Solution manual for microelectronic circuit design 5th edition by jaeger

CHAPTER 1 1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system A

CHAPTER 1

1.1

Answering 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

NiCad/Lithium Ion battery chargers

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

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

Receiver

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

Electrical timer

Iron, vacuum cleaner, toaster Oven, refrigerator, stove, etc

*These appliances are historically based only upon on-off

1.2

N =1327x10(2020−1970)/6.52= 61.9 x 109 transistors/chip

1.3

N = 2.233x10( 9)x10(2021−2014)/10.1=11.0 x 109 Transistors/Chip

1.4

B =19.97 x 100.1997 2021−1960 ( )= 30.3 x 1012 = 30.3 Tb/chip

1.5

(a)

1.6

N2

N1 =1327x10

Y2−1970

( ) /6.52

1327x10(Y1−1970 ) /6.52 =10(Y2−Y1 ) /6.52

(a) Y2−Y1= 6.52log2 =1.96 years

(b) Y2−Y1= 6.52log10 = 6.52 years

1.7

N2

N1 = 2.233x10

9

( )x10(Y2 −2014 ) /10.1

2.233x109

( )x10(Y1 −2014 ) /10.1 =10(Y2−Y1 ) /10.1

(a) Y2−Y1=10.1log2 = 3.07 years

(b) Y2−Y1=10.1log10 =10.1 years

Although this distance corresponds to the diameter of only a few atoms, ITRS projections are

on track to produce feature sizes in this range See the Intel website for example

P = 268x10( 6tubes) (1.5W tube)= 402 MW! I = 4.02 x 10 W

V = 268x10( 6tubes) (80cm3/ tube)= 21.4x109

cm3= 21400 m3

1.10 D, D, A, A, D, A, A, D, A, D, A

1.11

28

256bits=19.53mV

bit and

3.06V 19.53mV bit

=156.7 bits →157 LSB

15710 = 128+16 +8+ 4 +1( )10 =100111012

1.12

210 bits= 2.5V

1024 bits= 2.441mV

bit

01011001102= 28+ 26+ 25+ 22+ 21

( )10= 35810 V O = 358 2.5V

1024







= 0.874 V

1.13

212

4096bits = 2.441 mV VMSB=10V

2 = 5.000 V

1001001010012= 211+ 28+ 25+ 23+ 20 = 234510 V O = 2345 2.441mV( )= 5.724 V

1.14

215

bits= 0.3052mV

bit and

6.89V

15

bits

( )= 22577 bits

2257710 = 16384 + 4096 + 2048+ 32 +16 +1( )10

2257710 =1011000001100012

1.15 (a) A 4 digit readout ranges from 0000 to 2000 and has a resolution of 1 part in 2,000

The number of bits must satisfy 2B ≥ 2,000 where B is the number of bits Here B = 11 bits (b) 2B ≥ 106 yields B = 20 bits

1.16

1.17

IB = dc component = 7.50 mA, ib = signal component = 0.003 cos (1000t) A

1.18

VGS = 2.5 V, vgs = 0.5u(t-1) + 0.1 cos 2000πt Volts

1.19

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

1.20

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

1.21

V = 1 V, R1 = 24 kΩ, R2= 30 kΩ and R3 = 11 kΩ

24k Ω+ 30kΩ 11kΩ( )= 0.749 V V2 =1V 30k Ω 11kΩ

24k Ω+ 30kΩ 11kΩ( )= 0.251 V

Checking:V1+V2 = 0.749 + 0.251=1.00 V which is correct.

24k Ω+ 30kΩ 11kΩ( )= 31.2 μA I2= I1

R3

R2+ R3

= 31.2( μA)30k 11k Ω+11kΩΩ = 8.37 μA

I3= I1

R2

R2+ R3

= 31.2( μA)30k 30k Ω+11kΩΩ = 22.8 μA Checking: I2+ I3= 31.2 μA

V = 8 V, R1 = 30 kΩ, R2= 24 kΩ and R3 = 15 kΩ

30k Ω+ 24kΩ 15kΩ( )= 8V 30k Ω+ 9.23kΩ 30kΩ = 6.12 V

V2= 8V 24k Ω 15kΩ

30k Ω+ 24kΩ 15kΩ( )=1.88 V Checking: 6.12+1.88 = 8.00 V

I2= I1

R3

R2+ R3

30k Ω+ 9.23kΩ







24k Ω+15kΩ= 78.4 μA

I3= I1

R2

R2+ R3

30k Ω+ 9.23kΩ







24k Ω+15kΩ=125 μA

Checking: I1= 8V

30k Ω+ 9.23kΩ= 204 μA and I1=I2+ I3

1.23

I2= 200μA 150kΩ

150k Ω+150kΩ







=100 μA I3= 200μA 150kΩ

150k Ω+150kΩ







=100 μA

V3= 200μA 150k( Ω 150kΩ)68k Ω+82kΩ 82kΩ 



= 8.2V Checking: I1+ I2 = 200 μA and I2R2 =100μA 82k( Ω)= 8.2 V

1.24

I1= 4mA (3.9k Ω+ 5.6kΩ)

3.9k Ω+ 5.6kΩ

( )+ 2.4kΩ = 3.19 mA I2= 4mA 2.4kΩ

9.5k Ω+ 2.4kΩ = 0.807 mA

V3= 4mA 2.4kΩ 9.5kΩ( )3.9k 5.6k Ω+ 5.6kΩΩ = 4.52 V

Checking: I1+ I2 = 4.00 mA and I2R3= 0.807mA 5.6kΩ( )= 4.52 V

1.25

Summing currents at the open circuited output node yields:

v

104 +.025v= 0 so v = 0 and vth= v i − v = v i

To find the Thévenin equivalent resistance, we apply a test source to the output with vi set to zero:

Thévenin equivalent circuit:

Summing currents at the output node:

i x= − v

R1 − g m v = 0 but v = −v x

i x= v x

R1+ g m v x = 0 R th=v x

i x = 1 1

R1+ g m

10kΩ+ 0.025S

= 39.8 Ω

vi 39.8 Ω

Norton equivalent circuit:

The short circuit current is:

in = v

75kΩ+ 0.002v and v = v i → in = v i

75kΩ+ 0.002v i = 2.01x10−3v i

To find the Thévenin equivalent resistance, we apply a test source to the output with vi set to zero:

Summing currents at the output node:

i x= − v

R1 − g m v = 0 but v = −v x

i x= v x

R1+ g m v x = 0 R th=v x

i x = 1 1

R1+ g m

75kΩ+ 0.002S

= 467 Ω 2.01 x 10-3 vi

467 Ω

1.27

(a)

i n= −βi but i = −v i

R1 and i n= β

R1v i = 150

39kΩv i = 3.85 x 10−3v i

R th=v x

i x ; i x = v x

R2 +βi but i = 0 since v R1= 0 R th = R2=100 kΩ

Noton equivalent circuit:

v th = v oc= −βiR2 where i+βi +i i = 0 and v th = R2 β

β+1







i i =100kΩ 150

151





 i i = 99300i i

Rth is found in part (a)

ii

99300

100

(a)

v th = v oc= −β i R2 but i = −v i

R1 and v th =β v i R2

R1 =120 v i

56kΩ

75kΩ= 89.6 v i

R th=v x

i x ; i x = v x

R2 +βi but i= 0 since vR1= 0 R th = R2 = 75 kΩ

Thévenin equivalent circuit:

(b)

v th = v oc= −β i R2 where i+βi +i i = 0 and v th = R2 β

β+1







i i = 75kΩ 120

121





 i i = 74400 i i

89.6 vi

75 kΩ

ii

R th=v x

i x ; i x = v x

R2 +βi but i+βi = 0 so i = 0 and R th = R2= 75 kΩ

Thévenin equivalent circuit:

1.29

(a) i i= v i

R1−βi= v i

R1+β v i

R1 = v i

β+1

R1 R=v i

i i = R1

β+1=

100kΩ

76 =1.32 kΩ

(b) Source is ii in part (b)

v i = −iR1 and i i = −i −βi= −(β+1)i R=v i

i i = − −1

β+1R1=100kΩ

76 =1.32 kΩ

1.30

The open circuit voltage is v th = −g m v R2 where v = +i i R1

v th = −g m R1R2i i= − 0.0025( ) 2x105

( ) 2x106

( )i i =1.0 x 109

i i

For i i = 0, v = 0, and R th = R2 = 2 MΩ

74400 ii

75 kΩ

( ) R AB =10kΩ+10kΩ 10kΩ+ 10kΩ 10kΩ ( )=16 kΩ

b

( ) R CD =10kΩ 10kΩ+ 10kΩ 10kΩ ( )= 6 kΩ

c

( ) R EF =10kΩ 10kΩ 10kΩ+10kΩ( )= 4 kΩ

d

( ) Terminals B & D are the same as E & F R BD = 4 kΩ

1.32

1.33

Ω resistor was shorted, or the 82 kΩ resistor was open, then the output

kΩ resistor was open, the output would be -9 V Otherwise the voltage would be between –9 and +9 volts

1.35

1.36

1.37

10−4∠00 = 4x104∠56o A = 4x104 ∠A = 56 o

1.38

(a) A=10−1∠−12o

2x10−3∠0o = 50∠−12o A = 50 ∠A = -12o

(b)

1.39

(a) A v= −R2

R1 = −560kΩ

12kΩ = −46.7 (b) A v= −360kΩ

18kΩ = −20.0 (c) A v= −62kΩ

2kΩ = −31.0

1000 2000

f(Hz)

v

5V

3V

1.40

v o( )t = −R2

R1v i( )t = −7500

910 (0.01sin 750πt )= −82.4 sin 750( πt ) mV

i i = v i

R1= 0.01V

910Ω=11.0μA and i i( )t = 11.0 sin 750( πt) μA

1.41 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vi

Therefore Av = 1

1.42 Since the voltage across the op amp input terminals must be zero, v- = v+ = vi Also, i- = 0

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

v1− v−

R1 +v2− v−

R2 = i−+v−− v o

R3 but v-= v+= 0 and i- = 0

v o= −R3

R1v1−−R3

R2v2= −0.255sin3770t − 0.250sin10000t ( ) volts

1.44

b 1 b 2 b 3 v O (V)

000 0

001 -0.625

010 -1.250

011 -1.875

100 -2.500

101 -3.125

110 -3.750

111 -4.375

1.45 Low-pass amplifier

1.46 Band-pass amplifier

1.48 Refers to Prob 1.45

1.49

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

1.51

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)

(b)

(c)

-2 -1 0 1 2

-10 -5 0 5 10

-10 -5 0 5 10

1.52

(a) 4700 1( −.01)≤ R ≤ 4700 1+.01( ) or 4650Ω ≤ R ≤ 4750Ω

(b) 4700 1( −.05)≤ R ≤ 4700 1+.05( ) or 4460Ω ≤ R ≤ 4940Ω

(c) 4700 1( −.10)≤ R ≤ 4700 1+.10( ) or 4230Ω ≤ R ≤ 5170Ω

1.53

1.54

1.55

Yes, the resistor is within the allowable range of values

1.56

(a) 5V 1( −.05)≤ V ≤ 5V 1+.05( ) or 4.75V≤ V ≤ 5.25V

V = 5.30 V exceeds the maximum range, so it is out of the specification limits

(b) However, if the meter is reading 1.5% high, then the actual voltage would be

1.57

-10 -5 0 5

x10-3

1.58

At 30 o C, 7500Ω 1− 0.05( )≤ R ≤ 7500Ω 1+ 0.05( ) or 7120 Ω ≤ R ≤ 7880 Ω

Adding the effect of TC for ΔT = 45oC:

Rmin= 7120Ω 1+ 452200

106







= 7820 Ω Rmax= 7780Ω 1+ 45 2.2x10 ( −3)

7820 Ω ≤ R ≤ 8550 Ω with accumulated rounding

7830 Ω ≤ R ≤ 8650 Ω more exact calculation

1.59 I = 200 μA, R1 = 150 kΩ, R2 = 68 kΩ and R3 = 82 kΩ

I1= I R2+ R3

R1+ R2+ R3

1+ R1

R2+ R3

and similarly I2= I 1

1+R2+ R3

R1

I1max= 200 1.02( )

1+ 150kΩ 0.90( )

68kΩ 1.1( )+82kΩ 1.1( )

μA=112 μA I1min= 200 0.98( )

1+ 150kΩ 1.1( )

68kΩ 0.90( )+82kΩ 0.90( )

μA= 88.2 μA

I2max= 200 1.02( )

1+68kΩ 0.90( )+82kΩ 0.90( )

150kΩ 1.1( )

μA=112 μA I2min = 200 0.98( )

1+68kΩ 1.1( )+82kΩ 1.1( )

150kΩ 0.90( )

μA= 88.2 μA

V3= I2R3= 1 I

R1+ 1

R3+ R2

R1R3

1

150kΩ 1.1( )+82kΩ 1.11( )+

68kΩ 0.9( )

150kΩ 1.1( ) (82kΩ) ( )1.1

9.60 V

1

150kΩ 0.9( )+

1

82kΩ 0.9( )+

68kΩ 1.1( )

150kΩ 0.9( ) (82kΩ) ( )0.9

= 6.89 V

Let R X = R2 R3 then V1= V R1

R1+ R X

= V1

1+R X

R1

RXmin=30kΩ 0.9( ) (11kΩ) ( )0.9

30kΩ 0.9( )+11kΩ 0.9( )= 7.24kΩ RXmax=30kΩ 1.1( ) (11kΩ) ( )1.1

30kΩ 1.1( )+11kΩ 1.1( ) = 8.85kΩ

V1max= 1 1.05( )

1+ 7.24kΩ

24kΩ 1.1( )

= 0.824 V V1min= 1 0.95( )

1+ 8.85kΩ

24kΩ 0.9( )

= 0.674 V

R1+ R X

and I2= I1

R3

R2+ R3

R1+ R X







1+R2

R3

I2max= 1 1.05( )

24kΩ 0.9( )+ 7.24kΩ

1

1+(30kΩ) ( )0.9

11kΩ 1.1( )

=11.3 μA

I2min

= 1 0.95( )

24kΩ 1.1( )+8.85kΩ

1

1+(30kΩ) ( )1.1

11kΩ 0.9( )

= 6.22 μA

I3= I1

R2

R2+ R3

= I1

1+R3

R2

I3max= 1 1.05( )

24kΩ 0.9( )+ 7.24kΩ

1

1+(11kΩ) ( )0.9

30kΩ 1.1( )

= 28.0 μA

I3min= 1 0.95( )

24kΩ 1.1( )+8.85kΩ

1

1+(11kΩ) ( )1.1

30kΩ 0.9( )

=18.6 μA

1.61

From Prob 1.24: R th = 1

g m+ 1

R1

1.62 For one set of 200 cases using the Equations in Prob 1.59 (mA & kΩ):

I = 0.200* 0.98+ 0.04* RAND()( ) R1=150* 0.9 + 0.2* RAND()( )

R2= 68* 0.9 + 0.2* RAND()( ) R3= 82* 0.9 + 0.2* RAND()( )

1.63 For one set of 200 cases using the equations in Prob 1.60

V =1* 0.95+ 0.1* RAND()( ) R1= 24000* 0.9 + 0.2* RAND()( )

R2= 30000* 0.9 + 0.2* RAND()( ) R3=11000* 0.9 + 0.2* RAND()( )

1.64 3.29, 0.995, -6.16; 3.295, 0.9952, -6.155

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

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