Solutions Manual -
Microelectronic Circuit
Design - 4th Ed
By Richard C. Jaeger, Travis N. Blalock -
McGraw-Hill
,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
N = 1610x100.1548 (
2020−1970 )
= 8.85 x 1010 transistors/P
1.5
0.1548 ( Y2 −1970) 0.1548 ( Y −Y )
N2 1610x10
= = 10 2 1
N1 (1 ) 0.1548 Y −1970
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.
1-1 ©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
Microelectronic Circuit
Design - 4th Ed
By Richard C. Jaeger, Travis N. Blalock -
McGraw-Hill
,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
N = 1610x100.1548 (
2020−1970 )
= 8.85 x 1010 transistors/P
1.5
0.1548 ( Y2 −1970) 0.1548 ( Y −Y )
N2 1610x10
= = 10 2 1
N1 (1 ) 0.1548 Y −1970
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.
1-1 ©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
