mh00260 cosmos ch 11 ps - pennsylvania state...
TRANSCRIPT
Chapter 11 Problem Solutions 11.1 (a)
( )0.7 30.1 23 K
3 1.5 0.05 30 K
EE
CC
RR
RR
− − −= ⇒ =
− = ⇒ =
(b) ( ) ( )2 2 26 2 6 76CE C C E Cv i R R i= − + = −
(c) ( ) 2 2max 0 0.7 Vcm CB CEv v v⇒ = ⇒ =
So ( )2 20.7 6 76 69.74 AC Ci i μ= − ⇒ =
( )( ) ( ) ( ) ( )max 0.7 3
2 0.06974 max 0.908 V23
CMCM
vv
− − −= ⇒ =
( ) ( )min 3 V min 2.3 VCM S CMv V v⇒ = − ⇒ = −
11.2
180, 85 dB
18017,783 0.01012
d dB
dcm
cm cm
A C M RR
AC M RR AA A
= =
= = = ⇒ =
Assume the common-mode gain is negative.
( ) ( )( )
0
0
0
180 0.01012180 2sin mV 0.01012 2sin V0.36sin 0.02024sin
d d cm cm
d cm
v A v A vv v
v t tv t t
ω ωω ω
= += −= −= −
Ideal Output: ( )0 0.360sin Vv tω=
Actual Output: ( )0 0.340sin Vv tω= 11.3 a.
( )
( ) ( )( )
( ) ( )( )( ) ( )
1 1
12 2
4 4
2 2
4 4
10 2 0.71.01 mA
8.51.01 1.01 mA
2 21 11 100 101
100 1.01 0.50 mA101 20 0.7 5 4.3 V
5 0.5 2 0.7 4.7 V
C C
C C
CE CE
CE CE
I I
II I
I I
V V
V V
β β
−= ⇒ =
= = ⇒ ≅+ +
+
⎛ ⎞⎛ ⎞= ⇒ ≅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= − − − ⇒ =
= − − − ⇒ =⎡ ⎤⎣ ⎦
b.
( ) ( )( )
( )
4 4
4 4
2 4 2
1 2
1 1
For 2.5 V 0.7 2.5 1.8 V5 1.8 1.6 mA
21 1012 2 1.6 3.23 mA
1003.23 mA
10 2 0.72.66 k
3.23
CE C
C C
C C C
C
V V
I I
I I I
I I
R R
ββ
= ⇒ = − + =−= ⇒ =
⎛ ⎞+ ⎛ ⎞+ = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
≈ =
−= ⇒ = Ω
11.4 a. Neglecting base currents
1 3 1 1
1 1
30 0.7400 A R 73.25 k0.4
10 V V 9.3 V15 9.3 28.5 k
0.2
CE C
C C
I I R
V
R R
μ −= = ⇒ = ⇒ = Ω
= ⇒ =−= ⇒ = Ω
b. ( )( )
( )
( )( )( )
( )
( )
( )( )( )( )
0 3
0
10
100 0.02613 k
0.250 125 k0.4
We have100 28.5
622 2 13 10
12 1
1
100 28.5 1 0.1132 125 10113 10
113 10
6220log0.113
Cd d
B
Ccm
B
B
cm
dB
r
r Q
RA Ar R
RArr Rr R
A
C M RR C
π
π
π
π
β
ββ
= = Ω
= = Ω
= = ⇒ =+ +
⎧ ⎫⎪ ⎪⎪ ⎪= − ⎨ ⎬++ ⎪ ⎪+⎪ ⎪+⎩ ⎭
⎧ ⎫⎪ ⎪⎪ ⎪= − ⇒ = −⎨ ⎬+ ⎪ ⎪+⎪ ⎪+⎩ ⎭
⎛ ⎞= ⇒⎜ ⎟⎝ ⎠
54.8 dBdBM RR =
c.
( ) ( )
( )
( )( )
0
2 2 13 10 46 k
1 2 121 13 10 2 101 125 12.6 M2
id B id
icm B
icm
R r R R
R r R r
R
π
π β
= + = + ⇒ = Ω
= + + +⎡ ⎤⎣ ⎦
= + + ⇒ = Ω⎡ ⎤⎣ ⎦
11.5
(a) ( )max 0CM CBv V⇒ = so that ( ) ( ) ( ) ( )0.5max 5 5 8
2 2Q
CM C
Iv R= − = −
( )max 3 VCMv =
(b)
( ) ( )2
0.25 0.018 0.08654 mA2 2 0.026 2
0.08654 8 0.692 V
CQd dm
T
C C
IV VI g
VV I R
⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ = ⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
Δ = Δ ⋅ = =
(c)
( )( )2
0.25 0.010 0.04808 mA0.026 2
0.04808 8 0.385 VC
I
V
⎛ ⎞⎛ ⎞Δ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Δ = =
11.6
( )( )( )
( )
1 4
1 4 1 1 4
1 1
1 2
so 1.2 2 6 0.1
3 0.7 353
0.13 1For 1 1 40 0.05
C
C C
CM C C C C
P I I V V
I I I I I mA
R R k
v V V V V R R k
+ −= + −
≅ = ⇒ = =
− − −= ⇒ = Ω
−= + ⇒ = = ⇒ = ⇒ = Ω
One-sided output
( )( )
1 0.05 where 1.923 /2 0.026
Then1 1.923 40 38.52
d m C m
d d
A g R g mA V
A A
= = =
= ⇒ =
11.7 a.
( ) ( )
( ) ( )
1 2
1 2
1 2 1
1 2
0 0.7 2 85 52
5 0.7 0.050 mA85 1
100 0.0501 2 101 2
Or 0.0248 mA
5 100 0.7
So 3.22 V
EE
E E
EC C
C C
CE CE C
CE CE
I I
I I
II I
I I
V V I
V V
ββ
= + + −
−= ⇒ =+
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠= =
= = − − −⎡ ⎤⎣ ⎦= =
b. ( ) ( )1max for 0 and 5 100 2.52 Vcm CB C Cv V V I= = − =
( )So max 2.52 Vcmv =
( ) 1 2min for and cmv Q Q at the edge of cutoff ( )min 4.3 Vcmv⇒ = −
(c) Differential-mode half circuits
( )
( )( )
( )( )( )
( )( )( )( )
.2
11
Then2
1
1
12 1
100 0.026105 2
0.0248
Then100 1001 16.3
2 105 101 2
dm E
E
d
E
Co m C d
E
TE
CQ
d d
v VV g V R
r
V Rr
vV
Rr
Rv g V R Ar R
Vr k R kI
A A
ππ π
π
ππ
π
π
ππ
π
β
β
ββ
β
⎛ ⎞ ′− = + +⎜ ⎟⎝ ⎠
+⎡ ⎤′= +⎢ ⎥
⎣ ⎦
− /=
+⎡ ⎤′+⎢ ⎥
⎣ ⎦
= − ⇒ = ⋅′+ +
′= = = Ω = Ω
= ⋅ ⇒ =+
11.8 a. For 1 2 0v v= = and neglecting base currents
( )0.7 1062 k
0.15E ER R− − −
= ⇒ = Ω
b.
( )( )( )
( )( )( )
( )
( )( )( )( )
02
10
2
100 0.02634.7 k
0.075
100 5071.0
2 34.7 0.5
12 1
1
100 50 1 0.3982 62 10134.7 0.5
134.7 0.5
71.020log0.398
Cd
d B
T
CQ
d d
Ccm
EB
B
cm
dB dB
v RAv r R
VrI
A A
RARr Rr R
A
C M RR C M RR
π
π
π
π
β
β
ββ
= =+
= = = Ω
= ⇒ =+
⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥++
+⎢ ⎥+⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
= − ⎢ ⎥ ⇒ = −+ ⎢ ⎥+⎢ ⎥+⎣ ⎦
= ⇒ 45.0 dB=
c. ( )( )
22 34.7 0.5 70.4 k
id B
id id
R r RR R
π= += + ⇒ = Ω
Common-mode input resistance
( )
( )( )
1 2 121 34.7 0.5 2 101 62 6.28 M2
icm B E
icm
R r R R
R
π β= + + +⎡ ⎤⎣ ⎦
= + + ⇒ = Ω⎡ ⎤⎣ ⎦
11.9
(a)
( )( )
1 2
1 2
1 2
1 1.69 1.6 18.97 A
390
9.49 9.39 A2
9.39 0.51 9 4.21 V
E
E
EC C
C C
v v V V
I
I A I I
v v
μ
μ μ
= = ⇒ =−= ⇒
= = =
= = − = −
(b)
( )( )
( )( )
3
6 32
1
9.39 0.361 mA/V0.026
0.361 10 0.005 1.805 A2
1.805 10 510 10 0.921 4.21 0.921 3.29 V
4.21 0.921 5.13 V
m
dm
C C
C
g
VI g
v V v
v
μ−
−
= ⇒
Δ = = × =
Δ = × × = ⇒ = − + ⇒ −
= − − ⇒ −
11.10 (a)
( )( )
( )
1 2
1 2
1 2
1 2
1 2
06 A
605.90 A
5.90 0.360 30.875 V
0.6 0.8751.475 V
E E
C C
C C
EC EC
v vI I
I Iv v
V V
μβ
μ
= == ≅
== == = −= −= = + − −=
(b) (i)
( )( )
5.90 0.227 mA/V0.026
0.227 360 81.70
m
d m C
cm
g
A g RA
= ⇒
= = ==
(ii) ( )( )
( )( )( )( )
60 0.02640.8
2 0.0059264 K
0.227 3600.0442
2 61 40001
264
m Cd
cm
g RA r
A
π= = =
=−
= = −+
11.11
( )( )
( )( )
( )( )
1 2
1 2
1 2
6 3
2 2 2 1
2
For 0.20 V0.1 mA0.1 30 10
7 V0.1 3.846 mA/V
0.026
3.846 0.008 30.77 A2
30.77 10 30 10 0.923 V
7 0.9236.077 V
7 0.9237.923 V
C C
C C
m
dm
C C
C C C
C
v vI Iv v
g
vI g
v I R
v I v v
v
μ
−
= == == = −= −
= =
Δ = = ⇒
Δ = Δ ⋅ = × × =
↑⇒ ↓⇒ ↓⇒ = − += −
= − −= −
11.12
( )
( )( )
1 2
1 2
50 KFor 0
0.7 1075
0.124 mA0.0615 mA
0.0615 2.365 mA/V0.026120 0.026
50.7 K0.0615
C
E
C C
m
Rv v
I
I I
g
rπ
== =
− − −=
== =
= =
= =
Differential Input
1 22 2d dv Vv v= = −
Half-circuit.
( )( )
1
2
1 2
2 2
2
2 22
22
2.365 50 118.25
dm C C
C C
o C C C C
C
dm C
d m C
V RI g v I R
Rv I R
R Rv v v I R I R
IRVg R
A g R
Δ⎛ ⎞Δ = + ⇒ Δ = −Δ +⎜ ⎟⎝ ⎠
Δ⎛ ⎞Δ = +Δ −⎜ ⎟⎝ ⎠Δ Δ⎛ ⎞ ⎛ ⎞= Δ − Δ = −Δ + − Δ −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠= − Δ
⎛ ⎞= − ⎜ ⎟⎝ ⎠
= − = − = −
Common-mode input.
( )
( )
( ) ( )( )
( )( )
( )( )
[ ]
1 1
2 2
1 2
2
1 1 2
1 211 2
21 2
21 2
2
cm m E
cm
E
m cm cmm
EE
C cm
CE
C cm
CE
C cm C
o C C
Vv V g V RrvV
Rr
g v vI g Vr R
Rr
RR vv IR
r R
RR vv IR
r R
RR v Rv v v
ππ π
π
π
π
ππ
π
π
π
β
βββ
β
β
β
β
β β
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
=⎛ ⎞
+ +⎜ ⎟⎝ ⎠
Δ = = =+ +⎛ ⎞++ ⎜ ⎟
⎝ ⎠Δ⎛ ⎞− + ⋅⎜ ⎟
⎝ ⎠Δ = −Δ =+ +
Δ⎛ ⎞− −⎜ ⎟⎝ ⎠Δ = −Δ =
+ +
Δ⎛ ⎞− + + −⎜ ⎟⎝ ⎠= Δ − Δ = [ ]
( )( )
( )( )( )( )( )( )( )
2
22
1 2
120 0.51 2 50.7 121 2 75
0.0032966118.25 35,870.5
0.003296691.1 dB
cm
cm
E
cmE
dB
R v
R v
r R
RAr R
C M RR
C M R R
π
π
β
β
ββ
Δ⎛ ⎞⎜ ⎟⎝ ⎠
Δ⎛ ⎞− ⎜ ⎟⎝ ⎠=
+ +
−− Δ= =+ + +
= −
= =
=∫
11.13
( )
( ) ( )
1 2
1 2
1
2
00.7 10
750.124 mA
0.0615 mA0.0615 2.365 mA/V0.026
0.01
2.377 mA/V2.353 mA/V120 0.026
50.7 K0.0615
E
C C
m
m
m
m
m
v v
I
I I
g
gggg
rπ
= =− − −
=
== =
= =
Δ=
==
= =
( )
( ) ( )
1 1
2 2
1 2 1 2
1 2
1 2
2
2
2
2 2
250 2.377 2.353 118.25
2 2
dm
dC m c
dC m c
d do C C m C m C
dC m m
Cd m m d
VI g
Vv g R
Vv g R
V Vv v v g R g R
VR g g
RA g g A
Δ =
Δ = −
Δ = +
= Δ − Δ = − −
= − +
−= − + = + ⇒ = −
Common-Mode
( ) ( )
( )
( )
( ) ( )
( ) ( )
1 21 2
1 2
1 11 2 1 2
2.377 2.353 501211 1 2 751 2 50.7
1.2 0.003343358.99
91 dB
m C cm m C cmC C
E E
m m Cocm
cmE
cm
dB
g R v g R vv v
R Rr r
g g RvA
vR
r
A
C M R R
π π
π
β β
β
− −Δ = Δ =
⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− − − −= = =
⎛ ⎞ ⎛ ⎞+ ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠−= ⇒ = −
=∫
11.14 (a)
( ) ( )
1 2
1 2
1 2
00.7 V
5 0.7 4.3 mA1
2.132 mA2.132 1 5
2.87 V
E
E
C C
C C
v vv
I
I Iv v
= == +
−= =
= == = −= −
(b) 1 2 2
1
0.5, 0 on off
v v QQ
= =
( ) ( )
1 2
1 2
2
1200, 4.3 mA 4.264 mA121
5 V 4.264 1 5
0.736 V
C C
C C
C
I I
v v
v
⎛ ⎞= = =⎜ ⎟⎝ ⎠
= − = −
= −
(c) 2.1320.7 V 82.0 mA/V0.026E mv g≈ = =
( ) ( )82.01 41.0
2 2 2d d
m C C m C d dv VI g v I R g R V VΔ = Δ = Δ ⋅ = ⋅ = ⋅ =
0.015 0.615 Vd cV v= ⇒ Δ =
2 1C Cv v↓ ↑
1
2
2.87 0.615 2.255 V2.87 0.615 3.485 V
C
C
vv
= − + = −= − − = −
11.15
(a) 1 38.46 mA/V
0.0261 100
0.01
100 38.46 2.6 K
Cm
T
od
d
d m C
C
c
Ig
Vv
Av
A g RR
R
= = =
= = =
==
=
(b) With 1 2 0v v= =
( )( ) ( )1 2 10 1 2.6 7.4 max 7.4 C C cmv v V v V= = − = ⇒ = 11.16 a. i. ( )01 02 0v v− =
ii.
( ) ( )( )
1 2
01 02 1 1 2 2
2 1 01 02
1 mA
1 7.9 8 0.1 V
C C
C C C C
C C C
I I
v v V I R V I R
I R R v v
+ +
= =
⎡ ⎤ ⎡ ⎤− = − − −⎣ ⎦ ⎣ ⎦= − = − ⇒ − = −
b.
( )
( )( )
( )( )
0 1 2
3
13 13
9
13 91 1 1
13 92 2
exp
2 10So exp10 1.1 109.524 10
exp 10 9.524 10 0.952 mA
1.1 10 9.524 10 1.048 mA
BES S
T
BE
T
BEC S C
T
C C
vI I IV
vV
vI I IV
I I
−
− −
−
−
⎛ ⎞= + ⎜ ⎟
⎝ ⎠⎛ ⎞ ×=⎜ ⎟ + ×⎝ ⎠
= ×
⎛ ⎞= = × ⇒ =⎜ ⎟
⎝ ⎠
= × × ⇒ =
i. ( )( )01 02 2 2 1 1 01 02 01 021.048 0.952 8 0.768 VC C C Cv v I R I R v v v v− = − ⇒ − = − ⇒ − =
ii. ( )( ) ( )( )01 02
01 02 01 02
1.048 7.9 0.952 88.279 7.616 0.663 V
v vv v v v
− = −− = − ⇒ − =
11.17 From Equation (11.12(b))
( ) ( ) ( )
2 /
/
/
110.90
11So 1 0.111
0.90ln 0.111 0.026 ln 0.111 0.0571 V
d T
d T
d T
QC v V
v V
v V
d T d
Ii
e
e
e
v V v
=+
=+
= − =
= = ⇒ = −
11.18 From Example 11.2, we have
( )( ) ( )
( )( )
( )( ) ( )
( ) ( )
max / 0.026
max / 0.026
max / 0.026
max 10.54 0.026 1 0.02
max0.5
4 0.026
max 10.98 0.54 0.026 1
10.490 9.423 max1
d
d
d
dv
d
dv
d v
ve
v
ve
ve
−
−
−
+ −+ =
+
⎡ ⎤+ =⎢ ⎥
+⎢ ⎥⎣ ⎦
+ =+
By trial and error ( )max 23.7 mVdv =
11.19 a.
1 BE3
1 1
12 2
For 1 mA, 0.7 V20 0.7 19.3 k
10.026 1ln ln 0.599 k
0.1 0.1T
Q Q
I V
R R
V IR RI I
= =−= ⇒ = Ω
⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ ⇒ = Ω⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
b. ( )( )
4
04
180 0.02646.8 k
0.10.1 3.846 mA/V
0.026100 1 M0.1
m
r
g
r
π = = Ω
= =
= ⇒ Ω
From Chapter 10 ( )
( ) ( )( )
( ) ( )
( )( ) ( )( )
0 04 4
4
0
01
010
1
0.599 46.8 0.5911 1 3.846 0.591 3.27 M
100 2 M0.05
1 12
181 3.27 181 1
592 181 139 M
m E
E
icm
icm
R r g R r
R rR
r
rR R
R
π
π
β β
= +⎡ ⎤⎣ ⎦= =
= + = Ω⎡ ⎤⎣ ⎦
= ⇒ Ω
⎡ ⎤⎛ ⎞≅ + +⎡ ⎤ ⎢ ⎜ ⎟⎥⎣ ⎦ ⎝ ⎠⎣ ⎦= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= ⇒ = Ω
(c) From Eq. (11.32(b))
( )
( )( )
( )( )( )( )
2 11
0.05 1.923 /0.026180 0.026
93.6 0.05
0Then
1.923 50 0.00760
2 181 32701
93.6
m Ccm
o
B
m
B
cm cm
g RAR
r R
g mA V
r k
R
A A
π
π
β−
=+
++
= =
= = Ω
=
−= ⇒ = −
+
11.19 For 3.5 CMv V= and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be
3.5 1 4.5 CV V= + = Assume the bias is 10 ,V± and 0.5 .QI mA= Then 0.25 CI mA=
Now 10 4.5 22 0.25C CR R k−= ⇒ = Ω
In this case, ( ) ( )100 0.02610.4
0.25r kπ = = Ω
Then ( )( )( )100 22
1012 10.4 0.5dA = =
+ So gain specification is met.
( ) ( )( )( )( )
4
For 80
1 101 0.51 110 1 1 1.03 2 2 0.026 100
dB
Q o oo
T
CMRR dB
I R RCMRR R M
Vβ
β
= ⇒
⎡ ⎤+⎡ ⎤= = + = + ⇒ = Ω⎢ ⎥⎢ ⎥
⎢ ⎥⎣ ⎦ ⎣ ⎦
Need to use a Modified Widlar current source. ( )
( )( )
( )( )
1
1 1 1 1
2
1
100If 100 , then 200 0.5
100 0.0265.2
0.50.5 19.23 /
0.026Then 1030 200 1 19.23 0.216 5.2 225
Also let 225 and 0.5
o o m E
A o
m
E E E E
E REF
R r g R r
V V r k
r k
g mA V
R r R r k R R
R I mA
π
π
π π
= +⎡ ⎤⎣ ⎦
= = = Ω
= = Ω
= =
= + ⇒ = Ω = ⇒ = Ω⎡ ⎤⎣ ⎦= Ω ≅
11.20
(a) ( )0.7 10
37.2 0.25E ER R k
− − −= ⇒ = Ω
(b)
( ) ( )
( )
( ) ( ) ( )
( )
1 21 2 1 2
1 11 1
2 2
1 2
1 2
1 or 1
Then
11
From this, we find
11
e em m
E E
ee
B B
e
ee e
B E
B
eB
E
V V V Vg V g V V Vr r R r R
V V V rV V V
r R r r RV V V
r VV V V Vr r R R
r RV VrV
r R rR
π ππ π π π
π π π
π ππ
π π π
π
π
π π
π
π
π π
β
β
β
⎛ ⎞++ + + = + =⎜ ⎟⎝ ⎠
⎛ ⎞−= ⇒ = −⎜ ⎟+ +⎝ ⎠= −
⎛ ⎞ ⎡ ⎤+ − + − =⎜ ⎟ ⎢ ⎥+⎝ ⎠ ⎣ ⎦
++ ⋅=
+ + ++
( )
( )( )
2 2
Now
We have120 0.026 0.12525 , 4.81 /
0.125 0.026
B
o m C m C e
m
Rr
V g V R g R V V
r k g mA V
π
π
π
⎡ ⎤+⎢ ⎥⎣ ⎦
= − = − −
= ≅ Ω = =
(i)
( )( )
( )
( )( ) ( )
1 2Set and 2 2
Then
25 0.51 0.022 25 22.02625 0.5 25 0.51
37.2 121 25
So0.00494
Now
4.81 50 0.00494 1192
d d
d d
e
e d
d oo d d
d
V VV V
V V
V
V V
V VV V A
V
= = −
⎛ + ⎞⎛ ⎞− ⎜ ⎟ −⎜ ⎟⎝ ⎠⎝ ⎠= =⎡ ⎤+ ++ +⎢ ⎥⎣ ⎦
= −
⎛ ⎞= − − − − ⇒ = =⎜ ⎟⎝ ⎠
(ii)
( )( )
( )
( )
( )( ) ( )
1 2Set Then
25 0.51 2.02252.0256725 0.5 25 0.51
37.2 121 25
0.9972Then
4.81 50 0.9972
or 0.673
cm
cmcm
e
e cm
o cm cm
ocm
cm
V V V
V VV
V V
V V V
VAV
= =
+⎛ ⎞+⎜ ⎟ −⎝ ⎠= =⎡ ⎤+ ++ +⎢ ⎥⎣ ⎦
=
= − −⎡ ⎤⎣ ⎦
= = −
11.21 From Equation (11.18)
( ) ( )
( ) ( )
( )
0 2 1
For 2 mA, 1 mA1Then 38.46 mA/V
0.026Now 2 38.46 0.015So 3.47 k
Now 10 1 3.476.53 V
For 0 max 6.53 V
C C m C d
CQm
T
Q CQ
m
C
C
C C C
CB cm
v v v g R vI
gV
I I
g
RR
V V I R
V v
+
= − =
=
= =
= =
== Ω
= − = −=
= ⇒ =
11.22 The small-signal equivalent circuit is
1 1 2 2
1 2 1 2
A KVL equation: v V V vv v V V
π π
π π
= − +− = −
A KCL equation
( )
( ) ( )
1 21 2
1 2 1 2
1 2 1 1 1 2 2 1 2
0
1 0
1 1Then 2 and 2 2
m m
m
V Vg V g Vr r
V V g V Vr
v v V V v v V v v
π ππ π
π π
π π π ππ
π π π
+ + + =
⎛ ⎞+ + = ⇒ = −⎜ ⎟
⎝ ⎠
− = ⇒ = − = − −
At the 01v node:
01 01 021 0m
C L
v v v g VR R π
−+ + =
( )01 02 2 11 1 1 1
2 mC L L
v v g v vR R R
⎛ ⎞ ⎛ ⎞+ − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ (1)
At the 02v node:
02 02 012 0m
C L
v v v g VR R π
−+ + =
( )02 01 1 21 1 1 1
2 mC L L
v v g v vR R R
⎛ ⎞ ⎛ ⎞+ − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ (2)
From (1):
( )02 01 2 1112
Lm L
C
Rv v g R v vR
⎛ ⎞= + − −⎜ ⎟
⎝ ⎠
Substituting into (2)
( ) ( )
( )
( )
01 2 1/ 01 1 2
01 1 22
011 2
1 1 1 1 1 1 112 2
1 1 1 1 12
122
Lm L m
C C L C L L
L Lm
C C CC
L Lm
C C C
Rv g R v v v g v vR R R R R R
R Rv g v vR R RR
v R Rg v vR R R
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − − + − = −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞
+ + = − − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎣ ⎦
⎛ ⎞ ⎛ ⎞+ = − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1 2For dv v v− =
011
12
2
m L
vd L
C
g RvA
v RR
−= =
⎛ ⎞+⎜ ⎟
⎝ ⎠
From symmetry: 022
12
2
m L
vd L
C
g RvA
v RR
= =⎛ ⎞
+⎜ ⎟⎝ ⎠
Then 02 01
2
m Lv
d L
C
v v g RAv R
R
−= =
⎛ ⎞+⎜ ⎟
⎝ ⎠
11.23 The small-signal equivalent circuit is
KVL equation: 1 1 2 2 1 2 1 2 or v V V v v v V Vπ π π π= − + − = − KCL equation:
( )
( )( )
( )( )
( )
1 21 2
1 2 1 2
1 2 2 2 1 2
0 2
1 2
01 2
0
1 0
1Then 2 or 2
Now 12
1For 2
m m
m
m C L
m C L
d d m C Ld
V Vg V g V
r r
V V g V Vr
v v V V v v
v g V R R
g R R v v
vv v v A g R Rv
π ππ π
π π
π π π ππ
π π
π
+ + + =
⎛ ⎞+ + = ⇒ = −⎜ ⎟
⎝ ⎠
− = − = − −
= −
= −
− ≡ ⇒ = =
11.23 a.
10 7 6 k0.5D DR R−= ⇒ = Ω
1 2 1 mAQ D D QI I I I= + ⇒ =
b. ( )10 6D DS GSI V V= + −
and DGS TN
n
IV VK
= +
For 0.50.5 mA, 2 3.12 V0.4D GSI V= = + =
and 10.12DSV =
Load line is actually nonlinear. c. Maximum common-mode voltage when 1M and 2M reach the transition point, or
( ) 3.12 2 1.12DS GS TNV sat V V V= − = = = Then
( )02 sat 7 1.12 3.12cm DS GSv v v V= − + = − +
Or ( )max 9 Vcmv =
Minimum common-mode voltage, voltage across QI becomes zero.
( )( )
So min 10 3.12min 6.88 V
cm
cm
vv
= − +⇒ = −
11.24
( )
( )
( ) ( )( )
2 2 2
1 1 1
0 2 1
2 1
1 2
We have and
Then
C m C m b e C
C m C m b e C
C C
m b e C m b e C
m C b b
V g V R g V V R
V g V R g V V R
V V Vg V V R g V V R
g R V V
π
π
= − = − −
= − = − −
= −= − − − − −⎡ ⎤⎣ ⎦= −
Differential gain 0
1 2d m C
b b
VA g RV V
= =−
Common-mode gain 0cmA = 11.25 (a)
1 23 3 10 3Then 70
0.1
cm C C
C C
v V V V V
R R k
= ⇒ = =−= ⇒ = Ω
(b)
( )
( )( )( )( )
75 5623Now
11 12
151 0.215623 1 1.45 2 150 0.026
dB
Q o
T
oo
CMRR dB CMRR
I RCMRR
V
RR M
ββ
= ⇒ =
+⎡ ⎤= +⎢ ⎥
⎣ ⎦⎡ ⎤
= + ⇒ = Ω⎢ ⎥⎢ ⎥⎣ ⎦
Use a Widlar current source. [ ]1o o m ER r g R′= +
Let AV of current source transistor be 100 V.
( )( )
( )
( )( ) ( ) ( )
100 0.2Then 500 , 7.69 /0.2 0.026
150 0.02619.5
0.2So 1450 500 1 7.69 0.247
Now 0.247 19.5 250
Then ln
0.2 0.250 0.026 ln0.2
o m
E E
E E E E
REFQ E T
Q
REFRE
r k g mA V
r k
R R k
R R r R R
II R VI
I I
π
π
= = Ω = =
= = Ω
′ ′= + ⇒ = Ω⎡ ⎤⎣ ⎦′ = ⇒ = ⇒ = Ω
⎛ ⎞= ⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞
= ⇒⎜ ⎟⎜ ⎟⎝ ⎠
( )1 1
1.37
10 0.7 10Then 14.1
1.37
F mA
R R k
=
− − −= ⇒ = Ω
11.26 At terminal A.
( )( )
( )1 15
1 2 2THA A
R R R RR R R kR R
δ δδ δ
+ ⋅ += = = ≅ = Ω
+ + +
Variation in THR is not significant
( )( )( )
( )1 5 5 11 2
ATHA
A
RRV VR R R R
δ δδ δ
+ + +⎛ ⎞= = =⎜ ⎟+ + + +⎝ ⎠
At terminal B.
5 2
2.5
THB
THB
RR R R k
RV V VR R
+
= = = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠
From Eq. (11.27) ( )
( )( )( )
( )( )( )( ) ( )
( )
( ) ( )
2 12 1
2 12 1
2 1
where and 2
120 0.0265 , 12.5
0.25120 3
So 10.32 12.5 5
We can find
5 12.5
2
2.5 2 5 1 2.5 52 2
2.5 1.2
CO THB THA
B
B
O
THB THA
THB THA
R V VV V V V V
r R
R k r k
V VV V V
V V V V
V V
π
π
β
δδ
δ δ δ δδ δ
δ
− −= = =
+
= Ω = = Ω
− −= = − −
+− = −
+⎡ ⎤− = − ⎢ ⎥+⎣ ⎦
+ − + −= =+ +
−≅ = −
( )( )25
Then 10.3 1.25 12.9OV
δ
δ δ= − − =
So for 0.01 0.01δ− ≤ ≤ We have 20.129 0.129 OV V− ≤ ≤ 11.27 a.
( )( )2180 0.026
23.4 k0.2
So 46.8 k
id
id
R r
r
R
π
π
=
= = Ω
= Ω
b. Assuming ,rμ → ∞ then
( )( )( )
01
181 1
181 181 M
icm
icm
icm
R R
R
R
β≅ +⎡ ⎤⎣ ⎦= ⎡ ⎤⎣ ⎦= ⇒ = Ω
11.28 (a)
( )1 1
1
2 2
10 0.7 100.5 38.6 K
0.026 0.5ln 236 0.14 0.14
I RR
R R
− − −= = ⇒ =
⎛ ⎞= ⇒ = Ω⎜ ⎟⎝ ⎠
(b)
( )
( )
( )( )
( ) ( )
( )( )
4 4 4
4
4
10.141 5.385 mA/V
0.026180 0.026
33.4 K0.14
33.4 0.236 0.234 K100 714 K0.14
714 1 5.385 0.234
1614 K181 1614 292 MΩ
icm o
o o m E m
E
o
o
icm
R R
R r g R g
r
R
r
R
R
π
β≈ +
′= + = =
= =
′ = =
= =
= +⎡ ⎤⎣ ⎦== ≈
(c)
( )
( )( )
( )( )( )( )
11
1
1
0.07 2.692 mA/V2 1 0.026
1
180 0.02666.86 K
0.072.692 40
2 181 16141
66.860.0123
m Ccm m
o
cm
cm
g RA gR
r
r
A
A
π
π
β−
= = =+
+
= =
−=
+
= −
11.29
( )
( )( )
( )
1 1 1 3
11 1
32 2 2
23 22 3 2
22 2 2
/ 219.23
2 100 0.026 5.2/ 2
/ 2, 19.23
219.23
Then 30 3.12 2
d m
Qm Q
T
T
Q Q Q
Qmd m Q
T
A g R r
Ig I
V
VrI I I
Ig RA g IV
IR I R V
π
πβ
=
= =
= = =
= = =
= ⋅ ⇒ =
Maximum 2 1 18 o ov v mV− = ± for linearity ( ) ( )( )3
2 2
max 18 30 0.54 so 3.12 is OK.
o
Q
v mV VI R V
= ± ⇒ ±=
From 1dA :
( )
( )
( )( )
12
1 1 3 1
12
1 1
2 1
11 1 1
2 12 1
1 1 11
21
5.2
20 19.23 19.235.2
19.23 5.220
5.2
Let 5 10 2
19.23 10 5.2Then 20 44.8
5.2
10Now 10
10So
QQ Q
Q
Q
Q
RI
I R r IR
I
I RI R
IR V I R V
I R VI R
I R RI
II
π
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠= = ⎜ ⎟⎛ ⎞⎜ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=+
⋅ = ⇒ =
= ⇒ =+
= ⇒ =
⎛ ⎞⎜⎜⎝
2
1
1 2
2 2 2
1 1 1
44.8 4.48
Let 100 , 448
Then3.12 6.96
10 100
Q
Q
Q Q
Q
Q
II
I A I A
I R R k
I R R k
μ μ
⎛ ⎞= ⇒ =⎟ ⎜ ⎟⎟ ⎜ ⎟
⎠ ⎝ ⎠= =
= ⇒ = Ω
= ⇒ = Ω
11.30 a.
( )
( )
( ) ( )( )( )
( )( )
231 3
23 3 3
23 3
2
3 3
1 1
1 1
21 1
4
200.25 2
5020 12.5 4 4
12.5 49 30 0
49 49 4 12.5 303.16 V
2 12.520 3.16 0.337 mA
50
0.168 mA2
0.168 0.25 2 2.82 V2.82 10
GSGS
GS GS GS
GS GS
GS GS
Q
QD D
GS GS
DS DS
VI V
V V V
V V
V V
I I I
II I
V VV V
−= = −
− = − +
− + =
± −= ⇒ =
−= ⇒ = =
= ⇒ =
= − ⇒ == − − − ⇒
( )( )( )
4
1
1 1
7.18 V
10 0.168 24 5.97 V5.97 2.82 8.79 V
D
DS DS
VV V
=
= − == − − ⇒ =
(b)
(c)
( )
( )( )( ) ( )( )
( ) ( )( )( ) ( )( )
1 2 1
1
1
1
4 1
4 4
Max 2.82 2 0.82
Now 10 0.168 24 5.97 max 5.97 5.97 0.82max 5.15
max max 5.15 2.82max 7.97
min3.16 2 1.16
CM DS DS DS GS TN
D
S DS
S
CM S GS
CM
CM DS GS
DS GS TN
v V V V sat V VV
V VV V satV Vv V Vv V
v V V sat VV sat V V
−
⇒ = = = −− =
= − == − = −=
= + = +=
= + += − = − =( ) ( )
Then min 10 1.16 2.82 min 6.02 CM CM
Vv v V= − + + ⇒ = −
11.31 a.
( )
( )
21 2 1 1 2
1 2 1 2
1 2 1
21 3 3
1 1
120 A 100 V 1.2 2.30 VFor 5.4 V and 12 V 5.4 2.30 12 4.3 V
10 4.3 47.5 k0.12
240 A
240 200 1.2 2.30 V20 2.3 73.7
0.24
D D GS GS GS
DS DS D
D D
Q D D Q
GS GS
I I V Vv v V V V
R R
I I I I I
I V V
R R
μ
μ
= = = − ⇒ = == = − = = ⇒ − − + = =
−= ⇒ = Ω
= + ⇒ = =
= = − ⇒ =−= ⇒ = 5 kΩ
b.
( )( )04
04
1 1 416.7 k0.01 0.24
1 5.4 13 A416.7
Q
Q DS Q
rI
I V Ir
λ
μ
= = = Ω
Δ = ⋅ Δ = ⇒ Δ ≅
11.32 (a)
( )
( )( )
( )
( )
2
2
25
160 A
28080 4 0.52
80 160 0.5
80 0.5 1.207 V160
5 2 37.5 K 2 1.207 3.21 V0.08
Q
nD GS TN
OS
o
GS
D DS
Ik WI V V
L
V
V
V
R V
μ=′ ⎛ ⎞= −⎜ ⎟⎝ ⎠
= −
= −
= + =
−= = = − − =
(c) ( )
( )2
1 2
1.207 0.5 0.707 VThen 2 0.707 1.29 VAnd 1.207 1.29
2.50 V
DS GS TN
S O DS
cm GS S
cm
V sat V VV V V sat
v v v V Vv
= − = − == − = − = +
= = = + = +=
(b)
11.33
( )( )
( )
( )
5 0.2 8 3.4 V
0.2 0.8 1.694 V0.25
1.694 0.80.894 V3.4 0.8942.506
2.506 1.694 4.2 V
D
DGS TN
n
GS TNDS
S D DS
CM S GS CM
v
IV VK
V VV sat
V V V sat
v V V v
= − =
= +
= + =
= − = −== −= −=
= + = + ⇒ =
(b)
( )( )( )( )( )( )6 3
2
2 2
22
2 0.25 0.2 0.4472 mA/V
0.4472 0.05 22.36 A
22.36 10 8 10 0.179 V
3.43.4 0.179 3.58 V
dD D D D m m n D
D
D
D D
D D
Vv I R I g g K I
I
v
v vv v
μ−
Δ = Δ ⋅ Δ = ⋅ =
= =
Δ = ⇒
Δ = × × =
= + Δ= + ⇒ =
(c)
( )( )( )( )6 3
2 2
50 mV0.4472 0.025 11.18 A
11.18 10 8 10 0.0894 V
3.4 0.0894 3.31 V
d
D
D
D D
vI
v
v v
μ−
= −Δ = − ⇒ −
Δ = − × × = −
= − ⇒ =
11.34 a.
( )
1 2
01 02 1 1 2 2
01 02 2 2 1 1 2 1
0.5 mAD D
D D D D
D D D D D D D
I I
v v V I R V I R
v v I R I R I R R
+ +
= =
⎡ ⎤ ⎡ ⎤− = − − −⎣ ⎦ ⎣ ⎦− = − = −
i. 1 2 01 026 k , 0D DR R v v− = Ω − =
ii. 1 26 k , 5.9 kD DR R= Ω = Ω
( )( )01 02 01 020.5 5.9 6 0.05 Vv v v v− = − ⇒ − = − b.
( )( )( )( ) ( )
( )( )( )( )
2 21 2
1 22
1 22 2
1
2
0.4 / , 0.44 /
1 0.4 0.44 1.190.4 1.19 0.476 mA0.44 1.19 0.524 mA
n n
GS GS
Q n n GS TN
GS TN GS TN
D
D
K mA V K mA VV V
I K K V V
V V V VII
= ==
= + −
= + − ⇒ − == == =
i.
( )( )1 2
01 02 01 02
6 k0.524 0.476 6 0.288 V
D DR Rv v v v
= = Ω− = − ⇒ − =
ii.
( )( ) ( )( )1 2
01 02
01 02
6 k , 5.9 k0.524 5.9 0.476 6
3.0916 2.856 0.236 V
D DR Rv v
v v
= Ω = Ω− = −
= − ⇒ − =
11.35 (a) From Equation (11.69)
( ) ( )( ) ( )
( )
22
2
2
2
1 12 2 2
0.1 0.10.90 0.50 12 0.25 2 0.25
0.40 0.4472 1 0.2
0.8945 1 0.2
n nDd d
Q Q Q
d d
d d
d d
K Ki v vI I I
v v
v v
v v
⎛ ⎞= − ⋅ − ⎜ ⎟⎜ ⎟
⎝ ⎠
⎡ ⎤= − ⋅ − ⎢ ⎥
⎢ ⎥⎣ ⎦
+ = − −
= − −
Square both sides [ ]( )
( )( )( )( )( )
2 2
22 2
2 2 2
0.80 1 0.2
0.2 0.80 0
1 1 4 0.2 0.804 or 1
2 0.2
d d
d d
d
v v
v v
v V V
= −
− + =
± −= =
Then 2 V or 1 Vdv = ± ±
max
0.25But 1.580.1
So 1V, 1V
Qd
n
d d
Iv
kv v
= = =
= ± ⇒ = −
b. From part (a), ,max 1.58 Vdv = 11.36
( )
( ) ( ) ( )
( )
( )
0
1
2
1
2max max max
max
max
12 2
2
1So linear 2 2
1 1 12 2 2 2 2
Then 0.0212 2
1 10.982 2 2 2
d
D
Q n nd d v
d Q Q
n
Q
nDd
Q Q
n n nd d d
Q Q n
nd
Q
n nd
Q Q
idI K K v v
dv I I
KI
Ki vI I
K K Kv v vI I I
K vI
K Kv vI I
=
⎛ ⎞⎜ ⎟⎜ ⎟ ⎛ ⎞⎝ ⎠ = ⋅ − +⎜ ⎟⎜ ⎟
⎝ ⎠
=
= + ⋅
⎡ ⎤⎛ ⎞⎢ ⎥+ ⋅ − + ⋅ ⋅ − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ =
+ ⋅
⎡ ⎤+ ⋅ = + ⋅⎢ ⎥
⎢ ⎥⎣ ⎦( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2max max
2max max max
2max max
2max max
12
0.15 1 0.15 0.150.49 0.98 12 0.2 2 2 0.2 2 0.2
0.49 0.600 max 0.50 0.6124 1 0.6124
0.600 max 0.010 0.6124 1 0.6124
nd d
Q
d d d
d d d
d d d
K vI
v v v
v v v
v v v
⎡ ⎤⎛ ⎞⎢ ⎥⋅ − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞⎢ ⎥+ ⋅ = + ⋅ ⋅ − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
+ = + ⋅ −
= + ⋅ −
By trial and error ( )max 0.429 Vdv ≈ 11.37 (b)
( )( )
( )( )( )( )
2 2
1
2 2 0.05 0.008696
0.0417 mA/V
0.0417 0.05 0.002085 mA2
0.002085 510 1.0631.063 4.565 3.502 V
1.063 4.565 5.628 V
m p D
dm
D
D D
D
g K I
VI g
vv vv
= =
=
Δ = = =
Δ = =↑⇒ = − = −= − − = −
( )( )( )( )( )
( )
( )( )( )
( )( )
2
2
2
2
1 2
1 2
9 12
8 2
8 2 0.05 390 0.8
8 39 1.6 0.64
39 61.4 16.96 0
61.4 3769.96 4 39 16.962 39
1.217 V 2.2170.01739 mA 8.696 A
8.696 0.510 9
S S SG
S D
P S SG TP SG
SG SG
SG SG SG
SG SG
SG
S
S D D
D D
I R VI I
K R V V V
V V
V V V
V V
V
VI I Iv v
μ
= + +=
= + +
= − +
= − + +
− + =
± −=
= == = ⇒
= = − = −4.565 V
(b)
( )( )
( )( )( )( )
2 2 0.05 0.008696 0.0417 mA/V
0.0417 0.05 0.002085 mA2
0.002085 510 1.063 V
m P DQ
dD D D D m
D
g K I
Vv I R I g
v
= = =
Δ = Δ ⋅ Δ = ⋅ = =
Δ = =
1 1 1, ,D Dv I v↑ ↓ ↓
1
2
4.565 1.063 5.628V4.565 1.063 3.502V
D
D
vv
= − − = −= − + = −
11.38 (a)
( )
( )( )( )
1 22
0
6 A
6 0.430
0.847 V0.847 V
36 0.36 3 0.84 V
0.847 0.841.69 V
D n SG TP
D
SG
SG
S
D D D
SD S D
SD
v v
I K V VI
V
VVv I R
V V vv
μ
= =
= +=
+ =
== += −= − = −
= − = − −=
(b) (i)
( )( )( )( )
2
2 30 6 26.83 A/V
26.83 0.36 9.660
d m D m n D
d d
cm
A g R g K I
A AA
μ
= =
= =
= ⇒ ==
(ii) ( )( )
( )( )( )( )
26.83 0.364.83
2 226.83 0.36
0.04481 2 1 2 26.83 4
m Dd d
m Dcm
m O
g RA A
g RAg R
= = ⇒ =
−−= = = −
+ +
11.39
( )( )
( )( )
( )( )
( ) ( )( )
1 2
1 2
1 2
2 2 2
1 1
For 0.30 V0.1 mA
0.1 1 2 V0.1
0.1 30 107 V
2 2 0.1 0.1 0.2 mA/V
0.2 0.1 0.02 mA2
0.02 30 0.6 V7 0.6 6.4 V
7 0.6 7.6 V
D D
DSG TP
P
D D
m p D
dD m
D D D
D D D
D D
v vI I
IV VK
v v
g K I
VI g
v I Rv v vv v
= = −= =
= −
= + =
= = −= −
= = =
⎛ ⎞Δ = = =⎜ ⎟⎝ ⎠
Δ = Δ = =↑⇒ = − + ⇒ = −= − − ⇒ = −
11.40
( )( )( )( )
( )( )( )( )
1 2
2
2
2
2
For 00 2 10
10 2
2 0.15 75 1
22.5 44 12.5 0
So 1.61 and 0.15 1.61 1 55.9 A
2 2 0.15 0.0559
0.1831 mA/V
GS D S
GS n S GS TN
GS GS
GS GS
GS D
m n D
m
v vV I R
V K R V V
V V
V V
V V I
g K I
g
μ
= == + −
= + −
= + −
− + =
= = − ⇒
= =
=
Use Half-circuits – Differential gain
1
2
1 2
2 2
2 2
dD m D
do m D
o D D m d D
od m D
d
V Rv g R
V Rv g R
v v v g V Rv
A g RV
Δ⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠Δ⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= − = −
= = −
Now – Common-Mode Gain
( )
( )
( )
( )
( )( )
( )( )
( )( )( )( )
( )( )( )
1
2
1 2
2
1 2
21 2
21 2
So 1 2
1 2Then
0.1831 50 9.16
0.1831 0.50.0
1 0.1831 2 75
i gs m gs S cm
cmgs
m S
Dm D cm
Dm S
DD cm
Dm S
O D D
m D cmo
m D
m Docm
cm m S
d
cm
V V g V R VV
Vg R
Rg R Vv
g R
Rgm R Vv
g Rv v v
g R Vv
g R
g RvA
V g R
A
A
= + =
=+
Δ⎛ ⎞− +⎜ ⎟⎝ ⎠=+
Δ⎛ ⎞− −⎜ ⎟⎝ ⎠=+
= −− Δ
=+
− Δ= =
+
= − = −
−= = −
+03216
69.1 dBbB
C M R R =∫
11.41 a.
( )
( )( )
02 04
202
2
404
4
2
150 375 kΩ0.4100 250 kΩ0.40.4 15.38 mA/V
0.02615.38 375 250 2307
d m
A
C
A
C
Cm
T
d d
A g r r
VrIVrII
gV
A A
=
= = =
= = =
= = =
= ⇒ =
b.
02 04 375 250 150 kΩL LR r r R= = ⇒ =
11.41 From 11.40
1 2 55.9 A0.183 mA/V
D D
m
I Ig
μ= ==
( )
( ) ( )
( )
1 1 2 2
1 2 1 2
2 1
1 2
: 2 2
2 2
2 2 2 20.183 50 9.15
2 2:
1 2
d dd D m D D m D
d dO D D m D m D
d d m mO D m m D m m
d m D
M Mm D cm m
CM O D Dm S
V VA v g R v g R
V Vv v v g R g R
V V g gv R g g R g g
A g R
g gg R v g
A v v vg R
+⎛ ⎞Δ = − ⋅ Δ = + ⎜ ⎟⎝ ⎠
= Δ − Δ = − −
− − ⎛ Δ Δ ⎞⎛ ⎞= ⋅ + = ⋅ − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= − = − = −
Δ Δ⎛ ⎞ ⎛− + −⎜ ⎟⎝ ⎠ ⎝= Δ − Δ = +
+ ( )
( ) ( ) ( )
( ) ( )( ) ( ) ( )
1 2
0.01 0.183 0.001831 2
0.00183 500.003216
1 0.183 2 75
69.1 dB
D CM
m S
O m Dcm m
cm m S
cm
dB
R v
g Rv g R
A gv g R
A
C M R R
⎞⎜ ⎟
⎠+
−Δ= = Δ = =
+
−= = −
+
=∫
11.42 (a)
( )( )( )( )
( )( )( )
( )( )
1 2
2
2
2
2
1 2
1 2 1
05 2
5 2
5 2 0.5 2 1.6 0.64
5 2 2.2 1.282 2.2 3.72 0
2.2 4.84 4 2 3.722 2
2.02 V5 2.022.02 V, 1.49 mA
20.745 mA
0.745 1 5
D S SG
p S SG TP SG
SG SG SG
SG SG
SG SG
SG
SG
S S
D D
D D D D
v vI R V
K R V V V
V V V
V VV V
V
V
v I
I I
v v v v
= == +
= + +
= − + +
= − +− − =
± +=
=−= = =
= =
= = − ⇒ = 2 4.26 V= −
(b)
( )( ) ( )
( ) ( ) ( ) ( )
2
1 2 2
2 21 2 2
1 2
2 22 2 2
2 22 2 2 2 2
22 2
22 2
55
5
1
5 0.5 2 1.8 0.8
5 3.6 3.24 1.6 0.64
5 2 4.2 3.882 4.2 1.12 0
S S SG
D D S SG
p SG TP p SG TP S SG
SG SG
SG SG SG
SG SG SG SG SG
SG SG
SG SG
S
I R VI I R V
K V V K V V R V
V V
V V V
V V V V V
V VV V
V
= += + +
⎡ ⎤= + + + +⎣ ⎦= −
⎡ ⎤= − + − +⎣ ⎦⎡ ⎤= − + + − + +⎣ ⎦
= − +− − =
( ) ( )( )2
4.2 17.64 4 2 1.12
2 2G
± +=
( ) ( )
( ) ( ) ( ) ( )
2 1
2 21 2
1 2
1 2
1 2
2.339 V 1.339 V2.339 V
0.5 1.339 0.8 0.5 2.339 0.80.1453 mA 1.184 mA0.1453 1 5 1.184 1 54.855 V 3.816 V
SG SG
S
D D
D D
D D
D D
V Vv
I II Iv vv v
= ==
= − = −= == − = −= − = −
(c)
( )( )
( )( )( ) ( )( )
2 1
1 2
1 2
22
2.02 V 2 0.5 0.745
1.22 mA/V1.22 0.1 0.122 mA
0.122 1 0.122 V
4.26 0.122 4.26 0.1224.138 V 4.382 V
dm m p D
S
m
D D
D D
D D
D D
VI g g K I
v
gIv I R
v vv vv v
Δ = =
≈ =
=Δ = =Δ = Δ = =
↓ ↑= − + = − −= − = −
11.43
a. ( ) ( )( )( )4 8 4 0.0264
Qf Q f T
T
Ig I g V
V= ⇒ = =
0.832 mAQI⇒ =
Neglecting base currents.
1 130 0.7 35.2 k0.832
R R−= ⇒ = Ω
b. 04 02100 240 k
0.416A
CQ
Vr rI
= = = = Ω
( ) ( )02 04
0.416 16 mA / V0.026|| 16 240 || 240
1920
CQm
T
d m
d
Ig
VA g r r
A
= = =
= =⇒ =
( )( )
0 02 04 0
180 0.0262 , 11.25 k
0.41622.5 k
|| 120 k
id
id
R r r
R
R r r R
π π= = = Ω
⇒ = Ω
= ⇒ = Ω
c. Max. common-mode voltage when 0CBV = for 1Q and 2Q .
Therefore ( ) ( )( )
3max 15 0.7max 14.3 V
cm EB
cm
v V V Qv
+= − = −=
Min. common-mode voltage when 0CBV = for 5.Q
Therefore ( ) ( )min 0.7 0.7 15 13.6 V
So 13.6 14.3 Vcm
cm
vv
= + + − = −− ≤ ≤
( )( )
( )( )
0
0
1 1 22
100 120 k0.832
181 120 21.7 M
icm
A
Q
icm icm
R R
VRI
R R
β≅ +
= = = Ω
= ⇒ = Ω
11.43 (a)
( )( )
( )
2
2 0.4 1
1.265 mA/V1 10
0.1
10 1.2657.91 K
m n D
m
od
d
d m D
D
D
g K I
gvAv
A g RR
R
=
=
=
= = =
===
(b)
( )( )
( )( )
1 2
1 2
Quiescent 010 1 7.91 2.09 V
1 0.8 2.38 V0.4
2.38 0.8 1.58
So 2.09 1.58 2.38
2.89 V
D D
DGS TN
n
DS
cm D DS GS
cm
v vv v
IV VK
V sat
v v V sat V
v
= == = − =
= + = + =
= − =
= − += − +
=
11.44
( )
( )
1 2
1 2
1 2
2For 2.5
0.25 2
10 3Let 3 , then 28 0.25
28Then 100 7.14 /
2
And 22
0.0807.14 2 0.252
1274 (Extremely la
m Dd
CM
QD D
D D D D
mm
nm D
g RA
v VI
I I mA
V V V R R k
gg mA V
k Wg IL
WL
W WL L
=
=
= = =
−= = = ⇒ = Ω
= ⇒ =
′ ⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞⎛ ⎞= ⇒⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
rge transistors to meet the gain requirement.)
Need 0.10CMA =
From Eq. (11.64(b))
( )( )( )
1 27.14 28
So 0.10 140 1 2 7.14
m DCM
m o
oo
g RA
g R
R kR
=+
= ⇒ = Ω+
For the basic 2-transistor current source
( ) ( )1 1 200
0.01 0.5o oQ
R r kIλ
= = = = Ω
This current source is adequate to meet common-mode gain requirement. 11.45 Not in detail, Approximation looks good. a.
( ) ( )
( )( )
( )
( ) ( )( )( )
( )( )
211
211
21 1 1
21 1
2
1 1
1 2 1 2
02 02
5 and 2 2
5 2 0.050 120
5 2 2 1
2 3 3 0
3 3 4 2 32.186 V
2 25 2.186 0.141 mA
20
0.0704 mA2
5 0.0704 25 3.24
GSS S D n GS TN
S
GSGS
GS GS GS
GS GS
GS GS
S S
SD D D D
VI I I K V V
RV V
V V V
V V
V V
I I
II I I I
v v
− − −= = = −
−= −
− = − +
− − =
± += ⇒ =
−= ⇒ =
= = ⇒ = =
= − ⇒ = V
b. ( ) ( ) ( )
( ) ( )0
2 2 0.05 2.186 10.119 mA/V1 1 710 k
0.02 0.0704
m n GS TN
m
DQ
g K V Vg
rIλ
= − = −=
= = = Ω
1 1 2 2
01 011
0
,
0
gs S gs S
Sm gs
D
V v V V v Vv v V
g VR r
= − = −−
+ + =
( )01 10 0
1 1 0Sm S
D
Vv g v V
R r r⎛ ⎞
+ + − − =⎜ ⎟⎝ ⎠
(1)
02 022
0
0Sm gs
D
v v Vg VR r
−+ + =
( )02 20 0
1 1 0Sm S
D
Vv g v V
R r r⎛ ⎞
+ + − − =⎜ ⎟⎝ ⎠
(2)
( ) ( )
01 021 2
0 0
01 021 2
0 0 0
2
S S Sm gs m gs
S
S Sm S m S
S
v V v V Vg V g Vr r R
v v V Vg v V g v Vr r r R
− −+ + + =
− + + − + − =
( ) 01 021 2
0 0 0
2 12m S mS
v vg v v V g
r r r R⎧ ⎫
+ + + = + +⎨ ⎬⎩ ⎭
(3)
From (1)
10
01
0
1
1 1
S m m
D
V g g vr
v
R r
⎛ ⎞+ −⎜ ⎟
⎝ ⎠=⎛ ⎞
+⎜ ⎟⎝ ⎠
Then
( )1
0 021 2
0 00
0
12 12
1 1
S m m
m S mS
D
V g g vr v
g v v V gr r R
rR r
⎛ ⎞+ −⎜ ⎟ ⎧ ⎫⎝ ⎠+ + + = + +⎨ ⎬
⎛ ⎞ ⎩ ⎭+⎜ ⎟⎝ ⎠
(3)
( )
( )
1 2 0 1 02 00 0 0 0 0
0 01 2 1 02
0 0 0
01
1 1 1 1 1 2 1 1 12
1 1 2 1 11 2 1
m S m m S mD D S D
m m S m mD D S D
mD
g v v r V g g v v V g rR r r R r r R R r
r rg v v g v v V g g
R R r r R R r
rg vR
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧ ⎫ ⎛ ⎞+ + + + − + + = + + ⋅ +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎝ ⎠⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎪ ⎪+ + − + + = + + + − +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭
⋅ 0 0 02 2 02
0 0 0
1 1 2 1 2 12 2S m m mD D S D D S D
r r rv v v V g g gR R r r R R R R R r
⎛ ⎞ ⎧ ⎫⎛ ⎞+ + ⋅ + + = + + + ⋅ + + − −⎨ ⎬⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎩ ⎭
( )0 0 01 2 2 02 0
0 0
1 1 1 1 22 1 1m S m mD D D S D D
r r rg v v v v V g g r
R R R r r R R R⎧ ⎫⎛ ⎞⎛ ⎞ ⎛ ⎞⎪ ⎪⋅ + + ⋅ + + = + + + + +⎨ ⎬⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎩ ⎭
(4)
Then substituting into (2), 02 20 0
1 1 1m S m
D
v g v V gR r r
⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Substitute numbers: ( )
( )( )
1 2 2 02710 710 1 10.11925 25 25 710
1 1 710 20.119 1 1 0.119 710710 20 25 25S
v v v v
V
⎡ ⎤ ⎡ ⎤+ + + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎧ ⎫⎛ ⎞= + + + + +⎡ ⎤⎨ ⎬⎜ ⎟ ⎣ ⎦⎝ ⎠⎩ ⎭
(4)
( )[ ] ( ) ( )
1 2 02
1 2 02
0.119 28.4 29.4 0.0414 0.1204 1.470 6.83928.4296
or 0.4010 0.4150 0.00491
S
S
S
v v v VV
V v v v
+ + = + +=
= + +
( )02 21 1 1Then 0.119 0.11925 710 710Sv v V⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (2)
( ) ( ) ( )[ ]( ) ( ) ( )
( ) ( )
( ) ( )
02 2 1 2 02
02 1 2
02 1 2
1
2
02
02
0.0414 0.119 0.1204 0.401 0.4150 0.004910.0408 0.04828 0.0690
1.183 1.691
Now 2
2
So 1.183 1.6912 2
Or 1.437 0.508 1.437,
dcm
dcm
d dcm cm
d cm d
v v v v vv v vv v v
vv v
vv v
v vv v v
v v v A A
+ = + += −
= −
= +
= −
⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − ⇒ =
10
0.508
1.43720log 9.03 dB0.508
cm
dB dBC M R R C M R R
= −
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
11.46
KVL:
1 1 2 2
1 2 1 2So gs gs
gs gs
v V V vv v V V
= − +− = −
KCL:
( ) ( )1 2 1 2
1 1 2 2 1 2
01 1So ,2 2
m gs m gs gs gs
gs gs
g V g V V V
V v v V v v
+ = ⇒ = −
= − = − −
Now 02 02 01
2
0102
1 1
m gsD L
D L L
v v v g VR R
vvR R R
−+ = −
⎛ ⎞= + −⎜ ⎟
⎝ ⎠
(1)
01 01 021
0201
1 1
m gsD L
D L L
v v v g VR R
vvR R R
−+ = −
⎛ ⎞= + −⎜ ⎟
⎝ ⎠
(2)
From (1): 01 02 21 Lm L gs
D
Rv v g R VR
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
Substitute into (2):
( ) ( )
( )
021 02 2
1 2 1 2 02 2
02 021 2 2
1 2
1 1 1 11
1 1 112
11 222
2
Lm gs m L gs
D D L D L L
L Lm m
D D DD
m LL L
m dD D D L
D
vRg V v g R VR R R R R R
R Rg v v g v v vR R RR
g Rv vR Rg v v AR R R v v R
R
⎛ ⎞⎛ ⎞ ⎛ ⎞− = + + + + −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞− ⋅ − + + − = + +⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
⋅⎛ ⎞ ⎛ ⎞− = + ⇒ = =⎜ ⎟ ⎜ ⎟ − ⎛ ⎞⎝ ⎠ ⎝ ⎠ +⎜ ⎟
⎝ ⎠
From symmetry 011
1 2
12
2
m L
dL
D
g RvAv v R
R
− ⋅= =
− ⎛ ⎞+⎜ ⎟
⎝ ⎠
Then 02 01
1 2 2
m Lv
L
D
v v g RAv v R
R
−= =
− ⎛ ⎞+⎜ ⎟
⎝ ⎠
11.47
( )
( ) ( )( )
( )
1 2 1 2 1 2 1 2
1 2 2
2 1 2
0 2 1 2
and 0Then 2
1Or 2
2
Or 2
gs gs m gs m gs gs gs
gs
gs
mm gs D L D L
md D L
v v V V g V g V V Vv v V
V v v
gv g V R R R R v v
gA R R
− = − + = ⇒ = −− = −
= − −
= − = −
=
11.48
From Equation (11.64(a)), 2n Q
d D
K IA R= ⋅
We need 2 100.2dA = =
( )0.5Then 10 or 20
2n
D n D
KR K R= ⋅ ⋅ =
If we set 20 ,DR k= Ω then 21 /nK mA V=
For this case ( )( )10 0.25 20 5 VDV = − =
( )( ) ( )
( )
0.25 1 1.5 V1
1.5 1 0.5
Then max sat5 0.5 1.5
Or max 6 V
GS
DS GS TN
cm D DS GS
cm
V
V sat V V V
v V V V
v
= + =
= − = − =
= − += − +
=
11.49
( )( )
( ) ( )( )( )
1 1 1
2 2 2
2 1 2 1
1 2
1 2
Now
Define
Then and 0
d m gs D m D s
d m gs D m D s
o d d m D s m D s
o m D
d
od m D cm
d
V g V R g R V VV g V R g R V V
V V V g R V V g R V V
V g R V VV V V
VA g R AV
= − = − −= − = − −
= − = − − − − −
= −− ≡
= = =
11.49
( )( )( )
( )( )
( )( )( )( )
02 04
02
04
2 2 0.12 0.075
0.1897 mA/V1 1 889 kΩ
0.015 0.0751 1 667 kΩ
0.02 0.075
0.1897 889 667 72.3
d m
m n DQ
n DQ
p DQ
d d
A g r r
g k I
rI
rI
A A
λ
λ
=
= =
=
= = =
= = =
= ⇒ =
11.50 (a)
( )
( )( )
21 2
1 2
1
1 2 1 1
1 2
0.080 10 0.40 /2 2
0.1 1 1.5 0.4
1.5 1 0.5 For 3
3 1.5 0.5 2 10 2 80
0.1
nn n
DGS GS TN
n
DS
CM D D CM GS DS
D D
D D
K WK K mA VL
IV V V VK
V sat Vv V V V v V V sat
V V V
R R k
′⎛ ⎞⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= = + = + =
= − == + ⇒ = = − +
= − + ⇒ = =−= ⇒ = Ω
(b)
( )( )
( )( )
( )( )( )
1 and 2 0.4 0.1 0.4 /2
1Then 0.4 80 162
1645 177.8
So 0.090
1 20.4 80
0.090 443 1 2 0.4
d m D m
d
dBcm
cm
m Dcm
m o
oo
A g R g mA V
A
C M R R C M R RA
Ag RA
g R
R kR
= = =
= =
= ⇒ = =
=
=+
= ⇒ = Ω+
If we assume 10.01 Vλ −= for the current source transistor, then
( )( )1 1 500
0.01 0.2oQ
r kIλ
= = = Ω
So the CMRR specification can be met by a 2-transistor current source.
( )
( ) ( ) ( )
3 4
23 4 3
3
3 1 3 3
Let 1
0.080 0.2Then 1 0.040 / and 1 3.24 2 0.04
For 3 , 3 3 1.5 4.5 min 4.5 10 5.5
Qn n GS TN
n
CM D GS DS DS
W WL L
IK K mA V V V V
Kv V V V V V V V sat
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞= = = = + = + =⎜ ⎟⎝ ⎠
= − = − − = − − = − ⇒ = − − − = > So design is OK.
On reference side: For ( )1, max 3.24GSW V VL
⎛ ⎞ ≥ =⎜ ⎟⎝ ⎠
320 20 3.24 16.76 GSV V− = − =
Then 16.67 5.173.24
= ⇒ We need six transistors in series.
( )
( )
2
2
20 3.24 2.793 6
20.0800.2 2.793 1 1.56 for each of the 6 transistors.
2
GS
nREF GS TN
V V
K WI V VLW WL L
−= =
′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
11.51
( )( )
( )( )
122 2 0.25 0.25 0.50 /
1 0.50 3 0.752
d m D
m n D
d
A g R
g K I mA V
A
=
= = =
= =
From Problem 11.26
( )
( ) ( )( )
1 2 1 2
2 1 2
5 1, 2.5 and 1.25
2Then
0.75 1.25 0.9375
A B
o d
V V V V V V V
V A V V
δδ
δ
δ δ
+= = = = − =
+
= ⋅ − = =
So for 0.01 0.01δ− ≤ ≤ 29.375 9.375 oV mV− ≤ ≤
11.52 From previous results
( )
2 11 1 1 1 1 1
1 2
32 3 2 3 2 2
2 1
1 1 2 2
1 2
1 22
1
2 20
1 1and 2 302 2
Set 5 and 2.5 2 2
Let 0.1 Then 100 , 50
0.06 20Then 2 0.12 100
o od m n Q
od m n Q
o o
Q Q
Q Q
v vA g R K I Rv v
vA g R K I R
v vI R I R
V V
I I mAR k R k
W WL L
−= = = ⋅ =
−
= = = ⋅ =−
= =
= == Ω = Ω
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( ) ( )1 2
2
3 3 4
6.67
2 300.060and 2 0.1 2402 50
WL
W W WL L L
⎛ ⎞= =⎟ ⎜ ⎟⎝ ⎠
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
11.53
a. 2
11 1 GS
D DSSP
vi IV
⎛ ⎞= −⎜ ⎟
⎝ ⎠
2
22 1 GS
D DSSP
vi IV
⎛ ⎞= −⎜ ⎟
⎝ ⎠
( )
( )
( ) ( )
( ) ( )( )
( )( )
1 21 2
2 1
1 2 2 1
2 21 1 2
21 1 1 1 2
21 1 2
1 1
2
1Then 2
GS GSD D DSS DSS
P P
DSSGS GS
P
DSS DSSd d
P P
D D Q D Q D
DSSD Q D d
P
DSSD D Q D Q D d
P
DSSD Q D Q d
P
v vi i I I
V V
Iv v
V
I Iv v
V Vi i I i I i
Ii I i v
VI
i i I i I i vV
Ii I i I vV
⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= −
= − ⋅ = ⋅−
+ = ⇒ = −
− − = ⋅−
− − + − = ⋅−
⎡ ⎤− = − ⋅⎢
−⎢⎣ ⎦⎥⎥
Square both sides
( )
( )
( ) ( )
2
2 21 1 2
2
2 22
1
22 2
2 21 2 2
1 04
144
2
212 2
DSSD D Q Q d
P
DSSQ Q Q d
PD
Q Q DSS d DSS dD Q Q
P P
Ii i I I v
V
II I I vV
i
I I I v I vi I IV V
⎡ ⎤− + − ⋅ =⎢ ⎥
−⎢ ⎥⎣ ⎦
⎡ ⎤⎛ ⎞± − − ⋅⎢ ⎥⎜ ⎟⎝ ⎠ −⎢ ⎥⎣ ⎦=
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟= ± − − +
⎜ ⎟⎢ ⎥− −⎝ ⎠⎣ ⎦
Use + sign
( ) ( )
( )
2
2 21 2 2
2 2
1
2 2
1
2 1
2
212 2
212 2
Or
21 12 2
We had
Then
1 12 2
Q Q DSS DSSD d d
P P
Q Q DSS DSS dD d
P Q Q P
DSS DSS dDd
Q P Q Q P
D Q D
D
Q P
I I I Ii v v
V V
I I I I vi vV I I V
I I vi vI V I I V
i I i
iI V
⎛ ⎞⎜ ⎟= + ⋅ − ⋅⎜ ⎟− −⎝ ⎠
⎛ ⎞ ⎛ ⎞= + − ⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⋅ − ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠
= −
⎛ ⎞= − ⎜ ⎟−⎝ ⎠
2 22 DSS DSS d
dQ Q P
I I vv
I I V⎛ ⎞ ⎛ ⎞
⋅ − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
b. 1
2 2
2 2
If , then
21 112 2
2
D Q
DSS DSS dd
P Q Q P
DSS DSS dP d
Q Q P
i I
I I vvV I I V
I I vV vI I V
=
⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⋅ − ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
Square both sides
( )
( )
( )
2 22 2
2 22 22 2
2 2 22
22 2
22
2
21 0
2 2 14
2 12
DSS DSS dP d
Q Q P
DSS DSSd d P
Q P Q
DSS DSS DSSP
Q Q Q Pd
DSS
Q P
Qd P
DSS
I I vV vI I V
I Iv v V
I V I
I I I VI I I V
vII V
Iv V
I
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − ⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦
⎛ ⎞ ⎛ ⎞− ⋅ + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞± −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠=
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
⎛ ⎞= ⎜ ⎟
⎝ ⎠1/ 2
Or Qd P
DSS
Iv V
I⎛ ⎞
= ⎜ ⎟⎝ ⎠
c. For vd small,
( )
( )
1
10
212 2
212d
Q Q DSSD d
P Q
Q DSSDf v
vd P Q
I I Ii v
V I
I Idigd V I→
≈ + ⋅ ⋅−
= = ⋅ ⋅−
( ) 1Or max2
Q DSSf
P
I Ig
V⎛ ⎞
⇒ = ⎜ ⎟−⎝ ⎠
11.53
( )2d m o oA g r R=
Want 400dA = From Example 11.15, 2 1 or M= Ω Assuming that 0.283 /mg mA V= for the PMOS from Example 11.15, then 285 .oR M= Ω
( )
( )1
1 1 2
So 400 1000 285000 0.4014 / 22
0.0800.04028 0.1 10.12
nm m DQ
k Wg g mA V IL
W W WL L L
′⎛ ⎞⎛ ⎞= ⇒ = = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
11.54 a.
( )1 2
0
1 mA
7 10 0.5 6 kΩQ D D Q
D D
I I I I
v R R
= + ⇒ =
= = − ⇒ =
b.
( )
( ) ( )( ) ( )
1max2
1 21max max 0.25 mA/V4 2
Q DSSf
P
f f
I Ig
V
g g
⋅⎛ ⎞= ⎜ ⎟−⎝ ⎠
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
c.
( )( )( )
max2
0.25 6 1.5
m Dd f D
d d
g RA g R
A A
= = ⋅
= ⇒ =
11.55 a.
( ) ( )
( )( )( ) ( )
( )
( )( )( )
( )
2
2
2
2
1 2
2
52 1
5 2 0.8 20 12
15 2 16 14
8 33 27 0
33 1089 4 8 272 8
1.125 V5 1.125
200.306 mA
0.153 mA1.17 V
GS GSS DSS
S P
GSGS
GS GS GS
GS GS
GS
S
D D
o
V VI IR V
VV
V V V
V V
V
I
I Iv
− − − ⎛ ⎞= = −⎜ ⎟
⎝ ⎠
⎛ ⎞− = −⎜ ⎟⎜ ⎟−⎝ ⎠
⎛ ⎞− = + +⎜ ⎟⎝ ⎠
+ + =
− ± −=
= −− −
=
== ==
(b) 11.56 Equivalent circuit and analysis is identical to that in problem 11.36.
2
1
02 01
12
2
12
2
2
m L
dL
D
m L
dL
D
m Lv
d L
D
g RA
RR
g RA
RR
v v g RA
v RR
⋅=⎛ ⎞
+⎜ ⎟⎝ ⎠
− ⋅=⎛ ⎞
+⎜ ⎟⎝ ⎠
−= =
⎛ ⎞+⎜ ⎟
⎝ ⎠
11.57 (a)
( )
( )( )
2 4
2
4
0.1 3.846 mA/V0.026120 1200 K0.180 800 K0.13.846 1200 800
1846
d m o o
m
o
o
d
d
A g r r
g
r
r
A
A
=
= =
= =
= =
=
=
(b)
( )( )For 923 3.846 1200 800
480240 480 480 K480
d L
LL L
L
A R
RR RR
= =
= = ⇒ =+
11.58 (a)
( ) ( )1 1
2250 A 1
2250 1 252.8 A180
5 0.7 536.8 K
0.2528
Q REF QI I I
R R
μβ
μ
⎛ ⎞= = +⎜ ⎟
⎝ ⎠⎛ ⎞= + =⎜ ⎟⎝ ⎠
− − −= ⇒ =
(b)
( )
( )( )
2 4
2
4
0.125 4.808 mA/V0.026150 1200 K
0.1251004.808 1200 800 800 K
0.1252308
d m o o m
o
d o
d
A g r r g
r
A r
A
= = =
= =
= = =
=
(c) ( )( )
2 4
2 180 0.0262 74.9 K
0.1251200 800 480 K
id id
o o o o
R r R
R r r R
π= = ⇒ =
= = = =
(d) ( )( )max 5 0.7 4.3 Vmin 0.7 0.7 5 3.6 V
cm
cm
vv
= − == + − = −
11.59 a.
0 3 4
0 0
122
0.2 2 A100
QB B
Q
II I I
II I
β
μβ
⎛ ⎞⎛ ⎞= + ≈ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= = ⇒ =
b.
( ) ( )( )
02 04
02 04
100 1000 k0.1
0.1 3.846 mA/V0.026
3.846 1000 1000 1923
A
CQ
CQm
T
d m d
Vr rI
Ig
VA g r r A
= = = = Ω
= = =
= = ⇒ =
c. ( )
( )( )02 04
3.846 1000 1000 250 641d m L
d d
A g r r R
A A
=
= ⇒ =
11.60 a.
( )02 04
202
404
02 04
2125
80
If 2 mA, then 38.46 mA/V125 kΩ, 80 kΩ
So 38.46 125 80 200
Or 1508
d m L
CQ Qm
T T
A
CQ CQ
A
CQ CQ
Q m
d
d
A g r r R
I Ig
V VVrI IVrI I
I gr r
A
A
=
= =
= =
= =
= == =
= ⎡ ⎤⎣ ⎦=
For each gain of 1000. lower the current level
02
04
For 0.60 mA, 0.30 mA0.3 11.54 mA/V
0.026125 417 kΩ0.380 267 kΩ0.311.54 417 267 200 1036
So 0.60 mA is adequate
Q CQ
m
d
Q
I I
g
r
r
A
I
= =
= =
= =
= =
= ⎡ ⎤ =⎣ ⎦=
b.
( ) ( )( )
For 10 V, 0.6 VFor 0, max 2 10 2 0.6Or max 8.8 V
BE EB
CB cm EB
cm
V V VV v V V
v
+
+
= = == = − = −
=
11.61 a. From symmetry.
( )( )
3 4 3 4
3 4
1 2
1 2 1 3
1 2
0.1 10.1
Or 2 V
0.1 1 2 V0.1
102 2 10
Or 10 V
GS GS DS DS
DS DS
SG SG
SD SD SG DS
SD SD
V V V V
V V
V V
V V V V
V V
= = = = +
= =
= = + =
= = − −= − −
= =
b.
( )( )
( )( )( )
( )( )( ) ( )( )
0
0
1 1 1 MΩ0.01 0.1
1 1 0.667 MΩ0.015 0.1
22 0.1 2 1 0.2 /
0.2 1000 667 80
nn DQ
pP DQ
m p SG TP
d m on op d
rI
rI
g K V VmA V
A g r r A
λ
λ
= = ⇒
= = ⇒
= += − =
= = ⇒ =
(c)
( )( )
( )( )
2 1
44
22
2 4
0.1 2
1 1 1000 0.01 0.1
1 1 667 0.015 0.1
667 1000 400
QD D
on D
oP D
o o o
II I mA
r kI
r kI
R r r k
λ
λ
= = =
= = = Ω
= = = Ω
= = = Ω
11.62
( )
( )( )
( )( )
( )( )( )( )
4 2
4
2
0.082 2.5 0.052
0.1414 mA/V1 1000 K
0.02 0.051 1333 K
0.015 0.050.1414 1000 1333
80.8
d m o o
m
o
o
d
d
A g r r
g
r
r
A
A
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
= =
= =
=
=
11.63
( )
( )( )
( )( )
( )( )
04 04 4 4
04
4
4
4
3
3
04
0
0
1
80 800 K0.1
0.1 3.8460.026100 0.026
0.126 K
1 26 0.963 K
Assume 100100 0.026
26 kΩ0.1
0.1 3.846 mA/V0.026800 1 3.846 0.963 3.763 MΩ
3.763MThen
m
m
m
v m
R r g R r
r
g
r
R r
r
g
R
R
A g r
π
π
π
π
β
⎡ ⎤= +⎣ ⎦
= =
= =
=
=
= =
=
= =
= =
= + ⇒⎡ ⎤⎣ ⎦⇒ = Ω
= − ( )
( )
2 0
02120 1200 kΩ0.10.1 3.846 mA/V
0.0263.846 1200 3763 3499
m
v v
R
r
g
A A
= =
= =
= − ⎡ ⎤ ⇒ = −⎣ ⎦
b. For
( )( )
04
02 04
800, 800 kΩ0.1
3.846 1200 800 1846v m
v
R r
A g r r
A
= = =
= −
= − ⎡ ⎤ ⇒ = −⎣ ⎦
(c) For part (a), ( )3.763 1.2 0.910 oR M= = Ω
For part (b), ( )1.2 0.8 0.48 oR M= = Ω 11.64
( )
( )
( )
5 3 4 3 45
3 4
5
166
1 1 1Now
So 1
1 1
E B B C CB
C C Q
QB
QEB
I I I I II
I I II
I
III
β β β β
β β
β β β
+ += = =
+ + ++ ≈
≈+
= =+ +
6 5For balance, we want B BI I=
1So that Q QI I=
11.65 Resistance looking into drain of M4.
4 1
44 4
04
14 1
04 04
10 04 4 1
04
1
Or 1
sg X
X sgX m sg
XX m
m
V I RV V
I g Vr
R VI g Rr r
RR r g Rr
≅−
± =
⎡ ⎤+ + =⎢ ⎥
⎣ ⎦⎡ ⎤
= + +⎢ ⎥⎣ ⎦
a.
( )( )( )
( )( )( )( )
( )( )
( )( )
( )
2 2
2
2
4
4
0
2 2 0.080 0.1
0.179 /1 1 667
0.015 0.1
2 2 0.080 0.1
0.179 /1 1 500
0.02 0.1
1500 1 0.179 1 590.5 kΩ500
0.179 667 590.5 56.06
d m o o
m n DQ
on DQ
m P DQ
op DQ
d d
A g r R
g K I
mA V
r kI
g K I
mA V
r kI
R
A A
λ
λ
=
= =
=
= = = Ω
= =
=
= = = Ω
⎡ ⎤= + + =⎢ ⎥⎣ ⎦= ⎡ ⎤ ⇒ =⎣ ⎦
b.
( )1 0 04When 0, 500 kΩ
0.179 667 500 51.15d d
R R rA A
= = == ⎡ ⎤ ⇒ =⎣ ⎦
(c) For part (a), 2 667 590.5 313 o o o oR r R R k= = ⇒ = Ω
For part (b), 2 4 667 500 286 Ωo o o oR r r R k= = ⇒ = 11.66 Let 100, 100 AV Vβ = =
[ ]
( )( )
( )( )( ) ( )
2
4 4
4
2 4
100 1000 0.1
1 where Now
100 0.02626
0.10.1 3.846 /
0.02626 1 0.963
Then 1000 1 3.846 0.963 4704
3.846 1000 4704 3172
Ao
CQ
o o m E E E
m
E
o
d m o o d
Vr kI
R r g R R r R
r k
g mA V
R kR k
A g r R A
π
π
= = = Ω
′ ′= + =
= = Ω
= =
′ = = Ω= + = Ω⎡ ⎤⎣ ⎦
= = ⇒ =
11.67 (a) For Q2, Q4
(1) 42 2 4 4
2 4
x xx m m
o o
V V VI g V g Vr r
ππ π
−= + + +
(2) 4 42 2
2 4 2
xm
o
V V Vg V
r r rπ π
ππ π
−+ =
(3) 4 2V Vπ π= −
From (2) 4 22 4 2 2
1 1xm
o o
V V gr r r rπ
π π
⎡ ⎤= + +⎢ ⎥
⎢ ⎥⎣ ⎦
( )
( )( )
( )( )
( )
4
2 22
2
2
2
4
Now
120 0.5 0.496 1 2 121
1 1200.5 0.0041 2 1 1 121
So120 0.026
761 0.0041
0.0041 0.158 0.026100 24.4
0.0041120 0.
QC
QC C
m
o
II mA
II I mA
r k
g mA V
r M
r
π
π
ββ
ββ β
⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟= = ⇒ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠
= = Ω
= = /
= ⇒ Ω
=( )
( )
4
4
4 42 2
0266.29
0.4960.496 19.08 /0.026100 202
0.496Now
1 1 0.158 which yields 6.29 761 24400 0.318
m
o
x x
o o
k
g mA V
r k
V VV V
r rπ π
= Ω
= =
= = Ω
⎡ ⎤= + + ⇒ =⎢ ⎥
⎢ ⎥⎣ ⎦
From (1),
( )( )
4 4 22 4 2
2
6
2 6
1
119.08 0.1581 1 24400 which yields 135
24400 202 0.318 24400
80Now 160 0.5
Then 135 160 73.2
x xx m m
o o o
x xo
x x
o
o o o o
V VI V g gr r r
I VR kV I
r k
R R r R k
π⎛ ⎞
= + + − −⎜ ⎟⎝ ⎠
⎡ ⎤⎛ ⎞− −⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥= + + = = Ω⎢ ⎥⎢ ⎥⎣ ⎦
= = Ω
= = ⇒ = Ω
(b)
where / 2
c cd m o m
d
iA g R gv
Δ= =
( )
( ) ( )
1 1 3 3 1 3
11 1 3 3
1
1 3 31
1 3 1
1 1
1 3 1
and 2
Also
1So
121Or 6.29761
Then 22 4
So 0.158 19.08 9.624 2
So
dm m
m
d d
d dm m
cm
vi g V g V V V
Vg V r V
r
V r Vr
V V V
v vV V
v vi g g V
igv
π π π π
ππ π π
π
π π ππ
π π π
π π
π
β
Δ = + + =
⎛ ⎞+ =⎜ ⎟
⎝ ⎠⎛ ⎞+ =⎜ ⎟⎝ ⎠⎛ ⎞ = ≅⎜ ⎟⎝ ⎠
= ⇒ =
⎛ ⎞ ⎛ ⎞Δ = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Δ= ( )( )
( )( )( )
1 3
9.62 9.62 73.2 704/ 2
Now 2 where 1761 121 6.29 1522
Then 3.044
d dd
id i i
i
id
A A
R R R r rR k
R M
π πβ
= ⇒ = ⇒ =
= = + += + = Ω
= Ω
11.69 (a)
( )
( )( )
( )( )( )
( )
2 4
2
4
100
Let 0.5 1 1 200
0.02 0.251 1 160
0.025 0.25Then 100 200 160 1.125 /
22
0.0801.125 2 0.25 31.62
d m o o
Q
on D
oP D
m m
nm D
n
A g r r
I mA
r kI
r kI
g g mA V
K Wg IL
W WL L
λ
λ
= =
=
= = = Ω
= = = Ω
= ⇒ =
′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
Now P
WL
⎛ ⎞⎜ ⎟⎝ ⎠
somewhat arbitrary. Let 31.6P
WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
11.70
( )( )( )
( )( )
( )( )
( )( )
( )
2 4
2
4
Let Then 0.5 2 3 3 0.0417
1 1 3205 0.015 0.0208
1 1 2404 0.02 0.0208
Then
80 3205 2404 0.0582
22
0.00.0582 2
d m o o
Q REF
Q REF
Q Q REF
on D
oP D
d m m
nm D
n
A g r r
P I I V V
I II I I mA
r kI
r kI
A g g mA V
k Wg IL
λ
λ
+ −
=
= + −
== − − ⇒ = =
= = = Ω
= = = Ω
= = ⇒ = /
′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= ( )80 0.0208 1.022 n n
W WL L
⎛ ⎞⎛ ⎞ ⎛ ⎞⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
11.71
( )
( )( )( )
( )
2
2
2
1 2
1
1 666.7 K0.015 0.1
400 666.70.60 mA/V
22
0.080.60 2 0.12
0.090 0.004
22.5
d m o o
m o
on D
d m
m
nD
A g r R
g r
rI
A gg
k W IL
WL
WL
W WL L
λ
=
≈
=
= =
= ==
′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
11.72
( )
[ ][ ]
( )( )
( )( )
( )( )
( )( )
4 6
4 4 2 4 4
6 6 8 6 6
2 4
6 8
4
6
4
where11
We have1 1667
0.015 0.0401 1250
0.02 0.040
0.0602 15 0.040 0.268 2
0.0252 10 0.040 0.141 2
Then
1667 166
d m o o
o o o m o
o o o m o
o o
o o
m
m
o
A g R R
R r r g rR r r g r
r r k
r r k
g mA V
g mA V
R
=
= + += + +
= = = Ω
= = = Ω
⎛ ⎞= = /⎜ ⎟⎝ ⎠
⎛ ⎞= = /⎜ ⎟⎝ ⎠
= + ( )( )( )( )6
7 1 0.268 1667 748
1250 1250 1 0.141 1250 222.8 o
M
R M
+ ⇒ Ω⎡ ⎤⎣ ⎦= + + ⇒ Ω⎡ ⎤⎣ ⎦
(a)
4 6 748 222.8 172 o o o oR R R R M= = ⇒ = Ω
(b) ( ) ( )( )4 4 6 0.268 172000 46096d m o o dA g R R A= = ⇒ =
11.73
( )
( )( )( )( )
( )( )
2 4
2 4
2 4
1
1 500 K0.02 0.1
2 2 0.5 0.1
0.4472 mA/V0.4472 500 500 112
500 500 250 K
d m o o
o oD
m n D
d d
o o o o
A g r r
r rI
g K I
A A
R r r R
λ
=
= =
= =
= =
== ⇒ =
= = ⇒ =
11.74 (a)
( )
( )
( )( )
2
3
2
1
1 1
3 1 1
0.4 1 1.894 V0.5
0.4 1 1.894 V0.5
1.894 1 0.894 V
1.894 0.894 1.894 4 4.89 V
DP p SG TP
SG
DN n GS TN
GS
DS GS TN
SG DS GS CM
I K V V
V
I K V V
V
V sat V VV V V sat V vV V V
+
+ + −
= +
+ = =
= −
+ = =
= − = − == + − += + − + ⇒ = = −
(b)
( )
( )( )( )( )
( )( )
2 4
2 41 1 166.7 K
0.015 0.4
2 2 0.5 0.4 0.8944 mA/V
0.8944 166.7 166.7 74.5
d m o o
o oD
m n D
d d
A g r r
r rI
g K I
A A
λ
=
= = = =
= = =
= ⇒ =
11.75 (a) For 2V 2.7 Vcmv V += + ⇒ = If is a 2-transistor current source,
0.7 0.73.4 V 3.4 V
Q
cm
I
V vV V V
−
− + −
= − −= − ⇒ = − =
(b)
( )
( )( )
2 4 2
4
100 1000 K0.160 600 K0.1
0.1 3.846 mA/V0.026
3.846 1000 600 1442
d m o o o
o
m
d d
A g r r r
r
g
A A
= = =
= =
= =
= ⇒ =
11.76 (a) 3.4 VV V+ −= − = (b)
( )( )
2
4
75 1250 K0.0640 666.7 K
0.060.06 2.308 mA/V0.0262.308 1250 666.7
1004
o
o
m
d
d
r
r
g
A
A
= =
= =
= =
=
=
11.77
( )( )
( )( )
1 1
2
2
2 2 0.2 0.25 0.447 mA/V
0.75 28.85 mA/V0.026120 0.026
4.16 k0.75
m n Bias
CQm
T
T
CQ
g K I
Ig
V
VrIπβ
= = =
= = =
= = = Ω
( )
( )
( ) ( )( )( )( )
1 1 2 20
2 1 1 2 1 2
1 1 2 1 20
1 1 1 2 11 2
10
1 2
10
1 2
and
and 1
11
1 0.447 1211 1 0.447 4.16
18.9 mA/V
m gs m
m gs i gs
gs m m m
ii gs m gs gs
m
mi
m
mCm
i mCm
i g V g VV g V r v V Vi V g g g r
vv V g V r Vg r
gi v
g rgi
gv g r
g
π
π π π
π
ππ
π
π
β
β
= += = +
= + ⋅
= + =+
+= ⋅
++
= = =+ +
⇒ =
11.78
( ) ( )( )
( )
( ) ( )( )
( ) ( ) ( )
0 2
0 2
2
2 0 2 0 2
1 1 500 kΩ0.01 0.2
80 400 kΩ0.2
2 2 0.2 0.2
0.4 mA/V
0.4 500 400 88.9
n DQ
A
CQ
m n DQ
d m
d
r MI
Vr QI
g M K I
A g M r M r Q
A
λ= = =
= = =
= =
=
⎡ ⎤= ⎣ ⎦= ⎡ ⎤ ⇒ =⎣ ⎦
If the IQ current source is ideal, 0cmA = and dBC M RR = ∞ 11.79 a.
b. Assume RL is capacitively coupled. Then
( )( )
( )( )
1
1 1
2 2
2 2
0.7 0.0875 mA8
0.9 0.0875 0.8125 mA
2 2 1 0.0875 0.592 mA/V
0.8125 31.25 mA/V0.026100 0.026
3.2 k0.8125
CQ DQ Q
BEDQ
CQ
m P DQ m
CQm m
T
T
CQ
I I IVIR
I
g K I g
Ig g
V
Vr rIπ πβ
+ =
= = =
= − =
= = ⇒ =
= = ⇒ =
= = ⇒ = Ω
c. ( )
( )( )( )
( ) ( )( )
( )
( )
0 1 2 2
0 0
2 1 1 2
0 1 2 1 1 2
0 0 1 2 1 1 2
1 2 1 1 20
1 2 1 1 2
1 2 1 1 2
1
We find
0.592 31.2
m sg m L
i sg sg i
m sg
m sg m m sg L
i m m m L
m m m Lv
i m m m L
m m m
V g V g V R
V V V V V V
V g V R r
V g V g g V R r R
V V V g g g R r R
g g g R r RVAV g g g R r R
g g g R r
π
π π
π
π
π
π
π
= − −
+ = ⇒ = −
=
⎡ ⎤= − +⎣ ⎦⎡ ⎤= − − +⎣ ⎦
⎡ ⎤+⎣ ⎦= =⎡ ⎤+ +⎣ ⎦
+ = + ( )( )( )
( )( )( )( )
5 0.592 8 3.2
42.8842.88
Then 1 42.88
Lv
L
RA
R
=
=+
11.80 a. Assume RL is capacitively coupled.
( )( )
( )( )
1 1
2 2
2 2
0.7 0.0875 mA8
1.2 0.0875 1.11 mA
2 2 1 0.0875 0.592 mA/V
1.11 42.7 mA/V0.026100 0.026
2.34 kΩ1.11
DQ
CQ
m p DQ m
CQm m
T
T
CQ
I
I
g K I g
Ig g
V
Vr rIπ πβ
= =
= − =
= = ⇒ =
= = ⇒ =
= = ⇒ =
b.
( )( )( )
( )
( )( )( )
2 2 1
1 1 2 2
1 2 1 1 2
01 2 1 1 2
0
1
1 21.6 Ω0.592 0.592 42.7 8 2.34
sg X
X m m sg
m sg
X X m m m
X
X m m m
V VI g V g V
g V R r V
I V g g g R r
VRI g g g R r
R
π
π π
π
π
== +
=
⎡ ⎤= +⎣ ⎦
= =+
= ⇒ =+
11.81 (a)
(1) ( )
22
0om
o
V Vg V
rπ
π− −
+ =
(2) ( )
2 1 12 1 1
1 1 or 0om m i m i
o o o
V V V Vg V g V g V V
r r r r rπ π π
π ππ π
− − ⎛ ⎞− −+ = + + = − +⎜ ⎟
⎝ ⎠
Then 1
1
1 1m i
o
g VV
r r
π
π
=⎛ ⎞
+⎜ ⎟⎝ ⎠
From (1)
( )( )
22 2
22
2 2 2 12
1
1 2 22
1
1
2
1
1 0
11
1 1
1
1 1
Now
2 2 0.25 0.025 0.158 /
0.025 0.9615 /0.026
1
om
o o
mo
o o m o m io
o
m o moo
vi
o
m n Q
Qm
T
o
Vg Vr r
gr
V r g V r g Vr
r r
g r grV
AV
r r
g K I mA V
Ig mA V
V
rI
π
π
π
π
λ
⎛ ⎞+ + =⎜ ⎟
⎝ ⎠⎛ ⎞
+⎜ ⎟⎛ ⎞ ⎝ ⎠= − + = −⎜ ⎟⎛ ⎞⎝ ⎠ +⎜ ⎟⎝ ⎠
⎛ ⎞− +⎜ ⎟
⎝ ⎠= =⎛ ⎞
+⎜ ⎟⎝ ⎠
= = =
= = =
=( )( )
( )( )
( )( )
2
1 2000 0.02 0.025
50 2000 0.025
100 0.026104
0.025Then
10.158 2000 0.96152000 30039
1 12000 104
Q
Ao
Q
T
Q
v v
k
Vr kI
Vr kI
A A
πβ
= = Ω
= = = Ω
= = = Ω
⎛ ⎞− +⎜ ⎟⎝ ⎠= ⇒ = −
⎛ ⎞+⎜ ⎟⎝ ⎠
To find Ro; set 10 0i m iV g V= ⇒ =
( )
( )
( )( )
( )
( )
22
1
2 12 2
2 1 22
Then
1
Combining terms,
1 1
12000 1 2000 104 0.9615 192.2 2000
xx m
o
x o
xx m x o
o o
xo o o m
x o
o
V VI g V
rV I r r
VI g I r rr r
VR r r r gI r
R M
ππ
π π
π
π
− −= +
= −
⎛ ⎞= + − +⎜ ⎟⎝ ⎠
⎡ ⎤⎛ ⎞= = + +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞= + + ⇒ = Ω⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(b)
(1) ( )3
3 33
0o gsm gs
o
V Vg V
r
− −+ =
(2) ( ) ( )3 3 2 3
3 3 2 2 2 23 2 2 2
1 or 0o gs gs gsm gs m m
o o o o
V V V V Vg V g V V g
r r r rπ
π π
− − − − − ⎛ ⎞+ = + = + −⎜ ⎟
⎝ ⎠
(3) ( ) ( )3 2 22
2 2 12 2 1
gsm m i
o o
V V VV g V g Vr r r
π πππ
π
− − − −+ + = +
From (2), 32
2 22
1gs
o mo
VV
r gr
π =⎛ ⎞
+⎜ ⎟⎝ ⎠
Then
(3) 32 2 1
2 2 1 2
1 1 1 gsm m i
o o o
VV g g V
r r r rππ
⎛ ⎞+ + + = +⎜ ⎟
⎝ ⎠
or
3 32 1
2 2 1 22 2
2
1 1 11
gs gsm m i
o o oo m
o
V Vg g V
r r r rr g
rπ
⎡ ⎤+ + + = +⎢ ⎥⎛ ⎞ ⎣ ⎦+⎜ ⎟
⎝ ⎠
3 31 1 10.9615 0.96151 104 2000 2000 20002000 0.9615
2000
gs gsi
V VV⎡ ⎤+ + + = +⎢ ⎥⎛ ⎞ ⎣ ⎦+⎜ ⎟
⎝ ⎠
Then 53 1.83 10gs iV V= ×
From (1), ( )53 3
3 3
1 1 or 2000 0.158 1.83 102000
om gs o i
o o
Vg V V V
r r⎛ ⎞ − ⎛ ⎞+ = = − + ×⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
75.80 10ov
i
VAV
= = − ×
To find Ro
(1) ( )3
3 33
x gsx m gs
o
V VI g V
r
− −= +
(2) ( ) ( )3 3 2
3 3 2 23 2
x gs gsm gs m
o o
V V V Vg V g V
r rπ
π
− − − − −+ = +
(3) ( )2 1 2x oV I r rπ π= −
3 33 3
3
3
1From (1)
10.1582000 2000
2000So 0.1585
xx gs m
o o
xx gs
xx
gs
VI V gr r
VI V
VIV
⎛ ⎞= + +⎜ ⎟
⎝ ⎠⎛ ⎞= + +⎜ ⎟⎝ ⎠
−=
( ) ( )( )
3 3 2 23 2 3 2
3 2
10
From (2),
1 1 1
1 1 10.158 0.96152000 2000 2000 2000
/ 2000Then 0.159 2000 104 0.9620.1585 2000
We find 6.09 10
xgs m m
o o o o
xgs
x x xx
xo
x
VV g V g
r r r r
VV V
I V V I
VRI
π
π
⎡ ⎤ ⎛ ⎞+ + + = +⎜ ⎟⎢ ⎥
⎣ ⎦ ⎝ ⎠⎡ ⎤ ⎛ ⎞+ + + = +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
−⎡ ⎤ + = −⎢ ⎥⎣ ⎦
= = × Ω
11.82 Assume emitter of Q1 is capacitively coupled to signal ground.
( )( )
( )
( ) ( )( )( )
1
1
1
800.2 0.1975 81
0.2 0.00247 81
80 0.02610.5
0.19750.1975 7.60 /0.0262 2 0.2 0.00247
0.0445 /
CQ
DQ
m
m n D
m
I mA
I mA
r k
g Q mA V
g M K I
g M mA V
π
⎛ ⎞= =⎜ ⎟⎝ ⎠
= =
= = Ω
= =
= =
=
( ) ( )
( )( )
( ) ( )
( )( )( )
( )( )
11
1
1
11
1
and or
1Then 1 or 11
11
7.60 20Then 48.4
110.0445 10.5
i gs m gs gsm
ii
m
m
m Coo m C v
i
m
v v
VV V V V g M V r V
g M r
VV V Vg M r
g M r
g Q RVV g Q V R AV
g M r
A A
ππ π π
π
π ππ
π
π
π
= + = =
⎛ ⎞= + =⎜ ⎟⎜ ⎟ ⎛ ⎞⎝ ⎠ +⎜ ⎟⎜ ⎟
⎝ ⎠−
= − ⇒ = =⎛ ⎞
+⎜ ⎟⎜ ⎟⎝ ⎠
−= ⇒ = −⎛ ⎞
+⎜ ⎟⎜ ⎟⎝ ⎠
11.83 Using the results from Chapter 4 for the emitter-follower:
( )( )
( )( )
[ ]( )
9 07 0118
0 4
88
89
9
07
011
011 011
11
1||1
100 0.0262.6 k
11 0.01 mA
100100 0.026
260 k0.01100 500 k0.2
100 500 k0.2
0.21 , 7.690.026
100 0.0
T
C
CC
A
Q
A
Q
m E m
r r Rr
R R
VrIII
r
VrIVrI
R r g R g
r
ππ
π
π
π
ββ
β
β
⎡ + ⎤+⎢ ⎥+⎢ ⎥=
⎢ ⎥+⎢ ⎥⎣ ⎦
= = = Ω
≈ = =
= = Ω
= = = Ω
= = = Ω
′= + = =
=( )
( )( )011
0
0
2613 k
0.20.2 13 0.197 k500 1 7.69 0.197 1257 k
Then
260 500 12572.6
1015 ||101
5 0.0863 0.0848 K 84.8
ERR
R
R
= Ω
′ = = Ω= + = Ω⎡ ⎤⎣ ⎦
⎡ + ⎤+⎢ ⎥
⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
= ⇒ = ⇒ Ω
11.84
( )( )( )
( )( )( )
( ) ( )
( )( )( )( )
1 2
2
2
1
30 3
0 0
1100 0.026
5.2 k0.5
100 0.026 100 0.026520 k
0.5 /100 0.5520 101 5.2 1.05 M
100 0.026505 , 2.6 k101 1
2.6 505 5 0.521 0.472 k101
i
i i
R r r
r
r
R R
rR r
R R
π π
π
π
ππ
β= + +
= = Ω
= = = Ω
= + ⇒ ≅ Ω
+= = = Ω
+= = ⇒ = Ω
( )
( )
30 3 3
3
0 33
5
1 5
mVV g Vr
V Vr
ππ
π
ππ
β
⎛ ⎞= − +⎜ ⎟
⎝ ⎠⎛ ⎞+= − ⎜ ⎟⎝ ⎠
(1)
( )0 332 2
3
02 2 3
3
50
1 150 50
m
m
V VVg V
r
Vg V V
r
πππ
π
π ππ
−= +
⎛ ⎞= + −⎜ ⎟
⎝ ⎠ (2)
12 1 1 2
1
1 21
1
mVV g V rr
V rr
ππ π π
π
π ππ
β
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞+= ⎜ ⎟⎝ ⎠
(3)
and 1 2inV V Vπ π= + (4)
20.5 19.23 mA/V
0.026mg = =
Then
( ) ( )0 3 3 0101 5 0.0051492.6
V V V Vπ π⎛ ⎞= − ⇒ = −⎜ ⎟⎝ ⎠
(1)
( )
( )( )
02 0
0
2 0
And1 119.23 0.005149
2.6 50 500.02208
Or 0.001148
VV V
V
V V
π
π
⎛ ⎞= − + −⎜ ⎟⎝ ⎠
= −
= −
(2)
And ( )1 2 0 0.001148in inV V V V Vπ π= − = + (4)
So
( ) ( ) ( )0 01010.001148 0.001148 5.2520inV V V ⎛ ⎞− = +⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠
(3)
( ) ( ) ( ) 00 00.001148 0.001159 1.01 438in v
in
VV V V AV
− − = ⇒ = = −
11.85
( )
2
2
1
5 1 mA5
1 0.8 2.21 V0.5
2.21 50.206 mA
35
GS
I
V
I
= =
= + =
− −= =
( )( )( )( )
( )( )
( ) ( )( )( ) ( )
( )( )( )
( )( )
0 2 2 2 02
2 1 1 01 1 0 1
2 1 01 1 0
0 2 2 02 1 01 1 0
2 2 02 1 01 10
2 2 02
2 2 2
1 1 1
and
So
Then
1
2 2 0.5 1 1.414 /
2 2 0.2 0.206 0.406
m gs
gs m sg sg in
gs m in
m m in
m mv
in m
m n D
m p D
V g V R r
V g V r R V V V
V g V r R V
V g R r g V r R V
g R r g r RVAV g R r
g K I mA V
g K I mA
=
= − = −
= − −
⎡ ⎤= − −⎣ ⎦−
= =+
= = =
= = =
( )( )
( )( )
( )( )( )( )( )( )
011 1
022 2
2 02
1 01
/
1 1 485 k0.01 0.206
1 1 100 k0.01 1
5 100 4.76 k
35 485 32.6 k
1.414 4.76 0.406 32.6Then
1 1.414 4.76So 11.5
D
D
v
v
V
rI
rI
R r
R r
A
A
λ
λ
= = = Ω
= = = Ω
= = Ω
= = Ω
−=
+⇒ = −
Output Resistance—From the results for a source follower in Chapter 6.
0 2 022
0
1 1 5 1001.4140.707 4.76
So 0.616 k
m
R R rg
R
= =
== Ω
11.86 a.
( )
2 2
22 2
2
1 1
5 10 k0.5
0.5 1 2.41 0.25
5 2.4174.1 k
0.1
DSG TP
p
R R
IV V VK
R R
= ⇒ = Ω
= − = + =
− −= ⇒ = Ω
b.
( )( )( )( )
( )( )( )( )( )( )( )( )( )( )
( )( )
( )( )
0 2 2 02 2
2 0 1 1 01 1 1
2 02 2 1 01 10
2 02 2
1 1 1
2 2 2
011 1
022 2
and
1
2 2 0.1 0.1 0.2 /
2 2 0.25 0.5 0.707 /
1 1 1000 k0.01 0.1
1 1 200 0.01 0.5
m sg
sg m gs gs in
m mv
in m
m n D
m p D
D
D
V g V r R
V V g V r R V V
g r R g r RVAV g r R
g K I mA V
g K I mA V
rI
rI
λ
λ
= −
⎡ ⎤= − − =⎣ ⎦−
= =+
= = =
= = =
= = = Ω
= = =
( )( )( )( )( )( )
02 2
01 1
0 2 022
0
k
200 10 9.52 k
1000 74.1 69.0 k
0.707 9.52 0.2 69Then
1 0.707 9.52So 12.0
1 1 10 2000.707
1.414 9.52Or 1.23 k
v
v
m
r R
r R
A
A
R R rg
R
Ω
= = Ω
= = Ω
−=
+⇒ = −
= =
== Ω
11.87 a.
( )
( ) ( )
2
023 1 1
1
03
3
034
2
2 2
0.25 mA5 2 12 k0.25
on 2 0.7 2.6 k0.5
5 5 3 4 k0.5
on 5
3 0.7 5 2.43 k3
C
BEC E E
E
C CC
BEC
E
E E
I
R R
v VI R R
Rv
R RI
v VI
R
R R
=−= ⇒ = Ω
− −= ⇒ = ⇒ = Ω
− −= = ⇒ = Ω
− − −⎡ ⎤⎣ ⎦=
− += ⇒ = Ω
b. Input resistance to base of Q3, ( )
( )( )
( )( )
( )
( )( )( )( )( )
( )( )
3 3 1
3
3
021 2 3
2
1 1
403
02 3 1
4 4 2
20
03 4
1100 0.026
5.2 k0.5
5.2 101 2.6 267.8 k12
0.25 9.62 mA/V0.0261 9.62 12 267.8 55.22
Now 1
where 1
1and
1
i E
i
d m id
m
d d
C i
E
i E
E
E
R r R
r
Rv
A g R Rv
g
A A
R Rvv r R
R r R
Rvv r R
π
π
π
π
π
β
βββ
ββ
= + +
= = Ω
= + = Ω
= =
= =
= ⇒ =
−=
+ +
= + +
+=
+ +
( )( )
( )( )( )( )( )( )
( )( )( )( )
( )( )( )
2
4
0
03
4
3
03
02
0
100 0.0260.867 k
3101 2.43
0.99650.867 101 2.430.867 101 2.43 246.3 k5.2 k
100 4 246.3So 1.47
5.2 101 2.6
So 55.2 0.9965 1.47 80.9
i
d dd
r
vvRr
vv
vA Av
π
π
= = Ω
= =+
= + = Ω= Ω
−= = −
+
= = − ⇒ = −
c. Using Equation (11.32b) ( )
( )
( )( )
( )( )( )( )
( )( )( )
2 31
0
2
2
1 1
0 031
03 02
10
2 11
100 0.02610.4 k
0.259.62 12 267.8
0.05692 101 100
110.4
Then
0.9965 1.47 0.0569 0.08335
80.920log 590.08335
m icm
cm cm
cm cm
cm
dB dB
g R RA
Rr
r
A A
v vA A
v vA
C M RR C M RR
π
π
β−
=+
+
= = Ω
−= = − =
+
⎛ ⎞⎛ ⎞= ⋅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= − − ⇒ =
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
.7 dB
11.88 a.
011 1
1
042 2
4
10 10 2 80 k0.1
10 10 6 20 k0.2
C CC
C CC
vR RI
vR RI
− −= = ⇒ = Ω
− −= = ⇒ = Ω
b.
( )
( )( )
( ) ( )
( )( )( )( )
01 021 1 1 3
1
3
1 1
042 4 2
01 02
4
2
0.1 3.846 mA/V0.026180 0.026
23.4 k0.2
3.846 80 23.4 69.6
12
0.2 7.692 mA/V0.0261 7.692 20 76.92
Then 76.9 69.6 5352
d m Cd
m
d d
d m C
m
d
d d
v vA g R rv
g
r
A A
vA g Rv v
g
A
A A
π
π
−= = −
= =
= = Ω
= − ⇒ = −
= =−
= =
= =
= − ⇒ = −
11.89 a. Neglect the effect of r0 in determining the differential-mode gain.
( ) ( )
( )( )
( )
( )( )
( )( )
( )( )
021 2 3 3 3
22
3
1 51
2
33
02
3
3
1 where 12
1
12 0.7 12 23.3 1.94 mA12
1 1.942 37.3 mA/V
0.026200 0.026
112 1.94 8 4.24 V2
4.24 0.7 1.07 mA3.3
200 0.0264.86 k
1.07
d m C i i Ed
C
E
C
m
C
C
vA g R R R r Rv
RA
r R
I IR
g
rI
v
I
r
R
π
π
π
π
β
ββ
= = = + +
−=
+ +
− − −= = = ≈
⋅= =
=
= − =
−= =
= = Ω
( )( )
( )3
1
4.86 201 3.3 668 k1 37.3 8 668 147.42
i
dA
= + = Ω
= ⎡ ⎤ =⎣ ⎦
( )( )
( )( )
( )( )( )
1 2
0 055
2 31
0
2
2
Then 147.4 1.197 176
80 41.2 k1.94
2 11
200 0.0265.36 k
1 1.942
d d d
A
C
m C icm
A A A A
VR rI
g R RA
Rr
r
π
π
β
= ⋅ = − ⇒ = −
= = = = Ω
−=
++
= = Ω⋅
( )( )( )( )
( )( )
1
2
37.3 8 6680.09539
2 201 41.21
5.361.197
0.09539 1.197 0.114
cm
cm cm
A
AA A
−= = −
+
= −= − − ⇒ =
b.
( )
( )( ) ( )( )
1 2
1 2
03
2.015sin 1.985sin0.03sin V
2.0sin2
176 0.03 0.114 2
d
d
cm
d d cm cm
v v v t tv t
v vv t
v A v A v
ω ωω
ω
= − = −=
+= =
= += − +
Or 03 5.052sinv tω= −
Ideal, 0cmA =
( )( )03
03
So176 0.03
5.28sind dv A v
v tω= = −= −
c. ( )
( ) ( )
( )
( )( ) ( )( )
2
0 0
02
2 2 5.36 10.72 kΩ
2 2 1 1
80 82.5 k1 1.942
2 2 201 41.2 201 82.5
16.6 M 16.6 MSo 4.15 M
id id
icm
A
C
icm
icm
R r R
R R r
VrI
R
R
π
β β
= = ⇒ =
≅ + +
= = = Ω⋅
= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= Ω Ω
⇒ = Ω
11.90 a.
( )
( ) ( ) ( )( )
( ) ( ) ( )( )
( )
( )
241 4
12
4 4
24 4 4
24 4
2
4
1
02
202 33 3
5
24
24 55 0.2 2
24 11 4 4
11 43 20 0
43 43 4 11 203.37 V
2 1124 3.37 0.375 mA
550.37512 40 4.5 V
2
GSn GS Th
GS GS
GS GS GS
GS GS
GS
Q
GSD n GS Th
VI k V V
R
V V
V V V
V V
V
I I
v
v VI k V V
R
−= = −
− = −
− = − +
− + =
± −= =
−= = =
⎛ ⎞= − =⎜ ⎟⎝ ⎠
−= = −
( ) ( ) ( )
( ) ( ) ( )( )
23 3 3
23 3
2
3
3
4.5 0.2 6 4 4
1.2 3.8 0.3 0
3.8 3.8 4 1.2 0.33.09 V
2 1.24.5 3.09 0.235 mA
6
GS GS GS
GS GS
GS
D
V V V
V V
V
I
− = − +
− + =
± −= =
−= =
( )
( ) ( )
( ) ( )
( ) ( )( ) ( )
( ) ( )
( ) ( )
2 2
1 2 1
3 22
3 5
3 3
2
1 2
0 05
0.3752 2 0.22
0.387 mA/V1 1 0.387 40 7.742 2
1
2 2 0.2 0.235
0.434 mA/V0.434 4
0.4821 0.434 6
So 7.74 0.482 3.73
1 1 133 0.02 0.375
m n D
d m D d
m D
m
m n D
d d d
Q
g K I
A g R A
g RA
g R
g K I
A
A A A A
R rIλ
⎛ ⎞= = ⎜ ⎟⎝ ⎠
=
= = ⇒ =
−=
+
= =
=−
= = −+
= ⋅ = − ⇒ = −
= = = =
( ) ( )( ) ( )
( ) ( )
21
2 0
k
0.387 401 2 1 2 0.387 133
0.1490.149 0.482 0.0718
m Dcm
m
cm cm
g RA
g R
A A
Ω
−−= =
+ += −= − − ⇒ =
b.
( )( ) ( )( ) ( )
1 2
1 2
03
03
0.3sin
2sin2
3.73 0.3 0.0718 2 0.975sin V
d
cm
d d cm cm
v v v tv vv t
v A v A vv t
ω
ω
ω
= − =+= =
= += − + ⇒ = −
Ideal, 0cmA =
( )( )03 3.73 0.3d dv A v= = − Or ( )03 1.12sin Vv tω⇒ = −
11.91 The low-frequency, one-sided differential gain is
( )( )
( )( )
( )
( )( )
( )
( )
022
2 2
3 12
12
12
100 0.0265.2 k
0.51 100 102 87.7
5.2 0.51
0.5 19.23 mA/V0.0262 1 19.23 10 387 pF
12
1 2 5.2 0.5 10 8 387 10
v m Cd B
C
B
v v
M m C
m
M M
HB M
v rA g R
v r R
R
r R
r
A A
C C g R
g
C C
fr R C C
π
π
π
π
μ
π π
β
π
π −
⎛ ⎞= = ⎜ ⎟+⎝ ⎠
⋅=
+
= = Ω
⋅= ⇒ =
+= +
= =
= + ⇒ =⎡ ⎤⎣ ⎦
=⎡ ⎤ +⎣ ⎦
=⎡ ⎤× × + ×⎣ ⎦
So 883 kHzHf⇒ =
11.92
a. From Equation (11.117), ( ) ( )6 120 0
1 12 2 5 10 0.8 10Zf R Cπ π −
= =× ×
Or 39.8 kHzZf =
b. From Problem 11.69, 883 .Hf kHz= From Equation (11.116(b)), the low-frequency common- mode gain is
( )
( )( )( )( )
0
6
3
4
10 4
2 11
5.2 k , 19.23 mA/VSo
19.23 10
2 101 5 100.515.2 5.2 10
9.9 1087.720log 98.9 dB
9.9 10
m Ccm
B
m
cm
dB
g RARR
r r
r g
A
C M RR
π π
π
β
−
−
−=⎡ ⎤+⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
= Ω =
−=⎡ ⎤×⎛ ⎞⎢ ⎥+ +⎜ ⎟ ×⎝ ⎠⎢ ⎥⎣ ⎦
= − ×
⎛ ⎞= =⎜ ⎟×⎝ ⎠
11.93
a. From Equation (7.72), ( )2m
Tgf
C Cπ μπ=
+
( )
( ) ( )( )
( )( )( )
( )
36
12
3 12
1 38.46 mA/V0.026
38.46 10Then 800 102
Or 7.65 10 F 7.65 pFAnd 6.65 pF
1 1 1 38.46 10
386 pF1
2
120 0.0263.12 k
11
2 3.12 1 10 6.65 386 10
O
m
M m C
HB M
H
g
C C
C CC
C C g R
fr R C C
r
f
π μ
π μ
π
μ
π π
π
π
π
π
−
−
−
= =
×× =+
+ = × ==
= + = +⎡ ⎤⎣ ⎦=
=⎡ ⎤ +⎣ ⎦
= = Ω
=⎡ ⎤× × + ×⎣ ⎦
r 535 kHzHf =
b. From Equation (11.140), ( )( )6 120 0
1 12 2 10 10 10Zf R Cπ π −
= =×
Or 15.9 kHzZf = 11.94 The differential-mode half circuit is:
( )
( )( )
( )( )
( ) ( )
02
12 2 or
111
100 0.0265.2 k
0.51 100 10
50025.2 101 5.2 101
dm C C
vE
E
vE E
vg R Rv A
r RR
r
r
AR R
π
π
π
β
ββ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =
+ +⎛ ⎞++ ⎜ ⎟⎝ ⎠
= = Ω
⎛ ⎞⎜ ⎟⎝ ⎠= =
+ +