Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
_
Chapter 2
2.1
1000
(a) For I > 0.6 V, O =
( I 0.6 )
1020
For I < 0.6 V, O = 0
1000
(b) (ii) O = 0 =
[10 sin ( t )1 0.6]
1020
0. 6
Then sin (
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
_
Chapter 5
5.1
(a) i E = (1 + )i B 1 + =
325
= 116 = 115
2.8
115
=
= 0.9914
1 + 116
iC = i E i B = 325 2.8 = 322 A
=
(b) 1 + =
1.80
= 90 = 89
0.020
89
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
_
Chapter 8
8.1
(b) (i) R D =
PD , max
24
= 6
4
= (12 )(2 ) = 24 W
(ii) PD , max = 30 = (20 )I DQ I DQ = 1.5 A
I D , max = 2(1.5) = 3 A
RD =
(c) (i) I D
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
_
Chapter 6
6.1
I CQ
(a) (i) g m =
VT
VT
r =
ro =
I CQ
=
=
0.5
= 19.23 mA/V
0.026
(180)(0.026) = 9.36 k
0.5
V A 150
=
= 300 k
I CQ 0.5
2
= 76.92 mA/V
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
_
Chapter 7
7.1
a.
T ( s) =
T (s) =
V0 ( s )
Vi ( s )
=
1/ ( sC1 )
1/ ( sC1 ) + R1
1
1 + sR1C1
b.
fH =
1
1
=
f H = 159 Hz
3
2 R1C1 2 (10 )(106 )
c.
V0
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
_
Chapter 10
10.1
I1 = I 2 =
0 2V V
R1 + R2
2V + I 2 R2 = VBE + I C R3
2V +
IC =
a.
R2
( 2V V ) = VBE + IC R3
R1 + R2
1
R3
R2
2V ( 2V + V )
VBE
R1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
_
Chapter 13
13.1
0 ( 3)
= 15 k
0 .2
k W
p
2
I D3 =
2 L (V SG 3 + VTP )
3
0.04
2
0. 2 =
(40 )(V SG 3 0.4 ) V SG 3 = 0.9 V
2
(a) R D 2 =
0 .9
=
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
_
Chapter 11
11.1
(a) CMRR dB = o = Ad d = (250)(1.5 sin t ) (mV)
o = 0.375 sin t (V)
(b) CMRR dB = 80 dB CMRR = 10 4 =
250
Acm = 0.025
Acm
o = (250
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
_
Chapter 14
14.1
Ad =
vo
= 80
vi
vo (max) = 4.5 vi (max) = 56.25 mV
vi (max) rms =
56.25
= 39.77 mV
2
So
_
14.2
(a)
4.5
= 0.028125 mA
160
4.5
iL =
= 4
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
_
Chapter 15
15.1
(a) Noninverting amplifier
R
R
8 = 1 + 2 2 = 7 R 2 = 210 k , R1 = 30 k
R1
R1
At noninverting terminal
1
1
=
= 5.305 10 6
RC =
2 f 2
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Problem Solutions
_
Chapter 17
17.1
0 ( 0.2 )
= 2k
0.1
(b) (i) 1 = 1 V, Q1 off, Q 2 on
(a) R C =
O 2 = 0 (0.2 )(2) = 0.4 V
O1 = 0
(ii) 1 = 0.4 V, Q1 on, Q 2 off
O1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Problem Solutions
_
Chapter 16
16.1
k W
2
(a) O = V DD n R D 2( I VTN ) O O
2 L
0.1 W
2
0.1 = 3.3
(40) 2(3.3 0.5)(0.1) (0.1)
2 L
[
]
[
]
W
W
0.1 = 3.3 (1.1) =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
_
Chapter 12
12.1
(a) A f =
A
5 10 4
= 9.98 10 3
100 =
1 + A
1 + 5 10 4
(
)
A
A = 2000
1 + A(0.012 )
_
(b) 80 =
12.2
(a) A f =
A
10 5
80 =
= 0.0
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
_
Chapter 3
3.1
Kn =
k n W 120 10
2
=
0.75 mA/V
2L
2 0. 8
(a) (i) I D = 0
[
]
= (0.75)[2(2 0.4)(0.1) (0.1) ] = 0.2325 mA
= (0.75)[2(3 0.4)(0.1) (0.1)
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
_
Chapter 4
4.1
(a) (i) g m = 2
kn W
I DQ
2 L
0.1 W
W
0.5 = 2
(0.5) = 2.5
2 L
L
k W
(ii) I DQ = n (VGSQ VTN )2
2 L
0.1
2
0. 5 =
(2.5)(VG
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
_
Chapter 16
Exercise Solutions
EX16.1
k ' W
2
(a) O = V DD i D R D = VDD n R D 2( I VTN ) O O
2 L
0.1
2
0.1 = 3
(4)RD 2(3 0.5)(0.1) (0.1) R D =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
_
Chapter 5
Exercise Solutions
EX5.1
I E = (1 + )I B
IE
1.20
=
= 141.2 = 140.2
I B 0.0085
140.2
=
=
= 0.9929
1 + 141.2
I C = I E I B = 1.20 0.0085 = 1.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
_
Chapter 2
Exercise Solutions
EX2.1
V S V B V
12 4.5 0.6
= 27.6 mA
R
0.25
R (max ) = V S + V B = 12 + 4.5 = 16.5 V
Conduction cycle:
I = 12 sin t1 =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
_
Chapter 4
Exercise Solutions
EX4.1
k'
gm = 2 n
2
W
I DQ
L
W
0.1 W
1.8 = 2
= 20.25
(0.8)
L
2 L
_
EX4.2
k ' W 0.1
2
(a) K n = n =
2 L
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
_
Chapter 7
Exercise Solutions
EX7.1
(a) (i) f L =
1
S =
2 S
S = (RS + RP )C S
1
1
=
3.183 ms
2 f L 2 (50 )
3.183 10 3 = (2 + 8) 10 3 (C S ) C S = 0.
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
_
Chapter 6
Exercise Solutions
EX6.1
V BB V BE (on ) 0.85 0.7
=
I BQ = 0.833 A
RB
180
= I BQ = (120 )(0.000833 ) = 0.10 mA
(a) I BQ =
I CQ
VCEQ = VCC
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
_
Chapter 8
Exercise Solutions
EX8.1
1
VCC = 12 V
2
25
= 2 I CQ = 2 = 4.17 A
12
(a) PT = I CVCE ; At VCEQ =
25 = I CQ (12) I C ,max
24
= 5.76
4.16
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
_
Chapter 10
Exercise Solutions
EX10.1
I REF =
V + V BE (on ) V 3 0.7 ( 3)
=
= 0.1128 mA
47
R1
I REF
0.112766
=
= 0.1109 mA
2
2
1 + 1 +
120
I
= I B
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
_
Chapter 13
Exercise Solutions
EX13.1
5 0.6 0.6 ( 5)
= 0.352 mA
25
0.352
I C10 (5) = (0.026 ) ln
I
C10
I REF =
By trial and error
I C10 16 A
Then