Microelectronic circuit design 3rd edition by r jaeger solutions mmzzhh

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
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 =

(b)

log2
= 1.52 years
0.1977

Y2 − Y1 =

log10
= 5.06 years
0.1977

1.4
0.1548(2020−1970)

N = 1610x10
1.5

= 8.85 x 1010 transistors/μP
(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

−0.05806(2020−1970)

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
= 511 kA!
220V

D, D, A, A, D, A, A, D, A, D, A

1.9
1.10

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
mV
5V
=
= 19.53
bit
2 bits 256bits
8

2.77V
= 142 LSB
mV
19.53
bit

and

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

1.12
VLSB =

2.5V
2.5V
mV
=
= 2.44
bit
2 bits 1024 bits
10

(

01011011012 = 28 + 26 + 25 + 23 + 22 + 20

)

10

⎛ 2.5V ⎞
VO = 365 ⎜
⎟ = 0.891 V
⎝ 1024 ⎠

= 36510

1.13

(

)

mV
6.83V 14
10V
= 0.6104
and
2 bits = 11191 bits
14
10V
bit
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
5.12V
mV
V
5.12V
= 1.25
and VO = (1011101110112 )VLSB ± LSB

=
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

VLSB =

(

)

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

1

R2

V

I

+
V

3

R

2

3

-

V1 = 10V

22kΩ

(

)

22kΩ + 47kΩ 180kΩ

= 10V

22kΩ
= 3.71 V
22kΩ + 37.3kΩ

37.3kΩ
= 6.29 V Checking : 6.29 + 3.71 = 10.0 V
22kΩ + 37.3kΩ
⎛
⎞
180kΩ
180kΩ
10V
I2 = I1
=⎜
= 134 μA
⎟
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ
⎛
⎞
47kΩ
47kΩ
10V
I3 = I1
=⎜
= 34.9 μA
⎟
47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ

V2 = 10V

Checking : I1 =

10V
= 169μA and I1 = I2 + I3

22kΩ + 37.3kΩ

1-4

©R. C. Jaeger & T. N. Blalock
6/9/06


1.21

V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ.
V

+

1

R

I
R

V

I3

+

2

1

V

2

R

2

3

-

V1 = 18V

56kΩ

(

)

56kΩ + 33kΩ 11kΩ

= 15.7 V

V2 = 18V

33kΩ 11kΩ

(

)

56kΩ + 33kΩ 11kΩ

= 2.31 V

Checking :V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct.
I1 =

18V

(

)

56kΩ + 33kΩ 11kΩ

I3 = I1

= 280 μA

I2 = I1

11kΩ
11kΩ
= (280 μA)
= 70.0 μA
33kΩ + 11kΩ

33kΩ + 11kΩ

33kΩ
33kΩ
= (280 μA)
= 210 μA
33kΩ + 11kΩ
33kΩ + 11kΩ

1.22

I1 = 5mA

(5.6kΩ + 3.6kΩ) = 3.97 mA
(5.6kΩ + 3.6kΩ)+ 2.4kΩ

(

Checking : I2 + I3 = 280 μA

I2 = 5mA

2.4kΩ
= 1.03 mA
9.2kΩ + 2.4kΩ

= 3.72V
)5.6kΩ3.6kΩ
+ 3.6kΩ

V3 = 5mA 2.4kΩ 9.2kΩ

Checking : I1 + I2 = 5.00 mA

and

I2 R2 = 1.03mA(3.6kΩ)= 3.71 V

1.23

150kΩ
150kΩ
= 125 μA
I3 = 250μA
= 125 μA
150kΩ + 150kΩ
150kΩ + 150kΩ
82kΩ
V3 = 250μA 150kΩ 150kΩ
= 10.3V
68kΩ + 82kΩ
Checking : I1 + I2 = 250 μA and I2 R2 = 125μA(82kΩ)= 10.3 V
I2 = 250μA

(

)

1-5
6/9/06

1.24
+
R

v
v

1

s

+
g v

v

m

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

ix

g v
m

vx

Summing currents at the output node :
v
− 0.002v = 0 but v = −vx
5x10 4
vx
v
1
ix =
+ 0.002vx = 0 Rth = x =
= 495 Ω
4
1
ix
5x10
+ gm
R1

ix = −

Thévenin equivalent circuit:

495 Ω
v

s

1-6

©R. C. Jaeger & T. N. Blalock
6/9/06


1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24,
and
⎛ 1
⎞−1
Rth = ⎜
+ .025⎟ = 39.6 Ω
⎝ 4kΩ
⎠
+
v

R

1

-

vs

g v
m

in

The short circuit current is :
v
in =
+ 0.025v and v = vs
4kΩ
v
i n = s + 0.025vs = 0.0253vs
4kΩ
Norton equivalent circuit:

0.0253v

s

39.6 Ω

1-7
6/9/06


1.26 (a)

+

βi
R

vs

v th

R2

1

i

Vth = Voc = −β i R2

-

i =−

but

vs
R1

and

Vth = β vs

R2
39kΩ
= 120 vs
= 46.8 vs
R1
100kΩ

ix
βi
R

v

Rth

R2

1

x

i

Rth =

vx
;
ix

ix =

vx
+ βi
R2

but

i = 0 since VR 1 = 0.

Rth = R2 = 39 kΩ.

Thévenin equivalent circuit:

39 k Ω
58.5v

s

(b)
+

βi
i

s

R

R2

1

i

v th
-

⎛ i ⎞
Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β ⎜ − s ⎟ R2 = 38700 is
⎝ β + 1⎠

1-8

©R. C. Jaeger & T. N. Blalock
6/9/06


βi
R

Rth

R2

1

v

x

i

Rth =

vx
;
ix

ix =

vx
+ βi but
R2

i + βi = 0 so i = 0

and Rth = R2 = 39 kΩ

Thévenin equivalent circuit:

39 k Ω

38700i s
1.27

βi
R

vs

R2

1

in

i

in = −β i but i = −

vs
R1

and in =

β
R1

vs =

From problem 1.26(a), Rth = R2 = 56 kΩ.

0.00133v

s

100
vs = 1.33 x 10−3 vs
75kΩ
Norton equivalent circuit:

56 k Ω

1-9
6/9/06


1.28
is
v

βi
R

s

R2

1

i

is =

vs
v
v
β +1
− β i = s + β s = vs
R1

R1
R1
R1

R=

vs
R
100kΩ
= 1 =
= 1.24 Ω
is β + 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 MΩ

1.30

5V
3V
f (Hz)
0
0

500

1000

1.31

2V
f (kHz)
0
9

10

11

v = 4sin (20000πt )sin (2000πt )=

[

]

4
cos(20000πt + 2000πt )+ cos(20000πt − 2000πt )
2
v = 2cos(22000πt )+ 2cos(18000πt )
1.32

2∠36 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 =

10−1∠ −12 o
10−2 ∠ − 45o
o
A
=
=
5∠
−
45
= 100∠ −12 o
(b)
2x10−3 ∠0 o
10−3 ∠0 o

1.34

(a) Av = −

R2
620kΩ
180kΩ
=−
= −44.3 (b) Av = −
= −10.0
14kΩ
R1
18kΩ

(c) Av = −

62kΩ
= −38.8
1.6kΩ

1.35
vo (t ) = −

IS =

R2
v s (t )= (−90.1 sin 750πt ) mV
R1

VS 0.01V
=

= 11.0μA and
R1 910Ω

is = (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
v
+ 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

vO (V)

3

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 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

1.42

5 kHz

High-pass amplifier

Amplitude
16

f
10 kHz
1.43
vO (t ) = 10x5sin (2000πt )+ 10x3cos(8000πt )+ 0x3cos(15000πt )

[

]

vO (t ) = 50sin(2000πt )+ 30cos(8000πt ) volts

1.44
vO (t ) = 20x0.5sin (2500πt )+ 20x0.75cos(8000πt )+ 0x0.6cos(12000πt )

[

]

vO (t ) = 10.0sin (2500πt )+ 15.0cos(8000πt ) 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

1

2

3

4

5
x10-3

(a)
10
5
0
-5

-10
0

(b)

1-14

©R. C. Jaeger & T. N. Blalock
6/9/06


10
5
0
-5
-10
0

1

2

3

4

(c)

5
x10-3

10
5
0
-5
-10
0

1

2

3

4

5
x10-3

(d)
1.47

(a) 3000(1− .01)≤ R ≤ 3000(1+ .01) or 2970Ω ≤ R ≤ 3030Ω

(b) 3000(1− .05)≤ R ≤ 3000(1+ .05) or 2850Ω ≤ R ≤ 3150Ω

(c) 3000(1− .10) ≤ R ≤ 3000(1+ .10) or 2700Ω ≤ R ≤ 3300Ω
ΔV ≤ 0.05V

0.05

= 0.0200 or 2.00%
2.50

1.48

Vnom = 2.5V

1.49

20000μF (1− .5)≤ C ≤ 20000μF (1+ .2) or 10000μF ≤ R ≤ 24000μF

1.50

8200(1− 0.1)≤ R ≤ 8200(1+ 0.1) or 7380Ω ≤ R ≤ 9020Ω
The resistor is within the allowable range of values.

T=

1-15
6/9/06


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
5.30
= 5.22V which is within specifications limits.
Vmeter = 1.015Vact or Vact =

1.015
1.52

ΔR 6562 − 6066
Ω
=
= 4.96 o
ΔT
100 − 0
C
= R o + TCR (ΔT)= 6066 + 4.96(27)= 6200Ω

TCR =
R nom

0 C

1-16

©R. C. Jaeger & T. N. Blalock
6/9/06


1.53
V1

+

I2

R

1

R2

V

I3

+

R

V2

3

-

Let RX = R2 R3

R min
X =
V1max =

I1 =

I2min =

I3 = I1

I3max =

I3min =

R1
=
R1 + RX

47kΩ(0.9)(180kΩ)(0.9)

47kΩ(0.9)+ 180kΩ(0.9)
10(1.05)

33.5kΩ
1+
22kΩ(1.1)

V
R1 + RX

I2max =

then V1 = V

= 4.40V

and I2 = I1

V1
R
1+ X
R1

= 33.5kΩ R max
=
X
V1min =

R3
=
R2 + R3

47kΩ(1.1)(180kΩ)(1.1)

47kΩ(1.1)+ 180kΩ(1.1)

10(0.95)

41.0kΩ
1+
22kΩ(0.9)

= 3.09V

V
R1 + R2 +

R1 R2

R3

10(1.05)

22000(0.9)(47000)(0.9)

22000(0.9)+ 47000(0.9)+

10(0.95)
22000(1.1)+ 47000(1.1)+
R2
=
R2 + R3

= 41.0kΩ

= 158 μA

180000(1.1)

22000(1.1)(47000)(1.1)

= 114 μA

180000(0.9)

V
R1 + R3 +

R1 R3

R2

10(1.05)

22000(0.9)+ 180000(0.9)+

22000(0.9)(180000)(0.9)

10(0.95)
22000(1.1)+ 180000(1.1)+

= 43.1 μA

47000(1.1)

22000(1.1)(180000)(1.1)

= 28.3 μA

47000(0.9)

1-17
6/9/06


1.54
I1 = I

R2 + R3
=I

R1 + R2 + R3

1+

and similarly I2 = I

R1
R2 + R3

mA = 4.12 mA

2400(0.95)

1+

1
R + R3
1+ 2
R1

I1min =
1+

5600(1.05)+ 3600(1.05)
5(1.02)

I2max =

5600(0.95)+ 3600(0.95)

V3 = I2 R3 =

V3min =

1+

5(1.02)

I1max =

V3max =

1

mA = 1.14 mA

2400(1.05)

1+

mA = 3.80 mA

5600(0.95)+ 3600(0.95)
5(0.98)

I2min =

2400(1.05)

5(0.98)

5600(1.05)+ 3600(1.05)

mA = 0.936 mA

2400(0.95)

I
1
1
R
+
+ 2
R1 R3 R1 R3

5(1.02)

5600(0.95)
1
1
+
+
2400(1.05) 3600(1.05) 2400(1.05)(3600)(1.05)
5(0.98)

= 4.18 V

5600(1.05)
1
1

+
+
2400(0.95) 3600(0.95) 2400(0.95)(3600)(0.95)

= 3.30 V

1.55
Rth =

From Prob. 1.24 :

Rthmax =

1
gm +

1
1
0.002(0.8)+
5x10 4 (1.2)

1
R1
= 619 Ω

Rthmin =

1-18

1

1
0.002(1.2)+
5x10 4 (0.8)

= 412 Ω

©R. C. Jaeger & T. N. Blalock
6/9/06


1.56

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

V = 10 * (0.95 + 0.1* RAND())

R1 = 22000 * (0.9 + 0.2 * RAND())

R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND())
V1

I2

I3

Min

3.23 V

116 μA

29.9 μA

Max

3.71 V

151 μA

40.9 μA

Average

3.71 V

133 μA

35.1 μA

1.57

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

I = 0.005* (0.98 + 0.04 * RAND())

R1 = 2400 * (0.95 + 0.1* RAND())

R1 = 5600 * (0.95 + 0.1* RAND()) R3 = 3600 * (0.95 + 0.1* RAND())
I1

I2

V3

Min

3.82 mA

0.96 mA

3.46 V

Max

4.09 mA

1.12 mA

4.08 V

Average

3.97 mA

1.04 mA

3.73 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 A)(97.80 kΩ) = 9.99 V; 10 V

1-19
6/9/06


CHAPTER 2
2.1
Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor.
2.2
Based upon Table 2.1, a resistivity of 1015 Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator.
2.3
I max

⎛ 10−8 cm2 ⎞
⎛ 7 A ⎞
= ⎜10
⎟ = 500 mA
⎟(5μm)(1μm)⎜
2
cm 2 ⎠
⎝
⎝ μm ⎠

2.4

EG
⎛
⎞
ni = BT 3 exp⎜ −
⎟
−5
⎝ 8.62 x10 T ⎠
31
For silicon, B = 1.08 x 10 and EG = 1.12 eV:
-10

3

ni = 2.01 x10 /cm

9

3

13

6.73 x10 /cm

30

For germanium, B = 2.31 x 10 and EG = 0.66 eV:
3

13

ni = 35.9/cm

3

2.27 x10 /cm

15

function f=temp(T)
ni=1E14;
f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
14

3

16

ni = 10 /cm3 for T = 739 K

for T = 506 K

2.6

⎛
⎞
EG
ni = BT 3 exp⎜ −
−5 ⎟
⎝ 8.62x10 T ⎠

with

B = 1.27x1029 K −3cm−6
6

3

T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm
3

T = 100 K: ni = 6.03 x 10-19/cm

20

3

8.04 x 10 /cm .

2.5
Define an M-File:

ni = 10 /cm

3

8.36 x 10 /cm .

11

3

T = 500 K: ni = 2.79 x10 /cm


2.7

⎛
cm2 ⎞⎛
V ⎞
6 cm
vn = −μn E = ⎜ −700
⎟⎜ 2500 ⎟ = −1.75x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
V ⎞
cm2 ⎞⎛
5 cm
v p = +μ p E = ⎜ +250
⎟⎜ 2500 ⎟ = +6.25x10
cm ⎠
V − s ⎠⎝
s
⎝
⎛
1 ⎞⎛
cm ⎞

4 A
jn = −qnvn = −1.60x10−19 C ⎜1017 3 ⎟⎜ −1.75x106
⎟ = 2.80x10
s ⎠
cm ⎠⎝
cm2
⎝
⎛
1 ⎞⎛
cm ⎞
−10 A
j p = qnv p = 1.60x10−19 C ⎜103 3 ⎟⎜ 6.25x105
⎟ = 1.00x10
s ⎠
cm 2
⎝ cm ⎠⎝

(

)

(

)

2.8

⎛ E ⎞
ni2 = BT 3 exp⎜ − G ⎟
⎝ kT ⎠

B = 1.08x1031

⎛
⎞
1.12
T 3 exp⎜ −
⎟
⎝ 8.62x10−5 T ⎠
Using a spreadsheet, solver, or MATLAB yields T = 305.22K

(10 ) = 1.08x10
10

2

31

Define an M-File:
function f=temp(T)
f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
Then: fzero('temp',300) | ans = 305.226 K
2.9

v=

j − 1000 A / cm 2
cm
=
= − 105

2
Q
s
0.01C / cm

2.10
C ⎞⎛
cm ⎞
MA
⎛
6 A
j = Qv = ⎜ 0.4 3 ⎟⎜10 7
=4 2
⎟ = 4 x10
2
cm ⎠⎝
sec ⎠
cm
cm
⎝

21


2.11

⎛
V ⎞
cm2 ⎞⎛
6 cm

vn = −μn E = ⎜−1000
⎟⎜ −2000 ⎟ = +2.00x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
V ⎞
cm 2 ⎞⎛
5 cm
v p = +μ p E = ⎜ +400
⎟⎜ −2000 ⎟ = −8.00x10
V − s ⎠⎝
cm ⎠
s
⎝
⎛
1 ⎞⎛
cm ⎞
−10 A
jn = −qnvn = −1.60x10−19 C ⎜103 3 ⎟⎜ +2.00x106
⎟ = −3.20x10
s ⎠
cm2
⎝ cm ⎠⎝
⎛
1 ⎞⎛
cm ⎞
4 A
j p = qnv p = 1.60x10−19 C ⎜1017 3 ⎟⎜ −8.00x105

⎟ = −1.28x10
s ⎠
cm ⎠⎝
cm2
⎝

(

)

(

)

2.12

(a )

E=

V
5V
= 5000
−4
cm
10 x10 cm

(b )

(

)

V ⎞
⎛
−4
V = ⎜105
⎟ 10 x10 cm = 100 V
cm
⎝
⎠

2.13

⎛ 1019 ⎞⎛
cm ⎞
7 A
j p = qpv p = 1.60x10−19 C ⎜ 3 ⎟⎜10 7
⎟ = 1.60x10
s ⎠
cm2
⎝ cm ⎠⎝
⎛
A ⎞
i p = j p A = ⎜1.60x10 7 2 ⎟ 1x10−4 cm 25x10−4 cm = 4.00 A
cm ⎠
⎝

(

)

(

)(

)

2.14
For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p )

σ ≥ 1000(Ω − cm) for a conductor
−1

ni ≥

σ

q (μn + μ p )

1000(Ω − cm)

−1

=

cm 2
1.602x10−19 C (100 + 50)
v − sec
39

⎛
⎞
1.73x10
E
n 2i =
= BT 3 exp⎜ − G ⎟ with
6
cm
⎝ kT ⎠

=

4.16x1019
cm3

B = 1.08x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV

This is a transcendental equation and must be solved numerically by iteration. Using the HP
solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above the
melting temperature of silicon.

22


2.15
For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p )

σ ≤ 10−5 (Ω − cm) for an insulator
−1

ni ≥

σ

q (μn + μ p )

10−5 (Ω − cm)

−1

=

⎛ cm 2 ⎞
1.602x10 C (2000 + 750)⎜
⎟
⎝ v − sec ⎠
⎛ E ⎞
5.152x1020
n 2i =
= BT 3 exp⎜ − G ⎟ with
6
cm
⎝ kT ⎠

(

−19

)

=

2.270x1010
cm 3

B = 1.08x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV

Using MATLAB as in Problem 2.5 yields T = 316.6 K.
2.16

Si

Si

Si

P

B

Si

Si

Si

Si

Donor electron
fills acceptor

vacancy

No free electrons or holes (except those corresponding to ni).
2.17

(a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron
and will act as a donor impurity.
(b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in one
electron and will act as an acceptor impurity.
2.18
Since Ge is from column IV, acceptors come from column III and donors come from column
V. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi

23


2.19
(a) Germanium is from column IV and indium is from column III. Thus germanium has one
extra electron and will act as a donor impurity.
(b) Germanium is from column IV and phosphorus is from column V. Thus germanium has
one less electron and will act as an acceptor impurity.
2.20
A ⎞
V
⎛
= jρ = ⎜10000 2 ⎟(0.02Ω − cm ) = 200
, a small electric field.
cm ⎠
σ
cm

⎝
j

E=

2.21

⎛ C ⎞⎛
cm ⎞
A
jndrift = qnμn E = qnv n = 1.602x10−19 1016 ⎜ 3 ⎟⎜10 7
⎟ = 16000 2
s ⎠
cm
⎝ cm ⎠⎝

(

)( )

2.22
⎛ 1015 atoms ⎞
⎛ 10−4 cm ⎞3
N =⎜
⎟(1μm)(10μm)(0.5μm )⎜
⎟ = 5,000 atoms
3
⎝ cm
⎠
⎝ μm ⎠

2.23
N A > N D : N A − N D = 1015 −1014 = 9x1014 /cm3

If we assume N A − N D >> 2ni = 1014 / cm 3 :
p = N A − N D = 9x1014 /cm3 | n =
If we use Eq. 2.12 : p =

9x1014 ±

ni2 251026
=
= 2.78x1012 /cm3
p 9x1014

(9x10 ) + 4(5x10 ) = 9.03x10
14

2

13

2

14

2
and n = 2.77x10 /cm . The answers are essentially the same.
12

3

2.24
N A > N D: N A − N D = 5 x1016 − 1016 = 4 x1016 /cm 3 >> 2ni = 2 x1011 /cm 3
p = N A − N D = 4 x1014 /cm 3 | n =

ni2
10 22
=
= 2.50 x10 5 /cm 3
p 4 x1016

2.25
N D > N A: N D − N A = 3x1017 − 2x1017 = 1x1017 /cm3

2ni = 2x1017 /cm3 ; Need to use Eq. (2.11)
n=
p=

24

1017 ±

( ) ( ) = 1.62x10
2

1017 + 4 1017

2

2

2
i

34

n
10
=
= 6.18x1016 /cm3
n 1.62x1017

17

/cm3


2.26
N D − N A = −2.5x1018 / cm 3
Using Eq. 2.11: n =

−2.5x1018 ±

(−2.5x10 ) + 4(10 )
18

2

10

2

2

ni2
= ∞.
p
No, the result is incorrect because of loss of significant digits

Evaluating this with a calculator yields n = 0, and n =

within the calculator. It does not have enough digits.

2.27

(a) Since boron is an acceptor, NA = 6 x 1018/cm3. Assume ND = 0, since it is not specified.
The material is p-type.
At room temperature, ni = 1010 /cm3 and N A − N D = 6 x1018 / cm3 >> 2n i
ni2
10 20 /cm 6
So p = 6 x10 /cm and n =
=
= 16.7 /cm3
18
3
p 6 x10 /cm
(b)
⎛
⎞

3
1.12
⎟
At 200K, ni2 = 1.08x1031 (200) exp⎜⎜ −
= 5.28x109 /cm6
−5
⎟
⎝ 8.62x10 (200)⎠
18

3

ni = 7.27x10 4 /cm 3

N A − N D >> 2ni , so p = 6x1018 /cm3 and n =

5.28x109
= 8.80x10−10 /cm 3
18
6x10

2.28

(a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The
material is n-type.
At room temperature, n i = 1010 / cm 3 and N D − N A = 3 x1017 / cm 3 >> 2n i
10 20 /cm 6
ni2
=
= 333 /cm 3

17
3
n 3x10 /cm
⎛
⎞
3
1.12
⎟
= 4.53x1015 /cm6
(b) At 250K, ni2 = 1.08x1031 (250) exp⎜⎜−
−5
⎟
⎝ 8.62x10 (250)⎠

So n = 3 x1017 /cm 3 and p =

ni = 6.73x10 7 /cm 3

N D − N A >> 2ni , so n = 3x1017 / cm 3 and n =

4.53x1015
= 0.0151/ cm 3
17
3x10

2.29

(a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3.
Since NA > ND, the material is p-type.

25