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Solutions Manual Microelectronic Circuit Design 4th Ed

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Solutions Manual - Microelectronic Circuit Design - 4th Ed 
 
 
By Richard C. Jaeger, Travis N. Blalock - McGraw-Hill (2010) 
 
NOTE: these answers are for the International Edition (?) 
 
But they’re still very similar to the original (sometimes a, b, c, d answers will be switched around, and 
some numbers may be a little off. As a general rule, subtract 3 from the answer you are looking for 
and that should be the real one) 
 
 
Special thanks to Moser from NIU for the main files 
 
 
February 1, 2012 
 
 
Go to this website for book updates / corrections by the publisher: 
http://www.jaegerblalock.com/ 
 
 1-1 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
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. 
1.2 
 B =19.97 x 100.1997 2020−1960( ) =14.5 x 1012 =14.5 Tb/chip 
 
1.3 
(a) 
 
B2
B1
= 19.97x10
0.1977 Y2 −1960( )
19.97x100.1977 Y1 −1960( )
=100.1977 Y2 −Y1( ) so 2 =100.1977 Y2 −Y1( )
Y2 −Y1 = log20.1977 =1.52 years 
(b) 
 
Y2 −Y1 = log100.1977 = 5.06 years 
 
1.4
 N =1610x100.1548 2020−1970( ) = 8.85 x 1010 transistors/μP 
 
1.5
 
N2
N1
= 1610x10
0.1548 Y2 −1970( )
1610x100.1548 Y1 −1970( )
=100.1548 Y2 −Y1( )
(a) Y2 −Y1 = log20.1548 =1.95 years
(b) Y2 −Y1 = log100.1548 = 6.46 years
 
 
1.6 . F = 8.00x10−0.05806 2020−1970( )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 
 
 
 1-3 
 6/9/06 
1.8
 
P = 75x106 tubes( )1.5W tube( )=113 MW! I = 1.13 x 108W220V = 511 kA! 
 
1.9 D, D, A, A, D, A, A, D, A, D, A 
 
1.10 
 
VLSB = 10.24V212 bits =
10.24V
4096bits
= 2.500 mV VMSB = 10.24V2 = 5.120V
1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342 2.500mV( )= 5.855 V
 
 
1.11 
 
VLSB = 5V28 bits =
5V
256bits
=19.53 mV
bit
 and 2.77V
19.53 mV
bit
=142 LSB
14210 = 128 + 8 + 4 + 2( ) =100011102
 
10
 
1.12 
 
VLSB = 2.5V210 bits =
2.5V
1024 bits
= 2.44 mV
bit
01011011012 = 28 + 26 + 25 + 23 + 22 + 20( )10 = 36510 VO = 365 2.5V1024⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.891 V 
 
1.13 
 
VLSB = 10V214 bits = 0.6104
mV
bit
 and 6.83V
10V
214 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 
 
VLSB = 5.12V212 bits =
5.12V
4096 bits
=1.25 mV
bit
 and VO = 1011101110112( )VLSB ± VLSB2
VO = 211 + 29 + 28 + 27 + 25 + 24 + 23 + 2 +1( )101.25mV ± 0.0625V
VO = 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 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
+ -
V
1
V
2
+
-
R
1
R 2 R3
I3
I2
 
 
 
V1 =10V 22kΩ22kΩ + 47kΩ 180kΩ( )=10V
22kΩ
22kΩ + 37.3kΩ = 3.71 V
V2 =10V 37.3kΩ22kΩ + 37.3kΩ = 6.29 V Checking : 6.29 + 3.71 = 10.0 V
I2 = I1 180kΩ47kΩ +180kΩ =
10V
22kΩ + 37.3kΩ
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
180kΩ
47kΩ +180kΩ =134 μA
I3 = I1 47kΩ47kΩ +180kΩ =
10V
22kΩ + 37.3kΩ
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
47kΩ
47kΩ +180kΩ = 34.9 μA
Checking : I1 = 10V22kΩ + 37.3kΩ =169μA and I1 = I2 + I3
 
 
 1-4 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 
 1-5 
 6/9/06 
1.21 
V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ. 
V
+ -
V
1
V
2
+
-
R
1
R 2 R3
I3
I2
 
 
V1 =18V 56kΩ56kΩ + 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 = 18V56kΩ + 33kΩ 11kΩ( )= 280 μA I2 = I1
11kΩ
33kΩ +11kΩ = 280 μA( ) 11kΩ33kΩ +11kΩ = 70.0 μA
I3 = I1 33kΩ33kΩ +11kΩ = 280 μA( ) 33kΩ33kΩ +11kΩ = 210 μA Checking : I2 + I3 = 280 μA
 
 
1.22 
 
I1 = 5mA
5.6kΩ + 3.6kΩ( )
5.6kΩ + 3.6kΩ( )+ 2.4kΩ = 3.97 mA I2 = 5mA
2.4kΩ
9.2kΩ + 2.4kΩ =1.03 mA
V3 = 5mA 2.4kΩ 9.2kΩ( ) 3.6kΩ5.6kΩ + 3.6kΩ = 3.72V 
Checking : I1 + I2 = 5.00 mA and I2R2 =1.03mA 3.6kΩ( )= 3.71 V 
 
1.23 
 
I2 = 250μA 150kΩ150kΩ +150kΩ =125 μA I3 = 250μA
150kΩ
150kΩ +150kΩ =125 μA
V3 = 250μA 150kΩ 150kΩ( ) 82kΩ68kΩ + 82kΩ =10.3V
Checking : I1 + I2 = 250 μA and I2R2 =125μA 82kΩ( )=10.3 V 
 
1.24
 1-6 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 
R
1
+
-
v
gmv
vs
vth
+
-
 
 
 
Summing currents at the output node yields:
v
5x104
+ .002v = 0 so v = 0 and vth = vs − v = vs
 
 
 
R
1
vx
+
-
v
g v
m
ix
 
 
Summing currents at the output node :
ix = − v5x104 − 0.002v = 0 but v = −vx
ix = vx5x104 + 0.002vx = 0 Rth =
vx
ix
= 1
1
R1
+ gm
= 495 Ω
Thévenin equivalent circuit: 
vs
495 Ω
 
 
 
 1-7 
 6/9/06 
1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24, 
and 
 
 
Rth = 14kΩ + .025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
−1
= 39.6 Ω 
 
R
1
vs
+
-
v
gmv in
 
 
 
The short circuit current is :
in = v4kΩ + 0.025v and v = vs 
in = vs4kΩ + 0.025vs = 0.0253vs 
 
Norton equivalent circuit: 
 
39.6 Ω0.0253v s
 
 
1.26
 1-8 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 (a) 
R
1 R2
βi
vs
i
+
-
vth
 
 
 
Vth = Voc = −β i R2 but i = − vsR1 and Vth = β vs 
R2
R1
=120 vs 39kΩ100kΩ = 46.8 vs 
R
1 R2
βi
i
Rth vx
ix
 
 
 
Rth = vxix
 ; ix =
vx
R2
+ βi but i = 0 since VR1 = 0. Rth = R2 = 39 kΩ.
 
 
Thévenin equivalent circuit: 
 
 
58.5v s
39 k Ω
 
 
(b) 
R
1 R2
βi
is
i
+
-
vth
 
 
 
Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β − isβ +1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ R2 = 38700 is 
 
 1-9 
 6/9/06 
R
1 R2
βi
i
Rth vx
 
 
 
Rth = vxix
 ; ix =
vx
R2
+ βi but i + βi = 0 so i = 0 and Rth = R2 = 39 kΩ
 
Thévenin equivalent circuit: 
 
39 k Ω
38700i s
 
 
1.27
βi
R
1 R2vs
i
in
 
 
 
in = −β i but i = − vsR1 and in =
β
R1
vs = 10075kΩ vs =1.33 x 10
−3 vs
 
From problem 1.26(a), Rth = R2 = 56 kΩ. Norton equivalent circuit: 
56 k Ω0.00133v s
 
 
1.28
 1-10 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
R
1 R2
βi
vs
i
is
 
 
 
is = vsR1
− β i = vs
R1
+ β vs
R1
 = vs
β +1
R1
 R = vs
is
= R1β +1 =
100kΩ
81
=1.24 Ω
 
 
1.29 
 
The open circuit voltage is vth = −gmv R2 and v = +isR1.
vth = −gmR1R2is = − 0.0025( )105( )106( )i s= 2.5 x 108is
For i
 
= 0, v = 0 and R = R =1 MΩs th 2
 
1.30 
5 V
3 V
0
f (Hz)
500 10000 
 
1.31 
2 V
0
f (kHz)
9 10 11 
 
v = 4sin 20000πt( )sin 2000πt( )= 42 cos 20000πt + 2000πt( )+ cos 20000πt − 2000πt( )[ ]
v = 2cos 22000πt( )+ 2cos 18000πt( ) 
 
1.32 
 
A = 2∠36
o
10−5∠00 = 2x10
5∠36o A = 2x105 ∠A = 36o 
 
 
 1-11 
 6/9/06 
1.33 
(a) 
 
A = 10
−2∠ − 45o
2x10−3∠0o = 5∠ − 45
o (b) 
 
A = 10
−1∠ −12o
10−3∠0o =100∠ −12
o 
 
1.34 
 
(a) Av = − R2R1
= − 620kΩ
14kΩ = −44.3 (b) Av = −
180k = −10.0 (c) Av = − 62kΩ1.6kΩ
Ω = −38.8 
18kΩ
 
1.35 
 
vo t( )= − R2R1 vs t( )= −90.1 sin 750πt ( ) mV
IS = VSR1
= 0.01V
910Ω =11.0μA and 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
R2
+ i− + v−R1
= 0 or vs − vo
R2
+ vs
R1
= 0 and Av = vovs
=1+ R2
R1
 
 
1.38 Writing a nodal equation at the inverting input terminal of the op amp gives 
 
v1 − v−
R1
+ v2 − v−
R2
= i− + v− − voR3
 but v- = v+ = 0 and i- = 0
vo = − R3R1
v1 − − R3R2
v2 = −0.255sin3770t − 0.255sin10000t volts
 
 
1.39
vO = −VREF b12 +
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 
b1b2b
3
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 Low-pass amplifier 
 
Amplitude
f
10
6 kHz 
 
 1-12 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 
 1-13 
 6/9/06 
1.41 Band-pass amplifier 
 
f
20
1 kHz 5 kHz
Amplitude
 
 
 
1.42 High-pass amplifier 
 
f
10 kHz
16
Amplitude
 
 
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. 
 
1.46 
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 
(b)
-10
-5
0
5
10
0 1 2 3 4 5
x10-3 
 1-14 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 
 1-15 
 6/9/06 
(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) 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 
 
 
1.48 
 
Vnom = 2.5V ΔV ≤ 0.05V T = 0.052.50 = 0.0200 or 2.00% 
 
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. 
 
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 
 
Vmeter =1.015Vact or Vact = 5.301.015 = 5.22V which is within specifications limits. 
 
1.52 
 
TCR = ΔRΔT =
6562 − 6066
100 − 0 = 4.96
Ω
oC 
 1-16 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
Rnom = R 0oC + TCR ΔT( )= 6066 + 4.96 27( )= 6200Ω
 
 
 1-17 
 6/9/06 
1.53 
V2
+
-
V
+ -V1
R2
I3I2R1
R3
 
 
Let RX = R2 R3 then V1 = V R1R1 + RX
= V1
1+ RX
R1
RX
min = 47kΩ 0.9( )180kΩ( )0.9( )
47kΩ 0.9( )+180kΩ 0.9( )= 33.5kΩ RXmax =
47kΩ 1.1( )180kΩ( )1.1( )
47kΩ 1.1( )+180kΩ 1.1( )= 41.0kΩ
V1
max = 10 1.05( )
1+ 33.5kΩ
22kΩ 1.1( )
= 4.40V V1min =
10 0.95( )
1+ 41.0kΩ
22kΩ 0.9( )
= 3.09V
I1 = VR1 + RX
 and I2 = I1 R3R2 + R3
= V
R1 + R2 + R1R2R3 
 
I2
max = 10 1.05( )
22000 0.9( )+ 47000 0.9( )+ 22000 0.9( )47000( )0.9( )180000 1.1( )
=158 μA
 I2
min = 10 0.95( )
22000 1.1( )+ 47000 1.1( )+ 22000 1.1( )47000( )1.1( )180000 0.9( )
=114 μA
I3 = I1 R2R2 + R3
= V
R1 + R3 + R1R3R2
 I3
max = 10 1.05( )
22000 0.9( )+180000 0.9( )+ 22000 0.9( )180000( )0.9( )47000 1.1( )
= 43.1 μA
 I3
min = 10 0.95( )
22000 1.1( )+180000 1.1( )+ 22000 1.1( )180000( )1.1( )47000 0.9( )
= 28.3 μA
 
 
1.54
 
I1 = I R2 + R3R1 + R2 + R3
= I 1
1+ R1
R2 + R3
 and similarly I2 = I 1
1+ R2 + R3
R1
I1
max = 5 1.02( )
1+ 2400 0.95( )
5600 1.05( )+ 3600 1.05( )
mA = 4.12 mA I1min =
5 0.98( )
1+ 2400 1.05( )
5600 0.95( )+ 3600 0.95( )
mA = 3.80 mA
I2
max = 5 1.02( )
1+ 5600 0.95( )+ 3600 0.95( )
2400 1.05( )
mA =1.14 mA I2min =
5 0.98( )
1+ 5600 1.05( )+ 3600 1.05( )
2400 0.95( )
mA = 0.936 mA
V3 = I2R3 = I1
R1
+ 1
R3
+ R2
R1R3
V3
max = 5 1.02( )
1
2400 1.05( )+
1
3600 1.05(
)+
5600 0.95( )
2400 1.05( )3600( )1.05( )
= 4.18 V
V3
min = 5 0.98( )
1
2400 0.95( )+
1
3600 0.95( )+
5600 1.05( )
2400 0.95( )3600( )0.95( )
= 3.30 V
 
1.55 
 
From Prob. 1.24 : Rth = 1
gm + 1R1 
Rth
max = 1
0.002 0.8( )+ 15x104 1.2( )
= 619 Ω Rthmin = 1
0.002 1.2( )+ 15x104 0.8( )
= 412 Ω
 
 1-18 ©R. C. Jaeger & T. N. Blalock 
 6/9/06 
 
 
 1-19 
 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 
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 
 
Imax = 107 Acm2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 5μm( )1μm( ) 10
−8cm2
μm2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 500 mA 
 
2.4 
⎟⎠
⎞⎜⎝
⎛−= − Tx
EBTn Gi 5
3
1062.8
exp 
For silicon, B = 1.08 x 10
31
 and EG = 1.12 eV: 
 ni = 2.01 x10
-10
/cm
3
 6.73 x10
9
/cm
3
 8.36 x 10
13
/cm
3
. 
For germanium, B = 2.31 x 10
30
 and EG = 0.66 eV: 
ni = 35.9/cm
3
 2.27 x10
13
/cm
3
 8.04 x 10
15
/cm
3
. 
 
2.5 
Define an M-File: 
 
function f=temp(T) 
ni=1E14; 
f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); 
 
ni = 10
14
/cm
3
 for T = 506 K ni = 10
16
/cm3 for T = 739 K 
 
2.6 
 
ni = BT 3 exp − EG8.62x10−5T
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ with B = 1.27x10
29 K−3cm−6 
T = 300 K and EG = 1.42 eV: ni = 2.21 x10
6
/cm
3
T = 100 K: ni = 6.03 x 10
-19/cm
3
 T = 500 K: ni = 2.79 x10
11
/cm
3
 
 20 
2.7 
 
vn = −μnE = −700 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 2500
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = −1.75x10
6 cm
s
v p = +μ pE = +250 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 2500
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = +6.25x10
5 cm
s
jn = −qnvn = −1.60x10−19C( )1017 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1.75x106 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.80x104 Acm2
jp = qnv p = 1.60x10−19C( )103 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 6.25x105 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.00x10−10 Acm2 
 
2.8 
 
ni
2 = BT 3 exp − EG
kT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ B = 1.08x10
31
1010( )2 =1.08x1031T 3 exp − 1.128.62x10−5T⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 
Using a spreadsheet, solver, or MATLAB yields T = 305.22K 
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 
s
cm
cmC
cmA
Q
jv 52
2
10 
/01.0
/1000 −=−==
 
 
2.10 
22
67
3 4104sec
104.0
cm
MA
cm
Axcm
cm
CQvj ==⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛== 
 
 21 
 
2.11 
 
vn = −μnE = −1000 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −2000
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = +2.00x10
6 cm
s
v p = +μ pE = +400 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −2000
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = −8.00x10
5 cm
s
jn = −qnvn = −1.60x10−19C( )103 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +2.00x106 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −3.20x10−10 Acm2
jp = qnv p = 1.60x10−19C( )1017 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −8.00x105 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −1.28x104 Acm2
 
 
2.12 
( ) ( ) ( ) V 100101010 b 5000
1010
5 454 =⎟⎠
⎞⎜⎝
⎛=== −− cmxcm
VV
cm
V
cmx
VEa 
 
2.13 
 
jp = qpv p = 1.60x10−19C( )1019cm3⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 10
7 cm
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.60x10
7 A
cm2
ip = j p A = 1.60x107 Acm2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1x10
−4cm( )25x10−4cm( )= 4.00 A
 
 
2.14 
 
For intrinsic silicon, σ = q μnni + μ pni( )= qni μn + μ p( )
σ ≥1000 Ω − cm( )−1 for a conductor
ni ≥ σq μn + μ p( )=
1000 Ω − cm( )−1
1.602x10−19C 100 + 50( ) cm2v − sec
= 4.16x10
19
cm3
n i
2 = 1.73x10
39
cm6
= BT 3 exp − EG
kT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ with
B =1.08x1031 K−3cm−6, k = 8.62x10-5eV/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 μnni + μ pni( )= qni μn + μ p( )
σ ≤10−5 Ω − cm( )−1 for an insulator
ni ≥ σq μn + μ p( )=
10−5 Ω − cm( )−1
1.602x10−19C( )2000 + 750( ) cm2v − sec⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
= 2.270x10
10
cm3
n i
2 = 5.152x10
20
cm6
= BT 3 exp − EG
kT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ with
B =1.08x1031 K−3cm−6, k = 8.62x10-5eV/K and EG =1.12eV 
Using MATLAB as in Problem 2.5 yields T = 316.6 K. 
 
2.16 
 
Si Si Si
SiB
Si Si Si
P
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 
( ) field. electric small a ,20002.010000 2 cm
Vcm
cm
AjjE =−Ω⎟⎠
⎞⎜⎝
⎛=== ρσ 
 
2.21 
 
jn
drift = qnμnE = qnvn = 1.602x10−19( )1016( ) Ccm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 107 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =16000 Acm2 
 
2.22 
 
N = 10
15 atoms
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1μm( )10μm( )0.5μm( )
10−4cm
μm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
3
= 5,000 atoms 
 
2.23 
 
N A > N D: N A − N D =1015 −1014 = 9x1014 /cm3
If we assume N A − N D >> 2ni =1014 / cm3 :
p = N A − N D = 9x1014 /cm3 | n = ni
2
p
= 2510
26
9x1014
= 2.78x1012 /cm3
If we use Eq. 2.12 : p = 9x10
14 ± 9x1014( )2 + 4 5x1013( )2
2
= 9.03x1014
and n = 2.77x1012 /cm3. The answers are essentially the same. 
 
2.24 
35
16
222
314
3113161616
10502
104
10104
102210410105
/cmx.
xp
n | n/cmxNNp
/cmxn/cmxxN: NNN
i
DA
iDADA
====−=
=>>=−=−>
 
 
2.25 
 
N D > N A: ND − N A = 3x1017 − 2x1017 =1x1017 /cm3
2ni = 2x1017 /cm3; Need to use Eq. (2.11)
n = 10
17 ± 1017( )2 + 4 1017( )2
2
=1.62x1017 /cm3
p = ni
2
n
= 10
34
1.62x1017
= 6.18x1016 /cm3
 
 24 
 
2.26 
 
N D − N A = −2.5x1018 / cm3
Using Eq. 2.11: n = −2.5x10
18 ± −2.5x1018( )2 + 4 1010( )2
2
Evaluating this with a calculator yields n = 0, and n = ni
2
p
= ∞.
No, the result is incorrect because of loss of significant digits
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. 
3
318
6202
318
i
318310
7.16
106
10 and 106 So
 2n >> /106 and 10 re, temperaturoomAt 
/cm
/cmx
/cm
p
nn/cmxp
cmxNN/cmn
i
DAi
====
=−=
 
(b) 
 
At 200K, ni
2 =1.08x1031 200( )3 exp − 1.128.62x10−5 200( )
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 5.28x109 /cm6
ni = 7.27x104 /cm3 N A − N D >> 2ni, so p = 6x1018 /cm3 and n = 5.28x10
9
6x1018
= 8.80x10−10 /cm3
 
 
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. 
3
317
6202
317
i
317310
i
333
103
10 and 103 So
 2n >> /103 and /10n re, temperaturoomAt 
/cm
/cmx
/cm
n
np/cmxn
cmxNNcm
i
AD
====
=−=
 
 
(b) At 250K, ni
2 =1.08x1031 250( )3 exp − 1.128.62x10−5 250( )
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 4.53x1015 /cm6
ni = 6.73x107 /cm3 N D − N A >> 2ni , so n = 3x1017 / cm3 and n = 4.53x10
15
3x1017
= 0.0151/ cm3
 
 
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 
 
3
318
6202
318
i
318310
i
716
106
10 and 106 So
 2n >> /106 and /10n re, temperaturoomAt (b)
/cm.
/cmx
/cm
p
nn/cmxp
cmxNNcm
i
DA
====
=−=
 
 
2.30 
(a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3. 
Since NA > ND, the material is p-type. 
3
317
6202
317
i
317310
333
103
10 and 103 So
 2n >> /103 and 10 re, temperaturoomAt (b)
/cm
/cmx
/cm
p
nn/cmxp
cmxNN/cmn
i
DAi
====
=−=
 
 
2.31 
ND = 4 x 1016/cm3. Assume NA = 0, since it is not specified. 
 
N D > N A : material is n - type | N D − N A = 4x1016 / cm3 >> 2ni = 2x1010 / cm3
n = 4x1016 / cm3 | p = ni
2
n
= 10
20
4x1016
= 2.5x103 / cm3
N D + N A = 4x1016 / cm3 | Using Fig. 2.13, μn =1030 cm
2
V − s and μp = 310
cm2
V − s
ρ = 1
qμnn =
1
1.602x10−19C( )1030 cm2V − s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
4x1016
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
= 0.152 Ω − cm
 
 
 26 
2.32 
NA = 1018/cm3. Assume ND = 0, since it is not specified. 
 
N A > N D : material is p - type | N A − N D =1018 / cm3 >> 2ni = 2x1010 / cm3
p = 1018 / cm3 | n = ni
2
p
= 10
20
1018
=100 / cm3
N D + N A =1018 / cm3 | Using Fig. 2.13, μn = 375 cm
2
V − s and μp =100
cm2
V − s
ρ = 1
qμ p p =
1
1.602x10−19C 100 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1018
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
= 0.0624 Ω − cm
 
 
2.33 
Indium is from column 3 and is an acceptor. NA = 7 x 1019/cm3. Assume ND = 0, since it is not 
specified. 
 
N A > N D : material is p - type | N A − N D = 7x1019 /cm3 >> 2ni = 2x1010 /cm3
p = 7x1019 /cm3 | n= ni
2
p
= 10
20
7x1019
=1.43/cm3
N D + N A = 7x1019 / cm3 | Using Fig. 2.13, μn =120 cm
2
V − s and μp = 60
cm2
V − s
ρ = 1
qμ p p =
1
1.602x10−19C 60 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
7x1019
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
=1.49 mΩ − cm
 
 
2.34 
 
Phosphorus is a donor : N D = 5.5x1016 / cm3 | Boron is an acceptor : N A = 4.5x1016 / cm3
N D > N A : material is n - type | N D − N A =1016 / cm3 >> 2ni = 2x1010 / cm3
n =1016 /cm3 | p = ni
2
p
= 10
20
1016
=104 /cm3
N D + N A =1017 / cm3 | Using Fig. 2.13, μn = 800 cm
2
V − s and μp = 230
cm2
V − s
ρ = 1
qμnn =
1
1.602x10−19C 800 cm
2
V − s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1016
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
= 0.781 Ω − cm
 
 
 27 
 
2.35 
 
ρ = 1
qμ p p | μ p p =
1
1.602x10−19C( )0.054Ω − cm( )=
1.16x1020
V − cm − s
 
An iterative solution is required. Using the equations in Fig. 2.8: 
 
NA μp μp p 
1018 96.7 9.67 x 1020 
1.1 x1018 93.7 1.03 x 1020 
1.2 x 1017 91.0 1.09 x 1020 
1.3 x 1019 88.7 1.15 x 1020 
 
2.36 
 
ρ = 1
qμ p p | μ p p =
1
1.602x10−19C( )0.75Ω − cm( )=
8.32x1018
V − cm − s
 
An iterative solution is required. Using the equations in Fig. 2.8: 
 
NA μp μp p 
1016 406 4.06 x 
1018 
2 x 1016 363 7.26 x 
1018 
3 x 1016 333 1.00 x 
1019 
2.4 x 1016 350 8.40 x 1018 
 
2.37 
Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic 
because it contains impurities. Addition of the impurities has increased the resistivity. 
 
2.38 
 
ρ = 1
qμnn | μnn ≈ μnN D =
1
1.602x10−19C( )2Ω − cm( )=
3.12x1018
V − cm − s 
An iterative solution is required. Using the equations in Fig. 2.8: 
 
ND μn μnn 
1015 1350 1.35 x 1018 
2 x 1015 1330 2.67 x 1018 
2.5 x 1015 1330 3.32 x 1018 
 28 
2.3 x 1015 1330 3.06 x 1018 
 
 29 
 
2.39 (a) 
 
ρ = 1
qμnn | μnn ≈ μnN D =
1
1.602x10−19C( )0.001Ω − cm( )=
6.24x1021
V − cm − s
 
An iterative solution is required. Using the equations in Fig. 2.8: 
 
ND μn μnn 
1019 116 1.16 x 1021 
7 x 1019 96.1 6.73 x 1021 
6.5 x 1019 96.4 6.3 x 1021 
 
(b) 
ρ = 1
qμp p | μp p ≈ μpNA =
1
1.602x10−19C( )0.001Ω − cm( ) =
6.24x1021
V − cm − s
 
An iterative solution is required using the equations in Fig. 2.8: 
 
NA μp μp p 
1.3 x 1020 49.3 6.4 x 1021 
 
2.40 
Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but 
the hole and electron concentrations remain unchanged. See Problem 2.37 for example. 
However, it is physically impossible to add exactly equal amounts of the two impurities. 
 
2.41 
(a) For the 1 ohm-cm starting material: 
 
ρ = 1
qμ p p | μ p p ≈ μ pN A =
1
1.602x10−19C( )1Ω − cm( )=
6.25x1018
V − cm − s 
An iterative solution is required. Using the equations in Fig. 2.8: 
 
NA μp μp p 
1016 406 4.1 x 1018 
1.5 x 1016 383 5.7 x 1018 
1.7 x 1016 374 6.4 x 1019 
 
 30 
To change the resistivity to 0.25 ohm-cm: 
 
ρ = 1
qμ p p | μ p p ≈ μ pN A =
1
1.602x10−19C( )0.25Ω − cm( )=
2.5x1019
V − cm − s 
NA μp μp p 
6 x 1016 276 1.7 x 1019 
8 x 1016 233 2.3 x 1019 
1.1 x 1017 225 2.5 x 1019 
Additional acceptor concentration = 1.1x10
17
- 1.7x10
16
 = 9.3 x 10
16
/cm
3
(b) If donors are added: 
 
ND ND + NA μn ND - NA μnn 
2 x 1016 3.7 x 1016 1060 3 x 1015 3.2 x 1018 
1 x 1017 1.2 x 1017 757 8.3 x 1016 6.3 x 1019 
8 x 1016 9.7 x 1016 811 6.3 x 1016 5.1 x 1019 
4.1 x 1016 5.8 x 1016 950 2.4 x 1016 2.3 x 1019 
 
So ND = 4.1 x 10
16
/cm
3
 must be added to change achieve a resistivity of 0.25 ohm-cm. The 
silicon is converted to n-type material. 
 
2.42 
Phosphorus is a donor: ND = 1016/cm3 and μn = 1250 cm2/V-s from Fig. 2.8. 
 
σ = qμnn ≈ qμnN D = 1.602x10−19C( )1250( )1016( )= 2.00Ω − cm 
Now we add acceptors until σ = 5.0 (Ω-cm) -1: 
 
σ = qμ p p | μ p p ≈ μ p N A − N D( )= 5 Ω − cm( )
−1
1.602x10−19C
= 3.12x10
19
V − cm − s 
 
NA ND + NA μp NA - ND μp p 
1 x 1017 1.1 x 1017 250 9 x 1016 2.3 x 1019 
2 x 1017 2.1 x 1017 176 1.9 x 1017 3.3 x 1019 
1.8 x 1017 1.9 x 1017 183 1.7 x 1016 3.1 x 1019 
 
 31 
 
2.43 
Boron is an acceptor: NA = 1016/cm3 and μp = 405 cm2/V-s from Fig. 2.8. 
 
σ = qμ p p ≈ qμ pN A = 1.602x10−19C( )405( )1016( )= 0.649Ω − cm 
Now we add donors until σ = 5.5 (Ω-cm) -1: 
 
σ = qμnn | μnn ≈ μn N D − N A( )= 5.5 Ω − cm( )
−1
1.602x10−19C
= 3.43x10
19
V − cm − s 
 
ND ND + NA μn ND - NA μp p 
8 x 1016 9 x 1016 832 7
x 1016 5.8 x 1019 
6 x 1016 7 x 1016 901 5 x 1016 4.5 x 1019 
4.5 x 1016 5.5 x 1016 964 3.5 x 1016 3.4 x 1019 
 
2.44 
 
VT = kTq =
1.38x10−23T
1.602x10−19
= 8.62x10−5T
 
T (K) 50 75 100 150 200 250 300 350 400 
VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5 
 
2.45 
 
j = −qDn − dndx
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = qVTμn
dn
dx
j = 1.602x10−19C( )0.025V( ) 350 cm2V − s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
1018 − 0
0 −10−4
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
cm4
= −14.0 kA
cm2 
 
2.46 
 
j = −qDp dpdx = −1.602x10
−19C( )15 cm2s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ −
1019 / cm3
2x10−4cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ exp −
x
2x10−4cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
j =1.20x105 exp −5000 x
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
A
cm2
I 0( )= j 0( )A = 1.20x105 Acm2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 10μm
2( )10−8cm2μm2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =12.0 mA
 
 
 32 
2.47 
 
jp = qμp pE − qDp dpdx = qμp p E −VT
1
p
dp
dx
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0 → E = VT
1
p
dp
dx
E ≈ VT 1N A
dN A
dx
= 0.025 −10
22 exp −104 x( )
1014 +1018 exp −104 x( )
E 0( )= −0.025 10221014 +1018 = −250 Vcm
E 5x10−4cm( )= −0.025 1022 exp −5( )1014 +1018 exp −5( )= −246 Vcm
 
 
2.48 
 
jn
drift = qμnnE = 1.60x10−19C( )350 cm2V − s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
1016
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −20
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = −11.2
A
cm2
jp
drift = qμ p pE = 1.60x10−19C( )150 cm2V − s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
1.01x1018
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −20
V
cm
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = −484
A
cm2
jn
diff = qDn dndx = 1.60x10
−19C( )350 ⋅ 0.025cm2s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
104 −1016
2x10−4cm4
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = −70.0
A
cm2
jp
diff = −qDp dpdx = −1.60x10
−19C( )150 ⋅ 0.025cm2s⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ 
1018 −1.01x1018
2x10−4cm4
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 30.0
A
cm2
jT = −11.2 − 484 − 70.0 + 30.0 = −535 Acm2 
 
 
2.49 NA = 2ND 
E C
E A N A
Holes
E
V
NA
N
D
N A
N
D
N A
N DE D
 
 
 
2.50 
 
λ = hc
E
= 6.626x10
−34 J − s( )3x108 m / s( )
1.12eV( )1.602x10−19 J / eV( ) =1.108 μm 
 
 33 
 
2.51 
p-type silicon
n-type silicon
Si02
Al - Anode Al - Cathode
 
 
2.52 
An n-type ion implantation step could be used to form the n+ region following step (f) in Fig. 
2.17. A mask would be used to cover up the opening over the p-type region and leave the 
opening over the n-type silicon. The masking layer for the implantation could just be 
photoresist. 
 
n-type silicon
p-type silicon
Si02
Photoresist
Structure after exposure and 
development of photoresist layer
Mask
n-type silicon
p-type silicon
Structure following ion 
implantation of n-type impurity 
n+
Ion implantation
Side view
Top View
Mask for ion implantation
 
 
2.53 
 
(a) N = 8 1
8
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ + 6
1
2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ + 4 1()= 8 atoms
(b) V = l 3 = 0.543x10−9 m( )3 = 0.543x10−7cm( )3 =1.60x10−22cm3
(c) D = 8 atoms
1.60x1022cm3
= 5.00x1022 atoms
cm3
(d ) m = 2.33 g
cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1.60x10
22cm3 = 3.73x10−22 g
(e) From Table 2.2, silicon has a mass of 28.086 protons.
mp = 3.73x10
−22 g
28.082 8( )protons =1.66x10−24
g
proton
Yes, near the actual proton rest mass. 
 34 
CHAPTER 3 
 
3.1 
φ j = VT ln NANDni2 = 0.025V( )ln
1019 ⋅ cm−3( )1018 ⋅ cm−3( )
1020 ⋅ cm−6 = 0.979V
wdo = 2εsq
1
NA
+ 1
ND
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1019cm−3
+ 1
1018cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.979V( ) 
w do = 3.73 x 10−6cm = 0.0373μm
 
xn = wdo
1+ ND
NA
= 0.0373μm
1+ 10
18cm−3
1019cm−3
= 0.0339 μm | x p = wdo
1+ NA
ND
= 0.0373μm
1+ 10
19cm−3
1018cm−3
= 3.39 x 10-3 μm
E MAX = qNA xpεs =
1.60x10−19C( )1019cm−3( ) 3.39x10−7cm( )
11.7 ⋅ 8.854 x10−14 F /cm = 5.24 x 10
5 V
cm 
 
3.2 
ppo = NA = 10
18
cm3
 | npo = ni
2
ppo
= 10
20
1018
= 10
2
cm3
nno = ND = 10
15
cm3
 | pno = ni
2
nno
= 10
20
1015
= 10
5
cm3 
φ j = VT ln NANDni2 = 0.025V( )ln
1018cm−3( )1015cm−3( )
1020cm−6
= 0.748 V
wdo = 2εsq
1
NA
+ 1
ND
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1018cm−3
+ 1
1015cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.748V( )
wdo = 98.4 x 10−6cm = 0.984 μm 
 
3.3 
 
ppo = N A = 10
18
cm3
 | npo = ni
2
p po
= 10
20
1018
= 10
2
cm3
nno = N D = 10
18
cm3
 | pno = ni
2
nno
= 10
20
1018
= 10
2
cm3 
φ j = VT ln NANDni2 = 0.025V( )ln
1018 ⋅ cm−3( )1018 ⋅ cm−3( )
1020 ⋅ cm−6 = 0.921V
wdo = 2εsq
1
NA
+ 1
ND
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1018cm−3
+ 1
1018cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.921V( ) 
w do = 4.881x10−6cm = 0.0488 μm
 
 34 
 
 
3.4 
 
ppo = N A = 10
18
cm3
 | npo = ni
2
p po
= 10
20
1018
= 10
2
cm3
nno = N D = 10
18
cm3
 | pno = ni
2
nno
= 10
20
1018
= 10
2
cm3
φ j = VT ln NANDni2 = 0.025V( )ln
1018 ⋅ cm−3( )1020 ⋅ cm−3( )
1020 ⋅ cm−6 =1.04V
wdo = 2εsq
1
NA
+ 1
ND
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1018cm−3
+ 1
1020cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1.04V( ) 
w do = 0.0369 μm 
 
3.5 
 
ppo = N A = 10
16
cm3
 | npo = ni
2
p po
= 10
20
1016
= 10
4
cm3
nno = N D = 10
19
cm3
 | pno = ni
2
nno
= 10
20
1019
= 10
cm3
 
φ j = VT ln N A N Dni2
= 0.025V( )ln 1019 ⋅ cm−3( )1016 ⋅ cm−3( )1020 ⋅ cm−6 = 0.864V
wdo = 2εsq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1019cm−3
+ 1
1016cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.864V( ) 
wdo = 0.334 μm 
 
3.6 
 
wd = wdo 1+ VRφ j | (a) wd = 2wdo requires VR = 3φ j = 2.55 V | wd = 0.4μm 1+
5
0.85
= 1.05 μm
 
 
3.7 
 
wd = wdo 1+ VRφ j | (a) wd = 3wdo requires VR = 8φ j = 4.80 V | wd =1μm 1+
10
0.6
= 4.20 μm
 
 
3.8 
 
jn = σE , σ = 1ρ =
1
0.5 Ω⋅ cm =
2
Ω⋅ cm | E =
jn
σ =
1000A ⋅ cm−2
2 Ω⋅ cm( )−1 = 500
V
cm
 
 
 35
3.9 
 
jp = σE | E = jnσ = jnρ = 5000A ⋅ cm
−2( )2Ω⋅ cm( )=10.0 kVcm 
 
3.10 
 
j ≅ jn = qnv = 1.60x10−19C( ) 4x1015cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
107cm
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 6400
A
cm2 
 
3.11 
 
j ≅ jp = qpv = 1.60x10−19C( ) 5x1017cm3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
107cm
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 800
kA
cm2 
 
3.12 
 
jp = qμ p pE − qDp dpdx = 0 → E = −
Dp
μ p
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ 1p = −
kT
q
dp
dx
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
p
dp
dx
 
p(x) = N o exp − xL
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | 
1
p
 | E = −VT
L
= − 0.025V
10−4cm
= −250 V
cm
dp
dx
= 1
L
The exponential doping results in a constant electric field. 
 
3.13 
 
jp = qDn dndx = qμnVT
dn
dx
 | dn
dx
= 2000A / cm
2
1.60x10−19C( )500cm2 /V − s( )0.025V( )=
1.00 x 1021
cm4
 
 
3.14 
 10 =10
4 ⋅10−16 exp 40VD( )−1[ ]+ VD and the solver yields VD = 0.7464 V 
 
3.15 
 
f =10 −104 ID − 0.025ln ID + ISIS
 | f ' = −104 − 0.025
ID + IS
 | ID
' = ID − ff ' 
Starting the iteration process with ID = 100 μA and IS = 10-13A: 
 
ID f f' 
 
1.000E-04 8.482E+0
0 
-1.025E+04 
9.275E-04 1.512E-
01 
-1.003E+04 
9.426E-04 3.268E-
06 
-1.003E+04 
9.426E-04 9.992E-
16 
-1.003E+04
36 
3.16 
(a) Create the following m-file: 
function fd=current(id) 
 fd=10-1e4*id-0.025*log(1+id/1e-13); 
 Then: fzero('current',1) yields ans = 9.4258e-04 
(b) Changing IS to 10
-15
A: 
 function fd=current(id) 
 fd=10-1e4*id-0.025*log(1+id/1e-15); 
Then: fzero('current',1) yields ans = 9.3110e-04 
 
3.17 
 
T = qVT
k
= 1.60x10
−19C 0.025V( )
1.38x10−23 J / K
= 290 K 
 
3.18 
 
VT = kTq =
1.38x10−23 J / K( )T
1.60x10−19C
= 8.63x10−5T
For T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV
 
 
3.19 
 
Graphing ID = IS exp 40VDn
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ yields : 
1.21.00.80.60.40.20.0
0
1
2
3
4
5
6
(a)
(b)
(c)
 
 
 37
3.20 
 
nVT = n kTq =1.04
1.38x10−23 J / K( )300( )
1.60x10−19C
= 26.88 mV T = 26.88mV1.602x10
-19
1.38x10-23
= 312 K 
 
3.21 
 
iD = IS exp vDnVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ or 
vD
nVT
= ln 1+ iD
IS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
For iD >> IS , vDnVT
≅ ln iD
IS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ or ln ID( )= 1nVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ vD + ln IS( )
 
which is the equation of a straight line with slope 1/nVT and x-axis intercept at -ln (IS). The 
values of n and IS can be found from any two points on the line in the figure: e. g. iD = 10
-4
 A 
for vD = 0.60 V and iD = 10
-9
 A for vD = 0.20 V. Then there are two equations in two 
unknowns: 
 
ln 10-9( )= 40n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ .20 + ln IS( ) or 9.21= 8n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ln IS( )
ln 10-4( )= 40n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ .60 + ln IS( ) or 20.72 = 24n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ln IS( ) 
Solving for n and IS yields n = 1.39 and IS = 3.17 x 10
-12
 A = 3.17 pA. 
 
3.22 
 
VD = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | ID = IS exp
VD
nVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
(a) VD =1.05 0.025V( )ln 1+ 7x10−5 A10−18 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.837V | (b) VD =1.05 0.025V( )ln 1+
5x10−6 A
10−18 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.768V
 
(c) ID =10−18 A exp 01.05 ⋅ 0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = 0 A
(d) ID =10−18 A exp −0.075V1.05 ⋅ 0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −0.943x10
−19 A
(e) ID =10−18 A exp −5V1.05 ⋅ 0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −1.00x10
−18 A
 
 
3.23 
 
VD = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | ID = IS exp
VD
nVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
(a) VD = 0.025V ln 1+ 10
−4 A
10−17 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.748V | (b) VD = 0.025V ln 1+
10−5 A
10−17 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.691V 
 38 
(c) ID =10−17 A exp 00.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = 0 A
(d) ID =10−17 A exp −0.06V0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −0.909x10
−17 A
(e) ID =10−17 A exp −4V0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −1.00x10
−17 A
 
 
3.24 
 
ID = IS exp VDVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ =10
−17 A exp 0.675
0.025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = 5.32x10
−6 A = 5.32 μA
VD = VT ln IDIS
+1⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.025V( )ln 15.9x10
−6 A
10−17 A
+1⎛ ⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.703 V 
 
3.25 
 
VD = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 2 0.025V( )ln 1+ 40A10−10 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.34 V
VD = 2 0.025V( )ln 1+ 100A10−10 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.38 V 
 
3.26 
 
(a) IS = ID
exp VD
nVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
= 2mA
exp 0.82
0.025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
=1.14x10−17 A
(b) ID =1.14x10−17 A exp −50.025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −1.14x10
−17 A
 
 
3.27 
 
(a) IS = ID
exp VD
nVT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
= 300μA
exp 0.75
0.025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
= 2.81x10−17 A
(b) ID = 2.81x10−17 A exp −30.025
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = −2.81x10
−17 A
 
 
 39
3.28 
 
VD = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | 10
-14 ≤ IS ≤10−12 | VD = 0.025V( )ln 1+ 10−3 A10−12 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.518 V
VD = 0.025V( )ln 1+ 10−3 A10−14 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.633 V | So, 0.518 V ≤ VD ≤ 0.633 V 
 
3.29 
 
VT =
1.38x10−23 307( )
1.60x10−19
= 0.0264V | ID = IS exp VD0.0264n
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
Varying n and IS by trial-and-error with a spreadsheet: 
 
n IS 
 
1.039 7.606E-15 
 
VD ID-Measured ID-Calculated 
Error 
Squared 
 
0.500 6.591E-07 6.276E-07 9.9198E-16 
0.550 3.647E-06 3.885E-06 5.6422E-14 
0.600 2.158E-05 2.404E-05 6.0672E-12 
0.650 1.780E-04 1.488E-04 8.518E-10 
0.675 3.601E-04 3.702E-04 1.0261E-10 
0.700 8.963E-04 9.211E-04 6.1409E-10 
0.725 2.335E-03 2.292E-03 1.8902E-09 
0.750 6.035E-03 5.701E-03 1.1156E-07 
0.775 1.316E-02 1.418E-02 1.0471E-06 
 
 
Total Squared Error 
 
1.1622E-06 
 
3.30 
 
VT = kTq =
1.38x10−23 J / K( )T
1.60x10−19C
= 8.63x10−5T
For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV
 
 
3.31 
 
kT
q
= 1.38x10
−23 303( )
1.60x10−19
= 26.1 mV | VD = 0.0261V( )ln 1+ 10−32.5x10−16
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.757 V
ΔV = −1.8mV / K( )20K( )= −36.0 mV | VD = 0.757 − 0.036 = 0.721 V 
 
 40 
3.32 
 
kT
q
= 1.38x10
−23 298( )
1.602x10−19
= 25.67 mV | (a) VD = 0.02567V( )ln 1+ 10−410−15
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.650 V 
 
ΔV = −2.0mV / K( )25K( )= −50.0 mV
(b) VD = 0.650 − 0.050 = 0.600 V 
 
3.33 
 
kT
q
= 1.38x10
−23 298( )
1.602x10−19
= 25.67 mV | (a) VD = 0.02567V( )ln 1+ 2.5x10−410−14
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.615 V 
 
b( ) ΔV = −1.8mV / K( )60K( )= −50.0 mV VD = 0.615− 0.108 = 0.507 V
(c) ΔV = −1.8mV / K( )−80K( )= +144 mV VD = 0.615+ 0.144 = 0.758 V 
 
3.34 
 
dvD
dT
= vD −VG − 3VT
T
= 0.7 −1.21− 3 0.0259( )
300
= −1.96 mV
K 
 
 41
3.35 
 
IS2
IS1
= T2
T1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
3
exp − EG
k
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
T2
− 1
T1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ =
T2
T1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
3
exp EG
kT1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1−
T1
T2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ 
f x( )= x( )3 exp EGkT1
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1−
1
x
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ x =
T2
T1 
Using trial and error with a spreadsheet yields �T = 4.27 K, 14.6 K, and 30.7 K to increase the 
saturation current by 2X, 10X, and 100X respectively. 
 
x f(x) Delta T 
 
1.00000 1.00000 0.00000 
1.00500 1.27888 1.50000 
1.01000 1.63167 3.00000 
1.01500 2.07694 4.50000 
1.01400 1.97945 4.20000 
1.01422 2.00051 4.26600 
1.01922 2.54151 5.76600 
1.02422 3.22151 7.26600 
1.02922 4.07433 8.76600 
1.03422 5.14160 10.26600 
1.03922 6.47438 11.76600 
1.04422 8.13522 13.26600 
1.04922 10.20058 14.76600 
1.04880 10.00936 14.64000 
1.10000 90.67434 30.00000 
1.10239 100.0012 30.71610 
 
3.36 
 
wd = wdo 1+ VRφ j | (a) wd =1μm 1+
5
0.8
= 2.69 μm (b) wd =1μm 1+ 100.8 = 3.67 μm 
 
3.37 
 
φ j = VT ln N A N Dni2
= 0.025V( )ln 1016cm−3( )1015cm−3( )1020cm−6 = 0.633 V
wdo = 2εsq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1016cm−3
+ 1
1015cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.633V( )
wdo = 0.949 μm | wd = wdo 1+ VRφ j
wd = 0.949μm 1+ 10V0.633V = 3.89 μm | wd = 0.949μm 1+
100V
0.633V
= 12.0 μm
 
 
 42
3.38 
 
φ j = VT ln N A N Dni2 = 0.025V( )ln
1018cm−3( )1020cm−3( )
1020cm−6
=1.04 V
wdo = 2εsq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( )
1.602x10−19C
1
1018cm−3
+ 1
1020cm−3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 1.04V( )
wdo = 0.0368 μm | wd = wdo 1+ VRφ j
wd = 0.0368μm 1+ 51.04 = 0.0887 μm | wd = 0.0368μm 1+
25
1.04
= 0.184 μm
 
 
3.39 
 
Emax =
2 φ j + VR( )
wd
= 2 φ j + VR( )
wdo 1+ VRφ j
= 2φ j
wdo
1+ VRφ j
3x105 V
cm
= 2 0.6V( )
10−4cm
1+ VR
0.6
→ VR = 374 V
 
 
3.40 
 
E = 2φ j
wdo
= 2 0.748V( )
0.984x10−4cm
=15.2 kV
cm
 | φ j + VR = Emax2
wdo
φ j
=
3x105 V
cm
0.984x10−4cm( )
2 0.748V
VR = 291.3− 0.748 = 291 V 
 
3.41 
VZ = 4 V; RZ = 0 Ω since the reverse breakdown slope is infinite. 
 
3.42 
Since NA >> ND, the depletion layer is all on the lightly-doped side of the junction. Also, VR 
>> φj, so φj can be neglected. 
 
Emax = qN A x pεS =
qN Awd
εS =
qN A
εS
2εS
q
VR
N A
N A = Emax
2 εS
2qVR
= 3x10
5( )2 11.7( )8.854x10−14( )
2 1.602x10−19( )1000 = 2.91 x 1014 / cm3
 
 
 43
3.43 
 
φ j = VT ln N AN Dni2 = 0.025ln
10151020
1020
= 0.864V
wdo = 2εSq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7( )8.854x10−14( )
1.602x10−19
1
1015
+ 1
1020
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.864 =1.057x10
−4cm
Cjo
" = εS
wdo
=
11.7 8.854x10−14( )
1.057x10−4
= 9.80x10-9 F / cm2 | Cj = Cjo
" A
1+ VRφ j
= 9.80x10
-9 0.05( )
1+ 5
0.864
=188 pF
 
3.44 
 
φ j = VT ln N AN Dni2 = 0.025ln
10181015
1020
= 0.748V
wdo = 2εSq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7( )8.854x10−14( )
1.602x10−19
1
1018
+ 1
1015
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.748 = 0.984x10
−4cm
Cjo
" = εS
wdo
=
11.7 8.854x10−14( )
0.984x10−4
= 10.5x10-9 F / cm2 | Cj = Cjo
" A
1+ VRφ j
= 10.5x10
-9 0.02( )
1+ 10
0.748
= 55.4 pF
 
3.45 
 
(a) CD = IDτTVT =
10−4 A 10−10 s( )
0.025V
= 400 fF (b) Q = IDτT =10−4 A 10−10 s( )=10 fC
(c) CD =
25x10−3 A 10−10 s( )
0.025V
=100 pF | Q = IDτT = 5x10−3 A 10−10 s( )= 0.50 pC 
 
3.46 
 
(a) CD = IDτTVT =
1A 10−8 s( )
0.025V
= 0.400 μF (b) Q = IDτT =1A 10−8 s( )=10.0 nC
(c) CD =
100mA 10−8 s( )
0.025V
= 0.04 μF | Q = IDτT =100mA 10−8 s( )=1.00 nC 
 
 44 
3.47 
 
φ j = VT ln N AN Dni2 = 0.025ln
10191017
1020
= 0.921V
wdo = 2εSq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7( )8.854x10−14( )
1.602x10−19
1
1019
+ 1
1017
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.921 = 0.110 μm
Cjo = εS Awdo =
11.7 8.854x10−14( )10−4( )
0.110x10−4
= 9.42 pF / cm2 | Cj = Cjo
1+ VRφ j
= 9.42 pF
1+ 5
0.921
= 3.72 pF
 
 
3.48 
 
φ j = VT ln N AN Dni2 = 0.025ln
10191016
1020
= 0.864V
wdo = 2εSq
1
N A
+ 1
N D
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ φ j =
2 11.7( )8.854x10−14( )
1.602x10−19
1
1019
+ 1
1016
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0.864 = 0.334μm
Cjo = εS Awdo =
11.7 8.854x10−14( )0.25cm2( )
0.334x10−4
= 7750 pF | Cj = Cjo
1+ VRφ j
= 7750 pF
1+ 3
0.864
= 3670 pF
 
 
3.49 
VDC
RFC
C
10 μH 10 μH
C
L =
 
 
C = Cjo
1+ VRφ j
 (a) C = 39 pF
1+ 1V
0.75V
= 25.5 pF | fo = 1
2π LC =
1
2π 10−5 H( )25.5 pF = 9.97MHz
(b) C = 39 pF
1+ 10V
0.75V
=10.3pF | fo = 1
2π LC =
1
2π 10−5 H( )10.3pF =15.7 MHz
 
 
3.50 
 
(a) VD = 0.025V( )ln 1+ 50A10−7 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.501 V | (b) VD = 0.025V( )ln 1+ 50A10−15 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.961 V 
 
 45
3.51 
 
(a) VD = 0.025V( )ln 1+ 4x10−3 A10−11 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.495 V | (b) VD = 0.025V( )ln 1+
4x10−3 A
10−14 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.668 V 
 
3.52 
 
RS = Rp + Rn Rp = ρ p LpAp = 1Ω − cm( )
0.025cm
0.01cm2
= 2.5Ω
Rn = ρn LnAn = 0.01Ω − cm( )
0.025cm
0.01cm2
= 0.025Ω RS = 2.53 Ω
 
 
3.53 
 
(a) VD
' = VT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.025V( )ln 1+ 10
−3
5x10−16
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.708V
VD = VD' + ID RS = 0.708V +10−3 A 10Ω( )= 0.718 V
(b) VD = VD' + ID RS = 0.708V +10−3 A 100Ω( )= 0.808 V
 
 
3.54 
 
ρc =10Ω − μm2 Ac =1μm2 RC = ρcAc =
10Ω − μm2
1μm2 =10Ω / contact
5 anode contacts and 14 cathode contacts
Resistance of anode contacts = 10Ω
5
= 2Ω
Resistance of cathode contacts = 10Ω
14
= 0.71Ω
 
 
3.55 
(a) From Fig. 3.21a, the diode is approximately 10.5 μm long x 8 μm wide. Area = 84 μm2. 
(b) Area = (10.5x0.13 μm) x (8x0.13μm) = 1.42 μm2. 
 
 46 
3.56 
 
(a) 5 =104 ID + VD | VD = 0 ID = 0.500mA | ID = 0 VD = 5V
Forward biased - VD = 0.5 V ID = 4.5V104Ω = 0.450 mA
(b) − 6 = 3000ID + VD | VD = 0 ID = −2.00mA | ID = 0 VD = −6V
In reverse breakdown - VD = −4 V ID = −2V3kΩ = −0.667 mA
(c) − 3 = −3000ID + VD | VD = 0 ID = −1.00mA | ID = 0 VD = −3V
Reverse biased - VD = −3 V ID = 0 
1 2 3 4
1 mA
2 mA
(a) Q-point
vD
65
-1-2-3-4-5-6
-1 mA
-2 mA
(b) Q-point
iD
(c) Q-point
 
 
3.57 
 
(a) 10 = 5000ID + VD | VD = 0 ID = 2.00 mA | VD = 5 V ID =1.00 mA
Forward biased - VD = 0.5V ID = 9.5V5kΩ =1.90 mA
(b) -10 = 5000ID + VD | VD = 0 ID = −2.00 mA | VD = −5 V ID = −1.00 mA
In reverse breakdown - VD = −4V ID = −6V5kΩ = −1.20 mA
(c) − 2 = 2000ID + VD | VD = 0 ID = −1.00 mA | ID = 0 VD = −2 V
Reverse biased - VD = −2 V ID = 0 
 
1 2 3 4
1 mA
2 mA (a) Q-point
vD
65
-1-2-3-4-5-6
-1 mA
-2 mA
(b) Q-point
iD
(c) Q-point
 
 
 47
3.58 
*Problem 3.58 - Diode Circuit SPICE Results 
V 1 0 DC 5 
R 1 2 10K VD = 0.693 V 
D1 2 0 DIODE1 ID = 0.431 μA 
.OP 
.MODEL DIODE1 D IS=1E-15 
.END 
 
3.59 
 
(a) −10 =104 ID + VD | VD = 0 ID = −1.00 mA | VD = −5 V ID = −0.500 mA
In reverse breakdown - VD = −4 V ID = −10 − (−4)V10kΩ = −0.600 mA
(b) 10 =104 ID + VD | VD = 0 ID =1.00 mA | VD = 5 V ID = 0.500 mA
Forward biased - VD = 0.5 V ID = 10 − 0.5V10kΩ = 0.950 mA
(c) − 4 = 2000ID + VD | VD = 0 ID = −2.00 mA | ID = 0 VD = − 4 V 
Reverse biased - VD = −4 V ID = 0 
1 2 3 4
1 mA
2 mA
(b) Q-point
vD
65
-1-2-3-4-5-6
-1 mA
-2 mA
(c) Q-point
iD
(a) Q-point
 
 
 
 
 48 
1 2 3 4 5 6 7
0.002
0.001
-0.001
-0.002
iD (A)
vD (V)-7 -6 -5 -4 -3 -2 -1
 
 49
 
3.60 
+
-V
R
iD
vD
+
-
 
The load line equation: V = iD R + vD We need two points to plot the load line. 
(a) V = 6 V and R = 4 kΩ: For vD = 0, iD = 6V/4 kΩ = 1.5 mA and for iD = 0, vD = 6V. 
 Plotting this line on the graph yields the Q-pt: (0.5 V, 1.4 mA). 
(b) V = -6 V and R = 3 kΩ: For vD = 0, iD = -6V/3 kΩ = -2 mA and for iD = 0, vD = -6V. 
 Plotting this line on the graph yields the Q-pt: (-4 V, -0.67 mA). 
(c) V = -3 V and R = 3 kΩ: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) 
(d) V = +12 V and R = 8 kΩ: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) 
(e) V = -25 V and R = 10 kΩ: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA) 
1 2 3 4 5 6 7-1-2-3-4-5 -6 -7
.001
.002
-.001
-.002
i (A)
D
v (V)
D
Q-Point 
(-4V,-0.67 mA)
Q-Point 
(0.5V,1.4 mA) Load line for (a)
Load line for (b)
Q-Point 
(-4V,-2.1 mA)
(e)
(c)
Q-Point 
(-3V,0 mA)
(d)
Q-Point 
(0.5V,1.45
mA)
 
 
 50 
3.61 
Using the equations from Table 3.1, (f = 10-10
-9 exp ..., etc.) 
VD = 0.7 V requires 12 iterations, VD = 0.5 V requires 22 iterations, 
VD = 0.2 V requires 384 iterations - very poor convergence because the second iteration (VD = 
9.9988 V) is very bad. 
 
3.62 
Using Eqn. (3.28), 
 
V = iD R + VT ln iDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ or 10 =10
4 iD + 0.025ln 1013iD( ). 
We want to find the zero of the function f = 10 −10
4iD − 0.025ln 1013iD( ) 
 
iD f 
.001 -0.576 
.0001 8.48 
.0009 0.427 
.00094 0.0259 - converged 
 
3.63 
f =10 −104 ID − 0.025ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | f
' = −104 − 0.025
ID + IS 
 
x f(x) f'(x) 
 
1.0000E+00 -9.991E+03 -1.000E+04 
9.2766E-04 1.496E-01 -1.003E+04 
9.4258E-04 3.199E-06 -1.003E+04 
9.4258E-04 9.992E-16 -1.003E+04 
9.4258E-04 9.992E-16 -1.003E+04 
 
3.64 
Create the following m-file: 
function fd=current(id) 
 fd=10-1e4*id-0.025*log(1+id/1e-13); 
Then: fzero('current',1) yields ans = 9.4258e-04 + 1.0216e-21i 
 
 51
3.65 
The one-volt source will forward bias the diode. Load line: 
 1 =10
4 ID + VD | ID = 0 VD =1V | VD = 0 ID = 0.1mA → 50 μA, 0.5 V( ) 
Mathematical model: f =1−10
−9 exp 40VD( )−1[ ]+ VD → 49.9 μA, 0.501 V( ) 
Ideal diode model: ID = 1V/10kΩ = 100 μA; (100 μA, 0 V) 
Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0μA; (40.0 μA, 0.6 V) 
 
3.66 
Using Thévenin equivalent circuits yields and then combining the sources 
+
-
+-
+
-
V
I 1 k Ω1.2 k Ω
1.5 V1.2 V
 
0.3 V
+-
+
-
V
I 2.2 k Ω
 
(a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will 
assume it is "on". Substituting the ideal diode model for the forward region yields 
 
I =
2.2kΩ
0.3V = 0.136 mA. This current is greater than zero, which is consistent with the diode 
being "on". Thus the Q-pt is (0 V, +0.136 mA). 
Ideal Diode: 
0.3 V
+-
+
-
V
I
2.2 k Ω
 CVD: 
0.3 V
+
-
0.6 V
I
2.2 k Ω
+-
on
V
 
(b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it 
is "on". Substituting the CVD model with Von = 0.6 V yields 
I =
2.2kΩ
0.3V − 0.6V = −136 μA. 
This current is negative which is not consistent with the assumption that the diode is "on". 
Thus the diode must be off. The resulting Q-pt is: (0 mA, -0.3 V). 
0.3 V+
-
I=0 2.2 k Ω- +V
 
 52 
(c) The second estimate is more realistic. 0.3 V is not sufficient to forward bias the diode into 
significant conduction. For example, let us assume that IS = 10
-15
 A, and assume that the full 
0.3 V appears across the diode. Then 
 
iD =10−15 A exp 0.3V0.025V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ −1
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ =163 pA, a very small current. 
 
3.67 
The nominal values are: 
 
VA = 3V R2R1 + R2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 3V
2kΩ
2kΩ + 3kΩ
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.20V and RTHA =
R1R2
R1 + R2
= 2kΩ 3kΩ( )
2kΩ + 3kΩ =1.20kΩ
VC = 3V R4R3 + R4
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 3V
2kΩ
2kΩ + 2kΩ
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.50V and RTHC =
R3R4
R3 + R4
= 2kΩ 2kΩ( )
2kΩ + 2kΩ =1.00kΩ
ID
nom = 1.50 −1.20
1.20 +1.00
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
V
kΩ =136 μA
 
For maximum current, we make the Thévenin equivalent voltage at the diode anode as large as 
possible and that at the cathode as small as possible. 
 
VA = 3V
1+ R1
R2
= 3V
1+ 2kΩ 0.9( )
2kΩ 1.1( )
=1.65V and RTHA = R1R2R1 + R2
= 2kΩ 0.9( )2kΩ 1.1( )
2kΩ 0.9( )+ 2kΩ 1.1( )= 0.990kΩ
VC = 3V
1+ R3
R4
= 3V
1+ 3kΩ 1.1( )
2kΩ 0.9( )
=1.06V and RTHC = R3R4R3 + R4
= 3kΩ 1.1( )2kΩ 0.9( )
3kΩ 1.1( )+ 2kΩ 0.9( )=1.17kΩ
ID
max = 1.65−1.06
0.990 +1.17
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
V
kΩ = 274 μA
 
For minimum current, we make the Thévenin equivalent voltage at the diode anode as small as 
possible and that at the cathode as large as possible. 
 
VA = 3V
1+ R1
R2
= 3V
1+ 2kΩ 1.1( )
2kΩ 0.9( )
=1.350V and RTHA = R1R2R1 + R2
= 2kΩ 1.1( )2kΩ 0.9( )
2kΩ 1.1( )+ 2kΩ 0.9( )= 0.990kΩ
VC = 3V
1+ R3
R4
= 3V
1+ 3kΩ 0.9( )
2kΩ 1.1( )
=1.347V and RTHC = R3R4R3 + R4
= 3kΩ 0.9( )2kΩ 1.1( )
3kΩ 0.9( )+ 2kΩ 1.1( )=1.21kΩ
ID
min = 1.350 −1.347
0.990 +1.21
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
V
kΩ =1.39 μA ≅ 0
 
 
 53
3.68 
SPICE Input Results 
*Problem 3.68 NAME D1 
V1 1 0 DC 4 MODEL DIODE 
R1 1 2 2K ID 1.09E-10 
R2 2 0 2K VD 3.00E-01 
R3 1 3 3K 
R4 3 0 2K 
D1 2 3 DIODE 
.MODEL DIODE D IS=1E-15 RS=0 
.OP 
.END 
The diode is essentially off - VD = 0.3 V and ID = 0.109 nA. This result agrees with the CVD 
model results. 
 
3.69 (a) 
 
(a) Diode is forward biased :V = 3− 0 = 3 V | I = 3− −7( )
16kΩ = 0.625 mA
(b) Diode is forward biased :V = −5 + 0 = −5 V | I = 5− −5( )
16kΩ = 0.625 mA
(c) Diode is reverse biased : I = 0 | V = −5+16kΩ I( )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD = −10 V 
(b) 
 
(a) Diode is forward biased :V = 3 − 0.7 = 2.3 V | I = 2.3 − −7( )
16kΩ = 0.581 mA
(b) Diode is forward biased :V = −5 + 0.7 = −4.3 V | I = 5− −4.3( )
16kΩ = 0.581 mA
(c) Diode is reverse biased : I = 0 | V = −5+16kΩ I( )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD = −10 V
 
 
3.70 (a) 
 
(a) Diode is forward biased :V = 3 − 0 = 3 V | I = 3− −7( )
100kΩ =100 μA
(b) Diode is forward biased :V = −5 + 0 = −5 V | I = 5− −5( )
100kΩ =100 μA
(c) Diode is reverse biased : I = 0 A | V = −5+100kΩ I( )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 A | V = 7 −100kΩ I( )= 7 V | VD = −10 V 
(b) 
 54 
 
(a) Diode is forward biased :V = 3 − 0.6 = 2.4 V | I = 2.4 − −7( )
100kΩ = 94.0 μA
(b) Diode is forward biased :V = −5 + 0.6 = −4.4 V | I = 5− −4.4( )
100kΩ = 94.0 μA
(c) Diode is reverse biased : I = 0 | V = −5+100kΩ I( )= −5 V | VD = −10 V
(d ) Diode is reverse biased : I = 0 | V = 7 −100kΩ I( )= 7 V | VD = −10 V
 
 
3.71 (a) 
 
(a) D1 on, D2 on : ID2 =
0 − −9( )
22kΩ = 409μA | ID1 = 409μA −
6 − 0
43kΩ = 270μA
 D1 : 409 μA, 0 V( ) D2 : 270 μA, 0 V( )
(b) D1 on, D2 off : ID2 = 0 | ID1 = 6 − 043kΩ =140 μA | VD2 = −9 − 0 = −9V
 D1 : 140 μA, 0 V( ) D2 : 0 A, − 9 V( ) 
 
(c) D1 off, D2 on : ID1 = 0 | ID2 =
6 − −9( )
65kΩ = 230 μA | VD1 = 6 − 43x10
3 ID2 = −3.92 V
 D1 : 0 A,−3.92 V( ) D2 : 230 μA,0 V( )
(d ) D1 on, D2 on : ID2 =
0 − −6( )
43kΩ =140 μA | ID1 =
9 − 0
22kΩ −140 μA = 270 μA
 D1 : 140 μA,0 V( ) D2 : 270 μA,0 V( ) 
 
(b) 
 
 
(a) D1 on, D2 on :
ID2 =
-0.75− 0.75− −9( )
22kΩ = 341μA | ID1 = 341μA −
6 − −0.75( )
43kΩ =184μA
D1 : 184 μA, 0.75 V( ) D2 : 341 μA, 0.75 V( )
(b) D1 on, D2 off :
ID2 = 0 | ID1 = 6 − 0.7543kΩ =122μA | VD2 = −9 − 0.75 = −9.75V
D1 : 122 μA, 0.75 V( ) D2 : 0 A, − 9.75 V( )
 
 55
 
(c) D1 off, D2 on :
ID1 = 0 | ID2 =
6 − 0.75− −9( )
65kΩ = 219μA | VD1 = 6 − 43x10
3 ID2 = −3.43V
D1 : 0 A, − 3.43 V( ) D2 : 219 μA, 0.75 V( )
(d) D1 on, D2 on :
ID2 =
0.75− 0.75− −6( )
43kΩ =140μA | ID1 =
9 − 0.75
22kΩ − 400μA = 235μA
D1 : 235 μA, 0.75 V( ) D2 : 140 μA, 0.75 V( )
 
 
3.72 (a) 
 
(a) D1 and D2 forward biased
ID2 =
0 − −9( )
15
V
kΩ = 600μA ID1 = ID2 −
6 − 0( )
15
V
kΩ = 200μA
D1 : 0 V, 200 μA( ) D2 : 0 V, 600 μA( )
 
 
 
(b) D1 forward biased, D2 reverse biased 
ID1 = 6 − 015
V
kΩ = 400μA VD2 = −9 − 0 = −9 V
D1 : 0 V, 400 μA(
) D2 : -9 V, 0 A( )
 
 
 
(c) D1 reverse biased, D2 forward biased
ID2 =
6V − −9V( )
30kΩ = 500μA VD1 = 6 −15000ID2 = −1.50V
D1 : −1.50 V, 0 A( ) D2 : 0 V, 500 μA( )
 
 
 
(d) D1 and D2 forward biased
ID2 =
0 − −6( )
15
V
kΩ = 400μA ID1 =
9 − 0( )
15
V
kΩ − ID2 = 200μA
D1 : 0 V, 200 μA( ) D2 : 0 V, 400 μA( )
 
 
(b) 
 
 
(a) D1 on, D2 on :
ID2 =
-0.75 − 0.75 − −9( )
15kΩ = 500μA | ID1 = 500μA −
6 − −0.75( )
15kΩ = 50.0μA
D1 : 50.0 μA, 0.75 V( ) D2 : 500 μA, 0.75 V( ) 
 56 
 
 
(b) D1 on, D2 off :
ID2 = 0 | ID1 = 6 − 0.7515kΩ = 350μA | VD2 = −9 − 0.75 = −9.75V
D1 : 350 μA, 0.75 V( ) D2 : 0 A, − 9.75 V( )
 
 
 
(c) D1 off, D2 on :
ID1 = 0 | ID2 =
6 − 0.75 − −9( )
30kΩ = 475μA | VD1 = 6 −15x10
3 ID2 = −1.13V
D1 : 0 A, −1.13 V( ) D2 : 475 μA, 0.75 V( ) 
 
(d) D1 on, D2 on :
ID2 =
0.75 − 0.75 − −6( )
15kΩ = 400μA | ID1 =
9 − 0.75
15kΩ − 400μA =150μA
D1 : 150 μA, 0.75 V( ) D2 : 400 μA, 0.75 V( )
 
 
3.73 Diodes are labeled from left to right
 
 
(a) D1 on, D2 off, D3 on : ID2 = 0 | ID1 = 10 − 03.3kΩ + 6.8kΩ = 0.990mA
ID3 + 0.990mA =
0 − −5( )
2.4kΩ → ID3 =1.09mA | VD2 = 5− 10 − 3300ID1( )= −1.73V
D1 : 0.990 mA, 0 V( ) D2 : 0 mA, −1.73 V( ) D3 : 1.09 mA, 0 V( ) 
 
 
(b) D1 on, D2 off, D3 on : ID2 = 0 | ID3 = 0
ID1 =
10 − 0( )V
8.2kΩ +12kΩ = 0.495mA | VD2 = 5− 10 − 8200ID1( )= −0.941V
ID3 =
0 − −5V( )
10kΩ − ID1 = 0.005mA
D1 : 0.495 mA, 0 V( ) D2 : 0 A, − 0.941 V( ) D3 : 0.005 mA, 0 V( ) 
 
 57
 
(c) D1 on, D2 on, D3 on
ID1 =
0 − −10( )
8.2kΩ V =1.22mA > 0 | I12K =
0 − 2( )
12kΩ V = −0.167mA | ID2 = ID1 + I12K =1.05mA > 0
I10K =
2 − −5( )
10kΩ V = 0.700mA | ID3 = I10K − I12K = 0.533mA > 0
D1 : 1.22 mA, 0 V( ) D2 : 1.05 mA, 0 V( ) D3 : 0.533 mA, 0 V( ) 
 
(d) D1 off, D2 off, D3 on : ID1 = 0, ID2 = 0
ID3 =
12 − −5( )
4.7 + 4.7 + 4.7
V
kΩ =1.21mA > 0 | VD1 = 0 − −5+ 4700ID3( )= −0.667V < 0
VD2 = 5− 12 − 4700ID3( )= −1.33V < 0
D1 : 0 A, − 0.667 V( ) D2 : 0 A, −1.33 V( ) D3 : 1.21 mA, 0 V( ) 
 
3.74 Diodes are labeled from left to right 
 
(a) D1 on, D2 off, D3 on : ID2 = 0 | ID1 =
10 − 0.6 − −0.6( )
3.3kΩ + 6.8kΩ = 0.990mA
ID3 + 0.990mA =
−0.6 − −5( )
2.4kΩ → ID3 = 0.843mA | VD2 = 5− 10 − 0.6 − 3300ID1( )= −1.13V
D1 : 0.990 mA, 0.600 V( ) D2 : 0 A, −1.13 V( ) D3 : 0.843 mA, 0.600V( ) 
 
 
(b) D1 on, D2 off, D3 off : ID2 = 0 | ID3 = 0
ID1 =
10 − 0.6 − −5( )
8.2kΩ +12kΩ +10kΩV = 0.477mA | VD2 = 5− 10 − 0.6 − 8200ID1( )= −0.490V
VD3 = 0 − −5+10000ID1( )= +0.230V < 0.6V so the diode is off
D1 : 0.477 mA, 0.600 V( ) D2 : 0 A, − 0.490 V( ) D3 : 0 A, 0.230 V( ) 
 
(c) D1 on, D2 on, D3 on
ID1 =
−0.6 − −9.4( )
8.2
V
kΩ =1.07mA > 0 | I12K =
−0.6 − 1.4( )
12
V
kΩ = −0.167mA
ID2 = ID1 + I12K = 0.906mA > 0 | I10K =
1.4 − −5( )
10
V
kΩ = 0.640mA | ID3 = I10K − I12K = 0.807mA > 0
D1 : 1.07 mA, 0.600 V( ) D2 : 0.906 mA, 0.600 V( ) D3 : 0.807 mA, 0.600 V( )
 
 58 
 
(d) D1 off, D2 off, D3 on : ID1 = 0, ID2 = 0
ID3 =
11.4 − −5( )
4.7 + 4.7 + 4.7
V
kΩ =1.16mA > 0 | VD1 = 0 − −5+ 4700ID3( )= −0.452V < 0
VD2 = 5− 11.4 − 4700ID3( )= −0.948V < 0
D1 : 0 A, − 0.452 V( ) D2 : 0 A, − 0.948 V( ) D3 : 1.16 mA, 0.600 V( ) 
 
3.75 
*Problem 3.75(a) (Similar circuits are used for the other three cases.) 
V1 1 0 DC 10 
V2 4 0 DC 5 
V3 6 0 DC -5 
R1 2 3 3.3K 
R2 3 5 6.8K 
R3 5 6 2.4K 
D1 1 2 DIODE 
D2 4 3 DIODE 
D3 0 5 DIODE 
.MODEL DIODE D IS=1E-15 RS=0 
.OP 
.END 
 
NAME D1 D2 D3 
MODEL DIODE DIODE DIODE 
ID 9.90E-04 -1.92E-12 7.98E-04 
VD 7.14E-01 -1.02E+00 7.09E-01 
 
NAME D1 D2 D3 
MODEL DIODE DIODE DIODE 
ID 4.74E-04 -4.22E-13 2.67E-11 
VD 6.95E-01 -4.21E-01 2.63E-01 
 
NAME D1 D2 D3 
MODEL DIODE DIODE DIODE 
ID 8.79E-03 1.05E-03 7.96E-04 
VD 7.11E-01 7.16E-01 7.09E-01 
 
NAME D1 D2 D3 
MODEL DIODE DIODE DIODE 
ID -4.28E-13 -8.55E-13 1.15E-03 
VD -4.27E-01 -8.54E-01 7.18E-01 
 
For all cases, the results are very similar to the hand analysis. 
 
 
3.76 
 59
 
ID1 =
10 − −20( )
10kΩ +10kΩ =1.50mA | ID2 = 0
ID3 =
0 − −10( )
10kΩ =1.00mA | VD2 =10 −10
4 ID1 − 0 = −5.00V
D1 : 1.50 mA, 0 V( ) D2 : 0 A,−5.00 V( ) D3 : 1.00 mA, 0 V( )
 
 
3.77 
*Problem 3.77 
V1 1 0 DC -20 
V2 4 0 DC 10 
V3 6 0 DC -10 
R1 1 2 10K 
R2 4 3 10K 
R3 5 6 10K 
D1 3 2 DIODE 
D2 3 5 DIODE 
D3 0 5 DIODE 
.MODEL DIODE D IS=1E-14 RS=0 
.OP 
.END 
NAME D1 D2 D3 
MODEL DIODE DIODE DIODE 
ID 1.47E-03 -4.02E-12 9.35E-04 
VD 6.65E-01 -4.01E+00 6.53E-01 
The simulation results are very close to those given in Ex. 3.8. 
 
3.78 
 
VTH = 24V 3.9kΩ3.9kΩ +11kΩ = 6.28V | RTH =11kΩ 3.9kΩ = 2.88kΩ
IZ = 6.28 − 42.88kΩ = 0.792mA > 0 | IZ ,VZ( )= 0.792 mA,4 V( ) 
 
 60 
3.79 
 
−6.28 = 2880ID + VD | ID = 0,VD = −6.28V | VD = 0, ID = −6.282880 = −2.18mA 
-1 mA
-2 mA
Q-point
vD
i
D
-6 -5 -4 -3 -2 -1
 Q-Point: (-0.8 mA, -4 V) 
 
3.80 
 
IS = 27 − 915kΩ =1.20mA → IL <1.20 mA | RL >
9V
1.2mA
= 7.50 kΩ
 
 
3.81 
 
IS = 27 − 915kΩ =1.20mA | P = 9V( )1.20mA( )=10.8 mW 
 
3.82 
 
IZ = VS −VZRS
− VZ
RL
= VS
RS
−VZ 1RS
+ 1
RL
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | PZ = VZ IZ
IZ
nom = 30V
15kΩ − 9V
1
15kΩ +
1
10kΩ
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.500 mA | PZ
nom = 9V 0.500mA( )= 4.5 mW
IZ
max = 30V 1.05( )
15kΩ 0.95( )− 9V 0.95( )
1
15kΩ 0.95( )+
1
10kΩ(1.05)
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 0.796 mA
PZ
max = 9V .95( )0.796mA( )= 6.81 mW
IZ
min = 30V 0.95( )
15kΩ 1.05( )− 9V 1.05( )
1
15kΩ 1.05( )+
1
10kΩ(0.95)
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 0.215 mA
PZ
min = 9V 1.05( )0.215mA( )= 2.03 mW
 
 
3.83 
 
(a) VTH = 60V 100Ω150Ω +100Ω = 24.0V | RTH =150Ω 100Ω = 60Ω | IZ =
24 −15
60
=150 mA
P = 15IZ = 2.25 W | (b) IZ = 60 −15150 = 300 mA | P = 15IZ = 4.50 W
 
 
 61
3.84 
 
IZ = VS −VZRS
− VZ
RL
= VS
RS
−VZ 1RS
+ 1
RL
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | PZ = VZ IZ
IZ
nom = 60 −15( )V
150Ω −
15V
100Ω =150 mA | PZ
nom =15V 150mA( )= 2.25 W
IZ
max = 60V 1.1( )
150Ω 0.90( )−15V 0.90( )
1
150Ω 0.90( )+
1
100Ω(1.1)
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 266 mA
PZ
max =15V 0.90( )266mA( )= 3.59 W
IZ
min = 60V 0.90( )
150Ω 1.1( ) −15V 1.1( )
1
150Ω 1.1( )+
1
100Ω(0.9)
⎛ 
⎝ 
⎜ ⎜ 
⎞ 
⎠ 
⎟ ⎟ = 43.9 mA
PZ
min =15V 1.1( )43.9mA( )= 0.724 W
 
 
3.85 
Using MATLAB, create the following m-file with f = 60 Hz: 
function f=ctime(t) 
f=5*exp(-10*t)-6*cos(2*pi*60*t)+1; 
Then: fzero('ctime',1/60) yields ans = 0.01536129698461 
 and �T = (1/60)-0.0153613 = 1.305 ms. 
 
ΔT = 1
120π
2Vr
VP
 | Vr = ITC =
5
0.1 60( )= 0.8333V
ΔT = 1
120π
2 0.8333( )
6
 = 1.40 ms
 
 
3.86 
 
VD = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 2 0.025V( )ln 1+ 48.6A10−9 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1.230 V
62 
3.87 
 
Von = nVT ln 1+ IDIS
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ | VD = Von + ID RS
VD =1.6 0.025V( )ln 1+ 100A10−8 A
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ +100A 0.01Ω( )=1.92 V
Pjunction ≅ VonIDC = Von IPΔT2T = 0.92V
100A
2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1ms
16.7ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 2.75 W
PR ≅ 43
T
ΔT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ IDC
2 RS = 43
16.7ms
1ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 3A( )
2
0.01Ω = 2.00 W
Ptotal = 4.76 W 
 
3.88 
 
VDC = 1T v t( )dt0
T∫ = 1T VP −Von( )T − TVr2
⎡ 
⎣ ⎢ 
⎤ 
⎦ ⎥ = VP −Von( )−
0.05 VP −Von( )
2
⎡ 
⎣ 
⎢ 
⎢ 
⎤ 
⎦ 
⎥ 
⎥ = 0.975 VP −Von( )
VDC = 0.975 18V( )=17.6 V 
 
3.89 
 
PD = 1T iD
2 t( )RS dt
0
T∫ = 1T IP2 1− tΔT
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
2
RS dt
0
ΔT∫
PD = IP
2 RS
T
1− 2tΔT +
t 2
ΔT 2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
2
dt
0
ΔT∫ = IP2 RST t − t
2
ΔT +
t 3
3ΔT 2
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 0
ΔT
PD = IP
2 RS
T
ΔT − ΔT + ΔT
3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =
1
3
IP
2 RS
ΔT
T
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
 
 
3.90 
Using SPICE with VP = 10 V. 
0s 10ms 20ms 30ms 40ms 50ms
t
15V
10V
5V
0V
-5V
-10V
-15V
Voltage
 
 
 63
3.91 
 
(a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.55 10.5 160 =1.05F
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d) Isurge = ωCVP = 2π 60( )1.05( )6.3 2( )= 3530 A
(e) ΔT = 1ω
2Vr
VP
= 1
2π 60( )
2 .25( )
6.3 2
= 0.628ms | IP = Idc 2TΔT =
7.91
.5
2
60
1
.628ms
= 841A
 
 
3.92 
 
VO
nom = − VP −Von( )= − 6.3 2 −1( )= −7.91V
VO
max = − VPmax −Von( )= − 6.3 1.1( ) 2 −1[ ]= −8.80V
VO
min = − VPmin −Von( )= − 6.3 0.9( ) 2 −1[ ]= −7.02V 
 
3.93 
*Problem 3.93 
VS 1 0 DC 0 AC 0 SIN(0 10 60) 
D1 2 1 DIODE 
R 2 0 0.25 
C 2 0 0.5 
.MODEL DIODE D IS=1E-10 RS=0 
.OPTIONS RELTOL=1E-6 
.TRAN 1US 80MS 
.PRINT TRAN V(1) V(2) I(VS) 
.PROBE V(1) V(2) I(VS) 
.END 
 
V(2) *REAL(Rectifier)*
Time (s)Circuit3_93b-Transient-8
+0.000e+000 +10.000m +20.000m +30.000m +40.000m +50.000m +60.000m +70.000m
-10.000
-5.000
+0.000e+000
+5.000
+10.000
SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A 
 
Vdc = − VP −Von( )= − 10 −1( )= −9.00V | Vr = I TC = 9.00V0.25Ω 160s 10.5F =1.20V
ISC = ωCVP = 2π 60( )0.5( )10( )=1890A | ΔT = 1ω 2VrVP =
1
2π 60( )
2 1.2( )
10
=1.30ms
IP = Idc 2TΔT =
9
0.25
2
60
1
1.3ms
= 923A
 
 64 
V(1) V(2) *REAL(Rectifier)*
Time (s)Circuit3_93b-Transient-11
(Amp)
-10.000
-5.000
+0.000e+000
+5.000
+10.000
+0.000e+000 +20.000m +40.000m +60.000m +80.000m +100.000m +120.000m +140.000m
 
 
SPICE Graph Results: VDC = -6.55 V, Vr = 0.58 V, IP = 150 A, ISC = 370 A 
Note that a significant difference is caused by the diode series resistance. 
 
3.94 
 
(a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.25 10.5 1400 = 0.158F
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d ) Isurge = ωCVP = 2π 400( )0.158( )6.3 2( )= 3540 A
(e) ΔT = 1ω
2Vr
VP
= 1
2π 400( )
2 .25( )
6.3 2
= 94.3μs | IP = Idc 2TΔT =
7.91
.5
2
400
1
94.3μs = 839A 
 
3.95 
 
(a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.25 10.5 1105 = 633μF
(c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d ) Isurge = ωCVP = 2π 105( )633μF( )6.3 2( )= 3540A
(e) ΔT = 1ω
2Vr
VP
= 1
2π 105( )
2 .25( )
6.3 2
= 0.377μs | IP = Idc 2TΔT =
7.91
.5
2
105
1
0.377μs = 839A
 
3.96 
 65
 
(a) C = I T
Vr
= 1
3000 0.01( )
1
60
= 556 μF (b) PIV ≥ 2VP = 2 ⋅ 3000 = 6000V
(c) Vrms = 3000
2
= 2120 V (d ) ΔT = 1ω
2Vr
VP
= 1
2π 60( ) 2 0.01( )= 0.375ms
IP = Idc 2TΔT =1
2
60
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
0.375ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 88.9A (e) Isurge = ωCVP = 2π 60( )556μF( )3000( )= 629A
 
 
3.97 Assuming Von = 1 V: 
 
C = VP −Von
Vr
T 1
R
= 1
0.025
1
60
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
30
3.3
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 6.06 F | PIV = 2VP = 2 3.3 +1( )V = 8.6 V | Vrms = 3.3 +12 = 3.04 V
ΔT = 1ω
2T
RC
VP −Von
VP
= 1
2π 60( )
2
0.110Ω 6.06F( )
1
60
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
3.3V
4.3V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 0.520 ms
IP = Idc 2TΔT = 30
2
60
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
0.520ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ =1920 A | Isurge = ωCVP = 2π 60 / s( )6.06F( )4.3V( )= 9820 A
 
3.98 
0s 5ms 10ms 15ms 20ms 25ms 30ms
40V
20V
0V
-20V
v1
vS
vO
Time
 VDC = 2(VP - Von) = 2(17 - 1) = 32 V. 
 
3.99 
*Problem 3.99 
VS 2 1 DC 0 AC 0 SIN(0 1500 60) 
D1 2 3 DIODE 
D2 0 2 DIODE 
C1 1 0 500U 
C2 3 1 500U 
RL 3 0 3K 
.MODEL DIODE D IS=1E-15 RS=0 
.OPTIONS RELTOL=1E-6 
.TRAN 0.1MS 100MS 
.PRINT TRAN V(2,1) V(3) I(VS) 
 66 
.PROBE V(3) V(2,1) I(VS) 
.END 
0s 20ms 40ms 60ms 80ms 100ms
Time
4.0kV
3.0kV
2.0kV
1.0kV
0V
-1.0kV
-2.0kV
vO
vS
 
Simulation Results: VDC = 2981 V, Vr = 63 V 
The doubler circuit is effectively two half-wave rectifiers connected in series. Each capacitor is 
discharged by I = 3000V/3000� = 1 A for 1/60 second. The ripple voltage on each capacitor is 
33.3 V. With two capacitors in series, the output ripple should be 66.6 V, which is close to the 
simulation result. 
 
3.100 
 
(a) Vdc = − VP −Von( )= − 15 2 −1( )= −20.2 V (b) C = I Vr T2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 20.2V0.5Ω 10.25V⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1120s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.35 F
(c) PIV ≥ 2VP = 2 ⋅15 2 = 42.4 V (d ) Isurge = ωCVP = 2π 60( )1.35( )15 2( )=10800 A
(e) ΔT = 1ω
2Vr
VP
= 1
2π 60( )
2 .25( )
15 2
= 0.407 ms | IP = Idc TΔT =
20.2V
0.5Ω
1
60
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
0.407ms
=1650 A
 
 
3.101 
 
(a) Vdc = − VP −Von( )= − 9 2 −1( )= −11.7 V (b) C = I Vr T2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 11.7V0.5Ω 10.25V⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1120s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.780 F
(c) PIV ≥ 2VP = 2 ⋅ 9 2 = 25.5 V (d ) Isurge = ωCVP = 2π 60( )0.780( )9 2( )= 3740 A
(e) ΔT = 1ω
2Vr
VP
= 1
2π 60( )
2 .25( )
9 2
= 0.526 ms | IP = Idc TΔT =
11.7V
0.5Ω
1
60
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
0.407ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 958A 
 
 67
3.102 
*Problem 3.102 
VS1 1 0 DC 0 AC 0 SIN(0 14.14 400) 
VS2 0 2 DC 0 AC 0 SIN(0 14.14 400) 
D1 3 1 DIODE 
D2 3 2 DIODE 
C 3 0 22000U 
R 3 0 3 
.MODEL DIODE D IS=1E-10 RS=0 
.OPTIONS RELTOL=1E-6 
.TRAN 1US 5MS 
.PRINT TRAN V(1) V(2) V(3) I(VS1) 
.PROBE V(1) V(2) V(3) I(VS1) 
.END 
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms
Time
20V
10V
0V
-10V
-20V
vS
vO
 
Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 A 
 
VDC = VP −Von =10 2 − 0.7 =13.4 V | Vr = 13.43
1
800
1
22000μF = 0.254 V
ΔT = 1
120π
2Vr
VP
= 1
120π
2 0.254( )
14.1
= 0.504 ms
IP = Idc TΔT =
13.4V
3Ω
1
60
s 1
0.504 ms
=150 A
 
Simulation with RS = 0.02 Ω. 
V(1) V(2) *REAL(Rectifier)*
Time (s)Circuit3_102-Transient-15
-15.000
-10.000
-5.000
+0.000e+000
+5.000
+10.000
+15.000
+0.000e+000 +2.000m +4.000m +6.000m +8.000m +10.000m +12.000m +14.000m
 
Simulation Results: VDC = -12.9 V, Vr = 0.20 V, IP = 33.3 A, ISC = 362 A. RS results in a 
significant reduction in the values of IP and ISC. 
 
 68 
3.103 
 
(a) C = VP −Von
Vr
T 1
R
= 1
0.025
1s
120
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
30A
3.3V
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 3.03 F (b) PIV = 2VP = 2 3.3+1( )V = 8.6 V 
(c) Vrms = 3.3+1
2
= 3.04 V (d) ΔT = 1ω
2Vr
VP
= 1
2π 60( )
2 0.025( )3.3( )
4.3
= 0.520 ms
(e) IP = Idc TΔT = 30A
1
60
s
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ 
1
0.520ms
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 962 A | Isurge = ωCVP = 2π 60 / s( )3.03F( )4.3V( )= 4910 A

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