Answering 1.1 machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Modems and routers Calculator Camcorder Carbon monoxide detector Cash register CD and DVD players Ceiling fan (remote) Cellular phones Coffee maker Compass Copy machine Cordless phone Depth finder Digital Camera Digital watch Digital voice recorder Digital scale Digital thermometer Electronic dart board Electric guitar Electronic door bell Electronic gas pump Elevator Exercise machine Fax machine Fish finder Garage door opener GPS Hearing aid Invisible dog fences Laser pointer LCD projector Light dimmer Keyboard synthesizer Keyless entry system Laboratory instruments Metal detector Microwave oven
Model airplanes MP3 player Musical greeting cards Musical tuner Pagers Personal computer Personal planner/organizer (PDA) Radar detector Broadcast Radio (AM/FM/Shortwave) Razor Satellite radio receiver Security systems Sewing machine Smoke detector Sprinkler system Stereo system Amplifier CD/DVD player Receiver Tape player Stud sensor Talking toys Telephone Telescope controller Thermostats Toy robots Traffic light controller TV receiver & remote control Variable speed appliances Blender Drill Mixer Food processor Fan Vending machines Video game controllers Wireless headphones & speakers Wireless thermometer Workstations Electromechanical Appliances* Air conditioning and heating systems Clothes washer and dryer Dish washer Electrical timer Iron, vacuum cleaner, toaster Oven, refrigerator, stove, etc. *These appliances are historically based only upon on-off (bang-bang) control. However, many of the high end versions of these appliances have now added sophisticated electronic control.
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) 0.1977(Y2 -1960) B2 19.97x10 0.1977(Y2 -Y1 ) 0.1977(Y2 -Y1 ) = = 10 so 2 = 10 0.1977(Y1 -1960) B1 19.97x10
Y2 - Y1 = log2 = 1.52 years 0.1977
(b)
Y2 - Y1 =
log10 = 5.06 years 0.1977
1.4
N = 1610x10
1.5
0.1548(2020-1970)
= 8.85 x 1010 transistors/P
(2 ) N 2 1610x10 0.1548(Y2 -Y1 ) = = 10 0.1548(Y1 -1970) N1 1610x10 log2 (a) Y2 - Y1 = = 1.95 years 0.1548 log10 (b) Y2 - Y1 = = 6.46 years 0.1548
0.1548 Y -1970
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
1.5W tube)= 113 MW! P = 75x106 tubes (
1.9 1.10
(
)
I=
1.13 x 108W = 511 kA! 220V
D, D, A, A, D, A, A, D, A, D, A
10.24V 10.24V 10.24V = = 2.500 mV VMSB = = 5.120V 12 2 2 bits 4096bits 1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500mV )= 5.855 V VLSB =
1.11
VLSB = 5V 5V mV = = 19.53 bit 2 bits 256bits
8 10
and
128 14210 = ( + 8 + 4 + 2) = 100011102
2.77V = 142 LSB mV 19.53 bit
1.12
VLSB = 2.5V 2.5V mV = = 2.44 bit 2 bits 1024 bits
10
01011011012 = 28 + 26 + 25 + 23 + 22 + 20
(
)
10
= 36510
2.5V VO = 365 = 0.891 V 1024
1.13
10V mV 6.83V 14 = 0.6104 and 2 bits = 11191 bits 14 bit 10V 2 bits 1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1) 10 VLSB = 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
VLSB = 5.12V 5.12V mV V = = 1.25 and VO = ( 1011101110112 ) LSB LSB V 12 bit 2 2 bits 4096 bits 11 9 8 7 5 4 3 VO = 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
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 k, R2= 47 k and R3 = 180 k.
+
V
1
I2 R2
R V
1
+
V
I
3
2
R
3
-
V1 = 10V V2 = 10V
22k + 47k 180k
(
22k
)
= 10V
22k = 3.71 V 22k + 37.3k
37.3k = 6.29 V Checking : 6.29 + 3.71 = 10.0 V 22k + 37.3k 180k 180k 10V I2 = I1 = = 134 A 47k + 180k 22k + 37.3k 47k + 180k 47k 47k 10V I3 = I1 = = 34.9 A 47k + 180k 22k + 37.3k 47k + 180k 10V = 169A and I1 = I2 + I3 22k + 37.3k
Checking : I1 =
1-4
R. C. Jaeger & T. N. Blalock 6/9/06
1.21
V = 18 V, R1 = 56 k, R2= 33 k and R3 = 11 k.
+
V
1
I
2
R V
1
+
V
I3 R
R
2
2
3
-
V1 = 18V
56k
56k + 33k 11k 18V
(
)
= 15.7 V
V2 = 18V
33k 11k
56k + 33k 11k
(
)
= 2.31 V
Checking :V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct. I1 = 56k + 33k 11k
(
)
= 280 A
I2 = I1
11k 11k = (280 A) = 70.0 A 33k + 11k 33k + 11k Checking : I2 + I3 = 280 A
I3 = I1
33k 33k = (280 A) = 210 A 33k + 11k 33k + 11k
1.22
I1 = 5mA
(5.6k + 3.6k) = 3.97 mA (5.6k + 3.6k)+ 2.4k
I2 = 5mA
2.4k = 1.03 mA 9.2k + 2.4k
V3 = 5mA 2.4k 9.2k
(
3.6k )5.6k + 3.6k = 3.72V and
Checking : I1 + I2 = 5.00 mA
1.23
I2 R2 = 1.03mA(3.6k)= 3.71 V
150k 150k = 125 A I3 = 250A = 125 A 150k + 150k 150k + 150k 82k V3 = 250A 150k 150k = 10.3V 68k + 82k Checking : I1 + I2 = 250 A and I2 R2 = 125A(82k)= 10.3 V I2 = 250A
(
)
1-5 6/9/06
1.24
+
v v R
1
s
+ g v
m
v
th
-
Summing currents at the output node yields: v + .002v = 0 so v = 0 and v th = vs - v = vs 5x10 4
+
v R
1
g v
m
ix
vx
Summing currents at the output node : ix = - v - 0.002v = 0 but v = -vx 5x10 4 vx v 1 ix = + 0.002vx = 0 Rth = x = = 495 4 1 ix 5x10 + gm R1
Thvenin equivalent circuit:
495 v
s
1-6
R. C. Jaeger & T. N. Blalock 6/9/06
1.25 The Thvenin equivalent resistance is found using the same approach as Problem 1.24, and
1 -1 Rth = + .025 = 39.6 4k
+
v vs R
1
g v
m
in
The short circuit current is : v in = + 0.025v and v = vs 4k v i n = s + 0.025vs = 0.0253vs 4k Norton equivalent circuit:
0.0253v
s
39.6
1-7 6/9/06
1.26 (a)
i vs R
1
+ R2 v th -
i
Vth = Voc = - i R2
but
i =-
vs R1
and
ix
Vth = vs
R2 39k = 120 vs = 46.8 vs R1 R
1
R2 100k
i i
Rth
v
x
Rth =
vx ; ix
ix =
vx + i R2
but
i = 0 since VR 1 = 0.
Rth = R2 = 39 k.
Thvenin equivalent circuit:
39 k 58.5v
s
(b)
i i
s
+ R2 v th -
R
1
i
i Vth = Voc = - i R2 where i + bi + is = 0 and Vth = - - s R2 = 38700 is + 1
1-8
R. C. Jaeger & T. N. Blalock 6/9/06
i R
1
R2 i
Rth
v
x
Rth =
vx ; ix
ix =
vx + i but R2
i + i = 0 so i = 0
and Rth = R2 = 39 k
Thvenin equivalent circuit:
39 k
38700i s
1.27
i vs R
1
R2 i
in
in = - i but i = -
vs R1
and in =
R1
vs =
100 vs = 1.33 x 10-3 vs 75k Norton equivalent circuit:
From problem 1.26(a), Rth = R2 = 56 k.
0.00133v
s
56 k
1-9 6/9/06
1.28
is v
s
i R
1
R2 i
is =
vs v v +1 - i = s + s = vs R1 R1 R1 R1
R=
vs R 100k = 1 = = 1.24 is + 1 81
1.29 The open circuit voltage is vth = -g mv R2 and v = +is R1.
vth = -g m R1 R2is = -(0.0025) 105 106 i s= 2.5 x 108 is For is = 0, v = 0 and Rth = R2 = 1 M
( )( )
1.30
5V 3V f (Hz) 0
1.31
0
500
1000
2V 0 f (kHz) 9 10 11
4 cos(20000t + 2000t )+ cos(20000t - 2000t ) 2 v = 2cos(22000t )+ 2cos( 18000t ) v = 4sin (20000t )sin (2000t )=
1.32
[
]
236 o A = -5 0 = 2x105 36 o 10 0
A = 2x105
A = 36o
1-10
R. C. Jaeger & T. N. Blalock 6/9/06
1.33
(a) A =
1.34
10-1 -12 o 10-2 - 45o = 5 - 45o (b) A = = 100 -12 o 2x10-3 0 o 10-3 0 o R2 620k 180k =- = -44.3 (b) Av = - = -10.0 14k R1 18k 62k = -38.8 1.6k
(a) Av = -
1.35
vo (t ) = -
(c) Av = -
R2 v s (t )= (-90.1 sin 750t ) mV R1
IS =
VS 0.01V = = 11.0A and R1 910
is = ( 11.0 sin 750t ) A
1.36 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vs.
Therefore Av = 1.
1.37 Since the voltage across the op amp input terminals must be zero, v- = v+ = vs. Also, i- = 0.
v v- - vo + i- + - = 0 R2 R1
or
vs - v o vs + =0 R2 R1
and A v =
vo R = 1+ 2 vs R1
1.38 Writing a nodal equation at the inverting input terminal of the op amp gives
v -v v1 - v- v2 - v- + = i- + - o R1 R2 R3 vo = -
but v- = v+ = 0
and
i- = 0
R3 R v1 - - 3 v2 = -0.255sin 3770t - 0.255sin10000t volts R1 R2
1-11 6/9/06
1.39
b b b 0 1 1 1 0 0 vO = -VREF 1 + 2 + 3 (a) vO = -5 + + = -1.875V (b) vO = -5 + + = -2.500V 2 4 8 2 4 8 2 4 8
b1b2b
3
vO (V) 0 -0.625 -1.250 -1.875 -2.500 -3.125 -3.750 -4.375
000 001 010 011 100 101 110 111
1.40 Low-pass amplifier
Amplitude 10
f 6 kHz
1-12
R. C. Jaeger & T. N. Blalock 6/9/06
1.41 Band-pass amplifier
Amplitude 20
f 1 kHz 5 kHz
1.42
High-pass amplifier
Amplitude 16
f 10 kHz
1.43
15000t ) vO (t ) = 10x5sin (2000t )+ 10x3cos(8000t )+ 0x3cos( vO (t ) = 50sin(2000t )+ 30cos(8000t ) volts
[
]
1.44
12000t ) vO (t ) = 20x0.5sin (2500t )+ 20x0.75cos(8000t )+ 0x0.6cos( vO (t ) = 10.0sin (2500t )+ 15.0cos(8000t ) volts
[
]
1.45 The gain is zero at each frequency:
vo(t) = 0.
1-13 6/9/06
1.46
t=linspace(0,.005,1000); w=2*pi*1000; v=(4/pi)*(sin(w*t)+sin(3*w*t)/3+sin(5*w*t)/5); v1=5*v; v2=5*(4/pi)*sin(w*t); v3=(4/pi)*(5*sin(w*t)+3*sin(3*w*t)/3+sin(5*w*t)/5); plot(t,v) plot(t,v1) plot(t,v2) plot(t,v3)
2 1 0 -1 -2 0 1 2 3 4 5 x10-3
(a)
10 5 0 -5 -10 0 1 2 3 4
(b)
5 x10-3
1-14
R. C. Jaeger & T. N. Blalock 6/9/06
10 5 0 -5 -10 0 1 2 3 4 5 x10-3
(c)
10 5 0 -5 -10 0
(d)
1.47
1
2
3
4
5 x10-3
(c) 3000( .10) R 3000( .10) or 2700 R 3300 1- 1+ Vnom = 2.5V V 0.05V T= 0.05 = 0.0200 or 2.00% 2.50
(b) 3000( .05) R 3000( .05) or 2850 R 3150 1- 1+
(a) 3000( .01) R 3000( .01) or 2970 R 3030 1- 1+
1.48
1.49 1.50
1- 1+ 20000F ( .5) C 20000F ( .2) or 10000F R 24000F 8200( 0.1) R 8200( 0.1) or 7380 R 9020 1- 1+ The resistor is within the allowable range of values.
1-15 6/9/06
1.51
1- 1+ (a) 5V ( .05) V 5V ( .05) or 5.75V V 5.25V V = 5.30 V exceeds the maximum range, so it is out of the specification limits.
(b) If the meter is reading 1.5% high, then the actual voltage would be 5.30 Vmeter = 1.015Vact or Vact = = 5.22V which is within specifications limits. 1.015
1.52
TCR = R nom
R 6562 - 6066 = = 4.96 o 100 - 0 T C = R o + TCR (T)= 6066 + 4.96(27)= 6200
0 C
1-16
R. C. Jaeger & T. N. Blalock 6/9/06
1.53
+
R V
1
V1
I2 R2
+
V2
I3 R
3
-
Let RX = R2 R3
then V1 = V
R1 = R1 + RX
R min = X V1max =
47k(0.9)+ 180k(0.9) 10( ) 1.05 33.5k 1+ 22k( ) 1.1 = 4.40V
180k)(0.9) 47k(0.9)(
V1 R 1+ X R1
= 33.5k R max = X V1min =
10(0.95)
47k( )+ 180k( ) 1.1 1.1 = 3.09V
1.1 47k( )( 1.1 180k)( )
= 41.0k
41.0k 1+ 22k(0.9) V R1 R2 R3
I1 =
V R1 + RX
and I2 = I1
R3 = R2 + R3
R1 + R2 +
I2max =
10( ) 1.05 22000(0.9)+ 47000(0.9)+
22000(0.9)(47000)(0.9) 180000( ) 1.1
= 158 A
I2min =
10(0.95) 22000( )+ 47000( )+ 1.1 1.1 R2 = R2 + R3 V R1 + R3 + R1 R3 R2
1.1 22000( )(47000)( ) 1.1 180000(0.9)
= 114 A
I3 = I1
I3max =
10( ) 1.05
22000(0.9)+ 180000(0.9)+
180000)(0.9) 22000(0.9)( 47000( ) 1.1
= 43.1 A
I3min =
10(0.95) 22000( )+ 180000( )+ 1.1 1.1
1.1 22000( )( 1.1 180000)( ) 47000(0.9)
= 28.3 A
1-17 6/9/06
1.54
I1 = I
R2 + R3 =I R1 + R2 + R3
1 1+ R1 R2 + R3
and similarly I2 = I
I1max = 1+ I2max = 1+
5600( )+ 3600( ) 1.05 1.05 5600(0.95)+ 3600(0.95) 2400( ) 1.05 I 1 1 R + + 2 R1 R3 R1 R3 5( ) 1.02 mA = 1.14 mA I2min =
2400(0.95)
5( ) 1.02
1 R + R3 1+ 2 R1
mA = 4.12 mA
I1min = 1+
5600(0.95)+ 3600(0.95) 1.05 5600( )+ 3600( ) 1.05 2400(0.95) 5(0.98) mA = 0.936 mA
2400( ) 1.05
5(0.98)
mA = 3.80 mA
1+
V3 = I2 R3 =
V3max =
5600(0.95) 1 1 + + 2400( ) 3600( ) 2400( )(3600)( ) 1.05 1.05 1.05 1.05 5(0.98)
5( ) 1.02
= 4.18 V
V3min =
5600( ) 1.05 1 1 + + 2400(0.95) 3600(0.95) 2400(0.95)(3600)(0.95)
1 gm + 1 R1 = 619
min Rth =
= 3.30 V
1.55
From Prob. 1.24 : Rth = 1 1 0.002(0.8)+ 5x10 4 ( ) 1.2
max Rth =
1 1 0.002( )+ 1.2 5x10 4 (0.8)
= 412
1-18
R. C. Jaeger & T. N. Blalock 6/9/06
1.56
For one set of 200 cases using the equations in Prob. 1.53. R1 = 22000 * (0.9 + 0.2 * RAND())
V = 10 * (0.95 + 0.1* RAND())
R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND()) V1 Min Max Average 3.23 V 3.71 V 3.71 V I2 116 A 151 A 133 A I3 29.9 A 40.9 A 35.1 A
1.57
For one set of 200 cases using the Equations in Prob. 1.54: R1 = 2400 * (0.95 + 0.1* RAND())
R1 = 5600 * (0.95 + 0.1* RAND()) R3 = 3600 * (0.95 + 0.1* RAND()) I1 Min Max Average 3.82 mA 4.09 mA 3.97 mA I2 0.96 mA 1.12 mA 1.04 mA V3 3.46 V 4.08 V 3.73 V
I = 0.005* (0.98 + 0.04 * RAND())
1.58 1.59
3.29, 0.995, -6.16; 3.295, 0.9952, -6.155 (a) (1.763 mA)(20.70 k) = 36.5 V (b) 36 V (c) (0.1021 A)(97.80 k) = 9.99 V; 10 V
1-19 6/9/06