Fundamentals of Microelectronics (3rd Edition, 2026) – Solutions Manual – by Razavi

Student Solutions Manual – Microelectronic
Circuit Design, 6th Edition (Jaeger & Blalock) |
Complete Step-by-Step Answers




1-1 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.2

B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip

1.3
(a)
B 0.1977 (Y2 −1960) 0.1977 ( Y −Y ) 0.1977 ( Y −Y )
19.97x10
2
= 0.1977 (Y1 −1960)
= 10 2 1
so 2 = 10 2 1


B1 19.97x10
log2
Y2 − Y1 = = 1.52 years
0.1977
log10
(b) Y2 − Y1 = = 5.06 years
0.1977


1.4
(2020−1970 )
N = 1610x100.1548 = 8.85 x 1010 transistors/P

1.5
0.1548 ( Y2 −1970) 0.1548 ( Y −Y )
N2 1610x10
= = 10 2 1

0.1548 ( Y1 −1970)
N1 1610x10
log2
(a) Y2 − Y1 = = 1.95 years
0.1548
log10
(b) Y2 − Y1 = = 6.46 years
0.1548


1.6 F = 8.00x10−0.05806 (2020−1970 )m = 10 nm .

No, this distance corresponds to the diameter of only a few atoms. Also, the wavelength of the
radiation needed to expose such patterns during fabrication is represents a serious problem.

1.7
From Fig. 1.4, there are approximately 600 million transistors on a complex Pentium IV
microprocessor in 2004. From Prob. 1.4, the number of transistors/P will be 8.85 x 1010 . in
2020. Thus there will be the equivalent of 8.85x1010 /6x108 = 148 Pentium IV processors.




1-2 ©R. C. Jaeger & T. N. Blalock
6/9/06

,1.8 1.13 x 108 W
6


(
P = 75x10 tubes )(1.5W tube)= 113 MW! I=
220V
= 511 kA!


1.9 D, D, A, A, D, A, A, D, A, D, A

1.10
10.24V 10.24V 10.24V
VLSB = = = 2.500 mV VMSB = = 5.120V
12
2 bits 4096bits 2
1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500mV )= 5.855 V

1.11
5V 5V = 19.53 mV 2.77V
VLSB = = and mV = 142 LSB
28 bits 256bits bit 19.53
bit
14210 = (128 + 8 + 4 + 210
) = 100011102
1.12
2.5V 2.5V = 2.44 mV
VLSB = =
210 bits 1024 bits bit
 2.5V 
2
(
0101101101 = 28 + 26 + 25 + 23 + 22 + 20 ) = 365 10
V = 365
O  
= 0.891 V
10 1024

1.13
10V mV 6.83V
V = = 0.6104 and (214 bits)= 11191 bits
LSB
214 bits bit 10V
1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 110)
1119110 = 101011101101112

1.14

A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000. The
number of bits must satisfy 2B ≥ 10,000 where B is the number of bits. Here B = 14 bits.

1.15
V = 5.12V = 5.12V = 1.25 mV and V = (101110111011 )V  VLSB
LSB 2 LSB
212 bits 4096 bits bit O 2

O (
11 9 8 7 5 4 3
)
V = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV  0.0625V
10

VO = 3.754  0.000625 or 3.753V  VO  3.755V




1-3
6/9/06

, 1.16

I B = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A

1.17

VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000< t Volts

1.18

vCE = [5 + 2 cos (5000t)] V

1.19

vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V

1.20

V = 10 V, R1 = 22 k, R2 = 47 k and R3 = 180 k.
V
+ 1 -

R I2 + I3
1


V R2 V R
2 3


-


22k
V = 10V 22k = 3.71 V
= 10V
1
(
22k+ 47k 180k ) 22k+ 37.3k
37.3k
V = 10V = 6.29 V Checking : 6.29 + 3.71 = 10.0 V
2 22k+ 37.3k
I 2 = I1 180k  10V  180k
47k+ 180k = 22k+ 37.3k 47k+ 180k = 134 A
 
I 3 = I1 47k  10V  47k
47k+ 180k = 22k+ 37.3k 47k+ 180k = 34.9 A
 
10V
Checking : I1 = = 169A and I1 = I2 + I3
22k+ 37.3k




1-4 ©R. C. Jaeger & T. N. Blalock
6/9/06