Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 1
1.1
ni = BT 3 / 2 e
(a) Silicon
− Eg / 2 kT
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
= 2.067 × 1019 exp [ −25.58]
ni = 1.61× 108 cm −3
(i)
ni = ( 5.23 × 1015 ) ( 250 )
(ii)
ni = ( 5.23 × 1015 ) ( 350 )
(b)
GaAs
(i)
ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= 3.425 × 1019 exp [ −18.27 ]
ni = 3.97 ×1011 cm −3
3/ 2
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
______________________________________________________________________________________
1.2
a.
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
⎛
⎞
−1.1
1012 = 5.23 × 1015 T 3 / 2 exp ⎜
⎟
−6
⎝ 2(86 × 10 )(T ) ⎠
⎛ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.3
Silicon
(a)
ni = ( 5.23 × 1015 ) (100 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦
= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3
(b)
ni = ( 5.23 × 1015 ) ( 300 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦
= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3
(c)
ni = ( 5.23 × 1015 ) ( 500 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦
= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3
Germanium.
(a)
ni = (1.66 × 1015 ) (100 )
3/ 2
⎡
⎤
−0.66
⎥ = (1.66 × 1018 ) exp [ −38.37 ]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦
ni = 35.9 cm −3
(b)
ni = (1.66 × 1015 ) ( 300 )
3/ 2
⎡
⎤
−0.66
⎥ = ( 8.626 × 1018 ) exp [ −12.79]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦
ni = 2.40 × 1013 cm −3
(c)
ni = (1.66 × 1015 ) ( 500 )
3/ 2
⎡
⎤
−0.66
⎥ = (1.856 × 1019 ) exp [ −7.674]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦
ni = 8.62 ×1015 cm −3
______________________________________________________________________________________
1.4
(a) n-type; no = 10
15
(
) = 5.76 × 10 cm
(
)
n2
2.4 × 1013
cm ; po = i =
no
1015
−3
2
11
2
−3
ni2
1.5 × 1010
=
= 2.25 × 10 5 cm −3
no
1015
______________________________________________________________________________________
(b) n-type; no = 1015 cm −3 ; po =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.5
(a) p-type; p o = 1016 cm −3 ; no =
(
) = 3.24 × 10
(
)
ni2
1.8 × 10 6
=
po
1016
2
cm −3
−4
2
ni2
2.4 × 1013
=
= 5.76 × 1010 cm −3
po
1016
______________________________________________________________________________________
(b) p-type; p o = 1016 cm −3 ; no =
1.6
(a)
(b)
n-type
no = N d = 5 × 1016 cm −3
n 2 (1.5 × 10 )
po = i =
= 4.5 × 103 cm −3
no
5 × 1016
10 2
(c)
no = N d = 5 × 1016 cm −3
From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3
( 3.97 × 10 ) = 3.15 ×10 cm
p =
11 2
6
−3
5 × 1016
______________________________________________________________________________________
o
1.7
(a) p-type; p o = 5 × 1016 cm −3 ; no =
(
) = 4.5 × 10 cm
(
)
ni2
1.5 × 1010
=
po
5 × 1016
2
3
−3
2
ni2
1.8 × 10 6
=
= 6.48 × 10 −5 cm −3
po
5 × 1016
______________________________________________________________________________________
(b) p-type; p o = 5 × 1016 cm −3 ; no =
1.8
(a) Add boron atoms
(b) N a = po = 2 × 1017 cm −3
(
)
2
ni2
1.5 × 1010
=
= 1.125 × 10 3 cm −3
po
2 × 1017
______________________________________________________________________________________
(c) no =
1.9
(a)
no = 5 × 1015 cm −3
n 2 (1.5 × 10 )
po = i =
⇒ po = 4.5 × 104 cm −3
no
5 × 1015
10 2
(b)
n o > p o ⇒ n-type
(c) no ≅ N d = 5 × 1015 cm −3
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.10
a.
b.
Add Donors
N d = 7 × 1015 cm −3
Want po = 106 cm −3 = ni2 / N d
So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21
⎛ − 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
______________________________________________________________________________________
1.11
(a) I = Aσ Ε = 10 −5 (1.5)(10) ⇒ I = 0.15 mA
( )
(
)
Iρ 1.2 × 10 −3 (0.4)
= 2.4 V/cm
=
A
ρ
2 × 10 − 4
______________________________________________________________________________________
(b) I =
AΕ
⇒Ε=
(
)
1.12
J 120
−1
=
= 6.67 (Ω − cm)
Ε 18
σ
(6.67)
σ ≅ eμ n N d ⇒ N d =
=
= 3.33 × 1016 cm −3
eμ n
1.6 × 10 −19 (1250)
______________________________________________________________________________________
J =σΕ ⇒σ =
(
)
1.13
1
1
1
⇒ Nd =
=
= 7.69 × 1015 cm −3
eμ n N d
eμ n ρ 1.6 × 10 −19 (1250)(0.65)
Ε
(b) J = ⇒ Ε = ρ J = (0.65)(160 ) = 104 V/cm
ρ
______________________________________________________________________________________
(a) ρ ≅
(
)
1.14
σ
1.5
=
= 9.375 × 1015 cm −3
eμ n
1.6 × 10 −19 (1000)
σ
0.8
=
= 1.25 × 1016 cm −3
(b) N a =
−19
eμ p
1.6 × 10 (400)
______________________________________________________________________________________
(a) σ ≅ eμ n N d ⇒ N d =
(
(
)
)
1.15
(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 )
−1
(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 ×103 A / cm2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.16
Dn = (0.026)(1250) = 32.5 cm 2 /s; D p = (0.026 )(450 ) = 11.7 cm 2 /s
J n = eDn
⎛ 1016 − 1012 ⎞
dn
⎟⎟ = −52 A/cm 2
= 1.6 × 10 −19 (32.5)⎜⎜
dx
⎝ 0 − 0.001 ⎠
(
J p = −eD p
)
⎛ 1012 − 1016 ⎞
dp
⎟⎟ = −18.72 A/cm 2
= − 1.6 × 10 −19 (11.7 )⎜⎜
dx
⎝ 0 − 0.001 ⎠
(
)
Total diffusion current density
J = −52 − 18.72 = −70.7 A/cm 2
______________________________________________________________________________________
1.17
J p = −eD p
dp
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟
⎜ Lp ⎟
⎜ Lp ⎟
⎝ ⎠
⎝ ⎠
(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞
−19
Jp =
15
⎜⎜ ⎟⎟
⎝ Lp ⎠
10 × 10 −4
J p = 2.4 e
− x / Lp
J p = 2.4 A/cm2
(a)
x=0
(b)
x = 10 μ m
J p = 2.4 e−1 = 0.883 A/cm 2
x = 30 μ m
J p = 2.4 e−3 = 0.119 A/cm 2
______________________________________________________________________________________
(c)
1.18
a.
N a = 1017 cm −3 ⇒ po = 1017 cm −3
n 2 (1.8 × 10 )
⇒ no = 3.24 × 10−5 cm −3
no = i =
1017
po
6 2
b.
n = no + δ n = 3.24 × 10−5 + 1015 ⇒ n = 1015 cm −3
p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3
______________________________________________________________________________________
⎛N N ⎞
1.19 Vbi = VT ln⎜⎜ a 2 d ⎟⎟
⎝ ni ⎠
(a) (i)
(ii)
(
)(
(
(
)
)( )
(
)
⎡ (10 )(10 ) ⎤
V = (0.026 ) ln ⎢
⎥ = 0.937 V
⎣⎢ (1.5 × 10 ) ⎦⎥
⎡ 5 × 1017 1015 ⎤
Vbi = (0.026 ) ln ⎢
⎥ = 0.739 V
10 2
⎣⎢ 1.5 × 10
⎦⎥
18
(iii)
)
⎡ 5 × 1015 5 × 1015 ⎤
Vbi = (0.026 ) ln ⎢
⎥ = 0.661 V
2
⎢⎣ 1.5 × 1010
⎥⎦
bi
18
10 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)(
)
⎡ 5 × 1015 5 × 1015 ⎤
Vbi = (0.026 ) ln ⎢
⎥ = 1.13 V
2
⎢⎣
⎥⎦
1.8 × 10 6
(b) (i)
(
(
)
)( )
(
)
⎡ (10 )(10 ) ⎤
V = (0.026 ) ln ⎢
⎥ = 1.41 V
⎢⎣ (1.8 × 10 ) ⎥⎦
⎡ 5 × 1017 1015 ⎤
Vbi = (0.026 ) ln ⎢
⎥ = 1.21 V
2
⎢⎣ 1.8 × 10 6
⎥⎦
(ii)
18
(iii)
18
6 2
bi
______________________________________________________________________________________
1.20
⎛N N ⎞
Vbi = VT ln⎜⎜ a 2 d ⎟⎟
⎝ ni ⎠
or
Na =
(n ) exp⎛⎜ V ⎞⎟ = (1.5 × 10) exp⎛⎜ 0.712 ⎞⎟ = 1.76 × 10
2
i
2
16
−3
cm
⎜V ⎟
Nd
1016
⎝ 0.026 ⎠
⎝ T ⎠
______________________________________________________________________________________
1.21
bi
⎡ N a (1016 ) ⎤
⎛N N ⎞
⎥
Vbi = VT ln ⎜ a 2 d ⎟ = ( 0.026 ) ln ⎢
10 2
⎢⎣ (1.5 × 10 ) ⎥⎦
⎝ ni ⎠
For N a = 1015 cm −3 , Vbi = 0.637 V
For N a = 1018 cm −3 , Vbi = 0.817 V
______________________________________________________________________________________
1.22
⎛ T ⎞
kT = (0.026) ⎜
⎟
⎝ 300 ⎠
200
250
300
350
400
450
500
kT
0.01733
0.02167
0.026
0.03033
0.03467
0.0390
0.04333
(T)3/2
2828.4
3952.8
5196.2
6547.9
8000.0
9545.9
11,180.3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛
⎞
−1.4
⎟
ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜
⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝
⎠
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
T
ni
200
1.256
250
6.02 × 103
300
1.80 × 106
350
1.09 × 108
400
2.44 × 109
450
2.80 × 1010
500
2.00 × 1011
Vbi
1.405
1.389
1.370
1.349
1.327
1.302
1.277
______________________________________________________________________________________
1.23
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
−1/ 2
⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤
⎥ = 0.684 V
Vbi = ( 0.026 ) ln ⎢
⎢⎣
(1.5 ×10 10 ) 2 ⎥⎦
1 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
0.684
⎝
⎠
−1/ 2
(a)
3 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠
−1/ 2
(b)
= 0.255 pF
= 0.172 pF
−1/ 2
5 ⎞
⎛
= 0.139 pF
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠
______________________________________________________________________________________
(c)
1.24
(a)
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
−1 / 2
5 ⎞
⎛
For VR = 5 V, C j = (0.02) ⎜ 1 +
⎟
⎝ 0. 8 ⎠
−1 / 2
⎛ 1. 5 ⎞
For VR = 1.5 V, C j = (0.02) ⎜1 +
⎟
⎝ 0. 8 ⎠
= 0.00743 pF
−1 / 2
= 0.0118 pF
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.00743 + 0.0118
= 0.00962 pF
2
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ
C j (avg ) =
where
τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 )
or
τ = 4.52 ×10−10 s
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
______________________________________________________________________________________
1.25
⎛ V ⎞
C j = C jo ⎜⎜1 + R ⎟⎟
⎝ Vbi ⎠
−1 / 2
(
)( )
)
⎡ 5 × 1015 1017 ⎤
; Vbi = (0.026 ) ln ⎢
⎥ = 0.739 V
10 2
⎣⎢ 1.5 × 10
⎦⎥
(
For V R = 1 V,
Cj =
0.60
1
1+
0.739
= 0.391 pF
For VR = 3 V,
Cj =
0.60
3
1+
0.739
= 0.267 pF
For V R = 5 V,
0.60
Cj =
(a)
fo =
(b) f o =
1
2π LC
(
5
1+
0.739
=
(
= 0.215 pF
2π 1.5 × 10
2π 1.5 × 10
1
−3
)(0.391× 10 )
−12
1
−3
)(0.267 × 10 )
−12
⇒ f o = 6.57 MHz
⇒ f o = 7.95 MHz
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
fo =
1
⇒ f o = 8.86 MHz
2π 1.5 × 10 0.215 × 10 −12
______________________________________________________________________________________
(c)
(
−3
)(
)
1.26
a.
⎡
⎛V ⎞ ⎤
⎛V ⎞
I = I S ⎢ exp ⎜ D ⎟ − 1⎥ − 0.90 = exp ⎜ D ⎟ − 1
⎝ VT ⎠ ⎦
⎝ VT ⎠
⎣
⎛V ⎞
exp ⎜ D ⎟ = 1 − 0.90 = 0.10
⎝ VT ⎠
VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V
b.
⎡
⎛ VF ⎞ ⎤
⎛ 0.2 ⎞
⎢ exp ⎜ ⎟ − 1⎥ exp ⎜
⎟ −1
IS ⎣
IF
0.026 ⎠
⎝ VT ⎠ ⎦
⎝
= ⋅
=
IR
IS ⎡
⎛ VR ⎞ ⎤ exp ⎛ −0.2 ⎞ − 1
⎜
⎟
⎢exp ⎜ ⎟ − 1⎥
⎝ 0.026 ⎠
⎝ VT ⎠ ⎦
⎣
=
2190
−1
IF
= 2190
IR
______________________________________________________________________________________
⎡ ⎛V ⎞ ⎤
1.27 I D = I S ⎢exp⎜⎜ D ⎟⎟ − 1⎥
⎣⎢ ⎝ VT ⎠ ⎦⎥
(a) (i)
(ii)
(iii)
(iv)
(
)
(
)
(
)
(
)
(
)
⎛ 0.3 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 1.03 μ A
⎝ 0.026 ⎠
⎛ 0.5 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 2.25 mA
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 4.93 A
⎝ 0.026 ⎠
⎡ ⎛ − 0.02 ⎞ ⎤
−12
A
I D = 10 −11 ⎢exp⎜
⎟ − 1⎥ = −5.37 × 10
⎣ ⎝ 0.026 ⎠ ⎦
(v)
⎡ ⎛ − 0.20 ⎞ ⎤
−11
I D = 10 −11 ⎢exp⎜
⎟ − 1⎥ ≅ −10 A
0
.
026
⎠ ⎦
⎣ ⎝
(vi)
I D = − 10 −11 A
(b) (i)
(ii)
(iii)
(
)
(
)
(
)
(
)
⎛ 0.3 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 0.0103 μ A
⎝ 0.026 ⎠
⎛ 0.5 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 22.5 μ A
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 49.3 mA
⎝ 0.026 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
(iv)
⎡ ⎛ − 0.02 ⎞ ⎤
−14
I D = 10 −13 ⎢exp⎜
A
⎟ − 1⎥ = −5.37 × 10
0
.
026
⎝
⎠
⎣
⎦
(v)
I D ≅ −10 −13 A
I D ≅ −10 −13 A
(vi)
______________________________________________________________________________________
⎛I ⎞
1.28 V D = VT ln⎜⎜ D ⎟⎟
⎝ IS ⎠
⎛ 10 × 10 −6 ⎞
⎟⎟ = 0.359 V
(a) (i) V D = (0.026 ) ln⎜⎜
−11
⎠
⎝ 10
⎛ 100 × 10 −6 ⎞
⎟⎟ = 0.419 V
V D = (0.026 ) ln⎜⎜
−11
⎠
⎝ 10
⎛ 10 −3 ⎞
V D = (0.026) ln⎜⎜ −11 ⎟⎟ = 0.479 V
⎝ 10 ⎠
⎡ ⎛ V ⎞ ⎤
(ii) − 5 × 10 −12 = 10 −11 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.018 V
⎣ ⎝ 0.026 ⎠ ⎦
⎛ 10 × 10 −6 ⎞
⎟⎟ = 0.479 V
(b) (i) V D = (0.026 ) ln⎜⎜
−13
⎠
⎝ 10
⎛ 100 × 10 −6 ⎞
⎟⎟ = 0.539 V
V D = (0.026 ) ln⎜⎜
−13
⎠
⎝ 10
⎛ 10 −3 ⎞
V D = (0.026) ln⎜⎜ −13 ⎟⎟ = 0.599 V
⎝ 10 ⎠
⎡ ⎛ V ⎞ ⎤
(ii) − 10 −14 = 10 −13 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.00274 V
⎣ ⎝ 0.026 ⎠ ⎦
______________________________________________________________________________________
1.29
(a)
(b)
VD
⎛ 0.7 ⎞
10−3 = I S exp ⎜
⎟
⎝ 0.026 ⎠
I S = 2.03 × 10 −15 A
I D ( A ) ( n = 1)
I D ( A )( n = 2 )
0.1
9.50 ×10
1.39 ×10 −14
0.2
4.45 ×10 −12
9.50 ×10 −14
0.3
2.08 ×10 −10
6.50 ×10 −13
−
9
0.4
9.75 ×10
4.45 ×10 −12
0.5
4.56 ×10 −7
3.04 ×10 −11
−
5
0.6
2.14 ×10
2.08 ×10 −10
0.7
10 −3
1.42 ×10 −9
______________________________________________________________________________________
−14
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.30
(a)
I S = 10 −12 A
VD(v)
0.10
0.20
0.30
0.40
0.50
0.60
0.70
(b)
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
I S = 10 −14 A
VD(v)
ID(A)
log10ID
−13
−12.3
0.10
4.68 ×10
−
11
−10.66
0.20
2.19 ×10
−
9
−8.99
0.30
1.03 ×10
−8
−7.32
0.40
4.80 ×10
−
6
−5.65
0.50
2.25 ×10
−4
−3.98
0.60
1.05 ×10
−
3
−2.31
0.70
4.93 ×10
______________________________________________________________________________________
1.31
a.
⎛ V − VD1 ⎞
ID2
= 10 = exp ⎜ D 2
⎟
I D1
⎝ VT
⎠
ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV
b.
ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV
______________________________________________________________________________________
1.32
⎛ 2 ⎞
(a) (i) V D = (0.026) ln⎜
⎟ = 0.539 V
−9
⎝ 2 × 10 ⎠
⎛ 20 ⎞
(ii) V D = (0.026) ln⎜
⎟ = 0.599 V
−9
⎝ 2 × 10 ⎠
⎛ 0.4 ⎞
(b) (i) I D = 2 × 10 −9 exp⎜
⎟ ⇒ 9.60 mA
⎝ 0.026 ⎠
⎛ 0.65 ⎞
(ii) I D = 2 × 10 −9 exp⎜
⎟ ⇒ 144 A
⎝ 0.026 ⎠
______________________________________________________________________________________
(
)
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.33
⎛I ⎞
⎛ 2 × 10−3 ⎞
= 0.6347 V
VD = Vt ln ⎜ D ⎟ = (0.026) ln ⎜
−14 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
⎛ 2 × 10−3 ⎞
= 0.5150 V
VD = (0.026) ln ⎜
−12 ⎟
⎝ 5 × 10 ⎠
0.5150 ≤ VD ≤ 0.6347 V
______________________________________________________________________________________
1.34
⎛ 0.30 ⎞
−8
(a) 1.5 × 10 −3 = I S exp⎜
⎟ ⇒ I S = 1.46 × 10 A
⎝ 0.026 ⎠
⎛ 0.35 ⎞
(b) (i) I D = 1.462 × 10 −8 exp⎜
⎟ ⇒ I D = 10.3 mA
⎝ 0.026 ⎠
⎛ 0.25 ⎞
(ii) I D = 1.462 × 10 −8 exp⎜
⎟ ⇒ I D = 0.219 mA
⎝ 0.026 ⎠
______________________________________________________________________________________
(
)
(
)
1.35
(
)
(
)
(
)
(
)
⎛ 0.8 ⎞
(a) I D = 10 − 22 exp⎜
⎟ ⇒ 2.31 nA
⎝ 0.026 ⎠
⎛ 1.0 ⎞
I D = 10 − 22 exp⎜
⎟ ⇒ 5.05 μ A
⎝ 0.026 ⎠
⎛ 1.2 ⎞
I D = 10 − 22 exp⎜
⎟ ⇒ 11.1 mA
⎝ 0.026 ⎠
⎡ ⎛ − 0.02 ⎞ ⎤
− 23
I D = 10 − 22 ⎢exp⎜
⎟ − 1⎥ = −5.37 × 10 A
0
.
026
⎠ ⎦
⎣ ⎝
For V D = −0.20 V, I D = −10 −22 A
For V D = −2 V, I D = −10 −22 A
(b)
⎛ 0.8 ⎞
I D = 5 × 10 −24 exp⎜
⎟ ⇒ 115 pA
⎝ 0.026 ⎠
⎛ 1.0 ⎞
I D = 5 × 10 − 24 exp⎜
⎟ ⇒ 0.253 μ A
⎝ 0.026 ⎠
⎛ 1.2 ⎞
I D = 5 × 10 − 24 exp⎜
⎟ ⇒ 0.554 mA
⎝ 0.026 ⎠
(
)
(
)
(
)
(
)
⎡ ⎛ − 0.02 ⎞ ⎤
− 24
A
I D = 5 × 10 − 24 ⎢exp⎜
⎟ − 1⎥ = −2.68 × 10
0
.
026
⎠ ⎦
⎣ ⎝
For V D = −0.20 V, I D = −5 × 10 −24 A
For V D = −2 V, I D = −5 × 10 −24 A
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.36
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
______________________________________________________________________________________
1.37
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)
______________________________________________________________________________________
1.38
(a) V PS = I D R + V D
( )
2.8 = I D 10 6 + VD ;
(
)
⎛ V ⎞
I D = 5 × 10 −11 exp⎜ D ⎟
⎝ 0.026 ⎠
By trial and error,
V D = 0.282 V, I D = 2.52 μ A
(b)
I D ≅ −5 × 10 −11 A, VD = −2.8 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.39
⎛ 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)
−4
0.50
0.5194
4.75 ×10
−
4
0.517
0.5194
4.7415 ×10
−4
0.5194
0.5194
4.740 ×10
VD = 0.5194 V
I D = 0.4740 mA
______________________________________________________________________________________
1.40
I s = 5 × 10 −13 A
⎛ R2 ⎞
⎛ 30 ⎞
VTH = ⎜
⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V
R
R
80 ⎠
+
⎝
⎝ 1
2 ⎠
⎛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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.41
(a) I D1 = I D 2 = 1 mA
(i)
⎛ 10 −3 ⎞
V D1 = V D 2 = (0.026 ) ln⎜⎜ −13 ⎟⎟ = 0.599 V
⎝ 10 ⎠
(ii)
⎛ 10 −3 ⎞
⎟ = 0.617 V
V D1 = (0.026 ) ln⎜⎜
−14 ⎟
⎝ 5 × 10 ⎠
⎛ 10 −3 ⎞
⎟ = 0.557 V
V D 2 = (0.026 ) ln⎜⎜
−13 ⎟
⎝ 5 × 10 ⎠
(b) V D1 = V D 2
Ii
= 0.5 mA
2
⎛ 0.5 × 10 −3 ⎞
⎟⎟ = 0.581 V
V D1 = V D 2 = (0.026 ) ln⎜⎜
−13
⎝ 10
⎠
(i)
I D1 = I D 2 =
(ii)
I
I D1
5 × 10 −14
= S1 =
= 0.10
I D 2 I S 2 5 × 10 −13
So I D1 = 0.10 I D 2
I D1 + I D 2 = 1.1I D 2 = 1 mA
So I D 2 = 0.909 mA, I D1 = 0.0909 mA
Now
⎛ 0.0909 × 10 −3 ⎞
⎟⎟ = 0.554 V
V D1 = (0.026 ) ln⎜⎜
−14
⎝ 5 × 10
⎠
⎛ 0.909 × 10 −3 ⎞
⎟⎟ = 0.554 V
V D 2 = (0.026 ) ln⎜⎜
−13
⎝ 5 × 10
⎠
______________________________________________________________________________________
1.42
(
)
⎛ 0.635 ⎞
(a) I D 3 = 6 × 10 −14 exp⎜
⎟ ⇒ 2.426 mA
⎝ 0.026 ⎠
0.635
= 0.635 mA
IR =
1
I D1 = I D 2 = 2.426 + 0.635 = 3.061 mA
⎛ 3.061 × 10 −3 ⎞
⎟⎟ = 0.641 V
V D1 = V D 2 = (0.026 ) ln⎜⎜
−14
⎝ 6 × 10
⎠
V I = 2(0.641) + 0.635 = 1.917 V
(b) I D 3 = 2.426 mA
0.635
= 1.27 mA
IR =
0.5
I D1 = I D 2 = 2.426 + 1.27 = 3.696 mA
⎛ 3.696 × 10 −3 ⎞
⎟⎟ = 0.6459 V
V D1 = V D 2 = (0.026 ) ln⎜⎜
−14
⎝ 6 × 10
⎠
V I = 2(0.6459 ) + 0.635 = 1.927 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
1.43
(a) Assume diode is conducting.
Then, VD = Vγ = 0.7 V
0. 7
⇒ 23.3 μ A
30
1.2 − 0.7
⇒ 50 μ A
I R1 =
10
Then I D = I R1 − I R 2 = 50 − 23.3
So that I R 2 =
Or I D = 26.7 μ A
(b) Let R1 = 50 k Ω Diode is cutoff.
30
⋅ (1.2) = 0.45 V
30 + 50
Since VD < Vγ , I D = 0
VD =
______________________________________________________________________________________
1.44
At node VA:
5 − VA
V
= ID + A
(1)
2
2
At node V B = V A − Vγ
(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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5 − VA VA
−
= 2.5 − VA ⇒ I D Neg. ⇒ I D = 0
2
2
Both (a), (b) I D = 0
From (1) I D =
2
⋅ 5 = 2 V ⇒ VD = 0.50 V
5
______________________________________________________________________________________
VA = 2.5, VB =
1.45
(a) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 0.7 mA
VO = 0.7 V; I D = (I i − 0.7 ) mA; for I i ≥ 0.7 mA
(b) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 1.7 mA
VO = 1.7 V; I D = (I i − 1.7 ) mA; for I i ≥ 1.7 mA
(c) VO = 0.7 V; I D1 = I i ; I D 2 = 0 ; for 0 ≤ I i ≤ 2 mA
______________________________________________________________________________________
1.46
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
(2)
9.3 0.7
=
+ 14.3
R1
R2
Using Eq. (1),
9.3 4.3
=
− 2 + 14.3 ⇒
R1
R1
R1 = 0.41 kΩ
Then R2 = 82.5Ω 82.5Ω
______________________________________________________________________________________
1.47
5 − 0.7
= 0.215 mA, VO = 0.7 V
20
5 − 0.6
= 0.220 mA, VO = 0.6 V
(ii) I =
20
5 − 0.7 − (− 5)
= 0.2325 mA, VO = (0.2325)(20 ) − 5 = −0.35 V
(b) (i) I =
40
5 − 0.6 − (− 5)
I=
= 0.235 mA, VO = (0.235)(20 ) − 5 = −0.30 V
(iii)
40
(a) (i) I =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2 − 0.7 − (− 8)
= 0.372 mA, VO = 2 − (0.372 )(5) = 0.14 V
25
2 − 0.6 − (− 8)
= 0.376 mA, VO = 2 − (0.376 )(5) = 0.12 V
(ii) I =
25
(d) (i) I = 0 , VO = −5 V
(ii) I = 0 , VO = −5 V
______________________________________________________________________________________
(c) (i) I =
1.48
(a) I =
5 − VO
⎛ V ⎞
, I = 5 × 10 −14 exp⎜ D ⎟
20
⎝ 0.026 ⎠
(
)
By trial and error, V D = VO = 0.5775 V, I = 0.221 mA
10 − V D
, VO = 5 − I (20 ) − V D
40
I = 0.2355 mA, V D = 0.579 V, VO = −0.289
(b) I =
10 − V D
, VO = 2 − I (5)
25
I = 0.3763 mA, V D = 0.5913 V, VO = 0.1185
(c) I =
(d) I = −5 × 10 −14 A, VO ≅ −5 V
______________________________________________________________________________________
1.49
(a)
Diode forward biased VD = 0.7 V
5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V
(b) P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω
______________________________________________________________________________________
1.50
(a)
0.65
= 0.65 mA = I D1
1
I D 2 = 2(0.65) = 1.30 mA
I R 2 = I D1 =
ID2 =
(b)
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
IR2 =
______________________________________________________________________________________
1.51
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V
(0.026)
τd = T =
= 0.026 kΩ = 26Ω
a.
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
b.
(0.026)
= 260Ω
0. 1
id = 0.05 I DQ = 5 μ A peak-to-peak
For I DQ = 0.1 mA ⇒ τ d =
vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak
______________________________________________________________________________________
1.52
(a) rd =
VT
0.026
=
=1 kΩ
I DQ 0.026
0.026
⇒ 100 Ω
0.26
0.026
⇒ 10 Ω
(c) rd =
2.6
______________________________________________________________________________________
(b) rd =
1.53
a.
diode resistance rd = VT /I
⎛
⎞
⎜ VT /I ⎟
⎛ rd ⎞
vd = ⎜
⎟ vS
⎟ vS = ⎜ V
⎝ rd + RS ⎠
⎜⎜ T + RS ⎟⎟
⎝ I
⎠
⎛ VT
⎞
vd = ⎜
⎟ vs = vo
⎝ VT + IRS ⎠
b.
RS = 260Ω
I = 1 mA,
⎞
v0 ⎛ VT
v
0.026
=⎜
⇒ 0 = 0.0909
⎟=
vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26)
vS
I = 0.1 mA,
I = 0.01 mA.
v0
v
0.026
=
⇒ 0 = 0.50
vs 0.026 + ( 0.1)( 0.26 )
vS
v0
v
0.026
=
⇒ 0 = 0.909
vS 0.026 + (0.01)(0.26)
vS
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.54
pn junction diode
⎛ 0.72 × 10 −3 ⎞
⎟ = 0.548 V
V D = (0.026 ) ln⎜⎜
−13 ⎟
⎝ 5 × 10
⎠
Schottky diode
⎛ 0.72 × 10 −3 ⎞
⎟⎟ = 0.249 V
V D = (0.026 ) ln⎜⎜
−8
⎝ 5 × 10
⎠
______________________________________________________________________________________
1.55
⎛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
IS =
I
0.5 × 10−3
=
⎛V ⎞
⎛ 0.4796 ⎞
exp ⎜ a ⎟ exp ⎜
⎟
⎝ 0.026 ⎠
⎝ VT ⎠
I S = 4.87 × 10 −12 A
______________________________________________________________________________________
1.56
(a)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I1 + I 2 = 0.5 × 10 −3
⎛V ⎞
⎛V ⎞
5 × 10 −8 exp ⎜ D ⎟ + 10−12 exp ⎜ D ⎟ = 0.5 × 10 −3
⎝ VT ⎠
⎝ VT ⎠
⎛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)
⎛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 ⎜
V
⎝ T ⎠
⎝ VT ⎠
⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞
⎛ 0. 9 ⎞
exp ⎜ D1 ⎟ = ⎜
exp ⎜
⎟
−12 ⎟
V
10
⎝ 0.026 ⎠
⎠
⎝ T ⎠ ⎝
⎛ 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 ⎠
______________________________________________________________________________________
1.57
VZ = VZ 0 = 5.6 V at I Z = 0.1 mA
rZ = 10Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I Z rZ = ( 0.1)(10 ) = 1 mV
VZ0 = 5.599
a.
RL → ∞ ⇒
IZ =
10 − 5.599
4.401
=
= 8.63 mA
R + rZ
0.50 + 0.01
VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 )
VZ = V0 = 5.685 V
b.
11 − 5.599
= 10.59 mA
0.51
VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V
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
c.
I = IZ + IL
V
V − V0
V − VZ 0
, IZ = 0
I L = 0 , I = PS
RL
R
rZ
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
______________________________________________________________________________________
1.58
10 − 6.8
= 6.4 mA
0.5
P = I Z VZ = (6.4)(6.8) = 43.5 mW
(b) I Z = (0.1)(6.4) = 0.64 mA
I L = 6.4 − 0.64 = 5.76 mA
(a) I Z =
VZ
V
6.8
⇒ RL = Z =
= 1.18 k Ω
RL
I Z 5.76
______________________________________________________________________________________
IL =
1.59
I Z rZ = ( 0.1)( 20 ) = 2 mV
VZ 0 = 6.8 − 0.002 = 6.798 V
a.
RL = ∞
IZ =
10 − 6.798
⇒ I Z = 6.158 mA
0.5 + 0.02
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 )
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
______________________________________________________________________________________
1.60
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
______________________________________________________________________________________
1.61 V D = 0, I D = 0.2 A
V D = 0.60 V, I D = 0.1995 A
V D = 0.65 V, I D = 0.1964 A
V D = 0.70 V, I D = 0.1754 A
V D = 0.72 V, I D = 0.1468 A
V D = 0.74 V, I D = 0.0853 A
V D = 0.7545 V, I D = 0
______________________________________________________________________________________
1.62
⎡ ⎛ V ⎞ ⎤
(a) 0.16 = 0.20 − 5 × 10 −14 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = 0.7126 V
⎣ ⎝ 0.026 ⎠ ⎦
(b) P = (0.16 )(0.7126 ) = 0.114 W
______________________________________________________________________________________
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 2
2.1
⎛ 1000 ⎞
(a) For υ I > 0.6 V, υ O = ⎜
⎟(υ I − 0.6 )
⎝ 1020 ⎠
For υ I < 0.6 V, υ O = 0
⎛ 1000 ⎞
(b) (ii) υ O = 0 = ⎜
⎟[10 sin (ω t )1 − 0.6]
⎝ 1020 ⎠
0. 6
Then sin (ω t )1 =
= 0.06 ⇒ (ω t )1 = 3.44° ⇒ 0.01911π rad
10
Also (ω t )2 = 180 − 3.44 = 176.56° ⇒ 0.9809π rad
Now
υ O (avg ) =
=
T
2π
1
1
υ O (t )dt =
[10 sin x − 0.6]dx
T 0
2π 0
∫
∫
0.9809π
0.9809π ⎤
1 ⎡
− 0.6 x
⎢− 10 cos x
⎥
2π ⎣
0.01911π
0.01911π ⎦
1
[(− 10 )(− 0.9982 − 0.9982 ) − 0.6(0.9809π − 0.01911π )]
2π
υ O (avg ) = 2.89 V
=
⎤
⎛ 1000 ⎞ ⎡
⎛π ⎞
(iii) υ O ( peak ) = ⎜
⎟ ⎢10 sin ⎜ ⎟ − 0.6⎥ = 9.2157 V; i d (max ) = 9.2157 mA
1020
2
⎝
⎠⎣
⎝ ⎠
⎦
(iv) PIV = 10 V
______________________________________________________________________________________
2.2
v0 = vI − vD
⎛i ⎞
v
vD = VT ln ⎜ D ⎟ and iD = 0
I
R
⎝ S⎠
⎛ v ⎞
v0 = vI − VT ln ⎜ 0 ⎟
⎝ IS R ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.3
⎛ 1⎞
(a) υ S = 120 2 ⎜ ⎟ = 16.97 V (peak)
⎝ 10 ⎠
υ O ( peak ) = 16.27 V
16.27
= 8.14 mA
2
(c) υ O = 16.97 sin ω t − 0.7
(b) i D ( peak ) =
sin (ω t )1 =
0. 7
= 0.04125 ⇒ (ω t )1 = 2.364°
16.97
(ω t )2 = 180 − 2.364 = 177.64°
⎛ 177.64 − 2.364 ⎞
%=⎜
⎟ × 100% = 48.7%
360
⎝
⎠
(d)
υ O (avg ) =
=
0.9869π
1
[16.97 sin x − 0.7]dx
2π 0.01313π
∫
0.9869π
0.9869π
⎤
1 ⎡
− 0. 7 x
(− 16.97 ) cos x
⎢
⎥
2π ⎣
0.01313π
0.01313π ⎦
1
[(− 16.97 )(− 0.99915 − 0.99915) − 0.7(0.9738π )]
2π
υ O (avg ) = 5.06 V
=
υ O (avg )
5.06
= 2.53 mA
2
2
______________________________________________________________________________________
(e) i D (avg ) =
=
2.4
(a) υ R (t ) = 15 sin ω t − 0.7 − 9 = 15 sin ω t − 9.7
(ω t )1 = sin −1 ⎛⎜ 9.7 ⎞⎟ = 40.29° ⇒ 0.2238π rad
⎝ 15 ⎠
(ω t )2 = 180 − 40.29 = 139.71° ⇒ 0.7762π rad
υ R (avg ) =
=
0.7762π
1
[15 sin x − 9.7]dx
2π 0.2238π
∫
0.7762π
0.7762π
⎤
1 ⎡
1
(− 15) cos x
− 9.7 x
⎢
⎥ = 2π [(− 15)(− 0.7628 − 0.7628 ) − 9.7(0.5523π )]
2π ⎣
0.2238π
0.2238π ⎦
υ R (avg ) = 0.9628 V
i D (avg ) = 0.8 =
0.9628
⇒ R = 1.20 Ω
R
(b)
⎛ 139.71 − 40.29 ⎞
%=⎜
⎟ × 100% = 27.6%
360
⎝
⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.5
(a) i ( peak ) =
υ R ( peak )
⇒R=
R
(b) υ R (t ) = 15 sin ω t − 9.7
15 − 9.7
= 4.417 Ω
1. 2
(ω t )1 = 0.2238π ; (ω t )2 = 0.7762π
υ R (avg ) =
1
0.7762π
∫ [15 sin x − 9.7]dx
π 0.2238π
Or from Problem 2.4, υ R (avg ) = 2(0.9628) = 1.9256 V
υ (avg ) 1.9256
=
= 0.436 A
i D (avg ) = R
R
4.417
(c)
⎛ 139.71 − 40.29 ⎞
%=⎜
⎟ × 100% = 27.6%
360
⎝
⎠
______________________________________________________________________________________
2.6
(a) υ S ( peak ) = 12 + 0.7 = 12.7 V
N 1 120 2
=
= 13.4
N2
12.7
12
= 60 Ω
0 .2
VM
12
=
⇒ 6667 μ F
C=
2 fRV r 2(60 )(60 )(0.25)
(b) R =
(c) PIV = 2υ S (max ) − Vγ = 2(12.7 ) − 0.7 = 24.7 V
______________________________________________________________________________________
2.7
v0 = vS − 2Vγ ⇒ vS ( max ) = v0 ( max ) + 2Vγ
a.
b.
v ( max ) = 25 V ⇒ vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V
For 0
N1 160
N
=
⇒ 1 = 6.06
N 2 26.4
N2
v ( max ) = 100 V ⇒ vS ( max ) = 101.4 V
For 0
N1
N
160
=
⇒ 1 = 1.58
N 2 101.4
N2
PIV = 2vS ( max ) − Vγ = 2 ( 26.4 ) − 0.7
From part (a)
PIV = 2 (101.4 ) − 0.7
or PIV = 52.1 V or, from part (b)
or PIV = 202.1 V
______________________________________________________________________________________
2.8
(a)
vs (max) = 12 + 2(0.7) = 13.4 V
13.4
vs ( rms ) =
⇒ vs (rms) = 9.48 V
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
Vr =
VM
VM
⇒C =
2 f RC
2 f Vr R
C=
12
⇒ C = 2222 μ F
2 ( 60 )( 0.3)(150 )
(c)
id , peak =
VM ⎡
2VM ⎤
⎢1 + π
⎥
R ⎢⎣
Vr ⎥⎦
2 (12 ) ⎤
12 ⎡
⎢1 + π
⎥
150 ⎢
0.3 ⎥
⎣
⎦
id , peak = 2.33 A
=
______________________________________________________________________________________
2.9
(a)
vS ( max ) = 12 + 0.7 = 12.7 V
vS ( rms ) =
Vr =
(b)
vS ( max )
2
⇒ vS ( rms ) = 8.98 V
VM
V
12
⇒C = M =
fRC
fRVr ( 60 )(150 )( 0.3)
iD , max =
or
C = 4444 μ F
VM ⎛
VM ⎞ 12 ⎛
12 ⎞
⎜ 1 + 4π
⎟
⎜⎜ 1 + 4π
⎟⎟ =
2Vr ⎠ 150 ⎜⎝
2 ( 0.3) ⎟⎠
R ⎝
iD , max = 4.58 A
or
(c) For the half-wave rectifier
______________________________________________________________________________________
2.10
(a) υ O ( peak ) = 10 − 0.7 = 9.3 V
VM
9.3
=
⇒ 620 μ F
fRV r (60 )(500 )(0.5)
(c) PIV = 10 + 9.3 = 19.3 V
______________________________________________________________________________________
(b) C =
2.11
(a) 10.3 ≤ υ O ≤ 12.3 V
(b) Vr =
VM
12.3
=
= 0.586 V
fRC (60 )(1000 ) 350 × 10 − 6
(
)
10.3
= 0.490 V
(60)(1000) 350 × 10 −6
So 0.490 ≤ V r ≤ 0.586 V
Vr =
(
)
VM
12.3
=
⇒ 513 μ F
fRV r (60)(1000 )(0.4)
______________________________________________________________________________________
(c) C =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.12
( )
(a) υ S ( peak ) = 8.5 2 = 12.02 V
VO max = 12.02 − 0.7 = 11.32 V
VM
11.32
=
= 0.03773 F
2 f RV r 2(60 )(10 )(0.25)
(c) PIV = 2υ S ( peak ) − Vγ = 2(12.02 ) − 0.7 = 23.34 V
(b) C =
______________________________________________________________________________________
2.13
(a)
vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V
vs ( rms ) =
C=
16.4
2
= 11.6 V
VM
15
=
= 2857 μ F
2 f RVr 2 ( 60 )(125 )( 0.35 )
(b)
______________________________________________________________________________________
2.14
______________________________________________________________________________________
2.15
(a) υ S = 12.8 V
N 1 120 2
=
= 13.3
N2
12.8
12
= 24 Ω
0 .5
V r = 3% ⇒ V r = (0.03)(12 ) = 0.36 V
(b) R =
C=
VM
12
=
= 0.0116 F
2 fRV r 2(60 )(24 )(0.36 )
(c) i D ( peak ) =
2V M ⎞⎟ 12 ⎛⎜
V M ⎛⎜
2(12 ) ⎞⎟
=
1+ π
1+ π
⎜
0.36 ⎟⎠
R ⎜⎝
V r ⎟⎠ 24 ⎝
i D ( peak ) = 13.3 A
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d) i D (avg ) =
1
π
2V r V M ⎛⎜ π
⋅
1+
VM
R ⎜⎝
2
2V M ⎞⎟ 1
=
V r ⎟⎠ π
2(0.36 ) ⎛ 12 ⎞⎛⎜ π
⎜ ⎟ 1+
12 ⎝ 24 ⎠⎜⎝
2
2(12 ) ⎞⎟
0.36 ⎟⎠
i D (avg ) = 0.539 A
(e) PIV = 12.8 + 12 = 24.8 V
______________________________________________________________________________________
2.16
(a) υ S = 9 + 2(0.8) = 10.6 V
N 1 120 2
=
= 16
N2
10.6
9
= 90 Ω
0. 1
VM
9
=
⇒ 4167 μ F
C=
2(60)(90)(0.2)
2 fRV r
(b) R =
(c) i D ( peak ) =
2V M ⎞⎟ 9 ⎛⎜
V M ⎛⎜
2(9 ) ⎞⎟
=
1+ π
1+ π
= 3.08 A
0.2 ⎟⎠
R ⎜⎝
V r ⎟⎠ 90 ⎜⎝
(d) i D (avg ) =
1
π
2V r V M ⎛⎜ π
⋅
1+
VM
R ⎜⎝
2
2V M ⎞⎟ 1
=
V r ⎟⎠ π
i D (avg ) = 0.1067 A
(e) PIV = υ S (max ) − Vγ = 10.6 − 0.8 = 9.8 V
2(0.2 ) ⎛ 9 ⎞⎛⎜ π
⎜ ⎟ 1+
9 ⎝ 90 ⎠⎜⎝
2
2(9 ) ⎞⎟
0.2 ⎟⎠
______________________________________________________________________________________
2.17
For vi > 0
Vγ = 0
Voltage across RL + R1 = vi
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ RL ⎞
1
⇒ v0 = ⎜
⎟ vi = vi
R
+
R
2
⎝ L
1 ⎠
Voltage Divider
______________________________________________________________________________________
2.18
For
vi > 0, (Vγ = 0 )
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 ⎠
v0 ( rms ) =
v0 ( max )
⇒ v0 ( rms ) = 3.04 V
2
b.
______________________________________________________________________________________
2.19
3 .9
= 0.975 mA
4
20 − 3.9
II =
= 1.342 mA
12
I Z = I I − I L = 1.342 − 0.975 = 0.367 mA
(a) I L =
PZ = I Z V Z = (0.367 )(3.9) = 1.43 mW
3.9
= 0.39 mA
10
I Z = 1.342 − 0.39 = 0.952 mA
(b) I L =
PZ = (0.952 )(3.9 ) = 3.71 mW
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.20
(a)
40 − 12
= 0.233 A
120
P = ( 0.233)(12 ) = 2.8 W
IZ =
(b)
So
IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A
12
0.21 =
⇒ RL = 57.1Ω
RL
P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W
(c)
______________________________________________________________________________________
2.21
(a) PZ = I Z V Z
4 = I z (15.4 ) ⇒ I Z (max ) = 259.74 mA
So 15 ≤ I z ≤ 259.74 mA
60 − 15.4
= 297.33 mA
0.15
So I L (max ) = 297.33 − 15 = 282.33 mA
(b) I I =
I L (min ) = 297.33 − 259.74 = 37.59 mA
15.4
= 54.55 Ω
0.28233
15.4
R L (max ) =
= 410 Ω
0.03759
So 54.55 ≤ R L ≤ 410 Ω
______________________________________________________________________________________
Then R L (min ) =
2.22
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 =
400
= 40 mA
10
10
RL
⇒ RL = 2 kΩ
(c)
For Ri = 175Ω
I I = 57.1 mA I L = 26.3 mA
I Z = 30.8 mA
I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA
RL =
10
⇒ RL = 585Ω
17.1
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.23
a.
From Eq. (2.30)
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. (2.28(b))
b.
Ri =
20 − 10
⇒ Ri = 8.08Ω
1187.5 + 50
PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W
PL = I L ( max )V0 = ( 0.5 )(10 ) ⇒ PL = 5 W
______________________________________________________________________________________
2.24
(a) I L = 0
10 − 5.6
⇒ 83.0 mA
50 + 3
V Z = 5.6 + (0.083)(3) = 5.85 V = V L
IZ =
PZ = I Z V Z = (0.083)(5.85) = 0.486 W
(b)
10 − V L V L − 5.6 V L
=
+
50
3
200
0.20 + 1.867 = V L (0.02 + 0.3333 + 0.005)
So V L = 5.769 V
5.769
= 28.84 mA
0.2
10 − 5.769
II =
= 84.62 mA
0.050
And I Z = I I − I L = 55.8 mA
Then I L =
PZ = (0.0558)(5.769 ) = 0.322 W
(c) I L = 0
12 − 5.6
⇒ 120.8 mA
50 + 3
V Z = V L = 5.6 + (0.1208)(3) = 5.962 V
IZ =
PZ = (0.1208)(5.962 ) = 0.72 W
(d)
12 − V L V L − 5.6 V L
=
+
50
3
200
0.24 + 1.867 = V L (0.02 + 0.333 + 0.005)
So V L = 5.88 V
5.88
12 − 5.88
= 29.4 mA; I I =
= 122.4 mA
0.20
0.05
I Z = 122.4 − 29.4 = 93 mA
Then I L =
PZ = (0.093)(5.88) = 0.547 W
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.25
(a) Set I Z = 10 mA; I L =
V L 7.5
=
= 7.5 mA
1
RL
I I = 10 + 7.5 = 17.5 mA
I I = 17.5 =
12 − 7.5
⇒ Ri = 257 Ω
Ri
(b) 7.5 = V ZO + (0.01)(12 ) ⇒ V ZO = 7.38 V
For V I = (1.1)(12 ) = 13.2 V
13.2 − V L V L − 7.38 V L
=
+
257
12
1000
0.05136 + 0.615 = V L (0.00389 + 0.0833 + 0.001) ⇒ V L = 7.556 V
For V I = (0.9 )(12 ) = 10.8 V
10.8 − V L V L − 7.38 V L
=
+
257
12
1000
0.04202 + 0.615 = V L (0.08819 ) ⇒ V L = 7.450 V
⎛ 7.556 − 7.450 ⎞
Then, Source Reg = ⎜
⎟ × 100% = 4.42%
⎝ 13.2 − 10.8 ⎠
(c) For R L = 1 k Ω , V L = 7.50 V
12 − 7.38
⇒ 17.17 mA
257 + 12
V L = 7.38 + (0.01717 )(12 ) = 7.586 V
For R L = ∞ , I Z =
⎛ 7.586 − 7.50 ⎞
Then , Load Reg = ⎜
⎟ × 100% = 1.15%
7.50
⎠
⎝
______________________________________________________________________________________
2.26
% Reg =
=
So
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
I Z ( max ) − I Z ( min ) = 0.1 A
Now
I L ( max ) =
Ri =
Now
280 =
or
VL ( max ) − VL ( min )
6
6
= 0.012 A, I L ( min ) =
= 0.006 A
500
1000
VPS ( min ) − VZ
I Z ( min ) + I L ( max )
15 − 6
⇒ I Z ( min ) = 0.020 A
I Z ( min ) + 0.012
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
I Z ( max ) = 0.1 + 0.02 = 0.12 A
VPS ( max ) − 6
Ri =
and
VPS ( max ) − VZ
I Z ( max ) + I L ( min )
⇒ VPS ( max ) = 41.3 V
0.12 + 0.006
or
______________________________________________________________________________________
280 =
2.27
Using Figure 2.19
VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V
a.
For
VPS ( min ) :
I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA
Ri =
b.
For
VPS ( min ) − VZ
II
VPS ( max )
=
15 − 10
⇒ Ri = 200Ω
25
⇒ 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
% Reg =
ΔV0
× 100% ⇒ % Reg = 3.5%
V0 ( nom )
c.
______________________________________________________________________________________
2.28
From Equation (2.28(a))
VPS ( min ) − VZ
24 − 16
Ri =
=
I Z ( min ) + I L ( max ) 40 + 400
Also Vr =
or
Ri = 18.2Ω
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 )
______________________________________________________________________________________
2.29
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
% Reg =
Vr =
ΔVL
0.4
=
⇒ % Reg = 5.0%
V0 ( nom )
8
VM
VM
⇒C =
2 fRC
2 fRVr
R = Ri + rz = 3 + 0.5 = 3.5Ω
C=
12
⇒ C = 0.0357 F
2 ( 60 )( 3.5 )( 0.8 )
Then
______________________________________________________________________________________
2.30
For −6.3 ≤ υ I ≤ 3 V, υ O = υ I
υ −3
For υ I > 3 V, I = I
and υ O = υ I − I (0.5)
1.5
⎛υ − 3⎞
υ O = υ I − (0.5)⎜⎜ I
⎟⎟ = 0.667υ I + 1.0
⎝ 1. 5 ⎠
υ I + 6.3
and υ O = υ I − I (0.5)
2.5
⎛ υ + 6.3 ⎞
υ O = υ I − (0.5)⎜⎜ I
⎟⎟ = 0.8υ I − 1.26
⎝ 2. 5 ⎠
______________________________________________________________________________________
For υ I < −6.3 V, I =
2.31
For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0
i = 0, vO = 0
For 0 ≤ vI ≤ 3, Zener not in breakdown, so 1
(a)
vI − 3
mA
20
1
⎛ v −3⎞
vo = ⎜ I
⎟ (10 ) = vI − 1.5
2
⎝ 20 ⎠
At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA
For vI > 3 i1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
For
For vI < 0, both diodes forward biased
0 − vI
−i1 =
.
10 At vI = −10 V , i1 = −1 mA
vI > 3, i1 =
vI − 3
.
20 At vI = 10 V , i1 = 0.35 mA
______________________________________________________________________________________
2.32
(a)
For
1
V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI
3
vI > 5.7 V
vI − (V1 + 0.7 ) 15 − V1 V1
+
= , v0 = V1 + 0.7
1
2
1
vI − v0 15 − ( v0 − 0.7 ) v0 − 0.7
+
=
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
iD = 0 for 0 ≤ vI ≤ 5.7
Then for vI > 5.7 V
v −v
iD = I O =
1
⎛v
⎞
vI − ⎜ I + 3.42 ⎟
0.6vI − 3.42
2.5
⎝
⎠
iD =
1
1
or
For vI = 15, iD = 5.58 mA
______________________________________________________________________________________
2.33
(a) (i) V B = 1.8 V
For υ I ≥ 1.1 V, υ O = υ I
For υ I ≤ 1.1 V, υ O = 1.1 V
(ii) V B = −1.8 V
For υ I ≥ −2.5 V, υ O = υ I
For υ I ≤ −2.5 V, υ O = −2.5 V
(b) (i) V B = 1.8 V
For υ I ≥ 2.5 V, υ O = 2.5 V
For υ I ≤ 2.5 V, υ O = υ I
(ii) V B = −1.8 V
For υ I ≥ −1.1 V, υ O = −1.1 V
For υ I ≤ −1.1 V, υ O = υ I
______________________________________________________________________________________
2.34
For
vI = 30 V, i =
30 − 10.7
= 0.175 A
100 + 10
v0 = i(10) + 10.7 = 12.5 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
______________________________________________________________________________________
2.35
(a) (i) V B = 5 V
For υ I ≥ 5.7 V, υ O = υ I − 5.7
For υ I ≤ 5.7 V, υ O = 0
(ii) V B = −5 V
For υ I ≥ −4.3 V, υ O = υ I + 4.3
For υ I ≤ −4.3 V, υ O = 0
(b) (i) V B = 5 V
For υ I ≥ 4.3 V, υ O = 0
For υ I ≤ 4.3 V, υ O = υ I − 4.3
(ii) V B = −5 V
For υ I ≥ −5.7 V, υ O = 0
For υ I ≤ −5.7 V, υ O = υ I + 5.7
______________________________________________________________________________________
2.36
a.
Vγ = 0
Vγ = 0.6
b.
Vγ = 0
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vγ = 0.6
______________________________________________________________________________________
2.37
______________________________________________________________________________________
2.38
One possible example is shown.
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.
______________________________________________________________________________________
2.39
(a) Square wave between +40 V and 0.
(b) Square wave between +35 V and −5 V.
(c) Square wave between +5 V and −35 V.
______________________________________________________________________________________
2.40
a.
For
Vγ = 0 ⇒ Vx = 2.7 V
V = 0.7 V ⇒ Vx = 2.0 V
b.
For γ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.41
Circuit similar to Figure 2.31(a) with V B = −10 V.
______________________________________________________________________________________
2.42
In steady-state, υ O = (10 sin ω t + 5) V
______________________________________________________________________________________
2.43
(i) V B = 5 V, In steady-state,
υ O = (10 sin ω t − 5) V
(ii) V B = −5 V, In steady-state,
υ O = (10 sin ω t − 15) V
______________________________________________________________________________________
2.44
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.
ID2 = 0
Same as (a)
10 =
(I )
2
( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA
V0 = I ( 9.5 ) ⇒ V0 = 9.16 V
I
⇒ I D1 = I D 2 = 0.482 mA
2
______________________________________________________________________________________
I D1 = I D 2 =
2.45
a.
I = I D1 = I D 2 = 0 V0 = 10
b.
10 = I ( 9.5 ) + 0.6 + I ( 0.5) ⇒ I = I D 2 = 0.94 mA
I D1 = 0
V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V
c.
10 = I ( 9.5 ) + 0.6 + I ( 0.5) + 5 ⇒ I = I D 2 = 0.44 mA I D1 = 0
V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
d.
10 = I ( 9.5 ) + 0.6 +
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 =
______________________________________________________________________________________
2.46
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
I D1 = I D 2 =
⇒ I D1 = I D 2 = 7.6 mA
0.5
I D 3 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA
I=
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.
V = 4.4 V
V1 = 5 V, V2 = 0 D1 off, D2, D3 on 0
10 − 4.4
I=
I = 0.589 mA
⇒
9.5
4.4 − 0.6
ID2 =
I D 2 = 7.6 mA
⇒
0.5
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
10 − 4.4
⇒
9.5
4.4 − 0.6 − 2
ID2 =
⇒
0.5
I=
V0 = 4.4 V
I = 0.589 mA
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
______________________________________________________________________________________
2.47
(a) V1 = 4.4 V, I D1 = 0.2 =
10 − 0.6 − 4.4
⇒ R1 = 25 k Ω
R1
I R 2 = 0.2 + 0.3 = 0.5 mA
V 2 = −0.6 V, I R 2 = 0.5 =
4.4 − (− 0.6 )
⇒ R 2 = 10 k Ω
R2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I R 3 = 0.5 + 0.5 = 1.0 mA
− 0.6 − (− 5)
= 4.4 k Ω
1. 0
(b) Assume all diodes conducting
10 − 0.6 − 4.4
V1 = 4.4 V, I D1 =
= 0.5 mA
10
4.4 − (− 0.6 )
V 2 = −0.6 V, I R 2 =
= 1.25 mA
4
Then I D 2 = 1.25 − 0.5 = 0.75 mA
R3 =
− 0.6 − (− 5)
= 2 mA
2.2
Then I D 3 = 2 − 1.25 = 0.75 mA
I R3 =
(c) Diode D 2 cutoff ⇒ I D 2 = 0
V 2 = −0.6 V, I D1 =
10 − 0.6 − (− 0.6 ) 10
=
= 1.11 mA
R1 + R 2
9
V1 = 10 − 0.6 − (1.11)(3) = 6.07 V
− 0.6 − (− 5)
= 1.76 mA
2. 5
Then I D 3 = 1.76 − 1.11 = 0.65 mA
I R3 =
(d) Diode D3 cutoff ⇒ I D 3 = 0
10 − 0.6 − 4.4
= 0.833 mA
6
4.4 − (− 5) 9.4
I R2 =
=
= 1.044 mA
R 2 + R3
9
V1 = 4.4 V, I D1 =
V 2 = (1.044 )(6 ) − 5 = 1.27 V
Then I D 2 = 1.044 − 0.833 = 0.211 mA
______________________________________________________________________________________
2.48
(a) I D1 = I D 2 = 2.5 mA
0.7 − 0.7 − (− 2)
⇒ R = 0.8 k Ω
R
(b) I D1 = 0.2 I D 2 , I D1 + I D 2 = 5
I D 2 = 2.5 =
0.2 I D 2 + I D 2 = 5 ⇒ I D 2 = 4.167 mA
2
⇒ R = 0.48 k Ω
R
(c) I D1 = 5I D 2 , I D1 + I D 2 = 5
I D 2 = 4.167 =
5 I D 2 + I D 2 = 5 ⇒ I D 2 = 0.833 mA
2
⇒ R = 2.4 k Ω
R
______________________________________________________________________________________
I D 2 = 0.833 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.49
(a) D1 and D 2 on
5 − 0.7 − V A V A − 0.7 − (− 5)
5+
=
2
1. 1
⎛1 1 ⎞
5 + 2.15 − 3.909 = V A ⎜ +
⎟ ⇒ V A = 2.30 V
⎝ 2 1 .1 ⎠
5 − 0.7 − 2.3
= 1.0 mA
2
2.3 − 0.7 − (− 5)
I D2 =
= 6.0 mA
1.1
(b) D1 cutoff, I D1 = 0
Then I D1 =
I D 2 = 5 mA, V A = 0.7 + (5)(2.5) − 5 = 8.2 V
5 − 0.7 − 0
= 2.15 mA
2
Then I D 2 = 5 + 2.15 = 7.15 mA
(c) V A = 0 , I D1 =
0 − 0.7 − (− 5)
⇒ R 2 = 0.60 k Ω
R2
______________________________________________________________________________________
I D 2 = 7.15 =
2.50
(a) (i) υ I = 5 V, D1 and D 2 on
5 − (υ O + 0.6 ) 5 − υ O υ O υ O − 0.6
+
=
+
5
5
0. 5
0. 5
0.88 + 1.0 + 1.2 = υ O (0.20 + 0.20 + 2.0 + 2.0) ⇒ υ O = 0.7 V
(ii) υ I = −5 V
⎛ 0.5 ⎞
⎟υ I = −0.455 V
⎝ 0.5 + 5 ⎠
υO = ⎜
(b) (i) υ I = 5 V, υ O = 4.4 V
(ii) υ I = −5 V, υ O = −0.6 V
______________________________________________________________________________________
2.51
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For vI > 0. when D1 and D4 turn off
10 − 0.7
I=
= 0.465 mA
20
v0 = I (10 kΩ ) = 4.65 V
v0 = vI for − 4.65 ≤ vI ≤ 4.65
______________________________________________________________________________________
2.52
(a) All diodes on
15 − V A V A − 0.7 V A − 0.7 − (− 5) V A − 0.7 − (− 10 )
=
+
+
6.15
2
14
24
2.439 + 0.35 − 0.307 − 0.3875 = V A (0.1626 + 0.50 + 0.0714 + 0.0417 ) ⇒ V A = 2.70 V
2.70 − 0.7
= 1.0 mA
2
2.70 − 0.7 − (− 5)
I D2 =
= 0.50 mA
14
2.70 − 0.7 − (− 10 )
I D3 =
= 0.50 mA
24
(b) D1 cutoff, I D1 = 0
Then I D1 =
15 − V A V A − 0.7 − (− 5) V A − 0.7 − (− 10 )
=
+
6.15
3.3
5. 2
2.439 − 1.303 − 1.788 = V A (0.1626 + 0.303 + 0.1923) ⇒ V A = −0.991 V
− 0.991 − 0.7 − (− 5)
= 1.0 mA
3. 3
− 0.991 − 0.7 − (− 10 )
I D3 =
= 1.60 mA
5. 2
(c) D1 and D 2 cutoff, I D1 = I D 2 = 0
Then I D 2 =
I D3 =
15 − 0.7 − (− 10 )
24.3
=
= 3.25 mA
R1 + R 4
6.15 + 1.32
V A = 15 − (3.25)(6.15) = −5 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.53
a.
R1 = 5 kΩ, R2 = 10 kΩ
⇒ V0 = 0
D1 and D2 on
10 − 0.7 0 − ( −10 )
−
= 1.86 − 1.0
5
10
I D1 = 0.86 mA
I D1 =
b.
R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0
I=
10 − 0.7 − ( −10 )
= 1.287
15
V0 = IR2 − 10 ⇒ V0 = −3.57 V
______________________________________________________________________________________
2.54
If both diodes on
(a)
VA = −0.7 V, VO = −1.4 V
I R1 =
10 − ( −0.7 )
= 1.07 mA
10
−1.4 − ( −15 )
= 2.72 mA
IR2 =
5
I R1 + I D1 = I R 2 ⇒ I D1 = 2.72 − 1.07
I D1 = 1.65 mA
(b) 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
VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off ,
I D1 = 0
______________________________________________________________________________________
2.55
(a)
D1 on, D2 off
10 − 0.7
I D1 =
= 0.93 mA
10
VO = −15 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
D1 on, D2 off
10 − 0.7
I D1 =
= 1.86 mA
5
VO = −15 V
______________________________________________________________________________________
(b)
2.56
15 − (V0 + 0.7 )
V + 0.7 V0
= 0
+
10
20
20
15 0.7 0.7
1 ⎞
⎛ 4.0 ⎞
⎛ 1 1
−
−
= V0 ⎜ +
+ ⎟ = V0 ⎜
⎟
10 10 20
⎝ 20 ⎠
⎝ 10 20 20 ⎠
V0 = 6.975 V
V0
⇒ I D = 0.349 mA
20
______________________________________________________________________________________
ID =
2.57
(a) Diode is cutoff, I D = 0 , V D = 0
V A = VB = 3 V
(b) Diode is conducting, V D = 0.7 V
5 − V B V B V B − 0.7 V B − 0.7 − 2
=
+
+
10
10
10
10
0.50 + 0.07 + 0.27 = V B (0.10 + 0.10 + 0.10 + 0.10 ) ⇒ V B = 2.1 V and V A = 1.4 V
5 − VB VB
=
+ ID
10
10
5 − 2.1 2.1
So I D =
−
= 0.08 mA
10
10
(c) Diode is cutoff, I D = 0
1
(5) = 2.5 V, V B = 1 (4) = 2.0 V
2
2
V D = 2 − 2.5 = −0.5 V
VA =
(d) Diode is conducting, V D = 0.7 V
8 − V B V B V B − 0. 7 V B − 0 . 7 − 2
=
+
+
10
10
10
10
0.80 + 0.07 + 0.27 = V B (0.40 ) ⇒ V B = 2.85 V and V A = 2.15 V
8 −VB VB
1
Then I D =
−
= [8 − 2(2.85)] = 0.23 mA
10
10 10
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.58
vI = 0, D off, D on
1
I
vo
2
10 − 2.5
0.5 mA
=
15
= 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V
For vI > 7.5 V , Both D1 and D2 on
vI − vo vo − 2.5 vo − 10
=
+
vI = vo ( 5.5 ) − 33.75
15
10
5
or
When vo = 10 V, D2 turns off
vI = (10 )( 5.5 ) − 33.75 = 21.25 V
For vI > 21.25 V, vo = 10 V
______________________________________________________________________________________
2.59
(a) For υ I = 0.5 V, I D1 = I D 2 = I D 3 = 0 , υ O = 0.5 V
(b) For υ I = 1.5 V, D1 on; I D 2 = I D 3 = 0
1.5 − 0.7
= 0.0667 mA
4+8
υ O = 0.7 + (0.0667 )(8) = 1.23 V
I D1 =
(c) For υ I = 3 V, D1 and D 2 conducting, I D 3 = 0
3 − υ O υ O − 0. 7 υ O − 1. 7
=
+
4
8
6
0.75 + 0.0875 + 0.2833 = υ O (0.25 + 0.125 + 0.1667 ) ⇒ υ O = 2.069 V
2.069 − 0.7
= 0.171 mA
8
2.069 − 1.7
I D2 =
= 0.0615 mA
6
(d) For υ I = 5 V, all diodes conducting
Then I D1 =
5 − υ O υ O − 0.7 υ O − 1.7 υ O − 2.7
=
+
+
4
8
6
4
1.25 + 0.0875 + 0.2833 + 0.675 = υ O (0.25 + 0.125 + 0.1667 + 0.25)
So υ O = 2.90 V
2.90 − 0.7
= 0.275 mA
8
2.90 − 1.7
I D2 =
= 0.20 mA
6
2.90 − 2.7
I D3 =
= 0.05 mA
4
______________________________________________________________________________________
Then I D1 =
2.60
(a) I D 2 = 0 for υ I < 4.5 V
I D 2 = 100 mA for υ I > 4.5 V
(b) I D 2 = 0 for υ I < 9 V
I D 2 = 100 mA for υ I > 9 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.61
a.
b.
c.
V01 = V02 = 0
V01 = 4.4 V, V02 = 3.8 V
V01 = 4.4 V, V02 = 3.8 V
Logic “1” level degrades as it goes through additional logic gates.
______________________________________________________________________________________
2.62
a.
b.
c.
V01 = V02 = 5 V
V01 = 0.6 V, V02 = 1.2 V
V01 = 0.6 V, V02 = 1.2 V
Logic “0” signal degrades as it goes through additional logic gates.
______________________________________________________________________________________
2.63
(V1 AND V2 ) OR (V3 AND V4 )
______________________________________________________________________________________
2.64
10 − 1.5 − 0.2
= 12 mA = 0.012
R + 10
8.3
R + 10 =
= 691.7Ω
0.012
R = 681.7Ω
______________________________________________________________________________________
I=
2.65
10 − 1.7 − VI
=8
0.75
VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V
I=
______________________________________________________________________________________
2.66
5 − 1.7
= 0.220 k Ω
15
r f = 20 Ω ⇒ R = 200 Ω
R + rf =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.67
VR = 1 V, I = 0.8 mA
VPS = 1 + ( 0.8 )( 2 )
VPS = 2.6 V
______________________________________________________________________________________
2.68
I Ph = η eΦA
0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A
A = 3.75 × 10−2 cm 2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 3
3.1
Kn =
k n′ W 120 ⎛ 10 ⎞
2
⋅ =
⎜
⎟ ⇒ 0.75 mA/V
2 L
2 ⎝ 0. 8 ⎠
(a) (i) I D = 0
[
]
(iii) I = (0.75)[2(2 − 0.4)(0.1) − (0.1) ] = 0.2325 mA
(iv) I = (0.75)[2(3 − 0.4)(0.1) − (0.1) ] = 0.3825 mA
(ii) I D = (0.75) 2(1 − 0.4)(0.1) − (0.1) ⇒ 82.5 μ A
2
2
D
2
D
(b) (i) I D = 0
(ii) I D = (0.75)(1 − 0.4) = 0.27 mA
2
(iii) I D = (0.75)(2 − 0.4) = 1.92 mA
2
(iv) I D = (0.75)(3 − 0.4) = 5.07 mA
______________________________________________________________________________________
2
3.2
[
0.5 = K [2(0.6)V − V ]
1.0 = K [2(1.0)V − V ]
2
I D = K n 2(VGS − VTN )V DS − V DS
n
DS
2
DS
n
DS
2
DS
]
Take ratio
2
1.2V DS − V DS
2
2
⇒ V DS − 0.5V DS
= 1.2V DS − V DS
0.5 =
2
2V DS − V DS
or 1 − 0.5V DS = 1.2 − V DS
which yields V DS = 0.4 V
[
]
Then 0.5 = K n (1.2)(0.4) − (0.4) ⇒ K n = 1.56 mA/V 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
0.15 = K n ( 3 − 1.5 ) = 2.25 K n
K n = 0.0666
0.39 = K n ( 4 − 1.5 ) = 6.25 K n
K n = 0.0624
0.77 = K n ( 5 − 1.5 ) = 12.25 K n
K n = 0.0629
2
2
2
From last three, K n (Avg) = 0.0640 mA/V
iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V
2
(c) iD (sat) = 0.0640(4.5 − 1.5) ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V
2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.4
a.
i.
VGS = 0
VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V
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
ii.
iii.
b.
⇒ I D = 6.88 mA
VDS = 5 V Same as (ii)
VGS = 2 V
VDS ( sat ) = 2 − ( −2.5) = 4.5 V
VDS = 0.5 V ⇒ Nonsaturation
i.
ii.
I D = (1.1) ⎡⎣ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤⎦ ⇒ I D = 4.68 mA
VDS = 2.5 V ⇒ Nonsaturation
I D = (1.1) ⎡⎣ 2(2 − ( −2.5))(2.5) − (2.5) 2 ⎤⎦ ⇒ I D = 17.9 mA
VDS = 5 V ⇒ Saturation
I D = (1.1) ( 2 − ( −2.5) ) ⇒ I D = 22.3 mA
iii.
______________________________________________________________________________________
2
3.5
(a) V DS (sat ) = VGS − VTN = 2.2 − 0.4 = 1.8 V
2.2 = V DS > V DS (sat ) = 1.8 ⇒ Saturation
(b) V DS (sat ) = VGS − VTN = 1 − 0.4 = 0.6 V
V DS = −0.6 − (− 1) = 0.4 V < V DS (sat ) = 0.6 V ⇒ Nonsaturation
(c) VGS = 1 − 1 = 0 ⇒ Cutoff
______________________________________________________________________________________
3.6
(a) V SG = 2.2 − 2.2 = 0 ⇒ Cutoff
(b) V SG = 2 V, V SD = 2 − (− 1) = 3 V
V SD (sat ) = V SG + VTP = 2 + (− 0.4) = 1.6 V
So V SD = 3 > V SD (sat ) = 1.6 ⇒ Saturation
(c) V SG = 2 V, V SD = 2 − 1 = 1 V
V SD (sat ) = V SG + VTP = 2 + (− 0.4) = 1.6 V
So V SD = 1 < V SD (sat ) = 1.6 ⇒ Nonsaturation
______________________________________________________________________________________
3.7
ID =
k n′ ⎛ W ⎞
2
⎜ ⎟(VGS − VTN )
2 ⎝L⎠
0.12 ⎛ W ⎞
⎛W ⎞
2
⎜ ⎟[0 − (− 1.2 )] ⇒ ⎜ ⎟ = 5.79
2 ⎝L⎠
⎝L⎠
______________________________________________________________________________________
0. 5 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.8
kn′ = μ n Cox =
μ n ∈ox
=
tox
( 600 )( 3.9 ) (8.85 ×10−14 )
tox
(a)
500 A
k n′ = 41.4 μ A/V
(b)
250
kn′ = 82.8 μ A/V 2
(c)
100
k n′ = 207 μ A/V 2
(d)
50
k n′ = 414 μ A/V 2
=
2.071× 10 −10
tox
2
k n′ = 828 μ A/V 2
(e) 25
______________________________________________________________________________________
3.9
(a) K n =
(
)
(
)
Wμ n ∈ox
20 × 10 −4 (650 )(3.9 ) 8.85 × 10 −14
=
= 1.40 mA/V 2
2 L t ox
2 0.8 × 10 − 4 200 × 10 −8
(
)(
)
(b) I D = K n (VGS − VTN ) = (1.40)(2 − 0.4)
Or I D = 3.58 mA
(c) V DS (sat ) = VGS − VTN = 2 − 0.4 = 1.6 V
______________________________________________________________________________________
2
2
3.10
(a) I D =
k n′ ⎛ W ⎞
2
⎜ ⎟(VGS − VTN )
2 ⎝L⎠
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.6 = ⎜
⎟⎜ ⎟(1.4 − 0.8) ⇒ ⎜ ⎟ = 27.8
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
Or W = (27.8)(0.8) = 22.2 μ m
[
]
⎛ 0.12 ⎞
2
(b) I D = ⎜
⎟(27.8) 2(1.4 − 0.8)(0.4 ) − (0.4 ) = 0.534 mA
2
⎝
⎠
(c) V DS (sat ) = VGS − VTN = 1.4 − 0.8 = 0.6 V
______________________________________________________________________________________
3.11
k n′ = μ n C ox =
μ n ∈ox
t ox
=
(600 )(3.9)(8.85 ×10 −14 )
200 × 10 −8
k n′ = 0.1035 mA/V 2
ID =
k n′ ⎛ W ⎞
2
⎜ ⎟(VGS − VTN )
2 ⎝L⎠
⎛W ⎞
⎛ 0.1035 ⎞⎛ W ⎞
2
1. 2 = ⎜
⎟⎜ ⎟(3 − 0.6 ) ⇒ ⎜ ⎟ = 4.026
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
Then W = (4.026)(0.8) = 3.22 μ m
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.12
I D = WC ox (VGS − VTN )υ sat
C ox =
(
)
)(
)
∈ox (3.9 ) 8.85 × 10 −14
=
= 1.726 × 10 − 7 F/cm
−8
t ox
200 × 10
(
(
)
I D = 3.22 ×10 −4 1.726 ×10 −7 (3 − 0.6) 2 ×10 7
I D = 2.67 mA
______________________________________________________________________________________
3.13
ID =
k ′p ⎛ W ⎞
2
⎜ ⎟(V SG + VTP )
2 ⎝L⎠
⎛ 0.05 ⎞⎛ W ⎞
2
0.225 = ⎜
⎟⎜ ⎟(2 + VTP )
2
L
⎝
⎠⎝ ⎠
⎛ 0.05 ⎞⎛ W ⎞
2
0.65 = ⎜
⎟⎜ ⎟(3 + VTP )
2
L
⎝
⎠⎝ ⎠
Then
3 + VTP
0.65
=
= 1.70 ⇒ VTP = −0.571 V
0.225 2 + VTP
⎛ 0.05 ⎞⎛ W ⎞
⎛W ⎞
2
And 0.225 = ⎜
⎟⎜ ⎟(2 − 0.571) ⇒ ⎜ ⎟ = 4.41
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
______________________________________________________________________________________
3.14
VS = 5 V, VG = 0 ⇒ VSG = 5 V
VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V
VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation
I D = 2 ( 5 − 0.5) ⇒ I D = 40.5 mA
2
a.
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
2
I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA
⎣
⎦
VD = 5 V ⇒ VSD = 0 ⇒ I D = 0
d.
______________________________________________________________________________________
3.15
(a)
(b)
Enhancement-mode
From Graph VTP = + 0.5 V
2
0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p =
0.20
1.25 = k p ( 3 − 0.5 ) = 6.25 K p
0.20
2.45 = k p ( 4 − 0.5 ) = 12.25 K p
0.20
2
2
4.10 = k p ( 5 − 0.5 ) = 20.25 K p
2
0.202
Avg K p = 0.20 mA/V 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA
2
(c) iD (sat) = 0.20 (4.5 − 0.5) = 3.2 mA
______________________________________________________________________________________
3.16
(a)
(b)
(c)
VSD ( sat ) = VSG + VTP
VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V
VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V
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 ⎠
ID =
(a)
(b)
2
⎛ 0.040 ⎞
ID = ⎜
⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA
2
⎝
⎠
2
⎛ 0.040 ⎞
ID = ⎜
⎟ ( 6 )( 3) ⇒ I D = 1.08 mA
2
⎝
⎠
(c)
______________________________________________________________________________________
3.17
k ′p ⎛ W ⎞ 50 ⎛ 12 ⎞
2
⎜
⎟ ⇒ K p = 0.375 mA/V
⎜ ⎟=
2 ⎝ L ⎠ 2 ⎝ 0. 8 ⎠
(a) Nonsaturation
2
I D = (0.375) 2(2 − 0.5)(0.2) − (0.2) = 0.21 mA
(b) Nonsaturation
2
I D = (0.375) 2(2 − 0.5)(0.8) − (0.8) = 0.66 mA
(c) Nonsaturation
2
I D = (0.375) 2(2 − 0.5)(1.2) − (1.2) = 0.81 mA
(d) Saturation
2
I D = (0.375)(2 − 0.5) = 0.844 mA
(e) Saturation
2
I D = (0.375)(2 − 0.5) = 0.844 mA
______________________________________________________________________________________
Kp =
[
]
[
]
[
]
3.18
k ′p = μ p Cox =
μ p ∈ox
t0 x
=
( 250 )( 3.9 ) (8.85 × 10−14 )
(a)
tox = 500Å ⇒ k ′p = 17.3 μ A/V
(b)
250Å ⇒ k ′p = 34.5 μ A/V 2
(c)
100Å ⇒ k ′p = 86.3 μ A/V 2
(d)
50Å ⇒ k ′p = 173 μ A/V 2
(e)
25Å ⇒ k ′p = 345 μ A/V 2
t0 x
2
=
8.629 × 10 −11
t0 x
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
3.19
Cox =
∈ox ( 3.9 ) ( 8.85 × 10
=
t0 x
500 × 10−8
−14
) = 6.90 ×10 F/cm
−8
2
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:
ID =
k ′p ⎛ W ⎞
2
⎜ ⎟ (VSG + VTP )
2 ⎝ L ⎠p
2
⎛ 0.0259 ⎞ ⎛ W ⎞
⎛W ⎞
0.8 = ⎜
⎟ ⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19
L
2
⎝
⎠⎝ ⎠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
______________________________________________________________________________________
3.20
VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA
2
r0 =
1
1
=
⇒ r0 = 781 kΩ
λ I D ( 0.01)( 0.128 )
VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA
1
r0 =
⇒ r = 63.7 kΩ
( 0.01)(1.57 ) 0
2
VA =
1
=
1
⇒ VA = 100 V
λ ( 0.01)
______________________________________________________________________________________
3.21
ID =
ro =
k n′ ⎛ W ⎞
⎛ 0.12 ⎞
2
2
⎟(4 )(2 − 0.5) = 0.54 mA
⎜ ⎟(VGS − VTN ) = ⎜
2 ⎝L⎠
⎝ 2 ⎠
1
λ I DQ
Then V A =
1
⇒λ =
1
1
=
= 0.00926 V −1
ro I DQ
200 × 10 3 0.54 × 10 − 3
(
)(
)
= 108 V
λ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.22
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
______________________________________________________________________________________
3.23
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)
______________________________________________________________________________________
3.24
Ε ox = 6×10 6 V/cm
(
)(
)
(a) (i) VG = Ε ox t ox = 6 ×10 6 120 ×10 −8 = 7.2 V
(ii) VG =
7.2
= 2.4 V
3
(
)(
)
(b) (i) VG = 6 ×10 6 200 ×10 −8 = 12 V
12
= 4V
3
______________________________________________________________________________________
(ii) VG =
3.25
Want
(3)(24) = Ε ox t ox = (6 ×10 6 )t ox
t0 x = 1.2 × 10−5 cm = 1200 Angstroms
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.26
⎛ R2 ⎞
⎛ 18 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) = 3.6 V
⎝ 18 + 32 ⎠
⎝ R1 + R2 ⎠
Assume transistor biased in saturation region
V
V − VGS
2
ID = S = G
= K n (VGS − VTN )
RS
RS
3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 )
2
= VGS2 − 1.6VGS + 0.64
VGS2 − 0.6VGS − 2.96 = 0
0.6 ±
( 0.6 ) + 4 ( 2.96 )
2
⇒ VGS = 2.046 V
2
V − VGS 3.6 − 2.046
ID = G
=
⇒ I D = 0.777 mA
2
RS
VGS =
VDS = VDD − I D ( RD + RS )
= 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V
VDS > VDS ( sat )
______________________________________________________________________________________
3.27
(a)
VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V
If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56
2
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
− 2.6VDS + 4 = 0
0.25VDS
VDS =
ID =
2.6 ± 6.76 − 4
= 1.88 V
2 ( 0.25 )
4 − 1.88
= 2.12 mA
1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
Non-Sat region
2
⎤⎦ + VDS
5 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT ) VDS − VDS
2
⎤⎦ + VDS
5 = ( 0.25 )( 3) ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS
2
5 = 7.3VDS − 0.75VDS
0.75 VDS2 − 7.3VDS + 5 = 0
VDS =
7.3 ± 53.29 − 15
2 ( 0.75 )
VDS = 0.741 V
5 − 0.741
= 1.42 mA
3
______________________________________________________________________________________
ID =
3.28
⎛ 0.12 ⎞
2
0.8 = ⎜
⎟(80 )(VGS − 0.4 ) ⇒ VGS = 0.808 V
2
⎝
⎠
1
VGS =
⋅ Rin ⋅ V DD
R1
0.80825 =
1
(200)(1.8) ⇒ R1 = 445 k Ω
R1
R1 R2 = Rin = 200 k Ω ⇒ R 2 = 363 k Ω
______________________________________________________________________________________
3.29
(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
Biased in Non-Sat Region.
×
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
0.24 VSD
− 2.296 VSD + 3.5 = 0
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
(b)
VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V
I = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0
If Sat Region D
Non-Sat Region.
2
⎤⎦
5 = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD
2
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 =
7.72 ± 59.598 − 16
use − sign VSD = 0.698 V
2 ( 0.8 )
5 − 0.698
⇒ I D = 1.08 mA
4
______________________________________________________________________________________
ID =
3.30
⎛ 22 ⎞
VG = ⎜
⎟(6 ) − 3 = 1.40 V
⎝ 22 + 8 ⎠
3 = K p R S (V SG + VTP ) + V SG + VG
2
(
)
2
3 = (0.5)(0.5) V SG
− 1.6V SG + 0.64 + V SG + 1.40
0.25V
2
SG
+ 0.6V SG − 1.44 = 0 ⇒ V SG = 1.483 V
I D = (0.5)(1.483 − 0.8) = 0.2332 mA
V SD = 6 − (0.2332 )(0.5 + 5) = 4.72 V
______________________________________________________________________________________
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.31
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 )
______________________________________________________________________________________
3.32
ID =
V DD − V DSQ − V RS
RD
=
3.3 − 1.6 − 0.8
= 0.18 mA
5
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.18 = ⎜
⎟⎜ ⎟(0.8 − 0.4 ) ⇒ ⎜ ⎟ = 18.75
2
L
⎝
⎠⎝ ⎠
⎝L⎠
0.8
RS =
= 4.44 k Ω
0.18
⎛ R2 ⎞
⎟⎟(3.3)
VG = 0.8 + 0.8 = 1.6 = ⎜⎜
⎝ R1 + R 2 ⎠
Now I R = (0.05)(0.18) ⇒ 9 μ A
3.3
= 367 k Ω
0.009
⎛ R ⎞
Then 1.6 = ⎜⎜ 2 ⎟⎟(3.3) ⇒ R 2 = 178 k Ω and R1 = 189 k Ω
⎝ 367 ⎠
______________________________________________________________________________________
So R1 + R 2 =
3.33
0.2
= 0.2 mA
1
V DS = 1.8 − (0.2 )(4 + 1) = 0.8 V
ID =
Now V DS (sat ) = 0.8 − 0.4 = 0.4 V
VDS (sat ) = VGS − VTN ⇒ 0.4 = VGS − 0.4 ⇒ VGS = 0.8 V
ID =
k n′ ⎛ W ⎞
2
⎜ ⎟(VGS − VTN )
2 ⎝L⎠
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.2 = ⎜
⎟⎜ ⎟(0.8 − 0.4 ) ⇒ ⎜ ⎟ = 20.8
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
Now VG = VGS + I D R S = 0.8 + (0.2 )(1) = 1.0 V
VG = 1 =
1
1
⋅ Rin ⋅ V DD =
(200)(1.8) ⇒ R1 = 360 k Ω
R1
R1
R1 R2 = Rin = 200 k Ω ⇒ R 2 = 450 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
3.34
⎛ 0.12 ⎞
2
(a) 0.35 = ⎜
⎟(50 )(VGS − 0.4 ) ⇒ VGS = 0.742 V
⎝ 2 ⎠
(b) V DS = 1.8 − (0.35)(2 ) = 1.1 V
V DS (sat ) = VGS − VTN = 0.742 − 0.4 = 0.342 V
V DS > V DS (sat ) ⇒ Saturation
_____________________________________________________________________________________
3.35
⎛ 6 ⎞
VG = ⎜
⎟(10 ) − 5 = −2 V
⎝ 6 + 14 ⎠
⎛ k ′ ⎞⎛ W ⎞
2
VG = VGS + I D R S − 5 = VGS + ⎜⎜ n ⎟⎟⎜ ⎟ R S (VGS − VTN ) − 5
⎝ 2 ⎠⎝ L ⎠
⎛ 0.12 ⎞
2
5 − 2 = VGS + ⎜
⎟(25)(0.5) VGS − 0.8VGS + 0.16
2
⎝
⎠
(
)
2
+ 0.4VGS − 2.88 = 0 ⇒ VGS = 1.71 V
Or 0.75VGS
⎛ 0.12 ⎞
2
ID = ⎜
⎟(25)(1.71 − 0.4 ) = 2.58 mA
2
⎝
⎠
V DS = 10 − (2.58)(1.2 + 0.5) = 5.62 V
______________________________________________________________________________________
3.36
⎛W ⎞
Let ⎜ ⎟ = 20 for example, then
⎝L⎠
⎛ 0.05 ⎞
2
0.20 = ⎜
⎟(20 )(V SG − 0.6 ) ⇒ V SG = 1.232 V
2
⎝
⎠
Then V RS = 1.232 = (0.2)R S ⇒ R S = 6.16 k Ω
6 − 1.232 − 3
= 8.84 k Ω
0.2
VG = 3 − 1.232 − 1.232 = 0.536 V
RD =
IR =
6
= (0.1)(0.2 ) = 0.02 mA, ⇒ R1 + R 2 = 300 k Ω
R1 + R 2
⎛ R2 ⎞
⎛ R ⎞
⎟⎟(6 ) − 3 = ⎜⎜ 2 ⎟⎟(6) − 3
VG = 0.536 = ⎜⎜
⎝ 300 ⎠
⎝ R1 + R 2 ⎠
Or R 2 = 176.8 k Ω and R1 = 123.2 k Ω
______________________________________________________________________________________
3.37
I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V
2
V = 5 − ( −1.516 ) =⇒ VDS = 6.516 V
(a) (i) DS
I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V
2
(ii)
VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V = VDS = 1.516 V
(i) Same as (a) GS
V = VDS = 2.61 V
(ii) GS
______________________________________________________________________________________
(b)
3.38
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 =
______________________________________________________________________________________
3.39
(a)
RD =
5 −1
⇒ RD = 8 K
0.5
I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V
2
RS =
(b)
−2.81 − ( −5 )
⇒ RS = 4.38 K
0.5
Let RD = 8.2 K, RS = 4.3 K
Now
−VGS − ( −5 )
= I D = 0.25 (VGS − 1.4 )
2
4.3
5 − VGS = 1.075 (VGS2 − 2.8 VGS + 1.96 )
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(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 )
I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA
______________________________________________________________________________________
2
3.40
9 − 2 .5
⇒ R = 65 k Ω
R
⎛ k ′p ⎞⎛ W ⎞
⎟⎜ ⎟(V SG + VTP )2
I D = ⎜⎜
⎟⎝ L ⎠
2
⎝
⎠
I D = 0.10 =
Now V SG = V SD = 2.5 V
⎛ 0.05 ⎞⎛ W ⎞
⎛W ⎞
2
0.10 = ⎜
⎟⎜ ⎟(2.5 − 0.7 ) ⇒ ⎜ ⎟ = 1.235
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
And W = (1.235)(0.8) = 0.988 μ m
______________________________________________________________________________________
3.41
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 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.42
(a)
RD =
VD − ( −5 )
I DQ
=
5−2
⇒ RD = 12 K
0.25
2
⎛ W ⎞ ⎛ k ′p ⎞
I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP )
L
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
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
VSG =
6.46 ± 41.73 + 6.49
= 2.155 V
2 ( 3.11)
5 − 2.155
= 0.252 mA
11.3
VSD = 10 − ( 0.252 )(12 + 11.3) = 4.13 V
ID =
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
VSD = 10 − ( 0.248 )(12 + 11.3) = 4.22 V
______________________________________________________________________________________
ID =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.43
ID =
−VSD − ( −10 )
RD
⇒5=
−6 + 10
⇒ RD = 0.8 kΩ
RD
I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 )
2
2
5
+ 1.75 = 3.04 V ⇒ VG = −3.04
3
VSG =
⎛ 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
______________________________________________________________________________________
3.44
(a) Both M 1 and M 2 in saturation
I D1 = I D 2
k n′
k′
(4)(υ I − 0.4)2 = n (1)[0 − (− 0.6)]2 ⇒ υ I = 0.7 V
2
2
(b) (i) υ I = 0.6 V; M 1 in saturation, M 2 in nonsaturation
I D1 = I D 2
k n′
k′
(4)(0.6 − 0.4)2 = n (1) 2(0 − (− 0.6))(5 − υ O ) − (5 − υ O )2
2
2
2
We find (5 − υ O ) − 1.2(5 − υ O ) + 0.16 = 0 ⇒ (5 − υ O ) = 0.153
[
]
So υ O = 4.85 V
(ii) υ I = 1.5 V; M 1 in nonsaturation, M 2 in saturation
I D1 = I D 2
k n′
k′
(4) 2(1.5 − 0.4)υ O − υ O2 = n (1)(0 − (− 0.6))2
2
2
2
We find υ O − 2.2υ O + 0.09 = 0 ⇒ υ O = 0.0417 V
______________________________________________________________________________________
[
]
3.45
M 1 in nonsaturation, M 2 in saturation
I D1 = I D 2
[
]
k n′ ⎛ W ⎞
k n′
2
(1)[0 − (− 0.6)]2
⎜ ⎟ 2(3 − 0.4 )(0.025) − (0.025) =
2 ⎝ L ⎠1
2
⎛W ⎞
⎛W ⎞
⎜ ⎟ (0.1294 ) = 0.36 ⇒ ⎜ ⎟ = 2.78
L
⎝ ⎠1
⎝ L ⎠1
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.46
(a) Transistors matched
VGS1 = VGS 2 = 2.5 V, VO = 2.5 V
⎛ 0.12 ⎞
2
ID = ⎜
⎟(30 )(2.5 − 0.4) = 7.938 mA
⎝ 2 ⎠
(b) I D1 = I D 2
k n′ ⎛ W ⎞
k n′ ⎛ W ⎞
2
2
⎜ ⎟ (VGS 1 − VTN ) =
⎜ ⎟ (VGS 2 − VTN )
2 ⎝ L ⎠1
2 ⎝ L ⎠2
VGS 2 = 5 − VGS1
30
(VGS1 − 0.4) = (5 − VGS1 − 0.4)
15
Which yields VGS1 = 2.14 V, VGS 2 = VO = 2.86 V
Then
⎛ 0.12 ⎞
2
ID = ⎜
⎟(30 )(2.14 − 0.4 ) = 5.45 mA
2
⎝
⎠
(c) I D1 = I D 2
(15)(VGS1 − 0.4)2 = (30)(VGS 2 − 0.4)2
0.7071(VGS1 − 0.4 ) = 5 − VGS1 − 0.4
Which yields VGS1 = 2.86 V, VGS 2 = VO = 2.14 V
⎛ 0.12 ⎞
2
ID = ⎜
⎟(15)(2.86 − 0.4 ) = 5.45 mA
⎝ 2 ⎠
______________________________________________________________________________________
3.47
(a) V1 = VGS 3 = 2.5 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.8 = ⎜
⎟⎜ ⎟ (2.5 − 0.6 ) ⇒ ⎜ ⎟ = 3.69
⎝ 2 ⎠⎝ L ⎠ 3
⎝ L ⎠3
V 2 = 6 = VGS 2 + V1 = VGS 2 + 2.5 ⇒ VGS 2 = 3.5 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.8 = ⎜
⎟⎜ ⎟ (3.5 − 0.6 ) ⇒ ⎜ ⎟ = 1.59
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
VGS1 = 9 − V 2 = 9 − 6 = 3 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
0.8 = ⎜
⎟⎜ ⎟ (3 − 0.6 ) ⇒ ⎜ ⎟ = 2.31
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1
(b) (i) k n′ = (120)(1.05) = 126 μ A/V 2
No change; V1 = 2.5 V, V 2 = 6 V
(ii) k n′ = (120)(0.95) = 114 μ A/V 2
No change; V1 = 2.5 V, V 2 = 6 V
(c) k n′1 = 114 μ A/V 2 , k n′ 2 = k n′ 3 = 126 μ A/V 2
I D 2 = I D3
⎛ 0.126 ⎞
⎛ 0.126 ⎞
2
2
⎜
⎟(1.59 )(VGS 2 − 0.6 ) = ⎜
⎟(3.69 )(VGS 3 − 0.6 )
2
2
⎝
⎠
⎝
⎠
Now VGS 3 = V1 , VGS 2 = V 2 − V1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
So (V 2 − V1 − 0.6) = 1.523(V1 − 0.6)
And V 2 = 2.523V1 − 0.3138
Also I D1 = I D 3
⎛ 0.114 ⎞
⎛ 0.126 ⎞
2
2
⎜
⎟(2.31)(VGS 1 − 0.6 ) = ⎜
⎟(3.69 )(VGS 3 − 0.6)
⎝ 2 ⎠
⎝ 2 ⎠
We have VGS1 = 9 − V 2 = 9 − (2.523V1 − 0.3138) = 9.3138 − 2.523V1
Also VGS 3 = V1
Then (0.13167)(9.3138 − 2.523V1 − 0.6) = (0.23247)(V1 − 0.6)
8.7138 − 2.523V1 = 1.3287(V1 − 0.6)
Which yields V1 = 2.469 V and V 2 = 5.916 V
______________________________________________________________________________________
2
2
3.48
M L in saturation, M D in nonsaturation
I DD = I DL
[
]
⎛ k n′ ⎞⎛ W ⎞
⎛ k′ ⎞
2
2
⎜⎜ ⎟⎟⎜ ⎟ 2(5 − 0.6 )(0.15) − (0.15) = ⎜⎜ n ⎟⎟(2 )(5 − 0.15 − 0.6 )
L
2
2
⎝ ⎠⎝ ⎠ D
⎝ ⎠
⎛W ⎞
⎛W ⎞
⎜ ⎟ (1.2975) = 36.125 ⇒ ⎜ ⎟ = 27.8
⎝ L ⎠D
⎝ L ⎠D
______________________________________________________________________________________
3.49
M L in saturation, M D in nonsaturation
I DD = I DL
[
]
⎛ k n′ ⎞⎛ W ⎞
⎛ k′ ⎞
2
2
⎜⎜ ⎟⎟⎜ ⎟ 2(5 − 0.6 )(0.10 ) − (0.10 ) = ⎜⎜ n ⎟⎟(2 )[0 − (− 1.2 )]
⎝ 2 ⎠⎝ L ⎠ D
⎝ 2 ⎠
⎛W ⎞
⎛W ⎞
⎜ ⎟ (0.87 ) = 2.88 ⇒ ⎜ ⎟ = 3.31
L
⎝ ⎠D
⎝ L ⎠D
______________________________________________________________________________________
3.50
(a) Transistor in nonsaturation
3 − 0.08
ID =
= 0.09733 mA
30
⎛ 0.12 ⎞⎛ W ⎞
2
0.09733 = ⎜
⎟⎜ ⎟ 2(2.6 − 0.4 )(0.08) − (0.08)
2
L
⎝
⎠⎝ ⎠
[
]
⎛W ⎞
⎛W ⎞
0.09733 = ⎜ ⎟(0.020736 ) ⇒ ⎜ ⎟ = 4.69
L
⎝ ⎠
⎝L⎠
[
⎛ 0.12 ⎞⎛ W ⎞
2
(b) 0.09733 = ⎜
⎟⎜ ⎟ 2(3 − 0.4 )(0.08) − (0.08)
⎝ 2 ⎠⎝ L ⎠
]
⎛W ⎞
⎛W ⎞
0.09733 = ⎜ ⎟(0.024576 ) ⇒ ⎜ ⎟ = 3.96
⎝L⎠
⎝L⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.51
5 = I D R D + Vγ + V DS
5 = (12)R D + 1.6 + 0.15 ⇒ R D = 271 Ω
[
]
⎛ 0.08 ⎞ W
⎛W ⎞
2
I D = 12 = ⎜
2(5 − 0.6 )(0.15) − (0.15) ⇒ ⎜ ⎟ = 231
⎟
2
L
⎝
⎠
⎝L⎠
______________________________________________________________________________________
3.52
5 = V SD + I D R D + Vγ
5 = 0.20 + (15)R D + 1.6 ⇒ R D = 213 Ω
[
⎛ 0.04 ⎞⎛ W ⎞
2
I D = 15 = ⎜
⎟⎜ ⎟ 2(5 − 0.6)(0.20 ) − (0.20 )
⎝ 2 ⎠⎝ L ⎠
]
⎛W ⎞
⎛W ⎞
15 = ⎜ ⎟(0.0344 ) ⇒ ⎜ ⎟ = 436
L
⎝ ⎠
⎝L⎠
______________________________________________________________________________________
3.53
5 − 0.15
= 0.097 mA
50
0.097 ⎛ 0.12 ⎞⎛ W ⎞
2
ID =
=⎜
⎟⎜ ⎟ 2(5 − 0.6)(0.15) − (0.15)
2
⎝ 2 ⎠⎝ L ⎠
(a) I R =
[
]
⎛W ⎞
⎛W ⎞ ⎛W ⎞
0.0485 = ⎜ ⎟(0.07785) ⇒ ⎜ ⎟ = ⎜ ⎟ = 0.623
L
⎝ ⎠
⎝ L ⎠1 ⎝ L ⎠ 2
(b) M 1 nonsaturation, M 2 cutoff
IR = ID =
[
5 − VO ⎛ 0.12 ⎞
2
=⎜
⎟(0.623) 2(5 − 0.6 )VO − VO
50
⎝ 2 ⎠
]
We find 1.869VO2 − 17.45VO + 5 = 0 ⇒ VO = 0.297 V
______________________________________________________________________________________
3.54
(a) V DS 2 (sat ) = 0.5 = VGS 2 − VTN = VGS 2 − 0.4 ⇒ VGS 2 = 0.9 V
⎛ 120 ⎞⎛ W ⎞
⎛W ⎞
2
I Q1 = 125 = ⎜
⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 8.33
2
L
⎝
⎠⎝ ⎠ 2
⎝ L ⎠2
VGS 3 = VGS 2 = 0.9 V
⎛ 120 ⎞⎛ W ⎞
⎛W ⎞
2
I REF 1 = 225 = ⎜
⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 15
⎝ 2 ⎠⎝ L ⎠ 3
⎝ L ⎠3
⎛W ⎞
M 1 and M 2 matched, so ⎜ ⎟ = 8.33
⎝ L ⎠1
(b) VGS1 = 0.9 V ⇒ V D1 = V DS1 − VGS1 = 2 − 0.9 = 1.1 V
2.5 − 1.1
= 11.2 k Ω
0.125
______________________________________________________________________________________
RD =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.55
(a) V SDB (sat ) = 0.8 = V SGB + VTP = V SGB − 0.4 ⇒ V SGB = 1.2 V
⎛ 50 ⎞⎛ W ⎞
⎛W ⎞
2
I Q 2 = 200 = ⎜ ⎟⎜ ⎟ (1.2 − 0.4 ) ⇒ ⎜ ⎟ = 12.5
L
2
⎝ ⎠⎝ ⎠ B
⎝ L ⎠B
V SGC = V SGB = 1.2 V
⎛ 50 ⎞⎛ W ⎞
⎛W ⎞
2
I REF 2 = 125 = ⎜ ⎟⎜ ⎟ (1.2 − 0.4 ) ⇒ ⎜ ⎟ = 7.81
⎝ 2 ⎠⎝ L ⎠ C
⎝ L ⎠C
⎛W ⎞
M A and M B matched, so ⎜ ⎟ = 12.5
⎝ L ⎠A
(b) V SGA = 1.2 V
V DA = V SGA − V SDA = 1.2 − 4 = −2.8 V
− 2.8 − (− 5)
= 11 k Ω
0.20
______________________________________________________________________________________
RD =
3.56
V DS 2 (sat ) = 0.5 = VGS 2 − VTN = VGS 2 − 0.6 ⇒ VGS 2 = 1.1 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
I Q = 0.35 = ⎜
⎟⎜ ⎟ (1.1 − 0.6 ) ⇒ ⎜ ⎟ = 23.3
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
VGS 3 = VGS 2 = 1.1 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.15 = ⎜
⎟⎜ ⎟ (1.1 − 0.6 ) ⇒ ⎜ ⎟ = 10
2
L
⎝
⎠⎝ ⎠ 3
⎝ L ⎠3
VGS 4 = 5 − 1.1 = 3.9 V
⎛ 0.12 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.15 = ⎜
⎟⎜ ⎟ (3.9 − 0.6 ) ⇒ ⎜ ⎟ = 0.23
⎝ 2 ⎠⎝ L ⎠ 4
⎝ L ⎠4
⎛W ⎞
M 1 and M 2 matched, so ⎜ ⎟ = 23.3
⎝ L ⎠1
VGS1 = 1.1 V; V D1 = V DS1 − VGS1 = 3.5 − 1.1 = 2.4 V
5 − 2 .4
= 7.43 k Ω
0.35
______________________________________________________________________________________
RD =
3.57
VDS ( sat ) = VGS − VP
VDS > VDS ( sat ) = −VP , I D = I DSS
So
______________________________________________________________________________________
3.58
a.
VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat )
VGS = 0 ⇒ I D = I DSS = 6 mA
2
⎛ V ⎞
⎛ −1 ⎞
I D = I DSS ⎜1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA
⎝ −3 ⎠
⎝ VP ⎠
b.
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ −2 ⎞
I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA
⎝ −3 ⎠
2
c.
ID = 0
d.
______________________________________________________________________________________
3.59
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
⎛
1 ⎞
2.8 = I DSS ⎜1 − ⎟
⎝ VP ⎠
2
2
⎛
3 ⎞
0.30 = I DSS ⎜ 1 − ⎟
V
⎝
P ⎠
2
2
⎛
1 ⎞
⎜1 − ⎟
2.8 ⎝ VP ⎠
=
= 9.33
2
0.30 ⎛
3 ⎞
⎜1 − ⎟
⎝ VP ⎠
⎛
1 ⎞
⎜1 − ⎟
⎝ VP ⎠ = 3.055
⎛
3 ⎞
⎜1 − ⎟
⎝ VP ⎠
1
9.165
= 3.055 −
VP
VP
8.165
= 2.055 ⇒ VP = 3.97 V
VP
1−
2
1 ⎞
⎛
2.8 = I DSS ⎜ 1 −
⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA
⎝ 3.97 ⎠
______________________________________________________________________________________
3.60
VS = −VGS , VSD = VS − VDD
V ≥ VSD ( sat ) = VP − VGS
Want SD
VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP
V ≤ −2.5 V
So DD
⎛ V ⎞
I D = 2 = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
2
⎛ V ⎞
2 = 6 ⎜1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V
⎝ 2.5 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.61
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
______________________________________________________________________________________
3.62
I D = K (VGS − VTN )
2
250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2
2
______________________________________________________________________________________
3.63
2
⎛ V ⎞
V
V
I D = I DSS ⎜ 1 − GS ⎟ = S = − GS
RS
RS
⎝ VP ⎠
2
V
⎛ V ⎞
10 ⎜ 1 − GS ⎟ = − GS
−5 ⎠
0.2
⎝
2
⎛ 2V
V ⎞
2 ⎜1 + GS + GS ⎟ = −VGS
5
25 ⎠
⎝
2 2 9
VGS + VGS + 2 = 0
25
5
2
2VGS + 45VGS + 50 = 0
VGS =
−45 ±
( 45 ) − 4 ( 2 )( 50 )
⇒ VGS = −1.17 V
2 ( 2)
2
V
1.17
I D = − GS =
⇒ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.64
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
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
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R
R
R
+
2 ⎠
1
⎝ 1
1
1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ
R1
5R2
= 0.5 ⇒ R2 = 0.556 MΩ
5 + R2
______________________________________________________________________________________
3.65
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
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
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.66
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
2
⎛ V ⎞
5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V
3 ⎠
⎝
VSD = VDD − I D ( RS + RD )
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 ⎠
______________________________________________________________________________________
3.67
⎛ R2 ⎞
⎛ 60 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ ( 20 ) ⇒ VG = 6 V
R
+
R
⎝ 140 + 60 ⎠
⎝ 1
2 ⎠
2
⎛ V ⎞
V
V − VGS
I D = I DSS ⎜ 1 − GS ⎟ = S = G
RS
RS
⎝ VP ⎠
2
⎛
V ⎞
(8 )( 2 ) ⎜⎜1 − GS ⎟⎟ = 6 − VGS
⎝ ( −4 ) ⎠
⎛ V
V2 ⎞
16 ⎜ 1 + GS + GS ⎟ = 6 − VGS
2
16 ⎠
⎝
VGS2 + 9VGS + 10 = 0
VGS =
−9 ±
( 9 ) − 4 (10 )
2
2
⇒ VGS = −1.30
⎛ ( −1.30 ) ⎞
I D = 8 ⎜⎜ 1 −
⎟ ⇒ I D = 3.65 mA
( −4 ) ⎟⎠
⎝
VDS = VDD − I D ( RS + RD )
= 20 − ( 3.65 )( 2 + 2.7 )
VDS = 2.85 V
2
VDS > VDS ( sat ) = VGS − VP
= −1.30 − ( −4 )
= 2.7 V (Yes)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.68
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
⎝ 450 + 20 ⎠
⎝ R1 + R2 ⎠
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
3.69
2
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0
⎝ VP ⎠
I D = I DSS = 4 mA
VDD − VDS 10 − 3
=
⇒ RD = 1.75 kΩ
ID
4
______________________________________________________________________________________
RD =
3.70
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
2
⎛ V ⎞
1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V
2 ⎠
⎝
VG = VS + VGS = 10 + 0.586 = 10.586
⎛ R2 ⎞
VG = ⎜
⎟ VDD
⎝ R1 + R2 ⎠
⎛ R ⎞
10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ
⎝ 200 ⎠
R1 = 94 kΩ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.71
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
______________________________________________________________________________________
3.72
VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN
For 0.75 − 0.15 = 0.6
Biased in the saturation region
V − VDS 3 − 0.7
I D = DD
=
⇒ I D = 46 μ A
RD
50
I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2
2
2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 4
4.1
(a) (i) g m = 2
k n′ ⎛ W ⎞
⎜ ⎟ I DQ
2 ⎝L⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
0.5 = 2 ⎜
⎟⎜ ⎟(0.5) ⇒ ⎜ ⎟ = 2.5
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
⎛ k ′ ⎞⎛ W ⎞
(ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
0. 5 = ⎜
⎟(2.5)(VGSQ − 0.4 ) ⇒ VGSQ = 2.4 V
⎝ 2 ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
(b) (i) 0.5 = 2 ⎜
⎟⎜ ⎟(0.15) ⇒ ⎜ ⎟ = 8.33
L
2
⎝
⎠⎝ ⎠
⎝L⎠
⎛ 0.1 ⎞
2
(ii) 0.15 = ⎜
⎟(8.33)(VGSQ − 0.4 ) ⇒ VGSQ = 1.0 V
2
⎝
⎠
______________________________________________________________________________________
4.2
⎛ k ′p ⎞⎛ W ⎞
⎟⎜ ⎟ I DQ
(a) (i) g m = 2 ⎜⎜
⎟
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
⎛ 0.04 ⎞⎛ W ⎞
1. 2 = 2 ⎜
⎟⎜ ⎟(0.15) ⇒ ⎜ ⎟ = 120
2
L
⎝L⎠
⎠⎝ ⎠
⎝
⎛ k ′p ⎞⎛ W ⎞
2
(ii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGQ + VTP
⎝ 2 ⎠⎝ L ⎠
⎛ 0.04 ⎞
2
0.15 = ⎜
⎟(120 )(V SGQ − 0.6 ) ⇒ V SGQ = 0.85 V
2
⎠
⎝
(
)
⎛W ⎞
⎛ 0.04 ⎞⎛ W ⎞
(b) (i) 1.2 = 2 ⎜
⎟⎜ ⎟(0.50 ) ⇒ ⎜ ⎟ = 36
⎝L⎠
⎝ 2 ⎠⎝ L ⎠
⎛ 0.04 ⎞
2
(ii) 0.50 = ⎜
⎟(36 )(V SGQ − 0.6 ) ⇒ V SGQ = 1.43 V
⎝ 2 ⎠
______________________________________________________________________________________
4.3
I D = K n (VGS − VTN ) (1 + λVDS )
2
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
ΔVDS
5
=
= 12.5 kΩ
ΔI D
0.4
______________________________________________________________________________________
ro =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.4
1
r0 =
λ ID
1
1
=
⇒ I D = 0.833 mA
λ r0 ( 0.012 )(100 )
______________________________________________________________________________________
ID =
4.5
2
(a) I D = K n (VGS − VTN ) (1 + λV DS )
I D = I DO (1 + λV DS )
Then
0.258 (1 + 3.3λ )
=
0.250 (1 + 1.5λ )
Or 1.032(1 + 1.5λ ) = 1 + 3.3λ ⇒ λ = 0.01826 V −1
0.250 = I DO [1 + (0.01826 )(1.5)] ⇒ I DO = 0.2433 mA
ro =
1
1
=
= 225 k Ω
(0.01826)(0.2433)
(b) I D = (0.2433)[1 + (0.01826 )(5)] = 0.2655 mA
λ I DO
______________________________________________________________________________________
4.6
(a)
ro =
(i)
ro =
(ii)
(b)
1
λ ID
=
1
( 0.02 )( 0.05 )
= 1000 K
1
= 100 K
0.02
( )( 0.5 )
ΔI D =
(i)
ΔVDS
1
=
= 0.001 mA = 1.0 μ A
ro
1000
ΔI D 1.0
=
⇒ 2%
ID
50
ΔI D =
ΔVDS
1
=
= 0.01 ⇒ 10 μ A
100
ro
ΔI D
10
=
⇒ 2%
500
(ii) I D
______________________________________________________________________________________
4.7
I D = 1.0 mA
1
1
ro =
=
= 100 K
λ I D ( 0.01)(1)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.8
(a) V DD = I DQ R D + V DSQ
3.3 = I DQ (5) + 1.5 ⇒ I DQ = 0.36 mA
⎛ k ′ ⎞⎛ W ⎞
2
I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN
⎝ 2 ⎠⎝ L ⎠
⎛ 0. 1 ⎞
2
0.36 = ⎜
⎟(40 ) VGSQ − 0.4 ⇒ VGSQ = 0.824 V
2
⎠
⎝
(
(
)
)
⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞
2
(b) K n = ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜
⎟(40 ) = 2 mA/V
2
2
L
⎠
⎠
⎝
⎝
⎝ ⎠
g m = 2 K n I DQ = 2 (2)(0.36) = 1.697 mA/V
ro =
1
1
=
= 111.1 k Ω
λ I DQ (0.025)(0.36 )
Aυ = − g m (ro R D ) = −(1.697)(111.1 5) = −8.12
______________________________________________________________________________________
4.9
(a) Aυ = − g m R D
− 3.8 = − g m (10 ) ⇒ g m = 0.38 mA/V
⎛ k ′ ⎞⎛ W ⎞
g m = 2 K n I DQ = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
⎛ 0.1 ⎞⎛ W ⎞
0.38 = 2 ⎜
⎟⎜ ⎟(0.12 ) ⇒ ⎜ ⎟ = 6.02
2
L
⎝L⎠
⎠⎝ ⎠
⎝
(b) − 5 = − g m (10 ) ⇒ g m = 0.50 mA/V
⎛W ⎞
⎛ 0.1 ⎞⎛ W ⎞
0.50 = 2 ⎜
⎟⎜ ⎟(0.12 ) ⇒ ⎜ ⎟ = 10.4
2
L
⎝L⎠
⎠⎝ ⎠
⎝
______________________________________________________________________________________
4.10
(a) V DD = I DQ R D + V DSQ
5 = (0.5)R D + 3 ⇒ R D = 4 k Ω
⎛ k ′ ⎞⎛ W ⎞
2
I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
⎛ 0.08 ⎞⎛ W ⎞
2
0.5 = ⎜
⎟⎜ ⎟(1.2 − 0.6 ) ⇒ ⎜ ⎟ = 34.7
2
L
⎝L⎠
⎠⎝ ⎠
⎝
(
)
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.08 ⎞
(b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(34.7 )(0.5) = 1.666 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠
1
1
ro =
=
= 133.3 k Ω
λ I DQ (0.015)(0.5)
(c) Aυ = − g m (ro R D ) = −(1.666)(133.3 4) = −6.47
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.11
2
K n vgs2 = K n ⎡⎣Vgs sinω t ⎤⎦ = K nVgs2 sin 2ω t
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 )
______________________________________________________________________________________
4.12
0.01 =
Vgs
4 (VGSQ − VTN )
So Vgs = ( 0.01)( 4 )( 3 − 1) ⇒ Vgs = 0.08 V
______________________________________________________________________________________
4.13
⎛ R2 ⎞
⎛ 60 ⎞
⎟⎟ ⋅ V DD = ⎜
(a) VGS = ⎜⎜
⎟(3.3) = 0.66 V
R
+
R
⎝ 60 + 240 ⎠
2 ⎠
⎝ 1
⎛ k ′ ⎞⎛ W ⎞
⎛ 0. 1 ⎞
2
2
I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN ) = ⎜
⎟(80 )(0.66 − 0.4 ) = 0.270 mA
2
L
2
⎠
⎝
⎝ ⎠⎝ ⎠
V DSQ = V DD − I DQ R D = 3.3 − (0.270 )(8) = 1.14 V
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
(b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(80)(0.270) = 2.078 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠
1
1
=
= 185 k Ω
ro =
λ I DQ (0.02 )(0.27 )
⎛ R1 R 2 ⎞
⎟
(c) Aυ = − g m (ro R D )⎜
⎜R R +R ⎟
Si ⎠
⎝ 1 2
We find ro R D = 185 8 = 7.668 k Ω
R1 R2 = 60 240 = 48 k Ω
⎛ 48 ⎞
So Aυ = −(2.078)(7.668)⎜
⎟ = −15.3
⎝ 48 + 2 ⎠
______________________________________________________________________________________
4.14
Av = − g m ( r0 || RD )
−10 = − g m (100 || 5 ) ⇒ g m = 2.1 mA/V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.15
⎛ R2 ⎞
⎛ 175 ⎞
⎟⎟ ⋅ V DD = ⎜
(a) VG = ⎜⎜
⎟(5) = 2.1875 V
R
+
R
⎝ 175 + 225 ⎠
2 ⎠
⎝ 1
2.1875 = VGS + I D R S = VGS + K n R S (VGS − VTN )
(
2
2.1875 = VGS + (1)(1) VGS
− 1.6VGS + 0.64
or V
2
GS
2
)
− 0.6VGS − 1.5475 = 0 ⇒ VGS = 1.58 V
I DQ = K n (VGS − VTN ) = (1)(1.58 − 0.8) = 0.608 mA
2
2
V DSQ = V DD − I DQ (R S + R D ) = 5 − (0.608 )(1 + 4 ) = 1.96 V
(b) Aυ =
− g m RD
1 + g m RS
g m = 2 (1)(0.608 ) = 1.56 mA/V
− (1.56)(4)
= −2.44
1 + (1.56)(1)
− g m (R D R L ) − (1.56)(R D R L )
=
= −0.6094(R D R L )
(c) Aυ =
1 + g m RS
1 + (1.56)(1)
Aυ =
− (0.75)(2.44) = −(0.6094)(R D R L ) ⇒ R D R L = 3.0 k Ω
4 R L = 3 ⇒ R L = 12 k Ω
______________________________________________________________________________________
4.16
(a) V DSQ = V DD − I DQ (R S + R D )
5 = 12 − (2 )(R S + R D ) ⇒ R S + R D = 3.5 k Ω
R S = 0.5 k Ω , then R D = 3 k Ω
(
I DQ = K n VGSQ − VTN
(
2 = 1.5 VGSQ − 1.2
)
2
) ⇒V
2
GSQ
= 2.355 V
V G = V GSQ + I DQ R S = 2.355 + (2 )(0.5) = 3.355 V
⎛ R2 ⎞
1
⎟⎟ ⋅ V DD =
VG = ⎜⎜
⋅ Rin ⋅ V DD
R1
⎝ R1 + R 2 ⎠
1
3.355 =
(250)(12) ⇒ R1 = 894 k Ω
R1
R1 R2 = 894 R2 = 250 ⇒ R2 = 347 k Ω
(b) g m = 2 (1.5)(2 ) = 3.464 mA/V
Aυ =
− g m (R D R L )
=
− (3.464)(3 10)
= −2.93
1 + g m RS
1 + (3.464)(0.5)
______________________________________________________________________________________
4.17
(a) From Problem 4.16;
R S = 0.5 k Ω , R D = 3 k Ω ,
R1 = 894 k Ω , R 2 = 347 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) g m = 2 (1.5)(2 ) = 3.464 mA/V
Aυ = − g m (R D R L ) = −(3.464)(3 10) = −7.99
______________________________________________________________________________________
4.18
(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 )
⇒ RS = 1 K
1 + ( 2 ) RS
______________________________________________________________________________________
4.19
Av =
− g m RD
1 + g m RS
−8 =
− g m RD
1 + g m (1)
(a)
(1)
−16 = − g m RD
(2)
8=
16
⇒
1 + g m (1)
g m = 1 mA/V
RD = 16 K
Then
Av = −10 =
− (1)(16 )
1 + (1) RS
RS = 0.6 K
(b)
______________________________________________________________________________________
4.20
⎛ k ′ ⎞⎛ W ⎞
2
(a) I DQ = I Q = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
0. 5 = ⎜
⎟(50 )(VGSQ − 0.8) ⇒ VGSQ = 1.247 V
2
⎝
⎠
(
)
V DSQ = V + − I DQ R D − − VGSQ = 5 − (0.5)(6 ) + 1.247 = 3.25 V
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
(b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(50)(0.5) = 2.236 mA/V
L
2
⎝ 2 ⎠
⎝ ⎠⎝ ⎠
1
1
=
= 100 k Ω
ro =
λ I DQ (0.02 )(0.5)
Aυ = − g m (ro R D ) = −(2.236)(100 6) = −12.7
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
(
)
(d) A = − g (r R R ) = −(2.236 )(100 6 6 ) = −6.51
(c) Aυ = − g m ro R D R L = −(2.236 ) 100 6 20 = −9.86
υ
m
o
D
L
______________________________________________________________________________________
4.21
(a) V DSQ = 5 − (− 5) − I DQ (R S + R D )
5.5 = 10 − (0.1)(R S + R D ) ⇒ R S + R D = 45 k Ω
(
I DQ = K n VGSQ − VTN
)
2
(0.1) = (0.85)(VGSQ − 0.8)2 ⇒ VGSQ = 1.143 V
VGS + I D R S = 5
1.143 + (0.1)R S = 5 ⇒ R S = 38.6 k Ω
⇒ R D = 6.43 k Ω
(b) g m = 2 K n I DQ = 2 (0.85)(0.1) = 0.583 mA/V
ro =
1
1
=
= 500 k Ω
λ I DQ (0.02 )(0.1)
(
)
(
)
(c) Aυ = − g m R D ro R L = −(0.583) 6.43 500 40 = −3.19
______________________________________________________________________________________
4.22
(a) V DSQ = V DD − I DQ (R S + R D )
2 = 3.3 − (0.5)(R S + R D ) ⇒ R S + R D = 2.6 k Ω
(
)
− (− 0.8)) ⇒ V
I DQ = K n VGSQ − VTN
2
0.5 = 2 VGSQ
2
(
GSQ
= −0.3 V
V GSQ + I DQ R S = 0
− 0.3 + (0.5)R S = 0 ⇒ R S = 0.6 k Ω
⇒ RD = 2 k Ω
(b) g m = 2 K n I DQ = 2 (2)(0.5) = 2 mA/V
Aυ =
− g m (R D R L )
=
− (2)(2 10)
= −1.52
1 + g m RS
1 + (2)(0.6 )
______________________________________________________________________________________
4.23
(a) V DD = I DQ R D + V DSQ + I DQ R S and VGS + I DQ R S = 0
Then V DD = K n R D (VGS − VTN ) + V DSQ − VGS
2
5 = (2)(2)(VGS + 0.8) + 2.5 − VGS
Which yields
2
4VGS
+ 5.4VGS + 0.06 = 0 , ⇒ VGS = −0.0112 V
2
and I DQ = 2(− 0.0112 + 0.8) = 1.244 mA
2
5 = (1.244 )(2 ) + 2.5 + (1.244 )R S ⇒ R S = 8.99 Ω ≅ 9 Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) g m = 2 K n I DQ = 2 (2)(1.244) = 3.155 mA/V
Aυ =
− g m (R D R L )
− (3.155)(2 2)
=
= −3.07
1 + g m RS
1 + (3.155)(0.009)
______________________________________________________________________________________
4.24
a.
5 = I DQ RS + VSDQ + I DQ RD − 5
5 = I DQ RS + 6 + I DQ (10 ) − 5
I DQ =
1.
2.
3.
4
RS + 10
VS = VSDQ + I DQ RD − 5 = VSGQ
1 + I DQ (10 ) = VSGQ
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.
I DQ = ( 0.20 )( 3 − 2 ) = 0.20 mA
2
g m = 2 K P I DQ = 2 ( 0.2 )( 0.2 ) = 0.4 mA / V
Av = − g m ( RD || RL ) = − ( 0.4 )(10 || 10 ) ⇒ Av = −2.0
c.
4
= 0.133 mA
30
VSGQ = 1 + (0.133)(10) = 2.33 V
Choose RS = 20 kΩ ⇒ I DQ =
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.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.25
(a)
I DQ = K p (VSGQ + VTP )
2
0.25 = 0.8 (VSGQ − 0.5 )
2
VSGQ = 1.059 V
3 − 1.059
RS =
⇒ RS = 7.76 K
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 )
g m = 2 K p I DQ = 2 ( 0.8 )( 0.25 ) = 0.8944 mA/V
Av = − ( 0.8944 )(10.2 || 2 )
Av = −1.50
(c)
ΔVO = ΔI ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418
So ΔVO = 0.836 peak-to-peak
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.26
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
6
g m ( RD RL )
RS + RD =
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
( 0.47 ) + 4 ( 2.2 )( 3.93)
⇒ RD = 1.23 kΩ,
2 ( 2.2 )
VG = VGS + VS = 2.8 + ( 6 )( 0.1) = 3.4 V
−0.47 +
RD =
2
RS = 0.10 kΩ
1
1
⋅ Rin ⋅ VDD = (100 )(18 ) = 3.4 ⇒ R1 = 529 kΩ
R1
R1
VG =
529 R2
= 100 ⇒ R2 = 123 kΩ
529 + R2
______________________________________________________________________________________
4.27
(a) V S = V SGQ = V SDQ + I DQ R D − 9
(
V SGQ = 5 + I DQ (4) − 9 = K p (4) V SGQ + VTP
(
)
) −4
2
2
V SGQ = 8 V SGQ
− 2.4V SGQ + 1.44 − 4
Or 8V
2
SGQ
− 20.2V SGQ + 7.52 = 0 ⇒ V SGQ = 2.071 V
(
I DQ = I Q = K p V SGQ + VTP
) = 2(2.071 − 1.2) = 1.518 mA
2
2
(b) g m = 2 K p I DQ = 2 (2)(1.518) = 3.485 mA/V
ro =
1
1
=
= 22 k Ω
λ I DQ (0.03)(1.518)
Aυ = − g m (R D ro ) = −(3.485)(4 22) = −11.8
(
)
(
)
(c) Aυ = − g m R D ro R L = −(3.485 ) 4 22 8 = −8.29
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.28
(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 =
⇒ RS = 18.8 K
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
g m = 2 K p I DQ = 2 ( 0.25 )( 0.5 ) = 0.7071 mA/V
Av = − ( 0.7071)(15.2 )
Av = −10.7
______________________________________________________________________________________
4.29
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
Aυ = −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
VG = VGS + VS = 3.14 + 2 = 5.14
2
VG =
2
1
1
⋅ Rin ⋅ VDD ⇒ ( 200 )( 20 ) = 5.14 ⇒ R1 = 778 kΩ
R1
R1
778R2
= 200 ⇒ R2 = 269 kΩ
778 + R2
______________________________________________________________________________________
4.30
(a) Aυ =
Ro =
g m ro
(5)(100) = 0.998
=
1 + g m ro 1 + (5)(100 )
1
1
ro = 100 = 0.2 100 ⇒ R o ≅ 200 Ω
gm
5
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
g m (ro R S )
(b) Aυ =
1 + g m (ro R S )
We have ro R S = 100 5 = 4.762 k Ω
(5)(4.762) = 0.960
1 + (5)(4.762)
______________________________________________________________________________________
Aυ =
4.31
Av =
0.98 =
g m ( RL ro )
1 + g m ( RL ro )
g m ro
⇒ g m ro = 49
1 + g m ro
⎛ Rr ⎞
gm ⎜ L o ⎟
⎝ RL + ro ⎠
Also 0.49 =
=
1 + g m ( RL ro )
⎛ Rr ⎞
1 + gm ⎜ L o ⎟
⎝ RL + ro ⎠
g m ( RL ro )
0.49 =
0.49 =
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
______________________________________________________________________________________
4.32
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.33
0 − (− 1.5)
= 0.15 mA
10
⎛ k ′ ⎞⎛ W ⎞
2
I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN
⎝ 2 ⎠⎝ L ⎠
(a) I DQ =
(
)
(
)
⎛ 0.1 ⎞
2
0.15 = ⎜
⎟(80 ) VGSQ − 0.4 ⇒ VGSQ = 0.594 V
2
⎝
⎠
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
(b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(80)(0.15) = 1.549 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠
1
1
ro =
=
= 333 k Ω
λ I DQ (0.02)(0.15)
We find ro R S = 333 10 = 9.708 k Ω
Then Aυ =
(c) R o =
g m (ro R S )
1 + g m (ro R S )
=
(1.549)(9.708) = 0.938
1 + (1.549 )(9.708)
1
1
ro R S =
333 10 = 0.6456 9.708
gm
1.549
or R o = 605 Ω
______________________________________________________________________________________
4.34
(a) I DQ =
(b) Aυ =
V DD − V DSQ
RS
=
2.5 − 1.25
= 2.5 mA
0 .5
g m RS
1 + g m RS
0.85 =
g m (0.5)
⇒ g m = 11.33 mA/V
1 + g m (0.5)
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
11.33 = 2 ⎜
⎟⎜ ⎟(2.5) ⇒ ⎜ ⎟ = 257
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
⎛ k ′ ⎞⎛ W ⎞
(c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2
⎝ 2 ⎠⎝ L ⎠
⎛ 0. 1 ⎞
2
2. 5 = ⎜
⎟(257 )(VGSQ − 0.6 ) ⇒ VGSQ = 1.041 V
⎝ 2 ⎠
V IQ = V GSQ + V O = 1.041 + 1.25 = 2.291 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.35
(a) P = I Q V DD ⇒ 2.5 = I Q (2.5) ⇒ I Q = 1 mA
(b) R o =
1
ro
gm
1
1
=
= 50 k Ω
λ I DQ (0.02 )(1)
ro =
So 0.5 =
1
50 ⇒ g m = 1.98 mA/V
gm
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
1.98 = 2 ⎜
⎟⎜ ⎟(1) ⇒ ⎜ ⎟ = 19.6
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
g m ro
(1.98)(50) = 0.990
(c) Aυ =
=
1 + g m ro 1 + (1.98)(50 )
⎛ 0.1 ⎞
(d) g m = 2 ⎜
⎟(100 )(1) = 4.472 mA/V
⎝ 2 ⎠
1
1
ro =
50 = 0.2236 50 ⇒ R o = 223 Ω
gm
4.472
______________________________________________________________________________________
Ro =
4.36
⎛ R2 ⎞
⎛ 350 ⎞
⎟⎟ ⋅ V DD = ⎜
(a) VG = ⎜⎜
⎟(10 ) = 2.917 V
⎝ 350 + 850 ⎠
⎝ R1 + R 2 ⎠
10 = I DQ R S + V SGQ + V G
⎛ k ′p ⎞⎛ W ⎞ ⎛ 0.04 ⎞
2
Now K p = ⎜⎜ ⎟⎟⎜ ⎟ = ⎜
⎟(80 ) = 1.6 mA/V
2
2
L
⎝
⎠
⎝
⎠
⎝ ⎠
2
So 10 = (1.6 )(4) V SGQ − 1.2 + V SGQ + 2.917
(
We find 6.4V
2
SGQ
)
− 14.36V SGQ + 2.133 = 0 ⇒ V SGQ = 2.084 V
I DQ = 1.6(2.084 − 1.2 ) = 1.25 mA
2
V SDQ = 10 − (1.25 )(4 ) = 5 V
(b) g m = 2 K p I DQ = 2 (1.6)(1.25) = 2.828 mA/V
1
ro =
λ I DQ
=
1
(0.05)(1.25)
= 16 k Ω
ro R S R L = 16 4 4 = 1.778 k Ω
Aυ =
(
g m ro R S R L
(
)
1 + g m ro R S R L
(c) Ag =
io
υi
=
)
=
(2.828)(1.778) = 0.834
1 + (2.828)(1.778)
io υ o
1 υo 1
⋅
=
⋅
= (0.834 ) = 0.2085 mA/V
υ o υ i RL υ i 4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d) R o =
1
1
R S ro =
4 16 = 0.3536 3.2
gm
2.828
R o = 318 Ω
______________________________________________________________________________________
4.37
(a) (i) K n =
k n′ W ⎛ 0.1 ⎞
2
⋅ =⎜
⎟(20 ) = 1 mA/V
2 L ⎝ 2 ⎠
g m = 2 K n I DQ = 2 (1)(5) = 4.472 mA/V
ro =
1
λ I DQ
=
1
(0.02)(5)
= 10 k Ω
ro R L = 10 4 = 2.857 k Ω
Aυ =
(ii) R o =
g m (ro R L )
(4.472)(2.857 ) = 0.927
1 + g m (ro R L ) 1 + (4.472)(2.857 )
=
1
1
ro =
10
gm
4.472
R o = 219 Ω
(b) (i) g m = 2 (1)(2 ) = 2.828 mA/V
ro =
1
(0.02)(2)
= 25 k Ω
ro R L = 25 4 = 3.448 k Ω
Aυ =
(ii) R o =
(2.828)(3.448) = 0.907
1 + (2.828)(3.448)
1
1
ro =
25
gm
2.828
R o = 349 Ω
______________________________________________________________________________________
4.38
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 ⎟ ⎜ ⎟ I Q
⎝2
⎠⎝ L ⎠
W
⎛W ⎞
4.75 = 2 ( 0.030 ) ⎜ ⎟ ( 4 ) ⇒
= 47.0
L
⎝L⎠
⎛1
⎞⎛ W ⎞
g m = 2 ⎜ μ n Cox ⎟⎜ ⎟ I Q
⎝2
⎠⎝ L ⎠
4.75 = 2 ( 0.030 )( 60 ) I Q ⇒ I Q = 3.13 mA
b.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.39
I DQ = K n (VGS − VTN )
2
5 = 5 (VGS + 2 ) ⇒ VGS = −1 V ⇒ VS = −VGS = 1 V
2
I DQ =
VS − ( −5 )
RS
⇒ RS =
1+ 5
⇒ RS = 1.2 kΩ
5
g m = 2 K n I DQ = 2 ( 5 )( 5 ) = 10 mA / V
r0 =
Av =
=
1
1
=
= 20 kΩ
λ I DQ ( 0.01)( 5 )
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
gm
10
______________________________________________________________________________________
R0 =
4.40
(a) V S = I DQ R S − 5 = (5)(1) − 5 = 0 ⇒ VGS = 0
⎛ k ′ ⎞⎛ W ⎞
2
I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN
2
L
⎝ ⎠⎝ ⎠
0
.
⎛ 1 ⎞⎛ W ⎞
⎛W ⎞
2
5=⎜
⎟⎜ ⎟[0 − (− 2 )] ⇒ ⎜ ⎟ = 25
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
(
)
⎛ 0. 1 ⎞
(b) g m = 2 ⎜
⎟(25)(5) = 5 mA/V
⎝ 2 ⎠
1
1
=
= 20 k Ω
λ I DQ (0.01)(5)
ro =
ro R S = 20 1 = 0.9524 k Ω
g m (ro R S )
Aυ =
1 + g m (ro R S )
=
(5)(0.9524) = 0.826
1 + (5)(0.9524 )
1
1
ro R S = 20 1 ⇒ Ro = 165 Ω
gm
5
(c) Ro =
(d) ro R S R L = 20 1 2 = 0.6452 k Ω
Aυ =
(
g m ro R S R L
(
)
1 + g m ro R S R L
)
=
(5)(0.6452 ) = 0.763
1 + (5)(0.6452 )
______________________________________________________________________________________
4.41
R0 =
1
RS
gm
Output resistance determined primarily by gm
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
Set
= 0.2 kΩ ⇒ g m = 5 mA/V
gm
g m = 2 K n I DQ ⇒ 5 = 2 ( 4 ) I DQ ⇒ I DQ = 1.56 mA
I DQ = K n (VGS − VTN )
2
1.56 = 4 (VGS + 2 )
VGS = −1.38 V, VS = −VGS = 1.38 V
2
RS =
1.38 − ( −5 )
⇒ RS = 4.09 kΩ
1.56
5 ( 4.09 2 )
g m ( RS RL )
=
⇒ Av = 0.870
Av =
1 + g m ( RS RL ) 1 + 5 ( 4.09 2 )
______________________________________________________________________________________
4.42
(a) g m = 2 K p I DQ = 2 (5)(10) = 14.14 mA/V
ro =
1
=
1
= 10 k Ω
(0.01)(10)
g m ro
(14.14)(10) = 0.993
=
Aυ =
1 + g m ro 1 + (14.14 )(10 )
(b) R o =
(c) Aυ =
λ I DQ
1
1
ro =
10 = 0.07072 10 ⇒ R o = 70.2 Ω
gm
14.14
g m (ro R L )
1 + g m (ro R L )
0.90 =
(14.14)(ro R L )
⇒ (ro R L ) = 0.6365 k Ω
1 + (14.14 )(ro R L )
10 R L = 0.6365 ⇒ R L = 680 Ω
______________________________________________________________________________________
4.43
ΔiD = I DQ =
−1
⋅ Δv0
RS RL
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 )
I DQ
⎡1 + g m ( RS RL ) ⎤⎦
gm ⎣
______________________________________________________________________________________
vi ( min ) = −
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.44
(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
R1
⎝ R1 + R2 ⎠
1
2.607 = ( 300 )( 3) ⇒ R1 = 345.2 K ⇒ R2 = 2291 K
R1
(b)
Av =
g m RS
1 + g m RS
g m = 2 K n I DQ = 2 ( 0.5 )( 0.25 ) = 0.7071 mA/V
Av =
( 0.7071)( 6 )
⇒ Av = 0.809
1 + ( 0.7071)( 6 )
Ro =
1
1
6 = 1.414 || 6
RS =
gm
( 0.7071)
Ro = 1.14 K
______________________________________________________________________________________
4.45
Ri =
1
1
= 0.5 ⇒ g m =
= 2 mA/V
gm
0.5
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2=2 ⎜
⎟⎜ ⎟(0.25) ⇒ ⎜ ⎟ = 80
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
Aυ = g m R D
20 = (2 )R D ⇒ R D = 10 k Ω
______________________________________________________________________________________
4.46
(a) R o = R D = 500 Ω
(b) VGSQ = 1.2 V
V DD − V DS 2.2 − (V DS (sat ) + 0.3) 2.2 − (1.2 − 0.4 + 0.3)
=
=
RD
0.5
0.5
I DQ = 2.2 mA
I DQ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
(c) I DQ = K n VGSQ − VTN
)
2
2.2 = K n (1.2 − 0.4) ⇒ K n = 3.438 mA/V 2
2
g m = 2 K n I DQ = 2 (3.438)(2.2) = 5.5 mA/V
Ri =
1
1
=
⇒ Ri = 182 Ω
g m 5.5
(d) Aυ = g m R D = (5.5)(0.5) = 2.75
______________________________________________________________________________________
4.47
⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞
2
K n = ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜
⎟(80 ) = 4 mA/V
⎝ 2 ⎠⎝ L ⎠ ⎝ 2 ⎠
g m = 2 K n I DQ = 2 (4)(0.5) = 2.828 mA/V
Aυ = g m R D = (2.828)(4) = 11.3
Ri = 10
1
1
= 10
= 10 0.3536 ⇒ Ri = 342 Ω
gm
2.828
______________________________________________________________________________________
4.48
a.
VGS + I DQ RS = 5
5 − VGS
2
I DQ =
= K n (VGS − VTN )
RS
5 − VGS = (10 )( 3) (VGS2 − 2VGS + 1)
30VGS2 − 59VGS + 25 = 0
( 59 ) − 4 ( 30 )( 25)
⇒ VGS = 1.35 V
VGS =
2 ( 30 )
2
I DQ = ( 3)(1.35 − 1) ⇒ I DQ = 0.365 mA
VDSQ = 10 − ( 0.365 )( 5 + 10 ) ⇒ VDSQ = 4.53 V
59 ±
2
b.
g m = 2 K n I DQ = 2 ( 3)( 0.365 ) ⇒ g m = 2.093 mA / V
r0 =
1
=
1
⇒r =∞
( 0 )( 0.365 ) 0
A = g m ( RD RL ) = ( 2.093) ( 5 4 ) ⇒ Av = 4.65
c. v
λ I DQ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.49
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.
⎞
⎛ RD ⎞ ⎛
RS
i0 = ⎜
⎟⎟ ⋅ ii
⎟ ⎜⎜
⎝ RD + RL ⎠ ⎝ 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 )
v0 = i0 RL = (1.84 )( 2 ) sin ω t ⇒ v0 = 3.68sin ω t ( mV )
______________________________________________________________________________________
4.50
(a) VO = (V DS (sat ) + 0.25) − VGS
VO = VGS − VTN + 0.25 − VGS = −0.4 + 0.25 = −0.15 V
1.8 − (− 0.15)
= 0.975 k Ω
2
(b) Aυ = g m R D
6 = g m (0.975) ⇒ g m = 6.154 mA/V
RD =
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
6.154 = 2 ⎜
⎟⎜ ⎟(2 ) ⇒ ⎜ ⎟ = 94.7
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
⎛ k ′ ⎞⎛ W ⎞
(c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2
⎝ 2 ⎠⎝ L ⎠
⎛ 0. 1 ⎞
2
2=⎜
⎟(94.7 )(VGSQ − 0.4 ) ⇒ VGSQ = 1.05 V
2
⎝
⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.51
2
(a) I DQ = I Q = K n VGSQ − VTN
(
(
2 = 4 VGSQ − 0.6
) ⇒V
)
2
GSQ
= 1.307 V
V D = V DSQ − V GSQ = 3.5 − 1.307 = 2.193 V
RD =
3.3 − 2.193
= 0.554 k Ω
2
(b) g m = 2 K n I DQ = 2 (4)(2) = 5.657 mA/V
Ri =
1
1
=
⇒ Ri = 177 Ω
g m 5.657
(
)
(
)
(c) Aυ = g m R D R L = (5.657) 0.554 4 = 2.75
______________________________________________________________________________________
4.52
2
(a) I DQ = K p V SGQ + VTP
(
(
)
)
1.2 = 2.5 V SGQ − 0.8 ⇒ V SGQ = 1.493 V
2
+
RS =
V − V SGQ
I DQ
=
3.3 − 1.493
= 1.51 k Ω
1. 2
V SDQ = 6.6 − I DQ (R S + R D ) ⇒ 3 = 6.6 − 1.2(1.51 + R D ) ⇒ R D = 1.49 k Ω
(b) g m = 2 K p I DQ = 2 (2.5)(1.2) = 3.464 mA/V
Aυ = g m (R D R L ) = (3.464)(1.49 4) = 3.76
______________________________________________________________________________________
4.53
(a)
KD
=
KL
Aυ =
(W L )D
=5
(W L )L
⎛W ⎞
So ⎜ ⎟ = 25
⎝ L ⎠D
From Example 4.11,
(3.3 − 0.6) + (0.6)(1 + 5) = 1.05 V
υ GSDt =
1+ 5
1.05 − 0.6
υ GSDQ =
+ 0.6 = 0.825 V
2
⎛ k ′ ⎞⎛ W ⎞
⎛ 0 .1 ⎞
2
(b) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (VGSDQ − VTN )2 = ⎜
⎟(25)(0.825 − 0.6 )
2
L
2
⎝
⎠
⎝
⎠
⎝ ⎠
D
I DQ = 0.0633 mA
Now I DD = I DL
(
)
(
)
⎛W ⎞
⎛W ⎞
2
2
⎜ ⎟ VGSDQ − VTN = ⎜ ⎟ VGSLQ − VTN
L
L
⎝ ⎠D
⎝ ⎠L
(
)
25
VGSDQ − VTN = V DD − VO − VTN
1
5(0.825 − 0.6 ) = 3.3 − VO − 0.6
Or V DSDQ = VO = 1.575 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.54
(a) Neglect λ in dc analysis.
Transition points:
For point B,
VOtB = V DD − VTNL = 5 − 0.8 = 4.2 V
For point A,
I DD = I DL
(
(1.2)(V
) = K (V
K nD VGSDQ − VTND
GSDQ
− 0 .6
2
nL
GSLQ
− VTNL
) = (0.2)[0 − (− 0.8)]
2
)
2
2
0. 2
(0.8) + 0.6 = 0.9266 V
1. 2
Then VOtA = VGSDQ − VTND = 0.9266 − 0.6 = 0.3266 V
So VGSDQ =
For point A: VOtA = 0.3266 V, V GSDQ = 0.9266 V
For point B: VOtB = 4.2 V, V GSDQ = 0.9266 V
(b) V GSDQ = 0.9266 V,
4.2 − 0.3266
+ 0.3266 = 2.2633 V
2
2
2
(c) I DQ = K nD VGSDQ − VTND = (1.2 )(0.9266 − 0.6) = 0.128 mA
V DSDQ =
(
(d) A = − g (r
υ
mD
roD = roL =
oD
)
roL )
1
1
=
= 390.6 k Ω
λ I DQ (0.02 )(0.128)
g mD = 2 K nD I DQ = 2 (1.2)(0.128) = 0.7838 mA/V
Aυ = −(0.7838)(390.6 390.6) = −153
______________________________________________________________________________________
4.55
(a) VTN = 0.6 V
I D = K n (V DS − VTN )
2
0.5 = K n (1.5 − 0.6) ⇒ K n = 0.6173 mA/V 2
2
1 dI D
=
= 2 K n (V DS − VTN )
r dV DS
Then r =
1
2 K n (V DS − VTN )
=
1
⇒ r = 900 Ω
2(0.6173)(1.5 − 0.6 )
(b) I D = (0.6173)(3 − 0.6) = 3.56 mA
2
1
⇒ r = 337 Ω
2(0.6173)(3 − 0.6)
______________________________________________________________________________________
r=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.56
a.
I DQ = K nD (VGS − VTND ) = ( 0.5 ) ( 0 − ( −1) )
2
2
I DQ = 0.5 mA
I DQ = K nL (VGSL − VTNL ) = K nL (VDD − VO − VTNL )
2
0.5 = 0.030 (10 − V0 − 1)
2
2
0.5
= 9 − V0 ⇒ V0 = 4.92 V
0.030
b.
I DD = I DL
K nD (Vi − VTND ) = K nL (VDD − Vo − VTNL )
2
2
K nD
(Vi − VTND ) = VDD − Vo − VTNL
K nL
K nD
(Vi − VTND )
K nL
Vo = VDD − VTNL −
Av =
dVo
K nD
=−
=−
dVi
K nL
(W / L )D
(W / L )L
500
⇒ Av = −4.08
30
______________________________________________________________________________________
Av = −
4.57
(a)
I DQ = K L (VGSL − VTNL ) = K L (VDSL − VTNL )
2
2
I D = ( 0.1)( 4 − 1) = 0.9 mA
2
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
K D (VGSD − VTND ) = K L (VGSL − VTNL )
2
2
KD
(VGG + Vi − Vo − VTND ) = Vo − VTNL
KL
⎛
KD ⎞
Vo ⎜ 1 +
⎟=
⎜
K L ⎟⎠
⎝
Av =
KD
(VGG + Vi − VTND ) + VTNL
KL
KD / KL
dVo
=
⇒
dVi 1 + K D / K L
Av =
1
1 + KL / KD
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) From Problem 4.55.
1
RLD =
2 K L (VDSL − VTNL )
1
=
2 ( 0.1)( 4 − 1)
= 1.67 k Ω
g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V
Av =
g m ( RLD || RL )
L
(1.90 )(1.67 || 4 )
⇒ Av = 0.691
1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 || 4 )
=
______________________________________________________________________________________
4.58
a.
From Problem 4.57.
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.
R0 =
1
1
1.67 = 0.526 ||1.67
RLD =
1.90
gm
R0 = 0.40 kΩ
______________________________________________________________________________________
4.59
(a) Aυ = − g mD roD roL
(
)
roD =
1
1
=
= 100 k Ω
λ D I DQ (0.02 )(0.5)
roL =
1
1
=
= 50 k Ω
λ L I DQ (0.04 )(0.5)
roD roL = 100 50 = 33.33 k Ω
Then − 40 = − g mD (33.33) ⇒ g mD = 1.20 mA/V
⎛ k ′ ⎞⎛ W ⎞
g mD = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠ D
( )
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
1.20 = 2 ⎜
⎟⎜ ⎟ (0.5) ⇒ ⎜ ⎟ = 14.4
⎝ L ⎠D
⎝ 2 ⎠⎝ L ⎠ D
⎛ k ′p ⎞⎛ W ⎞
2
(b) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGQ + VTP
L
2
⎝
⎠⎝ ⎠ L
⎛ 0.04 ⎞
2
0.5 = ⎜
⎟(50 )(V SGQ − 0.4 ) ⇒ V SGQ = 1.107 V
2
⎝
⎠
(
)
V SGQ = V + − V B
1.107 = 2.5 − V B ⇒ V B = 1.393 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ k ′ ⎞⎛ W ⎞
(c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (VGSDQ − VTN )2
⎝ 2 ⎠⎝ L ⎠ D
⎛ 0.1 ⎞
2
0.5 = ⎜
⎟(14.4)(VGSDQ − 0.4) ⇒ VGSDQ = 1.233 V
2
⎝
⎠
______________________________________________________________________________________
4.60
(a) (i) roD =
1
1
=
= 100 k Ω
λ D I DQ (0.04)(0.25)
roL =
1
1
=
= 200 k Ω
λ L I DQ (0.02 )(0.25)
roD roL = 100 200 = 66.67 k Ω
Aυ = − g mD (roD roL )
− 25 = − g mD (66.67 ) ⇒ g mD = 0.375 mA/V
⎛ k ′p ⎞⎛ W ⎞
⎟⎜ ⎟ I DQ
g mD = 2 ⎜⎜
⎟
⎝ 2 ⎠⎝ L ⎠ D
⎛ 0.04 ⎞⎛ W ⎞
⎛W ⎞
0.375 = 2 ⎜
⎟⎜ ⎟ (0.25) ⇒ ⎜ ⎟ = 7.03
2
L
⎝
⎠⎝ ⎠ D
⎝ L ⎠D
⎛ k ′ ⎞⎛ W ⎞
(ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V B − VTN )2
⎝ 2 ⎠⎝ L ⎠ L
⎛ 0.1 ⎞
2
0.25 = ⎜
⎟(10 )(V B − 0.4 ) ⇒ V B = 1.107 V
⎝ 2 ⎠
⎛ k ′p ⎞⎛ W ⎞
2
(iii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGDQ + VTP
L
2
⎝
⎠
D
⎝ ⎠
⎛ 0.04 ⎞
2
0.25 = ⎜
⎟(7.03)(V SGDQ − 0.6) ⇒ V SGDQ = 1.933 V
⎝ 2 ⎠
1
1
(b) (i) roD =
=
= 250 k Ω
λ D I DQ (0.04 )(0.1)
(
roL =
)
1
1
=
= 500 k Ω
λ L I DQ (0.02 )(0.1)
roD roL = 250 500 = 166.7 k Ω
Aυ = − g mD (roD roL )
− 25 = − g mD (166.7 ) ⇒ g mD = 0.15 mA/V
⎛ k ′p ⎞⎛ W ⎞
⎟⎜ ⎟ I DQ
g mD = 2 ⎜⎜
⎟
⎝ 2 ⎠⎝ L ⎠ D
⎛ 0.04 ⎞⎛ W ⎞
⎛W ⎞
0.15 = 2 ⎜
⎟⎜ ⎟ (0.1) ⇒ ⎜ ⎟ = 2.81
⎝ 2 ⎠⎝ L ⎠ D
⎝ L ⎠D
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ k ′ ⎞⎛ W ⎞
(ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V B − VTN )2
⎝ 2 ⎠⎝ L ⎠ L
⎛ 0.1 ⎞
2
0.1 = ⎜
⎟(10 )(V B − 0.4 ) ⇒ V B = 0.874 V
2
⎝
⎠
⎛ k ′p ⎞⎛ W ⎞
2
(iii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGDQ + VTP
L
2
⎝ ⎠⎝ ⎠ D
⎛ 0.04 ⎞
2
0.1 = ⎜
⎟(2.81)(V SGDQ − 0.6 ) ⇒ V SGDQ = 1.934 V
2
⎝
⎠
______________________________________________________________________________________
(
)
4.61
⎛ 85 ⎞
K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2
⎝ 2⎠
g m1 = 2 K n1 I D1 = 2 ( 2.125 )( 0.1) = 0.9220
ro1 =
ro 2 =
1
1
=
= 200 K
λ1 I D1 ( 0.05 )( 0.1)
1
1
=
= 133.3 K
( 0.075)( 0.1)
Av = − g m1 ( ro1 || ro 2 ) = − ( 0.922 )( 200 || 133.3)
λ2 I D 2
Av = −73.7
______________________________________________________________________________________
4.62
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
______________________________________________________________________________________
4.63
(a) I DD = I DL
K nD (VGSD − VTND ) = K nL (VGSL − VTNL )
2
2
(V I − VO − 0.4) = VO − 0.4
0.5
2V I − 0.8 = 3VO − 0.4
⎛2⎞
⎛1⎞
VO = ⎜ ⎟V I − ⎜ ⎟(0.4 )
⎝3⎠
⎝3⎠
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For V I = 0.8 V, VO = 0.4 V
For V I = 2.5 V, V O= 1.533 V
(b) I D = K nL (VGSL − VTNL ) = (0.5)(VO − 0.4)
2
2
2
2
⎡⎛ 2 ⎞
⎡⎛ 2 ⎞
⎤
⎤
⎛1⎞
I D = (0.5)⎢⎜ ⎟V I − ⎜ ⎟(0.4 ) − 0.4⎥ = (0.5)⎢⎜ ⎟V I − 0.533⎥
⎝ 3⎠
⎣⎝ 3 ⎠
⎦
⎣⎝ 3 ⎠
⎦
For V I = 0.8 V, I D = 0 ; For V I = 2.5 V, I D = 0.642 mA
(c) From (a), voltage gain = constant = 2/3 = 0.667
______________________________________________________________________________________
4.64
(a) V SD (sat ) = V SG + VTP = (2.5 − 1) − 0.4 = 1.1 V
V SD = V + − VO ⇒ VO (max ) = V + − V SD (sat ) = 2.5 − 1.1 = 1.4 V
⎛ k ′p ⎞⎛ W ⎞
⎛ 0.04 ⎞
2
2
(b) I D = ⎜⎜ ⎟⎟⎜ ⎟ (V SGL + VTP ) = ⎜
⎟(5)(2.5 − 1 − 0.4 ) = 0.121 mA
2
L
2
⎝
⎠
⎝
⎠⎝ ⎠ L
′
k
⎛ p ⎞⎛ W ⎞
2
(c) I D = ⎜⎜ ⎟⎟⎜ ⎟ (V SGD + VTP )
L
2
⎝
⎠⎝ ⎠ D
⎛ 0.04 ⎞
2
0.121 = ⎜
⎟(50 )(V SGD − 0.4 ) ⇒ V SGD = 0.748 V
2
⎝
⎠
⎛ k ′p ⎞⎛ W ⎞
⎛ 0.04 ⎞
(d) g mD = 2 ⎜⎜ ⎟⎟⎜ ⎟ I D = 2 ⎜
⎟(50 )(0.121) = 0.6957 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠ D
1
1
=
= 330.6 k Ω
roD = roL =
λ I D (0.025)(0.121)
Aυ =
g mD (roD roL )
(0.6957 )(165.3) = 0.9914
1 + g mD (roD roL ) 1 + (0.6957 )(165.3)
=
______________________________________________________________________________________
4.65
2
(a) I DQ = K n VGSDQ − VTN
(
(
)
) ⇒V
I = K (V
+V )
1 = 0.5(V
− 0.6 ) ⇒ V
1 = 2 VGSDQ − 0.6
GSDQ
= 1.307 V
SGLQ
= 2.014 V
2
2
DQ
p
SGLQ
TP
2
SGLQ
VO = 3.3 − 2.014 = 1.286 V
V DSDQ = VO − V S = 1.286 − (− 1.307 ) = 2.593 V
(b) I d = I l
g mDVi = g mLVo
Aυ =
Vo g mD
=
=
Vi
g mL
Kn
Kp
2
=2
0. 5
______________________________________________________________________________________
(c) Aυ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.66
(a)
⎛ 85 ⎞
K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2
⎝ 2⎠
g m1 = 2 K n1 I D1 = 2 ( 2.125 )( 0.1) = 0.922 mA/V
ro1 =
1
λ1 I D1
=
1
= 200 K
0.05
( )( 0.1)
ro 2 =
1
1
=
= 133.3 K
λ2 I D 2 ( 0.075 )( 0.1)
Ri1 =
1
1
=
= 1.085 K
g m1 0.922
(b)
Ri1
⎛
⎞
1.085
⎛
⎞
Vgs1 = − ⎜
⎟ Vi = − ⎜
⎟ Vi = −0.956Vi
+
+
R
0.050
1.085
0.050
⎝
⎠
⎝ i1
⎠
Vgs1
= + ( 0.956 )( 0.922 )( 200 )(133.3)
Av = − g m1 ( ro1 ro 2 ) ⋅
Vi
Av = 70.5
Ri = 0.05 +
(c)
1
1
= 0.05 +
⇒ Ri = 1.135 K
0.922
g m1
Ro ≈ ro1 ro 2 = 200 133.7 ⇒ Ro ≈ 80 K
(d)
______________________________________________________________________________________
4.67
(a)
g m1 = 2 K n I D1 = 2 ( 2 )( 0.1) = 0.8944 mA/V
g m 2 = 2 K p I D 2 = 2 ( 2 )( 0.1) = 0.8944 mA/V
1
1
=
= 100 K
λ I D ( 0.1)( 0.1)
(b) The small-signal equivalent circuit
ro1 = ro 2 =
g m1Vi +
(1)
Vsg 2
ro1
+ g m 2Vsg 2 +
Vsg 2 − Vo
ro 2
=0
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(2)
Vo Vo − Vsg 2
+
= g m 2Vsg 2
ro
ro 2
⎛1 1 ⎞
⎛ 1
⎞
Vo ⎜ +
+ gm2 ⎟
⎟ = Vsg 2 ⎜
r
r
r
⎝ o o2 ⎠
⎝ o2
⎠
1 ⎞
⎛ 1
⎛ 1
⎞
Vo ⎜ +
+ 0.8944 ⎟ ⇒ Vsg 2 = Vo ( 0.03317 )
⎟ = Vsg 2 ⎜
50
100
100
⎝
⎠
⎝
⎠
(1)
⎛ 1
1 ⎞ V
g m1Vi + Vsg 2 ⎜ + g m 2 + ⎟ = o
ro 2 ⎠ ro 2
⎝ ro1
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.
g m 2Vsg 2 + I x =
Vx Vx − Vsg 2
+
ro
ro 2
Vsg 2
Vsg 2 − Vx
(1)
(2)
(2)
ro1
+ g m 2Vsg 2 +
ro 2
=0
⎛ 1
1 ⎞ V
Vsg 2 ⎜ + g m 2 + ⎟ = x
r
r
ro 2
o2 ⎠
⎝ o1
1 ⎞ Vx
⎛ 1
Vsg 2 ⎜
+ 0.8944 +
⎟=
100 ⎠ 100
⎝ 100
Vsg 2 = Vx ( 0.010936 )
(1)
⎛1 1 ⎞
⎛ 1
⎞
I x = Vx ⎜ + ⎟ − Vsg 2 ⎜ + g m 2 ⎟
⎝ ro ro 2 ⎠
⎝ ro 2
⎠
1 ⎞
⎛ 1
⎛ 1
⎞
I x = Vx ⎜ +
+ 0.8944 ⎟
⎟ − Vx ( 0.010936 ) ⎜
⎝ 50 100 ⎠
⎝ 100
⎠
I x = Vx ( 0.03 − 0.0098905 )
V
Ro = x = 49.7 K
Ix
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.68
2
(a) I DQ1 = K n1 VGSQ1 − VTN 1
(
(
)
) ⇒V
I
= K (V
+V )
0.5 = 1.0(V
− 0.6 ) ⇒ V
0.2 = 0.2 VGSQ1 − 0.6
GSQ1
= 1.6 V
SGQ 2
= 1.307 V
2
2
DQ 2
p2
SGQ 2
TP 2
2
SGQ 2
0.6
= 3kΩ
0.2
V D1 = 0.6 + 2 = 2.6 V
R S1 =
5 − 2 .6
= 12 k Ω
0. 2
V G1 = V GSQ1 + 0.6 = 1.6 + 0.6 = 2.2 V
R D1 =
⎛ R2 ⎞
1
1
⎟⎟(5) =
(400 )(5)
⋅ Rin ⋅ (5) =
VG1 = 2.2 = ⎜⎜
R1
R1
⎝ R1 + R 2 ⎠
Or R1 = 909 k Ω and R1 R2 = Rin = 400 k Ω ⇒ R 2 = 714 k Ω
V S 2 = V D1 + V SGQ 2 = 2.6 + 1.307 = 3.907 V
5 − 3.907
= 2.19 k Ω
0 .5
V D 2 = V S 2 − 3 = 3.907 − 3 = 0.907 V
RS 2 =
RD2 =
0.907
= 1.81 k Ω
0.5
(b) g m1 = 2 K n1 I DQ1 = 2 (0.2)(0.2) = 0.4 mA/V
g m 2 = 2 K p 2 I DQ 2 = 2 (1)(0.5) = 1.414 mA/V
Aυ = (− g m1 R D1 )(− g m 2 R D 2 ) = g m1 g m 2 R D1 R D 2
Aυ = (0.4 )(1.414 )(12 )(1.81) = 12.3
______________________________________________________________________________________
4.69
2
(a) I DQ1 = K n1 (VGS1 − VTN 1 )
0.1 = 0.2(VGS1 − 0.6 ) ⇒ VGSQ1 = 1.307 V
2
(
I DQ 2 = K p 2 V SGQ 2 + VTP
(
0.25 = 1.0 V SGQ 2 − 0.6
)
2
) ⇒V
2
SGQ 2
= 1.10 V
V G1 = V GSQ1 + I DQ1 R S 1 = 1.307 + (0.1)(1) = 1.407 V
⎛ R2 ⎞
1
⎟⎟ ⋅ V DD =
⋅ Rin ⋅ V DD
VG1 = ⎜⎜
R1
⎝ R1 + R 2 ⎠
1
1.407 =
(250 )(3.3) ⇒ R1 = 586 k Ω
R1
R1 R2 = Rin = 250 k Ω ⇒ R 2 = 436 k Ω
V D1 = I DQ1 R S 1 + V DSQ1 = (0.1)(1) + 1.2 = 1.3 V
R D1 =
3. 3 − 1. 3
= 20 k Ω
0.1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V S 2 = V D1 + V SGQ 2 = 1.3 + 1.1 = 2.4 V
3. 3 − 2. 4
= 3. 6 k Ω
0.25
V D 2 = V S 2 − V SDQ 2 = 2.4 − 1.8 = 0.6 V
RS 2 =
RD2 =
0. 6
= 2. 4 k Ω
0.25
(b) g m1 = 2 K n1 I DQ1 = 2 (0.2)(0.1) = 0.2828 mA/V
g m 2 = 2 K p 2 I DQ 2 = 2 (1)(0.25) = 1.0 mA/V
Aυ = g m1 g m 2 R D1 R D 2 = (0.2828 )(1.0 )(20 )(2.4 ) = 13.6
______________________________________________________________________________________
4.70
⎛ 0.04 ⎞
2
K p1 = ⎜
⎟(20 ) = 0.4 mA/V
2
⎝
⎠
⎛ 0.1 ⎞
2
K n2 = ⎜
⎟(80 ) = 4.0 mA/V
⎝ 2 ⎠
0.6
(a) R S 1 =
= 6kΩ
0.1
V D1 = 1.8 − 0.6 − 1 = 0.2 V
R D1 =
0.2 − (− 1.8)
= 20 k Ω
0.1
(
I DQ1 = K p1 V SGQ1 + VTP
(
)
2
) ⇒V
I
= K (V
−V )
0.3 = 4(V
− 0.4) ⇒ V
0.1 = 0.4 V SGQ1 − 0.4
2
SGQ1
= 0.90 V
2
DQ 2
n2
GSQ 2
TN
2
GSQ 2
GSQ 2
= 0.6739 V
V G1 = 1.8 − 0.6 − V SGQ1 = 1.8 − 0.6 − 0.9 = 0.3 V
⎛ R2 ⎞
⎟⎟(3.6 ) − 1.8
VG1 = ⎜⎜
⎝ R1 + R 2 ⎠
1
0.3 =
(200)(3.6) − 1.8 ⇒ R1 = 343 k Ω
R1
R1 R2 = 200 k Ω ⇒ R 2 = 480 k Ω
V D1 = 1.8 − 0.6 − 1.0 = 0.2 V
V S 2 = V D1 − V GSQ 2 = 0.2 − 0.6739 = −0.4739 V
RS 2 =
− 0.4739 − (− 1.8)
= 4.42 k Ω
0. 3
⎛ − g m1 R D1 ⎞⎛ g m 2 R S 2 ⎞
⎟⎜
⎟
(b) Aυ = ⎜⎜
⎟⎜
⎟
⎝ 1 + g m1 R S1 ⎠⎝ 1 + g m 2 R S 2 ⎠
g m1 = 2 K p1 I DQ1 = 2 (0.4)(0.1) = 0.4 mA/V
g m 2 = 2 K n 2 I DQ 2 = 2 (4)(0.3) = 2.191 mA/V
Aυ =
− (0.4 )(20) (2.191)(4.42)
⋅
= −2.13
1 + (0.4 )(6) 1 + (2.191)(4.42)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
(c) R o =
RS 2 =
4.42 = 0.4564 4.42 ⇒ Ro = 414 Ω
g m2
2.191
______________________________________________________________________________________
4.71
(a)
I DQ1 =
10 − VGS1
2
= K n1 (VGS1 − VTN 1 )
RS 2
10 − VGS 1 = ( 4 )(10 ) (VGS2 1 − 4VGS 1 + 4 )
40VGS2 1 − 159VGS 1 + 150 = 0
(159 ) − 4 ( 40 )(150 )
⇒ VGS 1 = 2.435 V
2 ( 40 )
2
I DQ1 = ( 4 )( 2.435 − 2 ) ⇒ I DQ1 = 0.757 mA
VDSQ1 = 20 − ( 0.757 )(10 ) ⇒ VDSQ1 = 12.4 V
VGS 1 =
159 ±
2
Also I DQ 2 = 0.757 mA
VDSQ 2 = 20 − ( 0.757 )(10 + 5 ) ⇒ VDSQ 2 = 8.65 V
(b)
c.
g m1 = g m 2 = 2 KI DQ = 2 ( 4 )( 0.757 ) ⇒ g m1 = g m 2 = 3.48 mA / V
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.72
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)
R
+
R
+
R
⎝ 500 ⎠
2
3 ⎠
⎝ 1
R2 + R3 = 332.5 ⇒ R2 = 250 kΩ
R1 = 500 − 250 − 82.5 ⇒ R1 = 167.5 kΩ
VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V
5 − 3.6
RD =
⇒ RD = 0.467 kΩ
3
b.
Av = − g m1 RD
g m1 = 2 K n I DQ = 2 ( 2 )( 3) = 4.90 mA / V
Av = − ( 4.90 )( 0.467 ) ⇒ Av = −2.29
______________________________________________________________________________________
4.73
a.
VS 1 = I DQ RS − 10 = ( 5 )( 2 ) − 10 ⇒ VS 1 = 0
I DQ = K1 (VGS1 − VTN )
2
5 = 4 (VGS 1 − 1.5 ) ⇒ VGS1 = 2.618 V
2
VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 ⇒ R3 = 26.2 kΩ
VD1 = VS 1 + 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 = 0.6 kΩ
5
10 − 6.12
R1 =
⇒ R1 = 38.8 kΩ
0.1
RD =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
Av = − g m1 RD
g m1 = 2 K n I DQ = 2 ( 4 )( 5 ) = 8.944 mA / V
Av = − ( 8.944 )( 0.6 ) ⇒ Av = −5.37
______________________________________________________________________________________
4.74
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.
gm =
r0 =
c.
2 I DSS ⎛ VGS ⎞ 2 ( 6 ) ⎛ −0.551 ⎞
1−
1−
=
⎟ ⇒ g m = 3.265 mA/V
−3 ⎠
( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝
1
λ I DQ
=
1
( 0.01)( 4 )
⇒ r0 = 25 kΩ
Aυ = − g m (ro R D ) = −(3.265)(25 1) ⇒ Aυ = −3.14
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.75
VGS + I DQ ( RS1 + RS 2 ) = 0
⎛ V ⎞
I DQ = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
2
⎛ V ⎞
VGS + I DSS ( RS1 + RS 2 ) ⎜1 − GS ⎟ = 0
⎝ VP ⎠
2
⎛ V ⎞
VGS + ( 2 )( 0.1 + 0.25 ) ⎜1 − GS ⎟ = 0
⎝ VP ⎠
⎛ 2
V2 ⎞
VGS + 0.7 ⎜ 1 − VGS + GS 2 ⎟ = 0
⎜
⎟
⎝ ( −2 ) ( −2 ) ⎠
0.175VGS2 + 1.7VGS + 0.7 = 0
(1.7 ) − 4 ( 0.175)( 0.7 )
⇒ VGS = −0.4314 V
VGS =
2 ( 0.175 )
2 I DSS ⎛ VGS ⎞ 2 ( 2 ) ⎛ −0.431 ⎞
−1.7 ±
gm =
Av =
2
⎜1 −
⎟=
−VP ⎝
VP ⎠
− g m ( RD RL )
1 + g m RS1
=
⎜1 −
2 ⎝
⎟ ⇒ g m = 1.569 mA/V
−2 ⎠
− (1.569 ) ( 8 4 )
1 + (1.569 )( 0.1)
⇒ Av = −3.62
(v / R ) v R
i
⎛ 50 ⎞
Ai = 0 = 0 L = 0 ⋅ G = ( −3.62 ) ⎜ ⎟ ⇒ Ai = −45.2
ii ( vi / RG ) vi RL
⎝ 4⎠
______________________________________________________________________________________
4.76
I DSS
= 4 mA
2
V
VDSQ = DD = 10 V
2
VDSQ = VDD − I DQ ( RS + RD )
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 DQ = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
2
⎛
⎞
⎛
V
4⎞
4 = 8 ⎜⎜ 1 − GS ⎟⎟ ⇒ VGS = ( −4.2 ) ⎜⎜1 −
⎟⎟ ⇒ VGS = −1.23 V
−
4.2
8
)⎠
⎝
⎠
⎝ (
VG = VS + VGS = 2 − 1.23
⎛ R2 ⎞
⎛ R2 ⎞
VG = 0.77 V = ⎜
⎟ ( 20 ) = ⎜
⎟ ( 20 ) ⇒ R2 = 3.85 kΩ, R1 = 96.2 KΩ
⎝ 100 ⎠
⎝ R1 + R2 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.77
a.
I DSS
= 5 mA
2
V
12
=6V
VDSQ = DD =
2
2
12 − 6
⇒ RS = 1.2 kΩ
RS =
5
I DQ =
⎛ V ⎞
I DQ = I DSS ⎜1 − GS ⎟
⎝ VP ⎠
2
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
⎛ R2 ⎞
1
VG = ⎜
⎟ VDD = ⋅ Rin ⋅ VDD
R
R
R
+
⎝ 1
2 ⎠
1
1
4.536 = (100 )(12 ) ⇒ R1 = 265 kΩ
R1
265R2
= 100 ⇒ R2 = 161 kΩ
265 + R2
b.
gm =
r0 =
2 I DSS ⎛ VGS ⎞ 2 (10 ) ⎛ −1.46 ⎞
=
1−
⎜1 −
⎟ ⇒ g m = 2.83 mA/V
−5 ⎠
5 ⎝
( −VP ) ⎜⎝ VP ⎟⎠
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 )
1
1
1.2 = 0.353 1.2 ⇒ R0 = 0.273 kΩ
RS =
2.83
gm
______________________________________________________________________________________
R0 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.78
a.
⎛ R2 ⎞
⎛ 110 ⎞
VG = ⎜
⎟ VDD = ⎜
⎟ (10 ) = 5.5 V
⎝ 110 + 90 ⎠
⎝ R1 + R2 ⎠
I DQ =
10 − (VG − VGS )
⎛ V ⎞
= I DSS ⎜1 − GS ⎟
⎝ VP ⎠
RS
⎛ 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 = 0.511 V
2 ( 3.265 )
2
2
⎛ 0.511 ⎞
I DQ = ( 2 ) ⎜1 −
⎟ ⇒ I DQ = 1.00 mA
1.75 ⎠
⎝
VSDQ = 10 − (1.00 )( 5 ) ⇒ VSDQ = 5.0 V
b.
gm =
2 I DSS ⎛ VGS ⎞ 2 ( 2 ) ⎛ 0.511 ⎞
⎜1 −
⎟=
⎜1 −
⎟ ⇒ g m = 1.618 mA/V
1.75 ⎠
VP ⎝
VP ⎠ 1.75 ⎝
(1.618 ) ( 5 10 )
⇒ Av = 0.844
1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 )
⎛R ⎞
(v / R )
i
Ai = 0 = 0 L = Av ⋅ ⎜ i ⎟
ii
( vi / Ri )
⎝ RL ⎠
Av =
g m ( RS RL )
=
Ri = R1 R2 = 90 110 = 49.5 kΩ
⎛ 49.5 ⎞
Ai = ( 0.844 ) ⎜
⎟ ⇒ Ai = 4.18
⎝ 10 ⎠
c.
Δid = 1.0 mA
vsd = ( 3.33)(1.0 ) = 3.33 V
Maximum swing in output voltage = 6.66 V peak-to-peak
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.79
⎛ 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
gm =
2 I DSS ⎛ VGS ⎞ 2 ( 8 ) ⎛ 1.17 ⎞
⎜1 −
⎟=
⎜1 −
⎟ ⇒ g m = 2.83 mA/V
4 ⎝
4 ⎠
VP ⎝ VP ⎠
RS = 3.125 − RD
Av =
− g m RD
1 + g m RS
−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
⎝ 400 ⎠
⎝ R1 + R2 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 5
5.1
(a) i E = (1 + β )i B ⇒ 1 + β =
325
= 116 ⇒ β = 115
2.8
β
115
=
= 0.9914
1 + β 116
iC = i E − i B = 325 − 2.8 = 322 μ A
α=
(b) 1 + β =
1.80
= 90 ⇒ β = 89
0.020
89
= 0.9889
90
iC = 1.80 − 0.02 = 1.78 mA
______________________________________________________________________________________
α=
5.2
i
726
= 0.9918
(a) α = C =
i E 732
α
0.9918
=
= 121
1 − α 1 − 0.9918
i B = i E − iC = 732 − 726 = 6 μ A
β=
2.902
= 0.9801
2.961
α
0.980074
β=
=
= 49.19
1 − α 1 − 0.980074
i B = 2.961 − 2.902 ⇒ i B = 59 μ A
______________________________________________________________________________________
(b) α =
5.3
(a)
For
For
(b)
or
β = 110:
β = 180:
α=
α=
β
1+ β
=
110
= 0.99099
111
180
= 0.99448
181
0.99099 ≤ α ≤ 0.99448
I C = β I B = 110 ( 50 μ A ) ⇒ I C = 5.50 mA
I C = 180 ( 50 μ A ) ⇒ I C = 9.00 mA
5.50 ≤ I C ≤ 9.0 mA
so
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.4
(a) i B =
iE
1.25
=
⇒ i B = 8.28 μ A
1 + β 151
⎛ β ⎞
⎛ 150 ⎞
⎟⎟ ⋅ i E = ⎜
iC = ⎜⎜
⎟(1.25) = 1.242 mA
⎝ 151 ⎠
⎝1+ β ⎠
150
α=
= 0.9934
151
4.52
(b) i B =
⇒ i B = 55.8 μ A
81
⎛ 80 ⎞
i C = ⎜ ⎟(4.52 ) = 4.46 mA
⎝ 81 ⎠
80
= 0.9877
81
______________________________________________________________________________________
α=
5.5
(a)
α
β=
0.9
0.95
0.98
0.99
0.995
0.999
9
19
49
99
199
999
α
1−α
(b)
β
α=
β
1+ β
20
0.9524
50
0.9804
100
0.9901
150
0.9934
220
0.9955
400
0.9975
______________________________________________________________________________________
5.6
(a) I B =
IE
1.2
=
⇒ I B = 14.8 μ A
1 + β 81
⎛ β ⎞
⎛ 80 ⎞
⎟⎟ ⋅ I E = ⎜ ⎟(1.2 ) = 1.185 mA
I C = ⎜⎜
+
1
β
⎝ 81 ⎠
⎝
⎠
80
α=
= 0.9877
81
VC = 5 − (1.185)(2 ) = 2.63 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.8
⇒ I B = 9.88 μ A
81
⎛ 80 ⎞
I C = ⎜ ⎟(0.8) = 0.790 mA
⎝ 81 ⎠
α = 0.9877
VC = 5 − (0.790)(2 ) = 3.42 V
(b) I B =
1.2
⇒ I B = 9.92 μ A
121
⎛ 120 ⎞
IC = ⎜
⎟(1.2 ) = 1.19 mA
⎝ 121 ⎠
120
α=
= 0.9917
121
VC = 5 − (1.19 )(2 ) = 2.62 V
(c) (i) I B =
0.8
⇒ I B = 6.61 μ A
121
⎛ 120 ⎞
IC = ⎜
⎟(0.8) = 0.7934 mA
⎝ 121 ⎠
α = 0.9917
VC = 5 − (0.7934 )(2 ) = 3.41 V
______________________________________________________________________________________
(ii) I B =
5.7
⎛V ⎞
I E = I Eo exp⎜⎜ BE ⎟⎟
⎝ VT ⎠
⎛V ⎞
0.80 × 10 − 3 = 5 × 10 −14 exp⎜⎜ BE ⎟⎟
⎝ VT ⎠
(
)
⎛ 0.80 × 10 −3 ⎞
⎟ = 0.6109 V
Then V BE = (0.026) ln⎜⎜
−14 ⎟
⎝ 5 × 10
⎠
0.9910
α
=
= 110
β=
1 − α 1 − 0.9910
I C = αI E = (0.9910)(0.80) = 0.7928 mA
IE
0.80
=
⇒ I B = 7.21 μ A
1 + β 111
VC = 5 − I C RC = 5 − (0.7928)(2) = 3.41 V
______________________________________________________________________________________
IB =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.8
(a)
0.75
⇒ 12.3 μ A
61
⎛ 60 ⎞
I C = ( 0.75 ) ⎜ ⎟ = 0.738 mA
⎝ 61 ⎠
60
α=
= 0.9836
61
VC = I C RC − 10 = ( 0.738 )( 5 ) − 10
VC = −6.31 V
(b)
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 =
IB =
(c) Yes, VC < 0 in both cases so that B-C junction is reverse biased.
______________________________________________________________________________________
5.9
(a) VC = I C (5) − 10
− 1.2 = I C (5) − 10 ⇒ I C = 1.76 mA
I
1.76
IE = C =
= 1.785 mA
α 0.986
I B = I E − I C = 1.785 − 1.76 ⇒ I B = 25 μ A
⎛V ⎞
(b) I E = I Eo exp⎜⎜ EB ⎟⎟
⎝ VT ⎠
⎛V ⎞
1.785 × 10 −3 = 2 × 10 −15 exp⎜⎜ EB ⎟⎟
⎝ VT ⎠
⎛ 1.785 × 10 −3 ⎞
⎟ = 0.7154 V
V EB = (0.026) ln⎜⎜
−15
⎟
⎠
⎝ 2 × 10
______________________________________________________________________________________
5.10
⎛υ ⎞
⎛ 0.615 ⎞
iC = I S exp⎜⎜ BE ⎟⎟ = 5 × 10 −15 exp⎜
⎟
⎝ 0.026 ⎠
⎝ VT ⎠
i C = 93.7 μ A
(
)
93.7
= 0.7495 μ A
125
i E = (126 )(0.7495) = 94.44 μ A
______________________________________________________________________________________
iB =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.11
vEB / VT
⇒ 0.5 × 10 −3 = I Eo1e 0.650 / 0.026
Device 1: iE = I Eo1e
So that
I EO1 = 6.94 × 10−15 A
−3
0.650 / 0.026
Device 2: 12.2 × 10 = I Eo 2 e
Or
I Eo 2 = 1.69 × 10−13 A
I Eo 2 1.69 × 10 −13
=
⇒ Ratio = 24.4
I Eo1 6.94 × 10 −15
______________________________________________________________________________________
Ratio of areas =
5.12
For transistor A:
⎛ 275 × 10 −6 ⎞
⎛ IC ⎞
⎟
⎟ = (0.026 ) ln⎜
⎟
⎜ 8 × 10 −16 ⎟ = 0.6906 V
⎠
⎝
⎝ I SA ⎠
υ BE ( A) = VT ln⎜⎜
For transistor B:
I SB = 4 I SA = 4 8 ×10 −16 = 3.2 × 10 −15 A
(
)
⎛ 275 × 10 −6 ⎞
⎟ = 0.6546 V
−15 ⎟
⎝ 3.2 × 10 ⎠
______________________________________________________________________________________
υ BE (B ) = (0.026 ) ln⎜⎜
5.13
⎛ υ ⎞
(a) i C = I Co ⎜⎜1 + CE ⎟⎟
VA ⎠
⎝
2 ⎞
⎛
0.6 = I Co ⎜1 + ⎟ ⇒ I Co = 0.58537 mA
⎝ 80 ⎠
At υ CE = 5 V
5 ⎞
⎛
i C = (0.58537 )⎜1 + ⎟ = 0.622 mA
⎝ 80 ⎠
Δυ CE
5−2
(b) ro =
=
⇒ ro = 137 k Ω
Δi C
0.621956 − 0.60
______________________________________________________________________________________
5.14
BVC E 0 =
BVC B 0
3
β
= 3
60
100
BVC E 0 = 12.9 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.15
BVC E 0 =
56 =
220
3
β
BVC B 0
3
β
⇒3β =
220
= 3.93
56
β = 60.6
______________________________________________________________________________________
5.16
BVC E 0 =
BVC B 0
3
β
BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50
BVC B 0 = 184 V
______________________________________________________________________________________
5.17
−0.7 − ( −10 )
IE =
= 1.86 mA
5
⎛ 75 ⎞
I C = (1.86 ) ⎜ ⎟ = 1.836 mA
⎝ 76 ⎠
VC = −0.7 + 4 = 3.3 V
(a)
RC =
10 − 3.3
⇒ RC = 3.65 K
1.836
0.5
= 0.00658 mA
76
VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V
IB =
(b)
⎛ 75 ⎞
I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA
⎝ 76 ⎠
−1 − ( −5 )
⇒ RC = 8.11 K
RC =
0.493
IE
(10 ) + 0.7 + I E ( 4 ) − 8
76
7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA
O=
⎛ 75 ⎞
I C = (1.767 ) ⎜ ⎟ = 1.744 mA
⎝ 76 ⎠
VCE = 8 − (1.744 )( 4 ) − ⎡⎣(1.767 )( 4 ) − 8⎤⎦
(c)
= 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V
⎛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
(d)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.18
For Fig. P5.17(a), R E = 5 + 5% = 5.25 k Ω
IE =
−0.7 − ( −10 )
5.25
I C = 1.75 mA
= 1.77 mA
10 − 3.3
= 3.83 K
1.75
RE = 5 − 5% = 4.75 K
RC =
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. P5.17(c), R E = 4 + 5% = 4.2 k Ω
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
IB =
8 − 0.7
= 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
______________________________________________________________________________________
5.19
(a) VCC = I C RC + VCE
2.5 − 1.1
= 0.35 mA
4
⎛V ⎞
I C = I S exp⎜⎜ BE ⎟⎟
⎝ VT ⎠
IC =
⎛ 0.35 × 10 −3 ⎞
⎟ = 0.7091 V
V BE = V BB = (0.026 ) ln⎜⎜
−16 ⎟
⎠
⎝ 5 × 10
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) I E =
VCC − VCE 2.5 − 1.1
=
= 0.7 mA
RE
2
⎛ β ⎞
⎛ 90 ⎞
⎟⎟ ⋅ I E = ⎜ ⎟(0.70 ) = 0.6923 mA
I C = ⎜⎜
+
1
β
⎝ 91 ⎠
⎝
⎠
⎛ 0.6923 × 10 −3 ⎞
⎟ = 0.7269 V
V BE = (0.026 ) ln⎜⎜
−16
⎟
⎠
⎝ 5 × 10
V BB = V BE + I E R E = 0.7269 + (0.7 )(2 ) = 2.127 V
______________________________________________________________________________________
5.20
(a) I C = 0 , VCE = 2 V
(b) I C = βI B = (120 )(2 ) ⇒ I C = 0.24 mA
VCE = 2 − (0.24 )(4 ) = 1.04 V
1.4 − 0.7
= 0.35 mA
2
⎛ 120 ⎞
IC = ⎜
⎟(0.35) = 0.3471 mA
⎝ 121 ⎠
VCE = 2 − (0.3471)(4 ) − (0.35)(2 ) = −0.088 V - Not possible
Transistor in Saturation
VCE = 0.2 V
(c) I E =
V E = 0.7 V ⇒ VC = 0.9 V
2 − 0.9
= 0.275 mA
4
______________________________________________________________________________________
IC =
5.21
2 − (0.7 + 0.2 )
= 0.7333 mA
1.5
⎛ β ⎞
⎛ 120 ⎞
⎟⎟ ⋅ I E = ⎜
I C = ⎜⎜
⎟(0.7333) = 0.7273 mA
⎝ 121 ⎠
⎝ 1+ β ⎠
(a) I E =
V EC = V E = 0.9 V
(b) I C = βI B = (120 )(15) ⇒ I C = 1.8 mA - Not possible
Transistor in saturation
V EC = 0.2 V
2 − 0.2
= 1.2 mA
1.5
I C = I E − I B = 1.2 − 0.015 = 1.185 mA
IE =
(c) Transistor cutoff
I C = 0 , V EC = 2 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.22
V BB − V BE (on )
1.3 − 0.7
⇒ RB =
= 120 k Ω
RB
0.005
I CQ = β I BQ = (100 )(0.005 ) = 0.5 mA
(a) I BQ =
3 − 1.5
= 3kΩ
0.5
(b) For β = 75 , I CQ = (75 )(0.005 ) = 0.375 mA
RC =
VCE = 3 − (0.375)(3) = 1.875 V
For β = 125 , I CQ = (125 )(0.005 ) = 0.625 mA
VCE = 3 − (0.625)(3) = 1.125 V
So 1.125 ≤ VCE ≤ 1.875 V
______________________________________________________________________________________
5.23
(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=
β
⇒ α = 0.9926
1+ β
I C = β I B ⇒ I C = 0.269 mA
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
I B + IC = I E
IE =
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 =
β
⇒ α = 0.9140
1+ β
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.24
(V − 0.7 ) + 5
5 − VB
⎛ β ⎞
⎛ 90 ⎞
⎟⎟ ⋅ I E = ⎜ ⎟ ⋅ I E
, IE = B
, I C = ⎜⎜
10
3
⎝ 91 ⎠
⎝ 1+ β ⎠
(a) I C =
Then
5 − V B ⎛ 90 ⎞⎛ V B + 4.3 ⎞
= ⎜ ⎟⎜⎜
⎟⎟ ⇒ V B = −2.136 V
10
3
⎝ 91 ⎠⎝
⎠
−2.136 − 0.7 + 5
= 0.721 mA
3
(b) VCE = 10 − I C (10) − I E (3)
IE =
⎡ ⎛ 90 ⎞ ⎤
2 = 10 − I E ⎢3 + ⎜ ⎟(10 )⎥ = 10 − I E (12.89 )
⎣ ⎝ 91 ⎠ ⎦
Then I E = 0.6206 mA
And V B = 0.7 + (0.6206)(3) − 5 = −2.438 V
______________________________________________________________________________________
5.25
3.3 − 0.85
= 0.245 mA
10
0.85 − 0.7
IB =
⇒ IB = 3μ A
50
I C = I E − I B = 0.245 − 0.003 = 0.242 mA
(a) I E =
β=
I C 0.242
=
= 80.67
I B 0.003
β
80.67
= 0.9878
1 + β 81.67
VC = (0.242 )(10 ) − 3.3 = −0.88 V
V EC = 0.85 − (− 0.88) = 1.73 V
(b) β = (80.67 )(1.10 ) = 88.73
α=
IB =
=
V E − 0.7
3.3 − V E
⎛ V − 0.7 ⎞
, IE =
= (89.73)⎜⎜ E
⎟⎟ , ⇒ V E = 0.8371 V
50
10
⎝ 50 ⎠
3.3 − 0.8371
= 0.2463 mA
10
⎛ 88.73 ⎞
IC = ⎜
⎟(0.2463) = 0.2435 mA
⎝ 89.73 ⎠
IE =
VC = (0.2435)(10 ) − 3.3 = −0.8645 V
V EC = 0.8371 − (− 0.8645) = 1.70 V
______________________________________________________________________________________
5.26
5 − 0.7
⇒ 17.2 μ A
250
I C = (120 )( 0.0172 ) = 2.064 mA
VC = ( 2.064 )(1.5 ) − 5 = −1.90 V
VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V
IB =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
5.27
⎛ 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 =
______________________________________________________________________________________
5.28
0.5
⎛ 50 ⎞
I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B =
= 0.0098 mA
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 )
P = 3.91 mW
Or S
______________________________________________________________________________________
5.29
I
⇒ I E1 = I E 2 = 0.5 mA
2
I C1 = I C 2 ≈ 0.5 mA
I E1 = I E 2 =
VC1 = VC 2 = 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V
______________________________________________________________________________________
5.30
RE = 0 I B =
(a)
2 − 0.7 1.3
=
RB
RB
⎛ 1.3 ⎞ 5 − 2
I C = ( 80 ) ⎜ ⎟ =
= 0.8 ⇒ RC = 3.75 K
⎝ RB ⎠ RC
RB = 130 K
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.8
⎛ 81 ⎞
RE = 1 K I B =
= 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA
80
⎝ 80 ⎠
2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K
(b)
(c)
5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K
IB =
For part (a)
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.
______________________________________________________________________________________
5.31
(a) RC =
VCC − VCEQ
I CQ
=
9 − 4.5
= 18 k Ω
0.25
0.25
⇒ I BQ = 3.125 μ A
80
9 − 0.7
RB =
⇒ R B = 2.656 M Ω
0.003125
(b) I CQ = (120 )(0.003125 ) = 0.375 mA
I BQ =
V CEQ = 9 − (0.375 )(18 ) = 2.25 V
______________________________________________________________________________________
5.32
(a) I C = I E = 0 , VC = 6 V
(b) I E =
0.9 − 0.7
⎛ 150 ⎞
= 0.2 mA, I C = ⎜
⎟(0.2 ) = 0.1987 mA
1
⎝ 151 ⎠
VC = 6 − (0.1987 )(10 ) = 4.013 V
1.5 − 0.7
= 0.8 mA
1
Transistor in saturation
VC = 1.5 − 0.7 + 0.2 = 1 V
(c) I E =
6 −1
= 0.5 mA
10
2.2 − 0.7
= 1.5 mA
(d) I E =
1
VC = 2.2 − 0.7 + 0.2 = 1.7 V
IC =
6 − 1 .7
= 0.43 mA
10
______________________________________________________________________________________
IC =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.33
VBB = 0.
⎛ RL ⎞
⎛ 10 ⎞
Cutoff V0 = ⎜
⎟ VCC = ⎜
⎟ (5)
⎝ 10 + 5 ⎠
⎝ RC + RL ⎠
V0 = 3.33 V
a.
VBB = 1 V
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 ⎠
b.
V0 = VCE ( sat ) = 0.2 V
c.
Transistor in saturation
______________________________________________________________________________________
5.34
β = 100
(a)
⎛ 100 ⎞
I Q = 0.1 mA I C = ⎜
⎟ ( 0.1) = 0.0990 mA
⎝ 101 ⎠
VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V
(i)
⎛ 100 ⎞
I Q = 0.5 mA I C = ⎜
⎟ ( 0.5 ) = 0.495 mA
⎝ 101 ⎠
VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V
(ii)
I Q = 2 mA Transistor is in saturation
(iii)
(b)
VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V
β = 150
⎛ 150 ⎞
I Q = 0.1 mA I C = ⎜
⎟ ( 0.1) = 0.09934 mA
⎝ 151 ⎠
VO = 5 − ( 0.09934 )( 5) ⇒ VO = 4.503 V
(i)
% change =
4.503 − 4.505
× 100% = −0.044%
4.503
⎛ 150 ⎞
I Q = 0.5 mA I C = ⎜
⎟ ( 0.5 ) = 0.4967 mA
⎝ 151 ⎠
VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V
(ii)
% change =
2.517 − 2.525
× 100% = −0.32%
2.525
I Q = 2 mA Transistor in saturation
V = −8.5 V
No change
(iii) o
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.35
5−4
= 0.20 mA
5
⎛ 121 ⎞
IQ = I E = ⎜
⎟(0.20 ) = 0.2017 mA
⎝ 120 ⎠
(a) I C =
5−2
= 0.60 mA
5
⎛ 121 ⎞
IQ = I E = ⎜
⎟(0.60 ) = 0.605 mA
⎝ 120 ⎠
(b) I C =
5−0
= 1.0 mA
5
⎛ 121 ⎞
IQ = I E = ⎜
⎟(1.0 ) = 1.008 mA
⎝ 120 ⎠
______________________________________________________________________________________
(c) I C =
5.36
For
I Q = 0,
then
PQ = 0
⎛ 50 ⎞
I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5) = 0.49 mA
⎝ 51 ⎠
For
0.5
= 0.0098 mA, VB = 0.490 V , VE = 1.19 V
51
VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V
IB =
P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒
For
For
I Q = 1.0 mA,
Using the same calculations as above, we find P = 5.95 mW
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
P = 3.87 mW
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
______________________________________________________________________________________
5.37
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
5.38
12 − 6
2.727
= 2.727 mA, I B =
= 0.03409 mA
2.2
80
0.7 − (− 12 )
I R2 =
= 0.127 mA
100
I R1 = I B + I R 2 = 0.1611 mA
(a) I C =
V I = I R1 R1 + 0.7 = (0.1611)(15) + 0.7 = 3.12 V
12 − 9
= 1.364 mA, I B = 0.01705 mA
2.2
I R1 = 0.01705 + 0.127 = 0.14405 mA
(b) For V CEQ = 9 V, I C =
V I = (0.14405)(15) + 0.7 = 2.86 V
12 − 3
= 4.0909 mA, I B = 0.05114 mA
2.2
I R1 = 0.05114 + 0.127 = 0.1781 mA
For V CEQ = 3 V, I C =
V I = (0.1781)(15) + 0.7 = 3.37 V
So 2.86 ≤ V I ≤ 3.37 V
______________________________________________________________________________________
5.39
For VCE = 4.5
5 − 4.5
= 0.5 mA
I CQ =
1
0.5
I BQ =
= 0.02 mA
25
0.7 − ( −5 )
IR2 =
= 0.057 mA
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
I CQ =
= 4 mA
1
4
I BQ =
= 0.16 mA
25
I R 2 = 0.057 mA
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
5.40
5 − 2.5
=5K
0.5
0.5
IB =
= 0.00417 mA
120
5 − 0.7
RB =
= 1032 K
0.00417
RC =
(a)
Choose RC = 5.1 K
RB = 1 MΩ
(b)
For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K
5 − 0.7
= 3.91 μ A ⇒ I CQ = 0.469 mA
1.1
VCEQ = 2.37 V
I BQ =
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
I BQ =
= 4.78 μ A ⇒ I C = 0.573 mA
0.90
VCEQ = 1.78 V
RB = 1 MΩ − 10% = 0.90 MΩ
I BQ = 4.78 μ A ⇒ I C = 0.573 mA
VCEQ = 2.37 V
RC = 5.1 k − 10% = 4.59 K
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
5.41
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 ⎠
kT
ln (10 )
e
______________________________________________________________________________________
VO =
5.42
5−4
0.25
= 0.25 mA, I B =
= 0.002083 mA
120
4
V I = (0.002083)(200 ) + 0.7 = 1.117 V
(a) (i) I C =
⎛ 121 ⎞
(ii) I C = 0.25 mA, I E = ⎜
⎟(0.25) = 0.252 mA
⎝ 120 ⎠
V I = (0.002083)(200 ) + 0.7 + (0.252)(1) = 1.369 V
5 − 2.5
0.625
= 0.625 mA, I B =
= 0.005208 mA
120
4
V I = (0.005208)(200 ) + 0.7 = 1.742 V
(b) (i) I C =
⎛ 121 ⎞
(ii) I E = ⎜
⎟(0.625) = 0.6302 mA
⎝ 120 ⎠
V I = 1.742 + (0.6302 )(1) = 2.372 V
(c) Transistor biased in saturation
3.5 = I B (200 ) + 0.7 + I E (1)
V − 0.2
5 − VO
, IE = O
IC =
4
1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V − 0.2 ⎛ 5 − VO ⎞
I B = I E − IC = O
− ⎜⎜
⎟⎟ = VO (1.25) − 1.45
1
⎝ 4 ⎠
Then 3.5 = [VO (1.25) − 1.45](200 ) + 0.7 + (VO − 0.2 )
VO = 1.167 V
______________________________________________________________________________________
5.43
For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0
VO = 0
If Q reaches saturation, VO = 4.8
4.8
= 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
IC =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.44
For VI ≥ 4.3, Q is off and VO = 0
⎛ 101 ⎞
5=⎜
⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA
⎝ 100 ⎠
When transistor enters saturation,
(a)
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
______________________________________________________________________________________
5.45
IC =
VCC − VCE (sat ) 5 − 0.2
=
= 24 mA
0.2
RC
IC
24
= 20 ⇒ I B =
= 1.2 mA
IB
20
V I − V BE (on )
5 − 0.7
⇒ RB =
= 3.58 k Ω
RB
1.2
______________________________________________________________________________________
IB =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.46
(a) V I = 0 , ⇒ I B = I C = I E = 0, VO = 0
(b) V I = 2.5 V,
V I = I B R B + V BE (on ) + I E R E , and I E = (1 + β )I B
Then I B =
V I − V BE (on )
2.5 − 0.7
=
⇒ I B = 50.7 μ A
R B + (1 + β )R E 10 + (51)(0.5)
I C = (50 )(0.0507 ) = 2.535 mA, I E = (51)(0.0507 ) = 2.586 mA
VO = I E R E = (2.586 )(0.5) = 1.293 V
(c) V I = 5 V, Transistor in saturation
2.8
= 5.6 mA
0 .5
V − V BE (on ) − VO 5 − 0.7 − 2.8
=
= 0.15 mA
IB = I
10
RB
VO = 2.8 V, I E =
I C = I E − I B = 5.6 − 0.15 = 5.45 mA
______________________________________________________________________________________
5.47
IC =
VO 8.8
=
= 17.6 mA
RC 0.5
IC
17.6
= 25 ⇒ I B =
= 0.704 mA
IB
25
9 − V EB (on ) − V I
9 − 0.7 − 5
⇒ RB =
= 4.69 k Ω
RB
0.704
______________________________________________________________________________________
IB =
5.48
3 − 1.6
0.7
= 0.7 mA, I BQ =
⇒ I BQ = 5.833 μ A
120
2
V − V BE (on )
1 − 0 .7
R B = BB
=
= 51.4 k Ω
I BQ
0.005833
(a) I CQ =
(b) ΔVO = 3.3 − 0.2 = 2.8 V, peak-to-peak
ΔVO
3 − 1.6
=
= Aυ = −4.67
(c)
ΔV I
0.7 − 1.0
ΔVO (max )
2.8
(d) ΔV I (max ) =
=
= 0.6 V
4.667
Aυ
So υ i = 0.6 V, peak-to-peak
______________________________________________________________________________________
5.49
I BQ =
I CQ
β
=
0.15
⇒ I BQ = 1.25 μ A
120
⎛1+ β ⎞
⎛ 121 ⎞
⎟⎟ = (0.15)⎜
I EQ = I CQ ⎜⎜
⎟ = 0.15125 mA
β
⎝ 120 ⎠
⎝
⎠
We have RTH = 200 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.00125 )(200 ) + 0.7 + (0.15125 )(2 ) = 1.2525 V
⎛ R2 ⎞
1
⎟⎟ ⋅ VCC =
VTH = ⎜⎜
⋅ RTH ⋅ VCC
R
+
R
R
2 ⎠
1
⎝ 1
1
So 1.2525 =
(200)(2.5)
R1
⇒ R1 = 399 k Ω and R 2 = 401 k Ω
VCEQ = 2.5 − I CQ R C − I EQ R E = 2.5 − (0.15 )(6 ) − (0.15125 )(2 ) = 1.30 V
______________________________________________________________________________________
5.50
I CQ 0.20
⎛ 1+ β ⎞
⎛ 151 ⎞
⎟⎟ ⋅ I CQ = ⎜
=
⇒ I BQ = 1.33 μ A
I EQ = ⎜⎜
⎟(0.20 ) = 0.2013 mA, I BQ =
β
150
⎝ 150 ⎠
⎝ β ⎠
V CC = I CQ R C + V CEQ + I EQ R E
2.5 = (0.20 )RC + 1.5 + (0.2013)(1) ⇒ RC = 4 k Ω
VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.00133 )(120 ) + 0.7 + (0.2013 )(1) = 1.061 V
VTH =
1
1
⋅ RTH ⋅ VCC ⇒ 1.061 =
(120)(2.5)
R1
R1
So R1 = 283 k Ω and R 2 = 208 k Ω
______________________________________________________________________________________
5.51
RTH = R1 R2 = 20 15 = 8.57 k Ω
⎛ R2 ⎞
⎛ 15 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(10 ) = 4.29 V
⎝ 15 + 20 ⎠
⎝ R1 + R 2 ⎠
I EQ
VCC = I EQ R E + V EB (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
=
⇒ I EQ = 4.62 mA
Then I EQ =
8.57
1.085
1+
101
I EQ
⎛ 4.62 ⎞
VB =
⋅ RTH + VTH = ⎜
⎟(8.57 ) + 4.29 ⇒ V B = 4.68 V
1+ β
⎝ 101 ⎠
______________________________________________________________________________________
5.52
(a)
RTH = 42 58 = 24.36 k Ω
⎛ 42 ⎞
VTH = ⎜
⎟(24 ) = 10.08 V
⎝ 100 ⎠
10.08 − 0.7
9.38
I BQ =
=
⇒ I BQ = 7.30 μ A
24.36 + (126)(10) 1284.36
I CQ = 0.913 mA, I EQ = 0.9202 mA
V CEQ = 14.8 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
R1 + 5% = 60.9, R2 + 5% = 44.1
RTH = 25.58 K
10.08 − 0.7
9.38
I BQ =
=
⇒ 7.30 μ A
25.58 + 126 (10 ) 1285.58
I CQ = 0.912 mA
VCEQ = 14.81
I EQ = 0.919
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
VTH = 10.08
VTH = 9.50
I EQ = 0.8635 mA
VCEQ = 15.37 V
R1 − 5% = 55.1 K
R2 + 5% = 44.1 K
10.67 − 0.7
9.97
I BQ =
=
= 7.76 μ A
24.50 + 1260 1284.5
I CQ = 0.970 mA
I EQ = 0.978 mA
VCEQ = 14.22 V
RTH = 24.50 K
VTH = 10.67 V
R1 − 5% = 55.1 K
R2 − 5% = 39.90
10.08 − 0.7
9.38
I BQ =
=
= 7.31 μ A
23.14 + 1260 1283.14
I CQ = 0.914 mA
I EQ = 0.9211 mA
VCEQ = 14.79 V
RTH = 23.14 K
VTH = 10.08
So we have 0.857 ≤ I CQ ≤ 0.970 mA
14.22 ≤ VCEQ ≤ 15.37 V
______________________________________________________________________________________
5.53
(a) RTH = R1 R2 = 96 24 = 19.2 k Ω
⎛ R2 ⎞
⎛ 24 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(9 ) = 1.80 V
⎝ 24 + 96 ⎠
⎝ R1 + R 2 ⎠
V − V BE (on )
1.80 − 0.7
I BQ = TH
=
⇒ I BQ = 10.98 μ A
RTH + (1 + β )R E 19.2 + (81)(1)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I CQ = β I BQ = (80 )(0.01098 ) = 0.8782 mA, I EQ = (81)(0.01098 ) = 0.8892 mA
V CEQ = 9 − (0.8782 )(5.25 ) − (0.8892 )(1) = 3.50 V
1.80 − 0.7
⇒ I BQ = 7.846 μ A
19.2 + (121)(1)
I CQ = (120 )(0.007846 ) = 0.9415 mA, I EQ = (121)(0.007846 ) = 0.9494 mA
(b) I BQ =
V CEQ = 9 − (0.9415 )(5.25 ) − (0.9494 )(1) = 3.108 V
⎛ 0.9415 − 0.8782 ⎞
For I CQ : ⎜
⎟ × 100% = 7.21%
0.8782
⎝
⎠
⎛ 3.108 − 3.500 ⎞
For VCEQ : ⎜
⎟ × 100% = −11.2%
3.500
⎝
⎠
______________________________________________________________________________________
5.54
(a)
I CQ ≅ I EQ = 0.4 mA
3
3
RC =
⇒ RC = 7.5 k Ω; RE =
⇒ RE = 7.5 k Ω
0.4
0.4
9
R1 + R2 ≅
= 112.5 k Ω
( 0.2 )( 0.4 )
⎛ 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) = 0.004R2 − 3.56 × 10−5 R22 + 3.73
R = 48 k Ω ⇒ R1 = 64.5 k Ω
From this quadratic, we find 2
(b) Standard resistor values:
Set R E = RC = 7.5 k Ω and R1 = 62 k Ω , R 2 = 47 k Ω
Now RTH = R1 R 2 = 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.55
(a)
RTH = R1 R2 = 12 2 = 1.714 k Ω
⎛ R2 ⎞
⎛ 2⎞
⎟⎟(10 ) − 5 = ⎜ ⎟(10 ) − 5 ⇒ VTH = −3.571 V
VTH = ⎜⎜
⎝ 14 ⎠
⎝ R1 + R 2 ⎠
(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
______________________________________________________________________________________
5.56
(a) RTH = R1 R2 = 40 40 = 20 k Ω
⎛ R 2 ⎞ + ⎛ 40 ⎞
⎟⎟ ⋅ V = ⎜
VTH = ⎜⎜
⎟(2.5) = 1.25 V
⎝ 40 + 40 ⎠
⎝ R1 + R 2 ⎠
V + = I EQ R E + V EB (on ) + I BQ RTH + VTH
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V + − V EB (on ) − VTH
2.5 − 0.7 − 1.25
=
⇒ I BQ = 6.57 μ A
RTH + (1 + β )R E
20 + (91)(0.7 )
I CQ = 0.5914 mA, I EQ = 0.5980 mA
So I BQ =
V ECQ = 2.5 − (0.5914 )(1.6 ) − (0.5980 )(0.7 ) = 1.135 V
2.5 − 0.7 − 1.25
⇒ I BQ = 4.375 μ A
20 + (151)(0.7 )
I CQ = 0.6563 mA, I EQ = 0.6607 mA
(b) I BQ =
V ECQ = 2.5 − (0.6563 )(1.6 ) − (0.6607 )(0.7 ) = 0.9874 V
⎛ 0.6563 − 0.5914 ⎞
For I CQ : ⎜
⎟ × 100% = 10.97%
0.5914
⎝
⎠
⎛ 0.9874 − 1.135 ⎞
For V ECQ : ⎜
⎟ × 100% = −13.0%
1.135
⎝
⎠
______________________________________________________________________________________
5.57
(a)
RTH = 36 68 = 23.5 k Ω
⎛ 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
VCEQ = 10 − (0.0888 )(42 ) − (0.0906 )(30 ) ⇒ V CEQ = 3.55 V
(b)
R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K
RTH = 7.85 k
I BQ =
VTH = 3.46
3.46 − 0.7
= 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.58
(a)
⎛ 68 ⎞
RTH = 36 68 = 23.5 k Ω ; VTH = ⎜
⎟(10 ) − 5 = 1.54 V
⎝ 36 + 68 ⎠
5 = (51)I BQ (30 ) + 0.7 + I BQ (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 Ω
VTH = 1.54 V, R E = 10 k Ω , RC = 14 k Ω
5 = (51)I BQ (10 ) + 0.7 + I BQ (7.85 ) + 1.54
2.76
⇒ 5.33 μ A, I CQ = 0.266 mA, I EQ = 0.272 mA
517.85
V ECQ = 10 − (0.272 )(10 ) − (0.266 )(14 ) = 3.56 V
I BQ =
______________________________________________________________________________________
5.59
(a)
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω
VTH =
I BQ =
Then
1
⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
R1
I CQ
β
=
0.8
= 0.008 mA
100
1
( 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 ⎠
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
β = 75, I BQ =
For
Then
1.145 − 0.7
= 0.0103 mA
5.05 + ( 76 )( 0.5 )
I CQ = ( 75 )( 0.0103) = 0.775 mA
β = 150, I BQ =
For
Then
1.145 − 0.7
= 0.00552 mA
5.05 + (151)( 0.5 )
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
RTH = ( 0.1)(101)(1) = 10.1 k Ω
1
1
VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1)
R1
R1
(c)
which yields R1 = 63.6 k Ω
63.6 R2
= 10.1 ⇒ R2 = 12.0 k Ω
63.6 + R2
And
⎛ R2 ⎞
⎛ 12 ⎞
VTH = ⎜
⎟ (VCC ) = ⎜
⎟ (10 ) = 1.587 V
R1 + R2 ⎠
12 + 63.6 ⎠
⎝
⎝
Now
β = 75, I BQ =
For
So
1.587 − 0.7
= 0.0103 mA
10.1 + ( 76 )(1)
I CQ = 0.773 mA
β = 150, I BQ =
For
Then
1.587 − 0.7
= 0.00551 mA
10.1 + (151)(1)
I CQ = 0.826 mA
% Change =
ΔI CQ
I CQ
=
0.826 − 0.773
× 100% ⇒ % Change = 6.63%
0.8
______________________________________________________________________________________
5.60
VCC ≅ I CQ ( RC + RE ) + VCEQ
10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω
Let RE = 0.875 k Ω
R = ( 0.1)(121)( 0.875 ) = 10.6 k Ω
Then, for bias stable TH
0.8
I BQ =
= 0.00667 mA
120
1
(10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 )
R1
71.8R2
= 10.6 ⇒ R2 = 12.4 k Ω
71.8
+ R2
R
=
71.8
k
Ω
1
So
and
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10
= 0.119 mA
+ 12.4
71.8
Then
This is close to the design specification.
______________________________________________________________________________________
IR ≅
5.61
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 + V BE (on ) + (1 + β )I BQ R E − 6
⎛ R2 ⎞
⎟⎟(12) − 6 , RTH = R1 R 2
VTH = ⎜⎜
⎝ R1 + R 2 ⎠
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
______________________________________________________________________________________
5.62
(a) RTH = (0.1)(1 + β )R E = (0.1)(121)(0.2 ) = 2.42 k Ω
⎡
⎛ 1+ β ⎞ ⎤
⎟⎟ R E ⎥ = 6 − I CQ (2.202 )
VCEQ = 6 − I CQ ⎢ RC + ⎜⎜
⎝ β ⎠ ⎦
⎣
2.8 = 6 − I CQ (2.202 ) ⇒ I CQ = 1.453 mA
Then I EQ = 1.465 mA, I BQ = 12.11 μ A
VTH = I BQ RTH + V BE (on ) + I EQ R E − 3 = (0.01211)(2.42 ) + 0.7 + (1.465 )(0.2 ) − 3
Then VTH = −1.978 V =
1
1
⋅ RTH (6) − 3 =
(2.42)(6) − 3
R1
R1
Which yields R1 = 14.2 k Ω and R 2 = 2.92 k Ω
(b) For R1 = (1.05)(14.2) = 14.91 k Ω
R 2 = (0.95)(2.92 ) = 2.774 k Ω
RTH = R1 R 2 = 2.34 k Ω
2.774
⎛
⎞
VTH = ⎜
⎟(6) − 3 = −2.059 V
⎝ 2.774 + 14.91 ⎠
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3 − 2.059 − 0.7
⇒ I BQ = 9.08 μ A
So I BQ =
2.34 + (121)(0.2)
I CQ = 1.090 mA, I EQ = 1.099 mA
V CEQ = 6 − (1.09 )(2 ) − (1.099 )(0.2 ) = 3.60 V
For R1 = (0.95)(14.2 ) = 13.49 k Ω
R 2 = (1.05)(2.92 ) = 3.066 k Ω
RTH = R1 R 2 = 2.50 k Ω
3.066
⎛
⎞
VTH = ⎜
⎟(6 ) − 3 = −1.889 V
⎝ 3.066 + 13.49 ⎠
3 − 1.889 − 0.7
I BQ =
⇒ I BQ = 15.39 μ A
2.50 + (121)(0.2)
I CQ = 1.847 mA, I EQ = 1.863 mA
V CEQ = 6 − (1.847 )(2 ) − (1.863)(0.2 ) = 1.933 V
So 1.09 ≤ I CQ ≤ 1.847 mA
1.933 ≤ VCEQ ≤ 3.60 V
______________________________________________________________________________________
5.63
Let
VCEQ ≅ VCC − I CQ ( RC + RE )
5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω
RE = 0.333 k Ω
R = 2 kΩ
and C
β = 100
Nominal value of
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 kΩ
3
I BQ =
= 0.03 mA
100
1
1
VTH = ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6
R1
R1
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
VTH + 6 − 0.7
−4.19 + 6 − 0.7
I BQ =
=
= 0.0387 mA ⇒ I C = 2.90 mA
RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333)
β = 150, I BQ =
For
−4.19 + 6 − 0.7
= 0.0207 mA
3.36 + (151)( 0.333)
I = 3.10 mA
Then C
Specifications are met.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.64
⎡
⎛ 1+ β ⎞ ⎤
⎟⎟ R E ⎥
(a) V + = V ECQ + I CQ ⎢ RC + ⎜⎜
⎝ β ⎠ ⎦
⎣
3.3 = 1.5 + I CQ (3.022 ) ⇒ I CQ = 0.5956 mA
I BQ = 6.618 μ A, I EQ = 0.6022 mA
V + = I EQ R E + V EB (on ) + I BQ RTH + VTH
3.3 = (0.6022 )(2 ) + 0.7 + (0.006618)(2.4 ) + VTH
So VTH = 1.380 V =
1
1
⋅ RTH ⋅ V + =
(2.4)(3.3)
R1
R1
Which yields R1 = 5.74 k Ω and R 2 = 4.12 k Ω
V + − V EB (on ) − VTH 3.3 − 0.7 − 1.38
=
⇒ I BQ = 4.61 μ A
RTH + (1 + β )R E
2.4 + (131)(2 )
I CQ = 0.60 mA, I EQ = 0.6045 mA
(b) I BQ =
V ECQ = 3.3 − (0.60 )(1) − (0.6045 )(2 ) = 1.49 V
______________________________________________________________________________________
5.65
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
1
1
VTH = ⋅ 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
______________________________________________________________________________________
5.66
For I EQ ≅ I CQ , RC + R E =
VCC − VCEQ
I CQ
=
2.5 − 1.6
= 4.5 k Ω
0.2
So R E = 0.5 k Ω
For β = 100
RTH = (0.1)(101)(0.5) = 5.05 k Ω
0.2
⇒ I BQ = 2 μ A, I EQ = 0.202 mA
100
VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.002 )(5.05 ) + 0.7 + (0.202 )(0.5) = 0.8111 V
I BQ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTH =
1
1
⋅ RTH ⋅ VCC ⇒ 0.8111 =
(5.05)(2.5)
R1
R1
So R1 = 15.6 k Ω and R 2 = 7.47 k Ω
For β = 80 ,
VTH − V BE (on )
0.8111 − 0.7
=
⇒ I BQ = 2.439 μ A, I CQ = 0.1951 mA
RTH + (1 + β )R E 5.05 + (81)(0.5)
For β = 120 ,
I BQ =
0.8111 − 0.7
⇒ I BQ = 1.695 μ A, I CQ = 0.2034 mA
5.05 + (121)(0.5)
Design is valid
______________________________________________________________________________________
I BQ =
5.67
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
60
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE
R1
I BQ =
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
β = 45
Check: For
⎛ 51.2 ⎞
VTH = ⎜
⎟ (15 ) = 6.21V
⎝ 51.2 + 72.5 ⎠
V − VBE ( on )
6.21 − 0.7
I BQ = TH
=
= 0.0215 mA
RTH + (1 + β ) RE 30 + ( 46 )( 4.92 )
I CQ = 0.967 mA,
Check: For
I BQ =
β = 75
ΔI C
= 3.27%
IC
6.21 − 0.7
= 0.0136 mA
30.0 + ( 76 )( 4.92 )
ΔI C
= 2.31%
IC
Design criterion is satisfied.
______________________________________________________________________________________
I CQ = 1.023 mA,
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.68
(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
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω
RC = 13.3 k Ω, I BQ =
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
______________________________________________________________________________________
5.69
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
R1
⎝ R1 + R2 ⎠
I EQ
= 0.0165 mA
I BQ =
1+ β
5 = I EQ RE + VEB ( on ) + I B RTH + VTH
5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) +
0.80 =
1
( 30.3)( 4 ) − 2
R1
1
( 30.3)( 4 ) ⇒ R1 = 152 kΩ
R1
152 R2
= 30.3 ⇒ R2 = 37.8 kΩ
152 + R2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.70
a.
RTH = R1 R2 = 10 20 = 6.67 k Ω
⎛ R2 ⎞
⎛ 20 ⎞
⎟⎟(10) − 5 = ⎜
VTH = ⎜⎜
⎟(10 ) − 5 = 1.67 V
R
+
R
⎝ 20 + 10 ⎠
2 ⎠
⎝ 1
b.
10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
I BQ =
10 − 0.7 − 1.67 7.63
=
⇒ 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
______________________________________________________________________________________
5.71
V + − V − ≅ I CQ ( RC + RE ) + VECQ
20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω
Let
Let
RE = 10 k Ω
RC = 14 k Ω
then
β = 60 from previous problem.
RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 )
Or RTH = 61 k Ω
0.5
I BQ =
= 0.00833 mA
60
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5
R1
⎝ R1 + R2 ⎠
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 kΩ
10
10
=
⇒ 18.4 μ A
R1 + R2 70 + 474
40 μ A
So the
current limit is met.
______________________________________________________________________________________
IR ≅
5.72
⎡
⎤
⎛ 1+ β ⎞
⎟⎟ ⋅ R E ⎥
(a) V ECQ = V + − V − − I CQ ⎢ RC + ⎜⎜
⎝ β ⎠
⎣
⎦
(
)
⎡
⎛ 81 ⎞ ⎤
2.7 = 5 − (0.15)⎢ RC + ⎜ ⎟(2 )⎥ ⇒ RC = 13.3 k Ω
⎝ 80 ⎠ ⎦
⎣
I EQ = 0.1519 mA, I BQ = 1.875 μ A
RTH = (0.1)(81)(2 ) = 16.2 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V + = I EQ R E + V EB (on ) + I BQ RTH + VTH
2.5 = (0.1519 )(2 ) + 0.7 + (0.001875)(16.2) + VTH ⇒ VTH = 1.466 V
VTH =
1
1
⋅ RTH (5) − 2.5 ⇒ 1.466 =
(16.2)(5) − 2.5
R1
R1
So R1 = 20.4 k Ω and R 2 = 78.7 k Ω
(b) For β = 60
V + − VTH − V EB (on ) 2.5 − 1.466 − 0.7
=
⇒ I BQ = 2.417 μ A
RTH + (1 + β )R E
16.2 + (61)(2 )
I CQ = 0.145 mA , I EQ = 0.1474 mA
I BQ =
V ECQ = 5 − (0.145 )(13.3) − (0.1474 )(2 ) = 2.777 V
For β = 100
2.5 − 1.466 − 0.7
⇒ I BQ = 1.531 μ A
16.2 + (101)(2)
I CQ = 0.1531 mA, I EQ = 0.1546 mA
I BQ =
V ECQ = 5 − (0.1531)(13.3) − (0.1546 )(2 ) = 2.655 V
⎛ 0.1531 − 0.145 ⎞
For I CQ : ⎜
⎟ × 100% = 5.4%
0.15
⎝
⎠
⎛ 2.655 − 2.777 ⎞
For V ECQ : ⎜
⎟ × 100% = −4.52%
2.70
⎝
⎠
____________________________________________________________________________________
5.73
a.
RTH = 500 500 70 = 250 70 = 54.7 k Ω
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.74
RE ≅
VE
1.5
=
= 1.87 k Ω
I CQ 0.8
10 = (0.8)RC + 4 + 1.5 ⇒ RC = 5.63 k Ω
RTH = (0.1)(1 + β )R E = (0.1)(121)(1.87 ) = 22.6 k Ω
I BQ =
0.8
⇒ I BQ = 6.67 μ A
120
VTH = I BQ RTH + V BE (on ) + I EQ R E =
1
⋅ RTH ⋅ VCC
R1
(0.00667 )(22.6) + 0.7 + (0.807 )(1.87 ) = 1 (22.6)(10)
R1
which yields R1 = 95.8 k Ω and R 2 = 29.6 k Ω
______________________________________________________________________________________
5.75
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 R 2 , Design bias stable circuit
RTH = (0.1)(51)(19.8) = 101 k Ω
⎛ R2 ⎞
1
⎟⎟(10 ) − 5 =
⋅ RTH ⋅ (10 ) − 5
VTH = ⎜⎜
R1
⎝ R1 + R 2 ⎠
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
573 R2
= 101
573 + R2
R2 = 123 K
(b)
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.76
0.7
0.5
= 1.4 k Ω , I BQ =
⇒ I BQ = 4.167 μ A
0.5
120
V CEQ = 6 − I CQ R C − 0.7
(a) R E ≅
2.5 = 6 − (0.5)RC − 0.7 ⇒ RC = 5.6 k Ω
RTH = (0.1)(121)(1.4 ) = 16.9 k Ω
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 3
VTH = (0.004167 )(16.9 ) + 0.7 + (121)(0.004167 )(1.4 ) − 3 = −1.5237 V
1
1
⋅ RTH (6 ) − 3 ⇒ −1.5237 =
(16.9)(6) − 3
R1
R1
VTH =
which yields R1 = 68.7 k Ω and R 2 = 22.4 k Ω
(b) For standard resistor values:
Let R E = 1.5 k Ω , RC = 5.6 k Ω , R1 = 68 k Ω , R 2 = 22 k Ω
RTH = R1 R 2 = 68 22 = 16.62 k Ω
⎛ R2 ⎞
⎛ 22 ⎞
⎟⎟(6) − 3 = ⎜
VTH = ⎜⎜
⎟(6 ) − 3 = −1.533 V
R
+
R
⎝ 22 + 68 ⎠
2 ⎠
⎝ 1
3 + VTH − V BE (on )
3 − 1.533 − 0.7
I BQ =
=
⇒ I BQ = 3.87 μ A
RTH + (1 + β )R E
16.62 + (121)(1.5)
I CQ = 0.4646 mA, I EQ = 0.4684 mA
V CEQ = 6 − (0.4646 )(5.6 ) − (0.4684 )(1.5) = 2.70 V
______________________________________________________________________________________
5.77
0.7
6−3
= 7 k Ω , RC + R E =
= 30 k Ω , ⇒ RC = 23 k Ω
0.1
0.1
RTH = (0.1)(1 + β )R E = (0.1)(111)(7 ) = 77.7 k Ω
(a) R E ≅
I BQ =
0 .1
⇒ I BQ = 0.909 μ A, I EQ = 0.1009 mA
110
V + = I EQ R E + V EB (on ) + I BQ RTH + VTH
3 = (0.1009 )(7 ) + 0.7 + (0.000909 )(77.7 ) + VTH , ⇒ VTH = 1.523 V
VTH =
1
1
⋅ RTH (6 ) − 3 ⇒ 1.523 =
(77.7 )(6) − 3
R1
R1
which yields R1 = 103 k Ω and R 2 = 316 k Ω
V + − V EB (on ) − VTH
3 − 0.7 − 1.523
=
⇒ I BQ = 0.685 μ A
RTH + (1 + β )R E
77.7 + (151)(7 )
I CQ = 0.1027 mA, I EQ = 0.1034 mA
I BQ =
V ECQ = 6 − (0.1027 )(23) − (0.1034 )(7 ) = 2.914 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.78
(a) V ECQ = 18 − I CQ R C − V RE
6 = 18 − (1.2 )RC − 1.5 ⇒ RC = 8.75 k Ω
1.2
⎛ 76 ⎞
I EQ = ⎜ ⎟(1.2 ) = 1.216 mA, I BQ =
⇒ I BQ = 16 μ A
75
75
⎝ ⎠
1.5
RE =
= 1.234 k Ω
1.216
RTH = (0.1)(1 + β )R E = (0.1)(76 )(1.234 ) = 9.375 k Ω
V + = I EQ R E + V EB (on ) + I BQ RTH + VTH
9 = (1.216 )(1.234 ) + 0.7 + (0.016 )(9.375) + VTH ⇒ VTH = 6.65 V
VTH =
1
(RTH )(18) − 9 ⇒ 6.65 = 1 (9.375)(18) − 9
R1
R1
which yields R1 = 10.78 k Ω and R 2 = 71.8 k Ω
Set RC = 9.1 k Ω , R E = 1.2 k Ω , R1 = 11 k Ω , R 2 = 68 k Ω
RTH = R1 R2 = 11 68 = 9.47 k Ω
⎛ R2 ⎞
⎛ 68 ⎞
⎟⎟(18) − 9 = ⎜
VTH = ⎜⎜
⎟(18) − 9 = 6.494 V
⎝ 68 + 11 ⎠
⎝ R1 + R 2 ⎠
V + − V EB (on ) − VTH
9 − 0.7 − 6.494
I BQ =
=
⇒ I BQ = 17.94 μ A
RTH + (1 + β )R E
9.47 + (76 )(1.2 )
I CQ = 1.345 mA, I EQ = 1.363 mA
V ECQ = 18 − (1.345 )(9.1) − (1.363)(1.2 ) = 4.12 V
______________________________________________________________________________________
5.79
RTH = R1 R2 = 100 40 = 28.6 k Ω
⎛ R2 ⎞
⎛ 40 ⎞
⎟⎟(10 ) = ⎜
VTH = ⎜⎜
⎟(10 ) = 2.86 V
R
+
R
⎝ 40 + 100 ⎠
2 ⎠
⎝ 1
V − V BE (on )
2.86 − 0.7
=
I B1 = TH
RTH + (1 + β )R E1 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.74 − 0.7 − ( −10 )
IE2 =
⇒ I E 2 = 2.808 mA
5
I B 2 = 0.0232 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
______________________________________________________________________________________
5.80
VE1 = −0.7
I R1 =
−0.7 − ( −5 )
= 0.215 mA
20
VE 2 = −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
______________________________________________________________________________________
5.81
V B1 = V RE + V BE (on ) = 0.5 + 0.7 = 1.2 V
1.2
= 60 k Ω
0.020
0.5
RE =
= 2.5 k Ω
0.2
V B 2 = V BE (on ) + VCE + V RE = 0.7 + 1.2 + 0.5 = 2.4 V
R3 =
R2 =
V B 2 − V B1 2.4 − 1.2
=
= 60 k Ω
I R2
0.020
R1 =
V + − V B 2 5 − 2.4
=
= 130 k Ω
0.020
I R1
VC 2 = 2VCE + V RE = 2(1.2) + 0.5 = 2.9 V
V + − VC 2 5 − 2.9
=
= 10.5 k Ω
0.20
I C2
______________________________________________________________________________________
RC =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.82
RTH = 40 80 = 26.67 k Ω
⎛ 40 ⎞
VTH = ⎜
⎟(9 ) = 3 V
⎝ 40 + 80 ⎠
VTH − V BE (on )
3 − 0 .7
=
⇒ I B1 = 8.56 μ A
I B1 =
RTH + (1 + β n )R E1 26.67 + (121)(2)
I C1 = 1.027 mA, I E1 = 1.036 mA
9 − V C1
+ I B 2 = I C1
2
9 = I E 2 (0.1) + V EB (on ) + VC1 ⇒ I B 2 =
So
9 − 0.7 − VC1 8.3 − VC1
=
1 + β p (0.1)
8.1
(
)
9 − VC1 8.3 − VC1
+
= 1.027 ⇒ VC1 = 7.214 V
2
8.1
8.3 − 7.214
I B2 =
= 0.134 mA
8.1
I C 2 = 10.73 mA, I E 2 = 10.86 mA
V E1 = I E1 R E1 = (1.036 )(2 ) = 2.072 V
VCE1 = VC1 − V E1 = 7.214 − 2.072 = 5.14 V
V EC 2 = 9 − (10.86 )(0.1) − (10.73)(0.2 ) = 5.77 V
______________________________________________________________________________________
5.83
RTH = R1 R 2 = 50 100 = 33.3 k Ω
⎛ R2 ⎞
⎛ 100 ⎞
⎟⎟(10 ) − 5 = ⎜
VTH = ⎜⎜
⎟(10 ) − 5 = 1.67 V
⎝ 100 + 50 ⎠
⎝ R1 + R 2 ⎠
5 = I E1 R E1 + V EB (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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA
−0.87 − ( −5 )
⇒ RC1 = 5.21 kΩ
0.792
______________________________________________________________________________________
RC1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 6
6.1
I CQ
(a) (i) g m =
VT
β VT
rπ =
ro =
I CQ
=
=
0.5
= 19.23 mA/V
0.026
(180)(0.026) = 9.36 k Ω
0.5
V A 150
=
= 300 k Ω
I CQ 0.5
2
= 76.92 mA/V
0.026
(180)(0.026) = 2.34 k Ω
rπ =
2
150
ro =
= 75 k Ω
2
0.25
(b) (i) g m =
= 9.615 mA/V
0.026
(80)(0.026) = 8.32 k Ω
rπ =
0.25
100
ro =
= 400 k Ω
0.25
0.08
(ii) g m =
= 3.077 mA/V
0.026
(80)(0.026) = 26 k Ω
rπ =
0.08
100
ro =
= 1250 k Ω
0.08
______________________________________________________________________________________
(ii) g m =
6.2
(a) g m =
rπ =
ro =
I CQ
VT
β VT
I CQ
⇒ I CQ = (95)(0.026 ) = 2.47 mA
=
(125)(0.026) = 1.32 k Ω
2.47
VA
200
=
= 81 k Ω
I CQ 2.47
(b) I CQ = g mVT = (120)(0.026) = 3.12 mA
β = g m rπ = (120)(1.2) = 144
______________________________________________________________________________________
6.3
0.8
1.2
= 30.77 mA/V; g m =
= 46.15 mA/V
0.026
0.026
So 30.77 ≤ g m ≤ 46.15 mA/V
gm =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
rπ =
(180)(0.026) = 5.85 k Ω ; r = (90)(0.026) = 1.95 k Ω
π
0.8
1.2
So 1.95 ≤ rπ ≤ 5.85 k Ω
______________________________________________________________________________________
6.4
(
)
⎛ 0.12 ⎞
−3
(a) iC = I CQ + g mυ be = 0.12 + ⎜
⎟ 5 × 10 sin ω t
0
.
026
⎝
⎠
iC = 0.12 + 0.0231 sin ω t (mA)
υ CE = VCEQ − i C RC = [3.3 − (0.12)(15)] − (15)(0.0231 sin ω t )
υ CE = 1.5 − 0.346 sin ω t (V)
υ
− 0.346
= −69.2
(b) Aυ = ce =
0.005
υ be
______________________________________________________________________________________
6.5
(a) I CQ =
β (V BB − V BE (on ))
RB
=
(120)(1.10 − 0.7 ) = 0.436 mA
110
0.436
= 16.78 mA/V
0.026
(120)(0.026) = 7.15 k Ω
rπ =
0.436
80
ro =
= 183 k Ω
0.436
⎞
⎛ rπ
7.15 ⎞
⎟ = −(16.78)(4 183)⎛⎜
(b) Aυ = − g m (RC ro )⎜⎜
⎟
⎟
+ 110 ⎠
r
R
+
7
.
15
⎝
B ⎠
⎝ π
Aυ = −4.0
gm =
0.5 sin (100t )
= −0.125 sin (100t ) (V)
Aυ
−4
______________________________________________________________________________________
(c) υ s =
υo
=
6.6
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
0.578
= 0.00482 mA
120
VBB = I BQ RB + VBE ( on )
= ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V
I BQ =
I CQ
β
=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
rπ =
gm =
r0 =
β VT
I CQ
I CQ
VT
=
=
(120 )( 0.026 )
0.578
= 5.40 kΩ
0.578
= 22.2 mA/V
0.026
VA
100
=
= 173 kΩ
I CQ 0.578
⎛ r
⎞
V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS
r
R
+
B ⎠
⎝ π
β ( r0 RC )
⎛ r
⎞
Av = − g m ⎜ π ⎟ ( r0 RC ) = −
rπ + RB
⎝ rπ + RB ⎠
Av = −
(120 ) ⎡⎣173 4.33⎤⎦
(120 )( 4.22 )
⇒ Av = −16.7
5.40 + 25
30.4
______________________________________________________________________________________
6.7
rπ =
=−
(120)(0.026) = 6.24 k Ω
0.5
(a) Ri = R B + rπ = 50 + 6.24 = 56.24 k Ω
(b) Ri = R B rπ = 100 6.24 = 5.87 k Ω
(c) Ri = r π = 6.24 k Ω
______________________________________________________________________________________
6.8
ro =
VA
80
=
= 400 k Ω
I CQ 0.2
(a) R o = RC ro = 4 400 = 3.96 k Ω
(b) R o = RC ro = 10 400 = 9.76 k Ω
(c) R o = RC R L ro = 10 5 400 = 3.333 400 = 3.31 k Ω
______________________________________________________________________________________
6.9
10 − 4
= 1.5 mA
4
1.5
= 0.015 mA
I BQ =
100
(100 )( 0.026 )
= 1.73 K
rπ =
1.5
5sin ω t ( mV )
v
= 2.89sin ω t ( μ A )
ib = be =
1.73 kΩ
rπ
I CQ =
So
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
i B (t ) = I BQ + i b (t ) = 15 + 2.89 sin ω t ( μ A)
iC1 (t ) = β i B (t ) = 1.5 + 0.289 sin ω t (mA)
υ C1 (t ) = 10 − iC1 (t )RC = 10 − [1.5 + 0.289 sin ω t ](4 ) (V)
υ C1 (t ) = 4 − 1.156 sin ω t (V)
υ (t ) − 1.156
Aυ = c
=
⇒ Aυ = −231
υ be (t ) 0.005
______________________________________________________________________________________
6.10
a.
1
VECQ = VCC = 5 V
2
VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ
0.5
= 0.005
β 100
VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V
I BQ =
I CQ
=
gm =
I CQ
=
b.
rπ =
r0 =
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
β RC
Av = −
=−
(100 )(10 ) ⇒ A = −18.1
5.2 + 50
rπ + RB
c.
______________________________________________________________________________________
6.11
v
vo = 1.2sin ω t ( V )
iC ( t ) RC + vo = 0 ⇒ iC ( t ) =
iC ( t ) = −0.60sin ω t ( mA )
−1.2sin ω t
2
i (t )
ib ( t ) = C
= −6sin ω t ( μ A )
β
vbe ( t ) = ib ( t ) ⋅ rπ g m rπ = β
100
rπ =
=2K
50
vbe ( t ) = −12sin ω t ( mV )
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.12
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 R 2 = RTH = (0.1)(1 + β )R E
I BQ =
= (0.1)(151)(0.2) = 3.02 k Ω
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.
rπ =
(150)(0.026) = 1.09 k Ω
3.57
3.57
gm =
= 137 mA/V
0.026
− β RC
− (150 )(1.2 )
=
⇒ Aυ = −5.75
rπ + (1 + β )R E 1.09 + (151)(0.2)
______________________________________________________________________________________
Aυ =
6.13
(a) RTH = R1 R 2 = 33 50 = 19.88 k Ω
⎛ R2 ⎞
⎛ 50 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(3.3) = 1.988 V
⎝ 50 + 33 ⎠
⎝ R1 + R 2 ⎠
VCC = I BQ (1 + β )R E + V EB (on ) + I BQ RTH + VTH
3.3 − 0.7 − 1.988
= 0.005063 mA
19.88 + (101)(1)
I CQ = 0.506 mA; I EQ = 0.511 mA
Then I BQ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V ECQ = VCC − I CQ RC − I EQ R E = 3.3 − (0.506 )(2 ) − (0.511)(1)
V ECQ = 1.78 V
(b) rπ =
Aυ =
β VT
I CQ
=
(100)(0.026) = 5.14 k Ω
0.506
− β RC
− (100 )(2 )
=
= −1.884
rπ + (1 + β )R E 5.14 + (101)(1)
(c) R1 = (1.05)(33) = 34.65 k Ω
R 2 = (0.95)(50 ) = 47.5 k Ω
RTH = R1 R 2 = 34.65 47.5 = 20.03 k Ω
47.5
⎛
⎞
VTH = ⎜
⎟(3.3) = 1.908 V
⎝ 47.5 + 34.65 ⎠
⎛ 3.3 − 0.7 − 1.908 ⎞
⎟⎟ = 0.5718 mA
I CQ = (100)⎜⎜
⎝ 20.03 + (101)(1) ⎠
(100 )(0.026) = 4.547 k Ω
rπ =
0.5718
− (100 )(2)
= −1.895
Aυ =
4.547 + (101)(1)
Also R1 = (0.95)(33) = 31.35 k Ω
R 2 = (1.05)(50 ) = 52.5 k Ω
RTH = 31.35 52.5 = 19.63 k Ω
52.5
⎞
⎛
VTH = ⎜
⎟(3.3) = 2.066 V
52
.
5
31
.
35
+
⎠
⎝
⎛ 3.3 − 0.7 − 2.066 ⎞
⎟⎟ = 0.4427 mA
I CQ = (100)⎜⎜
⎝ 19.63 + (101)(1) ⎠
rπ =
(100)(0.026) = 5.873 k Ω
0.4427
− (100 )(2 )
= −1.871
Aυ =
5.873 + (101)(1)
So 1.871 ≤ Aυ ≤ 1.895
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.14
(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
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω
I BQ =
⎛ 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
− β RC
− (100 )(2 )
=
⇒ Aυ = −1.95
rπ + (1 + β )R E 1.31 + (101)(1)
______________________________________________________________________________________
(b)
Aυ =
6.15
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.16
(a) I CQ ≅
VCC − VCEQ
RC + R E
=
9 − 5.20
= 0.905 mA
2.2 + 2
I BQ ≅ 0.00754 mA; I EQ ≅ 0.9123 mA
RTH = (0.1)(1 + β )R E = (0.1)(121)(2) = 24.2 k Ω
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E
= (0.00754 )(24.2 ) + 0.7 + (0.9123)(2 ) = 2.707 V
Now VTH = 2.707 =
1
1
⋅ RTH ⋅ VCC =
(24.2)(9)
R1
R1
Or R1 = 80.5 k Ω and then R 2 = 34.6 k Ω
(120)(0.026) = 3.448 k Ω
(b) rπ =
0.905
0.905
gm =
= 34.81 mA/V
0.026
100
ro =
= 110 k Ω
0.905
(
) (
Vπ = i s R1 R 2 rπ = i s 80.5 34.6 3.448
Vπ = i s (3.02 )
(
υ o = − g mVπ ro RC R L
Then R m =
υo
is
)
)
(
)
= − g m (3.02) 110 2.2 1 = −(34.81)(3.02 )(110 0.6875)
Or R m = −71.8 V/mA
(c) For β = 100 ,
2.707 − 0.7
= 0.008873 mA
24.2 + (101)(2)
I CQ = 0.8873 mA
I BQ =
(100)(0.026) = 2.93 k Ω , g
0.8873
100
= 34.13 mA/V, ro =
= 113 k Ω
m =
0.8873
0.026
0.8873
Vπ = i s (RTH rπ ) = i s (24.2 2.93) = i s (2.614 )
rπ =
υo
= −(34.13)(2.614 )(113 0.6875) = −61.0 V/mA
is
For β = 150 ,
Rm =
⎛ 2.707 − 0.7 ⎞
⎟⎟ = 0.923 mA
I CQ = (150 )⎜⎜
⎝ 24.2 + (151)(2 ) ⎠
rπ = 4.225 k Ω , g m = 35.5 mA/V, ro = 108 k Ω
Vπ = i s (24.2 4.225) = i s (3.597 )
Rm =
υo
is
= −(35.5)(3.597 )(108 0.6875) = −87.2 V/mA
So 61.0 ≤ R m ≤ 87.2 V/mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.17
0.8
= 0.009877 mA, I CQ = 0.790 mA
81
0.2
RB =
= 20.25 k Ω
0.009877
− 3 − (− 5)
RC =
= 2.53 k Ω
0.79
0.79
100
(ii) g m =
= 30.38 mA/V, ro =
= 127 k Ω
0.026
0.79
ro RC = 127 2.53 = 2.48 k Ω
(a) (i) I BQ =
⎛ 2.48 ⎞
i o = − g mυ s ⎜
⎟
⎝ 2.48 + 4 ⎠
i
⎛ 2.48 ⎞
G f = o = −(30.38)⎜
⎟ = −11.63 mA/V
υs
⎝ 2.48 + 4 ⎠
0.8
= 0.00661 mA, I CQ = 0.7934 mA
121
0.2
RB =
= 30.3 k Ω
0.00661
− 3 − (− 5)
RC =
= 2.52 k Ω
0.7934
0.7934
80
(ii) g m =
= 30.52 mA/V, ro =
= 101 k Ω
0.026
0.7934
(b) (i) I BQ =
ro RC = 101 2.52 = 2.459 k Ω
⎛ 2.459 ⎞
G f = −(30.52)⎜
⎟ = −11.62 mA/V
⎝ 2.459 + 4 ⎠
______________________________________________________________________________________
6.18
0.25
= 0.00208 mA, I EQ = 0.252 mA
120
5 = (0.00208)(2.5) + 0.7 + (0.252 )R E ⇒ R E = 17.0 k Ω
(a) (i) I BQ =
VCEQ = 3 = 10 − (0.252 )(17 ) − (0.25)RC ⇒ RC = 10.9 k Ω
(120)(0.026) = 12.48 k Ω
0.25
= 9.615 mA/V, rπ =
0.026
0.25
⎛ rπ
⎞
12.48 ⎞
⎟ = −(9.615)(10.9 5)⎛⎜
Aυ = − g m (RC R L )⎜⎜
⎟ = −27.5
⎟
⎝ 12.48 + 2.5 ⎠
⎝ rπ + R S ⎠
(ii) g m =
(
)
(iii) υ o = −(27.49 ) 5 × 10 −3 sin ω t = −0.137 sin ω t (V)
(b) (i) 5 = 0.7 + (0.252 )R E ⇒ R E = 17.1 k Ω
VCEQ = 3 = 10 − (0.252)(17.06) − (0.25)RC ⇒ RC = 10.8 k Ω
(ii) g m = 9.615 mA/V, rπ = 12.48 k Ω
Aυ = − g m (RC R L ) = −(9.615)(10.8 5) = −32.9
(
)
(iii) υ o = −(32.86 ) 5 × 10 −3 sin ω t = −0.164 sin ω t (V)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.19
5 − 0.7
= 0.005292 mA, ⇒ I CQ = 0.4234 mA
2.5 + (81)(10 )
(80)(0.026) = 4.91 k Ω
0.4234
gm =
= 16.28 mA/V, rπ =
0.026
0.4234
⎛ rπ ⎞
4.91 ⎞
⎟ = −(16.28)(5 5)⎛⎜
Aυ = − g m (RC R L )⎜⎜
⎟ = −26.97
⎟
+
4
.
91
+ 2.5 ⎠
r
R
⎝
S ⎠
⎝ π
υ
Aυ
io = o = υ s
RL
RL
(a) (i) I BQ =
or
Gf =
io
υs
=
(
− 26.97
= −5.39 mA/V
5
)
(ii) υ o = −(26.97 ) 4 × 10 −3 sin ω t = −0.108 sin ω t (V)
(
i o = − 5.39 × 10
−3
)(4 ×10 sin ω t ) ⇒ −21.6 sin ω t ( μ A)
−3
⎛
⎞
5 − 0.7
⎟⎟ = 0.4256 mA
(b) (i) I CQ = (120 )⎜⎜
⎝ 2.5 + (121)(10 ) ⎠
(120)(0.026) = 7.33 k Ω
0.4256
gm =
= 16.37 mA/V, rπ =
0.026
0.4256
⎛ 7.33 ⎞
Aυ = −(16.37 )(5 5)⎜
⎟ = −30.5
⎝ 7.33 + 2.5 ⎠
Gf =
−30.5
= −6.1 mA/V
5
(
)
(ii) υ o = −(30.5) 4 × 10 −3 sin ω t = −0.122 sin ω t (V)
i o = −24.4 sin ω t ( μ A)
______________________________________________________________________________________
6.20
RTH = R1 R2 = 27 15 = 9.64 K
⎛ R2 ⎞
⎛ 15 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ ( 9 ) = 3.214 V
⎝ 15 + 27 ⎠
⎝ R1 + R2 ⎠
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 =
ro =
(100 )( 0.026 )
1.92
= 73.9 mA/V rπ =
= 1.35 K
0.026
1.92
100
= 52.1 K
1.92
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
Vo = − g mVπ r0 RC RL
⎛
rπ RTH
⎝ π
TH
⎞
) V = ⎜⎜ r R + R ⎟⎟V
π
S
⎠
S
rπ RTH = 1.35 9.64 = 1.184 K
⎛ 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 ⎜⎜
⎟
ro RC + RL ⎟⎠
Io
⎝
=
AI =
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
______________________________________________________________________________________
6.21
a.
0.35
= 0.00347 mA
101
V B = − I B R B = −(0.00347 )(10 ) ⇒ V B = −0.0347 V
I E = 0.35 mA, I B =
V E = V B − V BE (on ) ⇒ V E = −0.735 V
b.
VC = VCEQ + V E = 3.5 − 0.735 = 2.77 V
⎛ β ⎞
⎛ 100 ⎞
⎟⎟ ⋅ I E = ⎜
I C = ⎜⎜
⎟(0.35) = 0.347 mA
⎝ 101 ⎠
⎝ 1+ β ⎠
V + − VC 5 − 2.77
RC =
=
⇒ RC = 6.43 k Ω
IC
0.347
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c)
⎛ R B rπ ⎞
⎟(R r )
Aυ = − g m ⎜
⎜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
rπ =
0.347
R B rπ = 10 7.49 = 4.28 k Ω
⎛ 4.28 ⎞
Aυ = −(13.3)⎜
⎟(6.43 288) ⇒ Aυ = −81.7
⎝ 4.28 + 0.1 ⎠
d.
⎛ R B rπ ⎞
⎟(R r )
Aυ = − g m ⎜
⎜R r +R ⎟ C o
S ⎠
⎝ B π
R B rπ = 10 7.49 = 4.28 k Ω
⎛ 4.28 ⎞
Aυ = −(13.3)⎜
⎟(6.43 288) ⇒ Aυ = −74.9
⎝ 4.28 + 0.5 ⎠
______________________________________________________________________________________
6.22
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
=
= 0.0155 mA
I BQ = TH
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.
rπ =
(180)(0.026) = 1.67 k Ω
2.80
2.80
= 108 mA/V
gm =
0.026
(c)
⎛ R1 R 2 rπ ⎞
⎟(R R )
Aυ = g m ⎜
⎜R R r +R ⎟ C L
S ⎠
⎝ 1 2 π
R1 R 2 rπ = 6 1.5 1.67 = 0.698 k Ω
⎛ 0.698 ⎞
Aυ = (108)⎜
⎟(1 1.2) = 45.8
⎝ 0.698 + 0.2 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.23
a.
9 = I EQ RE + VEB ( on ) + I BQ RS
I EQ = 0.75 mA, I BQ =
0.75
= 0.00926 mA
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.
(
⎛ rπ ⎞
⎟ RC R L ro
Aυ = − g m ⎜⎜
⎟
⎝ rπ + R S ⎠
(80)(0.026) = 2.81 k Ω
rπ =
0.741
80
= 108 k Ω
ro =
0.741
−80
Aυ =
3.71 10 108
2.81 + 2
Aυ = −43.9
(
)
)
d.
Ri = R S + rπ = 2 + 2.81 ⇒ Ri = 4.81 k Ω
______________________________________________________________________________________
6.24
⎡ 4 − 0.7 ⎤
(a) For β = 80 , I CQ = (80 )⎢
⎥ = 0.6439 mA
⎣ 5 + (81)(5) ⎦
(80)(0.026) = 3.23 k Ω , r = 50 = 77.7 k Ω
0.6439
gm =
= 24.77 mA/V, rπ =
o
0.026
0.6439
0.6439
(3.23 5)
rπ R B
(
)
⋅υ s =
Vπ =
(3.23 5) + 1 ⋅υ s = 0.6624 υ s
rπ R B + R S
Aυ = − g m ⋅
Vπ
υs
(
)
(
⎡ 4 − 0.7 ⎤
For β = 120 , I CQ = (120)⎢
⎥ = 0.6492 mA
⎣ 5 + (121)(5) ⎦
g m = 24.97 mA/V, rπ = 4.806 k Ω , ro = 77 k Ω
Vπ =
)
⋅ ro RC R L = −(24.77 )(0.6624) 77.7 4 4 = −32
(4.806 5)
(
)
(4.806 5) + 1 ⋅υ = 0.710 υ
s
(
)
s
Aυ = −(24.97 )(0.710 ) 77 4 4 = −34.6
So 32 ≤ Aυ ≤ 34.6
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Ri = R B rπ
For β = 80 , Ri = 5 3.23 = 1.96 k Ω
For β = 120 , Ri = 5 4.806 = 2.45 k Ω
So 1.96 ≤ Ri ≤ 2.45 k Ω
(c) R o = RC ro
For β = 80 , R o = 4 77.7 = 3.804 k Ω
For β = 120 , R o = 4 77 = 3.802 k Ω
So 3.802 ≤ R o ≤ 3.804 k Ω
______________________________________________________________________________________
6.25
Assume an npn transistor with β = 100 and V A = ∞ . Let VCC = 10 V.
0.5
= 50
0.01
I CQ = 1 mA
Av =
Bias at
and let RE = 1 k Ω
For a bias stable circuit
RTH = (0.1)(1 + β )R E = (0.1)(101)(1) = 10.1 k Ω
VTH =
1
1
⋅ RTH ⋅ VCC =
(10.1)(10) = 101
R1
R1
R1
1
= 0.01 mA
100
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E
I BQ =
101
= (0.01)(10.1) + 0.7 + (101)(0.01)(1)
R1
which yields R1 = 55.8 k Ω , R 2 = 12.3 k Ω
Now
(100)(0.026) = 2.6 k Ω
rπ =
1
1
gm =
= 38.46 mA/V
0.026
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V o = − g m Vπ R C
⎞
⎛ R1 R 2 rπ
⎟ ⋅ V = ⎛⎜ 10.1 2.6 ⎞⎟ ⋅ V = 0.674V
where Vπ = ⎜
s
⎜ R R r + R ⎟ s ⎜ 10.1 2.6 + 1 ⎟ s
S ⎠
⎝
⎠
⎝ 1 2 π
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 Ω , R E = 1 k Ω
R1 = 56 k Ω , R 2 = 12 k Ω
RTH = R1 R2 = 9.88 K
⎛ R2 ⎞
⎛ 12 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 1.765 V
⎝ 12 + 56 ⎠
⎝ R1 + R2 ⎠
1.765 − 0.7
= 9.60 μ A
I BQ =
9.88 + (101)(1)
I CQ = 0.9605 mA
rπ =
(100 )( 0.026 )
0.9605
RTH rπ = 2.125 K
= 2.707 K
gm =
0.9605
= 36.94
0.026
⎛ RTH rπ
⎞
⎟ ⋅ V = ⎛ 2.125 ⎞ ⋅ V = (0.680 ) ⋅ Vi
Vπ = ⎜
⎜ R r + R ⎟ i ⎜⎝ 2.125 + 1 ⎟⎠ i
S ⎠
⎝ TH π
(
)
Aυ = − 0.680 g m RC = −(0.680 )(36.94 )(2) = −50.2
Design specification met.
______________________________________________________________________________________
6.26
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
= 65 mA/V
0.026
(100)(0.026) = 1.54 k Ω , r = ∞
rπ =
o
1.69
gm =
(c)
Av =
− β ( RC RL )
⋅
RB Rib
rπ + (1 + β ) RE RB Rib + RS
Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω
RB Rib = 10 304.5 = 9.68 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
Aυ =
− (100 )(6.8 6.8) ⎛ 9.68 ⎞
⋅⎜
⎟ = −1.06
1.54 + (101)(3) ⎝ 9.68 + 0.5 ⎠
⎛ RC ⎞
⎟(− β i b )
i o = ⎜⎜
⎟
⎝ RC + R L ⎠
⎛
⎞
RB
⎟ ⋅ is
ib = ⎜⎜
⎟
⎝ R B + rπ + (1 + β )R E ⎠
⎛ RC ⎞⎛
⎞
RB
⎟⎜
⎟
Ai = −(β )⎜⎜
⎟⎜ R + r + (1 + β )R ⎟
R
R
+
L ⎠⎝
B
π
E ⎠
⎝ C
⎞
10
⎛ 6.8 ⎞⎛
⎟⎟ ⇒ Ai = −1.59
= −(100 )⎜
⎟⎜⎜
(
)(
)
6
.
8
6
.
8
10
1
.
54
101
3
+
+
+
⎝
⎠⎝
⎠
(d)
(e)
Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω
Aυ =
− β (RC R L )
rπ + (1 + β )R E
=
− (100 )(6.8 6.8)
1.54 + (101)(3)
Aυ = −1.12
Ai = −1.59 , same as part (c)
______________________________________________________________________________________
6.27
r=
υ ce
1
=
g mυ ce g m
So re = rπ
1
ro
gm
______________________________________________________________________________________
6.28
(a) Set
RC
= 12.5 ⇒ RC = 12.5 R E
RE
Set VCEQ = 1.5 ≅ 3.3 − I CQ (RC + R E ) = 3.3 − I CQ (13.5R E )
Set I CQ = 0.1 mA, ⇒ R E = 1.33 k Ω and RC = 16.7 k Ω
We find I BQ = 0.833 μ A
Set RTH = (0.1)(1 + β )R E = (0.1)(121)(1.33) = 16.1 k Ω
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E = (0.000833)(16.1) + 0.7 + (121)(0.000833)(1.33)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTH = 0.8475 =
1
1
⋅ RTH ⋅ VCC =
(16.1)(3.3)
R1
R1
So R1 = 62.7 k Ω , R 2 = 21.7 k Ω
Actual gain:
(120)(0.026) = 31.2 k Ω
rπ =
0.1
Rib = rπ + (1 + β )R E = 31.2 + (121)(1.33) = 192.1 k Ω
Ri = RTH Rib = 16.1 192.1 = 14.85 k Ω
Then Aυ =
− (120 )(16.7 ) ⎛ 14.85 ⎞
⋅⎜
⎟ = −10.4
31.2 + (121)(1.33) ⎝ 14.85 + 0.1 ⎠
(b) Ris = R S + Ri = 0.1 + 14.85 = 14.95 k Ω
R o = RC = 16.7 k Ω
______________________________________________________________________________________
6.29
100
= 20.
Need a voltage gain of 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
5 × 10 −3
= 25 × 10 3 ⇒ Ri = 25 k Ω
0.2 × 10 − 6
Ri = RTH Rib , Let RTH = 50 k Ω , then Rib = 50 k Ω
Ri =
Rib = rπ + (1 + β )R E ≅ (1 + β )R E
Rib
50
=
= 0.495 k Ω
1 + β 101
Let R E = 0.5 k Ω , VCC = 10 V, I CQ = 0.2 mA
For β = 100 , R E ≅
0.2
= 0.002 mA
100
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
Then I BQ =
1
1
⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002)( 0.5)
R1
R1
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
I R = ( 0.2 )(12.7 ) = 2.54 V.
So dc biasing is OK.]
[Note: CQ C
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.30
Set
RC
= 15 ⇒ RC = 15 R E
RE
Set RC = 5 k Ω , then R E = 0.333 k Ω
Set I CQ = 0.5 mA, then V ECQ ≅ 5 − (0.5)(5.333) = 2.33 V
rπ = 4.68 k Ω , Rib = rπ + (1 + β )R E = 4.68 + (91)(0.333) = 35 k Ω
Set Ri = 22 k Ω = RTH Rib = RTH 35 ⇒ RTH ≅ 60 k Ω
Now I BQ = 0.00556 mA, I EQ = 0.506 mA
VCC = I EQ R E + V EB (on ) + I BQ RTH + VTH
5 = (0.506 )(0.333) + 0.7 + (0.00556 )(60 ) + VTH
VTH = 3.798 =
1
1
⋅ RTH ⋅ VCC =
(60)(5)
R1
R1
So that R1 = 79 k Ω and R 2 = 249 k Ω
______________________________________________________________________________________
6.31
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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.
______________________________________________________________________________________
6.32
RTH = R1 R 2 = 33 50 = 19.88 k Ω
⎛ R2 ⎞
⎛ 50 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(3.3) = 1.988 V
⎝ 50 + 33 ⎠
⎝ R1 + R 2 ⎠
3.3 = (101)I BQ (1) + 0.7 + I BQ (19.88) + 1.988
So I BQ = 0.005063 mA, I CQ = 0.5063 mA, I EQ = 0.5114 mA
V ECQ = 3.3 − (0.5114)(1) − (0.5063)(2 ) = 1.776 V
Then ΔV EC = 1.776 − 0.5 = 1.276 V, or ΔV EC = 3 − 1.776 = 1.224 V
So ΔV EC = 2(1.224 ) = 2.448 V peak-to-peak
______________________________________________________________________________________
6.33
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
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Output signal swing determined by current:
= 3.24 V peak-to-peak
Max. output swing
______________________________________________________________________________________
6.34
(a) 5 = (81)I BQ (10) + 0.7 + I BQ (2.5)
So that I BQ = 0.005292 mA, I CQ = 0.4234 mA, I EQ = 0.4287 mA
Now V ECQ = 10 − (0.4234 )(5) − (0.4287 )(10 ) = 3.60 V
ΔV EC = ΔI C (RC R L ) = ΔI C (2.5)
For ΔI C = 0.4234 ⇒ ΔV EC = 1.06 V
So that ΔV EC = 2(1.06) = 2.12 V peak-to-peak
(b) ΔI C = 2(0.4234 ) = 0.847 mA peak-to-peak
______________________________________________________________________________________
6.35
I EQ = 0.8 mA, I CQ = 0.790 mA, I BQ = 0.009877 mA
V E = 0.7 + (0.009877 )(20 ) = 0.898 V
VC = (0.79 )(2.5) − 5 = −3.025 V
Then V ECQ = V E − VC = 3.923 V
ΔV EC = ΔI C (RC R L ) = ΔI C (2.5 4 ) = ΔI C (1.538)
For ΔI C = 0.79 − 0.08 = 0.71 mA, then ΔV EC = (0.71)(1.538) = 1.09 V
So, ΔI C = 2(0.71) = 1.42 mA peak-to-peak,
⎛ 2.5 ⎞
Δi o = ⎜
⎟ΔI C = 0.546 mA peak-to-peak
⎝ 2.5 + 4 ⎠
______________________________________________________________________________________
6.36
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
Δi c = −
1
Δυ ce
6.4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For υ ce (min ) = 1 V, Δυ ce = 5.38 − 1 = 4.38 V ⇒ Δi c =
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 =
6.37
ro =
100
I CQ
Neglect ro as 1st 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
= 0.01821
100
RTH = ( 0.1)(101)(1.2 ) = 12.12 K
I BQ =
VTH =
1
1
⋅ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.38
dc load line
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 R 2 = (0.1)(1 + β )R E
= (0.1)(181)(0.1) = 1.81 k Ω
VTH =
1
⋅ RTH ⋅ V + = I BQ RTH + V BE (on ) + (1 + β )I BQ R E
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
______________________________________________________________________________________
6.39
I CQ = 0.647 mA, VCEQ = 10 − (0.647 )(9) = 4.18 V
Δi c = I CQ = 0.647 mA
So Δυ ce = Δi c (4 4 ) = (0.647 )(2 ) = 1.294 V
Voltage swing is well within the voltage specification. Then Δυ ce = 2(1.294) = 2.59 V peak-to-peak.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.40
(a) Aυ =
(1 + β )R E
rπ + (1 + β )R E
0.92 =
(121)(0.5) ⇒ r = 5.261 = βVT = (120)(0.026) ⇒ I = 0.593 mA
π
CQ
rπ + (121)(0.5)
I CQ
I CQ
VA
20
=
= 33.7 k Ω
I CQ 0.593
(b) ro =
Aυ =
(1 + β )(ro R E )
rπ + (1 + β )(ro R E )
ro R E = 33.7 0.5 = 0.4927 k Ω
(121)(0.4927 ) = 0.919
5.261 + (121)(0.4927 )
Then Aυ =
(c) For (a), R o = R E
rπ
5.261
= 0.5
= 0.5 0.04348 ⇒ R o = 40 Ω
1+ β
121
rπ
= 0.04 33.7 ⇒ R o = 39.95 Ω
1+ β
______________________________________________________________________________________
For (b), Ro = R E ro
6.41
(a) Rib = rπ + (1 + β )R E
(1 + β )R E
(1 + β )R E
=
rπ + (1 + β )R E
Rib
( A )(R ) (0.95)(50)
R = υ ib =
= 0.586 k Ω
Aυ =
1+ β
81
50 = rπ + (81)(0.586 ) ⇒ rπ = 2.5 k Ω
E
rπ =
β VT
I CQ
(b) R o = R E
⇒ I CQ =
(80)(0.026) = 0.832 mA
2.5
rπ
2.5
= 0.586
= 0.586 0.03086
1+ β
81
R o = 29.3 Ω
______________________________________________________________________________________
6.42
(a) g m = 76.92 mA/V, rπ = 1.04 k Ω , ro = 25 k Ω
(1 + β )ro
(81)(25)
=
= 0.999
Aυ =
R S + rπ + (1 + β )ro 1 + 1.04 + (81)(25)
Ri = R S + rπ + (1 + β )ro = 1 + 1.04 + (81)(25) ⇒ Ri = 2.027 M Ω
Ro = ro
rπ + R S
⎛ 1.04 + 1 ⎞
= 25 ⎜
⎟ = 25 0.0252 ⇒ R o = 25.2 Ω
1+ β
⎝ 81 ⎠
(b) rπ = 10.4 k Ω , ro = 250 k Ω
Aυ =
(81)(250)
= 0.9994
1 + 10.4 + (81)(250)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Ri = 1 + 10.4 + (81)(250) ⇒ Ri = 20.26 M Ω
⎛ 10.4 + 1 ⎞
R o = 250 ⎜
⎟ = 250 0.1407 ⇒ R o = 141 Ω
⎝ 81 ⎠
______________________________________________________________________________________
6.43
r + RS
(a) R o = π
1+ β
r + 0.5
0.015 = π
⇒ rπ = 1.315 k Ω
121
β VT
(120)(0.026) = 2.37 mA
rπ =
⇒ I CQ =
I CQ
1.315
50
= 21.07 k Ω
2.373
(1 + β )ro
(121)(21.07 )
=
= 0.9993
Aυ =
R S + rπ + (1 + β )ro 0.5 + 1.315 + (121)(21.07 )
______________________________________________________________________________________
(b) ro =
6.44
a.
RTH = R1 R 2 = 10 10 = 5 k Ω
⎛ R2 ⎞
⎛ 10 ⎞
⎟⎟(18) − 9 = ⎜
VTH = ⎜⎜
⎟(18) − 9 = 0
⎝ 10 + 10 ⎠
⎝ R1 + R 2 ⎠
0 − 0.7 − (− 9 )
I BQ =
= 0.0869 mA
5 + (181)(0.5)
I CQ = 15.6 mA, I EQ = 15.7 mA
VCEQ = 18 − (15.7 )(0.5) = 10.1 V
b.
c.
rπ =
Aυ =
(180)(0.026) = 0.30 k Ω
15.6
(1 + β )(R E R L ) ⎛⎜ R1 R2 Rib ⎞⎟
⋅
rπ + (1 + β )(R E R L ) ⎜ R1 R 2 Rib + R S ⎟
⎝
⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Rib = rπ + (1 + β )(R E R L ) = 0.30 + (181)(0.5 0.3) = 34.2 k Ω
R1 R 2 Rib = 5 34.2 = 4.36 k Ω
(181)(0.5 0.3) ⎛ 4.36 ⎞
⋅⎜
⎟ ⇒ Aυ = 0.806
0.3 + (181)(0.5 0.3) ⎝ 4.36 + 1 ⎠
Aυ =
d.
Rib = rπ + (1 + β )(R E R L )
Rib = 0.30 + (181)(0.188) = 34.3 k Ω
⎛ rπ + R1 R 2 R S ⎞
⎟ = 0.5 ⎛⎜ 0.3 + 5 1 ⎞⎟ ⇒ R = 6.18 Ω
Ro = R E ⎜
o
⎜
⎟
⎜ 181 ⎟
1+ β
⎝
⎠
⎝
⎠
______________________________________________________________________________________
6.45
a.
RTH = R1 R 2 = 10 10 = 5 k Ω
⎛ R2 ⎞
⎟⎟(− 10 ) = −5 V
VTH = ⎜⎜
⎝ R1 + R 2 ⎠
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 10
− 5 − 0.7 − (− 10 )
= 0.0174 mA
5 + (121)(2)
I CQ = 2.09 mA, I EQ = 2.11 mA
I BQ =
VCEQ = 10 − (2.09 )(1) − (2.11)(2 ) = 3.69 V
b.
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,
R1 R2 Rib = 5 122.5 = 4.80 k Ω
Av =
(121) ( 2 2) ⎛ 4.80 ⎞
⋅⎜
⎟ ⇒ Av = 0.484
1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
d.
Rib = rπ + (1 + β )(R E R L )
Rib = 1.49 + (121)(2 2) ⇒ Rib = 122 k Ω
⎛ rπ + R1 R 2 R S ⎞
⎟ = 2 ⎛⎜ 1.49 + 5 5 ⎞⎟ ⇒ R = 32.4 Ω
Ro = R E ⎜
o
⎜
⎟
⎜ 121 ⎟
1+ β
⎝
⎠
⎝
⎠
______________________________________________________________________________________
6.46
(a) RTH = R1 R 2 = 585 135 = 109.7 k Ω
⎛ R2 ⎞
⎛ 135 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(3.3) = 0.61875 V
⎝ 135 + 585 ⎠
⎝ R1 + R 2 ⎠
3.3 = (1 + β )I BQ R E + V EB (on ) + I BQ RTH + VTH
3.3 − 0.7 − 0.61875
= 0.001649 mA
109.7 + (91)(12 )
Then I CQ = 0.1484 mA, I EQ = 0.150 mA
I BQ =
V ECQ = 3.3 − (0.15)(12 ) = 1.5 V
(c) rπ =
Aυ =
(90)(0.026) = 15.77 k Ω , r =
0.1484
o
(1 + β )(ro R E R L )
rπ + (1 + β )(ro R E R L )
60
= 404 k Ω
0.1484
ro R E R L = 404 12 4 = 2.978 k Ω
(91)(2.978) = 0.945
15.77 + (91)(2.978)
Rib = rπ + (1 + β )(ro R E R L ) = 15.77 + (91)(2.978) = 286.8 k Ω
Aυ =
⎛ RTH
⎞⎛ ro R E ⎞
109.7
⎞⎛⎜ 404 12 ⎞⎟
⎟ = (91)⎛⎜
⎟⎜
= 18.7
Ai = (1 + β )⎜⎜
⎟
⎟
⎜
⎟
⎝ 109.7 + 286.8 ⎠⎜⎝ 404 12 + 4 ⎟⎠
⎝ RTH + Rib ⎠⎝ ro R E + R L ⎠
(d) Rib = 286.8 k Ω
R o = R E ro
rπ
15.77
= 12 404
⇒ R o = 171 Ω
1+ β
91
______________________________________________________________________________________
6.47
(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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
⎛ 80 ⎞
I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA
⎝ 81 ⎠
I CQ 0.494
=
⇒ g m = 19 mA / V
gm =
0.026
VT
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)
For R S = 0
⎛V
⎞
V o = −⎜⎜ π + g mVπ ⎟⎟(R L ro )
⎝ rπ
⎠
−V o
so that Vπ =
⎛ 1+ β ⎞
⎜
⎟
⎜ r ⎟(R L ro )
⎝ π ⎠
Now V s + Vπ = V o
V s = V o − Vπ = V o +
We find
Aυ =
Vo
⎛ 1+ β ⎞
⎜
⎟
⎜ r ⎟(R L ro )
⎝ π ⎠
(1 + β )(R L ro )
(81)(0.5 304)
Vo
=
=
V s rπ + (1 + β )(R L ro ) 4.21 + (81)(0.5 304)
Aυ = 0.906
Rib = rπ + (1 + β )(R L ro ) ≅ 4.21 + (81)(0.5) = 44.7 k Ω
⎛ RB ⎞
⎛ ro ⎞
⎟ ⋅ I s and I o = ⎜
⎟
I b = ⎜⎜
⎜ r + R ⎟(1 + β )I b
⎟
+
R
R
ib ⎠
L ⎠
⎝ B
⎝ o
Then
Ai =
⎛ RB ⎞⎛ ro ⎞
Io
= (1 + β ) ⎜
⎟⎜
⎟
Is
⎝ RB + Rib ⎠⎝ ro + RL ⎠
⎛ 10 ⎞
Ai ≅ ( 81) ⎜
⎟ (1) ⇒ Ai = 14.8
⎝ 10 + 44.7 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d)
⎛ R B Rib ⎞
⎛ 10 44.7 ⎞
⎟ ⋅Vs = ⎜
⎟ ⋅ V = (0.803)V s
V s′ = ⎜
⎜R R +R ⎟
⎜ 10 44.7 + 2 ⎟ s
S ⎠
⎝ B ib
⎝
⎠
(
)(
)
Then Aυ = 0.803 0.906 ⇒ Aυ = 0.728
Ai = 14.8 (unchanged)
______________________________________________________________________________________
6.48
(a)
I CQ = 1.98 mA, rπ =
(100)(0.026) = 1.313 k Ω
1.98
VA
100
=
= 50.5 k Ω
I CQ 1.98
ro =
r + RS
1.31 + 10
Ro = π
ro =
50.5 ⇒ R o = 112 Ω
1+ β
101
(b) From Equation (6.68)
(1 + β ) ( ro RL )
100
Av =
ro =
= 50.5 K
1.98
rπ + (1 + β ) ( ro RL )
(i)
RL = 0.5 K
(101) ( 50.5 0.5)
1.31 + (101) ( 50.5 0.5 )
(101)( 0.4951)
Av =
⇒ Av = 0.974
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 )
______________________________________________________________________________________
6.49
(a)
5 − 0.7
= I EQ = 1.303 mA, I CQ = 1.29 mA
3.3
V ECQ = 0.7 − (− 5) = 5.7 V
(b) rπ =
Aυ =
(110)(0.026) = 2.217 k Ω , r = 50 = 38.76 k Ω
1.29
(1 + β )(ro R E R L )
rπ + (1 + β )(ro R E R L )
o
1.29
ro R E R L = 38.76 3.3 1 = 0.7525 k Ω
(111)(0.7525) = 0.974
2.217 + (111)(0.7525)
Rib = rπ + (1 + β )(ro R E R L ) = 2.217 + (111)(0.7525) = 85.7 k Ω
Aυ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
rπ
2.217
R E ro =
3.3 38.76 ⇒ R o = 19.8 Ω
1+ β
111
υ (t ) 2.8 sin ω t
(c) i s (t ) = s =
⇒ i s (t ) = 32.7 sin ω t ( μ A)
Rib
85.7
Ro =
⎛ R E ro ⎞
⎟ ⋅ i (t ) = (111)⎛⎜ 3.041 ⎞⎟(32.7 sin ω t )
i o (t ) = (1 + β )⎜
⎜R r +R ⎟ s
⎝ 3.041 + 1 ⎠
L ⎠
⎝ E o
or i o (t ) = 2.73 sin ω t (mA)
υ o (t ) = i o (t )R L = 2.73 sin ω t (V)
υ eb (t ) = −i s (t )rπ = −(32.7 sin ω t )(2.217 )
υ eb (t ) = −72.5 sin ω t (mV)
______________________________________________________________________________________
6.50
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
I BQ =
10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ
b.
rπ =
(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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Rib = rπ + (1 + β ) RE = 508 kΩ
RB Rib = 434 508 = 234 kΩ
vb =
RB Rib
RB Rib + RS
⋅ vS =
vb = 0.997vS ⇒ vS =
c.
234vS
234
=
vS
234 + 0.7 234.7
4.02
⇒ vS = 4.03 V peak-to-peak
0.997
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 ⎞
⎟ ⋅υ s = 0.99υ s = (0.99 )(4.03)
72
.
⎝ 3 + 0.7 ⎠
υb = ⎜
υ b = 3.99 V peak-to-peak
(1 + β )(R E R L )
⋅υ b
rπ + (1 + β )(R E R L )
(101)(0.833) ⋅ (3.99)
=
2.63 + (101)(0.833)
υo =
υ o = 3.87 V peak-to-peak
______________________________________________________________________________________
6.51
RTH = R1 R 2 = 40 60 = 24 k Ω
⎛ 60 ⎞
VTH = ⎜
⎟(10 ) = 6 V
⎝ 60 + 40 ⎠
6 − 0.7
= 0.0131 mA
For β = 75 , I BQ =
24 + (76 )(5)
I CQ = 0.984 mA
For β = 150 , I BQ =
6 − 0.7
= 0.00680 mA
24 + (151)(5)
I CQ = 1.02 mA
For β = 75 , rπ =
(75)(0.026) = 1.98 k Ω
0.984
(150)(0.026) = 3.82 k Ω
For β = 150 , rπ =
1.02
For β = 75 , Rib = rπ + (1 + β )(R E R L ) = 65.3 k Ω
For β = 150 , Rib = 130 k Ω
Aυ =
R1 R 2 Rib
(1 + β )(R E R L )
⋅
rπ + (1 + β )(R E R L ) R1 R 2 Rib + R S
For β = 75 , R1 R 2 Rib = 40 60 65.3 = 17.5 k Ω
Aυ =
(76)(0.833) ⋅ 17.5 = 0.789
1.98 + (76 )(0.833) 17.5 + 4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For β = 150 , R1 R 2 Rib = 40 60 130 = 20.3 k Ω
Aυ =
(151)(0.833) ⋅ 20.3 = 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
______________________________________________________________________________________
6.52
(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
= 5.543 mA
β = 200 I E =
⎛ 100 ⎞
+
1
⎜
⎟
⎝ 201 ⎠
β = 50 I E =
2.80 ≤ I E ≤ 5.54 mA
VE = I E RE , β = 50, VE = 2.80 V
β = 200, VE = 5.54 V
β = 50, I CQ = 2.748 mA, rπ = 0.473 K
(b)
β = 200, I CQ = 5.515 mA, rπ = 0.943 K
[
]
Ri = R B rπ + (1 + β )(R E R L )
β = 50 , Ri = 100 [0.473 + (51)(1 1)] = 100 25.97 = 20.6 k Ω
β = 200 , Ri = 100 [0.943 + (201)(1 1)] = 100 101.4 = 50.3 k Ω
From Fig. (6.68)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(1 + β ) ( RE RL ) ⎛ Ri ⎞
⋅⎜
⎟
rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠
( 51) (1 1)
⎛ 20.6 ⎞
=
⋅⎜
⎟
0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠
Av =
β = 50 ⇒ Av = 0.661
β = 200 ⇒ Av =
( 201) (1 1) ⎛ 50.3 ⎞
⎜
⎟
0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠
Av = 0.826
______________________________________________________________________________________
6.53
Vo = (1 + β ) I b RL
Vs
rπ + (1 + β ) RL
Ib =
so Av =
(1 + β ) RL
rπ + (1 + β ) RL
For β = 100, RL = 0.5 k Ω
rπ =
(100 )( 0.026 )
0.5
Then Av ( min ) =
= 5.2 k Ω
(101)( 0.5 )
= 0.9066
5.2 + (101)( 0.5 )
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 )
______________________________________________________________________________________
6.54
(a) VCEQ = VCC − I EQ R E
Let VCEQ = 2.5 V, then 2.5 = 5 − I EQ (0.5) ⇒ I EQ ≅ I CQ = 5 mA, I BQ = 0.04167 mA
rπ =
(120)(0.026) = 0.624 k Ω
5
Rib = rπ + (1 + β )(R E R L ) = 0.624 + (121)(0.5 0.5) = 30.87 k Ω
⎛ RTH
⎞⎛ R E ⎞
⎟⎜⎜
⎟⎟
Ai = (1 + β )⎜⎜
⎟
⎝ RTH + Rib ⎠⎝ R E + R L ⎠
⎛
⎞⎛ 0.5 ⎞
RTH
⎟⎟⎜
10 = (121)⎜⎜
⎟ ⇒ RTH = 6.113 k Ω
⎝ RTH + 30.87 ⎠⎝ 0.5 + 0.5 ⎠
VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.04167 )(6.113) + 0.7 + (5)(0.5) = 3.455 V
VTH = 3.455 =
1
1
⋅ RTH ⋅ VCC =
(6.113)(5)
R1
R1
So R1 = 8.85 k Ω and R 2 = 19.8 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Ro = R E
rπ
0.624
= 0.5
⇒ Ro = 5.10 Ω
1+ β
121
⎛ RTH
⎞⎛ R E ⎞
⎟⎜⎜
⎟⎟
(b) Ai = (1 + β )⎜⎜
⎟
⎝ RTH + Rib ⎠⎝ R E + R L ⎠
Rib = rπ + (1 + β )(R E R L ) = 0.624 + (121)(0.5 2 ) = 49.02 k Ω
6.113
⎛
⎞⎛ 0.5 ⎞
Ai = (121)⎜
⎟⎜
⎟ = 2.68
⎝ 6.113 + 49.02 ⎠⎝ 0.5 + 2 ⎠
______________________________________________________________________________________
6.55
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
I BQ =
= 0.00625 mA
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
______________________________________________________________________________________
6.56
a.
Let RE = 24 Ω and VCEQ = 12 VCC = 12 V ⇒ I EQ =
I CQ = 0.493 A, I BQ = 6.58 mA
rπ =
( 75)( 0.026 )
0.493
= 3.96 Ω
⎛ RE ⎞
I 0 = (1 + β ) I b ⎜
⎟
⎝ RE + RL ⎠
⎛ RTH ⎞
Ib = I S ⎜
⎟
⎝ RTH + Rib ⎠
12
= 0.5 A
24
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Rib = rπ + (1 + β )(R E R L )
= 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
⇒ RTH = 74.9 Ω (Minimum value)
0.140 =
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.
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.
rπ
3.96
24 = 0.0521 24 ⇒ R0 = 52 mΩ
RE =
1+ β
76
______________________________________________________________________________________
R0 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.57
The output of the emitter follower is
⎛ RL ⎞
vo = ⎜
⎟ ⋅ vTH
⎝ RL + Ro ⎠
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
which yields Ro = 0.364 k Ω
= ( 0.95 )
4 + Ro
10 + Ro
⎛ rπ + R1 R 2 R S ⎞
⎟R r
We have Ro = ⎜
⎟ E o
⎜
1+ β
⎠
⎝
The first term dominates
Let R1 R 2 R S ≅ R S , then
rπ + R S
r +4
⇒ 0.364 = π
1+ β
1+ β
rπ
β VT
4
4
or 0.364 =
+
=
+
1 + β 1 + β I CQ (1 + β ) 1 + β
Ro ≅
0.364 ≅
The factor
Or
VT
4
+
I CQ 1 + β
4
1+ β
V
4
4
Ro ≅ 0.32 = T
= 0.044
= 0.0305.
I
CQ
is in the range of 91
to 131
We can set
I CQ = 0.08125 mA.
To take into account other factors, set
0.15
= 0.00136 mA
110
5
RE =
= 33.3 k Ω
VCEQ ≅ 5 V ,
0.15
For
set
I BQ =
Design a bias stable circuit.
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10) − 5 = ( RTH )(10) − 5
+
R
R
R
⎝ 1
2 ⎠
1
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
I CQ = 0.15 mA,
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
(91)(33.3 4)
370 340.6
⋅
⇒ Aυ = 0.9332
15.6 + (91)(33.3 4 ) 370 340.6 + 4
Aυ =
For β = 90, RL = 10 k Ω
Rib = 715.4 k Ω
Aυ =
(91)(33.3 10)
370 715.4
⋅
⇒ Aυ = 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
Aυ =
(131)(33.3 4)
370 490
⋅
⇒ Aυ = 0.9360
22.5 + (131)(33.3 4 ) 370 490 + 4
For β = 130, RL = 10 k Ω
Rib = 1030 k Ω
Aυ =
(131)(33.3 10)
370 1030
⋅
⇒ Aυ = 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
ΔvO
= 3.5%
vO
______________________________________________________________________________________
6.58
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
Need a gain of
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
op-amp voltage follower (see Chapter 9) between RS and CC1 .
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
(12 − ( −12 ) ) = 8 V
3
24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω
Set I EQ = 0.5 A, VCEQ =
50
( 0.5 ) = 0.49 A
51
β VT ( 50 )( 0.026 )
rπ =
=
= 2.65 Ω
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
I BQ =
= 0.0098 A = 9.8 mA
50
24
Let I R ≅
≅ 10 I B = 98 mA
R1 + R2
So that R1 + R2 = 245 Ω
R2
VTH =
( 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
______________________________________________________________________________________
6.59
1
= 38.46 mA/V
0.026
Aυ = g m RC = (38.46 )(2 ) = 76.9
(a) g m =
β
120
=
= 0.9917
1 + β 121
r
(120)(0.026) = 3.12 k Ω
(c) Ri = π , rπ =
1+ β
1
3.12
Ri =
⇒ Ri = 25.8 Ω
121
(d) Ro = RC = 2 k Ω
______________________________________________________________________________________
(b) Ai =
6.60
⎛ 80 ⎞
(a) I CQ = ⎜ ⎟(2 ) = 1.975 mA
⎝ 81 ⎠
(80)(0.026) = 1.053 k Ω
1.975
= 75.97 mA/V, rπ =
gm =
0.026
1.975
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Aυ = g m (RC R L ) = (75.97 )(1.5 2.5) = 71.2
⎛ β ⎞⎛ RC ⎞ ⎛ 80 ⎞⎛ 1.5 ⎞
⎟ = ⎜ ⎟⎜
⎟⎟⎜⎜
(b) Ai = ⎜⎜
⎟ = 0.370
⎟
⎝ 1 + β ⎠⎝ RC + R L ⎠ ⎝ 81 ⎠⎝ 1.5 + 2.5 ⎠
r
1.053
(c) Ri = π =
⇒ Ri = 13 Ω
1+ β
81
80
= 40.5 k Ω
(d) ro =
1.975
(i)
R oc = ro = 40.5 k Ω
R o = ro RC R L = 40.5 1.5 2.5 = 0.916 k Ω
(ii)
______________________________________________________________________________________
6.61
⎛ 110 ⎞
(a) I CQ = ⎜
⎟(0.5) = 0.4955 mA
⎝ 111 ⎠
(110)(0.026) = 5.77 k Ω
0.4955
= 19.06 mA/V, rπ =
gm =
0.026
0.4955
⎛ R ⎞⎡ r
⎤
⎛ 4 ⎞ ⎡ 5.77 ⎤
Aυ = g m ⎜⎜ C ⎟⎟ ⎢ π R S ⎥ = (19.06 )⎜ ⎟ ⎢
1⎥ = (76.24 ) 0.05198 1
⎝ 1 ⎠ ⎣ 111 ⎦
⎦
⎝ R S ⎠ ⎣1 + β
Aυ = 3.77
[
β
]
110
= 0.991
111
r
5.77
(c) Ri = R S + π = 1 +
= 1.052 k Ω
1+ β
111
(d) Ro = RC = 4 k Ω
______________________________________________________________________________________
(b) Ai =
1+ β
=
6.62
0.7
= 0.25 mA
2.8
I CQ ≅ 1.50 − 0.25 = 1.25 mA
(a) I R 2 =
VC = (0.25)(5 + 2.8) = 1.95 V = VCEQ
(120)(0.026) = 2.5 k Ω
1.25
= 48.08 mA/V, rπ =
0.026
1.25
Aυ = g m (R1 R L ) = (48.08)(5 10 ) = 160.3
(b) g m =
⎛ β ⎞⎛ R1 ⎞ ⎛ 120 ⎞⎛ 5 ⎞
⎟⎟ = ⎜
⎟⎟⎜⎜
(c) Ai = ⎜⎜
⎟⎜
⎟ = 0.3306
⎝ 1 + β ⎠⎝ R1 + R L ⎠ ⎝ 121 ⎠⎝ 5 + 10 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.63
(a)
10 − 0.7
= 0.93 mA
10
I CQ = 0.921 mA
I EQ =
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
______________________________________________________________________________________
6.64
(a)
I EQ = 0.93 mA, I CQ = 0.921 mA
VECQ = 6.10 V
0.921
= 35.42 mA/V rπ = 2.82 K
0.026
(b)
From Eq. 6.90
gm =
Av = g m
=
Av =
( R R ) ⎡ rπ
RS
⎤
⎢1 + β RE RS ⎥
⎣
⎦
0.1
⎤
⎢ 101 10 0.1⎥
⎣
⎦
C
L
( 35.42 ) ( 50 5 ) ⎡ 2.82
( 35.42 )( 4.545)
0.1
[ 0.0218]
Av = 35.1
______________________________________________________________________________________
6.65
(a)
I EQ = 1 mA, I CQ = 0.9917 mA
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 =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.66
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
V o = − g mVπ (R B R L )
Vπ =
− (R E Rie )
R E Rie + R S
⋅ V s = −(0.4971)V s
V o = (19 )(0.4971)V s (100 1)
Aυ = 9.37
c.
IX =
VX VX
+
− g mVπ , Vπ = −V X
R E rπ
IX
1
1
1
=
=
+ + gm
VX
Ri R E rπ
or Ri = R E rπ
1
1
= 1 5.253
= 0.84 0.05252
gm
19
Ri = 49.4 Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.67
(a) RTH = R1 R 2 = 150 50 = 37.5 k Ω
⎛ R2 ⎞
⎛ 50 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(9) = 2.25 V
R
R
+
⎝ 50 + 150 ⎠
2 ⎠
⎝ 1
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E
2.25 − 0.7
= 0.00373 mA, I CQ = 0.4663 mA, I EQ = 0.470 mA
37.5 + (126)(3)
VCEQ = 9 − (0.4663)(6 ) − (0.470 )(3) = 4.79 V
or I BQ =
(125)(0.026) = 6.97 k Ω
0.4663
= 17.93 mA/V, rπ =
0.026
0.4663
⎛ R S R E ⎞⎛ β ⎞
υ
⎟⎜
⎟(RC R L )
Rm = o = ⎜
i s ⎜⎝ R S R E + Rie ⎟⎠⎜⎝ 1 + β ⎟⎠
(b) g m =
We find R S R E = 100 3 = 2.913 k Ω
r + RTH 6.97 + 37.5
=
= 0.3529 k Ω
Rie = π
1+ β
126
2.913
⎞⎛ 125 ⎞
⎛
Rm = ⎜
⎟(6 4 ) = 2.12 V/mA
⎟⎜
⎝ 2.913 + 0.3529 ⎠⎝ 126 ⎠
(
) (
)
(c) υ s = i s R S R E Rie = i s 100 3 0.3529 = i s (0.3148)
So i s =
υs
0.3148
Then Aυ =
υ o υ o is
2.12
=
⋅
=
= 6.73
υ s i s υ s 0.3148
______________________________________________________________________________________
6.68
(a) VCEQ ≅ VCC − I CQ (RC + R E )
Let VCEQ = 2.5 V and I CQ = 0.25 mA
Then 2.5 = 5 − (0.25)(RC + 0.5) ⇒ RC = 9.5 k Ω
⎛ β ⎞
⎟⎟ ⋅ i e ⋅ (RC R L )
⎝ 1+ β ⎠
υ o = i c (RC R L ) = ⎜⎜
ie =
υs
Rie
Aυ =
=
υs
⎛ rπ + RTH ⎞
⎟⎟
⎜⎜
⎝ 1+ β ⎠
⎞
υo ⎛
β
⎟(RC R L )
=⎜
υ s ⎜⎝ rπ + RTH ⎟⎠
Now rπ =
(100)(0.026) = 10.4 k Ω
0.25
⎛
⎞
530.2
100
⎟⎟(9.5 12 ) =
Aυ = 25 = ⎜⎜
⇒ RTH = 10.81 k Ω
10.4 + RTH
⎝ 10.4 + RTH ⎠
Also I BQ = 0.0025 mA
VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E = (0.0025)(10.81) + 0.7 + (101)(0.0025)(0.5)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTH = 0.853275 =
1
1
⋅ RTH ⋅ VCC =
(10.81)(5)
R1
R1
We obtain R1 = 63.3 k Ω and R 2 = 13 k Ω
(b) From part (a), I CQ = 0.25 mA, VCEQ = 2.5 V
(c) Aυ = g m (RC R L )
0.25
= 9.615 mA/V
0.026
Aυ = (9.615)(9.5 12 ) = 51
gm =
______________________________________________________________________________________
6.69
(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 R ) ⎡ rπ
B
RS
L
⎤
RS ⎥
⎢
⎣1 + β
⎦
0.984
= 37.85 mA/V
0.026
rπ = 1.59 K
gm =
( 37.85) (100 2 ) ⎡1.59
⎤
0.05⎥
⎢
0.05
⎣ 61
⎦
= 1484 ⎡⎣ 0.0261 0.05⎤⎦
Av =
Av = 25.4
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.70
i s ( peak ) = 2.5 μ A, V o ( peak ) = 5 mV
So we need R m =
υo
is
=
5 × 10 −3
= 2 × 10 3 Ω ⇒ R m = 2 k Ω
−6
2.5 × 10
We have
⎛ RS RE ⎞
RL ) ⎜
⎜ R R + R ⎟⎟
ie ⎠
⎝ S E
Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω
Now β = 120, so we have
Vo ⎛ β ⎞
=⎜
⎟ ( RC
Is ⎝ 1+ β ⎠
⎛ RS RE ⎞
⎛ RS RE ⎞
⎛ 120 ⎞
2=⎜
⎟⎟ = 2.204 ⎜⎜
⎟⎟
⎟ ( 4 5 ) ⎜⎜
⎝ 121 ⎠
⎝ RS RE + Rie ⎠
⎝ RS RE + Rie ⎠
RS RE
Then
RS RE + Re
= 0.9075
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
⇒ 0.196 =
Rie = π
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
VTH = ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 )
R1
R1
I BQ =
which yields R1 = 25.3 k Ω
and R2 = 33.2 k Ω
______________________________________________________________________________________
6.71
a.
20 − 0.7
= 1.93 mA
10
I CQ = 1.91 mA
I EQ =
VECQ = VCC + VEB ( on ) − I C RC
= 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
Neglect effect hoe
Assume
2.45 ≤ hie ≤ 3.7 kΩ
80 ≤ h fe ≤ 120
(
)
V o = h fe I b (RC R L )
Rie =
⎛ RE ⎞
hie
⎟⋅ Is
, I e = ⎜⎜
⎟
1 + h fe
⎝ R E + Rie ⎠
⎛ I
⎞
Vs
Ib = ⎜ e ⎟ , Is =
⎜ 1 + h fe ⎟
R S + (R E Rie )
⎝
⎠
⎛ h fe ⎞
⎞
⎛
⎞⎛
1
⎟(R R )⎜ R E ⎟⎜
⎟
Aυ = ⎜
⎜ 1 + h fe ⎟ C L ⎜ R E + Rie ⎟⎜ R + R R ⎟
S
E
ie
⎝
⎠
⎝
⎠
⎝
⎠
hie = 3.7 kΩ, h fe = 120
High gain device:
3.7
Rie =
= 0.0306 k Ω
121
R E Rie = 10 0.0306 = 0.0305 k Ω
10
1
⎛ 120 ⎞
⎛
⎞⎛
⎞
Aυ = ⎜
⎟(6.5 5)⎜
⎟⎜
⎟ ⇒ Aυ = 2.711
+
+
121
10
0
.
0306
1
0
.
0305
⎝
⎠
⎝
⎠⎝
⎠
h = 2.45 kΩ, h = 80
fe
Low gain device: ie
2.45
Rie =
= 0.03025 k Ω
81
R E Rie = 10 0.03025 = 0.0302 k Ω
10
1
⎛ 80 ⎞
⎛
⎞⎛
⎞
Aυ = ⎜ ⎟(6.5 5)⎜
⎟⎜
⎟ ⇒ Aυ = 2.70
⎝ 81 ⎠
⎝ 10 + 0.03025 ⎠⎝ 1 + 0.0302 ⎠
2.70 ≤ Av ≤ 2.71
So A υ ≅ constant
c.
Ri = R E Rie
We found
0.0302 ≤ Ri ≤ 0.0305 kΩ
h , R = RC = 6.5 kΩ
Neglecting oe o
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.72
a.
Small-signal voltage gain
Aυ = g m (RC R L ) ⇒ 25 = g m (RC 1)
For V ECQ = 3 V, VC = −V ECQ + V EB (on ) = −3 + 0.7 = −2.3 V
VCC − I CQ RC + VC = 0 ⇒ I CQ =
5 − 2.3 2.7
=
RC
RC
For I CQ = 1 mA, RC = 2.7 k Ω
1
= 38.5 mA/V
0.026
Aυ = (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 = ∞
______________________________________________________________________________________
6.73
1
= 38.46 mA/V
0.026
(120)(0.026) = 3.12 k Ω , r = (80)(0.026) = 2.08 k Ω
rπ 1 =
π2
1
1
⎞
⎛
r
(1 + β 1 )⎜⎜ R E π 2 ⎟⎟
1
β2 ⎠
+
V
⎝
Aυ1 = o1 =
Vi
⎞
⎛
r
rπ 1 + (1 + β 1 )⎜ R E π 2 ⎟
⎜
1 + β 2 ⎟⎠
⎝
(a) g m1 = g m 2 =
We find R E
Then Aυ1 =
(b) Aυ 2 =
rπ 2
2.08
=1
= 0.02504 k Ω
1+ β 2
81
(121)(0.02504) = 0.4927
3.12 + (121)(0.02504 )
Vo 2
= g m 2 RC = (38.46 )(4) = 153.8
Vo1
(c) Aυ = Aυ1 ⋅ Aυ 2 = (0.4927 )(153.8) = 75.8
______________________________________________________________________________________
6.74
0.5
2
= 19.23 mA/V, g m 2 =
= 76.92 mA/V
0.026
0.026
(100)(0.026) = 5.2 k Ω , r = (100)(0.026) = 1.3 k Ω
rπ 1 =
π2
0.5
2
(a) g m1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vo1
= − g m1 (RC1 Rib 2 )
Vi
Aυ1 =
where Rib 2 = rπ 2 + (1 + β 2 )R E 2 = 1.3 + (101)(4 ) = 405.3 k Ω
Aυ1 = −(19.23)(4 405.3) = −76.17
(b) Aυ 2 =
Vo 2
(1 + β 2 )(R E 2 )
(101)(4) = 0.9968
=
=
Vo1 rπ 2 + (1 + β 2 )R E 2 1.3 + (101)(4)
(c) Aυ = Aυ1 ⋅ Aυ 2 = (− 76.17 )(0.9968) = −75.93
______________________________________________________________________________________
6.75
a.
⎛ R2 ⎞
⎛ 20 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH 1 = ⎜⎜
⎟(10 ) = 2.0 V
R
+
R
⎝ 20 + 80 ⎠
2 ⎠
⎝ 1
RTH 1 = R1 R 2 = 20 80 = 16 k Ω
2 − 0.7
= 0.0111 mA
16 + (101)(1)
1.11
I C1 = 1.11 mA; g m1 =
= 42.74 mA/V
0.026
(100)(0.026) = 2.34 k Ω
rπ 1 =
1.11
∞
ro1 =
=∞
1.11
⎛ R4 ⎞
15 ⎞
⎟ ⋅ VCC = ⎛⎜
VTH 2 = ⎜⎜
⎟(10 ) = 1.50 V
⎟
⎝ 15 + 85 ⎠
⎝ R3 + R 4 ⎠
RTH 2 = R3 R 4 = 15 85 = 12.75 k Ω
I B1 =
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 R 4 rπ 2 = 15 85 2.06 = 12.75 2.06
Ri 2 = 1.773 k Ω
Aυ′1 = − g m1 (RC1 Ri 2 ) = −(42.7 )(2 1.773)
Aυ′1 = −40.17
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
6.76
a.
⎛ R2 ⎞
⎛ 12.7 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH 1 = ⎜⎜
⎟(12 ) = 1.905 V
⎝ 12.7 + 67.3 ⎠
⎝ R1 + R 2 ⎠
RTH 1 = R1 R 2 = 12.7 67.3 = 10.68 k Ω
1.905 − 0.7
= 0.00477 mA
10.68 + (121)(2 )
I C1 = 0.572 mA
I B1 =
0.572
= 22 mA/V
0.026
(120)(0.026) = 5.45 k Ω
rπ 1 =
0.572
∞
ro1 =
=∞
0.572
⎛ R4 ⎞
45 ⎞
⎟ ⋅ VCC = ⎛⎜
VTH 2 = ⎜⎜
⎟(12 ) = 9.0 V
⎟
R
+
R
45
⎝ + 15 ⎠
4 ⎠
⎝ 3
RTH 2 = R3 R 4 = 15 45 = 11.25 k Ω
g m1 =
9.0 − 0.7
= 0.0405 mA
11.25 + (121)(1.6)
I C 2 = 4.86 mA
I B2 =
4.86
= 187 mA/V
0.026
(120)(0.026) = 0.642 k Ω , r = ∞
rπ 2 =
o2
4.86
g m2 =
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Ri 2 = R3 R 4 Rib
Rib = rπ 2 + (1 + β )(R E 2 R L )
= 0.642 + (121)(1.6 0.25) = 26.8 k Ω
Ri 2 = 15 45 26.8 = 7.92 k Ω
c.
Aυ1 = − g m1 (RC1 Ri 2 ) = −(22 )(10 7.92 ) ⇒ Aυ1 = −97.2
Aυ 2 =
(1 + β )(R E 2 R L )
(121)(0.216)
=
rπ 2 + (1 + β )(R E 2 R L ) 0.642 + (121)(0.216)
Aυ 2 = 0.976
Overall gain, Aυ = −(97.2)(0.976 ) = −94.9
d.
RiS = R1 R 2 rπ 1 = 67.3 12.7 5.45 = 3.61 k Ω
r + RS
Ro = π 2
RE 2
1+ β
where R S = R3 R3 RC1 = 15 45 10 = 5.29 k Ω
Ro =
0.642 + 5.29
1.6 = 0.049 1.6 ⇒ R o = 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 =
______________________________________________________________________________________
6.77
5 − 2(0.7 )
0.7
= 72 mA, I R1 =
= 1.4 mA
0.050
0.5
70.6
⎛ 80 ⎞
I E 2 = 72 − 1.4 = 70.6 mA, I C 2 = ⎜ ⎟(70.6) = 69.73 mA, I B 2 =
= 0.8716 mA
81
⎝ 81 ⎠
(a) I R 2 =
⎛ 120 ⎞
I E1 = 1.4 + 0.8716 = 2.2716 mA, I C1 = ⎜
⎟(2.2716) = 2.253 mA
⎝ 121 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
2.253
69.73
= 86.65 mA/V, g m 2 =
= 2681.9 mA/V
0.026
0.026
(120)(0.026) = 1.385 k Ω , r = (80)(0.026) = 0.02983 k Ω
rπ 1 =
π2
2.253
69.73
V s = V π 1 + Vπ 2 + V o
g m1 =
⎡⎛ V
⎤
⎞
V o = ⎢⎜⎜ π 1 + g m1Vπ 1 ⎟⎟ + g m 2Vπ 2 ⎥ (0.05)
⎢⎣⎝ rπ 1
⎥⎦
⎠
⎛V
⎞
⎛ 1
⎞
Vπ 2 = ⎜⎜ π 1 + g m1Vπ 1 ⎟⎟(0.5 rπ 2 ) = Vπ 1 ⎜
+ 86.65 ⎟(0.5 0.02983) = Vπ 1 (2.4595)
r
1
.
385
⎝
⎠
⎝ π1
⎠
⎡ ⎛ 1
⎤
⎞
V o = ⎢Vπ 1 ⎜
+ 86.65 ⎟ + (2681.9 )Vπ 1 (2.4595)⎥ (0.05) = Vπ 1 (334.175)
1
.
385
⎠
⎣ ⎝
⎦
V s = Vπ 1 + Vπ 2 + V o = Vπ 1 + Vπ 1 (2.4595) + Vo
So Vπ 1 = (V s − Vo )(0.28906 )
And V o = (334.175)(V s − Vo )(0.28906 ) = 96.596(V s − Vo )
V
Aυ = o = 0.990
Vs
(c) For Rib :
We have V o = (0.989754)V s
Vπ 1 = (V s − V o )(0.28906 ) = V s (1 − 0.98975)(0.28906 ) = V s (0.0029618)
V
Vs
rπ 1
Rib = s =
=
⇒ Rib = 467.6 k Ω
I s ⎛Vπ 1 ⎞ 0.0029618
⎜
rπ 1 ⎟⎠
⎝
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For R o :
(1) Vπ 1 + Vπ 2 + V x = 0
⎛V
⎞ V
(2) I x + g m 2Vπ 2 + ⎜⎜ π 1 + g m1Vπ 1 ⎟⎟ = x
⎝ rπ 1
⎠ 0.05
We had Vπ 2 = Vπ 1 (2.4595)
(1) Vπ 1 + Vπ 1 (2.4595) + V x = 0 ⇒ Vπ 1 = −V x (0.28906 )
⎛ 1
⎞ V
(2) I x + g m 2Vπ 1 (2.4595) + Vπ 1 ⎜⎜
+ g m1 ⎟⎟ = x
r
⎝ π1
⎠ 0.05
⎡
⎛ 1
⎞⎤ V
I x + Vπ 1 ⎢(2681.9 )(2.4595) + ⎜
+ 86.65 ⎟⎥ = x
⎝ 1.385
⎠⎦ 0.05
⎣
V
I x − V x (0.28906 )[6683.5] = x
0.05
V
R o = x = 0.512 Ω
Ix
______________________________________________________________________________________
6.78
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
I E 2 = (1 + β ) I E1 = (1 + β ) I B1
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.717 − 1.40
I B1 =
⇒ I B1 = 0.128 μΑ
2
91.0 + (101) (1)
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
VCE1 = 5.14 V
I C 2 = 1.29 mΑ
VCE 2 = 5.84 V
b.
0.0128
= 0.492 mΑ / V
0.026
1.292
gm2 =
= 49.7 mΑ / V
0.026
g m1 =
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 ⎟
r
⎝ π 1 ⎠⎦
⎝ rπ 1 ⎠
⎣
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎧
⎛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
______________________________________________________________________________________
6.79
I E1 = I Bias + I B 2 = I Bias +
I C2
β2
⎛ β ⎞⎡
I ⎤
I C1 = ⎜⎜ 1 ⎟⎟ ⎢ I Bias + C 2 ⎥
β2 ⎦
⎝ 1 + β 1 ⎠⎣
1⎤
⎛ 120 ⎞ ⎡
(a) I C 2 = I Bias = 1 mA, I C1 = ⎜
⎟ ⎢1 + ⎥ = 1.004 mA
⎝ 121 ⎠ ⎣ 80 ⎦
g m1 = 38.62 mA/V, rπ 1 = 3.108 k Ω , ro1 = 79.68 k Ω
g m 2 = 38.46 mA/V, rπ 2 = 2.08 k Ω , ro 2 = 50 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V
V − Vπ 2
(1) I x = x + g m 2Vπ 2 + x
+ g m1V π 1
ro 2
ro1
⎡V
V − Vπ 2 ⎤
(2) Vπ 2 = ⎢ π 1 + g m1Vπ 1 + x
⎥ ⋅ rπ 2
ro1 ⎦
⎣ rπ 1
(3) Vπ 1 + Vπ 2 = 0 ⇒ Vπ 1 = −Vπ 2
V
V − Vπ 2
− g m1Vπ 2
Then (1) I x = x + g m 2Vπ 2 + x
ro 2
ro1
⎛ 1
⎛
⎞
1 ⎞
1
⎟ + Vπ 2 ⎜ g m 2 −
I x = V x ⎜⎜
+
− g m1 ⎟⎟
⎟
⎜
ro1
⎝ ro 2 ro1 ⎠
⎝
⎠
⎛ 1
⎛r ⎞
1 ⎞
⎟ ⋅ rπ 2 + V x ⎜ π 2 ⎟
(2) Vπ 2 = −Vπ 2 ⎜⎜
+ g m1 +
⎟
⎜r ⎟
r
r
o1 ⎠
⎝ π1
⎝ o1 ⎠
⎡ ⎛ 1
⎤
⎛ rπ ⎞
1 ⎞
⎟ ⋅ rπ 2 ⎥ = V x ⎜ 2 ⎟
+ g m1 +
Vπ 2 ⎢1 + ⎜⎜
⎟
⎜r ⎟
ro1 ⎠
⎢⎣ ⎝ rπ 1
⎥⎦
⎝ o1 ⎠
⎡ ⎛ 1
⎤
1 ⎞
⎛ 2.08 ⎞
+ 38.62 +
Now (2) Vπ 2 ⎢1 + ⎜
⎟(2.08)⎥ = V x ⎜
⎟
3
.
108
79
.
68
⎠
⎝ 79.68 ⎠
⎣ ⎝
⎦
Vπ 2 = V x (0.00031825)
1 ⎞
1
⎛ 1
⎛
⎞
Then (1) I x = V x ⎜ +
− 38.62 ⎟ = V x (0.32495)
⎟ + V x (0.00031825)⎜ 38.46 −
79.68
⎝ 50 79.68 ⎠
⎝
⎠
V
R o = x = 30.77 k Ω
Ix
1 ⎞
⎛ 120 ⎞⎛
(b) I C 2 = 1 mA, I Bias = 0.2 mA, I C1 = ⎜
⎟⎜ 0.2 + ⎟ = 0.2107 mA
121
80
⎝
⎠⎝
⎠
g m1 = 8.104 mA/V, rπ 1 = 14.81 k Ω , ro1 = 379.7 k Ω
g m 2 = 38.46 mA/V, rπ 2 = 2.08 k Ω , ro 2 = 50 k Ω
⎡ ⎛ 1
⎤
1 ⎞
⎛ 2.08 ⎞
+ 8.104 +
Now (2) Vπ 2 ⎢1 + ⎜
⎟(2.08)⎥ = V x ⎜
⎟
379.7 ⎠
⎝ 379.7 ⎠
⎣ ⎝ 14.81
⎦
Vπ 2 = V x (0.0003043)
1 ⎞
1
⎛ 1
⎛
⎞
(1) I x = V x ⎜ +
− 8.104 ⎟ = V x (0.031867 )
⎟ + V x (0.0003043)⎜ 38.46 −
379.7
⎝ 50 379.7 ⎠
⎝
⎠
Vx
= 31.38 k Ω
Ro =
Ix
⎛ 120 ⎞⎛ 2 ⎞
(c) I C 2 = 2 mA, I Bias = 0 , I C1 = ⎜
⎟⎜ ⎟ = 0.02479 mA
⎝ 121 ⎠⎝ 80 ⎠
g m1 = 0.9536 mA/V, rπ 1 = 125.9 k Ω , ro1 = 3327 k Ω
g m 2 = 76.92 mA/V, rπ 2 = 1.04 k Ω , ro 2 = 25 k Ω
⎡ ⎛ 1
⎤
1 ⎞
⎛ 1.04 ⎞
+ 0.9536 +
Now (2) Vπ 2 ⎢1 + ⎜
⎟(1.04 )⎥ = V x ⎜
⎟
125
.
9
3327
⎠
⎝ 3327 ⎠
⎣ ⎝
⎦
Vπ 2 = V x (0.00015627 )
1 ⎞
1
⎛ 1
⎛
⎞
(1) I x = V x ⎜ +
− 0.9536 ⎟ = V x (0.05217 )
⎟ + V x (0.00015627)⎜ 76.92 −
3327
⎝ 25 3327 ⎠
⎝
⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V
R o = x = 19.17 k Ω
Ix
______________________________________________________________________________________
6.80
(a) RTH = R1 R 2 = 250 75 = 57.69 k Ω
⎛ R2 ⎞
⎛ 75 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(5) = 1.1538 V
R
+
R
⎝ 75 + 250 ⎠
2 ⎠
⎝ 1
V − V BE (on )
1.1538 − 0.7
=
= 0.003483 mA
I BQ = TH
RTH + (1 + β )R E 57.69 + (121)(0.6)
⇒ I CQ = 0.418 mA, I EQ = 0.4214 mA
VCEQ = 5 − (0.418)(5.6 ) − (0.4214 )(0.6 ) = 2.406 V
For Q : PQ = I CQ VCEQ = (0.418)(2.406 ) = 1.01 mW
2
RC = (0.418) (5.6) = 0.978 mW
For RC : PRC = I CQ
2
2
R E = (0.4214 ) (0.6) = 0.107 mW
For R E : PRE = I EQ
2
(b) Rib = rπ + (1 + β )R E
rπ =
(120)(0.026) = 7.464 k Ω
0.418
Rib = 7.464 + (121)(0.6) = 80.06 k Ω
Ri = RTH Rib = 57.69 80.06 = 33.53 k Ω
RiS = R S + Ri = 0.5 + 33.53 = 34.03 k Ω
⎛ RTH
⎞
⎛ RTH
⎞⎛ υ s ⎞
⎟ ⋅ is = ⎜
⎟⎜
⎟
; ib = ⎜⎜
⎟
⎜
⎟⎜
⎟
RiS
⎝ RTH + Rib ⎠
⎝ RTH + Rib ⎠⎝ RiS ⎠
⎛ RTH
⎛ RTH
⎞⎛ υ s ⎞
⎞⎛ υ s ⎞
⎟⎜
⎟⎜
⎟ ; i e = (1 + β )⎜
⎟
i c = β ⎜⎜
⎜
⎟
⎟⎜
⎜
⎟
⎟
⎝ RTH + Rib ⎠⎝ RiS ⎠
⎝ RTH + Rib ⎠⎝ RiS ⎠
57.69
⎛
⎞⎛ 0.1 sin ω t ⎞
Now i c = (120)⎜
⎟ = 0.1477 sin ω t (mA)
⎟⎜
⎝ 57.69 + 80.06 ⎠⎝ 34.03 ⎠
is =
υs
i e = 0.1489 sin ω t (mA)
1 2
1
2
I c RC = 0.978 + (0.1477 ) (5.6) = 1.039 mW
2
2
1
1
2
2
R E + I e2 R E = 0.107 + (0.1489) (0.6) = 0.1137 mW
For R E : PRE = I EQ
2
2
1
1
For Q : PQ = I CQ VCEQ − I c2 RC − I e2 R E = 1.01 − 0.0611 − 0.00665 = 0.942 mW
2
2
______________________________________________________________________________________
2
RC +
For RC : PRC = I CQ
6.81
(a) I BQ R B + V BE (on ) + (1 + β )I BQ R E + V − = 0
5 − 0.7
= 0.007363 mA; I CQ = 0.8836 mA, I EQ = 0.8909 mA
100 + (121)(4)
VCEQ = 10 − (0.8836)(4 ) − (0.8909)(4) = 2.902 V
I BQ =
For Q : PQ = I CQ V CEQ= (0.8836)(2.902 ) = 2.564 mW
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2
For RC : PRC = I CQ
RC = (0.8836) (4 ) = 3.123 mW
2
2
R E = (0.8909 ) (4 ) = 3.175 mW
For R E : PRE = I EQ
2
(b) ΔVCE = ΔI C (RC ) = (0.8836)(4 ) = 3.534 V - Not possible
ΔVCE = 2.902 − 0.5 = 2.402 V
2.402
= 0.6005 mA
4
1
1
2
2
p RC = (ΔI C ) RC = (0.6005) (4 ) = 0.721 mW
2
2
______________________________________________________________________________________
So ΔI C =
6.82
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.
−1
⋅ Δvec
1.43 kΩ
ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V
ΔiC =
⎛ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.83
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
100
= 119 kΩ
0.838
Neglecting base currents:
a.
RL = 1 kΩ
−1
−1
slope =
=
10 1 119 0.902 kΩ
r0 =
−1
⋅ ΔVce
0.902 kΩ
ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V
ΔiC =
1 ( 0.756 )
⇒ PRL = 0.286 mW
PRL =
2
1
2
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 =
______________________________________________________________________________________
6.84
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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.
−1
⋅ Δvce
0.909 kΩ
For ΔiC = 0.838 ⇒ Δvce = ( 0.909 )( 0.838 ) = 0.762 V
ΔiC =
⎛ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 7
7.1
a.
T ( s) =
T (s) =
V0 ( s )
Vi ( s )
=
1/ ( sC1 )
⎡⎣1/ ( sC1 ) ⎤⎦ + R1
1
1 + sR1C1
b.
fH =
1
1
=
⇒ f H = 159 Hz
3
2π R1C1 2π (10 )(10−6 )
c.
V0 ( s ) = Vi ( s ) ⋅
1
1 + sR1C1
For a step function
Vi ( s ) =
1
s
K
K2
1
1
V0 ( s ) = ⋅
= 1+
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
− R1C1
1
V0 ( s ) = +
s 1 + sR1C1
1
1
= −
1
s
+s
R1C1
v0 ( t ) = 1 − e − t / R1C1
______________________________________________________________________________________
7.2
a.
T (s) =
T (s) =
V0 ( s )
Vi ( s )
=
R2
R2 + ⎡⎣1/ ( sC2 ) ⎤⎦
sR2 C2
1 + sR2 C2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
fL =
1
1
=
⇒ f L = 1.59 Hz
2π R2 C2 2π (104 )(10 × 10−6 )
c.
V0 ( s ) = Vi ( s ) ⋅
sR2 C2
1 + sR2 C2
Vi ( s ) =
1
s
V0 ( s ) =
R2 C2
1
=
1 + sR2 C2 s + 1
R2 C2
v0 ( t ) = e − t / R2C2
______________________________________________________________________________________
7.3
V
(a) T (s ) = o =
Vi
R2
R2
1
sC 2
1
+ R1
sC 2
⎛ 1 ⎞
⎟⎟
R 2 ⎜⎜
R2
1
⎝ sC 2 ⎠
Now R 2
=
=
1
sC 2
1 + sR 2 C 2
R2 +
sC 2
R2
1 + sR 2 C 2
R2
=
Then T (s ) =
R2
R1 + R 2 + sR1 R 2 C 2
+ R1
1 + sR 2 C 2
⎛ R2 ⎞
1
⎟⎟ ⋅
T (s ) = ⎜⎜
(
R
+
R
1
+
s
R
2 ⎠
1 R 2 )C 2
⎝ 1
(b) τ = (R1 R 2 )C 2 = (10 20 )× 10 3 × 10 × 10 −6 ⇒ τ = 66.7 ms
1
= 2.39 Hz
2πτ 2π 66.7 × 10 −3
______________________________________________________________________________________
(c)
f =
1
=
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.4
a.
τ S = (Ri + R P )C S = (30 + 10)×10 3 × (10 ×10 −6 ) ⇒ τ S = 0.40 s
τ P = (Ri R P )C P = (30 10)× 10 3 × (50 × 10 −12 ) ⇒ τ P = 0.375 μ s
b.
fL =
fH =
1
2π τ S
1
2π τ P
=
1
⇒ f L = 0.398 Hz
2π (0.4)
=
1
⇒ f H = 424 kHz
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
(a)
Vo
R2
20
=
=
= 0.667
Vi
R 2 + R1 20 + 10
(b)
Vo
=1
Vi
(c) T (s ) =
T (s ) =
Vo (s )
=
Vi (s )
R2
R 2 + R1
R2
=
1
sC1
R2 +
R1
1 + sR1C1
⎛ R2 ⎞
(1 + sR1C1 )
R 2 (1 + sR1C1 )
⎟⎟ ⋅
= ⎜⎜
R1 + R 2 + sR1 R 2 C1 ⎝ R1 + R 2 ⎠ 1 + s (R1 R 2 )C1
[
R
We have K =
, τ = R C , τ = (R R )C
R +R
]
2
A
1
1
1
B
1
2
1
2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.6
a.
T (s ) =
Vo (s )
=
Vi (s )
RP
RP
1
sC P
⎛
1
1 ⎞
⎟
+ ⎜⎜ R S +
sC P ⎝
sC S ⎟⎠
1
sC P
RP
1
RP
=
=
1
sC P
1 + sR P C P
RP +
sC P
Then
RP ⋅
RP
T (s) =
⎛
1 ⎞
RP + ⎜ RS +
⎟ (1 + sRP CP )
sCS ⎠
⎝
RP
=
RP CP
1
+
+ sRS RP CP
RP + RS +
CS
sCS
⎤⎞
⎛ RP ⎞ ⎛ ⎡
sRP RS
RP
C
1
⋅ P+
+
⋅ CP ⎥ ⎟
T (s) = ⎜
⎟ × ⎜⎜1/ ⎢1 +
⎟
+
+
+
+
R
R
R
R
C
s
R
R
C
R
R
(
)
S ⎠ ⎝
P
S
S
S
P
S
S
P
⎝ P
⎣⎢
⎦⎥ ⎠
b.
−11
⎛
⎤⎞
1
⎛ 10 ⎞ ⎜ ⎡ 10 10
3
−11 ⎟
⎥
+
×
⋅
5
10
10
T (s) = ⎜
s
(
)
⎟ × 1/ ⎢1 + ⋅ −6 +
s ( 2 × 104 ) ⋅10 −6
⎥⎦ ⎟
⎝ 10 + 10 ⎠ ⎜ ⎢⎣ 20 10
⎝
⎠
1
1
≅ ⋅
1
2 1+
+ s ( 5 × 10−8 )
s ( 0.02 )
s = jω
T ( jω ) =
1
⋅
2
1
⎡
⎤
1
1 + j ⎢ω ( 5 × 10 −8 ) −
⎥
ω ( 0.02 ) ⎦
⎣
1
1
For ω L =
=
= 50
( RS + RR ) CS ( 2 ×104 )(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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For
1
ωH =
1
(R R )C = (5 ×10 )(10 ) = 2 ×10
S
T ( jω ) =
P
7
−11
3
P
1
⋅
2
1
⎡
⎤
1
1 + j ⎢ 2 ×10 7 5 ×10 −8 −
⎥
7
2 × 10 (0.02) ⎦
⎣
1 1
1 1
T ( jω ) ≅ ⋅
⇒ T ( jω ) = ⋅
2 1+ j
2 2
(
)(
In each case, T ( jω ) =
c.
)
(
)
RP
R
2 S + RP
1
⋅
RS = RP = 10 kΩ, CS = CP = 0.1 μ F
T (s) =
⎛ ⎡
⎤⎞
1 ⎜ ⎢ 1 1
1
⋅ 1/ 1 + ⋅ +
+ s ( 5 × 103 )(10−7 ) ⎥ ⎟
⎥ ⎟⎟
2 ⎜⎜ ⎢ 2 1 s 2 × 104 (10−7 )
⎦⎠
⎝ ⎣
(
)
s = jω
T ( jω ) =
1
⋅
2
1
⎡
⎤
1
1
⎥
+ j ⎢ω ( 5 × 10−4 ) −
−3
2
ω ( 2 × 10 ) ⎥⎦
⎢⎣
1
For ω =
= 500
4
2
10
×
(
)(10−7 )
1+
T ( jω ) =
1
⋅
2
1
For ω =
( 5 ×103 )(10−7 )
⎡
⎤
1
⎥
1.5 + j ⎢( 500 ) ( 5 × 10−4 ) −
−3
⎢⎣
( 500 ) ( 2 ×10 ) ⎥⎦
1
1
= ⋅
⇒ T ( jω ) = 0.298
2 1.5 − j ( 0.75 )
1
= 2 × 103
⎧
⎡
⎤ ⎞ ⎫⎪
1 ⎪ ⎛⎜
1
⎥ ⎟⎬
⋅ ⎨1/ 1.5 + j ⎢( 2 × 103 )( 5 × 10−4 ) −
−3
3
2 ⎪ ⎜
2 × 10 )( 2 × 10 ) ⎥⎦ ⎟ ⎪
⎢⎣
(
⎝
⎠⎭
⎩
1
1
= ⋅
⇒ T ( jω ) = 0.298
2 1.5 + j ( 0.75 )
T ( jω ) =
In each case, T ( jω ) <
RP
2 RP + RS
1
⋅
______________________________________________________________________________________
7.7
(a) T =
1
2 ⎤
⎡
⎢ 1 + ⎛⎜ f ⎞⎟ ⎥
⎜ f ⎟ ⎥
⎢
⎝ T ⎠ ⎥
⎢⎣
⎦
At f = f T , T =
3
1
( 2)
3
= 0.35355
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Or T dB = 20 log 10 (0.35355) = −9.03 dB
3
⎛ f ⎞
⎛ f ⎞
⎟⎟ = −3 tan −1 ⎜⎜
⎟⎟ = −3 tan −1 (1) = −135°
f
f
⎝ F ⎠
⎝ T ⎠
(b) Slope = 3(− 6) = −18 dB/octave = −60 dB/decade
φ = 3(− 90 ) = −270°
______________________________________________________________________________________
φ = − tan −1 ⎜⎜
7.8
______________________________________________________________________________________
7.9
(a) (ii) ω1 = 1 rad/s; ω 2 = 10 rad/s; ω 3 = 100 rad/s; ω 4 = 1000 rad/s
(iii) T (0) = 10
(iv) T (∞ ) = 10
(b) (ii) ω = 5 rad/s
(iii) T (0) = 0
(iv) T (∞ ) =
8
(0.2)2
= 200
______________________________________________________________________________________
7.10
ω ⎞⎛
⎛
⎞
⎟
⎜ j 2 ⎟⎜
1
⎟
(a) T ( jω ) = 5⎜ 10 ⎟⎜
⎜
ω ⎟⎜
ω ⎟
⎜ 1 + j 2 ⎟⎜ 1 + j
⎟
10 ⎠⎝
5 × 10 4 ⎠
⎝
⎛
⎞
jω ⎞⎛
1
⎟⎜
⎟
or T ( jω ) = 2.5 × 10 5 ⎜⎜ 2
4
⎟
⎜
⎟
⎝ 10 + jω ⎠⎝ 5 × 10 + jω ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ ω ⎞
5⎜ 2 ⎟
⎝ 10 ⎠
(b) T =
1
⋅
⎛ ω ⎞
⎛ ω ⎞
1+ ⎜ 2 ⎟
1+ ⎜
⎟
4
10
⎝ 5 × 10 ⎠
⎠
⎝
(i) At ω = 50 rad/s
⎛ 50 ⎞
5⎜
⎟
1
⎝ 100 ⎠
T =
⋅
= 2.236
2
2
⎛ 50 ⎞
⎛ 50 ⎞
1+ ⎜
1+ ⎜
⎟
⎟
4
⎝ 100 ⎠
⎝ 5 ×10 ⎠
(ii) At ω = 150 rad/s
⎛ 150 ⎞
5⎜
⎟
1
⎝ 100 ⎠
= 4.16
⋅
T =
2
2
150
150
⎞
⎛
⎞
⎛
1+ ⎜
1+ ⎜
⎟
⎟
4
⎝ 5 ×10 ⎠
⎝ 100 ⎠
2
2
(iii) At ω = 10 5
⎛ 10 5 ⎞
5⎜⎜ 2 ⎟⎟
⎝ 10 ⎠
1
= 2.236
2
2
⎛ 10 5 ⎞
⎛ 10 5 ⎞
⎟
1 + ⎜⎜ 2 ⎟⎟
1 + ⎜⎜
4 ⎟
⎝ 5 × 10 ⎠
⎝ 10 ⎠
______________________________________________________________________________________
T =
⋅
7.11
a.
⎛ r
⎞
V0 = − g mVπ RL Vπ = ⎜ π ⎟ Vi
⎝ rπ + RS ⎠
⎛ r
⎞
⎛ 5.2 ⎞
T = g m RL ⎜ π ⎟ = ( 29 )( 6 ) ⎜
⎟
⎝ 5.2 + 0.5 ⎠
⎝ rπ + RS ⎠
Tmidband = 159
b.
τ S = (R S + rπ )C C
fL =
1
2π τ S
τP =
⇒τS =
1
1
=
⇒ τ S = 5.31 ms, Open-Circuit
2π f L 2π (30 )
1
1
=
⇒ τ P = 0.332 μ s, Short-Circuit
2π f H 2π 480 × 10 3
(
)
c.
CC =
τS
(R S + rπ )
=
5.31× 10 −3
⇒ C C = 0.932 μ F
(0.5 + 5.2)×10 3
τ P = RL C L
τP
0.332 × 10 −6
⇒ C L = 55.3 pF
RL
6 × 10 3
______________________________________________________________________________________
CL =
=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.12
Vo
R 2 + R3
10 + 40
=
=
(a)
= 0.833
Vi
R1 + R 2 + R3 10 + 10 = 40
(b)
Vo
R2
10
=
=
= 0.50
Vi
R1 + R 2 10 + 10
⎛ 1 ⎞
R3 ⎜
⎟
R3
1
⎝ sC ⎠
=
=
(c) R3
1
sC
1 + sR3 C
R3 +
sC
R3
R2 +
1 + sR3 C
R 2 + R3 + sR 2 R3 C
V (s )
=
=
T (s ) = o
R3
R1 + R 2 + R3 + s(R1 + R 2 )R3 C
Vi (s )
R1 + R 2 +
1 + sR3 C
[
]
1 + s (R 2 R3 )C
⎛ R 2 + R3 ⎞
⎟⋅
or T (s ) = ⎜⎜
⎟
⎝ R1 + R 2 + R3 ⎠ 1 + s ((R1 + R 2 ) R3 )C
R 2 + R3
, τ A = (R2 R3 )C , τ B = ((R1 + R2 ) R3 )C
where K =
R1 + R 2 + R3
______________________________________________________________________________________
[
]
7.13 Computer Analysis
______________________________________________________________________________________
7.14
(a) Aυ max = g m R D , g m = 2 K n I DQ = 2 (0.4)(0.8) = 1.131 mA/V
Aυ max = (1.131)(1) = 1.13
(b) f H =
1
1
=
3
2πR D C L 2π 10 10 −12
( )(
)
f H = BW = 159 MHz
______________________________________________________________________________________
7.15
(a)
fH =
1
1
1
⇒ RC =
=
2πRC C L
2π f H C L 2π 800 × 10 6 0.08 × 10 −12
(
)(
)
or RC = 2.49 k Ω
VCC − VCEQ 2.5 − 1.25
(b) I CQ =
=
= 0.503 mA
RC
2.487
(c)
0.5026
= 19.33 mA/V
0.026
Aυ max = (19.33)(2.487 ) = 48.1
Aυ max = g m RC , g m =
______________________________________________________________________________________
7.16
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1 ⎞
⎛
⎟
⎜ ro ⋅
⎛
⎞
⎞
⎛
Vo (s )
sC L ⎟
1 ⎟
1
⎜
⎜
⎟
= − g m ro ⎜⎜
= − g m ro
= −g m ⎜
(a) T (s ) =
⎟
⎜
⎟
1
Vi (s )
1 + sro C L ⎟⎠
⎝
⎝ sC L ⎠
⎟
⎜⎜ ro +
sC L ⎟⎠
⎝
(b) g m = 2 K n I DQ = 2 (0.05)(0.1) = 0.1414 mA/V
ro =
1
1
=
= 1000 k Ω
λ I DQ (0.01)(0.1)
Aυ max = g m ro = (0.1414 )(1000 ) = 141.4
1
1
=
⇒ f H = 318 kHz
6
2π ro C L 2π 10 0.5 × 10 −12
______________________________________________________________________________________
(c)
f H = BW =
( )(
)
7.17
a.
RTH = R1 R2 = 10 1.5 = 1.304 kΩ
⎛ R2 ⎞
⎛ 1.5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (12 ) = 1.565 V
+
R
R
⎝ 1.5 + 10 ⎠
⎝ 1
2 ⎠
1.565 − 0.7
I BQ =
= 0.0759 mA
1.30 + (101)( 0.1)
I CQ = 7.585 mA
rπ =
(100 )( 0.026 )
= 0.343 kΩ
7.59
7.59
gm =
= 292 mA/V
0.026
Ri = R1 R2 [rπ + (1 + β )R E ]
= 10 1.5 [0.343 + (101)(0.1)]
= 1.30 10.44 ⇒ Ri = 1.159 k Ω
τ = (R S + Ri )C C = (0.5 + 1.16)×10 3 × (0.1× 10 −6 )
τ = 1.659 × 10 −4 s
fL =
b.
1
2π τ
=
1
⇒ f L = 959 Hz
2π 1.66 × 10 − 4
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
(100 )(1)( 0.111) V0
V0
=
⇒
= 6.69
Vi midband
Vi midband
1.659
⇒
c.
______________________________________________________________________________________
7.18
(a) V DSQ = V DD − I DQ (R D + R S )
3.2 = 9 − (0.8)(R D + 0.5) ⇒ R D = 6.75 k Ω
(
I DQ = K n VGSQ − VTN
(
0.8 = 0.5 VGSQ − 1.2
)
2
) ⇒V
2
GSQ
= 2.465 V
VG = (0.8)(0.5) + 2.465 = 2.865 V
VG =
1
1
⋅ Rin ⋅ V DD ⇒ 2.865 =
(160)(9)
R1
R1
which yields R1 = 503 k Ω and R 2 = 235 k Ω
(b) g m = 2 K n I DQ = 2 (0.5)(0.8) = 1.265 mA/V
(c)
Aυ =
− g m RD
− (1.265)(6.75)
= −5.23
=
1 + g m R S 1 + (1.265)(0.5)
fL =
1
1
1
⇒ CC =
=
⇒ C C = 0.06217 μ F
2πRin C C
2π f L Rin 2π (16 ) 160 × 10 3
τ S = Rin C C = (160 ×10 )(0.06217 ×10
3
(
−6
)
) = 9.947 ×10 s
−3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Aυ = 5.23
sτ S
1 + sτ S
⎤
⎡
⎢ ⎛⎜ f ⎞⎟ ⎥
⎢ ⎜ f ⎟ ⎥
⎝ L⎠ ⎥
= (5.23)⎢
2 ⎥
⎢
⎢ 1 + ⎛⎜ f ⎞⎟ ⎥
⎜ f ⎟ ⎥
⎢
⎝ L⎠ ⎦
⎣
(i) For f = 5 Hz,
⎡
⎤
5
⎢
⎥
⎢
⎥
16
Aυ = (5.23)⎢
⎥ = 1.56
2
⎢
5
⎛ ⎞ ⎥
⎢ 1+ ⎜ ⎟ ⎥
⎝ 16 ⎠ ⎥⎦
⎢⎣
(ii) For f = 14 Hz,
⎡
⎤
14
⎢
⎥
⎢
⎥
16
Aυ = (5.23)⎢
⎥ = 3.44
2
⎢
14
⎛ ⎞ ⎥
⎢ 1+ ⎜ ⎟ ⎥
⎝ 16 ⎠ ⎥⎦
⎢⎣
(iii) For f = 25 Hz,
⎡
⎤
25
⎢
⎥
⎢
⎥
16
Aυ = (5.23)⎢
⎥ = 4.405
2
⎢
⎛ 25 ⎞ ⎥
⎢ 1+ ⎜ ⎟ ⎥
⎝ 16 ⎠ ⎥⎦
⎢⎣
______________________________________________________________________________________
7.19
I DQ = K n (VGS − VTN ) ⇒ VGS =
2
I DQ
Kn
VS = −2.414 V
RS =
−2.414 − ( −5 )
⇒ RS = 2.59 kΩ
1
VD = VDSQ + VS = 3 − 2.414 = 0.586 V
5 − 0.59
RD =
⇒ RD = 4.41 kΩ
1
b.
+ VTN =
1
+ 1 = 2.414 V
0.5
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛
⎞
⎜
⎟
RD
⎟
I 0 = − ( g mVgs ) ⎜
⎜R +R + 1 ⎟
L
⎜ D
sCC ⎟⎠
⎝
Vi
Vgs =
1 + g m RS
I0 ( s )
Vi ( s )
=
⎡
⎤
sCC
− gm
⋅ RD ⎢
⎥
1 + g m RS
⎣⎢1 + s ( RD + RL ) CC ⎦⎥
I0 ( s )
T (s) =
Vi ( s )
s ( RD + RL ) CC
− g m RD
1
⋅
⋅
1 + g m RS RD + RL 1 + s ( RD + RL ) CC
=
c.
fL =
1
2π τ L
⇒τ L =
1
1
=
⇒ τ L = 15.92 ms
2π f L 2π (10 )
τL
15.9 × 10 −3
⇒ C C = 1.89 μ F
R D + R L (4.41 + 4 )× 10 3
______________________________________________________________________________________
τ L = (R D + R L )C C ⇒ C C =
=
7.20
a.
9 − VSG
2
= I D = K P (VSG + VTP )
RS
2
9 − VSG = ( 0.5 )(12 ) (VSG
− 4VSG + 4 )
2
6VSG
− 23VSG + 15 = 0
( 23) − 4 ( 6 )(15 )
⇒ VSG = 3 V
2 (6)
g m = 2 K P (VSG + VTP ) = 2 ( 0.5 )( 3 − 2 ) ⇒ g m = 1 mA/V
VSG =
Ro =
23 ±
2
1
RS = 1 12 ⇒ Ro = 0.923 kΩ
gm
b. τ = (Ro + R L )C C
fL =
1
2π τ
τ
⇒τ =
1
1
=
⇒ τ = 7.96 ms
2π f L 2π (20 )
7.96 × 10 −3
⇒ C C = 0.729 μ F
R o + R L (0.923 + 10 )× 10 3
______________________________________________________________________________________
c. C C =
=
7.21
(a) RTH = (0.1)(1 + β )R E = (0.1)(121)(4) = 48.4 k Ω
I EQ
1.5
I BQ =
=
= 0.012397 mA
1 + β 121
1
⋅ RTH ⋅ VCC
VTH = I BQ RTH + V BE (on ) + I EQ R E =
R1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
(48.4)(12) = (0.012397 )(48.4) + 0.7 + (1.5)(4)
so
R1
which yields R1 = 79.6 k Ω and R 2 = 124 k Ω
⎛ 120 ⎞
(b) I CQ = ⎜
⎟(1.5) = 1.488 mA
⎝ 121 ⎠
(120 )(0.026 ) = 2.097 k Ω , r = 50 = 33.6 k Ω
rπ =
o
1.488
1.488
(1 + β ) ro R E R L
Aυ =
rπ + (1 + β ) ro R E R L
(
(
)
)
Now ro R E R L = 33.6 4 4 = 1.888 k Ω
Aυ =
(121)(1.888) = 0.991
2.097 + (121)(1.888)
(c) Ro = R E ro
(d)
fL =
rπ
2.097
= 4 33.6
⇒ R o = 17.25 Ω
1+ β
121
1
1
=
2π (R o + R L )C C 2 2π (17.25 + 4000 ) 2 × 10 − 6
(
)
f L = 19.8 Hz
______________________________________________________________________________________
7.22
⎡
⎛ 1 ⎞⎤
⎟⎥
⎢ (ro R D )⎜⎜
⎛
⎞
⎡
⎤
ro R D
sC L ⎟⎠ ⎥
1 ⎟
⎝
⎢
⎜
(a) V o (s ) = − g m ro R D
⋅ V gs = − g m ⎢
⋅ V gs = − g m ⎢
⎥ ⋅ V gs
⎥
⎜
⎟
sC L ⎠
⎢⎣1 + s (ro R D )C L ⎥⎦
⎝
⎢ ro R D + 1 ⎥
sC L ⎦⎥
⎣⎢
⎛ 1 ⎞
⎜
⎟
⎜ sC ⎟(Vi (s ))
Vi (s )
⎝ i⎠
V gs =
=
1
1 + sR Si C i
+ R Si
sC i
T (s ) =
⎞
ro R D
⎞⎛
⎛
V o (s )
1
⎟
⎟⎜
= − g m ⎜⎜
⎟⎜
⎟
Vi (s )
⎝ 1 + sR Si C i ⎠⎝ 1 + s (ro R D )C L ⎠
(b) τ = R Si C i
(c) τ = (ro R D )C L
______________________________________________________________________________________
7.23
⎛ 1
1 ⎞⎟
−⎜
⎜
⎟
V gs
⎝ g m sC i ⎠
(a)
=
Vi
⎛ 1
1 ⎞⎟
⎜
+ RS
⎜ g sC ⎟
i ⎠
⎝ m
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 1 ⎞⎛ 1 ⎞
1
⎜
⎟⎜
⎟
⎜ g ⎟⎜ sC ⎟
⎛ 1
⎞
gm
1 ⎟ ⎝ m ⎠⎝ i ⎠
=
=
Now ⎜
⎜ g sC ⎟
1
1
⎛
1 ⎞
i ⎠
⎝ m
+
⎟C i
1 + s⎜⎜
⎟
g m sC i
⎝ gm ⎠
1
1 ⎞
⎛
−
⎜ −
⎟
V gs
gm
g
1
⎜
⎟⋅
m
So
=
=⎜
⎟
Vi
⎛
⎛ 1
⎞ ⎤
⎛ 1 ⎞ ⎞ ⎜ 1 +R ⎟ ⎡
1
S ⎟ ⎢1 + s⎜
⎟C i ⎟ ⎜
R S ⎟⎟C i ⎥
+ R S ⎜1 + s⎜⎜
⎜
⎟
g
⎜
⎟
⎠ ⎢⎣
gm
⎝ gm
⎠ ⎥⎦
⎝ gm ⎠ ⎠ ⎝ m
⎝
We have
⎤
⎡
⎥
⎢
⎡ R D R L (sC C ) ⎤
RD
⎥ ⋅ R L = − g mV gs ⎢
Vo = − g mV gs ⎢
⎥
1
⎥
⎢
⎣1 + s (R D + R L )C C ⎦
R
R
+
+
L
⎢ D
sC C ⎥⎦
⎣
⎛ R R ⎞ ⎡ s (R D + R L )C C ⎤
V o = − g mV gs ⎜⎜ D L ⎟⎟ ⎢
⎥
⎝ R D + R L ⎠ ⎣1 + s (R D + R L )C C ⎦
⎡ s (R D + R L )C C ⎤
V (s ) + g m (R D R L )
1
⋅⎢
Then T (s ) = o
=
⋅
⎥
Vi (s )
1 + g m RS
⎡
⎛ 1
⎞ ⎤ ⎣1 + s (R D + R L )C C ⎦
⎟
⎜
R S ⎟C i ⎥
⎢1 + s⎜
⎢⎣
⎝ gm
⎠ ⎥⎦
⎛ 1
⎞
(b) τ = ⎜⎜
R S ⎟⎟C i
⎝ gm
⎠
(c) τ = (R D + R L )C C
______________________________________________________________________________________
7.24
(a)
5 − 0.7
= 1.075 mA I CQ = 1.064 mA
4
VCEQ = 10 − (1.064 )( 2 ) − (1.075 )( 4 )
VCEQ = 3.57 V
I EQ =
gm =
rπ =
I CQ
VT
β VT
I CQ
=
1.064
= 40.92 mA/V
0.026
=
(100 )( 0.026 )
1.064
= 2.44 K
(b)
rπ
2440
= 200 + 4000
1+ β
101
R eq1 = 224.0 Ω ; τ 1 = R eq1C C1 = 1.053 ms
For C C1 ; R eq1 = R S + R E
For C C 2 ; R eq 2 = R C + R L = 2 + 47 = 49 k Ω
τ 2 = R eq 2 C C 2 = 49 ms
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
f1 =
1
1
=
⇒ f1 = 151 Hz
2πτ 1 2π (1.053 × 10−3 )
(c)
______________________________________________________________________________________
7.25
(a)
τ H = ( RC RL ) CL = ( 2 47 ) × 103 × 10 × 10−12
= 1.918 × 10−8 s
fH =
1
1
=
⇒ f H = 8.30 MHz
2πτ H 2π (1.918 × 10−8 )
(b)
1
1 + (2π τ H f )
= 0.1
2
2
⎛ 1 ⎞
2
⎜
⎟ = 100 = 1 + (2π τ H f )
⎝ 0.1 ⎠
99
99
2π τ H 2π 1.918 × 10 −8
f = 82.6 MHz
______________________________________________________________________________________
f =
=
(
)
7.26
(a)
5 − VSG
2
= K P (VSG + VTP )
R1
2
5 − VSG = (1)(1.2 )(VSG − 1.5 ) = (1.2 ) (VSG
− 3VSG + 2.25 )
2
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
For
(c)
CC1 , τ s1 = ( 284 + 200 ) ( 4.7 × 10 −6 ) = 2.27 ms
CC 2 , τ s 2 = (1.2 x103 + 50 × 103 )(10 −6 ) = 51.2 ms
CC2 dominates,
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
f 3− dB =
=
= 3.1 Hz
2πτ s 2 2π ( 51.2 × 10−3 )
______________________________________________________________________________________
7.27
2
Assume VTN = 1V , kn′ = 80 μ A / V , 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 kΩ
0.2
We need g m =
Aυ
RD
=
⎛ k ′ ⎞⎛ W ⎞
10
= 0.5 mA/V and g m = 2 K n I DQ = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
20
⎝ 2 ⎠⎝ L ⎠
W
⎛ 0.080 ⎞⎛ W ⎞
0.5 = 2 ⎜
= 7.81
⎟⎜ ⎟ ( 0.2 ) ⇒
2
L
L
⎝
⎠⎝ ⎠
or
Let
R1 + R2 =
9
( 0.2 ) I DQ
=
9
( 0.2 )( 0.2 )
= 225 k Ω
⎛ R2 ⎞
2
⎛ R2 ⎞
⎛ 0.080 ⎞
I DQ = 0.2 = ⎜
⎟ (9) = ⎜
⎟ ( 7.81)(VGS − 1) ⇒ VGS = 1.80 = ⎜
⎟ (9) ⇒
+
R
R
⎝ 2 ⎠
⎝ 225 ⎠
⎝ 1
2 ⎠
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 =
1
2π f 2
=
1
2π ( 3x10 )
3
= 5.305 × 10−5 s = RD CL or CL =
5.31× 10 −5
⇒ CL = 2.65 nF
20 × 103
______________________________________________________________________________________
7.28
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I BQ =
10 − 0.7
= 0.00997 mA
430 + ( 201)( 2.5 )
I CQ = ( 200 ) I BQ = 1.995 mA
rπ =
( 200 )( 0.026 )
= 2.61 k Ω
1.99
Rib = 2.61 + ( 201)( 2.5 ) = 505 k Ω
1
1
=
= 0.0106 s
τs =
2π f L 2π (15 )
= Req CC = ( 0.5 + 505 430 ) × 103 CC = 232.7 × 103 CC
C = 4.55 × 10−8 F ⇒ 45.5 nF
Or C
______________________________________________________________________________________
7.29
(a) V + = VCEQ + I EQ R E
3.3 = 1.8 + (0.25)R E ⇒ R E = 6 k Ω
0.25
= 0.002066 mA
121
V + = I BQ R B + V BE (on ) + I EQ R E
I BQ =
3.3 = (0.002066 )(R B ) + 0.7 + (0.25)(6 ) ⇒ R B = 532 k Ω
(120)(0.026 ) = 12.59 k Ω
⎛ 120 ⎞
(b) I CQ = ⎜
⎟(0.25) = 0.2479 mA, rπ =
0.2479
⎝ 121 ⎠
Rib = rπ + (1 + β )R E = 12.59 + (121)(6 ) = 738.6 k Ω
Ri = R B Rib = 532 738.6 = 309.25 k Ω
τS =
1
1
=
= 0.007958 = (R S + Ri )C C
2π f L 2π (20 )
so C C =
0.007958
⇒ C C = 0.0257 μ F
(0.1 + 309.25)×10 3
(c) For R S << R B ,
(1 + β )R E
(121)(6) = 0.983
=
rπ + (1 + β )R E 12.59 + (121)(6)
______________________________________________________________________________________
Aυ ≅
7.30
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
I BQ =
= 0.319 mA
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 ) << f 3− dB ( CC 2 )
2πτ
f3− dB ( CC1 ) ⇒ CC 2
f3− dB ( CC 2 ) ⇒ CC1
τ C >> τ C and f =
C1
Then, for
C2
acts as an open and for
f 3 − dB ( CC 2 ) = 25 Hz =
1
2πτ 2
, so that
acts as a short circuit.
1
τ2 =
= 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
0.6366
⇒ CC1 = 760 μ F
837.2
______________________________________________________________________________________
So CC1 =
7.31
From Problem 7.30
RTH = 0.6 K, I CQ = 31.9 mA, rπ = 81.5 Ω
τ C 2 >> τ C1 and f =
Then
f3− dB ( CC 2 ) ⇒ CC1
1
2π τ
so f 3− dB (C C 2 ) << f 3− dB (C C1 )
acts as an open circuit and for
f3− dB ( CC1 ) ⇒ CC 2
acts as a short circuit.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
f3− dB ( CC1 ) = 20 Hz =
⇒ τ C1 = 0.007958 s
2πτ C1
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
τ C 2 = 100τ C1 = 0.7958 s
CC1 =
⎛ r + RTH ⎞
⎛ 81.5 + 600 ⎞
Req 2 = RL + RE ⎜ π
⎟ = 10 + 50 ⎜
⎟
⎝ 101 ⎠
⎝ 1+ β ⎠
Req 2 = 10 + 50 6.748 = 15.95 Ω
0.7958
⇒ CC 2 = 0.050 F
15.95
______________________________________________________________________________________
CC 2 =
7.32
0 .2
⎛ 120 ⎞
= 0.001653 mA, I CQ = ⎜
⎟(0.2 ) = 0.1983 mA
121
⎝ 121 ⎠
V E = − (I BQ R i + V BE (on )) = −[(0.001653 )(10 ) + 0.7 ] = −0.7165 V
(a) I EQ = 0.2 mA, I BQ =
V C = V E + V CEQ = −0.7165 + 2.2 = 1.483 V
3 − 1.483
= 7.65 k Ω
0.1983
(120)(0.026 ) = 15.73 k Ω , g = 0.1983 = 7.627 mA/V
(b) rπ =
m
0.1983
0.026
⎛ rπ ⎞
15.73 ⎞
⎟ = −(7.627 )(7.65 20 )⎛⎜
Aυ = − g m (RC R L )⎜⎜
⎟ = −25.8
⎟
⎝ 15.73 + 10 ⎠
⎝ rπ + Ri ⎠
(c) For C C : τ C = (RC + R L )C C
RC =
fC =
1
2π τ C
=
1
2π (RC + R L )C C
⎛ r + Ri ⎞
1
⎟⎟C E ⇒ f E =
For C E : τ E = ⎜⎜ π
+
1
β
2
π
τE
⎝
⎠
1
(d) f E = 10 =
⇒ τ E = 0.015915 s
2π τ E
⎛ 15.73 + 10 ⎞
3
0.015915 = ⎜
⎟ × 10 × C E ⇒ C E = 74.8 μ F
121
⎝
⎠
1
f C = 50 =
⇒ τ C = 0.003183 s
2π τ C
0.003183 = (7.65 + 20)× 10 3 × C C ⇒ C C = 0.115 μ F
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.33
a.
I D = K n (VGS − VTN )
2
ID
0.5
+ VTN =
+ 0.8 = 1.8 V
Kn
0.5
VGS =
−VGS − ( −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
RS =
(b)
g m = 2 K n I DQ = 2 (0.5)(0.5) = 1 mA/V
τ A = R S C S = (6.4 × 10 3 )(5 ×10 −6 )
= 3.2 × 10 −2 s
1
1
fA =
=
⇒ f A = 4.97 Hz
2π τ A 2π 3.2 × 10 − 2
(
)
⎡ 6.4 × 10 3 ⎤
⎞
RS
−6
⎟C S = ⎢
⎥ 5 × 10
⎟
⎣1 + (1)(6.4 ) ⎦
⎝ 1 + g m RS ⎠
⎛
(
τ B = ⎜⎜
)
= 4.32 × 10 −3 s
1
1
fB =
=
⇒ f B = 36.8 Hz
2π τ B 2π 4.32 × 10 −3
(
c.
Av =
)
g m RD (1 + sRS CS )
⎡
⎞ ⎤
RS
⎟ CS ⎥
⎝ 1 + g m RS ⎠ ⎦
⎛
(1 + g m RS ) ⎢1 + s ⎜
⎣
As RS becomes large
Av →
g m RD ( sRS CS )
⎡
⎛ RS ⎞ ⎤
⎟ CS ⎥
⎝ g m RS ⎠ ⎦
( g m RS ) ⎢1 + s ⎜
⎣
⎡ ⎛ 1 ⎞ ⎤
⎟ CS ⎥
⎣ ⎝ gm ⎠ ⎦
⎛ 1 ⎞
1+ s ⎜
⎟ CS
⎝ gm ⎠
( g m RD ) ⎢ s ⎜
Av =
fB =
The corner frequency
1
2π (1/ g m ) CS
and the corresponding
fA → 0
g m = 2 K n I D = 2 ( 0.5 )( 0.5 ) = 1 mA / V
fB =
1
⇒ f B = 31.8 Hz
⎛ 1 ⎞
2π ⎜ −3 ⎟ ( 5 × 10−6 )
⎝ 10 ⎠
______________________________________________________________________________________
7.34
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎡
⎛ 1 ⎞ ⎤
⎟ ⎥
⎢ (ro R D )⎜⎜
⎛
⎞
sC L ⎟⎠ ⎥
V o1
1
⎝
⎢
⎟ = −g
(a) (i) T1 (s ) =
= − g m1 ⎜ ro R D
m1 ⎢
⎜
⎟
Vi
sC
⎛
⎞⎥
L
⎝
⎠
⎢ (ro R D ) + ⎜ 1 ⎟ ⎥
⎜
⎟
⎢⎣
⎝ sC L ⎠ ⎥⎦
1
T1 (s ) = − g m1 (ro R D )⋅
1 + s (ro R D )C L
[
(ii) T2 (s ) =
(iii) T (s ) =
(b) (i) f 3− dB =
]
Vo
1
= − g m 2 (ro R D )⋅
V o1
1 + s (ro R D )C L
[
]
Vo
1
2
= g m1 g m 2 (ro R D ) ⋅
2
Vi
1 + s (ro R D )C L
[
]
1
2π (ro R D )C L
1
= 100 k Ω , ro R D = 100 5 = 4.762 k Ω
(0.02)(0.5)
1
f 3− dB =
⇒ f 3− dB = 2.785 MHz
2π 4.762 × 10 3 12 × 10 −12
(ii) f 3− dB = 2.785 MHz
Now ro =
(
)(
)
2
⎧
⎫
1
1
⎪
⎪
(iii) Want ⎨
=
⎬
2
⎪ 1 + (2π f )(ro R D )C L 2 ⎪
⎩
⎭
1
1
So
=
= 0.7071
2
2
f
⎛
⎞
1+ ⎜
⎟
6
⎝ 2.785 × 10 ⎠
[
]
2
f
1
⎛
⎞
− 1 = 0.4142
⎜
⎟ =
6
0.7071
⎝ 2.785 × 10 ⎠
which yields f = 1.792 MHz
______________________________________________________________________________________
7.35
a.
b.
Expression for the voltage gain is the same as Equation (7.59) with RS = 0.
τ A = RE C E
R E rπ C E
rπ + (1 + β )R E
______________________________________________________________________________________
τB =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.36
⎛ 91 ⎞
(a) I EQ = ⎜ ⎟(0.15) = 0.1517 mA
⎝ 90 ⎠
3 − 0.7
RE =
= 15.16 k Ω
0.1517
V C = 0.7 − V ECQ = 0.7 − 2.2 = −1.5 V
− 1.5 − (− 3)
= 10 k Ω
0.15
0.15
(b) g m =
= 5.769 mA/V
0.026
Aυ = − g m RC = −(5.769 )(10 ) = −57.7
RC =
(
)(
)
(c) τ A = R E C E = 15.16 ×10 3 3 ×10 −6 = 4.548 × 10 −2 s
fA =
τB =
τB =
1
1
=
= 3.5 Hz
2πτ A 2π 4.548 × 10 − 2
(
)
R E rπ C E
(90)(0.026) = 15.6 k Ω
, where rπ =
0.15
rπ + (1 + β )R E
(15.16 ×10 )(15.6 ×10 )(3 ×10 ) = 5.085 ×10 s
15.6 × 10 + (91)(15.16 × 10 )
3
−6
3
3
1
−4
3
1
= 313 Hz
2πτ B 2π 5.085 × 10 − 4
______________________________________________________________________________________
fB =
=
(
)
7.37
(a) I EQ =
10 − 0.7
⎛ 90 ⎞
= 0.93 mA, I CQ = ⎜ ⎟(0.93) = 0.9198 mA
10
⎝ 91 ⎠
0.9198
= 35.38 mA/V
0.026
Aυ = g m RC R L = (35.38) 5 10 = 118
gm =
(
(b) f =
1
2πτ
)
=
(
)
1
1
=
2π (RC R L )C L 2π (5 10)× 10 3 × 3 ×10 −12
(
)
f = 15.9 MHz
______________________________________________________________________________________
7.38
2
(a) I DQ = K p V SGQ + VTP
(
(
0.2 = 0.1 V SGQ − 0.6
)
) ⇒V
2
SGQ
= 2.014 V
3 − 2.014
= 4.93 k Ω
0.2
V D = V SGQ − V SDQ = 2.014 − 1.9 = 0.114 V
RS =
RD =
0.114 − (− 3)
= 15.6 k Ω
0.2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
(b) f H =
2π (R D R L )C L
or
CL =
1
⇒ C L = 4.54 pF
2π (15.6 20)× 10 3 × 4 × 10 6
______________________________________________________________________________________
7.39
⎛ 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
gm
1.55
R0 = RS
R0 = 0.282 kΩ
τ = (Ro R L )C L and f H =
BW ≅ f H = 5 MHz ⇒ τ =
CL =
τ
Ro R L
=
1
2π τ
1
= 3.18 × 10 −8 s
2π 5 × 10 6
(
)
−8
3.18 × 10
⇒ C L = 121 pF
(0.282 4)×10 3
______________________________________________________________________________________
7.40
(a) Low-frequency
Mid-Band
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
High-frequency
(b)
(c)
12 − 0.7
= 11.3 μ A
1 MΩ
I CQ = 1.13 mA
I BQ =
rπ =
(100 )( 0.026 )
= 2.3 k Ω
1.13
1.13
= 43.46 mA / V
gm =
0.026
Am =
⎛ R r
⎞
Vo
(midband ) = − g m (RC R L )⎜⎜ B π ⎟⎟
Vs
⎝ R B rπ + R S ⎠
⎛ 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
fL =
1
2π τ L
(
, τ L = (R S + R B rπ )C C = (1 + 1000 2.3)× 10 3 × 10 × 10 −6
)
⇒ τ L = 3.29 ×10 −2 s, f L = 4.83 Hz
fH =
1
2π τ H
(
, τ H = (RC R L )C L = (5.1 500 )× 10 3 × 10 × 10 −12
)
⇒ τ H = 5.05 ×10 −8 s, f H = 3.15 MHz
______________________________________________________________________________________
7.41
1 ⎤
⎡
⎢ (R D R L )⋅ sC ⎥
⎞
⎛
1
L ⎥
⎟ = −g ⎢
(a) Aυ = − g m ⎜ R D R L
m
⎟
⎜
1
sC
⎢
⎥
L ⎠
⎝
⎢ (R D R L ) + sC ⎥
L ⎦
⎣
⎡
⎤
1
Aυ = − g m (R D R L )⎢
⎥
⎢⎣1 + s (R D R L )C L ⎥⎦
(b) τ = (R D R L )C L
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) 5 = I D R S + V SG = K p R S (V SG + VTP ) + V SG
2
5 = (0.25)(3.2)(V SG − 2) + V SG
2
2
− 2.2V SG − 1.8 = 0 ⇒ V SG = 3.41 V
We find 0.8V SG
I DQ = (0.25)(3.41 − 2) = 0.497 mA
2
τ = (10 20 )× 10 3 × 10 × 10 −12 = 6.67 × 10 −8 s
1
fH =
2πτ
=
1
⇒ f H = 2.39 MHz
2π 6.67 × 10 −8
(
)
g m = 2 K p I DQ = 2 (0.25)(0.497) = 0.705 mA/V
Aυ = − g m (R D R L ) = −(0.705)(10 20) = −4.7
______________________________________________________________________________________
7.42 Computer Analysis
______________________________________________________________________________________
7.43 Computer Analysis
______________________________________________________________________________________
7.44 Computer Analysis
______________________________________________________________________________________
7.45
0.25
= 9.615 mA/V
0.026
f
4 × 10 9
fβ = T =
⇒ f β = 33.3 MHz
βo
120
gm =
fT =
gm
gm
9.615 × 10 −3
⇒ Cπ + C μ =
=
2π (C π + C μ )
2π f T
2π 4 × 10 9
(
)
or C π + C μ = 0.3826 pF
Then C π = 0.3826 − 0.08 = 0.303 pF
______________________________________________________________________________________
7.46
(a)
fβ =
fT
βo
=
2 × 10 9
⇒ f β = 16.67 MHz
120
0.4
= 15.38 mA/V
0.026
gm
15.38 × 10 −3
Cπ + C μ =
=
2π f T
2π 2 × 10 9
C π + C μ = 1.224 pF, C π = 1.224 − 0.075 = 1.15 pF
gm =
(
(b) h fe =
βo
⎛ f ⎞
⎟
1+ ⎜
⎜ f ⎟
β
⎝
⎠
2
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
120
= 103
(i)At f = 10 MHz, h fe =
2
⎛ 10 ⎞
1+ ⎜
⎟
⎝ 16.67 ⎠
120
(ii)At f = 20 MHz, h fe =
⎛ 20 ⎞
1+ ⎜
⎟
⎝ 16.67 ⎠
(iii) At f = 50 MHz, h fe =
= 76.8
2
120
= 38.0
2
⎛ 50 ⎞
1+ ⎜
⎟
⎝ 16.67 ⎠
______________________________________________________________________________________
7.47
(a)
fβ =
fT
βo
=
540 × 10 6
⇒ f β = 4.5 MHz
120
0. 2
= 7.692 mA/V
0.026
gm
7.692 × 10 −3
=
Cπ + C μ =
⇒ C π + C μ = 2.267 pF
2π f T 2π 540 × 10 6
C π = 2.267 − 0.4 = 1.87 pF
gm =
(
)
0. 8
= 30.77 mA/V
0.026
gm
30.77 ×10 −3
fT =
=
⇒ f T = 2.16 GHz
2π Cπ + C μ
2π 2.267 ×10 −12
(b) g m =
(
)
(
)
2.16 ×10 9
⇒ f β = 18.0 MHz
120
______________________________________________________________________________________
fβ =
7.48
(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 ⎟⎠
Av ( s ) =
So
V0 ( s )
⎞
⎛ r ⎞⎛
1
= − g m RL ⎜ π ⎟ ⎜
⎟
Vi ( s )
⎝ rπ + rb ⎠ ⎜⎝ 1 + s ( rb rπ ) C1 ⎟⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
rπ =
(100 )( 0.026 )
(b) Midband gain:
For rb = 100 Ω
(i)
(ii)
(c)
(i)
1
= 2.6 k Ω, g m =
1
= 38.46 mA / V
0.026
⎛ 2.6 ⎞
Av1 = − ( 38.46 )( 4 ) ⎜
⎟ ⇒ Av1 = −148.1
⎝ 2.6 + 0.1 ⎠
For rb = 500 Ω
⎛ 2.6 ⎞
Av 2 = − ( 38.46 )( 4 ) ⎜
⎟ ⇒ Av 2 = −129.0
⎝ 2.6 + 0.5 ⎠
f 3− dB =
1
, τ = ( rb rπ ) C1
2πτ
For rb = 100 Ω
τ 1 = ( 0.1 2.6 ) × 103 ( 2.2 × 10−12 ) = 2.12 × 10−10 s ⇒ f 3− db = 751 MHz
For rb = 500 Ω
τ 2 = ( 0.5 2.6 ) × 103 ( 2.2 × 10−12 ) = 9.23 × 10 −10 s f 3− dB = 173 MHz
(ii)
______________________________________________________________________________________
7.49
f = 10 kHz = 104
Z i = 200 +
(
2500 1 − j (104 )(1.333 × 10−6 )
1 + (10 ) (1.333 × 10
4 2
)
)
−6 2
= 200 + 2500 − j 33.3 = 2700 − j 33.3
(b)
f = 100 kHz = 105
Z i = 200 +
(
2500 1 − j (105 )(1.333 × 10−6 )
1 + (10 ) (1.333 × 10
5 2
)
)
−6 2
Z i = 200 + 2456 − j 327 = 2656 − j 327
(c)
f = 1 MHz = 106
Z i = 200 +
(
2500 1 − j (106 )(1.333 ×10−6 )
1 + (10 ) (1.333 × 10
6 2
)
)
−6 2
Z i = 200 + 900 − j1200 = 1100 − j1200
(d)
______________________________________________________________________________________
7.50
a.
b.
CM = Cμ (1 + g m RL )
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V0 = − g mVπ RL Let Cπ + CM = Ci
rπ
Vπ =
1
sCi
⎛ RB ⎞
⋅⎜
⎟ Vi
R + RS ⎠
1
rπ
+ RB RS + rb ⎝ B
sC1
Aυ (s ) =
Vo (s )
Vi (s )
1
⎤
⎡
rπ ⋅
⎥
⎢
sC i
⎥
⎢
1
⎥
⎢
rπ +
⎥
⎢
⎛ RB ⎞
sC i
⎥
⎟⎢
= − g m R L ⎜⎜
⎟
⎥
⎝ R B + RS ⎠⎢ r ⋅ 1
⎥
⎢ π sC
i
⎢
+ R B R S + rb ⎥
1
⎥
⎢
⎥
⎢ rπ + sC
i
⎦
⎣
⎤
⎛ RB ⎞ ⎡
rπ
⎟⋅⎢
= − g m R L ⎜⎜
⎥
⎟
⎝ R B + R S ⎠ ⎢⎣ rπ + (1 + srπ C i )(R B R S + rb )⎥⎦
Let Req = ( RB RS + rb )
⎡
⎤
⎛ RB ⎞ ⎢
1
⎥
×
Av ( s ) = − β RL ⎜
⎟ ⎢
R
R
+
⎡
⎤
r + Req ) 1 + s rπ Req Ci ⎥⎥
S ⎠
⎝ B
⎣
⎦⎦
⎣⎢ ( π
(
Av ( s ) =
)
− β RL ⎛ RB ⎞
1
⋅⎜
⎟⋅
rπ + Req ⎝ RB + RS ⎠ 1 + s rπ Req Ci
(
)
1
fH =
2π ( rπ Req ) Ci
c.
______________________________________________________________________________________
7.51
High Freq. ⇒ CC1 , CC 2 , CE → short circuits
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
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 )
5
= 1.04 kΩ
Req = R1 R2 rπ = 5 1.04 = 0.861 kΩ
τ = Req ⋅ C i = (0.861× 10 3 )(603 × 10 −12 )
= 5.19 × 10 −7 s
1
1
f =
=
⇒ f = 307 kHz
2π τ 2π 5.19 × 10 − 7
______________________________________________________________________________________
(
)
7.52
RTH = R1 R2 = 60 5.5 = 5.04 kΩ
⎛ R2 ⎞
⎛ 5.5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (15 ) = 1.26 V
R
R
+
⎝ 5.5 + 60 ⎠
2 ⎠
⎝ 1
1.26 − 0.7
I BQ =
= 0.0222 mA
5.04 + (101)( 0.2 )
I CQ = 2.22 mA
rπ =
(100 )( 0.026 )
= 1.17 kΩ
2.22
2.22
gm =
= 85.4 mA/V
0.026
Lower 3 – dB frequency:
τ L = R eq ⋅ C C1
Req = R S + R1 R2 rπ
= 2 + 60 5.5 1.17 = 2.95 k Ω
τ L = (2.95 ×10 3 )(0.1×10 −6 ) = 2.95 ×10 −4 s
1
⇒ f L = 540 Hz
2π τ L 2π 2.95 × 10 − 4
Upper 3 – dB frequency:
gm
85.4 × 10 −3
fT =
⇒ 400 × 10 6 =
2π (C π + C μ )
2π (C π + C μ )
fL =
1
=
(
)
C π + C μ = 34 pF; C μ = 2 pF; C π = 32 pF
C M = C μ (1 + g m R C ) = 2[1 + (85.4 )(4 )] ⇒ C M = 685 pF
C i = C π + C M = 32 + 685 = 717 pF
Req = R S R1 R 2 rπ = 2 60 5.5 1.17 ⇒ Req = 0.644 k Ω
τ = Req ⋅ C i = (0.644 × 10 3 )(717 × 10 −12 )
= 4.62 × 10 −7 s
1
⇒ f H = 344 kHz
2π τ
______________________________________________________________________________________
fH =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.53
RTH = R1 R2 = 600 55 = 50.38 K
⎛ R2 ⎞
⎛ 55 ⎞
VTH = ⎜
⎟ (15 ) = ⎜
⎟ (15 ) = 1.2595 V
R
+
R
⎝ 600 + 55 ⎠
⎝ 1
2 ⎠
1.26 − 0.7
I BQ =
= 0.00222 mA
50.4 + (101)( 2 )
I CQ = 0.2217 mA
(100 )( 0.026 )
rπ =
= 11.73 K
0.222
0.2217
gm =
= 8.527 mA/V
0.026
Lower – 3dB Freq
τ L = R e q1 Cc1 ; Req1 = RS + RTH r π
= 0.50 + 50.38 11.73 = 10.0 K
τ L = (10 × 10 )( 0.1× 10−6 ) = 10−3 s
3
1
1
=
⇒ f L = 159 Hz
2π τ L 2π 10 −3
Upper – 3dB Freq
fL =
(
gm
fT =
2π ( Cπ + Cμ )
=
)
8.527 × 10−3
= 400 × 106
2π ( Cπ + 2 ) × 10−12
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 )
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.54
(a) RTH = R1 R2 = 33 22 = 13.2 k Ω
⎛ R2 ⎞
⎛ 22 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(5) = 2 V
R
+
R
⎝ 22 + 33 ⎠
2 ⎠
⎝ 1
V − V BE (on )
2 − 0.7
=
= 0.002106 mA
I BQ = TH
RTH + (1 + β )R E 13.2 + (151)(4 )
I CQ = 0.3159 mA, I EQ = 0.3180 mA
V CEQ = 5 − (0.3159 )(5) − (0.3180 )(4 ) = 2.15 V
fT
(b) f β =
=
βo
800 × 10 6
⇒ f β = 5.33 MHz
150
0.3159
= 12.15 mA/V
0.026
gm
12.15 × 10 −3
Cπ + C μ =
=
⇒ C π + C μ = 2.417 pF
2π f T 2π 800 × 10 6
C π = 2.417 − 0.45 = 1.97 pF
C M = C μ [1 + g m R C ] = (0.45 )[1 + (12.15 )(5)] = 27.79 pF
gm =
(
(c) rπ =
)
(150 )(0.026 ) = 12.35 k Ω , R
f 3− dB =
rπ = 13.2 12.35 = 6.38 k Ω
TH
0.3159
1
1
=
2π (RTH rπ )(C π + C M ) 2π 6.38 × 10 3 (1.97 + 27.79 )× 10 −12
(
)
f 3− dB = 838 kHz
______________________________________________________________________________________
7.55
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.08 ⎞⎛ 4 ⎞
g m1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟⎜
⎟(0.6) = 0.6928 mA/V
⎝ 2 ⎠⎝ 0.8 ⎠
⎝ 2 ⎠⎝ L ⎠
gm
0.6928 × 10 −3
=
⇒ f T = 1.84 GHz
2π C gs + C gd
2π (50 + 10)× 10 −15
______________________________________________________________________________________
fT =
7.56
(
)
I DQ = K n (VGS − VTN )
2
0.12 = K n (0.2) ⇒ K n = 3 mA/V 2
2
g m = 2 K n I DQ = 2 (3)(0.12) = 1.2 mA/V
gm
1.2 × 10 −3
⇒ f T = 3.82 GHz
=
2π C gs + C gd
2π (40 + 10)× 10 −15
______________________________________________________________________________________
fT =
(
)
7.57
(a) g m = 2 (1.5)(0.05) = 0.5477 mA/V
fT =
0.5477 × 10 −3
⇒ f T = 1.25 GHz
2π (60 + 10 )× 10 −15
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) g m = 2 (1.5)(0.3) = 1.342 mA/V
1.342 × 10 −3
⇒ f T = 3.05 GHz
2π (60 + 10 )× 10 −15
gm
(c) 3 × 10 9 =
⇒ g m = 1.319 mA/V
2π (60 + 10 )× 10 −15
fT =
2
g m = 2 K n I DQ ⇒ I DQ =
2
1 ⎛ gm ⎞
1 ⎛ 1.319 ⎞
⎜
⎟ =
⎜
⎟ = 0.29 mA
K n ⎜⎝ 2 ⎟⎠
1.5 ⎝ 2 ⎠
(d) g m = 2 (1.5)(0.25) = 1.225 mA/V
1.225 × 10 −3
⇒ C gs = 70 fF
2π C gs + 8 × 10 −15
2.5 × 10 9 =
(
)
______________________________________________________________________________________
7.58
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
μ n (VGS − VTN )
2π L2
450 ( 0.5 )
fT =
⇒ fT = 2.49 GHz
2
2π (1.2 ×10−4 )
fT =
(a)
fT =
450 ( 0.5 )
⇒ fT = 111 GHz
2π ( 0.18 × 10−4 )
(b)
______________________________________________________________________________________
7.59
(a) C M = C gd 1 + g m ro R D = (12) 1 + (3) 120 10 = 344.3 fF
[
(b) f 3− dB =
(
)]
[
(
)]
1
2πτ
τ = ri (C gs + C M ) = (10 4 )(80 + 344.3)× 10 −15 = 4.243 × 10 −9 s
1
⇒ f 3− dB = 37.5 MHz
2π 4.243 × 10 −9
______________________________________________________________________________________
f 3− dB =
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.60
gm
(Eq. (7.97))
fT =
2π (C gsT + C gdT )
⎛ 2⎞
CgdT = 0 and CgsT = ⎜ ⎟ (WLCox )
⎝ 3⎠
Let
⎛ μ C ⎞ ⎡W ⎤
g m = 2 K n I D = 2 ⎜ n ox ⎟ ⎢ ⎥ I D
⎝ 2 ⎠⎣ L ⎦
⎛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
⎝
⎠⎝ L ⎠
=
⋅
W Cox
2π L
3
fT =
μn I D
3
⋅
2π L 2W Cox L
______________________________________________________________________________________
7.61
(
)
⎛ μ C ⎞⎛ W ⎞ (400 ) 6.9 × 10 −9
(a) K n = ⎜⎜ n ox ⎟⎟⎜ ⎟ =
(8) ⇒ K n = 1.104 mA/V 2
2
⎝ 2 ⎠⎝ L ⎠
I D = K n (VGS − VTN ) = (1.104)(3 − 0.4) = 7.463 mA
2
2
g m = 2 K n I D = 2 (1.104 )(7.463) = 5.741 mA/V
g m′ =
gm
1 + g m rs
(0.8)g m =
rs =
gm
1
⇒ g m rs =
− 1 = 0.25
1 + g m rs
0.8
0.25
⇒ rs = 43.5 Ω
5.741
(b) I D = (1.104)(1 − 0.4) = 0.3974 mA
2
g m = 2 (1.104 )(0.3974 ) = 1.325 mA/V
g m′ =
1.325
= 1.253 mA/V
1 + (1.325)(0.04355)
g m′
⇒ 94.5%
gm
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.62
a.
I 0 = g mVgs and Vgs = I i Ri − g mVgs rS
Vgs =
so
I i Ri
1 + g m rS
I
g R
Ai = 0 = m i
I i 1 + g m rS
Then
b.
As an approximation, consider
In this case
I
1
Ai = 0 = g m′ Ri ⋅
Ii
1 + sRi ( C gsT + CM )
CM = CgdT (1 + g m′ RL ) and g m′ =
where
gm
1 + g m rs
As rS increases, CM decreases, so the bandwidth increases, but the current gain magnitude
decreases.
______________________________________________________________________________________
c.
7.63
⎛ 225 ⎞
(b) VGS = ⎜
⎟(10 ) = 3.103 V
⎝ 225 + 500 ⎠
I DQ = (1)(3.103 − 2) = 1.218 mA
2
g m = 2 K n I DQ = 2 (1)(1.218) = 2.207 mA/V
C M = C gd (1 + g m R D ) = (8)[1 + (2.207 )(5)] = 96.28 fF
(c)
f 3− dB =
1
2πτ
(
)
, τ = Ri R1 R 2 (C gs + C M )
Now Ri R1 R 2 = 1 500 225 = 0.9936 k Ω
τ = (0.9936 ×10 3 )(50 + 96.28)×10 −15 = 1.453 ×10 −10 s
f 3− dB =
1
⇒ f 3− dB = 1.095 GHz
2π 1.453 × 10 −10
(
)
⎛ R1 R 2 ⎞
⎟ = −(2.207 )(5)⎛⎜ 155.2 ⎞⎟ = −10.96
Aυ = − g m R D ⎜
⎜R R +R ⎟
⎝ 155.2 + 1 ⎠
i ⎠
⎝ 1 2
______________________________________________________________________________________
7.64
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) C M = C gd (1 + Aυ ) = (0.04)(1 + 15) = 0.64 pF
(b) f H =
1
2πτ
, ⇒τ =
τ = R eq (C gs + C M )
τ
1
2π f
=
1
= 3.183 × 10 −8 s
2π 5 × 10 6
(
)
3.183 × 10 −8
⇒ Req = 37.9 k Ω
C gs + C M
(0.2 + 0.64)×10 −12
______________________________________________________________________________________
or Req =
(
)
=
7.65
RTH = R1 R2 = 33 22 = 13.2 kΩ
⎛ R2 ⎞
⎛ 22 ⎞
VTH = ⎜
⎟ ( 5) = ⎜
⎟ ( 5) = 2 V
R
+
R
⎝ 22 + 33 ⎠
2 ⎠
⎝ 1
2 − 0.7
I BQ =
= 0.00261 mA
13.2 + (121)( 4 )
I CQ = 0.3138
rπ =
(120 )( 0.026 )
= 9.94 kΩ
0.3138
0.3138
gm =
= 12.07 mA/V
0.026
100
r0 =
= 318 kΩ
0.3138
a.
fT =
Cπ + Cμ =
gm
2π ( 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.
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
τ = R eq (C π + C M )
Req = R1 R 2 R S rπ = 33 22 2 rπ
= 1.74 9.94 ⇒ Req = 1.48 k Ω
τ = (1.48 ×10 3 )(2.20 + 27.6)×10 −12
τ = 4.41× 10 −8 s
fH =
1
1
=
⇒ f H = 3.61 MHz
2π τ 2π 4.41× 10 −8
(
(
V o = − g mVπ ro RC R L
Vπ =
R1 R 2 rπ
R1 R 2 rπ + R S
)
)
⋅ Vi
R1 R 2 rπ = 33 22 9.94 = 5.67 k Ω
5.67
⋅ Vi = (0.739 )Vi
5.67 + 2
ro RC R L = 318 4 5 = 2.18 k Ω
Vπ =
Aυ = −(12.07 )(0.739 )(2.18)
Aυ = −19.7
______________________________________________________________________________________
7.66
RTH = R1 R2 = 40 5 = 4.44 kΩ
⎛ R2 ⎞
⎛ 5 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10 ) = 1.111 V
R
R
+
⎝ 5 + 40 ⎠
⎝ 1
2 ⎠
1.111 − 0.7
= 0.00633 mA
I BQ =
4.44 + (121)( 0.5 )
I CQ = 0.760 mA
rπ =
(120 )( 0.026 )
= 4.11 kΩ
0.760
0.760
= 29.23 mA/V
gm =
0.026
r0 = ∞
gm
fT =
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:
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
τ H = R eq (C π + C M )
R eq = rπ R1 R 2 R S = 4.11 40 5 0.5
R eq = 0.405 k Ω
τ H = (0.405 ×10 3 )(15.6 + 149)×10 −12
= 6.67 × 10 −8 s
1
fH =
⇒ f H = 2.39 MHz
2π τ H
For lower frequency:
τ L = R eq C C1
R eq = R S + R1 R 2 rπ = 0.5 + 40 5 4.11
R eq = 2.64 k Ω
τ L = (2.64 ×10 3 )(4.7 ×10 −6 ) = 1.24 ×10 −2 s
fL =
1
⇒ f L = 12.8 Hz
2π τ L
b.
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.67
9 − VSG
2
I D = K P (VSG + VTP ) =
RS
( 2 )(1.2 ) (VSG2 − 4VSG + 4 ) = 9 − VSG
2
2.4VSG
− 8.6VSG + 0.6 = 0
VSG =
8.6 ±
(8.6 ) − 4 ( 2.4 )( 0.6 )
2 ( 2.4 )
2
VSG = 3.512 V
g m = 2 K P (VSG + VTP ) = 2 ( 2 )( 3.512 − 2 )
g m = 6.049 mA / V
I D = ( 2 )( 3.512 − 2 ) = 4.572 mA
1
1
r0 =
=
⇒ r0 = 21.9 kΩ
λ I o ( 0.01)( 4.56 )
2
(
CM = CgdT 1 + g m ( ro RD )
a.
b.
)
CM = (1) ⎡⎣1 + ( 6.04 ) ( 21.9 1) ⎤⎦ ⇒ CM = 6.785 pF
τ H = (Ri RG )(C gsT + C M )
τ H = (2 100)× 10 3 × (10 + 6.78)× 10 −12
τ H = 3.29 × 10 −8 s
fH =
1
⇒ f H = 4.84 MHz
2π τ H
Vo = − g m (ro R D )⋅ V gs
⎛ RG ⎞
100 ⎞
⎟ ⋅ Vi = ⎛⎜
V gs = ⎜⎜
⎟ ⋅ Vi
⎟
R
+
R
⎝ 102 ⎠
i ⎠
⎝ G
⎛ 100 ⎞
Aυ = −(6.04 )⎜
⎟(21.9 1)
⎝ 102 ⎠
Aυ = −5.67
______________________________________________________________________________________
7.68
2
(a) I DQ = K p V SGQ + VTP
(
(
)
)
0.5 = 0.5 V SGQ − 0.5 ⇒ V SGQ = 1.5 V
2
3 − 1. 5
= 3kΩ
0. 5
V D = 1.5 − 2 = −0.5 V
RS =
RD =
− 0.5 − (− 3)
= 5kΩ
0.5
(b) g m = 2 K p I DQ = 2 (0.5)(0.5) = 1 mA/V
⎛ RG ⎞
200 ⎞
⎟ = −(1)(5)⎛⎜
Aυ = − g m R D ⎜⎜
⎟ = −4.90
⎟
R
R
+
200
+4⎠
⎝
i ⎠
⎝ G
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) C M = C gd (1 + g m R D ) = (0.08 )[1 + (1)(5)] = 0.48 pF
(d) f 3− dB =
1
2πτ
(
)
where τ = R eq ⋅ C eq = (Ri RG ) C gs + C M = (4 200 )× 10 3 × (0.8 + 0.48)× 10 −12
which yields τ = 5.02 × 10 −9 s
1
Then f 3− dB =
⇒ f 3− dB = 31.7 MHz
2π 5.02 × 10 −9
______________________________________________________________________________________
(
)
7.69
⎛ R2 ⎞
⎛ 22 ⎞
VG = ⎜
⎟ ( 20 ) − 10 = ⎜
⎟ ( 20 ) − 10
⎝ 22 + 8 ⎠
⎝ R1 + R2 ⎠
VG = 4.67 V
10 − VSG − 4.67
2
= K P (VSG + VTP )
ID =
RS
5.33 − VSG = (1)( 0.5 ) (VSG2 − 4VSG + 4 )
2
− VSG − 3.33 = 0
0.5VSG
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.
τ = R eq (C gsT + C M )
R eq = Ri R1 R 2 = 0.5 8 22 = 0.461 k Ω
τ = (0.461×10 3 )(15 + 18.2)× 10 −12
= 1.53 ×10 −8 s
1
fH =
⇒ f H = 10.4 MHz
2π τ
c.
Vo = − g mV gs (R D R L )
⎛ R1 R 2 ⎞
⎟ ⋅ V = ⎛ 5.87 ⎞ ⋅ V = (0.9215)Vi
V gs = ⎜
⎜ R R + R ⎟ i ⎜⎝ 5.87 + 0.5 ⎟⎠ i
i ⎠
⎝ 1 2
Aυ = −(3.54)(0.9215)(2 5) ⇒ Aυ = −4.66
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.70
⎛ 100 ⎞
I E = 0.5 mA ⇒ I CQ = ⎜
⎟ ( 0.5 ) = 0.495 mA
⎝ 101 ⎠
0.495
gm =
= 19.0 mA/V
0.026
(100 )( 0.026 )
rπ =
= 5.25 kΩ
0.495
a.
Input: From Eq. (7.114(b))
⎡ r
⎤
τ Pπ = ⎢ π R E R S ⎥ C π
⎣1 + β
⎦
(
⎡ 5.25
⎤
=⎢
0.5 0.05⎥ × 10 3 × 10 × 10 −12
⎣ 101
⎦
= 2.43 × 10 −10 s
1
f Hπ =
⇒ f Hπ = 656 MHz
2π τ Pπ
Output: From Eq. (7.115(b))
τ Pμ = (R B R L )C μ = (100 1)× 10 3 × 10 −12
(
−10
= 9.90 × 10 s
1
f Hμ =
⇒ f Hμ = 161 MHz
2π τ Pμ
b.
Vo = − g mVπ (R B R L )
g mVπ +
Vπ Vπ Vi − (− Vπ )
+
+
=0
rπ R E
RS
⎡
Vi
1
1
1 ⎤
+
Vπ ⎢ g m + +
⎥=−
r
R
R
R
π
E
S ⎦
S
⎣
1
1
1 ⎤ − Vi
⎡
Vπ ⎢19 +
+
+
=
5
.
25
0
.
5
0
.
05 ⎥⎦ 0.05
⎣
Vπ (41.19) = −Vi (20 )
Vπ = −(0.4856 )Vi
Vo
= −(19 )(− 0.4856)(100 1)
Vi
Aυ = 9.14
c.
)
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
τ = (R B R L )C L = (100 1)× 10 3 × (15 × 10 −12 )
τ = 1.485 × 10 −8 s
f =
1
2π τ
⇒ f = 10.7 MHz
Since f < f Hμ ⇒ 3 -dB frequency dominated by C L
______________________________________________________________________________________
7.71
20 − 0.7
= 1.93 mA
10
⎛ 100 ⎞
I CQ = ⎜
⎟ (1.93) = 1.91 mA
⎝ 101 ⎠
1.91
= 73.5 mA/V
gm =
0.026
(100 )( 0.026 )
= 1.36 kΩ
rπ =
1.91
a.
Input:
⎡ r
⎤
τ Pπ = ⎢ π R E R S ⎥ C π
1
+
β
⎣
⎦
I EQ =
(
⎡1.36
⎤
=⎢
10 1⎥ × 10 3 × 10 × 10 −12
⎣ 101
⎦
)
τ Pπ = 1.327 ×10 −10 s
f Pπ =
1
2π τ Pπ
⇒ f Pπ = 1.20 GHz
Output:
τ Pμ = (RC R L )C μ = (6.5 5)× 10 3 × (10 −12 )
τ Pμ = 2.826 × 10 −9 s
f Pμ =
b.
1
⇒ f Pμ = 56.3 MHz
2π τ Pμ
Vo = − g mVπ (RC R L )
g mVπ +
Vπ Vπ Vi − (− Vπ )
+
+
=0
rπ R E
RS
⎛
V
1
1
1 ⎞
⎟=− i
Vπ ⎜⎜ g m + +
+
⎟
r
R
R
R
π
E
S ⎠
S
⎝
−
V
1
1 1⎞
⎛
i
Vπ ⎜ 73.5 +
+ + ⎟=
1
.
36
10
1
1
⎝
⎠
Vπ (75.34) = −Vi ⇒ Vπ = −(0.01327 )Vi
Vo = −(73.5)(− 0.01327)(6.5 5)Vi
Aυ = 2.76
c.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
τ = (RC R L )C L = (6.5 5)× 10 3 × (15 × 10 −12 )
τ = 4.24 × 10 −8 s
f =
1
2π τ
⇒ f = 3.75 MHz
Since f < f Pμ , 3-dB frequency is dominated by C L
______________________________________________________________________________________
7.72
VGS + I D RS = 5
ID =
5 − VGS
2
= K n (VGS − VTN )
RS
5 − VGS = ( 3)(10 ) (V 2GS − 2VGS + 1)
30V 2GS − 59VGS + 25 = 0
( 59 ) − 4 ( 30 )( 25)
⇒ VGS = 1.349 V
2 ( 30 )
g m = 2 K n (VGS − VTN ) = 2 ( 3)(1.35 − 1)
VGS =
59 ±
2
g m = 2.093 mA / V
On the output:
τ Pμ = (R D R L )C gdT = (5 4)× 10 3 × 4 × 10 −12
(
)
τ Pμ = 8.89 × 10 −9 s
f Pμ =
1
2π τ Pμ
⇒ f Pμ = 17.9 MHz
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
V
Av = 0 = ( 2.093)( 0.1857 ) ( 5 4 )
Vi
Av = 0.864
______________________________________________________________________________________
7.73
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
( 5.4 ) + 4 ( 4 )( 2.44 )
= 1.707
2 ( 4)
g m = 2 K P (VSG + VTP ) = 2 (1)(1.707 − 0.8 )
VSG =
5.4 ±
2
g m = 1.81 mA / V
3 ⋅ dB frequency due to CgsT : Req =
fA =
1
RS Ri
gm
1
2π Req ⋅ CgsT
1
4 0.5 = 0.246 kΩ
1.81
1
fA =
= 162 MHz
2π ( 246 ) ( 4 × 10−12 )
Req =
3 − dB frequency due to C gdT
fB =
=
1
2π ( RD RL ) CgdT
1
2π ( 2 4 ) × 103 × 10−12
f = 119 MHz
Midband gain
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
−
RS
−
4
gm
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
______________________________________________________________________________________
7.74
rπ =
(120 )( 0.026 )
1.02
g m = 39.23 mA/V
= 3.059 kΩ
a.
f Hπ =
Input:
(
1
2π τ π
)
τ π = R S R 2 R3 rπ (C π + 2C μ )
Req = 0.1 20.5 28.3 3.06 = 0.096 k Ω
τ π = (96)[12 + 2(2)]×10 −12 = 1.537 × 10 −9 s
f Hπ =
Output:
1
⇒ f Hπ = 103.6 MHz
2π 1.537 × 10 −9
(
)
f Hμ =
1
2π τ μ
τ μ = (RC R L )C μ
(
= (15 10)× 10 3 × 2 × 10 −12
)
τ μ = 6.67 × 10 s
−9
f Hμ =
1
⇒ f Hμ = 23.9 MHz
2π 6.67 × 10 −9
(
)
b.
⎛ R 2 R3 rπ ⎞
⎟
Aυ = g m (RC R L )⎜
⎜R R r +R ⎟
S ⎠
⎝ 2 3 π
R 2 R3 rπ = 20.5 28.3 3.059 = 2.433 k Ω
⎛ 2.433 ⎞
Aυ = (39.23)(5 10 )⎜
⎟ ⇒ Aυ = 125.6
⎝ 2.433 + 0.1 ⎠
C = 15 pF > C ⇒ C
L
μ
L
dominates frequency response.
c.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 8
8.1
24
= 6Ω
4
PD , max = (12 )(2 ) = 24 W
(b) (i) R D =
(ii) PD , max = 30 = (20 )I DQ ⇒ I DQ = 1.5 A
I D , max = 2(1.5) = 3 A
40
= 13.3 Ω
3
(c) (i) I D , max = 4 A
RD =
(ii) I D , max = 3 A
______________________________________________________________________________________
8.2
(a) PQ , max = VCEQ ⋅ I CQ
⎛ 24 ⎞
25 = ⎜ ⎟ ⋅ I CQ ⇒ I CQ = 2.083 A
⎝ 2 ⎠
24 − 12
RL =
= 5.76 Ω
2.083
2.083
I BQ =
= 0.03472 A
60
24 − 0.7
RB =
= 671 Ω
0.03472
β VT (60 )(0.026 )
(b) rπ =
=
= 0.7489 Ω
I CQ
2.083
Ib =
Vp
rπ
=
12 mV
= 16.02 mA
0.7489
I c = βI b = (60 )(0.01602 ) = 0.9614 A
1 2
1
2
I c RC = (0.9614 ) (5.76 ) = 2.66 W
2
2
For the transistor,
PQ = 25 − 2.66 = 22.34 W
Pavg =
______________________________________________________________________________________
8.3
30
= 25 Ω
1. 2
I CQ 0.6
I BQ =
=
= 0.0075 A
80
β
V − V BE (on ) 30 − 0.7
R B = CC
=
⇒ R B = 3.91 k Ω
I BQ
0.0075
(a) R L =
PQ , max = I CQ V CEQ = (0.6 )(15 ) = 9 W
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ VCC − VCEQ ⎞
⎟ ⋅ VCEQ
(b) PQ , max = I CQ VCEQ = ⎜⎜
⎟
RL
⎝
⎠
1
VCEQ = VCC
2
1 ⎛ VCC ⎞⎛ V CC ⎞
22.36
5=
⎜⎜
⎟⎟⎜⎜
⎟⎟ ⇒ V CC = 22.36 V; I C , max =
= 0.8944 A
25
25 ⎝ 2 ⎠⎝ 2 ⎠
1
(ΔI C )2 R L = 1 (0.6)2 (25) = 4.5 W
2
2
0.8944
1
2
For (b): ΔI C = I CQ =
= 0.4472 A; PL = (0.4472) (25) = 2.5 W
2
2
______________________________________________________________________________________
(c) For (a): ΔI C = 0.6 A; PL =
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.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
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
2
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 ⎤⎦
=
40 − VD S
⇒ VD S = 1.88 V
10
I D = 3.81 A, P = 7.16 W
V = 7 V, P = 39.375 W > PD ,max = 35 W
c.
Yes, at GG
______________________________________________________________________________________
8.6
a.
VDD
= 25 V
2
50 − 25
I DQ =
= 1.25 A
20
Set VDSQ =
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.
c.
PD = I DQVDSQ = (1.25 )( 25 ) ⇒ PD = 31.25 W
I D ,max = 2 I DQ ⇒ I D ,max = 2.5 A
VDS ,max = VDD ⇒ VDS ,max = 50 V
PD ,max = 31.25 W
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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)
(b)
PD = PD ,max − ( Slope ) (T j − 25 )
At PD = 0, T j ,max =
60
+ 25 ⇒ T j ,max = 145°C
0.5
T j ,max − Tcase
145 − 25
⇒ θ dev − amb = 2°C/W
60
or
(c)
______________________________________________________________________________________
PD ,max =
8.8
PD ,rated =
θ dev − amb
θ dev − amb =
T j ,max − Tamb
θ dev − case
or θ dev − case =
T j ,max − Tamb
PD ,rated
150 − 25
= 2.5°C/W
50
Then Tdev − Tamb = PD (θ dev − case + θ case − amb )
150 − 25 = PD ( 2.5 + θ case − amb ) ⇒ 125 = PD ( 2.5 + θ case − amb )
=
______________________________________________________________________________________
8.9
(a) T j , max − Tamb = PT (θ dev − case + θ snk − amb + θ case − snk )
120 − 25 = PT (1.5 + 2.8 + 0.6) ⇒ PT = 19.39 W
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Tcase = 25 + (19.39 )(0.6 + 2.8) = 90.9°C
(c) Tsnk = 25 + (19.39)(2.8) = 79.3°C
______________________________________________________________________________________
8.10
(a) T j , max − Tamb = P (θ dev − case + θ case − amb )
150 − 25 = 30(2.8 + θ case − amb ) ⇒ θ case − amb = 1.37°C / W
(b) T j , max = 25 + 20(2.8 + 1.37 ) = 108°C
______________________________________________________________________________________
8.11
(a) 150 − 25 = PT (3.8 + 1.5 + 4 ) ⇒ PT = 13.4 W
(b) P = I CQ V CEQ
13.4 = (3) ⋅ V CEQ ⇒ V CEQ = 4.48 V
V CC = V CE , max = 2V CEQ = 8.96 V
______________________________________________________________________________________
8.12
η=
PL
PS
PS = VCC ⋅ I Q
⎛V ⎞
PL = VP ⋅ I P = ⎜ CC ⎟ ( I Q )
⎝ 2 ⎠
1
⋅ VCC ⋅ I Q
η= 2
⇒ η = 50%
VCC ⋅ I Q
______________________________________________________________________________________
8.13
(a) Aυ =
(1 + β )R L
(1 + β )R L
=
βVT
rπ + (1 + β )R L
+ (1 + β )R
IC
L
We have (1 + β ) ≅ β
I C RL
RL
RL
Aυ =
=
=
V
1
I C R L + VT
RL +
RL + T
gm
IC
8
(b) (i) 0.9 =
8+
1
gm
⇒ g m = 1.125 A/V, I C = 29.25 mA
8
(ii) 0.95 =
8+
1
gm
8
⇒ g m = 2.375 A/V, I C = 61.75 mA
⇒ g m = 41.54 A/V, I C = 108 mA
1
8+
gm
______________________________________________________________________________________
(iii) 0.997 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.14
2
2
1 Vp
1 Vp
⋅
⇒ 0.5 = ⋅
⇒ V p = 2.828 V
2 RL
2 8
V p 2.828
=
= 0.3536 A
Ip =
8
RL
(b) For V o = −V p = −2.828 V
(a) PL =
I L = 0.3536 = (0.9)I O ⇒ I O = 0.393 A
______________________________________________________________________________________
8.15
1.6
= 0.20 A
8
I C ≅ 0.2 + 0.25 = 0.45 A
(a) VO = 1.6 V, I L =
0.45
= 17.31 A/V
0.026
8
Aυ =
= 0.9928
1
8+
17.31
(b) VO = 0 , I L = 0 , so I C = 0.25 A
gm =
0.25
= 9.615 A/V
0.026
8
Aυ =
= 0.9872
1
8+
9.615
(c) VO = −1.6 V, I L = −0.2 A, I C ≅ 0.25 − 0.2 = 0.05 A
gm =
0.05
= 1.923 A/V
0.026
8
Aυ =
= 0.939
1
8+
1.923
______________________________________________________________________________________
gm =
8.16
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
= 4.3 mA
vS ( min )
1
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.17
I D 3 = K (VGS 3 − VTN ) =
2
0 − VGS 3 − ( −5 )
R
12 (VGS 3 − 0.5 ) = 5 − VGS 3
2
2VGS2 3 − 11VGS − 2 = 0
VGS 3 =
(11) + 4 (12 )( 2 )
2 (12 )
2
11 ±
VGS 3 = VGS 2 = 1.072 V
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
2
vo ( min ) : i2 ( max ) = −3.93 =
V0 ( min )
1
vI ( min ) = vo ( min ) + VTN = −3.93 + 0.5
vI ( min ) = −3.43 V
⇒ V0 ( min ) = −3.93 V
vo ( max ) = 5 − VDS ( sat ) = 5 − 0.572
vo ( max ) = 4.43 V
I D1 ( max ) = 3.93 +
4.43
= 8.36 mA
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
______________________________________________________________________________________
8.18
(a) For υ O = −12 + 0.7 = −11.7 V, I Q =
11.3
+ 50 = 615 mA
0.02
⎛
2⎞
2 ⎞
⎛
I REF = ⎜⎜1 + ⎟⎟ ⋅ I Q = ⎜1 + ⎟(615) = 645.75 mA
40
β
⎝
⎠
⎝
⎠
0 − 0.7 − (− 12 )
R=
⇒ R = 17.5 Ω
0.6475
11.3
For υ O = 12 − 0.7 = +11.3 V, i L =
= 565 mA
0.02
i E1 (max ) = I Q + i L = 615 + 565 ⇒ i E1 (max ) = 1.18 A
1 VO 1 (11.3)
⋅
= ⋅
= 3.19 W
2 RL 2
20
PS = I Q (24 ) = (0.615 )(24 ) = 14.76 W
2
2
(b) PL =
PL
3.19
=
× 100% = 21.6%
PS 14.76
______________________________________________________________________________________
Define η =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.19
−20
(a) VO = −20 V, i L =
= −0.10 A
200
I Q = 0.10 + i E1 (min ) = 0.10 + 0.02 = 0.12 A
⎛
2⎞
2 ⎞
⎛
I REF = ⎜⎜1 + ⎟⎟ ⋅ I Q = ⎜1 + ⎟(0.12 ) = 0.1248 A
⎝ 50 ⎠
⎝ β⎠
0 − 0.7 − (− 24 )
R=
= 187 Ω
0.1248
(b) PQ1 = I Q VCE1 = (0.12 )(24 ) = 2.88 W
P = I Q VCE 2 + I REF (24 ) = (0.12 )(24 ) + (0.1248 )(24 ) = 5.88 W
(c) PL =
( )
2
1 VO 1 20 2
⋅
= ⋅
=1W
2 R L 2 200
1
× 100% = 11.4%
2.88 + 5.88
______________________________________________________________________________________
η=
8.20
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
(b)
For
RL = 500 Ω vo ( max ) = 4.8 V
vo < 0, vo ( min ) = −3.2 V
For
I 2′ =
vo −3.2
=
= −6.4 mA
RL
0.5
−2.5 ≤ vI ≤ 5.5 V
(c)
For vo = −2V , I 2′ ( max ) = −38.9 mA
R2 ( min ) =
−2
⇒ RL ( min ) = 51.4 Ω
−38.9
1 v2 1 ( 2)
⇒ PL = 38.9 mW
PL = ⋅ o = ⋅
2 RL 2 51.4
2
38.9
= 10%
389
______________________________________________________________________________________
PL = 10 ( 38.9 ) = 389 mW % =
8.21
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V 2 (V )
PL = P =
RL
RL
+ 2
1 (V ) 1 (V )
, V − = −V +
+ ⋅
PS = ⋅
2 RL
2 RL
+ 2
− 2
(V )
So P =
+ 2
S
η=
RL
PL
⇒ η = 100%
PS
______________________________________________________________________________________
8.22
(a)
As maximum conversion efficiency
η=
π
4
,
VP
= 0.785
VCC
⎛4⎞
So V p ( max ) = ( 0.785 )( 5 ) ⎜ ⎟
⎝π ⎠
V p ( max ) = 5 V
(b)
Maximum power dissipation occurs when
V2
Pθ ( max ) = 2CC
π RL
2=
( 5)
Vp =
2VCC
π
=
2 ( 5)
π
= 3.183 V
2
π 2 RL
⇒ RL = 1.27 Ω
(c)
______________________________________________________________________________________
8.23
2
P=
1 Vp
⋅
2 RL
2
(a)
(b)
1 Vp
⋅
⇒ V p = 49 V ⇒ V + = 52 V, V − = −52 V
2 24
VP 49
IP =
=
= 2.04 A
RL 24
50 =
η=
π VP
⋅
4 VCC
=
π ⎛ 49 ⎞
⎜ ⎟
4 ⎝ 52 ⎠
η = 74.0%
(c)
______________________________________________________________________________________
8.24
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(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
Vo ( max )
So 10 − Vo ( max ) =
⎡⎣10 − V0 ( max ) ⎤⎦ =
2
RL ⋅ K n
=
2
V0 ( max )
V02 ( max ) − 20.5V0 ( max ) + 100 = 0
V0 ( max ) =
( 5 )( 0.4 )
V0 ( max )
100 − 20V0 ( max ) + V02 ( max ) =
20.5 ±
Vo ( max )
2
( 20.5 ) − 4 (100 )
2
2
⇒ V0 ( max ) = 8 V
8
⇒ iL = 1.6 mA
5
i
1.6
VGS = L =
= 2 V ⇒ VI = 10 V
0.4
Kn
iL =
b.
1 (8)
⋅
= 6.4 mW
2 5
20 (1.6 )
PS =
= 10.2 mW
2
PL =
π
η=
PL
=
6.4
⇒ η = 62.7%
10.2
PS
______________________________________________________________________________________
8.25
(b) υ I = υ GSn + υ o
Also υ o = i L R L = i dn R L = KR L (υ GSn ) , but υ GSn = υ I − υ o
2
So υ o = KR L (υ I − υ o )
2
⎛ dυ ⎞
dυ o
= 2 KR L (υ I − υ o )⎜⎜1 − o ⎟⎟
dυ I
⎝ dυ I ⎠
2 KR L (υ I − υ o )
dυ o
=
, also υ I − υ o =
dυ I 1 + 2 KR L (υ I − υ o )
Then
2 KR L ⋅ υ o
dυ o
= Aυ =
dυ I
1 + 2 KR L ⋅ υ o
υo
KR L
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
KR L =
We find
(4)(0.05) = 0.4472
(i)
For υ o = 0 , Aυ = 0
(ii)
For υ o = 1 V, Aυ =
2(0.4472)(1)
= 0.472
1 + 2(0.4472)(1)
For υ o = 10 V, Aυ =
2(0.4472 ) 10
= 0.739
1 + 2(0.4472 ) 10
______________________________________________________________________________________
(iii)
8.26
⎛ 10 −3 ⎞
⎛i ⎞
⎛V ⎞
⎟ = 0.7004 V
(a) iCn = I S exp⎜⎜ BE ⎟⎟ ⇒ V BE = VT ln⎜⎜ Cn ⎟⎟ = (0.026) ln⎜⎜
−15 ⎟
⎝ VT ⎠
⎝ 2 × 10 ⎠
⎝ IS ⎠
V BB = 2V BE = 1.40077 V
PQ = i C ⋅υ CE = (1)(5) = 5 mW
3.5
= 3.5 mA ≅ i L
1
⎛ 3.5 × 10 −3 ⎞
⎟ = 0.732957 V
υ EB = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 2 × 10
υ BE = 1.40077 − 0.732957 = 0.66781 V
(b) For υ o = −3.5 V, iCp ≅
(
)
⎛ 0.66781 ⎞
i Cn = 2 × 10 −15 exp⎜
⎟ ⇒ i Cn = 0.2857 mA
⎝ 0.026 ⎠
Then iCp ≅ 0.2857 + 3.5 = 3.7857 mA
⎛ 3.7857 × 10 −3 ⎞
⎟ = 0.734997 V
−15
⎟
⎠
⎝ 2 × 10
υ BE = 1.40077 − 0.734997 = 0.66577 V
υ EB = (0.026 ) ln⎜⎜
(
)
⎛ 0.66577 ⎞
i Cn = 2 × 10 −15 exp⎜
⎟ ⇒ i Cn = 0.2642 mA
⎝ 0.026 ⎠
i Cp = 3.5 + 0.2642 = 3.764 mA
V BB
= −3.5 − 0.735 + 0.7004 = −3.535 V
2
2
For R L : PRL = i L2 R L = (3.5) (1) = 12.25 mW
For Q n : PQn = i Cnυ CEn = (0.2642 )[5 − (− 3.5)] = 2.25 mW
υ I = υ o − υ EB +
For Q p : PQp = i Cpυ ECp = (3.764 )[− 3.5 − (− 5)] = 5.65 mW
______________________________________________________________________________________
8.27
2
(a) (i) i Dn = K n (υ GSn − VTN )
V BB
= 2 V, ⇒ V BB = 4 V
2
(ii) P = i Dn ⋅υ DSn = (1)(12 ) = 12 mW
1 = 4(υ GSn − 1.5) ⇒ υ GSn =
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V − υ DS (sat )
υ
2
(b) (i) i D = K [υ DS (sat )] = o = DD
RL
RL
KR L [υ DS (sat )] + υ DS (sat ) − V DD = 0
Now KR L = (4 )(1) = 4
2
We have 4[υ DS (sat )] + υ DS (sat ) − 12 = 0 ⇒ υ DS (sat ) = 1.612 V
υ o (max ) = 12 − 1.612 = 10.39 V
(ii) i Dn = i L = 10.39 mA
i Dp = 0
2
υ GSn = 3.112 V
−V BB
+ υ GSn + υ o = −2 + 3.112 + 10.39 = 11.5 V
2
2
(iii) For R L : PRL = i L2 R L = (10.39) (1) = 108 mW
For M n : PMn = i Dnυ DSn = (10.39)[12 − 10.39] = 16.7 mW
For M p : PMp = i Dpυ SDp = 0
υI =
______________________________________________________________________________________
8.28
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 ⎞
3
⎛
⎞
VBE = VT ln ⎜ N ⎟ = ( 0.026 ) ln ⎜
−12 ⎟
I
×
6
10
⎝
⎠
⎝ S⎠
= 0.7004 V
Then 98.2 =
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
iD ≈
30 − 0.6 30 − 0.6
=
⇒ iD ≈ 545 mA
R1
53.97
⎛ 0.545 ⎞
VD = ( 0.026 ) ln ⎜
= 0.656 V
−12 ⎟
⎝ 6 × 10 ⎠
Approximation for iD is okay.
⇒ iN = iP = 545 mA
Diodes and transistors matched
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.29
(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 )
VGS 1 =
iL
+ VTN =
K n1
⎛
So vI + 2 + 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 +
⎤
dv0 ⎡
1
⎢1 +
⎥
dvI ⎢⎣ 2 2.5v0 ⎥⎦
vO = 5 V :
1=
For
⎤ dv
dv0 ⎡
dv
1
⎢1 +
⎥ = 0 (1.1414 ) ⇒ 0 = 0.876
dvI ⎢ 2 2.5 ( 5 ) ⎥ dvI
dvI
⎣
⎦
______________________________________________________________________________________
1=
8.30
vO = vI +
VBB
− VGS and VGS =
2
vO ≈ 0, I Dn = I DQ + iL = I DQ +
For
vO = vI +
Then
I Dn
+ VTN
Kn
vO
RL
I DQ + ( vO / RL )
VBB
− VTN −
2
Kn
vO ≅ vI +
vO = vI +
or
I DQ
VBB
1 v
− VTN −
⋅ 1+ ⋅ O
2
2 I DQ RL
Kn
For vO small,
⎡ 1 I DQ
I DQ
V
1 ⎤
vO ⎢1 + ⋅
⋅
⎥ = vI + BB − VTN −
2
2
K
I
R
Kn
⎢⎣
n
DQ L ⎥
⎦
I DQ
v
VBB
− VTN −
⋅ 1+ O
2
Kn
I DQ RL
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
dvO
1
=
= 0.95
dvI ⎡ 1 I DQ
1 ⎤
⋅
⎢1 + ⋅
⎥
⎢⎣ 2 K n I DQ RL ⎥⎦
So
1 I DQ
1
1
⋅
⋅
=
− 1 = 0.0526
2 K n I DQ RL 0.95
1
RL = 0.1 k Ω, then
K n I DQ
For
Or
= 0.01052
K n I DQ = 95.1
g = 2 K n I DQ = 190 mA/V
We can write m
This is the required transconductance for the output transistor. This implies a very large transistor.
______________________________________________________________________________________
8.31
(a) RTH = R1 R2 = 14 10 = 5.833 k Ω
⎛ R2 ⎞
⎛ 10 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(12 ) = 5 V
⎝ 10 + 14 ⎠
⎝ R1 + R 2 ⎠
5 − 0.7
I BQ =
= 0.5619 mA, I CQ = 50.57 mA
5.833 + (91)(0.02)
V
12
(b) R L = CC =
⇒ R L = 237 Ω
I CQ 50.57
1 (11)
⋅
= 255 mW
2 0.237
(d) PS = I CQVCC = (50.57)(12) = 607 mW
(c) PL (max ) =
2
255
× 100% = 42%
607
______________________________________________________________________________________
η=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.32
V
15
I CQ = CC = = 15 mA
1
RL
I BQ =
15
= 0.15 mA
100
(15)
1 V2
PL ( max ) = ⋅ CC =
⇒ PL ( max ) = 112.5 mW
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
______________________________________________________________________________________
8.33
(a) RTH = R1 R 2 = 2.3 1.75 = 0.9938 k Ω
⎛ R2 ⎞
⎛ 1.75 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(12 ) = 5.185 V
R
R
+
⎝ 1.75 + 2.3 ⎠
2 ⎠
⎝ 1
5.185 − 0.7
I BQ =
= 2.473 mA, I CQ = 98.91 mA
0.9938 + (41)(0.02)
V
12
(b) Want R L′ = CC =
= 121.3 Ω = a 2 R L = a 2 (8) ⇒ a = 3.89
I CQ 0.09891
(9) = 333.9 mW
1
⋅
2 (0.1213)
2
(c) PL =
PS = I CQVCC = (98.91)(12) ⇒ 1.187 W
0.3339
× 100% = 28.1%
1.187
______________________________________________________________________________________
(d) η =
8.34
a.
b.
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
2
I rms =
c.
Secondary
Primary
d.
IP =
PL
2
=
⇒ I rms = 0.314 A
Vrms 6.36
0.314
⇒ I P = 78.6 mA
4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W
2
η=
⇒ η = 37%
5.4
______________________________________________________________________________________
8.35
a.
⎛V
⎞
⎛1
⎞
ve = ⎜ π + g mVπ ⎟ RE′ = Vπ ⎜ + g m ⎟ RE′
⎝ rπ
⎠
⎝ rπ
⎠
⎛ 1+ β ⎞
= Vπ ⎜
⎟ RE′
⎝ rπ ⎠
vi = Vπ + ve ⇒ Vπ = vi − ve
⎛ 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
1
1
n1
2
PL = ⋅ I P2 RL , a = , I CQ = P so PL = .a 2 I CQ
RL
2
2
n2
a
PS = I CQ .VCC
For η = 50% :
1 2 2
⋅ a I CQ RL a 2 I R
V
VCC
VCC
PL
CQ L
so a 2 =
= 0.5 = 2
=
=
⇒ a 2 = CC
5
2VCC
I CQ ⋅ VCC
I CQ ⋅ RL ( 0.1)( 50 )
PS
c.
R0 =
49 ( 0.026 )
rπ
β VT
=
=
⇒ R0 = 0.255 Ω
1 + β (1 + β ) I CQ ( 50 )( 0.1)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.36
a.
With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.
⎛ R1 R2 ⎞
⎛ R1 R2 ⎞
ie
ie = (1 + β ) ib and ib = ⎜
= 8 = (1 + β ) ⎜
i
⎜ R R + R ⎟⎟ i
⎜ R R + R ⎟⎟
ii
ib ⎠
ib ⎠
⎝ 1 2
⎝ 1 2
so we need
where
Rib = rπ + (1 + β ) RL′ ≈ (1 + β ) RL′ = (101)( 0.8 ) = 80.8
⎛
⎞
R1 R2
Then 8 = (101) ⎜
⎜ R R + 80.8 ⎟⎟
⎝ 1 2
⎠
R1 R2
= 0.0792 or R1 R2 = 6.95 kΩ
R1 R2 + 80.8
Set
2VCC
V
12
= RL′ ⇒ I CQ = CC =
= 15 mA
2 I CQ
RL′ 0.8
15
= 0.15 mA
100
VTH = I BQ RTH + VBE
I BQ =
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 e = 0.9 I CQ = 13.5 mA =
IL
⇒ I L = 135 mA
a
1
2
( 0.135) ( 8) ⇒ PL = 72.9 mW
2
PS = VCC I CQ = (12 )(15 ) ⇒ PS = 180 mW
PL =
η=
PL
⇒ η = 40.5%
PS
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.37
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 =
______________________________________________________________________________________
8.38
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 ⎟⎠
Set
2VCC
12
= 0.9 kΩ ⇒ I CQ =
= 13.3 mA
2 I CQ
0.9
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
I BQ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
PL 57.6
=
⇒ η = 36.1%
PS 159.6
______________________________________________________________________________________
η=
8.39
⎛ 10 −3 ⎞
⎞
⎛I
⎟ = 1.473 V
(a) V BB = 2VT ln⎜⎜ Bias ⎟⎟ = 2(0.026) ln⎜⎜
−16 ⎟
⎝ 5 × 10 ⎠
⎝ I SD ⎠
⎛ 1.473 2 ⎞
(b) I CQ = I SQ exp⎜
⎟ ⇒ I CQ = 14 mA
⎝ 0.026 ⎠
______________________________________________________________________________________
8.40
−3
⎛ I CQ ⎞
⎛
⎞
⎟ = (0.026) ln⎜ 4 × 10 ⎟ = 0.73643 = V BEn = V EBp = V D
(a) V BE = VT ln⎜
⎜ 2 × 10 −15 ⎟
⎜ I SQ ⎟
⎝
⎠
⎠
⎝
⎛V ⎞
⎛ 0.73643 ⎞
I Bias = I SD exp⎜⎜ D ⎟⎟ = 4 × 10 −16 exp⎜
⎟ ⇒ I Bias = 0.8 mA
V
⎝ 0.026 ⎠
⎝ T ⎠
(
)
(b) V BB = 2V D = 1.473 V
(c) υ I = −V EBp = −0.7364 V
______________________________________________________________________________________
8.41
⎛ 0.5 × 10 −3 ⎞
⎟ = 0.76025 V
(a) V D1 = (0.026 ) ln⎜⎜
−16
⎟
⎠
⎝ 10
⎛ 0.5 × 10 −3 ⎞
⎟ = 0.72421 V
V D 2 = (0.026) ln⎜⎜
−16 ⎟
⎠
⎝ 4 × 10
V BB = V D1 + V D 2 = 1.48446 V
⎛ υ EBp ⎞
⎛υ ⎞
⎟
(b) iCn = i Cp = I SQn exp⎜⎜ BEn ⎟⎟ = I SQp exp⎜⎜
⎟
⎝ VT ⎠
⎝ VT ⎠
⎛υ ⎞
⎛υ ⎞
exp⎜⎜ BEn ⎟⎟
exp⎜⎜ BEn ⎟⎟
V
⎛ 2υ − V BB ⎞
⎝ VT ⎠
⎝ T ⎠
⎟⎟
=
= exp⎜⎜ BEn
=
I SQn
VT
υ
⎛ V BB − υ BEn ⎞
⎛ EBp ⎞
⎝
⎠
⎟ exp⎜⎜
⎟⎟
exp⎜⎜
⎟
V
V
T
⎝
⎠
⎝ T ⎠
I SQp
⎛ I SQp ⎞
2υ BEn − V BB
⎟
= ln⎜
⎜ I SQn ⎟
VT
⎠
⎝
υ BEn =
−15 ⎤
⎛ I SQp ⎞⎤ 1 ⎡
⎛
⎞
1⎡
⎟⎥ = ⎢1.48446 + (0.026 ) ln⎜ 1.6 × 10 ⎟⎥ = 0.75124 V
⎢V BB + VT ln⎜
−
16
⎜
⎟
⎜ I SQn ⎟⎥ 2 ⎢
2 ⎣⎢
⎝ 8 × 10
⎠⎦⎥
⎣
⎝
⎠⎦
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
υ EBp = V BB − υ BEn = 0.73322 V
⎛υ ⎞
⎛ 0.75124 ⎞
(c) I CQ = I SQn exp⎜⎜ BEn ⎟⎟ = 8 × 10 −16 exp⎜
⎟
V
⎝ 0.026 ⎠
⎝ T ⎠
⇒ i Cn = i Cp = 2.828 mA
(
)
(d) υ I = −υ EBp = −0.73322 V
______________________________________________________________________________________
8.42
a.
All transistors are matched.
⎛1+ β ⎞
iC
3 mA = iE1 + iB 3 = ⎜
⎟ iC +
β
⎝ β ⎠
⎛ 61 1 ⎞
3 = ⎜ + ⎟ iC ⇒ iC = 2.90 mA
⎝ 60 60 ⎠
b.
R = 200 Ω.
For vo = 6 V , let L
6
= 0.03 A = 30 mA ≅ iE 3
200
30
iB 3 =
= 0.492 mA
61
iE1 = 3 − 0.492 = 2.508 mA
2.508
iB1 =
⇒ 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
io =
Current gain
30 × 10−3
⇒ Ai = 3.72 × 103
8.07 × 10−6
⎛i ⎞
⎛ 30 × 10−3 ⎞
VBE 3 = VT ln ⎜ E 3 ⎟ = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
Ai =
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
v
6
Av = 0 =
⇒ Av = 0.989
vI 6.0646
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.43
1
i0 = 1 A, I B 3 ≅
⇒ 20 mA
50
a.
For
We can then write
⎡10 − ( v0,max + VBE 3 )
⎤
10 − VEB1
= 2⎢
− 20 ⎥
R1
R1
⎣⎢
⎦⎥
10 − VBE 2vo ,max
=
+ 40
R1
R1
If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then
9.3 2 ( 4 )
=
+ 40
R = R2 = 32.5 Ω
v
= 4 V,
R
R1
which yields 1
If we assume 0,max
then 1
9.3
I E1 =
⇒ I E1 = 0.286 A = I E 2
v
=
0,
32.5
I
b.
For
I = I E 4 = 2.86 A
I
= 10 I S 1,2 ,
Since S 3,4
then E 3
c.
We can write
⎧
rπ 1 ⎫
⎪ rπ 3 + R1
⎪
1 + β1 ⎪
1⎪
R0 = ⎨
⎬
2⎪
1 + β3
⎪
⎪
⎪
⎩
⎭
( 50 )( 0.026 )
βV
Now rπ 3 = 3 T =
= 0.4545 Ω
IC3
2.86
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
⎪
⎪⎩
⎪⎭
32.5
10.91
= 32.5 0.0902 = 0.0900
121
1 ⎧ 0.4545 + 0.0900 ⎫
⎨
⎬ or R 0 = 0.00534 Ω
51
2⎩
⎭
______________________________________________________________________________________
Then R0 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.44
1
Ri = rπ 1 + (1 + β ) ⎡ R1 ( rπ 3 + (1 + β ) 2 RL ) ⎤
⎣
⎦
2
{
}
iC1 ≈ 7.2 mA and iC 3 ≈ 7.2 mA
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 =
{
}
8.45
I DQ
(b) For M n 3 ; υ GSn3 =
K n3
+ VTN =
5
+1 = 2 V
5
υ SGp1 = υ GSn 3 = 2 V
I DQ1 = 2(2 − 1) = 2 mA
2
10 − 2
= 4 kΩ
2
(c) I DQ1 = I DQ 2 = 2 mA
R1 = R 2 =
(d) For υ o = 3.5 V, Assume M p 4 cutoff, so I DQ 4 = 0
I Dn3 = i o =
υo
RL
=
3.5
= 23.33 mA
0.15
23.33
+ 1 = 3.160 V
5
10 − (υ o + υ GSn3 ) 10 − (3.16 + 3.5)
I Dp1 = I R1 =
=
= 0.835 mA
R1
4
υ GSn 3 =
0.835
+ 1 = 1.646 V
2
υ I = υ o + υ GSn 3 − υ SGp1 = 3.5 + 3.160 − 1.646 = 5.014 V
υ SGp1 =
υ I = υ GSn 2 + K n 2 R2 (υ GSn2 − VTN )2 + V −
(
)
2
15.014 = υ GSn2 + 8 υ GSn
2 − 2υ GSn 2 + 1
2
or 8υ GSn
2 − 15υ GSn 2 − 7.014 = 0 ⇒ υ GSn 2 = 2.2625 V
I Dn 2 = 2(2.2625 − 1) = 3.188 mA
υ G 4 = υ I − υ GSn 2 = 5.014 − 2.2625 = 2.75 V
υ SGp 4 = υ o − υ G 4 = 3.5 − 2.75 = 0.75 V, ⇒ M p 4 cutoff
2
PL =
υ o2
=
(3.5)2 = 81.7 mW
0.15
RL
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.46
For υ I = −1.5 V and υ O = 0 , ⇒ υ SG 2 = 1.5 V = υ GS1
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
i D1 = i D 2 = 0.5 = ⎜
⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 20.4
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1
⎛ 0.04 ⎞⎛ W ⎞
⎛W ⎞
2
0. 5 = ⎜
⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 51.0
2
L
⎝
⎠⎝ ⎠ 2
⎝ L ⎠2
⎛ 0.04 ⎞⎛ W ⎞
⎛W ⎞
2
⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 20.4
⎝ 2 ⎠⎝ L ⎠ 4
⎝ L ⎠4
υ SG 4 = 1.5 V, 0.2 = ⎜
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
0. 2 = ⎜
⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 8.16
⎝ 2 ⎠⎝ L ⎠ 3
⎝ L ⎠3
______________________________________________________________________________________
8.47
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 C1 = ⎜ 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 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Because of rπ 1 and Z, neglect effect of r . Then neglecting r , r and r , we find
0
01
02
03
VX
I X = g m 3Vπ 3 + g m 2Vπ 2 + g m1Vπ 1 +
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
VX
ββ
β1
⋅ V X + 1 2 ⋅ VX +
⋅ VX +
Then I X = 1
rπ 1 + Z
rπ 1 + Z
rπ 1 + Z
rπ 1 + Z
Then
R0 =
rπ 1 + Z
VX
=
I X 1 + β1 + β1 β 2 + ( β1 + β1 β 2 ) β 3
(10 )( 0.026 )
rπ 1 =
1.534
Z = 25 kΩ
= 0.169 MΩ
Then
R0 =
169 + 25
1 + (10 ) + (10 )( 50 ) + ⎡⎣10 + (10 )( 50 ) ⎤⎦ ( 50 )
194
= 0.00746 kΩ or Ro = 7.46 Ω
26, 011
______________________________________________________________________________________
R0 =
8.48
a
Neglect base currents.
⎛I ⎞
VBB = 2VD = 2VT ln ⎜ Bias ⎟
⎝ IS ⎠
⎛ 5 × 10−3 ⎞
= 2 ( 0.026 ) ln ⎜
⇒ VBB = 1.281 V
−13 ⎟
⎝ 10
⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VBE1 + VEB 3 = VBB
I E1 = I E 3 + I C 2
⎛ β ⎞
I B 2 = IC 3 = ⎜ P ⎟ I E 3
⎝ 1+ βP ⎠
⎛ β ⎞
IC 2 = β n I B 2 = β n ⎜ P ⎟ I E 3
⎝ 1+ βP ⎠
⎛ βP ⎞
I E1 = I E 3 + β n ⎜
⎟ IE3
⎝ 1+ βP ⎠
⎡
⎛ β P ⎞⎤
I E1 = I E 3 ⎢1 + β n ⎜
⎟⎥
⎝ 1 + β P ⎠⎦
⎣
⎡
⎛ 1+ βn ⎞
⎛ 1+ βP ⎞
⎛ β P ⎞⎤
⎜
⎟ I C1 = ⎜
⎟ I C 3 ⎢1 + β n ⎜
⎟⎥
⎝ βP ⎠
⎝ 1 + β P ⎠⎦
⎝ βn ⎠
⎣
⎡I ⎤
⎡I ⎤
VBE1 = VT ln ⎢ C1 ⎥ , VEB 3 = VT ln ⎢ C 3 ⎥
⎣ IS ⎦
⎣ IS ⎦
21 ⎞
⎡
⎛ 20 ⎞ ⎤
⎟ I C 3 ⎢1 + (100 ) ⎜ ⎟ ⎥
⎝ 20 ⎠
⎝ 21 ⎠ ⎦
⎣
(1.01) IC1 = ⎛⎜
⎡ 21
⎤
= I C 3 ⎢ + 100 ⎥ = 101.05 I C 3
20
⎣
⎦
I C1 = 100.05 I C 3
⎛ 100.05I C 3 ⎞
⎛ IC 3 ⎞
VT ln ⎜
⎟ + VT ln ⎜
⎟ = VBB
IS
⎝
⎠
⎝ IS ⎠
⎛ 100.05I C2 3 ⎞
VT ln ⎜
⎟ = VBB
I S2
⎝
⎠
2
⎛V ⎞
100.05 I C 3
= exp ⎜ BB ⎟
2
IS
⎝ VT ⎠
⎛V ⎞
exp ⎜ BB ⎟ = 0.4995 mA = I C3
100.05
⎝ VT ⎠
Then I E 3 = 0.5245 mA
IC 3 =
IS
Now I C1 = 100.05I C 3 = 49.97 mA = I C1
⎛ 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
10
v0 = 10 V ⇒ iE1 ≈
= 0.10 A = iC1
100
100
iB1 =
= 1 mA
100
⎛ 4 × 10−3 ⎞
VBB = 2 ( 0.026 ) ln ⎜
= 1.2694 V
−13 ⎟
⎝ 10
⎠
⎛ 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 ⎠
V02 (10 )
=
⇒ PL = 1 W
RL
100
2
PL =
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.49
a.
⎛ 10 × 10−3 ⎞
VBB = 3 ( 0.026 ) ln ⎜
⇒ VBB = 1.74195 V
−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 = ( 0.026 ) ln ⎜
⇒ VBE1 = 0.58065 V
−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 )
PL =
:
( 20 )
v
=
⇒ PL = 20 W
20
RL
2
0
2
20
= −1 A
20
iC 5 + iC 4 + iE 3 = −io ( max ) = 1 A
i0 ( max ) = −
i ⎛ β ⎞ i ⎛ 1+ β p ⎞
iC 5 + C 5 ⋅ ⎜ n ⎟ + C 4 ⎜
⎟ =1
β n ⎝ 1 + β n ⎠ β n ⎜⎝ β p ⎟⎠
i ⎛ β ⎞ i ⎛ β ⎞ ⎡ 1 ⎛ 1 + β p ⎞⎤
iC 5 + C 5 ⎜ n ⎟ + C 5 ⎜ n ⎟ ⎢ ⎜
⎟⎥ = 1
β n ⎝ 1 + β n ⎠ β n ⎝ 1 + β n ⎠ ⎢⎣ β n ⎝⎜ β p ⎟⎠ ⎥⎦
⎡
1 ⎛ 20 ⎞ ⎤ ⎛ 1 ⎞ ⎡ 1 ⎛ 6 ⎞ ⎤
iC 5 ⎢1 + ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ = 1
20
⎝ 21 ⎠ ⎦ ⎝ 21 ⎠ ⎣ 20 ⎝ 5 ⎠ ⎦
⎣
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
iC 5 (1.05048) = 1 iC 5 = 0.952 A
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
i ⎛ β ⎞ 93.9
= 4.47 mA
iC1 = C 2 ⎜ n ⎟ =
β n ⎝ 1 + β n ⎠ 21
PQ 2 = I C 2 ( 24 − ( −20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 W
PQ 5 = ( 0.952 ) ( −10 − ( −24 ) ) = 13.3 W
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 9
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)
(d)
(e)
5 = 500 (1 − v1 ) ⇒ 500v1 = 495
v1 = 0.990 V
vO = 0
− 3 = 500 ( v2 − ( −0.5 ) )
−250 − 3 = 500v2
v2 = −0.506 V
______________________________________________________________________________________
9.2
(a) υ 2 =
υO
Aod
=
−2
= −2 × 10 − 4 V
10 4
⎛
1
⎞
⎟ ⋅υ I
⎝ 1 + 2000 ⎠
υ2 = ⎜
⎛ 1 ⎞
− 2 × 10 − 4 = ⎜
⎟ ⋅υ I ⇒ υ I = −0.4002 V
⎝ 2001 ⎠
1
⎛
⎞
(b) υ 2 = ⎜
⎟ ⋅υ I
⎝ 1 + 2000 ⎠
⎛ 1 ⎞
−3
⎟(2 ) = 0.9995 × 10 V
⎝ 2001 ⎠
υ2 = ⎜
υ O = 1 = Aod υ 2 = Aod (0.9995 ×10 −3 ) ⇒ Aod = 1000.5
______________________________________________________________________________________
9.3
(a) υ O = Aod (υ 2 − υ1 ) = 5 ×10 3 (2.0000 − 2.0010) = −5 V
(b) υ O = Aod (υ 2 − υ1 )
(
(
)
)
− 3.000 = 2 × 10 4 (3.0025 − υ1 ) ⇒ υ1 = 3.00265 V
(c) υ O = Aod (υ 2 − υ1 )
1.80 = Aod (0.01 − (− 0.01))× 10 −3 ⇒ Aod = 9 ×10 4
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
(a) Aυ =
− R 2 − 200
=
= −10
R1
20
−120
= −3
40
−40
(c) Aυ =
= −1
40
______________________________________________________________________________________
(b) Aυ =
9.6
200
⎫
= −10 ⎪
20
⎪
and
⎬ for each case
⎪
Ri = 20 kΩ
⎪
⎭
______________________________________________________________________________________
Av = −
9.7
a.
100
= −10
10
Ri = R1 = 10 kΩ
Av = −
b.
Aυ = −
100 100
= −5
10
Ri = R1 = 10 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
c.
100
= −5
10 + 10
Ri = 10 + 10 = 20 K
______________________________________________________________________________________
Av = −
9.8
0 −υO
4
⇒ R2 =
⇒ R 2 = 200 k Ω
R2
20 × 10 − 6
(a) i 2 =
− R2
− 200
⇒ −12 =
⇒ R1 = 16.67 k Ω
R1
R1
Aυ =
(b) i 2 =
0 −υO
0 − 1.5
=
⇒ i1 = i 2 = −7.5 μ A
R2
200 × 10 3
υO
+ 1.5
⇒ υ I = −0.125 V
− 12
Aυ
______________________________________________________________________________________
υI =
=
9.9
Av = −
(a)
(b)
(c)
(d)
(e)
R2
R1
Av = −10
Av = −1
Av = −0.20
Av = −10
Av = −2
Av = −1
(f)
______________________________________________________________________________________
9.10
(a) − 3 =
− R 2 − 200
=
⇒ R 2 = 200 k Ω , R1 = 66.67 k Ω
R1
R1
(b) − 8 =
− R 2 − 200
=
⇒ R 2 = 200 k Ω , R1 = 25 k Ω
R1
R1
(c) − 20 =
− R 2 − 200
=
⇒ R 2 = 200 k Ω , R1 = 10 k Ω
R1
R1
− R2 − R2
=
⇒ R 2 = 100 k Ω , R1 = 200 k Ω
R1
200
______________________________________________________________________________________
(d) − 0.5 =
9.11
υI
0.25
⇒ R1 = 5 k Ω
R1
50 × 10 − 6
− R2
− R2
Aυ =
⇒ −6.5 =
⇒ R 2 = 32.5 k Ω
R1
5
(a) i1 =
⇒ R1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
υO
(b) υ I =
Aυ
=
−4
= 0.6154 V
− 6.5
4
= 0.123 mA
32.5
______________________________________________________________________________________
i1 = i 2 =
9.12
(a) Aυ =
− R2
R1
− R2
⇒ R1 = 25 k Ω , R 2 = 500 k Ω
25
−1000
⇒ R1 = 50 k Ω , R 2 = 1 M Ω
(b) − 20 =
R1
− 20 =
(c) For (a), i1 =
υI
R1
=
− 0.2
⇒ i1 = −8 μ A
25
υI
− 0.2
=
⇒ i1 = −4 μ A
R1
50
______________________________________________________________________________________
For (b), i1 =
9.13
a.
Av =
⎛R ⎞
R2
1.05 R2
⇒
= 1.105 ⎜ 2 ⎟
R1
0.95 R1
⎝ R1 ⎠
⎛R ⎞
0.95 R2
= 0.905 ⎜ 2 ⎟
1.05 R1
⎝ R1 ⎠
Deviation in gain is +10.5% and − 9.5%
b.
Av ⇒
⎛R ⎞
⎛R ⎞
1.01R2
0.99 R2
= 1.02 ⎜ 2 ⎟ ⇒
= 0.98 ⎜ 2 ⎟
0.99 R1
1.01R1
⎝ R1 ⎠
⎝ R1 ⎠
Deviation in gain = ±2%
______________________________________________________________________________________
9.14
(a) (i) υ O =
(ii) i 2 =
iL =
− R2
− 15
⋅υ I =
(− 0.20) = 3 V
R1
1
0 −υO − 3
=
= −0.20 mA
R2
15
υO
RL
=
3
= 0.75 mA
4
i O + i 2 = i L ⇒ i O = 0.75 − (− 0.20 ) = 0.95 mA
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
−15
(0.05) = −0.75 V
1
0 − (− 0.75)
(ii) i 2 =
= 0.05 mA
15
−0.75
iL =
= −0.1875 mA
4
i O = i L − i 2 = −0.1875 − 0.05 = −0.2375 mA
(b) (i) υ O =
−15
(8 sin ω t ) (mV) ⇒ υ O = −0.12 sin ω t (V)
1
0.12 sin ω t
(ii) i 2 =
⇒ i 2 = 8 sin ω t ( μ A)
15
−0.12 sin ω t
iL =
⇒ i L = −30 sin ω t ( μ A)
4
iO = i L − i 2 = −38 sin ω t ( μ A)
______________________________________________________________________________________
(c) (i) υ O =
9.15
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 kΩ
For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V
______________________________________________________________________________________
9.16
υ O1 =
− R2
− 80
⋅υ I =
(− 0.15) = 1.2 V
R1
10
υO =
− R4
− 100
⋅υ O1 =
(1.2) = −6 V
R3
20
i1 = i 2 =
i3 = i 4 =
υI
R1
υ O1
R3
=
− (0.15)
⇒ i1 = i 2 = −15 μ A
10
=
1.2
⇒ i 3 = i 4 = 60 μ A
20
At υ O1 : i 2 + i O1 = i 3 ⇒ i O1 = 60 − (− 15) = 75 μ A; Out of Op-Amp
At υ O : iO 2 = i 4 = 60 μ A; Into Op-Amp
______________________________________________________________________________________
9.17
υ O ⎛ − R 2 ⎞⎛ − R 4 ⎞
⎟ = 100
⎟⎜
=⎜
υ I ⎜⎝ R1 ⎟⎠⎜⎝ R3 ⎟⎠
For υ I = 50 mV, υ O = (100 )(0.05) = 5 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
If i 4 = 50 μ A, R 4 =
Set R1 = 10 k Ω
Then
5
⇒ R 4 = 100 k Ω
50 × 10 − 6
υO
R
⎛ R ⎞⎛ 100 ⎞
⎟ ⇒ 2 = 10
= 100 = ⎜⎜ 2 ⎟⎟⎜⎜
⎟
R3
υI
⎝ 10 ⎠⎝ R3 ⎠
Set R 2 = 100 k Ω , R3 = 10 k Ω
______________________________________________________________________________________
9.18
⎛ − R 2 ⎞⎛ − R 4 ⎞⎛ − R6 ⎞
⎟⎜
⎟ = −300
⎟⎟⎜
Aυ = ⎜⎜
⎟
⎜
⎟⎜
⎝ R1 ⎠⎝ R3 ⎠⎝ R5 ⎠
For υ O = 6 V, set i 6 =
υO
R6
⇒ 60 μ A =
6
⇒ R 6 = 100 k Ω
R6
Set R 6 = 200 k Ω so that i 6 = 30 μ A
Set R1 = 20 k Ω
⎛ R ⎞⎛ R ⎞⎛ 200 ⎞
⎟
Now 300 = ⎜⎜ 2 ⎟⎟⎜⎜ 4 ⎟⎟⎜⎜
⎟
⎝ 20 ⎠⎝ R3 ⎠⎝ R5 ⎠
For example, set R 2 = 100 k Ω and R5 = 20 k Ω
⎛ R4 ⎞
⎛ 100 ⎞⎛⎜ R 4 ⎞⎟⎛ 200 ⎞
⎟
Then 300 = ⎜
⎜
⎟ = 50⎜⎜
⎟⎜
⎟
⎟
⎝ 20 ⎠⎝ R3 ⎠⎝ 20 ⎠
⎝ R3 ⎠
R
Or 4 = 6 , set R3 = 20 k Ω and R 4 = 120 k Ω
R3
______________________________________________________________________________________
9.19
(a) υ O =
− R2
⎛ 22 ⎞
⋅υ I = −⎜ ⎟(− 0.40 ) = 8.8 V
R1
⎝ 1 ⎠
− R2
1
1
= −(22 ) ⋅
⋅
= −21.8993
R1 ⎡
1
⎡
⎤
1 ⎛ R 2 ⎞⎤
+
23
(
)
1
⎟⎥
⎜⎜1 +
⎢1 +
⎢ 5 × 10 3
⎥
⎣
⎦
R1 ⎟⎠⎥⎦
⎢⎣ Aod ⎝
υ O = (− 21.8993)(− 0.40) = 8.7597 V
(b) Aυ =
(c) Aυ = −(0.998)(22 ) = −21.956
− 21.956 = −(22) ⋅
1
⇒ Aod = 1.1477 × 10 4
⎡
⎤
1
(23)⎥
⎢1 +
A
od
⎣
⎦
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.20
(a)
R
1
Av = − 2 ⋅
R1 ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ Aod ⎝
100
1
=−
⋅
1
25 ⎡
⎤
⎢⎣1 + 5 × 103 ( 5 ) ⎥⎦
Av = −3.9960
(b)
(c)
(d)
vO = −3.9960 (1.00 ) ⇒ vO = −3.9960 V
4 − 3.9960
× 100% = 0.10%
4
vO = Aod ( v2 − v1 ) = − Aod v1
− ( −3.9960 )
vO
=
Aod
5 × 10+3
v1 = −
v1 = 0.7992 mV
______________________________________________________________________________________
9.21
1
−100
= −9.98431
⋅
10 ⎡
1 ⎛ 100 ⎞⎤
⎜1 +
⎟⎥
⎢1 +
3
10 ⎠⎦
⎣ 7 × 10 ⎝
υ
7
υI = O =
= −0.7011 V
Aυ − 9.9843
(a) Aυ =
υ1 =
−υ O
−7
=
⇒ υ 1 = −1 mV
Aυ
7 × 10 3
(b) Aod =
−υO
υ1
=
− (− 5)
= 2.5 × 10 4
0.2 × 10 −3
Aυ = (− 10) ⋅
1
= −9.9956
1
⎡
⎤
⎢1 + 2.5 × 10 4 (11)⎥
⎣
⎦
υ
−5
υI = O =
= 0.50022 V
Aυ − 9.9956
______________________________________________________________________________________
9.22
(a) Aυ =
− R 2 ⎛ R3 R3 ⎞ − 50 ⎛ 50 50 ⎞
⎜1 +
⎟=
+ ⎟ = −60
+
⎜1 +
5 50 ⎠
R1 ⎜⎝ R 4 R 2 ⎟⎠ 10 ⎝
⎛
50 50 ⎞
(b) (i) − 100 = −5⎜⎜1 +
+ ⎟⎟ ⇒ R 4 = 2.78 k Ω
⎝ R 4 50 ⎠
⎛ 50 50 ⎞
(ii) − 150 = −5⎜⎜1 +
+ ⎟⎟ ⇒ R 4 = 1.79 k Ω
⎝ R 4 50 ⎠
_____________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.23
a.
Av = −
R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
R1 ⎝ R4 R2 ⎠
R1 = 500 kΩ
80 =
R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
500 ⎝ R4 R2 ⎠
Set R2 = R3 = 500 kΩ
⎛ 500 ⎞
500
+ 1⎟ = 2 +
⇒ R4 = 6.41 kΩ
80 = 1⎜ 1 +
R4
R4
⎝
⎠
b.
For vI = −0.05 V
−0.05
i1 = i2 =
⇒ i1 = i2 = −0.1 μ A
500 kΩ
v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05
v
0.05
i4 = − X = −
⇒ i4 = −7.80 μ A
6.41
R4
i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 μ A
______________________________________________________________________________________
9.24
(a)
Av = −1000 =
− R2 −500
=
R1
R1
R1 = 0.5 K
(b)
Av =
− R2 ⎛ R3 R3 ⎞
+ ⎟
⎜1 +
R1 ⎝ R4 R2 ⎠
−1000 =
−250 ⎛ 500 500 ⎞ −1250
+
⎜1 +
⎟=
R1 ⎝ 250 250 ⎠
R1
R1 = 1.25 K
______________________________________________________________________________________
9.25
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
v
i1 = I = i2
R
⎛v ⎞
v A = −i2 R = − ⎜ I ⎟ R = −vI
⎝R⎠
v A vI
i3 = − =
R R
v
v
2v
2v
i4 = i2 + i3 = − A − A = − A = I
R R
R
R
⎛ 2v ⎞
vB = v A − i4 R = −vI − ⎜ I ⎟ ( R ) = −3vI
⎝ R ⎠
( −3vI ) 3vI
v
i5 = − B = −
=
R
R
R
2vI 3vI 5vI
i6 = i4 + i5 =
+
=
R
R
R
v0
⎛ 5vI ⎞
v0 = vB − i6 R = −3vI − ⎜
⎟ R ⇒ = −8
vI
⎝ R ⎠
From Figure 9.12 ⇒ Av = −3
______________________________________________________________________________________
9.26
− R2
− 200
1
1
⋅
=
⋅
= −9.9978
R1 ⎡
20 ⎡
1 ⎛ 200 ⎞⎤
1 ⎛ R 2 ⎞⎤
+
+
⎜⎜1 +
⎟⎥
⎜1
⎟⎥
⎢1 +
⎢1
4
20 ⎠⎦
R1 ⎟⎠⎥⎦
⎣ 5 × 10 ⎝
⎢⎣ Aod ⎝
υ
− 4.80
= 0.4801056 V
(b) υ I = O =
Aυ − 9.9978
(a) Aυ =
(c) υ1 =
− υ O − (− 4.80 )
=
⇒ υ 1 = 96 μ V
Aod
5 × 10 4
4.801056 − 4.8
× 100% = 0.022%
4.8
______________________________________________________________________________________
(d)
9.27
(a) Aυ =
(i)
− R2
1
1
⋅
= (− 1) ⋅
= −0.9992
R1 ⎡
1
⎡
⎤
1 ⎛ R 2 ⎞⎤
⎜⎜1 +
⎟⎥
⎢1 +
⎢1 + 2.5 × 10 3 (2 )⎥
⎣
⎦
R1 ⎟⎠⎦⎥
⎣⎢ Aod ⎝
υ O = (− 0.9992 )(− 0.8) = 0.7993605 V
0.8 − 0.7993605
× 100% ≅ 0.08%
0.8
1
= −0.990099
(b) Aυ = (− 1) ⋅
1
⎡
⎤
⎢1 + 2 × 10 2 (2 )⎥
⎣
⎦
(i)
υ O = (− 0.990099 )(− 0.8) = 0.79208 V
(ii)
0.8 − 0.79208
× 100% = 0.99%
0.9
______________________________________________________________________________________
(ii)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.28
vl
v
v
R
= i2 = − O ⇒ O = − 2
R1
R1
R2
vl
ii =
(a)
(b)
i2 = i1 =
⎞
vl
v
1 ⎛ R2
= i3 + O = i3 +
⎜ − ⋅ vl ⎟
R1
RL
RL ⎝ R1
⎠
Then i3 =
vl ⎛
R2 ⎞
⎜1 +
⎟
R1 ⎝ RL ⎠
______________________________________________________________________________________
9.29
⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞
VX .max = ⎜
10 ⇒ VX .max = 0.09008 V
⋅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
______________________________________________________________________________________
9.30
υO = −
RF
R
R
120
120
120
⋅υ I 1 − F ⋅ υ I 2 − F ⋅υ I 3 = −
⋅υ I 1 −
⋅υ I 2 −
⋅υ I 3
40
20
60
R1
R2
R3
υ O = −3υ I 1 − 6υ I 2 − 2υ I 3
(a) υ O = −3(− 0.25) − 6(0.10 ) − 2(1.5) = −2.85 V
(b) 0.5 = −3(υ I 1 ) − 6(0.25) − 2(− 1.2) ⇒ υ I 1 = 0.133 V
______________________________________________________________________________________
9.31
(a) υ O = −2.5(1.2υ I 1 + 2.5υ I 2 + 0.25υ I 3 ) = −3υ I 1 − 6.25υ I 2 − 0.625υ I 3
RF
R
R
= 3 , F = 6.25 , F = 0.625
R1
R2
R3
Then
R3 is the largest resistor, so set R3 = 400 k Ω
Then R F = 250 k Ω , R1 = 83.3 k Ω , R 2 = 40 k Ω
(b) υ O = −3(− 1) − 6.25(0.25) − 0.625(2) = 0.1875 V
υO
0.1875
⇒ i F = 0.75 μ A
RF
250
______________________________________________________________________________________
iF =
9.32
=
υ O = −2(υ I 1 + 3υ I 2 ) = −2υ I 1 − 6υ I 2
Then
RF
R
= 2, F = 6
R1
R2
For υ I 1 = −1 V, υ I 2 = −0.5 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then υ O = −2(− 1) − 6(− 0.5) = 5 V
Set i F = 80 μ A =
υO
RF
=
5
⇒ R F = 62.5 k Ω
RF
Then R1 = 31.25 k Ω , R 2 = 10.42 k Ω
______________________________________________________________________________________
9.33
vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft )
1
1
⇒ 1 ms vI 2 ⇒ T2 =
⇒ 10 ms
100
103
R
R
10
10
vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2
1
5
R1
R2
f = 1 kHz ⇒ T =
vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V )
vO = −0.707 sin ( 2π ft ) − ( ±2 V )
______________________________________________________________________________________
9.34
υO = −
RF
R
⋅υ I 1 − F ⋅υ I 2
R1
R2
− 0.5 sin ω t = −
100
(0.004 + 0.125 sin ω t ) − 100 (− 0.006)
R1
R2
Set − 0.5 sin ω t = −
We have 0 = −
100
(0.125 sin ω t ) ⇒ R1 = 25 k Ω
R1
100
(0.004) − 100 (− 0.006)
R1
R2
0.4 0.6
+
⇒ R 2 = 37.5 k Ω
25 R 2
______________________________________________________________________________________
0=−
9.35
1
⎡υ
⎤
(a) υ O = −2 ⎢ I 1 + 2υ I 2 + υ I 3 ⎥ = − υ I 1 − 4υ I 2 − 2υ I 3
2
⎣ 4
⎦
R
R
1 R
Then F = , F = 4 , F = 2
R1 2 R 2
R3
Set R1 = 250 k Ω , Then R F = 125 k Ω , R 2 = 31.25 k Ω , R3 = 62.5 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) For υ I 1 = −2 V, υ I 2 = 0 , υ I 3 = −1 V
1
(− 2) − 4(0) − 2(− 1) = 3 V
2
For υ I 1 = 2 V, υ I 2 = 0.5 V, υ I 3 = 0
υO = −
1
(2) − 4(0.5) − 2(0) = −3 V
2
Then −3 ≤ υ O ≤ +3 V
υO = −
υ O max
3
=
⇒ i F max = 24 μ A
RF
125
______________________________________________________________________________________
i F max =
9.36
RF
R
R
⋅υ I 1 − F ⋅ υ I 2 − F ⋅υ I 3
R1
R2
R3
υO = −
− 6 sin ω t = −
We have −
RF
R
R
(2 + 2 sin ω t ) − F (0.5 sin ω t ) − F (− 4)
R1
R2
R3
RF
R
(2) + F (4) = 0 ⇒ R3 = 2 R1
R1
R3
Also − 6 = −
RF
R
(2) − F (0.5)
R1
R2
For υ O = 6 V, i F max = 120 μ A =
6
⇒ R F = 50 k Ω
RF
For υ I 1 max = 4 V, i1 max = 120 μ A =
Now 6 =
4
⇒ R1 = 33.33 k Ω and R3 = 2 R1 = 66.66 k Ω
R1
(50)(2) + (50)(0.5) = 100 + 25 = 100 + 25 ⇒ R = 8.33 k Ω
2
R1
R2
R1 R 2 33.33 R 2
______________________________________________________________________________________
9.37
a.
v0 = −
RF
R
R
R
⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 )
R3
R2
R1
R0
RF ⎡ a3 a2 a1 a0 ⎤
+ + +
( 5)
10 ⎢⎣ 2 4 8 16 ⎥⎦
R 1
v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ
10 2
So v0 =
b.
c.
v0 =
10 1
⋅ ⋅ 5 ⇒ v0 = 0.3125 V
10 16
v0 =
10 ⎡ 1 1 1 1 ⎤
+ + +
( 5) ⇒ v0 = 4.6875 V
10 ⎢⎣ 2 4 8 16 ⎥⎦
i.
ii.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.38
⎛ 20 ⎞
⎛ − 10 ⎞⎛ − 20 ⎞
(a) υ O = ⎜
⎟ ⋅υ I 1 − ⎜ ⎟ ⋅υ I 2 = 200υ I 1 − 20υ I 2
⎟⎜
⎝ 1 ⎠
⎝ 1 ⎠⎝ 1 ⎠
(b) υ O = (200 )(5) − (20 )(− 25 − 50 sin ω t ) = 1000 + 500 + 1000 sin ω t (mV)
υ O = 1.5 + 1.0 sin ω t (V)
(c) For the 20 k Ω resistor:
2.5
i max =
⇒ i max = 0.125 mA
20
For the 10 k Ω resistor:
50 mV
⎛ 10 ⎞
υ O1 = ⎜ ⎟(5) = 50 mV, i max =
⇒ i max = 5 μ A
10kΩ
⎝1⎠
______________________________________________________________________________________
9.39
For one-input
υ1 = −
υO
Aod
υ −υO
υ I 1 − υ1
υ1
=
+ 1
R1
RF
R 2 R3
⎡1
1
1 ⎤ υO
= υ1 ⎢ +
+
⎥−
R1
⎣⎢ R1 R 2 R3 R F ⎦⎥ R F
υ I1
=−
1
1 ⎤ υO
+
⎢ +
⎥−
Aod ⎣⎢ R1 R 2 R3 R F ⎦⎥ R F
υO ⎡ 1
⎡ 1
1
1 ⎛⎜ 1
1 ⎞⎟⎤
= −υ O ⎢
+
+
+
⎥
⎜
⎢⎣ Aod R F R F Aod ⎝ R1 R 2 R3 ⎟⎠⎥⎦
⎤
υ ⎡ 1
RF
1
⎥
=− O ⎢
+1+
⋅
R F ⎢ Aod
Aod R1 R 2 R3 ⎥
⎣
⎦
RF
1
where R P = R1 R2 R3
υO = −
⋅υ I 1 ⋅
R1
⎡
1 ⎛ R F ⎞⎤
⎟⎟⎥
⎜⎜1 +
⎢1 +
⎣⎢ Aod ⎝ R P ⎠⎦⎥
Therefore, for three-inputs υ O =
⎛R
⎞
R
R
−1
⋅ ⎜⎜ F ⋅υ I 1 + F ⋅υ I 2 + F ⋅υ I 3 ⎟⎟
R2
R3
1 ⎛ R F ⎞ ⎝ R1
⎠
⎟⎟
⎜⎜1 +
1+
Aod ⎝ R P ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.40
⎛ R ⎞ ⎛ 150 ⎞
(a) Aυ = ⎜⎜1 + 2 ⎟⎟ = ⎜1 +
⎟ = 11
R1 ⎠ ⎝
15 ⎠
⎝
⎛ 150 ⎞
(b) Aυ = ⎜1 +
⎟=4
50 ⎠
⎝
⎛ 20 ⎞
(c) Aυ = ⎜1 + ⎟ = 1.4
⎝ 50 ⎠
⎛ 20 ⎞
(d) Aυ = ⎜1 + ⎟ = 2
⎝ 20 ⎠
______________________________________________________________________________________
9.41
⎛ R ⎞
R
(a) Aυ = 15 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 14
R1 ⎠
R1
⎝
For υ O = −7.5 V ⇒ υ1 = −0.5 V
i = 120 μ A =
7.5 − 0.5
⇒ R 2 = 58.33 k Ω , R1 = 4.17 k Ω
R2
(b) υ O = (15)(0.25) = 3.75 V
0.25
⇒ i1 = i 2 = 60 μ A
4.17
______________________________________________________________________________________
i1 = i 2 =
9.42
⎛ R ⎞
(a) Aυ = ⎜⎜1 + 2 ⎟⎟
R1 ⎠
⎝
⎛ R ⎞
R
3 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 2 , Set R 2 = 290 k Ω , R1 = 145 k Ω
R1 ⎠
R1
⎝
⎛ R ⎞
R
(b) 9 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 8 , Set R 2 = 290 k Ω , R1 = 36.25 k Ω
R
R1
1 ⎠
⎝
⎛ R ⎞
R
(c) 30 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 29 , Set R 2 = 290 k Ω , R1 = 10 k Ω
R
R1
1 ⎠
⎝
⎛ R ⎞
R
(d) 1 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 0 , Set R 2 = 0 , R1 = 290 k Ω
R1 ⎠
R1
⎝
______________________________________________________________________________________
9.43
⎛ 1 ⎞
vB = ⎜
⎟ vI
⎝ 1 + 500 ⎠
⎛ 1 ⎞
v0 = Aod ⎜
⎟ vi
⎝ 501 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
a.
⎛ 1 ⎞
2.5 = Aod ⎜
⎟ ( 5 ) ⇒ Aod = 250.5
⎝ 501 ⎠
⎛ 1 ⎞
v0 = 5000 ⎜
⎟ ( 5 ) ⇒ v0 = 49.9 V
⎝ 501 ⎠
b.
______________________________________________________________________________________
9.44
⎛ 50 ⎞ ⎡⎛ 20 ⎞
⎛ 40 ⎞ ⎤
v0 = ⎜ 1 + ⎟ ⎢⎜
⎟ vI 2 + ⎜
⎟ vI 1 ⎥
+
⎝ 50 ⎠ ⎣⎝ 20 40 ⎠
⎝ 20 + 40 ⎠ ⎦
v0 = 1.33vI 1 + 0.667vI 2
______________________________________________________________________________________
9.45
(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
vO =
(b)
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⎠
(c)
______________________________________________________________________________________
9.46
⎛ R4 ⎞
⎟ ⋅υ I
(a) υ 2 = ⎜⎜
⎟
⎝ R3 + R 4 ⎠
⎛
υ O = ⎜⎜1 +
⎝
⎛ R ⎞
R2 ⎞
1
⎟⎟ ⋅υ 2 = ⎜⎜1 + 2 ⎟⎟ ⋅
⋅υ I
R1 ⎠
R1 ⎠ ⎛ R3 ⎞
⎝
⎟⎟
⎜⎜1 +
⎝ R4 ⎠
⎛ R ⎞
⎛ R ⎞
R
1
(b) 6 = ⎜⎜1 + 2 ⎟⎟ ⋅
⇒ ⎜⎜1 + 2 ⎟⎟ = 9 ⇒ 2 = 8
R1 ⎠ ⎛ 25 ⎞
R1 ⎠
R1
⎝
⎝
⎜1 + ⎟
⎝ 50 ⎠
Set R 2 = 200 k Ω , R1 = 25 k Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.47
(a)
vO ⎛
50 x ⎞
= ⎜⎜ 1 +
⎟
vI ⎝ (1 − x ) 50 ⎟⎠
vO ⎛
x ⎞ 1− x + x
= ⎜1 +
⎟=
vI ⎝ 1 − x ⎠
1− x
vO
1
=
vI 1 − x
1 ≤ Av ≤ ∞
Av =
(b)
(c) If x = 1, gain goes to infinity.
______________________________________________________________________________________
9.48
⎛υ ⎞
(a) υ X = ⎜⎜ I ⎟⎟(2 R ) + υ I = 3υ I
⎝ R⎠
υ X −υ I υ X υ X −υO
+
+
=0
2R
R
2R
1
1 ⎞ υ I υO
⎛ 1
=
υX ⎜
+ +
⎟−
⎝ 2R R 2R ⎠ 2R 2R
υ
⎛2⎞ υ
3υ I ⎜ ⎟ − I = O
⎝ R ⎠ 2R 2R
υO
= 11
υI
(b) For υ I = 0.25 V, ⇒ υ O = 2.75 V
(c) R = 30 k Ω , υ I = −0.15 V
so
0.15
⇒ 5μ A
30
For R 2 : i = 5 μ A
For R1 : i =
υ X = 3υ I = −0.45 V
0.45
⇒ 15 μ A
30
υ O = (11)(− 0.15) = −1.65 V
For R 4 : i =
1.65 − 0.45
⇒ 20 μ A
60
______________________________________________________________________________________
For R3 : i =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.49
(a)
(b)
vO
=1
vI
From Exercise TYU9.7
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
vO
⎝
=
vI ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
A
R1 ⎠ ⎦
od ⎝
⎣
But R2 = 0, R1 = ∞
vO
v
1
1
=
=
⇒ O = 0.999993
1
vI 1 + 1
vI
1+
Aod
1.5 × 105
Want
vO
1
= 0.990 =
⇒ Aod = 99
1
vI
1+
Aod
(b)
______________________________________________________________________________________
9.50
(a) υ O = Aod (υ 2 − υ1 ) = Aod (υ I − υ O )
υ O (1 + Aod ) = Aod υ I
υ
1
1
Aυ = O =
=
= 0.9524
1
1
υI
1+
1+
Aod
20
(b) Aυ =
1
= 0.995
1
200
1
= 0.9995
(c) Aυ =
1
1+
2000
1
= 0.99995
(d) Aυ =
1
1+
20000
______________________________________________________________________________________
1+
9.51
(a) Aυ1 =
Aυ 2 =
υ O1 ⎛ R 2 ⎞
⎟
= ⎜1 +
υ I ⎜⎝ R1 ⎟⎠
⎛ R ⎞
υ O2
= −⎜⎜1 + 2 ⎟⎟
R1 ⎠
υI
⎝
⎛ 60 ⎞
(b) υ O1 = ⎜1 + ⎟(− 0.5) = −2 V
⎝ 20 ⎠
⎛
⎝
υ O 2 = −⎜ 1 +
60 ⎞
⎟(− 0.5) = 2 V
20 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ R ⎞
⎛ 60 ⎞
(c) υ O1 − υ O 2 = 2⎜⎜1 + 2 ⎟⎟ ⋅υ I = 2⎜1 + ⎟(0.8) = 6.4 V
R
⎝ 20 ⎠
1 ⎠
⎝
______________________________________________________________________________________
9.52
iL =
(a)
(b)
vI
R1
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
______________________________________________________________________________________
9.53
⎛ 20 ⎞
(a) υ O = ⎜
⎟ ⋅υ I = (0.3333)υ I
⎝ 20 + 40 ⎠
(i) υ O = 1 V
(ii) υ O = −1.67 V
⎛ 20 ⎞
(b) υ O = ⎜
⎟ ⋅υ I = (0.3333)υ I
⎝ 20 + 40 ⎠
(i) υ O = 1 V
(ii) υ O = −1.67 V
⎛ 10 ⎞⎛ 6 ⎞
(c) υ O = ⎜1 + ⎟⎜
⎟ ⋅υ I = (0.2222 )υ I
⎝ 10 ⎠⎝ 6 + 48 ⎠
(i) υ O = 0.667 V
(ii) υ O = −1.111 V
______________________________________________________________________________________
9.54
a.
v −v
v
Rin = 1 and 1 0 = i1 and v0 = − Aod v1
i1
RF
v − ( − Aod v1 ) v1 (1 + Aod )
So i1 = 1
=
RF
RF
v
RF
Then Rin = 1 =
i1 1 + Aod
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
⎛ RS ⎞
RF
i1 = ⎜
⋅ i1
⎟ iS and v0 = − Aod ⋅
1 + Aod
⎝ RS + Rin ⎠
⎛ A ⎞⎛ RS ⎞
So v0 = − RF ⎜ od ⎟⎜
⎟ iS
⎝ 1 + Aod ⎠⎝ RS + Rin ⎠
RF
10
=
= 0.009990
Rin =
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
______________________________________________________________________________________
9.55
vO = iC RF , 0 ≤ iC ≤ 8 mA
For vO ( max ) = 8 V, Then RF = 1 kΩ
______________________________________________________________________________________
9.56
vI
10
⇒ R = 10 kΩ
so 1 =
R
R
In the ideal op-amp, R1 has no influence.
⎛ R ⎞
v0 = ⎜ 1 + 2 ⎟ vI
R⎠
⎝
Output voltage:
i=
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.
______________________________________________________________________________________
9.57
(a) i L =
Set
−υ I
− (− 5)
; R2 =
=1k Ω
5
R2
RF
1
=
; For example, set R1 = R F = 10 k Ω , R3 = 1 k Ω
R1 R3 R 2
(b) υ L = (5)(0.2) = 1 V = υ 1
i1 = i 2 =
υ I − υ1
R1
=
− 5 −1
= −0.6 mA
10
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
υ O = υ1 − i 2 R F = 1 − (− 0.6)(10) = 7 V
υO −υ L 7 −1
i3 =
R3
=
1
= 6 mA
υL
1
= = 1 mA
R2 1
i4 =
For the op-amp: iO + i 2 = i 3 ⇒ i O = i 3 − i 2 = 6 − (− 0.6 ) = 6.6 mA
______________________________________________________________________________________
9.58
(a)
i1 = i2 and i2 =
vx
+ iD , vx = −i2 RF
R2
⎛R ⎞
Then i1 = −i1 ⎜ F ⎟ + iD
⎝ R2 ⎠
⎛
R ⎞
Or iD = i1 ⎜1 + F ⎟
R2 ⎠
⎝
(b)
R1 =
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
______________________________________________________________________________________
9.59
(1)
VX VX − vO
+
R2
R3
(2)
VX VX − vO
+
=0
R1
RF
IX =
⎛ 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 =
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
______________________________________________________________________________________
Note: If
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.60
(a) Rid = R1 + R3 = 30 k Ω ; R1 = R3 = 15 k Ω
R2 R4
=
= 15 ⇒ R 2 = R 4 = 225 k Ω
R1 R3
(b) υ O = i L R L = (0.25)(10) = 2.5 V
υ I 2 − υ I1 =
υO
Ad
=
2.5
= 0.1667 V
15
(c) υ O = Ad (υ I 2 − υ I 1 ) = 15(1.2 − 1.5) = −4.5 V
iL =
υO
RL
=
− 4.5
= −0.45 mA
10
(d) υ O = (0.5)(10) = 5 V
υ I 2 − υ I1 =
υO
Ad
=
5
= 0.333 V
15
υ I 1 = 2 − 0.333 = 1.667 V
______________________________________________________________________________________
9.61
(a)
R2 R4
=
= 40 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 6.25 k Ω
R1 R3
(b)
R2 R4
=
= 25 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 10 k Ω
R1 R3
(c)
R2 R4
=
= 5 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 50 k Ω
R1 R3
R2 R4
=
= 0.5 ; Set R 2 = R 4 = 125 k Ω , R1 = R3 = 250 k Ω
R1 R3
______________________________________________________________________________________
(d)
9.62
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
1+ x ⎞
⎢
⎥ vI 2 − ⎛⎜
vO = ⎢1 + ⎜
⎟⎥
⎟ vI 1
⎝ 1− x ⎠
⎣ ⎝ 1 − x ⎠ ⎦ ⎢1 + ⎛ 1 − x ⎞ ⎥
⎢ ⎜⎝ 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 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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+ x ⎞⎜ 1 ⎟
⎛ 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−
=
=
vcm
1− x
1− x
1− x
Set
R2 = 50 (1 − x )
R1 = 50 (1 + x )
R3 = 50 (1 − x ) R4 = 50 (1 + x )
⎛
⎞
⎛ 1− x ⎞⎜ 1 ⎟
⎛ 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
Acm =
=
1+ x
1+ x
Worst common-mode gain
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
1
−
x
−
x
2 ⎣ ⎝ 1 − x ⎠ ⎥⎦ 2 ⎣
1− x
⎦
1.010
C M R RdB = 20 log10
= 33.98 dB
For x = 0.01 Ad = 1.010
0.0202
For x = 0.01,
Acm =
For x = 0.02, Ad =
1
= 1.020
0.98
C M R RdB = 20 log10
1.020
= 27.96 dB
0.04082
1
1.0526
= 1.0526 C M R RdB = 20 log10
≅ 20 dB
0.95
0.1053
______________________________________________________________________________________
For x = 0.05 Ad =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.63
(a) υ O = 10(υ 2 − υ1 ) = 10(1.4 − 1.8) = −4 V
i3 = i 4 =
υ2
R + 10 R
=
1.4
(11)(10)
= 0.0127 mA
⎛ 10 ⎞
⎛ 10 ⎞
⎟ ⋅υ 2 = ⎜ ⎟(1.4 ) = 1.273 V
⎝ 11 ⎠
⎝ 11 ⎠
υ − υ X 1.8 − 1.273
=
= 0.0527 mA
i1 = i 2 = 1
10
R
(b) υ O = 10(υ 2 − υ1 ) = 10(3.6 − 3.2 ) = 4 V
υ X = υY = ⎜
⎛ 10 ⎞
⎟(3.6 ) = 3.273 V
⎝ 11 ⎠
υ2
3.6
i3 = i 4 =
=
= 0.0327 mA
(11)(10) 110
3.2 − 3.273
i1 = i 2 =
= −0.00727 mA
10
(c) υ O = 10(− 1.35 − (− 1.20 )) = −1.5 V
υ X = υY = ⎜
i3 = i 4 =
−1.35
= −0.0123 mA
(11)(10)
⎛ 10 ⎞
⎟(− 1.35) = −1.227 V
⎝ 11 ⎠
− 1.2 − (− 1.227 )
i1 = i 2 =
= 0.00273 mA
10
______________________________________________________________________________________
υ X = υY = ⎜
9.64
(a) I E = (1 + β ) ⋅ I B = (76)(1.2) = 91.2 mA
10
⇒ R = 109.6 Ω
91.2
(b) I E = (101)(0.2 ) = 20.2 mA
R=
10
= 0.495 k Ω
20.2
6
(c) I E =
= 54.74 mA
0.1096
54.74
IO =
= 0.72 mA
76
4
(d) I E =
= 8.08 mA
0.495
8.08
IO =
= 0.080 mA
101
______________________________________________________________________________________
R=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.65
(a) υ O1 =
− R2
⋅υ CM
R1
⎛
⎛ R ⎞
R4 ⎞
⎟ ⋅υ CM ⋅ ⎜⎜1 + 2 ⎟⎟
⎟
+
R
R
R1 ⎠
4 ⎠
⎝
⎝ 3
υ O 2 = ⎜⎜
⎤
⎡ R4
⎢ R
⎛
⎞
R
R ⎥
υ O = υ O1 + υ O 2 = ⎢ 3 ⋅ ⎜⎜1 + 2 ⎟⎟ − 2 ⎥ ⋅υ CM
⎢ R4 ⎝
R1 ⎠ R1 ⎥
⎥
⎢1 + R
3
⎦
⎣
⎛ R 4 ⎞⎛ R 2 ⎞ ⎛ R 2 ⎞⎛ R 4 ⎞ R 4 R 2
⎜
⎟⎜
⎟
⎟ ⎜
⎟⎜
⎜ R ⎟⎜1 + R ⎟ − ⎜ R ⎟⎜1 + R ⎟ R − R
υO
1 ⎠ ⎝ 1 ⎠⎝
3 ⎠
⎝ 3 ⎠⎝
1
ACM =
=
= 3
R4
υ CM
⎛ R4 ⎞
1+
⎜1 +
⎟
⎜ R ⎟
R3
3 ⎠
⎝
86.4 62.4
−
9−6
= 0.3
(b) ACM = 9.6 10.4 =
86.4
10
1+
9.6
80.8 79.2
−
(c) ACM = 19.8 20.2 = 0.03149
80.8
1+
19.8
or
79.2 80.8
−
ACM = 20.2 19.8 = −0.0325
79.2
1+
20.2
⇒ ACM max = 0.0325
______________________________________________________________________________________
9.66
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vI 1 − v A v A − vB v A − 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 ⎠
(3)
⎛ R1 ⎞
⎛ R2 ⎞
v+ = ⎜
⎟ vB + ⎜
⎟ vI 2
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
Now v− = v+ ⇒ R1v A + R2 vI 1 = R1vB + R2 vI 2
So that v A = vB +
(4)
R2
( vI 2 − vI 1 )
R1
⎛ 1
v
vI 1
1
1 ⎞ v
= vA ⎜
+
+ ⎟− B − 0
R1 + R2
⎝ R1 + R2 RV R2 ⎠ RV R2
(1)
⎛ 1
vI 2
1
1 ⎞ v
= vB ⎜
+
+ ⎟− A
R1 + R2
⎝ R1 + R2 RV R2 ⎠ RV
( 2)
Then
⎛ 1
v ⎛ R ⎞⎛ 1
vI 1
1 ⎞
1
1 ⎞ v
1
= vB ⎜
+
+ ⎟ − B − 0 + ⎜ 2 ⎟⎜
+
+ ⎟ ( v I 2 − vI 1 )
R1 + R2
⎝ R1 + R2 RV R2 ⎠ RV R2 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠
⎛ 1
⎤
vI 2
R2
1
1 ⎞ 1 ⎡
= vB ⎜
+
+ ⎟−
⎢ vB + ( vI 2 − vI 1 ) ⎥
R1 + R2
R1
⎦
⎝ R1 + R2 RV R2 ⎠ RV ⎣
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
R2 RV R1
⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠
(1)
(2)
⎧⎪⎛ 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⎬
+
+
R
R
R
R
R
R
R
2
1
2
V
V ⎭
⎝ 1 ⎠⎩ 1
v0 =
R2 ⎞
2 R2 ⎛
⎜1 +
⎟ ( vI 2 − vI 1 )
R1 ⎝ RV ⎠
______________________________________________________________________________________
9.67
(a) i1 =
υ I1 − υ I 2
R1
=
(1.2 − 0.08 sin ω t ) − (1.2 + 0.08 sin ω t ) ⇒ i = −16 sin ω t ( μ A)
10
1
υ O1 = (1.2 − 0.08 sin ω t ) − (0.016 sin ω t )(40 ) = 1.2 − 0.72 sin ω t (V)
υ O 2 = (1.2 + 0.08 sin ω t ) − (− 0.016 sin ω t )(40 ) = 1.2 + 0.72 sin ω t (V)
υO =
R4
(υ O 2 − υ O1 ) = ⎛⎜ 120 ⎞⎟(2)(0.72 sin ω t ) = 4.32 sin ω t (V)
R3
⎝ 40 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) ii =
(− 0.65 + 0.05 sin ω t ) − (− 0.60 − 0.05 sin ω t )
10
i1 = −5 + 10 sin ω t ( μ A)
υ O1 = −0.65 + 0.05 sin ω t + (40)(− 0.005 + 0.010 sin ω t ) = −0.85 + 0.45 sin ω t (V)
υ O 2 = −0.60 − 0.05 sin ω t − (40 )(− 0.005 + 0.01 sin ω t ) = −0.40 − 0.45 sin ω t (V)
⎛ 120 ⎞
⎟[(− 0.40 − 0.45 sin ω t ) − (− 0.85 + 0.45 sin ω t )] = 1.35 − 2.7 sin ω t (V)
⎝ 40 ⎠
______________________________________________________________________________________
υO = ⎜
9.68
(a)
(b)
(c)
⎛ 40 ⎞
vOB = ⎜1 + ⎟ vI = 2.1667sin ω t
⎝ 12 ⎠
vOC = −
30
vI = −1.25sin ω t
12
vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t )
vO = 3.417 sin ω t
vO 3.417
=
= 6.83
vI
0.5
(d)
______________________________________________________________________________________
9.69
(a) iO =
υ I1 − υ I 2
R
0.25 − (− 0.25)
(b) R =
⇒ R = 100 Ω
5
(c) υ O1 = υ I 1 + i O R L = 0.25 + (5)(1) = 5.25 V
υ O 2 = υ I 2 = −0.25 V
υ I 1 − υ I 2 1.25 − 1.75
(d) iO =
=
= −1 mA
R
0. 5
υ O1 = υ I 1 + i O R L = 1.25 − (1)(3) = −1.75 V
υ O 2 = υ I 2 = 1.75 V
______________________________________________________________________________________
9.70
Ad =
vO
R ⎛ 2R ⎞
= 4 ⎜1 + 2 ⎟
vI 2 − vI 1 R3 ⎝
R1 ⎠
vO =
200 ⎛ 2 (115 ) ⎞
⎜1 +
⎟ ( 0.06sin ω t )
50 ⎝
R1 ⎠
For vO = 0.5
230
= 1.0833 ⇒ R1 = 212.3 K
R1
230
= 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K
R1
______________________________________________________________________________________
vO = 8 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.71
For υ O = 10 V, Ad = 200 ⇒ υ I 2 − υ I 1 =
10
= 0.05 V
200
R1 ( fixed ) =
0.05
⇒ R1 ( fixed ) = 1 k Ω
50 × 10 − 6
R ⎛ 2R ⎞
Ad = 4 ⎜⎜1 + 2 ⎟⎟
R3 ⎝
R1 ⎠
⎛ 2R ⎞
200 = (2.5)⎜⎜1 + 2 ⎟⎟ ⇒ R 2 = 39.5 k Ω
1 ⎠
⎝
For Ad = 5
⎡ 2(39.5) ⎤
5 = (2.5)⎢1 +
⎥ ⇒ R1 = 79 = R1 (var ) + R1 ( fixed ) = R1 (var ) + 1
R1 ⎦
⎣
R1 (var ) = 78 k Ω
______________________________________________________________________________________
9.72
υ O (υ O1 ) = −
R4
⋅υ O1
R3′
⎛
R 4 ⎞⎛ R 4 ⎞
⎟⎜1 +
⎟ ⋅υ O 2
⎟⎜
⎟
⎝ R3 + R 4 ⎠⎝ R3′ ⎠
υ O (υ O 2 ) = ⎜⎜
υ O = υ O (υ O1 ) + υ O (υ O 2 ) and υ O1 = υ O 2 ≡ υ CM
Then
ACM =
υ O ⎛ R 4 ⎞⎛ R 4 ⎞ ⎛ R 4 ⎞
⎟⎜1 +
⎟−⎜
⎟
=⎜
υ CM ⎜⎝ R3 + R 4 ⎟⎠⎜⎝ R3′ ⎟⎠ ⎜⎝ R3′ ⎟⎠
R 4 = 2 R3 = 60 k Ω , R3 = 30 k Ω , R3′ = 30 k Ω ±5%
For R3′ = 30 k Ω −5% = 28.5 k Ω
60 ⎞ ⎛ 60 ⎞
⎛ 60 ⎞⎛
ACM = ⎜
⎟⎜1 +
⎟−⎜
⎟ = −0.03509
⎝ 60 + 30 ⎠⎝ 28.5 ⎠ ⎝ 28.5 ⎠
For R3′ = 30 k Ω +5% = 31.5 k Ω
60 ⎞ ⎛ 60 ⎞
⎛ 60 ⎞⎛
ACM = ⎜
⎟⎜1 +
⎟−⎜
⎟ = +0.03175
31
.5 ⎠ ⎝ 31.5 ⎠
60
30
+
⎝
⎠⎝
Then −0.03509 ≤ ACM ≤ +0.03175
______________________________________________________________________________________
9.73
(a) R1C 2 = 20 ×10 3 0.02 ×10 −6 = 4 × 10 −4 s
υO =
(
)(
)
−1
4 × 10 − 4
∫ (0.25) cos ω tdt = (4 ×10 )ω ⋅ sin ω t
−0.25
−4
(
)
For υ O = 0.25 ⇒ 4 × 10 −4 (2π f ) = 1 ⇒ f = 398 Hz
Phase = 90°
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) (i) υ O = 1.5 =
0.25
⇒ f = 66.3 Hz
2π f 4 × 10 − 4
(
)
0.25
⇒ f = 663 Hz
2π f 4 × 10 − 4
______________________________________________________________________________________
(ii) υ O = 0.15 =
(
)
9.74
t
1.2
−1
−1
υ I (t ′)dt ′ =
(0.25)t ′
R1C 2 0
R1C 2
0
∫
(a) υ O =
− (0.25)(1.2)
⇒ R1C 2 = 60 ms
R1C 2
−5 =
(b) (i) 0 = −5 +
(0.10) ⋅ t ′′ ⇒ t ′′ = 3 s, t = 4.2 s
0.06
(0.10) ⋅ t ′′ ⇒ t ′′ = 6 s, t = 7.2 s
(ii) 5 = −5 +
0.06
______________________________________________________________________________________
9.75
−Z 2
, where Z 2 = R 2
R1
(a) Aυ =
− R2
1
⋅
R1 1 + jωR 2 C 2
Aυ =
− R2
R1
(b) At ω = 0 , Aυ (0 ) =
(c)
⎛ 1 ⎞
⎟⎟
R 2 ⎜⎜
R2
1
⎝ jωC 2 ⎠
=
=
1
jωC 2
1 + j ωR 2 C 2
R2 +
j ωC 2
Aυ =
R2
1
⋅
R1 1 + (ωR C )2
2 2
Set 1 + (ωR 2 C 2 ) = 2 ⇒ ω =
1
1
⇒ f =
R2 C 2
2πR 2 C 2
______________________________________________________________________________________
2
9.76
(a) R1 = 20 k Ω
R2
= 15 ⇒ R 2 = 300 k Ω
R1
ω=
1
= 2π f
R2 C 2
C2 =
1
1
=
⇒ C 2 = 106 pF
3
2π fR 2 2π 5 × 10 300 × 10 3
(
)(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) R1 = 15 k Ω
R2
= 25 ⇒ R 2 = 375 k Ω
R1
C2 =
1
⇒ C 2 = 28.3 pF
2π 15 × 10 375 × 10 3
______________________________________________________________________________________
(
9.77
)(
− jωR 2 C1
− R2
jωR1C1
=
⋅
1 + jωR1C1
R1 1 + jωR1C1
(b) As ω ⇒ ∞ , Aυ =
(c)
Aυ =
Set
)
− R2
1 + jωR1C1
1
, where Z 1 = R1 +
=
Z1
j ωC 1
j ωC 1
(a) Aυ =
Aυ =
3
− R2
R1
R2
ωR1C1
⋅
R1 1 + (ωR C )2
1 1
ωR1C1
1 + (ωR1C1 )
=
2
1
⇒ω =
2
1
1
⇒ f =
R1C1
2π R1C1
______________________________________________________________________________________
9.78
(a) Set R 2 = 350 k Ω
R2
= 15 ⇒ R1 = 23.33 k Ω
R1
2π f =
1
1
1
⇒ C1 =
=
⇒ C1 = 341 pF
R1C1
2π R1 f 2π 23.33 × 10 3 20 × 10 3
(
)(
)
(b) Set R1 = 20 k Ω
R2
= 25 ⇒ R 2 = 500 k Ω
R1
1
⇒ C1 = 227 pF
2π 20 × 10 3 35 × 10 3
______________________________________________________________________________________
C1 =
(
)(
)
9.79
Assuming the Zener diode is in breakdown,
R
1
vO = − 2 ⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V
R1
1
i2 =
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
iz =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10 − 0
10
= i2 =
⇒ i2 = 1.52 mA
Rs + R1
5.6 + 1
vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V
iz = 0
______________________________________________________________________________________
9.80
⎡
⎤
⎛ v ⎞
v
⎛ v ⎞
vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −I 10 ⎟
I
R
⎝ 10 ⎠
⎝ s 1⎠
⎣⎢ (10 )(10 ) ⎥⎦
For vI = 20 mV , vO = 0.497 V
For vI = 2 V , vO = 0.617 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.81
⎛ 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 ⎟
i
⎝ C2 ⎠
⎝ 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 ⎠
______________________________________________________________________________________
9.82
(
)
vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT
vO = (10−10 ) evI / 0.026
For vI = 0.30 V ,
vo = 1.03 × 10−5 V
For vI = 0.60 V ,
vo = 1.05 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.83
From Figure 9.40
⎛ R ⎞⎡ R
⎤
R
R
R
υ O = − F ⋅υ I 2 − F ⋅υ I 4 + ⎜⎜1 + F ⎟⎟ ⎢ P ⋅υ I 1 + P ⋅υ I 3 ⎥
R1
R2
RB
⎦
⎝ R N ⎠⎣ R A
= −10υ I 2 − υ I 4 + 2υ I 1 + 3υ I 3
Then
⎛ R ⎞⎛ R ⎞
⎛ R ⎞⎛ R ⎞
RF
R
= 10 , F = 1 , ⎜⎜1 + F ⎟⎟⎜⎜ P ⎟⎟ = 2 , ⎜⎜1 + F ⎟⎟⎜⎜ P ⎟⎟ = 3
R1
R2
⎝ R N ⎠⎝ R A ⎠
⎝ R N ⎠⎝ R B ⎠
Set R F = 500 k Ω , R1 = 50 k Ω , R 2 = 500 k Ω
Now R N = R1 R 2 = 50 500 = 45.45 k Ω
⎛R ⎞
⎛R ⎞
500 ⎞⎛ R P ⎞
500 ⎞⎛ R P ⎞
⎛
⎛
⎟⎟ = 12⎜⎜ P ⎟⎟ = 2 , Also ⎜1 +
⎟⎟ = 12⎜⎜ P ⎟⎟ = 3
Then ⎜1 +
⎟⎜⎜
⎟⎜⎜
⎝ 45.45 ⎠⎝ R A ⎠
⎝ 45.45 ⎠⎝ R B ⎠
⎝ RA ⎠
⎝ RB ⎠
2
R A = 333.3 k Ω
3
Then R P = 83.33 k Ω = R A R B RC
Let R A = 500 k Ω , then R B =
We find R A R B = 500 333.3 = 200 k Ω
So 200 RC = 83.33 ⇒ RC = 142.8 k Ω
______________________________________________________________________________________
9.84
⎛
υ O = ⎜⎜1 +
⎤ R
R F ⎞⎡ R P
R
R
R
⎟⎢
⋅υ I 1 + P ⋅υ I 2 + P ⋅υ I 3 ⎥ − R ⋅υ I 4 − F ⋅υ I 5
⎟
R N ⎠⎣ R A
RB
RC
R2
⎦ R1
⎝
= 3υ I 1 + 1.5υ I 2 + 2υ I 3 − 4υ I 4 − 6υ I 5
We have
RF
R
= 4 , F = 6 ; Set R F = 250 k Ω , R1 = 62.5 k Ω , R 2 = 41.67 k Ω
R1
R2
Now R N = R1 R 2 = 62.5 41.67 = 25 k Ω
⎛ R ⎞ ⎛ 250 ⎞
Also ⎜⎜1 + F ⎟⎟ = ⎜1 +
⎟ = 11
25 ⎠
⎝ RN ⎠ ⎝
(11)R P
(11)R P
(11)R P
R
1 RA 2
Now
= 3,
= 1.5 ,
=2 ⇒ A = ,
=
RA
RB
RC
R B 2 RC 3
Set R B = 250 k Ω , R A = 125 k Ω , RC = 187.5 k Ω
This yields R P = 34.09 k Ω , We have R p = R A R B RC R D
We find R A R B RC = 125 250 187.5 = 57.69 k Ω
Then 57.69 R D = 34.09 ⇒ R D = 83.3 k Ω
______________________________________________________________________________________
9.85
VO ⎛ R 2 ⎞ 12
R
⎟⎟ =
= ⎜⎜1 +
⇒ 2 = 1.143
VZ ⎝
R1 ⎠ 5.6
R1
IF =
VO − V Z
; Set I F = I Z (min ) = 1.2 mA
RF
Then R F =
12 − 5.6
= 5.33 k Ω
1. 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Set I D1 = 0.15 mA
V Z′ = V Z + Vγ = 5.6 + 0.7 = 6.3 V
6. 3
= 31.5 k Ω
0.2
Then I 3 = 0.2 + 0.15 = 0.35 mA
V − V Z′ 10 − 6.3
So R3 = S
=
= 10.6 k Ω
I3
0.35
______________________________________________________________________________________
Let I 4 = 0.2 mA, ⇒ R 4 =
9.86
R1 =
VO − V Z 12 − 5.6
=
= 3.2 k Ω
IZ
2
VO ⎛ R 2 ⎞ 12
R
⎟=
= ⎜⎜1 +
⇒ 2 = 1.143
⎟
V Z ⎝ R3 ⎠ 5.6
R3
Let I R = 2 mA, ⇒ R 2 + R3 =
VO 12
=
= 6kΩ
IR
2
Then 1.143R3 + R3 = 6 , ⇒ R3 = 2.8 k Ω and R 2 = 3.2 k Ω
V − VO 15 − 12
Let I R 4 = 4 mA, R 4 = IN
=
= 0.75 k Ω
I R4
4
______________________________________________________________________________________
9.87
Let R1 = R 2 = R3 = 20 k Ω
Let RT = 20(1 + δ ) k Ω
⎛ R3 ⎞ + 1
⎟ ⋅ V = (10 ) = 5 V
Now υ O1 = υ A = ⎜⎜
⎟
2
⎝ R3 + R1 ⎠
RT ⎞ + ⎡ 20(1 + δ ) ⎤
10(1 + δ )
⎟⎟ ⋅ V = ⎢
⎥ (10 ) =
(
)
+
δ
+
2+δ
R
R
+
20
1
20
⎣
⎦
2 ⎠
⎝ T
⎛
υ O 2 = υ B = ⎜⎜
So υ OA = υ A − υ B = 5 −
10(1 + δ ) 5(2 + δ ) − 10(1 + δ )
5δ
≅−
= −2.5δ
=
2+δ
2+δ
2
⎛ T − 300 ⎞
We have δ = y⎜
⎟ ; At T = 350 , RT = 21 k Ω , ⇒ 21 = 20(1 + δ ) ⇒ δ = 0.05
⎝ 300 ⎠
⎛ 350 − 300 ⎞
Then 0.05 = y⎜
⎟ ⇒ y = 0.30
⎝ 300 ⎠
For δ = 0.05 , υ OA = (2.5)(0.05) = 0.125 V
For the instrumentation amplifier, υ O = 5 =
R4 ⎛ 2 R2 ⎞
⎜1 +
⎟(0.125)
R3 ⎜⎝
R1 ⎟⎠
R4
R
= 4 and 2 = 4.5
R3
R1
______________________________________________________________________________________
For example, set
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.88
R − ΔR
⎛
⎞ + ⎛ R − ΔR ⎞ +
(a) υ A = ⎜
⎟ ⋅V = ⎜
⎟ ⋅V
⎝ R − ΔR + R + ΔR ⎠
⎝ 2R ⎠
R + ΔR
⎞ + ⎛ R + ΔR ⎞ +
⎟ ⋅V = ⎜
⎟ ⋅V
R
+
Δ
R
+
R
−
Δ
R
⎝
⎠
⎝ 2R ⎠
⎛
υB = ⎜
9
ΔR +
⎛
⎞
⎡ R − ΔR R + ΔR ⎤ +
⋅V = −
⋅ V = − ΔR⎜
−
⎟
3
2 R ⎥⎦
R
⎝ 20 × 10 ⎠
⎣ 2R
υ O1 = υ A − υ B = ⎢
(
)
or υ O1 = − 4.5 ×10 −4 (ΔR )
(b) For an instrumentation amplifier,
R ⎛ 2R ⎞
υ O = 4 ⎜⎜1 + 2 ⎟⎟ ⋅υ O1
R3 ⎝
R1 ⎠
For ΔR = 200 Ω , υ O = −5 V
−5 =
R4 ⎛ 2R2 ⎞
⎜1 +
⎟ − 4.5 × 10 − 4 (200 )
R3 ⎜⎝
R1 ⎟⎠
or 55.55 =
(
)
R4 ⎛ 2R2 ⎞
⎜1 +
⎟
R3 ⎜⎝
R1 ⎟⎠
R4
R
= 6 and 2 = 4.13
R3
R1
______________________________________________________________________________________
For example, set
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 10
10.1
I1 = I 2 =
0 − 2Vγ − V −
R1 + R2
2Vγ + I 2 R2 = VBE + I C R3
2Vγ +
IC =
a.
R2
( −2Vγ − V − ) = VBE + IC R3
R1 + R2
⎫⎪
R2 ⎞
1 ⎧⎪
− ⎛
⎨2Vγ − ( 2Vγ + V ) ⎜
⎟ − VBE ⎬
R3 ⎪⎩
⎪⎭
⎝ R1 + R2 ⎠
Vγ = VBE and R1 = R2
IC =
1 ⎧
1
⎫
−
⎨2Vγ − ( 2Vγ + V ) − VBE ⎬
R3 ⎩
2
⎭
or I C =
b.
−V −
2 R3
I C = 2 mA =
− ( −10 )
2 R3
I1 = I 2 = 2 mA =
⇒ R3 = 2.5 kΩ
−2 ( 0.7 ) − ( −10 )
⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ
R1 + R2
c.
______________________________________________________________________________________
10.2
⎛ 50 × 10 −6 ⎞
⎟ = 0.7004 V
(a) (i) I O = 50 μ A, V BE1 = (0.026) ln⎜⎜
−16
⎟
⎝ 10
⎠
−6
⎛ 150 × 10 ⎞
⎟ = 0.7289 V
(ii) I O = 150 μ A, V BE1 = (0.026 ) ln⎜⎜
−16
⎟
⎝ 10
⎠
−3
⎛ 1.5 × 10 ⎞
⎟ = 0.7888 V
(iii) I O = 1.5 mA, V BE1 = (0.026) ln⎜⎜
−16
⎟
⎝ 10
⎠
⎛ 48.0769 × 10 −6 ⎞
50
⎟ = 0.6994 V
= 48.08 μ A, V BE1 = (0.026) ln⎜⎜
(b) (i) I O =
⎟
2
10 −16
⎝
⎠
1+
50
⎛ 144.23 × 10 −6 ⎞
150
⎟ = 0.7279 V
= 144.23 μ A, V BE1 = (0.026) ln⎜⎜
(ii) I O =
−16
⎟
2
10
⎝
⎠
1+
50
⎛ 1.4423 × 10 −3 ⎞
1.5
⎟ = 0.7878 V
= 1.4423 mA, V BE1 = (0.026) ln⎜⎜
(iii) I O =
−16
⎟
2
10
⎝
⎠
1+
50
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.3
I C1 ≅
I REF
200
=
= 195.12 μ A
2
2
1+
1+
β
80
⎛ 195.12 × 10 −6 ⎞
⎟ = 0.6341 V
V BE1 = V BE 2 = (0.026) ln⎜⎜
−15
⎟
⎝ 5 × 10
⎠
⎛V
⎞
⎛ 0.6341 ⎞
I O = I S 2 exp⎜⎜ BE 2 ⎟⎟ = 2 × 10 −15 exp⎜
⎟ ⇒ I O = 78.05 μ A
⎝ 0.026 ⎠
⎝ VT ⎠
______________________________________________________________________________________
(
)
10.4
I C1 =
I REF
150
=
= 147.54 μ A
2
2
1+
1+
β
120
⎛ 147.54 × 10 −6 ⎞
⎟ = 0.7285 V
V BE1 = V BE 2 = (0.026) ln⎜⎜
⎟
10 −16
⎝
⎠
⎛V
⎞
⎛ 0.7285 ⎞
I O = I S 2 exp⎜⎜ BE 2 ⎟⎟ = 3 × 10 −16 exp⎜
⎟ ⇒ I O = 0.4426 mA
⎝ 0.026 ⎠
⎝ VT ⎠
______________________________________________________________________________________
(
)
10.5
Approximation: I C1 ≅
I REF
200
=
= 190.48 μ A
2
2
1+
1+
β
40
⎛ 190.48 × 10 −6 ⎞
⎟ = 0.63345 V
V BE1 = (0.026) ln⎜⎜
−15
⎟
⎝ 5 × 10
⎠
I C1 190.48
I B2 ≅
=
= 4.762 μ A
β
40
V BE 2 = V BE1 − I B 2 R = 0.63345 − (0.004762 )(2) = 0.62393 V
(
)
⎛ 0.62393 ⎞
I O = 5 × 10 −15 exp⎜
⎟ ⇒ I O = 132.07 μ A
⎝ 0.026 ⎠
______________________________________________________________________________________
10.6
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+
60
β
I B1 = I B 2 = 4.03 μ A
______________________________________________________________________________________
I C1 = I C 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.7
I REF =
V + − VBE ( on ) − V −
=
R1
5 − 0.7 − ( −5 )
18.3
I REF = 0.5082 mA
I
0.5082
I C1 = I C 2 = REF =
⇒ I C1 = I C 2 = 0.4958 mA
2
2
1+
1+
80
β
I B1 = I B 2 = ( 6.198 μ A )
______________________________________________________________________________________
10.8
(a) P = (I O + I REF ) V + − V −
1.8 = (0.25 + I REF )(5) ⇒ I REF = 0.11 mA
(
)
⎛ 0.25 × 10 −3 ⎞
⎟ = 0.68236 V
V BE1 = V BE 2 = (0.026) ln⎜⎜
−15
⎟
⎝ 10
⎠
0
.
68236
⎛
⎞
−16
(b) 0.11 × 10 −3 = I S 1 exp⎜
⎟ ⇒ I S 1 = 4.4 × 10 A
⎝ 0.026 ⎠
5 − 0.68236
= 39.25 k Ω
0.11
______________________________________________________________________________________
(c) R1 =
10.9
⎛
2⎞
2 ⎞
⎛
(a) I REF = ⎜⎜1 + ⎟⎟ ⋅ I O = ⎜1 +
⎟(0.5) = 0.5083 mA
β
120
⎝
⎠
⎝
⎠
5 − 0.7 − (− 5)
R1 =
= 18.3 k Ω
0.5083
V
100
= 200 k Ω
(b) Ro = A =
0.5
IO
ΔI O =
ΔVCE 2 7 − 0.7
=
= 0.0315 mA
200
Ro
ΔI O 0.0315
=
⇒ 6.3%
0.5
IO
______________________________________________________________________________________
10.10
I 0 = nI C1
I
I
I REF = I C1 + I B1 + I B 2 = I C1 + C1 + 0
β
β
⎛
⎛ 1+ n ⎞
1 n⎞
I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 +
⎟
β
β
β ⎠
⎝
⎠
⎝
I ⎛ 1+ n ⎞
nI REF
= 0 ⎜1 +
⎟ or I 0 =
β ⎠
n⎝
⎛ 1+ n ⎞
⎜1 +
β ⎟⎠
⎝
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.11
IO =
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 =
0.21
I REF
______________________________________________________________________________________
10.12
⎛
2⎞
2 ⎞
⎛
(a) I REF = ⎜⎜1 + ⎟⎟ ⋅ I O = ⎜1 + ⎟(0.12 ) = 0.123 mA
β
80
⎝
⎠
⎝
⎠
5 − 0. 7
R1 =
= 34.96 k Ω
0.123
V
80
= 666.7 k Ω
(b) ro = A =
I O 0.12
(i) ΔI O =
ΔV EC 2 − 0.7
=
⇒ ΔI O = 1.95 μ A
ro
666.7
4 − 0. 7
⇒ ΔI O = 4.95 μ A
666.7
______________________________________________________________________________________
(ii ) ΔI O =
10.13
I REF = 1 =
a.
b.
5 − 0.7 − ( −5 )
R1
⇒ R1 = 9.3 kΩ
I 0 = 2 I REF ⇒ I 0 = 2 mA
5 − 0.7
⇒ RC 2 = 2.15 kΩ
2
c.
______________________________________________________________________________________
For VEC 2 ( min ) = 0.7 ⇒ RC 2 =
10.14
I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA
⎛
3⎞
3 ⎞
⎛
I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟
β
60
⎝
⎠
⎝
⎠
I REF = 0.2625 mA
2.5 − 0.7
⇒ R1 = 6.86 K
0.2625
______________________________________________________________________________________
R1 =
10.15
Similar to Figure P10.14 biased at V + and V − .
2.5 − 0.7 − (− 2.5)
R1 =
= 21.5 k Ω
0. 2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.16
I 2 = 2 I1 and I3 = 3I1
I
(a) 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
______________________________________________________________________________________
10.17
a.
I 0 = I C1 and I REF = I C1 + I B 3 = I C1 +
I E 3 = I B1 + I B 2 +
I REF = I C1 +
I REF −
VBE 2 I C1 VBE
=
+
R2
R2
β
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 +
⎟⎟ +
⎝ ( 80 )( 81) ⎠ ( 81)(10 )
I REF = 0.700216 + 0.000864
I REF = 0.7011 mA =
10 − 2 ( 0.7 )
⇒ R1 = 12.27 kΩ
R1
b.
______________________________________________________________________________________
10.18
a.
I ES
1+ β
I ES = I BR + I B1 + I B 2 + ... + I BN = (1 + N ) I BR
I 0i = I CR and I REF = I CR + I BS = I CR +
=
(1 + N ) ICR
β
Then I REF = I CR +
or I 0i =
(1 + N ) I CR
β (1 + β )
I REF
⎛
(1 + N ) ⎞
⎜⎜ 1 + β (1 + β ) ⎟⎟
⎝
⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎡
⎤
6
I REF = ( 0.5 ) ⎢1 +
⎥ = 0.5012 mA
⎣⎢ ( 50 )( 51) ⎦⎥
R1 =
5 − 2 ( 0.7 ) − ( −5 )
⇒ R1 = 17.16 kΩ
0.5012
b.
______________________________________________________________________________________
10.19
⎡
⎡
2 ⎤
2 ⎤
I REF = I O ⎢1 +
⎥ = (0.15)⎢1 +
⎥ = 0.15018 mA
⎣ β (1 + β ) ⎦
⎣ (40 )(41) ⎦
3 − 0.7 − 0.7 − (− 3)
R1 =
= 30.63 k Ω
0.15018
______________________________________________________________________________________
10.20
I 0 = I REF ⋅
1
⎛
⎞
2
⎜⎜1 +
⎟⎟
⎝ β (2 + β ) ⎠
For I 0 = 0.8 mA
⎛
2 ⎞
I REF = ( 0.8 ) ⎜⎜1 +
⎟⎟ ⇒ I REF = 0.8024 mA
⎝ 25 ( 27 ) ⎠
18 − 2 ( 0.7 )
R1 =
⇒ R1 = 20.69 kΩ
0.8024
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.21
The analysis is exactly the same as in the text. We have
1
I 0 = I REF ⋅
⎛
⎞
2
⎜⎜1 +
⎟⎟
⎝ β (2 + β ) ⎠
______________________________________________________________________________________
10.22
(a) I O = 0.5 mA, I B 2 =
0. 5
= 0.005556 mA
90
I B2
= 0.002778 mA
2
I E 3 = I B1 + I B 2 = 0.008334 mA
I B1 =
0.008334
= 0.0001366 mA
61
I
I C1 = O = 0.25 mA
2
I REF = I C1 + I B 3 = 0.2501366 mA
I B3 =
5 − 0.6 − 0.7 − (− 5)
= 34.78 k Ω
0.2501366
(b) I B1 = 0.002778 mA
R1 =
I B 2 = 0.005556 mA
I E 3 = 0.008334 mA
I B 3 = 0.0001366 mA
______________________________________________________________________________________
10.23
(a)
Assuming RO ≈
β ro3
2
VA
VA
100
=
=
= 400 K
rO 3 =
I O I REF 0.25
RO =
(100 )( 400 )
2
⇒ RO = 20 MΩ
(b)
RO =
ΔV
ΔV
5
⇒ ΔI O =
=
ΔI O
20 MΩ 20 MΩ
ΔI O = 0.25 μ A
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.24
I REF =
V + − VBE1 − V − 5 − 0.7
=
9.3
R1
I REF = 0.4624 mA
IO =
VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞
ln ⎜
ln ⎜
⎟
⎟=
1.5
RE ⎝ I O ⎠
⎝ IO ⎠
⎛ 0.4624 ⎞
I O = 0.01733ln ⎜
⎟
⎝ IO ⎠
By trial and error
I O = 41.7 μ A
V BE 2 = 0.7 − I O R E = 0.7 − (0.0417 )(1.5)
V BE 2 = 0.6375 V
______________________________________________________________________________________
10.25
⎛ 200 × 10 −6 ⎞
⎟ = 0.6347 V
(a) V BE1 = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 5 × 10
⎛ I REF ⎞
⎟
I O R E = VT ln⎜⎜
⎟
⎝ IO ⎠
⎛ 0.2 ⎞
⎟
I O (0.5) = (0.026 ) ln⎜⎜
⎟
⎝ IO ⎠
I O ≅ 61.4 μ A
⎛ 61.4 × 10 −6 ⎞
⎟ = 0.6040 V
V BE 2 = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 5 × 10
⎛ 200 × 10 −6 ⎞
⎟ = 0.6347 V
(b) V BE1 = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 5 × 10
V BE1 − V BE 2 = I O R E
⎡ ⎛I
⎞
⎛ I ⎞⎤
VT ⎢ln⎜⎜ REF ⎟⎟ − ln⎜⎜ O ⎟⎟⎥ = I O R E
⎝ I S 2 ⎠⎦⎥
⎣⎢ ⎝ I S1 ⎠
⎛I
I ⎞
VT ln⎜⎜ REF ⋅ S 2 ⎟⎟ = I O R E
⎝ I O I S1 ⎠
⎡⎛
⎞
⎤
(0.026) ln ⎢⎜⎜ 0.2 ⎟⎟⎛⎜ 7 ⎞⎟⎥ = I O (0.5)
⎣⎢⎝ I O ⎠⎝ 5 ⎠⎦⎥
I O ≅ 71.2 μ A
⎛ 71.2 × 10 −6 ⎞
⎟ = 0.5991 V
V BE 2 = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 7 × 10
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.26
⎛ 100 × 10 −6 ⎞
⎟ = 0.61669 V
(a) V BE1 = (0.026) ln⎜⎜
−15 ⎟
⎠
⎝ 5 × 10
V BE 2 = V BE1 + I REF R E = 0.61669 + (0.1)(0.7 ) = 0.68669 V
(
)
⎛ 0.68669 ⎞
I O = 5 × 10 −15 exp⎜
⎟ ⇒ I O = 1.477 mA
⎝ 0.026 ⎠
(b) V BE1 = 0.61669 V
V BE 2 = 0.68669 V
(
)
⎛ 0.68669 ⎞
I O = 2 × 10 −15 exp⎜
⎟ ⇒ I O = 0.5906 mA
⎝ 0.026 ⎠
______________________________________________________________________________________
10.27
(a) I REF =
5 − 0.7 − (− 5)
= 0.186 mA
50
⎞
⎛I
I O R E = VT ln⎜⎜ REF ⎟⎟
I
⎝ O ⎠
⎛ 0.186 ⎞
⎟
I O (3) = (0.026) ln⎜⎜
⎟
⎝ IO ⎠
I O ≅ 19.53 μ A
V BE 2 = V BE1 − I O R E = 0.7 − (0.01953)(3) = 0.6414 V
(b) ro 2 =
VA
80
=
⇒ ro 2 = 4.096 M Ω
I O 0.01953
0.01953
= 0.7512 mA/V
0.026
(120 )(0.026 ) = 159.8 k Ω
rπ 2 =
0.01953
R E rπ 2 = 3 159.8 = 2.945 k Ω
g m2 =
Ro = (4.096 )[1 + (0.7512 )(2.945)] = 13.16 M Ω
______________________________________________________________________________________
10.28
Ro = ro 2 1 + g m 2 (R E rπ 2 )
[
]
From 10.27, I O = 19.53 μ A
I
0.01953
= 0.7512 mA/V
g m2 = O =
0.026
VT
rπ 2 =
β VT
IO
=
(80)(0.026) = 106.5 k Ω
0.01953
V
80
⇒ 4.096 M Ω
ro 2 = A =
I O 0.01953
[
]
Then Ro = (4.096) 1 + (0.7512)(3 106.5) = 13.07 M Ω
ΔVO
5
=
= 0.382 μ A
13.07
Ro
______________________________________________________________________________________
ΔI O =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.29
I REF =
5 − 0.7 − ( −5 )
R1
= 0.50
R1 = 18.6 K
(a)
⎛I ⎞
I O RE = VT ln ⎜ REF ⎟
⎝ IO ⎠
0.026 ⎛ 0.50 ⎞
ln ⎜
RE =
⎟
0.050 ⎝ 0.050 ⎠
RE = 1.20 K
RO = rc 2 [1 + RE′ g m 2 ]
RE′ = RE rπ 2
rπ 2 =
( 75)( 0.026 )
= 39 K
0.050
V
100
ro 2 = A =
⇒ 2 MΩ
I O 0.05
(b)
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
ΔI O =
ΔV
5
=
= 0.772 μ A
RO 6.477
ΔI O
0.772
× 100% =
× 100 = 1.54%
IO
50
(c)
______________________________________________________________________________________
10.30
Let R1 = 10 k Ω
Then I REF =
RE =
3 − 0.7 − (− 3)
= 0.53 mA
10
VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.53 ⎞
⎟=
ln⎜
ln⎜
⎟ ⇒ R E = 1.228 k Ω
I O ⎜⎝ I O ⎟⎠ (0.05) ⎝ 0.05 ⎠
______________________________________________________________________________________
10.31
⎛I ⎞
VBE = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
−15
0.7 = ( 0.026 ) ln ⎜
⎟ ⇒ I S = 2.03 × 10 A
⎝ IS ⎠
⎛ 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
⋅ ln ⎜
RE = T ln ⎜ REF ⎟ =
⎟ ⇒ RE = 1.92 kΩ
I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠
______________________________________________________________________________________
R1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.32
(a) I REF =
(b) R E =
3 − 0.7 − (− 3)
= 0.265 mA
20
VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.265 ⎞
⎟=
ln⎜
ln⎜
⎟ ⇒ R E = 253 Ω
(0.1) ⎝ 0.1 ⎠
I O ⎜⎝ I O ⎟⎠
______________________________________________________________________________________
10.33
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
I
⎝ S ⎠
Now
⎛ 0.4825 × 10−3 ⎞
= 0.681 V
VBE = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 2.03 × 10
⎠
VBE1 = 0.681 V
So
I REF ≅
10 − 0.681 − ( −10 )
40
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⇒ I REF = 0.483 mA
⎛ 0.483 ⎞
I 0 (12 ) = ( 0.026 ) ln ⎜
⎟
⎝ I0 ⎠
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
______________________________________________________________________________________
10.34
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.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.35
⎛I
⎞
⎛ 0.5 × 10 −3 ⎞
⎟ = 0.70038 V
(a) V BE1 = VT ln⎜⎜ REF ⎟⎟ = (0.026 ) ln⎜⎜
−15
⎟
⎝ I S1 ⎠
⎝ 10
⎠
V + = I REF R1 + V BE1 + I REF R E1 + V −
Then R1 =
3 − 0.70038 − (0.5)(0.5) − (− 3)
= 10.1 k Ω
0.5
⎛ 0.2 × 10 −3 ⎞
⎟ = 0.67656 V
V BE 2 = (0.026) ln⎜⎜
−15
⎟
⎝ 10
⎠
V BE1 + I REF R E1 = V BE 2 + I O R E 2
0.70038 + (0.5)(0.5) − 0.67656
= 1.37 k Ω
0. 2
(b) R1 = 10.1 k Ω
Then R E 2 =
⎛ 0.2 × 10 −3 ⎞
⎟ = 0.65854 V
V BE 2 = (0.026) ln⎜⎜
−15 ⎟
⎝ 2 × 10
⎠
0.70038 + (0.5)(0.5) − 0.65854
RE 2 =
= 1.46 k Ω
0. 2
______________________________________________________________________________________
10.36
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
⎝ IS ⎠
⎝ IS ⎠
⎡ ⎛ I ⎞2
⎛ I ⎞⎤
VT ⎢ln ⎜ REF ⎟ − ln ⎜ 0 ⎟ ⎥ = I 0 RE
⎢⎣ ⎝ I S ⎠
⎝ I S ⎠ ⎥⎦
⎛ I 2 REF ⎞
VT ln ⎜
⎟ = I 0 RE
⎝ I0 I S ⎠
b.
⎛ 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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.37
(a) I REF =
5 − 0.7 − (− 5)
= 0.93 mA
10
⎛I
⎞
I O 2 R E 2 = VT ln⎜⎜ REF ⎟⎟
⎝ I O2 ⎠
⎛ 0.93 ⎞
⎟
I O 2 (1) = (0.026 ) ln⎜⎜
⎟
⎝ I O2 ⎠
I O 2 ≅ 68 μ A
⎛ 0.93 ⎞
⎟
I O 3 (2 ) = (0.026 ) ln⎜⎜
⎟
⎝ I O3 ⎠
I O 3 ≅ 40.7 μ A
VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.93 ⎞
⎟=
ln⎜
ln⎜
⎟ ⇒ R E 2 = 4.99 k Ω
I O 2 ⎜⎝ I O 2 ⎟⎠ (0.02 ) ⎝ 0.02 ⎠
(b) R E 2 =
RE3 =
(0.026 ) ln⎛ 0.93 ⎞ ⇒ R = 0.797 k Ω
⎜
⎟
E3
(0.08) ⎝ 0.08 ⎠
______________________________________________________________________________________
10.38
(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 ⎟
I
⎝ S ⎠
⎝ 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 )
Then
IO =
2 R2
⎛R ⎞
V −V
− IO ⎜ E ⎟ + IO
2 R2
⎝ R2 ⎠
+
V + −V −
2 Rε
−
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(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
______________________________________________________________________________________
10.39
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 REF =
I 03 = 3I REF = 4.65 mA
b.
VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V
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
______________________________________________________________________________________
10.40
10 − 0.7 − 0.7 − (− 10 )
= 0.775 mA
24
I O1 = 2 I REF = 1.55 mA
(a) I REF =
I O 2 = I REF = 0.775 mA
I O 3 = 3I REF = 2.325 mA
0 − 0.7 − (− 10 )
= 6kΩ
1.55
10 − 0.7
RC 2 =
= 12 k Ω
0.775
10 − 0.7
RC 3 =
= 4 kΩ
2.325
______________________________________________________________________________________
(b) RC1 =
10.41
I C1 = I C 2 =
10 − 0.7 − 0.7 − ( −10 )
10
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
= 1.86 mA
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
10.42
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 ⎜
⎟
I
⎝ C3 ⎠
⎝ 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
______________________________________________________________________________________
10.43
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 =
VRE1 3
= ⇒ RE1 = 3 kΩ
I 01 1
VRE 2 = 3 V ⇒ RE 2 =
VRE 2 3
= ⇒ RE 2 = 1.5 kΩ
2
I 02
VRE 3 = 3 V ⇒ RE 3 =
VRE 3 3
= ⇒ RE 3 = 0.75 kΩ
4
I 03
I 01 = 1 mA
I 02 = 2 mA
I 03 = 4 mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.44
2.5 − VGS ⎛ 0.08 ⎞
2
I REF =
=⎜
⎟ ( 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 )
15
I o = 92.0 μ A
I REF =
VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5
VDS 2 ( sat ) = 0.619 V
______________________________________________________________________________________
10.45
(a)
2
⎛ 80 ⎞⎛ W ⎞
I REF = 50 = ⎜ ⎟⎜ ⎟ (VGS − 0.5 )
2
L
⎝ ⎠⎝ ⎠1
2.0 − VGS
I REF = 0.050 =
R
Design such that
VGS = 0.75 V
So
0.050 =
VDS 2 ( sat ) = 0.25 = VGS − 0.5
2 − 0.75
⇒ R = 25 K
R
2
⎛ 80 ⎞ ⎛ W ⎞
⎛W ⎞
50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20
⎝ 2 ⎠ ⎝ L ⎠1
⎝ L ⎠1
(b)
⎛W ⎞
⎜ ⎟
20
50
⎛W ⎞
⎝ L ⎠1 I REF
=
⇒
=
⇒ ⎜ ⎟ = 40
W
W
I
100
⎛ ⎞
⎛ ⎞
⎝ L ⎠2
O
⎜ ⎟
⎜ ⎟
⎝ L ⎠2
⎝ L ⎠2
1
1
RO =
=
⇒ RO = 667 K
λ I O ( 0.015 )( 0.1)
ΔI O =
ΔV
1
=
⇒ 1.5 μ A
RO 666
ΔI O
⎛ 1.5 ⎞
× 100% = ⎜
⎟ × 100% ⇒ 1.5%
IO
⎝ 100 ⎠
(c)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.46
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 )
(a)
ΔV 3 − 2.44
ΔI O =
=
⇒ 2.8 μ A
RO
200
(i)
I O = 252.8 μ A
ΔI O =
(ii)
I O = 260.3 μ A
ΔI O =
(iii)
ΔV 4.5 − 2.44
=
⇒ 10.3 μ A
RO
200
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
RO =
=
= 133.3 K
λ I O ( 0.02 )( 0.375 )
IO =
(b)
ΔI O =
(i)
I O = 379.2 μ A
ΔI O =
(ii)
ΔV 3 − 2.44
=
⇒ 4.20 μ A
RO
133.3
ΔV 4.5 − 2.44
=
⇒ 15.5 μ A
RO
133.3
I O = 390.5 μ A
ΔI O =
ΔV 6 − 2.44
=
⇒ 26.7 μ A
RO
133.3
I = 401.7 μ A
(iii) O
______________________________________________________________________________________
10.47
2
(a) I REF = K n1 (VGS1 − VTN 1 )
0.2 = 0.2(VGS1 − 0.4) ⇒ VGS1 = VGS 2 = 1.4 V
2
I O = K n 2 (VGS 2 − VTN 2 )
2
Now I O = (0.2 − 0.01)(1.4 − 0.4) = 0.19 mA
2
I O = (0.2 + 0.01)(1.4 − 0.4) = 0.21 mA
So 0.19 ≤ I O ≤ 0.21 mA
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) I O = (0.2)[1.4 − (0.4 − 0.02)] = 0.2081 mA
2
I O = (0.2)[1.4 − (0.4 + 0.02)] = 0.1921 mA
So 0.1921 ≤ I O ≤ 0.2081 mA
______________________________________________________________________________________
2
10.48
2
(a) I REF = K n1 (VGS1 − VTN )
0.2 = 0.2(VGS1 − 0.5) ⇒ VGS1 = 1.5 V
2
VGS1 = VGS 2 + I O R S = VGS 2 + K n 2 R S (VGS 2 − VTN )
(
2
)
2
1.5 = VGS 2 + (0.2)(10) VGS
2 − VGS 2 + 0.25
2
or 2VGS
2 − VGS 2 − 1 = 0 ⇒ VGS 2 = 1.0 V
V − VGS 2 1.5 − 1.0
I O = GS1
=
⇒ I O = 50 μ A
RS
10
(b) I O = 0.5 I REF = 0.1 mA
0.1 = 0.2(VGS 2 − 0.5) ⇒ VGS 2 = 1.207 V
V − VGS 2 1.5 − 1.207
=
= 2.93 k Ω
R S = GS1
0.1
IO
______________________________________________________________________________________
2
10.49
Ix =
(1)
Vx − VA
+ g mVgs 2
ro
Ix =
(2)
So
Vgs1 = Vx , Vgs 2 = −VA
Ix =
⎛1
⎞
Vx
− VA ⎜ + g m ⎟
ro
⎝ ro
⎠
Ix =
VA
+ g mVx ⇒ VA = ro [ I x − g mVx ]
ro
(1)
(2)
VA
+ g mVgs1
ro
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
g m >>
ro
Since
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 )
Ro ≅
1
gm
Usually, g m ro >> 2, so that
______________________________________________________________________________________
10.50
VGS 2 = V DS 2 (sat ) + VTN = 1 + 0.5 = 1.5 V
⎛ k ′ ⎞⎛ W ⎞
2
I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN )
⎝ 2 ⎠⎝ L ⎠ 2
⎛ 0.08 ⎞⎛ W ⎞
⎛W ⎞
2
0.15 = ⎜
⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 3.75
L
2
⎝
⎠⎝ ⎠ 2
⎝ L ⎠2
⎛ 0.08 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.5 = ⎜
⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 12.5
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1
(
)
VGS 3 = V + − V − − VGS1 = 1.8 − (− 1.8) − 1.5 = 2.1 V
⎛ 0.08 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.5 = ⎜
⎟⎜⎜ ⎟⎟(2.1 − 0.5) ⇒ ⎜ ⎟ = 4.88
L
2
⎝
⎠⎝ 3 ⎠
⎝ L ⎠3
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.51
(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 =
1
1
=
= 183.4 K
λ I O ( 0.015 )( 0.3636 )
ΔI O =
ΔV 1.5 − 1.705
=
⇒ −1.12 μ A
RO
183.4
I O = 362.5 μ A
ΔI O =
ΔV 3 − 1.705
=
⇒ 7.06 μ A
RO
183.4
I O = 370.7 μ A
(c)
______________________________________________________________________________________
10.52
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.53
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
______________________________________________________________________________________
10.54
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
Then
− 1.2
(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
______________________________________________________________________________________
10.55
V SG 2 = V SD 2 (sat ) − VTP = 1 + 0.5 = 1.5 V
⎛ k ′p ⎞⎛ W ⎞
2
I O = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 2 + VTP )
⎝ 2 ⎠⎝ L ⎠ 2
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
2
80 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 2.67
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 250 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 8.33
L
2
⎝ ⎠⎝ ⎠ 1
⎝ L ⎠1
V SG 3 = V SG 4
Then 2V SG 3 = V + − V − − V SG1 = 3 − (− 3) − 1.5 = 4.5 V
So V SG 3 = 2.25 V
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
⎛W ⎞
2
250 = ⎜ ⎟⎜ ⎟ (2.25 − 0.5) ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.72
⎝ 2 ⎠⎝ L ⎠ 3, 4
⎝ L ⎠3 ⎝ L ⎠4
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.56
⎛ k ′p ⎞⎛ W ⎞
⎛ k ′p ⎞⎛ W ⎞
2
2
(a) I REF = ⎜⎜ ⎟⎟⎜ ⎟ (V SG1 + VTP ) = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP )
2
2
L
L
⎝
⎠
⎝
⎠
1
3
⎝ ⎠
⎝ ⎠
V SG 3 = 3 − V SG1
25 (V SG1 − 0.4 ) = 5 (3 − V SG1 − 0.4 )
3.236V SG1 = 3.4944 ⇒ V SG1 = 1.08 V and V SG 3 = 1.92 V
⎛ 60 ⎞
2
I REF = ⎜ ⎟(25)(1.08 − 0.4 ) ⇒ I REF = 0.347 mA
⎝ 2 ⎠
⎛ 60 ⎞
2
I O = ⎜ ⎟(15)(1.08 − 0.4 ) ⇒ I O = 0.208 mA
⎝ 2 ⎠
(
(b) V SD 2 sat ) = V SG 2 + VTP = 1.08 − 0.4 = 0.68 V
3 − 0.68
= 11.15 k Ω
0.208
______________________________________________________________________________________
R=
10.57
V SD 2 (sat ) = 0.35 = V SG 2 + VTP = V SG 2 − 0.4 ⇒ V SG 2 = 0.75 V
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 220 = ⎜ ⎟⎜ ⎟ (0.75 − 0.4 ) ⇒ ⎜ ⎟ = 59.9
L
2
⎝ ⎠⎝ ⎠ 1
⎝ L ⎠1
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
2
I O = 80 = ⎜ ⎟⎜ ⎟ (0.75 − 0.4 ) ⇒ ⎜ ⎟ = 21.8
L
2
⎝ ⎠⎝ ⎠ 2
⎝ L ⎠2
V SG 3 = 3 − 0.75 = 2.25 V
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
2
220 = ⎜ ⎟⎜ ⎟ (2.25 − 0.4 ) ⇒ ⎜ ⎟ = 2.14
⎝ 2 ⎠⎝ L ⎠ 3
⎝ L ⎠3
______________________________________________________________________________________
10.58
2
(a) I REF = 100 = 100(VGS1 − 0.5) ⇒ VGS1 = 1.5 V
For V D 4 = −2 V, V DS 4 + V DS 2 = 3 V = V DS 3 + V DS1
Then I O = 100 μ A
(b) Ro = ro 4 + ro 2 (1 + g m ro 4 )
g m = 2 K n I O = 2 (0.1)(0.1) = 0.2 mA/V
ro 2 = ro 4 =
1
1
=
= 500 k Ω
λ I O (0.02)(0.1)
R o = 500 + 500[1 + (0.2 )(500 )] ⇒ R o = 51 M Ω
ΔI O =
ΔV D 4
4
=
= 0.07843 μ A
Ro
51
ΔI O
⎛ 0.07843 ⎞
× 100% = ⎜
⎟ × 100% = 0.07843%
IO
⎝ 100 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.59
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
I X = g mVgs 6 +
IX =
⎛
VX − VS 6 VX
1 ⎞
=
− VS 6 ⎜ g m + ⎟
r06
r06
r06 ⎠
⎝
⎛
VX
1 ⎞
− I X ⎜ g m + ⎟ ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦
r06
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
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
λ I0
=
1
= 250 kΩ
0.02
( )( 0.2 )
{
}
R0 = 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ × 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ ( 250 )
R0 = 2575750 kΩ ⇒ R0 = 2.58 × 10 Ω
9
______________________________________________________________________________________
10.60
⎛ k ′p ⎞⎛ W ⎞
⎛ k ′ ⎞⎛ W ⎞
2
2
I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 3 − VTN ) = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 4 + VTP )
2
2
L
L
⎝ ⎠⎝ ⎠ 3
⎝ ⎠⎝ ⎠ 4
⎛ 60 ⎞
⎛ 100 ⎞
2
2
⎟(5)(VGS 3 − 0.4 ) = ⎜ ⎟(10 )(V SG 4 − 0.4 )
⎜
2
2
⎠
⎝ ⎠
⎝
We find V SG 4 = 0.91287VGS 3 + 0.03485
⎛ k ′ ⎞⎛ W ⎞
⎛ k ′ ⎞⎛ W ⎞
2
2
I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 1 − VTN ) = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 3 − VTN )
⎝ 2 ⎠⎝ L ⎠ 1
⎝ 2 ⎠⎝ L ⎠ 3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
20 (VGS1 − 0.4 ) = 5 (VGS 3 − 0.4 )
Now V SG 4 + VGS 3 + VGS1 = 6
0.91287VGS 3 + 0.03485 + VGS 3 + VGS 1 = 6
Then VGS 3 = 3.1184 − 0.52277VGS1
And 2(VGS1 − 0.4) = 3.1184 − 0.52277VGS 1 − 0.4
So VGS 1 = 1.395 V
VGS 3 = 2.389 V
V SG 4 = 2.216 V
⎛ 0. 1 ⎞
2
I REF = ⎜
⎟(20 )(1.395 − 0.4 ) = 0.99 mA
⎝ 2 ⎠
⎛ 0.1 ⎞
2
IO = ⎜
⎟(20 )(1.395 − 0.4 ) = 0.99 mA
2
⎠
⎝
V DS 2 (sat ) = 1.395 − 0.4 = 0.995 V
______________________________________________________________________________________
10.61
V DS 2 (sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.4 ⇒ VGS 2 = 0.9 V
⎛ k ′ ⎞⎛ W ⎞
2
I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN )
2
L
⎝ ⎠⎝ ⎠ 2
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
2
50 = ⎜
⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 4
⎝ L ⎠2
⎝ 2 ⎠⎝ L ⎠ 2
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
2
I REF = 500 = ⎜
⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 40
⎝ L ⎠1
⎝ 2 ⎠⎝ L ⎠ 1
VGS 3 = V SG 4
6 = VGS1 + VGS 3 + V SG 4 = 0.9 + 2VGS 3 ⇒ VGS 3 = V SG 4 = 2.55 V
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
2
500 = ⎜
⎟⎜ ⎟ (2.55 − 0.4 ) ⇒ ⎜ ⎟ = 2.16
⎝ L ⎠3
⎝ 2 ⎠⎝ L ⎠ 3
⎛W ⎞
⎛ 60 ⎞⎛ W ⎞
2
500 = ⎜ ⎟⎜ ⎟ (2.55 − 0.4 ) ⇒ ⎜ ⎟ = 3.61
2
L
⎝ L ⎠4
⎝ ⎠⎝ ⎠ 4
______________________________________________________________________________________
10.62
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 (VGS 1 − 1) ⎣⎡1 + ( 0.02 )( 4 ) ⎦⎤
2
Or
80
2
= (VGS 1 − 1) ⇒ VGS1 = 1.962
86.4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
I O = K n (VGS 1 − VTN ) (1 + λnVGS 1 )
2
= 80 (1.962 − 1) ⎡⎣1 + ( 0.02 )(1.962 ) ⎤⎦
2
Or I 0 = 76.94 μ A
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%.
b.
______________________________________________________________________________________
10.63
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 ) ⎤⎦
2
Or VGS 4 = 1.98 V = VGS 1
I O = K n (VGS 2 − VTN ) (1 + λVGS 2 )
2
To a very good approximation I 0 = 80 μ A
From a PSpice analysis, I 0 = 80.00 μ A for VD 3 = −1 V and the output resistance is
b.
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
______________________________________________________________________________________
10.64
VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0
ID =
(a)
(b)
k n′ ⎛ W ⎞
2
⎜ ⎟ (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
(c)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.65
(a)
k′ ⎛ W ⎞
K n1 = n ⎜ ⎟ = 50 ( 5 ) = 250 μ A / V 2
2 ⎝ L ⎠1
R=
=
⎛
⎜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
______________________________________________________________________________________
10.66
V GS 3 = VGS 4 = VGS 5 ⇒ VGS 3 =
5
V
3
2
⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5
⎞
⎛W ⎞
I REF = 0.1 = ⎜
= 2.68
⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟
2
L
3
⎠⎝ ⎠ 3 ⎝
⎠
⎝ L ⎠ 3, 4 , 5
⎝
2
⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5
⎞
⎛W ⎞
I O1 = 0.2 = ⎜
⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟ = 5.35
⎝ 2 ⎠⎝ L ⎠ 1 ⎝ 3
⎠
⎝ L ⎠1
2
⎞
⎛W ⎞
⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5
I O 2 = 0.3 = ⎜
⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟ = 8.03
⎝ 2 ⎠⎝ L ⎠ 2 ⎝ 3
⎠
⎝ L ⎠2
______________________________________________________________________________________
10.67
(
)
P = (I REF + I O1 + I O 2 + I O 3 ) V + − V −
5 = (I REF + 0.1 + 0.2 + 0.4 )[1.8 − (− 1.8)] ⇒ I REF = 0.689 mA
V DS 2 (sat ) = 0.4 = VGS 2 − 0.4 ⇒ VGS 2 = 0.8 V = VGS
⎛ k ′ ⎞⎛ W ⎞
2
I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 1 − VTN )
L
2
⎝ ⎠⎝ ⎠ 1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
0.689 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 86.1
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I O1 = 0.1 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 12.5
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I O 2 = 0.2 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 25
2
L
⎝
⎠⎝ ⎠ 3
⎝ L ⎠3
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I O 3 = 0.4 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 50
⎝ 2 ⎠⎝ L ⎠ 4
⎝ L ⎠4
______________________________________________________________________________________
10.68
24 = I REF R + V SG + VGS
V SG =
I REF
− VTP = 5.7735 I REF + 0.8
⎛ k ′p ⎞
⎜
⎟(1)
⎜ 2 ⎟
⎝
⎠
VGS =
I REF
+ VTN = 4.472 I REF + 0.8
⎛ k n′ ⎞
⎜⎜ ⎟⎟(1)
⎝ 2 ⎠
So 24 = I REF (100 ) + 5.7735 I REF + 0.8 + 4.4721 I REF + 0.8
I REF = x
Let
Then 100 x 2 + 10.2456x − 22.4 = 0 ⇒ x = 0.4248 ⇒ x 2 = I REF = 0.1805 mA
I 1 = (0.2)I REF = 0.0361 mA
I 2 = (1.25)I REF = 0.2256 mA
I 3 = (0.8)I REF = 0.1444 mA
I 4 = (4 )I REF = 0.722 mA
______________________________________________________________________________________
10.69
10 = I REF R + V SG + VGS
V SG =
I REF
+ 0.8 = 5.7735 I REF + 0.8
⎛ 0.06 ⎞
1
(
)
⎜
⎟
⎝ 2 ⎠
VGS =
I REF
+ 0.8 = 4.472 I REF + 0.8
⎛ 0.1 ⎞
1
(
)
⎜
⎟
⎝ 2 ⎠
8.4 = 100 I REF + 10.2456 I REF
Let
I REF = x
Then 100 x 2 + 10.2456x − 8.4 = 0 ⇒ x = 0.2431 ⇒ x 2 = I REF = 59.09 μ A
I 1 = (0.2)I REF = 11.82 μ A, I 2 = (1.25)I REF = 73.87 μ A
I 3 = (0.8)I REF = 47.27 μ A, I 4 = (4)I REF = 236.4 μ A
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.70
W
9
L 2
⋅ I REF = ( 200 ) ⇒ I D 2 = 120 μ A
ID2 =
W
15
L1
W
⎛ 20 ⎞
L 4
⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 μ A
IO =
W
⎝ 9 ⎠
L 3
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
(
(
(
(
)
)
)
)
______________________________________________________________________________________
10.71
⎛ k ′ ⎞⎛ W ⎞
2
For M 1 : I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 1 − VTN )
⎝ 2 ⎠⎝ L ⎠ 1
⎛ 100 ⎞
2
100 = ⎜
⎟(4 )(VGS 1 − 0.4) ⇒ VGS 1 = VGS 2 = 1.107 V
2
⎝
⎠
⎛ k ′ ⎞⎛ W ⎞
⎛ 100 ⎞
2
2
I D 2 = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN ) = ⎜
⎟(2.5)(1.107 − 0.4) ⇒ I D 2 = 62.5 μ A
2
L
2
⎝
⎠
⎝
⎠
⎝ ⎠
2
⎛ k ′p ⎞⎛ W ⎞
2
Also I D 2 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP )
2
L
⎝
⎠
3
⎝ ⎠
⎛ 60 ⎞
2
62.5 = ⎜ ⎟(6 )(V SG 3 − 0.4 ) ⇒ V SG 3 = V SG 4 = 0.9893 V
⎝ 2 ⎠
⎛ 60 ⎞
2
I O = ⎜ ⎟(4 )(0.9893 − 0.4 ) = 41.67 μ A
⎝ 2 ⎠
______________________________________________________________________________________
10.72
2
⎛ 40 ⎞ ⎛ W ⎞
I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 )
⎝ 2 ⎠ ⎝ L ⎠1
1.75 − VSG1
I REF =
= 50
R
VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V
1.75 − 0.95
⇒ R = 16 K
0.05
2
⎛ 40 ⎞⎛ W ⎞
⎛W ⎞
50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4
⎝ 2 ⎠⎝ L ⎠1
⎝ L ⎠1
R=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
W
I O1 120
⎛W ⎞
L 2
=
=
⇒ ⎜ ⎟ = 49
I REF
50
20.4
( ) ⎝ L ⎠2
( )
( )
W
I D3
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
⎝ 2 ⎠⎝ L ⎠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
______________________________________________________________________________________
( )
10.73
For
a.
vGS = 0, iD = I DSS (1 + λ vDS )
VD = −5 V, vDS = 5
iD = ( 2 ) ⎡⎣1 + ( 0.05 )( 5) ⎤⎦ ⇒ iD = 2.5 mA
VD = 0, vDS = 10
b.
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
c.
______________________________________________________________________________________
10.74
⎛ V ⎞
I 0 = I DSS ⎜1 − GS ⎟
VP ⎠
⎝
⎛ V ⎞
2 = 4 ⎜ 1 − GS ⎟
⎝ VP ⎠
2
2
VGS
2
= 1−
= 0.293
VP
4
So VGS = ( 0.293)( −4 ) = −1.17 V
V
Then I 0 = S and VS = −VGS
R
( −1.17 )
−V
⇒ R = 0.586 kΩ
R = GS = −
I0
2
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.75
1
1
−
−
VT
− 38.46
= 0.026 =
(a) Aυ =
⇒ Aυ = −1846
1
1
1
1
0.00833 + 0.0125
+
+
V AN V AP
120 80
⎡ ⎛ V ⎞⎤⎛ V ⎞
(b) I CQ = I S1 ⎢exp⎜⎜ I ⎟⎟⎥⎜⎜1 + CE1 ⎟⎟
⎢⎣ ⎝ VT ⎠⎥⎦⎝ V AN ⎠
⎡ ⎛ V ⎞⎤⎛ 1.25 ⎞
200 × 10 − 6 = 5 × 10 −16 ⎢exp⎜⎜ I ⎟⎟⎥⎜1 +
⎟
⎣⎢ ⎝ VT ⎠⎦⎥⎝ 120 ⎠
(
)
⎡
⎤
200 × 10 − 6
or V I = (0.026 ) ln ⎢
⎥ = 0.6943 V
−16
⎣ 5 × 10 (1 + 1.25 120) ⎦
⎡ ⎛ V ⎞⎤⎛ 1.25 ⎞
(c) 200 × 10 − 6 = 10 −15 ⎢exp⎜⎜ EB ⎟⎟⎥⎜1 +
⎟
80 ⎠
⎣⎢ ⎝ VT ⎠⎦⎥⎝
(
)
(
)
⎡
⎤
200 × 10 − 6
V EB = (0.026 ) ln ⎢ −15
⎥ = 0.6762 V
⎣ 10 (1 + 1.25 80 ) ⎦
(
)
+
V EB = 0.6762 = V − V B ⇒ V B = 1.824 V
______________________________________________________________________________________
10.76
(a) Aυ = − g m1 ro1 ro 2
(
ro1 =
)
1
1
1
1
=
= 250 k Ω , ro 2 =
=
= 166.7 k Ω
λ1 I DQ (0.02 )(0.2 )
λ 2 I DQ (0.03)(0.2)
− 100 = − g m1 (250 166.7 ) ⇒ g m1 = 1 mA/V
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
g m1 = 1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟⎜ ⎟ (0.2) ⇒ ⎜ ⎟ = 25
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1
⎝ 2 ⎠⎝ L ⎠ 1
⎛ k ′ ⎞⎛ W ⎞
(b) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V I − VTN )2 (1 + λ1V DS1 )
⎝ 2 ⎠⎝ L ⎠ 1
⎛ 0.1 ⎞
2
0.2 = ⎜
⎟(25)(V I − 0.5) [1 + (0.02 )(1.25)] ⇒ V I = 0.895 V
⎝ 2 ⎠
⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′p ⎞⎛ W ⎞
(c) K n1 = K p 2 ⇒ ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜⎜ ⎟⎟⎜ ⎟
⎝ 2 ⎠⎝ L ⎠ 1 ⎝ 2 ⎠⎝ L ⎠ 2
⎛W ⎞
⎛ 100 ⎞
Then ⎜ ⎟ = ⎜
⎟(25) = 41.67
⎝ L ⎠ 2 ⎝ 60 ⎠
⎛ k ′p ⎞⎛ W ⎞
2
I DQ = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 2 + VTP ) (1 + λ 2V SD 2 )
⎝ 2 ⎠⎝ L ⎠ 2
⎛ 0.06 ⎞
2
0.2 = ⎜
⎟(41.67 )(V SG 2 − 0.5) [1 + (0.03)(1.25)] ⇒ V SG 2 = 0.8927 V
⎝ 2 ⎠
V SG 2 = 0.8927 = V + − VG ⇒ VG = 1.607 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.77
⎛V ⎞
I REF = I S 1 exp ⎜ EB1 ⎟
⎝ VT ⎠
a.
b.
c.
⎛I ⎞
⎛ 1× 10−3 ⎞
⇒ VEB1 = 0.5568
or VEB1 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
−13 ⎟
⎝ 5 × 10 ⎠
⎝ I S1 ⎠
R1 =
5 − 0.5568
⇒ R1 = 4.44 kΩ
1
From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V
2.5 ⎞
⎛
1+
⎜
⎛
⎞
V
2.5
⎡
⎤
80 ⎟
= 10−3 ⎜
10−12 exp ⎜ I ⎟ ⎢1 +
⎟
⎥
⎝ VT ⎠ ⎣ 120 ⎦
⎜⎜ 1 + 0.5568 ⎟⎟
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
Av =
− (1/ VT )
(1/ VAN ) + (1/ VAP )
1
−38.46
Av = 0.026 =
1
1
0.00833 + 0.0125
+
120 80
Av = −1846
−
d.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.78
⎛I
⎞
⎛ 0.5 × 10−3 ⎞
VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
⎟ ⇒ VBE = 0.5208
I S1 ⎠
10−12 ⎠
⎝
⎝
a.
5 − 0.5208
R1 =
⇒ R1 = 8.96 kΩ
0.5
b.
c.
Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal
to each other.
⎛V ⎞⎛ V ⎞
⎛V ⎞⎛ V ⎞
I CO = I SO exp ⎜ EBO ⎟ ⎜ 1 + ECO ⎟ = I C 2 = I S 2 exp ⎜ BE ⎟ ⎜1 + CE 2 ⎟
VAP ⎠
⎝ VT ⎠ ⎝
⎝ VT ⎠ ⎝ VAN ⎠
⎛ V ⎞ ⎛ 2.5 ⎞
⎛ VBE ⎞ ⎛
2.5 ⎞
−12
5 × 10−13 exp ⎜ EBO ⎟ ⎜1 +
⎟ ⎜1 +
⎟ = 10 exp ⎜
⎟
V
V
80
120
⎝
⎠
⎝
⎠
⎝ 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 ⎟
⎝ VT ⎠
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 =
− (1/ VT )
(1/ VAN ) + (1/ VAP )
1
−38.46
0.026
=
Av =
1
1 0.00833 + 0.0125
+
120 80
Av = −1846
−
d.
______________________________________________________________________________________
10.79
⎛ k ′ ⎞⎛ W ⎞
(a) For M O : I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V I − VTN )2 (1 + λ nV DSO )
⎝ 2 ⎠⎝ L ⎠ O
⎛ 100 ⎞⎛ W ⎞
⎛W ⎞
2
100 = ⎜
⎟⎜ ⎟ (1.2 − 0.5) [1 + (0.02 )(1.5)] ⇒ ⎜ ⎟ = 3.96
2
L
⎝ L ⎠O
⎝
⎠⎝ ⎠ O
For M 1 , M 2 : For I REF = I O ⇒ V SD 2 = V SD1 = V SG = 1.5 V
⎛ 60 ⎞⎛ W ⎞
⎛W ⎞
⎛W ⎞
2
100 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) [1 + (0.02 )(1.5)] ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.24
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠ 2 ⎝ L ⎠1
⎛W ⎞
For M 3 : V SG 3 = 3 − 1.5 = 1.5 V = V SD 3 ⇒ ⎜ ⎟ = 3.24
⎝ L ⎠3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Aυ = − g mO roO ro 2
(
)
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
g mO = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(3.96)(0.1) = 0.2814 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠ O
1
1
roO = ro 2 =
=
= 500 k Ω
λ I DQ (0.02 )(0.1)
Aυ = −(0.2814)(500 500) = −70.35
______________________________________________________________________________________
10.80
2
(a) I REF = K p (V SG + VTP ) 1 + λ pV SD1
(
)
0.080 = 0.1(V SG − 0.5) [1 + (0.02)V SG ]
Approximation, for λ = 0 ⇒ V SG = 1.394 V
(b) I O = I 2
2
(
K n (V I − VTN ) (1 + λ nV DSO ) = K p (V SG + VTP ) 1 + λ pV SD 2
2
2
)
K n = K p , V DSO = V SD 2 , λ n = λ p
So V I = 1.394 V
(c) I O ≅ I REF = 80 μ A
Aυ = − g mo (roo ro 2 )
g mo = 2 K n I O = 2 (0.1)(0.08) = 0.1789 mA/V
roo = ro 2 =
1
1
=
= 625 k Ω
λ I O (0.02)(0.08)
Aυ = −(0.1789)(625 625) = −55.9
______________________________________________________________________________________
10.81
3 − 0.6
⇒ I CO = 51.06 μ A
47
V
120
roo = AN =
⇒ roo = 2.35 M Ω
I CO 0.05106
(a) I REF = I CO =
ro 2 =
V AP
90
=
⇒ ro 2 = 1.763 M Ω
I CO 0.05106
gm =
I CO 0.05106
=
= 1.964 mA/V
0.026
VT
Aυ = − g m (roo ro 2 ) = −(1.964)(2350 1763) = −1978
(
)
(
)
(b) Aυ = − g m roo ro 2 R L = −(1.964 ) 2350 1763 300 = −454
(
)
(c) Aυ = −(1.964 ) 2350 1763 150 = −256
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.82
5 − 0.6
I REF =
= 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 +
+
+
120
80
RL
RL
Av =
(a)
−9.6692
0.0052375 +
1
RL
RL = ∞ Av = −1846
(b) RL = 250 K, Av = −1047
______________________________________________________________________________________
10.83
(a)
To a good approximation, output resistance is the same as the widlar current source.
R0 = r02 ⎡⎣1 + g m 2 ( rπ 2 || RE ) ⎤⎦
Av = − g m 0 ( r0 || RL || R0 )
(b)
______________________________________________________________________________________
10.84
Output resistance of Wilson source
βr
R0 ≅ 03
2
Then
Av = − g m ( r0 || R0 )
V
80
= 400 kΩ
r03 = AP =
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
⎢⎣
⎥⎦
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.85
(a) I D 2 = I D 0 = I REF = 200 μ A
For M 2 ; ro 2 =
1
1
=
= 250 K
λP I D 2 ( 0.02 )( 0.2 )
⎛ 0.04 ⎞
gm2 = 2 K P I D 2 = 2 ⎜
⎟ ( 35 )( 0.2 )
⎝ 2 ⎠
g m 2 = 0.748 mA/V
For M 0 ; r∞ =
(b)
1
λn I Do
=
1
( 0.015)( 0.2 )
= 333 K
⎛ 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)
Av = −114.3
Want Av = −57.15 = −0.80 (142.8 || RL )
142.8 || RL = 71.375 =
142.8RL
⇒ RL = 143 K
142.8 + RL
(c)
______________________________________________________________________________________
10.86
Assume M1, M2 matched
I REF = I D 2 = I Do = 200 μ A
1
1
ro 2 =
=
= 250 K
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.87
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
g m1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I D1 = 2 ⎜
⎟(20)(0.1) = 0.6325 mA/V
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠1
1
1
ro1 =
=
= 500 k Ω
λ n I D1 (0.02 )(0.1)
R o 2 = ro 2 + ro3 (1 + g m 2 ro 2 )
⎛ 0.06 ⎞
g m2 = 2 ⎜
⎟(80 )(0.1) = 0.9798 mA/V
⎝ 2 ⎠
1
ro 2 = ro3 =
= 500 k Ω
(0.02)(0.1)
R o 2 = 500 + 500[1 + (0.9798)(500 )] = 245,949 k Ω
Aυ = − g m1 (ro1 Ro 2 ) = −(0.6325)(500 245,949) = −316
______________________________________________________________________________________
10.88
− g m2
Aυ =
1
1
+
ro3 ro 4 ro1 ro 2
⎛ 0.1 ⎞
gm = 2 ⎜
⎟(25)(0.08) = 0.6325 mA/V
⎝ 2 ⎠
ro =
1
1
=
= 312.5 k Ω
λ I D (0.04)(0.08)
− (0.6325)
− 0.40
=
= −19,531
1
1
2(0.00001024 )
+
(312.5)2 (312.5)2
______________________________________________________________________________________
2
Aυ =
10.89
g m1Vi =
(1)
(2)
V − ( −Vπ 2 )
Vπ 2 Vπ 2
+
+ g m 2Vπ 2 + O
rπ 2 ro1
ro 2
VO VO − ( −Vπ 2 )
+
+ g m 2Vπ 2 = 0
RO 3
ro 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(1)
⎛ 1
1
1 ⎞ V
+ + gm2 + ⎟ + O
g m1Vi = Vπ 2 ⎜
ro 2 ⎠ ro 2
⎝ rπ 2 ro1
(2)
⎛ 1
⎛ 1
⎞
1 ⎞
VO ⎜
+ ⎟ + Vπ 2 ⎜
+ gm2 ⎟ = 0
⎝ RO 3 ro 2 ⎠
⎝ ro 2
⎠
g m >>
1
ro
(1)
⎛ 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 ⎠
Vπ 2 = −
VO ⎛ 1
1 ⎞
+ ⎟
⎜
g m 2 ⎝ RO 3 ro 2 ⎠
Then
(1) g m1Vi = −
VO ⎛ 1
1 ⎞ ⎛ 1 + β ⎞ VO
+ ⎟⎜
⎜
⎟+
g m 2 ⎝ RO 3 ro 2 ⎠ ⎝ rπ 2 ⎠ ro 2
⎛ 1
1 ⎞ ⎛ 1 + β ⎞ VO
= −VO ⎜
+ ⎟⎜
⎟+
⎝ RO 3 ro 2 ⎠ ⎝ β ⎠ ro 2
≈−
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
V
− g m1ro3 β 2
0.25
Av = O =
gm =
= 9.615 mA/V
1+ β
0.026
Vi
ro3 =
Av =
− ( 9.615 )( 320 )(120 )
80
= 320 K
0.25
2
= −366,165
121
______________________________________________________________________________________
10.90 Design Problem
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.91
Let (W L )1, 2 = 50 and (W L ) = 25 for all other transistors
Let I REF = 80 μ A
ro =
1
λI D
=
1
(0.04)(0.08)
= 312.5 k Ω
⎛ 0.06 ⎞
gm = 2 ⎜
⎟(50 )(0.08) = 0.6928 mA/V
⎝ 2 ⎠
Aυ =
− g m2
1
1
+
ro 3 ro 4 ro1 ro 2
− (0.6928)
− 0.480
=
1
1
2(0.00001024 )
+
(312.5)2 (312.5)2
2
=
Aυ = −23,438
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 11
11.1
(a) CMRR dB = ∞ ⇒ υ o = Ad υ d = (250)(1.5 sin ω t ) (mV)
υ o = 0.375 sin ω t (V)
(b) CMRR dB = 80 dB ⇒ CMRR = 10 4 =
250
⇒ Acm = 0.025
Acm
υ o = (250 )(0.0015 sin ω t ) + (0.025)(3 sin ω t )
υ o = 0.45 sin ω t (V)
(c) CMRR dB = 50 dB ⇒ CMRR = 316.2 =
250
⇒ Acm = 0.791
Acm
υ o = (250)(0.0015 sin ω t ) + (0.791)(3 sin ω t )
υ o = 2.75 sin ω t (V)
______________________________________________________________________________________
11.2
(a) (i) υ o1 = − g m Rυ1 = −(1)(5)(0.7 + 0.1 sin ω t )
υ o1 = −3.5 − 0.5 sin ω t (V)
(ii) υ o 2 = − g m Rυ 2 = −(1)(5)(0.7 − 0.1 sin ω t )
υ o 2 = −3.5 + 0.5 sin ω t (V)
(iii) υ o1 − υ o 2 = −1.0 sin ω t (V)
(b) Δ(υ1 − υ 2 ) = (0.7 + 0.1 sin ω t ) − (0.7 − 0.1 sin ω t ) = 0.2 sin ω t
−0.5
= −2.5
0.2
+0.5
(ii) Ad 2 =
= +2.5
0.2
−1
(iii) Ad 3 =
= −5
0.2
______________________________________________________________________________________
(c) (i) Ad 1 =
11.3
(a) Neglect dc base currents
I E = I C1 + I C 2 = 0.2 mA =
− 0.7 − (− 3)
⇒ R E = 11.5 k Ω
RE
3 − 1.2
= 18 k Ω
0.1
(c) For VCB = 0 ⇒ VCE = 0.7 V
3 = I C (18) + 0.7 + 2 I C (11.5) − 3
So I C = 0.1293 mA
υ o1 = υ cm + = 3 − (0.1293)(18) = 0.673 V
υ cm − = −2.3 V
So −2.3 ≤ υ cm ≤ 0.673 V
______________________________________________________________________________________
υ o1 = 1.2 V, ⇒ RC =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.4
a.
I1 =
10 − 2 ( 0.7 )
IC 2 =
⇒ I1 = 1.01 mA
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
2
⎛ 1+ β ⎞
⎛ 101 ⎞
IC 2 + ⎜
⎟ ( 2 IC 4 ) = ⎜
⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA
⎝ 100 ⎠
⎝ β ⎠
IC 4 =
I1 ≈ I C 2 = 3.23 mA
R1 =
10 − 2 ( 0.7 )
3.23
⇒ R1 = 2.66 kΩ
______________________________________________________________________________________
11.5
a.
Neglecting base currents
30 − 0.7
⇒ R1 = 73.25 kΩ
0.4
VCE1 = 10 V ⇒ VC1 = 9.3 V
15 − 9.3
RC =
⇒ RC = 28.5 kΩ
0.2
I1 = I 3 = 400 μ A ⇒ R1 =
b.
rπ =
(100 )( 0.026 )
= 13 kΩ
0.2
50
r0 ( Q3 ) =
= 125 kΩ
0.4
We have
Ad =
(100 )( 28.5 )
⇒ Ad = 62
2 ( rπ + RB )
2 (13 + 10 )
β RC
=
⎧
⎫
⎪
⎪
1
⎪
Acm = −
⎨
⎬
rπ + RB ⎪ 2r0 (1 + β ) ⎪
1+
rπ + RB ⎭⎪
⎩⎪
⎧
⎫
(100 )( 28.5) ⎪⎪
1
⎪⎪
=−
⇒ Acm = −0.113
⎨
2
125
101
13 + 10 ⎪
( )( ) ⎬⎪
1+
13 + 10 ⎭⎪
⎩⎪
β RC ⎪
⎛ 62 ⎞
C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 54.8 dB
⎝ 0.113 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 =
______________________________________________________________________________________
11.6
(a)
vCM ( max ) ⇒ VCB = 0
vCM ( max ) = 3 V
so that
vCM ( max ) = 5 −
IQ
2
( RC ) = 5 −
( 0.5)
2
(8)
(b)
⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞
V
ΔI = g m ⋅ d = ⎜
=⎜
⎟⋅
⎟⎜
⎟ = 0.08654 mA
2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠
ΔVC 2 = ΔI ⋅ RC = ( 0.08654 )( 8 ) = 0.692 V
(c)
⎛ 0.25 ⎞ ⎛ 0.010 ⎞
ΔI = ⎜
⎟⎜
⎟ = 0.04808 mA
⎝ 0.026 ⎠ ⎝ 2 ⎠
ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V
______________________________________________________________________________________
11.7
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
Ad =
1
0.05
g m RC where g m =
= 1.923 mA / V
2
0.026
Then
Ad =
1
(1.923)( 40 ) ⇒ Ad = 38.5
2
______________________________________________________________________________________
11.8
a.
IE
( 2 ) + I E (85 ) − 5
2
5 − 0.7
IE =
⇒ I E = 0.050 mA
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
(c) Differential-mode half circuits
⇒ vcm ( min ) = −4.3 V
⎛V
⎞
v
− d = Vπ + ⎜ π + g mVπ ⎟ .RE′
2
⎝ rπ
⎠
⎡ (1 + β ) ⎤
= Vπ ⎢1 +
RE′ ⎥
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 )
0.0248
= 105 k Ω RE′ = 2 k Ω
Then
Ad =
1 (100 )(100 )
⋅
⇒ Ad = 16.3
2 105 + (101)( 2 )
______________________________________________________________________________________
11.9
a.
For v1 = v2 = 0 and neglecting base currents
RE =
b.
Ad =
rπ =
Ad =
−0.7 − ( −10 )
0.15
⇒ RE = 62 kΩ
v02
β RC
=
vd 2 ( rπ + RB )
β VT
I CQ
=
(100 )( 0.026 )
0.075
= 34.7 kΩ
(100 )( 50 )
⇒ Ad = 71.0
2 ( 34.7 + 0.5 )
⎡
⎢
⎤
⎥
⎥
rπ + RB ⎢ 2 RE (1 + β ) ⎥
⎢1 +
⎥
rπ + RB ⎦⎥
⎣⎢
β RC ⎢
Acm = −
1
⎡
⎤
⎥
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
(100 )( 50 ) ⎢
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
c.
Rid = 2 ( rπ + RB )
Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩ
Common-mode input resistance
1
⎡ rπ + RB + 2 (1 + β ) RE ⎤⎦
2⎣
1
= ⎡⎣34.7 + 0.5 + 2 (101)( 62 ) ⎤⎦ ⇒ Ricm = 6.28 MΩ
2
Ricm =
______________________________________________________________________________________
11.10
⎛ 81 ⎞
(a) I E1 = I E 2 = ⎜ ⎟(50 ) = 50.625 ⇒ I E = 101.25 μ A
⎝ 80 ⎠
3 − 0.6
RE =
= 23.7 k Ω
0.10125
− 1.5 − (− 3)
RC =
= 30 k Ω
0.05
3 − (1 + 0.6 )
(b) (i) I E =
⇒ I E = 59.07 μ A
23.7
1 ⎛ 80 ⎞
I C1 = I C 2 = ⎜ ⎟(59.07 ) = 29.17 μ A
2 ⎝ 81 ⎠
υ C1 = υ C 2 = I C RC − 3 = (0.02917 )(30 ) − 3 = −2.125 V
(ii) g m =
0.02917
= 1.122 mA/V
0.026
υ d = 12 mV,
υd
= 6 mV
2
υ C 2 = −2.125 − (1.122 )(30 )(0.006 ) = −2.327 V
υ C1 = −2.125 + (1.122)(30 )(0.006 ) = −1.923 V
______________________________________________________________________________________
11.11
(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
0.026
Ad = g m RC = ( 0.227 )( 360 ) = 81.7
Acm = 0
gm =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(ii)
Ad =
Acm =
( 60 )( 0.026 )
g m RC
= 40.8 rπ =
2
0.0059
= 264 K
− ( 0.227 )( 360 )
= −0.0442
2 ( 61)( 4000 )
1+
264
______________________________________________________________________________________
11.12
I Q = 0.2 mA, ⇒ I C1 = I C 2 =
1 ⎛ 80 ⎞
⎜ ⎟(0.2) = 0.09877 mA
2 ⎝ 81 ⎠
(a) For VCB = 0 , υ C1 = υ C 2 = υ CM = −2.5 = I C RC − 5
5 − 2.5
= 25.3 k Ω
0.09877
0.09877
(b) Ad = g m RC , g m =
= 3.799 mA/V
0.026
Ad = (3.799 )(25.3) = 96.1
So RC =
(c) υ d = 14 mV,
υd
2
= 7 mV
υd
= −2.5 − (3.799 )(25.3)(0.007 ) = −3.173 V
2
υ C 2 = −2.5 + (3.799)(25.3)(0.007 ) = −1.827 V
(d) CMRR dB = 60 dB ⇒ CMRR = 1000
υ C1 = −2.5 − g m RC ⋅
1 ⎡ (81)(0.2 )R o ⎤
⎢1 +
⎥ ⇒ R o = 257 k Ω
2 ⎣ (0.026)(80 ) ⎦
______________________________________________________________________________________
1000 =
11.13
(a) Neglect dc base currents
I E = I C1 + I C 2 = 240 μ A
− 0.7 − (− 5)
= 17.9 k Ω
0.24
5−3
RC =
= 16.7 k Ω
0.12
0.12
(b) g m =
= 4.615 mA/V
0.026
Ad = g m RC = (4.615)(16.667 ) = 76.9
ΔRC
0.5
(c) Acm ≅
=
= 0.0279
17.9
RE
RE =
⎛ 76.9 ⎞
CMRRdB = 20 log 10 ⎜
⎟ = 68.8 dB
⎝ 0.0279 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.14
v1 = v2 = 0
IE =
−0.7 − ( −10 )
75
= 0.124 mA
I C1 = I C 2 = 0.0615 mA
gm =
0.0615
= 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
V
ΔI = g m d
2
V
ΔvC1 = − g m1 d Rc
2
Vd
ΔvC 2 = + g m 2
Rc
2
= 50.7 K
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.15
(a)
v1 = v2 = 0
vE = +0.7 V
5 − 0.7
IE =
= 4.3 mA
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
Q1 off
(b)
⎛ 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
(c)
vE ≈ 0.7 V
ΔI = g m
vd
2
gm =
2.132
= 82.0 mA/V
0.026
ΔvC = ΔI ⋅ RC = g m
Vd = 0.015 ⇒ Δvc = 0.615 V
( 82.0 )
Vd
⋅ RC =
⋅ Vd (1) = 41.0Vd
2
2
vC 2 ↓ vC1 ↑
vC1 = −2.87 + 0.615 = −2.255 V
vC 2 = −2.87 − 0.615 = −3.485 V
______________________________________________________________________________________
11.16
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.17
(a) I O = I E1 + I E 2 ⇒ I E1 = I E 2 = 0.4 mA
[
] [
]
(i) υ O1 − υ O 2 = V + − I E1 RC1 − V + − I E 2 RC 2 = I E 2 RC 2 − I E1 RC1
υ O1 − υ O 2 = 0
(ii) υ O1 − υ O 2 = (0.4)(7.6 − 7.4) = 0.08 V
(b) V BE1 = V BE 2
⎛V ⎞
⎛V ⎞
0.8 = 2.9 × 10 −15 + 3.1× 10 −15 ⋅ exp⎜⎜ BE ⎟⎟ ⇒ exp⎜⎜ BE ⎟⎟ = 1.333 × 1014
⎝ VT ⎠
⎝ VT ⎠
(
)
(
)(
)
= (3.1× 10 )(1.333 × 10 ) = 0.4133 mA
Then I E1 = 2.9 × 10 −15 1.333 × 1014 = 0.3867 mA
I E2
−15
14
(i) υ O1 − υ O 2 = (0.4133 − 0.3867 )(7.5) = 0.1995 V
(ii) υ O1 − υ O 2 = (0.4133)(7.6) − (0.3867 )(7.4) = 0.2795 V
______________________________________________________________________________________
11.18
(a)
⎛ −υ d ⎞
1
⎟=4
⇒ exp⎜⎜
VT ⎟⎠
⎛ −υ d ⎞
⎝
⎟⎟
1 + exp⎜⎜
⎝ VT ⎠
− υ d = (0.026 ) ln (4 ) ⇒ υ d = −36.0 mV
i C1
= 0.20 =
IQ
so that
⎛ +υd ⎞
1
⎟ = 0.1111
⇒ exp⎜⎜
VT ⎟⎠
⎛ +υd ⎞
⎝
⎟⎟
1 + exp⎜⎜
⎝ VT ⎠
so that υ d = (0.026 ) ln (0.1111) = −57.1 mV
______________________________________________________________________________________
(b)
iC 2
= 0.90 =
IQ
11.19
(a)
⎡
⎤
IQ
⎛ IQ ⎞
⎟ ⋅υ d (max )⎥ −
⎢0.5 I Q + ⎜⎜
⎟
⎝ 4VT ⎠
⎣⎢
⎦⎥ 1 + exp(− υ d (max ) VT )
⎡
⎤
⎛ IQ ⎞
⎟ ⋅υ d (max )⎥
⎢0.5 I Q + ⎜⎜
⎟
⎢⎣
⎥⎦
⎝ 4VT ⎠
⎡
⎛
⎞
⎤
⎛
⎞
⎤
(0.995)⎢0.5 + ⎜⎜ 1 ⎟⎟ ⋅υ d (max )⎥ =
= 0.005
1
⎢⎣
⎥⎦ 1 + exp(− υ d (max ) VT )
⎝ 4VT ⎠
By trial and error, υ d (max ) ≅ 14 mV
(b)
⎡
(0.985)⎢0.5 + ⎜⎜ 1 ⎟⎟ ⋅υ d (max )⎥ =
⎝ 4VT ⎠
1
1 + exp(− υ d (max ) VT )
⎣⎢
⎦⎥
(
)
By trial and error, υ d max ≅ 21.2 mV
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.20
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
I = 0.5 mA.
Assume the bias is ±10 V , and Q
Then I C = 0.25 mA
Now
10 − 4.5
⇒ RC = 22 k Ω
0.25
RC =
In this case,
Then
Ad =
rπ =
(100 )( 0.026 )
0.25
(100 )( 22 )
2 (10.4 + 0.5 )
= 10.4 k Ω
= 101
So gain specification is met.
For CMRRdB = 80 dB ⇒
CMRR = 104 =
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.
R o = ro 1+ g m (R E1 rπ )
[
]
If V A = 100 V, then ro =
rπ =
100
= 200 k Ω
0.5
(100)(0.026) = 5.2 k Ω
0.5
0.5
gm =
= 19.23 mA/V
0.026
Then 1030 = 200 1 + (19.23)(R E1 rπ ) ⇒ R E1 rπ = 0.216 k Ω ⇒ R E1 5.2 = 0.216
[
]
So, R E1 = 225 Ω ; also I REF ≅ 0.5 mA
______________________________________________________________________________________
11.21
(a)
RE =
−0.7 − ( −10 )
0.25
⇒ RE = 37.2 k Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
⎛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
⎝ rπ + RB ⎠
Vπ 2 = V2 − Ve
Then
⎛ 1 + β ⎞ ⎡ rπ
⎤ V
(V1 − Ve ) + (V2 − Ve )⎥ = e
⎟⎢
⎝ rπ ⎠ ⎣ rπ + RB
⎦ RE
(1) ⎜
From this, we find
r + RB
⋅ V2
V1 + π
rπ
Ve =
⎡ rπ + RB
r + RB ⎤
+1+ π
⎢
⎥
rπ ⎦
⎣ RE (1 + β )
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
Vd ⎛ ⎛ 25 + 0.5 ⎞ ⎞
1− ⎜
⎟
2 ⎜⎝ ⎝ 25 ⎠ ⎟⎠
Vd
( −0.02 )
= 2
Ve =
2.026
⎡ 25 + 0.5
25 + 0.5 ⎤
+1+
⎢
⎥
25 ⎦
⎣ ( 37.2 )(121)
So
Ve = −0.00494Vd
Now
V
⎛ V
⎞
Vo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 ) Vd ⎟ ⇒ Ad = o = 119
V
2
⎝
⎠
d
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.22
(a) Ad = g m RC
1.2
= 75
0.016
0.125
gm =
= 4.808 mA/V
0.026
75
Then RC =
= 15.6 k Ω
4.808
(b) For VCB = 0 ⇒ υ C1 = υ C 2 = υ CM = 3 − (0.125)(15.6 ) = 1.05 V
Ad =
1 ⎡ (0.25)(4000 ) ⎤
= 19,231 ⇒ CMRR dB = 85.7 dB
⎢1 +
(0.026) ⎥⎦
2⎣
______________________________________________________________________________________
(c) CMRR =
11.23
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
r
⎝ π
⎠
(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
1 ⎞
v01 ⎜
+
⎟ − v02 ⎜ ⎟ = g m ( v2 − v1 )
R
R
L ⎠
⎝ RL ⎠ 2
⎝ C
(1)
At the v02 node:
v02 v02 − v01
+
+ g mVπ 2 = 0
RC
RL
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 1
⎛ 1 ⎞ 1
1 ⎞
v02 ⎜
+
⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 )
R
R
L ⎠
⎝ RL ⎠ 2
⎝ C
(2)
From (1):
⎛ R ⎞ 1
v02 = v01 ⎜1 + L ⎟ − g m RL ( v2 − v1 )
⎝ RC ⎠ 2
Substituting into (2)
⎛ R ⎞⎛ 1
⎛ 1
⎛ 1 ⎞ 1
1 ⎞ 1
1 ⎞
v01 ⎜1 + L ⎟ ⎜
+
+
⎟ − g m RL ( v2 − v1/ ) ⎜
⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 )
⎝ RL ⎠ 2
⎝ RC ⎠ ⎝ RC RL ⎠ 2
⎝ RC RL ⎠
⎡ ⎛ RL
⎛ 1 RL
⎞⎤
1 ⎞ 1
v01 ⎜
+ 2+
+ 1⎟ ⎥
⎟ = g m ( v1 − v2 ) ⎢1 − ⎜
R
R
R
2
R
C ⎠
C
⎠⎦
⎝ C
⎣ ⎝ C
v01 ⎛
RL ⎞
1 ⎛ RL ⎞
⎜2+
⎟ = − gm ⎜
⎟ ( v1 − v2 )
2 ⎝ RC ⎠
RC ⎝
RC ⎠
For v1 − v2 = vd
1
− g m RL
v01
Av1 =
= 2
vd ⎛
RL ⎞
⎜2+
⎟
R
C ⎠
⎝
From symmetry:
Av =
1
g m RL
v02
Av 2 =
= 2
vd ⎛
RL ⎞
⎜2+
⎟
R
C ⎠
⎝
v02 − v01
g m RL
=
vd
⎛
RL ⎞
⎜2+
⎟
RC ⎠
⎝
Then
______________________________________________________________________________________
11.24
The small-signal equivalent circuit is
KVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
KCL equation:
V
Vπ 1
+ g m Vπ 1 + g m Vπ 2 + π 2 = 0
rπ
rπ
⎛
⎞
⎝ rπ
⎠
(Vπ 1 + Vπ 2 )⎜⎜ 1 + g m ⎟⎟ = 0 ⇒ Vπ 1 = −Vπ 2
Then υ1 − υ 2 = −2Vπ 2 , or Vπ 2 = −
Now υ o = − g mVπ 2 (RC R L )
1
(υ1 − υ 2 )
2
1
g m (RC RL )(υ1 − υ 2 )
2
υ
1
For υ1 − υ 2 ≡ υ d ⇒ Ad = o = g m (RC R L )
υd 2
______________________________________________________________________________________
=
11.25
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 )
Ad =
Differential gain
V0
= g m RC
Vb1 − Vb 2
A =0
Common-mode gain cm
______________________________________________________________________________________
11.26
(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.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
100
0.2
= 500 k Ω , g m =
= 7.69 mA/V
0.2
0.026
(150)(0.026) = 19.5 k Ω
rπ =
0.2
So 1450 = 500[1 + (7.69)R E′ ] ⇒ R E′ = 0.247 k Ω
Now ro =
Now R E′ = R E rπ ⇒ 0.247 = R E 19.5 ⇒ R E = 250 Ω
⎛I
⎞
We have I Q R E = VT ln⎜ REF ⎟
⎜ IQ ⎟
⎝
⎠
I
(0.2)(0.250) = (0.026) ln ⎡⎢ REF ⎤⎥ ⇒ I REF = 1.37 mA
⎣ (0.2) ⎦
10 − 0.7 − (− 10 )
⇒ R1 = 14.1 k Ω
1.37
______________________________________________________________________________________
Then R1 =
11.27
R A ⎞ + ⎡ R(1 + δ ) ⎤ + ⎛ 1 + δ ⎞ +
⎟⎟ ⋅ V = ⎢
⎟ ⋅V
⎥ ⋅V = ⎜
⎝ 2+δ ⎠
⎣ R(1 + δ ) + R ⎦
⎝ RA + R ⎠
⎛
υ A = ⎜⎜
1
2
υ B = V + = 2.5 V
⎛ 1.01 ⎞
For δ = +0.01 , υ A = ⎜
⎟(5) = 2.5124 V
⎝ 2.01 ⎠
⎛ 0.99 ⎞
For δ = −0.01 , υ A = ⎜
⎟(5) = 2.4874 V
⎝ 1.99 ⎠
υ d ≅ 12.5 mV
R B ≅ R R = 20 k Ω
rπ =
β VT
I CQ
=
(120)(0.026) = 31.2 k Ω
0.1
(120)(15) = 17.58
2(rπ + R B ) 2(31.2 + 20 )
υ o = Ad ⋅υ d = (17.58)(0.0125) = 0.22 V
Ad =
β RC
=
−0.22 ≤ υ O 2 ≤ +0.22 V
______________________________________________________________________________________
11.28
(a) Rid = 2rπ =
(b) Ricm =
2 β VT 2(120 )(0.026 )
=
= 49.9 k Ω
I CQ
0.125
1
[rπ + (1 + β )(2 Ro )] = 1 ⎡⎢ (120)(0.026) + (121)(2)(8000)⎤⎥
2 ⎣ 0.125
2
⎦
Ricm = 968 M Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.29
(a)
I1 =
R2 =
(b)
10 − 0.7 − ( −10 )
R1
= 0.5 ⇒ R1 = 38.6 K
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 )
= 33.4 K
rπ 4 =
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
− ( 2.692 )( 40 )
2 (181)(1614 )
1+
66.86
Acm = −0.0123
= 66.86 K
Acm =
______________________________________________________________________________________
11.30
Ad 1 = g m1 ( R1 rπ 3 )
g m1 =
rπ 3 =
Ad 2 =
I Q1 / 2
VT
β VT
IQ 2 / 2
= 19.23I Q1
=
2 (100 )( 0.026 )
IQ 2
=
5.2
IQ 2
IQ2 / 2
g m 3 R2
, g m3 =
= 19.23I Q 2
2
VT
Then 30 =
Maximum
(19.23) IQ 2
⋅ R2 ⇒ I Q 2 R2 = 3.12 V
2
vo 2 − vo1 = ±18 mV
for linearity
vo 3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 V
so I Q 2 R2 = 3.12 V is OK.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From Ad 1 :
⎡
⎛ 5.2 ⎞ ⎤
⎟ ⎥
⎢ R1 ⎜
⎜ I Q2 ⎟ ⎥
⎢
⎠
⎝
20 = 19.23I Q1 (R1 rπ 3 ) = 19.23I Q1 ⎢
⎥
⎞⎥
⎛
5
.
2
⎢
⎟
⎜
⎢ R1 + ⎜ I ⎟ ⎥
⎝ Q2 ⎠ ⎦
⎣
19.23I Q1 R1 (5.2 )
20 =
I Q 2 R1 + 5.2
Let
I Q1
2
⋅ R1 = 5 V, ⇒ I Q1 R1 = 10 V
Then 20 =
19.23(10)(5.2)
⇒ I Q 2 R1 = 44.8 V
I Q 2 R1 + 5.2
Now I Q1 R1 = 10 ⇒ R1 =
10
I Q1
⎛ 10 ⎞
I
⎟ = 44.8 ⇒ Q 2 = 4.48
So I Q 2 ⎜
⎜ I Q1 ⎟
I Q1
⎝
⎠
Let I Q1 = 100 μ A, then I Q 2 = 448 μ A
Then
I Q 2 R 2 = 3.12 ⇒ R 2 = 6.96 k Ω
I Q1 R1 = 10 ⇒ R1 = 100 k Ω
______________________________________________________________________________________
11.31
a.
20 − VGS 3
2
= 0.25 (VGS 3 − 2 )
50
20 − VGS 3 = 12.5 (VGS2 3 − 4VGS 3 + 4 )
I1 =
12.5VGS2 3 − 49VGS 3 + 30 = 0
VGS 3 =
49 ±
( 49 ) − 4 (12.5)( 30 )
⇒ VGS 3 = 3.16 V
2 (12.5 )
2
20 − 3.16
⇒ I1 = I Q = 0.337 mA
50
IQ
I D1 =
⇒ I D1 = 0.168 mA
2
I1 =
0.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS 1 = 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
(c)
Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS 1 − 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 ) + VGS 1 = 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
______________________________________________________________________________________
11.32
2
(a) I D1 = K n1 (VGS1 − VTN )
60 = 100(VGS1 − 0.3) ⇒ VGS1 = 1.075 V
2
V D1 = V DS1 − VGS1 + V1 = 4 − 1.075 − 1.15 = 1.775 V
3 − 1.775
= 20.4 k Ω
0.06
(i) I Q = I 1 = I D1 + I D 2 = 120 μ A
RD =
I 1 = K 3 (VGS 3 − VTN )
2
120 = 200(VGS 3 − 0.3) ⇒ VGS 3 = 1.075 V
2
3 − 1.075 − (− 3)
= 41 k Ω
0.12
(iii) VGS1 = VGS 4 = 1.075 V
(ii) R1 =
(b) ro =
1
1
=
= 833.3 k Ω
λ 4 I Q (0.01)(0.12)
ΔV DS 4
2.3
=
⇒ ΔI Q = 2.76 μ A
ro
833.3
______________________________________________________________________________________
ΔI Q =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.33
(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
RD =
= 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V
0.08
VGS =
(c)
VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 V
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)
______________________________________________________________________________________
11.34
(a) υ CM (max ) = υ D 2 − υ DS 2 (sat ) + υ GS 2
⎛ k ′ ⎞⎛ W ⎞
2
i D 2 = ⎜⎜ n ⎟⎟⎜ ⎟(υ GS 2 − VTN )
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
0.09 = ⎜
⎟(4 )(υ GS 2 − 0.35) ⇒ υ GS 2 = 1.021 V
2
⎠
⎝
υ DS 2 (sat ) = 1.0208 − 0.35 = 0.6708 V
Then 2.25 = υ D 2 − 0.6708 + 1.021 ⇒ υ D 2 = 1.90 V
3 − 1.90
= 12.2 k Ω
0.09
(b) (i) υ D 2 = 1.90 V
RD =
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
(ii) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(4 )(0.09 ) = 0.2683 mA/V
2
L
⎝ 2 ⎠
⎝ ⎠⎝ ⎠
υ D 2 = 1.90 + g m ⋅
υd
2
⎛ 0.12 ⎞
⋅ R D = 1.90 + (0.2683)⎜
⎟(12.2) = 2.096 V
⎝ 2 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
υd
⎛ 0.05 ⎞
⋅ R D = 1.90 − (0.2683)⎜
⎟(12.2) = 1.818 V
2
⎝ 2 ⎠
______________________________________________________________________________________
(iii) υ D 2 = 1.90 − g m ⋅
11.35
2
(a) I Q = I 1 = K n3 (VGS 3 − VTN )
0.25 = 0.20(VGS 3 − 0.4) ⇒ VGS 3 = 1.518 V
2
5 − 1.518 − (− 5)
= 33.9 k Ω
0.25
0.25
(b) I D1 = I D 2 =
= 0.125 mA
2
υ O1 − υ O 2 = V + − I D1 R D1 − V + − I D 2 R D 2 = I D 2 R D 2 − I D1 R D1
(i) υ O1 − υ O 2 = 0
(ii ) υ O1 − υ O 2 = (0.125)(15.5 − 14.5) = 0.125 V
R1 =
(
) (
)
(c) I Q = I D1 + I D 2 = (K n1 + K n 2 )(VGS − VTN )
2
0.25 = (0.125 + 0.115)(VGS − VTN ) ⇒ (VGS − VTN ) = 1.04166
I D1 = (0.125)(1.04166) = 0.1302 mA
I D 2 = (0.115)(1.04166) = 0.1198 mA
(i) υ O1 − υ O 2 = (15)(0.1198 − 0.1302 ) = −0.156 V
(ii) υ O1 − υ O 2 = (0.1198)(15.5) − (0.1302 )(14.5) = −0.031 V
______________________________________________________________________________________
2
2
11.36
(a)
⎛ K ⎞
Kn
i D1 1
= +
⋅υ d 1 − ⎜ n ⎟ ⋅υ d2
⎜ 2I Q ⎟
IQ 2
2I Q
⎝
⎠
0.20 = 0.50 +
⎛ 0.20 ⎞ 2
0.20
⎟⎟ ⋅υ d
⋅υ d 1 − ⎜⎜
2(0.15)
⎝ 2(0.15) ⎠
− 0.3674 = υ d 1 − (0.6667 )υ d2
[
0.135 = υ d2 1 − (0.6667 )υ d2
]
0.6667υ − υ + 0.135 = 0 ⇒ υ d2 = 0.15 ⇒ υ d = −0.3873 V
4
d
2
d
(b) 0.80 = 0.50 −
⎛ 0.20 ⎞ 2
0.20
⎟⎟υ d
⋅υ d 1 − ⎜⎜
2(0.15)
⎝ 2(0.15) ⎠
0.3674 = −υ d 1 − (0.6667 )υ d2 ⇒ υ d = −0.3873 V
(c) υ d = υ d , max =
IQ
Kn
=
0.15
= +0.866 V
0.20
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.37
i D1
IQ
Lin
= 0.5 +
g f υd
IQ
, Now g f =
Kn IQ
2
=
(0.2)(0.10) = 0.10 mA/V
2
(0.1) ⋅υ = 0.5 + υ
d
(0.1) d
Kn
(0.2) = 1
=
We have
2 I Q 2(0.1)
= 0.5 +
0.5 + υ d − ⎡0.5 + υ d 1 − υ d2 ⎤
⎥⎦
⎢⎣
= 0.005
(a)
0.5 + υ d
0.0025 = υ d ⎡0.995 − 1 − υ d2 ⎤ , ⇒ υ d ≅ 0.19 V
⎢⎣
⎥⎦
0.5 + υ d − ⎡0.5 + υ d 1 − υ d2 ⎤
⎢⎣
⎥⎦
= 0.015
(b)
0.5 + υ d
0.0075 = υ d ⎡0.985 − 1 − υ d2 ⎤ , ⇒ υ d ≅ 0.285 V
⎢⎣
⎥⎦
______________________________________________________________________________________
11.38
(b)
g m = 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
vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 V
ΔI = g m
vD1 = −1.063 − 4.565 = −5.628 V
9 = I S RS + VSG + 1
IS = 2I D
8 = 2 K P RS (VSG + VTP ) + VSG
2
8 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG
2
2
8 = 39 (VSG
− 1.6VSG + 0.64 ) + VSG
2
39VSG
− 61.4VSG + 16.96 = 0
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
V
ΔI D = g m ⋅ d = ( 0.0417 )( 0.05 ) = 0.002085 mA
2
ΔvD = ( 0.002085 )( 510 ) = 1.063 V
ΔvD = ΔI D ⋅ RD
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
v1 ↑, I D1 ↓, vD1 ↓
vD1 = −4.565 − 1.063 = −5.628 V
vD 2 = −4.565 + 1.063 = −3.502 V
______________________________________________________________________________________
11.39
(a)
v1 = v2 = 0
I D = K n (VSG + VTP )
2
ID = 6 μA
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
gm = 2 Kn I D
= 2 ( 30 )( 6 ) = 26.83 μ A/V
Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66
Acm = 0
(ii)
g m RD ( 26.83)( 0.36 )
=
⇒ 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 )
Ad =
______________________________________________________________________________________
11.40
(a) g m = 2 K p I DQ = 2 (0.12 )(0.075) = 0.1897 mA/V
(i) υ D1 − υ D 2 = g m R Dυ d = (0.1897 )(30 )(0.1) = 0.5692 V
(ii) υ D1 − υ D 2 = g m R Dυ d = (0.1897 )(30 )(0.2) = 1.138 V
(b) (i) Δυ D 2 = g m R D ⋅
(ii) Acm =
υd
2
= (0.1897 )(30 )(0.1) = 0.5692 V
− g m RD
− (0.1897 )(30 )
=
= −0.003748
1 + 2 g m R o 1 + 2(0.1897 )(4000 )
g m R D (0.1897 )(30 )
=
= 2.846
2
2
Δυ D 2 = Ad υ d + Acmυ cm , let υ cm = 1 V, υ d = 0.2 V
Ad =
Then Δυ D 2 = (2.846 )(0.2) − (0.003748)(1) = 0.5655 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.41
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
g m = 2 K n 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 )
ΔR ⎞
⎛
− g m ⎜ RD + D ⎟ Vcm
2 ⎠
⎝
vD1 =
1 + g m ( 2 RS )
ΔR ⎞
⎛
− gm ⎜ RD − D ⎟ Vcm
2 ⎠
⎝
vD 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 )
= −0.003216
C M R R ∫ = 69.1 dB
bB
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.42
From 11.41
I D1 = I D 2 = 55.9 μ A
g m = 0.183 mA/V
⎛ +V ⎞
ΔvD 2 = + g m 2 ⎜ d ⎟ RD
⎝ 2 ⎠
Vd
Vd
vO = ΔvD1 − ΔvD 2 = − g m1
RD − g m 2
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 : ΔvD1 = − g m1
Vd
⋅ RD
2
Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15
Δg ⎞
Δg ⎞
⎛
⎛
− ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM
2
2 ⎠
⎝
⎠
⎝
+
ACM : vO = ΔvD1 − ΔvD 2 =
1 + g m ( 2 RS )
1 + g m ( 2 RS )
Acm =
Acm =
−Δg m RD
vO
=
vcm 1 + g m ( 2 RS )
− ( 0.00183) ( 50 )
1 + ( 0.183)( 2 ) ( 75 )
Δg m = ( 0.01) ( 0.183) = 0.00183
= −0.003216
C M R R ∫ = 69.1 dB
dB
______________________________________________________________________________________
11.43
2
(a) 5 = I S R S + V SG = 2 K p (V SG + VTP ) R S + V SG
(
)
2
5 = 2(1.2 )(2 ) V SG
− 1.2V SG + 0.36 + V SG
4.8V
2
SG
− 4.76V SG − 3.272 = 0 ⇒ V SG = 1.459 V = υ S
I S = 2(1.2)(1.459 − 0.6) = 1.77 mA
2
I D1 = I D 2 = 0.885 mA
υ D1 = υ D 2 = I D R D − 5 = (0.885)(1) − 5 = −4.115 V
(
)
2
− 1.2V SG + 0.36 + V SG + 1
(b) 5 = 2(1.2)(2 ) V SG
4.8V
2
SG
− 4.76V SG − 2.272 = 0 ⇒ V SG = 1.344 V
υ S = V SG + 1 = 2.344 V
I D1 = I D 2 = (1.2)(1.344 − 0.6 ) = 0.664 mA
2
υ D1 = υ D 2 = (0.664 )(1) − 5 = −4.336 V
(c) Using part (a), g m = 2 (1.2 )(0.885) = 2.061 mA/V
υd
= −4.115 + (2.061)(1)(0.1) = −3.909 V
2
υ D 2 = −4.115 − (2.061)(1)(0.1) = −4.321 V
υ D1 = −4.115 + g m R D ⋅
(d) Using part (b), g m = 2 (1.2 )(0.664) = 1.785 mA/V
υd
= −4.336 + (1.785)(1)(0.1) = −4.158 V
2
υ D 2 = −4.336 − (1.785)(1)(0.1) = −4.515 V
______________________________________________________________________________________
υ D1 = −4.336 + g m R D ⋅
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.44
(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
ID
1
+ VTN =
+ 0.8 = 2.38 V
Kn
0.4
VGS =
VDS ( sat ) = 2.38 − 0.8 = 1.58
So vcm = vD − VDS ( sat ) + VGS
= 2.09 − 1.58 + 2.38
vcm = 2.89 V
______________________________________________________________________________________
11.45
g m RD
2
For vCM = 2.5 V
Ad =
I D1 = I D 2 =
IQ
2
= 0.25 mA
Let VD1 = VD 2 = 3 V , then RD =
Then 100 =
g m ( 28 )
10 − 3
⇒ RD = 28 k Ω
0.25
⇒ g m = 7.14 mA / V
2
k′ ⎛ W ⎞
And g m = 2 n ⎜ ⎟ I D
2⎝L⎠
⎛ 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. (11.82(b))
ACM =
g m RD
1 + 2 g m Ro
So 0.10 =
( 7.14 )( 28 )
⇒ Ro = 140 k Ω
1 + 2 ( 7.14 ) Ro
For the basic 2-transistor current source
Ro = ro =
1
1
=
= 200 k Ω
λ I Q ( 0.01)( 0.5 )
This current source is adequate to meet common-mode gain requirement.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.46
Not in detail, Approximation looks good.
a.
IS =
−VGS 1 − ( −5 )
RS
and I S = 2 I D = 2 K n (VGS1 − VTN )
5 − VGS 1
2
= 2 ( 0.050 )(VGS 1 − 1)
20
5 − VGS1 = 2 (VGS2 1 − 2VGS 1 + 1)
2VGS2 1 − 3VGS 1 − 3 = 0
VGS 1 =
3±
( 3) + 4 ( 2 )( 3)
⇒ VGS 1 = 2.186 V
2 ( 2)
2
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
r0 =
1
λ I DQ
=
1
( 0.02 )( 0.0704 )
= 710 kΩ
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
R
r
r0
⎝ D 0⎠
(1)
v02
v − VS
+ g mVgs 2 + 02
=0
RD
r0
⎛ 1
V
1⎞
+ ⎟ + g m ( v2 − VS ) − S = 0
v02 ⎜
R
r
r0
⎝ D 0⎠
(2)
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
v − V v − VS
V
+ g mVgs 2 = S
g mVgs1 + 01 S + 02
r0
r0
RS
v
v
V
2V
g m ( v1 − VS ) + 01 + 02 − S + g m ( v2 − VS ) = S
r0
r0
r0
RS
⎧
v
v
2 1 ⎫
g m ( v1 + v2 ) + 01 + 02 = VS ⎨2 g m + + ⎬
r0
r0
r0 RS ⎭
⎩
(3)
From (1)
⎛
1⎞
VS ⎜ g m + ⎟ − g m v1
r0 ⎠
v01 = ⎝
⎛ 1
1⎞
+ ⎟
⎜
⎝ RD r0 ⎠
Then
⎛
1⎞
VS ⎜ g m + ⎟ − g m v1
r0 ⎠
⎧
v
2 1 ⎫
⎝
+ 02 = VS ⎨2 g m + + ⎬
g m ( v1 + v2 ) +
r0
r
RS ⎭
⎛ 1
1⎞
0
⎩
+ ⎟
r0 ⎜
⎝ RD r0 ⎠
(3)
⎛ 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
⎪
D ⎠
0
S ⎠⎝
D ⎠ ⎝
0 ⎠⎭
⎝
⎝ D 0⎠
⎩⎪⎝
⎛ 1
⎧
⎛
r ⎞
r
r
r
1⎞
2 1
2
1⎫
g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜
+ ⎟ = VS ⎨2 g m + +
+ 2 gm ⋅ 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 )⎪⎬
⎜1 +
⎟+
R
R
R
r
r
R
R
R
⎪⎩
⎪⎭ (4)
⎝
0
D
D ⎠
S ⎝
D ⎠
D
⎝ D 0⎠
⎛ 1
⎛
1⎞
1⎞
v02 ⎜
+ ⎟ + g m v2 = VS ⎜ g m + ⎟
R
r
r
0 ⎠
⎝
Then substituting into (2), ⎝ D 0 ⎠
( 0.119 ) ⎡⎢v1
⎣
710
710 ⎤
1 ⎤
⎡1
+ v2 + v2
+ v02 ⎢ +
⎥
25
25 ⎥⎦
⎣ 25 710 ⎦
⎧
⎫
1
1 ⎛ 710 ⎞ 2
= VS ⎨0.119 +
+ ⎜1 +
⎟ + ⎡⎣1 + ( 0.119 )( 710 ) ⎤⎦ ⎬
710 20 ⎝
25 ⎠ 25
⎩
⎭
Substitute numbers:
( 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
⎛
Then v02 ⎜ +
⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 +
⎟
710 ⎠
⎝ 25 710 ⎠
⎝
(2)
(4)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
v
v1 = vcm + d
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
⎛ 1.437 ⎞
C M R RdB = 20 log10 ⎜
⎟ ⇒ C M R RdB = 9.03 dB
⎝ 0.508 ⎠
______________________________________________________________________________________
11.47
KVL:
v1 = Vgs1 − Vgs 2 + v2
So v1 − v2 = Vgs1 − Vgs 2
KCL:
g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2
So Vgs1 =
1
1
( v1 − v2 ) , Vgs 2 = − ( v1 − v2 )
2
2
Now
v02 v02 − v01
+
= − g mVgs 2
RD
RL
⎛ 1
1 ⎞ v01
= v02 ⎜
+
⎟−
R
R
RL
L ⎠
⎝ D
v01 v01 − v02
+
= − g mVgs1
RD
RL
⎛ 1
1 ⎞ v02
= v01 ⎜
+
⎟−
R
R
RL
L ⎠
⎝ D
⎛
R ⎞
v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2
R
⎝
D ⎠
From (1):
(1)
(2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 ⎠⎝ D
L ⎠
L ⎠
⎛ 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 ⎞
gm ⎜
= 2
⎟ ( v1 − v2 ) =
⎜2+
⎟ ⇒ Ad 2 =
2 ⎝ RD ⎠
RD ⎝
RD ⎠
v1 − v2 ⎛
RL ⎞
⎜2+
⎟
R
⎝
D ⎠
1
− ⋅ g m RL
v01
= 2
Ad 1 =
v1 − v2 ⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
From symmetry
Av =
v02 − v01
g m RL
=
v1 − v2
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
Then
______________________________________________________________________________________
11.48
⎛ −υ ⎞
(a) υ o = − g m (R D R L )⎜⎜ d ⎟⎟
⎝ 2 ⎠
1
Ad = g m (R D R L )
2
(b) υ o = − g m (R D R L )υ gs
υ cm = υ gs + g mυ gs (2 Ro )
υ gs =
Acm =
υ cm
1+ 2 g m R o
− g m (R D R L )
1+ 2 g m R o
______________________________________________________________________________________
11.49
(a) Ad =
Δυ o 0.5
=
=5
Δυ d 0.1
g m RD
, g m = 2 K n I DQ = 2 (0.15)(0.1) = 0.2449 mA/V
2
2(5)
RD =
= 40.8 k Ω
0.2449
(b) υ CM (max ) = υ O − υ DS (sat ) + υ GS
Ad = 5 =
i D = K n (υ GS − VTN )
2
0.1 = 0.15(υ GS − 0.4 ) ⇒ υ GS = 1.216 V
2
υ DS (sat ) = 1.216 − 0.4 = 0.8165 V
υ O = 5 − (0.1)(40.8) = 0.92 V
υ CM (max ) = 0.92 − 0.8165 + 1.216 = 1.32 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.50
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
______________________________________________________________________________________
11.51
(a) υ CM = υ O − υ DS (sat ) + υ GS
i D = K n (υ GS − VTN )
2
⎛ 0.1 ⎞
2
0.1 = ⎜
⎟(10 )(υ GS − 0.4 ) ⇒ υ GS = 0.8472 V
⎝ 2 ⎠
υ DS (sat ) = 0.8472 − 0.4 = 0.4472 V
Then υ O = 1.5 + 0.4472 − 0.8472 = 1.1 V
υ O = 3 − (0.1)R D = 1.1 ⇒ R D = 19 k Ω
Now Ad =
Ad =
g m RD
⎛ 0.1 ⎞
, where g m = 2 ⎜
⎟(10 )(0.1) = 0.4472 mA/V
2
⎝ 2 ⎠
(0.4472)(19) = 4.248
2
CMRR dB = 50 dB ⇒ CMRR = 316.2
CMRR =
[
]
⎤
1
1⎡
⎛ 0.1 ⎞
1 + 2 K n I Q ⋅ R o ⇒ 316.2 = ⎢1 + 2 ⎜
⎟(10 )(0.2) ⋅ R o ⎥ ⇒ Ro = 998 k Ω
2
2 ⎢⎣
⎝ 2 ⎠
⎥⎦
(b) Use cascode current source similar to Figure 10.18 with υ DS 2 (sat ) = 0.3 V.
______________________________________________________________________________________
11.52
(a) From Problem 11.27, υ d ≅ 12.5 mV.
g R
Ad = m D , where g m = 2 (1)(0.1) = 0.6325 mA/V
2
(
0.6325)(20 )
Ad =
= 6.325
2
υ O = (6.325)(0.0125) = 0.0791 V
So −0.0791 ≤ υ O ≤ 0.0791 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.53
From previous results
Ad 1 =
vo 2 − vo1
= g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20
v1 − v2
and Ad 2 =
Set
I Q1 R1
2
vo 3
1
1
2 K n 3 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
2
⎛ 0.06 ⎞⎛ W ⎞
⎛ 20 ⎞
⎛W ⎞ ⎛W ⎞
Then 2 ⎜
⎟⎜ ⎟ ( 0.1) = ⎜
⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67
⎝ 2 ⎠⎝ L ⎠1
⎝ 100 ⎠
⎝ L ⎠1 ⎝ L ⎠ 2
⎛ 2 ( 30 ) ⎞
⎛ 0.060 ⎞⎛ W ⎞
⎛W ⎞ ⎛W ⎞
and 2 ⎜
⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240
⎟⎜ ⎟ ( 0.1) = ⎜
50
⎝ 2 ⎠⎝ L ⎠3
⎝ L ⎠3 ⎝ L ⎠ 4
⎝
⎠
2
______________________________________________________________________________________
11.54
a.
⎛ v ⎞
iD1 = I DSS ⎜ 1 − GS1 ⎟
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
VP
( vGS 2 − vGS1 )
I DSS
=−
VP
⋅ vd =
I DSS
( −VP )
⋅ vd
iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1
( i − I − i ) = ( −IV ) ⋅ v
2
D1
DSS
D1
Q
2
2
d
P
iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) =
Then iD1 ( I Q − iD1 ) =
I DSS
( −VP )
2
⋅ vd2
⎤
I
1⎡
⎢ I Q − DSS 2 ⋅ vd2 ⎥
2 ⎢⎣
( −VP ) ⎦⎥
Square both sides
2
⎤
I
1⎡
iD2 1 − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd2 ⎥ = 0
4 ⎣⎢
( −VP ) ⎦⎥
⎤
I
⎛1⎞⎡
I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd2 ⎥
4
⎝ ⎠ ⎢⎣
( −VP ) ⎥⎦
2
2
Q
iD1 =
2
2
I Q 1 2 ⎡ 2 2 I Q I DSS vd2 ⎛ I DSS vd2 ⎞ ⎤
⎢
⎟ ⎥
±
+⎜
iD1 =
IQ − IQ −
2
⎜ ( −V )2 ⎟ ⎥
⎢
2 2
−VP )
(
P
⎝
⎠ ⎦
⎣
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Use + sign
⎞
1 2 I Q I DSS 2 ⎛ I DSS
iD1 =
+
⋅ vd − ⎜
⋅ vd2 ⎟
2
2
⎜ ( −V )
⎟
2 2 ( −VP )
P
⎝
⎠
2
IQ
iD1 =
IQ
2
+
2
2
2
2
2
2
2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞
1 IQ
vd
−⎜
⎜ I Q ⎟⎟ ⎜ VP ⎟
2 ( −VP )
IQ
⎝
⎠ ⎝ ⎠
Or
2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞
iD1 1 ⎛ 1 ⎞
= +⎜
−⎜
⎟ ⋅ vd
⎜ I ⎟⎟ ⎜ V ⎟
I Q 2 ⎝ −2VP ⎠
IQ
⎝ Q ⎠ ⎝ P⎠
We had
iD 2 = I Q − iD1
Then
2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞
iD 2 1 ⎛ 1 ⎞
−⎜
= −⎜
⎟ ⋅ vd
⎜ I ⎟⎟ ⎜ V ⎟
IQ
I Q 2 ⎝ −2VP ⎠
⎝ Q ⎠ ⎝ P⎠
b.
If iD1 = I Q , then
2
1=
2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞
1 ⎛ 1 ⎞
+⎜
−⎜
⎟ ⋅ vd
⎜ I ⎟⎟ ⎜ V ⎟
2 ⎝ −2VP ⎠
IQ
⎝ Q ⎠ ⎝ P⎠
2
2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞
−⎜
⎜ I Q ⎟⎟ ⎜ VP ⎟
IQ
⎝
⎠ ⎝ ⎠
VP = vd
2
2
Square both sides
VP
2
2
2
⎡ 2I
⎛ I DSS ⎞ ⎛ vd ⎞ ⎤
DSS
⎥
=v ⎢
−⎜
⎟
⎜
⎟
⎜ I Q ⎟ VP ⎥
⎢ IQ
⎝
⎠ ⎝ ⎠ ⎦
⎣
2
d
2
2
⎛ I DSS ⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2
2
⋅ vd + VP = 0
⎜⎜
⎟⎟ ⎜ ⎟ ( vd ) −
IQ
⎝ I Q ⎠ ⎝ VP ⎠
vd2 =
2 I DSS
±
IQ
2
2
2
⎛ 2 I DSS ⎞
⎛I ⎞ ⎛ 1 ⎞
2
⎜⎜
⎟⎟ − 4 ⎜⎜ DSS ⎟⎟ ⎜ ⎟ (VP )
I
I
V
⎝ Q ⎠
⎝ Q ⎠ ⎝ P⎠
2
⎛ 2I ⎞ ⎛ 1 ⎞
2 ⎜ DSS ⎟ ⎜ ⎟
⎜ I Q ⎟ VP
⎝
⎠ ⎝ ⎠
2
2 ⎛ IQ ⎞
vd2 = (VP ) ⎜
⎟
⎝ I DSS ⎠
1/ 2
⎛ IQ ⎞
Or vd = VP ⎜
⎟
⎝ I DSS ⎠
c.
For vd small,
iD1 ≈
gf =
IQ
2
2 I DSS
1 IQ
+ ⋅
⋅ vd
IQ
2 ( −VP )
2 I DSS
diD1
1 IQ
⋅
⋅
vd → 0 =
d vd
IQ
2 ( −VP )
⎛ 1 ⎞ I Q I DSS
Or ⇒ g f ( max ) = ⎜
⎟
2
⎝ −VP ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.55
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⎞
g f ( max ) = ⎜ ⎟
⎝ 4⎠
(1)( 2 )
2
⇒ g f ( max ) = 0.25 mA/V
c.
g m RD
= g f ( max ) ⋅ RD
2
Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5
Ad =
______________________________________________________________________________________
11.56
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
⎝
⎠
8VGS2 + 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)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.57
Equivalent circuit and analysis is identical to that in problem 11.47.
1
⋅ g m RL
Ad 2 = 2
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
1
− ⋅ g m RL
Ad 1 = 2
⎛
RL ⎞
⎜2+
⎟
R
D ⎠
⎝
Av =
v02 − v01
g m RL
=
vd
⎛
RL ⎞
⎜2+
⎟
RD ⎠
⎝
______________________________________________________________________________________
11.58
(a) Ad = g m (ro 2 ro 4 )
I CQ
gm =
VT
=
0.2
= 7.692 mA/V
0.026
ro 2 =
V A2 120
=
= 600 k Ω
I CQ 0.2
ro 4 =
V A4
80
=
= 400 k Ω
I CQ 0.2
Ad = (7.692)(600 400) = 1846
(b) R o = ro 2 ro 4 = 600 400 = 240 k Ω
(
(c) Ad = g m ro 2 ro 4 R L
)
Ad = (0.75)(1846) = 1384.5 = (7.692)(240 R L ) ⇒ R L = 720 k Ω
______________________________________________________________________________________
11.59
(a)
⎛
2⎞
I Q = 250 μ A I REF = I Q ⎜1 + ⎟
β
⎝
⎠
2 ⎞
⎛
= 250 ⎜1 +
⎟ = 252.8 μ A
⎝ 180 ⎠
R1 =
5 − ( 0.7 ) − ( −5 )
0.2528
⇒ R1 = 36.8 K
(b)
Ad = g m ( ro 2 ro 4 )
Ad = ( 4.808 ) (1200 800 )
Ad = 2308
0.125
= 4.808 mA/V
0.026
150
ro 2 =
= 1200 K
0.125
100
ro 4 =
= 800 K
0.125
gm =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(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
______________________________________________________________________________________
11.60
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.
r02 = r04 =
gm =
I CQ
VA 100
=
= 1000 kΩ
I CQ 0.1
=
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
______________________________________________________________________________________
11.61
IQ
(a) Ad =
IQ
2V A 2
+
2VT
IQ
2V A4
+
1
RL
IQ
1000 =
2(0.026 )
IQ
1
+
+
2(90) 2(60) 250
IQ
( )
1000 I Q (0.005556 + 0.008333) + 4 = I Q (19.23) ⇒ I Q = 0.749 mA
(b) υ CM (max ) = V + − 2V EB (on ) = 5 − 2(0.6 ) = 3.8 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.62
⎛υ ⎞
(b) υ o1 = − g m1 (ro1 Ro 3 )⎜⎜ d ⎟⎟
⎝ 2 ⎠
υ
1
Ad 1 = o1 = − g m1 (ro1 R o3 )
υd
2
I CQ
0.2
=
= 7.692 mA/V
g m1 =
0.026
VT
ro1 =
V A1 120
=
= 600 k Ω
I CQ 0.2
R o3 = ro 3 rπ 3
1
, where g m3 = 7.692 mA/V,
g m3
rπ 3 =
Then Ro 3 = 400 15.6
(120)(0.026) = 15.6 k Ω , r
0.2
o3
= 400 k Ω
1
= 15.014 0.130 = 0.1289 k Ω
7.692
1
(7.692)(600 0.1289) = −0.4956
2
1
(c) Ad 2 = + g m 2 (ro 2 Ro 4 ) = +0.4956
2
(d) Ad 3 = + g m1 (ro1 Ro 3 ) = 2(0.4956) = 0.9912
Ad 1 = −
______________________________________________________________________________________
11.63
g m 4V sg 4 = g m3V sg 3 = g m1V gs1
V gs1 = V1 − V s , and V gs 2 = V 2 − V s
g m1V gs1 + g m1V gs 2 +
Vo − V s V s
=
ro 2
Ro
g m1 (V1 + V 2 − 2V s ) +
⎛ 1
Vo
1 ⎞
⎟
= V s ⎜⎜
+
⎟
ro 2
⎝ R o ro 2 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
g m1 (V1 + V 2 ) +
⎛
Vo
1
1 ⎞
⎟
= V s ⎜⎜ 2 g m1 +
+
ro 2
R o ro 2 ⎟⎠
⎝
Now g m1 = 2 K n I DQ = 2 (0.5)(0.1) = 0.4472 mA/V
ro 2 =
1
= 500 k Ω
(0.02)(0.1)
Then (0.4472 )(V1 + V 2 ) +
Vo
1
1 ⎤
⎡
= V s ⎢2(0.4472 ) +
+
⎥ = V s (0.8969)
500
2000
500
⎣
⎦
We have (1) V s = (0.4986 )(V1 + V 2 ) + Vo (0.0022299 )
V − V s Vo
Also (2) o
+
+ g m1V gs 2 − g m1V gs1 = 0
ro 2
ro 4
⎛ 1
1 ⎞ Vs
⎟−
V o ⎜⎜
+
⎟ r + g m1 (V 2 − V s − (V1 − V s )) = 0
r
r
o4 ⎠
o2
⎝ o2
⎛ 1
1 ⎞ Vs
⎟⎟ −
+ g m1 (V2 − V1 ) = 0
Vo ⎜⎜
+
⎝ ro 2 ro 4 ⎠ ro 2
We find ro 4 =
1
(0.03)(0.1)
= 333.3 k Ω
1 ⎞ 1
⎛ 1
[(0.4986)(V1 + V2 ) + Vo (0.0022299)] + (0.4472)(V2 − V1 ) = 0
Then Vo ⎜
+
⎟−
⎝ 500 333.3 ⎠ 500
V o [(0.005) − (0.00000446 )] − (0.0009972 )(V1 + V 2 ) + (0.4472 )(V 2 − V1 ) = 0
(a) Let V1 = V d and V 2 = 0
V o [0.00499554] − V d (0.0009972 ) − V d (0.4472 ) = 0
V
Ad = o = 89.72
Vd
(b) Let V1 = 0 and V 2 = −V d
V o [0.00499554] + V d (0.0009972 ) − V d (0.4472 ) = 0
V
Ad = o = 89.32
Vd
Vd
V
and V 2 = − d
2
2
V o [0.00499554] = V d (0.4472 )
(c) Let V1 =
Vo
= 89.52
Vd
______________________________________________________________________________________
A=
11.64
a.
From symmetry.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.1
+1
0.1
VGS 3 = VGS 4 = VDS 3 = VDS 4 =
Or VDS 3 = VDS 4 = 2 V
0.1
+1 = 2 V
0.1
VSD1 = VSD 2 = VSG1 − (VDS 3 − 10 )
VSG1 = VSG 2 =
= 2 − ( 2 − 10 )
Or VSD1 = VSD 2 = 10 V
b.
r0 n =
r0 p =
1
λn I DQ
1
=
1
( 0.01)( 0.1)
⇒ 1 MΩ
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 = 80
λP I DQ
(c)
I D 2 = I D1 =
ro 4 =
ro 2 =
1
λn I D 4
1
λP I D 2
IQ
2
=
=
= 0.1 mA
1
( 0.01)( 0.1)
1
= 1000 k Ω
( 0.015)( 0.1)
= 667 k Ω
Ro = ro 2 ro 4 = 667 1000 = 400 k Ω
______________________________________________________________________________________
11.65
(a) Ad = g m (ro 2 ro 4 )
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜
⎟(8)(0.06 ) = 0.3098 mA/V
2
L
⎝ 2 ⎠
⎝ ⎠⎝ ⎠ n
1
ro 2 =
= 666.7 k Ω
(0.025)(0.06)
1
ro 4 =
= 416.7 k Ω
(0.04)(0.06)
Ad = (0.3098)(666.7 416.7 ) = 79.44
⎛ k ′p ⎞⎛ W ⎞
2
(b) i D 3 = ⎜⎜ ⎟⎟⎜ ⎟ (υ SG 3 + VTP )
L
2
⎝ ⎠⎝ ⎠ p
⎛ 0.04 ⎞
2
0.06 = ⎜
⎟(10 )(υ SG 3 − 0.3) ⇒ υ SG 3 = 0.8477 V
2
⎝
⎠
υ O = 1.8 − 0.8477 = 0.9523
υ CM (max ) = υ O − υ DS (sat ) + υ GS
⎛ 0.1 ⎞
2
0.06 = ⎜
⎟(8)(υ GS − 0.3) ⇒ υ GS = 0.6873 V, υ DS (sat ) = 0.3873 V
⎝ 2 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
υ CM (max ) = 0.9523 − 0.3873 + 0.6873 = 1.25 V
______________________________________________________________________________________
11.66
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
80
= 800 kΩ
0.1
Av = − g m ( r02 r04 )
R = 0, r04 =
= − ( 3.846 ) ⎡⎣1200 800 ⎤⎦ ⇒ Av = −1846
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
R = ( 3.763 1.2 ) = 0.910 M Ω
For part (a), o
R = (1.2 0.8 ) = 0.48 M Ω
For part (b), o
______________________________________________________________________________________
(c)
11.67
I B5 =
I E5
I +I
I +I
= B3 B 4 = C 3 C 4
1+ β
1+ β
β (1 + β )
Now I C 3 + I C 4 ≈ I Q
So I B 5 ≈
I B6 =
IQ
β (1 + β )
I Q1
IE6
=
1 + β β (1 + β )
For balance, we want I B 6 = I B 5
So that I Q1 = I Q
______________________________________________________________________________________
11.68
Resistance looking into drain of M4.
V sg 4 ≅ I X R1
I X + g m 4V sg 4 =
V X − V sg 4
ro 4
⎡
R ⎤ V
I X ⎢1 + g m 4 R1 + 1 ⎥ = X
ro 4 ⎦ ro 4
⎣
⎡
R ⎤
Or R o = ro 4 ⎢1 + g m 4 R1 + 1 ⎥
ro 4 ⎦
⎣
a.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 ( 0.080 )( 0.1)
= 0.179 mA / V
ro 4 =
1
λ p I DQ
=
1
( 0.02 )( 0.1)
= 500 k Ω
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
R =r
r = 667 500 ⇒ R = 286 kΩ
o
For part (b), o o 2 o 4
______________________________________________________________________________________
11.69
Let β = 100, VA = 100 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VA 100
=
= 1000 k Ω
I CQ 0.1
ro 2 =
Ro 4 = ro 4 [1 + g m RE′ ] where RE′ = rπ RE
Now
rπ =
(100 )( 0.026 )
= 26 k Ω
0.1
0.1
= 3.846 mA / V
gm =
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
______________________________________________________________________________________
11.70
(a) For Q2, Q4
(1)
Ix =
Vx − Vπ 4
V
+ g m 2Vπ 2 + g m 4Vπ 4 + x
ro 2
ro 4
g m 2Vπ 2 +
(2)
(3)
Vx − Vπ 4
V
= π4
ro 2
rπ 4 rπ 2
Vπ 4 = −Vπ 2
From (2)
⎡ 1
⎤
Vx
1
= Vπ 4 ⎢
+
+ gm2 ⎥
ro 2
⎢⎣ rπ 4 rπ 2 ro 2
⎥⎦
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
⎛ β ⎞ ⎛ I Q ⎞ ⎛ 120 ⎞
IC 4 = ⎜
⎟⎜ ⎟ = ⎜
⎟ ( 0.5 ) = 0.496 mA
⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 121 ⎠
⎛ 120 ⎞
⎛ I Q ⎞ ⎛ 1 ⎞⎛ β ⎞
⎟ ⇒ I C 2 = 0.0041 mA
IC 2 = ⎜ ⎟ ⎜
⎟⎜
⎟ = ( 0.5 ) ⎜⎜
2
⎟
⎝ 2 ⎠ ⎝ 1 + β ⎠⎝ 1 + β ⎠
⎝ (121) ⎠
So
rπ 2 =
(120 )( 0.026 )
= 761 k Ω
0.0041
0.0041
= 0.158 mA/V
gm2 =
0.026
100
⇒ 24.4 M Ω
ro 2 =
0.0041
(120 ) ( 0.026 )
= 6.29 k Ω
rπ 4 =
0.496
0.496
gm4 =
= 19.08 mA / V
0.026
100
ro 4 =
= 202 k Ω
0.496
Now
⎡
⎤
Vx
Vx
1
1
= Vπ 4 ⎢
+
+ 0.158⎥ ⇒ which yields Vπ 4 =
ro 2
6.29
761
24400
0.318
(
) ro 2
⎣⎢
⎦⎥
From (1),
Ix =
⎛
Vx Vx
1 ⎞
+
+ 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
= 160 k Ω
Now ro 6 =
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Δi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 =
vd
2
⎛V
⎞
Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3
⎝ rπ 1
⎠
⎛1+ β ⎞
So Vπ 1 ⎜
⎟ rπ 3 = Vπ 3
⎝ rπ 1 ⎠
⎛ 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⎠
Δ
i
= 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
______________________________________________________________________________________
11.71 - Design Problem
______________________________________________________________________________________
11.72
Input: −8 ≤ V d ≤ 8 mV
Output: −0.8 ≤ Vo ≤ 0.8 V
V
0.8
= 100
Ad = o =
V d 0.008
Ad = g m (ro 2 ro 4 )
Let I Q = 0.5 mA, I DQ = 0.25 mA
1
= 160 k Ω
(0.025)(0.25)
1
ro 4 =
= 100 k Ω
(0.04)(0.25)
ro 2 =
100 = g m (160 100 ) ⇒ g m = 1.625 mA/V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠ n
⎛W ⎞
⎛ 0.1 ⎞⎛ W ⎞
1.625 = 2 ⎜
⎟⎜ ⎟ (0.25) ⇒ ⎜ ⎟ = 52.8
2
L
⎝
⎠⎝ ⎠ n
⎝ L ⎠n
______________________________________________________________________________________
11.73
For current source, I REF = I Q
( )(
)
0.8 = (2 I )(6 ) ⇒ I ≅ 66 μ A
A = g (r r )
P = 2I Q V + − V −
Q
d
o2
m
ro 2 =
ro 4 =
Q
o4
1
= 1515 k Ω
⎛ 0.066 ⎞
(0.02)⎜
⎟
⎝ 2 ⎠
1
= 1010 k Ω
⎛ 0.066 ⎞
(0.03)⎜
⎟
⎝ 2 ⎠
240 = g m (1515 1010 ) ⇒ g m = 0.396 mA/V
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ
⎝ 2 ⎠⎝ L ⎠ n
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
0.396 = 2 ⎜
⎟⎜ ⎟ (0.033) ⇒ ⎜ ⎟ = 23.76
⎝ 2 ⎠⎝ L ⎠ n
⎝ L ⎠n
______________________________________________________________________________________
11.74
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 ⎟ ⎜ ⎟ ID
⎝ 2 ⎠⎝ L ⎠
⎛ 0.08 ⎞⎛ W ⎞
0.60 = 2 ⎜
⎟⎜ ⎟ ( 0.1)
⎝ 2 ⎠⎝ L ⎠
⎛W ⎞
0.090 = 0.004 ⎜ ⎟
⎝L⎠
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 22.5
⎝ L ⎠1 ⎝ L ⎠ 2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.75
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
ro 2 = ro 4 =
ro 6 = ro8 =
1
( 0.015)( 0.040 )
1
( 0.02 )( 0.040 )
= 1667 k Ω
= 1250 k Ω
⎛ 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
______________________________________________________________________________________
11.76
(a) Ad = g m (ro 2 ro 4 )
g m = 2 K n I DQ = 2 (0.2)(0.06 ) = 0.2191 mA/V
ro 2 = ro 4 =
1
= 666.7 k Ω
(0.025)(0.06)
Ad = (0.2191)(666.7 666.7 ) = 73.0
(b) R o = ro 2 ro 4 = 333.3 k Ω
(c) i D 3 = K p (υ SG 3 + VTP )
2
0.06 = 0.2(υ SG 3 − 0.3) ⇒ υ SG 3 = 0.8477 V
2
υ O = V + − υ SG 3 = 2.8 − 0.8477 = 1.9523 V
i D1 = K n (υ GS1 − VTN )
2
0.06 = 0.2(υ GS1 − 0.3) ⇒ υ GS1 = 0.8477 V, ⇒ υ DS1 (sat ) = 0.5477 V
3
υ CM (max ) = υ O − υ DS1 (sat ) + υ GS1 = 1.9523 − 0.5477 + 0.8477 = 2.25 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.77
2
(a) i D 3 = K p (υ SG 3 + VTP )
0.25 = 0.25(υ SG 3 − 0.4) ⇒ υ SG 3 = 1.4 V
2
Then υ GS 1 = 1.4 V, υ DS1 (sat ) = 1.0 V
υ CM = υ O − υ DS1 (sat ) + υ GS1
(
)
3 = V + − 1.4 − 1.0 + 1.4 ⇒ V + = 4 V = −V −
(b) Ad = g m (ro 2 ro 4 )
g m = 2 K n I DQ = 2 (0.25)(0.25) = 0.50 mA/V
ro 2 = ro 4 =
1
(0.02)(0.25)
= 200 k Ω
Ad = (0.50 )(200 200 ) = 50
______________________________________________________________________________________
11.78
(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
0.1
60
ro 4 =
= 600 K
0.1
0.1
gm =
= 3.846 mA/V
0.026
Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 1442
Ad = g m ( ro 2 ro 4 )
ro 2 =
______________________________________________________________________________________
11.79
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.80
(a) I E 2 = 0.25 mA, I B 2 = 0.001656 mA, I C 2 = 0.2483 mA
I D1 = 0.25 + 0.001656 = 0.25166 mA
g m1 = 2 K n I DQ = 2 (0.2 )(0.25166) = 0.4487 mA/V
g m2 =
rπ 2 =
I C 2 0.2483
=
= 9.55 mA/V
0.026
VT
(150)(0.026) = 15.71 k Ω
0.2483
(0.4487 )[1 + (9.55)(15.71)] = 8.42 mA/V
g mC =
1 + (0.4487 )(15.71)
(b) I E 2 = 0.45 mA, I B 2 = 0.002980 mA, I C 2 = 0.4470 mA
I D1 = 0.05 + 0.00298 = 0.05298 mA
g m1 = 2 (0.2 )(0.05298) = 0.2059 mA/V
0.4470
= 17.19 mA/V
0.026
rπ 2 = 8.725 k Ω
g m2 =
g mC =
(0.2059)[1 + (17.19)(8.725)] = 11.12 mA/V
1 + (0.2059 )(8.725)
______________________________________________________________________________________
11.81
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
A =0
C M RR = ∞
dB
and
If the IQ current source is ideal, cm
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.82
(a)
0.7
= 0.0875 mA
8
I Q = 0.5 = I R1 + I E 2 = 0.0875 + I E 2 ⇒ I E 2 = 0.4125 mA
(b) I R1 =
I B 2 = 0.002279 mA, I C 2 = 0.41022 mA
I D1 = I R1 + I B 2 = 0.0875 + 0.002279 = 0.08978 mA
g m1 = 2 (0.7 )(0.08978) = 0.5014 mA/V
g m2 =
0.41022
= 15.78 mA/V
0.026
rπ 2 = 11.41 k Ω
(
)
(c) V o = − g m1V sg + g m 2Vπ R L
(
)
Vπ = g m1V sg (R1 rπ 2 )
[
]
V o = − g m1V sg + g m 2 g m1V sg (R1 rπ 2 ) R L
[
]
− V o = V sg g m1 + g m 2 g m1 (R1 rπ 2 ) R L
Now V sg = Vo − Vi
[
]
[
]
So V o 1 + g m1 (1 + g m 2 (R1 rπ 2 ))R L = Vi g m1 1 + g m 2 (R1 rπ 2 ) R L
Aυ =
[
]
g m1 1 + g m 2 (R1 rπ 2 ) R L
Vo
=
Vi 1 + g m1 1 + g m 2 (R1 rπ 2 ) R L
[
]
(0.5014)[1 + (15.78)(8 11.41)]R
37.71R
A =
=
1 + 37.71R
1 + (0.5014 )[1 + (15.78)(8 11.41)]R
L
L
υ
L
L
(i) For R L = 10 k Ω ,
Aυ =
(37.71)(10) = 0.99736
1 + (37.71)(10 )
(ii) For R L = 100 k Ω ,
Aυ =
(37.71)(100) = 0.99973
1 + (37.71)(100)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.83
0.7
(a) I R1 =
= 0.0875 mA
8
I Q = 0.8 = I R1 + I E 2 = 0.0875 + I E 2 ⇒ I E 2 = 0.7125 mA
I B 2 = 0.003936 mA, I C 2 = 0.7086 mA
I D1 = I R1 + I B 2 = 0.0875 + 0.003936 = 0.091436 mA
g m1 = 2 (0.7 )(0.091436 ) = 0.506 mA/V
0.7086
= 27.25 mA/V
0.026
rπ 2 = 6.605 k Ω
(b) I x = g m 2Vπ + g m1V sg
g m2 =
(
)
Vπ = g m1V sg (R1 rπ 2 )
Now V sg = V x
Ix
1
=
= g m1 + g m 2 g m1 (R1 rπ 2 ) = 0.506 + (0.506)(27.25)(8 6.605) = 50.39 mA/V
V x Ro
Or R o = 19.8 Ω
______________________________________________________________________________________
11.84
(a)
(1)
g m 2Vπ +
g m 2Vπ +
(2)
Vπ =
Then
From (1)
Vo − ( −Vπ )
ro 2
Vo − ( −Vπ )
ro 2
g m1Vi
⎛ 1 1⎞
⎜ + ⎟
⎝ ro1 rπ ⎠
=0
= g m1Vi +
⎛ 1 1⎞
−Vπ −Vπ
+
or 0 = g m1Vi − Vπ ⎜ + ⎟
ro1
rπ
⎝ ro1 rπ ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛
Vo
1 ⎞
=0
⎜ g m 2 + ⎟ Vπ +
r
ro 2
o2 ⎠
⎝
⎛
1 ⎞
⎜ gm2 + ⎟
ro 2 ⎠
⎛
1 ⎞
Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = − ro 2 g m1Vi ⎝
r
⎛
1
1⎞
o2 ⎠
⎝
⎜ + ⎟
r
r
⎝ o1 π ⎠
⎛
1 ⎞
− g m1ro 2 ⎜ g m 2 + ⎟
ro 2 ⎠
V
⎝
Av = o =
Vi
⎛ 1 1⎞
⎜ + ⎟
⎝ ro1 rπ ⎠
Now
g m1 = 2 K n I Q = 2 ( 0.25 )( 0.025 ) = 0.158 mA / V
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
IQ
=
(100 )( 0.026 )
0.025
= 104 k Ω
Then
1 ⎞
⎛
− ( 0.158 )( 2000 ) ⎜ 0.9615 +
⎟
2000 ⎠
⎝
Av =
⇒ Av = −30039
1 ⎞
⎛ 1
+
⎜
⎟
⎝ 2000 104 ⎠
Vi = 0 ⇒ g m1Vi = 0
To find Ro; set
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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)
(1)
g m 3Vgs 3 +
g m 3Vgs 3 +
(2)
(3)
Vo − ( −Vgs 3 )
ro3
Vo − ( −Vgs 3 )
= g m 2Vπ 2 +
−Vgs 3 − ( −Vπ 2 )
ro 2
−Vgs 3 − ( −Vπ 2 )
( −Vπ 2 )
Vπ 2
+ g m 2Vπ 2 +
= g m1Vi +
rπ 2
ro 2
ro1
Vπ 2 =
From (2),
Then
(3)
or
ro 3
=0
Vgs 3
⎛
1 ⎞
ro 2 ⎜ g m 2 + ⎟
r
o2 ⎠
⎝
Vgs 3
⎛ 1
1
1⎞
Vπ 2 ⎜
+ gm2 +
+ ⎟ = g m1Vi +
ro 2 ro1 ⎠
ro 2
⎝ rπ 2
⎛
1 ⎞ Vgs 3
or 0 = Vπ 2 ⎜ g m 2 + ⎟ −
ro 2 ⎠ ro 2
⎝
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vgs 3
⎡ 1
1
1⎤
+ ⎥ = g m1Vi +
⎢ + gm2 +
ro 2 ro1 ⎦
ro 2
⎛
1 ⎞ r
ro 2 ⎜ g m 2 + ⎟ ⎣ π 2
r
o2 ⎠
⎝
Vgs 3
Vgs 3
1
1 ⎤
⎡ 1
+ 0.9615 +
+
= 0.9615Vi +
1 ⎞ ⎢⎣104
2000 2000 ⎥⎦
2000
⎛
2000 ⎜ 0.9615 +
⎟
2000 ⎠
⎝
Vgs 3 = 1.83 × 105 Vi
Vgs 3
Then
⎛
−Vo
1 ⎞
1 ⎞
⎛
5
or Vo = −2000 ⎜ 0.158 +
⎜ g m 3 + ⎟ Vgs 3 =
⎟ (1.83 × 10 )Vi
2000 ⎠
ro3 ⎠
ro 3
⎝
⎝
From (1),
Av =
Vo
= −5.80 × 107
Vi
To find Ro
(1)
(2)
(3)
I x = g m 3Vgs 3 +
g m 3Vgs 3 +
Vx − ( −Vgs 3 )
ro 3
Vx − ( −Vgs 3 )
ro3
= g m 2Vπ 2 +
Vπ 2 = − I x ( ro1 rπ 2 )
⎛
1 ⎞ V
From (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x
r
ro 3
o3 ⎠
⎝
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From (2),
⎡
⎛
1
1 ⎤ V
1 ⎞
+ ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟
Vgs 3 ⎢ g m 3 +
r
r
r
r
o3
o2 ⎦
o3
o2 ⎠
⎣
⎝
1
1 ⎤ Vx
1 ⎞
⎡
⎛
+
+
= Vπ 2 ⎜ 0.9615 +
Vgs 3 ⎢ 0.158 +
⎟
2000 2000 ⎥⎦ 2000
2000 ⎠
⎣
⎝
V
⎡ I − Vx / 2000 ⎤
Then ⎢ x
( 0.159 ) + x = − I x ( 2000 104 ) ( 0.962 )
⎥
2000
⎣ 0.1585 ⎦
Vx
We find Ro =
= 6.09 × 1010 Ω
Ix
______________________________________________________________________________________
11.85
Assume emitter of Q1 is capacitively coupled to signal ground.
⎛ 80 ⎞
I CQ = 0.2 ⎜ ⎟ = 0.1975 mA
⎝ 81 ⎠
0.2
= 0.00247 mA
I DQ =
81
( 80 )( 0.026 )
= 10.5 k Ω
rπ =
0.1975
0.1975
= 7.60 mA / V
g m ( Q1 ) =
0.026
g m ( M 1 ) = 2 K n I D = 2 ( 0.2 )( 0.00247 )
g m ( M 1 ) = 0.0445 mA / V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vi = Vgs + Vπ and Vπ = g m ( M 1 )Vgs rπ or Vgs =
Vπ
g m ( M 1 ) rπ
⎛
⎞
Vi
1
Then Vi = Vπ ⎜⎜1 +
⎟⎟ or Vπ =
⎛
⎞
1
⎝ g m ( M 1 ) rπ ⎠
⎜⎜ 1 +
⎟⎟
g
M
r
(
)
m
1 π ⎠
⎝
− g m ( Q1 ) RC
Vo
=
Vo = − g m ( Q1 )Vπ RC ⇒ Av =
Vi ⎛
⎞
1
⎜⎜ 1 + g ( M ) r ⎟⎟
m
1 π ⎠
⎝
− ( 7.60 )( 20 )
⇒ Av = −48.4
Then Av =
⎛
⎞
1
+
1
⎜⎜ ( 0.0445 )(10.5 ) ⎟⎟
⎝
⎠
______________________________________________________________________________________
11.86
rπ 11 =
(120)(0.026) = 15.6 k Ω
0.2
R E′ = R3 rπ 11 = 0.2 15.6 = 0.1975 k Ω
0.2
= 7.692 mA/V
0.026
V
120
ro11 = A11 =
= 600 k Ω
I C11 0.2
RC11 = ro11 (1 + g m11 R E′ ) = 600[1 + (7.692 )(0.1975)] = 1512 k Ω
g m11 =
RC 7 = ro 7 =
V A7
60
=
= 300 k Ω
I C 7 0.2
Z = RC 7 RC11 = 300 1512 = 250 k Ω
rπ 8 =
(120)(0.026) = 3.12 k Ω
rπ 9 =
(120)(0.026) = 377.5 k Ω
1
1 ⎛ 120 ⎞
I C9 =
⎜
⎟ = 0.008264 mA
120 ⎝ 121 ⎠
0.008264
⎛r +Z ⎞
⎛ 377.5 + 250 ⎞
rπ 8 + ⎜⎜ π 9
⎟⎟
3.12 + ⎜
⎟
121
⎝ 121 ⎠
⎝
⎠
=5
Now R o = R 4
= 5 0.06864
121
121
Or R o = 67.7 Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.87
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 =
= 2.6 kΩ
1
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 ⎠
(V − V )
Vπ 3
= g m 2Vπ 2 + 0 π 3
rπ 3
50
⎛ 1
1 ⎞ V
g m 2Vπ 2 = Vπ 3 ⎜
+ ⎟− 0
r
50
⎝ π3
⎠ 50
(1)
(2)
⎛V
⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2
r
⎝ π1
⎠
⎛ 1+ β ⎞
= Vπ 1 ⎜
⎟ rπ 2
⎝ rπ 1 ⎠
and
Vin = Vπ 1 + Vπ 2
(3)
(4)
0.5
gm2 =
= 19.23 mA/V
0.026
Then
⎛ 101 ⎞
V0 = −Vπ 3 ⎜
⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 )
⎝ 2.6 ⎠
(1)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
And
1 ⎞ V
⎛ 1
+ ⎟− 0
19.23Vπ 2 = −V0 ( 0.005149 ) ⎜
⎝ 2.6 50 ⎠ 50
= −V0 ( 0.02208 )
Or Vπ 2 = −V0 ( 0.001148 )
(2)
And
Vπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 )
(4)
So
⎛ 101 ⎞
−V0 ( 0.001148 ) = ⎣⎡Vin + V0 ( 0.001148 ) ⎤⎦ ⎜
⎟ ( 5.2 )
⎝ 520 ⎠
(3)
−V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av =
V0
= −438
Vin
______________________________________________________________________________________
11.88
I2 =
5
= 1 mA
5
1
+ 0.8 = 2.21 V
0.5
2.21 − ( −5 )
I1 =
= 0.206 mA
35
VGS 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 )
g m 2 = 2 K n 2 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 =
r02 =
1
λ1 I D1
1
λ2 I D 2
=
1
( 0.01)( 0.206 )
=
1
( 0.01)(1)
= 485 kΩ
= 100 kΩ
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 4.
R0 =
1
R2 r02
gm2
=
1
5 100
1.414
= 0.707 4.76
So R0 = 0.616 kΩ
______________________________________________________________________________________
11.89
5−0
= 5kΩ
1
2
2
I 2 = K p (V SG + VTP ) ⇒ 1 = 1(V SG − 0.8) ⇒ V SG = 1.8 V
(a) R 2 =
R1 =
5 − (− 1.8)
= 27.2 k Ω
0.25
(b) g m1 = 2 (0.5)(0.25) = 0.7071 mA/V
g m 2 = 2 (1)(1) = 2 mA/V
ro1 =
1
(0.02)(0.25)
= 200 k Ω , ro 2 =
1
= 50 k Ω
(0.02)(1)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c)
V o1 = − g m1Vin (ro1 R1 )
V sg 2 = Vo − V o1
Vo Vo
+
+ g m 2V sg 2 = 0
R 2 ro 2
⎛ 1
1 ⎞
⎟ + g m 2 Vo + g m1Vin (ro1 R1 ) = 0
V o ⎜⎜
+
⎟
R
r
o2 ⎠
⎝ 2
[
]
⎞
⎛ 1
1
V o ⎜⎜
+
+ g m 2 ⎟⎟ = − g m1 g m 2 (ro1 R1 )Vin
⎠
⎝ R 2 ro 2
− g m1 g m 2 (ro1` R1 ) − (0.7071)(2)(200 27.2 )
V
=
Aυ = o =
= −15.25
Vin ⎛ 1
⎞
⎛1 1
⎞
1
+ 2⎟
⎜
⎟
⎜ +
⎜ R + r + g m2 ⎟
⎝ 5 50
⎠
o2
⎝ 2
⎠
V
V
(d) I x = x + x + g m 2V x
ro 2 R 2
R o = ro 2 R 2
1
1
= 50 5 = 4.545 0.5
2
g m2
R o = 0.450 k Ω
______________________________________________________________________________________
11.90
5−0
= 20 k Ω
0.25
− 0.7 − (− 5)
R E1 =
= 17.2 k Ω
0.25
5 − 0.7
= 17.2 k Ω
RC =
0.25
0 − (− 5)
RE 2 =
= 2.5 k Ω
2
υ
g
(b) Ad 1 = o 2 = m1 (R rπ 3 )
υd
2
(a) R =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.25
g m1 =
= 9.615 mA/V
0.026
(120)(0.026) = 12.48 k Ω
rπ 3 =
0.25
(
9.615)
(20 12.48) = 36.94
Ad 1 =
2
A3 = − g m3 (RC Ri 4 )
Ri 4 = rπ 4 + (1 + β )R E 2
0.25
= 9.615 mA/V
0.026
(120)(0.026) = 1.56 k Ω
rπ 4 =
2
(
)(
Ri 4 = 1.56 + 121 2.5) = 304 k Ω
g m3 =
A3 = −(9.615)(17.2 304 ) = −156.5
A4 =
Now
(1 + β )R E 2
(121)(2.5) = 0.995
=
rπ 4 + (1 + β )R E 2 1.56 + (121)(2.5)
Ad =
(c) Acm1 =
υo
= Ad 1 ⋅ A3 ⋅ A4 = (36.94 )(− 156.5)(0.995) = −5752
υd
− g m1 (R rπ 3 )
1+
2(1 + β )R o
rπ 1
rπ 1 =
Acm1 =
(120)(0.026) = 12.48 k Ω
0.25
− (9.615)(20 12.48)
= −0.01905
2(121)(200 )
12.48
Acm = Acm1 ⋅ A3 ⋅ A4 = (− 0.01905)(− 156.5)(0.995) = 2.966
1+
⎛ 5752 ⎞
CMRRdB = 20 log 10 ⎜
⎟ = 65.8 dB
⎝ 2.966 ⎠
______________________________________________________________________________________
11.91
a.
b.
RC1 =
10 − v01 10 − 2
=
⇒ RC1 = 80 kΩ
I C1
0.1
RC 2 =
10 − v04 10 − 6
=
⇒ RC 2 = 20 kΩ
IC 4
0.2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 =
______________________________________________________________________________________
11.92
a.
Neglect the effect of r0 in determining the differential-mode gain.
v
1
Ad 1 = 02 = 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 )
gm2 = 2
= 37.3 mA/V
0.026
( 200 )( 0.026 )
rπ 3 =
IC 3
1
(1.94 )(8) = 4.24 V
2
4.24 − 0.7
IC 3 =
= 1.07 mA
3.3
( 200 )( 0.026 )
rπ 3 =
= 4.86 kΩ
1.07
Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ
v02 = 12 −
Ad 1 =
1
( 37.3) ⎡⎣8 668⎤⎦ = 147.4
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 )
= 5.36 kΩ
1
⋅ (1.94 )
2
− ( 37.3) ( 8 668 )
= −0.09539
Acm1 =
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
r0 =
VA
80
=
= 82.5 kΩ
I C 2 1 ⋅ 1.94
( )
2
2 Ricm = ⎡⎣ 2 ( 201)( 41.2 ) ⎤⎦ ⎡⎣( 201)( 82.5 ) ⎤⎦
= 16.6 MΩ 16.6 MΩ
So ⇒ Ricm = 4.15 MΩ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.93
a.
I1 =
24 − VGS 4
2
= kn (VGS 4 − VTh )
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) − 4 (11) ( 20 )
= 3.37 V
2 (11)
2
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 ) − 4 (1.2 ) ( 0.3)
= 3.09 V
2 (1.2 )
2
4.5 − 3.09
= 0.235 mA
6
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 0.375 ⎞
g m 2 = 2 K n I D 2 = 2 ( 0.2 ) ⎜
⎟
⎝ 2 ⎠
= 0.387 mA/V
1
1
Ad 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
g m 3 = 2 K n I D 3 = 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 )
______________________________________________________________________________________
11.94
(a) Ad =
β RC
rπ + R B
Assuming I CQ ≅ I EQ ,
rπ =
(150)(10) = 146
A =
d
9.75 + 0.5
(150)(0.026) = 9.75 k Ω
0.4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) C M ≅ C μ (1 + g m RC )
0.4
= 15.38 mA/V
0.026
C M = (0.2)[1 + (15.38)(10 )] = 30.96 pF
gm =
(c)
fH =
1
1
=
2π (rπ R )(C π + C ) 2π (9.75 0.5)× 10 (1.2 + 30.96 )× 10
3
B
−12
M
f H = 10.4 MHz
______________________________________________________________________________________
11.95
(a)
fZ =
(b) rπ =
1
1
=
⇒ f Z = 39.8 kHz
6
2πR o C o 2π 10 × 10 0.4 × 10 −12
(
)(
)
(150)(0.026) = 9.75 k Ω
0.4
⎛ R ⎞
R o ⎜⎜1 + B ⎟⎟
(10)⎛⎜1 + 0.5 ⎞⎟
r
π
⎝
⎠
⎝ 9.75 ⎠
=
R eq =
2(1 + β )R o
0.5 2(151)(10,000)
R
1+
+
1+ B +
9.75
9.75
rπ
rπ
Or R eq = 33.94 Ω
fP =
1
1
=
2πReq C o 2π (33.94) 0.4 × 10 −12
(
)
f P = 11.7 GHz
______________________________________________________________________________________
11.96
fT =
a.
From Equation (7.73),
gm
2π ( Cπ + Cμ )
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 =
fH =
(120 )( 0.026 )
1
= 3.12 kΩ
1
2π ⎡⎣3.12 1⎤⎦ × 103 × ( 6.65 + 386 ) × 10 −12
Or f H = 535 kHz
fZ =
b.
From Equation (11.140),
1
1
=
2π R0 C0 2π (10 × 106 )(10−12 )
f Z = 15.9 kHz
Or
______________________________________________________________________________________
11.97
1
β RC
2
(a) Aυ =
rπ + (1 + β )R E
rπ =
(120)(0.026) = 12.48 k Ω
0.25
1
(120)(8)
2
= 19.5
Aυ =
12.48 + (121)(0.1)
1
(120)(8)
2
= 11.2
(b) Aυ =
12.48 + (121)(0.25)
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 12
12.1
(a) A f =
A
5 × 10 4
⇒ β = 9.98 × 10 − 3
⇒ 100 =
1 + Aβ
1 + 5 × 10 4 β
(
)
A
⇒ A = 2000
1 + A(0.012 )
______________________________________________________________________________________
(b) 80 =
12.2
(a) A f =
A
− 10 5
⇒ −80 =
⇒ β = −0.01249
1 + Aβ
1 + − 10 5 β
(
)
− 5 × 10
= −66.58
1 + − 5 × 10 4 (− 0.015)
______________________________________________________________________________________
4
(b) A f =
(
)
A
1 + Aβ
β = 0.15
12.3
Af =
(a)
T = Aβ
T =∞
(i)
4
3
(ii) A = 80 dB ⇒ A = 10 ⇒ T = 1.5 × 10
(iii) T = 15
Af =
(i)
(ii)
1
β
= 6.667
Af = 6.662
A = 6.25
(iii) f
(b)
(i) T = ∞
3
(ii) T = 2.5 × 10
(iii) T = 25
Af =
(i)
(ii)
1
β
= 4.00
Af = 3.9984
A = 3.846
(iii) f
______________________________________________________________________________________
12.4
(a)
A
1
≅ = 125
1 + Aβ β
β = 0.0080
Af =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
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
______________________________________________________________________________________
12.5
A
− 2 × 10 4
⇒ −80 =
⇒ β = −0.01245
1 + Aβ
1 + − 2 × 10 4 β
(a) A f =
(b)
dA f
(
=
Af
)
1
dA
⋅
(1 + Aβ ) A
1
dA
dA
⇒
= 2.5%
A
1 + − 2 × 10 (− 0.01245) A
______________________________________________________________________________________
0.01 =
[ (
dA f
1
dA
⋅
1 + βA A
4
]
)
⋅
12.6
Af
=
1
⋅ (0.10 ) ⇒ 1 + β A = 100
1 + βA
100 − 1
β=
⇒ β = 1.98 × 10 −3
5 × 10 4
5 × 10 4
A
Now A f =
=
⇒ A f = 500
1 + βA 1 + 1.98 × 10 − 3 5 × 10 4
______________________________________________________________________________________
0.001 =
(
)(
)
12.7
⎛ A1 ⎞⎛ A2 ⎞
⎟⎟⎜⎜
⎟⎟
(a) Aυf = ⎜⎜
⎝ 1 + A1 β 1 ⎠⎝ 1 + A2 β 1 ⎠
⎛ 200 ⎞⎛ 10 ⎞
⎟⎟⎜⎜
⎟⎟
50 = ⎜⎜
⎝ 1 + 200 β 1 ⎠⎝ 1 + 10 β 1 ⎠
(1 + 200β 1 )(1 + 10β 1 ) = (200)(10) = 40
50
Then 2000 β 12 + 210 β 1 − 39 = 0 ⇒ β 1 = 0.096685
Aυf =
50 =
A1 A2
1 + A1 A2 β 2
2000
⇒ β 2 = 0.0195
1 + 2000 β 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) From circuit (a):
⎛
⎞⎛
⎞
200
9
⎟⎟⎜⎜
⎟⎟ = 47.33 ⇒ (− 5.43% )
Aυf = ⎜⎜
⎝ 1 + (200 )(0.096685) ⎠⎝ 1 + (9 )(0.096685) ⎠
For circuit (b):
1800
Aυf =
= 49.86 ⇒ (− 0.28% )
1 + (1800)(0.0195)
______________________________________________________________________________________
12.8
(a)
VO = (−10)(−15)( −20)Vε = −3000Vε
Vε = β VO + VS
So VO = −3000( β VO + VS )
We find
Avf =
For
VO
−3000
=
VS 1 + 3000 β
Avf = −120 =
−3000
⇒ β = 0.008
1 + 3000 β
(b) Now VO = (−9)(−13.5)(−18)Vε = −2187Vε
Then
Avf =
−2187
−2187
=
= −118.24
1 + 2187 β 1 + 2187(0.008)
% change =
120 − 118.24
× 100 ⇒ 1.47% change
120
______________________________________________________________________________________
12.9
(a) Aυ f H = Aυ f f C
(5 ×10 )(10) = (25) f ⇒ f = 20 kHz
A f
(10 )(8) = 40
=
(b) A =
4
C
υ
υf
C
5
H
fC
20 × 10 3
______________________________________________________________________________________
12.10
(a)
fC =
(5 ×10 )(10) ⇒ f = 10 kHz
4
C
50
10 4 (10 )
⇒ f C = 2 kHz
(b) f C =
50
______________________________________________________________________________________
( )
12.11
(a) (i) Aυ =
(ii) 75 =
Aυf f C
fH
=
(75)(35 ×10 3 ) = 5.25 ×10 5
5
5.25 × 10
⇒ β = 0.01333
1 + 5.25 × 10 5 β
5
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
(b) A = 5.25 × 10 5 (0.9) = 4.725 × 10 5
Aυf =
4.725 × 10 5
= 74.99
1 + 4.725 × 10 5 (0.01333)
(
(
)
)
Aυ f H
4.725 × 10 5 (5)
=
⇒ f C = 31.5 kHz
Aυf
74.99
fC =
______________________________________________________________________________________
12.12
Low freq.
100 =
Af =
A0
1 + A0 β
5000
⇒ β = 0.0098
1 + (5000) β
Freq. response
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 + ⎜ ⎟⎜ ⎟ + 49
f1
f 2 ⎝ f1 ⎠⎝ f 2 ⎠
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 ⎞
j ⎟
1+ j
+j
+ j
fA
f B ⎜⎝ f A ⎟⎜
⎠⎝ f B ⎠
So
100
100
=
1 ⎛ jf ⎞⎛ jf ⎞
f
f ⎛ f ⎞⎛ f ⎞
f
f
1+ j
j ⎟ 1+ j
+j
+ j
+j
+ ⎜ ⎟⎜ ⎟
fA
f B ⎜⎝ f A ⎟⎜
f
f
f
50
50
50
1
2
⎠⎝ B ⎠
⎝ 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
fB
1
+
= 0.002010
106 f B
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 )
f B = 1.105 × 103 Hz
+ sign
f A = 9.05 × 10 Hz
+ sign
______________________________________________________________________________________
2
12.13
(a)
Fig. P12.7(a)
⎡
⎤
⎛ 200 ⎞
⎢
⎥
⎜
f ⎟
⎢
⎥
⎜⎜ 1 + j ⎟⎟
f1 ⎠
10
⎤
⎢
⎥⎡
⎝
Af = ⎢
⎥ ⎢1 + (10)(0.1126) ⎥
⎛ 200 ⎞
⎣
⎦
⎢1 + ⎜
⎟ (0.1126) ⎥
⎢ ⎜1+ j f ⎟
⎥
⎢⎣ ⎜⎝
⎥⎦
f1 ⎟⎠
200
⎡
⎤
=⎢
⎥ (4.704)
f
⎛
⎞
⎢ ⎜ 1 + j ⎟ + 22.52 ⎥
f1 ⎠
⎣⎢ ⎝
⎦⎥
=
=
940.73
f
23.52 + j
f1
=
40
jf
1+
(23.52) f1
940.73
1
⋅
f
23.52 1 + j
(23.52) f1
f −3dB = (23.52)(100) ⇒ 2.352 kHz
Fig P12.7(b)
(200)(10)
f
1+ j
f1
2000
Af =
=
f
(0.0245)(200)(10)
1+
1 + j + 49
f
f
1
1+ j
f1
=
2000
⋅
50
1
f −3dB = (50)(100) ⇒ 5 KHz
f
(50) f1
Overall feedback ⇒ wider bandwidth.
1+ j
(b)
______________________________________________________________________________________
12.14
v0 = A1 A2 vi + A1vn
v0 = (100)vi + (1)vn = (100)(10) + (1)(1) ⇒
S0 1000
=
= 1000
N0
1
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.15
(a)
(b)
Circuit (b) – less distortion
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.16
Aυ f =
Aυ
=
5 × 10 3
⇒ Aυf = 121.95
1 + (0.0080 ) 5 × 10 3
)
R = R (1 + β A ) = (10 )[1 + (0.0080 )(5 × 10 )] ⇒ R = 410 k Ω
if
1 + β υ Aυ
υ
i
(
3
υ
if
1× 10 3
R of =
=
⇒ R of = 24.4 Ω
1 + β υ Aυ 1 + (0.0080 ) 5 × 10 3
______________________________________________________________________________________
Ro
12.17
(
)
V fb = β υ Vo = (0.0096)(2.5) ⇒ V fb = 24 mV
V∈ = Vi − V fb = 25 − 24 = 1 mV
V o = Aυ V∈ ⇒ Aυ =
Vo
2.5
=
= 2.5 × 10 3 V/V
V∈ 0.001
Vo
2.5
=
= 100 V/V
Vi 0.025
______________________________________________________________________________________
Aυf =
12.18
⎛ R ⎞
R
Avf ≈ ⎜ 1 + 2 ⎟ = 20 ⇒ 2 = 19
R1 ⎠
R1
⎝
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 − vd )
1 ⎞ A v
v0 ⎜ + ⎟ = 0 L d + S
R
R
R
R2
2 ⎠
0
⎝ 0
A0 L vd (vS − vd )
+
R0
R2
v0 =
⎛ 1
1 ⎞
⎜ + ⎟
R
R
2 ⎠
⎝ 0
From (1):
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vS − vd vS − vd
+
−
R1
R2
iS =
1 ⎡ A0 L vd (vS − vd ) ⎤
⋅
+
R2 ⎢⎣ R0
R2 ⎥⎦
⎛ 1
1 ⎞
⎜ + ⎟
R
R
2 ⎠
⎝ 0
A0 L 1 ⎞
1 ⎞
⎛
⎛
−
⎜ 1
⎜ 1
R0 R2 ⎟
R2 ⎟
1
1
⎜
⎟
⎜
⎟
+
−
− vd
+
+
iS = vS
R2 ⎟
⎜ R1 R2
⎜ R1 R2 1 + R2 ⎟
1+
⎜
⎜
R0 ⎟⎠
R0 ⎟⎠
⎝
⎝
vd = iS Ri
⎧
⎡⎛ 1
⎡⎛ 1
1 ⎞⎛ R ⎞ A
1 ⎤⎫
1 ⎞ ⎛ R2 ⎞ 1 ⎤
⎪ Ri ⎢⎜ + ⎟ ⎜1 + 2 ⎟ + 0 L − ⎥ ⎪
⎢ ⎜ + ⎟ ⎜1 + ⎟ − ⎥
R
R
R
R
R
⎪
2 ⎠⎝
0 ⎠
0
2 ⎦⎪
⎝ 1
⎢ ⎝ R1 R2 ⎠ ⎝ R0 ⎠ R2 ⎥
iS ⎨1 + ⎣
⎬ = vS ⎢
⎥
R
R
2
⎪
⎪
1+
1+ 2
⎢
⎥
R0
R
⎪
⎪
0
⎣⎢
⎦⎥
⎩
⎭
⎡1 R 1
⎡ 1 R2 1
1 A0 L ⎤ ⎪⎫
1⎤
⎪⎧ R
iS ⎨1 + 2 + Ri ⎢ + 2 ⋅ +
+
⎥ ⎬ = vS ⎢ + ⋅ + ⎥
⎣ R1 R1 R0 R0 R0 ⎦ ⎭⎪
⎣ R1 R1 R0 R0 ⎦
⎩⎪ R0
Let
⎡R ⎛ R ⎞
⎤ ⎪⎫
⎡ R ⎛ R ⎞⎤
⎪⎧
iS ⎨ R0 + R2 + Ri ⎢ 0 + ⎜ 1 + 2 ⎟ + A0 L ⎥ ⎬ = vS ⎢ 0 + ⎜ 1 + 2 ⎟ ⎥ (1)
R
R
R1 ⎠ ⎦
1 ⎠
⎣ 1 ⎝
⎦ ⎭⎪
⎣ R1 ⎝
⎩⎪
R2 = 190 kΩ , R1 = 10 kΩ
⎧
⎡ 0.1
⎤⎫
⎡ 0.1
⎤
iS ⎨0.1 + 190 + 100 ⋅ ⎢
+ 20 + 105 ⎥ ⎬ = vS ⎢
+ 20 ⎥
⎣ 10
⎦⎭
⎣ 10
⎦
⎩
iS (1.000219 × 107 ) = vS (20.01)
Rif =
vS
≅ 5 × 105 kΩ ⇒ Rif ≅ 500 MΩ
iS
Output Resistance
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.19
a.
vS − vd v0 − (vS − vd )
=
R1
R2
and
vd =
v0
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 ⎟ ⎥
R
R
R2 ⎣ A ⎝
R1 ⎠ ⎦
2 ⎠
⎝ 1
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
v0
= ⎝
vS
1 ⎛ R2 ⎞
1 + ⎜1 + ⎟
A⎝
R1 ⎠
which can be written as
Avf =
v0
=
vS
β=
b.
1
R
1+ 2
R1
20 =
c.
A
⎡ ⎛ R2 ⎞ ⎤
1 + ⎢ A / ⎜1 + ⎟ ⎥
R1 ⎠ ⎦
⎣ ⎝
105
1 + (105 ) β
105
−1
β = 20 5 ⇒ β = 0.04999
10
So
R2 1
R
1
= −1 =
− 1 ⇒ 2 = 19.004
R1 β
0.04999
R1
Then
A → 9 × 104
d.
Af =
ΔAf
Af
9 × 104
= 19.99956
1 + (9 × 10 4 )(0.04999)
=
ΔAf
−4.444 × 10−4
= −2.222 × 10−3 % ⇒
= −0.005%
20
Af
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.20
I ∈ = I i − I fb = 20 − 19 = 1 μ A
I fb = β i I o ⇒ I o =
19
⇒ I o = 2 mA
0.0095
Ai =
I o 2 × 10 −3
=
= 2 × 10 3 A/A
I ∈ 1× 10 − 6
Aif =
Io
2
=
= 100 A/A
I i 0.020
Rif =
Ri
1 + β i Ai
=
500
= 25 Ω
1 + (0.0095) 2 × 10 3
(
[
)
(
)](
)
R of = (1 + β i Ai )Ro = 1 + (0.0095) 2 × 10 3 20 × 10 3 ⇒ R of = 400 k Ω
______________________________________________________________________________________
12.21
I fb = I i − I ∈ = 25 − 0.8 = 24.2 μ A
I o = Ai f I i = (125)(25) ⇒ I o = 3.125 mA
βi =
Aif =
I fb
=
Io
24.2
= 0.007744 A/A
3125
Ai
1 + β i Ai
Ai
⇒ Ai = 3906 A/A
1 + (0.007744 )Ai
______________________________________________________________________________________
125 =
12.22
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 =
so
I0
Ai
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I fb = I S −
I0
Ai
From above
⎛ R ⎞
I fb ⎜ 1 + F ⎟ = I 0
R3 ⎠
⎝
⎛
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
I
Aif = 0
IS
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
=
⎡
1 ⎛ RF ⎞ ⎤
⎢1 + ⎜1 +
⎟⎥
A
R3 ⎠ ⎦
i ⎝
⎣
=
βi =
b.
25 =
c.
Ai
= Aif
Ai
1+
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
1
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
⎝
105
1 + (105 ) β i
105
−1
β i = 25 5 ⇒ β i = 0.03999
10
so
so
RF
R
1
1
=
−1 =
− 1 ⇒ F = 24.0
R3 β i
R3
0.03999
Ai = 105 − (0.15)(105 ) = 8.5 × 104
d.
so
Aif =
ΔAif
A
8.5 × 104
= 24.9989
1 + (8.5 × 104 )(0.03999)
=−
1.10 × 10−3
= −4.41× 10−5 ⇒ −4.41× 10−3 %
25
so if
______________________________________________________________________________________
12.23
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I S = I ε + I fb , V1 = I ε Ri
I fb = I 0 +
V0
R3
and
I 0 = Ai I ε ⇒ I ε =
V0 = V1 − I fb RF
I0
Ai
Now
I fb = Ai I ε +
1
(V1 − I fb RF )
R3
⎡ R ⎤
V
I fb ⎢1 + F ⎥ = Ai I ε + 1
R3
⎣ R3 ⎦
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
R
R
3 ⎦
3 ⎠
⎣
⎣⎝
⎦ R3
V
Iε = 1
Ri
⎧⎪ 1 ⎡⎛ R ⎞
⎤ 1 ⎫⎪
⎡ R ⎤
I S ⎢1 + F ⎥ = V1 ⎨ ⋅ ⎢⎜ 1 + F ⎟ + Ai ⎥ + ⎬
R3 ⎠
⎣ R3 ⎦
⎪⎩ Ri ⎣⎝
⎦ R3 ⎪⎭
The
⎛ RF ⎞
⎜1 +
⎟
R3 ⎠
V
⎝
Rif = 1 =
I S ⎧⎪ 1 ⎡⎛ RF ⎞
⎤ 1 ⎫⎪
⎨ ⋅ ⎢⎜ 1 +
⎟ + Ai ⎥ + ⎬
R3 ⎠
⎪⎩ Ri ⎣⎝
⎦ R3 ⎪⎭
1/ R3
term in the denominator will be negligible. Then
25
⎧1 ⎡
5 ⎫
⎨ ⎣(25) + 10 ⎤⎦ ⎬
⎩2
⎭
Rif ≅ 5 × 10−4 kΩ ⇒ Rif = 0.5 Ω
Rif =
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1 105 + 1
= +
Rof 10 240 + 2
R0 f ≈
RF + Ri 240 + 2
= 5
⇒ R0 f ≈ 2.42 × 10−3 kΩ or R0 f ≈ 2.42 Ω
10 + 1
Ai + 1
so
______________________________________________________________________________________
12.24
V∈ = Vi − V fb = 0.2 − 0.195 ⇒ V∈ = 5 μ V
βz =
V fb
Io
=
0.195 × 10 −3
= 0.039 V/A
5 × 10 −3
I o 5 × 10 −3
=
= 1000 A/V
V∈ 5 × 10 −6
Ag
1000
Agf =
=
= 25 A/V
1 + Ag β z 1 + (1000)(0.039)
Ag =
(
) (
)
R = R (1 + β A ) = (10 × 10 )[1 + (0.039)(1000 )] ⇒ R = 400 k Ω
Rif = Ri 1 + β z Ag = 20 × 10 3 [1 + (0.039 )(1000 )] ⇒ Rif = 800 k Ω
3
of
o
z
g
of
______________________________________________________________________________________
12.25
Agf =
Ag
1 + β z Ag
=
(
2000
= 40 A/V
1 + (2000)(0.0245)
)
I o = Agf Vi = (40 ) 150 × 10 −6 ⇒ I o = 6 mA
(
)
V fb = β z I o = (0.0245) 6 × 10 −3 ⇒ V fb = 147 μ V
V∈ = Vi − V fb = 150 − 147 ⇒ V∈ = 3 μ V
______________________________________________________________________________________
12.26
IE =
Also
Then
V − Vε
(1 + hFE )
⋅ I0 = S
hFE
RE
I 0 = hFE ( AgVε )
Vε =
so
I0
hFE Ag
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V
I0
1 + hFE
⋅ I0 = S −
hFE
RE hFE Ag RE
⎡1 + hFE
⎤
VS
1
+
⎢
⎥ I0 =
hFE Ag RE ⎥⎦
RE
⎢⎣ hFE
⎡ Ag (1 + hFE ) RE + 1 ⎤
VS
⎢
⎥ I0 =
hFE Ag RE
RE
⎣⎢
⎦⎥
⎤
hFE Ag RE
hFE Ag
I0
I0
1 ⎡
=
⋅⎢
≈
⎥⇒
VS RE ⎣⎢1 + Ag (1 + hFE ) RE ⎦⎥ VS 1 + (hFE Ag ) RE
β z = RE
b.
10 =
c.
d.
5 × 105
1 + (5 × 105 ) β z
5 × 105
−1
β z = 10 5 ⇒ β z = RE = 0.099998 kΩ
5 × 10
Ag → 5.5 × 105
If
then
5.5 × 105
= 10.0000182
Agf =
1 + (5.5 × 105 )(0.099998)
ΔAgf
Agf
=
1.82 × 10 −5
⇒ 1.82 × 10−4 %
10
______________________________________________________________________________________
12.27
I E = (1 + hFE ) AgVε ,
Now
(1 + hFE ) Ag I S Ri =
IE =
Vε
− IS
RE
and Vε = I S Ri , Vε = VS − Vε = VS − I S Ri
1
⋅ (VS − I S Ri ) − I S
RE
⎡
⎤
V
Ri
+ 1⎥ I S = S
⎢(1 + hFE ) Ag Ri +
R
RE
E
⎣
⎦
Rif =
⎡
⎤
VS
R
= RE ⎢(1 + hFE ) Ag Ri + i + 1⎥
IS
RE ⎦
⎣
We have
(1 + hFE ) Ag ≈ hFE Ag = 5 ×105 mS
RE ≈ 0.1 kΩ
20 ⎤
⎡
Rif = (0.1) ⎢(5 × 105 )(20) +
+ 1⎥
0.1
⎣
⎦
so
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
Rif = 106 kΩ
Vπ
= AgVε
rπ
I X = g mVπ +
VX − (−Vε )
R0
(1)
Vε = −( I X + AgVε )( RE || Ri )
(2)
V ⎡1 + Ag ( Rε || Ri ) ⎦⎤ = − I X ( RE || Ri )
or ε ⎣
Now:
I X = g m Ag rπ Vε +
VX Vε
+
R0 R0
(1)
⎛
1 ⎞ ⎡ − I X ( RE || Ri ) ⎤ VX
I X = ⎜ g m Ag rπ + ⎟ ⎢
⎥+
R
⎢1 + Ag ( RE || Ri ) ⎦⎥ R0
0 ⎠⎣
⎝
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)
⎠⎣
⎦ ⎭⎪
⎩⎪ ⎝
R0 f = 5.04 MΩ
or
______________________________________________________________________________________
12.28
Azf =
Az
1 + β g Az
0.20 × 10 6 =
Rif =
Rof =
(
Az
)
1 + 4.25 × 10 − 6 Az
⇒ Az = 1.333 V/ μ A
Ri
500
=
⇒ Rif = 75 Ω
1 + β g Az 1 + 4.25 × 10 − 6 1.333 × 10 6
(
Ro
1 + β g Az
)(
)
= 75 Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.29
I ε = I i − I fb = 40 − 38 = 2 μA
Az =
Vo 8
V
= =4
Iε 2
μA
βg =
Azf =
I fb
Vo
=
38
μA
= 4.75
8
V
Vo
8
V
=
= 0.2
I i 40
μA
______________________________________________________________________________________
12.30
a.
Assuming V1 is at virtual ground
(−V0 ) = − I fb RF
and
(−V0 ) = − Az I ε ⇒ I ε =
I fb = I S − I ε
⎛V ⎞
V0 = ( I S − I ε ) RF = I S RF − ⎜ 0 ⎟ RF
⎝ Az ⎠
So
so
⎡ R ⎤
V0 ⎢1 + F ⎥ = I S RF
⎣ Az ⎦
V
RF
AR
= z F
Azf = 0 =
I S ⎡ RF ⎤ Az + RF
⎢1 + A ⎥
z ⎦
⎣
Azf =
or
b.
Az
Az
=
⎛ 1 ⎞ 1 + Az β g
1 + Az ⎜
⎟
⎝ RF ⎠
βg =
1
RF
5 × 104 =
c.
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
V0
Az
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
d.
Az = (0.9)(5 × 106 ) = 4.5 × 106
Azf =
ΔAzf
Azf
4.5 × 106
= 4.994 × 10 4
1 + (4.5 × 106 )(1.98 × 10 −5 )
=−
55.4939
= −1.11× 10−3 ⇒ −0.111%
5 × 104
______________________________________________________________________________________
12.31
V1 = Iε Ri , − V0 = − Az I ε ⇒ V0 = Ax Iε
I fb = I S − I ε
−V0 = V1 − I fb RF
and
− Az I ε = V1 − ( I S − Iε ) RF
⎛V ⎞
⎛V ⎞
− Az ⎜ 1 ⎟ = V1 − I S RF + ⎜ 1 ⎟ RF
R
⎝ i⎠
⎝ Ri ⎠
⎡ A R ⎤
I S RF = V1 ⎢1 + z + F ⎥
⎣ Ri Ri ⎦
Rif =
V1
RF
=
I S ⎡ Az RF ⎤
+
⎢1 +
⎥
⎣ Ri Ri ⎦
We have
Rif =
50.5 × 103
⎡
5 × 106 50.5 × 103 ⎤
⎢1 + 10 × 103 + 10 × 103 ⎥
⎣
⎦
50.5 × 103
=
⇒ Rif = 99.79 Ω
[1 + 500 + 5.05]
______________________________________________________________________________________
12.32
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.33
Ri (max) = Ri (1 + T ) = 10(1 + 104 ) ⇒ Ri (max) ≅ 105 k Ω
(a)
Ri
10
=
≅ 10−3 k Ω
1 + T 1 + 104
Ri (min) = 1Ω
Ri (min) =
Or
(b)
Ro (max) = Ro (1 + T ) = 1(1 + 104 ) ⇒ Ro (max) ≅ 104 k Ω
Ro
1
=
≅ 10−4 k Ω
1 + T 1 + 104
Ro (min) = 0.1Ω
Ro (min) =
Or
______________________________________________________________________________________
12.34
i
Ag = o
vi
Overall Transconductance Amplifier,
Series output = current signal and Shunt input = current
signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions
are shown.
______________________________________________________________________________________
12.35
I CQ = 0.2 mA, rπ =
(140)(0.026) = 18.2 k Ω , g
0.2
(a)
(1)
⎛ 1 + h FE ⎞
Vπ
V −VA
⎟
+ g m Vπ = o
= Vπ ⎜⎜
⎟
rπ
R2
⎝ rπ ⎠
(2)
Vπ
A V − Vo
= υ ∈
rπ
Ro + rπ
(3)
V A − V o V A V A − Vi
+
+
=0
R2
R1
Ri
(4) V∈ = Vi − V A
m
=
0.2
= 7.692 mA/V
0.026
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From (1),
⎛ Aυ V∈ − V o ⎞
V o − V A ⎛ 1 + h FE ⎞ ⎡ ⎛ Aυ V∈ − V o ⎞⎤
⎟
⎜
⎟
⎟ ⎢rπ ⎜
= ⎜⎜
⎟ ⎜ R + r ⎟⎥ = (1 + h FE )⎜ R + r ⎟
R2
r
⎝ o π ⎠
⎝ π ⎠ ⎣⎢ ⎝ o π ⎠⎦⎥
Using (4),
⎛ A V − Vo ⎞
V o − Vi + V∈
⎟
= (1 + h FE )⎜⎜ υ ∈
⎟
R2
⎝ R o + rπ ⎠
(Eq. 5)
⎛ 1
V
1
1 ⎞ V
From (3), V A ⎜⎜
+
+ ⎟⎟ = o + i
R
R
R
R
Ri
1
i ⎠
2
⎝ 2
⎞
⎛
(Vi − V∈ )⎜⎜ 1 + 1 + 1 ⎟⎟ =
⎝ R2
Ri ⎠
R1
V o Vi
+
R 2 Ri
(Vi − V∈ )⎛⎜ 1 + 1 + 1 ⎞⎟ = o + i
V
V
10
1
30
10
30
⎝
⎠
(Vi − V∈ )(1.1333) = Vo (0.10) + Vi (0.0333)
We find, V∈ = Vi (0.9706 ) − V o (0.08824 )
From Eq. (5) above,
Vo − Vi (1 + h FE )Vo (1 + h FE )
V
+
=
⋅ Aυ V∈ − ∈
R2
Ro + rπ
R o + rπ
R2
V o − Vi
V
141
141
+
⋅Vo =
10 5 V∈ − ∈
10
0.5 + 18.2
0.5 + 18.2
10
( )
7.640V o − Vi (0.1) = 7.540 × 10 5 V∈ = 7.540 × 10 5 [Vi (0.9706) − Vo (0.08824 )]
(
6.653705 × 10 4 V o = Vi 7.318235 × 10 5
)
V
Then o = 11.0
Vi
⎛ A V − Vo ⎞
⎟
(b) From (2), Vπ = rπ ⎜⎜ υ ∈
⎟
⎝ R o + rπ ⎠
⎛ 1 + h FE ⎞
⎟
From (1), V o = V A + R 2Vπ ⎜⎜
⎟
⎝ rπ ⎠
⎛ A V − Vo ⎞
⎟
V o = (Vi − V∈ ) + R 2 (1 + h FE )⎜⎜ υ ∈
⎟
⎝ R o + rπ ⎠
⎤
⎡ R (1 + h FE ) ⎤
⎡ R 2 (1 + h FE )Aυ
V o ⎢1 + 2
− 1⎥ (Eq. 6)
⎥ = Vi + V∈ ⎢
R o + rπ ⎦
⎦
⎣
⎣ R o + rπ
From (3),
⎛ 1
V
1
1 ⎞ V
V A ⎜⎜
+
+ ⎟⎟ = o + i
⎝ R 2 R1 Ri ⎠ R 2 Ri
⎞
⎛
(Vi − V∈ )⎜⎜ 1 + 1 + 1 ⎟⎟ =
⎝ R2
Then, using Eq. (6),
R1
Ri ⎠
V o Vi
+
R 2 Ri
⎧
⎡ R 2 (1 + h FE )Aυ
⎤⎫
− 1⎥ ⎪
⎪Vi + V∈ ⎢
⎛ 1
⎛ 1
1 ⎞
1
1 ⎞ 1 ⎪
⎣ R o + rπ
⎦⎪
⎟⎟ − V∈ ⎜
Vi ⎜⎜
+
+ ⎟⎟ =
+
⎨
⎬
⎜
(
1
)
R
h
+
FE
⎝ R 2 R1 ⎠
⎝ R 2 R1 Ri ⎠ R 2 ⎪
⎪
1+ 2
⎪
⎪
R o + rπ
⎩
⎭
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vi (1.10 ) − V∈ (1.1333) =
(
(
Vi (1.098674 ) = V∈ 1.1333 + 1.0 × 10 4
So
)
6
1 ⎧⎪Vi + V∈ 7.540106 × 10 ⎪⎫
⎨
⎬
10 ⎪⎩
75.401
⎪⎭
)
Vi
= 9.103 × 10 3
V∈
⎛V ⎞
Then Rif = Ri ⎜⎜ i ⎟⎟ = 30 × 10 3 9.103 × 10 3 ⇒ Rif = 273 M Ω
⎝ V∈ ⎠
(
(c) I x + g mVπ +
)(
)
Vπ V x − V A
=
rπ
R2
⎛ A V −Vx ⎞
⎟
From (2), Vπ = rπ ⎜⎜ υ ∈
⎟
⎝ R o + rπ ⎠
V −V
(1 + h FE )
( Aυ V∈ − V x ) = x A
Then, I x +
R o + rπ
R2
Now, V A = −V∈ and V∈ = −V x (0.08824)
So, I x +
V
V (0.08824)
141
10 5 (− 0.08824V x ) − V x = x − x
0.5 + 18.2
10
10
[
(
]
)
I x = V x 0.091176 + 6.654 × 10 4
V
R of = x = 15 μ Ω
Ix
______________________________________________________________________________________
12.36
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
0.5
I C1
0.5
= 19.23 mA / V
0.026
(100)(0.026)
rπ 3 =
= 1.3 kΩ
2
2
g m3 =
= 76.92 mA / V
0.026
g m1 = g m 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
r
⎝ π1
⎠
Vπ 1
( RS + rπ 1 ) − Vπ 2 + Vb 2
Vi =
rπ 1
(1)
⎛ R ⎞
Vi = Vπ 1 ⎜1 + S ⎟ − Vπ 2 + Vb 2
⎝ rπ 1 ⎠
or
But Vπ 2 = −Vπ 1
so
⎛
R ⎞
Vi = Vπ 1 ⎜ 2 + S ⎟ + Vb 2
rπ 1 ⎠
⎝
(2)
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 ⎞
⎛ 1
1 ⎞ V
(V02 − V0 ) ⎜
+ ⎟ − b2
⎟ = V0 ⎜
r
R
R
R2
2 ⎠
⎝ L
⎝ π3 ⎠
(4)
Vb 2 − V0 Vb 2 Vπ 2
+
+
=0
R2
R1 rπ 1
(5)
Substitute numbers into (2), (3), (4) and (5):
1 ⎞
⎛
Vi = −Vπ 2 ⎜ 2 +
⎟ + Vb 2
5.2 ⎠
⎝
Vi = −Vπ 2 (2.192) + Vb 2
(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
⎛ 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)
(3)
From (2): Vb 2 = Vi + Vπ 2 (2.192). Substitute in (4) and (5) to obtain:
V02 (77.69) = V0 (77.96) − [Vi + Vπ 2 (2.192)](0.02)
(4′)
[Vi + Vπ 2 (2.192)](0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0
(5′)
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′)
(0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 (5′)
From (3): V02 = V0 (0.9455) − Vπ 2 (23.64). Substitute for V02 in (4′) to obtain:
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(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)
(5′)
Vπ 2 :
Next, solve
for
(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
Avf =
V0 484.12
=
⇒ Avf = 5.68
Vi
85.18
______________________________________________________________________________________
12.37
RTH = R1 || R2 = 400 || 75 = 63.2 kΩ
a.
⎛ R2 ⎞
⎛ 75 ⎞
VTH = ⎜
⎟ VCC = ⎜
⎟ (10) = 1.579 V
⎝ 75 + 400 ⎠
⎝ R1 + R2 ⎠
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
VC 2 = 10 − (0.497)(13) = 3.54 V
IC 2 ≈
IC 3 ≈
3.54 − 0.7
= 2.03 mA
1.4
Then
(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
rπ 1 =
b.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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)
V +V V
g m 2Vπ 2 + π 3 0 + π 3 = 0
RC 2
rπ 3
(4)
Vπ 3
V
V −V
+ g m 3Vπ 3 = 0 + 0 ε 1
rπ 3
RE 3
RF
(5)
Substitute numbers in (2), (3), (4) and (5):
or
1⎞ V
⎛ 1
⎞
⎛ 1
Vπ 1 ⎜
+ 32.81⎟ = (Vi − Vπ 1 ) ⎜
+ ⎟− 0
⎝ 3.66
⎠
⎝ 0.5 10 ⎠ 10
Vπ 1 (35.18) = Vi (2.10) − V0 (0.10)
(2)
Vπ 2 = −(32.81)Vπ 1 (88 || 6.28)
V
or π 2 = −Vπ 1 (120.2)
(3)
(19.12)Vπ 2 +
or
Vπ 3 V0 Vπ 3
+ +
=0
13 13 1.54
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
Vπ 3 (78.73) = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10)
Now substituting
Vπ 2 = −Vπ 1 (120.2)
(5)
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)
Vπ 1 (2.49 × 105 ) − V0 (9.152) = −Vi (0.10)
or
Then
Vπ 1 = V0 (3.674 × 10−5 ) − Vi (4.014 × 10−7 )
−5
Now substituting Vπ 1 = V0 (3.674 × 10 )
−Vi (4.014 × 10−7 )
in (2):
(35.18)[V0 (3.674) × 10−5 ) − Vi (4.014 × 10−7 )]
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
= 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
Ii
and I i = I RB1 + I b1
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
Vi
V (7.60 × 10−4 )
+ i
63.2
3.66
1
=
0.01582 + 2.077 × 10 −4
Rif = 62.4 kΩ
Rif =
or
d.
To determine
R0 f :
Equation (1) is modified to Vπ 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to:
Vπ 3 (78.73) + I X = V0 (0.8143) + Vπ 1 (0.10)
(5)
Now
Vπ 1 (35.18) = −V0 (0.10)
Vπ 2 = −Vπ 1 (120.2)
(2)
(3)
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
Vπ 3 = −V0 (9.101)
or
So then
(4)
−V0 (9.101)(78.73) + I X
= V0 (0.8143) + (0.10)(−V0 )(0.002843)
or
I X = V0 (717.3)
(5)
or
R0 f =
V0
= 0.00139 kΩ ⇒ R0 f = 1.39 Ω
IX
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
12.38
(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)
⎛ 1
⎞ V
V −V
+ g m1 ⎟ = A + A O
RF
⎝ rπ 1
⎠ RE
(Vi − VA ) ⎜
(2)
⎛ 1
1 ⎞
VB ⎜
+
⎟ + g m1 (Vi − VA ) = 0
⎝ RC1 rπ 2 ⎠
(3)
⎛ 1
VO
1 ⎞
+
=0
VC ⎜
⎟ + g m 2VB −
R
r
rπ 3
2
3
π
C
⎝
⎠
⎛
(4)
(VC − VO ) ⎜ g m3 +
⎝
1 ⎞ VO − VA
⎟=
rπ 3 ⎠
RF
(100)(0.026)
14.3
rπ 1 =
= 0.182 K g m1 =
= 550 mA/V
14.3
0.026
(100)(0.026)
4.62
rπ 2 =
= 0.563 K g m 2 =
= 178 mA/V
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
(Vi − VA ) ⎛⎜
1.2
⎝ 0.182
⎠ 0.05
(1)
(2)
1 ⎞
⎛ 1
VB ⎜
+
⎟ + (550)(Vi − VA ) = 0
⎝ 0.3 0.563 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VO
1 ⎞
⎛ 1
VC ⎜
+
=0
⎟ + 178 VB −
0.582
⎝ 0.65 0.582 ⎠
(3)
(VC − VO ) ⎛⎜172 +
⎝
(4)
1 ⎞ VO − VA
⎟=
0.582 ⎠
1.2
(Vi − VA )(555.5) = VA (20) + (VA − VO )(0.8333)
(1)
VB (5.109) + 550(Vi − VA ) = 0
(2)
VC (3.257) + 178VB − VO (1.718) = 0
(3)
(4)
(VC − VO )(173.7) = (VO − VA )(0.8333)
(1)
Vi (555.5) + VO (0.8333) = VA (576.3)
VB (5.109) + 550Vi − VA (550) = 0
(2)
VC (3.257) + 178VB − VO (1.718) = 0
(3)
VC (173.7) + VA (0.8333) = VO (174.5)
(4)
VB = VA (107.7) − Vi (107.7)
From (2)
VC = VO (1.0046) − VA (0.004797)
From (4)
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
Avf =
RF + RE 1.2 + 0.05
=
= 25.0
0.05
RE
Rif =
(b)
We have
Vi
Ii
and
Ii =
Vπ 1 Vi − VA
=
rπ 1
rπ 1
VA = Vi (1.00) − Vo (0.00008106)
= Vi (1.00) − (23.6)Vi (0.00008106)
VA = Vi (0.99809)
Then
Ii =
Vi (1 − 0.99809)
= Vi (0.01051)
0.182
Vi
⇒ Rif = 95.1 K
Vi (0.01051)
Rof ,
Vi = 0
Rif =
To find
set
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
12.39
(
)
g m υ gs1 + υ gs 2 = 0 ⇒ υ gs 2 = −υ gs1
⎛ R2 ⎞
⎟⎟ ⋅ V o
Vi = υ gs1 − υ gs 2 + ⎜⎜
⎝ R1 + R 2 ⎠
1
Vi = −2υ gs 2 + Vo
2
Vo
V
+ g mυ gs 2 + o = 0
Also
R1 + R 2
RD
υ gs 2 = −
Then Vi =
Vo ⎛ 1
1 ⎞
⎟
⎜⎜
+
g m ⎝ R D R1 + R 2 ⎟⎠
2Vo ⎛ 1
1 ⎞ 1
⎟ + Vo
⎜⎜
+
g m ⎝ R D R1 + R 2 ⎟⎠ 2
We have R1 + R 2 >> R D
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 2
1⎞
Vi ≅ Vo ⎜⎜
+ ⎟⎟
⎝ g m RD 2 ⎠
V
g m RD
1
=
Aυf = o =
Vi ⎛ 2
1 ⎞ ⎛2 + 1 g R ⎞
⎟ ⎜
⎜
m D ⎟
⎜ g R + 2⎟ ⎝
2
⎠
⎠
⎝ m D
Now g m = 2 K n I DQ = 2 (0.5)(0.5) = 1.0 mA/V
Aυf =
(1)(7 )
= 1.273
1
⎡
⎤
(
)(
)
2
+
1
7
⎢
⎥
2
⎣
⎦
______________________________________________________________________________________
12.40
(a) Neglect base currents
VGG − VGS
= I D1 + I C 2
RL
I D1 =
I C2 =
5 − VD 5 − VD
=
5
R D1
5 − (V D + 0.7 ) 4.3 − V D
=
1.6
RE 2
V D = 5 − 5 I D1 ⇒ I C 2 =
Then
4.3 − (5 − 5 I D1 )
= −0.4375 + 3.125I D1
1.6
2.5 − VGS
= I D1 + (− 0.4375 + 3.125 I D1 )
1.2
3.025 = VGS + 4.95 I D1 = VGS + 4.95(K n )(VGS − 0.5)
2
We find 7.425VGS2 − 6.425VGS − 1.16875 = 0 ⇒ VGS = 1.0197 V
Then I D1 = (1.5)(1.0197 − 0.5) = 0.405 mA
I C 2 = −0.4375 + 3.125(0.405) = 0.828 mA
2
(b)
g m1 = 2 K n I D1 = 2 (1.5)(0.405) = 1.559 mA/V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I
0.828
= 31.85 mA/V
g m2 = C 2 =
0.026
VT
rπ 2 =
β VT
=
I C2
(120)(0.026) = 3.768 k Ω
0.828
(1) V o = ( g m1V gs + g m 2Vπ ) R L
(2) Vi = V gs + Vo ⇒ V gs = Vi − Vo
(3)
V
VA
+ g m1V gs = π
R D1
rπ
(4)
V A + Vπ Vπ
+
+ g m 2 Vπ = 0
RE 2
rπ
So
−VA
1
⎞
⎛ 1
= Vπ ⎜
+
+ 31.85 ⎟ = Vπ (32.74 )
1.6
⎠
⎝ 1.6 3.768
V A = −52.385Vπ
−52.385Vπ
V
+ 1.559V gs = π
5
3.768
or 1.559V gs = Vπ (0.2654 + 10.477 )
(3)
so Vπ = 0.1451V gs = 0.1451(Vi − V o )
(1) V o = [1.559(Vi − V o ) + 31.85(0.1451)(Vi − V o )](1.2)
V
Then Aυ = o = 0.8812
Vi
(c)Set Vi = 0
I x + g m 2Vπ + g m1V gs =
Vx
RL
V gs = −V x
Vπ = 0.1451V gs = −0.1451V x
⎡ 1
⎤
I x = Vx ⎢
+ (31.85)(0.1451) + (1.559 )⎥ = V x (7.014 )
⎣1.2
⎦
V
R o = x = 143 Ω
Ix
______________________________________________________________________________________
12.41
g m RS
1 + g m RS
(a) (i) Aυ f =
g m = 2 K n I DQ = 2 (1.5)(1.2) = 2.683 mA/V
Aυ f =
(2.683)(1.5) = 0.801
1 + (2.683)(1.5)
(ii) R o f =
1
1
RS =
1.5 = 0.3727 1.5
gm
2.683
Ro f = 299 Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) (i) g m = 2 (2.25)(1.2 ) = 3.286 mA/V
Aυ f =
ΔAυ f
Aυ f
(3.286)(1.5) = 0.8313 mA/V
1 + (3.286)(1.5)
× 100% = +3.78%
1
1.5 = 0.3043 1.5
3.286
R o f = 253 Ω
(ii) Ro f =
ΔR o f
Ro f
× 100% = −15.4%
______________________________________________________________________________________
12.42
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
I B1 =
= 0.0187 mA
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)
I C 2 = (50)(0.03705) = 1.85 mA
IB2 =
(50)(0.026)
= 1.39 kΩ;
0.935
(50)(0.026)
rπ 2 =
= 0.703 kΩ
1.85
0.935
g m1 =
= 35.96 mA / V
0.026
1.85
gm2 =
= 71.15 mA / V
0.026
rπ 1 =
VS = Vπ 1 + Ve
Vπ 1
V V − V0
+ g m1Vπ 1 = e + e
rπ 1
R1
RF
(1)
(2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
g m1Vπ 1 +
Vπ 2 Vπ 2 Vπ 2
+
+
=0
RC1 RB 2 rπ 2
(3)
V
V −V
g m 2Vπ 2 + 0 + 0 e = 0
RC 2
RF
(4)
Substitute numerical values in (2), (3) and (4):
Ve = VS − Vπ 1
(1)
Vπ 1
1 ⎞
⎛ 1
⎛ 1 ⎞
+ (35.96)Vπ 1 = (VS − Vπ 1 ) ⎜
+
⎟ − V0 ⎜
⎟
1.39
⎝ 0.1 4.7 ⎠
⎝ 4.7 ⎠
or
Vπ 1 (46.89) = VS (10.213) − V0 (0.2128)
(2)
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
(71.15)Vπ 2 + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0
Vπ 2 = −Vπ 1 (22.85)
(4)
From (3):
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
Vπ 1 = VS (0.2178) − V0 (0.004538)
From (2):
Then
−(1625.6)[VS (0.2178) − V0 (0.004538)] + V0 (0.4255) − VS (0.2128) = 0
−VS (354.3) + V0 (7.802) = 0
or
Finally
⇒
V0
= 45.4
VS
______________________________________________________________________________________
12.43
For example, use a 2-stage amplifier. Each stage is shown in Fig. 12.29.
______________________________________________________________________________________
12.44
⎛ k ′ ⎞⎛ W ⎞
2
I DQ1 = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN )
2
L
⎝
⎠
⎝ ⎠
⎛W ⎞
Let all ⎜ ⎟ = 20
⎝L⎠
⎛ 0.1 ⎞
2
0.5 = ⎜
⎟(20)(VGS − 1.5) ⇒ VGS1, 2 = 2.207 V
⎝ 2 ⎠
⎛ 0.1 ⎞
2
I DQ 3 = 2 = ⎜
⎟(20)(VGS 3 − 1.5) ⇒ VGS 3 = 2.914 V
2
⎝
⎠
Want VG 3 = 2.914 V, then
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12 − 2.914
RD =
= 18.2 k Ω
0.5
g m1υ gs1 + g m1υ gs 2 = 0 ⇒ υ gs 2 = −υ gs1
Vi = υ gs1 − υ gs 2 + V A = −2υ gs 2 + V A
⎛ R1 ⎞
⎟⎟ ⋅ V o
V A = ⎜⎜
⎝ R1 + R 2 ⎠
Vo
Vo
+
= g m3υ gs 3
R L R1 + R 2
υ gs 3 = −(g m1υ gs 2 R D ) − Vo
⎛ 1
1 ⎞
⎟⎟ = g m3 − g m1υ gs 2 R D − V o
V o ⎜⎜
+
⎝ R L R1 + R 2 ⎠
(
υ gs 2 =
)
⎛ R
⎞
1
(V A − Vi ) = 1 ⎜⎜ 1 ⎟⎟ ⋅Vo − 1 ⋅Vi
2
2 ⎝ R1 + R 2 ⎠
2
⎡ 1 ⎛ R1 ⎞
⎤
⎛ 1
1 ⎞
1
⎟⎟ ⋅ Vo − ⋅ Vi ⎥ − g m3Vo
⎟⎟ = − g m1 g m3 R D ⎢ ⎜⎜
V o ⎜⎜
+
2
⎢⎣ 2 ⎝ R1 + R 2 ⎠
⎥⎦
⎝ R L R1 + R 2 ⎠
⎡ 1
⎤ 1
⎛ R1 ⎞
1
1
⎟⎟ + g m3 ⎥ = g m1 g m3 R DVi
+
+ g m1 g m3 R D ⎜⎜
Vo ⎢
⎢⎣ R L R1 + R 2 2
⎥⎦ 2
⎝ R1 + R 2 ⎠
1
g m1 g m 3 R D
Vo
2
Aυ f =
=
Vi
⎡ 1
⎤
⎛ R1 ⎞
1
1
⎟⎟ + g m3 ⎥
+
+ g m1 g m3 R D ⎜⎜
⎢
⎢⎣ R L R1 + R 2 2
⎥⎦
⎝ R1 + R 2 ⎠
⎛ 0.1 ⎞
g m1 = 2 ⎜
⎟(20)(0.5) = 1.414 mA/V
⎝ 2 ⎠
⎛ 0.1 ⎞
g m3 = 2 ⎜
⎟(20 )(2 ) = 2.828 mA/V
⎝ 2 ⎠
1
(1.414)(2.828)(18.2)
2
Then 8 =
⎡1
⎤
⎛ 15 ⎞
1
1
⎟⎟ + 2.828⎥
+ (1.414 )(2.828)(18.2 )⎜⎜
⎢ +
⎝ R1 + R 2 ⎠
⎣⎢10 R1 + R 2 2
⎦⎥
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
36.389
8=
⎡
546.8 ⎤
⎥
⎢2.928 +
R1 + R 2 ⎦
⎣
⇒ R1 + R 2 = 337 k Ω ⇒ R 2 = 322 k Ω
______________________________________________________________________________________
12.45
Io
R
R
5
=
= 83.33 = 1 + 1 ⇒ 1 = 82.33
I s 0.06
R2
R2
For example, let R 2 = 3 k Ω , R 2 = 247 k Ω
______________________________________________________________________________________
12.46
(a) (1) V A = (I D1 + I D 2 )R D 2
(2) I D1 R D1 = V SG 2
(V − VGS1 ) − V A
= I D1
(3) G
RF
Now VG − VGS1 − (I D1 + I D 2 )R D 2 = I D1 R F
And VGS 2 =
I D2
− VTP = I D1 R D1
Kp
I D 2 (0.3162 ) + 1 = I D1 (0.525)
I D 2 = I D1 (1.660 ) − 3.162
I D 2 = [I D1 (1.660 ) − 3.162]
2
Then VG − VGS 1 − [I D1 (1.660 ) − 3.162] R D 2 = I D1 (R D 2 + R F )
2
I D1 = K n (VGS1 − VTN )
2
7.6 − VGS1 − [I D1 (1.660 ) − 3.162] (0.25) = I D1 (0.75)
2
I D1 = 10(VGS1 − 1)
By trial and error, I D1 ≅ 3.98 mA
Then V SG 2 = I D1 R D1 = (3.98)(0.525) = 2.0895 V
2
I D1 = (10)(2.0895 − 1) = 11.87 mA
2
(b)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I o = g m 2V sg 2
V sg 2 = g m1V gs1 R D1
(
)
V A V A − − V gs1
+
RD2
RF
V A + V gs1
=0
I i + g m1V gs1 +
RF
Io =
V gs1 ⎤
⎡
or V A = − R F ⎢ I i + g m1V gs1 +
⎥
RF ⎦
⎣
⎛ 1
1 ⎞ V gs1
⎟⎟ +
+
I o = V A ⎜⎜
⎝ RD2 RF ⎠ RF
V gs1 ⎞⎛ 1
⎛
1 ⎞ V gs1
⎟⎜
⎟+
+
I o = − R F ⎜⎜ I i + g m1V gs1 +
⎜
⎟
R F ⎠⎝ R D 2 R F ⎟⎠ R F
⎝
V sg 2
I
1
Now V gs1 =
= o ⋅
g m1 R D1 g m 2 g m1 R D1
⎡⎛
⎛ 1
1 ⎞
1 ⎞⎛ 1
1 ⎞ 1 ⎤
⎟⎟ − R F V gs1 ⎢⎜⎜ g m1 +
⎟⎟⎜⎜
⎟−
+
+
I o = − R F I i ⎜⎜
⎥
R F ⎠⎝ R D 2 R F ⎟⎠ R F2 ⎦⎥
⎝ RD2 RF ⎠
⎣⎢⎝
⎡
⎛ g m1 g m1
⎛
R ⎞
RF
1 ⎞⎤
⎜⎜
⎟⎟⎥ = − I i ⎜⎜1 + F ⎟⎟
Then I o ⎢1 +
+
+
⎝ RD2 ⎠
⎣⎢ g m1 g m 2 R D1 ⎝ R D 2 R F R F R D 2 ⎠⎦⎥
So Ai =
Io
=
Ii
− g m 2 R D1
1
1+
g m1 (R F + R D 2 )
+
g m 2 R D1 R D 2
RF + RD2
(c) g m1 = 2 K n I D1 = 2 (10 )(3.98) = 12.62 mA/V
g m 2 = 2 K p I D 2 = 2 (10 )(11.87 ) = 21.79 mA/V
− (21.79 )(0.525)
(21.79)(0.525)(0.25)
1
1+
+
(12.62)(0.75)
0.75
Ai = −2.33
______________________________________________________________________________________
Ai =
12.47
(a) I Q = I D1 + I D 2
V SG 2 = I D1 R D
I D 2 = K p (V SG 2 + VTP ) = K p (I D1 R D + VTP )
2
[
]
I Q = I D1 + K p I D2 1 R D2 − 2 I D1 R D + 1
We find 2.756 I
2
D1
− 9.5 I D1 − 6 = 0
⇒ I D1 = 3.99 mA and I D 2 = 12.01 mA
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
I i + g m1V gs1 + I o = 0
V sg 2 = g m1V gs1 R D ⇒ V gs1 =
V sg 2
g m1 R D
I o = g m 2V sg 2
Then V gs1 =
Io
g m1 g m 2 R D
⎛
⎞
Io
⎟ + Io = 0
I i + g m1 ⎜⎜
⎟
⎝ g m1 g m 2 R D ⎠
− g m2 R D
I
−1
=
So that Ai = o =
Ii ⎛
1 ⎞ 1 + g m2 R D
⎜1 +
⎟
⎜ g R ⎟
m2 D ⎠
⎝
We find g m 2 = 2 K p I D 2 = 2 (10 )(12.01) = 21.92 mA/V
− (21.92)(0.525)
= −0.920
1 + (21.92 )(0.525)
______________________________________________________________________________________
Ai =
12.48
(a) Neglect base currents
I Q = I C1 + I C 2
V EB 2 = I C1 RC
⎛I R ⎞
I Q = I C1 + I S 2 exp⎜⎜ C1 C ⎟⎟
⎝ VT ⎠
⎡ I (200 ) ⎤
16 × 10 −3 = I C1 + 10 −15 exp ⎢ C1
⎥
⎣ 0.026 ⎦
By trial and error, I C1 = 3.92 mA and I C 2 = 12.08 mA
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
I o = g m 2 Vπ 2
Vπ 2 = g m1Vπ 1 (rπ 2 RC )
Then I o = g m1 g m 2Vπ 1 (rπ 2 RC )
I i + I o + g m1Vπ 1 +
Vπ 1
=0
rπ 1
⎛ 1 + β1 ⎞
⎟=0
I i + I o + Vπ 1 ⎜⎜
⎟
⎝ rπ 1 ⎠
Io
Now Vπ 1 =
g m1 g m 2 (rπ 2 RC )
⎡
⎛ 1 + β 1 ⎞⎤
1
⎜
⎟⎥ = 0
Then I i + I o ⎢1 +
⎜
⎟
⎢⎣ g m1 g m 2 (rπ 2 RC ) ⎝ rπ 1 ⎠⎥⎦
⎛ β ⎞
− ⎜⎜ 1 ⎟⎟(g m 2 )(rπ 2 RC )
I
⎝ 1 + β1 ⎠
So Ai = o =
Ii
⎛ β ⎞
1 + ⎜⎜ 1 ⎟⎟(g m 2 )(rπ 2 RC )
⎝ 1 + β1 ⎠
I
12.08
= 464.6 mA/V
(c) g m 2 = C 2 =
0.026
VT
rπ 2 =
β 2 VT
I C2
=
(180)(0.026) = 0.3874 k Ω
12.08
rπ 2 RC = 387.4 200 = 131.9 Ω
⎛ 180 ⎞
−⎜
⎟(464.6 )(0.1319 )
⎝ 181 ⎠
Ai =
= −0.984
⎛ 180 ⎞
(
)(
)
1+ ⎜
464
.
6
0
.
1319
⎟
⎝ 181 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.49
⎛ h
⎞
⎛ 100 ⎞
I C1 = ⎜ FE ⎟ I E1 = ⎜
⎟ (0.2) = 0.198 mA
1
+
h
⎝ 101 ⎠
FE ⎠
⎝
VC1 = 10 − (0.198)(40) = 2.08 V
2.08 − 0.7
= 1.38 mA
1
⎛ 100 ⎞
IC 2 = ⎜
⎟ (1.38) = 1.37 mA
⎝ 101 ⎠
IE2 =
(a)
For Q1 :
(100)(0.026)
= 13.1 k Ω
0.198
0.198
= 7.62 mA / V
g m1 =
0.026
Q2 :
rπ 1 =
For
(100)(0.026)
= 1.90 k Ω
1.37
1.37
gm2 =
= 52.7 mA / V
0.026
rπ 2 =
(b)
IS =
Vπ 1 Vπ 1 Vπ 1 − Ve
+
+
RS rπ 1
RF
V + Ve Vπ 2
+
=0
g m1Vπ 1 + π 2
RC1
rπ 2
(1)
(2)
Vπ 2
V V −V
+ g m 2Vπ 2 = e + e π 1
rπ 2
RE
RF
(3)
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)
From (3),
Ve = Vπ 2 (48.39) + Vπ 1 (0.0909)
Substituting into (1),
(3)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I S = Vπ 1 (0.2763) − (0.10) [Vπ 2 (48.39) + Vπ 1 (0.0909) ]
or
I S = Vπ 1 (0.2672) − Vπ 2 (4.839)
(1′)
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) (2′)
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
⎛ 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
Aif =
IO
= 8.60
Is
Ri =
(c)
We had
Vπ 1
IS
and
Ri = RS || Rif
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
Ri =
Vπ 1
1
=
= 0.04713
21.22
IS
Finally
0.04713 =
10 Rif
10 + Rif
⇒ Rif = 47.4 Ω
______________________________________________________________________________________
12.50
(a)
Ii =
(1)
Vπ 1
V −V
+ π 1 e2
RF
RS RB1 rπ 1
g m1Vπ 1 +
(2)
(3)
VC1
V
+ π2 = 0
RC1 RB 2 rπ 2
Vπ 2
V
V −V
+ g m 2Vπ 2 = e 2 + e 2 π 1
rπ 2
RE 2
RF
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(4)
Now
⎛ RC 2 ⎞
I o = −( g m 2Vπ 2 ) ⎜
⎟
⎝ RC 2 + RL ⎠
Ii =
(1)′
Vπ 1
V
− e2
RS RB1 rπ 1 RF RF
⎛
⎞
RF
⎟V − I R
Ve 2 = ⎜
⎜ RS RB1 rπ 1 RF ⎟ π 1 i F
⎝
⎠
So
Now, from (2)
g m1Vπ 1 +
(2)′
Also
(3)′
And
Vπ 2 + Ve 2 Vπ 2
+
=0
Rc1 RB 2 rπ 2
⎛
Vπ 2
Ve 2
1 ⎞
+
=0
⎜ g m1 +
⎟ Vπ 1 +
r
R
R
R
r
π2 ⎠
C1 RB 2
⎝
C1
B2 π 2
⎛
⎛ 1
Vπ 1
1 ⎞
1 ⎞
= Ve 2 ⎜
+
⎜ gm2 +
⎟ Vπ 2 +
⎟
r
R
R
R
F
F ⎠
π2 ⎠
⎝ E2
⎝
⎛ I ⎞⎛ R + RL ⎞
Vπ 2 = − ⎜ O ⎟⎜ C 2
⎟
⎝ g m 2 ⎠⎝ RC 2 ⎠
(4)′
Substitute (1)′ into (2)′ and (3)′
(2)″
⎡
⎤
⎛
Vπ 2
RF
1 ⎞
1
⎢
+
− Vπ 1 − I i RF ⎥ = 0
⎜ g m1 +
⎟ Vπ 1 +
rπ 2 ⎠
RC1 RB 2 rπ 2 RC1 RB 2 ⎢ RS RB1 rπ 1 RF
⎥
⎝
⎣
⎦
(3)″
⎤
⎛
Vπ 1 ⎛ 1
RF
1 ⎞
1 ⎞⎡
=⎜
+
− Vπ 1 − I i RF ⎥
⎜ 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)′.
(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
I BQ 2 =
= 0.01265 mA
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
g m1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(100)(0.026)
= 2.34 K
1.11
(100)(0.026)
= 2.06 K
rπ 2 =
1.265
RC1 RB 2 = 2 12.75 = 1.729 K
rπ 1 =
Now
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)″
Vπ 2
1 ⎞
1 ⎡ 10
⎛
⎤
+
⎜ 42.69 +
⎟ Vπ 1 +
⎢1.695 ⋅ Vπ 1 − I i (10) ⎥
2.06
0.940
1.729
⎝
⎠
⎣
⎦
46.587Vπ 1 + 1.064Vπ 2 − 5.784 I i = 0
(3)″
Vπ 1 ⎛ 1
1 ⎞ ⎡ 10
⎛ 101 ⎞
⎤
=⎜
+ ⎟
⋅ Vπ 1 − I i (10) ⎥
⎜
⎟ Vπ 2 +
10 ⎝ 0.5 10 ⎠ ⎣⎢1.695
⎝ 2.06 ⎠
⎦
49.03Vπ 2 = 12.29Vπ 1 − 21I i
From (2)″ Vπ 1 = (0.1242) I i − (0.02284)Vπ 2
Then
(3)″
49.03Vπ 2 = 12.29 [ (0.1242) I i − (0.02284)Vπ 2 ] − 21I i
49.31Vπ 2 = −19.47 I i
From (4)′
Then
⎛ I ⎞⎛ 4 + 4 ⎞
Vπ 2 = − ⎜ o ⎟ ⎜
⎟ = −(0.0411) I o
⎝ 48.65 ⎠ ⎝ 4 ⎠
(49.31) [ −(0.0411) I o ] = −19.47 I i
Io
= Ai = 9.61
Ii
______________________________________________________________________________________
12.51
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)
I C1 =
= 1.75 mA
9.98 + (121)(1)
VC1 = 10 − (1.75 )( 3) = 4.75 V
4.75 − 0.7
= 0.50 mA
8.1
(120)(0.026)
rπ 1 =
= 1.78 kΩ
1.75
1.75
g m1 =
= 67.31 mA / V
0.026
IC 2 ≈
(120)(0.026)
= 6.24 kΩ
0.50
0.50
gm2 =
= 19.23 mA / V
0.026
rπ 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
and
VS − Vπ 1
Vπ 1
V −V
=
+ π 1 e2
RS
RB || rπ 1
RF
(1)
V + Ve 2 Vπ 2
g m1Vπ 1 + π 2
+
=0
RC1
rπ 2
(2)
Vπ 2
V
V −V
+ g m 2Vπ 2 = ε 2 + ε 2 π 1
rπ 2
RE 2
RF
(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)
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
(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
and
or
Vπ 1 (68.91) + Vπ 2 (0.4936) − VS (0.6666) = 0
(2′)
Vπ 2 (19.39) = (0.9568) [Vπ 1 (4.813) − VS (2.00) ] − Vπ 1 (0.8333)
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V0 = −(19.23)(4)(−VS )(0.09670)
Then
V0
= 7.44
VS
______________________________________________________________________________________
12.52
Using the circuit from Problem 12.51, we have
Rif =
Vπ 1
IS
.
V − Vπ 1
IS = S
RS
Where
From Problem 12.51
Vπ 1 = VS (0.009673) − Vπ 2 (0.007163)
= 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)
or
Rif = 6.29 Ω
______________________________________________________________________________________
12.53
RTH = 1.4 || 17.9 = 1.298 kΩ
⎛ 1.4 ⎞
VTH = ⎜
⎟ (10) = 0.7254 V
⎝ 1.4 + 17.9 ⎠
0.7254 − 0.7
I B1 =
= 0.0196 mA
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)
rπ 2 =
= 0.408 kΩ
3.19
3.19
gm2 =
= 123 mA / V
0.026
IE2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
IS =
Vπ 1
V −V
+ π1 1
R1 || R2 || rπ 1
RF
V
V + Ve 2
g m1Vπ 1 + π 2 + π 2
=0
rπ 2
RC1
(1)
(2)
Vπ 2
V −V
+ g m 2Vπ 2 = e 2 1
rπ 2
RE1
(3)
Ve 2 − Vπ 1
V −V
V
= 1 + 1 π1
RE1
RE 2
RF
(4)
Enter numerical values in (1), (2), (3) and (4):
IS =
or
Vπ 1
V −V
+ π1 1
17.9 ||1.4 || 1.33
5
I S = Vπ 1 (1.722) − V1 (0.20)
(1)
V
V + Ve 2
(37.7)Vπ 1 + π 2 + π 2
=0
0.408
7
or
or
Vπ 1 (37.7) + Vπ 2 (2.594) + Ve 2 (0.1429) = 0 (2)
Vπ 2
V −V
+ (123)Vπ 2 = e 2 1
0.408
0.25
Vπ 2 (125.5) = Ve 2 (4) − V1 (4)
(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)]
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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)
Vπ 2 (151.3) = −Vπ 1 (374.55)
or
Then
Vπ 1 = −Vπ 2 (0.4040) so that
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 ⎠
Vπ 2 = −(0.01552) I 0
or
Then
I0
I
1
=
⇒ 0 = 5.33
I S (0.01552)(12.10)
IS
______________________________________________________________________________________
12.54
For example, use the circuit shown in Figure P12.49
______________________________________________________________________________________
12.55
(a)
Vi − V L V L − V O
=
R1
RF
⎛ 1
VO
1 ⎞ Vi
⎟⎟ −
= V L ⎜⎜
+
RF
⎝ R1 R F ⎠ R1
⎛ R ⎞ ⎛R ⎞
So (1) VO = V L ⎜⎜1 + F ⎟⎟ − ⎜⎜ F ⎟⎟Vi
R1 ⎠ ⎝ R1 ⎠
⎝
VO − V L V L V L
=
+
R3
R L R2
⎛
R
R ⎞
So (2) VO = V L ⎜⎜1 + 3 + 3 ⎟⎟
⎝ R L R2 ⎠
Then, from (1) = (2)
⎛
⎛ R ⎞ ⎛R ⎞
R
R ⎞
V L ⎜⎜1 + F ⎟⎟ − ⎜⎜ F ⎟⎟Vi = V L ⎜⎜1 + 3 + 3 ⎟⎟
R1 ⎠ ⎝ R1 ⎠
⎝ R L R2 ⎠
⎝
⎛ R
R
R ⎞ ⎛R ⎞
V L ⎜⎜1 + F − 1 − 3 − 3 ⎟⎟ = ⎜⎜ F ⎟⎟Vi
R
R
R
L
1
2 ⎠
⎝ R1 ⎠
⎝
Now V L = I O R L
⎛R
R
R ⎞ ⎛R ⎞
I O R L ⎜⎜ F − 3 − 3 ⎟⎟ = ⎜⎜ F ⎟⎟Vi
R
R
R
2 ⎠
L
⎝ R1 ⎠
⎝ 1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
RF
R1
IO
=
Vi ⎛ R L R F
R R ⎞
⎜⎜
− R3 − L 3 ⎟⎟
R
R2 ⎠
1
⎝
R
R
(b) Set F = 3
R1
R2
Ag f =
RF
Then Ag f =
Or Ag f =
(c) For
R1 − R F
=
R1 R3
− R3
I O −1
=
Vi
R2
IO
−1
= −0.5 mA/V =
⇒ R2 = 2 k Ω
Vi
R2
R
RF
= 3 , set R3 = 2 k Ω and R1 = R F = 10 k Ω
R1
R2
______________________________________________________________________________________
For
12.56
(a) V D1 = 3 − (1)(1.6) = 1.40 V, V SG 3 = (1)(1.6) = 1.6 V
I DQ 3 = K p (V SG 3 + VTP ) = (10 )(1.6 − 0.5) = 12.1 mA
2
2
VG = (12.1)(0.248) − 3 ≅ 0
(b) g m1V gs1 + g m1V gs 2 = 0 ⇒ V gs 2 = −V gs1
V sg 3 = g m1V gs1 R D
I o = g m3V sg 3 = g m1 g m3 R DV gs1
Vi = V gs1 − V gs 2 + I o R L = 2V gs1 + I o R L
Vi − I o R L
2
⎛ V − I o RL ⎞
⎟⎟
I o = g m1 g m3 R D ⎜⎜ i
2
⎝
⎠
V gs1 =
⎛ g g R R ⎞ g g R
I o ⎜⎜1 + m1 m3 D L ⎟⎟ = m1 m3 D ⋅ Vi
2
2
⎝
⎠
I
g m1 g m3 R D
Ag f = o =
Vi 2 + g m1 g m 3 R D R L
(c) g m1 = 2 K n I DQ1 = 2 (2 )(1) = 2.828 mA/V
g m3 = 2 K p I DQ 3 = 2 (10 )(12.1) = 22 mA/V
(2.828)(22)(1.6)
= 3.73 mA/V
2 + (2.828)(22)(1.6)(0.248)
______________________________________________________________________________________
Ag f =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.57
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 e3
rπ 1
RE1
RF
(2)
Vπ 2 = − g m1Vπ 1 ( RC1 || rπ 2 )
(3)
V +V
V
g m 2Vπ 2 + π 3 e 3 + π 3 = 0
RC 2
rπ 3
(4)
Vπ 3
V
V −V
+ g m 3Vπ 3 = e3 + e3 e1
rπ 3
RE 2
RF
(5)
Enter numerical values in (2)-(5):
Vπ 1
1 ⎞
⎛ 1
⎛ 1 ⎞
+ (19.23)Vπ 1 = Ve1 ⎜
+
⎟ − Ve3 ⎜
⎟
6.24
⎝ 0.1 0.8 ⎠
⎝ 0.8 ⎠
or
Vπ 1 (19.39) = Ve1 (11.25) − Ve 3 (1.25)
Vπ 2 = −(19.23)Vπ 1 (5 || 3.12) = −(36.94)Vπ 1
(2)
(3)
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) + Ve3 (0.5) = 0
(4)
1 ⎞
⎛ 1
⎞
⎛ 1
⎛ 1 ⎞
+ 76.92 ⎟ = Ve 3 ⎜
+
Vπ 3 ⎜
⎟ − Ve3 ⎜
⎟
⎝ 1.56
⎠
⎝ 0.1 0.8 ⎠
⎝ 0.8 ⎠
or
Vπ 3 (77.56) = Ve3 (11.25) − Ve1 (1.25)
From (1)
Then
or
Vπ 1 = VS − Ve1
(VS − Ve1 )(19.39) = Ve1 (11.25) − Ve3 (1.25)
VS (19.39) = Ve1 (30.64) − Ve3 (1.25)
Vπ 2 = −VS (36.94) + Ve1 (36.94)
(5)
(2′)
(3′)
(38.46)[−VS (36.94) + Ve1 (36.94)] + Vπ 3 (1.141) + Ve3 (0.5) = 0
From (5):
Then
Ve 3 = Vπ 3 (6.894) + Ve1 (0.1111)
VS (19.39) = Ve1 (30.64) − (1.25)[Vπ 3 (6.894) + Ve1 (0.1111)]
or
VS (19.39) = Ve1 (30.50) − Vπ 3 (8.6175)
(2″)
(4′)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
12.58
Computer Analysis
______________________________________________________________________________________
12.59
(100)(0.026)
= 5.2 kΩ
0.5
0.5
g m1 = g m 2 =
= 19.23 mA / V
0.026
(100)(0.026)
rπ 3 =
= 1.3 kΩ
2
2
g m3 =
= 76.92 mA / V
0.026
rπ 1 = rπ 2 =
Vπ 1
V
+ g m1Vπ 1 + g m 2Vπ 2 + π 2 = 0
rπ 1
rπ 2
Since
rπ 1 = rπ 2
and
g m1 = g m 2 ,
(1)
then Vπ 1 = −Vπ 2
VS = Vπ 1 − Vπ 2 + Ve3 = −2Vπ 2 + Ve3
V
V +V
g m 2Vπ 2 + π 3 + π 3 e 3 = 0
rπ 3
RC 2
(3)
Vπ 3
V
V
+ g m 3Vπ 3 = e3 + π 2
rπ 3
RF rπ 2
(4)
(2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ RC 3 ⎞
I0 = − ⎜
⎟ ( g m 3Vπ 3 )
⎝ RC 3 + RL ⎠
From (2):
(19.23)
Ve 3 = VS + 2Vπ 2
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⎞
+ 76.92 ⎟ = ⎜ ⎟ (VS + 2Vπ 2 ) + π 2
Vπ 3 ⎜
5.2
⎝ 1.3
⎠ ⎝ 10 ⎠
or
(77.69)Vπ 3 = (0.3923)Vπ 2 + (0.1)VS
(4′)
⎛ 2 ⎞
I0 = − ⎜
⎟ (76.92)Vπ 3 = −(51.28)Vπ 3
⎝ 2 +1⎠
(5′)
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
______________________________________________________________________________________
12.60
Io
1
=
= 3 mA/V ⇒ R E = 333.3 Ω
Vi R E
h FE Ag
I
(80)(1000)
=
= 2.9998875 mA/V
(b) Ag f = o =
Vi 1 + (h FE )Ag R E 1 + (80)(1000 )(0.33333)
(a) Ag f =
I o = Ag f (5) = 14.9994377
14.9994377 − 15.0
× 100% = −0.00375%
15.0
______________________________________________________________________________________
Error =
12.61
(a) 5 = (1 + h FE )I BQ R E + V EB (on ) + I BQ R B
5 − 0.7
= 0.030605 mA
100 + (81)(0.5)
I CQ = 2.448 mA, and I EQ = 2.479 mA
I BQ =
V ECQ = 5 − (2.479)(0.5) − (2.448)(1) = 1.31 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
V − Vπ
Vo Vo
+
+ h FE I b + o
=0
RF
RC ro
⎛ 1
V
1
1 ⎞
⎟ + h FE I b − π = 0
+ +
V o ⎜⎜
⎟
RF
⎝ RC ro R F ⎠
Vπ = I b rπ
V
V
V − Vo
Is = π + π + π
R B rπ
RF
⎛
r
r ⎞ V
I s = I b ⎜⎜1 + π + π ⎟⎟ − o
⎝ RB RF ⎠ RF
V
Is + o
RF
Ib =
⎛
r
r ⎞
⎜⎜1 + π + π ⎟⎟
⎝ RB RF ⎠
V
⎡
⎤
Is + o ⎥
⎢
⎛ 1
r ⎞
RF ⎥
1
1 ⎞ ⎛
⎟ + ⎜⎜ h FE − π ⎟⎟ ⎢
+ +
=0
V o ⎜⎜
⎟
r
rπ ⎥
⎢
R
R
r
R
π
o
F ⎠ ⎝
F ⎠
⎝ C
+
⎢1 +
⎥
⎣ RB RF ⎦
⎛ 1
⎛
r ⎞ V
r ⎞
r
r ⎞ ⎛
1
1 ⎞⎛
⎟⎜⎜1 + π + π ⎟⎟ + ⎜⎜ h FE − π ⎟⎟ ⋅ I s + ⎜⎜ h FE − π ⎟ ⋅ o = 0
+ +
V o ⎜⎜
⎟
R F ⎟⎠ R F
RF ⎠
⎝
⎝ RC ro R F ⎠⎝ R B R F ⎠ ⎝
V
100
= 40.85 k Ω
Now ro = A =
I CQ 2.448
rπ =
h FE VT (80)(0.026)
=
= 0.8497 k Ω
2.448
I CQ
1
1 ⎞⎛ 0.8497 0.8497 ⎞ ⎛
0.8497 ⎞
0.8497 ⎞ V o
⎛
⎛1
=0
Then V o ⎜ +
+
+ ⎟⎜1 +
⎟⋅
⎟ ⋅ I s + ⎜ 80 −
⎟ + ⎜ 80 −
100
10 ⎠ ⎝
10 ⎠
10 ⎠ 10
⎝
⎝ 1 40.85 10 ⎠⎝
V o (9.22114 ) + I s (79.915) = 0
V
Az f = o = −8.666 V/mA
Is
V
⎛
1 ⎞
8.666
⎛
⎞
⎜ 1+ o ⋅
⎟
1−
⎜
⎟
I s RF ⎟
Vπ
I b rπ
⎜
10
⎜
⎟
=
= rπ ⎜
(c) Ri f =
(
)
=
0
.
8497
rπ
rπ ⎟
0.8497 0.8497 ⎟
⎜
Is
Is
+
⎜⎜ 1 +
⎟
+
⎜1+
⎟
100
10 ⎠
⎝
R B R F ⎟⎠
⎝
Ri f = 0.1037 k Ω
(d) I x =
Ib =
Then
Vx
Vx Vx
+
+ h FE I b +
RC ro
R F + R B rπ
⎛ R B rπ
⎞
Vπ
⎟ ⋅V x
, and Vπ = ⎜
⎜
⎟
rπ
⎝ R B rπ + R F ⎠
⎞
Ix
1
1
1 ⎛ h ⎞⎛ R B rπ
1
⎟+
=
=
+ + ⎜⎜ FE ⎟⎟⎜
⎜
⎟
V x Ro f
RC ro ⎝ rπ ⎠⎝ R B rπ + R F ⎠ R B rπ + R F
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
R B rπ = 100 0.8497 = 0.8425 k Ω
1
1
1
1
⎛ 80 ⎞⎛ 0.8425 ⎞
= +
+⎜
⎟+
⎟⎜
Ro f 1 40.85 ⎝ 0.8497 ⎠⎝ 0.8425 + 10 ⎠ 0.8425 + 10
Then Ro f = 0.1186 k Ω
______________________________________________________________________________________
12.62
2
(a) (i) V + = I DQ R D + VGS = K n R D (VGS − VTN ) + VGS
(
)
2
3 = (0.5)(5) VGS
− 0.8VGS + 0.16 + VGS
We find 2.5V
2
GS
− VGS − 2.6 = 0 ⇒ VGS = 1.239 V
I DQ = (0.5)(1.239 − 0.4) = 0.3522 mA
2
(ii) g m = 2 K n I DQ = 2 (0.5)(0.3522 ) = 0.8392 mA/V
Vo − V gs
Vo
+ g mV gs +
=0
RF
RD
V gs − Vo
(2) I s =
⇒ V gs = I s R F + V o
RF
(b) (1)
⎛ 1
1 ⎞ ⎛
1 ⎞
⎟⎟ + ⎜⎜ g m −
⎟( I s R F + V o ) = 0
+
Then V o ⎜⎜
R F ⎟⎠
⎝ RD RF ⎠ ⎝
⎛ 1
⎞
+ g m ⎟⎟ + I s (g m R F − 1) = 0
V o ⎜⎜
⎝ RD
⎠
V o 1 − g m R F 1 − (0.8392)(25)
Az f =
=
=
1
1
Is
+ gm
+ 0.8392
RD
5
Az f = −19.23 V/mA
(c) I x =
Vx
+ g mV gs and V gs = V x
RD
Ix
1
1
1
=
=
+ g m = + 0.8392
V x Ro f
RD
5
Ro f = 0.962 k Ω
______________________________________________________________________________________
12.63
1
− RF
1 − g m RF
gm
=
(a) Az f =
1
1
+ gm
+1
g m RD
RD
As g m → ∞ , Az f = − R F
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1 − g m RF
1
+ gm
RD
1 − g m (25)
− 23.75 =
0.2 + g m
(b) − 0.95 R F =
So that g m = 4.6 mA/V
______________________________________________________________________________________
12.64
dc analysis
RTH = 24 || 150 = 20.7 kΩ
so
⎛ 24 ⎞
VTH = ⎜
⎟ (12) = 1.655 V
⎝ 24 + 150 ⎠
1.655 − 0.7
I BQ =
= 0.00556 mA
20.7 + (151)(1)
I CQ = 0.834 mA
(150)(0.026)
= 4.68 kΩ
0.834
0.834
gm =
= 32.08 mA / V
0.026
rπ =
VS − Vπ
Vπ
V − V0
=
+ π
RS
RB || rπ
RF
(1)
V V −V
g mVπ + 0 + 0 π = 0
RC
RF
(2)
From (1):
⎡1
VS
1
1 ⎤ V0
= Vπ ⎢ +
+
⎥−
5
5
20.7
||
4.68
R
RF
F ⎦
⎣
or
⎛
1 ⎞ V0
VS (0.20) = Vπ ⎜ 0.4620 +
⎟−
R
RF
F ⎠
⎝
From (2):
⎛
⎛1 1 ⎞
1 ⎞
⎜ 32.08 −
⎟ Vπ + V0 ⎜ +
⎟=0
RF ⎠
⎝
⎝ 6 RF ⎠
so
⎛
1 ⎞
−V0 ⎜ 0.1667 +
⎟
RF ⎠
⎝
Vπ =
⎛
1 ⎞
⎜ 32.08 −
⎟
RF ⎠
⎝
(2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
⎡
⎛
1 ⎞⎤
⎢ −V0 ⎜ 0.1667 +
⎟⎥
R
⎛
⎞
V
1 ⎢
F ⎠⎥
⎝
VS (0.20) = ⎜ 0.4620 +
− 0
⎟⎢
⎥
RF ⎠
RF
⎛
1 ⎞
⎝
⎢ ⎜ 32.08 −
⎟ ⎥
RF ⎠ ⎥⎦
⎢⎣ ⎝
RF
Neglect the
in the denominator term. Now
V0
V
= −5 ⇒ VS = − 0 = −V0 (0.20)
VS
5
⎡ −V ( 0.1667 RF + 1) ⎤
−V0 (0.20)(0.20) RF = (0.4620 RF + 1) ⎢ 0
⎥ − V0
32.08 RF
⎣
⎦
−1.283RF2 = −(0.4620 RF + 1)(0.1667 RF + 1) − 32.08 RF
1.206 RF2 − 32.71RF − 1 = 0
RF =
32.71 ± (32.71) 2 + 4(1.206)(1)
2(1.206)
so that
RF = 27.2 kΩ
______________________________________________________________________________________
12.65
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)
Vπ 2
g mVπ 1 +
=0
RC || RB || rπ
(2)
Vπ 3
g mVπ 2 +
=0
RC || RB || rπ
(3)
V
V V −V
g mVπ 3 + 0 + 0 + 0 π 1 = 0
RC RL
RF
(4)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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):
(17.4)Vπ 1 +
Vπ 2
=0
1.86
or Vπ 2 = −(32.36)Vπ 1
and substituting in (3):
(17.4)Vπ 2 +
Vπ 3
=0
1.86
(17.4)[−(32.36)Vπ 1 ] +
Vπ 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 +
⎟−
RF ⎠ RF
⎝
Substitute numerical values in (4):
⎛ 1 1 1 ⎞ Vπ 1
(17.4)(1047.3)Vπ 1 + V0 ⎜ + +
=0
⎟−
⎝ 8 4 RF ⎠ RF
⎛
⎛
1 ⎞
1 ⎞
Vπ 1 ⎜1.822 × 104 −
⎟ + V0 ⎜ 0.375 +
⎟=0
R
R
F ⎠
F ⎠
⎝
⎝
⎛
1 ⎞
−V0 ⎜ 0.375 +
⎟
RF ⎠
⎝
Vπ 1 =
1
1.822 × 104 −
RF
so that
⎡
⎛
1 ⎞⎤
⎢ −V0 ⎜ 0.375 +
⎟⎥
R
⎛
V
1 ⎞⎢
F ⎠⎥
⎝
Vi (0.10) = ⎜ 0.513 +
− 0
⎟⎢
⎥
1
R
R
F ⎠ 1.822 × 10 4 −
F
⎝
⎢
⎥
RF ⎥
⎢⎣
⎦
We have
V0
= −80
Vi
or
Vi = −
V0
80
⎡ ⎛
1 ⎞⎤
⎢ − ⎜ 0.375 +
⎟⎥
RF ⎠ ⎥ 1
(0.10) ⎛
1 ⎞⎢ ⎝
−
= ⎜ 0.513 +
−
⎟
1 ⎥ RF
80
RF ⎠ ⎢
4
⎝
1.822
10
×
−
⎢
⎥
RF ⎥
⎣⎢
⎦
1/ RF
Neglect that
term in the denominator.
−(0.00125RF ) = −
(0.513RF + 1)(0.375 RF + 1)
−1
1.822 × 104 RF
22.775 RF2 = (0.513RF + 1)(0.375 RF + 1) + 1.822 × 104 RF
We find
22.58 RF2 − 1.822 × 104 RF − 1 = 0
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
RF =
1.822 × 104 ± (1.822 × 10 4 ) 2 + 4(22.58)(1)
2(22.58)
or
RF = 0.807 MΩ
______________________________________________________________________________________
12.66
VS − (−Vd ) −Vd − V1
=
RS
RF
a.
or
⎛ 1
1 ⎞ VS V1
+
+
=0
Vd ⎜
⎟+
R
R
RS RF
F ⎠
⎝ S
(1)
V0 − V1 V1 V1 − ( −Vd )
=
+
R1
R2
RF
or
⎛ 1
V0
1
1 ⎞ Vd
= V1 ⎜ +
+
⎟+
R1
⎝ R1 R2 RF ⎠ RF
(2)
V
Vd = 0
A0 L
and V0 = A0 LVd or
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
(1)
V0
⎛ 1 1 1⎞ V ⎛ 1⎞
= V1 ⎜ + + ⎟ + 04 ⋅ ⎜ ⎟
50
⎝ 50 10 10 ⎠ 10 ⎝ 10 ⎠
or
V0 (0.02 − 10−5 ) = V1 (0.22)
(2)
⎛ 0.02 − 10−5 ⎞
V1 = V0 ⎜
⎟
⎝ 0.22 ⎠
Then
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
V0
V
−0.20
=
⇒ 0 = −21.94
VS 9.115 × 10 −3
VS
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
−Vd
−V ⋅ R
= d S
VS − (−Vd ) VS + Vd
RS
Rif =
b.
Now
Then
or
Vd =
V0
21.94VS
=−
A0 L
104
Rif =
(21.94 × 10−4 )(5)
1 − 21.94 × 10−4
Rif = 1.099 × 10−2 kΩ ⇒ Rif = 10.99 Ω
Because of the A0LVd source,
c.
R0 f = 0
______________________________________________________________________________________
12.67
For example, use the circuit shown in Figure 12.41
______________________________________________________________________________________
12.68
Break the loop
It = Iε
Ai I t +
Now
V0
V0
+
=0
R1 RF + RS Ri
⎛ RS ⎞
V0
Ir = ⎜
⎟⋅
⎝ RS + Ri ⎠ RF + RS Ri
⎛ R + Ri ⎞
V0 = I r ⎜ S
⎟ ⋅ ( RF + RS Ri )
⎝ RS ⎠
or
Then
⎛ 1
⎞ ⎡ ⎛ RS + Ri ⎞
⎤
1
Ai I t + ⎜ +
× I
⎟ ( RF + RS Ri ) ⎥ = 0
⎜ R R + R R ⎟⎟ ⎢ r ⎜ R
F
S
i ⎠ ⎣
S
⎝
⎠
⎦
⎝ 1
Ai
I
T = − r ⇒T =
It
⎡1
⎤ ⎛ RS + Ri ⎞
1
⎢ +
⎥⎜
⎟ ( RF + RS Ri )
⎣ R1 RF + RS Ri ⎦ ⎝ RS ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.69
Vπ 1
V
V −V
+ g m1Vπ 1 = ε 1 + ε 1 0
rπ 1
RE1
RF
g m1Vπ 1 +
Vπ 2 = Vt
Vr
RC1 rπ 2
(1)
= 0 ⇒ Vr = −( g m1Vπ 1 )( RC1 rπ 2 )
(2)
so that
V +V V
g m 2Vt + π 3 0 + π 3 = 0
RC 2
rπ 3
(3)
Vπ 3
V
V −V
+ g m 3Vπ 3 = 0 + 0 ε 1
rπ 3
RE 3
RF
(4)
From (4):
But
⎛ 1
⎛ 1
⎞ V
1 ⎞
V0 ⎜
+
+ g m3 ⎟ + ε 1
⎟ = Vπ 3 ⎜
R
R
r
F ⎠
⎝ E3
⎝ π3
⎠ RF
Vε 1 = −Vπ 1
⎛ 1
⎞ V
Vπ 3 ⎜
+ g m3 ⎟ − π 1
rπ 3
⎝
⎠ RF
V0 =
⎛ 1
1 ⎞
+
⎜
⎟
⎝ RE 3 RF ⎠
so
Then
⎡⎛ 1
⎞ ⎛ 1
1 ⎞⎤
+
Vπ 1 ⎢⎜ + g m1 ⎟ − ⎜
⎟⎥ =
⎠ ⎝ RE1 RF ⎠ ⎦
⎣⎝ rπ 1
⎛ 1
⎞ V
−Vπ 3 ⎜
+ g m3 ⎟ + π 1
⎝ rπ 3
⎠ RF
⎛ 1
1 ⎞
RF ⋅ ⎜
+
⎟
R
R
F ⎠
⎝ E3
(1′)
and
⎛ 1
1 ⎞
+
g m 2Vt + Vπ 3 ⎜
⎟+
⎝ RC 2 rπ 3 ⎠
⎛ 1
⎞ V
Vπ 3 ⎜
+ g m3 ⎟ − π 1
r
⎝ π3
⎠ RF = 0
⎛ 1
1 ⎞
+
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).
T =−
Vr
.
Vt
Then
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.70
Vr Vr Vr − Ve
+
+
=0
RS rπ 1
RF
(1)
V + Ve Vπ 2
g m1Vt + π 2
+
=0
RC1
rπ 2
(2)
Vπ 2
V V − Vr
+ g m 2Vπ 2 = e + e
rπ 2
RE
RF
(3)
Assume rπ 1 = 15.8 k Ω , g m1 = 7.62 mA/V, rπ 2 = 2.28 k Ω , g m 2 = 52.7 mA/V
1
1⎞ V
⎛1
Vr ⎜ +
+ ⎟− e = 0
⎝ 10 15.8 10 ⎠ 10
or
Vr (0.2633) = Ve (0.10)
(1)
1 ⎞ Ve
⎛ 1
(7.62)Vt + Vπ 2 ⎜ +
=0
⎟+
⎝ 40 2.28 ⎠ 40
or
Vt (7.62) + Vπ 2 (0.4636) + Ve (0.025) = 0
(2)
⎛ 1
⎞
⎛1 1 ⎞ V
+ 52.7 ⎟ = Ve ⎜ + ⎟ − r
Vπ 2 ⎜
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
or
Now
T =−
Vr
⇒ T = 84.45
Vt
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.71
RTH = R1 R 2 = 24 150 = 20.69 k Ω
⎛ R2 ⎞
⎛ 24 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(12 ) = 1.655 V
+
R
R
⎝ 24 + 150 ⎠
2 ⎠
⎝ 1
VTH − V BE (on )
1.655 − 0.7
= 0.0133 mA
I BQ =
=
RTH + (1 + h FE )R E 20.69 + (51)(1)
I CQ = 0.666 mA
0.666
= 25.62 mA/V
0.026
(50)(0.026) = 1.951 k Ω
rπ =
0.666
100
ro =
= 150 k Ω
0.666
From Problem 12.64, let R F = 27.2 k Ω
gm =
We see that Vπ = Vt . Let R B = R1 R 2 R S
(1) g mVt +
Vo
Vo
+
=0
ro RC R F + rπ R B
⎛ R B rπ
⎞
⎟ ⋅V
(2) V r = ⎜
⎜R r +R ⎟ o
π
B
F
⎝
⎠
Now R B = 24 150 5 = 20.69 5 = 4.027 k Ω ; rπ R B = 1.951 4.027 = 1.314 k Ω
ro RC = 150 6 = 5.769 k Ω
1
⎤
⎡ 1
(1) 25.62Vt + Vo ⎢
+
⎥ = 0 ⇒ Vo = −122.93Vt
⎣ 5.769 27.2 + 1.314 ⎦
⎞
⎛ 1.314
(2) V r = ⎜
⎟ ⋅ V o = (0.04608)(− 122.93)Vt
⎝ 1.314 + 27.2 ⎠
V
T = − r = 5.66
Vt
______________________________________________________________________________________
12.72
⎛ f ⎞
⎛ f ⎞
(a) φ = − tan −1 ⎜ 3 ⎟ − 2 tan −1 ⎜
⎟
4
⎝ 5 × 10 ⎠
⎝ 10 ⎠
At f 180 = 5.1× 10 4 Hz, φ ≅ −180°
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
β (5 × 10 4 )
(b) T = 0.25 =
⎛ 5.1× 10 4 ⎞
⎟
1 + ⎜⎜
3
⎟
⎠
⎝ 10
0.25 =
2
β (5 × 10 4 )
⎡ ⎛ 5.1× 10 4 ⎞ 2 ⎤
⎟ ⎥
⎢1 + ⎜
⎢ ⎜⎝ 5 × 10 4 ⎟⎠ ⎥
⎣
⎦
(51.0098)(2.0404)
β = 5.2 × 10 −4
______________________________________________________________________________________
12.73
⎛ f ⎞
⎛ f ⎞
(a) φ = −2 tan −1 ⎜
⎟
⎟ − tan −1 ⎜
3
5
⎝ 5 × 10 ⎠
⎝ 5 × 10 ⎠
At f 180 = 7.1× 10 4 Hz, φ ≅ −180°
β (10 4 )
(b) T = 1 =
2
⎡ ⎛ 7.1× 10 4 ⎞ 2 ⎤
⎛ 7.1× 10 4 ⎞
⎥
⎜
⎟
⎢1 + ⎜
⎟
1
+
⎜ 5 × 10 5 ⎟
⎢ ⎜⎝ 5 × 10 3 ⎟⎠ ⎥
⎝
⎠
⎦
⎣
4
β 10
1=
⇒ β = 0.0205
(202.64)(1.01)
( )
(c) Aυ f (0 ) =
10 4
= 48.54
1 + (0.0205) 10 4
( )
(d) Smaller
______________________________________________________________________________________
12.74
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
− tan −1 ⎜
− tan −1 ⎜ 5 ⎟
4 ⎟
4 ⎟
⎝ 10 ⎠
⎝ 5 × 10 ⎠
⎝ 10 ⎠
f = 8.1× 104 Hz, φ = −180.28°
φ = − tan −1 ⎜
At
Determine
T( f )
T = β (10 ) ×
at this frequency.
1
3
⎛ 8.1× 10 ⎞
1+ ⎜
⎟
4
⎝ 10
⎠
4
=
a.
b.
2
1
×
⎛ 8.1× 10 ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
4
2
×
1
⎛ 8.1× 10 4 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
2
β (103 )
(8.161)(1.904)(1.287)
β = 0.005
For
T ( f ) = 0.250 < 1 ⇒ Stable
β = 0.05
For
T ( f ) = 2.50 > 1 ⇒ Unstable
______________________________________________________________________________________
12.75
(b) Phase margin = 80° ⇒ φ = −100°
⎛ f ⎞
⎛ f ⎞
− tan −1 ⎜
3 ⎟
4 ⎟
⎝ 10 ⎠
⎝ 5 × 10 ⎠
3
f = 1.16 × 10 Hz
φ = −100 = −2 tan −1 ⎜
By trial and error,
Then
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
β (5 × 103 )
T =1=
2
⎛
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)
______________________________________________________________________________________
12.76
For β = 0.005,
c.
T ( f ) = 1(0 dB)
at f ≈ 2.10 ×10 Hz
4
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 × 10 Hz
4
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
or
φ = −203.1° ⇒ System is unstable.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
12.77
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
(a) φ = − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 7 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
At f 180 = 10 6 Hz, φ ≅ −180°
⎛ f ⎞
⎛ f ⎞
⎛f ⎞
(b) φ = −135° = − tan⎜⎜ 1353 ⎟⎟ − tan −1 ⎜⎜ 1355 ⎟⎟ − tan −1 ⎜⎜ 1357 ⎟⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
f 135 ≅ 10 5 Hz
T ( f 135 ) = 1 =
1=
β (5 × 10 4 )
⎛ 10 5 ⎞
1 + ⎜⎜ 3 ⎟⎟
⎝ 10 ⎠
β (5 × 10 4 )
(100)(1.414)(1)
2
⎛ 10 5 ⎞
1 + ⎜⎜ 5 ⎟⎟
⎝ 10 ⎠
2
⎛ 10 5 ⎞
1 + ⎜⎜ 7 ⎟⎟
⎝ 10 ⎠
2
⇒ β = 2.83 × 10 −3
5 × 10 4
= 351
1 + 2.83 × 10 − 3 5 × 10 4
______________________________________________________________________________________
(c) Aυ f (0 ) =
(
)(
)
12.78
(a) 100 =
10 5
⇒ β = 9.99 ± 10 −3
1 + 10 5 β
( )
T = 1 = β Av =
1=
(9.99 × 10−3 )(105 )
⎛ f ⎞
1+ ⎜ 3 ⎟
⎝ 10 ⎠
999
2
2
⎛ f ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
⎛ f ⎞
⎛ f ⎞
1+ ⎜ 3 ⎟ 1+ ⎜ 5 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
f = 3.08 × 105 Hz
2
2
Phase
f
f
− tan −1 5
103
10
3.08 × 105
3.08 × 105
= − tan −1
− tan −1
3
10
105
= −89.81 − 72.01
φ = − tan −1
φ = −161.8
Stable
(b) Phase Margin = 180 − 161.8 = 18.2°
______________________________________________________________________________________
12.79
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
(a) φ = − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
At f 180 = 3.33 × 10 5 Hz, φ ≅ −180°
(0.019)(10 3 )
2
2
2
1 + (33.3) 1 + (3.33) 1 + (0.333)
(0.019)(10 3 )
T ( f 180 ) =
= 0.156
(33.315)(3.477 )(1.054)
(b) (i) T ( f 180 ) =
(0.019)(10 3 )
(ii) T = 1 =
⎛ f ⎞
1+ ⎜ 4 ⎟
⎝ 10 ⎠
2
⎛ f ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
2
⎛ f ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
2
⇒ f = 1.2 × 10 5 Hz
5
5
⎛ 1.2 × 10 5 ⎞
−1 ⎛ 1.2 × 10 ⎞
−1 ⎛ 1.2 × 10 ⎞
⎟
⎜
⎟
⎜
⎟
−
−
tan
tan
4
⎟
⎜ 10 5 ⎟
⎜ 10 6 ⎟
⎝ 10
⎠
⎝
⎠
⎝
⎠
= −85.236° − 50.194° − 6.843°
φ = −142.3°
φ = − tan −1 ⎜⎜
10 3
= 50
1 + (0.019) 10 3
______________________________________________________________________________________
(c) A f (0 ) =
( )
12.80
f180
f
− 2 tan −1 1805
5 × 103
10
f180 = 1.05 × 105 Hz
φ = −180 = − tan −1
(a)
(0.0045)(2 × 103 )
T =
=
2
2
⎛ 1.05 × 105 ⎞ ⎡ ⎛ 1.05 × 105 ⎞ ⎤
+
1+ ⎜
1
⎢
⎜
⎟ ⎥
3 ⎟
5
⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10
⎠ ⎥⎦
9
(21.02)(2.1025)
T = f 0 = 0.204
180
(b)
System is stable
T =1=
9
2
2
⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤
1+ ⎜
1
+
⎢
⎥
⎟
⎜
⎟
3
5
⎝ 5 × 10 ⎠ ⎣⎢ ⎝ 10 ⎠ ⎦⎥
f = 3.88 × 104 Hz
4
3.88 × 104
−1 3.88 × 10
−
2
tan
5 × 103
105
= −82.66 − 42.41
φ = −125.1°
φ = − tan −1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
T =
=
(0.15)(2 × 103 )
⎛ 1.05 × 105 ⎞ ⎡ ⎛ 1.05 × 105 ⎞ ⎤
1+ ⎜
⎢1 + ⎜
⎟ ⎥
3 ⎟
5
⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10
⎠ ⎥⎦
2
2
300
(21.02)(2.1025)
T = 6.79
(c)
System is unstable
T = 1=
300
2
2
⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤
1+ ⎜
1
+
⎜ 5⎟ ⎥
3 ⎟ ⎢
⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦
f = 2.33 × 105 Hz
2.33 × 105
2.33 × 105
− 2 tan −1
3
5 × 10
105
= −88.77 − 133.54
φ = −222.3°
φ = − tan −1
______________________________________________________________________________________
12.81
Phase Margin = 45° ⇒ φ = −135°
φ = −135°
At
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
= − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠
f = 104 Hz, φ = −135.6°
T =1
= β (103 ) ×
1
2
×
1
⎛ 104 ⎞
⎛ 104 ⎞
1+ ⎜ 3 ⎟
1+ ⎜ 4 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
1
1
×
×
2
2
⎛ 104 ⎞
⎛ 10 4 ⎞
1+ ⎜ 5 ⎟
1+ ⎜ 6 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
1=
or
2
×
β (103 )
(10.05)(1.414)(1.005)(1.00)
β = 0.01428
______________________________________________________________________________________
12.82
⎛ f 120 ⎞
⎛ f
⎞
⎛ f
⎞
⎛ f
⎞
⎟⎟ − tan −1 ⎜⎜ 120 5 ⎟⎟ − tan −1 ⎜⎜ 120 6 ⎟⎟ − tan −1 ⎜⎜ 120 7 ⎟⎟
f
⎝ 4 × 10 ⎠
⎝ 4 × 10 ⎠
⎝ 4 × 10 ⎠
⎝ PD ⎠
φ = −120° = − tan −1 ⎜⎜
f 120 ≅ 2.31× 10 5 Hz
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4000
T ( f 120 ) = 1 =
2
2
2
2
⎛ 2.31× 10 5 ⎞
⎛ 2.31× 10 5 ⎞
⎛ 2.31× 10 5 ⎞
⎛ 2.31× 10 5 ⎞
⎟
⎜
⎟
⎜
⎟
⎟ 1+ ⎜
1 + ⎜⎜
⎜ 4 × 10 5 ⎟ 1 + ⎜ 4 × 10 6 ⎟ 1 + ⎜ 4 × 10 7 ⎟
⎟
f PD
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
4000
1=
2
⎛ 2.31× 10 5 ⎞
⎟ (1.1548)(1.00167 )(1.0)
1 + ⎜⎜
⎟
f PD
⎝
⎠
2.31× 10 5
4000
=
(1.1548)(1.00167 )(1.09)
f PD
f PD = 66.8 Hz
______________________________________________________________________________________
12.83
(a)
⎛ f ⎞
⎛ f ⎞
f 180 ⇒ −180° = −2 tan −1 ⎜⎜ 1804 ⎟⎟ − tan −1 ⎜⎜ 1806 ⎟⎟
10
⎝
⎠
⎝ 10 ⎠
f 180 = 1.42 × 10 5 Hz
T ( f 180 ) =
10 3
[1 + (14.2) ] 1 + (0.142)
2
2
=
10 3
= 4.89
(202.64)(1.01)
T ( f 180 ) > 1 ⇒ Unstable
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
(b) φ = −135° = − tan −1 ⎜⎜ 135 ⎟⎟ − 2 tan −1 ⎜⎜ 1354 ⎟⎟ − tan −1 ⎜⎜ 1356 ⎟⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎝ f PD ⎠
f 135 ≅ 0.414 × 10 4 Hz
T ( f 135 ) = 1 =
10 3
⎛ 0.414 × 10 4 ⎞
⎟
1 + ⎜⎜
⎟
f PD
⎝
⎠
2
2
⎡ ⎛ 0.414 × 10 4 ⎞ 2 ⎤
⎛ 0.414 × 10 4 ⎞
⎢1 + ⎜
⎟
⎟ ⎥ 1+ ⎜
⎟
⎜
⎟ ⎥
10 4
10 6
⎢ ⎜⎝
⎝
⎠
⎠ ⎦
⎣
0.414 × 10 4
10 3
=
f PD
(1.171)(1)
f PD = 4.85 Hz
______________________________________________________________________________________
12.84
(a)
⎛ f ⎞
⎝ 10 ⎠
⎛ f180 ⎞
⎛ f ⎞
− tan −1 ⎜ 1805 ⎟
4 ⎟
⎝ 5 × 10 ⎠
⎝ 10 ⎠
φ = −180 = − tan −1 ⎜ 1804 ⎟ − tan −1 ⎜
f180 ≅ 8.06 × 104 Hz
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
500
T =
2
⎛ 8.06 × 10 ⎞
⎛ 8.06 × 104 ⎞
1+ ⎜
⎟ 1+ ⎜
⎟
4
4
⎝ 10
⎠
⎝ 5 × 10 ⎠
500
=
(8.122)(1.897)(1.284)
4
(b)
2
⎛ 8.06 × 104 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
2
T = 25.3
T=
500
⎛
f ⎞⎛
f ⎞⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟ ⎜1 + j 4 ⎟ ⎜1 + j
⎟⎜1 + j 5 ⎟
f PD ⎠ ⎝
10 ⎠ ⎝
5 × 104 ⎠⎝
10 ⎠
⎝
= 60° ⇒ φ = −120°
(c)
Phase Margin
−120 = − tan −1
tan −1
Assume
f
f PD
f
f
f
f
− tan −1 4 − tan −1
− tan −1 5
f PD
10
5 × 104
10
≅ 90°
Then f ≅ 4.2 × 10 Hz
3
500
T =1=
1=
2
⎛ 4.2 × 10 ⎞
⎛ 4.2 × 10 ⎞
1+ ⎜
⎟ 1+ ⎜
⎟
4
⎝ 10
⎠
⎝ f PD ⎠
500
3
3
2
⎛ 4.2 × 103 ⎞
1+ ⎜
4 ⎟
⎝ 5 × 10 ⎠
2
⎛ 4.2 × 103 ⎞
1+ ⎜
⎟
5
⎝ 10
⎠
2
2
⎛ 4.2 × 103 ⎞
1+ ⎜
⎟ (1.085)(1.004)(1.0)
⎝ f PD ⎠
4.2 × 103
500
≅
(1.0846)(1.0035)(1.0)
f PD
f PD = 9.14 Hz
______________________________________________________________________________________
12.85
50 =
(a)
T=
104
⇒ β = 0.0199
1 + (104 ) β
(0.0199)(104 )
⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟ ⎜1 + j 5 ⎟
f
10
⎝
⎠
PD ⎠
⎝
Phase margin = 45° ⇒ φ = −135°
−135 = − tan −1
f = 105 Hz
f
f
− tan −1 5
f PD
10
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(0.0199)(104 )
T =1=
⎛ 105 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
2
⎛ 105 ⎞
1+ ⎜ 5 ⎟
⎝ 10 ⎠
2
105 (0.0199)(104 )
=
1.414
f PD
f PD = 711 Hz
20 =
(b)
104
⇒ β = 0.0499
1 + (104 ) β
(0.0499)(104 )
f ⎞⎛
f ⎞
⎛
⎜1 + j
⎟⎜1 + j 5 ⎟
711 ⎠⎝
10 ⎠
⎝
(0.0499)(104 )
T =1=
2
2
⎛ f ⎞
⎛ f ⎞
1+ ⎜
⎟ 1+ ⎜ 5 ⎟
⎝ 711 ⎠
⎝ 10 ⎠
T=
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°
______________________________________________________________________________________
12.86
(a)
AO = 100 dB ⇒ AO = 105
20 =
T=
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°
−135 = − tan −1
f
f
f
− tan −1 6 − tan −1 7
f PD
10
10
f ≈ 106 Hz
(0.04999)(105 )
T =1=
⎛ 106 ⎞
1+ ⎜
⎟
⎝ f PD ⎠
1=
2
⎛ 106 ⎞
1+ ⎜ 6 ⎟
⎝ 10 ⎠
(0.04999)(105 )
2
⎛ 106 ⎞
1+ ⎜
⎟ (1.414)(1.005)
⎝ f PD ⎠
106 (0.04999)(105 )
=
f PD (1.414)(1.005)
f PD = 2.84 Hz
2
⎛ 106 ⎞
1+ ⎜ 7 ⎟
⎝ 10 ⎠
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5=
(b)
105
⇒ β = 0.19999
1 + (105 ) β
(0.19999)(105 )
T =1=
⎛ 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°
______________________________________________________________________________________
12.87
T (0) = 100 dB ⇒ T (0) = 105
a.
T( f ) =
105
f ⎞⎛
f ⎞⎛
f
⎛
⎞
1+ j
⎜1 + j ⎟⎜1 + j
⎟
6 ⎟⎜
10 ⎠⎝
5 × 10 ⎠⎝
10 × 106 ⎠
⎝
T =1=
1
= 105 ×
⎛ f ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
2
1
×
⎛ f ⎞
1+ ⎜
6 ⎟
⎝ 5 × 10 ⎠
2
1
×
f
⎛
⎞
1+ ⎜
6 ⎟
⎝ 10 × 10 ⎠
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°
= 180° − 106.6° = 73.4°
Phase Margin
b.
or
f P′1 ∝
1
CF
10 75
=
f P′1 20
so
f P′1 = 2.67 Hz
Now
T =1=
1
= 105 ×
⎛ 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
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 2.66 × 105 ⎞
−1 ⎛ 0.266 ⎞
−1 ⎛ 0.266 ⎞
⎟ − tan ⎜
⎟ − tan ⎜
⎟
⎝ 5 ⎠
⎝ 10 ⎠
⎝ 2.67 ⎠
φ = − tan −1 ⎜
= −90° − 3.045° − 1.524° = −94.57°
= 180° − 94.57° = 85.4°
Phase Margin
______________________________________________________________________________________
12.88
(a)
fP =
1
2π (R R )C
o1
i2
=
i
1
2π (2 0.75)× 10 6 × 1.2 × 10 −12
(
)
f P = 243 kHz
(b)
f PD = 6 =
1
1
=
2π (Ro1 Ri 2 )(C i + C M ) 2π (2 0.75)× 10 6 × (C i + C M )
C i + C M = 4.863 × 10 −8 = 1.2 × 10 −12 + C M
C M = 0.0486 μ F
(c) C M = C F (1 + A )
0.0486 × 10 −6 = C F (1001)
C F = 48.6 pF
______________________________________________________________________________________
12.89
⎛ f
⎞
⎛ f 120 ⎞
⎛ f
⎞
⎟ − tan −1 ⎜⎜ 120 7 ⎟⎟
6 ⎟
⎝ 4 × 10 ⎠
⎝ 4 × 10 ⎠
φ = −120° = − tan −1 ⎜⎜ 120 ⎟⎟ − tan −1 ⎜⎜
⎝ f PD ⎠
f 120 ≅ 2.05 × 10 6 Hz
T ( f 120 ) = 1 =
4000
⎛ 2.05 × 10 6 ⎞
⎟
1 + ⎜⎜
⎟
f PD
⎝
⎠
2
⎛ 2.05 × 10 6 ⎞
⎟
1 + ⎜⎜
6
⎟
⎝ 4 × 10 ⎠
2
⎛ 2.05 × 10 6 ⎞
⎟
1 + ⎜⎜
7
⎟
⎝ 4 × 10 ⎠
2
2.05 × 10 6
4000
=
f PD
(1.1237 )(1.0013)
f PD = 577 Hz
______________________________________________________________________________________
12.90
(a) 40 =
5 × 10 5
⇒ β = 0.024998
1 + β 5 × 10 5
(
)
⎛ f ⎞
⎛ f
⎛ f ⎞
⎞
(b) φ = −120° = − tan −1 ⎜⎜ 120 ⎟⎟ − tan −1 ⎜⎜ 120 5 ⎟⎟ − tan −1 ⎜⎜ 1207 ⎟⎟
⎝ 5 × 10 ⎠
⎝ 10 ⎠
⎝ f PD ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
f 120 ≅ 2.71× 10 5 Hz
T ( f 120 ) = 1 =
(0.024998)(5 ×10 5 )
⎛ 2.71× 10 5 ⎞
⎟
1 + ⎜⎜
⎟
f PD
⎝
⎠
(
2.71× 10 5 (0.024998) 5 × 10 5
=
f PD
(1.1374)(1)
2
)
⎛ 2.71× 10 5 ⎞
⎟
1 + ⎜⎜
5
⎟
⎝ 5 × 10 ⎠
2
⎛ 2.71× 10 5 ⎞
⎟
1 + ⎜⎜
7
⎟
⎝ 10
⎠
2
f PD = 24.66 Hz
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 13
13.1
0 − (− 3)
= 15 k Ω
0 .2
⎛ k ′p ⎞⎛ W ⎞
⎟⎜ ⎟ (V SG 3 + VTP )2
I D 3 = ⎜⎜
⎟⎝ L ⎠
2
3
⎝
⎠
⎛ 0.04 ⎞
2
0. 2 = ⎜
⎟(40 )(V SG 3 − 0.4 ) ⇒ V SG 3 = 0.9 V
⎝ 2 ⎠
(a) R D 2 =
0 .9
= 9kΩ
0. 1
υ
g
(b) (i) Ad = o1 = m1 ⋅ R D1
υd
2
R D1 =
⎛ 0. 1 ⎞
g m1 = 2 ⎜
⎟(20 )(0.1) = 0.6325 mA/V
⎝ 2 ⎠
⎛ 0.6325 ⎞
Ad = ⎜
⎟(9 ) = 2.846
⎝ 2 ⎠
(ii) A2 =
υo
= − g m3 R D 2
υ o1
⎛ 0.04 ⎞
g m3 = 2 ⎜
⎟(40 )(0.2 ) = 0.8 mA/V
⎝ 2 ⎠
A2 = −(0.8)(15) = −12
(c) A = Ad ⋅ A2 = (2.846)(− 12 ) = −34.15
______________________________________________________________________________________
13.2
3−0
= 7.5 k Ω
0.4
V B 3 = 0.7 + (0.4 )(0.5) − 3 = −2.1 V
(a) RC 2 =
− 2.1 − (− 3)
= 3. 6 k Ω
0.25
υ
g
(b) (i) Ad = o1 = m1 (RC1 Ri 3 )
υd
2
R C1 =
(180)(0.026 ) = 11.7 k Ω
0.25
= 9.615 mA/V, rπ 3 =
0.026
0.4
Ri 3 = rπ 3 + (1 + β n )R E = 11.7 + (181)(0.5) = 102.2 k Ω
g m1 =
⎛ 9.615 ⎞
Ad = ⎜
⎟(3.6 102.2 ) = 16.72
⎝ 2 ⎠
υ
− β n RC 2
− (180)(7.5)
(ii) A2 = o =
=
= −13.21
υ o1 rπ 3 + (1 + β n )R E 11.7 + (181)(0.5)
(c) A = Ad ⋅ A2 = (16.72 )(− 13.21) = −220.9
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.3 Computer Analysis
______________________________________________________________________________________
13.4 Computer Analysis
______________________________________________________________________________________
13.5
(
Ad = g m1 ro 2 ro 4 Ri 6
(a)
)
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
Ri 6 = rπ 6 + (1 + β n ) ⎡⎣ R1 rπ 7 ⎤⎦
(120)(0.026)
= 15.6 k Ω
0.2
V (on) 0.6
I C 6 ≅ BE
=
= 0.030 mA
R1
20
rπ 7 =
rπ 6 =
(120)(0.026)
= 104 k Ω
0.030
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 ⎜
⎟ Ic6
⎝ R1 + rπ 7 ⎠
⎛ R1 ⎞
Vo1
= − β n (1 + β n )ro 7 ⎜
⎟ I b 6 and I b 6 =
Ri 6
⎝ R1 + rπ 7 ⎠
Then
Av 2 =
Vo − β n (1 + β n ) ro 7 ⎛ R1 ⎞
=
⎜
⎟
Vo1
Ri 6
⎝ R1 + rπ 7 ⎠
ro 7 =
VA
80
=
= 400 k Ω
I C 7 0.2
So
−(120)(121)(400) ⎛ 20 ⎞
⎜
⎟ ⇒ Av 2 = −2813
1160
⎝ 20 + 15.6 ⎠
Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 × 106
Av 2 =
(b)
Rid = 2rπ 1 and
Rid = 208 k Ω
rπ 1 =
(80)(0.026)
= 104 k Ω
0.020
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
f PD =
(c)
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
______________________________________________________________________________________
13.6
Q3
a.
acts as the protection device.
b.
Same as part (a).
______________________________________________________________________________________
13.7
If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5
So breakdown voltage ≈ 56.4 V.
______________________________________________________________________________________
13.8
⎛ 0.5 × 10 −3 ⎞
⎟ = 0.7184 V
(a) V EB12 = V BE11 = (0.026) ln⎜⎜
−16 ⎟
⎝ 5 × 10
⎠
15 − 0.7184 − 0.7184 − (− 15)
R5 =
= 57.1 k Ω
0.5
(0.026 ) ln⎛ 0.5 ⎞ = 2.438 k Ω
R4 =
⎜
⎟
(0.03) ⎝ 0.03 ⎠
V BE10 = V BE11 − I C10 R 4 = 0.7184 − (0.03)(2.438) = 0.6453 V
(b) I REF =
15 − 0.6 − 0.6 − (− 15)
= 0.5044 mA
57.1
⎛ 0.5044 ⎞
⎟ ⇒ I C10 ≅ 30.1 μ A
I C10 (2.438) = (0.026) ln⎜⎜
⎟
⎝ I C10 ⎠
⎛ 0.5044 − 0.5 ⎞
(c) I REF : ⎜
⎟ × 100% = 0.88%
0.5
⎝
⎠
⎛ 30.1 − 30 ⎞
I C10 : ⎜
⎟ × 100% = 0.33%
⎝ 30 ⎠
______________________________________________________________________________________
13.9
5 − 0.7184 − 0.7184 − (− 5)
= 17.13 k Ω
0. 5
(0.026 ) ln⎛ 0.5 ⎞ = 2.438 k Ω
R4 =
⎜
⎟
(0.03) ⎝ 0.03 ⎠
(a) R5 =
V EB12 = V BE11 = 0.7184 V
V BE10 = 0.6453 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5 − 0.6 − 0.6 − (− 5)
(b) I REF =
= 0.5137 mA
17.13
⎛ 0.5137 ⎞
⎟ ⇒ I C10 ≅ 30.22 μ A
I C10 (2.438) = (0.026 ) ln⎜⎜
⎟
⎝ I C10 ⎠
⎛ 0.5137 − 0.5 ⎞
(c) I REF : ⎜
⎟ × 100% = 2.74%
0. 5
⎝
⎠
⎛ 30.22 − 30 ⎞
I C10 : ⎜
⎟ × 100% = 0.733%
30
⎝
⎠
______________________________________________________________________________________
13.10
3 − V EB 2 − V BE1 − (− 3)
80
⎛ I
⎞
V EB 2 = V BE1 = (0.026 ) ln⎜⎜ REF−15 ⎟⎟
⎝ 3 × 10 ⎠
By trial and error, I REF ≅ 59.6 μ A
(a) I REF =
⎛ 0.0596 ⎞
⎟ ⇒ I 3 ≅ 11.9 μ A
I 3 (3.5) = (0.026) ln⎜⎜
⎟
⎝ I3 ⎠
I
3 × 10 −15
I 4 = S 4 ⋅ I REF =
⋅ (59.6 ) = 35.76 μ A
I S2
5 × 10 −15
I5 =
I S5
⋅ I REF
I S2
(
)
(
)
(1×10 ) ⋅ (59.6) = 11.92 μ A
=
(5 ×10 )
−15
−15
(b) I REF = 59.6 μ A, I 3 = 11.9 μ A
I4 =
(
)
(
)
(2 ×10 ) ⋅ (59.6) = 23.84 μ A
=
(5 ×10 )
I S4
8 × 10 −15
⋅ I REF =
⋅ (59.6 ) = 95.36 μ A
I S2
5 × 10 15
−15
I S5
⋅ I REF
−15
I S2
______________________________________________________________________________________
I5 =
13.11
5 − 0.6 − 0.6 − (−5)
⇒ I REF = 0.22 mA
40
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
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 = 7.10μ A
2
I C17 = 0.75I REF ⇒ I C17 = 0.165 mA
IC 6 ≅
I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA
_____________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.12
I C8 = I C9
⎛
2 ⎞⎟
= I C 9 ⎜1 +
⎜ βp ⎟
βp
⎝
⎠
⎛ 1+ β n ⎞
⎟
I E1 = I E 2 = I ⎜⎜
⎟
⎝ βn ⎠
2I = I C8 +
2I C 9
⎛ 1 + β n ⎞⎛ 1 ⎞
⎟
⎟⎜
I B 3 = I B 4 = I ⎜⎜
⎟⎜
⎟
⎝ β n ⎠⎝ 1 + β p ⎠
⎛ 1 + β n ⎞⎛ 1 ⎞
⎟
⎟⎜
I C10 = I C 9 + 2 I ⎜⎜
⎟⎜
⎟
⎝ β n ⎠⎝ 1 + β p ⎠
⎛
2 ⎞⎟⎛ 1 + β n ⎞⎛⎜ 1 ⎞⎟
⎜
⎟
I C10 = I C 9 + I C 9 ⎜1 +
⎜ β p ⎟⎜ β n ⎟⎜ 1 + β p ⎟
⎠⎝
⎝
⎠⎝
⎠
⎡ ⎛
2 ⎞⎛ 91 ⎞⎛ 1 ⎞⎤
50 = I C 9 ⎢1 + ⎜1 + ⎟⎜ ⎟⎜ ⎟⎥ = I C 9 (1.0259 ) ⇒ I C 9 = 48.738 μ A
40
⎠⎝ 90 ⎠⎝ 41 ⎠⎦
⎣ ⎝
I C2 = I =
I C 9 ⎛⎜
2 ⎞⎟ 48.738 ⎛
2 ⎞
=
1+
⎜1 + ⎟ = 25.587 μ A
⎜
⎟
2 ⎝ βp ⎠
2 ⎝ 40 ⎠
⎛ β p ⎞⎛ 1 + β n ⎞
40 91
⎟⎜
⎟ = (25.587 )⎛⎜ ⎞⎟⎛⎜ ⎞⎟ = 25.240 μ A
I C4 = I C2 ⎜
⎜ 1 + β p ⎟⎜ β n ⎟
⎝ 41 ⎠⎝ 90 ⎠
⎠
⎝
⎠⎝
I
48.738
I B9 = C 9 =
= 1.218 μ A
40
βp
I C4
25.240
=
= 0.631 μ A
βp
40
______________________________________________________________________________________
I B4 =
13.13
VB 5 − V − = VBE (on) + I C 5 (1)
= 0.6 + (0.0095)(1) = 0.6095
0.6095
IC 7 =
⇒ I C 7 = 12.2 μ A
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.14
(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
______________________________________________________________________________________
13.15
I C13 A = (0.25)I REF = (0.25)(0.5) = 0.125 mA
0. 6
= 0.012 mA
50
I C19 ≅ I C13 A − I R10 = 0.125 − 0.012 = 0.113 mA
I R10 ≅
I B19 =
I C19
βn
=
0.113
⇒ I B19 = 0.565 μ A
200
I C18 = I R10 + I B19 = 12 + 0.565 = 12.565 μ A
⎛ 12.565 × 10 −6 ⎞
⎟ = 0.54474 V
V BE18 = (0.026 ) ln⎜⎜
⎟
10 −14
⎝
⎠
⎛ 0.113 × 10 −3 ⎞
⎟ = 0.60185 V
V BE19 = (0.026) ln⎜⎜
−14
⎟
⎝ 10
⎠
V BB V BE18 + V Be19
=
= 0.57329 V
2
2
⎛ 0.57329 ⎞
I C14 = 3 × 10 −14 exp⎜
⎟ ⇒ I C14 = 113 μ A
⎝ 0.026 ⎠
______________________________________________________________________________________
(
)
13.16
⎛I
⎞
⎛ 80 × 10 −6 ⎞
⎟ = 1.22178 V
(a) V BB = 2VT ln⎜⎜ Bias ⎟⎟ = 2(0.026) ln⎜⎜
−15 ⎟
⎝ I SD ⎠
⎝ 5 × 10 ⎠
⎡1.22178 2 ⎤
I CN = I CP = I SQ exp ⎢
⎥ ⇒ I CN = I CP = 128 μ A
⎣ 0.026 ⎦
(b) For υ I = 3 V, υ O ≅ 3 V, i L ≅
3
= 0.3 mA
10
First approximation:
0.3
I BN ≅
⇒ I BN = 2.5 μ A
120
I D = 80 − 2.5 = 77.5 μ A
⎛ 77.5 × 10 −6 ⎞
⎟ = 1.22013 V
V BB = 2(0.026 ) ln⎜⎜
−15 ⎟
⎝ 5 × 10
⎠
⎛ 0.3 × 10 −3 ⎞
⎟ = 0.63304 V
V BEN = (0.026) ln⎜⎜
−15 ⎟
⎝ 8 × 10
⎠
V EBP = V BB − V BEN = 0.58709 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
⎛ 0.58709 ⎞
I CP = 8 × 10 −15 exp⎜
⎟ ⇒ I CP = 51.24 μ A
⎝ 0.026 ⎠
Second approximation:
⎛ 120 ⎞
I CN = ⎜
⎟(0.3 + 0.05124 ) = 0.34834 mA, I BN = 2.903 μ A
⎝ 121 ⎠
I D = 80 − 2.903 = 77.1 μ A
⎛ 77.1× 10 −6 ⎞
⎟ = 1.219864 V
V BB = 2(0.026) ln⎜⎜
−15 ⎟
⎝ 5 × 10
⎠
⎛ 0.34834 × 10 −3 ⎞
⎟ = 0.636922 V
V BEN = (0.026) ln⎜⎜
−15
⎟
⎝ 8 × 10
⎠
V EBP = 1.219864 − 0.636922 = 0.582942 V
(
)
R1 + R 2 =
(0.1)I Bias
⎛ 0.582942 ⎞
I CP = 8 × 10 −15 exp⎜
⎟ ⇒ I CP = 43.7 μ A
⎝ 0.026 ⎠
____________________________________________________________________________________
13.17
V BE
V BB
1.160
=
= 96.67 k Ω
(0.1)(0.12)
⎛ I CQ ⎞
⎛ (0.9)(120 × 10 −6 ) ⎞
⎟ = (0.026) ln⎜
⎟ = 0.6187 V
= V ln⎜
⎜
⎟
⎜ I ⎟
5 × 10 −15
⎝
⎠
⎝ S ⎠
⎛ R2 ⎞
⎛ R ⎞
⎟⎟(V BB ) ⇒ 0.6187 = ⎜⎜ 2 ⎟⎟(1.16 )
V BE = ⎜⎜
⎝ 96.67 ⎠
⎝ R1 + R 2 ⎠
So R 2 = 51.56 k Ω , R1 = 45.11 k Ω
____________________________________________________________________________________
T
13.18
A = −g
(r r R )
d
m
o4 o6
(a)
From example 13.4
gm =
i2
9.5
= 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)
From Equation (13.20),
− β n (1 + β n ) R9 ( Ract 2 Ri 3 R017 )
Av 2 =
For
{
}
Ri 2 R9 + ⎡⎣ rπ 17 + (1 + β n ) Rg ⎤⎦
Rg = 0, Ri 2 = 1.95 M Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Using the results of Example 13.5
Av 2 =
(
−200(201)(50) 92.6 4050 92.6
(1950){50 + 9.63}
) ⇒ A = −792
v2
______________________________________________________________________________________
13.19
Let I C10 = 40 μ A, then I C1 = I C 2 = 20 μ A. Use the procedure in Example 13.4:
Ri 2 = 4.07 MΩ
(200)(0.026)
= 260 kΩ
0.020
0.020
gm6 =
= 0.769 mA/V
0.026
50
r06 =
⇒ 2.5 MΩ
0.02
rπ 6 =
Then
Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ
r06 =
50
⇒ 2.5 MΩ
0.02
Then
⎛ I CQ ⎞
Ad = − ⎜
⎟ (r04 Ract1 Ri 2 )
⎝ VT ⎠
⎛ 20 ⎞
= −⎜
⎟ (2.5 4.42 4.07)
⎝ 0.026 ⎠
So
Ad = −882
______________________________________________________________________________________
13.20
From Problem 13.11
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
(200)(0.026)
rπ 16 =
= 394 K
0.0132
Then
Ri 2 = rπ 16 + (1 + β ) RE1 = 394 + (201)(25.4) ⇒ 5.50 MΩ
Now
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(200)(0.026)
= 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
rπ 6 =
ro 4 =
50
⇒ 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
Now
Ract 2 =
50
⇒ 303 K
0.165
Ro17 =
50
= 303 K
0.165
From Eq. (13.20), assuming Ri 3 → ∞
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
______________________________________________________________________________________
13.21
Using results from Problem 13.20
⎛ 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
= 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
Ract 2 =
Overall gain
Av = ( −885)(−1090) = 964, 650
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.22
(a) I C18 + I C19 = I Bias
(0.25)I C19 + I C19 = 0.18 mA, ⇒ I C19 = 0.144 mA
I C18 = 0.036 mA
⎛ 0.144 × 10 −3 ⎞
⎟ = 0.60815 V
V BE19 = (0.026) ln⎜⎜
−14
⎟
⎝ 10
⎠
0.60815
R10 =
= 16.9 k Ω
0.036
(b) V BE19 = 0.60815 V
⎛ 0.036 × 10 −3 ⎞
⎟ = 0.5721 V
V BE18 = (0.026) ln⎜⎜
−14
⎟
⎝ 10
⎠
______________________________________________________________________________________
13.23
(b) I C19 = 0.168 mA, g m19 =
rπ 19 =
(200 )(0.026 ) = 30.95 k Ω , r
o19
0.168
I C18 = 12.84 μ A, g m18 =
rπ 18 =
0.168
= 6.462 mA/V,
0.026
50
= 298 k Ω
0.168
0.01284
= 0.4938 mA/V
0.026
(200 )(0.026 ) = 405 k Ω , r
0.01284
=
o18
=
50
= 3894 k Ω
0.01284
V x = Vπ 18 + Vπ 19
(1) I x =
(2)
Vπ 18
V
V
+ g m18Vπ 18 + π 18 + g m19Vπ 19 + x
rπ 18
ro18
ro19
Vπ 18
V
V
V
+ g m18Vπ 18 + π 18 = π 19 + π 19
rπ 18
ro18
R10
rπ 19
1 ⎞
1 ⎞
⎛ 1
⎛ 1
Then Vπ 18 ⎜
+ 0.4938 +
⎟ = Vπ 19 ⎜ +
⎟
405
3894
50
30
.95 ⎠
⎝
⎠
⎝
Vπ 18 (0.49653) = Vπ 19 (0.05231) ⇒ Vπ 18 = Vπ 19 (0.10535)
V
Then (1) I x = Vπ 18 (0.49653) + (6.462 )Vπ 19 + x
298
I x = Vπ 19 (0.10535)(0.49653) + (6.462)Vπ 19 + (0.003356)V x
I x = Vπ 19 (6.5143) + V x (0.003356)
Now V x = Vπ 18 + Vπ 19 = Vπ 19 (0.10535) + Vπ 19 = (1.10535)Vπ 19
Or Vπ 19 = (0.90469 )V x
Then I x = (0.90469 )V x (6.5143) + V x (0.003356 )
I
1
So x =
= 5.8968 ⇒ R eq = 170 Ω
V x R eq
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.24
Now
r + R01
Re14 = π 14
1+ βP
and R0 = R6 + Re14
Assume series resistance of Q18 and Q19 is small. Then
R01 = r013 A Re 22
where
r + R017 r013 B
Re 22 = π 22
1+ βP
R
= r [1 + g
( R r )]
m17
8 π 17
and 017 017
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
r013 B =
50
= 92.6 kΩ
0.54
Then
7.22 + 283 92.6
= 1.51 kΩ
51
R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ
(50)(0.026)
rπ 14 =
= 0.65 kΩ
2
Re 22 =
Then
Re14 =
0.65 + 1.50
= 0.0422 kΩ
51
or
Re14 = 42.2 Ω
Then
R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω
______________________________________________________________________________________
13.25
⎡
⎛ r
⎞⎤
Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3 ⎟ ⎥
⎝ 1 + β P ⎠⎦
⎣
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
rπ 1 =
Then
(201)(27.4) ⎤
⎡
Rid = 2 ⎢547 +
⎥
11
⎣
⎦
Rid ⇒ 2.095 MΩ
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
I C1 = 7.10 μ A
(200)(0.026)
= 732 K
0.0071
(10)(0.026)
rπ 3 =
= 36.6 K
0.0071
(201)(36.6) ⎤
⎡
Rid = 2 ⎢ 732 +
⎥
11
⎣
⎦
Rid ⇒ 2.80 MΩ
rπ 1 =
______________________________________________________________________________________
13.26
We can write
A0
⎛
f ⎞⎛
f ⎞
⎜1 + j
⎟ ⎜1 + j ⎟
f
f
1 ⎠
PD ⎠ ⎝
⎝
181, 260
=
f ⎞⎛
f ⎞
⎛
⎜1 + j
⎟ ⎜1 + j ⎟
f1 ⎠
10.7 ⎠ ⎝
⎝
A( f ) =
Phase:
⎛ f ⎞
−1 ⎛ f ⎞
⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠
⎝ f1 ⎠
φ
= −110°
=
70
°
,
For a Phase margin
φ = − tan −1 ⎜
So
⎛ f ⎞
−1 ⎛ f ⎞
−110° = − tan −1 ⎜
⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠
⎝ f1 ⎠
Assuming f >> 10.7 Hz, we have
⎛ f ⎞
f
tan −1 ⎜ ⎟ = 20° ⇒ = 0.364
f1
⎝ f1 ⎠
A( f ) = 1,
so
At this frequency,
1=
=
181, 260
2
⎛ f ⎞
2
1+ ⎜
⎟ ⋅ 1 + (0.364)
⎝ 10.7 ⎠
170,327
⎛ f ⎞
1+ ⎜
⎟
⎝ 10.7 ⎠
2
f
= 170,327 ⇒ f = 1.82 MHz
10.7
or
Then, second pole at
f1 =
f
⇒ f1 = 5 MHz
0.364
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.27
(a) 10 × 10 6 = f PD 10 6 ⇒ f PD = 10 Hz
( )
(b) f PD =
1
2πR eq C M
⇒ CM =
1
= 1.326 × 10 −8 F
2π 1.2 × 10 6 (10 )
(
)
−8
1.326 ×10
⇒ C F = 13.25 pF
1001
______________________________________________________________________________________
C M = C F (1 + A ) ⇒ C F =
13.28
⎛ f ⎞
⎛
⎞
⎛ f ⎞
⎟ = −90 − 2 tan −1 ⎜ f ⎟
⎟⎟ − 2 tan −1 ⎜
⎜f ⎟
⎜f ⎟
⎝ f PD ⎠
⎝ 2,3 ⎠
⎝ 2, 3 ⎠
φ = −110 = − tan −1 ⎜⎜
⎛ f ⎞
⎟ = 0.1763
So ⎜
⎜f ⎟
⎝ 2,3 ⎠
A =1=
200,000
⎛ f ⎞
⎟⎟
1 + ⎜⎜
⎝ f PD ⎠
2
⎡ ⎛ f ⎞ ⎤
⎟ ⎥
⎢1 + ⎜
⎢ ⎜⎝ f 2,3 ⎟⎠ ⎥
⎣
⎦
2
≅
200,000
[
⎛ f ⎞
2
⎜ ⎟ 1 + (0.1763)
⎝ 10 ⎠
]
⇒ f = 1.9397 MHz
1.9397 × 10 6
⇒ f 2,3 = 11.0 MHz
0.1763
______________________________________________________________________________________
Then f 2,3 =
13.29
⎛ k ′p ⎞⎛ W ⎞
2
(a) I D 3 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP )
L
2
⎝
⎠⎝ ⎠ 3
⎛ 40 ⎞
2
150 = ⎜ ⎟(50 )(V SG 3 − 0.4 ) ⇒ V SG 3 = 0.7873 V
⎝ 2 ⎠
0.7873
R D1 =
= 7.87 k Ω
0.1
⎛ k ′ ⎞⎛ W ⎞
2
I D 4 = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 4 − VTN )
⎝ 2 ⎠⎝ L ⎠ 4
⎛ 100 ⎞
2
200 = ⎜
⎟(40 )(VGS 4 − 0.4 ) ⇒ VGS 4 = 0.7162 V
2
⎝
⎠
0.7162 − (− 3)
RD2 =
= 24.8 k Ω
0.15
0 − (− 3)
RS =
= 15 k Ω
0.15
⎛g ⎞
⎛ 0. 1 ⎞
(b) (i) Ad 1 = ⎜⎜ m1 ⎟⎟ ⋅ R D1 , g m1 = 2 ⎜
⎟(20 )(0.1) = 0.6325 mA/V
2
⎝ 2 ⎠
⎝
⎠
⎛ 0.6325 ⎞
Ad 1 = ⎜
⎟(7.87 ) = 2.49
⎝ 2 ⎠
⎛ 0.04 ⎞
(ii) A2 = − g m3 R D 2 , g m 3 = 2 ⎜
⎟(50 )(0.15) = 0.7746 mA/V
⎝ 2 ⎠
A2 = −(0.7746 )(24.8) = −19.21
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(iii) A3 =
g m4 RS
⎛ 0. 1 ⎞
, g m4 = 2 ⎜
⎟(40 )(0.2 ) = 1.265 mA/V
1+ g m 4 R S
⎝ 2 ⎠
(1.265)(15) = 0.950
1 + (1.265)(15)
(c) A = Ad 1 ⋅ A2 ⋅ A3 = (2.49 )(− 19.21)(0.95) = −45.4
A3 =
______________________________________________________________________________________
13.30
⎛ 100 ⎞
2
(a) I D 3 = 100 = ⎜
⎟(25)(VGS 3 − 0.4 ) ⇒ VGS 3 = 0.6828 V
⎝ 2 ⎠
0.6828
R D1 =
= 13.66 k Ω
0.05
⎛ 40 ⎞
2
I D 4 = 200 = ⎜ ⎟(100 )(V SG 4 − 0.4 ) ⇒ V SG 4 = 0.7162 V
⎝ 2 ⎠
3 − (− 0.7162 )
= 37.16 k Ω
0. 1
3−0
RS =
= 15 k Ω
0.2
R D3 =
⎛g ⎞
⎛ 0.04 ⎞
(b) (i) Ad = ⎜⎜ m1 ⎟⎟ ⋅ R D1 , g m1 = 2 ⎜
⎟(80 )(0.05) = 0.5657 mA/V
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 0.5657 ⎞
Ad = ⎜
⎟(13.66 ) = 3.864
⎝ 2 ⎠
⎛ 0. 1 ⎞
(ii) A2 = − g m3 R D 3 , g m 3 = 2 ⎜
⎟(25)(0.1) = 0.7071 mA/V
⎝ 2 ⎠
A2 = −(0.7071)(37.16 ) = −26.28
(iii) A3 =
g m4 RS
⎛ 0.04 ⎞
, g m4 = 2 ⎜
⎟(100 )(0.2 ) = 1.265 mA/V
1+ g m 4 R S
⎝ 2 ⎠
(1.265)(15) = 0.950
1 + (1.265)(15)
(c) A = Ad ⋅ A2 ⋅ A3 = (3.864)(− 26.28)(0.95) = −96.5
A3 =
______________________________________________________________________________________
13.31
a.
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 ⎝ L ⎠ is increased to 50, then
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
13.32
υ CM (max) = V + − υ SD 6 (min ) − υ SG1
υ SD 6 (min ) = υ SD (sat ) + 0.2 = 0.9022 − 0.5 + 0.2 = 0.6022 V
i D1 = K p1 (υ SG1 + VTP )
2
20.2 = 250(υ SG1 − 0.5) ⇒ υ SG1 = 0.7843 V
υ CM (max ) = 5 − 0.6022 − 0.7843 = 3.61 V
2
υ CM (min ) = V − + υ GS 3 + υ SD1 (min ) − υ SG1
⎛ 0. 1 ⎞
2
2
i D 3 = K n (υ GS 3 − VTN ) , K n = ⎜
⎟(6.25) = 0.3125 mA/V
⎝ 2 ⎠
20.2 = 312.5(υ GS 3 − 0.5) ⇒ υ GS 3 = 0.7542 V
υ SD1 (min ) = 0.7843 − 0.5 + 0.2 = 0.4843 V
υ CM (min ) = −5 + 0.7542 + 0.4843 − 0.7843 = −4.55 V
So −4.55 ≤ υ CM ≤ 3.61 V
______________________________________________________________________________________
2
13.33
⎛ k ′p ⎞⎛ W ⎞
⎛ 0.04 ⎞
2
(a) K p 5 = ⎜⎜ ⎟⎟⎜ ⎟ = ⎜
⎟(50 ) = 1.0 mA/V
2
L
2
⎠
⎝ ⎠⎝ ⎠ 5 ⎝
+
V − V SG 5 − V −
2
K p 5 (V SG 5 + VTP ) =
R set
(1)(50)(V SG2 5 − 1.4VSG5 + 0.49) = 10 − VSG5
2
50V SG
5 − 69V SG 5 + 14.5 = 0 ⇒ V SG 5 = 1.121 V
I set = I Q =
10 − 1.121
= 0.1776 mA = I D 7
50
⎛ 0.04 ⎞
⎛ 0.1776 ⎞
(b) Ad = g m1 (ro 2 ro 4 ) , g m1 = 2 ⎜
⎟(50 )⎜
⎟ = 0.5960 mA/V
2
⎝
⎠
⎝ 2 ⎠
ro 2 =
1
1
= 281.5 k Ω , ro 4 =
= 563.1 k Ω
⎛ 0.1776 ⎞
⎛ 0.1776 ⎞
(0.04)⎜
(0.02)⎜
⎟
⎟
⎝ 2 ⎠
⎝ 2 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Ad = (0.5960)(281.5 563.1) = 111.9
⎛ 0. 1 ⎞
A2 = − g m 7 (ro 7 ro8 ) , g m 7 = 2 ⎜
⎟(30 )(0.1776 ) = 1.032 mA/V
⎝ 2 ⎠
ro 7 =
1
= 281.5 k Ω , r =
1
o8
(0.02)(0.1776)
(0.04)(0.1776)
A2 = −(1.032)(281.5 140.8) = −96.86
A = Ad ⋅ A2 = (111.9 )(− 96.86 ) = −10,839
= 140.8 k Ω
______________________________________________________________________________________
13.34
⎛ 0.04 ⎞
Ad = g m1 (ro 2 ro 4 ) , g m1 = 2 ⎜
⎟(50 )(0.1) = 0.6325 mA/V
⎝ 2 ⎠
ro 2 =
1
1
= 400 k Ω , ro 4 =
= 666.7 k Ω
(0.025)(0.1)
(0.015)(0.1)
Ad = (0.6325)(400 666.7 ) = 158.1
⎛ 0. 1 ⎞
A2 = − g m 7 (ro 7 ro8 ) , g m 7 = 2 ⎜
⎟(30 )(0.2 ) = 1.095 mA/V
⎝ 2 ⎠
ro 7 =
1
1
= 333.3 k Ω , ro8 =
= 200 k Ω
(0.015)(0.2)
(0.025)(0.2)
A2 = −(1.095)(333.3 200) = −136.9
A = Ad ⋅ A2 = (158.1)(− 136.9) = −21,644
______________________________________________________________________________________
13.35
f PD =
1
2π Req Ci
R =r
r
where eq 04 02 and
We can find that
Ci = C1 (1 + Av 2 )
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
f PD =
1
2π (2.51× 106 )(3024 × 10−12 )
or
f PD = 21.0 Hz
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.36
From Problem 13.33,
A2 = 96.86 , ro 2 = 281.5 k Ω , ro 4 = 563.1 k Ω
f PD =
1
2π (ro 2 ro 4 )C M
Then C M =
1
= 1.06 × 10 − 7 F
3
2π (8)(281.5 563.1)× 10
C M = C1 (1 + A2 ) ⇒ 1.06 × 10 −7 = C1 (97.86 )
Or C1 = 1.08 ×10 −9 F
______________________________________________________________________________________
13.37
R0 = r07 r08
We can find that
r07 = r08 = 2.52 MΩ
Then
R0 = 2.52 2.52
or
R0 = 1.26 MΩ
______________________________________________________________________________________
13.38
6
=2V
3
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I D = 0.5 = ⎜
= 3.91
⎟⎜ ⎟(2 − 0.4 ) ⇒ ⎜ ⎟
2
L
⎝
⎠⎝ ⎠
⎝ L ⎠ 2−5
(a) VGS 2 =
(b) Aυ =
g m1 (ro1 ro 2 )
1 + g m1 (ro1 ro 2 )
0.98 =
, ro1 = ro 2 =
1
(0.025)(0.5)
= 80 k Ω , ro1 ro 2 = 40 k Ω
g m1 (40 )
⇒ g m1 = 1.225 mA/V
1 + g m1 (40 )
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
g m1 = 1.225 = 2 ⎜
⎟⎜ ⎟ (0.5) ⇒ ⎜ ⎟ = 15
2
L
⎝
⎠⎝ ⎠ 1
⎝ L ⎠1
(c) R o =
1
1
ro1 ro 2 =
40 = 0.8163 40
g m1
1.225
R o = 800 Ω
______________________________________________________________________________________
13.39
(a)
2
⎛ 80 ⎞
I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ]
2
⎝ ⎠
I Q 2 = 180 μ A
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
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
L
2
⎝ ⎠⎝ ⎠8 P
⎝ L ⎠8 P
⎛ 80 ⎞⎛ W ⎞
⎛W ⎞
180 = ⎜ ⎟⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180
⎝ 2 ⎠⎝ L ⎠8 N
⎝ L ⎠8 N
______________________________________________________________________________________
13.40
⎛ 0.1 ⎞⎛ W ⎞
⎛ 0. 1 ⎞
2
2
I REF = I Q1= 0.150 = ⎜
⎟⎜ ⎟ (VGS 11 − 0.5) = ⎜
⎟(20 )(VGS 11 − 0.5) ⇒ VGS 11 = 0.8873 V
2
L
2
⎝
⎠⎝ ⎠ 11
⎝
⎠
V GS , REF = 5 − 0.8873 = 4.1127 V
For three NMOS transistors in series:
4.1127
= 1.3709 V
VGS =
3
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.15 = ⎜
⎟⎜ ⎟(1.3709 − 0.5) ⇒ ⎜ ⎟ = 3.96
⎝ 2 ⎠⎝ L ⎠
⎝L⎠
______________________________________________________________________________________
13.41
(a)
2
⎛ 80 ⎞
I Q 2 = 250 μ A = ⎜ ⎟ (5) (VGS 8 − 0.7 )
2
⎝ ⎠
⇒ VGS 8 = 1.818 V
⇒ VGS 6 = VSG 7 =
1.818
= 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 ) ( 800 533.3)
Ad 1 = 175
Second stage:
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
A2 = − g m 5 ( ro 5 ro9 )
⎛ 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
______________________________________________________________________________________
13.42
Ad = g m1 ( Ro 6 Ro8 )
(a)
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
1
1
ro1 = ro 6 = ro8 = ro10 =
=
= 2.67 M Ω
λ I DQ (0.015)(25)
ro 4 =
1
λ I D4
=
1
⇒ 1.33 M Ω
0.015
(
)( 50 )
Now
Ro 8 = 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
f PD =
(c)
1
1
=
⇒ f PD = 80 Hz
2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 )
GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz
______________________________________________________________________________________
13.43
(a)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
ro1 = ro8 = ro10 =
ro 6 =
ro 4 =
1
λn I D
1
λn I D 4
=
1
λp ID
=
1
= 2 MΩ
(0.02)(25)
1
= 2.67 M Ω
(0.015)(25)
=
1
= 1.33 M Ω
(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 ⎞
X1 = ⎜ ⎟
X6 = ⎜ ⎟
⎝ L ⎠1 and
⎝ L ⎠6
Define
Then
Ro = ⎣⎡63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦⎤ ⎡⎣ 41.8 X 1 ( 2 )( 2 ) ⎤⎦
= 134.8 X 6 167.2 X 1 =
Ad = g m1 Ro
22,539 X 1 X 6
134.8 X 6 + 167.2 X 1
⎛ 22,539 X 1 X 6
⎞
= (41.8 X 1 ) ⎜
⎟
134.8
167.2
X
+
X
6
1 ⎠
⎝
= 10, 000
⎛W ⎞
X6 = ⎜ ⎟ =
⎝ L ⎠6
Now
1 ⎛W ⎞
⎜ ⎟ = 0.674 X 1
2.2 ⎝ L ⎠1
We then find
⎛W ⎞
⎛W ⎞
X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟
L
⎝ ⎠1
⎝ L ⎠p
and
⎛W ⎞
⎜ ⎟ = 1.85
⎝ L ⎠n
______________________________________________________________________________________
13.44
+
−
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 ⎞
X1 = ⎜ ⎟
⎝ L ⎠1
where
Assume all width-to-length ratios are the same.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 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
X 12 =
W
= 16
L
for all transistors
So that
______________________________________________________________________________________
13.45
(a) Ad = Bg m1 ro 6 ro8
(
)
⎛ 0.1 ⎞
g m1 = 2 ⎜
⎟(20 )(0.06 ) = 0.4899 mA/V
⎝ 2 ⎠
1
= 138.9 k Ω
(0.04)(0.06)(3)
1
ro8 =
= 277.8 k Ω
(0.02)(0.06)(3)
ro 6 =
Ad = (3)(0.4899)(138.9 277.8) = 136.1
(b) Ro = ro 6 ro8 = 138.9 277.8 = 92.6 k Ω
(c)
f PD =
1
1
=
⇒ f PD = 343.7 kHz
2πR o C 2π 92.6 × 10 3 5 × 10 −12
(
(
)
)(
)
GBW = (136.1) 343.7 × 10 ⇒ GBW = 46.8 MHz
______________________________________________________________________________________
3
13.46
1
= 0.5 M Ω
(0.02)(2.5)(40)
1
ro8 =
= 0.667 M Ω
(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⎠
(a)
Assume all (W/L) ratios are the same except for
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛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 ⎞
⎜ ⎟ = ⎜ ⎟ = 49
Assume ⎝ 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 =
f 3− dB =
(c)
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 =
______________________________________________________________________________________
13.47
(a)
From previous results, we can write
Ro10 = g m10 ( ro10 ro 6 )
Ro12 = g m12 ( ro12 ro8 )
Ad = Bg m1 ( Ro10 Ro12 )
Now
1
λP B ( I Q / 2 )
=
1
= 0.5 M Ω
(0.02)(2.5)(40)
1
=
1
= 0.667 M Ω
(0.015)(2.5)(40)
ro10 = ro 6 =
ro12 = ro8 =
λn B ( I Q / 2 )
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Assume all transistors have the same width-to-length ratios except for M 5 and M 6 .
⎛W ⎞
2
⎜ ⎟= X
L
⎝
⎠
Let
Then
⎛ 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 ]
⎡ (20.9 X )(56.3 X ) ⎤
2
= 200 X ⎢
⎥ = 3048 X
⎣ 20.9 X + 56.3 X ⎦
Then
⎛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 ⎞
⎜ ⎟ = ⎜ ⎟ = 6.56
L
Assume ⎝ ⎠ A ⎝ L ⎠ B
⎛ 80 ⎞
I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V
⎝ 2⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Need 5 transistors in series
VGSC =
10 − 1.052
= 1.79 V
5
Then
⎛ 80 ⎞⎛ W ⎞
⎛W ⎞
I REF = 80 = ⎜ ⎟⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20
2
L
⎝ ⎠⎝ ⎠C
⎝ L ⎠C
f 3− dB =
(c)
Now
1
2π Ro C
where
Ro = Ro10 Ro12
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
f 3− dB =
= 1.36 kHz
2π (39 ×106 )(3 × 10−12 )
GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz
______________________________________________________________________________________
13.48
(a) ΔVO1 = 0.7 + (0.3)(0.4) = 0.82 V
0.82
= 5.47 k Ω
0.15
3−0
= 10 k Ω
RC =
0.3
RD =
(b) Ad =
g m1
(R D RiC ) , g m1 = 2 ⎛⎜ 0.04 ⎞⎟(50)(0.15) = 0.7746 mA/V
2
⎝ 2 ⎠
RiC = rπ + (1 + β )R E , rπ =
(120 )(0.026) = 10.4 k Ω
0. 3
RiC = 10.4 + (121)(0.4 ) = 58.8 k Ω
⎛ 0.7746 ⎞
Ad = ⎜
⎟(5.47 58.8) = 1.938
⎝ 2 ⎠
− β RC
− (120)(10)
(c) A2 =
=
= −20.41
rπ + (1 + β )R E 10.4 + (121)(0.4)
(d) Aυ = Ad ⋅ A2 = (1.938)(− 20.41) = −39.6
______________________________________________________________________________________
13.49
(a) R S =
0 − (− 3)
= 10 k Ω
0. 3
I D1 = K n (VGS1 − VTN )
2
0.3 = 3(VGS1 − 0.4) ⇒ VGS1 = 0.7162 V
2
0.7162 − (− 3)
= 12.39 k Ω
0. 3
ΔVO1 = (0.3)(0.5) + 0.7 = 0.85 V
RC 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.85
R C1 =
= 3.4 k Ω
0.25
(120 )(0.026 ) = 12.48 k Ω
0.25
(b)
g m1 =
= 9.615 mA/V, rπ 1 =
0.026
0.25
(
0.3
120 )(0.026 )
g m3 =
= 11.54 mA/V, rπ 3 =
= 10.4 k Ω
0.026
0. 3
g
Ad = m1 (RC1 Ri 3 )
2
Ri 3 = rπ 3 + (1 + β )R E = 10.4 + (121)(0.5) = 70.9 k Ω
9.615
(3.4 70.9) = 15.60
2
− βRC 2
− (120 )(12.39 )
=
= −20.97
(c) A2 =
rπ 3 + (1 + β )R E 10.4 + (121)(0.5)
Ad =
(d) A3 =
g mD R S
, g mD = 2 (3)(0.3) = 1.897 mA/V
1 + g mD R S
(1.897)(10) = 0.950
1 + (1.897)(10)
(e) Aυ = Ad ⋅ A2 ⋅ A3 = (15.6 )(− 20.97 )(0.95) = −310.8
A3 =
______________________________________________________________________________________
13.50
(a) For PMOS:
⎛ 0.04 ⎞
g m1 = g m 2 = 2 ⎜
⎟(40 )(0.125) = 0.6325 mA/V
⎝ 2 ⎠
roP =
1
1
=
= 228.6 k Ω
λ I D1 (0.035)(0.125)
For BJT:
ro 2 =
VA
150
=
= 1200 k Ω
I C 0.125
(b) Ad = g m 2 (roP ro 2 ) = (0.6325)(228.6 1200) = 121.5
______________________________________________________________________________________
13.51
(a) For NMOS:
⎛ 0. 1 ⎞
g mN = 2 ⎜
⎟(40 )(0.125) = 1.0 mA/V
⎝ 2 ⎠
roN =
1
= 400 k Ω
(0.02)(0.125)
For BJT:
100
= 800 k Ω
0.125
(b) Ad = g mN roN ro 2 = (1.0) 400 800 = 266.7
ro 2 =
(
)
(
)
______________________________________________________________________________________
13.52
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I REF = 200 μ A
K n = K p = 0.5 mA / V 2
λn = λ p = 0.015 V −1
A = g m1 ( Ro 6 Ro8 )
d
(a)
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 =
ro1 =
1
λn I D 4
=
1
= 333 k Ω
(0.015)(0.2)
1
1
=
= 667 k Ω
λ p I D1 (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
______________________________________________________________________________________
13.53
+
−
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 ⎞
⎜ ⎟ = 2.2 ⎜ ⎟
L
⎝ L ⎠n
Assume ⎝ ⎠ p
Ro8 = g m8 ( ro8 ro10 )
Ro 6 = g m 6 (ro 6 )(ro 4 ro1 )
ro8 =
ro10 =
1
λP I D 8
=
1
= 0.60 M Ω
(0.02)(83.3)
1
= 0.60 M Ω
λP I D 8
⎛ k ′p ⎞ ⎛ W ⎞
⎛ 35 ⎞
g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3)
L
2
⎝
⎠
⎝ 2⎠
8
⎝ ⎠
= 113.3 X
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛W ⎞
X2 =⎜ ⎟
⎝ L ⎠n
where
ro 6 =
ro 4 =
VA
80
=
= 0.960 M Ω
I C 6 83.3
1
λn I D 4
=
1
= 0.40 M Ω
(0.015)(167)
ro1 =
1
1
=
= 0.60 M Ω
λ p I D1 (0.02)(83.3)
gm6 =
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
L
⎝
⎠
⎝ 2⎠
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
______________________________________________________________________________________
13.54
For vcm (max), assume VCB (Q5 ) = 0. Then
VS = 15 − 0.6 − 0.6 = 13.8 V
I D 9 = I D10 =
0.236
= 0.118 mA
2
Using parameters given in Example 13.12
VSG =
I D9
0.118
− 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
13.55
I1 = I 2
K p (V SG + VTP ) =
2
I 2 = 0.15 =
V SG − V BE 7
R1
V SG − 0.6
⇒ V SG = 1.8 V
8
I 1 = 0.15 = K p (1.8 − 1) ⇒ K p = 0.234 mA/V 2
2
______________________________________________________________________________________
13.56
(a) K p (V SG + VTP ) =
V SG − V BE 7
R1
0.15(V SG − 1.2 ) =
V SG − 0.6
8
2
2
2
− 3.88V SG + 2.328 = 0 , ⇒ V SG = 2.437 V
We find 1.2V SG
I1 = I 2 =
2.437 − 0.6
= 0.2296 mA
8
(b) VC 7 = V + − 0.6 − 0.6
VC 6 = V − + 2.437
Set VC 6 = VC 7
Then V S − 1.2 = −V S + 2.437 ⇒ V S = 1.82 V
______________________________________________________________________________________
13.57
I C 5 = I C 4 = 300 μ A
We have
Ri 2 = rπ 13 =
β nVT
I C13
=
(200)(0.026)
= 17.3 kΩ
0.3
Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3)
or
Ad = 10.38
Now
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
_____________________________________________________________________________
13.58
Assuming the resistances looking into Q4 and into the output stage are very large, we have
β R013
| Av 2 | =
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)
rπ 13 =
= 17.3 kΩ
0.3
g m13 =
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
f PD =
1
2π Req Ci
Req = Ri 2 r012 r010
Neglecting R3 ,
r010 =
1
λ I D10
=
1
= 333 kΩ
(0.02)(0.15)
Neglecting R5 ,
50
= 333 kΩ
0.15
Ri 2 = rπ 13 + (1 + β ) RE13
r012 =
= 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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)
Ad = 56.6
or
Overall gain:
Av = (56.6)(1814) = 102, 672
Then unity-gain bandwidth = (77.4)(102, 672)
⇒ 7.95 MHz
______________________________________________________________________________________
13.59
Since VGS = 0 in J 6 , I REF = I DSS
⇒ I DSS = 0.8 mA
______________________________________________________________________________________
13.60
Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ]
a.
(100)(0.026)
= 13 kΩ
0.2
I
200 μ A
IC 5 ≅ C 6 =
= 2 μA
β
100
rπ 6 =
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 =
r04 =
Then
1
λ ID
=
1
= 500 kΩ
(0.02)(0.1)
VA 5.0
=
= 500 kΩ
I C 4 0.1
Av = (0.0943)[500 || 500 || 5670]
or
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Av = 22.6
______________________________________________________________________________________
13.61
a.
Set
Need
VSD (QE ) ≥ VSD ( sat ) = VP
VP = 3 V
I REF 2 =
so that
and
VZK = 3 V
For minimum bias ±3 V
VZK − VD1
R3
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,
I DSS = 0.2 mA
b.
Neglecting base currents
I 01 = I REF 1 = 0.5 mA =
12 − 0.6
R4
so that
R4 = 22.8 kΩ
______________________________________________________________________________________
13.62
a.
We have
gm2 =
2
⋅ I D ⋅ I DSS
| VP |
=
2
⋅ (0.5)(1)
4
= 0.354 mA/V
1
1
r02 =
=
= 100 kΩ
λ I D (0.02)(0.5)
r04 =
VA 100
=
= 200 kΩ
I D 0.5
0.5
= 19.23 mA/V
0.026
(200)(0.026)
rπ 4 =
= 10.4 kΩ
0.5
gm4 =
So
R04 = r04 ⎡⎣1 + g m 4 ( rπ 4 R2 ) ⎤⎦
= 200 ⎡⎣1 + (19.23) (10.4 0.5 ) ⎤⎦
= 2035 kΩ
Ad = g m 2 ( r02 R04 RL )
RL → ∞
For
Ad = 0.354 (100 || 2035 ) = 33.7
With these parameter values, gain can never reach 500.
b.
Similarly for this part, gain can never reach 700.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 14
14.1
Ad =
vo
= −80
vi
vo (max) = 4.5 ⇒ vi (max) = 56.25 mV
vi (max) rms =
56.25
= 39.77 mV
2
So
______________________________________________________________________________________
14.2
(a)
4.5
= 0.028125 mA
160
4.5
iL =
= 4.5 mA
1
Output Circuit = 4.528 mA
i2 =
v
−4.5
vi = − o =
⇒ vi = −0.05625 V
80
A
(b)
io ≈ 15 mA =
vo 4.5
=
RL RL
⇒ RL (min) = 300 Ω
______________________________________________________________________________________
14.3
(1)
vo = 2 V
(2)
v2 = 12.5 mV
(3)
AOL = 2 × 10 4
(4)
v1 = 8 μ V
(5) AOL = 1000
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.4
R
120
= −21.42857
(a) ACL (∞ ) = − 2 = −
R1
5.6
−21.42857
= −21.42376
22.42857
1+
10 5
− 21.42376 − (− 21.42857 )
× 100% = −0.0224%
− 21.42857
R
120
(b) ACL (∞ ) = − 2 = −
= −14.634146
R1
8.2
ACL =
−14.634146
= −14.63186
15.634146
1+
10 5
− 14.63186 − (− 14.634146 )
× 100% = −0.0156%
− 14.634146
______________________________________________________________________________________
ACL =
14.5
47
7.91176
6.8
=
= 7.90863
(a) (i) ACL =
7.91176
47 ⎞
⎛
1
+
⎜1 +
⎟
2 × 10 4
⎝ 6.8 ⎠
1+
2 × 10 4
7.90863 − 7.91176
(ii)
× 100% = −0.03956%
7.91176
7.91176
(b) (i) ACL =
= 7.84966
7.91176
1+
10 3
7.84966 − 7.91176
(ii)
× 100% = −0.785%
7.91176
______________________________________________________________________________________
1+
14.6
− R2
R
R1
R1
⇒ 2 = 15.12091
=
R1
⎛1 + R 2 ⎞ 1.0005 + 5 × 10 − 4 ⎛ R 2 ⎞
⎜ R ⎟
⎜
R1 ⎟⎠
1⎠
⎝
⎝
− R2
(a) − 15.0 =
1+
2 × 10 3
−15.12091
(b) ACL =
= −15.1160
16.12091
1+
5 × 10 4
______________________________________________________________________________________
14.7
90
⇒ AOL = 8.9991 × 10 5
90
1+
AOL
______________________________________________________________________________________
(1 − 0.0001)(90) =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.8
ACL = (1 − 0.0002 )(1) =
1
⇒ AOL = 4999
1
1+
AOL
______________________________________________________________________________________
14.9
A =+
(a)
R 2 210(1 ± 0.001)
=
R1
21(1 ± 0.001)
210.21
= 10.02
20.979
209.79
A min =
= 9.98
21.021
So 9.98 ≤ A ≤ 10.02
A max =
10.02
= 10.009
11.02
1+
10 4
9.98
A min =
= 9.969
10.98
1+
10 4
So 9.969 ≤ A ≤ 10.009
A max =
(b)
______________________________________________________________________________________
14.10
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
1 (1 + A0 L )
=
+
⋅
1
Rif
Ri R2
a.
For Ri = 1 kΩ
−(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
b.
For Ri = 10 kΩ
−(1/ 20)
⎡ 1
1 ⎛ 1
1
1 ⎞⎤
⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥
⎝
⎠⎦
⎣
−0.05
=
[0.01 + 1.6 × 10−4 ]
ACL =
or
⇒ ACL = −4.92
1
1 1 + 103
= +
⇒ Rif = 98.9 Ω
Rif 10 100
c.
For Ri = 100 kΩ
ACL =
=
−(1/ 20)
⎡ 1
1 ⎛ 1
1
1 ⎞⎤
⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥
⎝
⎠⎦
⎣
−0.05
[0.01 + 7 × 10−5 ]
or
⇒ ACL = −4.965
1
1 1 + 103
=
+
⇒ Rif = 99.8 Ω
Rif 100
100
______________________________________________________________________________________
14.11
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
vo
⎝
ACL = =
vi ⎡
1 ⎛ R2 ⎞ ⎤
⎢1 +
⎜1 + ⎟ ⎥
A
R1 ⎠ ⎦
OL ⎝
⎣
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
0.002 1 + 1 (50)
AOL
which yields
AOL = 1000
______________________________________________________________________________________
14.12
From Equation (14.18)
⎛A
1 ⎞
− ⎜ OL − ⎟
R
R
v
2 ⎠
⎝ o
Avf 1 = o1 =
v1 ⎛ 1
1
1 ⎞
+
+ ⎟
⎜
R
R
R
o
2 ⎠
⎝ 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
v1 =
vi
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
v1 =
We find
vi
vi
=
( 45.15495 )(10 ) + 1 452.5495
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vi
⎡
⎤
vo1 = −4.504495 × 103 ⎢
⎥
⎣ 452.5495 ⎦
Or
Avf 1 =
vo1
= −9.9536
vi
For the second stage, RL = ∞
⎛ 5 × 103
1 ⎞
−⎜
−
⎟
1
100
⎠ ⋅ v ′ = −4.950485 × 103 ⋅ v ′
vo 2 = ⎝
1
1
⎛1 1 ⎞
⎜ +
⎟
⎝ 1 100 ⎠
⎡
⎤
1
1 ⎢1 + 5 × 103 ⎥
K≡ +
= 49.61485
⎢
1 ⎥⎥
10 100 ⎢
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
Avf =
vo 2
= (−9.9536)(−9.95776) ⇒ Avf = 99.12
vi
______________________________________________________________________________________
14.13
a.
v1 − vI
v v −v
+ 1 + 1 0 =0
R3 + Ri R1
R2
⎡ 1
vI
1
1⎤ v
v1 ⎢
+ + ⎥= 0 +
R
R
R
R
R
R
+
i
1
2 ⎦
2
3 + Ri
⎣ 3
v0 v0 − A0 L vd v0 − v1
=0
+
+
RL
R0
R2
(1)
(2)
or
⎡ 1
A v
1
1⎤ v
v0 ⎢ +
+ ⎥ = 1 + 0L d
R0
⎣ RL R0 R2 ⎦ R2
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ Ri
⎝ R3 + Ri ⎠
So substituting numbers:
(3)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vI
1⎤ v
1
⎡ 1
v1 ⎢
+ + ⎥= 0 +
⎣10 + 20 10 40 ⎦ 40 10 + 20
(1)
or
v1[0.15833] = v0 [0.025] + vI [0.03333]
1 ⎤ v (10 4 )vd
⎡1 1
v0 ⎢ +
+ ⎥= 1 +
0.5
⎣1 0.5 40 ⎦ 40
or
So
or
(2)
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 )
⎝ 10 + 20 ⎠
(3)
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 )
(2)
v0 [3.025] = 1.333 ×104 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
ACL =
To find
v0
= 4.993
vI
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 ⎥⎥
1 ⎢⎢ 103 ⎥⎥
=
⋅
=
⋅
40 ⎥
R2 ⎥ 0.5 ⎢
R0′ f
R0 ⎢
1
+
⎢ R R ⎥
⎢1 + 10 20 ⎥
i ⎦
1
⎣
⎦
⎣
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
R0′ f = 3.5 Ω
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 ⎠
______________________________________________________________________________________
14.14
(a)
(b)
(i)
υ I −υO
Ri
υI
⎛1
υ I ⎜⎜
υ O − AOL (υ I − υ O )
Ro
⎛ 1
AOLυ I
A ⎞
1
= υ O ⎜⎜
+
+ OL ⎟⎟
Ro
Ro ⎠
⎝ Ri R o
+
Ri
=
⎝ 10
+
⎛ 1 1 5 × 10 3 ⎞
5 × 10 3 ⎞
⎟ = υO ⎜ + +
⎟
⎜ 10 1
1 ⎟⎠
1 ⎟⎠
⎝
υ I (5.0001×10 3 ) = υ O (5.0011×10 3 )
υO
= 0.9998
υI
(ii) I x =
V x − AOL (− V x ) V x
+
Ro
Ri
Ix
1 + AOL
1
1 1 + 5 × 10 3 1
=
=
+
=
+
V x Rof
Ro
Ri
1
10
R of ≅ 0.2 Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.15
and
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
v1 = − 0
A0L
v0 = − A0L v1
so that
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%
v0
v0
2
______________________________________________________________________________________
14.16
⎛ 40 ⎞
⎛ 4⎞
vB = ⎜
⎟ v2 = ⎜ ⎟ v2 = 0.8v2
⎝ 40 + 10 ⎠
⎝5⎠
(1)
v1 − v A v A − v0
=
10
40
v1 v0
1 ⎞
⎛1
+
= vA ⎜ + ⎟
10 40
10
40
⎝
⎠
v1 (0.1) + v0 (0.025) = v A (0.125)
(2)
v0 = A0 L vd = A0 L (vB − v A )
or
(3)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
v0 = A0 L [0.8v2 − v A ]
v0
− 0.8v2 = −v A
A0 L
⇒ v A = 0.8v2 −
v0
A0 L
Then
⎡
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
______________________________________________________________________________________
14.17
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 ⎥⎦
Rif 2 = 0.0109 kΩ
So
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
110.017
0.1109
= + ⋅
= 0.10 +
11.017
Rif 10 1 1 + 1 + 1
0.1109 1
so that
Rif = 99.1 Ω
R
:
b.
To determine 0 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 ⎥⎦
R
= 0.021 kΩ
which yields 0 f 1
For the second op-amp, then
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎡
⎤
⎢
⎥
A0 L
1
1 ⎢
⎥
=
⋅
R2
⎥
R0 f R0 ⎢
⎢1 + ( R + R ) || R ⎥
1
0 f1
i ⎦
⎣
⎡
⎤
⎥
1 ⎢
100
⎥
= ⋅⎢
0.10
1 ⎢1 +
⎥
⎢⎣ (0.121) ||10 ⎥⎦
or
c.
R0 f = 18.4 Ω
To find the gain, consider the second op-amp.
v01 − ( −vd 2 ) vd 2 −vd 2 − v02
+
=
0.1
Ri
0.1
(1)
v01
v02
1
1 ⎞
⎛ 1
+ vd 2 ⎜
+ +
⎟=−
0.1
0.1
10
0.1
0.1
⎝
⎠
or
v01 (10) + vd 2 (20.1) = −v02 (10)
v02 − A0 L vd 2 v02 − ( −vd 2 )
+
=0
R0
0.1
v02
⎛ 100 1 ⎞ v02
=0
− vd 2 ⎜
−
⎟+
1
0.1 ⎠ 0.1
⎝ 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.
(2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vI − ( −vd 1 ) vd 1 −vd 1 − v01
+
=
1
Ri
1
(1)
⎛1 1 1⎞
vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1)
⎝ 1 10 1 ⎠
or
vI (1) + vd 1 (2.1) = −v01 (1)
v01
v − A0 L vd 1 v01 − ( −vd 1 )
+ 01
+
=0
0.1109
R0
1
(2)
1 1⎞
⎛ 1
⎛ 100 1 ⎞
v01 ⎜
+ + ⎟ − vd 1 ⎜
− ⎟=0
⎝ 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
d.
Ideal
So ratio of actual to ideal = 0.650.
______________________________________________________________________________________
14.18
(a)
For the op-amp. A0 L ⋅ f 3dB = 10
f 3dB =
6
106
= 50 Hz
2 × 104
For the closed-loop amplifier.
f 3dB =
(b)
106
= 40 kHz
25
Open-loop amplifier.
A=
2 × 104
2 × 104
⇒| A | =
2
f
⎛ f ⎞
1+ j
1+ ⎜
f 3dB
⎟
⎝ f 3dB ⎠
f = 0.25 f 3dB ⇒ A =
f = 5 f 3− dB ⇒ A =
2 × 104
1 + (0.25) 2
2 × 104
1 + (5)
2
= 1.94 × 104
= 3.92 × 103
Closed-loop amplifier
f = 0.25 f 3dB ⇒ A =
f = 5 f 3− dB ⇒ A =
25
1 + (0.25) 2
25
1 + (5) 2
= 24.25
= 4.90
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
14.19
Ao = 100 dB, ⇒ Ao = 10 5
A = 38 dB, A = 79.43
10 5
Then 79.43 =
2
⎛ 10 4 ⎞
⎟
1 + ⎜⎜
⎟
⎝ f PD ⎠
10 4
10 5
≅
⇒ f PD = 7.94 Hz
f PD 79.43
( )
GBW = 10 5 (7.94) = 7.94 ×10 5 Hz
______________________________________________________________________________________
14.20
⎛ R ⎞ ⎛ 150 ⎞
(a) ACLO = ⎜⎜1 + 2 ⎟⎟ = ⎜1 +
⎟ = 11
R1 ⎠ ⎝
15 ⎠
⎝
f T = 1.2 × 10 6 = (11) f 3− dB ⇒ f 3− dB = 109 kHz
⎡ (150 )(1 ± 0.05) ⎤
(b) ACLO = ⎢1 +
(15)(1 ± 0.05) ⎥⎦
⎣
157.5
ACLO (max ) = 1 +
= 12.05
14.25
142.5
ACLO (min ) = 1 +
= 10.05
15.75
Then 10.05 ≤ ACLO ≤ 12.05
f T = 1.2 ×10 6 = (12.05) f 3− dB ⇒ f 3− dB = 99.6 kHz
f T = 1.2 × 10 6 = (10.05) f 3− dB ⇒ f 3− dB = 119.4 kHz
Then 99.6 ≤ f 3− dB ≤ 119.4 kHz
______________________________________________________________________________________
14.21
The open loop gain can be written as
A0
A0 L ( f ) =
⎛
f ⎞⎛
f ⎞
⎜1 + j ⋅
⎟ ⎜1 + j ⋅
⎟
×
5
106 ⎠
f
⎝
⎝
PD ⎠
5
where A0 = 2 × 10 .
The closed-loop response is
ACL =
A0 L
1 + β A0 L
At low frequency,
100 =
2 × 105
1 + β (2 × 105 )
So that β = 9.995 × 10 .
Assuming the second pole is the same for both the open-loop and closed-loop, then
−3
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ f ⎞
f ⎞
−1 ⎛
⎟ − tan ⎜
6 ⎟
⎝ 5 × 10 ⎠
⎝ f PD ⎠
80°, φ = −100°.
φ = − tan −1 ⎜
For a phase margin of
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
______________________________________________________________________________________
14.22
(a)
nd
1st stage
(10) f 3− dB = 1 MHz ⇒ f 3− dB = 100 kHz
2 stage
(50) f 3− dB = 1 MHz ⇒ f 3− dB = 20 kHz
Bandwidth of overall system ≅ 20 kHz
(b)
If each stage has the same gain, so
K 2 = 500 ⇒ K = 22.36
Then bandwidth of each stage
(22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz
______________________________________________________________________________________
14.23
−
(a) ACLO =
1+
R2
R1
− 10.0
=
= −9.9978
⎛1 + R 2 ⎞ 1 + 11
⎜
⎟
R1 ⎠
⎝
5 × 10 4
AO
f T = 1.5 ×10 = (9.9978) f 3− dB ⇒ f 3− dB = 150.033 kHz
6
(b) ACLO = (− 9.9978) = −999.34
3
At f 3− dB ; ⇒ ACL =
999.34
2
= 706.64
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
999.34
Then 706.64 =
3
2 ⎤
⎡
f
⎞
⎛
−dB
3
⎢ 1+ ⎜
⎟ ⎥
⎜
3 ⎟
⎢
⎝ 150.033 × 10 ⎠ ⎥
⎣
⎦
3
2
2
⎡ ⎛
f 3− dB
⎞ ⎤
⎛ 999.34 ⎞
⎢1 + ⎜⎜
⎟
⎥
=
⇒ f 3− dB = 76.49 kHz
⎜
⎟
3 ⎟
⎝ 706.64 ⎠
⎢⎣ ⎝ 150.033 × 10 ⎠ ⎥⎦
______________________________________________________________________________________
14.24
(5 × 10 4 ) f PD = 106 ⇒ f PD = 20 Hz
(25) f 3− dB 106 ⇒ f 3− dB = 40 kHz
Av =
Avo
1+ j
f
⇒ Av =
f 3− 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
= 11.18
______________________________________________________________________________________
14.25
(20 × 103 ) ⋅ Avf MAX = 106 ⇒ Avf MAX = 50
______________________________________________________________________________________
14.26
(a)
f max =
(b) f max =
SR
5 × 10 6
=
⇒ f max = 159 kHz
2πV PO
2π (5)
5 × 10 6
⇒ f max = 530.5 kHz
2π (1.5)
5 × 10 6
⇒ f max = 1.99 MHz
2π (0.4 )
______________________________________________________________________________________
(c)
14.27
a.
f max =
Using Equation (14.55),
VP 0 =
8 × 106
2π (250 × 103 )
or
VP 0 = 5.09 V
b.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
T=
1
1
=
= 4 × 10−6 s
f 250 × 103
Period
One-fourth period = 1 μ s
Slope =
VP 0
= SR = 8 V/μ s
1μ s
⇒ VP 0 = 8 V
______________________________________________________________________________________
14.28
f max =
SR
2πV PO
(
)
SR = 2π (10) 12 ×10 3 = 7.54 ×10 5 V/s
Or SR = 0.754 V/ μ s
______________________________________________________________________________________
14.29
(a)
f max = 20 × 10 3 =
0.63 × 10 6
⇒ V PO = 5.0 V
2πV PO
3 × 10 6
= 23.87 V
2π 20 × 10 3
______________________________________________________________________________________
(b) V PO =
(
)
14.30
For input (a), maximum output is 5 V.
S R = 1 V/μs
so
For input (b), maximum output is 2 V.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For input (c), maximum output is 0.5 V so the output is
______________________________________________________________________________________
14.31
For input (a),
Then
max v01 = 3 V.
v02 max = 3(3) = 9 V
For input (b),
max v01 = 1.5 V.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
v02 max = 3 (1.5) = 4.5V
______________________________________________________________________________________
14.32
⎛V ⎞
⎛V ⎞
I1 = I S1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟
V
⎝ T ⎠
⎝ VT ⎠
Want I1 = I 2 , so
I1
=1=
I2
=
⎛V ⎞
5 × 10−14 (1 + x) exp ⎜ BE1 ⎟
⎝ VT ⎠
⎛
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
T
⎝
⎠
⎝ VT ⎠
⎛ 0.0025 ⎞
= exp ⎜
⎟ = 1.10
⎝ 0.026 ⎠
Now
1 + x = (1 − x)(1.10) ⇒
x = 0.0476 ⇒ 4.76%
______________________________________________________________________________________
14.33
(a) Balanced circuit, I S 4 = 5 ×10 −15 A
(b) From Eq. (14.62), υ CE1 = 5 V, υ CE 2 = 5.6 − 1.2 = 4.4 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4 .4 ⎞
⎛
5
⎟
⎜1 +
120 = I S 3 ⋅ ⎝ 120 ⎠
0.6 I S 4 ⎛ 1.2 ⎞
1+
⎟
⎜1 +
80
⎝ 80 ⎠
1+
1.041667 I S 3 (1.036667 )
=
⋅
(1.015)
1.0075
I S4
I S3
= 1.0123 ⇒ I S 4 = 4.939 × 10 −15 A
I S4
(c) υ CE1 = 5 V, υ CE 2 = 5.6 − 2.5 = 3.1 V
3 .1 ⎞
⎛
5
⎜1 +
⎟
I
120
⎠
120 = S 3 ⋅ ⎝
0.6 I S 4 ⎛ 2.5 ⎞
1+
⎜1 +
⎟
80
80 ⎠
⎝
1+
1.041667 I S 3 (1.025833)
=
⋅
1.0075
I S 4 (1.03125)
I S3
= 1.03937 ⇒ I S 4 = 4.811× 10 −15 A
I S4
______________________________________________________________________________________
14.34
K n = 150 μ A/V 2
ΔK n = 150(1 + x ) − 150(1 − x ) = 300 x μ A/V 2
VOS =
1
2
I Q ⎛ ΔK n ⎞
⎜
⎟
2 K n ⎜⎝ K n ⎟⎠
1
200 ⎛ 300 x ⎞
⎜
⎟ = 0.8165x ⇒ x = 0.01837
2 2(150) ⎝ 150 ⎠
______________________________________________________________________________________
15 ×10 −3 =
14.35
(a) υ O = −30(10 ± 2)× 10 −3 = (− 300 ± 60)×10 −3 V
So −0.360 ≤ υ O ≤ −0.240 V
(b) υ O = −30(100 ± 2)×10 −3 = −3 ± 0.06 V
So −3.06 ≤ υ O ≤ −2.94 V
______________________________________________________________________________________
14.36
υ O = −30(25 sin ω t ± 2) mV
υ O = −0.75 sin ω t ± 0.06 V
So (− 0.75 sin ω t − 0.06 ) ≤ υ O ≤ (− 0.75 sin ω t + 0.06 ) V
______________________________________________________________________________________
14.37
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I=
0.5 × 10 −3
= 5 × 10−8 A
104
Also
I =C
i
dVo
I
1
⇒ Vo = ∫ Idt = ⋅ t
dt
C0
C
Then
5=
5 × 10−8
t ⇒ t = 1 03 s
10 × 10−6
______________________________________________________________________________________
14.38
⎛ 100 ⎞
(a) υ O1 = ⎜1 +
⎟(± 3) mV, −33 ≤ υ O1 ≤ 33 mV
10 ⎠
⎝
⎛ 50 ⎞
⎟(± 33 ± 3) mV, −180 ≤ υ O 2 ≤ 180 mV
⎝ 10 ⎠
υ O 2 = −⎜
(b) υ O1 = (11)(10 ± 3) mV, ⇒ 77 ≤ υ O1 ≤ 143 mV
υ O 2 = −5(143 + 3) = −730 mV
υ O 2 = −5(77 − 3) = −370 mV
So −0.73 ≤ υ O 2 ≤ −0.37 V
(c) υ O1 = (11)(100 ± 3) mV
1.067 ≤ υ O1 ≤ 1.133 V
υ O 2 = −5(1.133 + 0.003) = −5.68 V
υ O 2 = −5(1.067 − 0.003) = −5.32 V
So −5.68 ≤ υ O 2 ≤ −5.32 V
______________________________________________________________________________________
14.39
v0 due to vI
1 ⎞
⎛
v0 = (0.5) ⎜ 1 + ⎟ = 0.9545 V
⎝ 1.1 ⎠
+
Wiper arm at V = 10 V, (using superposition)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ R1 || R5 ⎞
⎛ 0.0909 ⎞
v1 = ⎜
⎟ (10) = ⎜
⎟ (10)
+
R
||
R
R
⎝ 0.0909 + 10 ⎠
4 ⎠
⎝ 1 5
= 0.090
⎛1⎞
v01 = − ⎜ ⎟ (0.090) = −0.090
⎝1⎠
Then
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
______________________________________________________________________________________
14.40
a.
or
R1′ = R2′ = 0.5 || 25 = 0.490 kΩ
R1′ = R2′ = 490 Ω
b.
From Equation (14.75),
⎛ 125 × 10−6 ⎞
+ (0.125) R1′
(0.026) ln ⎜
−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)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
14.41
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′
I
⎝ S3 ⎠
⎝ 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 ⎟ ⎥
i
⎝ C2 ⎠
⎝ iC 2 ⎠ ⎦
⎣
By trial and error:
iC1 = 252 μ A
i = 248 μ A
and C 2
or
iC1
= 1.0155
iC 2
______________________________________________________________________________________
14.42
(a) υ O ( 1 μA ) = 10 −6 200 × 10 3 = 0.2 V
(
)(
)
Insert resistor R3
υ O ( 2 μA ) = −0.2 = −(2 × 10 − 6 )R3 ⎜1 +
⎛
(
)(200 ×10 ) = 0.16 V
⎛ 200 ⎞
⎟ ⇒ R = 29.09 k Ω
) = −0.16 = − (0.5 × 10 )R ⎜1 +
20
(b) υ O ( 0.8 μA ) = 0.8 × 10
υ O ( 0.5 μA
⎝
200 ⎞
⎟ ⇒ R 3 = 9.09 k Ω
20 ⎠
−6
3
−6
3
3
⎠
⎝
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.43
(a) υ O = − I B R 2 = − 2 ×10 −6 150 × 10 3 = −0.3 V
(
)(
)
150
(0.02 ) − 0.3 = −0.5 V
15
150
(c) υ O = −
(− 0.02) − 0.3 = −0.1 V
15
150
(d) υ O = −
(0.1) − 0.3 = −1.3 V
15
______________________________________________________________________________________
(b) υ O = −
14.44
(a) υ O = 0.6 ×10 −6 250 ×10 3 = 0.15 V
(b) υ O = (41)(0.008) + 0.15 = 0.478 V
(c) υ O = (41)(− 0.0035) + 0.15 = 0.0065 V
(d) υ O = (41)(0.005 sin ω t ) + 0.15 = 0.205 sin ω t + 0.15 (V)
______________________________________________________________________________________
(
)(
)
14.45
a.
For I B 2 = 1 μ A, then
or
b.
v0 = − (10−6 )(104 )
v0 = −0.010 V
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 .
______________________________________________________________________________________
14.46
From Equation (14.83), we have
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
14.47
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 × 10 4 )
= −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.
______________________________________________________________________________________
14.48
a.
For a constant current through a capacitor.
1 t
I dt
C ∫0
0.1× 10−6
v0 =
⋅ t ⇒ v0 = (0.1)t
10 −6
or
v0 =
b.
c.
v0 = 1 V
At t = 10 s,
Then
100 × 10−12
⋅ t ⇒ v0 = (10 −4 )t
10−6
v0 = 1 mV
t = 10 s,
v0 =
At
______________________________________________________________________________________
14.49
(a) υ O1 = 3 ×10 −6 50 × 10 3 = 0.15 V
υ O 2 = 0.15 V
(
)(
)
(
)(
)
20
(0.15) + 3 ×10 − 6 20 ×10 3 = −0.09 V
20
(b) R A = 10 50 = 8.33 k Ω
υ O3 = −
R B = 20 20 = 10 k Ω
(
)(
)
(c) υ O1 = ± 50 × 10 3 0.3 × 10 −6 = ±0.015 V
υ O 2 = ±0.015 V
υ O 3 = ±(20 ×10 3 )(0.3 ×10 −6 ) ± 0.015 = ±0.021 V
______________________________________________________________________________________
14.50
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
v = −0.010 V
0
or
Circuit (b),
⎛ 50 ⎞
v0 = ( 0.7 × 10 −6 )( 50 × 103 ) − ( 0.9 × 10 −6 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 1.8
v = −1.765 V
0
or
______________________________________________________________________________________
14.51
⎛ 100 ⎞
(a) For V OS : υ O = ⎜1 +
⎟(± 3) = ±33 mV
10 ⎠
⎝
(
)(
)
υ (max ) = (0.37 ×10 )(100 ×10 ) = 0.037 V
For I B : υ O (max ) = 0.43 ×10 −6 100 ×10 3 = 0.043 V
−6
3
O
So 4 ≤ υ O ≤ 76 mV
(b) For V OS : υ O = ±33 mV
(
)(
)
For I OS : υ O = ± 0.06 × 10 −6 100 ×10 3 = ±0.006 V
So −39 ≤ υ O ≤ 39 mV
⎛ 100 ⎞
(c) υ O = ⎜1 +
⎟(0.2 ) ± 0.039
10 ⎠
⎝
So 2.161 ≤ υ O ≤ 2.239 V
______________________________________________________________________________________
14.52
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ Ri ⎞
⎜
⎟ (15) = 0.010 V
⎝ Ri + R2 ⎠
a.
15
= 0.0006667
15 + R2
15(1 − 0.0006667) = 0.0006667 R2
Then
R2 = 22.48 MΩ
R = R || R = 15 ||10 ⇒ R = 6 kΩ
1
i
F
1
b.
______________________________________________________________________________________
14.53
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
v01
v02
Worst case:
v01 = 0.31 V
b.
is positive and
v02 = −1.51 V
is negative, then
and
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 ⎠
R ≅ 4.15 MΩ
C
or
______________________________________________________________________________________
14.54
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) Offset voltage:
⎛ 50 ⎞
υ O1 = ⎜1 + ⎟(± 2 ) = ±12 mV
⎝ 10 ⎠
υ O 2 = ±12 ± 2 = ±14 mV
⎛ 20 ⎞
⎟(± 12 ) + (2 )(± 2 ) = ±16 mV
⎝ 20 ⎠
Bias current:
υ O1 = 0.21×10 −6 50 ×10 3 = 0.0105 V
υ O3 = ⎜ −
(
)(
)
or υ = (0.19 ×10 )(50 ×10 ) = 0.0095 V
−6
3
O1
υ O 2 = υ O1
υ O 3 = (− 1)(υ O1 ) + (0.21×10 −6 )(20 ×10 3 ) = −υ O1 + 0.0042
(
)(
)
or υ O3 = −υ O1 + 0.19 × 10 −6 20 × 10 3 = −υ O1 + 0.0038
By superposition
−2.5 ≤ υ O1 ≤ 22.5 mV
−4.5 ≤ υ O 2 ≤ 24.5 mV
−22.3 ≤ υ O 3 ≤ 10.7 mV
(b) Bias currents:
υ O1 = ± I OS 50 ×10 3 = ± 0.02 ×10 −6 50 ×10 3 ⇒ υ O1 = ±1 mV
υ O3
(
) (
)(
)
= ± I (20 × 10 ) = ± (0.02 × 10 )(20 × 10 ) ⇒ υ
3
−6
3
OS
By superposition: υ O 3 = ±υ O1 ± 2(2 ) ± 0.4
O3
= ±0.4 mV
−13 ≤ υ O1 ≤ 13 mV
−15 ≤ υ O 2 ≤ 15 mV
−17.4 ≤ υ O 3 ≤ 17.4 mV
______________________________________________________________________________________
14.55
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 ⎠
v0 = 9 mV
So net output voltage is
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
_____________________________________________________________________________________
14.56
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is
a.
v0 = 4 mV
For T = 50°C, the offset voltage for is
b.
V0 S = 2 mV + (0.0067)(25) = 2.1675 mV
so the output voltage for each circuit is
v0 = 4.335 mV
______________________________________________________________________________________
14.57
At T = 25°C,
a.
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
______________________________________________________________________________________
14.58
a.
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
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 ,
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ 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
I0S ,
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
______________________________________________________________________________________
14.59
a.
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
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 ⎠
= (0.1)(4) + (2 × 10−6 )(60 × 103 ) − (2 × 10 −6 )(60 × 103 )4
or
b.
v02 = 0.12 V
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 ⎠
= (0.105)(4) + (2.1× 10−6 )(60 × 103 ) − (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
or
⎛ 60 ⎞
⎛ 60 ⎞
v01 = v02 ⎜1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟
⎝ 20 ⎠
⎝ 20 ⎠
= (0.125)(4) + (2.5 × 10−6 )(60 × 103 ) − (2.5 × 106 )(50 × 103 (4)
v02 = 0.15 V
Due to I 0 S :
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
14.60
50
= 5.0
10
For common-mode, υ I 1 = υ I 2
From Chapter 9,
⎛ R2 ⎞
⎜⎜1 +
⎟
R1 ⎟⎠ R 2
⎝
Acm =
−
⎛ R3 ⎞ R1
⎜⎜1 +
⎟⎟
⎝ R4 ⎠
If R 2 = 50(1.015 ) = 50.75 , R1 = 10(1 − 0.015 ) = 9.85
(a) Ad =
R3 = 10(1 − 0.015) = 9.85 , R 4 = 50(1.015 ) = 50.75
50.75
9.85 − 50.75 = 6.15228 − 5.15228 = 5.046 × 10 − 6
Then Acm =
9.85
9.85 1.19409
1+
50.75
If
R3 = 10(1.015) = 10.15 , R 4 = 50(1 − 0.015) = 49.25
1+
50.75
9.85 − 50.75 = 6.15228 − 5.15228 = −0.051268
Acm =
Then
10.15 9.85 1.20609
1+
49.25
If
R 2 = 49.25 , R1 = 10.15
1+
49.25
10
.15 − 49.25 = 5.85222 − 4.85222 = +0.04877
Acm =
Then
9.85 10.15 1.19409
1+
50.75
5
⎛
⎞
Now CMRR dB (min ) = 20 log 10 ⎜
⎟ = 39.8 dB
⎝ 0.051268 ⎠
(b)
R 2 = 50(1.03) = 51.5 , R1 = 10(0.97 ) = 9.70
R 4 = 50(0.97 ) = 48.5 , R3 = 10(1.03) = 10.3
1+
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
51.5
1+
9.70 − 51.5 = 6.30928 − 5.30928 = −0.10519
Acm =
10.3 9.70 1.21237
1+
48.5
5
⎛
⎞
CMRR dB = 20 log 10 ⎜
⎟ = 33.5 dB
0
.
10519
⎝
⎠
______________________________________________________________________________________
14.61
(a) CMRR dB = 50 dB ⇒ CMRR = 316.2
5
= 0.01581
316.2
From Problem 14.60,
50(1 + x )
1+
10(1 − x ) 50(1 + x )
−
Acm = −0.01581 =
10(1 + x ) 10(1 − x )
1+
50(1 − x )
Acm =
1
≅ 1+ x
1− x
1 + 5(1 + x )(1 + x )
− 5(1 + x )(1 + x )
Then − 0.01581 =
1 + 0.2(1 + x )(1 + x )
x is small, so that
Neglect x 2 ,
1 + 5(1 + 2 x )
− 5(1 + 2 x )
1 + 0.2(1 + 2 x )
(− 0.01581)[1 + 0.2(1 + 2 x )] = 1 + 5(1 + 2 x ) − 5(1 + 2 x )[1 + 0.2(1 + 2 x )]
− 0.01581 ≅
We find (1 + 2 x ) − 0.003162(1 + 2 x ) − 1.01581 = 0 ⇒ (1 + 2 x ) = 1.009456
Then x = 0.004728 ⇒ x = 0.4728%
(b) CMRR dB = 75 dB, ⇒ CMRR = 5623.4
2
Acm =
5
= 0.00088914
5623.4
Then (− 0.00088914)[1 + 0.2(1 + 2 x )] = 1 − (1 + 2 x )
2
(1 + 2 x )2 − 0.000177828(1 + 2 x ) − 1.00088914 = 0 ⇒ (1 + 2 x ) = 1.0005334 ⇒ x = 0.0267%
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 15
15.1
(a) Noninverting amplifier
R
R
8 = 1 + 2 ⇒ 2 = 7 ⇒ R 2 = 210 k Ω , R1 = 30 k Ω
R1
R1
At noninverting terminal
1
1
=
= 5.305 × 10 − 6
RC =
2π f 2π 30 × 10 3
Let input C = 0.001 μ F, then R = 5.305 k Ω
(b) Set R1 = 15 k Ω , R 2 = 300 k Ω for inverting amplifier
(
R1C =
)
1
1
=
= 7.958 × 10 − 6
2π f 2π 20 × 10 3
(
)
Put C in series with R1
7.958 × 10 −6
⇒ C = 530.5 pF
15 × 10 3
______________________________________________________________________________________
C=
15.2
(a) T =
(b) T =
(c) T =
1
1 + (1.5)
= 0.4061 ⇒ −7.83 dB
4
1
1 + (1.5)
= 0.2841 ⇒ −10.93 dB
6
1
1 + (1.5)
= 0.1938 ⇒ −14.25 dB
8
1
(d) T =
= 0.1306 ⇒ −17.68 dB
10
1 + (1.5)
______________________________________________________________________________________
15.3
T = −6 dB, ⇒ T = 0.50
1
0.50 =
=
⎛ 1 ⎞
1+ ⎜
⎟
⎝ 0.9 ⎠
or (1.111)
2N
1
1 + (1.111)
2N
2
2N
⎛ 1 ⎞
=⎜
⎟ −1 = 3
⎝ 0.5 ⎠
For N=6, (1.111)
= 3.54 , ⇒ 6-pole filter
______________________________________________________________________________________
(2 )(6 )
15.4
(a) From Figure 15.8(a)
1
1
RC =
=
= 6.366 × 10 − 6
2π f 3− dB 2π 25 × 10 3
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Let C = 0.001 μ F, then R = 6.366 k Ω
And R3 = (0.707 )R = 4.50 k Ω
R 4 = (1.414 )R = 9.0 k Ω
(b) (i) T =
1
⎛ 25 ⎞
1+ ⎜ ⎟
⎝ 22 ⎠
(ii) T =
1
⎛ 25 ⎞
1+ ⎜ ⎟
⎝ 25 ⎠
(iii) T =
= 0.6123 ⇒ −4.26 dB
4
= 0.707 ⇒ −3 dB
4
1
= 0.7819 ⇒ −2.14 dB
4
⎛ 25 ⎞
1+ ⎜ ⎟
⎝ 28 ⎠
______________________________________________________________________________________
15.5
1
1
=
= 7.958 × 10 − 6
2π f 3− dB 2π 20 × 10 3
Let R = 20 k Ω , then C = 397.9 pF
And C1 = (3.546 )C = 1411 pF
(a) RC =
(
)
C 2 = (1.392 )C = 553.9 pF
C 3 = (0.2024 )C = 80.5 pF
(b) (i) T =
1
⎛ 10 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
(ii) T =
1
⎛ 15 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
(iii) T =
= 0.7071 ⇒ −3.0 dB
6
1
⎛ 25 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
(v) T =
= 0.9214 ⇒ −0.711 dB
6
1
⎛ 20 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
(iv) T =
= 0.9923 ⇒ −0.0673 dB
6
1
= 0.4557 ⇒ −6.83 dB
6
= 0.2841 ⇒ −10.9 dB
6
⎛ 30 ⎞
1+ ⎜ ⎟
⎝ 20 ⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.6
From Equation (15.7).
T (s) =
Y1Y2
Y1Y2 + Y4 (Y1 + Y2 + Y3 )
Y1 = Y2 = sC ,
For a high-pass filter, let
Y3 =
1
,
R3
Y4 =
and
1
R4
Then
T (s) =
s 2C 2
1 ⎛
1 ⎞
s 2 C 2 + ⎜ sC + sC + ⎟
R4 ⎝
R3 ⎠
1
1 ⎛
1 ⎞
1+
⎜2+
⎟
sR4 C ⎝
sR3C ⎠
Define τ 3 = R3 C and τ 4 = R 4 C
=
T (s ) =
Set s = jω
1
1 ⎛
1 ⎞
⎜2 +
⎟
1+
⎜
sτ 4 ⎝
sτ 3 ⎟⎠
T ( jω ) =
1
⎛
1
1 ⎞
⎟
⎜2 +
1+
⎜
jωτ 4 ⎝
jωτ 3 ⎟⎠
1
=
j ⎛
j ⎞
⎟
⎜2 −
1−
⎜
ωτ 4 ⎝ ωτ 3 ⎟⎠
1
=
⎛
1 ⎞⎟ 2 j
⎜1 −
⎜ ω 2τ τ ⎟ − ωτ
4
3 4 ⎠
⎝
2
⎧⎛
1 ⎞⎟
4 ⎫⎪
⎪⎜
T ( jω ) = ⎨⎜1 − 2
+
⎬
⎟
ω 2τ 42 ⎪
⎪⎩⎝ ω τ 3τ 4 ⎠
⎭
−1 / 2
For a maximally flat filter, we want
dT
=0
dω ω →∞
Taking the derivative, we find
d T ( jω )
dω
or
2
1 ⎞⎟
4 ⎫⎪
1 ⎧⎪⎛⎜
= − ⎨⎜1 − 2
+
⎬
2 ⎪⎝ ω τ 3τ 4 ⎟⎠
ω 2τ 42 ⎪
⎭
⎩
−3 / 2
⎡ ⎛
1 ⎞⎟⎛⎜ 2 ⎞⎟ 4(− 2) ⎤
× ⎢2⎜⎜1 − 2
⎟⎜ 3
⎟+ 3 2 ⎥
⎢⎣ ⎝ ω τ 3τ 4 ⎠⎝ ω τ 3τ 4 ⎠ ω τ 4 ⎥⎦
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
d T ( jω )
dω
ω →∞ = 0
⎡⎛ 4 ⎞⎛
1 ⎞⎟
8 ⎤
⎟⎜1 −
= ⎢⎜⎜ 3
− 3 2⎥
2
⎟
⎜
⎟
⎣⎢⎝ ω τ 3τ 4 ⎠⎝ ω τ 3τ 4 ⎠ ω τ 4 ⎦⎥
=
4 ⎡ 1 ⎛⎜
1 ⎞⎟ 2 ⎤
−
− ⎥
1
⎢
3
2
ω ⎣⎢τ 3τ 4 ⎜⎝ ω τ 3τ 4 ⎟⎠ τ 42 ⎦⎥
Then
⎡ 1 ⎛
1 ⎞⎟ 2 ⎤
⎜1 −
=0
⎢
2
⎜
⎟− 2 ⎥
⎢⎣τ 3τ 4 ⎝ ω τ 3τ 4 ⎠ τ 4 ⎥⎦ ω →∞
1
2
=
⇒ 2τ 3 = τ 4
So that
τ3
τ4
Then the transfer function can be written as:
2
⎫
⎧⎡
⎤
1
4
⎪
⎪
T ( jω ) = ⎨⎢1 − 2
+
2 ⎥
2
2 ⎬
ω
2
τ
ω
4
τ
⎥
3 ⎦
3 ⎪
⎪⎩⎣⎢
⎭
( )
−1 / 2
( )
⎧⎪
1
1
1 ⎫⎪
= ⎨1 − 2 2 +
+
⎬
2
ω 2τ 32 ⎪⎭
⎪⎩ ω τ 3 4 ω 2τ 32
(
⎫⎪
⎧⎪
1
= ⎨1 +
2 ⎬
⎪⎩ 4 ω 2τ 32 ⎪⎭
(
−1 / 2
)
−1 / 2
)
3 − dB frequency
2ω 2τ 32 = 1 or ω =
1
2 (τ 3 )
=
1
2 (R 3 C )
Define
ω=
1
RC
So that
R3 =
R
2
We had 2τ 3 = τ 4 or 2( R3C ) = R4 C ⇒ R4 = 2 R3
R = 2⋅R
So that 4
______________________________________________________________________________________
15.7
−14 dB ⇒ T = 0.1995
0.1995 =
1
⇒ (1.2 )
2N
2
2N
⎛ 1 ⎞
=⎜
⎟ − 1 = 24.1
⎝ 0.1995 ⎠
1 + (1.2)
N=9, 9 order filter
______________________________________________________________________________________
th
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.8
−12 dB ⇒ T = 0.2512
⇒ (1.333)
1
0.2512 =
2N
2
2N
⎛ 1 ⎞
=⎜
⎟ − 1 = 14.85
⎝ 0.2512 ⎠
⎛4⎞
1+ ⎜ ⎟
⎝3⎠
th
N=5, 5 order filter
______________________________________________________________________________________
15.9
1
T =
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
At f = 12 kHz,
2N
T = 0.9
1
0.9 =
⎛ f ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
Also
2N
=
⎛ 12 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
1
⎛ 14 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
2N
⎛ 14 ⎞
⎜
⎟
⎝ f3dB ⎠
=
2N
1
− 1 = 0.2346
(0.9) 2
0.01 =
⎛ 14 ⎞
⎜
⎟
⎝ f3dB ⎠
1
=
2N
2N
1
− 1 = 9999
(0.01) 2
2N
2N
9999
⎛ 14 ⎞
=⎜ ⎟ =
= 4.262 × 104
2N
0.2346
⎝ 12 ⎠
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
(1.16667) 2 N = 4.262 × 104
N = 35
Then
1
0.9 =
2N
⎛ 12 ⎞
1+ ⎜
⎟
⎝ f 3dB ⎠
⎛ 12 ⎞
⎜
⎟
⎝ f3dB ⎠
2N
= 0.2346
⎛
⎞
⎜
⎟
⎛ 12 ⎞
⎝ 2N ⎠
= (0.2346)0.014286
⎜
⎟ = (0.2346)
f
⎝ 3dB ⎠
= 0.9795
1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
So
f 3dB = 12.25 kHz
______________________________________________________________________________________
15.10
(a) T =
1
1
=
⎛ 1 ⎞
1+ ⎜
⎟
⎝ 0.8 ⎠
1 + (1.25)
2N
1
For N=3, T =
1 + (1.25)
= 0.4557 ⇒ −6.83 dB
6
1
(b) For N=5, T =
2N
= 0.3114 ⇒ −10.1 dB
1 + (1.25)
10
1
(c) For N=7, T =
= 0.2053 ⇒ −13.8 dB
14
1 + (1.25)
______________________________________________________________________________________
15.11
(a) T =
1
1 + (1.4)
2N
For N=3, T =
(b) For N=5, T =
1
1 + (1.4 )
= 0.3424 ⇒ −9.31 dB
6
1
1 + (1.4 )
= 0.1828 ⇒ −14.8 dB
10
(c) For N=7, T =
1
= 0.0944 ⇒ −20.5 dB
14
1 + (1.4)
______________________________________________________________________________________
15.12
Consider
For low-frequency:
vo
R2 + R3
=
vi R1 + R2 + R3
For high-frequency:
vo
R2
=
vi R1 + R2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
So we need
⎛ R2 ⎞
R2 + R3
= 25 ⎜
⎟
R1 + R2 + R3
⎝ R1 + R2 ⎠
R = 1.5 k Ω ⇒ R1 = 48.5 k Ω
Let R1 + R2 = 50 k Ω and 2
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
vo = 25.
Need to have
⎛ R ⎞
⎛ R ⎞
R
vo = 25 = ⎜1 + 5 ⎟ ⋅ vo1 = ⎜ 1 + 5 ⎟ (0.75)vi ⇒ 5 = 32.3
R
R
R
⎝
⎝
4 ⎠
4 ⎠
4
To check at high-frequency.
R2
1.5
vo1 =
vi =
vi = 0.03vi
R1 + R2
1.5 + 48.5
vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi
which meets the design specification
Consider the frequency response.
vo1
=
vi
R2 + R3
1
sC
R1 + R2 + R3
Now
R3
R3
1
=
sC 1 + sR3C
1
sC
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
=
=
2π ( R2 || R3 ) C 2π (1.5 ||144 ) × 103 C ( 9.33 × 103 ) C
fH ≅
1
1
=
π
2
144
||
50 ) × 103 C
2π ( R3 || ( R1 + R2 ) ) C
(
=
1
( 2.33 ×10 ) C
5
Set
25 kHz =
⎤
fL + fH 1 ⎡
1
1
⎥
= ⎢
+
2
2 ⎢ ( 9.33 × 103 ) C ( 2.33 × 105 ) C ⎥
⎣
⎦
Which yields
C = 2.23 nF
______________________________________________________________________________________
15.13
Av =
1
⎛ f ⎞
1+ ⎜
⎟
⎝ f3− dB ⎠
− 100 dB ⇒ 10−5
2N
So
10−5 =
1
⎛ 770 ⎞
1+ ⎜
⎟
⎝ 12 ⎠
2N
or
2
⎛ 1 ⎞
1 + (64.2) 2 N = ⎜ −5 ⎟ = 1010
⎝ 10 ⎠
or
(64.2) 2 N ≅ 1010
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
N
1
2
3
Left Side
4.112 × 103
1.7 × 107
7 × 1010
So, we need a 3rd order filter.
______________________________________________________________________________________
15.14
Low-pass: −50 dB ⇒ 3.16 × 10
Then
3.16 × 10−3 =
1
⎛ f ⎞
1+ ⎜ ⎟
⎝ fL ⎠
f L = 3.37 Hz
4
=
−3
1
⎛ 60 ⎞
1+ ⎜ ⎟
⎝ fL ⎠
4
We find
High Pass:
3.16 × 10−3 =
We find
1
⎛ f ⎞
1+ ⎜ H ⎟
⎝ f ⎠
f H = 1067 Hz
4
=
1
⎛f ⎞
1+ ⎜ H ⎟
⎝ 60 ⎠
4
Bandwidth: BW = f H − f L = 1067 − 3.37 ⇒
BW ≅ 1064 Hz
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.15
a.
v
vI
= − 02 −
R4
R3
v0
(1)
⎛ 1 ⎞
R1 ⎜
⎟
⎝ sC ⎠
v0
v
= − 01
R2
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
v01
v
= − 02 ⇒ v01 = −v02
R5
R5
Then
(2)
(3)
v0
v
= + 02
⎛ 1 ⎞
R2
⎛ 1 ⎞
v02 = v0 ⎜
⎟
⎜
⎟
⎝ sR2 C ⎠
⎝ sC ⎠ or
(2)
And
v ⎛ 1 ⎞
vI
= − 0 ⋅⎜
⎟−
R4
R3 ⎝ sR2 C ⎠
v0
⎛ 1 ⎞
R1 ⎜
⎟
⎝ sC ⎠
⎡
⎤
⎢
⎥
1
1
⎥
= −v0 ⎢
+
⎢ R3 ( sR2 C ) R1 ⋅ (1/ sC ) ⎥
⎢⎣
R1 + (1/ sC ) ⎥⎦
⎡
1 + sR1C ⎤
1
= −v0 ⎢
+
⎥
R1 ⎦
⎣ R3 ( sR2 C )
⎡ R + (1 + sR1C )( sR2 R3C ) ⎤
= −v0 ⎢ 1
⎥
( sC ) R1 R2 R3
⎣
⎦
Then
⎤
v0
( sC )( R1 R2 R3 )
1 ⎡
=− ⎢
⎥
vI
R4 ⎣ R1 + sR2 R3C + s 2 R1 R2 R3C 2 ⎦
or
1
−
v0
R4
Av ( s ) = =
1
1
vI
+ sC +
R1
sCR2 R3
−
Av ( jω ) =
b.
1
R4
1
1
+ jω C +
R1
jω CR2 R3
(1)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
−
Av ( jω ) =
1
R4
⎡
⎤
1
1
+ j ⎢ω C −
ω CR2 R3 ⎥⎦
R1
⎣
R1
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 ⎥⎦ ⎭⎪
⎩⎪ ⎣
=−
⎡
R1 ⎤
⎢ω R1C −
⎥=0
ω
CR
2 R3 ⎦
when ⎣
Av max
Then
R1 85
=
⇒ Av max = 28.3
3
R4
Av max =
Now
⎡
1
⎣
ω 2 C 2 R2 R3 ⎦
ω R1C ⎢1 −
⎤
⎥=0
ω=
or
1
C R2 R3
Then
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
ω=
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.16
a.
vI − v A
v
= A
R
⎛ 1 ⎞
⎜
⎟
⎝ sC ⎠
vI − vB vB − v0
=
R
R
(1)
(2)
and vA = vB
So
vI
⎛1
⎞
⎛ 1 + sRC ⎞
= v A ⎜ + sC ⎟ = v A ⎜
⎟
R
⎝R
⎠
⎝ R ⎠
(1)
or
vA =
vI
1 + sRC
Then
2vI
(2)
1 + sRC
⎡ 2
⎤
⎡1 − sRC ⎤
v0 = vI ⎢
− 1⎥ = vI ⎢
⎥
⎣1 + sRC ⎦
⎣1 + sRC ⎦
Now
v0
1 − jω RC
= A( jω ) =
1 + jω RC
vI
vI + v0 = 2vB = 2v A =
A=
1 + ω 2 R2C 2
1 + ω 2 R2C 2
⇒ A =1
Phase:
φ = −2 tan −1 (ω RC )
b.
RC = (104 )(15.9 × 10−9 ) = 1.59 × 10 −4
f
0
102
5 × 103
1/ 2π RC = 103 Hz
5 × 103
104
φ
0
−11.4
−53.1
−90°
−157
−169
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.17
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 =
1
2π R2 C
⇒ f 3dB2 =
1
2π ( R1 || R2 )C
b.
Vi
V − V0
+ i
=0
R1 || (1/ sC )
R2
Vi
+
Vi V0
=
R2 R2
⎛ R1 ⎞
⎜
⎟
⎝ 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 ] ⇒ f 3dB =
Vi
R1
Vi ⎝
R1 ⎠
2π ( R1 || R2 )C
______________________________________________________________________________________
15.18
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
⎥⎦
⎝
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
⎡
⎤
⎢
⎥
V0
R2 ⎢
1
⎥
=− ⋅
T (s) =
⎥
Vi
R1 ⎢ 1
⎛ R2 C2 ⎞
+ ⎜ 1 + ⋅ ⎟ + sR2 C2 ⎥
⎢
R1 C1 ⎠
⎣⎢ sR1C1 ⎝
⎦⎥
b.
R2
1
×
2
2 1/ 2
R1 ⎧
1 ⎞ ⎪⎫
⎪⎛ R2 C2 ⎞ ⎛
⎨⎜ 1 + . ⎟ + ⎜ ω R2 C2 −
⎟ ⎬
R1 C1 ⎠ ⎝
ω R1C1 ⎠ ⎪
⎪⎩⎝
⎭
⎛
1 ⎞
⎜ ω R2 C2 −
⎟ = 0,
ω R1C1 ⎠
⎝
T ( jω ) = −
when
we want
R
1
T ( jω ) = 50 = 2 ⋅
R1 ⎛ R2 C2 ⎞
⎜1 + ⋅ ⎟
R1 C1 ⎠
⎝
At the 3 − dB frequencies, we want
For
⎛
⎛ R2 C2 ⎞
1 ⎞
⎜ ω R2 C2 −
⎟ = ± ⎜1 + ⋅ ⎟
ω R1C1 ⎠
R1 C1 ⎠
⎝
⎝
f = 5 kHz,
f = 200 Hz,
+
use
sign and for
ω1 = 2π (200) = 1257
ω 2 = 2π (5 × 103 ) = 3.142 × 104
Define τ 2 = R 2 C 2 and τ 1 = R1C1
Then
50 =
R2
⋅
R1
1
(1)
τ
1+ 2
τ1
⎛
⎛ τ ⎞
1 ⎞
⎜⎜ ω 2τ 2 −
⎟⎟ = +⎜⎜1 + 2 ⎟⎟
ω 2τ 1 ⎠
⎝
⎝ τ1 ⎠
⎛
⎛ τ ⎞
1 ⎞
⎜⎜ ω1τ 2 −
⎟⎟ = −⎜⎜1 + 2 ⎟⎟
ω1τ 1 ⎠
⎝
⎝ τ1 ⎠
From (2)
ω 22τ 1τ 2 − 1 τ 1 + τ 2
=
ω 2τ 1
τ1
or
ω 2τ 1τ 2 −
1
ω2
= τ 1 +τ 2
τ 1 (ω 2τ 2 − 1) =
1
ω2
+τ 2
So
1
+τ 2
ω2
τ1 =
ω 2τ 2 − 1
Substituting into (3), we find
(2)
(3)
use − sign.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎡
⎤
⎢
τ (ω τ − 1) ⎥
1
⎥
= − ⎢1 + 2 2 2
ω1τ 2 −
1
⎢
⎥
⎛ 1
⎞
+τ 2 ⎥
⎜⎜
+ τ 2 ⎟⎟
⎢
ω2
ω2
⎣
⎦
⎠
ω1 ⋅ ⎝
(ω 2τ 2 − 1)
⎡⎛
⎤
⎞
⎤ 1
(ω 2τ 2 − 1) = − ⎢⎜⎜ 1 + τ 2 ⎟⎟ + τ 2 (ω 2τ 2 − 1)⎥
+τ 2 ⎥ −
⎠
⎣ω2
⎦ ω1
⎣⎢⎝ ω 2
⎦⎥
⎡ 1
ω1τ 2 ⎢
ω1
ω
1
1
⋅τ 2 + ω1τ 22 − 2 ⋅τ 2 +
=−
− τ 2 − ω 2τ 22 + τ 2
ω2
ω1
ω1
ω2
⎛ω
(ω1 + ω 2 )τ 22 + ⎜⎜ 1 −
⎝ ω2
⎛ 1
ω2 ⎞
1 ⎞
⎟⎟ = 0
⎟τ 2 + ⎜⎜
+
ω1 ⎟⎠
ω
ω
2 ⎠
⎝ 1
(3.2677 ×10 )τ − 24.96τ + 8.273 ×10 = 0
24.96 ± (24.96) − 4(3.2677 × 10 )(8.273 × 10 )
τ =
2(3.2677 × 10 )
4
−4
2
2
2
2
2
4
−4
4
Since ω 2 is large, τ 2 should be small so use minus sign:
τ 2 = 3.47 × 10 −5 s
Then
τ1 =
3.18 × 10 −5 + 3.47 × 10 −5
⇒ τ 1 = 7.32 × 10 − 4 s
9.09 × 10 − 2
Now
50 =
R2
⋅
R1
1
3.47 × 10−5
1+
7.32 × 10−4
Then
R2
= 52.37
R1
or
R2 = 524 kΩ
Also
τ 1 = R1C1 so that C1 = 0.0732 μ F
τ 2 = R 2 C 2 so that C 2 = 66.3 pF
______________________________________________________________________________________
15.19
Two noninverting amplifiers,
2
⎛ R2 ⎞
⎜⎜1 +
⎟ = 30 dB ⇒= 31.62
R1 ⎟⎠
⎝
R
which gives 2 = 4.62 , then R 2 = 250 k Ω , R1 = 54.1 k Ω
R1
For high-pass filter:
1
RC =
= 7.958 × 10 − 6
2π 20 × 10 3
Set R = 250 k Ω , then C = 31.8 pF
For low-pass filter:
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
= 1.061× 10 −3
2π (150 )
Set R = 250 k Ω , then C = 0.00424 μ F
______________________________________________________________________________________
RC =
15.20
1
⇒ R eq = 40 M Ω
50 × 10 0.5 × 10 −12
(a) R eq =
(
(b) R eq =
1
⇒ R eq = 10 M Ω
50 × 10 2 × 10 −12
3
(
3
)(
)
)(
)
1
⇒ R eq = 2 M Ω
(c) R eq =
3
50 × 10 10 × 10 −12
______________________________________________________________________________________
(
15.21
a.
)(
)
From Equation (15.28),
Q=
and
Now
V1 − V2
⋅ TC
Req
f C = 100 kHz
Req =
so that
TC =
1
⇒ 10 μ s
100 × 103
1
1
=
⇒ 1 MΩ
f C C (100 × 103 )(10 × 10−12 )
So
Q=
(2 − 1)(10 × 10−6 )
= 10 × 10−12 C
106
or
Q = 10 pC
I eq =
b.
c.
Q 10 × 10 −12
=
TC 10 × 10−6
Q = CV
or
I eq = 1 μ A
so find the time that V0 reaches 99% of its full value.
(
V o = V1 1 − e − t / τ
) where τ = RC
Then 0.99 = 1 − e − t / τ or e − t / τ = 0.01
or t = τ ln (100 )
τ = RC = (10 3 )(10 × 10 −12 ) = 10 −8 s
Then
t = 4.61× 10−8 s
______________________________________________________________________________________
15.22
gain = −10 ⇒
Low frequency
f 3dB = 10 × 103 Hz =
Set
f C C2
2π CF
C1
= 10
C2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
f C = 10 f 3dB = 100 kHz
Then
C2 2π (10 × 103 )
=
= 0.628
100 × 103
CF
C1 ,
The largest capacitor is
C1 = 30 pF
so let
Then
C2 = 3 pF
and
CF = 4.78 pF
______________________________________________________________________________________
15.23
a.
Time constant = τ = Req C F where
Req =
1
1
=
= 2 × 106 Ω
3
fC C1 (100 × 10 )(5 × 10−12 )
Then
τ = 2 × 10 6 30 × 10 −12
or
τ = 60 μ s
(
b.
)(
vo = −
or
Δv o =
1
τ
)
∫ v dt
I
(1)(TC )
τ
and TC =
1
fC
So
Δv0 =
1
(60 × 10−6 )(100 × 103 )
or
Δv0 = 0.167 V
c.
or
Now Δv0 = 13 = N (0.167)
N = 78 clock pulses
______________________________________________________________________________________
15.24
fo =
1
(b) C =
1
(a)
=
1
(
)(
2π 3 RC 2π 3 20 × 10 0.001× 10 − 6
f o = 4.59 kHz, R 2 = 8 R = 160 k Ω
2π 3 R f o
=
(
3
1
)(
2π 3 20 × 10 3 25 × 10 3
)
)
C = 184 pF, R 2 = 160 k Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.25
a.
v1 =
⎛ sRCV ⎞
R
⋅ v0 = ⎜
⎟ ⋅ v0
R + (1/ sCV )
⎝ 1 + sRCV ⎠
⎛ sRC ⎞
⋅v =
⋅v
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 ⎠
Set s = jω
R
v2 =
⎞ ⎛ jω RCV ⎞
R2 ⎛
−ω 2 R 2 C 2
⎟
⎜
2 2 2 ⎟⎜
R ⎝ 1 + 2 jω RC − ω R C ⎠ ⎝ 1 + jω RCV ⎠
1= −
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 )
b.
f 0,max =
1
2π (10 ) (10
4
−11
)(10−11 + 2[10−11 ])
f 0,max = 919 kHz
f 0,min =
1
2π (10 ) (10
4
−11
)(10−11 + 2[50 × 10−12 ])
f 0,min = 480 kHz
______________________________________________________________________________________
15.26
fo =
1
(b) R =
1
(a)
=
(
1
)(
2π 6 RC 2π 6 12 × 10 3 150 × 10 −12
f o = 36.1 kHz, R 2 = 29 R = 348 k Ω
2π 6C f o
=
(
1
)(
)
2π 6 0.001× 10 − 6 22 × 10 3
)
R = 2.95 k Ω , R 2 = 29 R = 85.6 k Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.27
v0 − v1 v1 v1 − v2
= +
1
1
R
sC
sC
v
or (v0 − v1 ) sC = 1 + (v1 − v2 ) sC
R
v1 − v2 v2
v2
= +
1
1
R
+R
sC
sC
v
v ( sC )
or (v1 − v2 ) sC = 2 + 2
R 1 + sRC
v0
v2
=−
1
R2
+R
sC
v
v sC
=− 0
or 2
1 + sRC
R2
(1)
(2)
(3)
so
v2 =
−v0
(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
v0
1 ⎤ ⎡ −v0 (1 + sRC ) ⎤ ⎡1 + sRC
1 ⎤
⎡
+
v0 = ⎢ 2 +
⎥×⎢
⎥⎢
⎥ + sR C ⋅ (1 + sRC )
+
sRC
sR
C
sRC
sRC
1
⎣
⎦⎣
⎦
⎦ ⎣
2
2
⎡1 + 2sRC ⎤ ⎡1 + sRC ⎤ ⎡ (1 + sRC ) + sRC ⎤ 1 + sRC
−1 = ⎢
⎥⎢
⎥−
⎥⎢
⎣ sRC ⎦ ⎣ sR2 C ⎦ ⎣ ( sRC )(1 + sRC ) ⎦ sR2 C
2
−1 =
(1 + 2sRC )(1 + 2 sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2
−
( sRC ) 2 ( sR2 C )
( sRC ) 2 ( sR2 C )
Set s = jω , then − 1 =
(1 + 2 jωRC )(1 + 3 jωRC + ω 2 R 2 C 2 ) − (1 + jωRC )(− ω 2 R 2 C 2 )
(− ω 2 R 2 C 2 )( jωR2 C )
(− ω 2 R 2 C 2 )( jωR2 C )
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:
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
−1 =
1=
2 jω RC + 3 jω RC − 2 jω 3 R3 C 3 + jω 3 R 3 C 3
(−ω 3 R 2 C 2 )( jω R2 C )
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−
15.28
Let R F 1 = R F 2 = R F 3 ≡ R F
⎛
υ o1 = ⎜⎜1 +
R F ⎞⎛ 1 ⎞
⎟⎜
⎟ ⋅υ o
R ⎟⎠⎝ 1 + sRC ⎠
⎝
⎛ R ⎞⎛ 1 ⎞
υ o 2 = ⎜⎜1 + F ⎟⎟⎜
⎟ ⋅υ o1
R ⎠⎝ 1 + sRC ⎠
⎝
υ o3 − υ o 2 υ o3 υ o3
+
+
=0
R
1 sC
R
⎛2
⎞ υ
+ sC ⎟ = o 2
R
⎝R
⎠
υ o3 ⎜
⎛
1
⎞
⎟ ⋅υ o 2
⎝ 2 + sRC ⎠
υ o3 = ⎜
RF
⋅υ o 3
R
R ⎛
1
⎞⎛ R F ⎞⎛ 1 ⎞⎛ R F ⎞⎛ 1 ⎞
υo = − F ⎜
⎟⎜
⎟⎜
⎟ ⋅υ o
⎟⎜1 +
⎟⎜1 +
R ⎝ 2 + sRC ⎠⎜⎝
R ⎟⎠⎝ 1 + sRC ⎠⎜⎝
R ⎟⎠⎝ 1 + sRC ⎠
υo = −
2
R ⎛ R ⎞ ⎛
1
⎞⎛ 1 ⎞⎛ 1 ⎞
1 = − F ⎜⎜1 + F ⎟⎟ ⎜
⎟⎜
⎟⎜
⎟
R ⎝
R ⎠ ⎝ 2 + sRC ⎠⎝ 1 + sRC ⎠⎝ 1 + sRC ⎠
Let s = jω
2
1= −
RF ⎛ RF ⎞ ⎛
⎞
⎞⎛
⎞⎛
1
1
1
⎟
⎟⎜
⎟⎜
⎜1 +
⎟ ⎜
R ⎜⎝
R ⎟⎠ ⎜⎝ 2 + jωRC ⎟⎠⎜⎝ 1 + jωRC ⎟⎠⎜⎝ 1 + jωRC ⎟⎠
=−
⎞
RF ⎛ RF ⎞ ⎛
⎞⎛
1
1
⎟
⎟⎟⎜⎜
⎟⎟ ⎜⎜
⎜⎜1 +
2 2
2 ⎟
R ⎝
R ⎠ ⎝ 2 + jωRC ⎠⎝ 1 + 2 jωRC − ω R C ⎠
=−
⎞
RF ⎛ RF ⎞ ⎛
1
⎟
⎟ ⎜
⎜1 +
R ⎜⎝
R ⎟⎠ ⎜⎝ 2 + 4 jωRC − 2ω 2 R 2 C 2 + jωRC − 2ω 2 R 2 C 2 − jω 3 R 3 C 3 ⎟⎠
2
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎤
RF ⎛ RF ⎞ ⎡
1
1+
⎢
⎥
R ⎜⎝
R ⎟⎠ ⎣ 2 − 4ω 2 R 2 C 2 + 5 jω RC − jω 3 R 3C 3 ⎦
2
1= −
Imaginary Term must be zero
5 jω 0 RC − jω 03 R 3 C 3 = 0
5 − jω 02 R 2 C 2 = 0
ω0 =
5
RC
Then
⎡
⎤
⎢
⎥
R ⎛ R ⎞
1
⎥
1 = − F ⎜1 + F ⎟ ⎢
2 2
R ⎝
R ⎠ ⎢
4R C − 5 ⎥
⎢⎣ 2 − R 2 C 2 ⎥⎦
2
2
1= −
RF ⎛ RF ⎞ ⎡ 1 ⎤ 1 RF ⎛ RF ⎞
= ⋅
1+
1+
R ⎜⎝
R ⎟⎠ ⎢⎣ 2 − 20 ⎥⎦ 18 R ⎜⎝
R ⎟⎠
18 =
RF ⎛ RF ⎞
RF
=2
⎜1 +
⎟ ⇒
R ⎝
R ⎠
R
2
2
______________________________________________________________________________________
15.29
(a) 1st stage:
1
jωC
⎞⎛ R F ⎞
⎛
1
⎛ R ⎞
⎟⎜⎜1 +
⋅ ⎜⎜1 + F ⎟⎟ ⋅υ O = ⎜⎜
⎟ ⋅υ O
⎟
1
R ⎠
R ⎟⎠
⎝ 1 + jωRV C ⎠⎝
+ RV ⎝
j ωC
nd
2 stage:
⎞⎛ R F ⎞
⎛
1
⎟⎟⎜⎜1 +
υ O 2 = ⎜⎜
⎟ ⋅υ O1
R ⎟⎠
⎝ 1 + jωRC ⎠⎝
υ O1 =
At node of C 3 :
υ A −υ O2
υ O2
υ
+ υ A ( jωC ) + A = 0 ⇒ υ A =
R
R
2 + jωRC
3rd stage:
υ O2
R
R
υ O = − F ⋅υ A = − F ⋅
R
R (2 + jωRC )
Now
⎞⎛ R F ⎞
R
⎞⎛ R F ⎞⎛
⎛
1
1
1
⎟⎜1 +
⎟⎟⎜⎜1 +
⎜⎜
υO = − F ⋅
⎟ ⋅υ O
⎟⎟⎜⎜
R (2 + jωRC ) ⎝ 1 + jωRC ⎠⎝
R ⎠⎝ 1 + jωRV C ⎟⎠⎜⎝
R ⎟⎠
2
RF ⎛ RF ⎞
1
1
1
⋅
⋅
⎟ ⋅
⎜1 +
R ⎜⎝
R ⎟⎠ (2 + jωRC ) (1 + jωRC ) (1 + jωRV C )
From the denominator, we have:
(2 + jωRC )(1 + jωRC )(1 + jωRV C ) = 2 + 3 jωRC − ω 2 R 2 C 2 (1 + jωRV C )
1= −
(
)
[(2 − ω R C )+ 3 jωRC ](1 + jωR C )
= (2 − ω R C ) + 3 jωRC + jωR C (2 − ω R C ) − 3ω RR C
2
2
2
V
2
2
2
2
V
For oscillation, imaginary part must be zero.
2
2
2
V
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
3ω o RC + ω o RV C 2 − ω o2 R 2 C 2 = 0
3RC + 2 RV C − ω o2 RV R 2 C 3 = 0
So that ω o =
1
RC
3R + 2 RV
RV
2
(b) 1 = −
RF ⎛ RF ⎞
1
⎟ ⋅
⎜1 +
R ⎜⎝
R ⎟⎠ 2 − ω o2 R 2 C 2 − 3ω o2 RRV C 2
(
)
Consider the term:
(
)
2 − ω o2 R 2 C 2 + 3RRV C 2 = 2 −
1 ⎛ 3R + 2 RV ⎞ 2 2
⎟ R C + 3RRV C 2
⎜
2 ⎜
⎟
R
R C ⎝
V
⎠
(
2
⎛ 3R + 2 RV ⎞⎛ 3RV ⎞ 2 RRV − (3R + 2 RV )(R + 3RV )
⎟⎜⎜1 +
⎟=
= 2 − ⎜⎜
⎟
RV
R ⎟⎠
RRV
⎠⎝
⎝
(
)
(
2 RRV − 3R 2 + 9 RRV + 2 RRV + 6 RV2
3R 2 + 9 RRV + 6 RV2
=−
RRV
RRV
Then
=
)
)
2
1=
RRV
RF ⎛ RF ⎞
⎜⎜1 +
⎟⎟ ⋅
2
R ⎝
R ⎠ 3R + 9 RRV + 6 RV2
(
)
For R = RV
2
1=
RF ⎛ RF ⎞
R2
⎟⎟ ⋅
⎜⎜1 +
R ⎝
R ⎠ 18 R 2
2
R ⎛ R ⎞
R
Or 18 = F ⎜⎜1 + F ⎟⎟ ⇒ F = 2
R ⎝
R ⎠
R
(c) For RV = 15 k Ω ,
1
2π 25 × 10 3 0.001× 10 − 6
For RV = 30 k Ω ,
fo =
(
)(
)
3(25) + 2(15)
⇒ f o = 16.84 kHz
15
1
3(25) + 2(30 )
⇒ f o = 13.5 kHz
−6
30
2π 25 × 10 0.001× 10
So 13.5 ≤ f o ≤ 16.84 kHz
______________________________________________________________________________________
fo =
15.30
(a)
υ O1 − υ O
(
3
)(
+ υ O1 ( jωC ) +
)
υ O1 − υ O 2
R
R
υ
υ
2
⎛
⎞
υ O1 ⎜ + jωC ⎟ = O + O 2
R
⎝R
⎠ R
=0
Then (1) υ O1 (2 + jωRC ) = υ O + υ O 2
υ O 2 − υ O1
υ − υ O3
+ υ O 2 ( jωC ) + O 2
=0
R
R
So (2) υ O 2 (2 + jωRC ) = υ O1 + υ O 3
υ O3 − υ O 2
υ
+ υ O 3 ( j ωC ) + O 3 = 0
R
R
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
And (3) υ O 3 (2 + jωRC ) = υ O 2
Also (4) υ O = −
RF
⋅υ O 3
R
υO
+υ
(2 + jωRC ) O3
υO
2
2
From (2) υ O 3 (2 + jωRC ) = υ O1 + υ O 3 ⇒ υ O 3 (2 + jωRC ) = 2υ O 3 +
(2 + jωRC )
υ
O
υ O 3 (4 + 4 jωRC − ω 2 R 2 C 2 − 2 ) =
(2 + jωRC )
υO
υ O3 =
(2 + jωRC )(2 − ω 2 R 2 C 2 + 4 jωRC )
υO
R
Using (4) υ O = − F ⋅
R (2 + jωRC )(2 − ω 2 R 2 C 2 + 4 jωRC )
From (1) υ O1 (2 + jωRC ) = υ O + υ O 3 (2 + jωRC ) ⇒ υ O1 =
RF
1
⋅
2 2
2
R (2) 2 − ω R C + 8 jωRC + jωRC 2 − ω 2 R 2 C 2 − 4ω 2 R 2 C 2
For oscillation, denominator must be real, so
10
8ω o RC + ω o RC 2 − ω o2 R 2 C 2 = 0 ⇒ ω o =
RC
R
1
(b) 1 = − F ⋅
R (2) 2 − ω o2 R 2 C 2 − 4ω o2 R 2 C 2
1= −
[ (
)
(
[ (
1= −
(
)
]
)
]
)
R
RF
R
1
1
1
=− F ⋅
⋅
=− F ⋅
(
−
56 )
R
R 4 − 2ω o2 R 2 C 2 − 4ω o2 R 2 C 2
R ⎡
⎤
⎛ 10 ⎞ 2 2
⎢4 − 6⎜ 2 2 ⎟ R C ⎥
⎝R C ⎠
⎦
⎣
[
]
RF
= 56
R
10
10
(c) C =
=
⇒ C = 0.00114 μ F
2π f o R 2π 22 × 10 3 20 × 10 3
Or
(
)(
)
R F = (56 )R ⇒ R F = 1.12 M Ω
______________________________________________________________________________________
15.31
(a)
υ O1 − υ O
RV
⎛ 1
υ O1 ⎜⎜
+ υ O1 ( jωC ) +
+
υ O1 − υ O 2
R
=0
⎞ υ
υ
1
+ jωC ⎟⎟ = O + O 2
R
R
⎠ RV
⎝ RV
⎛ 1
⎞ υ
υ
+ jωC ⎟ = O + O 2
υ O1 ⎜
⎜R R
⎟ RV
R
V
⎝
⎠
⎛υ
υ ⎞
Then (1) υ O1 (1 + jω (R RV )C ) = (R RV )⎜⎜ O + O 2 ⎟⎟
R ⎠
⎝ RV
From Problem 15.30,
(2) υ O 2 (2 + jωRC ) = υ O1 + υ O 3
(3) υ O 3 (2 + jωRC ) = υ O 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
RF
⋅υ O 3
R
2
2
From (2), υ O 3 (2 + jωRC ) = υ O1 + υ O 3 ⇒ υ O1 = υ O 3 (2 + jωRC ) − 1
(4) υ O = −
[
[
]
]
⎤
⎡υ
υ
2
Then (1) υ O 3 (2 + jωRC ) − 1 (1 + jω (R RV )C ) = (R RV )⎢ O + O 3 (2 + jωRC )⎥
R
⎦
⎣ RV
⎫
R RV
υ
⎪⎧
(2 + jωRC )⎪⎬ = (R RV )⋅ O
υ O 3 ⎨ (2 + jωRC )2 − 1 (1 + jω (R RV )C ) −
R
RV
⎪⎩
⎪⎭
R (R RV )
1
Then,
υO = − F ⋅
⋅υ O ⋅
R
RV
⎧⎪
⎫
R RV
2
(2 + jωRC )⎪⎬
⎨ (2 + jωRC ) − 1 (1 + jω (R RV )C ) −
R
⎪⎭
⎩⎪
[
]
[
]
Consider the denominator:
[4 + 4 jωRC − ω R C − 1](1 + jω (R R )C ) − R RR (2 + jωRC )
2
2
V
2
V
(3 − ω R C + 4 jωRC )(1 + jω (R R )C ) − R RR (2 + jωRC )
2
2
V
2
V
For oscillation, the denominator must be real, so
R RV
4 jω o RC + jω o (R RV )C 3 − ω o2 R 2 C 2 −
⋅ jω o RC = 0
R
4 R + (R RV ) 3 − ω o2 R 2 C 2 − R RV = 0
[
[
]
]
4 R + 2(R RV ) = ω (R RV )R 2 C 2
2
o
So that ω o =
⎛ R ⎞
⎟+2
4⎜
⎜R R ⎟
V ⎠
⎝
1
RC
(b) For RV = 15 k Ω ,
fo =
1
2π 25 × 10 0.001× 10 − 6
(
3
)(
)
4(25)
+ 2 ⇒ f o = 22.66 kHz
25 15
For RV = 30 k Ω ,
fo =
1
2π 25 × 10 3 0.001× 10 − 6
(
)(
4(25)
) 25 30
+ 2 ⇒ f o = 19.45 kHz
So 19.45 ≤ f o ≤ 22.66 kHz
______________________________________________________________________________________
15.32
a.
We can write
⎛ Zp ⎞
⎛ R1 ⎞
vB = ⎜
v
vA = ⎜
⎟ v0
⎜ Z + Z ⎟⎟ 0
s ⎠
⎝ R1 + R2 ⎠ and
⎝ p
Z p = RB
where
Z s = RA +
and
RB
1
=
sCB 1 + sRB CB
1 + sRAC A
1
=
sC A
sC A
Setting vA = vB , we have
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
R1
=
R1 + R2
RB
1 + sRB CB
1 + sRAC A
RB
+
1 + sRB CB
sC A
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 RA C A )(1 + jω RB CB )
or
jω RB C A + 1 + jω RA C 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
1
f0 =
2π RA RB C ACB
b.
(2)
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 RA C A + jω RB CB )
or
1+
R2
R
C
= 1+ A + B
R1
RB C A
Then
R2 RA CB
=
+
R1 RB C A
______________________________________________________________________________________
15.33
a.
We can write
⎛ R1 ⎞
vA = ⎜
⎟ v0
⎝ R1 + R2 ⎠
and
⎛
⎞
R || sL
vB = ⎜
⎟ v0
||
+
+
R
sL
R
sL
⎝
⎠
Setting vA = vB , we have
sRL
⎡
⎤
⎢
⎥
R1
R + sL
=⎢
⎥ ⋅ v0
sRL
R1 + R2 ⎢
+ R + sL ⎥
⎢⎣ R + sL
⎥⎦
R1
sRL
=
R1 + R2 sRL + ( R + sL) 2
(1)
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:
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 =
⋅
L
2π
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
1+
R2
=3
R1
so that
R2
=2
R1
______________________________________________________________________________________
15.34
1
2πRC
1
1
RC =
=
= 4.547 × 10 − 6
2π f o 2π 35 × 10 3
Let C = 0.001 μ F, then R = 4.55 k Ω
fo =
(
)
R2
=2
R1
______________________________________________________________________________________
Set
15.35
g m = 2 K n I DQ = 2 (0.7 )(0.8) = 1.497 mA/V
C2
= g m R = (1.497 )(2) = 2.993
C1
C 2 ≅ 3(0.02 ) = 0.06 μ F
2π f o =
1
⎛ CC ⎞
L⎜⎜ 1 2 ⎟⎟
⎝ C1 + C 2 ⎠
⎞
⎡ (0.02 )(0.06 )× 10 − 6 ⎤ ⎛
1
⎟ ⇒ L = 13.8 μ H
Then L ⎢
⎥ = ⎜⎜
3 ⎟
0.02 + 0.06
⎣
⎦ ⎝ 2π 350 × 10 ⎠
______________________________________________________________________________________
2
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.36
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
⎠
V ( sC1 + g m )
V1 = 0
⎛ 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 ) +
R
sL
⎝
⎠
⎣
⎦
L
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.37
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
(
L1 = ω 2 L1 L2C + gm RL1 ω 2 L2C
Now ω 2 =
)
1
( L1 + L2 )
1=
1
[ L2 C + g m RL2 C ]
C ( L1 + L2 )
1=
L2
L
(1 + g m R ) ⇒ 1 = (1 + g m R)
L1 + L2
L2
or
L1
= gm R
L2
______________________________________________________________________________________
15.38
(a) (1) g mVπ +
VC
V
+ C + (VC − Vπ )( jωC ) = 0
R
jωL1
(2) Vπ = (VC − Vπ )( jωC )( jωL2 )
(
)
(
Vπ 1 − ω 2 CL 2 = −VC ω 2 CL 2
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Vπ =
(
VC ω 2 L 2 C
)
ω L2 C − 1
2
⎞
⎛1
1
Then (1) Vπ (g m − jωC ) + VC ⎜⎜ +
+ jωC ⎟⎟ = 0
⎠
⎝ R jωL1
(ω L C )(g − jωC ) + V ⋅ ⎛⎜ 1 − j + jωC ⎞⎟ = 0
2
VC ⋅
2
m
ω 2 L2 C − 1
g m ω 2 L2 C
+
C
⎜R
⎝
(
ωL1
⎟
⎠
)
⎡
ωC ω 2 L 2 C ⎤
1
1
+ j ⎢ωC −
− 2
⎥=0
ωL1 ω L 2 C − 1 ⎦
R
⎣
ω L2 C − 1
Set the imaginary part equal to zero.
ω 2 L1C − 1 ωC ω 2 L 2 C
= 2
ωL1
ω L2 C − 1
2
(
)
(ω L C − 1)(ω L C − 1) = (ω L C )(ω L C )
2
2
2
1
2
2
1
2
ω L1 L 2 C − ω L 2 C − ω L1C + 1 = ω L1 L 2 C 2
4
2
2
2
4
ω o2 (L1 + L 2 )C = 1
1
So ω o =
(L1 + L2 )C
(b) Set the real part equal to zero,
g m ω o2 L 2 C 1
+ =0
ω o2 L2 C − 1 R
g m Rω o2 L2 C = 1 − ω o2 L 2 C
ω o2 L 2 C (g m R + 1) = 1
Then
L 2 C (g m R + 1)
(L + L2 )C
= 1 ⇒ gm R +1 = 1
(L1 + L2 )C
L2 C
We find g m R =
(c) C =
1
L1
L2
(2π f o ) (L1 + L2 )
2
=
1
[2π (750 ×10 )] (100 ×10 )
3
2
−6
⇒ C = 450 pF
L1
1
1
=1⇒ R =
=
⇒ R = 33.3 Ω
L2
g m 30
______________________________________________________________________________________
gmR =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.39
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
⎤
T ( s ) = ⎜1 + 2 ⎟ ⎢
= ⎜1 + ⎟ ⎢
⎥
2 2 2
⎥
R
sRC
sRC
R
+
+
−
(1
)(2
)
1
sRC
s
R
C
+
+
−
2
3
1
⎣
⎦
⎦ ⎝
⎝
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
R
−
R
C
+
j
RC
ω
ω
1
3
⎦
⎝
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
______________________________________________________________________________________
15.40
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vb − vo vb vb − va
+
+
=0
1
1
R
sC
sC
vb − vo
+ 2vb ⋅ sC − va ⋅ sC = 0
R
(1)
Va − Vb Va
+
=0
1
R
sC
1⎞
⎛
⎛ 1 + sRC ⎞
Va ⎜ sC + ⎟ = vb ⋅ sC ⇒ vb = va ⎜
⎟
R⎠
⎝
⎝ sRC ⎠
(2)
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
− sC ⎥ = o
va ⎢
( sRC ) ⋅ R
⎣
⎦ R
⎡ (1 + sRC )(1 + 2 sRC )
⎤
− sRC ⎥ = vo
va ⎢
sRC
⎣
⎦
vo (1 + sRC )(1 + 2 sRC ) − s 2 R 2 C 2
=
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 ⎟ ⋅
2
R
⎝
1 ⎠ 1 + 3sRC + ( sRC )
⎛ R ⎞⎡
⎤
jω RC
T ( jω ) = ⎜ 1 + 2 ⎟ ⎢
⎥
2 2 2
R1 ⎠ ⎣1 − ω R C + 3 jω RC ⎦
⎝
So 1 − ω 0 R C = 0
1
fO =
2π RC
So
2
2
2
⎛ R ⎞⎛ 1 ⎞
1 = ⎜1 + 2 ⎟ ⎜ ⎟
R1 ⎠ ⎝ 3 ⎠
⎝
Also
R2
=2
R1
So
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.41
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 ⎭
(1)
Then
vB =
v0
1
⋅
sL ⎧ ( R + sL)( R + 2 sL) − ( sL) 2 ⎫
⋅⎬
⎨
( sL)( sRL)
⎩
⎭
Now
⎛ R ⎞⎛
sRL
⎞
T ( s ) = ⎜1 + 2 ⎟ ⎜ 2
2 2
2 2 ⎟
R1 ⎠ ⎝ R + 3sRL + 2 s L − s L ⎠
⎝
or
⎛ R ⎞⎛
sRL
⎞
T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2
2 ⎟
R1 ⎠ ⎝ s L + 3sRL + R ⎠
⎝
And
⎛ R ⎞⎛
⎞
jω RL
T ( jω ) = ⎜1 + 2 ⎟ ⎜ 2
⎟
2 2
R1 ⎠ ⎝ R − ω L + 3 jω RL ⎠
⎝
R
f0 =
2π L
Frequency of oscillation:
Condition for oscillation:
⎛ R ⎞⎛ 1 ⎞
1 = ⎜1 + 2 ⎟ ⎜ ⎟
R1 ⎠ ⎝ 3 ⎠
⎝
or
R2
=2
R1
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.42
(
)
R2 f + R 2v
R2
⋅ V REF = −
⋅ V REF
R1
R1
υI = −
Set R 2 f = 20 k Ω
For υ I = −2 V, set R 2 v = 0
−2 = −
Then
20
(5) ⇒ R1 = 50 k Ω
R1
(20 + R2v )
(5) ⇒ R 2v = 20 k Ω
50
______________________________________________________________________________________
−4 = −
15.43
For υ O = 10 V, υ I = −5 V
i = 0.2 mA =
10 − (− 5)
⇒ R1 + R 2 = 75 k Ω
R1 + R 2
(VTH − VTL ) =
R1
(V H − V L )
R2
0.4 =
R1
[10 − (− 10)] ⇒ R2 = 50 R1
R2
R1 + R 2 = R1 + 50 R1 = 75 ⇒ R1 = 1.47 k Ω , R 2 = 73.53 k Ω
______________________________________________________________________________________
15.44
⎛ R1 ⎞
⎟⎟(V H − V L )
(a) VTH − VTL = ⎜⎜
⎝ R1 + R 2 ⎠
⎛ R1 ⎞
⎛ R1 ⎞
⎟⎟
⎟⎟[9 − (− 9)] = 18⎜⎜
0.2 = ⎜⎜
R
R
+
2 ⎠
⎝ R1 + R 2 ⎠
⎝ 1
Set R1 = 2 k Ω
Then 2 + R 2 =
(18)(2) ⇒ R = 178 k Ω
2
0.2
9
9
(b) i =
=
⇒ i = 50 μ A
R1 + R 2 2 + 178
______________________________________________________________________________________
15.45
⎛ R1 ⎞
⎛ 2 ⎞
⎟⎟ ⋅ V H = ⎜
(a) VTH = ⎜⎜
⎟(10 ) = 0.4 V
R
R
+
⎝ 2 + 48 ⎠
2 ⎠
⎝ 1
⎛ R1 ⎞
⎛ 2 ⎞
⎟⎟ ⋅ V L = ⎜
VTL = ⎜⎜
⎟(− 10 ) = −0.4 V
⎝ 2 + 48 ⎠
⎝ R1 + R 2 ⎠
(b) For 33.33 ≤ t ≤ 41.67 ms, sin[2π (60 )t ] ⇒ positive half cycle
At υ I = 0 , υ O = +10 V
At υ I = +0.4 = 10 sin [2π (60 )t1 ] ⇒ t1 = +0.106 + 33.333 = 33.439 ms
So, for 33.333 ≤ t ≤ 33.439 ms, υ O = +10 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
33.439 ≤ t ≤ 41.77 ms, υ O = −10 V
41.77 ≤ t ≤ 50 ms, υ O = +10 V
______________________________________________________________________________________
15.46
a.
Upper crossover voltage when v0 = +VP ,
Now
⎛ R1 ⎞
vB = ⎜
⎟ (+VP )
⎝ R1 + R2 ⎠
and
⎛ RA ⎞
⎛ RB ⎞
vA = ⎜
⎟ VREF + ⎜
⎟ VTH
⎝ RA + RB ⎠
⎝ RA + RB ⎠
v A = vB so that
⎛ R1 ⎞
⎛ RA ⎞
⎛ RB ⎞
⎜
⎟ VP = ⎜
⎟ VREF + ⎜
⎟ VTH
+
+
R
R
R
R
2 ⎠
B ⎠
⎝ 1
⎝ A
⎝ RA + RB ⎠
or
⎛ R + RB ⎞ ⎛ R1 ⎞
⎛ RA ⎞
VTH = ⎜ A
⎟⎜
⎟ VP − ⎜
⎟ VREF
+
R
R
R
⎝ 1
⎝ RB ⎠
2 ⎠⎝ B ⎠
Lower crossover voltage when v0 = −VP
So
⎛ R + RB ⎞ ⎛ R1 ⎞
⎛ RA ⎞
VTL = − ⎜ A
⎟⎜
⎟ VP − ⎜
⎟ VREF
R
+
R
R
2 ⎠⎝ B ⎠
⎝ 1
⎝ 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 ⎠
______________________________________________________________________________________
15.47
a.
vB VREF − vB v0 − vB
=
+
R1
R3
R2
⎛ 1
v
1
1 ⎞ V
vB ⎜ +
+ ⎟ = REF + 0
R3
R2
⎝ R1 R2 R3 ⎠
VTH = vB
So
when v0 = +VP and VTL = vB when v0 = −VP
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VREF VP
+
R3
R2
VTH =
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
⎝ R1 R2 R3 ⎠
and
VTL =
VREF VP
−
R3
R2
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
R
R
R
2
3 ⎠
⎝ 1
b.
VREF
VS =
⎛ 1
1
1 ⎞
+ ⎟
R3 ⎜ +
⎝ R1 R2 R3 ⎠
−10
−5 =
⎛ 1
1
1⎞
+ ⎟
10 ⎜ +
⎝ R1 R2 10 ⎠
1
1 1 1
+
= − = 0.10
R1 R2 5 10
ΔVT = VTH − VTL =
0.2 =
2VP
R2
⎛ 1
1
1 ⎞
+ ⎟
⎜ +
⎝ R1 R2 R3 ⎠
2(12)
R2 (0.10 + 0.10)
R = 600 kΩ
So 2
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
______________________________________________________________________________________
15.48
a.
If the saturated output voltage is
where
then the circuit behaves as a comparator
v0 < 6.2 V.
If the saturated output voltage is
no control.
b.
VP < 6.2 V,
VP > 6.2 V,
the output will flip to either +VP or −VP and the input has
Same as part (a) except the curve at vI ≈ 0 will have a finite slope.
Circuit works as a comparator as long as v01 < 8.7 V and v02 > −3.7 V. Otherwise the input has no
control.
______________________________________________________________________________________
c.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.49
⎛ R2 ⎞
⎟⎟ ⋅ V REF
(a) When υ O = 0 , υ + = V S = ⎜⎜
⎝ R1 + R 2 ⎠
When υ O = V H , υ + = VTH
V REF − VTH VTH − V H
=
R1
R2
⎛ 1
⎛ R + R2 ⎞
V REF V H
1 ⎞
⎟⎟ = VTH ⎜⎜ 1
⎟⎟
+
= VTH ⎜⎜
+
R1
R2
R
R
2 ⎠
⎝ 1
⎝ R1 R 2 ⎠
⎛ R2 ⎞
⎛ R1 ⎞
⎛ R1 ⎞
⎟⎟ + V H ⎜⎜
⎟⎟ = V S + V H ⎜⎜
⎟⎟
VTH = V REF ⎜⎜
⎝ R1 + R 2 ⎠
⎝ R1 + R 2 ⎠
⎝ R1 + R 2 ⎠
⎛ R1 ⎞
⎟⎟
VTL = V S + V L ⎜⎜
⎝ R1 + R 2 ⎠
(b) V S = −1.75 V, R1 = 4 k Ω
⎛ 4 ⎞
⎟⎟ ⇒ R 2 = 188 k Ω
VTH = −1.5 = −1.75 + (12 )⎜⎜
⎝ 4 + R2 ⎠
⎛ 188 ⎞
− 1.75 = ⎜
⎟ ⋅ V REF ⇒ V REF = −1.787 V
⎝ 188 + 4 ⎠
12 − (− 1.787 ) 13.787
(c) (i) For υ O = 12 V, i =
=
⇒ i = 71.8 μ A
R1 + R 2
4 + 188
12 − 1.787
⇒ i = 53.2 μ A
192
______________________________________________________________________________________
(ii) For υ O = −12 V, i =
15.50
a.
Switching point when v0 = 0.
Now
⎛ R2 ⎞
v+ = VREF = ⎜
⎟ vI
⎝ R1 + R2 ⎠ where vI = VS .
Then
⎛ R + R2 ⎞
⎛
R1 ⎞
VS = ⎜ 1
⎟ VREF = ⎜ 1 + ⎟ VREF
R
R
2
2 ⎠
⎝
⎠
⎝
v1
Now upper crossover voltage for
VTH − VREF VREF − VL
=
R1
R2
or VTH = −
occurs when v0 = VL and v+ = VREF . Then
⎛
R1
R ⎞
⋅ VL + VREF ⎜ 1 + 1 ⎟
R2
R
⎝
2 ⎠
or VTH = VS −
R1
⋅ VL
R2
Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VH − VREF VREF − VTL
=
R2
R1
or VTL = −
⎛
R1
R ⎞
⋅ VH + VREF ⎜ 1 + 1 ⎟
R2
R
⎝
2 ⎠
or VTL = VS −
b.
R1
⋅ VH
R2
VTH = −1
For
and
R1 = 0.833 kΩ
so that
Now
⎛
R ⎞
VS = ⎜ 1 + 1 ⎟ VREF
R
⎝
2 ⎠
VTL = −2, VS = −1.5 V.
VTL = VS −
Then
R1
R
⋅ VH ⇒ −2 = −1.5 − 1 (12)
R2
20
⎛ 0.833 ⎞
−1.5 = ⎜ 1 +
⎟ VREF
20 ⎠
⎝
which gives
VREF = −1.44 V
______________________________________________________________________________________
15.51
⎛ R1 ⎞
25 ⎞
⎟ ⋅ Vγ = ⎛⎜
(a) VTH = ⎜⎜
⎟(0.7 ) = 0.175 V
⎟
⎝ 25 + 75 ⎠
⎝ R1 + R3 ⎠
⎛ R1 ⎞
⎟(− Vγ ) = −0.175 V
VTL = ⎜⎜
⎟
⎝ R1 + R3 ⎠
(c) (i) υ I = 2 V, υ O = −0.7 V
I D1 = 0
⎛ − 0.7 − (− 10) ⎞
I R 2 = −⎜
⎟ = −0.465 mA
20
⎝
⎠
⎛ − 0.175 − (− 0.7 ) ⎞
I R 3 = +⎜
⎟ ⇒ I R3 = 7 μ A
75
⎝
⎠
I R 3 + I R 2 + I D 2 = 0 ⇒ I D 2 = − I R 3 − I R 2 = −0.007 − (− 0.465) = 0.458 mA
(ii) υ I = −2 V, υ O = +0.7 V
I D2 = 0
⎛ 10 − 0.7 ⎞
I R2 = ⎜
⎟ = 0.465 mA
⎝ 20 ⎠
⎛ 0.175 − (0.7 ) ⎞
I R3 = ⎜
⎟ ⇒ I R 3 = −7 μ A
75
⎝
⎠
I D1 = I R 2 + I R 3 = 0.465 − 0.007 = 0.458 mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.52
(a) Vγ + V Z = 0.7 + 5.6 = 6.3 V
VTH 0 − (− 6.3)
R
=
⇒ VTH = 1 (6.3)
R1
R2
R2
VTL = −
VTH − VTL = 0.6 =
Then
R1
(6.3)
R2
R1
[6.3 − (− 6.3)] = (12.6) R1
R2
R2
R2
= 21 , Set R1 = 4 k Ω , then R 2 = 84 k Ω
R1
(b) Maximum current in R 2 ,
iR2 =
6.3 6.3
=
= 0.075 mA
R2
84
10 − 6.3
= 0.8 + 0.075 = 0.875 mA
R
10 − 6.3
R=
= 4.23 k Ω
0.875
______________________________________________________________________________________
iR =
15.53
a.
v0 = VREF + 2Vγ
5 = VREF + 2(0.7)
or
VREF = 3.6 V
b.
⎛ R1 ⎞
VTH = ⎜
⎟ (VREF + 2Vγ )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
0.5 = ⎜
⎟ (5)
⎝ R1 + R2 ⎠
or 1 +
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.
For
v1 − (v1 + 0.7) VREF − v1 v1 − v0
+
=
100
1
1
v1 = −0.7,
then
10 − 0 3.6 − (−0.7) −0.7 − v0
+
=
100
1
1
or
v0 = −5.1 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.54
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
⎝ R1 + R2 ⎠ and v0 = −(VREF + 2Vγ )
so
⎛ R1 ⎞
VTL = − ⎜
⎟ (VREF + 2Vγ )
⎝ R1 + R2 ⎠
______________________________________________________________________________________
15.55
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
R1
⋅ (VREF + 2Vγ )
R2
By symmetry
R
VTL = − 1 ⋅ (VREF + 2vγ )
R2
______________________________________________________________________________________
or VTH =
15.56
⎛ R1 ⎞
⎛ 10 ⎞
⎟⎟ ⋅υ O = ⎜
(a) υ + = ⎜⎜
⎟(5) = 1.667 V
⎝ 10 + 20 ⎠
⎝ R1 + R 2 ⎠
⎛
⎛ t ⎞
t ⎞
⎟⎟ = 5 − 6.667 exp⎜⎜ −
⎟⎟ , for 0 < t < t1
⎝ τX ⎠
⎝ τX ⎠
υ X = 5 + (− 1.667 − 5) exp⎜⎜ −
⎛ ⎛ t − t1 ⎞ ⎞
⎛ (t − t1 ) ⎞
⎟⎟ ⎟ = −5 + 6.667 exp⎜⎜ −
⎟⎟ , for t1 < t < T
⎟
τ
τ
X
X
⎝
⎝
⎠
⎠
⎠
⎝
υ X = −5 + (1.667 − (− 5)) exp⎜⎜ − ⎜⎜
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ t ⎞
(b) 1.667 = 5 − 6.667 exp⎜⎜ − 1 ⎟⎟
⎝ τX ⎠
⎛ t ⎞
6.667 exp⎜⎜ − 1 ⎟⎟ = 3.333
⎝ τX ⎠
⎛ t ⎞
exp⎜⎜ + 1 ⎟⎟ = 2 ⇒ t1 = τ X ln (2)
⎝ τX ⎠
By symmetry, T − t1 = τ X ln (2 )
Then T = 2τ X ln (2 )
(
)(
)
f =
1
1
=
, τ X = R X C X = 40 × 10 3 0.02 × 10 −6 = 8 × 10 −4 s
T 2τ X ln (2)
f =
2(8 × 10 ) ln (2 )
1
−4
= 902 Hz
Duty cycle = 50%
______________________________________________________________________________________
15.57
⎛ R1 ⎞
⎛ 10 ⎞
⎟⎟(5) = ⎜
(a) υ + (+ ) = ⎜⎜
⎟(5) = 1.667 V
⎝ 10 + 20 ⎠
⎝ R1 + R 2 ⎠
⎛
R1 ⎞
⎛ 10 ⎞
⎟⎟(− 10 ) = ⎜
⎟(− 10 ) = −3.333 V
+
R
R
⎝ 10 + 20 ⎠
2 ⎠
⎝ 1
υ + (− ) = ⎜⎜
⎛
υ X = 5 + (− 3.333 − 5) exp⎜⎜ −
⎛ t ⎞
t ⎞
⎟ = 5 − 8.333 exp⎜⎜ −
⎟⎟ ,
⎝ τX ⎠
⎟
⎝ τX ⎠
⎛
υ X = −10 + (1.667 − (− 10 )) exp⎜⎜ −
⎝
for 0 < t < t1
(t − t1 ) ⎞⎟
τX
⎛ (t − t1 ) ⎞
⎟⎟ , for t1 < t < T
⎟ = −10 + 11.667 exp⎜⎜ − τ
X
⎠
⎝
⎠
⎛ t ⎞
(b) 1.667 = 5 − 8.333 exp⎜⎜ − 1 ⎟⎟
⎝ τX ⎠
⎛ −t ⎞
⎛ t ⎞
8.333 exp⎜⎜ 1 ⎟⎟ = 3.333 ⇒ exp⎜⎜ + 1 ⎟⎟ = 2.5
⎝τX ⎠
⎝ τX ⎠
t1 = τ X ln (2.5)
⎛ (T − t1 ) ⎞
⎟
Also − 3.333 = −10 + 11.667 exp⎜⎜ −
τ X ⎟⎠
⎝
⎛ (T − t1 ) ⎞
⎛ (T − t1 ) ⎞
⎟⎟ = 6.667 ⇒ exp⎜⎜ +
⎟ = 1.75
11.667 exp⎜⎜ −
τ
τ X ⎟⎠
X
⎝
⎠
⎝
T − t1 = τ X ln (1.75)
Now T = τ X [ln(2.5) + ln (1.75)] = τ X ln[(2.5)(1.75)]
1
, τ X = R X C X = 8 × 10 −4 s
T
1
f =
= 847 Hz
8 × 10 − 4 ln[(2.5)(1.75)]
______________________________________________________________________________________
f =
(
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.58
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 =
RX = 56 K
For example, Let C X = 680 pF
1
Within 2 of 1% of design specification.
______________________________________________________________________________________
15.59
( )(
)
t1 = 1.1R X C X = (1.1) 10 4 0.1× 10 −6 ⇒ t1 = 1.1 ms
(
0 < t < t1 , υ Y = 10 1 − e
−t / τ Y
)
τ Y = RY C Y = (2 × 10 )(0.02 × 10 −6 )
3
= 4 × 10 −5 s
Now
t1
τY
= 2.75
⇒ CY
completely charges during each cycle.
______________________________________________________________________________________
15.60
a.
Switching voltage
⎛ R1 + R3 ⎞
⎛ 10 + 10 ⎞
vX = ⎜
⎟ ⋅ VP = ⎜
⎟ (±10)
R
+
R
+
R
⎝ 10 + 10 + 10 ⎠
3
2 ⎠
⎝ 1
So vX = ±6.667 V
Using Equation (15.83(a))
2
⎛ 2
⎞
v X = V P + ⎜ − V P − V P ⎟e −t1 / τ X = V P
3
⎝ 3
⎠
5
2
Then 1 − ⋅ e −t1 / τ X =
3
3
1 5 − t1 / τ X
= ⋅e
or t1 = τ X ln (5)
3 3
T
1
1
t1 = =
=
⇒ t1 = 0.001 s
2 2 f 2(500 )
So
b.
(
10 −3 = τ X ln (5) ⇒ τ X = 6.21× 10 −4 = R X 0.01× 10 −6
RX = 62.1 kΩ
Switching voltage
⎛
⎞
R1
vX = ⎜
⎟ (±VP )
⎝ R1 + R3 + R2 ⎠
10
1
⎛
⎞
=⎜
⎟ (±VP ) = ⋅ (±VP )
10
+
10
+
10
3
⎝
⎠
Using Equation (15.83(a))
)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
⎞
⎛ 1
v X = V P + ⎜ − V P − V P ⎟e −t1 / τ X = V P
3
⎝ 3
⎠
4
1
Then 1 − e −t1 / τ X =
3
3
2 4 −t1 / τ X
= ⋅e
3 3
(
)
t1 = τ X ln (2 ) = 6.21× 10 −4 ln (2 ) = 4.30 × 10 −4 s
−4
T = 2t1 = 8.6 × 10 s
1
⇒ f = 1.16 kHz
T
______________________________________________________________________________________
f =
15.61
(a) β =
R1
20
=
= 0.625
R1 + R 2 20 + 12
0.7 ⎤
⎡
1+
⎢
10 ⎥ = τ (1.0485)
T = 250 × 10 − 6 = τ X ln ⎢
⎥
X
⎢1 − 0.625 ⎥
⎣⎢
⎦⎥
So τ X = 2.384 × 10 −4 = R X C X
Set C X = 0.01 μ F, then R X = 23.84 k Ω
(b) For t < 0 , υ X = 0.7 V
So υ Y must be < υ X ⇒ υ I < 0
⎛ (T ′ − T ) ⎞
⎟
(c) υ X = Vγ = 10 + [− β (10) − 10] exp⎜⎜ −
τ X ⎟⎠
⎝
⎛ (T ′ − T ) ⎞
⎛ (T ′ − T ) ⎞
⎟ = 9.3
⎟⎟ ⇒ 16.26 exp⎜⎜ −
0.7 = 10 − 16.25 exp⎜⎜ −
τX ⎠
τ X ⎟⎠
⎝
⎝
⎛ (T ′ − T ) ⎞
⎟ = 1.747
exp⎜⎜ +
τ X ⎟⎠
⎝
T ′ − T = τ X ln (1.747 ) = 2.384 × 10 −4 ln (1.747 )
T ′ − T = 133 μ s
______________________________________________________________________________________
15.62
⎡ Vγ ⎤
⎢1 +
⎥
VP ⎥
⎢
(a) T = τ X ln
, Vγ = 0.7 V, V P = 5 V, β = 0.5
⎢ 1− β ⎥
⎢
⎥
⎣⎢
⎦⎥
τ X = R X C X = (20 × 10 3 )(1.2 × 10 −6 ) = 2.4 × 10 −2 s
(
T = 2.4 × 10
−2
)
⎡ 0.7 ⎤
⎢1 + 5 ⎥
ln ⎢
⎥ ⇒ T = 19.78 ms
⎢ 1 − 0.5 ⎥
⎣⎢
⎦⎥
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) T ′ − T ≅ 0.4τ X = 9.6 ms
______________________________________________________________________________________
15.63
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.
+
If V = 5 V, then
IB ≅
5 − 0.7
= 0.043 mA
100
If β = 100, I C = 4.3 mA
1
Ic
VC = ∫ I C dt = ⋅ t
C
C
2 +
⋅ V = 3.33 V
Capacitor has charged to 3
t=
So that
VC ⋅ C (3.33)(50 × 10−6 )
=
4.3 × 10−3
IC ⋅
So recovery time t ≈ 38.7 ms
______________________________________________________________________________________
15.64
T = 1.1 RC
5 × 10−6 = 1.1 RC
−6
so RC = 4.545 ×10 s
For example, let
C = 100 pF
and R = 45.5 kΩ
From Problem (15.63), recovery time
t≅
VC ⋅ C (3.33)(100 × 10−12 )
=
IC
4.3 × 10−3
or
t = 77.4 ns
______________________________________________________________________________________
15.65
Duty cycle = 60% =
0.60 =
R A + RB
× 100%
R A + 2R B
25 + R B
⇒ R B = 50 k Ω
25 + 2 R B
f = 80 × 10 3 =
1
(0.693)(25 + 2(50))×10 3 C
⇒ C = 144.3 pF
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.66
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
f max =
= 4.81 kHz
(0.693)(10 + 2(10)) × 103 × (0.01× 10−6 )
f min =
So 627 Hz ≤ f ≤ 4.81 kHz
R + RB
Duty cycle = A
× 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%
______________________________________________________________________________________
15.67
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
f max =
= 2.72 kHz
(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%
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.68
a.
V + − 3VEB
R1 A + R1B
VEB = 0.7
IE3 = IE 4 =
Assume
IE3 = IE 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
b.
⎛I ⎞
⎛ 0.398 × 10−3 ⎞
VBB = 2VD = 2VT ln ⎜ D ⎟ = 2(0.026) ln ⎜
⎟
−13
⎝ 10
⎠
⎝ IS ⎠
I C1 = I C 2 =
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 ⎟
⎝ IS ⎠
⎝ IS ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
P = (0.398 + 0.398 + 4.16)(22) ⇒ P = 109 mW
c.
______________________________________________________________________________________
15.69
a.
b.
From Figure 15.44, 3.7 W to the load
V + ≈ 19 V
P=
c.
or
1 VP2
2 RL
VP = 2 RL P = 2(10)(3.7) ⇒ VP = 8.6 V
______________________________________________________________________________________
15.70
⎛ R ⎞
R
R
(a) ⎜⎜1 + 2 ⎟⎟ = 12 ⇒ 2 = 11 , and 4 = 12
R1 ⎠
R1
R3
⎝
2
V
(b) PL = L ⇒ V L = 2 R L PL = 2(12)(15) = 18.97 V
2R L
υ O1 max = υ O 2 max = 9.49 V
IL =
V L 18.97
=
= 1.58 A
RL
12
⎛ I ⎞⎛ V ⎞ ⎛ 0.8 ⎞⎛ 24 ⎞
⎟⎜
⎟ = 9.6 W
(c) (i) PL = ⎜⎜ L ⎟⎟⎜⎜ L ⎟⎟ = ⎜⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
V
24
= 30 Ω
(ii) R L = L =
I L 0.8
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.71
a.
Then
v01 = iR2 + vI where
i=
vI
R1
⎛ 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 ⎠ ⎦
v
R R
Av = L = 1 + 2 + 3
vI
R1 R1
Av = 10 ⇒
b.
Want
R2 R3
+
=9
R1 R1
⎛ R2 ⎞ R3
⎜1 + ⎟ =
R1 ⎠ R1
Also want ⎝
R2
=4
R1
so
R
=
200
kΩ
R1 = 50 kΩ, 2
Then
For
and
R2 ⎛ R2 ⎞
+ ⎜1 + ⎟ = 9
R1 ⎝
R1 ⎠
R3
=5
R1
c.
or
R = 250 kΩ
so 3
1 VP2
P=
2 RL
VP = 2 RL P = 2(20)(10) = 20 V
So peak values of output voltages are
v01 = v02 = 10 V
20
=1 A
20
______________________________________________________________________________________
Peak load current =
15.72
⎛ R ⎞
(a) υ O1 = ⎜⎜1 + 2 ⎟⎟ ⋅υ I
R1 ⎠
⎝
υO2 = −
R4
R ⎛ R ⎞
⋅υ O1 = − 4 ⎜⎜1 + 2 ⎟⎟ ⋅υ I
R3
R3 ⎝
R1 ⎠
υ L = υ O1 − υ O 2
Aυ =
υ L ⎛ R 2 ⎞⎛ R 4 ⎞
⎟
⎟⎜1 +
= ⎜1 +
υ I ⎜⎝ R1 ⎟⎠⎜⎝ R3 ⎟⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ R ⎞
R
(b) ⎜⎜1 + 2 ⎟⎟ = 12 ⇒ 2 = 11 , Set R 2 = 120 k Ω , then R1 = 10.9 k Ω
R1 ⎠
R1
⎝
R4
= 1 , Set R3 = R 4 = 120 k Ω
R3
(c) (i) PL =
V L2
(16)2 = 5.12 W
=
2 R L 2(25)
V L 16
=
= 0.64 A
R L 25
______________________________________________________________________________________
(ii) I L =
15.73
(a) From Problem 15.72
R
R
υ O 2 = − 4 ⋅υ O1 , Set 4 = 1
R3
R3
⎛ R ⎞⎛ R ⎞
Aυ = ⎜⎜1 + 2 ⎟⎟⎜⎜1 + 4 ⎟⎟
R1 ⎠⎝ R3 ⎠
⎝
⎛ R ⎞
R
25 = ⎜⎜1 + 2 ⎟⎟(2 ) ⇒ 2 = 11.5
R1 ⎠
R1
⎝
Set R 2 = R3 = R 4 = 100 k Ω , then R1 = 8.69 k Ω
⎛ V ⎞⎛ I ⎞ ⎛ 24 ⎞⎛ 1.2 ⎞
⎟ = 14.4 W
⎟⎜
(b) (i) PL = ⎜⎜ L ⎟⎟⎜⎜ L ⎟⎟ = ⎜⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
V
24
= 20 Ω
(ii) R L = L =
I L 1.2
V L2
(24)2 = 7.2 W
=
2 R L′
2(40 )
______________________________________________________________________________________
(c) PL =
15.74
Line regulation =
ΔV0
ΔV +
Now
ΔI =
ΔV +
R1
and ΔVZ = rZ ⋅ ΔI and ΔV0 = 10ΔVZ
So
ΔV0 = 10 ⋅ rZ ⋅
ΔV +
R1
So
ΔV0 10(15)
=
⇒ 1.61%
ΔV +
9300
______________________________________________________________________________________
Line regulation =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.75
(a) R of =
ΔVO
ΔI O
=
8
⇒ R of = 4 m Ω
2
(b) ΔVO = Rof ⋅ ΔI O = (10 )(1.2 ) ⇒ ΔVO = 12 mV
______________________________________________________________________________________
15.76
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
______________________________________________________________________________________
15.77
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 ⎠
IC 3 = IC 5 =
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
⎛I ⎞
VEB 4 = VT ln ⎜ Z 2 ⎟
⎝ I5 ⎠
⎛I ⎞
VEB 5 = VT ln ⎜ C 5 ⎟
⎝ 2I S ⎠
⎛ I ⎞
I Z 2 R4 = VT ln ⎜ C 5 ⎟
⎝ 2I Z 2 ⎠
Then
R4 =
⎛ 0.677 ⎞
0.026
⋅ ln ⎜
⎟
0.25
⎝ 2(0.25) ⎠
or
R4 = 31.5 Ω
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
15.78
regulation =
Line
Now
ΔV0
ΔV +
ΔVB 7 = ΔI C 3 ⋅ R1
⎛ R13 ⎞
⎜
⎟ (ΔV0 ) = ΔVB 7 = ΔI C 3 R1
R + R13 ⎠
and ⎝ 12
ΔI C 3 =
ΔVZ
ΔI Z ⋅ rZ
=
R1 + R2 + R3 R1 + R2 + R3
ΔI Z =
ΔV +
r0
and
and
Then
r0 =
where
VA
IZ
⎛ 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 +
______________________________________________________________________________________
15.79
(a) R1 + rz =
V + 25
=
= 2.0833 k Ω
IZ
12
R1 = 2.0833 − 0.012 = 2.0713 k Ω
(b) V + = 5.6 + (0.012 )(12 ) = 5.744 V
For x = 0 ,
⎛ R3 + R 4 ⎞
1+ 2 +1⎞
⎟ ⋅ VO ⇒ VO = ⎛⎜
5.744 = ⎜⎜
⎟(5.744) = 7.659 V
⎟
⎝ 2 +1 ⎠
⎝ R 2 + R3 + R 4 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For x = 1 ,
⎛
⎞
R4
1+ 2 +1⎞
⎟ ⋅ VO ⇒ VO = ⎛⎜
5.744 = ⎜⎜
⎟(5.744) = 22.976 V
⎟
R
+
R
+
R
⎝ 1 ⎠
3
4 ⎠
⎝ 2
So 7.659 ≤ VO ≤ 22.976 V
(c) For x = 1 , V1 =
1
VO
4
1
4
υ d = V REF − V1 = V REF − VO
⎛I ⎞
VO = AOLυ d − V BE = AOLυ d − VT ln⎜⎜ O ⎟⎟
⎝ IS ⎠
⎛I ⎞
1
⎛
⎞
VO = AOL ⎜V REF − VO ⎟ − VT ln⎜⎜ O ⎟⎟
4 ⎠
⎝
⎝ IS ⎠
⎛I ⎞
⎛ 1
⎞
VO ⎜1 + AOL ⎟ = AOLV REF − VT ln⎜⎜ O ⎟⎟
⎝ 4
⎠
⎝ IS ⎠
⎛I ⎞
AOLV REF − VT ln⎜⎜ O ⎟⎟
⎝ IS ⎠
VO =
1
1 + AOL
4
V (NL ) − VO (FL )
Load regulation = O
VO (NL )
AOLV REF − VT ln[I O (NL ) I S ] ⎡ AOLV REF − VT ln[I O (FL ) I S ]⎤
−⎢
⎥
1 + (1 4 )AOL
1 + (1 4 )AOL
⎣
⎦
=
AOLV REF − VT ln[I O (NL ) I S ]
1 + (1 4)AOL
⎡ ⎛ I (FL ) ⎞
⎛ I (NL ) ⎞⎤
⎛ I (FL ) ⎞
⎟ − ln⎜ O
⎟⎥
VT ⎢ln⎜⎜ O
⎟
VT ln⎜⎜ O
⎟
⎜ I
⎟
⎟
I
S
S
⎠
⎝
⎠⎦⎥
⎣⎢ ⎝
⎝ I O (NL ) ⎠
=
=
⎛ I (NL ) ⎞
⎛ I (NL ) ⎞
⎟
⎟
AOLV REF − VT ln⎜⎜ O
AOLV REF − VT ln⎜⎜ O
⎟
⎟
⎝ IS ⎠
⎝ IS ⎠
Let I O (FL ) = 5 A, I O (NL ) = 1% = 0.05 A
⎛ I (NL ) ⎞
⎟ ≅ 0.7 V
Let VT ln⎜⎜ O
⎟
⎝ IS ⎠
(0.026) ln⎛⎜ 5 ⎞⎟
Load regulation =
⎝ 0.05 ⎠
(5 ×10 )(5.744) − 0.7
3
=
0.11973
2.8719 × 10 4
Load regulation = 4.169 × 10 −4 %
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.80
(a) I E =
V Z 6.8
=
= 1.36 mA
R2
5
⎛ β ⎞
⎛ 80 ⎞
⎟⎟ ⋅ I E = ⎜ ⎟(1.36 ) = 1.343 mA
I O = ⎜⎜
β
1
+
⎝ 81 ⎠
⎝
⎠
For V BC = 0 , VO = 20 − V Z − V EB = 20 − 6.8 − 0.6 = 12.6 V
V
12.6
R L (max ) = O =
= 9.38 k Ω
I O 1.343
So 0 ≤ R L ≤ 9.38 k Ω
⎛ 10 4 ⎞ +
⎛ R1 ⎞ +
⎟
⎟⎟ V − V ZO = ⎜ 4
(b) V + = ⎜⎜
⎜ 10 + 20 ⎟ V − 6.8
⎝ R1 + rZ ⎠
⎝
⎠
(
)
(
)
⎛ 10 4 ⎞
⎟(13.2) = 13.17365 V
For V + = 20 V, V + = ⎜⎜ 4
⎟
⎝ 10 + 20 ⎠
⎛ 80 ⎞⎛ 20 − 13.17365 ⎞
I O = ⎜ ⎟⎜
⎟ = 1.3484 mA
5
⎝ 81 ⎠⎝
⎠
⎛ 10 4 ⎞
⎟(16 − 6.8) = 9.181637 V
For V + = 16 V, V + = ⎜⎜ 4
⎟
⎝ 10 + 20 ⎠
⎛ 80 ⎞⎛ 16 − 9.181637 ⎞
I O = ⎜ ⎟⎜
⎟ = 1.3468 mA
5
⎝ 81 ⎠⎝
⎠
So 1.3468 ≤ I O ≤ 1.3484 mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 16
16.1
⎛ k ′ ⎞⎛ W ⎞
(a) υ O = V DD − ⎜⎜ n ⎟⎟⎜ ⎟ R D 2(υ I − VTN )υ O − υ O2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞⎛ W ⎞
2
0.1 = 3.3 − ⎜
⎟⎜ ⎟(40) 2(3.3 − 0.5)(0.1) − (0.1)
2
L
⎝
⎠⎝ ⎠
[
]
[
]
⎛W ⎞
⎛W ⎞
0.1 = 3.3 − ⎜ ⎟(1.1) ⇒ ⎜ ⎟ = 2.91
⎝L⎠
⎝L⎠
⎛ k ′ ⎞⎛ W ⎞
(b) ⎜⎜ n ⎟⎟⎜ ⎟ R DVO2t + VO t − V DD = 0
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
⎜
⎟(2.91)(40 )VO t + VO t − 3.3 = 0
⎝ 2 ⎠
5.82VO2t + VO t − 3.3 = 0 ⇒ VO t = 0.672 V
VO t = V I t − VTN ⇒ V I t = 1.172 V
3.3 − 0.1
⇒ i D , max = 80 μ A
40
PD , max = (80 )(3.3) = 264 μ W
(c) i D , max =
______________________________________________________________________________________
16.2
(a) (i) K n R DVO2t + VO t − V DD = 0
(0.05)(100)VO2t + VO t − 3.3 = 0 ⇒ VO t = 0.7185 V
[
⇒ V I t = 1.219 V
(ii) υ O = 3.3 − (0.05)(100 ) 2(3.3 − 0.5)υ O − υ O2
]
We find 5υ − 29υ O + 3.3 = 0 ⇒ υ O = 0.116 V
2
O
(b) (i) (0.05)(30)VO2t + VO t − 3.3 = 0 ⇒ VO t = 1.187 V
[
⇒ V I t = 1.687 V
(ii) υ O = 3.3 − (0.05)(30 ) 2(3.3 − 0.5)υ O − υ O2
]
Or 1.5υ − 9.4υ O + 3.3 = 0 ⇒ υ O = 0.373 V
2
O
(c) (i) (0.05)(5)VO2t + VO t − 3.3 = 0 ⇒ VO t = 2.147 V
[
⇒ V I t = 2.647 V
(ii) υ O = 3.3 − (0.05)(5) 2(3.3 − 0.5)υ O − υ O2
]
Or 0.25υ − 2.4υ O + 3.3 = 0 ⇒ υ O = 1.663 V
______________________________________________________________________________________
2
O
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.3
(a) P = IV
0.25 = I (3.3) ⇒ I = 75.76 μ A
3.3 − 0.15
= 41.6 k Ω
0.07576
⎛ k ′ ⎞⎛ W ⎞
2
I = ⎜⎜ n ⎟⎟⎜ ⎟ 2(VGS − VTN )V DS − V DS
⎝ 2 ⎠⎝ L ⎠
R=
[
]
[
]
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
2
75.76 = ⎜
⎟⎜ ⎟ 2(3.3 − 0.5)(0.15) − (0.15) ⇒ ⎜ ⎟ = 1.85
2
L
⎝L⎠
⎠⎝ ⎠
⎝
V − V DS (sat )
⎛ k ′ ⎞⎛ W ⎞
2
(b) I D = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN ) = DD
R
⎝ 2 ⎠⎝ L ⎠
3.3 − V DS (sat )
⎛ 0.1 ⎞
2
⎜
⎟(1.85)V DS (sat ) =
41.6
⎝ 2 ⎠
2
(sat ) + V DS (sat ) − 3.3 = 0 ⇒ V DS (sat ) = 0.805 V
Or 3.848V DS
V DS (sat ) = VGS − VTN
0.805 = VGS − 0.5 ⇒ VGS = 1.305 V
Then 0.5 ≤ VGS ≤ 1.305 V
______________________________________________________________________________________
16.4
(a) From Equation (16.21)
(V DD − υ O − VTNL )2
K D (W L ) D
=
=
(W L )L 2(υ I − VTND )υ O − υ O2
KL
=
(W L )D
= 11.34
(W L )L
(1.8 − 0.08 − 0.4)2
1.7424
=
2
0.1536
2(1.4 − 0.4)(0.08) − (0.08)
PD , max = i D , max ⋅ V DD
0.3 = i D , max (1.8) ⇒ i D , max = 0.1667 mA
[
⎛ 0.1 ⎞⎛ W ⎞
2
i D , max = 0.1667 = ⎜
⎟⎜ ⎟ 2(1.4 − 0.4 )(0.08) − (0.08)
⎝ 2 ⎠⎝ L ⎠ D
]
⎛W ⎞
⎛W ⎞
Which yields ⎜ ⎟ = 21.7 and ⎜ ⎟ = 1.91
L
⎝ ⎠D
⎝ L ⎠L
(b) V I t =
(
1.8 − 0.4 + (0.4 ) 1 + 11.34
) = 0.7206 V
1 + 11.34
0.4 ≤ V I ≤ 0.7206 V
______________________________________________________________________________________
16.5
(a)
[
]
KD
2
2
2(3 − 0.5)(0.25) − (0.25) = [3 − 0.25 − 0.5]
KL
KD
K
(1.1875) = (5.0625) ⇒ D = 4.26
KL
KL
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
[
]
KD
2
2
2(2.5 − 0.5)(0.25) − (0.25) = [3 − 0.25 − 0.5]
KL
KD
K
(0.9375) = (5.0625) ⇒ D = 5.4
KL
KL
⎛ 0.1 ⎞
2
2
(c) i D = K L (VGSL − VTN ) = ⎜
⎟(1)(3 − 0.25 − 0.5) = 0.253 mA
⎝ 2 ⎠
P = i D ⋅ V DD = (0.253)(3) = 0.759 mW
______________________________________________________________________________________
16.6
(V DD − υ O − VTNL )
KD
=
K L 2(υ I − VTND )υ O − υ O2
2
(a)
υ I = Logic 1 = V DD − VTNL = 3 − 0.5 = 2.5 V
K D (W L ) D
(3 − 0.1 − 0.5)2
5.76
=
=
=
(W L )L 2(2.5 − 0.5)(0.1) − (0.1)2 0.39
KL
(W L )D
= 14.77
(W L )L
PD , max = i D , max ⋅ V DD
0.4 = i D , max (3) ⇒ i D , max = 0.1333 mA
[
⎛ 0.1 ⎞⎛ W ⎞
2
i D , max = 0.1333 = ⎜
⎟⎜ ⎟ 2(2.5 − 0.5)(0.1) − (0.1)
⎝ 2 ⎠⎝ L ⎠ D
]
⎛W ⎞
⎛W ⎞
Which yields ⎜ ⎟ = 6.84 and ⎜ ⎟ = 0.463
L
⎝ ⎠D
⎝ L ⎠L
(b) V I t =
(
3 − 0.5 + (0.5) 1 + 14.77
1 + 14.77
VO t = 1.016 − 0.5 = 0.516 V
) = 1.016 V
______________________________________________________________________________________
16.7
We have
KD
2
⎡ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = (VDD − vO − VTNL )
KL ⎣
(W / L )D ⎡
2
2
2 (V − V − V )( 0.08VDD ) − ( 0.08VDD ) ⎤ = (VDD − 0.08VDD − VTN )
⎦
(W / L )L ⎣ DD TN TN
(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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.8
VOH = VB − VTN = Logic 1
So
VB = 4 V ⇒ VOH = 3 V
(a)
VB = 5 V ⇒ VOH = 4 V
(b)
VB = 6 V ⇒ VOH = 5 V
(c)
VB = 7 V ⇒ VOH = 5 V ,since VDS = 0
(d)
For vI = VOH
K D ⎡⎣ 2 ( vI − VT ) vO − vO2 ⎤⎦ = K L [VB − vO − VT ]
2
Then
(a)
(1) ⎡⎣ 2 ( 3 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [ 4 − VOL − 1] ⇒ VOL = 0.657 V
(b)
(1) ⎡⎣ 2 ( 4 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [5 − VOL − 1] ⇒ VOL = 0.791 V
2
2
(1) ⎡⎣ 2 ( 5 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [6 − VOL − 1] ⇒ VOL = 0.935 V
2
(c)
(d)
Load in non-sat region
iDD = iOL
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
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
______________________________________________________________________________________
16.9
(a)
(
)
KD
V I t − VTND = −VTNL
KL
(
)
500
V I t − 0.5 = −(− 0.8) ⇒ V I t = 0.8578 V
100
For Driver: VO t = V I t − VTND = 0.8578 − 0.5 = 0.3578 V
For Load:
(b)
[
VO t = V DD + VTNL = 3.3 + (− 0.8) = 2.5 V
]
KD
2
2(υ I − VTND )υ O − υ O2 = (− VTNL )
KL
[
]
⎛ 500 ⎞
2
2
⎟ 2(3.3 − 0.5)υ O − υ O = [− (− 0.8)]
⎜
⎝ 100 ⎠
We find 5υ O2 − 28υ O + 0.64 = 0 ⇒ υ O = 0.0230 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) i D , max = K L (− VTNL ) = 100[− (− 0.8)] = 64 μ A
2
2
PD , max = i D , max ⋅ V DD = (64)(3.3) = 211 μ W
______________________________________________________________________________________
16.10
2
2
⎛ 500 ⎞ ⎡
⎜
⎟ 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤⎦ = ( −VTNL )
⎝ 50 ⎠ ⎣
So
( −VTNL ) = 4.9 ⇒ VTNL = −2.21 V
2
______________________________________________________________________________________
16.11
(a) PD , max = i D , max ⋅ V DD
80 = i D , max (1.8) ⇒ i D , max = 44.44 μ A
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
2
i D , max = 44.44 = ⎜
⎟⎜ ⎟ [− (− 0.6)] ⇒ ⎜ ⎟ = 2.47
⎝ L ⎠L
⎝ 2 ⎠⎝ L ⎠ L
KD
2
2
2(1.8 − 0.3)(0.06 ) − (0.06 ) = [− (− 0.6 )]
KL
(W L )D
KD
K
(0.1764) = (0.36) ⇒ D = 2.04 =
(W L )L
KL
KL
[
]
⎛W ⎞
Then ⎜ ⎟ = 5.04
⎝ L ⎠D
(
)
KD
V I t − VTND = −VTNL
KL
(b)
(
)
5.04
V I t − 0.3 = [− (− 0.6 )] ⇒ V I t = 0.720 V
2.47
For Driver: VO t = 0.720 − 0.3 = 0.420 V
For Load:
VO t = 1.8 − 0.6 = 1.2 V
(c) PD , max = 80 μ W
[
]
⎛ 5.04 ⎞
2
2
2⎜
⎟ 2(1.8 − 0.3)υ O − υ O = [− (− 0.6)]
2
.
47
⎠
⎝
We find 4.08υ O2 − 12.24υ O + 0.36 = 0 ⇒ υ O = 0.0297 V
______________________________________________________________________________________
16.12
a.
From Equation (16.27(b)):
2
⎛W ⎞ ⎡
⎛W ⎞
2
⎜ ⎟ ⎣ 2 ( 2.5 − 0.5 )( 0.05 ) − ( 0.05 ) ⎤⎦ = ⎜ ⎟ [− ( −1)]
⎝ L ⎠D
⎝ L ⎠L
⎛W ⎞
⎜ ⎟ =1
⎝ L ⎠L
Then
⎛W ⎞
⎜ ⎟ = 5.06
⎝ L ⎠D
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
b.
2
⎛ 80 ⎞
iD = ⎜ ⎟ (1) ⎡⎣ − ( −1) ⎤⎦
⎝ 2⎠
i = 40 μ A
or D
P = iD ⋅ VDD = ( 40 )( 2.5 ) ⇒ P = 100 μ W
______________________________________________________________________________________
16.13
a.
vI = 0.5 V ⇒ iD = 0 ⇒ P = 0
i.
vI = 5 V, From Equation (16.11),
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.
vI = 0.25 V ⇒ iD = 0 ⇒ P = 0
i.
vI = 4.3 V, From Equation (16.21),
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
iD = 10 [5 − 0.237 − 0.7 ] = 165 μ A
2
P = iD ⋅ VDD = (165 )( 5 ) ⇒ P = 825 μ W
c.
vI = 0.03 V ⇒ iD = 0 ⇒ P = 0
i.
vI = 5 V
iD = K L ( −VTNL ) = (10 ) ⎣⎡ − ( −2 ) ⎦⎤ = 40 μ A
2
2
P = iD ⋅ VDD = ( 40 )( 5 ) ⇒ P = 200 μ W
ii.
______________________________________________________________________________________
16.14
(a) υ I = V DD − VTNL = 5 − 0.5 = 4.5 V
[
]
K D 2(υ I − VTND )υ O1 − υ O2 1 = K L [V DD − υ O1 − VTNL ]
[
]
2
10 2(4.5 − 0.5)υ O1 − υ O2 1 = (1)[5 − υ O1 − 0.5]
2
We find 11υ O21 − 89υ O1 + 20.25 = 0 ⇒ υ O1 = 0.234 V
υ O 2 = V DD − VTNL = 4.5 V
(b) υ I = 0.234 V, ⇒ υ O1 = 4.5 V
From part (a), υ O 2 = 0.234 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.15
(a)
[
]
KD
2
2(υ I − VTND )υ O − υ O2 = (− VTNL )
KL
[
]
⎛4⎞
2
2
⎜ ⎟ 2(5 − 0.8)υ O1 − υ O1 = [− (− 1.2 )]
1
⎝ ⎠
We find 4υ O2 1 − 33.6υ O1 + 1.44 = 0 ⇒ υ O1 = 0.0431 V
⇒ υ O2 = 5 V
(b) For υ I = 0.0431 V, υ O1 = 5 V
From part (a), υ O 2 = 0.0431 V
______________________________________________________________________________________
16.16
(a) υ O = V DD − VTNLO = 2.5 − 0.5 = 2.0 V
[
( 2φ + υ − 2φ )]
υ = 2.5 − [0.5 + 0.25( 0.7 + υ − 0.7 )]
(b) υ O = V DD − VTNLO + γ
fp
O
O
fp
O
υ O − 2.209 = −0.25 0.7 + υ O
υ O2 − 4.418υ O + 4.88 = 0.0625(0.7 + υ O )
υ O2 − 4.4805υ O + 4.836 = 0 ⇒ υ O = 1.81 V
______________________________________________________________________________________
16.17
(a)
KD
(υ I − VTND ) = (− VTNL )
KL
Or υ I = [− (VTNL )]
KL
20
+ VTND = [− (− 0.6)]
+ 0.4 = 0.6683 V
KD
100
( 2φ + υ − 2φ )
= −0.6 + 0.25( 0.7 + 1.25 − 0.7 ) = −0.460 V
(b) VTNL = VTNLO + γ
fp
O
fp
20
+ 0.4 = 0.6057 V
100
______________________________________________________________________________________
υ I = [− (− 0.460)]
6.18
(a)
[
]
KD
2
2(υ I − VTND )υ O − υ O2 = (− VTNL )
KL
[
]
KD
2
2
2(1.8 − 0.4 )(0.1) − (0.1) = [− (− 0.6 )]
KL
KD
K
(0.27 ) = 0.36 ⇒ D = 1.33
KL
KL
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
[
]
2(1.333)[2(1.8 − 0.4)υ − υ ] = [− (− 0.6 )]
⎛K ⎞
2
(b) 2⎜⎜ D ⎟⎟ 2(υ I − VTND )υ O − υ O2 = [− VTNL ]
K
⎝ L ⎠
2
2
O
O
We find 2.667υ O2 − 7.467υ O + 0.36 = 0 ⇒ υ O = 49.1 mV
⎛ k ′ ⎞⎛ W ⎞
⎛ 0.1 ⎞
2
2
(c) i D , max = ⎜⎜ n ⎟⎟⎜ ⎟ (− VTNL ) = ⎜
⎟(1)[− (− 0.6)] ⇒ i D , max = 18 μ A
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠ L
P = i D , max ⋅ V DD = (18)(1.8) = 32.4 μ W
______________________________________________________________________________________
6.19
(a) One input high,
KD
K
2
2
2(3 − 0.5)(0.1) − (0.1) = [− (− 1)] ⇒ D = 2.04
KL
KL
[
]
(b) P = i D ⋅ V DD
0.1 = i D (3) ⇒ i D = 33.33 μ A
⎛ k ′ ⎞⎛ W ⎞
2
i D = ⎜⎜ n ⎟⎟⎜ ⎟ [− VTNL ]
⎝ 2 ⎠⎝ L ⎠ L
⎛W ⎞
⎛ 100 ⎞⎛ W ⎞
33.33 = ⎜
⎟⎜ ⎟ (1) ⇒ ⎜ ⎟ = 0.667
⎝ L ⎠L
⎝ 2 ⎠⎝ L ⎠ L
⎛W ⎞
Then ⎜ ⎟ = (2.04)(0.667 ) = 1.36
⎝ L ⎠D
[
]
(c) 3(2.04 ) 2(3 − 0.5)υ O − υ O2 = [− (− 1)]
2
6.12υ O2 − 30.6υ O + 1 = 0 ⇒ υ O = 0.0329 V
______________________________________________________________________________________
16.20
(a)
[
]
KD
2
2
2(2.5 − 0.4)(0.05) − (0.05) = [− (− 0.6 )]
KL
KD
K
(0.2075) = 0.36 ⇒ D = 1.735
KL
KL
(b) P = i D , max ⋅ V DD
50 = i D , max (2.5) ⇒ i D , max = 20 μ A
⎛ 100 ⎞⎛ W ⎞
2
i D , max = 20 = ⎜
⎟⎜ ⎟ [− (− 0.6 )]
L
2
⎠⎝ ⎠ L
⎝
⎛W ⎞
⎛W ⎞
We find ⎜ ⎟ = 1.11 and ⎜ ⎟ = 1.93
⎝ L ⎠L
⎝ L ⎠D
[
]
(c) (i) 2(1.735) 2(2.5 − 0.4)υ O − υ O2 = 0.36
3.47υ − 14.574υ O + 0.36 = 0 ⇒ υ O = 24.9 mV
2
O
[
]
(ii) 3(1.735) 2(2.5 − 0.4 )υ O − υ O2 = 0.36
5.205υ − 21.861υ O + 0.36 = 0 ⇒ υ O = 16.5 mV
2
O
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
[
]
(iii) 4(1.735) 2(2.5 − 0.4)υ O − υ O2 = 0.36
6.94υ O2 − 29.148υ O + 0.36 = 0 ⇒ υ O = 12.4 mV
______________________________________________________________________________________
16.21
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
So that
⎛W ⎞
⎜ ⎟ = 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 vX = vY = 0 ⇒ v01 = 5 and v03 = 4.2
Then
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 ]
2
K D 2 ∝ 8, K D 3 ∝ 8, K L 2 ∝ 1
2
2
⎤⎦ + 8 ⎡⎣ 2 ( 4.2 − 0.8 ) v02 − v02
⎤⎦ = (1) ⎡⎣ − ( −2 ) ⎤⎦
8 ⎡⎣ 2 ( 5 − 0.8 ) v02 − v02
2
2
2
+ 54.4v02 − 8v02
=4
67.2v02 − 8v02
Then
16v02 − 121.6v0 + 4 = 0
v02 =
121.6 ±
(121.6 ) − 4 (16 )( 4 )
2 (16 )
2
v = 0.0330 V
So 02
______________________________________________________________________________________
16.22
⎛ 100 ⎞
2
2
2
(a) i D , max = ⎜
⎟(1)[V DD − υ O − VTN ] = 50[3.3 − υ O − 0.4] = 50[2.9 − υ O ]
⎝ 2 ⎠
The output is small, so neglect υ O2 .
Then i D , max ≅ 50[8.41 − 5.8υ O ] (Eq. 1)
Also,
[
⎛ 100 ⎞
2
i D , max = ⎜
⎟(12 ) 2(υ GSX − VTN )υ DSX − υ DSX
⎝ 2 ⎠
[
]
]
2
= 600 2(2.9 − 0.4)υ DSX − υ DSX
≅ 3000υ DSX
(Eq. 2)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
And
[
]
⎛ 100 ⎞
2
i D , max = ⎜
⎟(12 ) 2(υ GSY − VTN )υ DSY − υ DSY
⎝ 2 ⎠
We have υ GSY = υ Y − υ DSX ; υ DSY = υ O − υ DSX
[
i D , max = 600 2(2.9 − υ DSX − 0.4 )(υ O − υ DSX ) − (υ O − υ DSX )
≅ 600[2(2.5 − υ DSX )(υ O − υ DSX )]
2
]
(Eq.3)
Now 3000υ DSX = 50[8.41 − 5.8υ O ]
υ DSX = 0.016667[8.41 − 5.8υ O ]
(Eq. 4)
Also 3000υ DSX = 600[2(2.5 − υ DSX )(υ O − υ DSX )] ≅ 1200[2.5υ O − 2.5υ DSX ]
2.5υ DSX = 2.5υ O − 2.5υ DSX
Or υ DSX ≅ 0.5υ O
Then from Eq. 4,
0.5υ O = 0.140 − 0.09667υ O ⇒ υ O ≅ 0.2346 V
(b) υ DSX ≅ 0.5υ O ≅ 0.1173 V
υ GSX = 2.9 V
υ GSY ≅ 2.9 − 0.1173 ≅ 2.783 V
υ DSY = υ O − υ DSX ≅ 0.1173 V
______________________________________________________________________________________
16.23
a.
We can write
2
2
⎤⎦ = K y ⎡⎣ 2 ( vY − vDSX − VTNY ) vDSY − vDSY
⎤⎦ = K L [ −VTNL ]
K x ⎡⎣ 2 ( v X − VTNX ) vDSX − vDSX
2
2
From the first and third terms, (neglect vDSX ),
4 ⎡⎣ 2 ( 5 − 0.8 ) vDSX ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦
or
2
vDSX = 0.067 V
2
From the second and third terms, (neglect vDSY ),
4 ⎡⎣ 2 ( 5 − 0.067 − 0.8 ) vDSY ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦
v
or DSY
Now
and
2
= 0.068 V
vGSX = 5, vGSY = 5 − 0.067 ⇒ vGSY = 4.933 V
v0 = vDSX + vDSY ⇒ v0 = 0.135 V
Since v0 is close to ground potential, the body-effect has little effect on the results.
______________________________________________________________________________________
16.24
(a)
[
]
1 ⎛ KD ⎞
⎜
⎟ 2(υ I − VTND )υ O − υ O2 = (− VTNL )2
4 ⎜⎝ K L ⎟⎠
[
]
1 ⎛ KD ⎞
⎜
⎟ 2(3.3 − 0.4 )(0.1) − (0.1)2 = [− (− 0.6)]2
4 ⎜⎝ K L ⎟⎠
⎛ KD ⎞
⎛K ⎞
⎜⎜
⎟⎟(0.1425) = 0.36 ⇒ ⎜⎜ D ⎟⎟ = 2.53
K
⎝ L ⎠
⎝ KL ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) P = i D , max ⋅ V DD
100 = i D , max (3.3) ⇒ i D , max = 30.3 μ A
⎛ 100 ⎞⎛ W ⎞
2
i D , max = 30.3 = ⎜
⎟⎜ ⎟ [− (− 0.6 )]
⎝ 2 ⎠⎝ L ⎠ L
⎛W ⎞
⎛W ⎞
Which yields ⎜ ⎟ = 1.68 and ⎜ ⎟ = 4.26
⎝ L ⎠L
⎝ L ⎠D
______________________________________________________________________________________
16.25
Y = [A OR (B AND C)] AND D
______________________________________________________________________________________
16.26
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.
______________________________________________________________________________________
16.27
( A + B)(C + D )
______________________________________________________________________________________
16.28
(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 ⎞
⎜ ⎟ = 1,
⎜ ⎟ = 1.37
then ⎝ L ⎠ D
For ⎝ L ⎠ L
So, for
⎛W ⎞
M 6 : ⎜ ⎟ = 1.37
⎝ L ⎠6
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
To achieve the required composite conduction parameter,
⎛W ⎞
M 1 − M 5 : ⎜ ⎟ = 2.74
⎝ L ⎠1−5
For
______________________________________________________________________________________
16.29
(a)
(b)
(W L ) A
[
]
2(2.5 − 0.4)(0.05) − (0.05) = [− (− 0.6 )]
1
⎛W ⎞
⎛W ⎞
⎜ ⎟ (0.2075) = 0.36 ⇒ ⎜ ⎟ = 1.735
⎝ L ⎠A
⎝ L ⎠A
2
2
⎛W ⎞
⎛W ⎞
⇒⎜ ⎟
= 2⎜ ⎟ = 3.47
⎝ L ⎠ B ,C , D
⎝ L ⎠A
______________________________________________________________________________________
16.30 Design Problem
______________________________________________________________________________________
16.31
V DD 2.5
=
= 1.25 V
2
2
VOP t = 1.25 − (− 0.4 ) = 1.65 V
(a) By symmetry, V I t =
VON t = 1.25 − 0.4 = 0.85 V
(c) For υ I = 1.1 V, NMOS in saturation, PMOS in nonsaturation
[
K n [υ I − VTN ] = K p 2(V DD − υ I + VTP )(V DD − υ O ) − (V DD − υ O )
2
2
(1.1 − 0.4) = 2(2.5 − 1.1 − 0.4)(V DD − υ O ) − (V DD − υ O )
(V DD − υ O )2 − 2(V DD − υ O ) + 0.49 = 0 ⇒ (V DD − υ O ) = 0.2859 V
2
2
Or υ O = 2.5 − 0.2859 = 2.214 V
For υ I = 1.4 V, NMOS in nonsaturation, PMOS in saturation
[
]
K n 2(υ I − VTN )υ O − υ O2 = K p (V DD − υ I + VTP )
2(1.4 − 0.4 )υ O − υ O2 = (2.5 − 1.4 − 0.4 )
2
υ O2 − 2υ O + 0.49 = 0 ⇒ υ O = 0.286 V
2
]
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
16.32
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(a) K n = ⎜
⎟(2 ) = 100 μ A/V ; K p = ⎜ ⎟(5) = 100 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
(i) By symmetry
V
3.3
V I t = DD =
= 1.65 V
2
2
⇒ VOP t = 1.65 − (− 0.4) = 2.05 V
⇒ VON t = 1.65 − 0.4 = 1.25 V
(iii) For υ O = 0.25 V; NMOS in nonsaturation, PMOS in saturation
2(υ I − VTN )υ O − υ O2 = (V DD − υ I + VTP )
2
2(υ I − 0.4 )(0.25) − (0.25) = (3.3 − υ I − 0.4 )
2
2
0.5υ I − 0.2 − 0.0625 = 8.41 − 5.8υ I + υ I2
υ I2 − 6.3υ I + 8.6725 = 0 ⇒ υ I = 2.03 V
For υ O = 3.05 V; NMOS in saturation, PMOS in nonsaturation
(υ I − VTN )2 = 2(V DD − υ I + VTP )(V DD − υ O ) − (V DD − υ O )2
(υ I − 0.4)2 = 2(3.3 − υ I − 0.4)(3.3 − 3.05) − (3.3 − 3.05)2
υ I2 − 0.8υ I + 0.16 = 0.5(2.9 − υ I ) − 0.0625
υ I2 − 0.3υ I − 1.2275 = 0 ⇒ υ I = 1.27 V
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(b) K n = ⎜
⎟(4) = 200 μ A/V ; K p = ⎜ ⎟(5) = 100 μ A/V
2
2
⎝
⎝ ⎠
⎠
3.3 − 0.4 +
(i) V I t =
200
(0.4)
100
= 1.436 V
200
1+
100
⇒ VOP t = 1.436 + 0.4 = 1.836 V
⇒ VON t = 1.436 − 0.4 = 1.036 V
(iii) For υ O = 0.25 V; NMOS in nonsaturation, PMOS in saturation
[
]
200 2(υ I − 0.4)(0.25) − (0.25) = 100(3.3 − υ I − 0.4 )
2
2
2(0.5υ I − 0.2 − 0.0625) = 8.41 − 5.8υ I + υ I2
υ I2 − 6.8υ I + 8.935 = 0 ⇒ υ I = 1.78 V
For υ O = 3.05 V ; NMOS in saturation, PMOS in nonsaturation
[
200(υ I − 0.4) = 100 2(3.3 − υ I − 0.4 )(3.3 − 3.05) − (3.3 − 3.05)
2
(
)
2 υ I2 − 0.8υ I + 0.16 = 0.5(2.9 − υ I ) − 0.0625
2
]
2υ − 1.1υ I − 1.0675 = 0 ⇒ υ I = 1.06 V
______________________________________________________________________________________
2
I
16.33
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(a) K n = ⎜
⎟(4) = 200 μ A/V ; K p = ⎜ ⎟(12 ) = 240 μ A/V
2
2
⎝
⎠
⎝ ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.3 − 0.4 +
200
(0.4)
240
= 1.707 V
200
1+
240
⇒ VOP t = 1.707 + 0.4 = 2.107 V
(i) V I t =
⇒ VON t = 1.707 − 0.4 = 1.307 V
(ii) For υ O = 3.1 V ; NMOS in saturation, PMOS in nonsaturation
[
200(υ I − 0.4) = 240 2(3.3 − υ I − 0.4 )(3.3 − 3.1) − (3.3 − 3.1)
2
2
]
υ I2 − 0.8υ I + 0.16 = 1.2[0.4(2.9 − υ I ) − 0.04]
υ I2 − 0.32υ I − 1.184 = 0 ⇒ υ I = 1.26 V
(iii) For υ O = 0.2 V ; NMOS in nonsaturaton, PMOS in saturation
[
]
200 2(υ I − 0.4 )(0.2) − (0.2) = 240(3.3 − υ I − 0.4 )
2
(
0.4υ I − 0.2 = 1.2 8.41 − 5.8υ I + υ I2
2
)
1.2υ I2 − 7.36υ I + 10.292 = 0 ⇒ υ I = 2.157 V
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(b) K n = ⎜
⎟(6 ) = 300 μ A/V ; K p = ⎜ ⎟(4 ) = 80 μ A/V
2
2
⎝
⎠
⎝ ⎠
3.3 − 0.4 +
300
(0.4)
80
= 1.25 V
300
1+
80
(ii) For υ O = 3.1 V; NMOS in saturation, PMOS in nonsaturation
(i) V I t =
[
300(υ I − 0.4) = 80 2(3.3 − υ I − 0.4)(3.3 − 3.1) − (3.3 − 3.1)
2
(
)
3.75 υ − 0.8υ I + 0.16 = 0.4(2.9 − υ I ) − 0.04
2
I
2
]
3.75υ I2 − 2.6υ I − 0.52 = 0 ⇒ υ I = 0.855 V
(iii) For υ O = 0.2 V; NMOS in nonsaturation, PMOS in saturation
[
]
300 2(υ I − 0.4)(0.2) − (0.2) = 80(3.3 − υ I − 0.4 )
2
2
3.75(0.4υ I − 0.2 ) = 8.41 − 5.8υ I + υ I2
υ I2 − 7.3υ I + 9.16 = 0 ⇒ υ I = 1.61 V
______________________________________________________________________________________
16.34
a.
For
N1
vO1 = 0.6 < VTN ⇒ vO 2 = 5 V
in nonsaturation and P1 in saturation. From Equation (16.43),
⎡ 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 =
or
9.6 ±
( 9.6 ) − 4 (1)(18.96 )
2
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
1.7 ≤ v ≤ 3.3 V
02
So
______________________________________________________________________________________
16.35
V0 Nt ≤ v01 ≤ V0 Pt
a.
By symmetry, VIt = 2.5 V
and
So
b.
V0 Pt = 2.5 + 0.8 = 3.3 V
V0 Nt = 2.5 − 0.8 = 1.7 V
1.7 ≤ v01 ≤ 3.3 V
For
vO 2 = 0.6 < VTN ⇒ vO 3 = 5 V
N 2 in nonsaturation and P2 in saturation. From Equation (16.43),
⎡ 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
So
vI22 − 9.6vI 2 + 18.96 = 0
vI 2 = v01 = 2.78 V
For v01 = 2.78, both N1 and P1 in saturation. Then
vI = 2.5 V
______________________________________________________________________________________
16.36
2.5
= 1.25 V
2
2
For 0.4 ≤ υ I ≤ 1.25 V; i D = 120(υ I − 0.4) μ A
(a) V I t =
i D , peak = 120(1.25 − 0.4) = 86.7 μ A
2
For 1.25 ≤ υ I ≤ 2.1 V; i D = 120(2.5 − υ I − 0.4) μ A
2
⎛ 1.8 ⎞
(b) V I t = ⎜
⎟ = 0.9 V
⎝ 2 ⎠
For 0.4 ≤ υ I ≤ 0.9 V; i D = 120(υ I − 0.4 ) μ A
2
i D , peak = 120(0.9 − 0.4 ) = 30 μ A
2
For 0.9 ≤ υ I ≤ 1.4 V; i D = 120(1.8 − υ I − 0.4 ) μ A
______________________________________________________________________________________
2
16.37
⎛ 80 ⎞
⎛ 40 ⎞
(a) K n = ⎜ ⎟(2 ) = 80 μ A/V 2 , K p = ⎜ ⎟(4 ) = 80 μ A/V 2
⎝ 2 ⎠
⎝ 2 ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.8
VI t =
= 0.9 V
2
2
i D , peak = 80(0.9 − 0.35) = 24.2 μ A
⎛ 80 ⎞
⎛ 40 ⎞
(b) K n = ⎜ ⎟(2 ) = 80 μ A/V 2 , K p = ⎜ ⎟(6) = 120 μ A/V 2
2
⎝ ⎠
⎝ 2 ⎠
1.8 − 0.35 +
VI t =
1+
80
(0.35)
120
80
120
= 0.9556 V
i D , peak = 80(0.9556 − 0.35) = 29.34 μ A
2
⎛ 80 ⎞
⎛ 40 ⎞
(c) K n = ⎜ ⎟(4 ) = 160 μ A/V 2 , K p = ⎜ ⎟(4 ) = 80 μ A/V 2
⎝ 2 ⎠
⎝ 2 ⎠
1.8 − 0.35 +
VI t =
160
(0.35)
80
160
1+
80
= 0.8057 V
i D , peak = 160(0.8057 − 0.35) = 33.23 μ A
2
______________________________________________________________________________________
16.38
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(a) K n = ⎜
⎟(3) = 150 μ A/V , K p = ⎜ ⎟(7.5) = 150 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
3.3
VI t =
= 1.65 V
2
2
i D , peak = 150(1.65 − 0.4 ) = 234 μ A
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(b) K n = ⎜
⎟(4) = 200 μ A/V , K p = ⎜ ⎟(4 ) = 80 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
3.3 − 0.4 +
VI t =
200
(0.4)
80
200
1+
80
= 1.369 V
i D , peak = 200(1.369 − 0.4 ) = 188 μ A
2
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(c) K n = ⎜
⎟(3) = 150 μ A/V , K p = ⎜ ⎟(12 ) = 240 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
3.3 − 0.4 +
VI t =
150
(0.4)
240
150
1+
240
= 1.796 V
i D , peak = 150(1.796 − 0.4 ) = 292 μ A
2
______________________________________________________________________________________
16.39
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
( )(
)
= (10 )(0.2 × 10 )(1.8) ⇒ P = 6.48 μ W
2
(a) P = fC LV DD
= 10 7 0.2 × 10 −12 (2.5) ⇒ P = 12.5 μ W
2
−12
2
7
(b) P = fC LV DD
______________________________________________________________________________________
2
16.40
2
(a) P = 4 × 10 6 150 × 10 6 0.12 × 10 −12 (1.8) ⇒ P = 233 W
(
)(
)(
)
(b) P = 233.28 = (4 × 10 )(300 × 10 )(0.12 × 10 )V
−12
⇒ V DD = 1.27 V
______________________________________________________________________________________
6
6
2
DD
16.41
(a)
(b)
P=
3
= 3 × 10−7 W
107
2
⇒ CL =
P = fCLVDD
CL =
(i)
CL =
(ii)
CL =
3 × 10−7
( 5 ×10 ) ( 5)
6
⇒ CL = 0.0024 pF
2
3 × 10−7
2
⇒ CL = 0.00551pF
2
⇒ CL = 0.0267 pF
( 5 ×10 ) ( 3.3)
6
3 × 10−7
( 5 ×10 ) (1.5)
6
P
2
fVDD
(iii)
______________________________________________________________________________________
16.42
(a)
(b)
P=
10
= 2 × 10−6 W
5 × 106
CL =
CL =
(i)
CL =
(ii)
CL =
P
2
fVDD
2 × 10−6
(8 ×10 ) ( 5)
6
2
⇒ CL = 0.01 pF
2 × 10−6
2
⇒ CL = 0.023 pF
2
⇒ CL = 0.111 pF
(8 ×10 ) ( 3.3)
6
2 × 10−6
(8 ×10 ) (1.5)
6
(iii)
_____________________________________________________________________________
16.43
(a) For υ I ≅ V DD , NMOS in nonsaturation
[
2
i D = K n 2(υ I − VTN )υ DS − υ DS
] and υ
di D
1
=
≅ K n [2(υ I − VTN )]
So
rds dυ DS
DS
very small
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
Or rds ≅
=
⎛ k n′ ⎞⎛ W ⎞
⎛W ⎞
⎜⎜ ⎟⎟⎜ ⎟ [2(V DD − VTN )] k n′ ⎜ ⎟ (V DD − VTN )
2
L
⎝ L ⎠n
⎝ ⎠⎝ ⎠ n
For υ I ≅ 0 , PMOS in nonsaturation
[
2
i D = K p 2(V DD − υ I + VTP )υ SD − υ SD
So
] and υ
SD
very small
⎛ k ′p ⎞⎛ W ⎞
di D
1
⎛W ⎞
=
= ⎜⎜ ⎟⎟⎜ ⎟ [2(V DD − υ I + VTP )] ≅ k ′p ⎜ ⎟ (V DD + VTP )
rsd dυ SD ⎝ 2 ⎠⎝ L ⎠ p
⎝ L ⎠p
Or rsd =
1
⎛W ⎞
k ′p ⎜ ⎟ (V DD + VTP )
⎝ L ⎠p
⎛W ⎞
⎛W ⎞
(b) Let ⎜ ⎟ = 2 , ⎜ ⎟ = 4 , and V DD = 5 V
⎝ L ⎠n
⎝ L ⎠p
For NMOS:
1
⇒ r = 1.11 k Ω
rds =
(0.1)(2)(5 − 0.5) ds
υ
0.5
i d = ds =
= 0.45 mA
rds 1.11
For PMOS:
1
⇒ r = 1.39 k Ω
rsd =
(0.04)(4)(5 − 0.5) sd
υ
0.5
= 0.36 mA
i d = sd =
rsd 1.39
______________________________________________________________________________________
16.44
3
(V DD + VTP − VTN ) = 0.5 + 3 (3.3 − 0.5 − 0.5)
8
8
⇒ V IL = 1.3625 V
(a) V IL = VTN +
5
(V DD + VTP − VTN ) = 0.5 + 5 (3.3 − 0.5 − 0.5)
8
8
⇒ V IH = 1.9375 V
V IH = VTN +
1
[2(1.3625) + 3.3 − 0.5 + 0.5] = 3.0125 V
2
1
VOLU = [2(1.9375) − 3.3 − 0.5 + 0.5] = 0.2875 V
2
Then NM L = 1.3625 − 0.2875 = 1.075 V
(b) VOHU =
NM H = 3.0125 − 1.9375 = 1.075 V
______________________________________________________________________________________
16.45
(a)
K n 100
=
=2
Kp
50
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V IL = 0.35 +
(2.5 − 0.35 − 0.35) ⎡2 2 − 1⎤
⎥
⎢
(2 − 1)
⎣⎢ 2 + 3 ⎦⎥
⇒ V IL = 0.8268 V
V IH = 0.35 +
(2.5 − 0.35 − 0.35) ⎡⎢ 2(2) − 1⎤⎥
(2 − 1)
⎣⎢ 3(2 ) + 1 ⎦⎥
⇒ V IH = 1.2713 V
1
[(1 + 2)(0.8268) + 2.5 − (2)(0.35) + 0.35]
2
⇒ VOHU = 2.3152 V
(b) VOHU =
VOLU =
(1.2713)(1 + 2) − 2.5 − (2)(0.35) + 0.35
2(2)
⇒ VOLU = 0.2410 V
Then NM L = 0.8268 − 0.2410 = 0.5858 V
NM H = 2.3152 − 1.2713 = 1.0439 V
______________________________________________________________________________________
16.46
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(a) K n = ⎜
⎟(2 ) = 100 μ A/V , K p = ⎜ ⎟(5) = 100 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
3
V IL = 0.4 + (3.3 − 0.4 − 0.4) = 1.3375 V
8
5
V IH = 0.4 + (3.3 − 0.4 − 0.4 ) = 1.9625 V
8
1
VOHU = [2(1.3375) + 3.3 − 0.4 + 0.4] = 2.9875 V
2
1
VOLU = [2(1.9625) − 3.3 − 0.4 + 0.4] = 0.3125 V
2
Then NM L = 1.3375 − 0.3125 = 1.025 V
NM H = 2.9875 − 1.9625 = 1.025 V
⎛ 100 ⎞
⎛ 40 ⎞
2
2
(b) K n = ⎜
⎟(4) = 200 μ A/V , K p = ⎜ ⎟(12) = 240 μ A/V
⎝ 2 ⎠
⎝ 2 ⎠
K n 200
=
= 0.8333
K p 240
V IL = 0.4 +
(3.3 − 0.4 − 0.4) ⎡2 0.8333 − 1⎤ = 1.4127 V
(0.8333 − 1) ⎢⎢⎣ 3.8333 ⎥⎥⎦
V IH = 0.4 +
(3.3 − 0.4 − 0.4) ⎡⎢ 2(0.8333) − 1⎤⎥ = 2.0370 V
(0.8333 − 1) ⎢⎣ 3(0.8333) + 1 ⎥⎦
1
[(1 + 0.8333)(1.4127 ) + 3.3 − (0.8333)(0.4) + 0.4]
2
⇒ VOHU = 2.9783 V
Now VOHU =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.0370(1 + 0.8333) − 3.3 − (0.8333)(0.4) + 0.4
VOLU =
2(0.8333)
⇒ VOLU = 0.3007 V
Then NM L = 1.4127 − 0.3007 = 1.1112 V
NM H = 2.9783 − 2.0370 = 0.9413 V
______________________________________________________________________________________
16.47
v A = vB = 5 V
a.
N1
and N 2 on, so vDS 1 ≈ 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).
For vA = vB = vC ≡ vI
b.
Want
K n ,eff = K p , eff
kn′ ⎛ W ⎞
k ′ ⎛ 3W ⎞
⋅⎜ ⎟ = P ⋅⎜
⎟
2 ⎝ 3L ⎠ n 2 ⎝ L ⎠ P
kn′ = 2k P′ ,
With
then
2 1 ⎛W ⎞
1 ⎛W ⎞
⋅ ⋅⎜ ⎟ = ⋅3⋅⎜ ⎟
2 3 ⎝ L ⎠n 2 ⎝ L ⎠ P
Or
c.
9 ⎛W ⎞
⎛W ⎞
⎜ ⎟ = ⋅⎜ ⎟
⎝ L ⎠n 2 ⎝ L ⎠ P
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 (16.41)
5 + ( −0.8 ) +
VIt =
1+
Kn
⋅ ( 0.8 )
Kp
Kn
Kp
Now
Kn
⎛9⎞
= ( 2) ⎜ ⎟ = 9
Kp
⎝ 2⎠
Then
VIt =
5 + ( −0.8 ) + 3 ( 0.8 )
1+ 3
⇒ VIt = 1.65 V
______________________________________________________________________________________
16.48
By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V.
N1
N2
N3
N4
N5
State
v0
1
0
off
on
off
on
on
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2
off
off
on
on
off
0
3
on
on
off
off
on
5
4
on
on
off
on
on
0
Logic function ( X
v OR vY ) ⊗ ( v X AND vZ )
Y )
Z )
with ( X
Exclusive OR of ( X
______________________________________________________________________________________
v OR v
v AND v
16.49
NMOS in Parallel
4-PMOS in series
(b)
CL
⎛W ⎞
⇒⎜ ⎟ =2
⎝ L ⎠n
⎛W ⎞
⇒ ⎜ ⎟ = 4 ( 4 ) = 16
⎝ L ⎠p
doubles ⇒ current must double to maintain switching speed.
⎛W ⎞
⇒⎜ ⎟ =4
⎝ L ⎠n
⎛W ⎞
⎜ ⎟ = 32
⎝ L ⎠p
______________________________________________________________________________________
16.50
⎛W ⎞
⎜ ⎟ = 4 ( 2) = 8
4-NMOS in series ⎝ L ⎠n
⎛W ⎞
⎜ ⎟ =4
L
4-PMOS in parallel ⎝ ⎠ p
⎛W ⎞
⎜ ⎟ = 16
⎝ L ⎠n
⎛W ⎞
⎜ ⎟ =8
⎝ L ⎠p
(b)
______________________________________________________________________________________
16.51
(a)
⎛W ⎞
⇒⎜ ⎟ =2
NMOS in parallel ⎝ L ⎠ n
⎛W ⎞
⇒ ⎜ ⎟ = 3 ( 4 ) = 12
3-PMOS in series ⎝ L ⎠ P
⎛W ⎞
⎜ ⎟ =4
⎝ L ⎠n
⎛W ⎞
⎜ ⎟ = 24
⎝ L ⎠p
(b)
______________________________________________________________________________________
16.52
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a)
⎛W ⎞
⎜ ⎟ = 3( 2) = 6
3-NMOS in series ⎝ L ⎠ n
⎛W ⎞
⎜ ⎟ =4
L
3-PMOS in parallel ⎝ ⎠ p
⎛W ⎞
⎜ ⎟ = 12
⎝ L ⎠n
⎛W ⎞
⎜ ⎟ =8
⎝ L ⎠p
(b)
______________________________________________________________________________________
16.53
(a) Y = ABC + DE
(b)
(c)
(W L ) An, Bn,Cn = 6 , (W L )Dn, En = 4
All (W L ) p = 8
______________________________________________________________________________________
16.54
(a)
(b)
Y = A( BD + CE )
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) NMOS: 3 transistors in series for pull down mode.
⎛W ⎞
⎜ ⎟ = 2 ( 3)( 2 ) = 12
For twice the speed: ⎝ L ⎠ n
⎛W ⎞
⎜ ⎟ = 2 ( 4) = 8
L
PMOS: ⎝ ⎠ P , A
⎛W ⎞
= 2 ( 2 )( 4 ) = 16
⎜ ⎟
⎝ L ⎠ P , B ,C , D , E
______________________________________________________________________________________
16.55
(a)
(b)
(c)
Y = A + BC + DE
⎛W ⎞
⎜ ⎟ =2
L
NMOS: ⎝ ⎠ n , A
⎛W ⎞
=4
⎜ ⎟
⎝ L ⎠ n , B ,C , D , E
PMOS: 3 transistors in series for the pull-up mode
⎛W ⎞
⎜ ⎟ = 3 ( 4 ) = 12
⎝ L ⎠p
______________________________________________________________________________________
16.56
(a) Y = A[B + CD ]
(b)
(c) (W L ) An, Bn = 4 , (W L )Cn , Dn = 8
(W L ) Ap = 4 , (W L )Bp ,Cp , Dp = 8
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.57
(a) For Y = AB C + A B C + A BC
(
)(
)(
We have Y = AB C + A B C + A BC = A + B + C A + B + C A + B + C
)
(b) All (W L )n = 3
All (W L ) p = 6
______________________________________________________________________________________
16.58
(a)
(b)
(W L ) An, Bn,Cn = 2 , (W L )Dn, En = 1
(W L ) Ap, Dp , Ep = 6 , (W L )Bp,Cp = 12
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.59
(a) Y = C ( A + B )
(b) All (W L )n = 4
(W L )Cp = 4 , (W L ) Ap, Bp = 8
______________________________________________________________________________________
16.60
(a) All (W L )n = 2
All (W L ) p = 20
(b) All (W L )n = 10
All (W L ) p = 4
______________________________________________________________________________________
16.61
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
3
4
5
6
N1
P1
NA
NB
NC
v01
N2
P2
v02
off
on
off
on
off
on
on
off
on
off
on
off
off
on
off
off
off
off
off
off
off
off
off
on
off
off
off
on
off
on
5
5
5
5
5
0
on
on
on
on
on
off
off
off
off
off
off
on
0
0
0
0
0
5
Logic function is
v02 = ( v A OR vB ) AND vC
______________________________________________________________________________________
16.62
State
v01
v02
v03
1
2
3
4
5
6
5
0
5
5
5
0
5
0
5
0
5
5
0
5
0
5
0
0
Logic function:
v03 = ( v X OR vZ ) AND vY
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.63
______________________________________________________________________________________
16.64
______________________________________________________________________________________
16.65
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.66
2 I = −C
dVC
dt
So
1
( 2I ) ⋅ t
C
ΔVC = −0.5 V
ΔVC = −
For
−0.5 = −
2 ( 2 x 10−12 ) ⋅ t
25 x 10−15
⇒ t = 3.125 ms
______________________________________________________________________________________
16.67
(a) (i) υ O = 0
(ii) υ O = φ − VTN = 3.3 − 0.4 = 2.9 V
(iii) υ O = 2.5 V
(b) (i) υ O = 0
(ii) υ O = φ − VTN = 1.8 − 0.4 = 1.4 V
(iii) υ O = φ − VTN = 1.8 − 0.4 = 1.4 V
______________________________________________________________________________________
16.68
(a) (i) υ O = 0
(ii) υ O = φ − VTN = 2.5 − 0.5 = 2 V
(iii) υ O = 1.8 V
(b) (i) υ O = 0
(ii) υ O = φ − VTN = 2 − 0.5 = 1.5 V
(iii) υ O = φ − VTN = 2 − 0.5 = 1.5 V
______________________________________________________________________________________
16.69
(a) υ O1 = 2.5 − 0.4 = 2.1 V
υ O 2 = 2.5 V
(b) υ I′1 = 2.5 − 0.4 = 2.1 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(W L )1
2
2
2(2.1 − 0.4 )(0.1) − (0.1) = (2.5 − 0.1 − 0.4 )
(W L )2
[
]
(W L )1 [0.33] = 4 ⇒ ⎛⎜ W ⎞⎟ = 12.1
⎝ L ⎠1
υ I′ 3 = 2.1 V
(W L )3
2
2
[
2(2.1 − 0.4 )(0.1) − (0.1) ] = [− (− 0.6)]
(W L )4
(W L )3 [0.33] = 0.36 ⇒ ⎛⎜ W ⎞⎟ = 1.09
⎝ L ⎠3
______________________________________________________________________________________
16.70
A
B
Y
0
0
1
0
1
0
1
0
0
⇒ indeterminate
1
1
0.1
Without the top transistor, the circuit performs the exclusive-NOR function.
______________________________________________________________________________________
16.71
A
0
0
1
1
A
1
1
0
0
B
0
1
0
1
B
1
0
1
0
Y
0
1
1
1
Z
1
0
0
0
Y = A + AB = A + B
Z = Y or Z = AB
______________________________________________________________________________________
16.72
(a)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
______________________________________________________________________________________
16.73
(a) (i) Y = 0
(ii) Y = 2.5 V
(iii) Y = 0
(iv) Y = 2.5 V
(b) (i) Y = 0
(ii) Y = 0
(iii) Y = 2.5 V
(iv) Y = 2.5 V
(c) For φ = 1 , φ = 0 ; then Y = B
For φ = 0 , φ = 1 ; then Y = A
A multiplexer
______________________________________________________________________________________
16.74
Y = AC + BC
______________________________________________________________________________________
16.75
(a) (i) Y = 0
(ii) Y = 2.5 V
(iii) Y = 2.5 V
(iv) Y = 0
(b) Y = AB + A B = A ⊗ B
______________________________________________________________________________________
16.76
A
0
1
0
1
B
0
0
1
1
Y
0
1
1
0
Exclusive-OR function.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.77
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.
______________________________________________________________________________________
16.78
Want Q to be the transition point of M 5 − M 6 .
From Equation (16.26(b)),
(
)
K5
V I t − VTN 5 = −VTN 6
K6
(
)
100
V I t − 0.4 = [− (− 0.6)]⇒ V I t = Q = 0.7795 V
40
This is region where M 1 and M 3 are biased in
the saturation region.
(
)
K1
V I t − VTN 1 = −VTN 3
K3
(
)
150
V I t − 0.4 = [− (− 0.6 )] ⇒ V I t = S = 0.7098 V
40
This analysis neglected the effect of M 2 starting to conduct.
______________________________________________________________________________________
16.79
vIt =
3.3 + ( −0.4 ) + 0.5
vI = 1.5 V
1+1
= 1.7 V
NMOS Sat; PMOS Non Sat
( vI − 0.5) = ⎡⎣2 ( 3.3 − vI − 0.4 )( 3.3 − vo1 ) − ( 3.3 − vo1 ) ⎤⎦ ⇒ vo1 = 2.88 V
2
2
vI = 1.6 V vo1 = 2.693 V
vI = 1.7 V vo1 = variable (switching region)
vI = 1.8 V NMOS Non Sat; PMOS Sat
( 3.3 − VI − 0.4 ) = ⎡⎣ 2 ( vI − 0.5 ) vo1 − vo21 ⎤⎦ ⇒ vo1 = 0.607 V
2
Now
vI = 1.5 V, vo1 = 2.88 V ⇒ vo ≈ 0V
vI = 1.6 V, vo1 = 2.693 V
NMOS Non Sat; PMOS Sat
( 3.3 − vo1 − 0.4 ) = ⎡⎣ 2 ( vo1 − 0.5) vo − vo2 ⎤⎦
2
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
2
( v01 − 0.5) = ⎡⎣ 2 ( 3.3 − v01 − 0.4 )( 3.3 − v0 ) − ( 3.3 − v0 ) ⎤⎦ ⇒ v0 = 3.298 V
2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.80
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.
______________________________________________________________________________________
16.81
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.
______________________________________________________________________________________
16.82
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.
______________________________________________________________________________________
16.83
For J = vX = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0.
For J = vX = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1.
Now consider J = K = CLK = 1. With vX = 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.
______________________________________________________________________________________
16.84
(a) 256K ⇒ 262,144 cells ⇒ 512× 512
Each decoder ⇒ 9 inputs
(b) (i) Row 52, address = 000110011
(ii) Row 129, address = 010000000
(iii) Row 241, address = 011110000
(c) (i) Column 24, address = 000010111
(ii) Column 165, address = 010100100
(iii) Column 203, address = 011001010
______________________________________________________________________________________
16.85
(a)
1-Megabit memory ⇒
= 1, 048,576 ⇒ 1024 × 1024
= 10
Number of input row and column decodes lines necessary
(b) 250K × 4 bits ⇒ 262,144 × 4 bits ⇒ 512 × 512
For 512 lines ⇒ 9 row and column decoder lines necessary.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.86
32 rows × 16 columns
Each column contains 8 bits
______________________________________________________________________________________
16.87
Assume the address line is initially uncharged, then
dVC
1
I
VC = ∫ Idt = ⋅ t
C
C
dt or
−12
V ⋅ C ( 2.7 ) ( 5.8 × 10 )
t= C
=
⇒
250 × 10−6
I
Then
I =C
t = 6.26 × 10−8 s ⇒ 62.6 ns
______________________________________________________________________________________
16.88
(a)
V DD − Q ⎛ k n′ ⎞⎛ W ⎞
= ⎜⎜ ⎟⎟⎜ ⎟ 2(VGSD − VTN )Q − Q 2
R
⎝ 2 ⎠⎝ L ⎠ D
[
[
]
2.5 − 0.02 ⎛ 80 ⎞⎛ W ⎞
2
= ⎜ ⎟⎜ ⎟ 2(2.5 − 0.4 )(0.02) − (0.02)
1
⎝ 2 ⎠⎝ L ⎠ D
]
⎛W ⎞
⎛W ⎞
0.062 = ⎜ ⎟ (0.0836 ) ⇒ ⎜ ⎟ = 0.74
⎝ L ⎠D
⎝ L ⎠D
(b) 16K ⇒ 16,384 cells
V
1.2
i D ≅ DD =
= 1.2 μ A
2
1
P = (1.2 )(1.2 )(16,384 ) ⇒ P = 23.6 mW
______________________________________________________________________________________
16.89
16 K ⇒ 16,384 cells
200
⇒ 12.2 μW
16,384
V
P
12.2
2.5
iD =
=
= 4.88 μ A ≅ DD =
⇒ R = 0.512 M Ω
2.5
VDD
R
R
PT = 200 mW ⇒ Power per cell =
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 ⎞
⎜ ⎟ = 0.797
L
So ⎝ ⎠
______________________________________________________________________________________
16.90
D = V DD = 2.5 V
Assume M P 3 in saturation; M NA , M N 1 in nonsaturation
I DP 3 = I DNA = I DN 1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛ k ′p ⎞⎛ W ⎞
⎛ 35 ⎞
2
2
I DP 3 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP ) = ⎜ ⎟(1)(2.5 − 0.4 ) = 77.175 μ A
L
2
2
⎝ ⎠
⎝ ⎠⎝ ⎠ 3
I DP 3 = I DN 1
[
⎛ 80 ⎞
77.175 = ⎜ ⎟(2) 2(2.5 − 0.4 )Q − Q 2
⎝ 2 ⎠
]
We find Q 2 − 4.2Q + 0.9647 = 0 ⇒ Q = 0.244 V
I DP 3 = I DNA
[
⎛ 80 ⎞
2
77.175 = ⎜ ⎟(1) 2(2.5 − 0.244 − 0.4 )(D − 0.244) − (D − 0.244)
⎝ 2 ⎠
[
1.9294 = 3.712(D − 0.244 ) − (D − 0.244 )
2
]
]
= 3.712 D − 0.9057 − D + 0.488 D − 0.05954
2
We find D − 4.2 D + 2.895 = 0 ⇒ D = 0.869 V
2nd approximation, M P 3 in nonsaturation
Assume V DS = V DD − D ≅ 2.5 − 0.869 = 1.631 V
2
[
]
⎛ 35 ⎞
2
I DP 3 = ⎜ ⎟(1) 2(2.5 − 0.4)(1.631) − (1.631) = 73.33 μ A
⎝ 2 ⎠
I DP 3 = I DN 1
[
⎛ 80 ⎞
73.33 = ⎜ ⎟(2) 2(2.5 − 0.4)Q − Q 2
⎝ 2 ⎠
]
Q 2 − 4.2Q + 0.9166 = 0 ⇒ Q = 0.231 V
I DP 3 = I DNA
[
⎛ 80 ⎞
2
73.33 = ⎜ ⎟(1) 2(2.5 − 0.231 − 0.4)(D − 0.231) − (D − 0.231)
⎝ 2 ⎠
[
(
1.833 = 3.738(D − 0.231) − D 2 − 0.462 D + 0.05336
]
)]
We find D − 4.2 D + 2.75 = 0 ⇒ D = 0.812 V
______________________________________________________________________________________
2
16.91
Approximation, M N 2 cutoff
I DP 2 = I DB , assume D = 0
Both M P 2 and M B in nonsaturation
[
(
) (
)]
2
⎛ 35 ⎞
⎜ ⎟(4) 2(2.5 − 0.4 ) 2.5 − Q − 2.5 − Q
⎝ 2 ⎠
[
[ (
⎛ 80 ⎞
= ⎜ ⎟(1) 2(2.5 − 0.4 )Q − Q 2
⎝ 2 ⎠
) (
)]
[
]
70 4.2 2.5 − Q − 6.25 − 5Q + Q 2 = 40 4.2Q − Q 2
]
We find 0.75Q + 2.8Q − 7.4375 = 0 ⇒ Q = 1.794 V
Approximation, M P1 cutoff; assume D = 2.5 V
I DNA = I DN 1
Both M A and M N 1 in nonsaturation
2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
[
⎛ 80 ⎞
2
⎜ ⎟(1) 2(2.5 − Q − 0.4 )(2.5 − Q ) − (2.5 − Q )
⎝ 2 ⎠
]
[
⎛ 80 ⎞
= ⎜ ⎟(2 ) 2(2.5 − 0.4 )Q − Q 2
⎝ 2 ⎠
(
2(2.1 − Q )(2.5 − Q ) − (2.5 − Q ) = 2 4.2Q − Q 2
2
(
2 5.25 − 2.1Q − 2.5Q + Q
2
]
)
) − (6.25 − 5Q + Q ) = 8.4Q − 2Q
2
2
We find 3Q 2 − 12.6Q + 4.25 = 0 ⇒ Q = 0.370 V
______________________________________________________________________________________
16.92
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
______________________________________________________________________________________
16.93
Design Problem
______________________________________________________________________________________
16.94
Design Problem
______________________________________________________________________________________
16.95
Design Problem
______________________________________________________________________________________
16.96
(a) Quantization error
Or LSB ≤ 0.10 V
For a
(b)
6-bit word , LSB =
1 − LSB =
=
1
LSB ≤ 1% ≤ 0.05 V
2
5
= 0.078125 V
64
5
= 0.078125 V
64
3.5424
× 64 = 45.34 ⇒ n = 45
5
(c)
Digital Output = 101101
45 × 5
= 3.515625
64
Δ = 3.5424 − 3.515625 = 0.026775 <
1
LSB.
2
.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
16.97
(a) Quantization error =
1 − LSB = 0.033 V
1
LSB ≤ 0.5% ≤ 0.0165 V
2
For a 7-bit word, LSB =
3.3
= 0.0258 V
128
(b) 1-LSB = 0.02578125 V
2.5321
× 128 = 98.2 ⇒ n = 98
(c)
3.3
Digital output = 1100010
(98)(3.3) = 2.5265625 V
Now
128
1
LSB
2
______________________________________________________________________________________
Δ = 2.5321 − 2.5265625 = 0.0055375 V <
16.98
⎛0 1 1 0 ⎞
(a) υ O = ⎜ + + + ⎟(5) = 1.875 V
⎝ 2 4 8 16 ⎠
⎛1 0 0 1 ⎞
(b) υ O = ⎜ + + + ⎟(5) = 2.8125 V
⎝ 2 4 8 16 ⎠
______________________________________________________________________________________
16.99
(a)
⎛1⎞
LSB = ⎜ ⎟ ( 5 ) = 0.3125 V
⎝ 16 ⎠
1
LSB = 0.15625 V
2
⎛ 10 ⎞
vo = ⎜
⎟ (5)
⎝ 20 + ΔR1 ⎠
Now
For vo = 2.5 + 0.15625 = 2.65625 V
(10 )( 5 )
⇒ ΔR1 = −1.176 K
2.65625
vo = 2.5 − 0.15625 = 2.34375 V
20 + ΔR1 =
For
20 + ΔR1 =
(10 )( 5)
2.34375 V
ΔR1 = +1.333 K
For
ΔR1 = 1.176 K ⇒ ΔR1 = 5.88%
(b)
⎛
⎞
10
R4 : vo = ⎜
⎟ ( 5)
⎝ 160 + ΔR4 ⎠
For
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
vo = 0.3125 + 0.15625 = 0.46875 V
(10 )( 5)
⇒ ΔR4 = −53.33K
0.46875
Or vo = 0.3125 − 0.15625 = 0.15625 V
160 + ΔR4 =
160 + ΔR4 =
(10 )( 5)
0.15625
⇒ ΔR4 = 160 K
ΔR = 53.33K ⇒ ΔR = 33.33%
4
4
For
______________________________________________________________________________________
16.100
R5 = 320 kΩ
R6 = 640 kΩ
R7 = 1280 kΩ
(a)
R8 = 2560 kΩ
⎛ 10 ⎞
vo = ⎜
⎟ ( 5 ) = 0.01953125 V
⎝ 2560 ⎠
(b)
______________________________________________________________________________________
16.101
(a)
VREF −5
=
⇒ I1 = −0.50 mA
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
I5
I 6 = = −0.015625 mA
2
Δvo = I 6 RF = ( 0.015625 )( 5 )
I1 =
(b)
(c)
(d)
Δvo = 0.078125 V
vo = − [ I 2 + I 5 + I 6 ] RF = [ 0.25 + 0.03125 + 0.015625] ( 5 )
vo = 1.484375 V
For 101010;
vo = ( 0.50 + 0.125 + 0.03125 )( 5 ) = 3.28125 V
v = ( 0.25 + 0.0625 + 0.015625 )( 5 ) = 1.640625 V
For 010101; o
Δvo = 1.640625 V
______________________________________________________________________________________
16.102
1
⎛V
⎞⎛ R ⎞ 3.3
LSB = ⎜⎜ REF ⎟⎟⎜ ⎟ =
= 0.20625 V
2
⎝ 8 R ⎠⎝ 2 ⎠ 16
Ideal
5
⎛ 5V
⎞
υ A for 101 ⇒ ⎜⎜ REF ⎟⎟(R ) = (3.3) = 2.0625 V
R
8
8
⎝
⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
Range of υ A = 2.0625 ± LSB
2
1.85625 ≤ υ A ≤ 2.26875 V
______________________________________________________________________________________
16.103
6
6-bits ⇒ 2 = 64 resistors
26 − 1 = 63 comparators
______________________________________________________________________________________
16.104
(a) 10- bit output ⇒ 1024 clock periods
=
1 clock period
1
1
=
= 1 μS
f 106
Maximum 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
v A = v′A ± LSB
2
So range of
≤ v A ≤ 0.715332031 V
0.710449219
⇒ 256 + 32 + 4 = 292 clock pulses
(c) 0100100100
______________________________________________________________________________________
16.105
(a)
3.125 =
N ×5
⇒ N = 640 ⇒ 512 + 128
1024
Output = 1010000000
(b)
N ×5
⇒ N = 381.19 ⇒ N = 381 ⇒ 256 + 64 + 32 + 16 + 8 + 4 + 1
1024
Output = 0101111101
1.8613 =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 17
17.1
0 − (− 0.2 )
= 2kΩ
0.1
(b) (i) υ1 = −1 V, Q1 off, Q 2 on
(a) R C =
υ O 2 = 0 − (0.2 )(2) = −0.4 V
υ O1 = 0
(ii) υ1 = −0.4 V, Q1 on, Q 2 off
υ O1 = 0 − (0.2 )(2) = −0.4 V
υ O2 = 0
(c) For (i) and (ii)
P = I Q 0 − V − = (0.2 )(1.8) = 0.36 mW
( )(
)
______________________________________________________________________________________
17.2
(a) i E =
− 1 − 0.7 − (− 2.5)
= 0.08 mA, ⇒ R E = 10 k Ω
RE
0 − (− 0.25)
= 6.25 k Ω
0.04
(b) (i) υ1 = −1.3 V, Q1 off, Q 2 on
RC =
− 1 − 0.7 − (− 2.5)
= 0.08 mA
10
υ O 2 = 0 − (0.08)(6.25) = −0.50 V
iE =
υ O1 = 0
(ii) υ1 = −0.7 V, Q1 on, Q 2 off
− 0.7 − 0.7 − (− 2.5)
= 0.11 mA
10
υ O1 = 0 − (0.11)(6.25) = −0.6875 V
iE =
υ O2 = 0
(c) (i) i E = 0.08 mA, P = (0.08)(2.5) = 0.2 mW
(ii) i E = 0.11 mA, P = (0.11)(2.5) = 0.275 mW
______________________________________________________________________________________
17.3
iC 2 = I Q = 0.5 =
3−0
⇒ RC 2 = 6 K
RC 2
iC1 = I Q = 0.5 =
3 −1
⇒ RC1 = 4 K
RC1
(a)
(b)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
iC1
=
IQ
=
(c)
So
⎛V ⎞
I S exp ⎜ BE1 ⎟
⎝ VT ⎠
⎡
⎛V ⎞
⎛ V ⎞⎤
I S ⎢exp ⎜ BE1 ⎟ + exp ⎜ BE 2 ⎟ ⎥
V
⎝ T ⎠
⎝ VT ⎠ ⎦
⎣
1
⎛ VBE 2 − VBE1 ⎞
1 + exp ⎜
⎟
VT
⎝
⎠
vI = VBE1 − VBE 2
iC1
=
IQ
1
⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠
0.1
=
0.5
1
= 0.2
⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠
⎛ −v ⎞ 1
−1 = 4
exp ⎜ I ⎟ =
⎝ VT ⎠ 0.2
( −vI ) = ( 0.026 ) ln (4)
vI = −0.0360 V
______________________________________________________________________________________
17.4
(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
______________________________________________________________________________________
17.5
(a)
Q2 on, vE = −1.2 − 0.7 = −1.9 V
iE = iC 2 =
−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
(b)
Q1 on, vE = −0.7 − 0.7 = −1.40 V
iE = iC1 =
−1.4 − ( −5.2 )
= 1.52 mA
2.5
v1 = −1V = −iC1 RC1 = − (1.52 )( RC1 )
RC1 = 0.658 k Ω
(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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d) (i) For vin = −0.7V , iE = 1.52 mA
iC 4 =
iC 3 =
−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 )
or
P = 21.8 mW
(ii) For vin = −1.7V , iE = 1.32 mA
iC 4 =
iC 3 =
−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 )
P = 20.7 mW
or
______________________________________________________________________________________
17.6
a.
I3 =
3.7 − 0.7
= 1.5 mA
0.67 + 1.33
VR = I 3 R4 + Vγ = (1.5)(1.33) + 0.7
or
VR = 2.70 V
b.
logic 1 level
= 3.7 − 0.7 ⇒ 3.0 V
For v X = vY = logic 1.
3 − 0.7
= 2.875 mA = iRC1
0.8
vB 3 = 3.7 − ( 2.875 )( 0.21) = 3.10 V
⇒ v01 ( logic 0 ) = 2.4 V
iE =
For v X = vY = logic 0, QR on
2.7 − 0.7
= 2.5 mA = iRC 2
0.8
vB 4 = 3.7 − ( 2.5 )( 0.24 ) = 3.1 V
⇒ v02 ( logic 0 ) = 2.4 V
iE =
______________________________________________________________________________________
17.7
0.7 − 0.7 − (− 2.1)
= 10.5 k Ω
0.20
0 − (− 2.1)
(b) R5 = R 6 =
= 17.5 k Ω
0.12
(c) I Q = I REF = 0.20 mA
(a) R1 =
υ O1 = −0.7 V, υ C 2 = 0
R C1 =
0. 7 − 0
= 3. 5 k Ω
0. 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d) I Q = I REF = 0.20 mA
υ O 2 = −0.7 V, υ CR = 0
0.7 − 0
= 3.5 k Ω
0.20
______________________________________________________________________________________
RC 2 =
17.8
V R = −0.5 V
iE =
R5 =
− 0.5 − 0.7 − (− 3)
= 0.4 mA, R E = 4.5 k Ω
RE
− 0.5 − (− 3)
= 6.25 k Ω
0. 4
V B 2 = −0.5 + 0.7 = 0.2 V, R1 =
0. 7 − 0. 2
= 1.25 k Ω
0. 4
0.2 − 0.7 − 0.7 − (− 3)
= 4. 5 k Ω
0.4
0 − (− 3)
R3 = R 4 =
= 3.75 k Ω
0. 8
υ OR = −1 V, ⇒ υ CR = −0.3 V
R2 =
0.7 − (− 0.3)
= 2. 5 k Ω
0.4
− 0.7 − (− 3)
υ I = 0 , iE =
= 0.511 mA
4.5
υ C1 = −1 + 0.7 = −0.3 V
RC 2 =
0.7 − (− 0.3)
= 1.957 k Ω
0.511
______________________________________________________________________________________
RC1 =
17.9
υ O = logic 1 = 1.8 V, logic 0 = 1.2 V
For υ I = logic 1 = 1.8 V
i E = 0.8 =
1.8 − 0.7
⇒ R E = 1.375 k Ω
RE
2.5 − 1.9
= 0.75 k Ω
0.8
1.5 − 0.7
0.8
For υ I = logic 0, Q R on; i E =
=
= 0.5818 mA
1.375
RE
υ C1 = 1.2 + 0.7 = 1.9 V, ⇒ RC1 =
υ CR = 1.2 + 0.7 = 1.9 V
2. 5 − 1. 9
= 1.031 k Ω
0.5818
1 .8
R 2 = R3 =
= 2.25 k Ω
0 .8
______________________________________________________________________________________
RC 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.10
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
IE5 =
5 − 0.7
⇒ I E1 = 0.239 mA
18
IE3 = 0
I E1 =
5 − 0.7
⇒ I E 5 = 1.72 mA
2.5
Y = 0.7 V
IE5 =
(b)
(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).
______________________________________________________________________________________
17.11
(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
A = B = Logic 0 = −2.7 V , QR on,
(c)
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)
iC 2 =
= 3 mA
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
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.12
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 =
5 − 0.7
⇒ iE1 = 1.65 mA
2.6
5 − (0.7 + 0.7)
iE 2 =
⇒ iE 2 = 3 mA
1.2
iC 3 = 0, iC 2 = iE 2 = 3 mA
iE1 =
V2 = 0
c.
5 − (1.8 + 0.7)
⇒ iE1 = 0.962 mA
2.6
5 − (1.6 + 0.7)
iE 2 =
⇒ iE 2 = 2.25 mA
1.2
iC 2 = 0, iC 3 = iE 2 = 2.25 mA
iE1 =
V = 1.8 V
2
d.
______________________________________________________________________________________
17.13
(a) υ R = (logic 0 + logic 1)/2 − 0.7 =
(b) For υ X = υ Y = logic 1 = 3.5 V
υ E1 = 3.5 − 0.7 − 0.7 = 2.1 V
3.5 + 3.1
− 0.7 = 2.6 V
2
2.1 − 0
= 0.175 mA
12
0.4
1
⇒ RC1 = 6.86 k Ω
i RC1 = ⋅ i E1 = 0.05833 mA =
3
R C1
i E1 =
(c) For υ X = υ Y = logic 0 = 3.1 V
υ E1 = V R − 0.7 = 2.6 − 0.7 = 1.9 V
1.9
= 0.1583 mA
12
0.4
1
⇒ RC 2 = 7.58 k Ω
i RC 2 = ⋅ i E 2 = 0.05277 mA =
3
RC 2
iE2 =
(d) For υ X = υ Y = logic 0 = 3.1 V
i E = 0.1583 mA
3.1 − 0.7
= 0.3 mA
8
P = (0.1583 + 0.3)(3.5) = 1.60 mW
______________________________________________________________________________________
i R1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.14
Assume Vγ = 0.4 V
(a) Logic 1 = 0.2 V, Logic 0 = −0.2 V
0 − 0.7 − (− 3.10 )
(b) i E =
= 0.25 ⇒ R E = 9.6 k Ω
RE
(c) i D1 + i R1 = i E
2i R1 + i R1 = 3i R1 = 0.25 ⇒ i R1 = 0.08333 mA
0. 4
= 4. 8 k Ω
0.08333
0.2 − 0.7 − (− 3.10 )
(d) i E =
= 0.2708 mA
9. 6
0.4 0.4
i R2 =
=
= 0.0833 mA
R 2 4.8
R1 =
i D 2 = i E − i R 2 = 0.2708 − 0.0833 = 0.1875 mA
(e) i E = 0.2708 mA
− 0.2 − (− 3.10 )
= 0.8788 mA
3.3
0.2 − (− 3.10 )
i R3 =
= 1.0 mA
3. 3
P = (i E + i R 3 + i R 4 )[0.9 − (− 3.10 )] = (0.2708 + 1.0 + 0.8788)(4 ) = 8.6 mW
______________________________________________________________________________________
iR4 =
17.15
a.
i1 =
i3 =
i4 =
( −0.9 − 0.7 ) − ( −3)
1
( −0.2 − 0.7 ) − ( −3)
15
( −0.2 − 0.7 ) − ( −3)
⇒ i1 = 1.4 mA
⇒ i3 = 0.14 mA
⇒ i4 = 0.14 mA
15
i2 + iD = i1 + i3 = 1.4 + 0.14 = 1.54 mA
0.4
i2 =
⇒ i2 = 0.8 mA
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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
d.
(0 − 0.7 − 0.7) − (−3)
⇒ i1 = 1.6 mA
1
(0 − 0.7) − ( −3)
i3 =
⇒ i3 = 0.153 mA
15
i4 = i3 ⇒ i4 = 0.153 mA
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
______________________________________________________________________________________
17.16
(a) (i) A = B = C = D = 0 V, Q1 , Q2 , Q3 , Q 4 off; Q5 , Q6 on
2.5 = i R1 (2 ) + 0.7 +
1 ⎛ i R1 ⎞
⎜
⎟(15)
2 ⎜⎝ 91 ⎟⎠
1.8 = i R1 (2 + 0.0824) ⇒ i R1 = 0.8644 mA
Y = 2.5 − i R1 (2 ) ⇒ Y = 0.771 V
(ii) A = B = 0 V, C = D = 2.5 V; Q1 , Q 2 , Q6 off; Q3 , Q 4 , Q5 on
⎛i ⎞
2.5 = i R1 (2 ) + 0.7 + ⎜⎜ R1 ⎟⎟(15)
⎝ 91 ⎠
1.8 = i R1 (2 + 0.1648) ⇒ i R1 = 0.8315 mA
Y = 2.5 − i R1 (2) ⇒ Y = 0.837 V
(iii) A = C = 2.5 V, B = D = 0 V; Q1 , Q3 on, Q5 , Q6 off
i R1 = 0 ⇒ Y = 2.5 V
(b) Y=(A OR B) AND (C OR D)
(c) (i) P = i R1 (2.5) = (0.8644 )(2.5) = 2.16 mW
2.5 − 0.7
= 0.12 mA
15
P = (i R1 + i R 3 )(2.5) = (0.8315 + 0.12 )(2.5) = 2.38 mW
(ii) i R 3 =
2.5 − 0.7
= 0.12 mA
15
P = (i R 2 + i R 3 )(2.5) = (0.24 )(2.5) = 0.60 mW
______________________________________________________________________________________
(iii) i R 2 = i R 3 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.17
a.
b.
logic 1 = 0 V
logic 0 = −0.4 V
v01 = A OR B
v02 = C OR D
v03 = v01 OR v02
or
v03 = ( A OR B ) AND (C OR D)
______________________________________________________________________________________
17.18
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.
b.
P = ( I DC + 0.5I DC + 0.1I DC + 0.1 I DC )( 3)
P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 μ W
______________________________________________________________________________________
17.19
(a) (i) For υ I = 0.1 V
υ1 = 0.8 V
2.5 − 0.8
= 0.1417 mA
12
i 2 = i 3 = 0 , υ O = 2 .5 V
i1 =
(ii) For υ I = 2.5 V
υ1 = 0.7 + 0.8 = 1.5 V
2.5 − 1.5
= 0.0833 mA
12
υ O = 0.1 V
i1 = i 2 =
2.5 − 0.1
= 0.20 mA
12
(b) (i) υ1 = 0.7 + 0.7 = 1.4 V
i3 =
υ I = υ1 − 0.7 = 0.7 V
(ii) υ1 = 0.7 + 0.8 = 1.5 V
υ I = υ1 − 0.7 = 0.8 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.20
(a)
vI = 0 ⇒ V1 = 0.7 V
3.3 − 0.7
= 0.433 mA
6
iB = iC = 0
vo = 3.3 V
i1 =
(b)
vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V
3.3 − 1.5
= 0.3 mA
i1 =
6
0.8
= 0.04 mA
iR =
20
iB = 0.3 − 0.04 = 0.26 mA
3.3 − 0.1
iC =
= 0.8 mA
4
vo = 0.1 V
______________________________________________________________________________________
17.21
i.
For
i1 =
ii.
vX = vY = 0.1 V ⇒ v′ = 0.8 V
5 − 0.8
⇒ i1 = 0.525 mA
8
i3 = i4 = 0
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 =
______________________________________________________________________________________
17.22
(i) υ X = υ Y = 0.1 V
υ ′ = 0 .8 V
3.3 − 0.8
i1 =
= 0.3125 mA
8
i3 = i 4 = 0
(ii) υ X = υ Y = 3.3 V
υ ′ = 0 . 8 + 0 .7 + 0 . 7 = 2 . 2 V
3. 3 − 2. 2
i1 =
= 0.1375 mA
8
0. 8
i 4 = 0.1375 −
= 0.08417 mA
15
3.3 − 0.1
i3 =
= 1.333 mA
2. 4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
17.23
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 =
⇒ i2 = 1.7 mA
2
i4 = iB1 + i2 ⇒ i4 = 2.375 mA
i1 =
0.8
⇒ i5 = 0.08 mA
10
iB 2 = i4 − i5 ⇒ iB 2 = 2.295 mA
i5 =
i3 =
5 − 0.1
⇒ i3 = 1.225 mA
4
v0 = 0.1V
iL′
5 − (0.1 + 0.7)
= 1.05 mA
4
iC (max) = β iB 2 = NiL′ + i3
=
b.
(20)(2.295) = N (1.05) + 1.225
So
N = 42
______________________________________________________________________________________
17.24
(a) υ1 = 0.8 + 0.7 + 0.8 = 2.3 V
3.3 − 2.3
= 0.25 mA = i B1
4
υ C1 = 0.8 + 0.7 + 0.1 = 1.6 V
i1 =
3.3 − 1.6
= 0.85 mA
2
i 4 = i B1 + i 2 = 0.25 + 0.85 = 1.10 mA
i2 =
0. 8
= 0.08 mA
10
i B 2 = i 4 − i 5 = 1.10 − 0.08 = 1.02 mA
i5 =
3.3 − 0.1
= 0.80 mA
4
(b) iCo (max ) = β i B 2 = i 3 + N i L′
i3 =
3.3 − (0.1 + 0.7 )
= 0.625 mA
4
(20)(1.02) = 0.8 + N (0.625) ⇒ N = 31.36 ⇒ N = 31
______________________________________________________________________________________
i L′ =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.25
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
2[
So
β
= 25
Assume
5 − 2.2 = i (1 + β ) R1 + R2 ]
5 − 2.2
⇒ i2 = 0.0589 mA
(26)(1.75) + 2
i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA
i2 =
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
______________________________________________________________________________________
17.26
(a)
vI = 0 V, Q1 forward active
5 − 0.7
= 0.717 mA
6
iC = (25)(0.71667) = 17.9 mA
iE = (26)(0.71667) = 18.6 mA
iB =
(b)
VI = 0.8 V
iB =
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
6
iE = − β R iE = −(0.5)(0.583) = −0.292 mA
iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA
iB =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.27
(a) (i) For υ I = 0.1 V, υ1 = 0.1 + 0.8 = 0.9 V, and υ O = 2.5 V
2.5 − 0.9
= 0.1333 mA
12
i 2 = i3 = 0
i1 =
(ii) For υ I = 2.5 V, υ1 = 0.8 + 0.7 = 1.5 V, and υ O = 0.1 V
2.5 − 1.5
= 0.0833 mA
12
i 2 = i1 (1 + 0.1) = (0.0833)(1.1) = 0.09167 mA
i1 =
2.5 − 0.1
= 0.20 mA
12
(b) (i) υ1 = 0.7 + 0.7 = 1.4 V
i3 =
υ I = 1.4 − 0.8 = 0.6 V
(ii) υ1 = 0.8 + 0.7 = 1.5 V
υ I = 1.5 − 0.8 = 0.7 V
______________________________________________________________________________________
17.28
a.
i.
i1 =
ii.
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
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 =
⇒ i4 = 0.533 mA
1.5
i1 =
iB 3 = ( iB 2 + i2 ) − i4 = 0.45 + 2.05 − 0.533
or
iB 3 = 1.97 mA
5 − 0.1
⇒ i3 = 2.23 mA
2.2
For Q3 :
i3 =
b.
i3
2.23
=
= 1.13 < β
iB 3 1.97
For Q2 :
i2
2.05
=
= 4.56 < β
iB 2 0.45
I /I < β
, then each transistor is in saturation.
Since ( C B )
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.29
(a) (i) υ X = 0.1 V, υ Y = 3.3 V
υ ′ = 0 . 1 + 0 .7 = 0 . 8 V
3.3 − 0.8
ii =
= 0.156 mA
16
i3 = i 4 = 0
(ii) υ X = υ Y = 3.3 V
υ ′ = 0 . 8 + 0 .7 + 0 . 7 = 2 . 2 V
3. 3 − 2. 2
i1 =
= 0.06875 mA
16
0. 8
i 4 = 0.06875 −
= 0.02875 mA
20
3.3 − 0.1
i3 =
= 0.5333 mA
6
(b) i C1 (max ) = β i 4 = i 3 + N i L′
3.3 − (0.1 + 0.7 )
= 0.15625 mA
16
(50)(0.02875) = 0.5333 + N (0.15625) ⇒ N = 5.8 ⇒ N = 5
i L′ =
(c) iC1 (max ) = β i 4 = (50)(0.02875) = 1.44 mA < 5 mA, ⇒ N = 5
______________________________________________________________________________________
17.30
For v X = vY = 5 V, Q, in inverse active mode.
a.
5 − ( 0.8 + 0.8 + 0.7 )
iB1 =
= 0.45 mA
6
iB 2 = iB1 + 2 β R iB1 = 0.45(1 + 2 [ 0.1]) = 0.54 mA
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
iB3 = 2.06 mA
Now
iL′ =
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
′
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 ) ⎜
⎟ =
2
⎝ R2 ⎠
= 1.05 mA
So
I L (max) = NI L′
or 1.05 = N (0.045)
⇒ N = 23
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
17.31
(a) (i) Vin = 0.1 V
5 − (0.1 + 0.8)
= 1.025 mA
4
i RCP = i Bo = 0 , Vout = 5 V
i RB =
(ii) Vin = 5 V
5 − (0.7 + 0.8 + 0.7 )
= 0.70 mA
4
V out = 0.7 + 0.1 = 0.8 V
i RB =
5 − 0.8
= 4.2 mA
1
i BS = (1.1)(0.7 ) = 0.77 mA
i RCP =
i Co = β i Bo , iCS = 4.2 − β i Bo , i ES = 0.77 + (4.2 − β i Bo )
0.7
1
i Bo = 0.77 + 4.2 − β i Bo − 0.7
i Bo = i ES −
(1 + β )i Bo = 4.27 ⇒ i Bo = 4.27 = 0.0837 mA
51
(b) (i) Vin = 0.1 V, V out = High,
i L′ = 5β R i RB = 5(0.1)(0.7 ) = 0.35 mA
V out = 5 − (0.35)(1) = 4.65 V
P = (5 − 0.1)(1.025) + (0.35)(5 − 4.65) = 5.145 mW
(ii) i L = 5(1.025) = 5.125 mA
P = (0.77 + 4.2 )(5) + (5.125)(0.1) = 25.4 mW
______________________________________________________________________________________
17.32
v X = vY = vZ = 0.1 V
a.
iB1 =
5 − (0.1 + 0.8)
⇒ iB1 = 1.05 mA
3.9
Then
iC1 = iB 2 = iC 2 = iB 3 = iC 3 = 0
v X = vY = vZ = 5 V
b.
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 3 = iB 2 + iC 2 −
= 1.73 + 2.05 − 1.0
0.8
⇒ iB 3 = 2.78 mA
iC 2 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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 =
______________________________________________________________________________________
17.33
5
= 0.098 μ A
51
V B 5 = 5 − (0.000098)(2) ≅ 5 V
(a) (i) I L′ = 5 μ A, i B 4 =
υ O = 3 .6 V
5
= 0.098 mA
51
V B 5 = 5 − (0.098)(2) = 4.804 V
(ii) I L′ = 5 mA, i B 4 =
υ O = 4.804 − 1.4 = 3.404 V
(iii) Q 4 in saturation
5 − V B 5 − (V E + 0.8)
=
2
2
5 − VC 5 − (V E + 0.1)
IC =
=
0.13
0.13
5 − (V E + 0.7 ) 5 − (V E + 0.1)
I B + I C = I E = I L = 25 =
+
2
0.13
25 = 2.10 − V E (0.5) + 37.69 − V E (7.692) ⇒ V E = 1.81 V
IB =
υ O = 1.81 − 0.7 = 1.11 V
(b) V B 4 = 0.7 + 0.8 = 1.5 V
I B4 =
5 − 1.5
= 1.75 mA
2
VC 4 = 0.7 + 0.1 = 0.8 V, I C 4 =
5 − 0.8
= 32.31 mA
0.13
I L = 1.75 + 32.31 = 34.06 mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.34
v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode.
a.
2.8 − (0.8 + 0.8 + 0.7)
⇒ iB1 = 0.25 mA
2
iB 2 = iB1 (1 + 3β R ) = 0.25(1 + 3 [0.3])
iB1 =
⇒ iB 2 = 0.475 mA
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
⇒ iB 3 = 4.235 mA
iB 3 = iB 2 + iC 2 −
v X = vY = vZ = 0.1 V
b.
iB1 =
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
iB 4 =
5iL′
5(0.075)
=
⇒ iB 4 = 0.00371 mA
1+ βF
101
______________________________________________________________________________________
17.35
v X = vY = vZ = 0.1 V
a.
iB1 =
2 − (0.1 + 0.8)
+ iB 3
RB1
where
iB 3 =
(2 − 0.7) − (0.9) 0.4
=
RB 2
1
⇒ 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
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
17.36
IB + ID =
(a)
IC − I D =
5.8 − 0.7
= 0.51 mA
10
5 − (0.7 − 0.3)
= 4.6 mA
1
Now
I
I
I D = 0.51 − I B = 0.51 − C = 0.51 − C
50
β
Then
So
I ⎞
1 ⎞
⎛
⎛
I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜1 + ⎟ − 0.51 = 4.6
50
50
⎝
⎠
⎝
⎠
I C = 5.01 mA
IC
5.01
⇒ I B = 0.1002 mA
50
I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA
IB =
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
IC =
⇒ I C = 4.9 mA
1
IB =
______________________________________________________________________________________
17.37
(a) (i) υ I = 0 , υ1 = 0.3 V
1 .5 − 0 .3
= 1.2 mA
1
i B = i C = 0 , υ O = 1.5 V
i1 =
(ii) υ I = 1.5 V, υ1 = 0.7 + 0.3 = 1.0 V
1.5 − 1.0
= 0.5 mA
1
0. 7
i B = 0.5 −
= 0.465 mA
20
i1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.5 − 0.4
= 0.9167 mA, υ O = 0.4 V
1.2
(b) (i) υ1 = 0.7 + 0.3 = 1.0 V, υ I = 0.7 V
iC =
i B = iC = 0
(ii) υ1 = 1.0 V, υ I = 0.7 V
1.5 − 0.4
= 0.9167 mA
1.2
i
0.9167
iB = C =
= 0.03667 mA
25
β
(c) iCo (max ) = β i B = iC + N i L′
iC =
1.5 − (0.4 + 0.3) 0.4
−
= 0.78 mA
1
20
(25)(0.465) = (0.9167) + N (0.78) ⇒ N = 13.7 ⇒ N = 13
______________________________________________________________________________________
i L′ =
17.38
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
vB 2 = 0.4 + 0.4 ⇒ vB 2 = 0.8 V
iB1 =
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 =
⇒ iB 4 = 0.0394 mA
0.76 + (31)(3.5)
iC 4 = β F iB 4 ⇒ iC 4 = 1.18 mA
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
iB 2 = iB1 (1 + 2 β R ) = 1.18(1 + 2 [0.1])
iB 2 = 1.41 mA
iB1 =
1.1 − 0.7
⇒ iB 4 = 0.00369 mA
(31)(3.5)
5 − 1.1
iR 2 =
= 5.13 mA ⇒ iC 2 ≈ 5.13 mA
0.76
iB 0 ≈ iB 2 + iC 2
iB 0 = 6.54 mA
iB 4 =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.39
(a) For the load, i RB1 =
2.5 − (0.4 + 0.3) 1
= (0.2 ) ⇒ R B1 = 18 k Ω
R B1
2
For υ X = υ Y = υ Z = logic 1
2.5 − (0.7 + 0.8)
= 0.05556 mA
18
2.5 − (0.7 + 0.1)
i RC1 =
R C1
i RB1 =
i B 2 = 0.1 = 0.05556 +
1.7 0.7
−
⇒ RC1 = 1.63 k Ω
RC1 0.7
(b) υ X = 0.4 V, υ B1 = 0.7 V, υ B 2 = 0
υ O ≅ 2.5 − 0.7 = 1.8 V
All i B = 0 , All iC = 0
(c) υ B1 = 1.5 V, υ B 2 = 0.7 V
2.5 − 1.5
= 0.0556 mA
18
2.5 − (0.7 + 0.1)
i C1 =
= 1.043 mA
1.63
i B 2 = 0.10 mA
i B1 =
4[2.5 − (0.4 + 0.3)]
= 0.40 mA
18
υ O = 0 .4 V
(d) iC 2 (max ) = β i B 2 = N i L′
i L′ = 0.1 mA
iC 2 =
(20)(0.1) = N (0.1) ⇒ N = 20
______________________________________________________________________________________
17.40
a.
For v X = vY = 3.6 V
5 − 2.1
= 0.29 mA
10
5 − 1.8
= 0.32 mA
vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 =
10
1.4
iB 2 = iB1 + iC1 −
= 0.29 + 0.32 − 0.0933
15
vB1 = 3(0.7) = 2.1 ⇒ iB1 =
So
iB 2 = 0.517 mA
vC 2 = 0.7 + 0.4 = 1.1 V
iC 2 =
5 − 1.1
= 0.951 mA
4.1
iB 5 = iB 2 + iC 2 −
or iB 5 = 1.293 mA
′
0.7
= 0.517 + 0.951 − 0.175
4
For v0 = 0.4 V, vB1 = 0.4 + 0.7 = 1.1 V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
1.1 − 0.7
= 0.00086 mA
(31)(15)
5 − 1.1
− 0.00086 or iL′ ≈ 0.39 mA
iL′ =
10
iC 5 (max) = β iΒ 5 = NiL′
iB′ 1 =
So
(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.)
______________________________________________________________________________________
17.41
a.
Assume no load. For v X = logic 0 = 0.4 V
iE1 =
5 − (0.4 + 0.7)
= 0.0975 mA
40
Essentially all of this current goes to ground from VCC .
P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW
b.
iR1 =
5 − (3)(0.7)
= 0.0725 mA
40
5 − (0.7 + 0.7 + 0.4)
= 0.064 mA
50
5 − (0.7 + 0.4)
iR 3 =
= 0.26 mA
15
P = (0.0725 + 0.064 + 0.26)(5)
iR 2 =
P = 1.98 mW
c.
For v0 = 0, vC 7 = 0.7 + 0.4 = 1.1 V
iR 7 =
5 − 1.1
⇒ iR 7 = 78 mA ≈ iSC
0.050
______________________________________________________________________________________
17.42
3 − (0.7 + 0.3)
= 1.0 mA
2
2.4 − 0.7
(b) i E =
= 0.85 mA
2
3 − 2. 4
RC =
= 0.706 k Ω
0.85
(c) (i) P = (1.0 )(3) = 3.0 mW
(a) i E =
(ii) P = (0.85)(3) = 2.55 mW
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
17.43
(a)
v I = v O = 2.5 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
VC =
i ⋅t
(V )(C )
1
iC1dt = C1 ⇒ t = C
∫
C
C
iC1
(5)(15 × 10−12 )
⇒ t = 5.19 ns
14.45 × 10−3
(5)(15 × 10−12 )
t=
⇒ t = 260 ns
0.289 × 10−3
t=
(c)
______________________________________________________________________________________
17.44
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) Assume R1 = R2 = 10 kΩ; β = 50
Then
iR1 = iR 2 =
0.7
= 0.07 mA
10
NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V
iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 )
2
2
iDN = 0.10 mA
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
i
So DP = 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
v=
(b)
1
i ⋅t
idt =
C∫
C
Cv (15 × 10−12 )(5)
=
i
1.53 × 10−3
t = 49 ns
t=
(c)
Cv (15 × 10−12 )(5)
=
i
0.1× 10−3
t = 0.75 μ s
t=
______________________________________________________________________________________