Neamen D. Microelectronics: Circuit Analysis and Design
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1348 Appendix E Answers to Selected Problems
9.85
R
F
= 5.33
k
,
R
4
= 31.5
k
,
R
3
= 10.6
k
,
R
2
R
1
= 1.143
9.87 For the bridge circuit,
R
T
= 20
(
1 + δ
)
k
,
R
1
= R
2
= R
3
= 20
k
For the instrumentation amplifier,
Let
R
4
R
3
= 4
,
R
2
R
1
= 4.5
CHAPTER 10
10.1 (a)
I
C
=
1
R
3
2V
γ
−(2V
γ
+ V
−
)
R
2
R
1
+ R
2
−V
BE
(c)
R
3
= 2.5k
,
R
1
= R
2
= 2.15 k
10.3
V
BE1
= V
BE2
= 0.6341
V,
I
O
= 78.05 μ
A
10.5
V
BE1
= 0.63345
V,
V
BE2
= 0.62393
V,
I
O
= 132.07 μ
A
10.7
I
REF
= 0.5082
mA,
I
C1
= I
C2
= 0.4958
mA,
I
B1
= I
B2
= 6.198 μ
A
10.9 (a)
R
1
= 18.3k
(b)
6.3%
10.11
I
REF
= 0.210
mA,
R
1
= 20.5
k
10.13 (a)
R
1
= 9.3
k
(b)
I
O
= 2
mA
(c)
R
C2
= 2.15
k
10.15 Similar to Figure P10.14, biased
at
V
+
and
V
−
.
R
1
= 21.5
k
10.17 (a)
I
O
=
I
REF
−
V
BE
1 + β
R
2
1 +
2
β(1 +β)
(b)
R
1
= 12.27
k
,
I
REF
= 0.7011
mA
10.19
R
1
= 30.63
k
10.21
I
O
=
I
REF
1 +
2
β(2 +β)
10.23 (a)
R
o
= 20
M
(b)
I
O
= 0.25 μ
A
10.25 (a)
V
BE1
= 0.6347
V,
V
BE2
= 0.6040
V,
I
O
∼
=
61.4 μ
A
(b)
V
BE1
= 0.6347
V,
V
BE2
= 0.5991
V,
I
O
∼
=
71.2 μ
A
10.27 (a)
I
REF
= 0.186
mA,
I
O
∼
=
19.53 μ
A,
V
BE2
= 0.6414
V
(b)
R
o
= 13.16
M
10.29 (a)
R
1
= 18.6
k
,
R
E
= 1.20
k
(b)
R
o
= 6.477
M
(c)
1.54%
10.31
V
BE1
= 0.718
V at 2 mA;
R
1
= 7.14
k
,
R
E
= 1.92
k
10.33
V
BE1
= 0.681
V,
I
REF
= 0.483
mA,
I
O
∼
=
8.7 μ
A,
V
BE2
= 0.5766
V
10.35 (a)
R
1
= 10.1
k
,
R
E2
= 1.37
k
,
V
BE1
= 0.70038
V,
V
BE2
= 0.67656
V
(b)
R
1
= 10.1
k
,
R
E2
= 1.46
k
,
V
BE2
= 0.65854
V
10.37 (a)
I
REF
= 0.93
mA,
I
O2
∼
=
68 μ
A,
I
O3
∼
=
40.7 μ
A
(b)
R
E2
= 4.99
k
,
R
E3
= 0.797
k
10.39 (a)
I
O1
= 3.1
mA,
I
O2
= 1.55
mA,
I
O3
= 4.65
mA
(b)
V
CE1
= 3.8
V,
V
EC2
= 5.35
V,
V
EC3
= 5.35
V
10.41
I
C1
= I
C2
= 1.86
mA,
I
C3
= I
C4
= 1.86
mA,
I
C5
= I
C6
= I
C7
= 0.136
mA,
V
CE3
= 7.02
V,
V
CE5
= 14.2
V,
V
EC7
= 4.89
V
10.43
R
E1
= 3
k
,
R
E2
= 1.5
k
,
R
E3
= 0.75
k
,
I
O1
= 1
mA,
I
O2
= 2
mA,
I
O3
= 4
mA
10.45 (a) For
V
GS
= 0.75
V,
R = 25
k
,
W
L
1
= 20,
W
L
2
= 40
(b)
R
o
= 667
k
(c)
1.5%
10.47 (a)
0.19 ≤ I
O
≤ 0.21
mA
(b)
0.1921 ≤ I
O
≤ 0.2081
mA
10.49
R
o
=
2 + g
m
r
o
g
m
(1 + g
m
r
o
)
∼
=
1
g
m
10.51 (a)
I
REF
= 606 μA
(b)
I
O
= 362.5 μA
(c)
I
O
= 370.7 μA
10.53
W
L
1
= 5.56,
W
L
2
= 2.78
,
W
L
3
= 1.81
nea80644_appE_1337-1358.qxd 8/6/09 10:09 AM Page 1348 pmath DATA-DISK:Desktop Folder:UDAYVEER/Neamen:
Appendix E Answers to Selected Problems 1349
10.55
W
L
1
= 8.33
,
W
L
2
= 2.67
,
W
L
3
=
W
L
4
= 2.72
10.57
W
L
1
= 59.9,
W
L
2
= 21.8
,
W
L
3
= 2.14
10.59
R
o
= 2.58 ×10
9
10.61
W
L
1
= 40,
W
L
2
= 4
,
W
L
3
= 2.16,
W
L
4
= 3.61
10.63 (a)
I
O
= 80 μA
(b)
I
O
= 80.05 μA
10.65 (a)
R = 6.12 k
(b)
(V
−
− V
−
)
min
= 1.66 V
(c)
W
L
5
= 2.5,
W
L
6
= 7.5
10.67
W
L
1
= 86.1
,
W
L
2
= 12.5
,
W
L
3
= 25,
W
L
4
= 50
10.69
I
REF
= 59.09 μ
A,
I
1
= 11.82 μ
A,
I
2
= 73.87 μ
A,
I
3
= 47.27 μ
A,
I
4
= 236.4 μ
A
10.71
I
D2
= 62.5 μ
A,
I
O
= 41.67 μ
A,
V
GS1
= V
GS2
= 1.107
V,
V
SG3
= V
SG4
= 0.9893
V
10.73 (a)
I
O
= 2.5mA
(b)
I
O
= 3mA
(c)
I
O
= 3.5mA
10.75 (a)
A
v
=−1846
(b)
V
I
= 0.6943
V
(c)
V
B
= 1.824
V
10.77 (a)
V
EB
= 0.5568 V
(b)
R
1
= 4.44 k
(c)
V
I
= 0.5389 V
(d)
A
v
=−1846
10.79 (a)
W
L
0
= 3.96
,
W
L
1
=
W
L
2
= 3.24
,
W
L
3
= 3.24
(b)
A
v
=−70.35
10.81 (a)
A
v
=−1978
(b)
A
v
=−454
(c)
A
v
=−256
10.83 (a) To a good approximation,
R
o
= r
o2
[1 + g
m2
(r
π2
R
E
)]
(b)
A
v
=−g
m0
(r
o
R
L
R
o
)
10.85 (a)
g
m0
= 0.80 mA/V, r
o
= 333 k
g
m2
= 0.748 mA/V, r
o2
= 250 k
(b)
A
v
=−114.3
(c)
R
L
= 143 k
10.87
A
v
=−316
10.89
A
v
=−366,165
CHAPTER 11
11.1 (a)
v
O
= 0.375 sin ωt
(V)
(b)
v
O
= 0.45 sin ω t
(V)
(c)
v
O
= 2.75 sin ω t
(V)
11.3 (a)
R
E
= 11.5
k
,
R
C
= 18
k
(c)
−2.3 ≤ v
cm
≤ 0.673
V
11.5 (a)
R
1
= 73.25
k
,
R
C
= 28.5
k
(b)
A
d
= 62
,
A
cm
=−0.113
,
CMRR
dB
= 54.8
dB
(c)
R
id
= 46
k
,
R
icm
= 12.6
M
11.7
R
1
= 53
k
,
R
C
= 40
k
,
A
d
= 38.5
11.9 (a)
R
E
= 62
k
(b)
A
d
= 71.0
,
A
cm
=−0.398
,
CMRR
dB
= 45.0
dB
(c)
R
id
= 70.4
k
,
R
icm
= 6.28
M
11.11 (a)
I
C1
= I
C2
= 5.90 μ
A,
V
EC1
= V
EC2
= 1.475
V
(b) (i)
A
d
= 81.7
,
A
cm
= 0
(ii)
A
d
= 40.8
,
A
cm
=−0.0442
11.13 (a)
R
E
= 17.9
k
,
R
C
= 16.7
k
(b)
|
A
d
|
= 76.9
(c)
A
d
= 76.9
,
A
cm
= 0.0279
,
CMRR
dB
= 68.8
dB
11.15 (a)
v
E
=+0.7
V,
v
C1
= v
C2
=−2.87
V
(b)
v
E
=+0.7
V,
v
C1
=−5
V,
v
C2
=−0.736
V
(c)
v
E
∼
=
+0.7
V,
v
C1
=−2.255
V,
v
C2
=−3.485
V
11.17 (a) (i)
v
O1
−v
O2
= 0
,
(ii)
v
O1
−v
O2
= 0.08
V
(b) (i)
v
O1
−v
O2
= 0.1995
V,
(ii)
v
O1
−v
O2
= 0.2795
V
11.19 (a)
v
d
(
max
)
∼
=
14
mV
(b)
v
d
(
max
)
∼
=
21.2
mV
nea80644_appE_1337-1358.qxd 8/6/09 10:09 AM Page 1349 pmath DATA-DISK:Desktop Folder:UDAYVEER/Neamen:
1350 Appendix E Answers to Selected Problems
11.21 (a)
R
E
= 37.2
k
(b) (i)
A
d
= 119
, (ii)
A
cm
=−0.673
11.23
A
v1
=
−
1
2
g
m
R
L
2 +
R
L
R
C
,
A
v2
=
+
1
2
g
m
R
L
2 +
R
L
R
C
A
v
=
g
m
R
L
2 +
R
L
R
C
11.25
A
d
= g
m
R
C
,
A
cm
= 0
11.27
−0.22 ≤ v
O2
≤ 0.22
V
11.29 (a)
R
1
= 38.6
k
,
R
2
= 236
(b)
R
icm
= 292
M
(c)
A
cm
=−0.0123
11.31 (a)
I
1
= I
Q
= 0.337
mA,
I
D1
= 0.168
mA,
V
DS1
= 8.79
V,
V
DS4
= 7.18
V
(c)
v
CM
(
max
)
= 7.97
V,
v
CM
(
min
)
=−6.02
V
11.33 (a)
I
Q
= 160 μ
A,
R
D
= 37.5
k
(c)
v
CM
= 2.50
V
11.35 (a)
R
1
= 33.9
k
(b) (i)
v
O1
−v
O2
= 0
,
(ii)
v
O1
−v
O2
= 0.125
V
(c) (i)
v
O1
−v
O2
=−0.156
V
(ii)
v
O1
−v
O2
=−0.031
V
11.37 (a)
v
d
(
max
)
∼
=
0.19
V
(b)
v
d
(
max
)
∼
=
0.285
V
11.39 (a)
I
D
= 6 μ
A,
V
SD
= 1.69
V
(b) (i)
A
d
= 9.66
,
A
cm
= 0
(ii)
A
d
= 4.83
,
A
cm
=−0.0448
11.41
A
d
=−9.16
,
A
cm
=−0.003216
,
CMRR
dB
= 69.1
dB
11.43 (a)
v
S
= 1.459
V,
v
D1
= v
D2
=−4.115
V
(b)
v
S
= 2.344
V,
v
D1
= v
D2
=−4.336
V
(c)
v
S
∼
=
1.459
V,
v
D1
=−3.909
V,
v
D2
=−4.321
V
(d)
v
S
∼
=
2.344
V,
v
D1
=−4.158
V,
v
D2
=−4.515
V
11.45 Let
I
Q
= 0.50
mA; For
V
D1
= V
D2
= 3
V,
then
R
D
= 28
k
;
W
L
1
=
W
L
2
= 1274
Need
R
o
= 200
k
11.47
A
d1
=
−
1
2
g
m
R
L
2 +
R
L
R
D
,
A
d2
=
+
1
2
g
m
R
L
2 +
R
L
R
D
A
d
=
g
m
R
L
2 +
R
L
R
D
11.49 (a)
R
D
= 40.8
k
(b)
v
cm
(
max
)
= 1.32
V
11.51 (a)
R
D
= 19
k
,
R
o
= 998
k
(b) Use cascode current source similar
to Figure 10.18 with
v
DS2
(
sat
)
= 0.3
V.
11.53 Let
I
Q1
= I
Q2
= 0.1
mA,
then
R
1
= 100
k
,
R
2
= 50
k
W
L
1
=
W
L
2
= 6.67
,
W
L
3
=
W
L
4
= 240
11.55 (a)
R
D
= 6
k
,
I
Q
= 1
mA
(b)
g
f
(
max
)
= 0.25
mA/V
(c)
A
d
= 1.5
11.57
A
d
=
g
m
R
L
2 +
R
L
R
D
11.59 (a)
R
1
= 36.8
k
(b)
A
d
= 2308
(c)
R
id
= 74.9
k
,
R
o
= 480
k
(d)
−3.6 ≤ v
cm
≤ 4.3
V
11.61 (a)
I
Q
= 0.749
mA
(b)
v
cm
(
max
)
= 3.8
V
11.63 (a)
A
v
= 89.72
(b)
A
v
= 89.32
(c)
A
v
= 89.52
11.65 (a)
A
d
= 79.44
(b)
v
CM
(
max
)
= 1.25
V
11.67
I
Q1
= I
Q
11.69 (a)
R
o
= 1
4.704 = 0.8247
M
(b)
A
d
= 3172
11.73
I
Q
∼
=
66 μ
A,
W
L
n
= 23.76
11.75 (a)
R
o
= 172
M
(b)
A
d
= 46096
11.77 (a)
V
+
=−V
−
= 4
V
(b)
A
d
= 50
11.79 (a)
V
+
=−V
−
= 3.4
V
(b)
A
d
= 1004
11.81 (a)
A
d
= 88.9
11.83 (a)
g
m1
= 0.506
mA/V,
g
m2
= 27.25
mA/V,
r
π2
= 6.605
k
(b)
R
o
= 19.8
11.85
A
v
=−48.4
11.87
R
i
∼
=
1.05
M
,
R
o
= 0.472
k
,
A
v
=−438
nea80644_appE_1337-1358.qxd 8/6/09 10:09 AM Page 1350 pmath DATA-DISK:Desktop Folder:UDAYVEER/Neamen:
Appendix E Answers to Selected Problems 1351
11.89 (a)
R
1
= 27.2
k
,
R
2
= 5
k
(b)
g
m1
= 0.7071
mA/V,
g
m2
= 2
mA/V,
r
o1
= 200
k
,
r
o2
= 50
k
(c)
A
v
=−15.25
(d)
R
o
= 0.450
k
11.91 (a)
R
C1
= 80
k
,
R
C2
= 20
k
(b)
A
d1
=−69.6
,
A
d
=−5352
11.93 (a)
A
d
=−3.73
,
A
cm
= 0.0718
(b)
v
O3
=−0.975 sin ω t
(V),
v
O3
(
ideal
)
=−1.12 sin ω t
(V)
11.95 (a)
f
z
= 39.8
kHz
(b)
f
p
= 11.7
GHz
11.97 (a)
A
v
= 19.5
(b)
A
v
= 11.2
CHAPTER 12
12.1 (a)
β = 9.98 ×10
−3
(b)
A = 2000
12.3 (a) (i)
T =∞
,
A
f
= 6.667
(ii)
T = 1.5 ×10
3
,
A
f
= 6.662
(iii)
T = 15
,
A
f
= 6.25
(b) (i)
T =∞
,
A
f
= 4.0
(ii)
T = 2.5 ×10
3
,
A
f
= 3.9984
(iii)
T = 25
,
A
f
= 3.846
12.5 (a)
β =−0.01245
(b)
dA
A
= 2.5%
12.7 (a)
β
1
= 0.096685
,
β
2
= 0.0195
(b) For (a),
A
v f
= 47.33 (−5.34%)
,
For (b),
A
v f
= 49.86 (−0.28%)
12.9 (a)
f
C
= 20
kHz
(b)
A
v f
= 40
12.11 (a) (i)
A
v
= 5.25 ×10
5
(ii)
β = 0.01333
(b) (i)
A
v f
= 74.99
(ii)
f
C
= 31.5
kHz
12.13 (a) For (a),
f
3-dB
= 2.352
kHz,
For (b),
f
3-dB
= 5
kHz
(b) Overall feedback
⇒
wider bandwidth
12.15 For (a), at low input, gain
=
81; at high
input, gain
=
73.4. For (b), at low input,
gain
=
81.1, at high input, gain
=
80.1
12.17
V
fb
= 24
mV,
V
ε
= 1
mV,
A
v
= 2.5 ×10
3
V/V,
A
v f
= 100
V/V
12.19 (a)
A
v f
=
A
1 +
A
1 +
R
2
R
1
(b)
β =
1
1 +
R
2
R
1
(c)
β = 0.04999
,
R
2
R
1
= 19.004
(d)
−2.222 × 10
−3
%
12.21
I
fb
= 24.2 μ
A,
I
o
= 3.125
mA,
β
i
= 0.007744
A/A,
A
i
= 3906
A/A
12.23
R
if
= 0.5
,
R
of
∼
=
2.42
12.25
A
gf
= 40
A/V,
I
o
= 6
mA,
V
fb
= 147 μ
V,
V
ε
= 3 μ
V
12.27
R
if
= 10
6
k
,
R
of
= 5.04 M
12.29
A
z
= 4V/μA
,
A
zf
= 0.2V/μA
,
β
g
= 4.75 μA/V
12.31
R
if
= 99.79
12.33 (a)
1 ≤ R
i
≤ 10
5
k
(b)
0.1 ≤ R
o
≤ 10
4
k
12.35 (a)
A
v f
= 11.0
(b)
R
if
= 273
M
(c)
R
of
= 15
m
12.37 (a)
g
m1
= 32.81
mA/V,
r
π1
= 3.66
k
g
m2
= 19.12
mA/V,
r
π2
= 6.28
k
g
m3
= 78.08
mA/V,
r
π3
= 1.54
k
(b)
A
v f
= 20.7
(c)
R
if
= 62.4
k
(d)
R
of
= 1.39
12.39
A
v f
= 1.273
12.41 (a) (i)
A
v f
= 0.801
(ii)
R
of
= 299
(b) (i)
+3.78%
(ii)
−15.4%
12.45 For example,
R
2
= 3
k
,
R
1
= 247
k
12.47 (a)
I
DQ1
= 3.99
mA,
I
DQ2
= 12.01
mA
(b)
A
i
=−0.920
12.49 (a)
g
m1
= 7.62
mA/V,
r
π1
= 13.1
k
g
m2
= 52.7
mA/V,
r
π2
= 1.90
k
(b)
A
if
= 8.60
(c)
R
if
= 47.4
12.51 (a)
g
m1
= 67.31
mA/V,
r
π1
= 1.78
k
g
m2
= 19.23
mA/V,
r
π2
= 6.24
k
(b)
A
v
= 7.44
12.53
A
if
= 5.33
12.55 (a)
A
gf
=
R
F
R
1
R
L
R
F
R
1
− R
3
−
R
L
R
3
R
2
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1352 Appendix E Answers to Selected Problems
(c)
R
2
= 2
k
, set
R
3
= 2
k
,
R
1
= R
F
= 10
k
12.57
A
gf
= 98.06
mA/V
12.59
A
gf
=−0.0653
mA/V
12.61 (a)
I
CQ
= 2.448
mA,
V
ECQ
= 1.31
V
(b)
A
zf
=−8.666
V/mA
(c)
R
if
= 0.1037
k
(d)
R
of
= 0.1186
k
12.63 (a)
A
zf
=−R
F
(b)
g
m
= 4.6
mA/V
12.65
R
F
= 0.807
M
12.71
T = 4.22
12.73 (a)
f
180
= 7.1 ×10
4
Hz
(b)
β = 0.0205
(c)
A
vf
(
0
)
= 48.54
(d) Smaller
12.75 (b)
β = 4.7 ×10
−4
12.77 (a)
f
180
= 10
6
Hz
(b)
β = 2.83 ×10
−3
(c)
A
vf
(0) = 351
12.79 (a)
f
180
= 3.33 ×10
5
Hz
(b) (i)
|
T ( f
180
)
|
= 0.156
(ii)
φ =−142.3
◦
(c)
A
f
(0) = 50
12.81
β = 0.01428
12.83 (a)
f
180
= 1.42 ×10
5
Hz,
|
T ( f
180
)
|
= 4.89
(b)
f
PD
= 4.85
Hz
12.85 (a)
β = 0.0199
,
f
PD
= 711
Hz
(b) Phase margin
= 29.8
◦
12.87 (a)
f = 0.976
MHz, phase margin
= 73.4
◦
(b)
f
P1
= 2.67
Hz, phase margin
= 85.4
◦
12.89
f
PD
= 577
Hz
CHAPTER 13
13.1 (a)
R
D2
= 15
k
,
R
D1
= 9
k
(b) (i)
A
d
= 2.846
, (ii)
A
2
=−12
(c)
A =−34.15
13.5 (a)
A
v
=−1.59 × 10
6
(b)
R
id
= 208 k
(c)
GBW = 12.3 MHz
13.7
v
d
∼
=
56.4V
13.9 (a)
R
5
= 17.13
k
,
R
4
= 2.438
k
,
V
EB12
= V
BE11
= 0.7184
V,
V
BE10
= 0.6453
V
(b)
I
REF
= 0.5137
mA,
I
C10
∼
=
30.22 μ
A
(c)
I
REF
:
2.74%
;
I
C10
:
0.733%
13.11
I
REF
= 0.22
mA,
I
C10
∼
=
14.2 μ
A,
I
C6
= 7.10 μ
A,
I
C17
= 0.165
mA,
I
C13A
= 0.055
mA
13.13
P = 48.8
mW,
I = 1.63
mA
13.15
I
C13A
= 0.125
mA,
I
R10
∼
=
0.012
mA,
I
C19
= 0.113
mA,
I
C18
= 12.565 μ
A,
V
BE19
= 0.60185
V,
V
BE18
= 0.54474
V,
I
C14
= 113 μ
A
13.17
R
1
= 45.11
k
,
R
2
= 51.56
k
13.19
A
d
=−882
13.21
A
v
= 964,650
13.23
R
eq
= 170
13.25 (a)
R
id
= 2.095
M
(b)
R
id
= 2.80
M
13.27 (a)
f
PD
= 10
Hz
(b)
C
F
= 13.25
pF
13.29 (a)
R
D1
= 7.87
k
,
R
D2
= 24.8
k
,
R
S
= 15
k
(b) (i)
A
d
= 2.49
(ii)
A
2
=−19.21
(iii)
A
3
= 0.950
(c)
A =−45.4
13.31 (a) Original:
g
m1
= g
m2
= 0.09975
mA/V,
New:
g
m1
= g
m2
= 0.1995
mA/V
(b)
A
d
= 250
,
A
v2
= 125
,
A
v
= 31,250
13.33 (a)
I
set
= I
Q
= I
D7
= 0.1776
mA
(b)
A
d
= 111.9
,
A
2
=−96.86
,
A =−10,839
13.35
f
PD
= 83.9
Hz
13.37
R
o
= 0.63
M
13.39 (a)
I
Q2
= 180 μ
A
(b)
W
L
8P
= 360
,
W
L
8N
= 180
13.41 (a)
I
D6
= I
D7
= 43.7 μ
A
(b)
A
v
=−35,350
13.43
W
L
1
=
W
L
P
= 4.06
,
W
L
N
= 1.85
13.45 (a)
A
d
= 136.1
(b)
R
o
= 92.6
k
(c)
f
PD
= 343.7
kHz,
GBW = 46.8
MHz
13.47 (a)
W
L
5
=
W
L
6
= 16.4
(c)
GBW = 27.2
MHz
13.49 (a)
R
S
= 10
k
,
R
C1
= 3.4
k
,
R
C2
= 12.39
k
(b)
A
d
= 15.60
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Appendix E Answers to Selected Problems 1353
(c)
A
2
=−20.97
(d)
A
3
= 0.95
(e)
A =−310.8
13.51 (a) For NMOS:
g
mN
= 1.0
mA/V,
r
oN
= 400
k
For BJT:
r
o2
= 800
k
(b)
A
d
= 266.7
13.53
W
L
N
= 6.16
,
W
L
P
= 13.6
13.55
K
p
= 0.234
mA/V
2
13.57
A
d
= 10.38
,
|
A
v2
|
= 1917
,
|
A
v
|
= 19,895
13.59
I
DSS
= 0.8
mA
13.61 (a) Set
V
P
= 3
V,
V
ZK
= 3
V,
R
3
= 24
k
, Set
I
DSS
= 0.2
mA
(b)
R
4
= 22.8
k
CHAPTER 14
14.1
v
i
(max)
rms
= 39.77 mV
14.3 (1)
v
O
= 2V
, (2)
v
2
= 12.5mV
,
(3)
A
OL
= 2 ×10
4
, (4)
v
1
= 8 μV
,
(5)
A
OL
= 1000
14.5 (a) (i)
A
CL
= 7.90863
(ii)
−0.03956%
(b) (i)
A
CL
= 7.84966
(ii)
−0.785%
14.7
A
OL
= 8.9991 ×10
5
14.9 (a)
9.98 ≤
|
A
|
≤ 10.02
(b)
9.969 ≤
|
A
|
≤ 10.009
14.11
A
OL
= 49,950
14.13 (a)
A
CL
= 4.927
,
R
if
= 2.687
M
,
R
of
= 3.16
(b)
−0.05%
14.15
v
O
=−1.993v
I1
−3.986v
I2
;
−0.35%
14.17 (a)
R
if
= 99.1
(b)
R
of
= 18.4
(c)
A
CL
= 0.650
(d) 0.650
14.19
f
PD
= 7.94
Hz,
GBW = 7.94 ×10
5
Hz
14.21
f
PD
= 448
Hz
14.23 (a)
A
CLO
=−9.9978
,
f
3-dB
= 150.033
kHz
(b)
A
CLO
=−999.34
,
f
3-dB
= 76.49
kHz
14.25
|
A
v
|
max
= 50
14.27 (a)
V
PO
= 5.09
V
(b)
V
PO
= 8
V
14.29 (a)
V
PO
= 5.0
V
(b)
V
PO
= 23.87
V
14.33 (a)
I
S4
= 5 ×10
−15
A
(b)
I
S4
= 4.939 ×10
−15
A
(c)
I
S4
= 4.811 ×10
−15
A
14.35 (a)
−0.360 ≤ v
O
≤−0.240
V
(b)
−3.06 ≤ v
O
≤−2.94
V
14.37
V
o
=
V
i
RC
· t
,
t = 10
3
s
14.39 At
V
+
node,
v
O
= 0.8645
V,
At center,
v
O
= 0.9545
V,
At
V
−
node,
v
O
= 1.0445
V
14.41
i
C1
i
C2
= 1.0155
14.43 (a)
v
O
=−0.30
V
(b)
v
O
=−0.50
V
(c)
v
O
=−0.10
V
(d)
v
O
=−1.30
V
14.45 (a)
v
O
=−0.010 V
14.47 (a)
v
O1
= 0.10 V
,
v
O2
=−0.45 V
14.49 (a)
v
O1
= 0.15
V,
v
O2
= 0.15
V,
v
O3
=−0.09
V
(b)
R
A
= 8.33
k
,
R
B
= 10
k
(c)
v
O1
=±0.015
V,
v
O2
=±0.015
V,
v
O3
=±0.021
V
14.51 (a)
4 ≤ v
O
≤ 76
mV
(b)
−39 ≤ v
O
≤ 39
mV
(c)
2.161 ≤ v
O
≤ 2.239
V
14.53 (a) For offset voltage,
|
v
O1
|
= 110 mV,
|
v
O2
|
= 610 mV
For bias currents,
v
O1
= 0.31 V,v
O2
=−1.51 V
14.55 For circuit (a),
v
O
= 9mV;
For circuit (b),
v
O
=−1.0815 V
14.57 (a)
v
O1
= 6mV,v
O2
= 28 mV;
(b)
v
O1
= 6.495 mV,
v
O2
= 30.31 mV
14.59 (a)
v
O1
= 0.10 V,v
O2
= 0.12 V;
(b)
v
O1
= 0.105 V,
v
O2
= 0.166 V;
(c) Due to
I
B
,v
O1
= 0.125 V,
v
O2
= 0.15 V;
Due to
I
OS
,v
O1
= 0.133 V,
v
O2
= 0.224 V
14.61 (a)
x = 0.4728%
(b)
x = 0.0267%
nea80644_appE_1337-1358.qxd 8/6/09 10:09 AM Page 1353 pmath DATA-DISK:Desktop Folder:UDAYVEER/Neamen:
1354 Appendix E Answers to Selected Problems
CHAPTER 15
15.1 (a) Noninverting amplifier:
Set
R
1
= 30
k
,
R
2
= 210
k
At noninverting terminal, let input
C = 0.001 μ
F, then
R = 5.305
k
(b) Inverting amplifier:
Set
R
1
= 15
k
,
R
2
= 300
k
;
Capacitor in series with
R
1
,
C = 530.5
pF
15.3 6-pole filter
15.5 (a) Let
R = 20
k
, then
C = 397.9
pF;
C
1
= 1411
pF,
C
2
= 553.9
pF,
C
3
= 80.5
pF
(b) (i)
|
T
|
=−0.0673
dB,
(ii)
|
T
|
=−0.711
dB,
(iii)
|
T
|
=−3.0
dB,
(iv)
|
T
|
=−6.83
dB,
(v)
|
T
|
=−10.9
dB
15.7 9-pole filter
15.9
N = 35, f
3-dB
= 12.25 kHz
15.11 (a)
|
T
|
=−9.31
dB
(b)
|
T
|
=−14.8
dB
(c)
|
T
|
=−20.5
dB
15.13 3-pole filter
15.15 (b) (i)
|
A
v
|
max
= 28.3
(ii)
f
o
= 5.305
kHz
(iii)
f
1
= 5.296
kHz,
f
2
= 5.315
kHz
15.17 (a)
T (s) =
1 +
R
2
R
1
1 + s(R
1
R
2
)C
1 + sR
2
C
,
f
3-dB1
=
1
2π R
2
C
,
f
3-dB2
=
1
2π(R
1
R
2
)C
(b)
T (s) =
1 +
R
2
R
1
[
1 + s(R
1
R
2
)C
]
,
f
3-dB
=
1
2π(R
1
R
2
)C
15.19 Two noninverting amplifiers:
Let
R
2
= 250
k
,
R
1
= 54.1
k
For high-pass filter: Set
R = 250
k
,
C = 31.8
pF
For low-pass filter: Set
R = 250
k
,
C = 0.00424 μ
F
15.21 (a)
Q = 10
pC
(b)
I
eq
= 1 μ
A
(c)
t = 4.61 ×10
−8
s
15.23 (a)
τ = 60 μ
s
(b)
v
O
= 0.167
V
(c) 78 clock pulses
15.25 (a)
f
o
=
1
2πR
√
C
(
C +2C
V
)
(b)
480 ≤ f
o
≤ 919
kHz
15.29 (a)
f
o
=
1
2π RC
·
3R + 2R
V
R
V
(b)
R
F
R
= 2
(c)
13.5 ≤ f
o
≤ 16.84
kHz
15.31 (a)
f
o
=
1
2πRC
·
4
R
R
R
V
+2
(b)
19.45 ≤ f
o
≤ 22.66
kHz
15.33 (a)
R
1
R
1
+ R
2
=
sRL
sRL+
(
R + sL
)
2
(b)
f
o
=
R
2πL
(c)
R
2
R
1
= 2
15.35
C
2
∼
=
0.06 μF, L = 13.8 μ
H
15.37 (a)
f
o
=
1
2π
√
C
(
L
1
+ L
2
)
(b)
L
1
L
2
= g
m
R
15.39
T(s) =
1 +
R
2
R
1
sRC
s
2
R
2
C
2
+3sRC +1
,
f
o
=
1
2π RC
,
R
2
R
1
= 2
15.41
T(s) =
1 +
R
2
R
1
sRL
s
2
L
2
+3sRL+ R
2
,
f
o
=
R
2πL
,
R
2
R
1
= 2
15.43
R
1
= 1.47
k
,
R
2
= 73.53
k
15.45 (a)
V
TH
= 0.4
V,
V
TL
=−0.4
V
(b) For
33.333 ≤ t ≤ 33.439
ms,
v
O
=+10
V,
For
33.439 ≤ t ≤ 66.667
ms,
v
O
=−10
V
15.47 (a)
V
TH
=
V
REF
R
3
+
V
P
R
2
1
R
1
+
1
R
2
+
1
R
3
,
V
TL
=
V
REF
R
3
−
V
P
R
2
1
R
1
+
1
R
2
+
1
R
3
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Appendix E Answers to Selected Problems 1355
(b)
R
1
= 10.17
k
,
R
2
= 600
k
(c)
V
TH
=−4.9
V,
V
TL
=−5.1
V
15.49 (a)
V
S
=
R
2
R
1
+ R
2
· V
REF
,
V
TH
= V
S
+
R
1
R
1
+ R
2
· V
H
V
TL
= V
S
+
R
1
R
1
+ R
2
· V
L
(b)
R
1
= 4
k
,
R
2
= 188
k
,
V
REF
=−1.787
V
(c) (i)
|
i
|
= 71.8 μ
A, (ii)
|
i
|
= 53.2 μ
A
15.51 (a)
V
TH
= 0.175
V,
V
TL
=−0.175
V
(c) (i)
I
D1
= 0
,
I
D2
= 0.458
mA,
I
R2
=−0.465
mA,
I
R3
= 7 μ
A
(ii)
I
D1
= 0.458
mA,
I
D2
= 0
,
I
R2
= 0.465
mA,
I
R3
=−7 μ
A
15.53 (a)
V
REF
= 3.6
V
(b) Let
R
2
= 90
k
,
R
1
= 10
k
(c)
v
O
=−5.1
V
15.55 (a)
v
O
=
V
REF
+2V
γ
,
v
O
=−
V
REF
+2V
γ
,
V
TH
=
R
1
R
2
V
REF
+2V
γ
,
V
TL
=−
R
1
R
2
V
REF
+2V
γ
15.57 (a)
v
X
= 5 −8.333e
−t/τ
X
for
0 ≤ t ≤ t
1
;
v
X
=−10 + 11.667e
−
(
t−t
1
)
/τ
X
for
t
1
≤ t ≤ T
(b)
f = 847
Hz
15.59 For
0 ≤ t ≤ t
1
,
t
1
= 1.1
ms,
v
Y
= 10(1 −e
−t/τ
Y
), τ
Y
= 4 ×10
−5
s
15.61 (a) Set
C
X
= 0.01 μ
F,
R
X
= 23.84
k
(b)
v
Y
<v
X
(c)
133 μ
s
15.63 (a) Let
C = 50 μ
F,
R = 1.09
M
(b)
t
∼
=
38.7
ms
15.65
R
B
= 50
k
,
C = 144.3
pF
15.67
1.40 ≤ f
o
≤ 2.72
kHz
50.5% ≤
duty cycle
≤ 98.1%
15.69 (a)
P = 3.7
W
(b)
V
+
∼
=
19
V
(c)
V
P
= 8.6
V
15.71 (a)
A
v
= 1 +
R
2
R
1
+
R
3
R
1
(b)
R
1
= 50
k
,
R
2
= 200
k
,
R
3
= 250
k
(c)
|
v
O1
|
=
|
v
O2
|
= 10
V,
i
L
= 1
A
15.73 (a) Set
R
2
= R
3
= R
4
= 100
k
,
R
1
= 8.69
k
(b) (i)
¯
P
L
= 14.4
W, (ii)
R
L
= 20
(c)
¯
P
L
= 7.2
W
15.75 (a)
R
of
= 4
m
(b)
V
O
= 12
mV
15.77 (a)
R
4
= 31.5
(b)
R
12
= 5.57
k
15.79 (a)
7.659 ≤ V
O
≤ 22.976
V
(b)
4.17 × 10
−4
%
CHAPTER 16
16.1 (a)
W
L
= 2.91
(b)
V
It
= 1.172
V,
V
Ot
= 0.672
V
(c)
i
D,max
= 80 μ
A,
P
D,max
= 264 μ
W
16.3 (a)
R = 41.6
k
,
W
L
= 0.193
(b)
0.5 ≤ V
GS
≤ 2.38
V
16.5 (a)
K
D
K
L
= 4.26
(b)
K
D
K
L
= 5.4
(c)
P = 0.759
mW
16.7
(W/L)
D
(W/L)
L
= 5.4
16.9 (a) Driver:
V
It
= 0.8578
V,
V
Ot
= 0.3578
V
Load:
V
It
= 0.8578
V,
V
Ot
= 2.5
V
(b)
v
O
= 0.0230
V
(c)
i
D,max
= 64 μ
A,
P
D,max
= 211 μ
W
16.11 (a)
W
L
L
= 2.47
,
W
L
D
= 5.04
(b) Driver:
V
It
= 0.720
V,
V
Ot
= 0.420
V
Load:
V
It
= 0.720
V,
V
Ot
= 1.20
V
(c)
P
D,max
= 80 μ
W,
v
O
= 0.0297
V
16.13 (a) (i)
P = 0
, (ii)
P = 1.16
mW
(b) (i)
P = 0
, (ii)
P = 0.825
mW
(c) (i)
P = 0
, (ii)
P = 0.20
mW
16.15 (a)
v
O1
= 0.0431
V,
v
O2
= 5
V
(b)
v
O1
= 5
V,
v
O2
= 0.0431
V
16.17 (a)
v
I
= 0.6683
V
(b)
v
I
= 0.6057
V
16.19 (a)
K
D
K
L
= 2.04
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1356 Appendix E Answers to Selected Problems
(b)
W
L
L
= 0.667
,
W
L
D
= 1.36
(c)
v
O
= 0.0329
V
16.21 (a)
W
L
ML1
= 0.417
,
W
L
MD1
= 1.35
(b)
v
O2
= 0.0330
V
16.23 (a)
v
O
∼
=
0.135
V
(b)
v
DSX
∼
=
0.067
V,
v
DSY
∼
=
0.068
V,
v
GSX
= 5
V,
v
GSY
∼
=
4.933
V
16.25
¯
Y =
[
A
OR (
B
AND
C
)] AND
D
16.27
Y =
(
A + B
)(
C + D
)
16.29 (a) For NMOS circuit,
M
NA
in parallel with [
M
NB
in series
with (
M
NC
in parallel with
M
ND
)
]
(b)
W
L
A
= 1.735
,
W
L
B,C,D
= 3.47
16.31 (a)
V
It
= 1.25
V,
V
OPt
= 1.65
V,
V
ONt
= 0.85
V
(c) For
v
I
= 1.1
V,
v
O
= 2.214
V
For
v
I
= 1.4
V,
v
O
= 0.286
V
16.33 (a) (i)
V
It
= 1.707
V
(ii)
v
I
= 1.26
V
(iii)
v
I
= 2.157
V
(b) (i)
V
It
= 1.25
V
(ii)
v
I
= 0.855
V
(iii)
v
I
= 1.61
V
16.35 (a)
1.7 ≤ v
O1
≤ 3.3
V
(b)
v
O3
= 5
V,
v
O1
= 2.78
V,
v
I
= 2.5
V
16.37 (a)
i
D,peak
= 24.2 μ
A
(b)
i
D,peak
= 29.34 μ
A
(c)
i
D,peak
= 33.23 μ
A
16.39 (a)
P = 12.5 μ
W
(b)
P = 6.48 μ
W
16.41 (a)
P = 3 ×10
−7
W
(b) (i)
C
L
= 0.0024
pF
(ii)
C
L
= 0.00551
pF
(iii)
C
L
= 0.0267
pF
16.43 (b) Let
W
L
N
= 2
,
W
L
P
= 4
,
V
DD
= 5
V
For NMOS:
i
d
= 0.45
mA
For PMOS:
i
d
= 0.36
mA
16.45 (a)
V
IL
= 0.8268
V,
V
IH
= 1.2713
V
(b)
NM
L
= 0.5858
V,
NM
H
= 1.0439
V
16.47 (a)
v
C
= 2.5
V
(b)
W
L
N
=
9
2
·
W
L
P
(c)
V
It
= 1.65
V
16.49 (a)
W
L
N
= 2
,
W
L
P
= 16
(b)
W
L
N
= 4
,
W
L
P
= 32
16.51 (a)
W
L
N
= 2
,
W
L
P
= 12
(b)
W
L
N
= 4
,
W
L
P
= 24
16.53 (a)
Y = ABC + DE
(b)
M
NA
, M
NB
, M
NC
in series and
are in parallel with
M
ND
, M
NE
in series
(c)
W
L
AN,BN,CN
= 6
,
W
L
DN,EN
= 4
,
W
L
P
= 8
16.55 (a)
Y = A + BC + DE
(b)
M
NA
in parallel with (
M
NB
in series
with
M
NC
)
in parallel with (
M
ND
in
series with
M
NE
)
(c) NMOS:
W
L
A
= 2
,
W
L
B,C,D,E
= 4
PMOS:
W
L
A,B,C,D,E
= 12
16.57 (b) All
W
L
N
= 3
, All
W
L
P
= 6
16.59 (a)
¯
Y = C
(
A + B
)
(b) All
W
L
N
= 4
,
W
L
CP
= 4
,
W
L
AP,BP
= 8
16.61 (a) State
v
O1
v
O2
150
250
350
450
550
605
(b)
v
O2
= (v
A
OR
v
B
)
AND
v
C
nea80644_appE_1337-1358.qxd 8/6/09 10:09 AM Page 1356 pmath DATA-DISK:Desktop Folder:UDAYVEER/Neamen:
16.63
(M
NA
in series with
M
M
¯
B
)
and in parallel
with (
M
N
¯
A
in series with
M
NB
)
16.65
M
NA
in series with (
M
NB
in parallel with
M
NC
)
in series with (
M
ND
in parallel with
M
NE
)
16.67 (a) (i)
v
O
= 0
, (ii)
v
O
= 2.9
V,
(iii)
v
O
= 2.5
V
(b) (i)
v
O
= 0
, (ii)
v
O
= 1.4
V,
(iii)
v
O
= 1.4
V
16.69 (a)
v
O1
= 2.1
V,
v
O2
= 2.5
V
(b)
W
L
1
= 12.1
,
W
L
3
= 1.09
16.71
Y = A +
¯
AB
Z =
¯
A
¯
B
16.73 (a) (i)
Y = 0
, (ii)
Y = 2.5
V,
(iii)
Y = 0
, (iv)
Y = 2.5
V
(b) (i)
Y = 0,
(ii)
Y = 0
,
(iii)
Y = 2.5
V, (iv)
Y = 2.5
V
(c) Multiplexer
16.75 (a) (i)
Y = 0
, (ii)
Y = 2.5
V,
(iii)
Y = 2.5
V, (iv)
Y = 0
(b) Exclusive OR
16.79 For
v
I
= 1.5V,v
O1
= 2.88 V,
v
O
∼
=
0
v
I
= 1.6V,v
O1
= 2.693 V,
v
O
= 0.00979 V
v
I
= 1.7V,
Switching point,
v
I
= 1.8V,v
O1
= 0.607 V,
v
O
= 3.298 V
16.81 (a) Positive edge when
CLK = 1,
Q =
D = D
16.83 Not actually a J-K flip-flop
16.85 (a) 1 Megabit memory
= 1,048,576 cells
⇒ 1024 ×1024
memory
Then 10 input row and
column decoder lines necessary.
(b)
250 K × 4 bits ⇒
262,144 × 4 bits ⇒
512 × 512
memory
Then 9 input row and
column decoder lines
necessary.
16.87
t = 62.6ns
16.89
R = 0.512 M,
W
L
= 0.797
16.91
¯
Q = 1.794
V,
Q = 0.370
V
16.97 (a) 7 bits
(b) 0.02578125 V
(c) 1100010
16.99 (a)
R
1
= 5.88%
(b)
R
4
= 33.3%
16.101 (a)
I
1
=−0.50 mA
,
I
2
=−0.25 mA
,
I
3
=−0.125 mA
,
I
4
=−0.0625 mA
,
I
5
=−0.03125 mA
,
I
6
=−0.015625 mA
(b)
v
O
= 0.078125 V
(c)
v
O
= 1.484375 V
(f)
v
O
= 1.640625 V
16.103 64 resistors, 63 comparators
16.105 (a) 1010000000
(b) 0101111101
CHAPTER 17
17.1 (a)
R
C
= 2
k
(b) (i)
v
O1
= 0
,
v
O2
=−0.4
V
(ii)
v
O1
=−0.4
V,
v
O2
= 0
(c)
P = 0.36
mW
17.3 (a)
R
C2
= 6
k
(b)
R
C1
= 4
k
(c)
v
I
=−0.0360
V
17.5 (a)
R
C2
= 0.758
k
(b)
R
C1
= 0.658
k
(c) For
v
in
=−0.7
V,
v
O1
=−0.7
V,
v
O2
=−1.7
V
For
v
in
=−1.7
V,
v
O1
=−1.7
V,
v
O2
=−0.7
V
(d) (i)
P = 21.8
mW, (ii)
P = 20.7
mW
17.7 (a)
R
1
= 10.5
k
(b)
R
5
= R
6
= 17.5
k
(c)
I
Q
= 0.20
mA,
R
C1
= 3.5
k
(d)
I
Q
= 0.20
mA,
R
C2
= 3.5
k
17.9
R
E
= 1.375
k
,
R
C1
= 0.75
k
,
R
C2
= 1.031
k
,
R
2
= R
3
= 2.25
k
17.11 (a)
V
R
=−1.7
V
(b) Logic 1
=−0.7
V, logic 0
=−2.7
V
(c)
V
E
=−2.4
V for logic 0,
V
E
=−1.4
V for logic 1
(d)
P = 39.9
mW for logic 0 and logic 1
17.13 (a)
V
R
= 2.6
V
(b)
R
C1
= 4.57
k
Appendix E Answers to Selected Problems 1357
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