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CHAPTER 1

1.1

Answering machine

Alarm clockAutomatic door

Automatic lights

ATM

Automobile:

Engine controller

Temp. control

ABS

Electronic dash

Navigation system

Automotive tune-up equip.

Baggage scannerBar code scanner

Battery charger

Cable/DSL Modems and routers

Calculator

Camcorder

Carbon monoxide detector

Cash register

CD and DVD players

Ceiling fan (remote)

Cellular phone

Coffee maker

Compass

Copy machine

Cordless phone

Depth finder

Digital watch

Digital recorder

Digital scale

Digital thermometer

Electronic dart board

Electric guitar

Electronic door bell

Electronic gas pumpElevator

Exercise machine

Fax machine

Fish finder

Garage door opener

GPS

Hearing aid

Invisible dog fences

Laser pointerLCD projector

Light dimmer

Keyboard synthesizer

Keyless entry system

Laboratory instruments

Metal detector

Microwave oven

Model airplanes

MP3 player

Musical greeting cards

Musical tunerPagers

Personal computer

Personal planner/organizer

Radar detector

Radio

Razor

Satellite radio and TV receivers

Security systems

Sewing machine

Smoke detector

Sprinkler system

Stereo system

Amplifier

CD player

Receiver

Tape player

Stud sensor

Talking toys

Telephone

Telescope controller

Toy robots

Traffic light controller

TV & remote controlVariable speed appliances

Blender

Drill

Mixer

Food processor

Fan

Vending machines

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Video games

Wireless headphones & speakers

Wireless thermometer

Workstations

Electromechanical Appliances*

Air conditioning and heating systems

Clothes washer and dryer

Dish washer

Electrical t imer

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.

1.2 B =19.97 x 100.1997 2020−1960( ) =14.5 x 10

12 =14.5 Tb/chip

1.3 (a)

B2

B1

=19.97 x100.1977 Y 2−1960( )

19.97 x100.1977 Y 1−1960( ) =100.1977 Y 2 −Y 1( ) so 2 =100.1977Y 2−Y 1( )

Y 2 −Y 1 =log 2

0.1977 =1.52 years

(b) Y 2 −Y 1 =log10

0.1977= 5.06 years

1.4 N =1027 x100.1505 2020−1970( ) = 34.4 x 10

9transistors / µP

1.5

N 2

N 1=

1027 x100.1505 Y 2 −1970( )

1027 x100.1505 Y 1−1970( ) =100.1505Y

2−Y

1( )

(a) Y 2 −Y 1 =log2

0.1505= 2.00 years

(b) Y 2 −Y 1 =log10

0.1505= 6.65 years

1.6 F = 8.214 x10−0.060792020−1970( )µm = 7.5 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. 1.5, there are approximately 3.5 million transistors on a Pentium IV

microprocessor. From Prob. 1.4, the number of transistors/µP will be 34.4 x 109.

Thus there will be the equivalent of 34.4x109/3.5x106 = 9830 Pentium IV processors.

1.8 P = (75 x 106 tubes)(1.5 W/tube) = 1.13 x 108 W = 113 MW

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I = 1.13 x 108W

220V = 511 kA!

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

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1.10

V LSB =10V

212bits=

10V

4096bits= 2.44 mV VMSB =

10V

2= 5V

1001001001002 = 211 + 28 + 25 + 22 = 234010 V O = 234010V

4096

= 5.71 V

1.11

V LSB =2.5V

210 bits=

2.5V

1024 bits= 2.44

mV

bit

01011001012 = 28 + 26 + 25 + 22 + 20( )10= 35710 V O = 357

2.5V

1024

= 0.872 V

1.12

V LSB =5V

28bits=

5V

256bits=19.53

mV

bitand

2.77V

19.53mV

bit

=142 bits

14210 = 128+ 8 + 4 + 2( )10=100011102

1.13

V LSB =10V

215bits= 0.3052

mV

bitand

6.83V

10V 2

15bits( )= 22381 bits

2238110 = 16384 + 4096 +1024 + 512+ 256 + 64 + 32 + 8+ 4 +1( )10

2238110 =1010111011011012

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 LSB =5.12V

212 bits=

5.12V

4096 bits=1.25

mV

bit and V O = 1011101110112( )V LSB ±

V LSB

2

V O = 211 + 29 + 28 + 27 + 25 + 24 + 23 + 2 +1( )10

1.25mV ± 0.0625V

V O = 3.754± 0.000625 or 3.753V ≤ VO ≤ 3.755V

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 2000πt Volts

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

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

1.20 V = 10 V, R1 = 4.7 kΩ, R2= 2.2 kΩ and R3 = 18 kΩ.

V

+ -V

1

V2

+

-

R1

R2 R

3

I3

I2

V 1 =10V 4.7k Ω

4.7k Ω+ 2.2k Ω 18k Ω( )=10V

4.7k Ω4.7k Ω+1.96k Ω

= 7.06 V

V 2 =10V 1.96k Ω

4.7k Ω+1.96k Ω= 2.94 V

I 2 = I 118k Ω

2.2k Ω+18k Ω=

10V

4.7k Ω+1.96k Ω

18k Ω2.2k Ω+18k Ω

=1.34 mA

I 3 = I 12.2k Ω

2.2k Ω+18k Ω=

10V

4.7k Ω+1.96k Ω

2.2k Ω2.2k Ω+18k Ω

= 0.164 mA

1.21 V = 18 V, R1 = 39 kΩ, R2= 43 kΩ and R3 = 11 kΩ.

V 1 =18V 39k Ω

39k Ω+ 43k Ω 11k Ω( )=14.7 V

V 2 =18V 43k Ω 11k Ω

39k Ω+ 43k Ω 11k Ω( )= 3.30 V

I 2 = I 111k Ω

43k Ω+11k Ω=

18V

39k Ω+ 43k Ω 11k Ω( )

11k Ω43k Ω+11k Ω

= 76.7 µ A

I 3 = I 143k Ω

43k Ω+11k Ω=

18V

39k Ω+ 43k Ω 11k Ω( )

43k Ω43k Ω+11k Ω

= 0.300 mA

1.22

I 1 = 5mA2.2k Ω+ 3.6k Ω( )

2.2k Ω+ 3.6k Ω( )+ 4.7k Ω= 2.76 mA

I 2 = 5mA4.7k Ω

5.8k Ω+ 4.7k Ω= 2.24 mA

V 3 = 5mA 4.7k Ω 5.8k Ω( ) 3.6k Ω2.2k Ω+ 3.6k Ω

= 8.06V

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1.23

I 2 = 250µ A150k Ω

150k Ω+100k Ω=150 µ A

I 3 = 250µ A100k Ω

150k Ω+100k Ω=100 µ A

V 3 = 250µ A 100k Ω 150k Ω( ) 82k Ω68k Ω+ 82k Ω

= 8.20V

1.24

R1

vs

+

-

v

gm

vv

th

+

-

Summing currents at the output node yields :

v

105 + .002v = 0 so v = 0 and vth = vs − v = vs

R1

vx

+

-

v

g vm

ix

Summing currents at the output node :

i x = −v

105− 0.002v = 0 but v = −vx

i x =vx

105+ 0.002v x = 0 Rth =

v x

i x=

1

1

R1

+ gm

= 498Ω

Thévenin equivalent circuit:

vs

498 Ω

1.25 The Thévenin equivalent resistance is found using the same approach as Problem

1.24, and Rth=

1

40k Ω+ .025

−1

= 40.0 Ω

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R1

vs

+

-

v

g mv in

The short circuit current is :

in =v

40k Ω+ 0.025v and v = vs

in =vs

40k Ω+ 0.025vs = 0.025vs

Norton equivalent circuit

40 Ω0.025vs

1.26 (a)

R1 R2

βi

vs

i

+

-

vth

V th =V oc = −β i R2 but i = −vs

R1

and

V th = β vs R2

R1

=150 vs 39k Ω

100k Ω= 58.5 vs

R1 R2

βi

i

Rthv

x

ix

Rth =v x

i x; ix =

v x

R2

+ βi

but i = 0. Rth = R2 = 39 k Ω.


58.5vs

39 k Ω

(b)

R1 R

2

βi

is

i

+

-

vth

V th =V oc = −β i R2 where i +bi +is = 0 and

V th =−β −is

β +1

R2 = 38700 is

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R1 R2

βi

i

Rth vx

Rth =v x

i x; ix =

v x

R2

+ βi

but i + βi = 0 so i = 0

and Rth = R2 = 39 k Ω


39 k Ω

38700is

1.27

R1 R2

βi

vs

i

in

in = −β i but i = −vs

R1

and

in =β

R1

vs =β

R1

vs

80

75k Ω=1.07 x 10−3 vs

From problem 1.22, Rth = R2 = 56 kΩ.

Norton equivalent circuit

56 k Ω0.00107v s

1.28

R1 R2

βi

vs

i

is

is =vs

R1

−β i =vs

R1

+ β vs

R1

= vs

β +1

R1

R =vs

is

=R1

β +1=

100k Ω151

= 662 Ω

1.29

The open circuit voltage is v th = −gmv R2 and v = +is R1.

v th = −gm R1 R2is = − 0.002( )105( )10

6( )is= 2 x 108is

For is = 0, v = 0 and Rth = R2 =1 M Ω

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1.30 1.31

f (Hz)

500 10000

5 V

3 V

0

f (kHz)

9 10 11

2 V

0

v = 4sin 20000πt ( )sin 2000πt ( )= 4

2cos 20000πt + 2000πt ( )+ cos 20000πt − 2000πt ( )[ ]

v = 2 cos 22000πt ( )+ 2cos 18000πt ( )

1.32 A = 2∠36o

10−5

∠00= 2 x105∠36o A = 2 x105 ∠A = 3 6o

1.33 (a) A = 10−2∠− 45

o

2 x10−3∠0

o= 5∠− 45

o(b) A = 10

−1∠−12o

10−3∠0

o=100∠−12

o

1.34

(a) Av = − R2

R1

= −620k Ω15k Ω

= −41.3 (b) Av = −180k Ω18k Ω

= −10.0 (c) Av = −62k Ω1.5k Ω

= −41.3

1.35

vo t ( )= − R2

R1

vs t ( )= −90.1 sin 750πt( )mV

IS =V S

R1

=0.01V

910Ω=11.0µ A and i s = 11.0 sin 750πt( ) µA

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− −vo

R2

+ i− +v−

R1

= 0 orvs − vo

R2

+vs

R1

= 0 and Av =vo

vs

=1+R2

R1

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1.38 Writing a nodal equation at the inverting input terminal of the op amp gives

v1 − v−

R1

+v2 − v−

R2

= i− +v− − vo

R3

but v - = v+ = 0 and i- = 0

vo = −R3

R1

v1 − −R3

R2

v2 = −0.510sin 3770t −1.02sin10000t volts

1.39

vO = −V REF

b1

2+

b2

4+

b3

8

(a) vO = −5

0

2+

1

4+

1

8

= −1.875V (b) vO = −5

1

2+

0

4+

0

8

= −2.500V

b1b2b3 vO (V)

000 0

001 -0.625

010 -1.250

011 -1.875

100 -2.500

101 -3.125

110 -3.750

111 -4.375

1.40 1.41 1.42

Amplitude

f

5

5 kHz

f

10

1 kHz 5 kHz

f

10 kHz

32

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Low-pass Amplifier Band-pass amplifier High-pass amplifier

1.43

vO t ( )= 5 x5sin 2000πt ( )+ 5 x3cos 8000πt ( )+ 0 x3cos 15000πt ( )vO t ( )= 25.0sin 2000πt ( )+15.0cos 8000πt ( ) volts

1.44

vO t ( )=10 x0.5sin 2500πt ( )+10 x0.75cos 8000πt ( )+ 0 x0.6cos 12000πt ( )vO t ( )= 5.00sin 2500πt ( )+ 7.50cos 8000πt ( ) volts

1.45 The gain is zero at each frequency: vo(t) = 0.

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)

(a)

-2

-1

0

1

2

0 1 2 3 4 5

x10-3

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(b)

-10

-5

0

5

10

0 1 2 3 4 5

x10-3

(c)

-10

-5

0

5

10

0 1 2 3 4 5

x10-3

(d)

-10

-5

0

5

10

0 1 2 3 4 5

x10-3

1.47

(a) 2000 1− .01( )≤ R ≤ 2000 1+ .01( ) or 1980Ω ≤ R ≤ 2020Ω

(b) 2000 1− .05( )≤ R ≤ 2000 1+ .05( ) or 1900Ω ≤ R ≤ 2100Ω

(c) 2000 1− .10( )≤ R ≤ 2000 1+ .10( ) or 1800Ω ≤ R ≤ 2200Ω

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1.48 V nom = 3.3V ∆V ≤ 0.05V T =0.05

3.30 x 100%=1.52%

1.49 10000µF 1− .5( )≤ C ≤ 10000µF 1+ .2( ) or 5000µF ≤ R ≤ 12000µF

1.50 8200 1− .05( )≤ R ≤ 8200 1+ .05( ) or 7790Ω≤ R ≤ 8610Ω

The resistor i s within the allowable range of values.

1.51 (a) 5V 1− .05( )≤ V ≤ 5V 1+ .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

V meter =1.015V act or Vact =5.30

1.015= 5.22V which is within specifications limits.

1.52

TCR = ∆ R∆T

= 6562 −6066

100−0= 4.96

ΩoC

Rnom = R0

oC + TCR ∆T( )= 6066 + 4.96 27( )= 6200Ω

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1.53

V

+ -V

1

V2

+

-

R1

R2 R

3

I3

I2

Let R X = R2 R3 then V1 =V R1

R1 + R X

=V 1

1+R X

R1

RX

min =2.2k Ω 0.9( )18k Ω( ) 0.9( )2.2k Ω 0.9( )+18k Ω 0.9( )

=1.78k Ω RX

max =2.2k Ω 1.1( )18k Ω( )1.1( )2.2k Ω 1.1( )+18k Ω 1.1( )

= 2.18k Ω

V 1max =

10 1.02( )1+

1.78k Ω4.7k Ω 1.1( )

= 7.59V V 1min =

10 0.98( )1+

2.18k Ω4.7k Ω 0.9( )

= 6.47V

I 1 =V

R1 + R X

and I 2 = I 1 R3

R2 + R3

=V

R1 + R2 +R1 R2

R3

I 2max =

10 1.02( )4700 0.9( )+ 2200 0.9( )+

4700 0.9( ) 2200( ) 0.9( )180001.1( )

=1.54 mA

I 2min =

10 0.98( )4700 1.1( )+ 2200 1.1( )+ 4700 1.1( ) 2200( )1.1( )

18000 0.9( )

=1.17 mA

I 3 = I 1 R2

R2 + R3

=V

R1 + R3 +R1 R3

R2

I 3max =

10 1.02( )

4700 0.9( )+18000 0.9( )+ 4700 0.9( )18000( ) 0.9( )2200 1.1

( )

= 0.209 mA

I 3min =

10 0.98( )

4700 1.1( )+18000 1.1( )+ 4700 1.1( )18000( )1.1( )2200 0.9( )

= 0.128 mA

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1.54

I 1 = I R2 + R3

R1 + R2 + R3

= I 1

1+R1

R2 + R3

and similarly I 2 = I 1

1+R2 + R3

R1

I 1max

=

5 1.01( )1+

4700 0.95( )2200 1.05( )+ 3600 1.05( )

mA = 2.92 mA I 1min

=

5 0.98( )1+

4700 1.05( )2200 0.95( )+ 3600 0.95( )

mA = 2.59 mA

I 2max =

5 1.01( )1+

2200 0.95( )+ 3600 0.95( )4700 1.05( )

mA = 2.39 mA I 2min =

5 0.99( )1+

2200 1.05( )+ 3600 1.05( )4700 0.95( )

mA = 2.09 mA

V 3 = I 2 R3 =I

1

R1

+1

R3

+R2

R1 R3

V 3max =5 1.01

( )1

4700 1.05( )+

1

3600 1.05( )+

2200 0.95( )4700 1.05( ) 3600( )1.05( )

= 8.72 V

V 3min =

5 0.99( )1

4700 0.95( )+

1

3600 0.95( )+

2200 1.05( )4700 0.95( ) 3600( ) 0.95( )

= 7.41 V

1.55

From Prob. 1.24 : Rth = 1gm +

1

R1

Rth

max =1

0.002 0.8( )+ 1

105 1.2( )

= 622 Ω Rth

min =1

0.002 1.2( )+ 1

105 0.8( )

= 415 Ω

1.56 For one set of 200 cases using the equations in Prob. 1.53.

V =10* 0.98+ 0.04 * RAND()( ) R1 = 4700* 0.9+ 0.2* RAND()( ) R1 = 2200* 0.9 + 0.2* RAND()( ) R3 =18000* 0.9 + 0.2* RAND()( )

V1 I2 I3

Min 6.65 V 1.20 mA 0.138 mA

Max 7.47 V 1.47 mA 0.201 mA

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Average 7.05 V 1.34 mA 0.165 mA

1.57 For one set of 200 cases using the Equations in Prob. 1.54:

I = 0.005* 0.99 + 0.02* RAND()( ) R1 = 4700* 0.95 + 0.1* RAND()( ) R1 = 2200* 0.95+ 0.1* RAND()( ) R3 = 3600* 0.95 + 0.1* RAND()( )

I1 I2 V3

Min 2.63 mA 2.14 mA 7.53 V

Max 2.88 mA 2.34 mA 8.49 V

Average 2.76 mA 2.24 mA 8.03 V

1.58 3.29, 0.995, -6.16; 3.295, 0.9952, -6.155

1.59 (a) (1.763 mA)(20.70 kΩ) = 36.5 V (b) 36 V

(c) (0.1021 µA)(97.80 kΩ) = 9.99 V; 10 V

ch01 net

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