,1.2
B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip
1.3
(a)
0.1977(Y2 −1960)
B2 19.97x10 0.1977(Y2 −Y1 ) 0.1977(Y2 −Y1 )
= 0.1977(Y1 −1960)
= 10 so 2 = 10
B1 19.97x10
log2
Y2 − Y1 = = 1.52 years
0.1977
log10
(b) Y2 − Y1 = = 5.06 years
0.1977
1.4
0.1548(2020−1970)
N = 1610x10 = 8.85 x 1010 transistors/μP
1.5
(2 )
0.1548 Y −1970
N 2 1610x10 0.1548(Y2 −Y1 )
= 0.1548(Y1 −1970)
= 10
N1 1610x10
log2
(a) Y2 − Y1 = = 1.95 years
0.1548
log10
(b) Y2 − Y1 = = 6.46 years
0.1548
−0.05806(2020−1970)
1.6 F = 8.00x10 μm = 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/μP 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
( )
P = 75x106 tubes (1.5W tube)= 113 MW! I=
1.13 x 108W
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 = 12
= = 2.500 mV VMSB = = 5.120V
2 bits 4096bits 2
1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500mV )= 5.855 V
1.11
5V 5V mV 2.77V
VLSB = 8
= = 19.53 and = 142 LSB
2 bits 256bits bit mV
19.53
bit
14210 = (128 + 8 + 4 + 2) = 100011102
10
1.12
2.5V 2.5V mV
VLSB = 10
= = 2.44
2 bits 1024 bits bit
⎛ 2.5V ⎞
(
01011011012 = 28 + 26 + 25 + 23 + 22 + 20 )
10
= 36510 VO = 365 ⎜ ⎟ = 0.891 V
⎝ 1024 ⎠
1.13
VLSB =
10V
14
2 bits
= 0.6104
mV
bit
and
6.83V 14
10V
( )
2 bits = 11191 bits
1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1)
10
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
and VO = (1011101110112 )VLSB ± LSB
5.12V 5.12V mV V
VLSB = 12
= = 1.25
2 bits 4096 bits bit 2
( 8 7 5 4 3
)
VO = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV ± 0.0625V
11 9
10
VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V
1-3
6/9/06
, 1.16
IB = 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 kΩ, R2= 47 kΩ and R3 = 180 kΩ.
V
+ 1 -
R I2 I
1
+ 3
V R2 V R
2 3
-
22kΩ 22kΩ
V1 = 10V = 10V = 3.71 V
(
22kΩ + 47kΩ 180kΩ ) 22kΩ + 37.3kΩ
37.3kΩ
V2 = 10V = 6.29 V Checking : 6.29 + 3.71 = 10.0 V
22kΩ + 37.3kΩ
180kΩ ⎛ 10V ⎞ 180kΩ
I2 = I1 =⎜ ⎟ = 134 μA
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
47kΩ ⎛ 10V ⎞ 47kΩ
I3 = I1 =⎜ ⎟ = 34.9 μA
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
10V
Checking : I1 = = 169μA and I1 = I2 + I3
22kΩ + 37.3kΩ
1-4 ©R. C. Jaeger & T. N. Blalock
6/9/06
B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip
1.3
(a)
0.1977(Y2 −1960)
B2 19.97x10 0.1977(Y2 −Y1 ) 0.1977(Y2 −Y1 )
= 0.1977(Y1 −1960)
= 10 so 2 = 10
B1 19.97x10
log2
Y2 − Y1 = = 1.52 years
0.1977
log10
(b) Y2 − Y1 = = 5.06 years
0.1977
1.4
0.1548(2020−1970)
N = 1610x10 = 8.85 x 1010 transistors/μP
1.5
(2 )
0.1548 Y −1970
N 2 1610x10 0.1548(Y2 −Y1 )
= 0.1548(Y1 −1970)
= 10
N1 1610x10
log2
(a) Y2 − Y1 = = 1.95 years
0.1548
log10
(b) Y2 − Y1 = = 6.46 years
0.1548
−0.05806(2020−1970)
1.6 F = 8.00x10 μm = 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/μP 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
( )
P = 75x106 tubes (1.5W tube)= 113 MW! I=
1.13 x 108W
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 = 12
= = 2.500 mV VMSB = = 5.120V
2 bits 4096bits 2
1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500mV )= 5.855 V
1.11
5V 5V mV 2.77V
VLSB = 8
= = 19.53 and = 142 LSB
2 bits 256bits bit mV
19.53
bit
14210 = (128 + 8 + 4 + 2) = 100011102
10
1.12
2.5V 2.5V mV
VLSB = 10
= = 2.44
2 bits 1024 bits bit
⎛ 2.5V ⎞
(
01011011012 = 28 + 26 + 25 + 23 + 22 + 20 )
10
= 36510 VO = 365 ⎜ ⎟ = 0.891 V
⎝ 1024 ⎠
1.13
VLSB =
10V
14
2 bits
= 0.6104
mV
bit
and
6.83V 14
10V
( )
2 bits = 11191 bits
1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1)
10
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
and VO = (1011101110112 )VLSB ± LSB
5.12V 5.12V mV V
VLSB = 12
= = 1.25
2 bits 4096 bits bit 2
( 8 7 5 4 3
)
VO = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV ± 0.0625V
11 9
10
VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V
1-3
6/9/06
, 1.16
IB = 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 kΩ, R2= 47 kΩ and R3 = 180 kΩ.
V
+ 1 -
R I2 I
1
+ 3
V R2 V R
2 3
-
22kΩ 22kΩ
V1 = 10V = 10V = 3.71 V
(
22kΩ + 47kΩ 180kΩ ) 22kΩ + 37.3kΩ
37.3kΩ
V2 = 10V = 6.29 V Checking : 6.29 + 3.71 = 10.0 V
22kΩ + 37.3kΩ
180kΩ ⎛ 10V ⎞ 180kΩ
I2 = I1 =⎜ ⎟ = 134 μA
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
47kΩ ⎛ 10V ⎞ 47kΩ
I3 = I1 =⎜ ⎟ = 34.9 μA
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
10V
Checking : I1 = = 169μA and I1 = I2 + I3
22kΩ + 37.3kΩ
1-4 ©R. C. Jaeger & T. N. Blalock
6/9/06
