Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
0
Chapter 9
9.1 (a)
( )( )( )
2 1
3 31 10 10 50O d
d d
v A v v
A A− −
= −
= − − ⇒ =
(b)
( )32 2
2
1 500 10 1 0.5 500
3 mV
v v
v
−= − = + =
=
(c)
( )1 1
1
5 500 1 500 4950.990 V
v vv
= − ⇒ ==
(d) 0Ov =
(e)
( )( )2
2
2
3 500 0.5
250 3 5000.506 V
v
vv
− = − −
− − == −
______________________________________________________________________________________ 9.2
(a) 442 102
102 −×−=
−==
od
O
Aυ
υ V
Iυυ ⋅⎟⎠⎞
⎜⎝⎛+
=200011
2
4002.02001
1102 4 −=⇒⋅⎟⎠⎞
⎜⎝⎛=×− −
II υυ V
(b) Iυυ ⋅⎟⎠⎞
⎜⎝⎛+
=200011
2
( ) 32 109995.02
20011 −×=⎟
⎠⎞
⎜⎝⎛=υ V
( ) 5.1000109995.01 32 =⇒×=== −
odododO AAA υυ______________________________________________________________________________________ 9.3
(a) V ( ) ( )( ) 50010.20000.2105 312 −=−×=−= υυυ odO A
(b) ( )12 υυυ −= odO A
( )( ) 00265.30025.3102000.3 114 =⇒−×=− υυ V
(c) ( )12 υυυ −= odO A
( )( ) 43 1091001.001.080.1 ×=⇒×−−= −odod AA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.4
( )
( )
25
0.790 0.8025
0.9875 2524.6875 0.0125
1975 K
iid I
i
i
i
i i
i
i
Rv v
R
RR
R RR
R
⎛ ⎞= ⎜ ⎟+⎝ ⎠
⎛ ⎞= ⎜ ⎟+⎝ ⎠
+ ==
=
______________________________________________________________________________________ 9.5
(a) 1020200
1
2 −=−
=−
=RR
Aυ
(b) 340120
−=−
=υA
(c) 14040
−=−
=υA
______________________________________________________________________________________ 9.6
200 1020
and for each case20 k
⎫= − = − ⎪⎪⎬⎪= Ω ⎪⎭
v
i
A
R
______________________________________________________________________________________ 9.7 a.
1
100 1010
10 k
= − = −
= = Ω
v
i
A
R R b.
510
100100−=−=υA
k 101 == RRi Ω
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ c.
100 510 10
10 10 20 K
v
i
A
R
= − = −+
= + = ______________________________________________________________________________________ 9.8
(a) 200102040
2622
2 =⇒×
=⇒−
=−
RRR
i Oυ kΩ
67.1620012 111
2 =⇒−
=−⇒−
= RRR
RAυ kΩ
(b) μυ
5.7102005.100
2132
2 −==⇒×−
=−
= iiR
i O A
125.012
5.1−=⇒
−+
== IO
I Aυ
υυ
υ
V
______________________________________________________________________________________ 9.9
2
1v
RA
R= −
(a) 10vA = −
(b) 1vA = −
(c) 0.20vA = −
(d) 10vA = −
(e) 2vA = −
(f) 1vA = −
______________________________________________________________________________________ 9.10
(a) 2002003 211
2 =⇒−
=−
=− RRR
RkΩ , 67.661 =R kΩ
(b) 2002008 211
2 =⇒−
=−
=− RRR
RkΩ , 251 =R kΩ
(c) 20020020 211
2 =⇒−
=−
=− RRR
RkΩ , 101 =R kΩ
(d) 100200
5.0 22
1
2 =⇒−
=−
=− RR
RR
kΩ , 2001 =R kΩ
______________________________________________________________________________________ 9.11
(a) 5105025.0
1611
1 =⇒×
=⇒=−
RRR
i Iυ kΩ
5.325
5.6 22
1
2 =⇒−
=−⇒−
= RR
RR
Aυ kΩ
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) 6154.05.6
4=
−−
==υ
υυ
AO
I V
123.05.32
421 === ii mA
______________________________________________________________________________________ 9.12
(a) 1
2
RR
A−
=υ
2525
20 12 =⇒
−=− R
Rk , Ω 5002 =R kΩ
(b) 50100020 11
=⇒−
=− RR
kΩ , 12 =R MΩ
(c) For (a), μυ
825
2.01
11 −=⇒
−== i
Ri I A
For (b), μυ
450
2.01
11 −=⇒
−== i
Ri I A
______________________________________________________________________________________ 9.13 a.
2 2 2
1 1 1
2 2
1 1
1.05 1.1050.95
0.95 0.9051.05Deviation in gain is +10.5% and 9.5%
⎛ ⎞= ⇒ = ⎜ ⎟
⎝ ⎠⎛ ⎞
= ⎜ ⎟⎝ ⎠
−
vR R RAR R R
R RR R
b.
2 2 2
1 1 1
1.01 0.991.02 0.980.99 1.01
Deviation in gain 2%
vR R RAR R R
⎛ ⎞ ⎛ ⎞⇒ = ⇒ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠= ±
2
1
RR
______________________________________________________________________________________ 9.14
(a) (i) ( ) 320.0115
1
2 =−−
=⋅−
= IO RR
υυ V
(ii) 20.015
30
22 −=
−=
−=
Ri Oυ mA
75.043===
L
OL R
iυ
mA
( ) 95.020.075.02 =−−=⇒=+ OLO iiii mA
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) (i) ( ) 75.005.0115
−=−
=Oυ V
(ii) ( ) 05.015
75.002 =
−−=i mA
1875.0475.0
−=−
=Li mA
2375.005.01875.02 −=−−=−= iii LO mA
(c) (i) ( tO ωυ sin8115−
= ) (mV) ⇒ tO ωυ sin12.0−= (V)
(ii) tit
i ωω
sin815sin12.0
22 =⇒= ( μ A)
tit
i LL ωω
sin304sin12.0
−=⇒−
= ( μ A)
tiii LO ωsin382 −=−= ( μ A) ______________________________________________________________________________________ 9.15
2
1 5
30 2.5% 29.25 30.75
v
v v
RA
R RA A
= −+
= − ± ⇒ ≤ ≤
2 2
1 1
So 29.25 and 30.752 1
R RR R
= =+ +
( ) (1 1We have 29.25 2 30.75 1R R+ = + )
1 2Which yields 18.5 and 599.6 R k R= Ω = kΩ
For 25 , then 0.731 0.769 I Ov mV v= ≤ ≤ V
______________________________________________________________________________________ 9.16
( ) 2.115.01080
1
21 =−
−=⋅
−= IO R
Rυυ V
( ) 62.120100
13
4 −=−
=⋅−
= OO RR
υυ V
( ) 1510
15.021
121 −==⇒
−=== ii
Rii Iυ μ A
6020
2.143
3
143 ==⇒=== ii
Rii Oυ μ A
At 1Oυ : ( ) 7515601312 =−−=⇒=+ OO iiii μ A; Out of Op-Amp At Oυ : 6042 == iiO μ A; Into Op-Amp ______________________________________________________________________________________ 9.17
1003
4
1
2 =⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
RR
RR
I
O
υυ
For 50=Iυ mV, ( )( ) 505.0100 ==Oυ V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
If μ504 =i A, 10010505
464 =⇒×
=−
RR kΩ
Set k 101 =R Ω
Then 1010010
1003
2
3
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛==
RR
RR
I
O
υυ
Set k , k 1002 =R Ω 103 =R Ω______________________________________________________________________________________ 9.18
3005
6
3
4
1
2 −=⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
RR
RR
RR
Aυ
For 6=Oυ V, set μυ
606
6 ⇒=R
i O A 10066
6
=⇒= RR
kΩ
Set kΩ so that 2006 =R 306 =i μ A Set kΩ 201 =R
Now ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
53
42 20020
300RR
RR
For example, set kΩ and 1002 =R 205 =R kΩ
Then ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
3
4
3
4 5020
20020
100300RR
RR
Or 63
4 =RR
, set k and 203 =R Ω 1204 =R kΩ
______________________________________________________________________________________ 9.19
(a) ( ) 8.840.0122
1
2 =−⎟⎠⎞
⎜⎝⎛−=⋅
−= IO R
Rυυ V
(b) ( )( )
8993.2123
10511
122
111
1
31
21
2 −=
⎥⎦⎤
⎢⎣⎡
×+
⋅−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++
⋅−
=
RR
A
RR
A
od
υ
( )( ) 7597.840.08993.21 =−−=Oυ V (c) ( )( ) 956.2122998.0 −=−=υA
( )( )
4101477.12311
122956.21 ×=⇒
⎥⎦
⎤⎢⎣
⎡+
⋅−=− od
od
A
A
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.20 (a)
( )
2
1 2
1
3
111 1
100 1125 1 5
5 103.9960
v
od
v
RA
R RA R
A
= − ⋅⎡ ⎤⎛ ⎞+ +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
= − ⋅⎡ ⎤+⎢ ⎥×⎣ ⎦
= −
(b) ( )3.9960 1.00 3.9960 VO Ov v= − ⇒ = −
(c) 4 3.9960 100% 0.10%
4−
× =
(d)
( )( )
2 1 1
1 3
1
3.99605 10
0.7992 mV
O od od
O
od
v A v v A v
vv
Av
+
= − = −
− −= − =
×=
______________________________________________________________________________________ 9.21
(a) 98431.9
101001
10711
110100
3
−=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +
×+
⋅−
=υA
7011.09843.97
−=−
==υ
υυ
AO
I V
11077
131 −=⇒×−
=−
= υυ
υυAO mV
(b) ( ) 43
1
105.2102.05
×=×−−
=−
=−υ
υOodA
( )( )
9956.911
105.211
110
4
−=
⎥⎦⎤
⎢⎣⎡
×+
⋅−=υA
50022.09956.95
=−
−==
υ
υυ
AO
I V
______________________________________________________________________________________ 9.22
(a) 605050
5501
10501
2
3
4
3
1
2 −=⎟⎠⎞
⎜⎝⎛ ++
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛++
−=
RR
RR
RR
Aυ
(b) (i) 78.250505015100 4
4
=⇒⎟⎟⎠
⎞⎜⎜⎝
⎛++−=− R
RkΩ
(ii) 79.150505015150 4
4
=⇒⎟⎟⎠
⎞⎜⎜⎝
⎛++−=− R
RkΩ
_____________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.23 a.
3 32
1 4 2
1
3 32
4 2
2 3
44 4
1
500 k
80 1500
Set 500 k
500 50080 1 1 1 2 6.41 k
vR RRA
R R RR
R RRR R
R R
RR R
⎛ ⎞= − + +⎜ ⎟
⎝ ⎠= Ω
⎛ ⎞= + +⎜ ⎟
⎝ ⎠= = Ω
⎛ ⎞= + + = + ⇒ = Ω⎜ ⎟
⎝ ⎠ b.
( )( )1 2 1 2
6 32 2
4 44
3 2 4 3
For 0.05 V0.05 0.1 A
500 k0.1 10 500 10 0.05
0.05 7.80 A6.410.1 7.80 7.90 A
I
X
X
v
i i i i
v i R
vi iR
i i i i
μ
μ
μ
−
= −−
= = ⇒ = = −Ω
= − = − − × × =
= − = − ⇒ = −
= + = − − ⇒ = −
______________________________________________________________________________________ 9.24 (a)
2
1 1
1
5001000
0.5 K
vRAR R
R
− −= − = =
=
(b)
3 32
1 4 2
1 1
1
1
250 500 500 12501000 1250 250
1.25 K
vR RRA
R R R
R RR
⎛ ⎞−= + +⎜ ⎟
⎝ ⎠− −⎛ ⎞− = + + =⎜ ⎟
⎝ ⎠=
______________________________________________________________________________________ 9.25
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
1 2
2
3
I
IA I
A I
vi iR
vv i R R vR
v viR R
= =
⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠
= − =
( )
( )
4 2 3
4
5
6 4 5
2 2
2 3
3 3
2 3 5
A A A I
IB A I
IB I
I I I
v v v vi i iR R R R
vv v i R v R vR
vv viR R R
v v vi i iR R R
= + = − − = − =
⎛ ⎞= − = − − = −⎜ ⎟⎝ ⎠
−= − = − =
= + = + =
I
00 6
53 8
From Figure 9.12 3
IB I
I
v
vvv v i R v RR v
A
⎛ ⎞= − = − − ⇒ = −⎜ ⎟⎝ ⎠
⇒ = −
______________________________________________________________________________________ 9.26
(a) 9978.9
202001
10511
120200
111
1
41
21
2 −=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +
×+
⋅−
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++
⋅−
=
RR
A
RR
A
od
υ
(b) 4801056.09978.9
80.4=
−−
==υ
υυ
AO
I V
(c) ( ) μυυ
υ 96105
80.4141 =⇒
×−−
=−
=od
O
AV
(d) %022.0%1008.4
8.4801056.4=×
−
______________________________________________________________________________________ 9.27
(a) ( )( )
9992.02
105.211
11
111
1
31
21
2 −=
⎥⎦⎤
⎢⎣⎡
×+
⋅−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++
⋅−
=
RR
A
RR
A
od
υ
(i) ( )( ) 7993605.08.09992.0 =−−=Oυ V
(ii) %08.0%1008.0
7993605.08.0≅×
−
(b) ( )( )
990099.02
10211
11
2
−=
⎥⎦⎤
⎢⎣⎡
×+
⋅−=υA
(i) ( )( ) 79208.08.0990099.0 =−−=Oυ V
(ii) %99.0%1009.079208.08.0
=×−
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.28
(a)
22
1 2
l O Oi
l
v v v Ri iR R v
= = = − ⇒ = −1R
(b)
22 1 3 3
1 1
23
1
1
Then 1
l Ol
L L
l
L
v v Ri i i i vR R R R
v RiR R
⎛ ⎞= = = + = + − ⋅⎜ ⎟
⎝ ⎠⎛ ⎞
= +⎜ ⎟⎝ ⎠
______________________________________________________________________________________ 9.29
( )
( )
3 1.max .max
3 1 4
2.max
1
2 2
1 1
2
0.1 110 0.09008
0.1 1 10
10 0.09008 111
So 111
+⎛ ⎞ ⎛ ⎞= ⋅ = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
= ⋅
= ⇒ =
= Ω
X X
O X
R RV V V
R R R
Rv VR
R RR R
R k
V
______________________________________________________________________________________ 9.30
33
22
11
IF
IF
IF
O RR
RR
RR
υυυυ ⋅−⋅−⋅−= 321 60120
20120
40120
III υυυ ⋅−⋅−⋅−=
321 263 IIIO υυυυ −−−= (a) ( ) ( ) ( ) 85.25.1210.0625.03 −=−−−−=Oυ V (b) ( ) ( ) ( ) 133.02.1225.0635.0 11 =⇒−−−−= II υυ V
______________________________________________________________________________________ 9.31
(a) ( ) 321321 625.025.6325.05.22.15.2 IIIIIIO υυυυυυυ −−−=++−=
Then 31
=RRF , 25.6
2
=RRF , 625.0
3
=RRF
3R is the largest resistor, so set 4003 =R kΩ Then k , k250=FR Ω 3.831 =R Ω , 402 =R kΩ
(b) ( ) ( ) ( ) 1875.02625.025.025.613 =−−−−=Oυ V
μυ
75.02501875.0
=⇒== FF
OF i
Ri A
______________________________________________________________________________________ 9.32 ( ) 2121 6232 IIIIO υυυυυ −−=+−=
Then 21
=RRF , 6
2
=RRF
For 11 −=Iυ V, 5.02 −=Iυ V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then ( ) ( ) 55.0612 =−−−−=Oυ V
Set μ80=Fi A 5.625=⇒== F
FF
O RRR
υkΩ
Then k , k25.311 =R Ω 42.102 =R Ω ______________________________________________________________________________________ 9.33
( ) ( ) ( )
( ) ( )( ) ( )( )( ) ( )
1
2 23
1 2 1 21 2
0.05 2 sin 2 0.0707sin 21 11 1 10
1001010 101 5
10 0.0707sin 2 2 1
0.707sin 2 2
I
I
F FO I I I I
O
O
v ft ft
f kHz T ms v T ms
R Rv v v v vR R
v ft V
v ft V
π π
π
π
= =
= ⇒ = ⇒ ⇒ = ⇒
= − ⋅ − ⋅ = − ⋅ − ⋅
= − − ±
= − − ±
______________________________________________________________________________________ 9.34
22
11
IF
IF
O RR
RR
υυυ ⋅−⋅−=
( ) ( )006.0100sin125.0004.0100sin5.021
−−+−=−R
tR
t ωω
Set ( ) 25sin125.0100sin5.0 11
=⇒−=− RtR
t ωω kΩ
We have ( ) ( )006.0100004.0100021
−−−=RR
5.376.025
4.00 22
=⇒+−= RR
kΩ
______________________________________________________________________________________ 9.35
(a) 321321 24
212
42 IIIII
IO υυυυυ
υυ −−−=⎥
⎦
⎤⎢⎣
⎡++−=
Then 21
1
=RRF , 4
2
=RRF , 2
3
=RRF
Set k2501 =R Ω , Then k125=FR Ω , 25.312 =R kΩ , 5.623 =R kΩ
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) For 21 −=Iυ V, 02 =Iυ , 13 −=Iυ V
( ) ( ) ( ) 31204221
=−−−−−=Oυ V
For 21 =Iυ V, 5.02 =Iυ V, 03 =Iυ
( ) ( ) ( ) 3025.04221
−=−−−=Oυ V
Then 33 +≤≤− Oυ V
μυ
24125
3max
maxmax
=⇒== FF
OF i
Ri A
______________________________________________________________________________________ 9.36
33
22
11
IF
IF
IF
O RR
RR
RR
υυυυ ⋅−⋅−⋅−=
( ) ( ) ( )4sin5.0sin22sin6321
−−−+−=−RR
tRR
tRR
t FFF ωωω
We have ( ) ( ) 1331
2042 RRRR
RR FF =⇒=+−
Also ( ) ( )5.02621 R
RRR FF −−=−
For 6=Oυ V, μ120max
=Fi A 506=⇒= F
F
RR
kΩ
For 4max1 =Iυ V, μ120
max1 =i A 33.3341
1
=⇒= RR
kΩ and 66.662 13 == RR k Ω
Now ( )( ) ( )( ) 33.82533.33
100251005.0502506 222121
=⇒+=+=+= RRRRRR
kΩ
______________________________________________________________________________________ 9.37 a.
( ) ( ) ( ) ( )
( )
0 3 2 1 03 2 1 0
3 02 10
5 5 5
So 510 2 4 8 16
F F F F
F
R R R Rv a a a aR R R R
a aR a av
= − ⋅ − − ⋅ − − ⋅ − − ⋅ −
⎡ ⎤= + + +⎢ ⎥⎣ ⎦
5
b. 0
12.5 5 10 k10 2
FF
Rv R= = ⋅ ⋅ ⇒ = Ω
c.
i. 0 0
10 1 5 0.3125 V10 16
v v= ⋅ ⋅ ⇒ =
ii. ( )0 0
10 1 1 1 1 5 4.68710 2 4 8 16
v v⎡ ⎤= + + + ⇒ =⎢ ⎥⎣ ⎦5 V
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.38
(a) 2121 20200120
120
110
IIIIO υυυυυ −=⋅⎟⎠⎞
⎜⎝⎛−⋅⎟
⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −=
(b) ( )( ) ( )( ) ttO ωωυ sin10005001000sin5025205200 ++=−−−= (mV) tO ωυ sin0.15.1 += (V) (c) For the 20 k resistor: Ω
125.020
5.2maxmax
=⇒= ii mA
For the 10 kΩ resistor:
( ) 5051
101 =⎟
⎠⎞
⎜⎝⎛=Oυ mV, μ5
1050
maxmax=⇒
Ω= i
kmV
i A
______________________________________________________________________________________ 9.39 For one-input
od
O
Aυ
υ −=1
F
OI
RRRRυυυυυ −
+=− 1
32
1
1
11
F
O
F
I
RRRRRRυ
υυ
−⎥⎥⎦
⎤
⎢⎢⎣
⎡++=
111
3211
1
1
F
O
Fod
O
RRRRRAυυ
−⎥⎥⎦
⎤
⎢⎢⎣
⎡++−=
111
321
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++−=
321
11111RRRARRA odFFod
Oυ
⎥⎥⎦
⎤
⎢⎢⎣
⎡⋅++−=
321
111RRR
RAAR
F
ododF
Oυ
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++
⋅⋅−=
P
F
od
IF
O
RR
A
RR
111
11
1
υυ where 321 RRRRP =
Therefore, for three-inputs ⎟⎟⎠
⎞⎜⎜⎝
⎛⋅+⋅+⋅⋅
⎟⎟⎠
⎞⎜⎜⎝
⎛++
−= 3
32
21
1111
1I
FI
FI
F
P
F
od
O RR
RR
RR
RR
A
υυυυ
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.40
(a) 1115
150111
2 =⎟⎠⎞
⎜⎝⎛ +=⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
RR
Aυ
(b) 450
1501 =⎟⎠⎞
⎜⎝⎛ +=υA
(c) 4.150201 =⎟
⎠⎞
⎜⎝⎛ +=υA
(d) 220201 =⎟
⎠⎞
⎜⎝⎛ +=υA
______________________________________________________________________________________ 9.41
(a) 141151
2
1
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+==
RR
RR
Aυ
For 5.7−=Oυ V 5.01 −=⇒υ V
μ120=i A 33.585.05.72
2
=⇒−
= RR
kΩ , 17.41 =R kΩ
(b) ( )( ) 75.325.015 ==Oυ V
μ6017.425.0
2121 ==⇒== iiii A
______________________________________________________________________________________ 9.42
(a) ⎟⎟⎠
⎞⎜⎜⎝
⎛+=
1
21RR
Aυ
2131
2
1
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
, Set 2902 =R kΩ , 1451 =R kΩ
(b) 8191
2
1
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
, Set 2902 =R kΩ , 25.361 =R kΩ
(c) 291301
2
1
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
, Set 2902 =R kΩ , 101 =R kΩ
(d) 0111
2
1
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
, Set 02 =R , 2901 =R kΩ
______________________________________________________________________________________ 9.43
0
1 11 500 501B I odv v v A⎛ ⎞ ⎛= =⎜ ⎟ ⎜+⎝ ⎠ ⎝
iv⎞⎟⎠
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
a. ( )12.5 5 250.5
501od odA A⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
b. ( )0 0
15000 5 49.9 V501
v v⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
______________________________________________________________________________________ 9.44
0 2
0 1 2
50 20 40150 20 40 20 40
1.33 0.667
I I
I I
v v
v v v
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦= +
1v
______________________________________________________________________________________ 9.45 (a)
1 2 2 2 2
2 2
1 2 2 2
1 2 2
1 2
20 40 101001 350
2Now 2 2 4
2 7 73
6 3So 7 7
I I
O
I I
oI I
O I I
v v v v v
v v v
v v v v vv
v v v
v v v
− −+ =
⎛ ⎞= + =⎜ ⎟⎝ ⎠
− + − =
⎛ ⎞+ = = ⎜ ⎟⎝ ⎠
= ⋅ + ⋅
(b) ( ) ( )6 30.2 0.3 0.3 V
7 7O Ov v⎛ ⎞= + ⇒ =⎜ ⎟⎝ ⎠
(c) ( ) ( )6 30.25 0.4 42.86 mV
7 7O Ov v⎛ ⎞ ⎛ ⎞= + − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
______________________________________________________________________________________ 9.46
(a) IRRR
υυ ⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
=43
42
IO
RRR
RRR
υυυ ⋅
⎟⎟⎠
⎞⎜⎜⎝
⎛+
⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
4
31
22
1
2
1
111
(b) 891
50251
1161
2
1
2
1
2 =⇒=⎟⎟⎠
⎞⎜⎜⎝
⎛+⇒
⎟⎠⎞
⎜⎝⎛ +⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
RR
Set k2002 =R Ω , k 251 =R Ω______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.47 (a)
( )501
1 50
111 1
11
O
I
O
I
Ov
I
v xv x
v x x xv x
vAv x
⎛ ⎞= +⎜ ⎟⎜ ⎟−⎝ ⎠
− +⎛ ⎞= + =⎜ ⎟− −⎝ ⎠
= =−
x
(b) 1 vA≤ ≤ ∞
(c) If x = 1, gain goes to infinity. ______________________________________________________________________________________ 9.48
(a) ( ) III
X RR
υυυ
υ 32 =+⎟⎟⎠
⎞⎜⎜⎝
⎛=
022
=−
++−
RRROXXIX υυυυυ
RRRRROI
X 22211
21 υυ
υ =−⎟⎠⎞
⎜⎝⎛ ++
RRROI
I 2223
υυυ =−⎟
⎠⎞
⎜⎝⎛
so 11=I
O
υυ
(b) For 25.0=Iυ V, 75.2=⇒ Oυ V (c) k , 30=R Ω 15.0−=Iυ V
For : 1R μ53015.0
⇒=i A
For : 2R μ5=i A 45.03 −== IX υυ V
For : 4R μ153045.0
⇒=i A
( )( ) 65.115.011 −=−=Oυ V
For : 3R μ2060
45.065.1⇒
−=i A
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.49
(a) 1O
I
vv
=
(b) From Exercise TYU9.7
2
1
2
1
2 1
5
1
11 1
But 0, 1 1 0.999993
1 11 11.5 10
O
I
od
O O
I I
od
RRv
v RA R
R Rv vv v
A
⎛ ⎞+⎜ ⎟
⎝ ⎠=⎡ ⎤⎛ ⎞+ +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦= = ∞
= = ⇒ =+ +
×
(b)
1Want 0.990 9911
Ood
I
od
vA
vA
= = ⇒ =+
______________________________________________________________________________________ 9.50
(a) ( ) ( OIododO AA )υυυυυ −=−= 12 ( ) IododO AA υυ =+1
9524.0
2011
111
1=
+=
+==
od
I
O
A
Aυυ
υ
(b) 995.0
20011
1=
+=υA
(c) 9995.0
200011
1=
+=υA
(d) 99995.0
2000011
1=
+=υA
______________________________________________________________________________________ 9.51
(a) ⎟⎟⎠
⎞⎜⎜⎝
⎛+==
1
211 1
RR
AI
O
υυ
υ
⎟⎟⎠
⎞⎜⎜⎝
⎛+−==
1
222 1
RR
AI
O
υυ
υ
(b) ( ) 25.0206011 −=−⎟
⎠⎞
⎜⎝⎛ +=Oυ V
( ) 25.0206012 =−⎟
⎠⎞
⎜⎝⎛ +−=Oυ V
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(c) ( ) 4.68.020601212
1
221 =⎟
⎠⎞
⎜⎝⎛ +=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+=− IOO R
Rυυυ V
______________________________________________________________________________________ 9.52
(a) 1
= IL
viR
(b)
( ) ( )( )
( ) ( )( ) ( )
1 1
1
max 10 V 1 9 10So max 1 mA
Then max 1 9 max 9 V
O L L I L L L
OI L L
L
I L I
v i R v i R i Rv i i
i
v i R v
= + = +≅ = + =≅
≅ = ⇒ ≅
______________________________________________________________________________________ 9.53
(a) ( ) IIO υυυ 3333.04020
20=⋅⎟
⎠⎞
⎜⎝⎛
+=
(i) 1=Oυ V (ii) 67.1−=Oυ V
(b) ( ) IIO υυυ 3333.04020
20=⋅⎟
⎠⎞
⎜⎝⎛
+=
(i) 1=Oυ V (ii) 67.1−=Oυ V
(c) ( ) IIO υυυ 2222.0486
610101 =⋅⎟
⎠⎞
⎜⎝⎛
+⎟⎠⎞
⎜⎝⎛ +=
(i) 667.0=Oυ V (ii) 111.1−=Oυ V
______________________________________________________________________________________ 9.54 a.
( ) ( )
1 011 0
1
1 1 11
1
1
and and
1So
Then 1
in odF
od od
F F
Fin
od
v vvR i vi R
v A v v Ai
R Rv RRi A
−= = =
− − += =
= =+
1A v−
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.
1 0
0
0
and 1
So 1
10 0.0099901 1001
10001001 0.009990
1000Want 0.9901001 0.009990
which y
S FS od
S in od
od SF S
od S in
Fin
od
SF S
S
S
S
R Ri i v AR R A
A Rv R iA R R
RRA
Rv R iR
RR
⎛ ⎞= = − ⋅⎜ ⎟+ +⎝ ⎠
⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
= = =+
⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ≤⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠
1i⋅
ields 1.099 kSR ≥ Ω
______________________________________________________________________________________ 9.55
( ), 0 8 mA
For max 8 V, Then 1 O C F C
O F
v i R iv R= ≤ ≤
= = kΩ
______________________________________________________________________________________ 9.56
10 so 1 10 kIv
i RR R
= = ⇒ = Ω
In the ideal op-amp, R1 has no influence.
Output voltage: 2
0 1 IRv vR
⎛ ⎞= +⎜ ⎟⎝ ⎠
v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. ______________________________________________________________________________________ 9.57
(a) 2R
i IL
υ−= ; ( ) 1
55
2 =−−
=R kΩ
Set 231
1RRR
RF = ; For example, set 101 == FRR kΩ , 13 =R kΩ
(b) ( )( ) 12.05 ==Lυ V 1υ=
6.010
15
1
121 −=
−−=
−==
Rii I υυ
mA
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
( )( ) 7106.0121 =−−=−= FO Riυυ V
61
17
33 =
−=
−=
Ri LO υυ
mA
111
24 ===
Ri Lυ mA
For the op-amp: ( ) 6.66.062332 =−−=−=⇒=+ iiiiii OO mA ______________________________________________________________________________________ 9.58 (a)
1 2 2 22
1 12
12
and ,
Then
Or 1
= = + = −
⎛ ⎞= − +⎜ ⎟
⎝ ⎠⎛ ⎞
= +⎜ ⎟⎝ ⎠
xD x F
FD
FD
vi i i i v i RR
Ri i iR
Ri iR
(b)
( )
1 11
2 2
2
5 5 1
12 1 1 11
For example, 5 , 55
= = ⇒ = Ω
⎛ ⎞= + ⇒ =⎜ ⎟
⎝ ⎠= Ω = Ω
I
F F
F
vR R ki
R RR R
R k R k ______________________________________________________________________________________ 9.59
(1) 2 3
X OXX
V vVIR R
−= +
(2) 1
0X OX
F
V vVR R
−+ =
1
2 3 3 1
0 2 3 3 1 3 2 1 3
1 3 2
1 2 3
1 2 3
1 3 2
2 1 31 3 2
From (2) 1
1 1 1Then (1) 1
1 1 1 1 1
or
1Note: If then , which corresp
FO X
FX X X
X F F
X
F
oF
FF o
Rv VR
RI V VR R R R
I R RV R R R R R R R R R
R R R RR R R
R R RRR R R R
R R R R R RR R R
⎛ ⎞= +⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞
= + − ⋅ +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
= = + − − = −
−=
=−
= ⇒ = = ∞ onds to an ideal current source.
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.60
(a) kΩ ; 3031 =+= RRRid 1531 == RR kΩ
22515 423
4
1
2 ==⇒== RRRR
RR
kΩ
(b) ( )( ) 5.21025.0 === LLO Riυ V
1667.015
5.212 ===−
d
OII A
υυυ V
(c) ( ) ( ) 5.45.12.11512 −=−=−= IIdO A υυυ V
45.010
5.4−=
−==
L
OL R
iυ
mA
(d) ( )( ) 5105.0 ==Oυ V
333.0155
12 ===−d
OII A
υυυ V
667.1333.021 =−=Iυ V ______________________________________________________________________________________ 9.61
(a) 403
4
1
2 ==RR
RR
; Set k25042 == RR Ω , 25.631 == RR kΩ
(b) 253
4
1
2 ==RR
RR
; Set k25042 == RR Ω , 1031 == RR kΩ
(c) 53
4
1
2 ==RR
RR
; Set k25042 == RR Ω , 5031 == RR kΩ
(d) 5.03
4
1
2 ==RR
RR
; Set k12542 == RR Ω , 25031 == RR kΩ
______________________________________________________________________________________ 9.62
( ) ( )( ) ( )
4 32 2 22 1 2 1
1 4 3 1 1 3 4 1
2 1
3 4
2
We have
/ 11 or 11 / 1 /
Set 50 1 , 50 150 1 , 50 1
1 1111 11
O I I O I
O I
R RR R R Rv v v v vR R R R R R R R
R x R xR x R x
xv vxxx
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
= + = −= − = +
⎡ ⎤⎢ ⎥⎡ + ⎤⎛ ⎞ ⎢ ⎥= + ⎜ ⎟⎢ ⎥ −− ⎛ ⎞⎢ ⎥⎝ ⎠⎣ ⎦ + ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦
2Iv
( )( )
1
2 1
2 1
11
1 1 1 11 1 1 1
1 11 1
I
O I I
I I
x vx
x x x xv v vx x x x
x xv vx x
+⎛ ⎞− ⎜ ⎟−⎝ ⎠
⎡ ⎤− + +⎡ ⎤ + +⎛ ⎞= ⋅ −⎢ ⎥⎢ ⎥ ⎜ ⎟− + + − −⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦+ +⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
( ) ( )( ) ( )
( )
1 2
2 1
3 4
2 1
2 1
1 2
For 0Set 50 1 50 1
50 1 50 1
1 1 1111 111
11
1 11 211 1 1
I I O
O I
I I
I I cm
O
cm
v v vR x R xR x R x
x xv vxx xx
xv vx
v v vx xv x x
v x x x
= ⇒ =
= + = −= + = −
⎛ ⎞⎜ ⎟+ +⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟+− −⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟
−⎝ ⎠+⎛ ⎞= − ⎜ ⎟−⎝ ⎠
= =
− − ++ −= − = =
− − −
Iv
( ) ( )( ) ( )
( )
2 1
3 4
2 1
Set 50 1 50 150 1 50 1
1 1 1111 111
111
1 1 21 1
O I
cm
cm
R x R xR x R x
x xv vxx xx
x vx
x x xAx x
= − = += − = +
⎛ ⎞⎜ ⎟− −⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟−+ +⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟+⎝ ⎠
−⎛ ⎞= −⎜ ⎟+⎝ ⎠+ − −
= =+ +
Iv
Worst common-mode gain2
1cmxAx
−=
− (b)
( )
( )
( )
( )
2 1
2 0.012For 0.01, 0.02021 1 0.01
2 0.02For 0.02, 0.04082
1 0.022 0.05
For 0.05, 0.10531 0.05
1 1For this condition, set , 1 V2 2
1 11 1 112 1 2 1
cm
cm
cm
I I d
d
xx Ax
x A
x A
v v v
x xxAx
−−= = = = −
− −−
= = = −−
−= = = −
−
= + = − ⇒ =
− + +⎡ + ⎤⎛ ⎞= + =⎜ ⎟⎢ ⎥− −⎝ ⎠⎣ ⎦
10
10
10
1 2 12 1 1
1.010For 0.01 1.010 20log 33.98 dB0.0202
1 1.0For 0.02, 1.020 20log 27.96 dB0.98 0.040821 1.0526For 0.05 1.0526 20log 20 dB
0.95 0.1053
d dB
d dB
d dB
x x x
x A C M R R
x A C M R R
x A C M R R
⎡ ⎤= ⋅ =⎢ ⎥ − −⎣ ⎦
= = = =
= = = = =
= = = = ≅
20
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.63
(a) ( ) =−= 1210 υυυO ( ) 48.14.110 −=− V
( )( ) 0127.010114.1
102
43 ==+
==RR
iiυ
mA
( ) 273.14.11110
1110
2 =⎟⎠⎞
⎜⎝⎛=⋅⎟
⎠⎞
⎜⎝⎛== υυυ YX V
0527.010
273.18.1121 =
−=
−==
Rii Xυυ
mA
(b) ( ) ( ) 42.36.31010 12 =−=−= υυυO V
( ) 273.36.31110
=⎟⎠⎞
⎜⎝⎛== YX υυ V
( )( ) 0327.0110
6.310112
43 ====υ
ii mA
00727.010
273.32.321 −=
−== ii mA
(c) ( )( ) 5.120.135.110 −=−−−=Oυ V
( )( ) 0123.0101135.1
43 −=−
== ii mA
( ) 227.135.11110
−=−⎟⎠⎞
⎜⎝⎛== YX υυ V
( ) 00273.010
227.12.121 =
−−−== ii mA
______________________________________________________________________________________ 9.64
(a) ( ) ( )( ) 2.912.1761 ==⋅+= BE II β mA
Ω=⇒= 6.1092.91
10 RR
(b) mA ( )( ) 2.202.0101 ==EI
495.02.20
10==R kΩ
(c) 74.541096.06
==EI mA
72.076
74.54==OI mA
(d) 08.8495.04
==EI mA
080.0101
08.8==OI mA
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.65
(a) CMO RR
υυ ⋅−
=1
21
⎟⎟⎠
⎞⎜⎜⎝
⎛+⋅⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=1
2
43
42 1
RR
RRR
CMO υυ
CMOOO RR
RR
RR
RR
υυυυ ⋅
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−⎟⎟⎠
⎞⎜⎜⎝
⎛+⋅
+=+=
1
2
1
2
3
4
3
4
21 11
3
4
1
2
3
4
3
4
3
4
1
2
1
2
3
4
11
11
RR
RR
RR
RR
RR
RR
RR
RR
ACM
OCM
+
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
==υυ
(b) 3.010
69
6.94.861
4.104.62
6.94.86
=−
=+
−=CMA
(c) 03149.0
8.198.801
2.202.79
8.198.80
=+
−=CMA
or
0325.0
2.202.791
8.198.80
2.202.79
−=+
−=CMA
0325.0max
=⇒ CMA ______________________________________________________________________________________ 9.66
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
01
1 2 2
2
1 2 2
1 21
1 2 1 2
1 22
1 2 1 2
(1)
(2)
(3)
(4)
AI A A B
v
I B B A B
v
A I
B I
v vv v v vR R R Rv v v v vR R R R
R Rv v vR R R R
R Rv v vR R R R
−
+
−− −= +
+− −
= ++
⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞
= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
( )
( )
( )
1 2 1 1 2 2
22 1
1
01
1 2 1 2 2 2
2
1 2 1 2 2
01 2
1 2 1 2 2 2 1 1 2
Now
So that
1 1 1 1
1 1 1 2
Then
1 1 1 1 1
A I B I
A B I I
I BA
V V
I AB
V V
I BB
V V
v v R v R v R v R vRv v v vR
vv vvR R R R R R R R
v vvR R R R R R R
vv v RvR R R R R R R R R R R R
− += ⇒ + = +
= + −
⎛ ⎞= + + − −⎜ ⎟+ +⎝ ⎠
⎛ ⎞= + + −⎜ ⎟+ +⎝ ⎠
⎛ ⎞ ⎛ ⎞= + + − − + +⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠⎝ ⎠
( )
( )
2 12
2 22 1
1 2 1 2 2 1
1 (1)
1 1 1 1 (2)
I IV
IB B I I
V V
v vR
v Rv v v vR R R R R R R R
⎛ ⎞+ −⎜ ⎟
⎝ ⎠⎛ ⎞ ⎡ ⎤
= + + − + −⎜ ⎟ ⎢ ⎥+ + ⎣ ⎦⎝ ⎠ Subtract (2) from (1)
( ) ( ) ( )
( )
( )
02 21 2 2 1 2 1
1 2 1 1 2 2 2 1
0 2 22 1
2 1 1 2 2 1 2 1
2 2 2 1 20 2 1
1 1 2 1 2
20
1 1 1 1 1
1 1 1 1 1
1
2
I I I I I IV V
I IV V
I IV V
vR Rv v v v v vR R R R R R R R R R
v R Rv vR R R R R R R R R R
R R R R Rv v vR R R R R R R
RvR
⎛ ⎞⎛ ⎞− = + + − − + ⋅ −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
⎧ ⎫⎛ ⎞⎛ ⎞⎪ ⎪= − + + + + ⋅⎨ ⎬⎜ ⎟⎜ ⎟ + +⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭⎧ ⎫⎛ ⎞
= − + + + +⎨ ⎬⎜ ⎟ + +⎝ ⎠⎩ ⎭
= ( )22 1
1
1 I IV
R v vR
⎛ ⎞+ −⎜ ⎟
⎝ ⎠ ______________________________________________________________________________________ 9.67
(a) ( ) ( )
titt
Ri II ω
ωωυυsin16
10sin08.02.1sin08.02.1
11
211 −=⇒
+−−=
−= ( μ A)
( ) ( )( ) tttO ωωωυ sin72.02.140sin016.0sin08.02.11 −=−−= (V) ( ) ( )( ) tttO ωωωυ sin72.02.140sin016.0sin08.02.12 +=−−+= (V)
( ) ( )( ) ttRR
OOO ωωυυυ sin32.4sin72.0240
12012
3
4 =⎟⎠⎞
⎜⎝⎛=−= (V)
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
)
(b) ( ) (10
sin05.060.0sin05.065.0 ttii
ωω −−−+−=
ti ωsin1051 +−= ( μ A) ( )( )ttO ωωυ sin010.0005.040sin05.065.01 +−++−= tωsin45.085.0 +−= (V) ( )( ) tttO ωωωυ sin45.040.0sin01.0005.040sin05.060.02 −−=+−−−−= (V)
( ) ( )[ ] tttO ωωωυ sin7.235.1sin45.085.0sin45.040.040
120−=+−−−−⎟
⎠⎞
⎜⎝⎛= (V)
______________________________________________________________________________________ 9.68
(a)
401 2.1667si12OB Iv v n tω⎛ ⎞= + =⎜ ⎟
⎝ ⎠
(b) 30 1.25sin12OC Iv v tω= − = −
(c)
( )2.1667sin 1.25sin3.417sin
O OB OC
O
v v v t tv t
ω ωω
= − = − −=
(d)
3.417 6.830.5
O
I
vv
= =
______________________________________________________________________________________ 9.69
(a) R
i IIO
21 υυ −=
(b) ( )Ω=⇒
−−= 100
525.025.0 RR
(c) ( )( ) 25.51525.011 =+=+= LOIO Riυυ V 25.022 −== IO υυ V
(d) 15.0
75.125.121 −=−
=−
=R
i IIO
υυmA
( )( ) 75.13125.111 −=−=+= LOIO Riυυ V 75.122 == IO υυ V
______________________________________________________________________________________ 9.70
( ) ( )
4 2
2 1 3 1
1
11
1 1 11
21
2 115200 1 0.06sin50
230For 0.5 1.0833 212.3 K
2308 V 32.33 7.11 K 7.11 K, (potentiometer) 205.2 K
Od
I I
O
O
O f
v R RAv v R R
v tR
v RR
v R R RR
ω
⎛ ⎞= = +⎜ ⎟− ⎝ ⎠
⎛ ⎞= +⎜ ⎟
⎝ ⎠
= = ⇒ =
= = ⇒ = ⇒ = =
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.71
For 10=Oυ V, 05.020010200 12 ==−⇒= IIdA υυ V
( ) ( ) 1105005.0
161 =⇒×
=−
fixedRfixedR kΩ
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
1
2
3
4 21
RR
RR
Ad
( ) 5.391
215.2200 2
2 =⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+= R
RkΩ
For 5=dA
( ) ( ) ( ) ( ) ( ) 1varvar795.39215.25 11111
+=+==⇒⎥⎦
⎤⎢⎣
⎡+= RfixedRRR
R
kΩ ( ) 78var1 =R______________________________________________________________________________________ 9.72
( ) 13
41 OOO R
Rυυυ ⋅
′−=
( ) 23
4
43
42 1 OOO R
RRR
Rυυυ ⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛′
+⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
( ) ( 21 OOOOO )υυυυυ += and CMOO υυυ ≡= 21 Then
⎟⎟⎠
⎞⎜⎜⎝
⎛′
−⎟⎟⎠
⎞⎜⎜⎝
⎛′
+⎟⎟⎠
⎞⎜⎜⎝
⎛+
==3
4
3
4
43
4 1RR
RR
RRR
ACM
OCM υ
υ
k , k602 34 == RR Ω 303 =R Ω , 303 =′R kΩ %5± For kΩ k 303 =′R %5− 5.28= Ω
03509.05.28
605.28
6013060
60−=⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛
+=CMA
For kΩ k 303 =′R 5.31%5 =+ Ω
03175.05.31
605.31
6013060
60+=⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛
+=CMA
Then 03175.003509.0 +≤≤− CMA______________________________________________________________________________________ 9.73
(a) s ( )( ) 46321 1041002.01020 −− ×=××=CR
( ) ( ) ttdtO ωω
ωυ sin104
25.0cos25.01041
44⋅
×−
=×−
=−− ∫
For ( )( ) 3981210425.0 4 =⇒=×⇒= − ffO πυ Hz Phase °= 90
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) (i) ( ) 3.661042
25.05.14
=⇒×
==−
ffO π
υ Hz
(ii) ( ) 6631042
25.015.04
=⇒×
==−
ffO π
υ Hz
______________________________________________________________________________________ 9.74
(a) ( ) ( )2.1
021021
25.011 tCR
tdtCR
t
IO ′−=′′−
= ∫υυ
( )( ) 602.125.05 2121
=⇒−
=− CRCR
ms
(b) (i) ( ) 306.010.050 =′′⇒′′⋅+−= tt s, 2.4=t s
(ii) ( ) 606.010.055 =′′⇒′′⋅+−= tt s, 2.7=t s
______________________________________________________________________________________ 9.75
(a) 1
2
RZ
A−
=υ , where 22
2
22
22
222 11
11
CRjR
CjR
CjR
CjRZ
ωω
ωω +
=+
⎟⎟⎠
⎞⎜⎜⎝
⎛
==
221
2
11
CRjRR
Aωυ +
⋅−
=
(b) At 0=ω , ( )1
20RR
A−
=υ
(c) ( )2
221
2
1
1
CRRR
Aω
υ+
⋅=
Set ( )2222
222 2
1121CR
fCR
CRπ
ωω =⇒=⇒=+
______________________________________________________________________________________ 9.76
(a) kΩ 201 =R
30015 21
2 =⇒= RRR
k Ω
fCR
πω 21
22
==
( )( ) 106103001052
12
1233
22 =⇒
××== C
fRC
ππpF
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
(b) kΩ 151 =R
37525 21
2 =⇒= RRR
kΩ
( )( ) 3.281037510152
12332 =⇒
××= CC
πpF
______________________________________________________________________________________ 9.77
(a) 1
2
ZR
A−
=υ , where 1
11
111
11Cj
CRjCj
RZωω
ω+
=+=
11
11
1
2
11
12
11 CRjCRj
RR
CRjCRj
Aω
ωωω
υ +⋅
−=
+−
=
(b) As ∞⇒ω , 1
2
RR
A−
=υ
(c) ( )2
11
11
1
2
1 CR
CRRR
Aω
ωυ
+⋅=
Set ( ) 11112
11
11
211
21
1 CRf
CRCR
CRπ
ωω
ω=⇒=⇒=
+
______________________________________________________________________________________ 9.78
(a) Set kΩ 3502 =R
33.2315 11
2 =⇒= RRR
kΩ
( )( ) 34110201033.232
12
112 1331
111
=⇒××
==⇒= CfR
CCR
fππ
π pF
(b) Set k 201 =R Ω
50025 21
2 =⇒= RRR
kΩ
( )( ) 227103510202
11331 =⇒
××= CC
πpF
______________________________________________________________________________________ 9.79 Assuming the Zener diode is in breakdown,
( )
( )
2
1
2 22
2
1 6.8 6.8 1
0 6.80 6.8 1
10 10 6.8 6.8 6.2 !!!5.6
O z O
O
zz z
s
Rv V v VR
vi i mAR
Vi i iR
= − ⋅ = − ⇒ = −
− −−= = ⇒ =
− −= − = − ⇒ = − mA
Circuit is not in breakdown. Now
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
( )( )
2 21
2 2
10 0 10 1.52 5.6 11.52 1 1.52
0
s
O O
z
i i mR Rv i R v
i
−= = ⇒ =
+ += − = − ⇒ = −
=
A
V
______________________________________________________________________________________ 9.80
( ) ( )( ) 1014 41
ln 0.026 ln 0.026ln1010 10
For 20 , 0.497 For 2 , 0.617
I IO T O
s
I O
I O
v vv V vI R
v mV v Vv V v V
−−
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − = − ⇒ = −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎣ ⎦
= == =
Iv
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.81
( ) ( )
( )( )
0 01 02 01 0
101 1
202 2
1 201 02
2 1
2 12 1
2 1
2 101 02
2 1
20
1
333 16.6520
ln
ln
ln ln
,
So ln
Then
16.65 0.026 ln
CBE T
S
CBE T
S
C CT T
C C
C C
T
v v v v v
iv v VI
iv v VI
i iv v V V
i iv vi iR R
v Rv v VR v
v Rvv
⎛ ⎞= − = −⎜ ⎟⎝ ⎠
⎛ ⎞= − = − ⎜ ⎟
⎝ ⎠⎛ ⎞
= − = − ⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞− = − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= =
⎛ ⎞− = ⋅⎜ ⎟
⎝ ⎠
= ⋅
2
( ) ( ) ( ) ( )( )
( )
1
2
2 10
1 2
10
10
2 10 10
1 2
0.4329ln
ln log log log 10
2.3026log
Then 1.0 log
e e
R
v Rvv R
x x x
x
v Rvv R
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞= ⋅⎜ ⎟
⎝ ⎠= = ⋅⎡ ⎤ ⎡⎣ ⎦ ⎣=
⎛ ⎞≅ ⋅⎜ ⎟
⎝ ⎠
⎤⎦
______________________________________________________________________________________ 9.82
( ) ( )( )( )
/ /14 4
/ 0.02610
5
10 10
10
For 0.30 , 1.03 10 For 0.60 , 1.05
I T I
I
v V v VO s
vO
I o
I o
v I R e e
v e
v V v Vv V v V
−
−
−
= − = −
=
= = ×= =
T
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.83 From Figure 9.40
⎥⎦
⎤⎢⎣
⎡⋅+⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛++⋅−⋅−= 314
22
1
1 IB
PI
A
P
N
FI
FI
FO R
RRR
RR
RR
RR
υυυυυ
3142 3210 IIII υυυυ ++−−=
Then 101
=RRF , 1
2
=RRF , 21 =⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+
A
P
N
F
RR
RR
, 31 =⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+
B
P
N
F
RR
RR
Set k , kΩ , 500=FR Ω 501 =R 5002 =R kΩ Now 45.455005021 === RRRN kΩ
Then 21245.45
5001 =⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
A
P
A
P
RR
RR
, Also 31245.45
5001 =⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
B
P
B
P
RR
RR
Let k , then 500=AR Ω 3.33332
== AB RR kΩ
Then kΩ33.83=PR CBA RRR=
We find 2003.333500 ==BA RR kΩ
So 8.14233.83200 =⇒= CC RR kΩ ______________________________________________________________________________________ 9.84
52
41
3211 IF
IR
IC
PI
B
PI
A
P
N
FO R
RRR
RR
RR
RR
RR
υυυυυυ ⋅−⋅−⎥⎦
⎤⎢⎣
⎡⋅+⋅+⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
54321 6425.13 IIIII υυυυυ −−++=
We have 41
=RRF , 6
2
=RRF ; Set 250=FR kΩ , 5.621 =R kΩ , 67.412 =R k Ω
Now 2567.415.6221 === RRRN kΩ
Also 1125
25011 =⎟⎠⎞
⎜⎝⎛ +=⎟⎟
⎠
⎞⎜⎜⎝
⎛+
N
F
RR
Now ( )
311
=A
P
RR
, ( )
5.111
=B
P
RR
, ( )
211
=C
P
RR
21
=⇒B
A
RR
, 32
=C
A
RR
Set k , k , 250=BR Ω 125=AR Ω 5.187=CR kΩ This yields k , We have 09.34=PR Ω DCBAp RRRRR =
We find 69.575.187250125 ==CBA RRR kΩ
Then 3.8309.3469.57 =⇒= DD RR kΩ ______________________________________________________________________________________ 9.85
143.16.5
1211
2
1
2 =⇒=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
VV
Z
O
F
ZOF R
VVI
−= ; Set ( ) 2.1min == ZF II mA
Then 33.52.1
6.512=
−=FR kΩ
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Set mA 15.01 =DI V 3.67.06.5 =+=+=′ γVVV ZZ
Let mA, 2.04 =I 5.312.03.6
4 ==⇒ R kΩ
Then mA 35.015.02.03 =+=I
So 6.1035.0
3.610
33 =
−=
′−=
IVV
R ZS kΩ
______________________________________________________________________________________ 9.86
2.32
6.5121 =
−=
−=
Z
ZO
IVV
R kΩ
143.16.5
1213
2
3
2 =⇒=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
RR
RR
VV
Z
O
Let mA, 2=RI 62
1232 ===+⇒
R
O
IV
RR kΩ
Then , k6143.1 33 =+ RR 8.23 =⇒ R Ω and 2.32 =R kΩ
Let mA, 44 =RI 75.04
1215
44 =
−=
−=
R
OIN
IVV
R kΩ
______________________________________________________________________________________ 9.87 Let k 20321 === RRR Ω Let ( )δ+= 120TR kΩ
Now ( ) 51021
13
31 ==⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
== +VRR
RAO υυ V
( )( ) ( ) ( )
δδ
δδυυ
++
=⎥⎦
⎤⎢⎣
⎡++
+=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛+
== +
211010
20120120
22 V
RRR
T
TBO
So ( ) ( ) ( ) δδδ
δδδδυυυ 5.2
25
211025
21105 −=−≅
++−+
=++
−=−= BAOA
We have ⎟⎠⎞
⎜⎝⎛ −
=300
300Tyδ ; At , 350=T 21=TR kΩ , ( ) 05.012021 =⇒+=⇒ δδ
Then 30.0300
30035005.0 =⇒⎟⎠⎞
⎜⎝⎛ −
= yy
For 05.0=δ , ( )( ) 125.005.05.2 ==OAυ V
For the instrumentation amplifier, ( )125.02
151
2
3
4⎟⎟⎠
⎞⎜⎜⎝
⎛+==
RR
RR
Oυ
For example, set 43
4 =RR
and 5.41
2 =RR
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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.88
(a) ++ ⋅⎟⎠⎞
⎜⎝⎛ Δ−
=⋅⎟⎠⎞
⎜⎝⎛
Δ++Δ−Δ−
= VR
RRVRRRR
RRA 2
υ
++ ⋅⎟⎠⎞
⎜⎝⎛ Δ+
=⋅⎟⎠⎞
⎜⎝⎛
Δ−+Δ+Δ+
= VR
RRVRRRR
RRB 2
υ
⎟⎠⎞
⎜⎝⎛
×Δ−=⋅
Δ−=⋅⎥⎦
⎤⎢⎣⎡ Δ+
−Δ−
=−= ++31 1020
922
RVRRV
RRR
RRR
BAO υυυ
or ( )( )RO Δ×−= −41 105.4υ
(b) For an instrumentation amplifier,
11
2
3
4 21 OO R
RRR
υυ ⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+=
For , Ω=Δ 200R 5−=Oυ V
( )( )200105.42
15 4
1
2
3
4 −×−⎟⎟⎠
⎞⎜⎜⎝
⎛+=−
RR
RR
or ⎟⎟⎠
⎞⎜⎜⎝
⎛+=
1
2
3
4 2155.55
RR
RR
For example, set 63
4 =RR
and 13.41
2 =RR
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