Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
Chapter 4 4.1
(a) (i) DQn
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛′=
22
( ) 5.25.021.025.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(ii) ( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 4.24.05.221.05.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
(b) (i) ( ) 33.815.021.025.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(ii) ( )( ) 0.14.033.821.015.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
______________________________________________________________________________________ 4.2
(a) (i) DQp
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 12015.0204.022.1 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(ii) ( )22 TPSGQ
pDQ VV
LWk
I +⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 85.06.0120204.015.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGQSGQ VV V
(b) (i) ( ) 3650.0204.022.1 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(ii) ( )( ) 43.16.036204.050.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGQSGQ VV V
______________________________________________________________________________________ 4.3
( ) ( )( )( )
[ ] [ ]( )( )
2
11
2 2
11 101 3.4
1 3.0 1 53.4 1 5 3.0 1 103.4 3.0 3 10 3.4 5 0.0308
5 12.5 k0.4
D n GS TN DS
DSD
D DS
DSo
D
I K V V V
VII V
VrI
λλλ
λ λλ λλ λ
= − +
++= ⇒ =
+ +
+ = +− = ⋅ − ⋅ ⇒ =
Δ= = = Ω
Δ ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.4
( ) ( )
0
0
1
1 1 0.833 mA0.012 100
D
D D
rI
I Ir
λ
λ
=
= = ⇒ =
______________________________________________________________________________________ 4.5
(a) ( ) ( DSTNGSnD VVVKI λ+−= 12 ) ( )DSDOD VII λ+= 1
Then ( )( )λ
λ5.113.31
250.0258.0
++
=
Or ( ) 01826.03.315.11032.1 =⇒+=+ λλλ V 1−
( )( )[ ] 2433.05.101826.01250.0 =⇒+= DODO II mA
( )( ) 2252433.001826.0
11===
DOo I
rλ
kΩ
(b) ( ) ( )( )[ ] 2655.0501826.012433.0 =+=DI mA ______________________________________________________________________________________ 4.6 (a)
(i) ( )( )1 1 1000 K
0.02 0.05oD
rIλ
= = =
(ii) ( )( )1 100 K
0.02 0.5or = =
(b)
(i)
1 0.001 mA 1.0 A1000
DSD
o
VIr
μΔ
Δ = = = =
1.0 2%50
D
D
IIΔ
= ⇒
(ii)
1 0.01 10 A100
10 2%500
DSD
o
D
D
VI
rI
I
μΔ
Δ = = = ⇒
Δ= ⇒
______________________________________________________________________________________ 4.7
( )( )
1.0 mA1 1 100 K
0.01 1
D
oD
I
rIλ
=
= = =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.8
(a) DSQDDQDD VRIV +=
mA ( ) 36.05.153.3 =⇒+= DQDQ II
( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 824.04.04021.036.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
(b) ( ) 24021.0
2=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
LWk
K nn mA/V 2
( )( ) 697.136.0222 === DQnm IKg mA/V
( )( ) 1.11136.0025.0
11===
DQo I
rλ
kΩ
( ) ( )( ) 12.851.111697.1 −=−=−= Dom RrgAυ ______________________________________________________________________________________ 4.9
(a) Dm RgA −=υ
mA/V ( ) 38.0108.3 =⇒−=− mm gg
DQnm IKg 2= DQn I
LWk
⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 02.612.021.0238.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(b) mA/V ( ) 50.0105 =⇒−=− mm gg
( ) 4.1012.021.0250.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
______________________________________________________________________________________ 4.10
(a) DSQDDQDD VRIV +=
k ( ) 435.05 =⇒+= DD RR Ω
( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( ) 7.346.02.1208.05.0 2 =⎟
⎠⎞
⎜⎝⎛⇒−⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(b) ( )( ) 666.15.07.34208.02
22 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= DQ
nm I
LWk
g mA/V
( )( ) 3.1335.0015.0
11===
DQo I
rλ
kΩ
(c) ( ) ( )( ) 47.643.133666.1 −=−=−= Dom RrgAυ ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.11
[ ]
[ ]
22 2
2
22
sin sin
1sin 1 cos 22
So 1 cos 22
2ω ω
ω ω
ω
⎡ ⎤= =⎣ ⎦
= −
= −
n gs n gs n gs
n gsn gs
K v K V t K V
t t
K VK v t
t
( )
( )
2
cos 22Ratio of signal at 2 to that at :
2 s
The coefficient of this expression is then:4
n gs
n GSQ TN gs
gs
GSQ TN
K Vt
inK V V V
V
V V
ωω ω
tω
⋅
− ⋅
−
______________________________________________________________________________________ 4.12
( )( )( )( )
0.014
So 0.01 4 3 1 0.08 V
=−
= − ⇒ =
gs
GSQ TN
gs gs
VV V
V V
______________________________________________________________________________________ 4.13
(a) ( ) 66.03.324060
60
21
2 =⎟⎠⎞
⎜⎝⎛
+=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
= DDGS VRR
RV V
( ) ( )( ) 270.04.066.08021.0
222 =−⎟
⎠⎞
⎜⎝⎛=−⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= TNGS
nDQ VV
LWk
I mA
( )( ) 14.18270.03.3 =−=−= DDQDDDSQ RIVV V
(b) ( )( ) 078.2270.08021.02
22 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= DQ
nm I
LWk
g mA/V
( )( ) 18527.002.0
11===
DQo I
rλ
kΩ
(c) ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
+−=
SiDom RRR
RRRrgA
21
21υ
We find 668.78185 ==Do Rr kΩ
482406021 ==RR kΩ
So ( )( ) 3.15248
48668.7078.2 −=⎟⎠⎞
⎜⎝⎛
+−=υA
______________________________________________________________________________________ 4.14
( )( )0 ||
10 100 || 5 2.1 mA/Vv m D
m m
A g r Rg g
= −− = − ⇒ =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.15
(a) ( ) 1875.25225175
175
21
2 =⎟⎠⎞
⎜⎝⎛
+=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
= DDG VRR
RV V
( )21875.2 TNGSSnGSSDGS VVRKVRIV −+=+=
( )( )( )64.06.1111875.2 2 +−+= GSGSGS VVV
or V 58.105475.16.02 =⇒=−− GSGSGS VVV
mA ( ) ( )( ) 608.08.058.11 22 =−=−= TNGSnDQ VVKI ( ) ( )( ) 96.141608.05 =+−=+−= DSDQDDDSQ RRIVV V
(b) Sm
Dm
RgRg
A+−
=1υ
( )( ) 56.1608.012 ==mg mA/V
( )( )( )( ) 44.2
156.11456.1
−=+−
=υA
(c) ( ) ( )( )
( )( ) ( )LDLD
Sm
LDm RRRR
RgRRg
A 6094.0156.11
56.11
−=+
−=
+
−=υ
( )( ) ( )( ) 0.36094.044.275.0 =⇒−=− LDLD RRRR kΩ
1234 =⇒= LL RR kΩ ______________________________________________________________________________________ 4.16
(a) ( )DSDQDDDSQ RRIVV +−=
( )( ) 5.32125 =+⇒+−= DSDS RRRR kΩ k , then k 5.0=SR Ω 3=DR Ω
( )2TNGSQnDQ VVKI −=
V ( ) 355.22.15.12 2 =⇒−= GSQGSQ VV ( )( ) 355.35.02355.2 =+=+= SDQGSQG RIVV V
DDinDDG VRR
VRR
RV ⋅⋅=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=121
2 1
( )( ) 894122501355.3 11
=⇒= RR
kΩ
347250894 2221 =⇒== RRRR kΩ
(b) ( )( ) 464.325.12 ==mg mA/V
( ) ( )( )
( )( ) 93.25.0464.31
103464.31
−=+
−=
+
−=
Sm
LDm
RgRRg
Aυ
______________________________________________________________________________________ 4.17
(a) From Problem 4.16; kΩ , kΩ , 5.0=SR 3=DR kΩ , kΩ 8941 =R 3472 =R
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) ( )( ) 464.325.12 ==mg mA/V
( ) ( )( ) 99.7103464.3 −=−=−= LDm RRgAυ ______________________________________________________________________________________ 4.18 (a)
15 2 7.5 Kv m D
D D
A g RR R
= −− = − ⇒ =
(b)
( )( )( )
12 7.5
5 11 2
m Dv
m S
SS
g RAg R
RR
−=
+
−− = ⇒ =
+ K
______________________________________________________________________________________ 4.19
(a) 1m D
vm S
g RA
g R−
=+
(1) ( )8
1 1m D
m
g Rg
−− =
+
(2) 16 m Dg R− = −
Then
( )168 1
1 116 K
mm
D
gg
R
= ⇒ =+
=
mA/V
(b)
( )( )( )1 16
101 1
0.6 K
vS
S
AR
R
−= − =
+
=
______________________________________________________________________________________ 4.20
(a) ( )22 TNGSQn
QDQ VVL
WkII −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′==
( )( ) 247.18.05021.05.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
( ) ( )( ) 25.3247.165.05 =+−=−−−= +GSQDDQDSQ VRIVV V
(b) ( )( ) 236.25.05021.02
22 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= DQ
nm I
LWk
g mA/V
( )( ) 1005.002.0
11===
DQo I
rλ
kΩ
( ) ( )( ) 7.126100236.2 −=−=−= Dom RrgAυ
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(c) ( ) ( )( ) 86.9206100236.2 −=−=−= LDom RRrgAυ
(d) ( ) ( )( ) 51.666100236.2 −=−=−= LDom RRrgAυ ______________________________________________________________________________________ 4.21
(a) ( ) ( DSDQDSQ RRIV + )−−−= 55 ( )( ) 451.0105.5 =+⇒+−= DSDS RRRR kΩ
( )2TNGSQnDQ VVKI −=
( ) ( )( ) 143.18.085.01.0 2 =⇒−= GSQGSQ VV V 5=+ SDGS RIV
( ) 6.3851.0143.1 =⇒=+ SS RR kΩ k43.6=⇒ DR Ω
(b) ( )( ) 583.01.085.022 === DQnm IKg mA/V
( )( ) 5001.002.0
11===
DQo I
rλ
kΩ
(c) ( ) ( )( ) 19.34050043.6583.0 −=−=−= LoDm RrRgAυ ______________________________________________________________________________________ 4.22
(a) ( )DSDQDDDSQ RRIVV +−=
( )( ) 6.25.03.32 =+⇒+−= DSDS RRRR kΩ
( )2TNGSQnDQ VVKI −=
V ( )( ) 3.08.025.0 2 −=⇒−−= GSQGSQ VV 0=+ SDQGSQ RIV k( ) 6.005.03.0 =⇒=+− SS RR Ω k2=⇒ DR Ω
(b) ( )( ) 25.0222 === DQnm IKg mA/V
( ) ( )( )
( )( ) 52.16.021
10221
−=+
−=
+
−=
Sm
LDm
RgRRg
Aυ
______________________________________________________________________________________ 4.23
(a) and SDQDSQDDQDD RIVRIV ++= 0=+ SDQGS RIV
Then ( ) GSDSQTNGSDnDD VVVVRKV −+−= 2
( )( )( ) GSGS VV −++= 5.28.0225 2
Which yields , 006.04.54 2 =++ GSGS VV 0112.0−=⇒ GSV V
and mA ( ) 244.18.00112.02 2 =+−=DQI ( )( ) ( ) Ω≅Ω=⇒++= 999.8244.15.22244.15 SS RR
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) ( )( ) 155.3244.1222 === DQnm IKg mA/V
( ) ( )( )
( )( ) 07.3009.0155.3122155.3
1−=
+
−=
+
−=
Sm
LDm
RgRRg
Aυ
______________________________________________________________________________________ 4.24 a.
( )5 55 6 10
DQ S SDQ DQ D
DQ S DQ
I R V I RI R I
= + + −= + + − 5
1.
410
5
DQS
S SDQ DQ D SGQ
IR
V V I R V
=+
= + − =
2. ( )1 10DQ SI V+ = GQ
3. ( )22DQ p SGQI K V= −
4Choose 10 k 0.20 mA20S DQR I= Ω ⇒ = =
2 2
1 (0.2)(10) 3 V0.20 (3 2) 0.20 /
SGQ
P P
VK K mA
= + == − ⇒ = V
b.
( )( )( )( )
( ) ( )( )
20.20 3 2 0.20 mA
2 2 0.2 0.2 0.4 /
|| 0.4 10 ||10 2.0
DQ
m P DQ
v m D L v
I
g K I mA V
A g R R A
= − =
= = =
= − = − ⇒ = −
c.
2 2
4Choose 20 k 0.133 mA30
1 (0.133)(10) 2.33 V0.133 (2.33 2) 1.22 /
2 (1.22)(0.133) 0.806 mA/V(0.806)(10 10) 4.03
S DQ
SGQ
p p
m
v v
R I
VK K mA V
gA A
= Ω⇒ = =
= + == − ⇒ =
= == − ⇒ = −
A larger gain can be achieved. ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.25 (a)
( )( )
( )
2
20.25 0.8 0.5
1.059 V3 1.059 7.76 K
0.251.059 1.5 0.441 V
0.441 310.2 K
0.25
DQ p SGQ TP
SGQ
SGQ
S S
D S SDQ
D D
I K V V
V
V
R R
V V V
R R
= +
= −
=−
= ⇒ =
= − = − = −
− − −= ⇒ =
(b)
( )( )( )
( )( )2 2 0.8 0.25 0.8944 mA/V
0.8944 10.2 || 21.50
v m D L
m p DQ
v
v
A g R R
g K I
AA
= −
= = =
= −= −
(c)
( ) ( )|| 0.25 10.2 || 2 0.418So 0.836 peak-to-peak
O D L
O
V I R RV
Δ = Δ = =Δ =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.26
( )
( )( )
( )
( )
( )( )
2
2
2
2
2.2 2 6 0.202 /
6 0.202 2.8 2.65
188 10 1.33 k 1.33
6
1
12.21
11 2.2 1.33
2.21 2.93 2.21
3.93 2.2
DQ n GSQ TN
m n DQ
n n
TN TN
DSQ DQ S D
S D S D
m D Lv
m S
D
D
D
DD
D
I K V V
g K I
K K mA V
V V V
V I R R
R R R R
g R RA
g R
RR
RRRR
= −
=
= ⇒ =
= − ⇒ = −
= − +1 −
+ = = Ω⇒ = −
= −+
⎛ ⎞⋅− ⎜ ⎟+⎝ ⎠− =
+ −
+ − =+
−( )( )
( ) ( )( )( )( )( )
( )( )
2
2
2
11 1
22
2
1 2.23.93 1.73 2.2 2.22.2 0.47 3.93 0
0.47 0.47 4 2.2 3.931.23 k , 0.10 k
2 2.22.8 6 0.1 3.4 V
1 1 100 18 3.4 529 k
529 100 123 k529
D D D
D D D
D D
D D
G GS S
G in DD
R R RR R R
R R
R R
V V V
V R V RR RR RR
+ =+ − =+ − =
− + += ⇒ = Ω
= + = + =
= ⋅ ⋅ = = ⇒ = Ω
= ⇒ = Ω+
SR = Ω
______________________________________________________________________________________ 4.27
(a) 9−+== DDQSDQSGQS RIVVV
( ) ( )( ) 44945 2 −+=−+= TPSGQpDQSGQ VVKIV
( ) 444.14.28 2 −+−= SGQSGQSGQ VVV
Or V 071.2052.72.208 2 =⇒=+− SGQSGQSGQ VVV
( ) ( ) 518.12.1071.22 22 =−=+== TPSGQpQDQ VVKII mA
(b) ( )( ) 485.3518.1222 === DQpm IKg mA/V
( )( ) 22518.103.0
11===
DQo I
rλ
kΩ
( ) ( )( ) 8.11224485.3 −=−=−= oDm rRgAυ
(c) ( ) ( )( ) 29.88224485.3 −=−=−= LoDm RrRgAυ ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.28 (a)
( )( )
( )
2
20.5 0.25 0.8
0.614 10 0.614 18.8 K
0.50.614 3 2.386 V
2.386 1015.2 K
0.5
DQ p SGQ TP
SGQ
SGQ S
S S
D S SDQ
D D
I K V V
V
V V V
R R
V V V
R R
= +
= +
= =−
= ⇒ =
= − = − = −
− − −= ⇒ =
(b)
( )( )( )( )
2 2 0.25 0.5 0.7071 mA/V
0.7071 15.210.7
v m D
m p DQ
v
v
A g R
g K I
AA
= −
= = =
= −
= −
______________________________________________________________________________________ 4.29
( )( )
( )( )10 20 1 10 k
v m D L
DSQ DD DQ S D
S D S D
A g R R
V V I R RR R R R
= −
= − += − + ⇒ + = Ω
Let 8 k , 2 kD SR R= Ω = Ω
( )20810 mgA −=−=υ
mA/V75.1=mg ( ) 766.0122 =⇒== nnDQn KKIK mA/V 2
( )( )( ) ( )
( )( )
2 2
11 1
22
2
1 2 2 V
1 0.766 2 3.14 V3.14 2 5.14
1 1 200 20 5.14 778 k
778 200 269 k778
S DQ S
DQ n GS TN GS GS
G GS S
G in DD
V I R
I K V V V VV V V
V R V RR RR RR
= = =
= − ⇒ = − ⇒ == + = + =
= ⋅ ⋅ ⇒ = ⇒ = Ω
= ⇒ = Ω+
______________________________________________________________________________________ 4.30
(a) ( )( )( )( ) 998.0
100511005
1=
+=
+=
om
om
rgrg
Aυ
Ω≅⇒=== 2001002.0100511
oom
o Rrg
R
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) ( )( )Som
Som
RrgRrg
A+
=1υ
We have 762.45100 ==So Rr kΩ
( )( )( )( ) 960.0
762.451762.45
=+
=υA
______________________________________________________________________________________ 4.31
( )( )1
0.98 491
m L ov
m L o
m om o
m o
g R rA
g R r
g rg r
g r
=+
= ⇒ =+
( )( )
( )( )
( )( )( )( )
Also 0.491
1
0.49
49 1 490.491 49 1 50
L om
m L o L o
m L o L om
L o
m L o
L o m L o
o o
R rgg R r R r
g R r R rgR r
g R rR r g R r
r r
⎛ ⎞⎜ ⎟+⎝ ⎠= =
+ ⎛ ⎞+ ⎜ ⎟+⎝ ⎠
=+ +
= =+ + +
50 K
0.98 mA/Vo
m
r
g
=
=
______________________________________________________________________________________ 4.32 (a)
( )( )( )( )2 25
1 1 2 250.981 1 25 0.5 || 25
20.49 K
m ov
m o
v
o om
o
g rA
g rA
R rg
R
= =+ +
=
= = =
=
(b)
( )( )
( )( )
( )( )
|| 2 25 || 2 2 1.8521 || 1 2 25 || 2 1 2 1.8520.787
m o Lv
m o L
v
g r RA
g r RA
= = =+ + +
=
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.33
(a) ( ) 15.010
5.10=
−−=DQI mA
( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 594.04.08021.015.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
(b) ( )( ) 549.115.08021.02
22 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= DQ
nm I
LWk
g mA/V
( )( ) 33315.002.0
11===
DQo I
rλ
kΩ
We find 708.910333 ==So Rr kΩ
Then ( )( )
( )( )( )( ) 938.0
708.9549.11708.9549.1
1=
+=
+=
Som
Som
RrgRrg
Aυ
(c) 708.96456.010333549.111
=== Som
o Rrg
R
or Ω= 605oR______________________________________________________________________________________ 4.34
(a) 5.25.0
25.15.2=
−=
−=
S
DSQDDDQ R
VVI mA
(b) Sm
Sm
RgRg
A+
=1υ
( )( ) 33.11
5.015.0
85.0 =⇒+
= mm
m gg
gmA/V
DQn
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 2575.221.0233.11 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(c) ( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 041.16.025721.05.2 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
291.225.1041.1 =+=+= OGSQIQ VVV V ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.35
(a) ( ) 15.25.2 =⇒=⇒= QQDDQ IIVIP mA
(b) om
o rg
R 1=
( )( ) 50102.0
11===
DQo I
rλ
kΩ
So 98.15015.0 =⇒= mm
gg
mA/V
DQn
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 6.19121.0298.1 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(c) ( )( )( )( ) 990.0
5098.115098.1
1=
+=
+=
om
om
rgrg
Aυ
(d) ( )( ) 472.4110021.02 =⎟⎠⎞
⎜⎝⎛=mg mA/V
Ω=⇒=== 223502236.050472.411
oom
o Rrg
R
______________________________________________________________________________________ 4.36
(a) ( ) 917.210850350
350
21
2 =⎟⎠⎞
⎜⎝⎛
+=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
= DDG VRR
RV V
GSGQSDQ VVRI ++=10
Now ( ) 6.180204.0
2=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
LWk
K pp mA/V 2
So ( )( )( ) 917.22.146.110 2 ++−= SGQSGQ VV
We find V 084.20133.236.144.6 2 =⇒=+− SGQSGQSGQ VVV
mA ( ) 25.12.1084.26.1 2 =−=DQI V ( )( ) 5425.110 =−=SDQV
(b) ( )( ) 828.225.16.122 === DQpm IKg mA/V
( )( ) 1625.105.0
11===
DQo I
rλ
kΩ
778.14416 ==LSo RRr kΩ
( )( )
( )( )( )( ) 834.0
778.1828.21778.1828.2
1=
+=
+=
LSom
LSom
RRrg
RRrgAυ
(c) ( ) 2085.0834.0411
==⋅=⋅==i
o
Li
o
o
o
i
og R
iiA
υυ
υυ
υυmA/V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(d) 2.33536.0164828.211
=== oSm
o rRg
R
Ω= 318oR______________________________________________________________________________________ 4.37
(a) (i) ( ) 12021.0
2=⎟
⎠⎞
⎜⎝⎛=⋅
′=
LWk
K nn mA/V 2
( )( ) 472.45122 === DQnm IKg mA/V
( )( ) 10502.0
11===
DQo I
rλ
kΩ
857.2410 ==Lo Rr k Ω
( )( )Lom
Lom
RrgRrg
A+
=1υ
( )( )( )( ) 927.0
857.2472.41857.2472.4
=+
=
(ii) 10472.411
== om
o rg
R
Ω= 219oR
(b) (i) ( )( ) 828.2212 ==mg mA/V
( )( ) 25202.0
1==or kΩ
448.3425 ==Lo Rr k Ω
( )( )( )( ) 907.0
448.3828.21448.3828.2
=+
=υA
(ii) 25828.211
== om
o rg
R
Ω= 349oR______________________________________________________________________________________ 4.38 a.
( )( )
( )
( ) ( )
40.95
1 10.95 4 1 0.95 4.75 mA/V
122
4.75 2 0.030 4 47.0
mm Lv
m L m
m m
m n ox Q
gg RA
g R gg g
Wg C IL
W WL L
μ
= ⇒ =+ +
= − ⇒ =
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞= ⇒⎜ ⎟⎝ ⎠
4
=
b. ( )( )
122
4.75 2 0.030 60 3.13 mA
m n ox Q
Q Q
Wg C IL
I I
μ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= ⇒ =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.39
( )( )
( )
( )( )
( ) ( )
2
2
0
5 5 2 1 V 1 V5 1 5 1.2 k
5
2 2 5 5 10 /
1 1 20 k0.01 5
DQ n GS TN
GS GS S GS
SDQ S S
S
m n DQ
DQ
I K V V
V V V VV
I R RR
g K I mA V
rIλ
= −
= + ⇒ = − ⇒ = − =
− − += ⇒ = ⇒ = Ω
= = =
= = = Ω
( )( )
( )( )( )( )
0
01
10 20 1.2 20.878
1 10 20 1.2 2
m S Lv
m S L
v
g r R RA
g r R R
A
=+
= ⇒+
=
0 0
1 1|| || || 20 ||1.2 91.9 10S o
m
R r R Rg
= = ⇒ = Ω
______________________________________________________________________________________ 4.40
(a) ( )( ) ⇒=−=−= 05155SDQS RIV 0=GSV
( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )[ ] 252021.05 2 =⎟
⎠⎞
⎜⎝⎛⇒−−⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(b) ( )( ) 552521.02 =⎟⎠⎞
⎜⎝⎛=mg mA/V
( )( ) 20501.0
11===
DQo I
rλ
k Ω
9524.0120 ==So Rr kΩ
( )( )
( )( )( )( ) 826.0
9524.0519524.05
1=
+=
+=
Som
Som
RrgRrg
Aυ
(c) Ω=⇒== 165120511
oSom
o RRrg
R
(d) 6452.02120 ==LSo RRr kΩ
( )( )
( )( )( )( ) 763.0
6452.0516452.05
1=
+=
+=
LSom
LSom
RRrg
RRrgAυ
______________________________________________________________________________________ 4.41
0
1S
m
R Rg
=
Output resistance determined primarily by gm
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
( )( )( )
( )
( )( )
( )( )
2
2
1Set 0.2 k 5 mA/V
2 5 2 4 1.56 mA
1.56 4 21.38 V, 1.38 V
1.38 54.09 k
1.565 4.09 2
0.8701 1 5 4.09 2
mm
m n DQ DQ DQ
DQ n GS TN
GS
GS S GS
S S
m S Lv v
m S L
gg
g K I I I
I K V V
VV V V
R R
g R RA A
g R R
= Ω⇒ =
= ⇒ = ⇒ =
= −
= += − = − =
− −= ⇒ = Ω
= = ⇒ =+ +
______________________________________________________________________________________ 4.42
(a) ( )( ) 14.1410522 === DQpm IKg mA/V
( )( ) 101001.0
11===
DQo I
rλ
k Ω
( )( )( )( ) 993.0
1014.1411014.14
1=
+=
+=
om
om
rgrg
Aυ
(b) Ω=⇒=== 2.701007072.01014.14
11oo
mo Rr
gR
(c) ( )( )Lom
Lom
RrgRrg
A+
=1υ
( )( )( )( )Lo
Lo
RrRr
14.14114.14
90.0+
= ( ) 6365.0=⇒ Lo Rr kΩ
Ω=⇒= 6806365.010 LL RR ______________________________________________________________________________________ 4.43
( )
0
0
0
1
min1
D DQS L
DQ S LDQ S L
S L
DQ S
S
L
i I vR R
I R Rv I R R
R RI R
vRR
−Δ = = ⋅Δ
⋅Δ = − ⋅ = −
+
= −+
( )( )( ) ( )
( )
( ) ( )
0
1
1
min 1
m S Lv
im S L
DQ S L m S Li
m S L
DQi m S L
m
g R R vA
vg R R
I R R g R Rv
g R R
Iv g R R
g
= =+
⎡ ⎤− +⎣ ⎦=
⎡ ⎤= − +⎣ ⎦
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.44 (a)
( )3 1.5 0.25 6 K
DD DSQ DQ S
S S
V V I RR R
= += + ⇒ =
( )( )
2
2
1.5 V
0.25 0.5 0.4
1.107 V
S
DQ n GSQ TN
GSQ
GSQ
V
I K V V
V
V
=
= −
= −
=
( )( )
2
1 2 1
1 21
1.107 1.5 2.607 V
1
12.607 300 3 345.2 K 2291 K
G GSQ S
G DD L DD
V V V
RV V R VR R R
R RR
= + + + =
⎛ ⎞= = − −⎜ ⎟+⎝ ⎠
= ⇒ = ⇒ =
(b)
( )( )
( )( )( )( )
( )
2 2 0.5 0.25 0.7071 m1
0.7071 60.809
1 0.7071 6
1 1 6 1.414 || 60.7071
1.14 K
m Sv m n DQ
m S
v v
o Sm
o
g RA g K I
g R
A A
R Rg
R
= = = =+
= ⇒ =+
= = =
=
A/V
______________________________________________________________________________________ 4.45
25.0
15.01==⇒== m
mi g
gR mA/V
DQn
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 8025.021.022 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
Dm RgA =υ
kΩ ( ) 10220 =⇒= DD RR______________________________________________________________________________________ 4.46
(a) 500== Do RR Ω(b) V 2.1=GSQV
( )( ) ( )
5.03.04.02.12.2
5.03.02.2 +−−=
+−=
−=
satVR
VVI DS
D
DSDDDQ
mA 2.2=DQI
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(c) ( )2TNGSQnDQ VVKI −=
mA/V ( ) 438.34.02.12.2 2 =⇒−= nn KK 2
( )( ) 5.52.2438.322 === DQnm IKg mA/V
Ω=⇒== 1825.5
11i
mi R
gR
(d) ( )( ) 75.25.05.5 === Dm RgAυ
______________________________________________________________________________________ 4.47
( ) 48021.0
2=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
LWk
K nn mA/V 2
( )( ) 828.25.0422 === DQnm IKg mA/V
( )( ) 3.114828.2 === Dm RgAυ
Ω=⇒=== 3423536.010828.2110110 i
mi R
gR
______________________________________________________________________________________ 4.48 a.
( )
( )( )( )
( ) ( )( )( )
( )( )( )( )
2
2
2
2
2
55
5 10 3 2 1
30 59 25 0
59 59 4 30 251.35 V
2 30
3 1.35 1 0.365 mA
10 0.365 5 10 4.53 V
GS DQ S
GSDQ n GS TN
S
GS GS GS
GS GS
GS GS
DQ DQ
DSQ DSQ
V I RV
I K V VR
V V V
V V
V V
I I
V V
+ =−
= = −
− = − +
− + =
± −= ⇒
= − ⇒ =
= − + ⇒ =
=
b.
( )( )
( ) ( )0 0
2 2 3 0.365 2.093 /
1 10 0.365
m n DQ m
DQ
g K I g mA V
r rIλ
= = ⇒ =
= = ⇒ = ∞
c. ( ) ( )( )2.093 5 4 4.65v m D L vA g R R A= = ⇒ =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.49 a.
( )( )( )
( )( )( )
2
20.75 0.5 1 2.225 V5 2.2255 3
0.7510
6 10 0.75 3.70 1.63 k
DQ p SG TP
SG SG
DQ S SG S S
SDQ DQ S D
D D
I K V V
V V
I R V R R
V I R RR R
= +
= − ⇒ =−
= + ⇒ = ⇒ =
= − += − + ⇒ = Ω
.70 kΩ
b.
( )( )
1
2 2 0.5 0.75 1.225 /
1 0.816 k1.225
1.63 k
im
m p DQ
i i
o D o
Rg
g K I mA V
R R
R R R
=
= = =
= ⇒ = Ω
= ⇒ = Ω
c.
[ ]
( )( )( ) ( )
0
0
0 0
0 0 0
1/
1.63 3.701.63 2 3.70 0.816
0.368 1.84sin
1.84 2 sin 3.68sin mV
SDi
D L S m
i
i
L
RRi iR R R g
i i
i i i t A
v i R t v t
ω μ
ω ω
⎛ ⎞⎛ ⎞= ⋅⎜ ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
= = =
= = ⇒ =
______________________________________________________________________________________ 4.50
(a) ( )( ) GSDSO VsatVV −+= 25.0 15.025.04.025.0 −=+−=−+−= GSTNGSO VVVV V
( ) 975.02
15.08.1=
−−=DR k Ω
(b) Dm RgA =υ
mA/V ( ) 154.6975.06 =⇒= mm gg
DQn
m IL
Wkg ⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 7.94221.02154.6 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
LW
LW
(c) ( )22 TNGSQn
DQ VVL
WkI −⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 05.14.07.9421.02 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSQGSQ VV V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.51
(a) ( )2TNGSQnQDQ VVKII −==
V ( ) 307.16.042 2 =⇒−= GSQGSQ VV 193.2307.15.3 =−=−= GSQDSQD VVV V
554.02
193.23.3=
−=DR k Ω
(b) ( )( ) 657.52422 === DQnm IKg mA/V
Ω=⇒== 177657.511
im
i Rg
R
(c) ( ) ( )( ) 75.24554.0657.5 === LDm RRgAυ ______________________________________________________________________________________ 4.52
(a) ( )2TPSGQpDQ VVKI +=
V ( ) 493.18.05.22.1 2 =⇒−= SGQSGQ VV
51.12.1
493.13.3=
−=
−=
+
DQ
SGQS I
VVR kΩ
( ) ( ) 49.151.12.16.636.6 =⇒+−=⇒+−= DDDSDQSDQ RRRRIV kΩ
(b) ( )( ) 464.32.15.222 === DQpm IKg mA/V
( ) ( )( ) 76.3449.1464.3 === LDm RRgAυ ______________________________________________________________________________________ 4.53
(a) ( )( ) 5===
L
D
L
D
LWLW
KK
Aυ
So 25=⎟⎠⎞
⎜⎝⎛
DLW
From Example 4.11,
( ) ( )( ) 05.151
516.06.03.3=
+++−
=GSDtυ V
825.06.02
6.005.1=+
−=GSDQυ V
(b) ( ) ( )( )22 6.0825.02521.0
2−⎟
⎠⎞
⎜⎝⎛=−⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= TNGSDQ
D
nDQ VV
LWk
I
mA 0633.0=DQI Now DLDD II =
( ) ( )22TNGSLQ
LTNGSDQ
D
VVL
WVVL
W−⎟
⎠⎞
⎜⎝⎛=−⎟
⎠⎞
⎜⎝⎛
( ) TNODDTNGSDQ VVVVV −−=−125
( ) 6.03.36.0825.05 −−=− OV Or V 575.1== ODSDQ VV
______________________________________________________________________________________
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, 2.48.05 =−=−= TNLDDOtB VVV V For point A, DLDD II =
( ) ( 22TNLGSLQnLTNDGSDQnD VVKVVK −=− )
( )( ) ( ) ( )[ ]22 8.002.06.02.1 −−=−GSDQV
So ( ) 9266.06.08.02.12.0
=+=GSDQV V
Then 3266.06.09266.0 =−=−= TNDGSDQOtA VVV V For point A: V, 3266.0=OtAV 9266.0=GSDQV V For point B: V, 2.4=OtBV 9266.0=GSDQV V
(b) V, 9266.0=GSDQV
2633.23266.02
3266.02.4=+
−=DSDQV V
(c) ( ) ( )( ) 128.06.09266.02.1 22 =−=−= TNDGSDQnDDQ VVKI mA
(d) ( )oLoDmD rrgA −=υ
( )( ) 6.390128.002.0
11====
DQoLoD I
rrλ
kΩ
( )( ) 7838.0128.02.122 === DQnDmD IKg mA/V
( )( ) 1536.3906.3907838.0 −=−=υA ______________________________________________________________________________________ 4.55
(a) V 6.0=TNV
( )2TNDSnD VVKI −=
mA/V ( ) 6173.06.05.15.0 2 =⇒−= nn KK 2
( )TNDSnDS
D VVKdVdI
r−== 21
Then ( ) ( )( ) Ω=⇒−
=−
= 9006.05.16173.02
12
1 rVVK
rTNDSn
(b) mA ( )( ) 56.36.036173.0 2 =−=DI
( )( ) Ω=⇒−
= 3376.036173.02
1 rr
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.56 a.
( ) ( ) ( )( )
( ) (( )
)
22
2 2
20
0 0
0.5 0 1
0.5
0.5 0.030 10 1
0.5 9 4.92 V0.030
DQ nD GS TND
DQ
DQ nL GSL TNL nL DD O TNL
I K V V
I mA
I K V V K V V V
V
V V
= − = − −
=
= − = − −
= − −
= − ⇒ =
b.
( ) ( )
( )
( )
( )( )
2 2
//
500 4.0830
DD DL
nD i TND nL DD o TNL
nDi TND DD o TNL
nL
nDo DD TNL i TND
nL
o nD Dv
i nL L
v v
I I
K V V K V V V
K V V V V VK
KV V V V V
K
W LdV KA
dV K W L
A A
=
− = − −
− = − −
= − − −
= = − = −
= − ⇒ = −
______________________________________________________________________________________ 4.57 (a)
( ) (( )( )
( )( )( )
)2 2
2
2
2
0.1 4 1 0.9
0.9 1 1 1.95 V1.95 4 5.95 V
DQ L GSL TNL L DSL TNL
D
DQ D GSD TND
GSD GSD
GG GSD DSL GG
I K V V K V V
I mA
I K V V
V VV V V V
= − = −
= − =
= −
= − ⇒ == + = + ⇒ =
b.
( ) ( )
( )
( )
2 2
1
/ 11 / 1 /
DD DL
D GSD TND L GSL TNL
DGG i o TND o TNL
L
D Do GG i TND
L L
D Lov v
i
TNL
D L L
I I
K V V K V V
K V V V V V VK
K KV V V VK K
K KdVA AdV D
V
K K K
=
− = −
+ − − = −
⎛ ⎞+ = + − +⎜ ⎟⎜ ⎟
⎝ ⎠
= = ⇒ =+ + K
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) From Problem 4.55.
( )
( )( )( )( )
( )( )
( )( )( )( )
12
1 1.67 2 0.1 4 1
2 2 1 0.9 1.90 /
|| 1.90 1.67 || 40.691
1 || 1 1.90 1.67 || 4
LDL DSL TNL
m D DQ
m LD Lv v
m LD L
RK V V
k
g K I mA V
g R RA A
g R R
=−
= = Ω−
= = =
= = ⇒ =+ +
L
______________________________________________________________________________________ 4.58 a. From Problem 4.57.
( )( )
( )( )( )( )
|| 1.90 1.67 ||101 || 1 1.90 1.67 ||100.731
m LD Lv
m LD L
v
g R RA
g R RA
= =+ +
=
b.
0
0
1 1 1.67 0.526 ||1.671.90
0.40 k
LDm
R Rg
R
= = =
= Ω
______________________________________________________________________________________ 4.59
(a) ( )oLoDmD rrgA −=υ
( )( ) 1005.002.0
11===
DQDoD I
rλ
kΩ
( )( ) 505.004.0
11===
DQLoL I
rλ
kΩ
33.3350100 ==oLoD rr kΩ Then ( ) 20.133.3340 =⇒−=− mDmD gg mA/V
( )DQD
nmD I
LWk
g ⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 4.145.021.0220.1 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
DD LW
LW
(b) ( )22 TPSGQ
L
pDQ VV
LWk
I +⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 107.14.050204.05.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGQSGQ VV V
BSGQ VVV −= +
V 393.15.2107.1 =⇒−= BB VV
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(c) ( )22 TNGSDQ
D
nDQ VV
LWk
I −⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 233.14.04.1421.05.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= GSDQGSDQ VV V
______________________________________________________________________________________ 4.60
(a) (i) ( )( ) 10025.004.0
11===
DQDoD I
rλ
kΩ
( )( ) 20025.002.0
11===
DQLoL I
rλ
kΩ
67.66200100 ==oLoD rr k Ω
( )oLoDmD rrgA −=υ ( )⇒−=− 67.6625 mDg 375.0=mDg mA/V
DQD
pmD I
LWk
g ⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 03.725.0204.02375.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
DD LW
LW
(ii) ( )2
2 TNBL
nDQ VV
LWk
I −⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 107.14.01021.025.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= BB VV V
(iii) ( )22 TPSGDQ
D
pDQ VV
LWk
I +⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 933.16.003.7204.025.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGDQSGDQ VV V
(b) (i) ( )( ) 2501.004.0
11===
DQDoD I
rλ
kΩ
( )( ) 5001.002.0
11===
DQLoL I
rλ
kΩ
7.166500250 ==oLoD rr kΩ
( )oLoDmD rrgA −=υ ( ) 15.07.16625 =⇒−=− mDmD gg mA/V
DQD
pmD I
LWk
g ⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
22
( ) 81.21.0204.0215.0 =⎟
⎠⎞
⎜⎝⎛⇒⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
DD LW
LW
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(ii) ( )2
2 TNBL
nDQ VV
LWk
I −⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 874.04.01021.01.0 2 =⇒−⎟⎠⎞
⎜⎝⎛= BB VV V
(iii) ( )22 TPSGDQ
D
pDQ VV
LWk
I +⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 934.16.081.2204.01.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGDQSGDQ VV V
______________________________________________________________________________________ 4.61
( )
( ) ( )
21
1 1 1
85 50 2.125 mA/V2
2 2 2.125 0.1 0.9220
n
m n D
K
g K I
⎛ ⎞= ⇒⎜ ⎟⎝ ⎠
= = =
( ) ( )
( )( )
11 1
22 2
1 1 200 K0.05 0.1
1 1 133.3 K0.075 0.1
oD
oD
rI
rI
λ
λ
= = =
= = =
( ) ( )( )1 1 2|| 0.922 200 ||133.373.7
v m o o
v
A g r rA
= − = −= −
______________________________________________________________________________________ 4.62
( )
( )( )
21
1 1 1
40 50 1.0 mA/V2 2
2 2 1 0.1 0.6325 mA/V
pp
m p D
k wKL
g K I
′ ⎛ ⎞ ⎛ ⎞= = ⇒⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= = =
( )( )
( )( )
11 1
22 2
1 1 133.3 K0.075 0.1
1 1 200 K0.05 0.1
oD
oD
rI
rI
λ
λ
= = =
= = =
( ) ( )( )1 1 2 0.6325 133.3 || 200
50.6v m o o
v
A g r r
A
= − = −
= −
______________________________________________________________________________________ 4.63
(a) DLDD II =
( ) ( 22TNLGSLnLTNDGSDnD VVKVVK −=− )
( ) 4.04.05.0
2−=−− OOI VVV
4.038.02 −=− OI VV
( )4.031
32
⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛= IO VV
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
)
For V, V 8.0=IV 4.0=OV For V, V 5.2=IV 533.1=OV
(b) ( ) ( )( 22 4.05.0 −=−= OTNLGSLnLD VVVKI
( ) ( ) ( )22
533.0325.04.04.0
31
325.0 ⎥
⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛=⎥
⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛= IID VVI
For V, ; For 8.0=IV 0=DI 5.2=IV V, 642.0=DI mA (c) From (a), voltage gain = constant = 2/3 = 0.667
______________________________________________________________________________________ 4.64
(a) ( ) ( ) 1.14.015.2 =−−=+= TPSGSD VVsatV V
( ) ( ) 4.11.15.2max =−=−=⇒−= ++ satVVVVVV SDOOSD V
(b) ( ) ( )( ) 121.04.015.25204.0
222 =−−⎟
⎠⎞
⎜⎝⎛=+⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= TPSGL
L
pD VV
LWk
I mA
(c) ( )2
2 TPSGDD
pD VV
LWk
I +⎟⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′=
( )( ) 748.04.050204.0121.0 2 =⇒−⎟
⎠⎞
⎜⎝⎛= SGDSGD VV V
(d) ( )( ) 6957.0121.050204.02
22 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ ′= D
D
pmD I
LWk
g mA/V
( )( ) 6.330121.0025.0
11====
DoLoD I
rrλ
kΩ
( )( )
( )( )( )( ) 9914.0
3.1656957.013.1656957.0
1=
+=
+=
oLoDmD
oLoDmD
rrgrrg
Aυ
______________________________________________________________________________________ 4.65
(a) ( )2TNGSDQnDQ VVKI −=
V ( ) 307.16.021 2 =⇒−= GSDQGSDQ VV
( )2TPSGLQpDQ VVKI +=
V ( ) 014.26.05.01 2 =⇒−= SGLQSGLQ VV V 286.1014.23.3 =−=OV ( ) 593.2307.1286.1 =−−=−= SODSDQ VVV V (b) ld II = omLimD VgVg =
p
n
mL
mD
i
o
KK
gg
VV
A ===υ
(c) 25.0
2==υA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.66 (a)
( )
( )( )
( )( )
( ) ( )
21
1 1 1
11 1
22 2
85 50 2.125 mA/V2
2 2 2.125 0.1 0.922 mA/V
1 1 200 K0.05 0.1
1 1 133.3 K0.075 0.1
n
m n D
oD
oD
K
g K I
rI
rI
λ
λ
⎛ ⎞= ⇒⎜ ⎟⎝ ⎠
= = =
= = =
= = =
(b)
( ) ( )( )( )( )
11
11
1
11 1 2
1 1 1.085 K0.922
1.085 0.9560.050 1.085 0.050
0.956 0.922 200 133.3
70.5
im
igs i i
i
gsv m o o
i
v
Rg
RV V V
RV
A g r rV
A
= = =
⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
= − ⋅ = +
=
iV
(c) 1
1 10.05 0.05 1.135 K0.922i i
m
R Rg
= + = + ⇒ =
(d) 1 2 200 133.7 80 Ko o o oR r r R≈ = ⇒ ≈ ______________________________________________________________________________________ 4.67 (a)
( )( )( )( )
( )( )
1 1
2 2
1 2
2 2 2 0.1 0.8944 mA/V
2 2 2 0.1 0.8944 mA/V
1 1 100 K0.1 0.1
m n D
m p D
o oD
g K I
g K I
r rIλ
= = =
= = =
= = = =
(b) The small-signal equivalent circuit
(1)
2 21 2 2
1 2
0sg sg om i m sg
o o
V Vg V g V
r r−
+ + + =V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (2)
( )
22 2
2
2 22 2
2 2
1 1 1
1 1 1 0.8944 0.0331750 100 100
o sgom sg
o o
o sg mo o o
o sg sg o
V VVg V
r r
V V gr r r
V V V V
−+ =
⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞+ = + ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(1)
( )
( )
1 2 21 2 2
1 1
1 10.8944 0.03317 0.8944100 100 100
0.8944 0.01 0.03033
44
om i sg m
o o o
oi o
i o
o
i
Vg V V g
r r r
VV V
V VVV
⎛ ⎞+ + + =⎜ ⎟
⎝ ⎠⎛ ⎞+ + +⎜ ⎟⎝ ⎠
= −
= −
=
(c) For output resistance, set 0.iV =
(1)
22 2
2
x sgxm sg x
o o
V VVg V I
r r−
+ = +
(2)
2 22 2
1 2
0sg sg xm sg
o o
V Vg V
r r−
+ +V
=
(2)
( )
2 21 2 2
2
2
1 1
1 10.8944100 100 100
0.010936
xsg m
o o o
xsg
sg x
VV g
r r r
VV
V V
⎛ ⎞+ + =⎜ ⎟
⎝ ⎠⎛ ⎞+ + =⎜ ⎟⎝ ⎠=
(1)
( )
( )
2 22 2
1 1 1
1 1 10.010936 0.894450 100 1000.03 0.0098905
49.7 K
x x sg mo o o
x x x
x x
xo
x
I V V gr r r
I V V
I VV
RI
⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛= + − +⎜ ⎟ ⎜⎝ ⎠ ⎝
= −
= =
⎞⎟⎠
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.68
(a) ( )21111 TNGSQnDQ VVKI −=
V ( ) 6.16.02.02.0 12
1 =⇒−= GSQGSQ VV
( )22222 TPSGQpDQ VVKI +=
V ( ) 307.16.00.15.0 22
2 =⇒−= SGQSGQ VV
32.06.0
1 ==SR kΩ
V 6.226.01 =+=DV
122.0
6.251 =
−=DR k Ω
V 2.26.06.16.011 =+=+= GSQG VV
( ) ( ) ( )(540015152.21121
21 R
RRRR
RV inG =⋅⋅=⎟⎟
⎠
⎞⎜⎜⎝
⎛+
== )
Or k and 9091 =R Ω 40021 == inRRR kΩ 7142 =⇒ R kΩ 907.3307.16.2212 =+=+= SGQDS VVV V
19.25.0907.35
2 =−
=SR kΩ
V 907.03907.3322 =−=−= SD VV
81.15.0
907.02 ==DR k Ω
(b) ( )( ) 4.02.02.022 111 === DQnm IKg mA/V
( )( ) 414.15.0122 222 === DQpm IKg mA/V
( )( ) 21212211 DDmmDmDm RRggRgRgA =−−=υ
( )( )( )( ) 3.1281.112414.14.0 ==υA______________________________________________________________________________________ 4.69
(a) ( )21111 TNGSnDQ VVKI −=
V ( ) 307.16.02.01.0 12
1 =⇒−= GSQGS VV
( )2222 TPSGQpDQ VVKI +=
V ( ) 10.16.00.125.0 22
2 =⇒−= SGQSGQ VV ( )( ) 407.111.0307.11111 =+=+= SDQGSQG RIVV V
DDG VRR
RV ⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=21
21 DDin VR
R⋅⋅=
1
1
( )( ) 5863.32501407.1 11
=⇒= RR
kΩ
25021 == inRRR kΩ k4362 =⇒ R Ω ( )( ) 3.12.111.01111 =+=+= DSQSDQD VRIV V
201.0
3.13.31 =
−=DR kΩ
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
V 4.21.13.1212 =+=+= SGQDS VVV
6.325.0
4.23.32 =
−=SR k Ω
V 6.08.14.2222 =−=−= SDQSD VVV
4.225.06.0
2 ==DR kΩ
(b) ( )( ) 2828.01.02.022 111 === DQnm IKg mA/V
( )( ) 0.125.0122 222 === DQpm IKg mA/V
( )( )( )( ) 6.134.2200.12828.02121 === DDmm RRggAυ ______________________________________________________________________________________ 4.70
( ) 4.020204.0
1 =⎟⎠⎞
⎜⎝⎛=pK mA/V 2
( ) 0.48021.0
2 =⎟⎠⎞
⎜⎝⎛=nK mA/V 2
(a) 61.06.0
1 ==SR k Ω
V 2.016.08.11 =−−=DV
( ) 201.0
8.12.01 =
−−=DR kΩ
( )2111 TPSGQpDQ VVKI +=
V ( ) 90.04.04.01.0 12
1 =⇒−= SGQSGQ VV
( )2222 TNGSQnDQ VVKI −=
V ( ) 6739.04.043.0 22
2 =⇒−= GSQGSQ VV 3.09.06.08.16.08.1 11 =−−=−−= SGQG VV V
( ) 8.16.321
21 −⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=RR
RVG
( )( ) 3438.16.320013.0 11
=⇒−= RR
kΩ
20021 =RR k kΩ 4802 =⇒ R Ω V 2.00.16.08.11 =−−=DV 4739.06739.02.0212 −=−=−= GSQDS VVV V
( ) 42.43.0
8.14739.02 =
−−−=SR kΩ
(b) ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
=22
22
11
11
11 Sm
Sm
Sm
Dm
RgRg
RgRg
Aυ
( )( ) 4.01.04.022 111 === DQpm IKg mA/V
( )( ) 191.23.0422 222 === DQnm IKg mA/V
( )( )( )( )
( )( )( )( ) 13.2
42.4191.2142.4191.2
64.01204.0
−=+
⋅+−
=υA
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(c) Ω=⇒=== 41442.44564.042.4191.211
22
oSm
o RRg
R
______________________________________________________________________________________ 4.71 (a)
( )
( )( )( )
( ) ( )( )( )
211 1 1 1
2
21 1 1
21 1
2
1 1
10
10 4 10 4 4
40 159 150 0
159 159 4 40 1502.435 V
2 40
GSDQ n GS TN
S
GS GS GS
GS GS
GS GS
VI K V VR
V V V
V V
V V
−= = −
− = − +
− + =
± −= ⇒ =
( )( )( )( )
( )( )
21 1
1 1
2
2 2
4 2.435 2 0.757 mA
20 0.757 10 12.4 V
Also 0.757 mA
20 0.757 10 5 8.65 V
DQ DQ
DSQ DSQ
DQ
DSQ DSQ
I I
V V
I
V V
= − ⇒ =
= − ⇒ =
=
= − + ⇒ =
(b) ( )( )1 2 1 22 2 4 0.757 3.48 /m m DQ m mg g KI g g mA= = = ⇒ = = V
c.
( )( )( )( )
( ) ( )( )( ) ( ) ( )
( )( ) ( )
0 2 2
2 1 1 2 2 1 2
1 2 1 2
2 2 2 1 2 1 2 1 2
2 2 2 1 2 1 2 1 2 1 1 2
1 1 22
2 1 2 1 1 2
1 20
1
m gs D L
gs m gs m gs S S
i gs gs gs i gs
gs m gs S S m i gs S S
gs m gs S S m gs S s m i S S
m i S Sgs
m S S m S S
m mv
i
V g V R R
V g V g V R R
V V V V V V
V g V R R g V V R R
V g V R R g V R R g V R R
g V R RV
g R R g R R
g gVA
V
= −
= − −
= − ⇒ = +
+ = − +
+ + = −
−=
+ +
= =( )( )
( )( )( ) ( )( )( )( )
1 2
1 2 1 2
2
1
3.48 10 10 5 22.42
1 3.48 3.48 10 10
S S D L
m m S S
v v
R R R Rg g R R
A A
+ +
= ⇒ =+ +
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.72 a.
( )( )( )
( )
( ) ( )
12
12
1 1
3 31 3
1 2 3
3 mA5 3 1.2 5 1.4 V
3 2 1 2.225 V2.225 1.4 0.825 V
5 0.825 5 82.5 k500
DQ
S DQ S
DQ GS TN
GS GS
G GS S
G
IV I R
I K V V
V VV V V
R RV R
R R R
== − = − = −
= −
= − ⇒ == + = − =
⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟+ + ⎝ ⎠⎝ ⎠Ω
( ) ( )
1 1 1
2 1
2 3 2 32
1 2 3
2 3 2
1.4 2.5 1.1 V1.1 2.225 3.325 V
5 3.325 5500
332.5 250 k
D S DSQ
G D GS
G
V V VV V V
R R R RVR R R
R R R
= + = − + == + = + =
⎛ ⎞+ +⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟+ + ⎝ ⎠⎝ ⎠+ = ⇒ = Ω
1 1
2 1 2
500 250 82.5 167.5 k
1.1 2.5 3.6 V5 3.6 0.467 k
3
D D DSQ
D D
R R
V V V
R R
= − − ⇒ =
= + = + =−
= ⇒ = Ω
Ω
b.
( )( )( )( )
1
1 2 2 2 3 4.90 /
4.90 0.467 2.29
v m D
m n DQ
v v
A g R
g K I mA V
A A
= −
= = =
= − ⇒ = −
______________________________________________________________________________________ 4.73 a.
( )( )( )
( )( )
( )( )
1 12
1 12
1 1
1 1 1 3 3 3
1 1 1
2 1
2 3
2 3 2
2 1 2
10 5 2 10 0
5 4 1.5 2.618 V2.618 V 0.1 26.2 k
0 3.5 3.5V3.5 2.62 6.12V
0.161.2 k 35 k
S DQ S S
DQ GS TN
GS GS
G GS S
D S DSQ
G D GS
D D DSQ
V I R V
I K V V
V VV V V IR R R
V V VV V V
R RR R R
V V V
= − = − ⇒ =
= −
= − ⇒ == + = = = ⇒ =
= + = + == + = + =
= ++ = Ω⇒ = Ω
= + =
Ω
1 1
3.5 3.5 7.0 V10 7 0.6 k
510 6.12 38.8 k
0.1
D DR R
R R
+ =−
= ⇒ = Ω
−= ⇒ = Ω
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.
( )( )( )( )
1
1 2 2 4 5 8.944 /
8.944 0.6 5.37
v m D
m n DQ
v v
A g R
g K I mA V
A A
= −
= = =
= − ⇒ = −
______________________________________________________________________________________ 4.74 a.
( )
( )
( )
2
2
1
4 6 13
43 1 0.551 V6
6 10 4 1 k
GSDQ DSS
P
GS
GS GS
DSQ DD DQ D
D D
VI IV
V
V V
V V I RR R
⎛ ⎞= −⎜ ⎟
⎝ ⎠
⎛ ⎞= −⎜ ⎟⎜ ⎟−⎝ ⎠
⎡ ⎤= − − ⇒ = −⎢ ⎥
⎣ ⎦= −
= − ⇒ = Ω
b.
( )( )
( )( )0 0
2 62 0.5511 1 3.2653 3
1 1 25 k0.01 4
DSS GSm m
P P
DQ
I Vg gV V
r rIλ
⎛ ⎞ −⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠
= = ⇒ = Ω
mA/V
c. ( ) ( )( ) 14.3125265.3 −=⇒−=−= υυ ARrgA Dom ______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.75
( )
( )
( )( )
( ) ( )
( ) ( ) ( )( )
1 22
2
1 2
2
2
2
2
2
0
1
1 0
2 0.1 0.25 1 0
20.7 1 0
2 2
0.175 1.7 0.7 0
1.7 1.7 4 0.175 0.70.4314 V
2 0.175
2
GS DQ S S
GSDQ DSS
P
GSGS DSS S S
P
GSGS
P
VGS GSGS
GS GS
GS GS
m
V I R R
VI IV
VV I R RV
VV
V
VV
V V
V V
g
+ + =
⎛ ⎞= −⎜ ⎟
⎝ ⎠
⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠⎛ ⎞⎜ ⎟+ − + =⎜ ⎟− −⎝ ⎠+ + =
− ± −= ⇒ =
=
−
( )
( ) ( )( )( )( )
( )( ) ( )
1
00 0
2 2 0.4311 1 1.569 mA/V2 2
1.569 8 43.62
1 1 1.569 0.1
/ 503.62 45.2/ 4
DSS GSm
P P
m D Lv v
m S
L Gi i
i i G i L
I V gV V
g R RA A
g R
v Ri v RA Ai v R v R
⎛ ⎞ −⎛ ⎞− = − ⇒ =⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠− −
= = ⇒ = −+ +
⎛ ⎞= = = ⋅ = − ⇒ = −⎜ ⎟⎝ ⎠
______________________________________________________________________________________ 4.76
( )( )( )
( ) ( )
2
2
4 mA2
10 V2
10 20 4 2.5 k2 V 4 0.5 k , 2.0 k
1
44 8 1 4.2 1 1.23 V4.2 8
2 1.23
DSSDQ
DDDSQ
DSQ DD DQ S D
S D S D
S DQ S S S D
GSDQ DSS
P
GSGS GS
G S GS
G
II
VV
V V I R RR R R R
V I R R R R
VI IV
V V V
V V V
V
= =
= =
= − += − + ⇒ + = Ω= = = ⇒ = Ω = Ω
⎛ ⎞= −⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞= − ⇒ = − − ⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠= + = −
( ) ( )2 22 1
1 2
0.77 V 20 20 3.85 k , 96.2 K100
R R R RR R
⎛ ⎞ ⎛ ⎞= = = ⇒ = Ω =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.77 a.
( ) ( )
( )( )
2
2
2
1 2 1
11
5 mA2
12 6 V2 2
12 6 1.2 k5
1
55 10 1 5 1 1.464 V5 10
6 1.464 4.536 V
1
14.536 100 12 26
DSSDQ
DDDSQ
S S
GSDQ DSS
P
GSGS GS
G S GS
G DD in DD
II
VV
R R
VI IV
V V V
V V V
RV V R VR R R
RR
= =
= = =
−= ⇒ = Ω
⎛ ⎞= −⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞= − ⇒ = − − ⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠= + = − =
⎛ ⎞= = ⋅ ⋅⎜ ⎟+⎝ ⎠
= ⇒ =
22
2
5 k
265 100 161 k265
R RR
Ω
= ⇒ = Ω+
b.
( )( )
( )( )( )( )
( )( )( )( )
0
0
0
0 0
2 102 1.461 1 2.83 mA/V5 5
1 1 20 k0.01 5
1
2.83 20 1.2 0.50.495
1 2.83 20 1.2 0.5
1 1 1.2 0.353 1.2 0.273 k2.83
DSS GSm m
P P
DQ
m s Lv
m S L
v v
Sm
I Vg g
V V
rI
g r R RA
g r R R
A A
R R Rg
λ
⎛ ⎞ −⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠
= = = Ω
=+
= ⇒ =+
= = = ⇒ = Ω
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.78 a.
( )
( )
( )( )
( )
( ) ( )( )( )
2
1 2
2
2
2
2
2
110 10 5.5 V110 90
101
10 5.5 2 5 11.75
4.5 10 1 1.143 0.3265
3.265 12.43 5.5 0
12.43 12.43 4 3.265 5.50.
2 3.265
G DD
G GS GSDQ DSS
S P
GSGS
GS GS GS
Gs GS
GS GS
RV VR R
V V VI I
R V
VV
V V V
V V
V V
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
− − ⎛ ⎞= = −⎜ ⎟
⎝ ⎠
⎛ ⎞− + = −⎜ ⎟⎝ ⎠
+ = − +
− + =
± −= ⇒
( )
=
( )( )
2
511 V
0.5112 1 1.00 mA1.75
10 1.00 5 5.0 V
DQ DQ
SDQ SDQ
I I
V V
⎛ ⎞= − ⇒ =⎜ ⎟⎝ ⎠
= − ⇒ =
b.
( )
( )( )
( )( )( )( )
( )( )
( )
00
1 2
2 22 0.5111 1 1.618 mA/V1.75 1.75
1.618 5 100.844
1 1 1.618 5 10
//90 110 49.5 k
49.50.844 4.1810
DSS GSm m
P P
m S Lv v
m S L
L ii v
i i i L
i
i i
I Vg gV V
g R RA A
g R R
v Ri RA A
i v R RR R R
A A
⎛ ⎞ ⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
= = ⇒ =+ +
⎛ ⎞= = = ⋅⎜ ⎟
⎝ ⎠= = = Ω
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
c.
( )( )1.0 mA3.33 1.0 3.33 V
d
sd
ivΔ =
= = 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
( )( )
( )
( )
2
2
1
44 8 1 4 1 1.17 V4 8
7.5 20 4 3.125 k2 82 1.171 1 2.83 mA/V
4 43.125
13 1
GSDQ DSS
P
GSGS GS
SDQ DD DQ S D
S D S D
DSS GSm m
P P
S D
m Dv
m S
m S
VI IV
V V V
V V I R RR R R R
I Vg gV V
R Rg RAg R
g R
⎛ ⎞= −⎜ ⎟
⎝ ⎠⎛ ⎞⎛ ⎞= − ⇒ = − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − += − + ⇒ + = Ω
⎛ ⎞ ⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
= −−
=+
− + = −
( )( ) ( )
( )( )
( )2 22 1
1 2
3 1 2.83 3.125 2.83
9.844 2.83 0.9433 2.61 k 0.516 k
20 4 0.516 17.94 V17.94 1.17 16.77 V
20 335 k . 65 k400
m D
D D
D D D S
S S
G S GS
G DD
g RR R
R R R R
V VV V V
R RV V R RR R
+ − =⎡ ⎤⎣ ⎦− = ⇒ = Ω =
= − ⇒ == − = − =
⎛ ⎞ ⎛ ⎞= = ⇒ = Ω =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
Ω
Ω
______________________________________________________________________________________
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