0.026
50
⇒ 2.5 MΩ
r06 =
0.02
Then
Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ
50
r06 =
⇒ 2.5 MΩ
0.02
Then
rπ 6 =
⎛ I CQ ⎞
Ad = − ⎜
⎟ (r04 Ract1 Ri 2 )
⎝ VT ⎠
⎛ 20 ⎞
= −⎜
⎟ (2.5 4.42 4.07)
⎝ 0.026 ⎠
So
Ad = −882
13.17
From Problem 13.8
I1 = I 2 = 7.10 µ A, I C17 = 0.165 mA, I C13 A = 0.055 mA
I C16 ≈ I B17 +
I E17 R8 + VBE17
R9
=
0.165 (0.165)(0.1) + 0.6
+
200
50
= 0.000825 + 0.01233
I C16 = 0.0132 mA
(200)(0.026)
= 31.5 K
0.165
RE1 = R9 [ rπ 17 + (1 + β ) R8 ] = 50 [31.5 + (201)(0.1)]
rπ 17 =
= 50 51.6 = 25.4 K
rπ 16 =
(200)(0.026)
= 394 K
0.0132
Then
Ri 2 = rπ 16 + (1 + β ) RE1 = 394 + (201)(25.4) ⇒ 5.50 MΩ
Now
(200)(0.026)
rπ 6 =
= 732 K
0.0071
0.0071
gm6 =
= 0.273 mA/V
0.026
50
ro 6 =
⇒ 7.04 MΩ
0.0071
Ract1 = ro 6 [1 + g m 6 ( R rπ 6 )]
= 7.04[1 + (0.273)(1 732)] = 8.96 MΩ
50
ro 4 =
⇒ 7.04 MΩ
0.0071
Then
Ad = − g m1 (ro 4 Ract1 Ri 2 )
⎛ 7.10 ⎞
= −⎜
⎟ (7.04 8.96 5.5)
⎝ 0.026 ⎠
Ad = −627
50
50
⇒ 303 K Ro17 =
= 303 K
0.165
0.165
From Eq. (13.20), assuming Ri 3 → ∞
Now Ract 2 =
Av 2 ≅ −
=
β (1 + β ) R9 ( Ract 2 R017 )
Ri 2 { R9 + [rπ 17 + (1 + β ) R8 ]}
−(200)(201)(50)(303 303)
(5500)[50 + 31.5 + (201)(0.1)]
Av 2 = −545
=
−3.045 × 108
5.588 × 105
Overall gain Av = (−627)(−545) = 341, 715
13.18
Using results from 13.17
⎛ 100 ⎞
Ri 2 = 5.50 MΩ, Ract1 ⎜
⎟ [1 + (0.273)(1 732)] ⇒ 17.93 MΩ
⎝ 0.0071 ⎠
100
ro 4 =
⇒ 14.08 MΩ
0.0071
⎛ 7.10 ⎞
Ad = − ⎜
⎟ (14.08 17.93 5.50)
⎝ 0.026 ⎠
Ad = −885
Now
100
100
Ract 2 =
= 606 K Ro17 =
= 606 K
0.165
0.165
−(200)(201)(50)(606 606)
−6.09 × 108
=
Av 2 =
(5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105
Av 2 = −1090
Overall gain
Av = (−885)(−1090) = 964, 650
13.19
Now
Re14 =
rπ 14 + R01
and R0 = R6 + Re14
1+ βP
Assume series resistance of Q18 and Q19 is small. Then
R01 = r013 A Re 22
where Re 22 =
rπ 22 + R017 r013 B
1+ βP
and R017 = r017 [1 + g m17 ( R8 rπ 17 )]
Using results from Example 13.6,
rπ 17 = 9.63 kΩ
rπ 22 = 7.22 kΩ
g m17 = 20.8 mA/V r017 = 92.6 kΩ
Then
R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 kΩ
50
r013 B =
= 92.6 kΩ
0.54
Then
7.22 + 283 92.6
= 1.51 kΩ
Re 22 =
51
R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ
(50)(0.026)
= 0.65 kΩ
rπ 14 =
2
Then
0.65 + 1.50
Re14 =
= 0.0422 kΩ
51
or
Re14 = 42.2 Ω
Then
R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω
13.20
⎡
⎛ r
Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3
⎝ 1+ βP
⎣
β n = 200, β P = 10
(a)
I C1 = 9.5 µ A
⎞⎤
⎟⎥
⎠⎦
(200)(0.026)
= 547 K
0.0095
(10)(0.026)
rπ 3 =
= 27.4 K
0.0095
Then
(201)(27.4) ⎤
⎡
Rid = 2 ⎢547 +
⎥
11
⎣
⎦
Rid ⇒ 2.095 MΩ
rπ 1 =
(b)
I C1 = 7.10 µ A
(200)(0.026)
= 732 K
rπ 1 =
0.0071
(10)(0.026)
= 36.6 K
rπ 3 =
0.0071
(201)(36.6) ⎤
⎡
Rid = 2 ⎢ 732 +
⎥
11
⎣
⎦
Rid ⇒ 2.80 MΩ
13.21
We can write
A0
⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟⎜ 1 + j ⎟
f PD ⎠⎝
f1 ⎠
⎝
181, 260
=
f ⎞⎛
f ⎞
⎛
⎜1 + j
⎟ ⎜1 + j ⎟
10.7 ⎠ ⎝
f1 ⎠
⎝
Phase:
⎛ f ⎞
−1 ⎛ f ⎞
φ = − tan −1 ⎜
⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠
⎝ f1 ⎠
A( f ) =
For a Phase margin = 70°, φ = −110°
So
⎛ f ⎞
−1 ⎛ f ⎞
−110° = − tan −1 ⎜
⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠
⎝ f1 ⎠
Assuming f 10.7, we have
⎛ f ⎞
f
tan −1 ⎜ ⎟ = 20° ⇒ = 0.364
f1
⎝ f1 ⎠
At this frequency, A( f ) = 1, so
1=
181, 260
⎛ f ⎞
2
1+ ⎜
⎟ ⋅ 1 + (0.364)
⎝ 10.7 ⎠
170,327
2
⎛ f ⎞
1+ ⎜
⎟
⎝ 10.7 ⎠
f
= 170,327 ⇒ f = 1.82 MHz
or
10.7
Then, second pole at
f
f1 =
⇒ f1 = 5 MHz
0.364
=
13.22
a.
2
Original g m1 and g m 2
⎛ W ⎞⎛ µ C ⎞
K p1 = K p 2 = ⎜ ⎟⎜ P ox ⎟ = (12.5)(10)
⎝ L ⎠⎝ 2 ⎠
= 125 µ A / V 2
So
⎛ IQ ⎞
g m1 = g m 2 = 2 K p1 ⎜ ⎟
⎝ 2⎠
= 2 (0.125)(10)
= 0.09975 mA/V
⎛W ⎞
If ⎜ ⎟ is increased to 50, then
⎝L⎠
K p1 = K p 2 = (50)(10) = 500 µ A / V 2
So
g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V
b.
Gain of first stage
Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025)
or
Ad = 501
Voltage gain of second stage remains the same, or
Av 2 = 251
Then Av = Ad ⋅ Av 2 = (501)(251)
or
Ad = 125, 751
13.24
a.
K p = (10)(20) = 200 µ A / V 2 = 0.2 mA / V 2
I REF = I SET
10 − VSG − (−10)
200
= k P (VSG − 1.5) 2
=
2
20 − VSG = (0.2)(200)(VSG
− 3VSG + 2.25)
2
40VSG
− 119VSG + 70 = 0
VSG =
Then
119 ± (119) 2 − 4(40)(70)
⇒ VSG = 2.17 V
2(40)
20 − 2.17
⇒ I REF = 89.2 µ A
200
M 5 , M 6 , M 8 matched transistors so that
I REF =
I Q = I D 7 = I REF = 89.2 µ A
b.
Small-signal voltage gain of input stage:
Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 )
1
= 1.12 MΩ
⎛ 89.2 ⎞
(0.02) ⎜
⎟
⎝ 2 ⎠
1
1
r04 =
=
= 2.24 MΩ
λn I D
⎛ 89.2 ⎞
(0.01) ⎜
⎟
⎝ 2 ⎠
Then
Ad = 2(200)(89.2) ⋅ (1.12 2.24)
r02 =
1
=
λP I D
or
Ad = 141
Small-signal voltage gain of second stage:
Av 2 = g m 7 (r07 r08 )
K n 7 = (20)(20) = 400 µ A / V 2
So
g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.0892) = 0.378 mA/V
1
1
r08 =
=
= 561 kΩ
λP I D 7 (0.02)(0.0892)
r07 =
1
λn I D 7
=
1
= 1121 kΩ
(0.01)(0.0892)
So
Av 2 = (0.378)(1121 561) ⇒ Av 2 = 141
Then overall voltage gain
Av = Ad ⋅ Av 2 = (141)(141) ⇒ Av = 19,881
13.25
Small-signal voltage gain of input stage:
Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 )
K p1 = (10)(10) = 100 µ A / V 2
1
1
=
= 1000 kΩ
I
⎛ ⎞
⎛ 0.2 ⎞
λP ⎜ Q ⎟ (0.01) ⎜
⎟
⎝ 2 ⎠
⎝ 2⎠
1
1
r04 =
=
= 2000 kΩ
⎛ IQ ⎞
⎛ 0.2 ⎞
(0.005)
λn ⎜ ⎟
⎜
⎟
⎝ 2 ⎠
⎝ 2⎠
r02 =
Then
Ad = 2(0.1)(0.2) ⋅ (1000 2000)
or
Ad = 133
Small-signal voltage gain of second stage:
Av 2 = g m 7 ( r07 r08 )
K n 7 = (20)(20) = 400 µ A / V 2
So
g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.2) = 0.566 mA/V
1
1
=
= 500 kΩ
r08 =
λP I D 7 (0.01)(0.2)
r07 =
1
λn I D 7
=
1
= 1000 kΩ
(0.005)(0.2)
So
Av 2 = (0.566)(1000 500) ⇒ Av 2 = 189
Then overall voltage gain is
Av = Ad ⋅ Av 2 = (133)(189) ⇒ Av = 25,137
13.26
f PD =
1
2π Req Ci
where Req = r04 r02 and Ci = C1 (1 + Av 2 )
We can find that
Av 2 = 251 and r04 = r02 = 5.025 MΩ
Now
Req = 5.025 5.025 = 2.51 MΩ
and
Ci = 12(1 + 251) = 3024 pF
So
1
f PD =
2π (2.51× 106 )(3024 × 10−12 )
or
f PD = 21.0 Hz
13.27
f PD =
1
2π Req Ci
where Req = r04 r02
From Problem 13.22,
r02 = 1.12 MΩ, r04 = 2.24 MΩ and Av 2 = 141
So
1
8=
2π (1.12 2.24) × 106 × Ci
or
Ci = 2.66 × 10−8 = C1 (1 + Av 2 ) = C1 (142)
or
C1 = 188 pF
13.28
R0 = r07 r08
We can find that
r07 = r08 = 2.52 MΩ
Then
R0 = 2.52 2.52
or
R0 = 1.26 MΩ
13.29
a.
V0 = ( g m1Vgs1 )(r01 r02 )
VI = Vgs1 + V0
Then V0 = g m1 (r01 r02 )(VI − V0 )
or
Av =
g m1 (r01 r02 )
1 + g m1 (r01 r02 )
b.
I X + g m1Vgs1 =
VX VX
and Vgs1 = −VX
+
r02 r01
R0 =
13.30
(a)
(b)
1
r r
g m1 01 02
2
⎛ 80 ⎞
I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ]
⎝ 2⎠
I Q 2 = 180 µ A
2
⎛ 80 ⎞
I D 6 = ⎜ ⎟ (25) (VGS 6 − 0.7 ) = 25 ⇒ VGS 6 = 0.8581 V
⎝ 2⎠
2
⎛ 40 ⎞
I D 7 = ⎜ ⎟ (50) (VSG 7 − 0.7 ) = 25 ⇒ VSG 7 = 0.8581 V
2
⎝ ⎠
Set
VSG 8 P = VGS 8 N = 0.8581 V
⎛ 40 ⎞ ⎛ W ⎞
⎛W ⎞
180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 360
⎝ 2 ⎠ ⎝ L ⎠8 P
⎝ L ⎠8 P
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180
⎝ 2 ⎠ ⎝ L ⎠8 N
⎝ L ⎠8 N
13.31
VGS11 ⇒
2
⎛ 80 ⎞
200 = ⎜ ⎟ (20) (VGS 11 − 0.7 )
⎝ 2⎠
VGS 11 = 1.20 V
Let M 12 = 2 transistors in series. Than
5 − 1.20
= 1.90 V
2
2
⎛ 80 ⎞⎛ W ⎞
⎛W ⎞
⎛W ⎞
200 = ⎜ ⎟⎜ ⎟ (1.90 − 0.7 ) ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.47
L
L
2
⎝ ⎠⎝ ⎠12
⎝ ⎠12 A ⎝ L ⎠12 B
VGS12 =
13.32
(a)
2
⎛ 80 ⎞
I Q 2 = 250µ A = ⎜ ⎟ (5) (VGS 8 − 0.7 )
⎝ 2⎠
⇒ VGS 8 = 1.818 V
1.818
⇒ VGS 6 = VSG 7 =
= 0.909 V
2
⎛ 80 ⎞
I D 6 = I D 7 = ⎜ ⎟ (25)(0.909 − 0.7) 2 = 43.7 µ A
⎝ 2⎠
(b)
⎛ 80 ⎞
⎛ 250 ⎞
g m1 = 2 ⎜ ⎟ (15) ⎜
⎟ ⇒ 0.5477 mA/V
⎝ 2⎠
⎝ 2 ⎠
1
= 800 K
ro 2 =
( 0.01)( 0.125)
r04 =
1
= 533.3K
( 0.015)( 0.125)
Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800
Ad 1 = 175
Second stage:
533.3)
A2 = − g m 5 (ro 5 ro 9 )
⎛ 40 ⎞
g m 5 = 2 ⎜ ⎟ (80)(250) ⇒ 1.265 mA/V
⎝ 2 ⎠
1
r05 =
= 266.7 K
(0.015)(0.25)
1
r09 =
= 400 K
(0.01)(0.25)
A2 = −(1.265)(266.7 400)
A2 = −202
Assume the gain of the output stage ≈ 1, then
Av = Ad 1 ⋅ A2 = (175)(−202)
Av = −35,350
13.33
(a)
Ad = g m1 ( Ro 6 Ro8 )
g m1 = 2 K n I DQ = 2 (0.5)(0.025) ⇒ 224 µ A / V
g m1 = g m8
g m 6 = 2 (0.5)(0.025) ⇒ 224 µ A / V
ro1 = ro 6 = ro8 = ro10 =
ro 4 =
1
1
=
= 2.67 M Ω
λ I DQ (0.015)(25)
1
1
=
⇒ 1.33 M Ω
λ I D 4 ( 0.015 )( 50 )
Now
Ro8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ω
Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) ⇒ Ro 6 = 531 M Ω
Then
Ad = (224)(531 1597) ⇒ Ad = 89, 264
(b)
Ro = Ro 6 Ro8 = 531 1597 ⇒ Ro = 398 M Ω
(c)
f PD =
1
1
=
⇒ f PD = 80 Hz
2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 )
GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz
13.34
(a)
ro1 = ro8 = ro10 =
ro 6 =
ro 4 =
1
λn I D
=
1
1
=
= 2 MΩ
λ p I D (0.02)(25)
1
= 2.67 M Ω
(0.015)(25)
1
1
=
= 1.33 M Ω
λn I D 4 (0.015)(50)
⎛ 35 ⎞ ⎛ W ⎞
⎛W ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ (25) = 41.8 ⎜ ⎟ = g m8
⎝ 2 ⎠ ⎝ L ⎠1
⎝ L ⎠1
⎛ 80 ⎞⎛ W ⎞
⎛W ⎞
g m 6 = 2 ⎜ ⎟⎜ ⎟ (25) = 63.2 ⎜ ⎟
⎝ 2 ⎠⎝ L ⎠6
⎝ L ⎠6
Ro = Ro 6 Ro8 = [ g m 6 (ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )]
⎛W ⎞
⎛W ⎞
Define X 1 = ⎜ ⎟ and X 6 = ⎜ ⎟
L
⎝ ⎠1
⎝ L ⎠6
Then
Ro = ⎣⎡ 63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦⎤ ⎡⎣ 41.8 X 1 ( 2 )( 2 ) ⎤⎦
22,539 X 1 X 6
= 134.8 X 6 167.2 X 1 =
134.8 X 6 + 167.2 X 1
Ad = g m1 Ro
⎛ 22,539 X 1 X 6
⎞
= (41.8 X 1 ) ⎜
⎟
134.8
167.2
+
X
X
6
1 ⎠
⎝
= 10, 000
1 ⎛W ⎞
⎛W ⎞
Now X 6 = ⎜ ⎟ =
⎜ ⎟ = 0.674 X 1
L
2.2
⎝ ⎠6
⎝ L ⎠1
We then find
⎛W ⎞
⎛W ⎞
X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟
⎝ L ⎠1
⎝ L ⎠p
and
⎛W ⎞
⎜ ⎟ = 1.85
⎝ L ⎠n
13.35
Let V + = 5V , V − = −5V
P = IT (10) = 3 ⇒ IT = 0.3 mA ⇒ I REF = 0.1 mA = 100 µ A
1
ro1 = ro8 = ro10 =
= 1 MΩ
(0.02)(50)
1
ro 6 =
= 1.33 MΩ
(0.015)(50)
1
ro 4 =
= 0.667 M Ω
(0.015)(100)
⎛ 35 ⎞ ⎛ W ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 59.2 X 1 = g m8
⎝ 2 ⎠ ⎝ L ⎠1
⎛W ⎞
where X 1 = ⎜ ⎟
⎝ L ⎠1
Assume all width-to-length ratios are the same.
⎛ 80 ⎞ ⎛ W ⎞
g m 6 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 89.4 X 1
⎝ 2 ⎠⎝ L ⎠
Now
Ro = Ro 6 Ro8 = ⎣⎡ g m 6 ( ro 6 ) ( ro 4 ro1 ) ⎦⎤ ⎡⎣ g m8 ( ro8 ro10 ) ⎤⎦
= ⎣⎡89.4 X 1 (1.33) ( 0.667 1) ⎦⎤ ⎡⎣59.2 X 1 (1)(1) ⎤⎦
( 47.6 X 1 )( 59.2 X 1 )
= [ 47.6 X 1 ] [59.2 X 1 ] =
47.6 X 1 + 59.2 X 1
So Ro = 26.4 X 1
Now
Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000
So that X 12 =
13.36
W
= 16 for all transistors
L
(a)
Ad = Bg m1 (ro 6 ro8 )
ro 6 = ro8 =
1
1
=
= 0.741 M Ω
λ I DQ (0.015)(90)
⎛ k ′ ⎞⎛ W ⎞
g m1 = 2 ⎜ n ⎟ ⎜ ⎟ I D1 = 2 (500)(30) = 245 µ A / V
⎝ 2 ⎠⎝ L ⎠
Ad = (3)(245)(0.741 0.741) ⇒ Ad = 272
(b)
Ro = ro 6 ro8 = 0.741 0.741 ⇒ Ro = 371 k Ω
(c)
f PD =
1
1
=
⇒ f PD = 85.8 kHz
2π Ro C 2π (371× 103 )(5 × 10−12 )
GBW = (272)(85.8 × 103 ) ⇒ GBW = 23.3 MHz
13.37
(a)
1
= 0.5 M Ω
(0.02)(2.5)(40)
1
= 0.667 M Ω
ro8 =
(0.015)(2.5)(40)
Ad = Bg m1 ( ro 6 ro8 )
ro 6 =
400 = (2.5) g m1 ( 0.5 0.667 ) ⇒ g m1 = 560 µ A / V
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
g m1 = 560 = 2 ⎜ ⎟ ⎜ ⎟ (40) ⇒ ⎜ ⎟ = 49
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
Assume all (W/L) ratios are the same except for
⎛W ⎞ ⎛W ⎞
M 5 and M 6 . ⎜ ⎟ = ⎜ ⎟ = 122.5
⎝ L ⎠5 ⎝ L ⎠ 6
(b)
Assume the bias voltages are
V + = 5V , V − = −5V .
⎛W ⎞
⎛W ⎞
Assume ⎜ ⎟ = ⎜ ⎟ = 49
⎝ L ⎠ A ⎝ L ⎠B
⎛ 80 ⎞
I Q = ⎜ ⎟ (49)(VGSA − 0.5) 2 = 80 ⇒ VGSA = 0.702 V
⎝ 2⎠
Then
⎛ 80 ⎞ ⎛ W ⎞
I REF = 80 = ⎜ ⎟ ⎜ ⎟ (VGSC − 0.5) 2
⎝ 2 ⎠ ⎝ L ⎠C
For four transistors
10 − 0.702
= 2.325 V
4
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
80 = ⎜ ⎟ ⎜ ⎟ (2.325 − 0.5) 2 ⇒ ⎜ ⎟ = 0.60
⎝ 2 ⎠ ⎝ L ⎠C
⎝ L ⎠C
VGSC =
(c)
f 3− dB =
1
2π Ro C
Ro = 0.5 0.667 = 0.286 M Ω
1
= 185 kHz
2π (286 × 103 )(3 × 10−12 )
GBW = (400)(185 × 103 ) ⇒ 74 MHz
f 3− dB =
13.38
(a)
From previous results, we can write
Ro10 = g m10 (ro10 ro 6 )
Ro12 = g m12 (ro12 ro8 )
Ad = Bg m1 ( Ro10 Ro12 )
Now
ro10 = ro 6 =
1
1
=
= 0.5 M Ω
(0.02)(2.5)(40)
λP B ( I Q / 2 )
ro12 = ro8 =
1
1
=
= 0.667 M Ω
(0.015)(2.5)(40)
λn B ( I Q / 2 )
Assume all transistors have the same width-to-length ratios except for M 5 and M 6 .
⎛W
Let ⎜
⎝L
Then
⎞
2
⎟= X
⎠
⎛ k ′p ⎞ ⎛ W ⎞
⎛ 35 ⎞
g m10 = 2 ⎜ ⎟ ⎜ ⎟ ( I DQ10 ) = 2 ⎜ ⎟ X 2 (2.5)(40)
⎝ 2⎠
⎝ 2 ⎠ ⎝ L ⎠10
= 83.67 X
⎛ k′ ⎞⎛ W ⎞
⎛ 80 ⎞
g m12 = 2 ⎜ n ⎟ ⎜ ⎟ ( I DQ12 ) = 2 ⎜ ⎟ X 2 (2.5)(40)
⎝ 2⎠
⎝ 2 ⎠ ⎝ L ⎠12
= 126.5 X
⎛ 80 ⎞
g m1 = 2 ⎜ ⎟ X 2 (40) = 80 X
⎝ 2⎠
Then
Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ω
Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M Ω
We want
20, 000 = (2.5)(80 X )[20.9 X 56.3 X ]
Then
⎡ (20.9 X )(56.3 X ) ⎤
2
= 200 X ⎢
⎥ = 3048 X
⎣ 20.9 X + 56.3 X ⎦
⎛W ⎞
X 2 = 6.56 = ⎜ ⎟
⎝L⎠
Then
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = (2.5)(6.56) = 16.4
⎝ L ⎠ 6 ⎝ L ⎠5
(b)
Assume bias voltages are V + = 5V , V − = −5V
⎛W ⎞
⎛W ⎞
Assume ⎜ ⎟ = ⎜ ⎟ = 6.56
⎝ L ⎠ A ⎝ L ⎠B
⎛ 80 ⎞
I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V
⎝ 2⎠
Need 5 transistors in series
10 − 1.052
VGSC =
= 1.79 V
5
Then
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
I REF = 80 = ⎜ ⎟ ⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20
⎝ 2 ⎠ ⎝ L ⎠C
⎝ L ⎠C
(c)
f 3− dB =
1
where Ro = Ro10 Ro12
2π Ro C
Now
Ro10 = 20.9 6.56 = 53.5 M Ω
Ro12 = 56.3 6.56 = 144 M Ω
Then
Ro = 53.5 144 = 39 M Ω
1
= 1.36 kHz
2π (39 × 106 )(3 × 10−12 )
GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz
f 3− dB =
13.39
Ad = g m ( M 2 ) ⋅ ⎡⎣ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤⎦
⎛ 40 ⎞
g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 447 µ A / V
⎝ 2 ⎠
1
1
ro 2 ( M 2 ) =
=
= 500 k Ω
λ I DQ (0.02)(0.1)
ro 2 (Q2 ) =
VA 120
=
= 1200 k Ω
I CQ 0.1
Then
Ad = 447(0.5 1.2) ⇒ Ad = 158
13.40
Ad = g m ( M 2 ) ⋅ ⎡⎣ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤⎦
⎛ 80 ⎞
g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 632 µ A / V
⎝ 2⎠
1
1
ro 2 ( M 2 ) =
=
= 667 k Ω
λ I DQ (0.015)(0.1)
VA
80
=
= 800 k Ω
I CQ 0.1
ro 2 (Q2 ) =
Ad = (632) ( 0.667 0.80 ) ⇒ Ad = 230
13.41
(a)
I REF = 200 µ A
K n = K p = 0.5 mA / V 2
λn = λ p = 0.015 V −1
Ad = g m1 ( Ro 6 Ro8 )
where
Ro8 = g m8 (ro8 ro10 )
Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 )
Now
g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V
1
1
ro8 =
=
= 667 k Ω
λP I D 8 (0.015)(0.1)
ro10 =
1
= 667 k Ω
λP I D 8
gm6 =
IC 6
0.1
=
= 3.846 mA/V
VT
0.026
ro 6 =
VA
80
=
= 800 k Ω
I C 6 0.1
ro 4 =
1
1
=
= 333 k Ω
λn I D 4 (0.015)(0.2)
ro1 =
1
λ p I D1
=
1
= 667 k Ω
(0.015)(0.1)
g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V
So
Ro8 = (0.447)(667)(667) ⇒ 198.9 M Ω
Ro 6 = (3.846)(800)(333 667) ⇒ 683.4 M Ω
Then
Ad = 447(198.9 683.4) ⇒ Ad = 68,865
13.42
Assume biased at V + = 10V , V − = −10V .
P = 3I REF (20) = 10 ⇒ I REF = 167 µ A
Ad = g m1 ( Ro 6 Ro8 ) = 25, 000
kn′ = 80 µ A / V 2 , k ′p = 35 µ A / V 2
λn = 0.015V −1 , λ p = 0.02 V −1
⎛W ⎞
⎛W ⎞
Assume ⎜ ⎟ = 2.2 ⎜ ⎟
L
⎝ ⎠p
⎝ L ⎠n
Ro8 = g m8 ( ro8 ro10 )
Ro 6 = g m 6 (ro 6 )(ro 4 ro1 )
ro8 =
ro10 =
1
λP I D 8
1
λP I D 8
=
1
= 0.60 M Ω
(0.02)(83.3)
= 0.60 M Ω
⎛ k ′p ⎞ ⎛ W ⎞
⎛ 35 ⎞
g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3)
⎝ 2⎠
⎝ 2 ⎠ ⎝ L ⎠8
= 113.3 X
⎛W ⎞
where X 2 = ⎜ ⎟
⎝ L ⎠n
VA
80
=
= 0.960 M Ω
ro 6 =
I C 6 83.3
ro 4 =
ro1 =
gm6 =
1
λn I D 4
=
1
= 0.40 M Ω
(0.015)(167)
1
1
=
= 0.60 M Ω
λ p I D1 (0.02)(83.3)
IC 6
83.3
=
= 3204 µ A / V
0.026
VT
⎛ k p′ ⎞ ⎛ W ⎞
⎛ 35 ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ I D1 = 2 ⎜ ⎟ (2.2) X 2 (83.3)
⎝ 2⎠
⎝ 2 ⎠ ⎝ L ⎠1
= 113.3 X
Now
Ro 6 = (3204)(0.960) ⎡⎣0.40 0.60 ⎤⎦ = 738 M Ω
Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Ω
Then
Ad = 25, 000 = (113.3 X ) ⎡⎣ 738 40.8 X ⎤⎦
⎡ 30,110 X ⎤
= (113.3 X ) ⎢
⎥
⎣ 738 + 40.8 X ⎦
which yields X = 2.48
or
⎛W ⎞
X 2 = 6.16 = ⎜ ⎟
⎝ L ⎠n
and
⎛W ⎞
⎜ ⎟ + (2.2)(6.16) = 12.3
⎝ L ⎠P
13.43
For vcm (max), assume VCB (Q5 ) = 0. Then
VS = 15 − 0.6 − 0.6 = 13.8 V
0.236
I D 9 = I D10 =
= 0.118 mA
2
Using parameters given in Example 13.11
I
0.118
VSG = D 9 − VTP =
+ 1.4 = 2.17 V
0.20
KP
Then
vcm (max) = 13.8 − 2.17 ⇒ vcm (max) = 11.6 V
For
vcm (min) , assume
VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 − 1.4 = 0.77 V
Now
VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) − 15
= 0.118 + 0.6 − 15 ⇒ VD10 = −14.28 V
Then
vcm (min) = −14.28 + VSD (sat) − VSG
= −14.28 + 0.77 − 2.17 = −15.68 V
Then, common-mode voltage range
−15.68 ≤ vcm ≤ 11.6
Or, assuming the input is limited to ±15 V, then
−15 ≤ vcm ≤ 11.6 V
13.44
For I1 = I 2 = 300 µ A,
VSG = VBE + (0.3)(8) = 0.6 + 2.4 = 3.0 V
Then
I 2 = K P (VSG + VTP ) 2
0.3 = K P (3 − 1.4)2 ⇒ K P = 0.117 mA / V 2
13.45
For VCB = 0 for both Q6 and Q7 , then
VS = 0.6 + 0.6 + VSG + (−VS )
So 2VS = 1.2 + VSG
Now
I1
0.6 + I 2 R1 = VSG =
+ VTP and I1 = I 2
KP
Also I1 = I 2 = K P (VSG + VTP ) 2 so
0.6 + (0.25)(8)(VSG − 1.4) 2 = VSG
2
− 2.8VSG + 1.96) = VSG
0.6 + 2(VSG
2
2VSG − 6.6VSG + 4.52 = 0
6.6 ± (6.6) 2 − 4(2)(4.52)
= 2.33 V
2(2)
Then
2VS = 1.2 + 2.33 = 3.53 and
VS = 1.765 V
VSG =
13.46
I C 5 = I C 4 = 300 µ A
Using the parameters from Examples 13.12 and 13.13, we have
βV
(200)(0.026)
Ri 2 = rπ 13 = n T =
= 17.3 kΩ
I C13
0.3
Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3)
or
Ad = 10.38
Now
I C13
0.3
=
= 11.5 mA/V
0.026
VT
g m13 =
r013 =
VA
50
=
= 167 kΩ
I C13 0.3
Then
| Av 2 | = g m13 ⋅ r013 = (11.5)(167)
or
Av 2 = 1917
Overall gain:
Av = (10.38)(1917) = 19,895
Assuming the resistances looking into Q4 and into the output stage are very large, we have
13.47
| Av 2 | =
β R013
rπ 13 + (1 + β ) RE13
where R013 = r013 ⎡⎣1 + g m13 ( RE13 rπ 13 ) ⎤⎦
I C13 = 300 µ A, r013 =
50
= 167 kΩ
0.3
0.3
= 11.5 mA / V
0.026
(200)(0.026)
=
= 17.3 kΩ
0.3
g m13 =
rπ 13
So
R013 = (167) ⎡⎣1 + (11.5) (1 17.3) ⎤⎦ ⇒ 1.98 MΩ
Then
| Av 2 | =
(200)(1980)
= 1814
17.3 + (201)(1)
Now
Ci = C1 (1 + Av 2 ) = 12 [1 + 1814]
⇒ Ci = 21, 780 pF
1
f PD =
2π Req Ci
Req = Ri 2 r012 r010
Neglecting R3 ,
r010 =
1
1
=
= 333 kΩ
λ I D10 (0.02)(0.15)
Neglecting R5 ,
50
= 333 kΩ
r012 =
0.15
Ri 2 = rπ 13 + (1 + β ) RE13
= 17.3 + (201)(1)
= 218 kΩ
Then
f PD =
1
2π ⎡⎣ 218 333 333⎤⎦ × 103 × ( 21, 780 ) × 10−12
or
f PD = 77.4 Hz
Unity-Gain Bandwidth
Gain of first stage:
Ad = 2 K n I Qs ⋅ ( R12 ro12 ro10 )
= 2(0.6)(0.3) ⋅ (218 333 333)
= (0.6)(218 333 333)
or
Ad = 56.6
Overall gain:
Av = (56.6)(1814) = 102, 672
Then unity-gain bandwidth = (77.4)(102, 672)
⇒ 7.95 MHz
13.48
Since VGS = 0 in J 6 , I REF = I DSS
⇒ I DSS = 0.8 mA
13.49
a.
Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ]
(100)(0.026)
= 13 kΩ
0.2
I
200 µ A
≅ C6 =
= 2 µA
100
β
rπ 6 =
IC 5
So
rπ 5 =
(100)(0.026)
= 1300 kΩ
0.002
Then
Ri 2 = 1300 + (101) [13 + (101)(0.3) ]
or
Ri 2 = 5.67 MΩ
Av = g m 2 ( r02 r04 Ri 2 )
b.
gm2 =
2
⋅ I D ⋅ I DSS
VP
=
2
⋅ (0.1)(0.2)
3
= 0.0943 mA / V
r02 =
1
1
=
= 500 kΩ
λ I D (0.02)(0.1)
r04 =
VA 5.0
=
= 500 kΩ
I C 4 0.1
Then
Av = (0.0943)[500 || 500 || 5670]
or
Av = 22.6
13.50
a.
Need VSD (QE ) ≥ VSD ( sat ) = VP For minimum bias ±3 V
Set VP = 3 V and VZK = 3 V
I REF 2 =
VZK − VD1
R3
3 − 0.6
⇒ R3 = 24 kΩ
0.1
Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA
Therefore,
so that R3 =
I DSS = 0.2 mA
b.
Neglecting base currents
12 − 0.6
I 01 = I REF 1 = 0.5 mA =
R4
so that
R4 = 22.8 kΩ
13.51
a.
gm2
We have
2
=
⋅ I D ⋅ I DSS
| VP |
=
2
⋅ (0.5)(1)
4
= 0.354 mA/V
r02 =
r04 =
1
λ ID
=
1
= 100 kΩ
(0.02)(0.5)
VA 100
=
= 200 kΩ
I D 0.5
0.5
= 19.23 mA/V
0.026
(200)(0.026)
= 10.4 kΩ
rπ 4 =
0.5
So
R04 = r04 ⎡⎣1 + g m 4 ( rπ 4 R2 ) ⎤⎦
gm4 =
= 200 ⎡⎣1 + (19.23) (10.4 0.5 ) ⎤⎦
= 2035 kΩ
Ad = g m 2 ( r02 R04 RL )
For RL → ∞
Ad = 0.354 (100 || 2035 ) = 33.7
b.
With these parameter values, gain can never reach 500.
Similarly for this part, gain can never reach 700.
Chapter 14
Exercise Solutions
EX14.1
a.
ACL =
b.
dACL
ACL
ACL =
−50
⇒ ACL = −49.949
⎡ ⎛ 1 ⎞
⎤
⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥
⎠
⎣ ⎝
⎦
dA
51
= 10 ×
⇒ CL = 0.0102%
5 × 104
ACL
−50
⇒ ACL = −49.943
51 ⎤
⎡
⎢⎣1 + 4.5 × 104 ⎥⎦
EX14.2
a.
For R0 = 0
1
1 1
= + (1 + 104 ) = 0.1 + 103
Ri f 10 10
⇒ Ri f = 10−3 kΩ = 1 Ω
b.
For R0 = 10 kΩ
1
1 1 ⎡1 + 104 + 1 ⎤
104
= + ×⎢
⎥ ≅ 0.1 +
Ri f 10 10 ⎣ 1 + 1 + 1 ⎦
3 (10 )
Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω
EX14.3
⎛ 40 ⎞
40 (1 + 104 ) + 99 ⎜1 + ⎟
1 ⎠
⎝
Ri f =
99
1+
1
5
4 × 10 + 4.059 × 103
≅
100
Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ
EX14.4
R
1 + 2 = 100
R1
a.
1
1 ⎡ 105 ⎤
=
⎢
⎥ = 10 ⇒ R0 f = 0.1 Ω
R0 f 100 ⎣100 ⎦
b.
1
1 ⎡ 105 ⎤
2
= ⎢
⎥ = 10
R0 f 10 ⎣100 ⎦
R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω
EX14.5
From Equation (14.43)
ACL ( f ) =
=
ACL 0
1+ j ⋅
f
f PD ( A0 /ACL 0 )
25
f
1+ j ⋅
( 50 ) (104 / 25)
a.
f = 2 kHz
b.
f = 20 kHz
25
=
1+ j ⋅
f
2 × 104
v0
= 25 ⇒ v0 ( peak ) = 1.25 mV
vI
v0
1
=
⋅ 25 ⇒ v0 ( peak ) = 0.884 mV
vI
2
c.
f = 100 kHz
v0
25
25
=
=
= 4.90
2
5.099
vI
1 + (100 / 20 )
⇒ v0 = 0.245 mV
EX14.6
Full-scale response = 1× 5 = 5 V
t=
5
⇒ t = 2.5 µ s
2
EX14.7
a.
FPBW =
SR
0.63 × 106
=
2π V0 ( max )
2π (1)
FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz
b.
FPBW =
0.63 × 106
= 1.0 × 104
2π (10 )
⇒ FPBW = 10 kHz
EX14.8
⎛I ⎞
⎛ 1.85 × 10−14 ⎞
V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜
−14 ⎟
⎝ 2 × 10
⎠
⎝ I S1 ⎠
⇒ V0 S = 2.03 mV
EX14.9
We need
iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V
By Equation (14.60(a))
⎡
⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟
⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎣
⎡
⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 +
⎟
50 ⎠
⎝ VT ⎠ ⎦ ⎝
⎣
By Equation (14.60(b))
⎡
⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC 2 = I S 2 ⎢exp ⎜ BE 2 ⎟ ⎥ ⎜1 + ⎟
⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎣
⎡
⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 4 ⎢exp ⎜ EB 4 ⎟ ⎥ ⎜1 +
⎟
50 ⎠
⎝ VT ⎠ ⎦ ⎝
⎣
I S1 = I S 2, take the ratio:
⎛v −v ⎞ I
exp ⎜ BE1 BE 2 ⎟ = S 3
VT
⎝
⎠ IS 4
vBE1 − vBE 2 = V0 S
⎛I ⎞
= VT ln ⎜ S 3 ⎟
⎝ IS 4 ⎠
= 0.026 ⋅ ln (1.05 )
⇒ V0 S = 1.27 m V
EX14.10
VOS =
0.020 =
I Q ⎛ ΔK n ⎞
1
⋅
⋅⎜
⎟
2 2Kn ⎝ Kn ⎠
1
150 ⎛ ΔK n ⎞
⋅
⋅
2 2 ( 50 ) ⎜⎝ 50 ⎟⎠
⇒ ΔK n = 1.63µ A / V 2
⇒
ΔK n 1.63
=
⇒ 3.26%
50
Kn
EX14.11
⎛ R5 ⎞ +
Want ⎜
⎟V = 5 m V
⎝ R5 + R4 ⎠
R5 R4 so
R5 =
R5
× V + = 0.005
R4
( 0.005)(100 )
= 0.05 kΩ
10
⇒ R5 = 50 Ω
EX14.12
R1′ = 25 1 = 0.9615 kΩ
R2′ = 75 1 = 0.9868 kΩ
For I Q = 100 µ A ⇒ iC1 = iC 2 = 50 µ A
From Equation (14.75)
⎛ 50 × 10−6 ⎞
( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 )
⎝ 10
⎠
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 )
⎝ IS 4 ⎠
0.58065 + 0.048075
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934
⎝ IS 4 ⎠
⎛i ⎞
ln ⎜ C 2 ⎟ = 22.284
⎝ IS 4 ⎠
50 × 10−6
= 4.7625 × 109
IS 4
I S 4 ≅ 1.05 × 10−14 A
EX14.13
From Equation (14.79)
⎛ R ⎞
v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
R1 ⎠
⎝
For v0 = 0
⎛ 100 ⎞
0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 +
⎟
10 ⎠
⎝
R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ
TYU14.1
v1CM ( max ) = V + − VSD1 ( sat ) − VSG1
v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1
We have:
I REF = 100 µ A, kn′ = 80 µ A / V 2 , k ′p = 40 µ A / V 2 ,
⎛W
⎜
⎝L
For
⎞
⎟ = 25
⎠
M1 :
2
⎛ 40 ⎞
I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP )
⎝ 2 ⎠
So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V
2
VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V
Then
vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V
For M 4 :
2
⎛ 80 ⎞
I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN )
⎝ 2⎠
So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V
2
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184
So
V − − 0.184 ≤ vCM ≤ V + − 1.13 V
TYU14.2
vo ( max ) = V + − VSD 8 ( sat ) − VSG10
vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat )
Now
50
VSG 8 = VSG10 =
+ 0.5 = 0.816 V
( 40/ 2 )( 25)
VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V
So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13
Also
VGS 6 =
50
+ 0.5 = 0.724 V
(80/ 2 )( 25)
VGS 4 =
100
+ 0.5 = 0.816 V
(80/ 2 )( 25)
VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
So
vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54
Then
V − + 0.54 ≤ vo ≤ V + − 1.13 V
TYU14.3
500
= −25
20
Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975
ACL ( ideal ) = −
−24.975 =
−25
⎡
26 ⎤
⎢1 + A ⎥
0L ⎦
⎣
26
−25
=
− 1 = 0.0010
A0 L −24.975
A0 L = 25.974
TYU14.4
ACL ( ∞ )
⎡ A (∞) ⎤
1 + ⎢ CL
⎥
⎣ A0 L ⎦
R
495
ACL ( ∞ ) = 1 + 2 = 1 +
= 100
R1
5
100
ACL =
⇒ ACL = 99.90
100
1+ 5
10
ACL ( ∞ ) = 100
a.
ACL =
b.
dACL
100
= 10 × 5 = 0.01%
10
ACL
ACL = 99.90 − ( 0.0001)( 99.90 )
⇒ ACL = 99.89
TYU14.5
ACL ( ∞ ) − ACL
= 1−
ACL ( ∞ )
ACL
= 1−
ACL ( ∞ )
1
A (∞)
1 + CL
A0 L
ACL ( ∞ )
ACL ( ∞ )
−1
A0 L
A0 L
So 0.001 =
=
A (∞)
A (∞)
1 + CL
1 + CL
A0 L
A0 L
1+
0.001 = 0.999 ⋅
ACL ( ∞ ) =
ACL ( ∞ )
A0 L
0.001
0.001
⋅ A0 L =
⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010
0.999
0.999
or
ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0
TYU14.6
ii ⎛ Rif ⎞
=⎜
⎟
I1 ⎝ Ri ⎠
iI 0.1
a.
=
= 1× 10−5
i1 104
iI
10
=
= 1× 10−3
i1 104
b.
TYU14.7
Voltage follower R2 = 0, R1 = ∞
Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 )
≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ
TYU14.8
f3dB =
(105 ) (10 ) ⇒ 20 kHz
fT
=
50
ACL 0
f max = f 3dB =
V0 ( max ) =
SR
2π V0 ( max )
0.8 × 106
SR
=
2π f3dB 2π ( 20 × 103 )
⇒ V0 ( max ) = 6.37 V
TYU14.9
v0 = I B1 R3 = (10−6 )( 200 × 103 )
a.
⇒ v0 = 0.20 V
b.
R4 = R1 R2 R3 = 100 50 200
⇒ R4 = 28.6 kΩ
Chapter 14
Problem Solutions
14.1
Ad =
vo
= −80
vi
vo (max) = 4.5 ⇒ vi (max) = 56.25 mV
So vi (max)rms =
56.25
2
= 39.77 mV
14.2
(a)
4.5
= 0.028125 mA
160
4.5
iL =
= 4.5 mA
1
Output Circuit = 4.528 mA
v
−4.5
⇒ vi = −0.05625 V
vi = − o =
A
80
(b)
v
4.5
io ≈ 15 mA = o =
RL RL
i2 =
⇒ RL (min) = 300 Ω
14.3
(1)
(2)
v2 = 12.5 mV
(3)
(4)
(5)
AOL = 2 × 104
v1 = 8 µ V
AOL = 1000
vo = 2 V
14.4
From Eq. (14.4)
ACL =
−15 =
− R2 / R1
1 ⎛ R2 ⎞
1+
⎜1 + ⎟
AoL ⎝
R1 ⎠
− R2 / R1
1 ⎛ R2 ⎞
1+
⎜1 + ⎟
R1 ⎠
2 × 103 ⎝
⎛R ⎞
15 ⎜ 2 ⎟
R
R
1 ⎞
⎛
+ ⎝ 1 3⎠ = 2
15 ⎜ 1 +
3 ⎟
R1
×
×
2
10
2
10
⎝
⎠
15.0075 =
R2
(0.9925)
R1
R2
= 15.12
R1
14.5
vI − v1 v1 − v0 v1
=
+ and v0 = − A0 L v1
R1
R2
Ri
so that v1 = −
v0
A0 L
⎛ 1
vI v0
1
1⎞
+
= v1 ⎜ +
+ ⎟
R1 R2
⎝ R1 R2 Ri ⎠
So
⎡1
vI
1 ⎛ 1
1
1 ⎞⎤
= −v0 ⎢ +
+ ⎟⎥
⎜ +
R1
⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦
Then
v0
−(1/ R1 )
=
= ACL
vI ⎡ 1
1 ⎛ 1
1
1 ⎞⎤
+ ⎟⎥
⎢ +
⎜ +
⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦
From Equation (14.20) for RL = ∞ and R0 = 0
1
1
1 (1 + A0 L )
= + ⋅
Rif Ri R2
1
For Ri = 1 kΩ
a.
−(1/ 20)
1 ⎛ 1
1 1 ⎞⎤
⎡ 1
⎢100 + 103 ⎜ 20 + 100 + 1 ⎟ ⎥
⎝
⎠⎦
⎣
−0.05
=
[0.01 + 1.06 × 10−3 ]
ACL =
or
⇒ ACL = −4.52
1 1 1 + 103
= +
⇒ Rif = 90.8 Ω
Rif 1 100
For Ri = 10 kΩ
b.
ACL =
=
−(1/ 20)
1 ⎛ 1
1
1 ⎞⎤
⎡ 1
⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥
⎝
⎠⎦
⎣
−0.05
[0.01 + 1.6 × 10−4 ]
or
⇒ ACL = −4.92
1
1 1 + 103
= +
⇒ Rif = 98.9 Ω
Rif 10
100
For Ri = 100 kΩ
c.
ACL =
=
or
−(1/ 20)
1 ⎛ 1
1
1 ⎞⎤
⎡ 1
⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥
⎝
⎠⎦
⎣
−0.05
[0.01 + 7 × 10−5 ]
⇒ ACL = −4.965
1
1 1 + 103
=
+
⇒ Rif = 99.8 Ω
Rif 100
100
14.6
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
v
⎝
ACL = o =
vi ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ AOL ⎝
For the ideal:
⎛ R2 ⎞ 0.10
= 50
⎜1 + ⎟ =
R1 ⎠ 0.002
⎝
vo (actual ) = (0.10)(1 − 0.001) = 0.0999
So
0.0999
50
=
= 49.95
1
0.002 1 +
(50)
AOL
which yields
AOL = 1000
14.7
From Equation (14.18)
⎛A
1 ⎞
− ⎜ OL − ⎟
v
⎝ Ro R2 ⎠
Avf 1 = o1 =
v1 ⎛ 1
1
1 ⎞
+
+ ⎟
⎜
R
R
R
2 ⎠
o
⎝ L
Or
⎛ 5 × 103
1 ⎞
−⎜
−
⎟
3
1
100
⎠ ⋅ v = −(4.99999 × 10 ) ⋅ v
vo1 = ⎝
1
1
1.11
⎛1 1 1 ⎞
⎜ + +
⎟
⎝ 10 1 100 ⎠
vo1 = −4.504495 × 103 ⋅ v1
Now
i1 vi − v1
=
≡K
v1
R1v1
Then
vi − v1 = KR1v1
which yields
vi
v1 =
KR1 + 1
Now, from Equation (14.20)
1 ⎤
⎡
1 + 5 × 103 + ⎥
1
1 ⎢
10
K= +
⎢
⎥
10 100 ⎢ 1 + 1 + 1 ⎥
10 100 ⎦⎥
⎣⎢
⎡ 5.0011× 103 ⎤
= (0.1) + (0.01) ⎢
⎥ = 45.15495
1.11
⎣
⎦
Then
vi
vi
v1 =
=
( 45.15495)(10 ) + 1 452.5495
vi
⎡
⎤
vo1 = −4.504495 × 103 ⎢
⎥
⎣ 452.5495 ⎦
Or
v
Avf 1 = o1 = −9.9536
vi
We find
For the second stage, RL = ∞
⎛ 5 × 103
1 ⎞
−⎜
−
⎟
1
100 ⎠
vo 2 = ⎝
⋅ v1′ = −4.950485 × 103 ⋅ v1′
⎛1 1 ⎞
⎜ +
⎟
⎝ 1 100 ⎠
⎡
⎤
1
1 ⎢1 + 5 × 103 ⎥
K≡ +
⎢
⎥ = 49.61485
10 100 ⎢ 1 + 1 ⎥
⎢⎣
100 ⎥⎦
vo1
vo1
vo1
v1′ =
=
=
KR1 + 1 (49.61485)(10) + 1 497.1485
Then
vo 2 −4.950485 × 103
=
= −9.95776
497.1485
vo1
So
v
Avf = o 2 = (−9.9536)(−9.95776) ⇒ Avf = 99.12
vi
14.8
a.
v1 − vI
v v −v
+ 1 + 1 0 =0
R3 + Ri R1
R2
⎡ 1
vI
1
1⎤ v
+ + ⎥= 0 +
v1 ⎢
⎣ R3 + Ri R1 R2 ⎦ R2 R3 + Ri
v0 v0 − A0 L vd v0 − v1
+
+
=0
RL
R0
R2
or
⎡1
A v
1
1⎤ v
+ ⎥ = 1 + 0L d
v0 ⎢ +
R
R
R
R
R0
0
2 ⎦
2
⎣ L
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ Ri
⎝ R3 + Ri ⎠
So substituting numbers:
vI
1
1⎤ v
⎡ 1
+ + ⎥= 0 +
v1 ⎢
⎣10 + 20 10 40 ⎦ 40 10 + 20
or
v1[0.15833] = v0 [0.025] + vI [0.03333]
1 ⎤ v (104 )vd
⎡1 1
v0 ⎢ +
+ ⎥= 1 +
0.5
⎣1 0.5 40 ⎦ 40
(1)
(2)
(3)
(1)
(2)
or
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 )
⎝ 10 + 20 ⎠
So
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 )
(3)
(2)
or
v0 [3.025] = 1.333 ×10 4 vI − 1.333 × 104 v1
From (1):
v1 = v0 ( 0.1579 ) + vI ( 0.2105 )
Then
v0 [3.025] = 1.333 × 104 vI − 1.333 × 104 ⎡⎣v0 ( 0.1579 ) + vI ( 0.2105 ) ⎤⎦
v0 ⎡⎣ 2.1078 ×103 ⎤⎦ = vI ⎡⎣1.0524 ×104 ⎤⎦
or
v
ACL = 0 = 4.993
vI
To find Rif : Use Equation (14.27)
⎛ 0.5 0.5 ⎞
iI ⎜ 1 +
+
⎟
1
40 ⎠
⎝
3
⎧⎛ 1
1 ⎞ ⎛ 0.5 0.5 ⎞ 0.5 ⎫ (10 ) vd
= v1 ⎨⎜ + ⎟ ⎜ 1 +
+
−
−
⎬
⎟
1
40 ⎠ (40) 2 ⎭
40
⎩⎝ 10 40 ⎠ ⎝
iI (1.5125) = v1{(0.125)(1.5125) − 0.0003125} − 25vd
or
iI (1.5125) = vI {0.18875} − 25vd
Now
vd = iI Ri = iI (20) and v1 = vI − iI (20)
So
iI (1.5125) = [vI − iI (20)] ⋅ [0.18875] − 25iI (20)
iI [505.3] = vI (0.18875)
or
vI
= 2677 kΩ
iI
Now Rif = 10 + 2677 ⇒ Rif = 2.687 MΩ
To determine R0 f : Using Equation (14.36)
⎡
1
1 ⎢⎢ A0 L
=
⋅
R2
R0′ f R0 ⎢
⎢1 + R R
1
i
⎣
or R0′ f = 3.5 Ω
⎤
⎡
⎤
⎥ 1 ⎢ 103 ⎥
⎥=
⎥
⋅⎢
⎥ 0.5 ⎢1 + 40 ⎥
⎥
⎢ 10 20 ⎥
⎣
⎦
⎦
Then R0 f = 1 kΩ 3.5 Ω
⇒ R0 f = 3.49 Ω
b.
Using Equation (14.16)
dACL
dA
⎛ 5 ⎞
= (−10) ⎜ 3 ⎟ ⇒ CL = −(0.05)%
ACL
ACL
⎝ 10 ⎠
14.9
v0 − A0 L vd v0 − vI
+
= 0 and vd = vI − v0
R0
Ri
So
v0 A0 L
v
v
−
⋅ (vI − v0 ) + 0 − I = 0
R0 R0
Ri Ri
⎡1 A
⎡1 A ⎤
1⎤
v0 ⎢ + 0 L + ⎥ = vI ⎢ + 0 L ⎥
⎣ R0 R0 Ri ⎦
⎣ Ri R0 ⎦
⎡ 1 (104 ) 1 ⎤
⎡ 1
(104 ) ⎤
+
+
+
v0 ⎢
⎥ = vI ⎢
⎥
⎣ 0.2 0.2 100 ⎦
⎣100 0.2 ⎦
v0 [5.000501× 104 ] = vI [5.000001× 10 4 ]
So ACL =
Set vI = 0
b.
i0 =
v0
= 0.9999
vI
v0 − A0 L vd v0
+ and vd = −v0
R0
Ri
⎡1 A
1⎤
i0 = v0 ⎢ + 0 L + ⎥
R
R
R
0
i ⎦
⎣ 0
Then
1
1 A0 L 1
=
+
+
R0 f R0 R0 Ri
or
1
1 (104 ) 1
=
+
+
R0 f 0.2 0.2 100
which yields
R0 f ≅ 0.02 Ω
14.10
vI 1 − v1 vI 2 − v1 v1 − v0
+
=
20
10
40
vI 1 vI 2 v0
1
1⎤
⎡1
+
+
= v1 ⎢ + + ⎥
20 10 40
⎣ 20 10 40 ⎦
v
and v0 = − A0 L v1 so that v1 = − 0
A0L
Then
⎧1
1
⎛ 7 ⎞⎫
⋅
vI 1 (0.05) + vI 2 (0.10) = −v0 ⎨ +
⎟⎬
3 ⎜
⎩ 40 2 × 10 ⎝ 40 ⎠ ⎭
= −v0 [2.50875 × 10−2 ]
⇒ v0 = −1.993vI 1 − 3.986vI 2
Δv0 2 − 1.993
Δv
=
⇒ 0 = 0.35%
2
v0
v0
14.11
⎛ 40 ⎞
⎛ 4⎞
vB = ⎜
⎟ v2 = ⎜ ⎟ v2 = 0.8v2
⎝ 40 + 10 ⎠
⎝5⎠
v1 − v A v A − v0
=
10
40
v1 v0
1 ⎞
⎛1
+
= vA ⎜ + ⎟
10 40
⎝ 10 40 ⎠
v1 (0.1) + v0 (0.025) = vA (0.125)
v0 = A0 L vd = A0 L (vB − v A )
or
v0 = A0 L [0.8v2 − v A ]
v0
− 0.8v2 = −v A
A0 L
⇒ v A = 0.8v2 −
Then
v0
A0 L
(1)
(2)
(3)
⎡
v ⎤
v1 (0.1) + v0 (0.025) = (0.125) ⎢0.8v2 − 0 ⎥
A
0L ⎦
⎣
0.125
⎡
⎤
v1 (0.1) − v2 (0.1) = −v0 ⎢0.025 +
103 ⎥⎦
⎣
= −v0 [2.5125 × 10−2 ]
⇒ Ad =
⇒
v0
= 3.9801
v2 − v1
ΔAd 0.0199
=
⇒ 0.4975%
Ad
4
14.12
a.
Considering the second op-amp and Equation (14.20), we have
⎡
⎤
1
1
1 ⎢1 + 100 ⎥
101
= +
⋅⎢
⎥ = 0.10 +
(0.1)(11)
Rif 2 10 0.1 ⎢ 1 + 1 ⎥
⎣⎢ 0.1 ⎦⎥
So Rif 2 = 0.0109 kΩ
The effective load on the first op-amp is then
RL1 = 0.1 + Rif 2 = 0.1109 kΩ
Again using Equation (14.20), we have
1
1 + 100 +
1
1 1
0.1109 = 0.10 + 110.017
= + ⋅
11.017
Rif 10 1 1 + 1 + 1
0.1109 1
so that
Rif = 99.1 Ω
b.
To determine R0 f :
For the first op-amp, we can write, using Equation (14.36)
⎡
⎤
⎡
⎤
⎥ 1 ⎢ 100 ⎥
1
1 ⎢ A0 L
⎥ = ⋅⎢
⎥
=
⋅⎢
40 ⎥
R2 ⎥ 1 ⎢
R0 f 1 R0 ⎢
1+
1+
⎢ R1 || Ri ⎥
⎢⎣ 1 || 10 ⎥⎦
⎣
⎦
which yields R0 f 1 = 0.021 kΩ
For the second op-amp, then
⎡
⎢
A0 L
1
1 ⎢
=
⋅
R2
R0 f R0 ⎢
⎢1 + ( R + R ) || R
1
0f1
i
⎣
⎡
⎤
⎥
1 ⎢
100
⎥
= ⋅⎢
0.10
1 ⎢1 +
⎥
⎢⎣ (0.121) ||10 ⎥⎦
or R0 f = 18.4 Ω
c.
⎤
⎥
⎥
⎥
⎥
⎦
To find the gain, consider the second op-amp.
v01 − (−vd 2 ) vd 2 −vd 2 − v02
+
=
0.1
Ri
0.1
(1)
v02 − A0 L vd 2 v02 − (−vd 2 )
+
=0
R0
0.1
(2)
vI − (−vd 1 ) vd 1 −vd 1 − v01
+
=
1
1
Ri
(1)
v01
v − A0 L vd 1 v01 − (−vd 1 )
+ 01
+
=0
0.1109
1
R0
(2)
v01
v02
1
1 ⎞
⎛ 1
+ vd 2 ⎜
+ +
⎟=−
0.1
0.1
⎝ 0.1 10 0.1 ⎠
or
v01 (10) + vd 2 (20.1) = −v02 (10)
v02
⎛ 100 1 ⎞ v02
− vd 2 ⎜
−
=0
⎟+
1
⎝ 1 0.1 ⎠ 0.1
v02 (11) − vd 2 (90) = 0
or
vd 2 = v02 (0.1222)
Then Equation (1) becomes
v01 (10) + v02 (0.1222)(20.1) = −v02 (10)
or
v01 = −v02 (1.246)
Now consider the first op-amp.
⎛1 1 1⎞
vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1)
⎝ 1 10 1 ⎠
or
vI (1) + vd 1 (2.1) = −v01 (1)
1 1⎞
⎛ 1
⎛ 100 1 ⎞
+ + ⎟ − vd 1 ⎜
− ⎟=0
v01 ⎜
⎝ 0.1109 1 1 ⎠
⎝ 1 1⎠
v01 (11.017) − vd 1 (99) = 0
or
vd 1 = v01 (0.1113)
Then Equation (1) becomes
vI (1) + v01 (0.1113)(2.1) = −v01
or vI = −v01 (1.234)
We had v01 = −v02 (1.246)
So vI = v02 (1.246)(1.234)
or
v02
= 0.650
vI
v02
=1
vI
So ratio of actual to ideal = 0.650.
d.
Ideal
14.13
(a)
For the op-amp. A0 L ⋅ f 3dB = 106
106
= 50 Hz
2 × 104
For the closed-loop amplifier.
106
f 3dB =
= 40 kHz
25
(b)
Open-loop amplifier.
2 × 104
2 × 104
A=
⇒| A | =
2
f
⎛ f ⎞
1+ j
1
+
f3dB
⎜
⎟
⎝ f3dB ⎠
f 3dB =
f = 0.25 f 3dB ⇒ A =
2 × 104
1 + (0.25)
f = 5 f 3− dB ⇒ A =
2 × 104
f = 5 f 3− dB ⇒ A =
25
2
= 1.94 × 104
= 3.92 × 103
1 + (5) 2
Closed-loop amplifier
25
f = 0.25 f 3dB ⇒ A =
= 24.25
1 + (0.25) 2
1 + (5) 2
= 4.90
14.14
The open loop gain can be written as
A0
A0 L ( f ) =
⎛
f ⎞⎛
f ⎞
⎜1 + j ⋅
⎟⎜1 + j ⋅
⎟
5
106 ⎠
f
×
PD ⎠ ⎝
⎝
where A0 = 2 × 105.
The closed-loop response is
A0 L
ACL =
1 + β A0 L
At low frequency,
2 × 105
100 =
1 + β (2 × 105 )
So that β = 9.995 × 10−3.
Assuming the second pole is the same for both the open-loop and closed-loop, then
⎛ f ⎞
f ⎞
−1 ⎛
φ = − tan −1 ⎜
⎟ − tan ⎜
⎟
×
5
106 ⎠
f
⎝
⎝ PD ⎠
For a phase margin of 80°, φ = −100°.
So
⎛ f ⎞
−100 = −90 − tan −1 ⎜
6 ⎟
⎝ 5 × 10 ⎠
or
f = 8.816 × 105 Hz
Then
A0 L = 1
=
2 × 105
⎛ 8.816 × 105 ⎞
1+ ⎜
⎟
f PD
⎝
⎠
2
⎛ 8.816 × 105 ⎞
1+ ⎜
⎟
6
⎝ 5 × 10 ⎠
2
or
8.816 × 105
≅ 1.9696 × 105
f PD
or
f PD = 4.48 Hz
14.15
(a)
1st stage
(10) f3− dB = 1 MHz ⇒ f 3− dB = 100 kHz
2nd stage
(50) f3− dB = 1 MHz ⇒ f 3− dB = 20 kHz
Bandwidth of overall system ≅ 20 kHz
(b)
If each stage has the same gain, so
2
K = 500 ⇒ K = 22.36
Then bandwidth of each stage
(22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz
14.16
Ao
A=
1+ j
A=
f
f3− dB
Ao
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
2
⇒ 200 =
5 × 104
⎛ 104 ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
Then
fT = (5 × 104 )(40) ⇒ fT = 2 MHz
14.17
2
⇒ f3− dB = 40 Hz
(5 × 104 ) f PD = 106 ⇒ f PD = 20 Hz
(25) f 3− dB 106 ⇒ f3− dB = 40 kHz
Avo
Av =
1+ j
f
⇒ Av =
f3− dB
25
f
⎛
⎞
1+ ⎜
3 ⎟
⎝ 40 × 10 ⎠
2
At f = 0.5 f 3− dB = 20 kHz
Av =
25
1 + (0.5) 2
= 22.36
At f = 2 f 3− dB = 80 kHz
Av =
25
1 + (2) 2
14.18
(20 × 103 ) ⋅ Avf
= 11.18
MAX
= 106 ⇒ Avf
MAX
= 50
14.19
From Equation (14.55),
10 × 106
SR
=
F P BW =
2π VP 0 2π (10)
or
F P BW = f max = 159 kHz
14.20
a.
VP 0
Using Equation (14.55),
8 × 106
=
2π (250 × 103 )
or
VP 0 = 5.09 V
b.
1
1
=
= 4 × 10−6 s
f 250 × 103
One-fourth period = 1 µ s
Period T =
Slope =
VP 0
= SR = 8 V/µ s
1µ s
⇒ VP 0 = 8 V
14.21
For input (a), maximum output is 5 V.
S R = 1 V/μs
so
For input (b), maximum output is 2 V.
For input (c), maximum output is 0.5 V so the output is
14.22
For input (a), max v01 = 3 V.
Then v02
max
= 3(3) = 9 V
For input (b), max v01 = 1.5 V.
Then v02
max
= 3 (1.5 ) = 4.5 V
14.23
f MAX = 20 kHz, SR = 0.8 V / µ s
V po =
0.8 × 106
SR
=
⇒ V po = 6.37 V
2π f MAX 2π (20 × 103 )
14.24
⎛V ⎞
⎛V ⎞
I1 = I S 1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟
⎝ VT ⎠
⎝ VT ⎠
Want I1 = I 2 , so
⎛V ⎞
5 × 10−14 (1 + x) exp ⎜ BE1 ⎟
I1
⎝ VT ⎠
=1=
I2
⎛V ⎞
5 × 10−14 (1 − x) exp ⎜ BE 2 ⎟
⎝ VT ⎠
=
⎛ V −V ⎞
(1 + x)
exp ⎜ BE1 BE 2 ⎟
VT
(1 − x)
⎝
⎠
Or
⎛ V − VBE1 ⎞
⎛ VOS ⎞
1+ x
= exp ⎜ BE 2
⎟ = exp ⎜
⎟
1− x
V
⎝
⎠
⎝ VT ⎠
T
⎛ 0.0025 ⎞
= exp ⎜
⎟ = 1.10
⎝ 0.026 ⎠
Now
1 + x = (1 − x )(1.10) ⇒
x = 0.0476 ⇒ 4.76%
14.25
From Equation (14.62),
vCE 2
⎛ vCE1 ⎞
⎛
⎜1+ V ⎟ I ⎜ 1+ V
AN ⎟
AN
⎜
= S3 ⋅⎜
⎜ 1 + vEB ⎟ I S 4 ⎜ 1 + vEC 4
⎜ V ⎟
⎜
VAP
AP ⎠
⎝
⎝
⎞
⎟
⎟
⎟
⎟
⎠
For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V so
5 ⎞
⎛
⎛ 0.6 ⎞
⎜ 1 + 80 ⎟ I S 3 ⎜ 1 + 80 ⎟
⋅⎜
⎜
⎟=
⎟
⎜⎜ 1 + 0.6 ⎟⎟ I S 4 ⎜⎜ 1 + 5 ⎟⎟
80 ⎠
80 ⎠
⎝
⎝
or
I S 3 (1.0625) 2
=
= 1.112
I S 4 (1.0075) 2
So
I S 3 = (10−14 ) (1.112 )
or
I S 3 = 1.112 × 10−14 A
14.26
By superposition:
R
vo (vi ) = − 2 ⋅ vi = −50vi
R1
⎛ R ⎞
vo (vos ) = ⎜ 1 + 2 ⎟ ⋅ vos = 51vos
R1 ⎠
⎝
So
vo = vo ( vi ) + vo ( vos ) = −50vi + 51vos
For vi = 20 mV and vos + 2.5 mV
vo = −50(0.02) + 51(0.0025) = −0.8725 V
For vi = 20 mV and vos = −2.5 mV
vo = −50(0.02) + 51(−0.0025) = −1.1275 V
So
−1.1275 ≤ vo ≤ −0.8725 V
14.27
vo = −50vi = −50 ⎡⎣ ±2.5 mV + sin ω t ( mV ) ⎤⎦
vo = [ ±0.125 − 0.25sin ω t ] ( v )
14.28
0.5 × 10−3
= 5 × 10−8 A
104
Also
i
dV
I
1
I = C o ⇒ Vo = ∫ Idt = ⋅ t
dt
C0
C
Then
5 × 10−8
5=
t ⇒ t = 103 s
10 × 10−6
I=
14.29
a.
⎛ 100 ⎞
| v01 | = 10 ⎜1 +
⎟ or | v01 | = 110 mV
10 ⎠
⎝
Then
⎛ 50 ⎞
| v02 | = | v01 | (5) + 10 ⎜1 + ⎟ = (110)(5) + (10)(6)
⎝ 10 ⎠
or
| v02 | = 610 mV
14.30
v0 due to vI
1 ⎞
⎛
v0 = (0.5) ⎜1 + ⎟ = 0.9545 V
⎝ 1.1 ⎠
Wiper arm at V + = 10 V, (using superposition)
⎛ R1 || R5 ⎞
⎛ 0.0909 ⎞
v1 = ⎜
⎟ (10) = ⎜
⎟ (10)
+
R
R
R
||
⎝ 0.0909 + 10 ⎠
4 ⎠
⎝ 1 5
= 0.090
⎛ 1⎞
Then v01 = − ⎜ ⎟ (0.090) = −0.090
⎝ 1⎠
Wiper arm in center, v1 = 0 and v02 = 0
Wiper arm at V − = −10 V, v1 = −0.090
So
v03 = 0.090
Finally, total output v0 : (from superposition)
Wiper arm at V + ,
v0 = 0.8645 V
Wiper arm in center,
v0 = 0.9545 V
Wiper arm at V − ,
v0 = 1.0445 V
14.31
a.
R1′ = R2′ = 0.5 || 25 = 0.490 kΩ
or
R1′ = R2′ = 490 Ω
b.
From Equation (14.75),
⎛ 125 × 10−6 ⎞
(0.026) ln ⎜
+ (0.125) R1′
−14 ⎟
⎝ 2 × 10
⎠
⎛ 125 × 10−6 ⎞
= (0.026) ln ⎜
+ (0.125) R2′
−14 ⎟
⎝ 2.2 × 10 ⎠
0.586452 + (0.125) R1′ = 0.583974 + (0.125) R2′
0.002478 = (0.125)( R2′ − R1′)
So R2′ − R1′ = 0.0198 kΩ ⇒ 19.8 Ω
Then
R2 (1 − x) Rx
R × Rx
− 1
= 0.0198
R2 + (1 − x) Rx R1 + xRx
(0.5)(1 − x)(50)
(0.5)(50) x
−
= 0.0198
(0.5) + (1 − x)(50) (0.5) + x(50)
25(1 − x)
25 x
−
= 0.0198
50.5 − 50 x 0.5 + 50 x
(0.5 + 50 x)(25 − 25 x) − (25 x)(50.5 − 50 x)
= 0.0198
(50.5 − 50 x )(0.5 + 50 x)
25 {0.5 − 0.5 x + 50 x − 50 x 2 − 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x − 25 x − 2500 x 2 }
25 {0.5 − x} = 0.0198 {25.25 + 2500 x − 2500 x 2 }
0.5 − x = 0.019998 + 1.98 x − 1.98 x 2
1.98 x 2 − 2.98 x + 0.48 = 0
x=
2.98 ± (2.98) 2 − 4(1.98)(0.48)
2(1.98)
So
x = 0.183
and
1 − x = 0.817
14.32
R1′ = R1 ||15 = 0.5 ||15 = 0.4839 kΩ
R2′ = R2 || 35 = 0.5 || 35 = 0.4930 kΩ
From Equation (14.75),
⎛i ⎞
⎛i ⎞
(0.026) ln ⎜ C1 ⎟ + iC1 R1′ = (0.026) ln ⎜ C 2 ⎟ + iC 2 R2′
⎝ IS3 ⎠
⎝ IS 4 ⎠
⎛i ⎞
(0.026) ln ⎜ C1 ⎟ = iC 2 R2′ − iC1 R1′
⎝ iC 2 ⎠
⎛i ⎞
⎡ i
R′ ⎤
(0.026) ln ⎜ C1 ⎟ = iC 2 R2′ ⎢1 − C1 ⋅ 1 ⎥
′
i
i
R
2 ⎦
C2
⎝ C2 ⎠
⎣
⎡
⎛i ⎞
⎛ i ⎞⎤
(0.026) ln ⎜ C1 ⎟ = iC 2 (0.4930) ⎢1 − (0.9815) ⎜ C1 ⎟ ⎥
⎝ iC 2 ⎠ ⎦
⎝ iC 2 ⎠
⎣
By trial and error:
iC1 = 252 µ A and iC 2 = 248 µ A
or
iC1
= 1.0155
iC 2
14.33
From Eq. (14.79), we have
⎛ R ⎞
vo = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
R1 ⎠
⎝
I B1 = 1 µ A I B 2 = 2 µ A
Setting vo = 0, we have
⎛ 200 ⎞
0 = (10−6 )( 200 × 103 ) − ( 2 × 10−6 ) R3 ⎜ 1 +
⎟
20 ⎠
⎝
200 × 10−3
⇒ R = 9.09 K
R3 =
( 2 ×10−6 ) (11) 3
14.34
1+
R2
= 80
R1
R1 = 6.329 K
V f = vI = 5sin ω t ( mV )
I1 + I B = I 2
(a)
(b)
VI = 0 ⇒ VX = 0 I 2 = I B
VO = (10−6 )(500 × 103 ) ⇒ vo = 0.50 V
v − VX
VX
+ IB = o
R1
R2
⎛ 1
⎛ 1
v
1 ⎞
1 ⎞
VX ⎜ + ⎟ + I B = 0 ⇒ vo = R2 ⎜ + ⎟ vI + I B R2
R1
⎝ R1 R2 ⎠
⎝ R1 R2 ⎠
vo = 80 ⎡⎣5sin ω t ( mV ) ⎤⎦ + (10−6 )( 500 × 103 )
vo = [ 0.5 + 0.4sin ω t ] ( v )
14.35
a.
For I B 2 = 1 µ A, then v0 = − (10−6 )(104 )
or
v0 = −0.010 V
b.
If a 10 kΩ resistor is included in the feedback loop
Now v0 = − I B 2 (10) + I B1 (10) = 0
Circuit is compensated if I B1 = I B 2 .
14.36
From Equation (14.83), we have
v0 = R2 I 0S
where R2 = 40 kΩ and I 0 S = 3 µ A.
Then
v0 = ( 40 × 103 )( 3 × 10−6 )
or
v0 = 0.12 V
14.37
a.
Assume all bias currents are in the same direction and into each op-amp.
v01 = I B1 (100 kΩ ) = (10−6 )(105 ) ⇒ v01 = 0.1 V
Then
v02 = v01 ( −5 ) + I B1 ( 50 kΩ )
= ( 0.1)( −5 ) + (10−6 )( 5 × 104 )
= −0.5 + 0.05
or
v02 = −0.45 V
b.
Connect R3 = 10 ||100 = 9.09 kΩ resistor to noninverting terminal of first op-amp, and
R3 = 10 || 50 = 8.33 kΩ resistor to noninverting terminal of second op-amp.
14.38
a.
For a constant current through a capacitor.
1 t
v0 = ∫ I dt
C 0
0.1× 10−6
⋅ t ⇒ v0 = (0.1)t
or v0 =
10−6
b.
At t = 10 s,
v0 = 1 V
c.
Then
100 × 10−12
v0 =
⋅ t ⇒ v0 = (10−4 )t
−6
10
At t = 10 s,
v0 = 1 mV
14.39
a.
Assume all bias currents are into the op-amp.
v01 = I B1 ( 50 kΩ ) = (10 ×10−6 )( 50 ×103 )
or
v01 = v02 = 0.5 V
v03 = ( −1)( v01 ) + (10 × 10−6 )( 20 × 103 )
or
v03 = −0.3 V
b.
RA = 10 || 50 ⇒ RA = 8.33 kΩ
RB = 20 || 20 ⇒ RB = 10 kΩ
c.
Assume the worst case offset current, that is, I 0 S = I B1 − I B 2 or I 0 S = I B 2 − I B1 .
From Equation (14.83),
v01 = R2 I 0 S = ( 50 × 103 )( 2 × 10−6 )
or
v01 = v02 = 0.1 V
v03 = ( −1) v01 − I 0 S R2
= ( −1)( 0.1) − ( 2 × 10−6 )( 20 × 103 )
or
v03 = −0.14 V
14.40
a.
Using Equation (14.79),
Circuit (a),
⎛ 50 ⎞
v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 ×10−6 )( 25 × 103 ) ⎜1 + ⎟
⎝ 50 ⎠
or
v0 = 0
Circuit (b),
⎛ 50 ⎞
v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 × 10 −6 )(103 ) ⎜1 + ⎟
⎝ 50 ⎠
−2
= 4 × 10 − 1.6
or
v0 = −1.56 V
b.
Assume I B1 = 0.7 µ A and I B 2 = 0.9 µ A, then using Equation (14.79):
Circuit (a),
⎛ 50 ⎞
v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )( 25 × 103 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 0.045
or
v0 = −0.010 V
Circuit (b),
⎛ 50 ⎞
v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 1.8
or
v0 = −1.765 V
14.41
a.
If R = 0,
⎛ 100 ⎞
v0,max = ⎜1 +
⎟ V0 S + I B (100 kΩ)
10 ⎠
⎝
= (11)(10 × 10−3 ) + (2 × 10−6 )(100 × 103 )
v0,max = 0.110 + 0.20
⇒ v0,max = 0.310 V
b.
R = 10.1|| 100 = 9.17 kΩ = R
14.42
a.
⎛ Ri ⎞
⎜
⎟ (15) = 0.010 V
⎝ Ri + R2 ⎠
15
= 0.0006667
15 + R2
15(1 − 0.0006667) = 0.0006667 R2
Then
R2 = 22.48 MΩ
b.
R1 = Ri || RF = 15 || 10 ⇒ R1 = 6 kΩ
14.43
a.
Assume the offset voltage polarities are such as to produce the worst case values, but the bias
currents are in the same direction.
Use superposition:
Offset voltages
⎛ 100 ⎞
| v01 | = ⎜1 +
⎟ (10) = 110 mV =| v01 |
10 ⎠
⎝
⎛ 50 ⎞
| v02 | = (5)(110) + ⎜ 1 + ⎟ (10)
⎝ 10 ⎠
⇒ | v02 | = 610 mV
Bias Currents:
v01 = I B (100 kΩ) = (2 ×10−6 )(100 × 103 ) = 0.2 V
Then
v02 = (−5)(0.2) + (2 × 10−6 )(50 × 103 ) = −0.9 V
Worst case: v01 is positive and v02 is negative, then
v01 = 0.31 V and v02 = −1.51 V
b.
Compensation network:
If we want
⎛ RB ⎞ +
+
⎜
⎟ V = 20 mV and V = 10 V
+
R
R
C ⎠
⎝ B
⎛ 8.33 ⎞
⎜
⎟ (10) = 0.020
⎝ 8.33 + RC ⎠
or
RC ≅ 4.15 MΩ
14.44
Assume bias currents are in same direction, but assume polarity of offset voltages are such as to
produce the worst case output.
a.
Let I B1 = 5.5 µ A, I B 2 = 4.5 µ A
Bias Current Effects:
v01 = I B1 (50 kΩ) = 0.275 V ⇒ v02 = 0.275 V
v03 = I B1 (20 kΩ) − v01 ⇒ v03 = −0.165 V
Offset Voltage Effects:
⎛ 50 ⎞
v01 = (5) ⎜1 + ⎟ = 30 mV ⇒ v02 = 30 mV
⎝ 10 ⎠
⎛ 20 ⎞
v03 = −v01 − 5 ⎜ 1 + ⎟ ⇒ v03 = −40 mV
⎝ 20 ⎠
Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = −0.205 V
14.45
For circuit (a), effect of bias current:
v0 = (50 × 103 )(100 × 10−9 ) ⇒ 5 mV
Effect of offset voltage
⎛ 50 ⎞
v0 = (2) ⎜1 + ⎟ = 4 mV
⎝ 50 ⎠
So net output voltage is v0 = 9 mV
For circuit (b), effect of bias current:
Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79),
⎛ 50 ⎞
v0 = (450 × 10−9 )(50 × 103 ) − (550 × 10−9 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
−2
= 2.25 × 10 − 1.1
or
v0 = −1.0775 V
If the offset voltage is negative, then
v0 = (−2)(2) = −4 mV
So the net output voltage is
v0 = −1.0815 V
14.46
a.
At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is
v0 = 4 mV
b.
For T = 50°C, the offset voltage for is
V0 S = 2 mV + (0.0067)(25) = 2.1675 mV
so the output voltage for each circuit is
v0 = 4.335 mV
14.47
a.
At T = 25°C,
V0 S = 1 mV, then
⎛ 50 ⎞
v01 = (1) ⎜1 + ⎟ ⇒ v01 = 6 mV
⎝ 10 ⎠
and
⎛ 60 ⎞
⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + (1) ⎜ 1 + ⎟
⎝ 20 ⎠
⎝ 20 ⎠
= 6(4) + (1)(4) ⇒ v02 = 28 mV
b.
At T = 50°C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then
v01 = (1.0825)(6) ⇒ v01 = 6.495 mV
and
v02 = (6.495)(4) + (1.0825)(4)
or
v02 = 30.31 mV
14.48
25°C; I B = 500 nA, I 0 S = 200 nA
50°C, I B = 500 nA + (8 nA / °C)(25°C) = 700 nA
I 0 S = 200 nA + (2 nA / °C)(25°C) = 250 nA
a.
Circuit (a): For I B , bias current cancellation, v0 = 0
Circuit (b): For I B , Equation (14.79),
⎛ 50 ⎞
v0 = (500 × 10−9 )(50 × 103 ) − (500 × 10 −9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.025 − 1.00 ⇒ v0 = −0.975 V
b.
Due to offset bias currents.
Circuit (a):
v0 = (200 × 10−9 )(50 × 103 ) ⇒ v0 = 0.010 V
Circuit (b):
Let I B 2 = 600 nA
I B1 = 400 nA
Then
⎛ 50 ⎞
v0 = (400 × 10−9 )(50 × 103 ) − (600 × 10−9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.020 − 1.20 ⇒ v0 = −1.18 V
c.
Circuit (a): Due to I B , v = 0
Circuit (b): Due to I B ,
⎛ 50 ⎞
v0 = (700 × 10−9 )(50 × 103 ) − (700 × 10−9 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 1.40 ⇒ v0 = −1.365 V
Circuit (a): Due to I 0 S ,
v0 = (250 × 10−9 )(50 × 103 ) ⇒ v0 = 0.0125 V
Circuit (b): Due to I 0 S ,
Let I B 2 = 825 nA
I B1 = 575 nA
Then
⎛ 50 ⎞
v0 = (575 × 10−9 )(50 × 103 ) − (825 × 10−9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.02875 − 1.65 ⇒ v0 = −1.62 V
14.49
25°C; I B = 2 µ A, I 0 S = 0.2 µ A
50°C, I B = 2 µ A + (0.020 µ A / °C)(25°C ) = 2.5 µ A
I 0 S = 0.2 µ A + (0.005 µ A / °C)(25°C) = 0.325 µ A
a.
Due to I B : (Assume bias currents into op-amp).
v01 = I B (50 kΩ) = (2 × 10−6 )(50 × 103 )
⇒ v01 = 0.10 V
⎛ 60 ⎞
⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟
20
⎝
⎠
⎝ 20 ⎠
−6
−6
3
= (0.1)(4) + (2 × 10 )(60 × 10 ) − (2 × 10 )(60 × 103 )4
or
v02 = 0.12 V
b.
Due to I 0 S :
1st op-amp. Let I B1 = 2.1 µ A
2nd op-amp. Let I B1 = 2.1 µ A
I B 2 = 1.9 µ A
v01 = I B1 (50 kΩ) = (2.1× 10−6 )(50 × 103 )
⇒ v01 = 0.105 V
⎛ 60 ⎞
⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟
20
⎝
⎠
⎝ 20 ⎠
3
−6
= (0.105)(4) + (2.1× 10 )(60 × 10 ) − (1.9 × 10 −6 )(50 × 103 )(4)
or
v02 = 0.166 V
c.
Due to I B :
v01 = (2.5 × 10 −6 )(50 × 103 ) ⇒ v01 = 0.125 V
⎛ 60 ⎞
⎛ 60 ⎞
v01 = v02 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜1 + ⎟
20
⎝
⎠
⎝ 20 ⎠
−6
3
= (0.125)(4) + (2.5 × 10 )(60 × 10 ) − (2.5 × 106 )(50 × 103 (4)
or
v02 = 0.15 V
Due to I 0 S :
Let I B1 = 2.625 µ A
I B 2 = 2.3375 µ A
v01 = I B1 (50 kΩ) = (2.6625 × 10−6 )(50 × 103 )
⇒ v01 = 1.133 V
⎛ 60 ⎞
⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜ 1 + ⎟
⎝ 20 ⎠
⎝ 20 ⎠
= (0.133)(4) + (2.6625 × 10−6 )(60 × 103 ) − (2.3375 × 10−6 )(50 × 103 )(4)
or
v02 = 0.224 V
14.50
⎛ R4 ⎞
⎛ R2 ⎞
vB = ⎜
⎟ vI 1 and v0 (vI 2 ) = vB ⎜ 1 + ⎟
R1 ⎠
⎝
⎝ R3 + R4 ⎠
or
⎛ R4 ⎞ ⎛ R2 ⎞
v0 (vI 2 ) = ⎜
⎟ ⎜ 1 + ⎟ vI 2
R1 ⎠
⎝ R3 + R4 ⎠ ⎝
or vI 1 .
v0 (vI 1 ) = −
R2
⋅ vI
R1
Then
⎛ R4 ⎞⎛ R2 ⎞
R2
v0 = ⎜
⎟⎜ 1 + ⎟ vI 2 − ⋅ vI 1
+
R
R
R
R1
4 ⎠⎝
1 ⎠
⎝ 3
V
V
we can write vI 2 = vcm + d and vI 1 = Vcm − d ⋅ Then
2
2
⎛ R4 ⎞⎛ R2 ⎞ ⎛
Vd ⎞ R2 ⎛
Vd ⎞
v0 = ⎜
⎟⎜ 1 + ⎟ ⎜ Vcm + ⎟ − ⋅ ⎜ Vcm − ⎟
R1 ⎠ ⎝
2 ⎠ R1 ⎝
2⎠
⎝ R3 + R4 ⎠ ⎝
Common-mode gain
⎛ R4 ⎞ ⎛ R2 ⎞ R2
v
Acm = 0 = ⎜
⎟ ⎜1 + ⎟ −
Vcm ⎝ R3 + R4 ⎠ ⎝
R1 ⎠ R1
Differential mode gain
v
1 ⎡⎛ R4 ⎞⎛ R2 ⎞ R2 ⎤
Ad = 0 = ⎢⎜
⎟⎜ 1 + ⎟ + ⎥
Vd 2 ⎣⎝ R3 + R4 ⎠ ⎝
R1 ⎠ R1 ⎦
Then
A
CM RR = d
Acm
1 ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤
⋅ ⎢⎜
⎟ ⎜1 + ⎟ + ⎥
R1 ⎠ R1 ⎦
2 ⎣⎝ R3 + R4 ⎠ ⎝
=
⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤
⎢⎜
⎟ ⎜1 + ⎟ − ⎥
R1 ⎠ R1 ⎦
⎣⎝ R3 + R4 ⎠ ⎝
⎡
⎢
1 ⎢ R4
⋅
2 ⎢ R3
⎢
⎣⎢
⎤
⎥
⎛ R2 ⎞ R2 ⎥
1
⋅
⋅ ⎜1 + ⎟ +
R1 ⎠ R1 ⎥
⎛ R4 ⎞ ⎝
+
1
⎥
⎜
⎟
⎝ R3 ⎠
⎦⎥
CM RR =
⎛ R2 ⎞ R2
R4
1
⋅
⋅ ⎜1 + ⎟ −
R3 ⎛ R4 ⎞ ⎝
R1 ⎠ R1
⎜1 + ⎟
⎝ R3 ⎠
Minimum CMRR ⇒ Maximum denominator
R
R
⇒ maximum 4 and minimum 2 . Then
R3
R1
R4 (1.02)(50)
=
= 5.204
R3 (0.98)(10)
R2 (0.98)(50)
=
= 4.804
R1 (1.02)(10)
Then
1 ⎡ 5.204
⎤
⋅ (5.804) + (4.804) ⎥
2 ⎢⎣ 6.204
⎦
CMRR =
⎡ 5.204
⎤
⎢⎣ 6.204 ⋅ (5.804) − (4.804) ⎥⎦
1
⋅ (9.6725)
= 2
(0.06447)
CMRR = 75.0 ⇒ CMRRdB = 20 log10 (75.0)
⇒ CMRRdB = 37.5 dB
14.51
Use the results of Problem 14.50:
R 1 + x ⎛ 50 ⎞
Let 4 =
⋅ ⎜ ⎟ ≈ (1 + 2 x)(5)
R3 1 − x ⎝ 10 ⎠
R 1 − x ⎛ 50 ⎞
Let 2 =
⋅ ⎜ ⎟ ≈ (1 − 2 x)(5)
R1 1 + x ⎝ 10 ⎠
Then
⎤
1 ⎡ (1 + 2 x ) 5
⋅ ( 6 − 10 x ) + (1 − 2 x )( 5 ) ⎥
⎢
2 ⎣ 6 + 10 x
⎦
⎡ (1 + 2 x ) 5
⎤
⋅ ( 6 − 10 x ) − (1 − 2 x )( 5 ) ⎥
⎢
⎣ 6 + 10 x
⎦
1
2
⎡30 + 10 x − 100 x + 30 − 10 x − 100 x 2 ⎤⎦
⎣
= 2
⎡30 + 10 x − 100 x 2 − ( 30 − 10 x − 100 x 2 ) ⎤
⎣
⎦
CMRR =
a.
20 x =
b.
20 x =
1
⋅ [60 − 200 x 2 ]
30 − 100 x 2
2
=
=
20 x
20 x
For CMRRdB = 90 dB ⇒ CMRR = 31, 623 x will be small, neglect the x 2 term. Then
30
⇒ x = 0.0000474 = 0.00474%
31, 623
For CMRRdB = 60 dB ⇒ CMRR = 1000. Then
30
⇒ x = 0.0015 = 0.15%
1000
Chapter 15
Exercise Solutions
EX15.1
For the circuit shown in Figure 15.7
1
f 3dB =
2π RC
or
1
1
=
= 3.979 × 10−6
RC =
2π f3dB 2π ( 40 × 103 )
For R = 75 K
Then
C = 5.31×10−11 = 53.1 pF
We have C3 = 1.414C = 75.1 pF
C4 = 0.707C = 37.5 pF
EX15.2
1
fC =
CReq
or
C=
1
1
=
f c Req (105 )( 20 × 106 )
C = 0.5 pF
EX15.3
Low-frequency gain: T = −
f 3dB =
C1
30
=−
= −6
C2
5
3
−12
fC C2 (100 × 10 )( 5 × 10 )
=
⇒ f 3dB = 6.63 kHz
2π CF
2π (12 × 10−12 )
EX15.4
1
f0 =
2π 3RC
1
1
=
= 6.13 × 10−6
RC =
3
2π f 0 3 2π (15 × 10 ) 3
Let C = 0.001 μF = 1 nF
Then R = 6.13 kΩ so R2 = 8R = 49 kΩ
EX15.5
f0 =
C=
1
1
⇒C =
2π RC
2π f 0 R
1
2π ( 800 ) (104 )
⇒ C ≅ 0.02 µ F
R2 = 2 R1 = 2 (10 ) ⇒ R2 = 20 kΩ
EX15.6
⎛ R1 ⎞
VTH = ⎜
⎟ VH
⎝ R1 + R2 ⎠
⎛ R1 ⎞
2=⎜
⎟ (12)
⎝ R1 + 20 ⎠
2 ( R1 + 20 ) = 12 R1
40 = 10 R1 ⇒ R1 = 4 kΩ
⎛ R1 ⎞
VTH − VTL = ⎜
⎟ (VH − VL )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
0.10 = ⎜
⎟ (10 − [ −10])
⎝ R1 + R2 ⎠
EX15.7
1+
R2
R
20
=
= 200 ⇒ 2 = 199
R1 0.10
R1
⎛ R2 ⎞
VS = ⎜
⎟ VREF
⎝ R1 + R2 ⎠
⎛
R ⎞
1 ⎞
⎛
VREF = ⎜1 + 1 ⎟ VS = ⎜1 +
⎟ (1) ⇒ VREF = 1.005 V
R
199
⎝
⎠
2 ⎠
⎝
I=
VH − VBE ( on ) − Vγ
R + 0.1
10 − 0.7 − 0.7
R + 0.1 =
= 43 kΩ
0.2
R = 42.9 kΩ
EX15.8
At t = 0− , let v0 = −5 so v X = −2.5. For t > 0
⎛ −t ⎞
v X = 10 + ( −2.5 − 10 ) exp ⎜ ⎟
⎝ rX ⎠
When v X = 5.0, output switches
⎛ t ⎞
5.0 = 10 − 12.5 exp ⎜ − 1 ⎟
⎝ rX ⎠
⎛ t ⎞ 10 − 5 5.0
=
exp ⎜ − 1 ⎟ =
⎝ rX ⎠ 12.5 12.5
⎛ t ⎞ 12.5
⎛ 12.5 ⎞
exp ⎜ + 1 ⎟ =
⇒ t1 = rX ⋅ ln ⎜
⎟ ⇒ t1 = rX ( 0.916 )
⎝ 5.0 ⎠
⎝ rX ⎠ 5.0
During the next part of the cycle
⎛ t ⎞
v X = −5 + ( 5 − [ −5]) exp ⎜ − ⎟
⎝ rX ⎠
When v X = −2.5, output switches
⎛ t ⎞
−2.5 = −5 + 10 exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 5 − 2.5 2.5
=
exp ⎜ − 2 ⎟ =
10
10
⎝ rX ⎠
⎛ t ⎞ 10
⎛ 10 ⎞
⇒ t2 = rX ⋅ ln ⎜
exp ⎜ + 2 ⎟ =
⎟ ⇒ t2 = rX (1.39 )
r
2.5
⎝ 2.5 ⎠
⎝ X⎠
1
Period = t1 + t2 = T = ⎡⎣( 0.916 ) + (1.39 ) ⎤⎦ rX = 2.31rX ⇒ Frequency =
2.31rX
rX = ( 50 × 103 )( 0.01× 10 −6 ) = 5 × 10 −4 s ⇒ f = 866 Hz
Duty cycle =
( 0.916 )
t1
× 100% =
× 100% ⇒ Duty cycle = 39.7%
t1 + t2
( 0.916 ) + (1.39 )
EX15.9
a.
rX = RX C X
⎛ R1 ⎞
⎛ 10 ⎞
vY = ⎜
⎟ v0 = ⎜
⎟ (12 ) = 1.2 V
⎝ 10 + 90 ⎠
⎝ R1 + R2 ⎠
R1
= 0.10
β=
R1 + R2
0.7 ⎤
⎡
⎢ 1 + 12 ⎥
⎡1 + Vγ /VP ⎤
T = rX ln ⎢
⎥
⎥ = rX ln ⎢
⎣ 1− β ⎦
⎢1 − (0.10) ⎥
⎣⎢
⎦⎥
T = 50 × 10−6 = rX ln [1.18] = (0.162) rX
50 × 10−6
⇒ RX = 3.09 kΩ
(0.1× 10−6 )(0.162)
b.
Recovery time
⎛ t ⎞
v X = VP + (−1.2 − VP ) exp ⎜ − ⎟
⎝ rX ⎠
RX =
When v X = Vγ , t = t2
⎛ t ⎞
0.7 = 12 + ( −1.2 − 12 ) exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 12 − 0.7
= 0.856
exp ⎜ − 2 ⎟ =
13.2
⎝ rX ⎠
⎛ 1 ⎞
t2 = rX ln ⎜
⎟ = ( 0.155 ) rX
⎝ 0.856 ⎠
rX = ( 3.09 × 103 )( 0.1× 10−6 ) = 3.09 × 10−4 ⇒ t2 = 48.0 µ s
EX15.10
T = 1.1 RC
T = 75 × 10−6
Let C = 10 nF
Then
75 × 10−6
R=
= 6.82 K
(1.1) (10 ×10−9 )
EX15.11
1
1
=
⇒ f = 802 Hz
0.693 ( RA + 2 RB ) C ( 0.693) ⎡⎣ 20 + 2 ( 80 ) ⎤⎦ × 103 × ( 0.01× 10−6 )
R + RB
20 + 80
Duty cycle = A
× 100% =
× 100% ⇒ Duty cycle = 55.6%
20 + 2 ( 80 )
RA + 2 RB
f =
EX15.12
P=
a.
1 VP2
⋅
2 RL
VP = 2 RL P = 2 ( 8 )(1) ⇒ VP = 4 V
IP =
VP 4
= ⇒ I P = 0.5 A
RL 8
VCE = 12 − 4 = 8 V
b.
I C ≈ 0.5 A
So P = I C ⋅ VCE = ( 0.5 )( 8 ) ⇒ P = 4 W
EX15.13
VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V
⎛ V ⎞
PS = VS ⎜ P ⎟
⎝ π RL ⎠
VS =
PS π R2 (10 ) π ( 8 )
=
VP
12.65
VS = 19.9 V
EX15.14
Line regulation =
Now
dV0
dV dV
= 0 ⋅ Z+
+
dV
dVZ dV
dV0 ⎛ 10 ⎞
= ⎜1 + ⎟ = 2
dVZ ⎝ 10 ⎠
dVZ ⎛ rZ ⎞
10
=⎜
= 0.00227
⎟=
+
dV
⎝ rZ + R1 ⎠ 10 + 4400
So Line regulation = ( 2 )( 0.00227 ) = 0.00454
0.454%
EX15.15
V1 V0 − V1
⎛1 1⎞ V
=
⇒ V1 ⎜ + ⎟ = 0
10
10
⎝ 10 10 ⎠ 10
V
⎛ 2⎞ V
V1 ⎜ ⎟ = 0 ⇒ V0 = 2V1 ⇒ V1 = 0
10
10
2
⎝ ⎠
V0 − V1 V0 V0 − A0 L (VZ − V1 )
+
+
=0
10
RL
R0
V0 V0 V0 A0 LVZ V1 A0 LV1
+
+
−
= −
10 RL R0
10
R0
R0
=
V0
A V
− 0L 0
2(10) 2 R0
V0
V 1000 ( 6.3) V0 (1000 ) V0
+ I0 + 0 −
=
−
10
0.5
0.5
20
2 ( 0.5 )
V0 [0.10 + 2.0 − 0.05 + 1000] + I 0 = 12, 600
V0 (1002.05) + I 0 = 12, 600
For I 0 = 1 mA ⇒ V0 = 12.5732
For I 0 = 100 mA ⇒ V0 = 12.4744
Load reg =
V0 ( NL ) − V0 ( FL )
V0 ( NL )
× 100%
12.5732 − 12.4744
× 100%
12.5732
Load reg = 0.786%
=
EX15.16
a.
IC 3 =
VZ − 3VBE ( on )
IC 3 =
R1 + R2 + R3
5.6 − 3 ( 0.6 )
3.9 + 3.4 + 0.576
⎛I ⎞
I C 4 R4 = VT ln ⎜ C 3 ⎟
⎝ IC 4 ⎠
=
3.8
⇒ I C 3 = 0.482 mA
7.88
⎛ 0.482 ⎞
I C 4 (0.1) = (0.026) ln ⎜
⎟
⎝ IC 4 ⎠
By trial and error
I C 4 = 0.213 mA
VB 7 = 2(0.6) + (0.482)(3.9) ⇒ VB 7 = 3.08 V
b.
⎛ R13 ⎞
⎜
⎟ V0 = VB 8 = VB 7
⎝ R13 + R12 ⎠
⎛ 2.23 ⎞
⎜
⎟ (5) = 3.08
⎝ 2.23 + R12 ⎠
( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R12
11.15 = 6.868 = 3.08R12 ⇒ R12 = 1.39 kΩ
TYU15.1
1
2π RC
1
1
=
= 1.59 × 10−5
RC =
2π f3dB 2π (104 )
f 3dB =
Let C = 0.01 µ F ⇒ R = 1.59 kΩ
Then
C1 = 0.03546 µ F
C2 = 0.01392 µ F
C3 = 0.002024 µ F
1
1
⎛ f ⎞
⎛ 20 ⎞
1+ ⎜ ⎟
1+ ⎜
⎟
⎝ 10 ⎠
⎝ f3d ⎠
T = 0.124 or T = −18.1 dB
T =
6
=
6
TYU15.2
f 3dB =
RC =
1
1
⇒ RC =
2π RC
2π f3dB
1
2π ( 50 × 103 )
= 3.18 × 10−6
Let C = 0.001 µ F = 1 nF ⇒ R = 3.18 kΩ
Then
R1 = 2.94 kΩ
R2 = 3.44 kΩ
R3 = 1.22 kΩ
R4 = 8.31 kΩ
T = 0.01 =
1
⎛ f
⎞
1 + ⎜ 3− dB ⎟
⎝ f ⎠
8
⎛ f
⎞ ⎛ 1 ⎞
4
1 + ⎜ 3− dB ⎟ = ⎜
⎟ = 10
f
0.01
⎝
⎠
⎝
⎠
8
2
⎛ f3− dB ⎞
f 3dB
⇒ f ≅ 15.8 kHz
⎜
⎟ ≅ 10 ⇒ f =
10
⎝ f ⎠
2
TYU15.3
1-pole
T =
2-pole
T =
3-pole
T =
4-pole
T =
2
⇒ −3.87 dB
4
⇒ −4.88 dB
6
⇒ −6.0 dB
8
⇒ −7.24 dB
1
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
TYU15.4
1
Req =
fC C
or f C C =
1
1
=
= 2 × 10−7
Req 5 × 106
If C = 10 pF ⇒ fC = 20 kHz
TYU15.5
1
1
f0 =
=
⇒ f 0 ≅ 65 kHz
4
2π 6 RC 2π 6 (10 )(100 × 10−12 )
R2 = 29 R = 29 (104 ) ⇒ R2 = 290 kΩ
TYU15.6
f0 =
1
⎛ CC ⎞
2π L ⋅ ⎜ 1 2 ⎟
⎝ C1 + C2 ⎠
=
1
⎡ (10−9 ) 2
2π (10 ) ⎢
−9
⎣ 2 × 10
−6
⎤
⋅⎥
⎦
⇒ f 0 = 7.12 MHz
C2
= gm R
C1
C2 1
1
⋅ =
⇒ g m = 0.25 mA / V
C1 R 4 × 103
We have
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜ ⎟⎜ ⎟ (VGS − VTh )
⎝ 2 ⎠⎝ L ⎠
k ′ ≅ 20 µ A / V 2 , VGS − VTh ≅ 1 V
gm =
So
W
0.25 × 10−3
=
= 12.5
L ( 20 × 10−6 ) (1)
and a value of W / L = 12.5 is certainly reasonable.
TYU15.7
⎛R ⎞
VTH = − ⎜ 1 ⎟ VL
⎝ R2 ⎠
⎛R ⎞
R
0.10 = − ⎜ 1 ⎟ ( −10 ) ⇒ 1 = 0.010
R2
⎝ R2 ⎠
Let R1 = 0.10 kΩ then R2 = 10 kΩ
TYU15.8
a.
⎛ R2 ⎞
⎛ 10 ⎞
VS = ⎜
⎟ VREF = ⎜
⎟ ( 2)
⎝ 1 + 10 ⎠
⎝ R1 + R2 ⎠
VS = 1.82 V
⎛ R1 ⎞
⎛ 1 ⎞
VTH = VS + ⎜
⎟ VH = 1.82 + ⎜
⎟ (10 )
+
R
R
1 + 10 ⎠
⎝
⎝ 1
2 ⎠
VTH = 2.73 V
VTL =
b.
⎛ R1 ⎞
⎛ 1 ⎞
VS + ⎜
⎟ VL = 1.82 + ⎜
⎟ ( −10 )
+
R
R
⎝ 1 + 10 ⎠
⎝ 1
2 ⎠
VTL = 0.91 V
TYU15.9
⎛
R ⎞
VS = ⎜ 1 + 1 ⎟ VREF
⎝ R2 ⎠
⎛R ⎞
⎛R ⎞
VTH = VS − ⎜ 1 ⎟ VL and VTL = VS − ⎜ 1 ⎟ VH
⎝ R2 ⎠
⎝ R2 ⎠
⎛R ⎞
Hysteresis Width = VTH − VTL = ⎜ 1 ⎟ (VH − VL )
⎝ R2 ⎠
⎛R ⎞
⎛R ⎞
2.5 = ⎜ 1 ⎟ ( 5 − [ −5]) = 10 ⎜ 1 ⎟
R
⎝ 2⎠
⎝ R2 ⎠
R
So 1 = 0.25
R2
Then
⎛
R ⎞
VS = −1 = ⎜ 1 + 1 ⎟ VREF = (1 + 0.25)VREF ⇒ VREF = −0.8 V
R
⎝
2 ⎠
Then
VTH = −1 − ( 0.25 )( −5 ) ⇒ VTH = 0.25 V
VTL = −1 − ( 0.25 )( 5 ) ⇒ VTL = −2.25 V
TYU15.10
⎛ R1 ⎞
1
⎛ 10 ⎞
vX = ⎜
⎟ v0 = ⎜
⎟ v0 = v0
3
⎝ 10 + 20 ⎠
⎝ R1 + R2 ⎠
10
t = 0,
vX = −
3
⎛ t ⎞
⎛ 10
⎞
v X = 10 + ⎜ − − 10 ⎟ exp ⎜ − ⎟
⎝ 3
⎠
⎝ rX ⎠
Output switches when v X =
10
3
10
= 10 − 13.33 exp
3
⎛ t1 ⎞
⎜− ⎟
⎝ rX ⎠
⎛ t ⎞ 10 − 3.33 6.67
exp ⎜ − 1 ⎟ =
=
13.33
13.33
⎝ rX ⎠
⎛ t ⎞ 13.33
exp ⎜ + 1 ⎟ =
≅2
⎝ rX ⎠ 6.67
t1 = rX ln (2) = (0.693)rX
T = 2(0.693)rX
f =
1
2(0.693)rX
rX = RX C X = (104 )( 0.1×10 −6 ) = 1×10 −3 ⇒ f = 722 Hz ⇒ Duty cycle = 50%
TYU15.11
⎛ R1 ⎞
20
= 0.333
β =⎜
⎟=
+
+ 40
20
R
R
⎝ 1
2 ⎠
rX = RX C X = (104 )( 0.01× 10−6 ) = 1× 10−4
⎛ 1 + Vγ / VP
T = rX ln ⎜
⎝ 1− β
Recovery time
0.7 ⎤
⎡
⎢ 1+ 8 ⎥
⎞
−4
⎥ ⇒ T = 48.9 µ s
⎟ = (1× 10 ) ln ⎢
⎠
⎢1 − 0.333 ⎥
⎣⎢
⎦⎥
⎛ R1 ⎞
⎛ 20 ⎞
vY = ⎜
⎟ v0 = ⎜
⎟ (8) = 2.667 V
+
R
R
⎝ 20 + 40 ⎠
2 ⎠
⎝ 1
⎛ t ⎞
0.7 = 8 + ( −2.667 − 8 ) exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 8 − 0.7
exp ⎜ − 2 ⎟ =
= 0.6844
⎝ rX ⎠ 10.66
⎛ 1 ⎞
t2 = rX ln ⎜
⎟ ⇒ t2 = 37.9 µ s
⎝ 0.685 ⎠
TYU15.12
f =
1
( 0.693)( RA + RB ) C
RA + RB =
1
( 0.693) fC
Let C = 0.01 µ F,
RA + RB =
f = 1kHz
1
( 0.693) (103 )( 0.01×10−6 )
Duty cycle = 55 =
= 1.443 × 105
RA + RB
× 100%
RA + 2 RB
(1.443 ×10 ) (100 )
(1.443 ×10 ) + R
(1.443 ×10 ) (100 − 55) ⇒ R
=
5
55 =
5
B
5
RB
55
B
= 118 kΩ so RA = 26.2 kΩ
TYU15.13
v01 ⎛ R2 ⎞ ⎛ 30 ⎞
a.
= ⎜ 1 + ⎟ = ⎜1 + ⎟ = 2.5
vI ⎝
R1 ⎠ ⎝ 20 ⎠
v02
R
50
=− 4 =−
= −2.5
20
vI
R3
(b)
P=
1 VL2 1 [12 − (−12)]2
⋅
= ⋅
= 240 mW
2 RL 2
1.2
Or
P = 0.24 W
c.
12
= V pi = 4.8 V
2.5
Chapter 15
Problem Solutions
15.1
(a)
For example:
vo ⎛ R2 ⎞
R
= ⎜1 + ⎟ = 10 ⇒ 2 = 9
vi ⎝
R1 ⎠
R1
Corner Frequency:
1
f =
= 5 × 103 ⇒ RC = 3.18 × 10−5
2π RC
(b)
For Example:
Low-Frequency:
1
− R2
jω C
vo
R
1
=
=− 2⋅
vi
R1
R1 1 + jω R2 C
So, set
R2
= 15 ⇒ For example, R1 = 10 k Ω, R2 = 150 k Ω
R1
R2 C =
1
1
=
= 1.06 × 10−5
2π f3− dB 2π (15 × 103 )
Then C = 70.7 pF
15.2
(a)
Av =
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
2
=
1
1 + (2) 2
= 0.447 ⇒ Av = −7 dB
(b)
Av =
(c)
Av =
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
2
2
=
=
1
1 + (2) 4
1
1 + (2)6
= 0.2425 ⇒ Av = −12.3 dB
= 0.1240 ⇒ Av = −18.1 dB
15.3
(a)
Figure 15.6
1
f =
2π RC
1
1
=
= 7.958 × 10−6
RC =
2π f 2π (20 × 103 )
Let R1 = R2 = R = 10 K
C = 795.8 pF
So C4 = 0.707C = 562.6 pF
C3 = 1.414C = 1.125 nF
(b)
1
T =
= 0.777
(i)
4
⎛ 18 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
1
T =
= 0.707
(ii)
4
⎛ 20 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
1
T =
= 0.637
(iii)
4
⎛ 22 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
15.4
Use Figure 15.10(b)
1
f3− dB =
2π RC
or
1
RC =
= 3.18 × 10−6
2π (50 × 103 )
For example, let C = 100 pF Then R = 31.8 kΩ
And R1 = 8.97 kΩ
R2 = 22.8 kΩ
R3 = 157 kΩ
From Equation (15.26)
1
T =
6
⎛ f ⎞
1 + ⎜ 3dB ⎟
⎝ f ⎠
We find
f kHz
30
35
40
45
|T|
0.211
0.324
0.456
0.589
15.5
From Equation (15.7).
Y1Y2
T (s) =
Y1Y2 + Y4 (Y1 + Y2 + Y3 )
For a high-pass filter, let Y1 = Y2 = sC ,
Y3 =
1
1
, and Y4 =
R3
R4
Then
s 2C 2
1 ⎛
1 ⎞
s 2 C 2 + ⎜ sC + sC + ⎟
R4 ⎝
R3 ⎠
1
=
1 ⎛
1 ⎞
1+
⎜2+
⎟
sR4 C ⎝
sR3C ⎠
T (s) =
Define r3 = R3C and r4 = R4 C
1
1 ⎛
1 ⎞
1+
⎜2+
⎟
sr4 ⎝
sr3 ⎠
Set s = jω
T (s) =
1
1 ⎛
1 ⎞
1+
⎜2+
⎟
jω r4 ⎝
jω r3 ⎠
1
=
j ⎛
j ⎞
1−
⎜2−
⎟
ω r4 ⎝ ω r3 ⎠
T ( jω ) =
=
1
⎛
1 ⎞ 2j
⎜1 − 2
⎟−
ω
r3 r4 ⎠ ω r4
⎝
2
⎧⎪⎛
1 ⎞
4 ⎫⎪
T ( jω ) = ⎨⎜1 − 2
⎟ + 2 2⎬
⎪⎩⎝ ω r3 r4 ⎠ ω r4 ⎪⎭
For a maximally flat filter, we want
dT
=0
dω ω →∞
−1/ 2
Taking the derivative, we find
d T ( jω )
dω
2
1 ⎧⎪⎛
1 ⎞
4 ⎫⎪
= − ⎨⎜1 − 2
⎟ + 2 2⎬
2 ⎪⎝ ω r3 r4 ⎠ ω r4 ⎪
⎩
⎭
−3 / 2
⎡ ⎛
1 ⎞ ⎛ 2 ⎞ 4(−2) ⎤
× ⎢ 2 ⎜1 − 2
⎟⎜ 3
⎟+ 3 2 ⎥
⎣⎢ ⎝ ω r3 r4 ⎠ ⎝ ω r3 r4 ⎠ ω r4 ⎦⎥
or
d T ( jω )
dω
=0
ω →∞
⎡⎛ 4 ⎞⎛
1 ⎞
8 ⎤
= ⎢⎜ 3
⎟⎜ 1 − 2
⎟− 3 2 ⎥
⎢⎣⎝ ω r3 r4 ⎠⎝ ω r3 r4 ⎠ ω r4 ⎥⎦
=
4 ⎡ 1
⎢
ω 3 ⎣⎢ r3 r4
Then
⎡ 1 ⎛
1 ⎞ 2⎤
⎢
⎜1 − 2
⎟− 2 ⎥
⎢⎣ r3 r4 ⎝ ω r3 r4 ⎠ r4 ⎥⎦
⎛
1 ⎞ 2⎤
⎜1 − 2
⎟− 2 ⎥
⎝ ω r3 r4 ⎠ r4 ⎦⎥
=0
ω →∞
1 2
= ⇒ 2r3 = r4
r3 r4
Then the transfer function can be written as:
So that
2
⎧⎪ ⎡
⎤
1
4 ⎫⎪
T ( jω ) = ⎨ ⎢1 − 2 2 ⎥ + 2 2 ⎬
ω (2r3 ) ⎦ ω (4r3 ) ⎪
⎩⎪ ⎣
⎭
−1/ 2
⎧
1
1
1 ⎫
= ⎨1 − 2 2 +
+ 2 2⎬
2 2 2
4(
)
ω
ω
ω
r
r
r3 ⎭
3
3
⎩
⎧
⎫
1
= ⎨1 +
2 2 2 ⎬
⎩ 4(ω r3 ) ⎭
3 − dB frequency
1
2ω 2 r32 = 1 or ω =
=
2r3
Define
1
ω=
RC
So that
R
R3 =
2
−1/ 2
−1/ 2
1
2 R3C
We had 2r3 = r4 or 2( R3C ) = R4 C ⇒ R4 = 2 R3
So that R4 = 2 ⋅ R
15.6
From Equation (15.25)
1
− 25 dB ⇒ T = 0.0562
T =
2N
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3− dB ⎠
f
f 3− dB
So
=
20
=2
10
0.0562 =
1
1 + (2) 2 N
1 + (2) 2 N = 316.6 ⇒ (2)2 N = 315.6
2 N ⋅ ln (2) = ln (315.6) ⇒ N = 4.15 ⇒ N = 5 A 5-pole filter
15.7
T =
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
At f = 12 kHz, T = 0.9
1
0.9 =
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
Also
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
=
2N
⎛ 14 ⎞
⎜
⎟
⎝ f3dB ⎠
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
=
⎛ 12 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
1
− 1 = 0.2346
(0.9) 2
1
0.01 =
⎛ 14 ⎞
⎜
⎟
⎝ f3dB ⎠
1
=
2N
⎛ 14 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
1
− 1 = 9999
(0.01) 2
2N
2N
⎛ 14 ⎞
=⎜ ⎟
⎝ 12 ⎠
2N
=
9999
= 4.262 × 104
0.2346
(1.16667) 2 N = 4.262 × 104
N = 35
Then
0.9 =
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
1
⎛ 12 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
2N
= 0.2346
⎛
⎞
⎜
⎟
⎛ 12 ⎞
⎝ 2N ⎠
= (0.2346)0.014286
⎜
⎟ = (0.2346)
⎝ f3dB ⎠
= 0.9795
So
f 3dB = 12.25 kHz
1
15.8
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
(a)
For N = 3
1
T =
= 0.2841
1 + (1.5)6
T =
2N
(b)
For N = 5
1
T =
= 0.1306
1 + (1.5)10
(c)
For N = 7
1
T =
= 0.05843
1 + (1.5)14
15.9
Consider
For low-frequency:
vo
R2 + R3
=
vi R1 + R2 + R3
For high-frequency:
vo
R2
=
vi R1 + R2
So we need
⎛ R2 ⎞
R2 + R3
= 25 ⎜
⎟
R1 + R2 + R3
⎝ R1 + R2 ⎠
Let R1 + R2 = 50 k Ω and R2 = 1.5 k Ω ⇒ R1 = 48.5 k Ω
Then
1.5 + R3
⎛ 1.5 ⎞
= 25 ⎜ ⎟ ⇒ R3 = 144 k Ω
50 + R3
⎝ 50 ⎠
Connect the output of this circuit to a non-inverting op-amp circuit.
At low-frequency:
vo1 =
R2 + R3
1.5 + 144
⋅ vi =
⋅ vi = 0.75vi
R1 + R2 + R3
48.5 + 1.5 + 144
Need to have vo = 25.
⎛ R ⎞
⎛ R ⎞
R
vo = 25 = ⎜1 + 5 ⎟ ⋅ vo1 = ⎜ 1 + 5 ⎟ (0.75)vi ⇒ 5 = 32.3
R4
⎝ R4 ⎠
⎝ R4 ⎠
To check at high-frequency.
R2
1.5
vo1 =
vi =
vi = 0.03vi
1.5 + 48.5
R1 + R2
vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi
which meets the design specification
Consider the frequency response.
1
R2 + R3
vo1
sC
=
1
vi
R1 + R2 + R3
sC
Now
R3
1
=
R3
sC 1 + sR3C
Then, we find
vo1
R3 + R2 (1 + sR3C )
=
vi
R3 + ( R1 + R2 )(1 + sR3C )
which can be rearranged as
( R2 + R3 ) (1 + s ( R2 || R3 )C )
vo1
=
vi
( R1 + R2 + R3 ) 1 + s ( R3 || ( R1 + R2 ) ) C
(
)
So
fL ≅
1
1
1
=
=
3
2π ( R2 || R3 ) C 2π (1.5 || 144 ) × 10 C ( 9.33 × 103 ) C
fH ≅
1
1
=
2π ( R3 || ( R1 + R2 ) ) C 2π (144 || 50 ) × 103 C
=
Set
1
( 2.33 ×10 ) C
5
⎤
fL + fH 1 ⎡
1
1
⎥
= ⎢
+
3
5
2
2 ⎢ ( 9.33 × 10 ) C ( 2.33 × 10 ) C ⎥
⎣
⎦
Which yields
C = 2.23 nF
25 kHz =
15.10
Av =
So
10−5 =
or
− 100 dB ⇒ 10−5
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
2N
1
⎛ 770 ⎞
1+ ⎜
⎟
⎝ 12 ⎠
2N
⎛ 1 ⎞
1 + (64.2) 2 N = ⎜ −5 ⎟ = 1010
⎝ 10 ⎠
or
(64.2) 2 N ≅ 1010
Now
2
N
1
2
3
Left Side
4.112 × 103
1.7 × 107
7 × 1010
So, we need a 3rd order filter.
15.11
Low-pass: −50 dB ⇒ 3.16 × 10−3
Then
1
1
3.16 × 10−3 =
=
4
4
⎛ f ⎞
⎛ 60 ⎞
1+ ⎜ ⎟
1+ ⎜ ⎟
⎝ fL ⎠
⎝ fL ⎠
We find f L = 3.37 Hz
High Pass:
1
1
=
3.16 × 10−3 =
4
4
⎛ fH ⎞
⎛ fH ⎞
1
+
1+ ⎜
⎜ 60 ⎟
⎟
⎝
⎠
⎝ f ⎠
We find f H = 1067 Hz
Bandwidth: BW = f H − f L = 1067 − 3.37 ⇒
BW ≅ 1064 Hz
15.12
a.
v
vI
= − 02 −
R4
R3
v0
⎛ 1 ⎞
R1 ⎜
⎟
⎝ sC ⎠
(1)
v0
v
= − 01
(2)
R2
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
v01
v
(3)
= − 02 ⇒ v01 = −v02
R5
R5
Then
⎛ 1 ⎞
v0
v
= + 02 or v02 = v0 ⎜
⎟ (2)
R2
⎛ 1 ⎞
⎝ sR2 C ⎠
⎜
⎟
⎝ sC ⎠
And
⎛ 1 ⎞
v0
⋅⎜
⎟−
sR
C
⎝ 2 ⎠ R ⎛ 1 ⎞
1 ⎜
⎟
⎝ sC ⎠
v
vI
=− 0
R4
R3
⎡
⎤
⎢
⎥
1
1
⎥
= −v0 ⎢
+
⎢ R3 ( sR2 C ) R1 ⋅ (1/ sC ) ⎥
⎢⎣
R1 + (1/ sC ) ⎥⎦
(1)
⎡
1 + sR1C ⎤
1
= −v0 ⎢
+
⎥
R1 ⎦
⎣ R3 ( sR2 C )
⎡ R + (1 + sR1C )( sR2 R3C ) ⎤
= −v0 ⎢ 1
⎥
( sC ) R1 R2 R3
⎣
⎦
Then
⎤
( sC )( R1 R2 R3 )
v0
1 ⎡
=− ⎢
⎥
vI
R4 ⎣ R1 + sR2 R3C + s 2 R1 R2 R3C 2 ⎦
or
−
Av ( s ) =
1
R4
v0
=
1
1
vI
+ sC +
R1
sCR2 R3
−
b.
Av ( jω ) =
1
R4
1
1
+ jω C +
R1
jω CR2 R3
or
−
1
R4
⎡
⎤
1
1
+ j ⎢ω C −
⎥
ω
R1
CR
R
2 3 ⎦
⎣
R
1
=− 1⋅
R4 ⎪⎧
⎡
R1 ⎤ ⎪⎫
⎨1 + j ⎢ω R1C −
⎬
ω CR2 R3 ⎥⎦ ⎪⎭
⎪⎩
⎣
R
1
Av ( jω ) = 1 ⋅
2 −1/ 2
R4 ⎧
R1 ⎤ ⎫⎪
⎪ ⎡
⎨1 + ⎢ω R1C −
⎬
ω CR2 R3 ⎥⎦ ⎭⎪
⎩⎪ ⎣
Av ( jω ) =
Av
⎡
R1 ⎤
when ⎢ω R1C −
⎥=0
ω CR2 R3 ⎦
⎣
max
Then
Av
max
Now
=
⎡
R1 85
=
⇒ Av
R4
3
ω R1C ⎢1 −
Then
⎣
max
= 28.3
⎤
1
1
⎥ = 0 or ω =
2
ω C R2 R3 ⎦
C R2 R3
2
f =
1
2π C R2 R3
=
1
2π (0.1× 10−6 ) (300) 2
So
f = 5.305 kHz
To find the two 3 − dB frequencies,
⎡
R1 ⎤
⎢ω R1C −
⎥ = ±1
ω CR2 R3 ⎦
⎣
ω 2 R1 R2 R3 C 2 − R1 = ±ω R2 R3 C
ω 2 (85 × 103 )(300) 2 (0.1× 10−6 ) 2 − 85 × 103 = ±ω (300) 2 (0.1× 10−6 )
ω 2 (7.65× 10−5 ) − 85 × 103 = ±ω (9 × 10−3 )
ω 2 (7.65× 10−5 ) ± ω (9 × 10−3 ) − 85 × 103 = 0
(9 × 10−3 ) 2 + 4(7.65 × 10−5 )(85 × 10−3 )
± (9 × 10−3 )
±
2(7.65 × 10−5 )
2(7.65 × 10−5 )
We find f = 5.315 kHz and f = 5.296 kHz
ω=
15.13
a.
vI − v A
v
= A
(1)
R
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
vI − vB vB − v0
(2)
=
R
R
and v A = vB
So
vI
⎛1
⎞
⎛ 1 + sRC ⎞
= v A ⎜ + sC ⎟ = v A ⎜
⎟ (1)
R
R
⎝
⎠
⎝ R ⎠
or
vI
vA =
1 + sRC
Then
2vI
vI + v0 = 2vB = 2vA =
(2)
1 + sRC
⎡ 2
⎤
⎡1 − sRC ⎤
v0 = vI ⎢
− 1⎥ = vI ⎢
⎥
sRC
1
+
⎣
⎦
⎣1 + sRC ⎦
Now
v0
1 − jω RC
= A( jω ) =
1 + jω RC
vI
A=
1 + ω 2 R2C 2
1 + ω 2 R2C 2
⇒ A =1
Phase:
φ = −2 tan −1 (ω RC )
b.
f
0
RC = (104 )(15.9 × 10−9 ) = 1.59 × 10−4
φ
0
−11.4
−53.1
−90°
−157
−169
10 2
5 × 103
1/ 2π RC = 103 Hz
5 × 103
10 4
15.14
a.
Vi
Vi − V0
+
=0
R1 R2 || (1/ sC )
Vi
Vi − V0
+
=0
R1 ⎡ R2 ⎤
⎢1 + sR C ⎥
2
⎣
⎦
R2
1
(Vi ) + Vi = V0
⋅
R1 1 + sR2 C
V0 R2 + R1 (1 + sR2 C ) ( R2 + R1 ) [1 + s ( R1 || R2 )C ]
=
=
Vi
R1 (1 + sR2 C )
R1 (1 + sR2 C )
⇒
V0 ⎛ R2 ⎞ ⎡1 + s ( R1 || R2 )C ⎤
= ⎜1 + ⎟ ⎢
⎥
Vi ⎝
R1 ⎠ ⎣ (1 + sR2 C ) ⎦
⇒ f 3dB1 =
⇒ f 3dB2 =
1
2π R2 C
1
2π ( R1 || R2 )C
b.
Vi
V − V0
+ i
=0
R1 || (1/ sC )
R2
Vi
V
V
+ i = 0
⎛ R1 ⎞ R2 R2
⎜
⎟
⎝ 1 + sR1C ⎠
⎡R
⎤
Vi ⎢ 2 ⋅ (1 + sR1C ) + 1⎥ = V0
⎣ R1
⎦
Vi
⋅ [ R2 + R1 + sR1 R2 C ] = V0
R1
V0 R2 + R1
V ⎛ R ⎞
1
=
⋅ [1 + s ( R1 || R2 )C ] ⇒ 0 = ⎜1 + 2 ⎟ [1 + s ( R1 || R2 )C ] ⇒ f3dB =
2π ( R1 || R2 )C
Vi
R1
Vi ⎝
R1 ⎠
15.15
a.
−V0
Vi
=
R1 + (1/ sC1 ) R2 || (1/ sC2 )
⎛ sC1 ⎞
⎛ 1 + sR2 C2 ⎞
Vi ⎜
⎟ = −V0 ⎜
⎟
⎝ 1 + sR1C1 ⎠
⎝ sC2 ⎠
V0
− sR2 C1
− sR2 C1
=
=
Vi (1 + sR1C1 )(1 + sR2 C2 ) 1 + sR1C1 + sR2 C2 + s 2 R1 R2 C1C2
⎡
⎤
⎢
⎥
V0
R
sC1
⎥
= − 2 ×⎢
⎥
Vi
R1 ⎢ 1
⎛ R2 C2 ⎞ 2
⎢ + sC1 ⎜ 1 + ⋅ ⎟ + s R2 C1C2 ⎥
R1 C1 ⎠
⎝
⎣⎢ R1
⎦⎥
or
⎡
⎤
⎢
⎥
V0
R2 ⎢
1
⎥
=− ⋅
T (s) =
⎥
Vi
R1 ⎢ 1
⎛ R2 C2 ⎞
+ ⎜1 + ⋅ ⎟ + sR2 C2 ⎥
⎢
R1 C1 ⎠
⎣⎢ sR1C1 ⎝
⎦⎥
b.
T ( jω ) = −
R2
1
×
2
2 1/ 2
R1 ⎧
1 ⎞ ⎫⎪
⎪⎛ R2 C2 ⎞ ⎛
⎨⎜ 1 + . ⎟ + ⎜ ω R2 C2 −
⎟ ⎬
R1 C1 ⎠ ⎝
ω R1C1 ⎠ ⎪
⎩⎪⎝
⎭
⎛
1 ⎞
when ⎜ ω R2 C2 −
⎟ = 0, we want
ω R1C1 ⎠
⎝
R
1
T ( jω ) = 50 = 2 ⋅
R1 ⎛ R2 C2 ⎞
⎜1 + ⋅ ⎟
R1 C1 ⎠
⎝
At the 3 − dB frequencies, we want
⎛
⎛ R2 C2 ⎞
1 ⎞
⎜ ω R2 C2 −
⎟ = ± ⎜1 + ⋅ ⎟
R1 C1 ⎠
ω R1C1 ⎠
⎝
⎝
For f = 5 kHz, use + sign and for f = 200 Hz, use − sign.
ω1 = 2π (200) = 1257
ω 2 = 2π (5 × 103 ) = 3.142 × 104
Define r2 = R2 C2 and r1 = R1C1
Then
50 =
R2
⋅
R1
1
r
1+ 2
r1
⎛
⎛ r2 ⎞
1 ⎞
⎜ ω 2 r2 −
⎟ = + ⎜1 + ⎟
ω 2 r1 ⎠
⎝
⎝ r1 ⎠
⎛
⎛ r2 ⎞
1 ⎞
⎜ ω1 r2 −
⎟ = − ⎜1 + ⎟
r
ω
⎝
⎝ r1 ⎠
1 1 ⎠
From (2)
ω 22 r1r2 − 1 r1 + r2
=
r1
ω 2 r1
(1)
(2)
(3)
or
ω 2 r1 r2 −
1
ω2
= r1 + r2
r1 (ω 2 r2 − 1) =
1
ω2
+ r2
So
1
+r
2
ω
r1 = 2
ω 2 r2 − 1
Substituting into (3), we find
⎡
⎤
⎢ r (ω r − 1) ⎥
1
⎥
ω1r2 −
= − ⎢1 + 2 2 2
1
⎡ 1
⎤
⎢
⎥
+
r
r
+
2
2 ⎥
⎢ω
⎢⎣
⎥⎦
ω
2
⎥
ω1 ⎢ 2
⎢ ω 2 r2 − 1 ⎥
⎢⎣
⎥⎦
⎡⎛ 1
⎤
⎡1
⎤ 1
⎞
ω1r2 ⎢ + r2 ⎥ − (ω 2 r2 − 1) = − ⎢⎜ + r2 ⎟ + r2 (ω 2 r2 − 1) ⎥
⎣ω2
⎦ ω1
⎠
⎣⎝ ω 2
⎦
ω1
ω
1
1
⋅ r2 + ω1r22 − 2 ⋅ r2 +
=−
− r2 − ω 2 r22 + r2
ω2
ω1
ω1
ω2
⎛ 1
⎛ω ω ⎞
1 ⎞
(ω1 + ω 2 )r22 + ⎜ 1 − 2 ⎟ r2 + ⎜ +
⎟=0
⎝ ω1 ω 2 ⎠
⎝ ω 2 ω1 ⎠
(3.2677 × 104 )r22 − 24.96r2 + 8.273 × 10−4 = 0
(24.96) 2 − 4(3.2677 ×10 4 )(8.273 × 10−4 )
24.96
±
2(3.2677 ×10 4 )
2(3.2677 × 104 )
Since ω 2 is large, r2 should be small so use minus sign:
r2 =
r2 = 3.47 × 10−5
Then
3.18 × 10−5 + 3.47 × 10−5
⇒ r1 = 7.32 × 10−4
r1 =
9.09 × 10−2
Now
R
1
50 = 2 ⋅
R1
3.47 × 10−5
1+
7.32 × 10−4
Then
R2
= 52.37 or R2 = 524 kΩ
R1
Also
r1 = R1C1 so that C1 = 0.0732 µ F
r2 = R2 C2 so that C2 = 66.3 pF
15.16
Gain = 10 dB ⇒ Gain = 3.162
For example, we may have
Want
R4
= 2.162
R3
For example, let
R3 = 50 kΩ,
R4 = 108 kΩ
f1 =
1
= 200
2π R1C1
So
R1C1 =
1
= 0.796 × 10−3
2π (200)
For example, let R1 = 200 kΩ ⇒ A large input resistance
C1 = 0.00398 µ F
1
1
f2 =
= 50 × 103 ⇒ R2 C2 =
= 3.18 × 10−6
2π R2 C2
2π (50 × 103 )
For example, let
R2 = 10 kΩ and C2 = 318 pF
15.17
f C = 100 kHz
1
Req =
fC C
a.
For C = 1 pF, Req = 10 MΩ
b.
For C = 10 pF, Req = 1 MΩ
c.
For C = 30 pF, Req = 333 kΩ
15.18
a.
From Equation (15.28),
V1 − V2
⋅ TC
Q=
Req
and f C = 100 kHz so that TC =
Now
Req =
So
1
⇒ 10 µ s
100 × 103
1
1
=
⇒ 1 MΩ
3
f C C (100 × 10 )(10 × 10−12 )
Q=
(2 − 1)(10 × 10−6 )
= 10 × 10−12 C
106
or
Q = 10 pC
Q 10 × 10−12
=
or I eq = 1 µ A
TC 10 × 10−6
b.
I eq =
c.
Q = CV so find the time that V0 reaches 99% of its full value.
V0 = V1 (1 − e − t / r ) where r = RC
Then 0.99 = 1 − e − t / r or e − t / r = 0.01
or t = r ln (100)
r = RC = (103 )(10 × 10−12 ) = 10−8 s
Then
t = 4.61×10−8 s
15.19
Low frequency gain = −10 ⇒
f 3dB = 10 × 103 Hz =
C1
= 10
C2
fC C2
2π CF
Set
f C = 10 f3dB = 100 kHz
Then
C2 2π (10 × 103 )
=
= 0.628
CF
100 × 103
The largest capacitor is C1 , so let
C1 = 30 pF
Then
C2 = 3 pF
and
CF = 4.78 pF
15.20
a.
Req =
Time constant = Req ⋅ CF = r where
1
1
=
= 2 × 106 Ω
3
−12
f C C1 (100 × 10 )(5 × 10 )
Then
r = (2 × 106 )(30 × 10−12 )
or
r = 60 µ s
b.
or
Δv0 =
v0 = −
1
vI ⋅ dt
r∫
(1)TC
1
,TC =
r
fC
So
Δv0 =
1
(60 × 10 )(100 × 103 )
−6
or
Δv0 = 0.167 V
c.
Now Δv0 = 13 = N (0.167)
or
N = 78 clock pulses
15.21
1
1
=
2π 3RC 2π 3(104 )(0.10 × 10−12 )
f O = 91.9 MHz
fO =
R2 = 8R = 80 kΩ
15.22
a.
⎛ sRCV ⎞
R
⋅ v0 = ⎜
⎟ ⋅ v0
R + (1/ sCV )
⎝ 1 + sRCV ⎠
R
⎛ sRC ⎞
⋅v =
⋅v
v2 =
1 1 ⎜⎝ 1 + sRC ⎟⎠ 1
R+
sC
R
⎛ sRC ⎞
v3 =
⋅v =
⋅v
1 2 ⎜⎝ 1 + sRC ⎟⎠ 2
R+
sC
R2
v0 = − ⋅ v3
R
Then
2
R ⎛ sRC ⎞ ⎛ sRCV ⎞
v0 = − 2 ⎜
⎟ v0
⎟ ⎜
R ⎝ 1 + sRC ⎠ ⎝ 1 + sRCV ⎠
v1 =
Set s = jω
⎛
⎞ ⎛ jω RCV ⎞
−ω 2 R 2 C 2
⎟
⎜
2 2 2 ⎟⎜
1
2
j
RC
R
C
+
−
ω
ω
⎝
⎠ ⎝ 1 + jω RCV ⎠
The real part of the denominator must be zero.
1 − ω 2 R 2 C 2 − 2ω 2 R 2 CCV = 0
so
1
ω0 =
R C (C + 2CV )
1= −
R2
R
b.
f 0,max =
1
4
2π (10 ) (10
−11
)(10−11 + 2[10−11 ])
f 0,max = 919 kHz
f 0,min =
1
4
2π (10 ) (10
f 0,min = 480 kHz
−11
)(10−11 + 2[50 × 10−12 ])
15.23
1
fO =
RC =
2π 6 RC
1
2π 6 fO
RC = 2.32 × 10
=
1
2π 6(28 × 103 )
−6
2.32 × 10−6
⇒ R = 18.56 kΩ
125 × 10−12
R2 = 29 R ⇒ R2 = 538.4 kΩ
R=
15.24
v0 − v1 v1 v1 − v2
(1)
= +
1
1
R
sC
sC
v
or (v0 − v1 ) sC = 1 + (v1 − v2 ) sC
R
v1 − v2 v2
v2
(2)
= +
1
1
R
+R
sC
sC
v
v ( sC )
or (v1 − v2 ) sC = 2 + 2
R 1 + sRC
v0
v2
(3)
=−
1
R
2
+R
sC
v
v sC
or 2
=− 0
1 + sRC
R2
so
−v0
v2 =
(1 + sRC )
sR2 C
From (2)
1
sC ⎤
⎡
v1 ( sC ) = v2 ⎢ sC + +
R 1 + sRC ⎥⎦
⎣
or
v (1 + sRC ) ⎡
1
1 ⎤
⋅ ⎢1 +
+
v1 = − 0
⎥
sR2 C
⎣ sRC 1 + sRC ⎦
From (1)
1
⎡
⎤
v0 ( sC ) = v1 ⎢ sC + + sC ⎥ − v2 ( sC )
R
⎣
⎦
Then
v
1 ⎤ ⎡ −v0 (1 + sRC ) ⎤ ⎡1 + sRC
1 ⎤
⎡
+
+ 0 ⋅ (1 + sRC )
v0 = ⎢ 2 +
⎢
⎥×⎢
⎥
⎥
1 + sRC ⎦ sR2 C
sRC ⎦ ⎣
sR2 C
⎣
⎦ ⎣ sRC
2
⎡1 + 2sRC ⎤ ⎡1 + sRC ⎤ ⎡ (1 + sRC ) + sRC ⎤ 1 + sRC
−1 = ⎢
⎥⎢
⎥−
⎥⎢
⎣ sRC ⎦ ⎣ sR2 C ⎦ ⎣ ( sRC )(1 + sRC ) ⎦ sR2 C
−1 =
(1 + 2 sRC )(1 + 2sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2
−
( sRC ) 2 ( sR2 C )
( sRC ) 2 ( sR2 C )
Set s =
(1 + 2 jω RC ) (1 + 3 jω RC + ω 2 R 2 C 2 ) (1 + jω RC ) ( −ω 2 R 2C 2 )
−
jω − 1 =
( −ω 2 R 2C 2 ) ( jω R2C )
( −ω 2 R 2C 2 ) ( jω R2C )
The real part of the numerator must be zero.
1 − ω 2 R 2 C 2 − 6ω 2 R 2 C 2 + ω 2 R 2 C 2 = 0
6ω 2 R 2 C 2 = 1
so that
1
ω0 =
6 RC
Condition for oscillation:
2 jω RC + 3 jω RC − 2 jω 3 R 3 C 3 + jω 3 R 3 C 3
−1 =
(−ω 3 R 2 C 2 )( jω R2 C )
1=
5 − ω 2 R2 C 2
(ω RC )(ω R2 C )
But
ω = ω0 =
1
6RC
Then
1⎞
⎛
2 2
1
⎜ 5 − ⎟ (6 R C )
6
⎠
6
1=
=⎝
( RC )( R2 C )
RR2 C 2
6 R 2C 2
⎛ 29 ⎞
⎜ ⎟ (6 R )
R
6
1= ⎝ ⎠
or 2 = 29
R2
R
5−
15.25
Let RF 1 = RF 2 = RF 3 ≡ RF
⎞⎛ 1 ⎞
⎟ ⎜ 1 + 5R ⎟ vo
⎠⎝
C ⎠
⎛ R ⎞⎛ 1 ⎞
Vo 2 = ⎜ 1 + F ⎟ ⎜
⎟ vo1
R ⎠ ⎝ 1 + 5 RC ⎠
⎝
vo3 + vo 2
v
v
+ o3 + o3 = 0
R
1/ sC R
2
⎛
⎞ v
vo3 ⎜ + sC ⎟ = o 2
⎝R
⎠ R
⎛ R
Vo1 = ⎜1 + F
R
⎝
⎛ 1 ⎞
vo3 = ⎜
⎟ vo 2
⎝ 2 + 5RC ⎠
R
vo = − F ⋅ vo 3
R
R ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞
vo = − F ⋅ ⎜
⎟ 1+
⎜
⎟ 1+
⎜
⎟ vo
R ⎝ 2 + 5 RC ⎠ ⎝⎜
R ⎠⎟ ⎝ 1 + 5 RC ⎠ ⎝⎜
R ⎟⎠ ⎝ 1 + 5 RC ⎠
1= −
RF
R
⎛ RF ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
⎜1 + R ⎟ ⎜ 2 + 5R ⎟ ⎜ 1 + 5R ⎟ ⎜ 1 + 5R ⎟
⎝
⎠ ⎝
C ⎠⎝
C ⎠⎝
C ⎠
2
Let S = jω
1= −
RF
R
=−
RF
R
=−
RF
R
⎞⎛
⎞⎛
⎞
1
1
1
⎛ RF ⎞ ⎛
⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎜ 1 + jω R ⎟ ⎜ 1 + jω R ⎟
⎝
⎠ ⎝
C ⎠⎝
C ⎠⎝
C ⎠
2
⎤
⎞⎡
1
1
⎛ RF ⎞ ⎛
⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎢
2 2 2 ⎥
⎝
⎠ ⎝
C ⎠ ⎣1 + 2 jω RC − ω R C ⎦
2
⎤
1
⎛ RF ⎞ ⎡
⎜1 + R ⎟ ⎢
2 2 2
2 2 2
3 3 3 ⎥
⎝
⎠ ⎣ 2 + 4 jω RC − 2ω R C + jω RC − 2ω R C − jω R C ⎦
2
⎤
RF ⎛ RF ⎞ ⎡
1
1+
⎢
⎜
⎟
2 2 2
3 3 3⎥
R ⎝
R ⎠ ⎣ 2 − 4ω R C + 5 jω RC − jω R C ⎦
Imaginary Term must be zero
5 jω 0 RC − jω 03 R 3 C 3 = 0
2
1= −
5 − jω 02 R 2 C 2 = 0
ω0 =
Then
R
1= − F
R
1= −
RF
R
5
RC
⎡
⎤
2
⎥
1
⎛ RF ⎞ ⎢
⎢
⎥
+
1
⎜
R ⎠⎟ ⎢
4R2 C 2 − 5 ⎥
⎝
⎢⎣ 2 − R 2 C 2 ⎥⎦
⎛ RF ⎞ ⎡ 1 ⎤ 1 RF
⎜ 1 + R ⎟ ⎢ 2 − 20 ⎥ = 18 ⋅ R
⎦
⎝
⎠ ⎣
2
R ⎛ R ⎞
R
18 = F ⎜ 1 + F ⎟ ⇒ F = 2
R ⎝
R ⎠
R
⎛ RF ⎞
⎜1 + R ⎟
⎝
⎠
2
2
15.26
a.
v0 − v01
v
v −v
= 01 + 01 02
R
R
⎛ 1 ⎞
⎜
⎟
sC
⎝
⎠
v01 − v02
v02
v −v
=
+ 02 03
R
R
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
v02 − v03
v
v
= 03 + 03
R
⎛ 1 ⎞ R
⎜
⎟
⎝ sC ⎠
R
v0 = − F ⋅ v03
R
We can write the equations as
v0 − v01 = v01 ( sRC ) + v01 − v02
v01 − v02 = v02 ( sRC ) + v02 − v03
v02 − v03 = v03 ( sRC ) + v03
and
R
v0 = − F ⋅ v03
R
(1)
(2)
(3)
(4)
(1)
(2)
(3)
(4)
Combining terms, we find
v0 = v01 (2 + sRC ) − v02
(1)
v01 = v02 (2 + sRC ) − v03
(2)
v02 = v03 (2 + sRC )
(3)
and
R
v0 = − F ⋅ v03
(4)
R
Combining Equations (3) and (2)
v01 = v03 (2 + sRC ) 2 − v03 = v03 ⎡⎣(2 + sRC ) 2 − 1⎤⎦
(2)
Then Equation (1) is
v0 = v03 ⎡⎣(2 + sRC ) 2 − 1⎤⎦ (2 + sRC ) − v03 (2 + sRC )
Using Equation (4), we find
R
− F ⋅ v03 = V03 ⎡⎣(2 + sRC ) 2 − 1⎤⎦ (2 + sRC ) − (2 + sRC )
R
To find the frequency of oscillation, set s = jω and set the imaginary part of the right side of the equation
to zero.
We will have
R
− F = (2 + jω RC )[4 + 4 jω RC − ω 2 R 2 C 2 − 1 − 1]
R
Then
jω RC (2 − ω 2 R 2 C 2 ) + 8 jω RC = 0
or
jω RC ⎣⎡ 2 − ω 2 R 2 C 2 + 8⎦⎤ = 0
{
}
Then the frequency of oscillation is
1
10
f0 =
⋅
2π RC
The condition to sustain oscillations is determined from
R
− F = 2[2 − ω 2 R 2 C 2 ] − 4ω 2 R 2 C 2
R
or
R
− F = 4 − 6ω 2 R 2 C 2
R
10
Setting ω 2 = 2 2 , we have
RC
RF
−
= 4 − 6(10)
R
or
RF
= 56
R
b.
For R = 5 kΩ and f 0 = 5 kHz, we find
10
⇒ C = 0.02 µ F
2π (5 × 103 )(5 × 103 )
and RF = 56(5) ⇒ RF = 280 kΩ
C=
15.27
a.
We can write
⎛ R1
vA = ⎜
⎝ R1 + R2
⎛ Zp
⎞
⎟ v0 and vB = ⎜⎜
⎠
⎝ Z p + Zs
where Z p = RB
and Z s = RA +
⎞
⎟⎟ v0
⎠
RB
1
=
sCB 1 + sRB CB
1 + sRA C A
1
=
sC A
sC A
Setting v A = vB , we have
RB
R1
1 + sRB CB
=
RB
1 + sRAC A
R1 + R2
+
sC A
1 + sRB CB
R1
sRB C A
=
(1)
R1 + R2 sRB C A + (1 + sRAC A )(1 + sRB CB )
To find the frequency of oscillation, set s = jω and set the real part of the denominator on the right side of
Equation (1) equal to zero.
The denominator term is
jω RB C A + (1 + jω RAC A )(1 + jω RB CB )
or
(2)
jω RB C A + 1 + jω RAC A + jω RB CB − ω 2 RA RB C ACB
Then from (2), we must have
1 − ω 02 RA RB C ACB = 0
or
f0 =
1
2π RA RB C ACB
b.
To find the condition for sustained oscillation, combine Equations (1) and (2). Then
R1
jω RB C A
=
R1 + R2
jω RB C A + jω RAC A + jω RB CB )
or
R2
R
C
= 1+ A + B
R1
RB C A
Then
R2 RA CB
=
+
R1 RB C A
1+
15.28
a.
We can write
⎛ R1 ⎞
vA = ⎜
⎟ v0
⎝ R1 + R2 ⎠
and
⎛
⎞
R || sL
vB = ⎜
⎟ v0
⎝ R || sL + R + sL ⎠
Setting v A = vB , we have
sRL
⎡
⎤
⎢
⎥
R1
R + sL
=⎢
⎥ ⋅ v0
sRL
R1 + R2 ⎢
+ R + sL ⎥
⎣⎢ R + sL
⎦⎥
R1
sRL
=
(1)
R1 + R2 sRL + ( R + sL) 2
To find the frequency of oscillation, set s = jω and se the real part of the denominator on the right side of
Equation (1) equal to zero.
The denominator term is:
jω RL + ( R + jω L) 2
or
(2)
jω RL + R 2 + 2 jω RL − ω 2 L2
Then
R 2 − ω 02 L2 = 0
or
R
1
f0 =
⋅
2π
L
b.
To find the condition for sustained oscillations, combine Equations (1) and (2).
R1
1
jω RL
=
=
R1 + R2
jω RL + 2 jω RL 3
Then
R
1+ 2 = 3
R1
so that
R2
=2
R1
15.29
1
2π RC
1
1
=
RC =
2π fO 2π (28 × 103 )
fO =
RC = 5.684 × 10−6
For example, Let R = 20 K
Then C = 284.2 pF
Also
R2
=2
R1
15.30
From Equation (15.59)
1
f0 =
⎛ CC ⎞
2π L ⎜ 1 2 ⎟
⎝ C1 + C2 ⎠
and from Equation (15.61)
C2
= gm R
C1
Now,
g m = 2 kn I DQ = 2 (0.5)(1) = 1.414 mA / V
We have C1 = 0.01 µ F, R = 4 kΩ, f 0 = 400 kHz
So
C2 = g m RC1 = (1.414)(4)(0.01)
or
C2 = 0.0566 µ F
and
400 × 103 =
1
⎡ (0.01)(0.0566) ⎤
2π L ⎢
× 10−6
⎥
0.01
0.0566
+
⎣
⎦
⎡
⎤
1
L(8.5 × 10−9 ) = ⎢
= 1.58 × 10−13
3 ⎥
⎣ 2π (400 × 10 ) ⎦
Then
L = 18.6 µ H
2
15.31
Vπ = −V0
V0
⎛ 1 ⎞
⎜
⎟
⎝ sC2 ⎠
+
V0 V0 − V1
+
= g mVπ = − g mV0
RL ⎛ 1 ⎞
⎜
⎟
⎝ sC1 ⎠
⎡
⎤
1
V0 ⎢ sC2 + sC1 +
+ g m ⎥ = V1 ( sC1 )
RL
⎣
⎦
V1 V0 − V1
+
+ g mVπ = 0
sL ⎛ 1 ⎞
⎜
⎟
⎝ sC1 ⎠
(1)
(2)
⎛ 1
⎞
V1 ⎜ + sC1 ⎟ = V0 ( sC1 + g m )
⎝ sL
⎠
V0 ( sC1 + g m )
V1 =
⎛ 1
⎞
⎜ + sC1 ⎟
sL
⎝
⎠
Then
⎡
⎤ V ( sC1 )( sC1 + g m )
1
V0 ⎢ s (C1 + C2 ) +
+ gm ⎥ = 0
RL
⎛ 1
⎞
⎣
⎦
⎜ + sC1 ⎟
⎝ sL
⎠
⎡
⎤⎛ 1
1
⎞
+ g m ⎥ ⎜ + sC1 ⎟ = sC1 ( sC1 + g m )
⎢ s (C1 + C2 ) +
RL
⎠
⎣
⎦ ⎝ sL
g
C1 + C2
sC
1
+ s 2 C1 (C1 + C2 ) +
+ 1 + sg m C1 + m = s 2 C12 + sg m C1
L
sRL L RL
sL
C1 + C2
sC g
1
+ s 2 C1C2 +
+ 1 + m =0
L
sRL L RL sL
Set s = jω
g
C1 + C2
jω C1
1
− ω 2 C1C2 +
+
+ m =0
L
jω RL L
RL
jω L
Then
ω2 =
C1 + C2
C1 + C2
⇒ ω0 =
C1C2 L
C1C2 L
and
gm
ω C1
1
+
=
ω L ω RL L RL
Then
gm
(C + C2 )C1
1
+
= 1
L RL L
C1C2 LRL
gm +
1 C1 + C2
=
RL
C2 RL
g m RL + 1 =
C1
C
+ 1 or 1 = g m RL
C2
C2
15.32
a.
V0 V0
V0
+ + g mVπ +
=0
(1)
1
sL1 R
+ sL2
sC
⎛
⎞
⎜ sL2 ⎟
Vπ = ⎜
(2)
⎟ V0
⎜ 1 + sL ⎟
⎜
2 ⎟
⎝ sC
⎠
Then
g m ( s 2 L2 C ) ⎪⎫
1
sC
⎪⎧ 1
+ +
+
V0 ⎨
⎬=0
2
2
⎪⎩ sL1 R 1 + s L2 C 1 + s L2 C ⎪⎭
2
2
2
2
⎪⎧ R (1 + s L2 C ) + ( sL1 )(1 + s L2 C ) s RL1C + g m ( sRL1 )( s L2 C ) ⎪⎫
+
⎨
⎬=0
( sRL1 )(1 + s 2 L2 C )
( sRL1 )(1 + s 2 L2 C )
⎩⎪
⎭⎪
Set s = jω . Both real and imaginary parts of the numerator must be zero.
R(1 − ω 2 L2 C ) + jω L1 (1 − ω 2 L2 C ) − ω 2 RL1C + ( jω g m RL1 )(−ω 2 L2 C ) = 0
Real part:
R (1 − ω 2 L2 C ) − ω 2 RL1C = 0
R = ω 2 RC ( L1 + L2 )
or
1
ω0 =
C ( L1 + L2 )
b.
Imaginary part:
jω L1 1 − ω 2 L2C − jω gm RL1 ω 2 L2C = 0
(
)
2
(
(
2
L1 = ω L1 L2C + gm RL1 ω L2C
)
)
Now ω 2 =
1=
1=
1
( L1 + L2 )
1
[ L2 C + g m RL2 C ]
C ( L1 + L2 )
L2
L
(1 + g m R ) ⇒ 1 = (1 + g m R )
L1 + L2
L2
or
L1
= gm R
L2
15.33
ω 0 = 2π (800 × 103 ) =
1
C ( L1 + L2 )
or
C ( L1 + L2 ) = 3.96 × 10−14
Also
L1
= gm R
L2
For example, if R = 1 kΩ, then
L1
= (20)(1) = 20
L2
So
L1 = 20 L2
Then
C (21L2 ) = 3.96 × 10−14 or CL2 = 1.89 × 10−15
C = 0.01 µ F
If
then
L2 = 0.189 µ H
and
L1 = 3.78 µ H
15.34
v0 − v1 v1 v1 − vB
= +
R
⎛ 1 ⎞ R
⎜
⎟
sC
⎝
⎠
and
vB
v −v
+ B 1 =0
R
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
or
(1)
(2)
1⎞ v
⎛
vB ⎜ sC + ⎟ = 1 ⇒ v1 = vB (1 + sRC )
R⎠ R
⎝
From (1)
2⎞ v
⎛
v0 ( sC ) = v1 ⎜ sC + ⎟ − B
R
⎝
⎠ R
or
v0 ( sRC ) = vB (1 + sRC )(2 + sRC ) − vB = vB [ (1 + sRC )(2 + sRC ) − 1]
Now
⎛ R ⎞⎡
⎤ ⎛ R2 ⎞ ⎡
sRC
sRC
⎤
= ⎜1 + ⎟ ⎢
T ( s ) = ⎜1 + 2 ⎟ ⎢
⎥
2 2 2
⎥
+
+
−
(1
)(2
)
1
R
sRC
sRC
R
+
+
−
2
3
1
sRC
s
R
C
⎣
⎦
⎦ ⎝
⎝
1 ⎠⎣
1 ⎠
or
⎛ R ⎞⎡
sRC
⎤
T ( s ) = ⎜1 + 2 ⎟ ⎢ 2 2 2
⎥
R1 ⎠ ⎣ s R C + 3sRC + 1 ⎦
⎝
⎛ R ⎞⎡
⎤
jω RC
T ( jω ) = ⎜ 1 + 2 ⎟ ⎢
⎥
2 2 2
ω
ω
−
+
1
3
R
R
C
j
RC
⎦
⎝
1 ⎠⎣
Frequency of oscillation:
1
f0 =
2π RC
Condition for oscillation:
⎛ R ⎞ ⎡ jω RC ⎤
1 = ⎜1 + 2 ⎟ ⎢
R1 ⎠ ⎣ 3 jω RC ⎥⎦
⎝
or
R2
=2
R1
15.35
vb − vo vb vb − va
+
+
=0
1
1
R
sC
sC
vb − vo
+ 2vb ⋅ sC − va ⋅ sC = 0
(1)
R
Va − Vb Va
(2)
+
=0
1
R
sC
1⎞
⎛
⎛ 1 + sRC ⎞
Va ⎜ sC + ⎟ = vb ⋅ sC ⇒ vb = va ⎜
⎟
R⎠
⎝
⎝ sRC ⎠
From (1)
⎛1
⎞ v
vb ⎜ + 2 sC ⎟ = o + va ⋅ sC
⎝R
⎠ R
Substitute (2) into (1)
⎛ 1 + sRC ⎞ ⎛ 1 + 2 sRC ⎞ vo
va ⎜
⎟⎜
⎟ = + va ⋅ sC
R
⎝ sRC ⎠ ⎝
⎠ R
⎡ (1 + sRC )(1 + 2 sRC )
⎤ v
va ⎢
− sC ⎥ = o
(
)
sRC
R
⋅
⎣
⎦ R
⎡ (1 + sRC )(1 + 2 sRC )
⎤
va ⎢
− sRC ⎥ = vo
sRC
⎣
⎦
2 2 2
vo (1 + sRC )(1 + 2 sRC ) − s R C
=
va
sRC
va
sRC
=
vo 1 + 3sRC + 2( sRC ) 2 − s 2 R 2 C 2
va
sRC
=
vo 1 + 3sRC + ( sRC ) 2
⎛ R ⎞
sRC
T ( s ) = ⎜1 + 2 ⎟ ⋅
R
+
+ ( sRC ) 2
sRC
1
3
1 ⎠
⎝
⎛ R ⎞⎡
⎤
jω RC
T ( jω ) = ⎜1 + 2 ⎟ ⎢
⎥
2 2 2
−
+
1
3
ω
ω
R
R
C
j
RC
⎦
1 ⎠⎣
⎝
So 1 − ω 02 R 2 C 2 = 0
So f O =
1
2π RC
⎛ R ⎞⎛ 1 ⎞
Also 1 = ⎜1 + 2 ⎟ ⎜ ⎟
R1 ⎠ ⎝ 3 ⎠
⎝
R
So 2 = 2
R1
15.36
v0 − v1 v1 v1 − vB
= +
sL
R
R
⎛ sL ⎞
vB = ⎜
⎟ v1
⎝ R + sL ⎠
or
(1)
(2)
⎛ R + sL ⎞
v1 = ⎜
⎟ vB
⎝ sL ⎠
Then
v0
⎛ 1 2⎞ v
= v1 ⎜ + ⎟ − B
sL
⎝ sL R ⎠ R
or
v0 ⎛ R + sL ⎞ ⎛ 1 2 ⎞
vB
=⎜
⎟ ⎜ + ⎟ vB −
sL ⎝ sL ⎠ ⎝ sL R ⎠
R
⎧⎛ R + sL ⎞⎛ R + 2 sL ⎞ 1 ⎫
= vB ⎨⎜
⎟⎜
⎟− ⎬
⎩⎝ sL ⎠⎝ sRL ⎠ R ⎭
Then
(1)
vB =
Now
v0
1
⋅
sL ⎧ ( R + sL)( R + 2 sL) − ( sL) 2
⎨
( sL)( sRL)
⎩
⎫
⋅⎬
⎭
⎛ R ⎞⎛
sRL
⎞
T ( s ) = ⎜1 + 2 ⎟ ⎜ 2
2 2
2 2 ⎟
+
+
−
3
2
R
R
sRL
s
L
s
L
⎠
⎝
1 ⎠⎝
or
⎛ R ⎞⎛
sRL
⎞
T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2
2 ⎟
R
+
+
3
s
L
sRL
R
⎝
⎠
⎝
1 ⎠
And
⎛ R ⎞⎛
⎞
jω RL
T ( jω ) = ⎜1 + 2 ⎟ ⎜ 2
⎟
2 2
R
−
+
ω
ω
3
R
L
j
RL
⎠
⎝
1 ⎠⎝
R
Frequency of oscillation: f 0 =
2π L
Condition for oscillation:
⎛ R ⎞⎛ 1 ⎞
1 = ⎜1 + 2 ⎟ ⎜ ⎟
R1 ⎠ ⎝ 3 ⎠
⎝
or
R2
=2
R1
15.37
From Equation (15.65(b)), the crossover voltage is
R
vI = − 2 ⋅ VREF
R1
Let R2 = RVAR + RF where RVAR is the potentiometer and RF is the fixed resistor.
Let VREF = −5 V, RF = 10 kΩ, and RVAR = 40 kΩ
Then we have
R
⎛ 10 ⎞
vI = − F ⋅ VREF = − ⎜ ⎟ (−5) = 1 V
R1
⎝ 50 ⎠
and
⎛ 50 ⎞
vI = − ⎜ ⎟ (−5) = 5 V
⎝ 50 ⎠
15.38
⎛ R1 ⎞
VTH − VTL = ⎜
⎟ (VH − VL )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
⎛ R1 ⎞
0.2 = ⎜
⎟ (13 − (−13) ) = ⎜
⎟ 26
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
R1
= 0.007692
So
R1 + R2
13
= 0.25 ⇒ R1 + R2 = 52
R1 + R2
Then
I=
R1 = (0.007692)(52) = 0.4 kΩ = R1
So R2 = 51.6 kΩ
15.39
a.
⎛ R1 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ VH = ⎜
⎟ (10)
+
R
R
⎝ 10 + 40 ⎠
⎝ 1
2 ⎠
so
VTH = 2 V
⎛ R1 ⎞
⎛ 10 ⎞
VTL = ⎜
⎟ VL = ⎜
⎟ (−10)
⎝ 10 + 40 ⎠
⎝ R1 + R2 ⎠
so
VTL = −2 V
b.
vI = 5sin ω t
15.40
a.
Upper crossover voltage when v0 = +VP ,
Now
⎛ R1 ⎞
vB = ⎜
⎟ (+VP )
⎝ R1 + R2 ⎠
and
⎛ RA ⎞
⎛ RB ⎞
vA = ⎜
⎟ VREF + ⎜
⎟ VTH
+
R
R
⎝ A
⎝ RA + RB ⎠
B ⎠
v A = vB so that
⎛ R1 ⎞
⎛ RA
⎜
⎟ VP = ⎜
⎝ R1 + R2 ⎠
⎝ RA + RB
or
⎞
⎛ RB
⎟ VREF + ⎜
⎠
⎝ RA + RB
⎞
⎟ VTH
⎠
⎛ R + RB ⎞ ⎛ R1 ⎞
⎛ RA ⎞
VTH = ⎜ A
⎟⎜
⎟ VP − ⎜
⎟ VREF
⎝ R1 + R2 ⎠ ⎝ RB ⎠
⎝ RB ⎠
Lower crossover voltage when v0 = −VP
So
⎛ R + RB ⎞ ⎛ R1 ⎞
⎛ RA ⎞
VTL = − ⎜ A
⎟ ⎜ ⎟ VP − ⎜
⎟ VREF
⎝ R1 + R2 ⎠ ⎝ RB ⎠
⎝ RB ⎠
b.
⎛ 10 + 20 ⎞ ⎛ 5 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ ⎜ ⎟ (10) − ⎜ ⎟ (2)
5
20
20
+
⎝
⎠⎝ ⎠
⎝ 20 ⎠
or
VTH = 2 V
and
⎛ 10 + 20 ⎞ ⎛ 5 ⎞
VTL = − ⎜
⎟ ⎜ ⎟ (10) − 1 ⇒ VTL = −4 V
⎝ 5 + 20 ⎠ ⎝ 20 ⎠
15.41
a.
vB VREF − vB v0 − vB
=
+
R1
R3
R2
⎛ 1
v
1
1 ⎞ V
+ ⎟ = REF + 0
vB ⎜ +
R3
R2
⎝ R1 R2 R3 ⎠
VTH = vB when v0 = +VP and VTL = vB when v0 = −VP
So
VREF VP
+
R3
R2
VTH =
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
⎝ R1 R2 R3 ⎠
and
VTL =
b.
VREF VP
−
R3
R2
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
⎝ R1 R2 R3 ⎠
VREF
⎛ 1
1
1 ⎞
+ ⎟
R3 ⎜ +
⎝ R1 R2 R3 ⎠
−10
−5 =
⎛ 1
1
1⎞
10 ⎜ +
+ ⎟
10
R
R
⎝ 1
⎠
2
VS =
1
1 1 1
+
= − = 0.10
R1 R2 5 10
ΔVT = VTH − VTL =
0.2 =
2VP
R2
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
R
R
R
2
3 ⎠
⎝ 1
2(12)
R2 (0.10 + 0.10)
So R2 = 600 kΩ
Then
1
1
+
= 0.10
R1 R2
1
1
+
= 0.10 ⇒ R1 = 10.17 kΩ
R1 600
c.
VTH = −5 + 0.1 = −4.9
VTL = −5 − 0.1 = −5.1
15.42
a.
If the saturated output voltage is VP < 6.2 V, then the circuit behaves as a comparator
where v0 < 6.2 V.
If the saturated output voltage is VP > 6.2 V, the output will flip to either +VP or −VP and the input has
no control.
b.
Same as part (a) except the curve at vI ≈ 0 will have a finite slope.
c.
Circuit works as a comparator as long as v01 < 8.7 V and v02 > −3.7 V. Otherwise the input has
no control.
15.43
a.
Switching point is when v0 = 0. Then
⎛ R2 ⎞
v+ = vI ≡ VS = ⎜
⎟ VREF
⎝ R1 + R2 ⎠
VTH occurs when vo = VH , then by superposition
⎛ R1 ⎞
⎛ R2 ⎞
v+ = VTH = ⎜
⎟ VH + ⎜
⎟ VREF
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
or
⎛ R1 ⎞
VTH = VS + ⎜
⎟ VH
⎝ R1 + R2 ⎠
VTL occurs when v0 = VL , then by superposition
⎛ R1 ⎞
⎛ R2 ⎞
v+ = VTL = ⎜
⎟ VL + ⎜
⎟ VREF
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
or
⎛ R1 ⎞
VTL = VS + ⎜
⎟ VL
⎝ R1 + R2 ⎠
b.
Now
For VTH = 2 V and VTL = 1 V, then VS = 1.5 V
⎛ 10 ⎞
2 = 1.5 + ⎜
⎟ (10)
⎝ 10 + R2 ⎠
0.5
10
=
⇒ R2 = 190 kΩ
10 10 + R2
⎛ 190 ⎞
Now VS = 1.5 = ⎜
⎟ VREF
⎝ 10 + 190 ⎠
so
VREF = 1.579 V
15.44
a.
Now
Switching point when v0 = 0.
⎛ R2 ⎞
v+ = VREF = ⎜
⎟ vI where vI = VS .
⎝ R1 + R2 ⎠
Then
⎛ R + R2 ⎞
⎛
R1 ⎞
VS = ⎜ 1
⎟ VREF = ⎜ 1 + ⎟ VREF
⎝ R2 ⎠
⎝ R2 ⎠
Now upper crossover voltage for v1 occurs when v0 = VL and v+ = VREF . Then
VTH − VREF VREF − VL
=
R1
R2
or VTH = −
⎛
R1
R ⎞
⋅ VL + VREF ⎜1 + 1 ⎟
R2
R
2 ⎠
⎝
R1
⋅ VL
R2
Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then
VH − VREF VREF − VTL
=
R2
R1
or VTH = VS −
or VTL = −
⎛
R1
R ⎞
⋅ VH + VREF ⎜1 + 1 ⎟
R2
⎝ R2 ⎠
or VTL = VS −
b.
R1
⋅ VH
R2
For VTH = −1 and VTL = −2, VS = −1.5 V. Then VTL = VS −
so that R1 = 0.833 kΩ
Now
⎛
R ⎞
VS = ⎜1 + 1 ⎟ VREF
⎝ R2 ⎠
⎛ 0.833 ⎞
−1.5 = ⎜1 +
⎟ VREF
20 ⎠
⎝
which gives
VREF = −1.44 V
15.45
(a)
vo (max) = 4.7 + 0.7 = 5.4 V
vo (min) = −5.4 V
⎛ R1 ⎞
VTH − TTL = ⎜
⎟ ( 5.4 − (−5.4) )
⎝ R1 + R2 ⎠
⎛ 2 ⎞
0.8 = ⎜
⎟ (10.8)
⎝ 2 + R2 ⎠
2 + R2 = 27 ⇒ R2 = 25 K
(b)
R=
Neglecting current in R1 and R2 ,
13.0 − 5.4
⇒ R = 15.2 K
0.5
15.46
a.
v0 = VREF + 2Vγ
5 = VREF + 2(0.7)
or
VREF = 3.6 V
b.
R1
R
⋅ VH ⇒ −2 = −1.5 − 1 (12)
R2
20
⎛ R1 ⎞
VTH = ⎜
⎟ (VREF + 2Vγ )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
0.5 = ⎜
⎟ (5)
⎝ R1 + R2 ⎠
R2
R
= 10 ⇒ 2 = 9
R1
R1
For example, let R2 = 90 kΩ and R1 = 10 kΩ
c.
For vI = 10 V, and v0 is in its low state. D1 is on and D2 is off.
or 1 +
v1 − (v1 + 0.7) VREF − v1 v1 − v0
+
=
100
1
1
For v1 = −0.7, then
10 − 0 3.6 − (−0.7) −0.7 − v0
+
=
100
1
1
or
v0 = −5.1 V
15.47
For v0 = High = (VREF + 2Vγ ). Then switching point is when.
⎛ R1 ⎞
vI = vB = ⎜
⎟ v0
⎝ R1 + R2 ⎠
⎛ R1 ⎞
or VTH = ⎜
⎟ (VREF + 2Vγ )
⎝ R1 + R2 ⎠
Lower switching point is when
⎛ R1 ⎞
v1 = vB = ⎜
⎟ v0 and v0 = −(VREF + 2Vγ )
⎝ R1 + R2 ⎠
so
⎛ R1 ⎞
VTL = − ⎜
⎟ (VREF + 2Vγ )
⎝ R1 + R2 ⎠
15.48
By symmetry, inverting terminal switches about zero.
Now, for v0 low, upper diode is on.
VREF − v1 = v1 − v0
v0 = 2v1 − VREF where v1 = −Vγ
so
v0 = −(VREF + 2Vγ )
Similarly, in the high state
v0 = (VREF + 2Vγ )
Switching occurs when non-inverting terminal is zero.
So for v0 low.
VTH − 0 0 − ⎣⎡ − (VREF + 2Vγ ) ⎦⎤
=
R1
R2
or VTH =
R1
⋅ (VREF + 2Vγ
R2
)
By symmetry
R
VTL = − 1 ⋅ (VREF + 2vγ )
R2
15.49
f =
1
2.2 RX C X
1
1
=
2.2 f (2.2)(12 × 103 )
RX C X = 3, 788 × 10−5
RX C X =
For example, Let RX = 56 K
C X = 680 pF
Within
1
of 1% of design specification.
2
15.50
(a)
⎛ R1 ⎞
⎛ 20 ⎞
vx1 = ⎜
⎟ vo = ⎜
⎟ vo
⎝ 20 + 5 ⎠
⎝ R1 + R2 ⎠
So vH = ±9.6
Also vx2 = 12 + (−9.6 − 12)e − t / τ x
= 12 − 21.6e − t / τ x
Set vx1 = vx2
9.6 = 12 − 21.6e − t / τ x
e+ t / τ x = 9 ⇒ t1 =
T
= τ x ln (9)
2
1
1
f = =
T 2τ x ln (9)
1
2(22 × 10 )(0.2 × 10−6 ) ln (9)
f = 51.7 Hz
Duty cycle = 50%
f =
3
15.51
(a)
⎛ R1 ⎞
⎛ 20 ⎞
vx+1 = ⎜
⎟ VH = ⎜
⎟ (15) = 12 V
⎝ 20 + 5 ⎠
⎝ R1 + R2 ⎠
⎛ 20 ⎞
vx−1 = ⎜
⎟ (−10) = −8 V
⎝ 20 + 5 ⎠
vx+2 = 15 + (−8 − 15)e − t / τ x = 15 − 23e− t / τ x
Then
12 = 15 − 23e − t1 / τ x
e − t1 / τ x = 7.667 ⇒ t1 − τ x ln (7.667)
⎛ − t2 − t1 ⎞
⎜⎜ τ
⎟⎟
x
⎠
vx−2 = −10 + (12 + 10)e⎝
Then
−8 = −10 + 22e
− ( t2 − t1 )
e
τx
⎛ − t2 − t1 ⎞
⎜⎜ τ
⎟⎟
x
⎝
⎠
= 11 ⇒ t2 − t1 = τ x ln (11)
Period = t2 = T = τ x [ ln (7.667) + ln (11) ] = τ x (4.435)
τ x = (22 × 103 )(0.2 × 10−6 ) = 4.4 × 10−3
T = 1.95 × 10−2
1
f = = 51.2 Hz
T
t1 = τ x ln(7.667) = (4.4 ×10 −3 ) ln (7.667)
t1 = 8.962 × 10−3
t1 8.962 × 10−3
=
T 1.951× 10−2
Duty Cycle = 45.9%
Duty cycle =
15.52
t1 = 1.1RX C X = (1.1)(10 4 )(0.1× 10−6 ) ⇒ t1 = 1.1 ms
0 < t < t1 , vY = 10(1 − e− t / rY )
rY = RY CY
Now
= (2 × 103 )(0.02 × 10−6 )
= 4 × 10−5 s
t1
= 2.75
rY
⇒ CY completely charges during each cycle.
15.53
a.
Switching voltage
⎛ R1 + R3 ⎞
⎛ 10 + 10 ⎞
vX = ⎜
⎟ ⋅ VP = ⎜
⎟ (±10)
⎝ 10 + 10 + 10 ⎠
⎝ R1 + R3 + R2 ⎠
So v X = ±6.667 V
Using Equation (15.83(b))
2
⎛ 2
⎞
v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP
3
⎝ 3
⎠
5
2
Then 1 − ⋅ e− t1 / rX =
3
3
1 5 − t1 / rX
or t1 = rX ln (5)
= ⋅e
3 3
T
1
1
t1 = =
=
⇒ t1 = 0.001 s
2 2 f 2(500)
10−3 = rX ln (5) ⇒ rX = 6.21× 10−4 = RX (0.01× 10−6 )
So RX = 62.1 kΩ
b.
Switching voltage
⎛
⎞
R1
vX = ⎜
⎟ (±VP )
⎝ R1 + R3 + R2 ⎠
10
1
⎛
⎞
=⎜
⎟ (±VP ) = ⋅ (±VP )
3
⎝ 10 + 10 + 10 ⎠
Using Equation (15.83(b))
1
⎛ 1
⎞
v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP
3
⎝ 3
⎠
4 − t1 / rX 1
Then 1 − ⋅ e
=
3
3
2 4 − t1 / rX
= ⋅e
3 3
t1 = rX ln (2) = (6.21× 10−4 ) ln (2) = 4.30 × 10−4 s
T = 2t1 = 8.6 × 10−4 s
1
f = ⇒ f = 1.16 kHz
T
15.54
From Equation (15.92)
⎛ ⎛ Vγ ⎞ ⎞
⎜1+ ⎜ ⎟ ⎟
⎝ VP ⎠ ⎟
⎜
T = rX ln
⎜ 1− β ⎟
⎜⎜
⎟⎟
⎝
⎠
R1
10
where β =
=
= 0.2857
R1 + R2 10 + 25
so
0.7 ⎤
⎡
⎢ 1+ 5 ⎥
100 = rX ln ⎢
⎥
⎢1 − 0.2857 ⎥
⎢⎣
⎥⎦
so τ X = 213.9 µ s = RX C X
For example, RX = 10 kΩ, C X = 0.0214 µ F
⎛ R1 ⎞
⎛ 10 ⎞
vY = ⎜
⎟ VP = ⎜
⎟ (5) = 1.43 V
⎝ 10 + 25 ⎠
⎝ R1 + R2 ⎠
and v X = 0.7 V
To trigger the circuit, vY must be brought to a voltage less than v X .
Therefore minimum triggering pulse is −0.73 V.
Using Equation (15.82) for T < t < T ′
v X = VP + (−0.2857VP − VP )e − t ′ / rX
Recovery period is when v X = Vγ = 0.7 V.
0.7 = 5 + (−6.43)e − t ′ / rX
6.43e− t ′ / rX = 4.3
or t ′ = rX ln (1.495)
rX = 213.9 µ s
so
t ′ = T ′ − T = 86.1 µ s
15.55
(a)
⎡1 + (Vr / VP ) ⎤
T =τx ⎢
⎥
⎣ 1− β ⎦
τ x = Rx Cx β = 1 2
T 0.69τ x = (0.69)(47 × 103 )(0.2 × 10−6 )
T = 6.49 ms
Recovery Time 0.4τ x
(b)
= 3.76 ms
15.56
a.
From Equation (15.95)
T = 1.1 RC
For T = 60 s = 1.1 RC
then RC = 54.55 s
For example, let
C = 50 µ F and R = 1.09 MΩ
b.
Recovery time: capacitor is discharged by current through the discharge transistor.
5 − 0.7
If V + = 5 V, then I B ≅
= 0.043 mA
100
If β = 100, I C = 4.3 mA
VC =
1
Ic
I C dt = ⋅ t
∫
C
C
2 +
⋅ V = 3.33 V
3
V ⋅ C (3.33)(50 × 10−6 )
So that t = C
=
IC ⋅
4.3 × 10−3
Capacitor has charged to
So recovery time t ≈ 38.7 ms
15.57
T = 1.1 RC
5 × 10−6 = 1.1 RC
so RC = 4.545 × 10−6 s
For example, let
C = 100 pF and R = 45.5 kΩ
From Problem (15.53), recovery time
V ⋅ C (3.33)(100 × 10−12 )
=
t≅ C
IC
4.3 × 10−3
or
t = 77.4 ns
15.58
From Equation (15.102),
1
f =
(0.693)(20 + 2(20)) × 103 × (0.1× 10−6 )
or f = 240.5 Hz
Duty cycle =
20 + 20
× 100% = 66.7%
20 + 2(20)
15.59
f =
1
(0.693)( RA + 2 RB )C
RA = R1 = 10 kΩ, RB = R2 + xR3
So 10 kΩ ≤ RB ≤ 110 kΩ
1
= 627 kHz
(0.693)(10 + 2(110)) × 103 × (0.01× 10−6 )
1
=
= 4.81 kHz
(0.693)(10 + 2(10)) × 103 × (0.01× 10−6 )
f min =
f max
So 627 Hz ≤ f ≤ 4.81 kHz
Duty cycle =
RA + RB
× 100%
RA + 2 RB
Now
10 + 10
× 100% = 66.7%
10 + 2(10)
and
10 + 110
× 100% = 52.2%
10 + 2(110)
So 52.2 ≤ Duty cycle ≤ 66.7%
15.60
1 kΩ ≤ RA ≤ 51 kΩ
1 kΩ ≤ RB ≤ 51 kΩ
1
= 1.40 Hz
(0.693)(1 + 2(51)) × 103 × (0.01× 10−6 )
1
= 2.72 kHz
f max =
(0.693)(51 + 2(1)) × 103 × (0.01× 10−6 )
or 1.40 kHz ≤ f ≤ 2.72 kHz
f min =
Duty cycle =
RA + RB
× 100%
RA + 2 RB
1 + 51
× 100% = 50.5%
1 + 2(51)
or
51 + 1
× 100% = 98.1%
51 + 2(1)
or
50.5% ≤ Duty cycle ≤ 98.1%
15.61
a.
I E3 = IE 4 =
V + − 3VEB
R1 A + R1B
Assume VEB = 0.7
I E3 = I E 4 =
22 − 3(0.7)
= 0.398 mA
25 + 25
Now
⎛ 20 ⎞
I C 3 = I C 4 = I C 5 = I C 6 = ⎜ ⎟ (0.398)
⎝ 21 ⎠
I C 3 = I C 4 = I C 5 = I C 6 = 0.379 mA
0.398 ⎛ 20 ⎞
⎜ ⎟ ⇒ I C1 = I C 2 = 0.018 mA
21 ⎝ 21 ⎠
I D = 0.398 mA, current in D1 and D2
I C1 = I C 2 =
b.
⎛I ⎞
⎛ 0.398 × 10−3 ⎞
VBB = 2VD = 2VT ln ⎜ D ⎟ = 2(0.026) ln ⎜
⎟
−13
⎝ 10
⎠
⎝ IS ⎠
or
VBB = 1.149 V = VBE 7 + VEB 8
Now
IC 7 ≈ IC 4 + IC 9 + I E8
I C 4 = 0.379 mA
⎛ 20 ⎞
I B9 = IC 8 = ⎜ ⎟ I E 8
⎝ 21 ⎠
So
⎛I ⎞
I E 8 = 1.05 I B 9 = 1.05 ⎜ C 9 ⎟
⎝ 100 ⎠
⎛ 100 ⎞
⎛ 21 ⎞
IC 7 = IC 4 + ⎜
⎟ I E 8 + I E 8 = I C 4 + (96.24) ⎜ ⎟ I C 8
⎝ 1.05 ⎠
⎝ 20 ⎠
So I C 7 = 0.379 mA + 101I C 8
and
⎛I ⎞
⎛I ⎞
VBE 7 = VT ln ⎜ C 7 ⎟ ; VEB 8 = VT ln ⎜ C 8 ⎟
I
⎝ S ⎠
⎝ IS ⎠
Then
⎡ ⎛I ⎞
⎛ I ⎞⎤
1.149 = 0.026 ⎢ ln ⎜ C 7 ⎟ + ln ⎜ C 8 ⎟ ⎥
⎝ I S ⎠⎦
⎣ ⎝ IS ⎠
⎡ I (0.379 × 10−3 ) + 101I C 8 ⎤
44.19 = ln ⎢ C 8
⎥
(10−13 ) 2
⎣
⎦
(10−13 ) 2 exp (44.19) = 101I C2 8 + 3.79 × 10 −4 I C 8
1.554 × 10−7 = 101I C2 8 + 3.79 × 10−4 I C 8
IC 8 =
(3.79 × 10−4 ) 2 + 4(101)(1.554 × 10−7 )
−3.79 × 10−4
±
2(101)
2(101)
I C 8 = 37.4 µ A
I C 7 = 0.379 + 101(0.0374) ⇒ I C 7 = 4.16 mA
⎛ 21 ⎞
I C 9 = 4.16 − 0.379 − 0.0374 ⎜ ⎟
⎝ 20 ⎠
I C 9 = 3.74 mA
c.
P = (0.398 + 0.398 + 4.16)(22) ⇒ P = 109 mW
15.62
a.
b.
From Figure 15.47, 3.7 W to the load
V + ≈ 19 V
c.
P=
1 VP2
2 RL
or
VP = 2 RL P = 2(10)(3.7) ⇒ VP = 8.6 V
15.63
1 VP2
P=
2 RL
so VP = 2 RP = 2(10)(20) ⇒ 20 V
peak-to-peak output voltage
Maximum output voltage of each op-amp = ±10 V. Current is (20 /10) = 2 A. Bias op-amps at ±12 V.
For A1 ,
v01 ⎛ R2 ⎞
R
= ⎜1 + ⎟ = 15 ⇒ 2 = 14
vI ⎝
R1 ⎠
R1
For A2 ,
v02
R
= 4 = 15
vI
R3
For example, let R1 = R3 = 10 kΩ, and R2 = 140 kΩ and R4 = 150 kΩ.
15.64
a.
v01 = iR2 + vI where i =
vI
R1
Then
⎛ R ⎞
v01 = vI ⎜ 1 + 2 ⎟
R1 ⎠
⎝
Now
⎛R ⎞
v02 = −iR3 = −vI ⎜ 3 ⎟
⎝ R1 ⎠
So
⎛ R ⎞ ⎡
⎛ R ⎞⎤
vL = v01 − v02 = vI ⎜1 + 2 ⎟ − ⎢ −vI ⎜ 3 ⎟ ⎥
R1 ⎠ ⎣
⎝
⎝ R1 ⎠ ⎦
Av =
b.
vL
R R
= 1+ 2 + 3
vI
R1 R1
Want Av = 10 ⇒
⎛ R ⎞ R
Also want ⎜1 + 2 ⎟ = 3
R1 ⎠ R1
⎝
R2 R3
+
=9
R1 R1
R
R2 ⎛ R2 ⎞
+ ⎜1 + ⎟ = 9 so 2 = 4
R1 ⎝
R1 ⎠
R1
For R1 = 50 kΩ, R2 = 200 kΩ
Then
and
R3
= 5 so R3 = 250 kΩ
R1
c.
P=
1 VP2
2 RL
or
VP = 2 RL P = 2(20)(10) = 20 V
So peak values of output voltages are
v01 = v02 = 10 V
Peak load current =
20
=1 A
20
15.65
(a)
⎛ R ⎞
vo1 = ⎜1 + 2 ⎟ vI
R1 ⎠
⎝
⎛ R2 ⎞
⎜ 1 + ⎟ vI
R1 ⎠
⎝
vL = vo1 − vo 2
vo 2 = −
Av =
R4
R3
vL ⎛ R2 ⎞ ⎛ R4 ⎞
= ⎜1 + ⎟ ⎜1 + ⎟
vI ⎝
R1 ⎠ ⎝ R3 ⎠
For v01 − 12 V, vo 2 = −12 V when R3 = R4
(b)
So vL = 24sin ω + (V )
⎛ R ⎞⎛ R ⎞
10 = ⎜ 1 + 2 ⎟ ⎜1 + 4 ⎟
R1 ⎠ ⎝ R3 ⎠
⎝
Let R3 = R4
(c)
Then
R2
=4
R1
15.66
(a)
From Problem 15.65
vO ⎛ R2 ⎞ ⎛ R4 ⎞
= ⎜1 + ⎟ ⎜1 + ⎟
vI ⎝
R1 ⎠⎝ R3 ⎠
For vo1 = vo 2 ⇒ R3 = R4
Then
⎛ R ⎞
vO
= 2 ⎜1 + 2 ⎟
vI
R1 ⎠
⎝
⎛ R ⎞ ⎛R ⎞
15 = 2 ⎜1 + 2 ⎟ ⇒ ⎜ 2 ⎟ = 6.5
R1 ⎠ ⎝ R1 ⎠
⎝
(b)
Peak output voltage
VP = 2 RL PL = 2(16)(20) = 25.3 V
Then Vo1 = Vo 2 = 12.65 V
iL (peak) =
25.3
= 1.58 A
16
15.67
Line regulation =
ΔV0
ΔV +
Now
ΔI =
ΔV +
and ΔVZ = rZ ⋅ ΔI and ΔV0 = 10ΔVZ
R1
So
ΔV0 = 10 ⋅ rZ ⋅
ΔV +
R1
So
Line regulation =
ΔV0 10(15)
=
⇒ 1.61%
ΔV +
9300
15.68
R0 f = −
ΔV0
ΔI 0
So R0 f =
−(−10 × 10−3 )
1
or
R0 f = 10 mΩ
15.69
For V0 = 8 V
V + (min) = V0 + I 0 (max) R11 + VBE11 + VBE10 + VEB 5
This assumes VBC 5 = 0.
Then
V + (min) = 8 + (0.1)(1.9) + 0.6 + 0.6 + 0.6
V + (min) = 9.99 V
15.70
a.
IC 3 = IC 5 =
VZ − 3VBE (npn)
R1 + R2 + R3
6.3 − 3(0.6)
= 0.571 mA
0.576 + 3.4 + 3.9
1 ⎛ 0.6 ⎞
IC 8 = ⎜
⎟ = 0.106 mA
2 ⎝ 2.84 ⎠
Neglecting current in Q9 , total collector current and emitter current in Q5 is
0.571 + 0.106 = 0.677
Now
I Z 2 R4 + VEB 4 = VEB 5
IC 3 = IC 5 =
⎛I ⎞
VEB 4 = VT ln ⎜ Z 2 ⎟
⎝ I5 ⎠
⎛I ⎞
VEB 5 = VT ln ⎜ C 5 ⎟
⎝ 2I S ⎠
⎛ I ⎞
Then I Z 2 R4 = VT ln ⎜ C 5 ⎟
⎝ 2I Z 2 ⎠
⎛ 0.677 ⎞
0.026
R4 =
⋅ ln ⎜
⎟
0.25
⎝ 2(0.25) ⎠
or
R4 = 31.5 Ω
b.
From Example 15.16, VB 7 = 3.43 V . Then
⎛ R13 ⎞
⎜
⎟ V0 = VB 8 = VB 7
⎝ R12 + R13 ⎠
or
⎛ 2.23 ⎞
⎜
⎟ (12) = 3.43
⎝ 2.23 + R12 ⎠
3.43(2.23 + R12 ) = (2.23)(12)
which yields
R12 = 5.57 kΩ
15.71
Line regulation =
ΔV0
ΔV +
Now
ΔVB 7 = ΔI C 3 ⋅ R1
⎛ R13 ⎞
and ⎜
⎟ (ΔV0 ) = ΔVB 7 = ΔI C 3 R1
⎝ R12 + R13 ⎠
ΔVZ
ΔI Z ⋅ rZ
and ΔI C 3 =
=
R1 + R2 + R3 R1 + R2 + R3
and ΔI Z =
V
ΔV +
where r0 = A
IZ
r0
Then
⎛ 0.015 ⎞
(0.4288)( ΔV0 ) = ΔI C 3 (3.9) = (3.9) ΔI Z ⎜
⎟
⎝ 7.876 ⎠
50
r0 =
= 87.6 kΩ
0.571
Then
⎛ ΔV + ⎞
(0.4288)(ΔV0 ) = (0.00743) ⎜
⎟
⎝ 87.6 ⎠
So
ΔV0
= 0.0198%
ΔV +
15.72
a.
IZ =
25 − 5
= 10
R1 + rZ
20
= 2 kΩ = R1 + 0.01 ⇒ R1 = 1.99 kΩ
10
b.
In the ideal case;
⎛ R3 + R4 ⎞
⎜
⎟ V0 = VZ
⎝ R2 + R3 + R4 ⎠
So R1 + rZ =
⎛ 2 +1 ⎞
⎜
⎟ V0 = 5 ⇒ V0 = 6.67 V
⎝ 2 +1+1⎠
and
⎛
⎞
R4
⎜
⎟ V0 = VZ
⎝ R2 + R3 + R4 ⎠
⎛ 1 ⎞
⎜
⎟ V0 = 5 ⇒ V0 = 20 V
⎝ 2 +1+1⎠
So
6.67 ≤ V0 ≤ 20 V
c.
3
3
⋅ V0 so vd = 5 − ⋅ V0
4
4
and V0 = A0 L vd − VBE
V1 =
⎛I ⎞
and VBE = VT ln ⎜ 0 ⎟
⎝ IS ⎠
Now
3
⎛
⎞
V0 = A0 L ⎜ 5 − ⋅ V0 ⎟ − VBE
4
⎝
⎠
⎛ 3
⎞
V0 ⎜ 1 + ⋅ A0 L ⎟ = 5 A0 L − VBE
4
⎝
⎠
5 A0 L − VBE
V0 =
3
1 + ⋅ A0 L
4
Load regulation =
V0 (NL) − V0 (FL)
V0 (NL)
5 A0 L − VBE (NL) 5 A0 L − VBE (FL)
−
(1 + 34 A0 L )
(1 + 34 A0 L )
=
5 A0 L − VBE (NL)
(1 + 34 A0 L )
⎛ I (FL) ⎞
VT ln ⎜ 0
⎟
I 0 (NL) ⎠
VBE (FL) − VBE (NL)
⎝
=
=
5 A0 L − VBE (NL)
5 A0 L − VBE (NL)
V0
6.67
=
= 1.67 mA
4 k Ω 4 kΩ
1
⎛
⎞
(0.026) ln ⎜
−3 ⎟
⎝ 1.67 × 10 ⎠
⇒ 3.33 × 10−4 %
Load regulation =
4
5(10 ) − 0.7
I 0 (FL) = 1 A, I 0 (NL) =
15.73
V
5.6
IE = Z =
= 1.12 mA
5
R2
I0 =
β
1+ β
⎛ 100 ⎞
⋅ IE = ⎜
⎟ (1.12) ⇒ I 0 = 1.109 mA Load current
⎝ 101 ⎠
For
VBC = 0 ⇒ V0
= 20 − VZ − 0.6
= 20 − 5.6 − 0.6
or
V0 = 13.8 V
Then
V
13.8
⇒ RL = 12.4 kΩ
RL = 0 =
I 0 1.109
So
0 ≤ RL ≤ 12.4 kΩ
Chapter 16
Exercise Solutions
EX16.1
COX =
γ =
( 3.9 ) (8.85 ×10−14 )
200 × 10−8
= 1.726 ×10−7 F / cm 2
2 (1.6 × 10−19 ) (11.7 ) ( 8.85 ×10 −14 )(1015 )
1.726 × 10−7
1
= 0.1055 V 2
ΔVTN = r ⎡⎣ 0.576 + VSB − 0.576 ⎤⎦ = ( 0.1055 ) ⎡⎣ 0.576 + 5 + 0.576 ⎤⎦
ΔVTN = 0.169 V
EX16.2
(a)
vo = VDD − I D RD
⎛ k′ ⎞⎛ W
vo = 3 − ⎜ n ⎟ ⎜
⎝ 2 ⎠⎝ L
vo = 0.1
⎞
2
⎟ ⎣⎡ 2 ( 3 − 0.5 ) vo − vo ⎦⎤ RD
⎠
2
⎛ 0.06 ⎞ ⎡
0.1 = 3 − ⎜
⎟ ( 5 ) ⎣( 5 )( 0.1) − ( 0.1) ⎤⎦ RD
⎝ 2 ⎠
0.1 = 3 − 0.0735RD
RD = 39.5 K
(b)
2
⎛ 0.06 ⎞
⎜
⎟ ( 5 )( 39.5 )(VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0
2
⎝
⎠
2
5.925 (VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0
(VIt − 0.5 ) = VOt
=
−1 ± 1 + 4 ( 5.925 )( 3)
2 ( 5.925 )
VOt = 0.632 V
VIt = 1.132 V
EX16.3
(a)
(i)
vo = VDD − VTNL = 3 − 0.4
vo = 2.6 V
(ii)
2
⎛W ⎞
⎛W ⎞
2
⎜ ⎟ ⎡⎣ 2 ( vI − 0.4 ) vo − vo ⎤⎦ = ⎜ ⎟ [VDD − vo − 0.4]
⎝ L ⎠D
⎝ L ⎠L
16 ⎡⎣ 2 ( 2.6 − 0.4 ) vo − vo2 ⎤⎦ = 2 [3 − vo − 0.4]
2
35.2vo − 8vo2 = 6.76 − 5.2vo + vo2
9vo2 − 40.4vo + 6.76 = 0
vo =
40.4 ± 1632.16 − 4 ( 9 )( 6.76 )
vo = 0.174 V
2 (9)
(b)
2
⎛ 60 ⎞
iD = ⎜ ⎟ (2) [3 − 0.174 − 0.4]
⎝ 2 ⎠
iD = 353.1 µ A
P = iD ⋅ VDD = 1.06 mW
EX16.4
(a)
2
⎛W ⎞
⎛W ⎞
2
⎜ ⎟ ⎣⎡ 2 ( vI − 0.4 ) vo − vo ⎦⎤ = ⎜ ⎟ ( − ( −0.8 ) )
L
L
⎝ ⎠D
⎝ ⎠L
6 ⎡⎣ 2 ( 3 − 0.4 ) vo − vo2 ⎤⎦ = 2 ( 0.64 )
6vo2 − 31.2vo + 1.28 = 0
vo =
31.2 ± 973.44 − 4 ( 6 )(1.28 )
2(6)
vo = 41.4 mV
(b)
2
2
⎛W ⎞
⎛W ⎞
⎜ ⎟ ( vIt − 0.4 ) = ⎜ ⎟ ( −(−0.8) )
L
L
⎝ ⎠D
⎝ ⎠L
2
6 ( vIt − 0.4 ) = 2 ( 0.64 )
⇒ vIt = 0.862 V
⎫
⎬ Driver
vOt = 0.862 − 0.4 = 0.462 V ⎭
vIt = 0.862 V
⎫
⎬ Load
vOt = VDD + VTNL = 3 − 0.8 = 2.2 V ⎭
(c)
2
⎛ 60 ⎞
iD = ⎜ ⎟ (2) ( − ( −0.8 ) ) = 38.4 µ A
⎝ 2 ⎠
P = iD ⋅ VDD = 115.2 µ W
EX16.5
We have
VOH = VDD − VTNLO + r ⎡⎣ 2φ fP + VSB − 2φ fP ⎤⎦
{
VOH = 5 − 0.8 + 0.35 ⎡⎣ 0.73 + VOH −
{
}
0.73 ⎤⎦}
VOH − 4.499 = −0.35 0.73 + VOH
Squaring both sides
2
VOH
− 8.998VOH + 20.241 = 0.1225(0.73 + VOH )
2
VOH − 9.1205VOH + 20.15 = 0
9.1205 ± 83.1835 − 4(20.15)
2
= 3.76 V
VOH =
VOH
EX16.6
a.
i.
A = logic 1 = 10 V, B = logic 0 “A” driver in nonsaturation. “B” driver off
2
⎛ kn′ ⎞ ⎛ W ⎞
⎛ k′ ⎞⎛ W ⎞
2
⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = ⎜ 2 ⎟ ⎜ L ⎟ ⎡⎣ 2 ( vI − vTND ) VOL − VOL ⎤⎦
⎝ ⎠ ⎝ ⎠D
⎝ ⎠ ⎝ ⎠L
2 ( 3) = (10 ) ⎡⎣ 2 (10 − 1.5 ) V0 L − V02L ⎤⎦
2
9 = 5 (17V0 L − V02L )
5V02L − 85V0 L + 9 = 0
⇒ V0 L = 0.107 V
2(5)
ii.
A = B = logic 1
2
⎛ kn′ ⎞ ⎛ W ⎞
⎛ k′ ⎞⎛ W ⎞
2
⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎡⎣ 2 ( vI − vTND )VOL − VOL ⎤⎦
⎝ ⎠ ⎝ ⎠D
⎝ ⎠ ⎝ ⎠L
V0 L =
85 ±
(85 )
2
− 4 ( 5 )( 9 )
2 ( 3) = ( 2 )(10 ) ⎣⎡ 2 (10 − 1.5 ) V0 L − V02L ⎦⎤
2
9 = 10 (17V0 L − V02L )
10V02L − 170V0 L + 9 = 0
V0 L =
170 ±
(170 ) − 4 (10 )( 9 )
⇒ V0 L = 0.0531 V
2 (10 )
2
b.
Both cases.
35
2
iD = ⋅ ( 2 )( 3) = 315 µ A ⇒ P = iD ⋅ VDD ⇒ P = 3.15 mW
2
EX16.7
P = iD ⋅ VDD ⇒ iD =
800
= 160 µ A
5
35 ⎛ W ⎞
2
⎛W ⎞
⎛W ⎞
iD = 160 = ⋅ ⎜ ⎟ (1.4 ) = 34.3 ⎜ ⎟ ⇒ ⎜ ⎟ = 4.66
L
2 ⎝ L ⎠L
⎝ ⎠L
⎝ L ⎠L
iD = 160 =
35 ⎛ W ⎞ ⎡
2
⎛W ⎞
⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 9.20
⎦ ⎝ L ⎠D
2 ⎝ L ⎠D ⎣
EX16.8
(a)
VOPt
VONt
(b)
VDD 2.1
=
= 1.05 V
2
2
= VIt − VTD = 1.05 − (−0.4) = 1.45 V
= VIt − VTN = 1.05 − 0.4 = 0.65 V
VIt =
VIt =
VOPt
2.1 + (−0.4) + 0.5(0.4)
1 + 0.5
= 1.16 + 0.4 = 1.56 V
= 1.16 V
VONt = 1.16 − 0.4 = 0.76 V
(c)
VOPt
VONt
EX16.9
2.1 + (−0.4) + 2(0.4)
= 0.938 V
1+ 2
= 0.938 + 0.4 = 1.338 V
= 0.538 V
VIt =
2
P = f ⋅ CL ⋅ VDD
( 0.10 ×10 ) = f ( 0.5 ×10 ) ( 3)
−6
−12
2
f = 2.22 × 104 Hz ⇒ f = 22.2 kHz
EX16.10
a.
K n / K p = 200 / 80 = 2.5 ⇒ VIt =
10 − 2 + 2.5(2)
1 + 2.5
⇒ VIt = 4.32 V
V0 Pt = 6.32 V
V0 Nt = 2.32 V
b.
VIL = 2 +
⎤
10 − 2 − 2 ⎡
2.5
⋅ ⎢2
− 1⎥ ⇒ VIL = 3.39 V
2.5 − 1 ⎣ 2.5 + 3 ⎦
1
{(1 + 2.5)(3.39) + 10 − (2.5)(2) + 2}
2
= 9.43 V
V0 HU =
V0 HU
VIH = 2 +
⎤
10 − 2 − 2 ⎡ 2(2.5)
⋅⎢
− 1⎥ ⇒ VIH = 4.86 V
2.5 − 1 ⎣⎢ 3(2.5) + 1 ⎦⎥
(4.86)(1 + 2.5) − 10 − (2.5)(2) + 2
2(2.5)
= 0.802 V
V0 LU =
V0 LU
c.
NM L = VIL − V0 LU = 3.39 − 0.802 ⇒ NM L = 2.59 V
NM H = V0 HU − VIH = 9.43 − 4.86 ⇒ NM H = 4.57 V
EX16.11
3 PMOS in series and 3 NMOS in parallel.
Worst Case: Only one NMOS is ON in Pull-down mode ⇒ same as the CMOS inverter ⇒ Wn = W .
All 3 PMOS are on during pull-up mode ⇒ W p = 3(2W ) = 6W .
EX16.12
NMOS: Worst Case, M NA , M NB on, Wn = 2(W ) or M NC , M ND or M NC , M NE on ⇒ Wn = 2(W ).
PMOS: M PA and M PC on or M PA and M PB on ⇒ WP = 2(2W ) = 4W
If M PD and M PE on, need WP = 2(4W ) = 8W
EX16.13
a.
vI = φ = 5 V ⇒ v0 = 4 V
b.
c.
d.
vI = 3 V, φ = 5 V ⇒ v0 = 3 V
vI = 4.2 V, φ = 5 V ⇒ v0 = 4 V
vI = 5 V, φ = 3 V ⇒ v0 = 2 V
EX16.14
(a)
vI = 8V , φ = 10V ⇒ vGSD = 8 V
M D in nonsaturation
K D ⎡⎣ 2(vGSD − VTND )vO − vO2 ⎤⎦
2
K L [VDD − vO − VTNL ]
KD ⎡
K
2
2
2 ( 8 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 9.78
⎦
KL ⎣
KL
(b)
vI = φ = 8V ⇒ vGSD = 6 V
KD ⎡
K
2
2
2 ( 6 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 15
⎣
⎦
KL
KL
EX16.15
16 K ⇒ 16384 cells
Total Power = 125 mW = (2.5) IT ⇒ IT = 50 mA
50 mA
⇒ I = 3.05 µ A
16384
V
2.5
⇒ R = 0.82 M Ω
or R = DD =
I
3.05
Then, for each cell, I =
Now, I ≅
VDD
R
TYU16.1
P = iD ⋅ VDD ⇒ iD =
750
= 150 µ A
5
35 ⎛ W ⎞
2
⎜ ⎟ (5 − 0.2 − 0.8)
2 ⎝ L ⎠L
⎛W ⎞
⎛W ⎞
150 = 280 ⎜ ⎟ ⇒ ⎜ ⎟ = 0.536
L
⎝ ⎠L
⎝ L ⎠L
150 =
⎛ k′ ⎞⎛ W ⎞
iD = ⎜ n ⎟ ⎜ ⎟ ⎡⎣ 2(vI − VTND )vO − vO2 ⎤⎦
⎝ 2 ⎠ ⎝ L ⎠D
35 ⎛ W ⎞
150 = ⎜ ⎟ ⎡⎣ 2(4.2 − 0.8)(0.2) − (0.2) 2 ⎤⎦
2 ⎝ L ⎠D
⎛W ⎞
⎛W ⎞
150 = 23.1 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 6.49
L
⎝ ⎠D
⎝ L ⎠D
TYU16.2
P = iD ⋅ VDD ⇒ I D =
350
= 70 µ A
5
2
⎛ k′ ⎞⎛ W ⎞
iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL )
⎝ 2 ⎠ ⎝ L ⎠L
35 ⎛ W ⎞
2
⎛W ⎞
70 = ⋅ ⎜ ⎟ ( 2 ) ⇒ ⎜ ⎟ = 1
2 ⎝ L ⎠L
⎝ L ⎠L
35 ⎛ W ⎞ ⎡
2
⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.05 ) − ( 0.05 ) ⎤
⎦
2 ⎝ L ⎠D ⎣
⎛W ⎞
⎛W ⎞
70 = 7.31 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 9.58
⎝ L ⎠D
⎝ L ⎠D
iD =
TYU16.3
P = iD ⋅ VDD ⇒ iD =
800
= 160 µ A
5
35 ⎛ W ⎞
2
⎛W ⎞
iD = 160 = ⋅ ⎜ ⎟ (1.4 ) ⇒ ⎜ ⎟ = 4.66
2 ⎝ L ⎠L
⎝ L ⎠L
iD = 160 µ A =
35 1 ⎛ W ⎞ ⎡
2
⎛W ⎞
⋅ ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 27.6
⎦ ⎝ L ⎠D
2 3 ⎝ L ⎠D ⎣
TYU16.4
a.
From the load transistor:
k
35
2
2
⎛ ′ ⎞⎛W ⎞
I DL = ⎜ n ⎟ ⎜ ⎟ (VGSL − VTNL ) = ( 0.5 )( 5 − 0.15 − 0.7 )
2
⎝ 2 ⎠ ⎝ L ⎠L
or
I DL = 150.7 µ A
Maximum v0 occurs when either A or B is high and C is high. For the two NMOS is series, the effective
k N is cut in half, so
I DL =
or
1 ⎡⎛ kn′ ⎞ ⎛ W ⎞ ⎤
2
⎢
⎜ ⎟ ⎥ ⎡ 2 (VGSD − VTND ) VDS − VDS ⎤⎦
2 ⎣⎜⎝ 2 ⎟⎠ ⎝ L ⎠ D ⎦ ⎣
1 ⎡ 35 ⎛ W ⎞ ⎤ ⎡
2
⎢ ⋅ ⎜ ⎟ ⎥ 2 ( 5 − 0.7 )( 0.15 ) − ( 0.15 ) ⎤⎦
2 ⎣ 2 ⎝ L ⎠D ⎦ ⎣
which yields
⎛W ⎞
⎜ ⎟ = 13.6
⎝ L ⎠D
150.7 =
b.
P = iD ⋅ VDD = (150.7 )( 5 ) ⇒ P = 753 µ W
TYU16.5
a.
v0 (max) occurs when A = B = 1 and C = D = 0 or A = B = 0 and C = D = 1
1 ⎛W ⎞
⎛W ⎞
2
2
⎜ ⎟ (−VTNL ) = ⋅ ⎜ ⎟ ⎡⎣ 2(vI − VTND )vO − vO ⎤⎦
2 ⎝ L ⎠D
⎝ L ⎠L
1 ⎛W ⎞
(0.5)(1.2)2 = ⋅ ⎜ ⎟ ⎣⎡ 2(5 − 0.7)(0.15) − (0.15) 2 ⎦⎤
2 ⎝ L ⎠D
⎛W ⎞
⎛W ⎞
0.72 = (0.634) ⎜ ⎟ ⇒ ⎜ ⎟ = 1.14
⎝ L ⎠D
⎝ L ⎠D
b.
2
⎛ k′ ⎞⎛ W ⎞
⎛ 35 ⎞
iD = ⎜ n ⎟ ⎜ ⎟ (−VTNL ) 2 = ⎜ ⎟ (0.5) [ −(−1.2) ]
⎝ 2⎠
⎝ 2 ⎠ ⎝ L ⎠L
iD = 12.6 µ A
P = iD ⋅ VDD = (12.6)(5) ⇒ P = 63 µ W
TYU16.6
a.
K n = K p = 50 µ A / V 2
VIt = 2.5 V
iD (max) = K n (VIt − VTN ) 2 = 50(2.5 − 0.8) 2 ⇒ iD (max) = 145 µ A
b.
K n = K p = 200 µ A / V 2
VIt = 2.5 V
iD (max) = (200)(2.5 − 0.8) 2 ⇒ iD (max) = 578 µ A
TYU16.7
a.
5 − 2 + (1)(0.8)
VIt =
1+1
VIt = 1.9 V
V0 Pt = 3.9 V
V0 Nt = 1.1 V
b.
3
VIL = 0.8 + ⋅ [5 − 2 − 0.8] ⇒ VIL = 1.63 V
8
1
V0 HU = {2(1.63) + 5 − 0.8 + 2}
2
V0 HU = 4.73 V
5
VIH = 0.8 + (5 − 2 − 0.8) ⇒ VIH = 2.18 V
8
1
V0 LU = {2(2.18) − 5 − 0.8 + 2}
2
V0 LU = 0.275 V
c.
NM L = VIL − V0 LU = 1.63 − 0.275 ⇒ NM L = 1.35 V
NM H = V0 HU − VIH = 4.73 − 2.18 ⇒ NM H = 2.55 V
TYU16.8
TYU16.9
NMOS − 2 transistors in series
Wn = 2 (W ) = 2W
PMOS − 2 transistors in series
W p = 2 ( 2W ) = 4W
TYU16.10
The NMOS part of the circuit is:
TYU16.11
The NMOS part of the circuit is:
TYU16.12
Exclusive-OR
A
0
1
0
1
B
0
0
1
1
f
0
1
1
0
TYU16.13
NMOS conducting for 0 ≤ vI ≤ 4.2 V
⇒ NMOS Conducting: 0 ≤ t ≤ 8.4 s
NMOS Cutoff: 8.4 ≤ t ≤ 10 s
PMOS cutoff for 0 ≤ vI ≤ 1.2 V
⇒ PMOS Cutoff: 0 ≤ t ≤ 2.4 s
PMOS Conducting: 2.4 ≤ t ≤ 10 s
TYU16.14
(a)
1 K ⇒ 32 × 32 array
Each row and column requires a 5-bit word ⇒ 6 transistors per row and column, ⇒ 32 × 6 + 32 × 6 = 384
transistors plus buffer transistors.
(b)
4 K ⇒ 64 × 64 array
Each row and column requires a 6-bit word ⇒ 7 transistors per row and column ⇒ 64 × 7 + 64 × 7 = 896
transistors plus buffer transistors.
(c)
16 K ⇒ 128 × 128 array
Each row and column requires a 7-bit word ⇒ 8 transistors per row and column
⇒ 128 × 8 + 128 × 8 = 2048 transistors plus buffer transistors.
TYU16.15
From Equation (16.84)
(W / L )nA 2 (VDDVTN ) − 3VTN2 2(2.5)(0.4) − 3(0.4) 2
=
=
= 0.526
2
2
(W / L )n1
(VDD − 2VTN )
( 2.5 − 2(0.4) )
From Equation (16.86)
(W / L ) p
(W / L )nB
=
2
⎡ 2(2.5)(0.4) − 3(0.4) 2 ⎤
kn′ 2 (VDDVTN ) − 3VTN
⋅
= (2.5) ⎢
⎥ = 0.862
2
k ′p
(2.5 − 0.4) 2
(VDD + VTP )
⎣
⎦
⎛W ⎞
⎛W ⎞
So ⎜ ⎟ of transmission gate device must be < 0.526 times the ⎜ ⎟ of the NMOS transistors in the
L
⎝ ⎠
⎝L⎠
⎛W ⎞
⎛W ⎞
inverter cell. The ⎜ ⎟ of the PMOS transistors must be < 0.862 times the ⎜ ⎟ of the transmission gate
⎝L⎠
⎝L⎠
W
W
⎛ ⎞
⎛ ⎞
devices. Then the ⎜ ⎟ of the PMOS devices must be < 0.453 times ⎜ ⎟ of NMOS devices in cell.
⎝L⎠
⎝L⎠
TYU16.16
Initial voltage across the storage capacitor = VDD − VTN = 3 − 0.5 = 2.5 V .
Now
dV
I
−I = C
or V = − ⋅ t + K
dt
C
2.5
= 1.25 V , and C = 0.05 pF . Then
where K = 2.5 V , t = 1.5 ms, V =
2
I (1.5 × 10−3 )
⇒
1.25 = 2.5 −
(0.05 × 10−12 )
I = 4.17 × 10−11 A ⇒ I = 41.7 pA
Chapter 16
Problem Solutions
16.1
(a)
ΔVTN =
Cax =
2e ∈s N a
Cax
⎡
⎤
⎣ 2φ fp + VSB − 2φ fp ⎦
∈ax (3.9)(8.85 × 10 −14 )
=
= 7.67 × 10−8
450 × 10−8
tax
2e ∈s N a = ⎡⎣ 2 (1.6 × 10 −19 ) (11.7 ) ( 8.85 × 10−14 )( 8 × 1015 ) ⎤⎦
Then
5.15 × 10−8 ⎡
ΔVTN =
⋅ 2(0.343) + VSB − 2(0.343) ⎤⎦
7.67 × 10−8 ⎣
For VSB = 1 V :
ΔVTN = 0.671 ⎣⎡ 1.686 − 0.686 ⎦⎤ ⇒ ΔVTN = 0.316 V
For VSB = 1 V :
1/ 2
= 5.15 × 10 −8
ΔVTN = 0.671 ⎣⎡ 2.686 − 0.686 ⎦⎤ ⇒ ΔVTN = 0.544 V
(b)
For VGS = 2.5 V, VDS = 5 V, transistor biased in the saturation region.
I D = K n (VGS − VTN ) 2
For VSB = 0,
I D = 0.2(2.5 − 0.8) 2 = 0.578 mA
For VSB = 1,
2
I D = 0.2 ( 2.5 − [ 0.8 + 0.316]) = 0.383 mA
For VSB = 2,
2
I D = 0.2 ( 2.5 − [ 0.8 + 0.544]) = 0.267 mA
16.2
(a)
ID =
VDD − vO
= K n ⎡⎣ 2(VGS − VTN )vO − vO2 ⎤⎦
RD
5 − (0.1)
2
= K n ⎡ 2 ( 5 − 0.8 )( 0.1) − ( 0.1) ⎤
3
⎣
⎦
40 × 10
−5
8 × 10 ⎛ W ⎞
or K n = 1.476 × 10−4 A / V 2 =
⎜ ⎟
2 ⎝L⎠
⎛W ⎞
So that ⎜ ⎟ = 3.69
⎝L⎠
b.
From Equation (16.10).
2
K n RD [VIt − VTN ] + [VIt − VTN ] − VDD = 0
2
(0.1476)(40) [VIt − 0.8] + [VIt − 0.8] − 5 = 0
−1 ± (1) 2 + 4(0.1476)(40)(5)
2(0.1476)(40)
or [VIt − 0.8] = 0.839
or [VIt − 0.8] =
So that VIt = 1.64 V
P = I D (max) ⋅ VDD
5 − (0.1)
= 0.1225 mA
40
or P = 0.6125 mW
and I D (max) =
16.3
a.
From Equation (16.10), the transistor point is found from
K n RD (VIt − VTN ) 2 + (VIt − VTN ) − VDD = 0
K n = 50 µ A / V 2 , RD = 20 k Ω, VTN = 0.8 V
(0.05)(20)(VIt − VTN ) 2 + (VIt − VTN ) − 5 = 0
−1 ± 1 + 4(0.05)(20)(5)
= 1.79 V So VIt = 2.59 V
2(0.05)(20)
V0t = 1.79 V
Output voltage for vI = 5 V is determined from Equation (16.12):
VIt − VTN =
v0 = 5 − (0.05)(20) ⎡⎣ 2(5 − 0.8)v0 − v02 ⎤⎦
v02 − 9.4v0 + 5 = 0
So v0 =
b.
9.4 ± (9.4) 2 − 4(1)(5)
= 0.566 V
2(1)
For RD = 200 kΩ,
−1 ± 1 + 4(0.05)(200)(5)
= 0.659 V So VIt = 1.46 V
2(0.05)(200)
V0t = 0.659 V
(VIt − VTN ) =
v0 = 5 − (0.05)(200) ⎡⎣ 2 ( 5 − 0.8 ) v0 − v02 ⎤⎦
or 10v02 − 85v0 + 5 = 0
v0 =
16.4
(a)
85 ±
2
(85 ) − 4 (10 )( 5 )
= 0.0592 V
2 (10 )
P = IV
0.25 = I (33) ⇒ I = 75.76 μA
3.3 − 0.15
⇒ R = 41.6 K
R=
0.07576
2
⎛ k ′ ⎞⎛ W ⎞
I = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN )
2
L
⎝ ⎠⎝ ⎠
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.8 ) ⇒ ⎜ ⎟ = 0.303
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
(b)
VDS (sat) = VGS − VTN
V − VDS (sat)
2
I D = K n (VGS − VTN ) = DD
R
3.3 − (VGS − 0.8 )
⎛ 0.08 ⎞
2
⎜
⎟ ( 0.303) (VGS − 1.6VGS + 0.64 ) =
41.6
⎝ 2 ⎠
0.504 (VGS2 − 1.6VGS + 0.64 ) = 4.1 − VGS
0.504VGS2 + 0.1936VGS − 3.777 = 0
−0.1936 ± 0.03748 + 4(0.504)(3.777)
2(0.504)
= 2.55 V
VGS =
VGS
For 0.8 ≤ VGS ≤ 2.55 V
Transistor biased in saturation region
16.5
(a)
P = I ⋅ VDD
0.25 = I (3.3) ⇒ I = 75.76 μA
For Sat Load
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
I = 75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.15 − 0.8 ) ⇒ ⎜ ⎟ = 0.343
2
L
⎝ ⎠ ⎝ ⎠L
⎝ L ⎠L
2
2
⎛ 80 ⎞
⎛ 80 ⎞ ⎛ W ⎞
⎜ ⎟ ( 0.343)( 3.3 − 0.15 − 0.8 ) = I = 75.76 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ 2 ( 2.5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤⎦
2
2
L
⎝ ⎠
⎝ ⎠ ⎝ ⎠D
W
⎛ ⎞
⎜ ⎟ = 3.89
⎝ L ⎠D
Eq 16.21
⎛
3.89 ⎞
3.3 − 0.8 + 0.8 ⎜⎜1 +
⎟⎟
0.343
⎝
⎠ = 5.994
VIt =
4.3677
3.89
1+
0.343
0.8 ≤ VGS ≤ 1.372 V
16.6
(a)
From Equation (16.23)
KD ⎡
K
2
2
2 ( 3 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 4.26
⎣
⎦
KL
KL
KD ⎡
K
2
2
2 ( 2.5 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 5.4
⎣
⎦
KL
KL
(b)
2
iD = K L (VGSL − VTNL ) = K L (VDD − vO − VTNL ) 2
⎛ 0.080 ⎞
2
=⎜
⎟ (1)(3 − 0.25 − 0.5) ⇒ iD = 0.203 mA
⎝ 2 ⎠
P = iD ⋅ VDD = (0.203)(3) ⇒ P = 0.608 mW
(c)
for both parts (a) and (b).
16.7
P = 0.4 mW = iD ⋅ VDD = iD (3) ⇒ iD = 0.1333 mA
iD = K L (VDD − vO − VTNL ) 2
2
⎛ 0.080 ⎞ ⎛ W ⎞
⎛W ⎞
0.1333 = ⎜
⎟ ⎜ ⎟ ( 3 − 0.1 − 0.5 ) = ( 0.2304 ) ⎜ ⎟
2
L
⎝
⎠ ⎝ ⎠L
⎝ L ⎠L
⎛W ⎞
So ⎜ ⎟ = 0.579
⎝ L ⎠L
KD ⎡
2
2
2 ( 2.5 − 0.5 )( 0.1) − ( 0.1) ⎤ = ( 3 − 0.1 − 0.5 )
⎣
⎦
KL
⇒
KD
⎛W ⎞
= 14.8 so that ⎜ ⎟ = 8.55
KL
⎝ L ⎠D
(
3 − 0.5 + 0.5 1 + 14.8
VIt =
)
1 + 14.8
or
VIt = 1.02 V , VOt = 0.52 V
16.8
We have
KD
2
⎡⎣ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = (VDD − vO − VTNL )
KL
(W / L )D ⎡
2
2
2 (VDD − VTN − VTN )( 0.08VDD ) − ( 0.08VDD ) ⎤ = (VDD − 0.08VDD − VTN )
⎣
⎦
(W / L )L
(W / L )D ⎡
2
2
2
⎤⎦ = ⎡⎣( 0.92 − 0.2 )VDD ⎤⎦ = 0.5184VDD
2 (VDD − 2 ( 0.2 )VDD ) ( 0.08VDD ) − 0.0064VDD
⎣
(W / L )L
(W / L )D
(W / L )D
= 5.4
[0.096] = 0.5184 ⇒
(W / L )L
(W / L )L
16.9
VOH = VB − VTN = Logic 1
So
(a)
VB = 4 V ⇒ VOH
(b)
VB = 5 V ⇒ VOH
(c)
VB = 6 V ⇒ VOH
(d)
VB = 7 V ⇒ VOH
For vI = VOH
= 3V
= 4V
= 5V
= 5 V ,since VDS = 0
K D ⎡⎣ 2 ( vI − VT ) vO − vO2 ⎤⎦ = K L [VB − vO − VT ]
2
Then
(1) ⎣⎡ 2 ( 3 − 1)VOL − VOL2 ⎦⎤ = ( 0.4 ) [ 4 − VOL − 1]
2
(a)
(1) ⎣⎡ 2 ( 4 − 1)VOL − VOL2 ⎦⎤ = ( 0.4 ) [5 − VOL − 1]
2
(b)
(1) ⎣⎡ 2 ( 5 − 1)VOL − VOL2 ⎦⎤ = ( 0.4 ) [6 − VOL − 1]
2
(c)
⇒ VOL = 0.657 V
⇒ VOL = 0.791 V
⇒ VOL = 0.935 V
(d)
Load in non-sat region
iDD = iOL
2
2
⎤⎦ = ( 0.4 ) ⎡ 2 ( 7 − VOL − 1)( 5 − VOL ) − ( 5 − VOL ) ⎤
(1) ⎡⎣ 2 ( 5 − 1) VOL − VOL
⎣
⎦
2
2
8VOL − VOL
= ( 0.4 ) ⎡⎣ 2 ( 6 − VOL )( 5 − VOL ) − ( 25 − 10VOL + VOL
)⎤⎦
2
= ( 0.4 ) ⎣⎡ 2 ( 30 − 11VOL + VOL
) − 25 + 10VOL − VOL2 ⎦⎤
2
2
⎤⎦
= ( 0.4 ) ⎡⎣60 − 22VOL + 2VOL
− 25 + 10VOL − VOL
2
2
8VOL − VOL
= 14 − 4.8VOL + 0.4VOL
2
1.4VOL
− 12.8VOL + 14 = 0
VOL =
12.8 ± 163.84 − 4 (1.4 )(14 )
2 (1.4 )
VOL = 1.27V
For load
VDS ( sat ) = 7 − 1.27 − 1 = 4.73V
VDS = 5 − 1.27 = 3.73 non-sat
16.10
a.
For load VOt = VDD + VTNL = 5 − 2 = 3 V
KD
⋅ (VIt − VTND ) = −VTNL
KL
500
(VIt − 0.8 ) = − ( −2 )
100
⇒ VIt = 1.69 V ⎫
⎬ Load
VOt = 3 V ⎭
Driver: VOt = VIt − VTND = 1.69 − 0.8 = 0.89 V
VIt = 1.69 V ⎫
⎬ Driver
V0t = 0.89 V ⎭
From Equation (16.29(b)):
b.
2
500
⋅ ⎡⎣ 2(5 − 0.8)v0 − v02 ⎤⎦ = ⎣⎡ − ( −2 ) ⎦⎤
100
5v02 − 42v0 + 4 = 0
v0 =
c.
42 ±
( 42 ) − 4 ( 5)( 4 )
⇒ v0 = 0.0963 V
2 (5)
2
iD = K L ( −VTNL ) = 100 ⎡⎣ − ( −2 ) ⎤⎦ ⇒ iD = 400 µ A
2
2
16.11
2
2
⎛ 500 ⎞ ⎡
⎜
⎟ ⎣ 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎦⎤ = ( −VTNL )
⎝ 50 ⎠
So
( −VTNL )
2
= 4.9 ⇒ VTNL = −2.21 V
16.12
(a)
P = iD ⋅ VDD
150 = iD ⋅ ( 3) ⇒ iD = 50 µ A
iD = K L (−VTNL ) 2
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
50 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎦⎤ ⇒ ⎜ ⎟ = 1.25
2
L
⎝ ⎠ ⎝ ⎠L
⎝ L ⎠L
2
KD ⎡
2
2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡⎣ − ( −1) ⎤⎦
⎦
KL ⎣
K D (W / L ) D
⎛W ⎞
=
= 2.04 ⇒ ⎜ ⎟ = 2.55
K L (W / L ) L
⎝ L ⎠D
For the Load:
VOt = VDD + VTNL = 3 − 1 ⇒ VOt = 2 V
2.04 (VIt − 0.5 ) = ⎡⎣ − ( −1) ⎤⎦ ⇒ VIt = 1.20 V
For the Driver:
VOt = VIt − VTND = 1.20 − 0.5 ⇒ VOt = 0.70 V
VIt = 1.20 V
(b)
NM L = VIL − VOLU
NM H = VOHU − VIH
VIL = 0.5 +
⎡⎣ − ( −1) ⎤⎦
= 0.902 V
( 2.04 )(1 + 2.04 )
2 ⎡ − ( −1) ⎤⎦
VIH = 0.5 + ⎣
= 1.31 V
3 ( 2.04 )
Then VOHU = ( 3 − 1) + ( 2.04 )( 0.902 − 0.5 ) = 2.82 V
VOLU =
(1.31 − 0.5)
= 0.405 V
2
NM L = 0.902 − 0.405 ⇒ NM L = 0.497 V
NM H = 2.82 − 1.31 ⇒ NM H = 1.51 V
16.13
a.
From Equation (16.29(b)):
2
⎛W ⎞ ⎡
⎛W ⎞
2
⎜ ⎟ ⎣ 2 ( 2.5 − 0.5 )( 0.05 ) − ( 0.05 ) ⎤⎦ = ⎜ ⎟ [− ( −1)]
L
L
⎝ ⎠D
⎝ ⎠L
W
⎛ ⎞
⎜ ⎟ =1
⎝ L ⎠L
⎛W ⎞
⎜ ⎟ = 5.06
⎝ L ⎠D
Then
2
⎛ 80 ⎞
iD = ⎜ ⎟ (1) ⎡⎣ − ( −1) ⎤⎦
⎝ 2⎠
or iD = 40 µ A
b.
P = iD ⋅ VDD = ( 40 )( 2.5 ) ⇒ P = 100 µ W
16.14
a.
vI = 0.5 V ⇒ iD = 0 ⇒ P = 0
vI = 5 V, From Equation (16.12),
i.
ii.
v0 = 5 − ( 0.1)( 20 ) ⎡⎣ 2 ( 5 − 1.5 ) v0 − v02 ⎤⎦
2v02 − 15v0 + 5 = 0
v0 =
15 ±
(15 ) − 4 ( 2 )( 5 )
⇒ v0 = 0.35 V
2 ( 2)
2
5 − 0.35
= 0.2325 mA
20
P = iD ⋅ VDD = ( 0.2325 )( 5 ) ⇒ P = 1.16 mW
iD =
b.
ii.
i.
vI = 0.25 V ⇒ iD = 0 ⇒ P = 0
vI = 4.3 V, From Equation (16.23),
100 ⎣⎡ 2 ( 4.3 − 0.7 ) v0 − v02 ⎦⎤ = 10 [5 − v0 − 0.7 ]
2
10 ⎡⎣7.2v0 − v02 ⎤⎦ = 18.49 − 8.6v0 + v02
Then
11v02 − 80.6v0 + 18.49 = 0
v0 =
80.6 ±
(80.6 ) − 4 (11)(18.49 )
⇒ v0 = 0.237 V
2 (11)
2
Then
2
iD = 10 [5 − 0.237 − 0.7 ] = 165 µ A
P = iD ⋅ VDD = (165 )( 5 ) ⇒ P = 825 µ W
c.
i.
ii.
vI = 0.03 V ⇒ iD = 0 ⇒ P = 0
vI = 5 V
iD = K L ( −VTNL ) = (10 ) ⎣⎡ − ( −2 ) ⎦⎤ = 40 µ A
2
2
P = iD ⋅ VDD = ( 40 )( 5 ) ⇒ P = 200 µ W
16.15
vo1 = 3.8V
Load & Driver in Sat, region, ML1, MD1
iDL = iDD
2
2
⎛W ⎞
⎛W ⎞
⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ( vGSD − VTND )
⎝ L ⎠L
⎝ L ⎠D
2
(1) ( 5 − vo1 − 0.8 ) = (10 )( vI − 0.8 )
0.16 = 10 ( vI − 0.8 )
2
2
vI = 0.9265V
Now MD2: Non Sat and ML2: Sat
iDL = iDD
2
⎛W ⎞
⎛W ⎞
2
⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ⎡⎣ 2 ( vo1 − VTND ) vo 2 − vo 2 ⎤⎦
⎝ L ⎠L
⎝ L ⎠D
(1)( 5 − vo 2 − 0.8 )
( 4.2 − vo 2 )
2
2
= (10 ) ⎡⎣ 2 ( 3.8 − 0.8 ) vo 2 − vo22 ⎤⎦
= 10 ⎡⎣ 6vo 2 − vo22 ⎤⎦
17.64 − 8.4vo 2 + vo22 = 60vo 2 − 10vo22
11vo22 − 68.4vo 2 + 17.64 = 0
vo 2 =
68.4 ± 4678.56 − 4 (11)(17.64 )
2 (11)
vo 2 = 0.270 V
16.16
a.
VIH
From Equation (16.41),
2 ⎡ − ( −2 ) ⎤⎦
= 0.8 + ⎣
⇒ VIH = 1.95 V = v01
3( 4)
M D 2 in non-saturation and M L 2 in saturation.
2
⎤⎦ = K L ( −VTNL )
K D ⎡⎣ 2 ( v01 − VTND ) v02 − v02
2
⎤⎦ = (1) ⎡⎣ − ( −2 ) ⎤⎦
4 ⎡⎣ 2 (1.95 − 0.8 ) v02 − v02
2
2
2
− 9.2v02 + 4 = 0
4v02
v02 =
9.2 ±
( 9.2 ) − 4 ( 4 )( 4 )
⇒ v02 = 0.582 V
2 ( 4)
2
Both M D1 and M L1 in saturation region. From Equation (16.28(b)).
4 ⋅ ( vI − 0.8 ) = − ( −2 )
or vI = 1.8 V
b.
VIL = 0.8 +
( +2 )
= 1.25 V = v01
4 (1 + 4 )
M D 2 in saturation, M L 2 in non-saturation
2
2
K D [ vO1 − VTND ] = K L ⎡ 2 ( −VTNL )( 5 − vO 2 ) − ( 5 − vO 2 ) ⎤
⎣
⎦
2
4 (1.25 − 0.8 ) = 2 ( 2 )( 5 − v02 ) − ( 5 − v02 )
( 5 − v02 )
5 − v02 =
2
2
− 4 ( 5 − v02 ) + 0.81 = 0
( 4)
4±
2
− 4 (1)( 0.81)
2 (1)
= 0.214 V
so
v02 = 4.786 V
To find vI :
2
4 ( v01 − 0.8 ) = (1) ( − ( −2 ) )
2
v01 − 0.8 = 1
v01 − 1.8 = V
c.
VIH = 1.95 V, VIL = 1.25 V
16.17
a.
i.
Neglecting the body effect,
v0 = VDD − VTN
Assume VDD = 5 V, then v0 = 4.2 V
ii.
Taking the body effect into account: From Problem 16.1.
VTN = VTN 0 + 0.671 ⎡⎣ 0.686 + VSB − 0.686 ⎤⎦
and VSB = v0
Then
v0 = 5 − ⎡0.8 + 0.671 0.686 + v0 − 0.686 ⎤
⎣
⎦
(
)
v0 = 4.756 − 0.671 0.686 + v0
0.671 0.686 + v0 = 4.756 − v0
0.450 ( 0.686 + v0 ) = 22.62 − 9.51v0 + v02
v02 − 9.96v0 + 22.3 = 0
v0 =
b.
⇒ v0 = 3.40 V
2
PSpice results similar to Figure 16.13(a).
9.96 ±
( 9.96 )
2
− 4 ( 22.3)
16.18
Results similar to Figure 16.13(b).
16.19
M X on, M Y cutoff.
a.
From Equation (16.29(b)):
2
KD ⎡
2
2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎣⎡ − ( −2 ) ⎦⎤
⎣
⎦
KL
or
b.
KD
= 2.44
KL
For v X = vY = .5 V
2 ( 2.44 ) ⎣⎡ 2 ( 5 − 0.8 ) v0 − v02 ⎦⎤ = ⎡⎣ − ( −2 ) ⎤⎦
2
4.88v02 − 41.0v0 + 4 = 0
v0 =
41 ±
2
( 41) − 4 ( 4.88 )( 4 )
2 ( 4.88 )
or
v0 = 0.0987 V
c.
2
⎛ 80 ⎞
iD = ⎜ ⎟ (1) ⎡⎣ − ( −2 ) ⎤⎦ = 160 µ A
2
⎝ ⎠
P = (160 )( 5 ) ⇒ P = 800 µ W
for both parts (a) and (b).
16.20
(a)
Maximum value of vO in low state- when only one input is high, then,
2
KD ⎡
2
2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡⎣ − ( −1) ⎤⎦
⎦
KL ⎣
KD
= 2.04
KL
(b)
P = iD ⋅ VDD
0.1 = iD (3) ⇒ iD = 33.3 µ A
2
⎛ k′ ⎞⎛ W ⎞
iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL )
2
L
⎝ ⎠ ⎝ ⎠L
2
⎛ 80 ⎞⎛ W ⎞
⎛W ⎞
33.3 = ⎜ ⎟⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 0.8325
⎝ 2 ⎠⎝ L ⎠ L
⎝ L ⎠L
⎛W ⎞
Then ⎜ ⎟ = 1.70
⎝ L ⎠D
(c)
3 ( 2.04 ) ⎡⎣ 2 ( 3 − 0.5 ) vO − vO2 ⎤⎦ = ⎡⎣ − ( −1) ⎤⎦ ⇒ vO = 0.0329 V
16.21
(a)
2
One driver in non-sat,
2
2
⎤⎦
I D = K L ( −VTNL ) = K D ⎡⎣ 2 (VGS − VTND ) VDSD − VDSD
2
KD ⎡
2
⎤
⎣⎡ − ( −1) ⎦⎤ = K ⎣ 2 ( 3.3 − 0.5 )( 0.1) − ( 0.1) ⎦
L
KD
= 1.82
KL
(b)
P = ± VDD
0.1 = ± ( 3.3) ⇒ I = 30.3 µ A
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
30.3 = I = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 0.7575
⎝ 2 ⎠ ⎝ L ⎠L
⎝ L ⎠L
⎛W ⎞
⎜ ⎟ = 1.38
⎝ L ⎠D
(c)
0.1
= 0.05 V
2
0.1
= 0.0333 V
Three inputs High, vo ≈
3
0.1
= 0.025 V
Four inputs High, vo ≈
4
Two inputs High, vo ≈
(i)
(ii)
(iii)
16.22
a.
P = iD ⋅ VDD
250 = iD ( 5 ) ⇒ iD = 50 µΑ
⎛ k' ⎞⎛W ⎞
2
iD = ⎜ n ⎟ ⎜ ⎟ [ −VTNL1 ]
2
L
⎝ ⎠ ⎝ ⎠ ML1
2
⎛ 60 ⎞ ⎛ W ⎞
50 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −2 ) ⎤⎦
⎝ 2 ⎠ ⎝ L ⎠ ML1
⎛W ⎞
So that ⎜ ⎟ = 0.417
⎝ L ⎠ ML1
KD
2
⎡ 2 ( vI − VTND ) vO − vO2 ⎦⎤ = [ −VTNL ]
⎣
KL
2
KD ⎡
2
2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = ⎡⎣ − ( −2 ) ⎤⎦
⎣
⎦
KL
or
KD
⎛W ⎞
= 3.23 ⇒ ⎜ ⎟
= 1.35
KL
⎝ L ⎠ MD1
b.
For v X = vY = 0 ⇒ v01 = 5 and v03 = 4.2
Then
2
K D 2 ⎣⎡ 2 ( vO1 − VTND ) vO 2 − vO2 2 ⎦⎤ + K D 3 ⎣⎡ 2 ( vO 3 − VTND ) vO 2 − vO2 2 ⎦⎤ = K L 2 [ −VTNL 2 ]
K D 2 ∝ 8, K D 3 ∝ 8, K L 2 ∝ 1
2
2
8 ⎣⎡ 2 ( 5 − 0.8 ) v02 − v02
⎦⎤ + 8 ⎣⎡ 2 ( 4.2 − 0.8 ) v02 − v02 ⎦⎤ = (1) ⎡⎣ − ( −2 ) ⎤⎦
2
2
+ 54.4v02 − 8v02
=4
67.2v02 − 8v02
Then
16v02 − 121.6v0 + 4 = 0
v02 =
121.6 ±
So v02 = 0.0330 V
16.23
2
(121.6 ) − 4 (16 )( 4 )
2 (16 )
2
a.
We can write
2
2
K x ⎣⎡ 2 ( v X − VTN ) vDSX − vDSX
⎦⎤ = K y ⎣⎡ 2 ( vY − vDSX − VTN ) vDSY − vDSY ⎦⎤ = K L [VDD − vO − VTN ]
2
where v0 = vDSX + vDSY
We have
v X = vY = 9.2 V , VDD = 10 V , VTN = 0.8 V
2
2
and vDSY
terms. Let v0 ≈ 2vDSX . Then from the first and
As a good first approximation, neglect the vDSX
third terms in the above equation.
9 ⎡⎣ 2 ( 9.2 − 0.8 ) vDSX ⎤⎦ ≅ (1)(10 − 2vDSX − 0.8 )
(151.2 ) vDSX
2
≅ 84.64 − 36.8vDSX
So that vDSX = 0.450 V
From the first and second terms of the above equation.
9 ⎡⎣ 2 ( 9.2 − 0.8 ) vDSX ⎤⎦ ≅ 9 ⎡⎣ 2 ( 9.2 − vDSX − 0.8 ) vDSY ⎤⎦
or
(16.8)( 0.45) = 2 ( 9.2 − 0.45 − 0.8) vDSY
which yields vDSY = 0.475 V
Then v0 = vDSX + vDSY = 0.450 + 0.475
or v0 = 0.925 V
We have vGSX = 9.2 V
and vGSY = 9.2 − vDSX = 9.2 − 0.45
or vGSY = 8.75 V
b.
Since v0 is close to ground potential, the body effect will have minimal effect on the results. From
a PSpice analysis:
For part (a):
vDSX = 0.462 V, vDSY = 0.491 V, v0 = 0.9536 V, vGSX = 9.2 V, and vGSY = 8.738 V
For part (b):
vDSX = 0.441 V, vDSY = 0.475 V, v0 = 0.9154 V, vGSX = 9.2 V, and vGSY = 8.759 V
16.24
a.
We can write
2
2
2
⎤⎦ = K y ⎡⎣ 2 ( vY − vDSX − VTNY ) vDSY − vDSY
⎤⎦ = K L [ −VTNL ]
K x ⎡⎣ 2 ( v X − VTNX ) vDSX − vDSX
2
From the first and third terms, (neglect vDSX
),
4 ⎣⎡ 2 ( 5 − 0.8 ) vDSX ⎦⎤ = (1) ⎣⎡ − ( −1.5 ) ⎦⎤
2
or vDSX = 0.067 V
2
From the second and third terms, (neglect vDSY
),
4 ⎡⎣ 2 ( 5 − 0.067 − 0.8 ) vDSY ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦
2
or vDSY = 0.068 V
Now
vGSX = 5, vGSY = 5 − 0.067 ⇒ vGSY = 4.933 V
and v0 = vDSX + vDSY ⇒ v0 = 0.135 V
Since v0 is close to ground potential, the body-effect has little effect on the results.
16.25
(a)
0.2
= 0.05 V
4
2
⎤⎦
− VDSD
We have VDS of each driver ≈
K L [ −VTNL ] = K D ⎡⎣ 2 (VGSD − VTN ) VDSD
2
2
KD ⎡
2
⎤
⎣⎡ − ( −1) ⎦⎤ = K ⎣ 2 ( 3.3 − 0.4 )( 0.05 ) − ( 0.05 ) ⎦
L
KD
= 3.478
KL
(b)
P = I VDD
0.15 = I ( 3.3) ⇒ I = 45.45 µ A
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
45.45 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 1.14
⎝ 2 ⎠ ⎝ L ⎠L
⎝ L ⎠L
⎛W ⎞
⎜ ⎟ = 3.95
⎝ L ⎠D
16.26
Complement of (B AND C) OR A ⇒ ( B ⋅ C ) + A
16.27
Considering a truth table, we find
A
B
Y
0
0
0
0
1
1
1
0
1
1
1
0
which shows that the circuit performs the exclusive-OR function.
16.28
( A + B )(C + D)
16.29
(a)
Carry-out = A • ( B + C ) + B • C
(b)
For vO1 = Low = 0.2 V
2
KD ⎡
2
2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎣⎡ − ( −1.5 ) ⎦⎤ ⇒
⎦
KL ⎣
⎛W ⎞
⎛W ⎞
For ⎜ ⎟ = 1, then ⎜ ⎟ = 1.37
⎝ L ⎠L
⎝ L ⎠D
⎛W ⎞
So, for M 6 : ⎜ ⎟ = 1.37
⎝ L ⎠6
To achieve the required composite conduction parameter,
⎛W ⎞
For M 1 − M 5 : ⎜ ⎟ = 2.74
⎝ L ⎠1− 5
16.30
AE: VDS ≈ 0.075
2
2
⎛W ⎞
(1) ⎡⎣ − ( −1) ⎤⎦ ≅ ⎜ ⎟ ⎡ 2 ( 3.3 − 0.4 )( 0.075 ) − ( 0.075 ) ⎤
⎣
⎦
L
⎝ ⎠D
⎛W ⎞
⎛W ⎞
⎛W ⎞
⎛W ⎞
⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 2.33 ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 4.66
⎝ L ⎠ A ⎝ L ⎠E
⎝ L ⎠ B ⎝ L ⎠C ⎝ L ⎠ D
16.31
Not given
16.32
a.
From Equation (16.43),
5 − 0.8 + 0.8
= VIt = 2.5 V
vI = VIt =
1+1
p − channel, V0 Pt = 2.5 − ( −0.8 ) ⇒ V0 Pt = 3.3 V
n − channel, V0 Nt = 2.5 − 0.8 ⇒ V0 Nt = 1.7 V
c
For vI = 2 V, NMOS in saturation and PMOS in nonsaturation. From Equation (16.49),
( 2 − 0.8)
2
2
= ⎡ 2 ( 5 − 2 − 0.8 )( 5 − v0 ) − ( 5 − v0 ) ⎤
⎣
⎦
1.44 = 4.4(5 − v0 ) − (5 − v0 ) 2
2
So ( 5 − v0 ) − 4.4 ( 5 − v0 ) + 1.44 = 0
( 5 − v0 ) =
4.4 ±
( 4.4 )
2
− 4 (1)(1.44 )
2
or
5 − v0 = 0.356 ⇒ v0 = 4.64 V
By symmetry, for vI = 3 V, v0 = 0.356 V
16.33
(a)
⎛ 80 ⎞
K n = ⎜ ⎟ ( 2 ) = 80 µ A / V 2
⎝ 2⎠
⎛ 40 ⎞
K p = ⎜ ⎟ ( 4 ) = 80 µ A / V 2
⎝ 2 ⎠
VDD + VTP +
(i)
VIt =
Kn
⋅ VTN
Kp
Kn
Kp
1+
=
3.3 − 0.4 + (1)(0.4)
1+1
VIt = 1.65 V
PMOS:
VOt = VIt − VTP = 1.65 − ( −0.4 ) ⇒ VOt = 2.05 V
NMOS:
VOt = VIt − VTN = 1.65 − ( 0.4 ) ⇒ VOt = 1.25 V
(iii) For vO = 0.4 V : NMOS: Non-sat: PMOS:Sat
2
⎤⎦ = K p [VSGP + VTP ]
K n ⎡⎣ 2 (VGSN − VTN )VDS − VDS
2
2 ( vI − 0.4 )( 0.4 ) − ( 0.4 ) = ( 3.3 − vI − 0.4 ) ⇒ vI = 1.89 V
2
For vO = 2.9 V , By symmetry
vI = 1.65 − (1.89 − 1.65 ) ⇒ vI = 1.41 V
(b)
⎛ 80 ⎞
K n = ⎜ ⎟ ( 2 ) = 80 µ A/V 2
⎝ 2⎠
⎛ 40 ⎞
K p = ⎜ ⎟ ( 2 ) = 40 µ A/V 2
⎝ 2 ⎠
2
80
⋅ ( 0.4 )
40
⇒ VIt = 1.44 V
80
1+
40
3.3 − 0.4 +
VIt =
(i)
PMOS:
VOt = 1.44 − ( −0.4 ) ⇒ VOt = 1.84 V
NMOS:
VOt = 1.44 − 0.4 ⇒ VOt = 1.04 V
For vO = 0.4 V
(iii)
(80 ) ⎣⎡ 2 ( vI − 0.4 )( 0.4 ) − ( 0.4 )
⎤ = 40 3.3 − vI − 0.4]2 ⇒ vI = 1.62 V
⎦ ( )[
For vO = 2.9 V : NMOS: Sat, PMOS: Non-sat
2
(80 ) [vI − 0.4]
16.34
(a)
(i)
2
= ( 40 ) ⎡ 2 ( 3.3 − vI − 0.4 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ vI = 1.18 V
⎣
⎦
From Eq. (16.43), switching voltage
VDD + VTP +
vI t =
(ii)
2
Kn
⋅ VTN
Kp
Kn
1+
Kp
2 ( 4)
( 0.4 ) 3.2266
12
=
⇒ vIt = 1.776 V
1.8165
2 ( 4)
1+
12
3.3 + ( −0.4 ) +
=
v0 = 3.1 V , PMOS, non-sat; NMOS, sat
⎛ k p′ ⎞ ⎛ W ⎞
2
⎛ kn′ ⎞ ⎛ W ⎞
2
⎜ ⎟ ⎜ ⎟ ⎡⎣ 2 (VSG + VTP ) VSD − VSD ⎤⎦ = ⎜ ⎟ ⎜ ⎟ (VGS − VTN )
L
L
2
2
⎝ ⎠ ⎝ ⎠n
⎝ ⎠⎝ ⎠p
2
2
⎛ 40 ⎞
⎛ 80 ⎞
⎜ ⎟ (12 ) ⎣⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − 3.1) − ( 3.3 − 3.1) ⎦⎤ = ⎜ ⎟ ( 4 ) [ vI − 0.4]
⎝ 2 ⎠
⎝ 2⎠
12 [1.16 − 0.4vI − 0.04] = 8 ⎣⎡ vI2 − 0.8vI + 0.16 ⎦⎤
8vI2 − 1.6vI − 12.16 = 0
vI =
(iii)
1.6 ± 2.56 + 4 ( 8 ) (12.16 )
2 (8)
⇒ vI = 1.337 V
v0 = 0.2 V PMOS: sat, NMOS, non-sat.
2
2
⎛ 40 ⎞
⎛ 80 ⎞
⎜ ⎟ (12 ) [3.3 − vI − 0.4] = ⎜ ⎟ ( 4 ) ⎣⎡ 2 ( vI − 0.4 )( 0.2 ) − ( 0.2 ) ⎦⎤
⎝ 2 ⎠
⎝ 2⎠
12 ⎡⎣8.41 − 5.8vI + vI2 ⎤⎦ = 8 [ 0.4vI − 0.2]
12vI2 − 72.8vI + 102.52 = 0
vI =
72.8 ± 5299.84 − 4 (12 )(102.52 )
vI = 2.222 V
(b)
2 (12 )
2 ( 6)
( 0.4 ) 3.5928
4
=
2.732
2 (6)
1+
4
3.3 + ( −0.4 ) +
vIt =
(i)
vIt = 1.315 V
(ii)
From (a), (ii)
4 [1.16 − 0.4vI − 0.04] = 12 ⎡⎣vI2 − 0.8vI + 0.16 ⎤⎦
12vI2 − 8vI − 2.56 = 0
vI =
8 ± 64 + 4 (12 )( 2.56 )
2 (12 )
⇒ vI = 0.903 V
(iii)
From (a), (iii)
⎡
4 ⎣8.41 − 5.8vI + vI2 ⎤⎦ = 12 [ 0.4vI − 0.2]
4vI2 − 28vI + 36.04 = 0
vI =
28 ± 784 − 4 ( 4 )( 36.04 )
2 ( 4)
16.35
a.
⇒ vI = 1.70 V
For vO1 = 0.6 < VTN ⇒ vO 2 = 5 V
N1 in nonsaturation and P1 in saturation. From Equation (16.45),
⎡ 2 ( vI − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI − 0.8]2
⎣
⎦
1.2vI − 1.32 = 17.64 − 8.4vI + vI2
or
vI2 − 9.6vI + 18.96 = 0
vI =
9.6 ±
( 9.6 )
2
− 4 (1)(18.96 )
2
or
vI = 2.78 V
V0 Nt ≤ v02 ≤ V0 Pt
b.
From symmetry, VIt = 2.5 V
V0 Pt = 2.5 + 0.8 = 3.3 V
and V0 Nt = 2.5 − 0.8 = 1.7 V
So 1.7 ≤ v02 ≤ 3.3 V
16.36
a.
V0 Nt ≤ v01 ≤ V0 Pt
By symmetry, VIt = 2.5 V
V0 Pt = 2.5 + 0.8 = 3.3 V
and V0 Nt = 2.5 − 0.8 = 1.7 V
So 1.7 ≤ v01 ≤ 3.3 V
b.
For vO 2 = 0.6 < VTN ⇒ vO 3 = 5 V
N 2 in nonsaturation and P2 in saturation. From Equation (16.57),
⎡ 2 ( vI 2 − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI 2 − 0.8]2
⎣
⎦
1.2vI 2 − 1.32 = 17.64 − 8.4vI 2 + vI22
or
vI22 − 9.6vI 2 + 18.96 = 0
So vI 2 = v01 = 2.78 V
For v01 = 2.78, both N1 and P1 in saturation. Then
vI = 2.5 V
16.37
a.
iPeak = K n ( vI − VTN )
1/ 2
iPeak = 0.1 ⋅ ( 2.5 − 0.8 ) = 0.538 ( mA )
b.
16.38
(a)
1/ 2
iPeak = 0.1 ⋅ (1.65 − 0.8 ) = 0.269 ( mA )
⎛ 50 ⎞
K n = ⎜ ⎟ ( 2 ) = 50 µ A / V 2
⎝ 2⎠
⎛ 25 ⎞
K p = ⎜ ⎟ ( 4 ) = 50 µ A / V 2
⎝ 2 ⎠
2
I D , peak = K n ( v1 − VTN ) = 50 ( 2.5 − 0.8 )
or I D , peak = 144.5 µ A
(b)
K n = 50 µ A / V 2 , K p = 25 µ A/V 2
From Equation (16.55),
50
5 − 0.8 +
(0.8)
25
vIt =
= 2.21 V
50
1+
25
Then
I D , peak = K n (VIt − VTN ) 2 = 50 ( 2.21 − 0.8 )
or I D , peak = 99.4 µ A
16.39
2
2
(a)
Switching Voltage, Eq. (16.43)
2 ( 4)
( 0.4 )
8
vIt =
= 1.65 V = vIt
2 ( 4)
1+
8
80
2
⎛ ⎞
iD, peak = ⎜ ⎟ ( 4 )(1.65 − 0.4 ) ⇒ iD , peak = 250 μA
⎝ 2⎠
(b)
3.3 − 0.4 +
2 ( 4)
( 0.4 )
4
= 1.436 V = vIt
vIt =
2 ( 4)
1+
4
2
⎛ 80 ⎞
iD , peak = ⎜ ⎟ ( 4 )(1.436 − 0.4 ) ⇒ iD , peak = 172 μA
⎝ 2⎠
(c)
3.3 − 0.4 +
2 ( 4)
( 0.4 )
12
vIt =
⇒ vIt = 1.776 V
2 ( 4)
1+
12
2
⎛ 80 ⎞
iD , peak = ⎜ ⎟ ( 4 )(1.776 − 0.4 ) ⇒ iD , peak = 303 μA
⎝ 2⎠
3.3 − 0.4 +
16.40
2
a. P = fCLVDD
For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2
or P = 50 µ W
For VDD = 15 V, P = (10 × 106 )(0.2 × 10−12 )(15) 2
or P = 450 µ W
b.
For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2
or P = 50 µ W
16.41
(a)
2
P = fCLVDD
= (150 × 106 )( 0.4 ×10−12 ) ( 5 ) = 1.5 × 10 −3 W / inverter
2
Total power: PT = ( 2 × 106 )(1.5 × 10−3 ) ⇒ PT = 3000 W !!!!
(b)
For f = 300 MHz
2
1.5 ×10 = ( 300 ×106 )( 0.4 × 10−12 ) VDD
⇒ VDD = 3.54 V
−3
16.42
3
= 3 × 10−7 W
107
(a)
P=
(b)
2
P = fCLVDD
⇒ CL =
P
2
fVDD
(i)
CL =
(ii)
CL =
(iii)
CL =
⇒ CL = 0.0024 pF
3 × 10−7
( 5 ×10 ) ( 5)
6
2
2
⇒ CL = 0.00551pF
2
⇒ CL = 0.0267 pF
3 × 10−7
( 5 ×10 ) ( 3.3)
6
3 × 10−7
( 5 ×10 ) (1.5)
6
16.43
10
= 2 × 10−6 W
5 × 106
P
CL =
2
fVDD
P=
(a)
(b)
2 × 10−6
⇒ CL = 0.01 pF
(i)
CL =
(ii)
CL =
(iii)
CL =
16.44
(a)
For vI ≅ VDD , NMOS in nonsaturation
(8 ×10 ) ( 5)
6
2
2
⇒ CL = 0.023 pF
2
⇒ CL = 0.111 pF
2 × 10−6
(8 ×10 ) ( 3.3)
6
2 × 10−6
(8 ×10 ) (1.5)
6
2
⎤⎦ and vDS ≅ 0
iD = K n ⎡⎣ 2 ( vI − VTN ) vDS − vDS
So
di
1
= D ≅ K n ⎡⎣ 2 (VDD − VTN ) ⎤⎦
rds dvDS
Or
rds =
1
⎛ kn′ ⎞ ⎛ W ⎞
⎜ 2 ⎟ ⎜ L ⎟ ⋅ 2 (VDD − VTN )
⎝ ⎠ ⎝ ⎠n
or
⎛W
kn′ ⎜
⎝L
For vI ≅ 0,
rds =
1
⎞
⎟ ⋅ (VDD − VTN )
⎠n
PMOS in nonsaturation
2
⎤⎦
iD = K p ⎡⎣ 2 (VDD − vI + VTP ) vSD − vSD
and vSD ≅ 0 for vI ≅ 0.
So
⎛ k ′p ⎞ ⎛ W ⎞
di
1
= D ≅ ⎜ ⎟ ⎜ ⎟ ⋅ 2 (VDD + VTP )
rsd dvSD ⎝ 2 ⎠ ⎝ L ⎠ p
or
rsd =
1
k ′p
⎛W
⎜⎜
⎝ Lp
1
⎞
⎟⎟ ⋅ (VDD + VTP )
⎠
⎛W ⎞
⎛W ⎞
(b) For ⎜ ⎟ = 2, ⎜ ⎟ = 4
⎝ L ⎠n
⎝ L ⎠p
rds =
rsd =
1
⇒ r = 2.38 k Ω
50
2
( )( )( 5 − 0.8) ds
1
( 25 )( 4 )( 5 − 0.8 )
⇒ rsd = 2.38 k Ω
⎛W ⎞
For ⎜ ⎟ = 2,.
⎝ L ⎠p
rsd =
1
⇒ r = 4.76 k Ω
25
2
( )( )( 5 − 0.8 ) sd
Now, for NMOS:
vds = id rds or id =
vds
0.5
=
⇒ id = 0.21 mA
rds 2.38
For PMOS:
For rsd = 2.38 k Ω,
id =
vsd
0.5
=
⇒ id = 0.21 mA
rsd 2.38
For rsd = 4.76 k Ω ,
id =
vsd
0.5
=
⇒ id = 0.105 mA
rsd 4.76
16.45
From Equation (16.63)
3
VIL = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIL = 4.125 V
8
and Equation (16.62)
1
V0 HU = ⋅ ⎡⎣ 2 ( 4.125 ) + 10 − 1.5 + 1.5⎤⎦
2
or V0 HU = 9.125 V
From Equation (16.69)
5
VIH = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIH = 5.875 V
8
and Equation (16.68)
1
V0 LU = ⋅ ⎡⎣ 2 ( 5.875 ) − 10 − 1.5 + 1.5⎤⎦
2
or V0 LU = 0.875 V
Now
NM L = VIL − V0 LU = 4.125 − 0.875 ⇒ NM L = 3.25 V
NM H = V0 HU − VTH = 9.125 − 5.875 ⇒ NM H = 3.25 V
16.46
From Equation (16.71)
VIL = 1.5 +
⎡
⎤
100
⎥
50 − 1 ⎥ = 1.5 + 7 ⎡ 2 ( 0.632 ) − 1⎤
2
⎣
⎦
⎢ 100 + 3 ⎥
⎢⎣
⎥
50
⎦
(10 − 1.5 − 1.5 ) ⎢⎢
⎛ 100 ⎞
− 1⎟
⎜
⎝ 50
⎠
or
VIL = 3.348 V
From Equation (16.70)
1 ⎧⎛ 100 ⎞
⎫
⎛ 100 ⎞
V0 HU = ⋅ ⎨⎜1 +
⎟ ( 3.348 ) + 10 − ⎜
⎟ (1.5 ) + 1.5⎬
2 ⎩⎝
50 ⎠
⎝ 50 ⎠
⎭
or V0 HU = 9.272 V
⎡
⎤
⎛ 100 ⎞
⎢ 2⎜
⎥
⎟
(10 − 1.5 − 1.5 ) ⎢ ⎝ 50 ⎠
= 1.5 +
− 1⎥ = 1.5 + 7 [1.51 − 1]
⎥
⎛ 100 ⎞ ⎢ ⎛ 100 ⎞
− 1⎟ ⎢ 3 ⎜
+1 ⎥
⎜
⎟
⎝ 50
⎠ ⎣⎢ ⎝ 50 ⎠
⎦⎥
From Equation (16.77)
VIH
or
VIH = 5.07 V
From Equation (16.76)
100
100
( 5.07 ) ⎛⎜1 + ⎞⎟ − 10 − ⎛⎜ ⎞⎟ (1.5 ) + 1.5
50
⎝
⎠
⎝ 50 ⎠
V0 LU =
⎛ 100 ⎞
2⎜
⎟
⎝ 50 ⎠
or V0 LU = 0.9275 V
Now NM L = VIL − V0 LU = 3.348 − 0.9275
or NM L = 2.42 V
NM H = V0 HU − VIH = 9.272 − 5.07
or NM H = 4.20 V
16.47
(a)
Kn = KP
3
(VDD + VTP − VTN )
8
3
= 0.4 + ( 3.3 − 0.4 − 0.4 ) ⇒ VIL = 1.3375 V
8
1
VOHu = {2 (1.3375 ) + 3.3 − 0.4 + 0.4}
2
VOHu = 2.9875 V
VIL = VTN +
VIH = 0.4 +
5
( 3.3 − 0.4 − 0.4 ) ⇒ VIH = 1.9625 V
8
1
{2 (1.9625) − 3.3 − 0.4 + 0.4}
2
= 0.3125 V
VOLu =
VOLu
NM H = VOHu − VIH = 2.9875 − 1.9625 ⇒ NM H = 1.025 V
NM L = VIL − VOLu = 1.3375 − 0.3125 ⇒ NM L = 1.025 V
(b)
⎡
⎤
2 ( 4)
⎢
⎥
( 3.3 − 0.4 − 0.4 ) ⎢
2.5
12
2
VIL = 0.4 +
− 1⎥ = 0.4 +
⎡( −0.147 ) ⎦⎤
2
4
⎢
⎥
( ) +3
⎛ ( 2 )( 4 ) ⎞
( −0.333) ⎣
1
−
⎥
⎜
⎟ ⎢
12
⎦
⎝ 12
⎠ ⎣
VIL = 1.505 V
VOHu =
⎫⎪ 1
⎛ 2 ( 4) ⎞
1 ⎧⎪⎛ ( 2 )( 4 ) ⎞
⎟ (1.505 ) + 3.3 − ⎜
⎟ ( 0.4 ) + 0.4 ⎬ = {2.5083 + 3.3 − 0.2667 + 0.4}
⎨⎜1 +
2 ⎪⎩⎝
12 ⎠
⎪⎭ 2
⎝ 12 ⎠
VOHu = 2.9708 V
⎡
⎤
⎛ 2 ( 4) ⎞
⎢ 2⎜
⎥
⎟
( 3.3 − 0.4 − 0.4 ) ⎢ ⎝ 12 ⎠
2.5
− 1⎥ = 0.4 +
VIH = 0.4 +
[ −0.2302] ⇒ VIH = 2.1282 V
⎢
⎥
−0.333)
⎛ 2 ( 4) ⎞
(
2 )( 4 )
(
− 1⎟ ⎢ 3
+1 ⎥
⎜
⎥⎦
12
⎝ 12
⎠ ⎢⎣
⎛ 2 ( 4) ⎞
⎛ 2 ( 4) ⎞
( 2.1282 ) ⎜1 +
⎟ − 3.3 − ⎜
⎟ ( 0.4 ) + 0.4
12 ⎠
12 ⎠
3.547 − 3.3 − 0.2667 + 0.4
⎝
⎝
VOLu =
=
⇒ VOLu = 0.2853 V
1.333
⎛ 2 ( 4) ⎞
2⎜
⎟
⎝ 12 ⎠
NM H = VOHu − VIH = 2.9708 − 2.1282 ⇒ NM H = 0.8426 V
NM L = VIL − VOLu = 1.505 − 0.2853 ⇒ NM L = 1.22 V
16.48
a.
v A = vB = 5 V
N1 and N 2 on, so vDS1 ≈ vDS 2 ≈ 0 V
P1 and P2 off
So we have a P3 − N 3 CMOS inverter. By symmetry, vC = 2.5 V (Transition Point).
b.
For v A = vB = vC ≡ vI
Want K n ,eff = K p ,eff
kn′ ⎛ W ⎞
k ′ ⎛ 3W ⎞
⋅⎜ ⎟ = P ⋅⎜
⎟
2 ⎝ 3L ⎠ n 2 ⎝ L ⎠ P
With kn′ = 2k P′ , then
2 1 ⎛W ⎞
1 ⎛W ⎞
⋅ ⋅⎜ ⎟ = ⋅3⋅⎜ ⎟
2 3 ⎝ L ⎠n 2 ⎝ L ⎠ P
9 ⎛W ⎞
⎛W ⎞
Or ⎜ ⎟ = ⋅ ⎜ ⎟
L
⎝ ⎠n 2 ⎝ L ⎠ P
c.
We have
⎛ k ′ ⎞ ⎛ W ⎞ ⎛ 2k ′p ⎞ ⎛ 9 ⎞ ⎛ W ⎞
Kn = ⎜ n ⎟ ⎜ ⎟ = ⎜
⎟⎜ ⎟⎜ ⎟
⎝ 2 ⎠ ⎝ L ⎠n ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ L ⎠ p
⎛ k ′p ⎞ ⎛ W ⎞
Kp = ⎜ ⎟⎜ ⎟
⎝ 2 ⎠⎝ L ⎠p
Then from Equation (16.55)
Kn
⋅ ( 0.8 )
Kp
5 + ( −0.8 ) +
VIt =
1+
Kn
Kp
Now
Kn
⎛9⎞
= ( 2) ⎜ ⎟ = 9
Kp
⎝ 2⎠
Then
VIt =
5 + ( −0.8 ) + 3 ( 0.8 )
1+ 3
⇒ VIt = 1.65 V
16.49
By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V.
State
N1
N2
N3
N4
N5
v0
1
2
3
4
0
0
5
0
off
off
on
on
on
off
on
on
off
on
off
off
on
on
off
on
Logic function ( v X OR vY ) ⊗ ( v X AND vZ )
Exclusive OR of ( vX OR vY ) with ( vX AND vZ )
16.50
⎛W ⎞
NMOS in Parallel ⇒ ⎜ ⎟ = 2
⎝ L ⎠n
⎛W ⎞
4-PMOS in series ⇒ ⎜ ⎟ = 4 ( 4 ) = 16
⎝ L ⎠p
(b)
CL doubles ⇒ current must double to maintain switching speed.
⎛W ⎞
⇒⎜ ⎟ =4
⎝ L ⎠n
⎛W ⎞
⎜ ⎟ = 32
⎝ L ⎠p
16.51
⎛W ⎞
4-NMOS in series ⎜ ⎟ = 4 ( 2 ) = 8
⎝ L ⎠n
⎛W ⎞
4-PMOS in parallel ⎜ ⎟ = 4
⎝ L ⎠p
(b)
16.52
⎛W
⎜
⎝L
⎛W
⎜
⎝L
⎞
⎟ = 16
⎠n
⎞
⎟ =8
⎠p
on
off
on
on
⎛W ⎞
NMOS in parallel ⇒ ⎜ ⎟ = 2
⎝ L ⎠n
⎛W ⎞
3-PMOS in series ⇒ ⎜ ⎟ = 3 ( 4 ) = 12
⎝ L ⎠P
(a)
(b)
⎛W
⎜
⎝L
⎛W
⎜
⎝L
⎞
⎟ =4
⎠n
⎞
⎟ = 24
⎠p
16.53
⎛W ⎞
3-NMOS in series ⎜ ⎟ = 3 ( 2 ) = 6
⎝ L ⎠n
⎛W ⎞
3-PMOS in parallel ⎜ ⎟ = 4
⎝ L ⎠p
(a)
(b)
16.54
(a)
(b)
⎛W
⎜
⎝L
⎛W
⎜
⎝L
⎞
⎟ = 12
⎠n
⎞
⎟ =8
⎠p
Y = A( B + C )( D + E )
⎛W ⎞
For NMOS in pull down mode, 3 in series ⇒ ⎜ ⎟ = 3 ( 2 ) = 6
⎝ L ⎠n
For PMOS
⎛W ⎞
⎜ ⎟ =4
⎝ L ⎠P, A
(c)
⎛W ⎞
= 2 ( 4) = 8
⎜ ⎟
⎝ L ⎠ P , B ,C , D , E
16.55
(a)
(b)
Y = A( BD + CE )
(c)
NMOS: 3 transistors in series for pull down mode.
⎛W ⎞
For twice the speed: ⎜ ⎟ = 2 ( 3)( 2 ) = 12
⎝ L ⎠n
⎛W ⎞
PMOS: ⎜ ⎟ = 2 ( 4 ) = 8
⎝ L ⎠P, A
⎛W ⎞
= 2 ( 2 )( 4 ) = 16
⎜ ⎟
⎝ L ⎠ P , B ,C , D , E
16.56
(a)
(b)
Y = A + BC + DE
⎛W ⎞
⎛W ⎞
NMOS: ⎜ ⎟ = 2
=4
⎜ ⎟
⎝ L ⎠n, A
⎝ L ⎠ n , B ,C , D , E
PMOS: 3 transistors in series for the pull-up mode
(c)
⎛W ⎞
⎜ ⎟ = 3 ( 4 ) = 12
⎝ L ⎠p
16.57
(a)
(b)
Y = A + ( B + D)(C + E )
(c)
⎛W ⎞
NMOS: ⎜ ⎟ = 2 ( 2 ) = 4
⎝ L ⎠n, A
PMOS:
⎛W ⎞
⎜ ⎟ = ( 2 ) 3 ( 4 ) = 24
⎝ L ⎠p
16.58
(a) A classic design is shown:
⎛W ⎞
= ( 2 )( 4 ) = 8
⎜ ⎟
⎝ L ⎠ n , B ,C , D , E
A, B, C signals supplied through inverters.
⎛W ⎞
⎛W ⎞
(b)
For Inverters, ⎜ ⎟ = 1 and ⎜ ⎟ = 2
⎝ L ⎠p
⎝ L ⎠n
⎛W ⎞
⎛W ⎞
For PMOS in Logic function, let ⎜ ⎟ = 1 , then for NMOS in Logic function, ⎜ ⎟ = 2.25
⎝ L ⎠p
⎝ L ⎠n
16.59
(a)
A classic design is shown:
(b)
⎛W ⎞
⎛W ⎞
=2
⎜ ⎟ = 1, ⎜ ⎟
L
⎝ ⎠ ND
⎝ L ⎠ NA, NB , NC
⎛W ⎞
⎛W ⎞
= 8, ⎜ ⎟
=4
⎜ ⎟
⎝ L ⎠ PA, PB
⎝ L ⎠ PC , PD
16.60
( A OR B )
AND C
16.61
⎛W ⎞
5-NMOS in series ⇒ ⎜ ⎟ = 5 ( 2 ) = 10
⎝ L ⎠n
⎛W ⎞
5-PMOS in parallel ⇒ ⎜ ⎟ = 4
⎝ L ⎠p
16.62
By definition:
NMOS off if gate voltage = 0
NMOS on if gate voltage = 5 V
PMOS off if gate voltage = 5 V
PMOS on if gate voltage = 0
State
1
2
N1
off
on
P1
on
off
NA
off
on
NB
off
off
NC
off
off
v01
5
5
N2
on
on
P2
off
off
v02
0
0
3
4
5
6
off
on
off
on
on
off
on
off
off
off
off
off
off
off
off
on
off
on
off
on
5
5
5
0
on
on
on
off
off
off
off
on
Logic function is
v02 = ( v A OR vB ) AND vC
16.63
State
1
2
3
4
5
6
Logic function:
v03 = ( vX OR vZ ) AND vY
16.64
v01
v02
v03
5
0
5
5
5
0
5
0
5
0
5
5
0
5
0
5
0
0
0
0
0
5
16.65
16.66
16.67
2 I = −C
dVC
dt
So
1
( 2I ) ⋅ t
C
For ΔVC = −0.5 V
ΔVC = −
−0.5 = −
16.68
(a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
(iii)
16.69
(a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
(iii)
2 ( 2 x 10−12 ) ⋅ t
25 x 10−15
⇒ t = 3.125 ms
vO = 0
vO = 4.2 V
vO = 2.5 V
vO = 0
vO = 3.2 V
vO = 2.5 V
vo = 0
vo = 2.9 V
vo = 2.4 V
vo = 0
vo = 2.0 V
vo = 2.0 V
16.70
Neglect the body effect.
v01 (logic 1) = 4.2 V , v02 (logic 1) = 5 V
a.
vI = 5 V ⇒ vGS 1 = 4.2 V
b.
M 1 in nonsaturation and M 2 in saturation. From Equation (16.23)
⎛W
⎜
⎝L
⎛W
⎜
⎝L
Or
⎛W
⎜
⎝L
2
⎞
⎛W ⎞
2
⎟ ⎡⎣ 2 ( vGS 1 − VTND ) vO1 − vO1 ⎤⎦ = ⎜ ⎟ (VDD − vO1 − VTNL )
L
⎠D
⎝ ⎠L
2
2
⎞ ⎡
⎟ ⎣ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤⎦ = (1) [5 − 0.1 − 0.8]
⎠D
⎞
⎛W ⎞
⎟ ( 0.67 ) = 16.81 ⇒ ⎜ ⎟ = 25.1
⎠D
⎝ L ⎠D
Now
v01 = 4.2 V ⇒ vGS 3 = 4.2 V
M 3 in nonsaturation and M 4 in saturation. From Equation (16.29(b)).
⎛W
⎜
⎝L
⎛W
⎜
⎝L
2
⎞
⎛W ⎞
2
⎟ ⎣⎡ 2 ( vGS 3 − VTND ) vO 2 − vO 2 ⎦⎤ = ⎜ ⎟ [ −VTNL ]
⎠D
⎝ L ⎠L
2
2
⎞ ⎡
⎟ ⎣ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤⎦ = ( 2 ) ⎡⎣ − ( −1.5 ) ⎤⎦
⎠D
⎛W ⎞
⎜ ⎟ (0.67) = 2.25
⎝ L ⎠D
⎛W ⎞
Or ⎜ ⎟ = 3.36
⎝ L ⎠D
16.71
A
B
Y
0
0
1
0
1
0
1
0
0
1
1
⇒ indeterminate
0.1
Without the top transistor, the circuit performs the exclusive-NOR function.
16.72
B
A
A
0
1
0
0
1
1
1
0
0
1
0
1
Y = A + AB = A + B
Z = Y or Z = AB
B
1
0
1
0
Y
0
1
1
1
Z
1
0
0
0
16.73
16.74
For φ = 1, φ = 0, then Y = B. And for φ = 0, φ = 1, then Y = A .
A multiplexer.
16.75
Y = AC + BC
16.76
Y = AB + AB = A ⊗ B
16.77
A
0
1
0
1
B
0
0
1
1
Y
0
1
1
0
Exclusive-OR function.
16.78
This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver
and load transistors must be maintained as discussed previously with the enhancement load inverter.
When φ1 is high, v01 becomes the complement of vI . When φ2 goes high, then v0 becomes the
complement of v01 or is the same as vI . The circuit is a shift register.
16.79
Let Q = 0 and Q = 1 ; as S increases, Q decreases. When Q reaches the transition point of the M 5 − M 6
inverter, the flip-flop with change state. From Equation (16.28(b)),
KL
VIt =
⋅ ( −VTNL ) + VTND
KD
where K L = K 6 and K D = K 5 .
Then
30
VIt =
⋅ ⎡ − ( −2 ) ⎤⎦ + 1 ⇒ VIt = Q = 2.095 V
100 ⎣
This is the region where both M 1 and M 3 are biased in the saturation region. Then
S=
K3
⋅ ( −VTNL ) + VTND =
K1
30
⋅ ⎡ − ( −2 ) ⎤⎦ + 1
200 ⎣
or S = 1.77 V
This analysis neglects the effect of M 2 starting to turn on at the same time.
16.80
Let vY = R, v X = S , v02 = Q, and v01 = Q. Assume VThN = 0.5 V and VThP = −0.5 V. For S = 0, we have
the following:
If we want the switching to occur for R = 2.5 V, then because of the nonsymmetry between the two
circuits, we cannot have Q and Q both equal to 2.5 V.
Set R = Q = 2.5 V and assume Q goes low.
For the M 1 − M 5 inverter, M 1 in nonsaturation and M 5 in saturation. Then
2
2
K n ⎡⎢ 2 ( 2.5 − 0.5 ) Q − Q ⎤⎥ = K p [ 2.5 − 0.5]
⎣
⎦
Or
2
⎛ Kp ⎞
4Q − Q = 4 ⎜
⎟
⎝ Kn ⎠
For the other circuit, M 2 − M 4 in saturation and M 6 in nonsaturation. Then
2
K n ( 2.5 − 0.5 ) + K n (Q − 0.5) 2 = K p ⎡ 2 5 − Q − 0.5 ( 2.5 ) − ( 2.52 ) ⎤
⎣
⎦
(
)
Combining these equations and neglecting the Q 3 term, we find
Kp
Q = 1.4 V and
kn
= 0.9
16.81
3.3 + ( −0.4 ) + 0.5
vIt =
= 1.7 V
1+1
vI = 1.5 V NMOS Sat; PMOS Non Sat
2
= ⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − vo1 ) − ( 3.3 − vo1 ) ⎤ ⇒ vo1 = 2.88 V
⎣
⎦
vI = 1.6 V vo1 = 2.693 V
vI = 1.7 V vo1 = variable (switching region)
( vI − 0.5)
2
vI = 1.8 V
NMOS Non Sat; PMOS Sat
( 3.3 − VI − 0.4 )
2
= ⎡⎣ 2 ( vI − 0.5 ) vo1 − vo21 ⎤⎦ ⇒ vo1 = 0.607 V
Now
vI = 1.5 V, vo1 = 2.88 V ⇒ vo ≈ 0V
vI = 1.6 V, vo1 = 2.693 V
NMOS Non Sat; PMOS Sat
2
( 3.3 − vo1 − 0.4 ) = ⎡⎣ 2 ( vo1 − 0.5 ) vo − vo2 ⎤⎦
vo = 0.00979 V
vI = 1.7 V, v o1 = Switching Mode ⇒ v0 = Switching Mode.
vI = 1.8 V, vo1 = 0.607 V NMOS Sat; PMOS Non Sat
( v01 − 0.5)
2
2
= ⎡ 2 ( 3.3 − v01 − 0.4 )( 3.3 − v0 ) − ( 3.3 − v0 ) ⎤ ⇒ v0 = 3.298 V
⎣
⎦
16.82
For R = φ = VDD and S = 0 ⇒ Q = 0, Q = 1
For S = φ = VDD and R = 0 ⇒ Q = 1, Q = 1
The signal φ is a clock signal.
For φ = 0, The output signals will remain in their previous state.
16.83
a.
Positive edge triggered flip-flop when CLK = 1, output of first inverter is D and then Q = D = D .
b.
For example, put a CMOS transmission gate between the output and the gate of M 1 driven by a
CLK pulse.
16.84
For J = 1, K = 0, and CLK = 1; this makes Q = 1 and Q = 0 .
For J = 0, K = 1, and CLK = 1 , and if Q = 1, then the circuit is driven so that Q = 0 and Q = 1.
If initially, Q = 0, then the circuit is driven so that there is no change and Q = 0 and Q = 1.
J = 1, K = 1, and CLK = 1, and if Q = 1, then the circuit is driven so that Q = 0.
If initially, Q = 0 , then the circuit is driven so that Q = 1.
So if J = K = 1, the output changes state.
16.85
For J = v X = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0.
For J = v X = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1.
Now consider J = K = CLK = 1. With v X = vZ = 1, the output is always v0 = 0, So the output does not
change state when J = K = CLK = 1. This is not actually a J − K flip-flop.
16.86
64 K ⇒ 65,536 transistors arranged in a 256 × 256 array.
(a)
Each column and row decoder required 8 inputs.
(b)
(i)
Address = 01011110 so input = a7 a6 a5 a4 a3 a2 a1a0
(ii)
Address = 11101111 so input = a7 a6 a5 a4 a3 a2 a1a0
(c)
(i)
(ii)
Address = 00100111 so input = a7 a6 a5 a4 a3 a2 a1a0
Address = 01111011 so input = a7 a6 a5 a4 a3 a2 a1a0
16.87
(a)
1-Megabit memory ⇒
= 1, 048,576 ⇒ 1024 × 1024
Nuclear & input row and column decodes lines necessary = 10
(b)
250K × 4 bits ⇒ 262,144 × 4 bits ⇒ 512 × 512
For 512 lines ⇒ 9 row and column decoder lines necessary.
16.88
Put 128 words in a 8 × 16 array, which means 8 row (or column) address lines and 16 column (or row)
address lines.
16.89
Assume the address line is initially uncharged, then
dV
1
I
I = C C or VC = ∫ Idt = ⋅ t
C
C
dt
−12
VC ⋅ C ( 2.7 ) ( 5.8 × 10 )
⇒
=
I
250 × 10−6
t = 6.26 × 10 −8 s ⇒ 62.6 ns
Then t =
16.90
(a)
⎛W
or ⎜
⎝L
(b)
5 − 0.1 ⎛ 35 ⎞⎛ W
= ⎜ ⎟⎜
1
⎝ 2 ⎠⎝ L
2
⎞
⎟ ⎣⎡ 2 ( 5 − 0.7 )( 0.1) − ( 0.1) ⎦⎤
⎠
⎞
⎟ = 0.329
⎠
16 K ⇒ 16,384 cells
2
= 2 µA
1
Power per cell = (2 µ A)(2 V ) = 4 µW
Total Power = PT = (4 µW )(16,384) ⇒ PT = 65.5 mW
iD ≅
Standby current = (2 µ A)(16,384) ⇒ IT = 32.8 mA
16.91
16 K ⇒ 16,384 cells
PT = 200 mW ⇒ Power per cell =
200
⇒ 12.2 µW
16,384
V
P
12.2
2.5
=
= 4.88 µ A ≅ DD =
⇒ R = 0.512 M Ω
iD =
VDD
R
R
2.5
If we want vO = 0.1 V for a logic 0, then
⎛ k′ ⎞⎛ W ⎞
iD = ⎜ n ⎟ ⎜ ⎟ ⎣⎡ 2 (VDD − VTN ) vO − vO2 ⎦⎤
⎝ 2 ⎠⎝ L ⎠
2
⎛ 35 ⎞ ⎛ W ⎞
4.88 = ⎜ ⎟ ⎜ ⎟ ⎡ 2 ( 2.5 − 0.7 )( 0.1) − ( 0.1) ⎤
⎣
⎦
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
So ⎜ ⎟ = 0.797
⎝L⎠
16.92
Q = 0, Q = 1
So D = Logic 1 = 5 V
A very short time after the row has been addressed, D remains charged at VDD = 5 V . Then M p 3, M A, and
M N 1 begin to conduct and D decreases. In steady-state, all three transistors are biased in the nonsaturation
region. Then
2
2
2
⎤
⎡
⎤
⎡
⎤
K p 3 ⎡⎣ 2 (VSG 3 + VTP 3 ) VSD 3 − VSD
3 ⎦ = K nA ⎣ 2 (VGSA − VTNA ) VDSA − VDSA ⎦ = K n1 ⎣ 2 (VGS 1 − VTN 1 ) VDS 1 − VDS 1 ⎦
Or
2
2
K p 3 ⎡ 2 (VDD + VTP 3 )(VDD − D ) − (VDD − D ) ⎤ = K nA ⎡ 2 (VDD − Q − VTNA )( D − Q ) − ( D − Q ) ⎤
⎣
⎦
⎣
⎦
= K n1 ⎡⎣ 2 (VDD − VTN 1 ) Q − Q 2 ⎤⎦ (1)
Equating the first and third terms:
2
⎛ 20 ⎞ ⎡
⎛ 40 ⎞
2
⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎤⎦ = ⎜ ⎟ ( 2 ) ⎡⎣ 2 ( 5 − 0.8 ) Q − Q ⎤⎦ (2)
⎝ 2 ⎠
⎝ 2 ⎠
2
As a first approximation, neglect the ( 5 − D ) and Q 2 terms. We find
Q = 1.25 − 0.25 D
(3)
Then, equating the first and second terms of Equation (1):
2
2
⎛ 20 ⎞ ⎡
⎛ 40 ⎞
⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎤⎦ = ⎜ ⎟ (1) ⎡⎣ 2 ( 5 − Q − 0.8 )( D − Q ) − ( D − Q ) ⎤⎦
2
2
⎝ ⎠
⎝ ⎠
Substituting Equation (3), we find as a first approximation: D = 2.14 V
Substituting this value of D into equation (2), we find
2
8.4 ( 5 − 2.14 ) − ( 5 − 2.14 ) = 4 ⎡⎣8.4Q − Q 2 ⎤⎦
We find Q = 0.50 V
Using this value of Q, we can find a second approximation for D by equating the second and third terms of
equation (1). We have
2
20 ⎡ 2 ( 4.2 − Q )( D − Q ) − ( D − Q ) ⎤ = 40 ⎡⎣ 2 ( 4.2Q ) − Q 2 ⎤⎦
⎣
⎦
Using Q = 0.50 V , we find D = 1.79 V
16.93
Initially M N 1 and M A turn on.
M N 1, Nonsat; M A , sat.
K nA [VDD − Q − VTN ] = K n1 ⎡⎣ 2 (VDD − VTN 1 ) Q − Q 2 ⎤⎦
2
2
⎛ 40 ⎞
⎛ 40 ⎞
2
⎜ ⎟ (1) [5 − Q − 0.8] = ⎜ ⎟ ( 2 ) ⎡⎣ 2 ( 5 − 0.8 ) Q − Q ⎤⎦
2
2
⎝ ⎠
⎝ ⎠
which yields
Q = 0.771 V
Initially M P 2 and M B turn on
Both biased in nonsaturation reagion
2
2
K P 2 ⎡ 2 (VDD + VTP 3 ) VDD − Q − VDD − Q ⎤ = K nB ⎡⎢ 2 (VDD − VTNB ) Q − Q ⎤⎥
⎥⎦
⎣
⎦
⎣⎢
2
2
⎛ 20 ⎞
⎡
⎤ ⎛ 40 ⎞ ⎡
⎤
⎜ ⎟ ( 4 ) ⎣⎢ 2 ( 5 − 0.8 ) 5 − Q − 5 − Q ⎦⎥ = ⎜ ⎟ (1) ⎢⎣ 2 ( 5 − 0.8 ) Q − Q ⎥⎦
⎝ 2 ⎠
⎝ 2 ⎠
(
) (
(
) (
)
)
which yields Q = 3.78 V
Note: (W / L) ratios do not satisfy Equation (16.86)
16.94
For Logic 1, v1:
( 5 )( 0.05 ) + ( 4 )(1) = (1 + 0.05 ) v1 ⇒ v1 = 4.0476 V
v2 :
(5)(0.025) + (4)(1) = (1 + 1.025)v2 ⇒ v2 = 4.0244 V
For Logic 0, v1:
(0)(0.05) + (4)(1) = (1 + 0.05)v1 ⇒ v1 = 3.8095 V
v2 :
(0)(0.025) + (4)(1) = (1 + 0.025)v2 ⇒ v2 = 3.9024 V
16.95
Not given
16.96
Not given
16.97
Not given
16.98
Quantization error =
(a)
1
LSB ≤ 1% ≤ 0.05 V
2
Or LSB ≤ 0.10 V
For a 6-bit word , LSB =
1 − LSB =
(b)
5
= 0.078125 V
64
5
= 0.078125 V
64
3.5424
× 64 = 45.34 ⇒ n = 45
5
Digital Output = 101101
45 × 5
= 3.515625
64
.
1
Δ = 3.5424 − 3.515625 = 0.026775 < LSB.
2
(c)
16.99
Quantization error =
(a)
1
LSB ≤ 0.5% ≤ 0.05 V
2
1 − LSB = 0.10 V
10
= 0.078125 V
128
1 − LSB = 0.078125 V
For a 7-beit word, LSB =
(b)
3.5424
×128 = 45.34272 ⇒ n = 45
10
Digital output = 0101101
(c)
Now
45 × 10
= 3.515625
128
Δ = 3.5424 − 3.515625 = 0.026775 V <
16.100
(a)
(b)
16.101
⎛0 1 0 1 ⎞
vo = ⎜ + + + ⎟ ( 5 )
⎝ 2 4 8 16 ⎠
vo = 1.5625 V
⎛1 0 1 0 ⎞
vo = ⎜ + + + ⎟ ( 5 )
⎝ 2 4 8 16 ⎠
vo = 3.125 V
⎛1⎞
LSB = ⎜ ⎟ ( 5 ) = 0.3125 V
⎝ 16 ⎠
1
LSB = 0.15625 V
2
⎛ 10 ⎞
Now vo = ⎜
⎟ (5)
⎝ 20 + ΔR1 ⎠
(a)
1
LSB
2
For vo = 2.5 + 0.15625 = 2.65625 V
⇒ ΔR1 = −1.176 K
2.65625
For vo = 2.5 − 0.15625 = 2.34375 V
20 + ΔR1 =
(10 )( 5)
20 + ΔR1 =
(10 )( 5 )
2.34375 V
ΔR1 = +1.333 K
For ΔR1 = 1.176 K ⇒ ΔR1 = 5.88%
⎛
⎞
10
For R4 : vo = ⎜
⎟ ( 5)
⎝ 160 + ΔR4 ⎠
vo = 0.3125 + 0.15625 = 0.46875 V
(b)
⇒ ΔR4 = −53.33K
0.46875
Or vo = 0.3125 − 0.15625 = 0.15625 V
160 + ΔR4 =
(10 )( 5 )
⇒ ΔR4 = 160 K
0.15625
For ΔR4 = 53.33K ⇒ ΔR4 = 33.33%
160 + ΔR4 =
16.102
(a)
(10 )( 5 )
R5 = 320 kΩ
R6 = 640 kΩ
R7 = 1280 kΩ
R8 = 2560 kΩ
(b)
⎛ 10 ⎞
vo = ⎜
⎟ ( 5 ) = 0.01953125 V
⎝ 2560 ⎠
16.103
(a)
V
−5
⇒ I1 = −0.50 mA
I1 = REF =
2 R 10
I
I 2 = 1 = −0.25 mA
2
I2
I 3 = = −0.125 mA
2
I3
I 4 = = −0.0625 mA
2
I4
I 5 = = −0.03125 mA
2
I
I 6 = 5 = −0.015625 mA
2
Δvo = I 6 RF = ( 0.015625 )( 5 )
(b)
Δvo = 0.078125 V
(c)
vo = − [ I 2 + I 5 + I 6 ] RF = [ 0.25 + 0.03125 + 0.015625] ( 5 )
vo = 1.484375 V
(d)
For 101010; vo = ( 0.50 + 0.125 + 0.03125 )( 5 ) = 3.28125 V
For 010101; vo = ( 0.25 + 0.0625 + 0.015625 )( 5 ) = 1.640625 V
Δvo = 1.640625 V
16.104
1
5
⎛V ⎞⎛ R ⎞ V
= 0.3125 V
LSB = ⎜ REF ⎟ ⎜ ⎟ = REF =
2
8
2
16
16
R
⎝
⎠⎝ ⎠
Ideal
3
⎛ 3V ⎞
v A for 011 ⇒ ⎜ REF ⎟ ( R ) = VREF = 1.875 V
8
8
R
⎝
⎠
1
Range of v A = 1.875 ± LSB
2
or 1.5625 ≤ vA ≤ 2.1875 V
16.105
6-bits ⇒ 26 = 64 resistors
26 − 1 = 63 comparators
16.106
(a)
10- bit output ⇒ 1024 clock periods
1
1
1 clock period = = 6 = 1 μS
f 10
May conversion time = 1024 μS = 1.024 mS
(b)
1
1⎛ 5 ⎞
LSB = ⎜
⎟ = 0.002441406 V
2
2 ⎝ 1024 ⎠
⎛ 5 ⎞
v′A = (128 + 16 + 2 ) ⎜
⎟ = 0.712890625 V
⎝ 1024 ⎠
1
So range of v A = v′A ± LSB
2
0.710449219 ≤ v A ≤ 0.715332031 V
(c)
0100100100 ⇒ 256 + 32 + 4 = 292 clock pulses
16.107
N ×5
⇒ N = 640 ⇒ 512 + 128
1024
Output = 1010000000
(b)
N ×5
⇒ N = 381.19 ⇒ N = 381 ⇒ 256 + 64 + 32 + 16 + 8 + 4 + 1
1.8613 =
1024
Output = 0101111101
(a)
3.125 =
Chapter 17
Exercise Solutions
EX17.1
a.
−0.7 − (−5)
iE =
= 1 mA ⇒ RE = 4.3 kΩ
RE
iC1 = iC 2 = 0.5 mA =
5 − 3.5
⇒ RC1 = RC 2 = 3 kΩ
RC1
b.
i.
v1 = 1 V
(1 − 0.7) − (−5)
⇒ iE = 1.23 mA
iE =
4.3
iC1 = iE = v01 = 5 − (1.23)(3) ⇒ v01 = 1.31 V
v02 = 5 V
ii.
v1 = −1 V
iE = 1 mA ⇒ v02 = 5 − (1)(3) ⇒ v02 = 2 V
v01 = 5 V
EX17.2
P (iCXY + iCR + i3 + i4 )(5.2)
v X = vY = logic 1 ⇒ iCXY = 3.22 mA
a.
iCR = 0
−0.7 + 5.2
i3 =
= 3 mA
1.5
−1.4 + 5.2
i4 =
= 2.53 mA
1.5
P = (3.22 + 0 + 3 + 2.53)(5.2) ⇒ P = 45.5 mW
v X = vY = logic 0 ⇒ iCXY = 0
b.
iCR = 2.92 mA
i3 = 2.53 mA
i4 = 3 mA
P = (0 + 2.92 + 2.53 + 3)(5.2) ⇒ P = 43.9 mW
EX17.3
3.5 − (2.45 + 0.7)
= 1.4 K
0.25
3.5 − 2(0.7)
R1 + R2 =
= 8.4 k ⇒ R2 = 7 K
0.25
2.45
R5 =
= 9.8 K
0.25
R1 =
EX17.4
I MAX = 1 =
−0.7 − (−5.2)
⇒ R3 = R4 = 4.5 K
R3
−0.7 − 0.7 − (−5.2)
= 3.22 mA
1.18
i5 = i1 = 1.40 mA
i3 = 1.0 mA
−1.4 − (−5.2)
i4 =
= 0.844 mA
4.5
P = (3.22 + 1.4 + 1.4 + 1.0 + 0.844)(5.2) = 40.8 mW
iCXY =
EX17.5
⎛ v − 0.7 − (−5.2) ⎞
iL = ⎜ oR
⎟ (10)
(1.18)(51)
⎝
⎠
voR − (−5.2)
i3 =
1.5
v + 5.2 ⎛ voR + 4.5 ⎞
⎡ 0 − (VoR + 0.7) ⎤
+⎜
(51) = oR
⎟ (10)
⎢
⎥
0.24
1.5
⎣
⎦
⎝ (1.18)(51) ⎠
⎡⎛ 51 ⎞ 1
1 − (6) ⎤ (0.7)(51) 5.2
4.5(10)
−voR ⎢⎜
+
=
+
+
⎟+
⎥
0.24
1.5 (1.18)(51)
⎣⎝ 0.24 ⎠ 1.5 (1.18)(51) ⎦
−voR [212.50 + 0.6667 + 0.166168] = 148.75 + 3.4666 + 0.747757
−voR [213.3328] = 152.9644
voR = −0.7170
EX17.6
(100)(0.026)
= 2.6 K
rπ 3 =
1
1
= 38.46 mA / V
g m3 =
0.026
Vn
Vn
=
Ib3 =
RC 2 + rπ 3 + (1 + β ) R3 0.24 + 2.6 + (101)(4.5)
Ib3 =
Vn
457.34
⎛ Vn ⎞
VO = − I b 3 ( RC 2 + rπ 3 ) = − ⎜
⎟ (0.24 + 2.6)
⎝ 457.34 ⎠
VO = −0.00621 Vn
⎛ Vn ⎞
VO′ = (1 + β ) I b 3 R3 = (101) ⎜
⎟ (4.5)
⎝ 457.34 ⎠
VO′ = 0.9938 Vn
EX17.7
P = I Q ⋅ VCC ⇒ 0.2 = I Q (1.7) ⇒ I Q = 117.6 µ A
QR on ⇒ v0 = 1.7 − I Q RC = 1.7 − 0.4 ⇒ RC =
VR =
1.7 + 1.3
⇒ VR = 1.5 V
2
0.4
⇒ RC = 3.40 kΩ
0.1176
EX17.8
(a)
v X = vY = 5 V
v1 = VBE ( sat ) + 2VY = 0.8 + 2(0.7) = 2.2 V
5 − 2.2
i1 =
= 0.70 mA
4
V − VCE ( sat ) 5 − 0.1
iRC = CC
=
= 1.225 mA
RC
4
P = (i1 + iRC )VCC = (0.70 + 1.225)(5)
or P = 9.625 mW
v X = vY = 0 ⇒ v1 = 0.70 V
(b)
V − v 5 − 0.70
= 1.075 mA
i1 = CC 1 =
4
R1
P = i1 ⋅ VCC = (1.075)(5) ⇒ P = 5.375 mW
EX17.9
vB1 = 0.1 + 0.8 = 0.9 V
(a)
5 − 0.9
= 0.683 mA
i1 =
6
All other currents ≈ 0
vB1 = 0.8 + 0.8 + 0.7 = 2.3 V
(b)
5 − 2.3
= 0.45 mA
i1 =
6
iB 2 = (1 + 0.2)(0.45) = 0.54 mA
5 − (0.8 + 0.1)
i2 =
= 2.733 mA
1.5
iE 2 = 2.733 + 0.54 = 3.273 mA
0.8
i4 =
= 0.533 mA
1.5
iB 0 = 3.273 − 0.533 = 2.74 mA
5 − 0.1
i3 =
= 2.23 mA
2.2
i
2.733
For Q2 : 2 =
= 5.06 < β F
0.54
iB 2
For Q0 :
i3
2.23
=
= 0.81 < β F
iB 0 2.74
EX17.10
v X = vY = 3.6 V, Output low:
5 − 2.3
= 0.45 mA
6
= (1 + 2 β R )i1
i1 =
iB 2
iB 2 = 0.54 mA
vB 3 = 0.8 + 0.1 = 0.9
5 − 0.9
= 2.05 mA
2
0.8
⇒ iB 0 = 2.06 mA
iB 0 = 2.05 + 0.54 −
1.5
5 − 0.9
iL′ =
= 0.683 mA
6
β F ⋅ iB 0 = N ⋅ iL′ ⇒ (20)(2.06) = N (0.683)
i2 =
⇒ N = 60
EX17.11
a.
iRC =
5 − 0.4
= 2.04 mA
2.25
2 + 2.04
⇒ iC′ = 3.67 mA
1
1+
10
3.67
iB′ =
= 0.367 mA
10
iD = 2 − 0.367 ⇒ iD = 1.63 mA
iC′ =
b.
iD = 0 ⇒ iB′ = iB = 2 mA
iC′ = β iB′ = (10)(2) = 20 mA = iRC + iL
iL = 20 − 2.04 ⇒ iL (max) ≈ 18 mA
EX17.12
(a)
v1 = 0.4 + 0.3 = 0.7, i1 =
5 − 0.7
= 0.1075 mA
40
All transistor currents are zero.
P = (0.1075)(5 − 0.4) ⇒ 495 µ W
5 − 1.4
= 0.090 mA
40
5 − 1.1
vC 2 = 0.7 + 0.4 = 1.1 V, i2 = iC 2 =
= 0.325 mA
12
iB 0 ≈ iB 2 + iC 2 = 0.09 + 0.325 = 0.415 mA
(b)
v1 = 1.4 V, i1 = iB 2 =
iC 0 ≈ 0
P = (i1 + i2 )(5) = (0.09 + 0.325)(5) = 2.08 mW
TYU17.1
a.
QR on
iE =
1.5 − 0.7 − ( −3.5)
= 2 mA ⇒ RE = 2.15 kΩ
RE
iCR ≈ iE = 2 mA =
b.
iE =
3.5 − 2
⇒ RC 2 = 0.75 kΩ
RC 2
v X = vY = 2 V ⇒ Q1 and Q2 on
2 − 0.7 − (−3.5) 4.8
=
⇒ iE = 2.233 mA
RE
2.15
RC1 =
3.5 − 2
1.5
=
⇒ RC1 = 0.672 kΩ
2.233
iCXY
TYU17.2
logic 1 = −0.7 V
Q1 and Q2 on when v X = vY = −0.7 V
−0.7 − 0.7 − (−5.2)
= 2.5 ⇒ RE = 1.52 kΩ
RE
iE =
vNOR = −1.5 ⇒ RC1 =
0 − (−1.5 + 0.7)
⇒ RC1 = 320 Ω
2.5
−1.5 − 0.7
⇒ VR = −1.1 V
2
−1.1 − 0.7 − (−5.2)
= 2.237 mA
QR on ⇒ iE =
1.52
0 − (−1.5 + 0.7)
⇒ RC 2 = 358 Ω
RC 2 =
2.237
−0.7 − (−5.2)
⇒ R3 = R4 = 1.8 kΩ
R3 = R4 =
2.5
VR =
TYU17.3
N M H = −0.70 − (−0.93) ⇒ N M H = 0.23 V
N M L = −1.17 − ( −1.40) ⇒ N M L = 0.23 V
TYU17.4
State
1
2
3
4
5
6
7
8
A
0
1
0
0
1
1
0
1
B
0
0
1
0
1
0
1
1
C
0
0
0
1
0
1
1
1
( A AND C ) OR ( B AND C )
true for
true for
states 6 and 8 states 3 and 5
Output goes high for these 4 states
TYU17.5
A
0
B
0
C
0
v0
0
Q01
off
“on”
off
off
on
on
off
on
Q02
off
off
on
off
on
off
on
on
Q03
off
off
off
on
off
on
on
on
Q1
off
off
off
on
off
on
on
on
Q2
on
on
on
off
on
off
off
off
QR
on
on
“off”
on
off
“off”
on
off
v0
0
0
1
0
1
1
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
1
0
1
1
1
1
1
1
0
0
0
1
⇒ ( A ⊕ B) ⊕ C
TYU17.6
v X = vY = 0.1 ⇒ v1 = 0.8
a.
5 − 0.8
⇒ i1 = 0.7 mA
i1 =
6
i2 = iR = iB = iRC = 0
v0 = 5 V
b.
c.
Same as part (a).
v X = vY = 5 V ⇒ v1 = 0.8 + 0.7 + 0.7 = 2.2 V
i1 = i2 =
5 − 2.2
⇒ i1 = i2 = 0.467 mA
6
0.8
⇒ iR = 0.053 mA
15
iB = i2 − iR ⇒ iB = 0.414 mA
iR =
iRC =
5 − 0.1
⇒ iRC = 0.98 mA
5
vC = 0.1 V
TYU17.7
(a)
From Tyu17.6, iB = 0.414 mA, iRC = 0.98 mA
v1′ = 0.1 + 0.7 = 0.8, iL′ =
5 − 0.8
= 0.70 mA
6
iC ,max = β iB = iRC + NiL′
(25)(0.414) = 0.98 + N (0.70)
N = 13.4 → N = 13
(b)
iC , max = β iB = (25)(0.414) = 10.35 mA < 15 mA
So from part (a), N = 13
TYU17.8
From EX17.9, iB 0 = 2.74 mA, i3 = 2.23 mA
vo = 0.1 V, so vB′ 1 = 0.1 + 0.8 = 0.9 V
5 − 0.9
= 0.6833 mA
6
iC 0, max = β iB 0 = i3 + Ni1′
(20)(2.74) = 2.23 + N (0.6833)
N = 76.9 ⇒ N = 76
i1′ =
TYU17.9
Q1 in saturation
5 − 0.9
⇒ iB1 = 0.683 mA
iB1 =
6
iC1 = iB 2 = iC 2 = 0
iB 0 = iC 0 = 0
vB 4 = 0.1 + 0.7 = 0.8 V
0.1
iB 4 =
⇒ iB 4 = 1.19 µ A
(21)(4)
iC 4 = 23.8 µ A
iB 3 = iC 3 = 0
TYU17.10
(a)
v X = vY = 0.4, vB1 = 0.4 + 0.7 = 1.1 V
5 − 1.1
= 1.393 mA
2.8
P = i1 (5 − 0.4) = (1.393)(5 − 0.4) = 6.41 mW
i1 =
(b)
v X = vY = 3.6 V
5 − 2.1
= 1.036 mA
2.8
5 − 1.1
= 5.132 mA
vC 2 = 0.7 + 0.4 = 1.1 V, i2 =
0.76
0.4
vE 4 = 1.1 − 0.7 = 0.4 V, iR 4 =
= 0.1143 mA
3.5
⎛ β ⎞
⎛ 25 ⎞
iR 3 = ⎜
⎟ iR 4 = ⎜ ⎟ (0.1143) = 0.1099 ≈ 0.11 mA
⎝ 26 ⎠
⎝1+ β ⎠
vB1 = 2.1, i1 =
P = (i1 + i2 + iR 3 )(5) = (1.036 + 5.132 + 0.11)(5)
or
P = 31.4 mW
TYU17.11
(a)
v X = 0.4 V, vC1 = 0.4 + 0.7 = 1.1 V
5 − 1.1
⇒ 97.5 µ A
40
(b)
v X = 3.6 V, vC1 = 3(0.7) = 2.1 V
i1 =
i1 =
5 − 2.1
⇒ 72.5 µ A
40
Chapter 17
Problem Solutions
17.1
a.
vI = −1.5 Vi Q1 off , Q2 on
−0.7 − (−3.5)
⇒ iE = 0.56 mA
iE =
5
iC1 = 0 ⇒ v01 = 3.5 V
iC 2 = iE ⇒ v02 = 3.5 − iE RC 2 = 3.5 − ( 0.56 )( 2 )
or
v02 = 2.38 V
b.
vI = 1.0 Vi Q1 on, Q2 off
iE =
iC 2
(1 − 0.7 ) − ( −3.5)
5
= 0 ⇒ v02 = 3.5 V
⇒ iE = 0.76 mA
c.
logic 0 at v02 (low level) = 2.38 V
Then
v01 = 2.38 = 3.5 − (0.76) RC1
or
RC1 = 1.47 kΩ
17.2
(a)
iC 2 = I Q = 0.5 =
(b)
iC1 = I Q = 0.5 =
(c)
3−0
⇒ RC 2 = 6 K
RC 2
3 −1
⇒ RC1 = 4 K
RC1
⎛V ⎞
I S exp ⎜ BE1 ⎟
iC1
⎝ VT ⎠
=
IQ
⎡
⎛V ⎞
⎛ V ⎞⎤
I S ⎢exp ⎜ BE1 ⎟ + exp ⎜ BE 2 ⎟ ⎥
V
⎝ T ⎠
⎝ VT ⎠ ⎦
⎣
1
⎛ VBE 2 − VBE1 ⎞
1 + exp ⎜
⎟
VT
⎝
⎠
vI = VBE1 − VBE 2
=
So
iC1
=
IQ
1
⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠
0.1
1
=
= 0.2
0.5
⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠
⎛ −v ⎞ 1
exp ⎜ I ⎟ =
−1 = 4
⎝ VT ⎠ 0.2
( −vI ) = ( 0.026 ) ln (4)
vI = −0.0360 V
17.3
(a)
vI = 0.5 V, Q1 on, Q2 off = v02 = 3 V
v01 = 3 − (1)(0.5) = 2.5 V
(b)
vI = −0.5 V Q1 off, Q2 on ⇒ v01 = 3 V
v02 = 3 − (1)(0.5) = 2.5 V
17.4
(a)
Q2 on, vE = −1.2 − 0.7 = −1.9V
−1.9 − ( −5.2 )
= 1.32 mA
2.5
v2 = −1V = −iC 2 RC 2 = − (1.32 )( RC 2 )
RC 2 = 0.758 k Ω
iE = iC 2 =
(b)
Q1 on, vE = −0.7 − 0.7 = −1.40 V
−1.4 − ( −5.2 )
= 1.52 mA
2.5
v1 = −1V = −iC1 RC1 = − (1.52 )( RC1 )
RC1 = 0.658 k Ω
iE = iC1 =
(c)
For vin = −0.7 V , Q1 on, Q2 off
⇒ vO1 = −0.70V
vO 2 = −1 − 0.7 ⇒ vO 2 = −1.7 V
For vin = −1.7 V , Q1 off , Q2 on
⇒ vO 2 = −0.7 V
vO1 = −1 − 0.7 ⇒ vO1 = −1.7 V
(d) (i) For vin = −0.7V , iE = 1.52 mA
iC 4 =
−1.7 − ( −5.2 )
3
−0.7 − ( −5.2 )
= 1.17 mA
= 1.5 mA
3
P = ( iE + iC 4 + iC 3 )( 5.2 ) = (1.52 + 1.17 + 1.5 )( 5.2 )
iC 3 =
or P = 21.8 mW
(ii) For vin = −1.7V , iE = 1.32 mA
iC 4 =
−0.7 − ( −5.2 )
3
−1.7 − ( −5.2 )
= 1.5 mA
= 1.17 mA
3
P = (1.32 + 1.5 + 1.17 )( 5.2 )
iC 3 =
or P = 20.7 mW
17.5
3.7 − 0.7
= 1.5 mA
0.67 + 1.33
VR = I 3 R4 + Vγ = (1.5 )(1.33) + 0.7
a.
I3 =
or
VR = 2.70 V
logic 1 level = 3.7 − 0.7 ⇒ 3.0 V
b.
For v X = vY = logic 1.
3 − 0.7
iE =
= 2.875 mA = iRC1
0.8
vB 3 = 3.7 − ( 2.875 )( 0.21) = 3.10 V
⇒ v01 ( logic 0 ) = 2.4 V
For v X = vY = logic 0, QR on
2.7 − 0.7
iE =
= 2.5 mA = iRC 2
0.8
vB 4 = 3.7 − ( 2.5 )( 0.24 ) = 3.1 V
⇒ v02 ( logic 0 ) = 2.4 V
17.6
(a)
I REF =
1.1 − 0.7 − ( −1.9 )
I Q = I REF
(b)
20
= 0.115 mA
= 0.115 mA
vo1 (max) = 1.1 − 0.7 = 0.4 V
i5 (max) =
vo1 (max) − ( −1.9 )
R5
0.4 + 1.9
= 11.5 K
0.2
vo1 = −0.4 V ⇒ vB 5 = +0.3 V
1.1 − 0.3
= 6.96 K
RC1 =
0.115
vo 2 = −0.4 V ⇒ vB 6 = +0.3V
1.1 − 0.3
= 6.96 K
RC 2 =
0.115
R5 = R6 =
(c)
(d)
17.7
logic 1 + logic 0 1 + 0
=
= 0.5 V
2
2
For i2 = 1 mA
VR =
R5 =
0.5 − ( −2.3)
1
For QR on,
iE =
or
⇒ R5 = 2.8 kΩ
VR − VBE − ( −2.3)
RE
0.5 − 0.7 + 2.3
⇒ RE = 2.1 kΩ
1
= VR + VBE = 0.5 + 0.7 = 1.2 V
RE =
VB 2
i1 =
1.2 − 1.4 − ( −2.3)
R2
or
1.2 − 1.4 + 2.3
⇒ R2 = 2.1 kΩ
1
1.7 − 1.2
R1 =
⇒ R1 = 0.5 kΩ
1
1 − (−2.3)
i3 =
= 3 ⇒ R3 = 1.1 kΩ
R3
R2 =
i4 =
0 − (−2.3)
= 3 ⇒ R4 = 0.767 kΩ
R4
For QR on,
v0 R = logic 0 = 0 V ⇒ vB 3 = 0.7 V
iE = iCR = 1 mA
So
1.7 − 0.7
RC 2 =
⇒ RC 2 = 1 kΩ
1
For vI = logic 1 = 1 V,
iE =
1 − 0.7 − ( −2.3)
For vNOR
= 1.238 mA
2.1
= logic 0 = 0 V,
vB 4 = 0.7 V
Then
1.7 − 0.7
RC1 =
⇒ RC1 = 0.808 kΩ
1.238
17.8
Maximum iE for vI = logic 1 = 3.3 V
Then iE = 5 mA =
3.3 − 0.7
⇒ RE = 0.52 kΩ
RE
For v02 = logic 1 = 3.3 V
iE 3 =
3.3 − 0
= 5 mA
R3
or
R3 = 0.66 kΩ
By symmetry,
R2 = 0.66 kΩ
For Q1 on,
iE = iRC1 = 5 mA =
So
RC1 = 0.12 kΩ
4 − ( 2.7 + 0.7 )
RC1
For QR on,
3 − 0.7
= 4.423 mA = iRC 2
0.52
4 − ( 2.7 + 0.7 )
and iRC 2 = 4.423 =
RC 2
iE =
⇒ RC 2 = 0.136 kΩ
17.9
Neglecting base currents:
(a)
I E1 = 0, I E 3 = 0
5 − 0.7
⇒ I E 5 = 1.72 mA
2.5
Y = 0.7 V
I E5 =
(b)
5 − 0.7
⇒ I E1 = 0.239 mA
18
=0
I E1 =
I E3
5 − 0.7
⇒ I E 5 = 1.72 mA
2.5
Y = 0.7 V
I E5 =
(c)
I E1 = I E 3 =
5 − 0.7
⇒ I E1 = I E 3 = 0.239 mA
18
I E 5 = 0, Y = 5 V
(d)
Same as (c).
17.10
(a)
(b)
VR = −(1)(1) − 0.7 ⇒ VR = −1.7 V
QR off , then vO1 = Logic 1 = −0.7 V
QR on, then vO1 = −(1)(2) − 0.7 ⇒
vO1 = Logic 0 = −2.7 V
QA / QB − off , then vO 2 = Logic 1 = −0.7 V
QA / QB − on, then vO 2 = −(1)(2) − 0.7 ⇒
vO 2 = Logic 0 = −2.7 V
(c)
A = B = Logic 0 = −2.7 V , QR on,
VE = −1.7 − 0.7 ⇒ VE = −2.4 V
A = B = Logic 1 = −0.7 V , QA / QB on,
VE = −0.7 − 0.7 ⇒ VE = −1.4 V
(d)
A = B = Logic 1 = −0.7 V , QA / QB on,
−2.7 − (−5.2)
= 1.67 mA
1.5
−0.7 − (−5.2)
= 3 mA
iC 2 =
1.5
P = (1.67 + 1 + 1 + 1 + 3)(5.2) ⇒ P = 39.9 mW
iC 3 =
A = B = Logic 0 = −2.7 V
iC 3 = 3 mA, iC 2 = 1.67 mA
P = 39.9 mW
17.11
a.
b.
AND logic function
logic 0 = 0 V
5 − (1.6 + 0.7)
= 2.25 mA
1.2
V2 = (2.25)(0.8) ⇒ logic 1 = 1.8 V
Q3 on, i =
c.
5 − 0.7
⇒ iE1 = 1.65 mA
2.6
5 − (0.7 + 0.7)
=
⇒ iE 2 = 3 mA
1.2
iC 3 = 0, iC 2 = iE 2 = 3 mA
iE1 =
iE 2
V2 = 0
d.
5 − (1.8 + 0.7)
⇒ iE1 = 0.962 mA
2.6
5 − (1.6 + 0.7)
=
⇒ iE 2 = 2.25 mA
1.2
iC 2 = 0, iC 3 = iE 2 = 2.25 mA
iE1 =
iE 2
V2 = 1.8 V
17.12
a.
b.
3.5 + 3.1
− 0.7 ⇒ VR = 2.6 V
2
For Q1 on, v X = vY = logic 1 = 3.5 V
VR =
3.5 − (0.7 + 0.7)
= 0.175 mA
12
0.175 0.4
=
⇒ RC1 = 4.57 kΩ
Want iRC1 =
RC1
2
iE =
c.
For Q2 on, iE =
Want iRC 2 =
d.
2.6 − 0.7
= 0.158 mA
12
0.158 0.4
=
⇒ RC 2 = 5.06 kΩ
RC 2
2
For vY = logic 1 = 3.5 V
3.5 − 0.7
= 0.35 mA, i E = 0.175 mA
iR1 =
8
P = ( iR1 + iE )(VCC ) = ( 0.35 + 0.175 )( 3.5 )
⇒ P = 1.84 mW
17.13
a.
logic 1 = 0.2 V
logic 0 = −0.2 V
b.
iE =
( 0 − 0.7 ) − ( −3.1)
RE
So
RE = 3 kΩ
= 0.8
0.8 0.4
=
⇒ R1 = 1 kΩ
2
R1
c.
Want iR1 =
d.
For v X = vY = logic 1 = 0.2 V
iE =
( 0.2 − 0.7 ) − ( −3.1)
iR 2 =
e.
3
= 0.867 mA
0.4
⇒ iR 2 = 0.4 mA
1
iD 2 = 0.467 mA
iE = 0.867 mA
0.2 − (−3.1)
i3 =
= 1 mA
3.3
−0.2 − (−3.1)
i4 =
= 0.879 mA
3.3
P = (0.867 + 1 + 0.879)(0.9 − (−3.1))
or P = 11.0 mW
17.14
a.
( −0.9 − 0.7 ) − ( −3)
i1 =
⇒ i1 = 1.4 mA
1
( −0.2 − 0.7 ) − ( −3)
⇒ i3 = 0.14 mA
i3 =
15
( −0.2 − 0.7 ) − ( −3)
⇒ i4 = 0.14 mA
i4 =
15
i2 + iD = i1 + i3 = 1.4 + 0.14 = 1.54 mA
0.4
⇒ i2 = 0.8 mA
i2 =
0.5
iD = 0.74 mA
v0 = −0.4 V
b.
i1 = 1.4 mA
(0 − 0.7) − (−3)
⇒ i3 = 0.153 mA
15
i4 = i3 ⇒ i4 = 0.153 mA
i3 =
i2 + iD = i4 ⇒ i2 = 0.153 mA
iD = 0
v0 = −(0.153)(0.5) ⇒ v0 = −0.0765 V
c.
i1 =
( 0 − 0.7 − 0.7 ) − ( −3)
1
⇒ i1 = 1.6 mA
(−0.2 − 0.7) − (−3)
⇒ i3 = 0.14 mA
15
i4 = i3 ⇒ i4 = 0.14 mA
i3 =
i2 + iD = i3 ⇒ i2 = 0.14 mA
iD = 0.0
v0 = −(0.14)(0.5) ⇒ v0 = −0.07 V
(0 − 0.7 − 0.7) − (−3)
⇒ i1 = 1.6 mA
1
(0 − 0.7) − (−3)
⇒ i3 = 0.153 mA
i3 =
15
i4 = i3 ⇒ i4 = 0.153 mA
d.
i1 =
i2 + iD = i1 + i4 = 1.6 + 0.153 = 1.753 mA
0.4
i2 =
⇒ i2 = 0.8 mA
0.5
iD = 0.953 mA
v0 = −0.40 V
17.15
a.
i.
A = B = C = D = 0 ⇒ Q1 − Q4 cutoff
So
VDD = 2 I E R 1 + VEB + I B R2
and
I
I
IB = E = E
1 + β P 51
Then
I
2.5 = 2 I E (2) + 0.7 + E ⋅ (15)
51
15
⎛
⎞
2.5 = 0.7 = I E ⋅ ⎜ 4 + ⎟
51 ⎠
⎝
So
I E = 0.419 mA
and
Y = 2.5 − 2(0.4192)(2) ⇒ Y = 0.823 V
ii.
A = C = 0, B = D = 2.5 V
Now
vB 5 = vB 6 = 2.5 − 0.7 = 1.8 V
and
Y = vB 5 + 0.7 ⇒ Y = 2.5 V
b.
Y = ( A OR B ) AND (C OR D)
17.16
a.
logic 1 = 0 V
logic 0 = −0.4 V
b.
v01 = A OR B
v02 = C OR D
v03 = v01 OR v02
or
v03 = ( A OR B ) AND (C OR D)
17.17
a.
For CLOCK = high, I DC flows through the left side of the circuit.. If D is high, I DC flows through
the left R resistor pulling Q low. If D is low. I DC flows through the right R resistor pulling Q low.
For CLOCK = low, I DC flows through the right side of the circuit maintaining Q and Q in their previous
state.
P = ( I DC + 0.5 I DC + 0.1I DC + 0.1 I DC )( 3)
b.
P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 µ W
17.18
(a)
vI = 0 ⇒ V1 = 0.7 V
3.3 − 0.7
= 0.433 mA
6
iB = iC = 0
i1 =
vo = 3.3 V
(b)
vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V
3.3 − 1.5
i1 =
= 0.3 mA
6
0.8
iR =
= 0.04 mA
20
iB = 0.3 − 0.04 = 0.26 mA
3.3 − 0.1
iC =
= 0.8 mA
4
vo = 0.1 V
17.19
i.
i1 =
For v X = vY = 0.1 V ⇒ v ′ = 0.8 V
5 − 0.8
⇒ i1 = 0.525 mA
8
i3 = i4 = 0
ii.
For v X = vY = 5 V,
v′ = 0.8 + 0.7 + 0.7 ⇒⇒ v′ = 2.2 V
5 − 2.2
⇒ i1 = 0.35 mA
8
0.8
i4 = i1 −
⇒ i4 = 0.297 mA
15
5 − 0.1
i3 =
⇒ i3 = 2.04 mA
2.4
i1 =
17.20
a.
For v X = vY = 5 V , both Q1 and Q2 driven into saturation.
v1 = 0.8 + 0.7 + 0.8 ⇒ v1 = 2.3 V
5 − 2.3
⇒ i1 = iB1 = 0.675 mA
4
5 − (0.8 + 0.7 + 0.1)
⇒ i2 = 1.7 mA
i2 =
2
i4 = iB1 + i2 ⇒ i4 = 2.375 mA
i1 =
0.8
⇒ i5 = 0.08 mA
10
= i4 − i5 ⇒ iB 2 = 2.295 mA
i5 =
iB 2
i3 =
5 − 0.1
⇒ i3 = 1.225 mA
4
v0 = 0.1V
b.
iL′
5 − (0.1 + 0.7)
= 1.05 mA
4
iC (max) = β iB 2 = NiL′ + i3
=
(20)(2.295) = N (1.05) + 1.225
So
N = 42
17.21
DX and DY off, Q1 forward active mode
v1 = 0.8 + 0.7 + 0.7 = 2.2 V
5 = i1 R1 + i2 R2 + v1 and i1 = (1 + β )i2
So 5 − 2.2 = i2 [ (1 + β ) R1 + R2 ]
Assume β = 25
5 − 2.2
⇒ i2 = 0.0589 mA
i2 =
(26)(1.75) + 2
i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA
i3 = β i2 ⇒ i3 = 1.47 mA
0.8
= 0.0589 + 1.47 − 0.16 ⇒
5
iBo = 1.37 mA
Qo in saturation
iBo = i2 + i3 −
iCo =
5 − 0.1
⇒ iCo = 0.817 mA
6
17.22
(a)
vI = 0 V, Q1 forward actions
5 − 0.7
iB =
= 0.717 mA
6
iC = (25)(0.71667) = 17.9 mA
iE = (26)(0.71667) = 18.6 mA
(b)
VI = 0.8 V
5 − (0.8 + 0.7)
= 0.583 mA
6
Because of the relative doping levels of the Emitter and collector, and because of the difference in B-C and
B-E areas, we have −iC ≈ iB = 0.583 mA and iE = small value.
(c)
vI = 3.6 Q1 inverse active.
5 − (0.8 + 0.7)
= 0.583 mA
iB =
6
iE = − β R iE = −(0.5)(0.583) = −0.292 mA
iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA
iB =
17.23
a.
i1 =
i.
v X = vY = 0.1 V, so Q1 in saturation.
5 − (0.1 + 0.8)
⇒ i1 = 0.683 mA
6
⇒ iB 2 = i2 = i4 = iB 3 = i3 = 0
ii.
v X = vY = 5 V, so Q1 in inverse active mode.
Assume Q2 and Q3 in saturation.
5 − (0.8 + 0.8 + 0.7)
⇒ i1 = iB 2 = 0.45 mA
6
5 − (0.8 + 0.1)
i2 =
⇒ i2 = 2.05 mA
2
0.8
⇒ i4 = 0.533 mA
i4 =
1.5
i1 =
iB 3 = ( iB 2 + i2 ) − i4 = 0.45 + 2.05 − 0.533
or
iB 3 = 1.97 mA
i3 =
b.
5 − 0.1
⇒ i3 = 2.23 mA
2.2
For Q3 :
i3
2.23
=
= 1.13 < β
iB 3 1.97
For Q2 :
i2
2.05
=
= 4.56 < β
iB 2 0.45
Since ( I C / I B ) < β , then each transistor is in saturation.
17.24
(a)
v X = vY = Logic 1
v ′ = 0.8 + 2 ( 0.7 ) = 2.2 V
5 − 2.2
i1 =
= 0.35 mA
8
0.8
i4 = i1 −
= 0.35 − 0.0533 = 0.2967 mA
15
5 − 0.1
i3 =
= 2.04 mA
2.4
5 − ( 0.1 + 0.7 )
iL′ =
= 0.525 mA
8
Assume β = 25
Then ( 25 )( 0.2967 ) = 2.04 + N ( 0.525 )
So N = 10.2 ⇒ N = 10
(b)
Now
5 = 2.04 + N ( 0.525 )
So N = 5.64 ⇒ N = 5
17.25
a.
5 − ( 0.8 + 0.8 + 0.7 )
= 0.45 mA
6
= iB1 + 2 β R iB1 = 0.45(1 + 2 [ 0.1]) = 0.54 mA
iB1 =
iB 2
For v X = vY = 5 V, Q, in inverse active mode.
iC 2 =
5 − ( 0.8 + 0.1)
2
iB 3 = ( iB 2 + iC 2 ) −
= 2.05 mA
0.8
= 0.54 + 2.05 − 0.533
1.5
or
iB 3 = 2.06 mA
Now
5 − (0.1 + 0.8)
= 0.683 mA
6
Then
iC 3 (max) = β F iB 3 = NiL′
or (20)(2.06) = N (0.683)
⇒ N = 60
iL′ =
b.
From above, for v0 high, I L′ = (0.1)(0.45) = 0.045 mA. Now
⎛ 5 − 4.9 ⎞
(21)(0.1)
I L′ (max) = (1 + β F ) ⎜
⎟ =
R
2
2
⎝
⎠
= 1.05 mA
So
I L (max) = NI L′
or 1.05 = N (0.045)
⇒ N = 23
17.26
(a)
Vin = 0.1V : QI , Sat : Qs , Qo , Cutoff
5 − (0.1 + 0.8)
= 1.025 mA
4
P = iI (5 − 0.1) = (1.025)(4.9) ⇒
iI =
P = 5.02 mW
(b)
Vin = 5V , QI , Inverse Active; QS , Qo , Saturation
vBI = 0.7 + 0.8 + 0.7 = 2.2V
5 − 2.2
= 0.7 mA
4
iEI = β R ⋅ iI = (0.1)(0.7) = 0.07 mA
iI =
Vout = 0.7 + 0.1 = 0.8 V
5 − 0.8
= 4.2 mA
i2 =
1
P = (iI + iEI + i2 )(5) = (0.7 + 0.07 + 4.2)(5) ⇒
P = 24.9 mW
17.27
a.
iB1 =
v X = vY = vZ = 0.1 V
5 − (0.1 + 0.8)
⇒ iB1 = 1.05 mA
3.9
Then
iC1 = iB 2 = iC 2 = iB 3 = iC 3 = 0
b.
v X = vY = vZ = 5 V
iB1 =
5 − (0.8 + 0.8 + 0.7)
⇒ iB1 = 0.692 mA
3.9
Then
iC1 = iB 2 = iB1 (1 + 3β R ) = (0.692)(1 + 3[0.5])
⇒ iC1 = iB 2 = 1.73 mA
5 − (0.1 + 0.8)
⇒ iC 2 = 2.05 mA
2
0.8
= iB 2 + iC 2 −
= 1.73 + 2.05 − 1.0
0.8
⇒ iB 3 = 2.78 mA
iC 2 =
iB 3
5 − 0.1
= 2.04 mA
2.4
5 − (0.1 + 0.8)
iL′ =
= 1.05 mA
3.9
iC 3 = iR 3 + 5iL′ = 2.04 + (5)(1.05)
⇒ iC 3 = 7.29 mA
iR 3 =
17.28
a.
v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode.
2.8 − (0.8 + 0.8 + 0.7)
⇒ iB1 = 0.25 mA
2
= iB1 (1 + 3β R ) = 0.25(1 + 3 [0.3])
⇒ iB 2 = 0.475 mA
iB1 =
iB 2
vC 2 = 0.8 + 0.1 = 0.9 V
0.9 − (0.7 + 0.1)
0.1
=
(1 + β F )(0.5)
(101)(0.5)
iB 4
=
iR 2
= 0.00198 mA (Negligible)
5 − 0.9
=
= 4.56 mA
0.9
⇒ iC 2 = 4.56 mA
0.8
= 0.475 + 4.56 − 0.8
1
= 4.235 mA
iB 3 = iB 2 + iC 2 −
⇒ iB 3
b.
v X = vY = vZ = 0.1 V
5 − (0.1 + 0.8)
⇒ iB1 = 2.05 mA
2
From part (a),
iL′ = β R ⋅ iB1 = (0.3)(0.25) = 0.075 mA
Then
5iL′
5(0.075)
iB 4 =
=
⇒ iB 4 = 0.00371 mA
1+ βF
101
iB1 =
17.29
a.
v X = vY = vZ = 0.1 V
2 − (0.1 + 0.8)
+ iB 3
RB1
where
(2 − 0.7) − (0.9) 0.4
iB 3 =
=
RB 2
1
iB1 =
⇒ iB 3 = 0.4 mA
Then
iB1 =
1.1
+ 0.4 ⇒ iB1 = 1.5 mA
1
iB 2 = 0 = iC 2
Q3 in saturation iC 3 = 5iL′ For v0 high,
vB′ 1 = 0.8 + 0.7 = 1.5 V ⇒ Q3′ off
2 − 1.5
= 0.5 mA
1
iL′ = β R iB′ 1 = (0.2)(0.5) = 0.1 mA
iB′ 1 =
Then
iC 3 = 0.5 mA
v X = vY = vZ = 2 V
b.
From part (a),
⇒ iB1 = 0.5 mA
iB 3 = 0 = iC 3
iB 2 = iB1 (1 + 3β R ) = (0.5)(1 + 3 [0.2])
iB 2 = 0.8 mA
iC 2 = 5iL′ , and from part (a), iL′ = 1.5 mA
So
iC 2 = 7.5 mA
17.30
5.8 − 0.7
= 0.51 mA
10
5 − (0.7 − 0.3)
IC − I D =
= 4.6 mA
1
Now
I
I
I D = 0.51 − I B = 0.51 − C = 0.51 − C
50
β
Then
I ⎞
1 ⎞
⎛
⎛
I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜ 1 + ⎟ − 0.51 = 4.6
50 ⎠
⎝ 50 ⎠
⎝
So I C = 5.01 mA
(a)
IB + ID =
5.01
⇒ I B = 0.1002 mA
50
I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA
IB =
IC
β
=
VCE = 0.4 V
(b)
I D = 0, VCE = VCE ( sat ) = 0.1 V
5.8 − 0.8
⇒ I B = 0.5 mA
10
5 − 0.1
⇒ I C = 4.9 mA
IC =
1
IB =
17.31
a.
v X = vY = 0.4 V
vB1 = 0.4 + 0.7 ⇒ vB1 = 1.1 V
5 − 1.1
⇒ iB1 = 1.39 mA
2.8
= 0.4 + 0.4 ⇒ vB 2 = 0.8 V
iB1 =
vB 2
iB 2 = iC 2 = iB 0 = iC 0 = iB 5 = iC 5
= iB 3 = iC 3 = 0 ( No load )
5 = iB 4 R2 + VBE + (1 + β )iB 4 R4
5 − 0.7
⇒ iB 4 = 0.0394 mA
0.76 + (31)(3.5)
= β F iB 4 ⇒ iC 4 = 1.18 mA
iB 4 =
iC 4
vB 4 = 5 − (0.0394)(0.76) ⇒ vB 4 = 4.97 V
b.
v X = vY = 3.6 V
vB1 = 0.7 + 0.7 + 0.3 ⇒ vB1 = 1.7 V
vB 2 = 1.4 V
vB 0 = 0.7 V
vC 2 = 1.1 V
5 − 1.7
⇒ iB1 = 1.1786 mA
2.8
= iB1 (1 + 2 β R ) = 1.18(1 + 2 [0.1])
iB1 =
iB 2
iB 2 = 1.41 mA
1.1 − 0.7
⇒ iB 4 = 0.00369 mA
(31)(3.5)
5 − 1.1
=
= 5.13 mA ⇒ iC 2 ≈ 5.13 mA
0.76
≈ iB 2 + iC 2
iB 4 =
iR 2
iB 0
iB 0 = 6.54 mA
17.32
(a)
vI = 0, v1 = 0.3 V
1.5 − 0.3
= 1.2 mA
1
iB = iC = 0, vO = 1.5 V
i1 =
(b)
vI = 1.5 V v1 = 0.7 + 0.3 = 1 V
1.5 − 1
= 0.5 mA
1
0.7
iR =
= 0.035 mA
20
iB = 0.5 − 0.035 = 0.465 mA
i1 =
1.5 − 0.4
= 0.917 mA
1.2
vO = 0.4 V
iC =
17.33
a.
Assuming the output transistor Q2 is a Schottky transistor, then
v0 = 0.4 V, iL′ =
Then
2.5 − (0.4 + 0.3)
= 0.5
RB1
RB1 = 3.6 kΩ
Then
iB1 =
2.5 − (0.7 + 0.8) 1.0
=
= 0.278 mA
3.6
RB1
0.7
= 1.50 mA
0.7
iE1 = iB1 + iC1 ⇒ iC1 = 1.50 − 0.278 = 1.222 mA
iB 2 = 0.5 mA, iE1 = 0.5 +
and iC1 =
2.5 − (0.7 + 0.1)
= 1.222 mA
RC1
b.
v X = vY = 0.4 V, vB1 = 0.7 V
⇒ RC1 = 1.39 kΩ
vC 2 = 2.5 − 0.7 ⇒ vC 2 = 1.8 V
All transistor currents are zero.
c.
vB1 = 1.5 V, vC1 = 0.8 V
Currents calculated in part (a).
iB 2 = 0.5 mA, iL′ = 0.5 mA
d.
iC 2 (max) = β iB 2 = NiL′ or (50)(0.5) = N (0.5)
So
N = 50
17.34
a.
For v X = vY = 3.6 V
vB1 = 3(0.7) = 2.1 ⇒ iB1 =
5 − 2.1
= 0.29 mA
10
5 − 1.8
vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 =
= 0.32 mA
10
1.4
iB 2 = iB1 + iC1 −
= 0.29 + 0.32 − 0.0933
15
So
iB 2 = 0.517 mA
vC 2 = 0.7 + 0.4 = 1.1 V
5 − 1.1
= 0.951 mA
iC 2 =
4.1
0.7
= 0.517 + 0.951 − 0.175
4
or iB 5 = 1.293 mA
For v0 = 0.4 V, vB′ 1 = 0.4 + 0.7 = 1.1 V
Then
1.1 − 0.7
= 0.00086 mA
iB′ 1 =
(31)(15)
5 − 1.1
− 0.00086 or iL′ ≈ 0.39 mA
iL′ =
10
So iC 5 (max) = β iΒ 5 = NiL′
iB 5 = iB 2 + iC 2 −
(30)(1.293) = N (0.39) ⇒ N = 99
b.
P = (0.29 + 0.32 + 0.951)(5) + (99)(0.39)(0.4)
P = 7.805 + 15.444 or P = 23.2 mW
(Assumming 99 load circuits which is unreasonably large.)
17.35
a.
Assume no load. For v X = logic 0 = 0.4 V
5 − (0.4 + 0.7)
iE1 =
= 0.0975 mA
40
Essentially all of this current goes to ground from VCC .
P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW
5 − (3)(0.7)
= 0.0725 mA
40
5 − (0.7 + 0.7 + 0.4)
= 0.064 mA
iR 2 =
50
5 − (0.7 + 0.4)
= 0.26 mA
iR 3 =
15
P = (0.0725 + 0.064 + 0.26)(5)
P = 1.98 mW
b.
iR1 =
c.
For v0 = 0, vC 7 = 0.7 + 0.4 = 1.1 V
iR 7 =
17.36
(a)
5 − 1.1
⇒ iR 7 = 78 mA ≈ iSC
0.050
vl = vO = 25 V ; A transient situation
vDS ( M N ) = 2.5 − 0.7 = 1.8 V
vGS ( M N ) = 2.5 − 0.7 = 1.8 V ⇒ M N in saturation
vSD ( M P ) = 5 − (2.5 + 0.7) = 1.8 V
vSG ( M P )5 − 2.5 = 2.5 V ⇒ M P in saturation
iDN = K n (vGSN − VTN ) 2 = (0.1)(1.8 − 0.8) 2 ⇒ iDN = 0.1 mA
iDP = K P (vSGP + VTP ) 2 = (0.1)(2.5 − 0.8) 2 ⇒ iDP = 0.289 mA
iC1 = β iDP = (50)(0.289) ⇒ iC1 = 14.45 mA
iC 2 = β iDN = (50)(0.1) ⇒ iC 2 = 5 mA
Difference between iE1 and iDN + iC 2 is a load current.
(b)
Assume iC1 = 14.45 mA is a constant
(V )(C )
i ⋅t
1
VC = ∫ iC1dt = C1 ⇒ t = C
C
C
iC1
t=
(c)
(5)(15 × 10−12 )
⇒ t = 5.19 ns
14.45 × 10−3
t=
(5)(15 × 10−12 )
⇒ t = 260 ns
0.289 × 10−3
17.37
(a)
Assume R1 = R2 = 10 kΩ; β = 50
0.7
= 0.07 mA
10
NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V
Then iR1 = iR 2 =
2
iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 )
iDN = 0.10 mA
2
Then iB 2 = 0.03 ⇒ iC 2 = (50)(0.03) = 1.5 mA
iE1 = 1.53 mA ⇒ iB1 = 0.03 mA ⇒ iC1 = 1.5 mA
So iDP = 0.10 mA
Now, M P biased in non-saturation region
vSGP = 2.5 V
2
⎤⎦
iDP = 0.10 = 0.10 ⎡⎣ 2(2.5 − 0.8)vSD − vSD
2
0.10 vSD
− 0.34 vSD + 0.10 = 0
vSD =
0.34 ± (0.34) 2 − 4(0.10)(0.10)
2(0.10)
Or
vSD = 0.325 V
Then vo = 5 − 0.325 − 0.7
vo = 3.975 V
1
i ⋅t
idt =
∫
C
C
Cv (15 × 10−12 )(5)
t=
=
i
1.53 × 10−3
t = 49 ns
(c)
Cv (15 × 10−12 )(5)
t=
=
i
0.1× 10−3
t = 0.75 µ s
(b)
v=
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