Solutions Manual Microelectronic Circuit Design 4th Ed

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 theyre 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 20201960( ) =14.5 x 1012 =14.5 Tb/chip

1.3 (a)

B2B1

= 19.97x100.1977 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 20201970( ) = 8.85 x 1010 transistors/P

1.5

N2N1

= 1610x100.1548 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 = 8.00x100.05806 20201970( )m =10 nmNo, this distance corresponds to the diameter of only a few atoms. Also, the wavelength of the radiation needed to expose such patterns during fabrication is represents a serious problem.

1.7 From Fig. 1.4, there are approximately 600 million transistors on a complex Pentium IV microprocessor in 2004. From Prob. 1.4, the number of transistors/P will be 8.85 x 1010. in 2020. Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors.


1-3 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 =10.24V4096bits

= 2.500 mV VMSB = 10.24V2 = 5.120V1001001001102 = 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 mVbit

=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 = 0.6104mVbit

and 6.83V10V

214 bits( )=11191 bits1119110 = 8192 + 2048 + 512 + 256 +128 + 32 +16 + 4 + 2 +1( )101119110 =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 0.0625VVO = 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 2000t Volts

1.18

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

1.19

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

1.20

V = 10 V, R1 = 22 k, R2= 47 k and R3 = 180 k.

V

+ -V

1

V2

+

-

R1

R 2 R3

I3I2

V1 =10V 22k22k + 47k 180k( )=10V22k

22k + 37.3k = 3.71 V

V2 =10V 37.3k22k + 37.3k = 6.29 V Checking : 6.29 + 3.71 = 10.0 V

I2 = I1 180k47k +180k =10V

22k + 37.3k

180k47k +180k =134 A

I3 = I1 47k47k +180k =10V

22k + 37.3k

47k47k +180k = 34.9 A

Checking : I1 = 10V22k + 37.3k =169A and I1 = I2 + I3


1-5 6/9/06

1.21

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

V

+ -V

1

V2

+

-

R1

R 2 R3

I3I2

V1 =18V 56k56k + 33k 11k( )=15.7 V V2 =18V33k 11k

56k + 33k 11k( )= 2.31 VChecking :V1 + V2 =15.7 + 2.31=18.0 V which is correct.I1 = 18V56k + 33k 11k( )= 280 A I2 = I1

11k33k +11k = 280 A( ) 11k33k +11k = 70.0 A

I3 = I1 33k33k +11k = 280 A( ) 33k33k +11k = 210 A Checking : I2 + I3 = 280 A

1.22

I1 = 5mA5.6k + 3.6k( )

5.6k + 3.6k( )+ 2.4k = 3.97 mA I2 = 5mA2.4k

9.2k + 2.4k =1.03 mA

V3 = 5mA 2.4k 9.2k( ) 3.6k5.6k + 3.6k = 3.72V Checking : I1 + I2 = 5.00 mA and I2R2 =1.03mA 3.6k( )= 3.71 V

1.23

I2 = 250A 150k150k +150k =125 A I3 = 250A150k

150k +150k =125 A

V3 = 250A 150k 150k( ) 82k68k + 82k =10.3VChecking : I1 + I2 = 250 A and I2R2 =125A 82k( )=10.3 V

1.24


R1

+

-

v

gmv

vs

vth

+

-

Summing currents at the output node yields:v

5x104+ .002v = 0 so v = 0 and vth = vs v = vs

R1

vx

+

-

v

g vm

ix

Summing currents at the output node :

ix = v5x104 0.002v = 0 but v = vxix = vx5x104 + 0.002vx = 0 Rth =

vxix

= 11R1

+ gm= 495

Thvenin equivalent circuit:

vs

495

1-7 6/9/06

1.25 The Thvenin equivalent resistance is found using the same approach as Problem 1.24, and

Rth = 14k + .025

1= 39.6

R1

vs

+

-

v

gmv in

The short circuit current is :

in = v4k + 0.025v and v = vs

in = vs4k + 0.025vs = 0.0253vs Norton equivalent circuit:

39.6 0.0253v s

1.26


(a)

R1 R2

ivs

i

+

-

vth

Vth = Voc = i R2 but i = vsR1 and Vth = vs

R2R1

=120 vs 39k100k = 46.8 vs

R1 R2

i

i

Rth vx

ix

Rth = vxix

; ix =vxR2

+ i but i = 0 since VR1 = 0. Rth = R2 = 39 k.


58.5v s

39 k

(b)

R1 R2

iis

i

+

-

vth

Vth = Voc = i R2 where i + bi + is = 0 and Vth = is +1

R2 = 38700 is

1-9 6/9/06

R1 R2

i

i

Rth vx

Rth = vxix

; ix =vxR2

+ i but i + i = 0 so i = 0 and Rth = R2 = 39 k


39 k

38700i s

1.27

iR

1 R2vs

i

in

in = i but i = vsR1 and in =

R1

vs = 10075k vs =1.33 x 103 vs

From problem 1.26(a), Rth = R2 = 56 k. Norton equivalent circuit:

56 k 0.00133v s

1.28


R1 R2

ivs

i

is

is = vsR1

i = vsR1

+ vsR1

= vs +1

R1 R = vs

is= R1 +1 =

100k81

=1.24

1.29

The open circuit voltage is vth = gmv R2 and v = +isR1.vth = gmR1R2is = 0.0025( )105( )106( )i s= 2.5 x 108isFor i

= 0, v = 0 and R = R =1 Ms th 2

1.30

5 V

3 V

0f (Hz)

500 10000

1.31

2 V

0f (kHz)

9 10 11

v = 4sin 20000t( )sin 2000t( )= 42 cos 20000t + 2000t( )+ cos 20000t 2000t( )[ ]v = 2cos 22000t( )+ 2cos 18000t( )

1.32

A = 236

o

10500 = 2x10536o A = 2x105 A = 36o

1-11 6/9/06

1.33

(a) A = 10

2 45o2x1030o = 5 45

o (b) A = 10

1 12o1030o =100 12

o

1.34

(a) Av = R2R1

= 620k14k = 44.3 (b) Av =

180k = 10.0 (c) Av = 62k1.6k = 38.8

18k

1.35

vo t( )= R2R1 vs t( )= 90.1 sin 750t ( ) mVIS = VSR1

= 0.01V910 =11.0A and is = 11.0 sin 750t( ) A

1.36 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vs. Therefore Av = 1.

1.37 Since the voltage across the op amp input terminals must be zero, v- = v+ = vs. Also, i- = 0.

v voR2

+ i + vR1= 0 or vs vo

R2+ vs

R1= 0 and Av = vovs

=1+ R2R1

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

v1 vR1

+ v2 vR2

= i + v voR3 but v- = v+ = 0 and i- = 0

vo = R3R1v1 R3R2

v2 = 0.255sin3770t 0.255sin10000t volts

1.39

vO = VREF b12 +b24

+ b38

(a) vO = 5

02

+ 14

+ 18

= 1.875V (b) vO = 5

12

+ 04

+ 08

= 2.500V

b1b2b3

vO (V)

000 0

001 -0.625

010 -1.250

011 -1.875

100 -2.500

101 -3.125

110 -3.750

111 -4.375

1.40 Low-pass amplifier

Amplitude

f

10

6 kHz


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 2000t( )+10x3cos 8000t( )+ 0x3cos 15000t( )vO t( )= 50sin 2000t( )+ 30cos 8000t( )[ ] volts

1.44

vO t( )= 20x0.5sin 2500t( )+ 20x0.75cos 8000t( )+ 0x0.6cos 12000t( )vO t( )= 10.0sin 2500t( )+15.0cos 8000t( )[ ] 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 5x10-3

(b)

-10

-5

0

5

10

0 1 2 3 4 5x10-3


1-15 6/9/06

(c)

-10

-5

0

5

10

0 1 2 3 4 5x10-3

(d)

-10

-5

0

5

10

0 1 2 3 4 5x10-3

1.47

(a) 3000 1 .01( ) R 3000 1+ .01( ) or 2970 R 3030 (b) 3000 1 .05( ) R 3000 1+ .05( ) or 2850 R 3150(c) 3000 1 .10( ) R 3000 1+ .10( ) or 2700 R 3300

1.48 Vnom = 2.5V V 0.05V T = 0.052.50 = 0.0200 or 2.00%

1.49 20000F 1 .5( ) C 20000F 1+ .2( ) or 10000F R 24000F

1.50 8200 1 0.1( ) R 8200 1+ 0.1( ) or 7380 R 9020 The resistor is within the allowable range of values.

1.51

(a) 5V 1 .05( ) V 5V 1+ .05( ) or 5.75V V 5.25VV = 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 = RT =6562 6066

100 0 = 4.96oC


Rnom = R 0oC + TCR T( )= 6066 + 4.96 27( )= 6200

1-17 6/9/06

1.53

V2

+

-

V

+ -V1

R2

I3I2R1

R3

Let RX = R2 R3 then V1 = V R1R1 + RX= V1

1+ RXR1

RXmin = 47k 0.9( )180k( )0.9( )

47k 0.9( )+180k 0.9( )= 33.5k RXmax =47k 1.1( )180k( )1.1( )47k 1.1( )+180k 1.1( )= 41.0k

V1max = 10 1.05( )

1+ 33.5k22k 1.1( )

= 4.40V V1min =10 0.95( )

1+ 41.0k22k 0.9( )

= 3.09V

I1 = VR1 + RX and I2 = I1 R3R2 + R3

= VR1 + R2 + R1R2R3

I2max = 10 1.05( )

22000 0.9( )+ 47000 0.9( )+ 22000 0.9( )47000( )0.9( )180000 1.1( )=158 A

I2min = 10 0.95( )

22000 1.1( )+ 47000 1.1( )+ 22000 1.1( )47000( )1.1( )180000 0.9( )=114 A

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 A

I3min = 10 0.95( )

22000 1.1( )+180000 1.1( )+ 22000 1.1( )180000( )1.1( )47000 0.9( )= 28.3 A

1.54

I1 = I R2 + R3R1 + R2 + R3= I 1

1+ R1R2 + R3

and similarly I2 = I 11+ R2 + R3

R1

I1max = 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

I2max = 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 = I1R1

+ 1R3

+ R2R1R3

V3max = 5 1.02( )

12400 1.05( )+

13600 1.05( )+

5600 0.95( )2400 1.05( )3600( )1.05( )

= 4.18 V

V3min = 5 0.98( )

12400 0.95( )+

13600 0.95( )+

5600 1.05( )2400 0.95( )3600( )0.95( )

= 3.30 V

1.55

From Prob. 1.24 : Rth = 1gm + 1R1

Rthmax = 1

0.002 0.8( )+ 15x104 1.2( )= 619 Rthmin = 1

0.002 1.2( )+ 15x104 0.8( )= 412


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 A 29.9 A Max 3.71 V 151 A 40.9 A Average 3.71 V 133 A 35.1 A

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 k) = 36.5 V (b) 36 V (c) (0.1021 A)(97.80 k) = 9.99 V; 10 V

CHAPTER 2

2.1

Based upon Table 2.1, a resistivity of 2.6 -cm < 1 m-cm, and aluminum is a conductor.

2.2

Based upon Table 2.1, a resistivity of 1015 -cm > 105 -cm, and silicon dioxide is an insulator.

2.3

Imax = 107 Acm2

5m( )1m( ) 10

8cm2

m2

= 500 mA

2.4

= TxEBTn Gi 5

3

1062.8exp

For silicon, B = 1.08 x 1031

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 1030

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 EG8.62x105T

with B = 1.27x10

29 K3cm6

T = 300 K and EG = 1.42 eV: ni = 2.21 x106/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 = nE = 700 cm2

V s

2500

Vcm

= 1.75x10

6 cms

v p = + pE = +250 cm2

V s

2500

Vcm

= +6.25x10

5 cms

jn = qnvn = 1.60x1019C( )1017 1cm3 1.75x106 cms = 2.80x104 Acm2jp = qnv p = 1.60x1019C( )103 1cm3 6.25x105 cms =1.00x1010 Acm2

2.8

ni2 = BT 3 exp EG

kT

B = 1.08x10

31

1010( )2 =1.08x1031T 3 exp 1.128.62x105T 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

scm

cmCcmA

Qjv 52

2

10 /01.0/1000 ===

2.10

2267

3 4104sec104.0

cmMA

cmAxcm

cmCQvj ==

==

21

2.11

vn = nE = 1000 cm2

V s

2000

Vcm

= +2.00x10

6 cms

v p = + pE = +400 cm2

V s

2000

Vcm

= 8.00x10

5 cms

jn = qnvn = 1.60x1019C( )103 1cm3 +2.00x106 cms = 3.20x1010 Acm2jp = qnv p = 1.60x1019C( )1017 1cm3 8.00x105 cms = 1.28x104 Acm2

2.12

( ) ( ) ( ) V 100101010 b 500010105 454 =

=== cmxcm

VVcmV

cmxVEa

2.13

jp = qpv p = 1.60x1019C( )1019cm3 10

7 cms

=1.60x10

7 Acm2

ip = j p A = 1.60x107 Acm2

1x10

4cm( )25x104cm( )= 4.00 A

2.14

For intrinsic silicon, = q nni + pni( )= qni n + p( ) 1000 cm( )1 for a conductorni q n + p( )=

1000 cm( )11.602x1019C 100 + 50( ) cm2v sec

= 4.16x1019

cm3

n i2 = 1.73x10

39

cm6= BT 3 exp EG

kT

with

B =1.08x1031 K3cm6, 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, = q nni + pni( )= qni n + p( ) 105 cm( )1 for an insulatorni q n + p( )=

105 cm( )11.602x1019C( )2000 + 750( ) cm2v sec

= 2.270x1010

cm3

n i2 = 5.152x10

20

cm6= BT 3 exp EG

kT

with

B =1.08x1031 K3cm6, 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 cmVcm

cmAjjE =

===

2.21

jn

drift = qnnE = qnvn = 1.602x1019( )1016( ) Ccm3 107 cms =16000 Acm2

2.22

N = 10

15 atomscm3

1m( )10m( )0.5m( )

104cmm

3

= 5,000 atoms

2.23

N A > N D: N A N D =1015 1014 = 9x1014 /cm3If 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 = 9x1014 9x1014( )2 + 4 5x1013( )2

2= 9.03x1014

and n = 2.77x1012 /cm3. The answers are essentially the same.

2.24

3516

222314

3113161616

10502104

10104

102210410105

/cmx.xp

n | n/cmxNNp

/cmxn/cmxxN: NNN

iDA

iDADA

======>>==>

2.25

N D > N A: ND N A = 3x1017 2x1017 =1x1017 /cm32ni = 2x1017 /cm3; Need to use Eq. (2.11)

n = 1017 1017( )2 + 4 1017( )2

2=1.62x1017 /cm3

p = ni2

n= 10

34

1.62x1017= 6.18x1016 /cm3

24

2.26

N D N A = 2.5x1018 / cm3

Using Eq. 2.11: n = 2.5x1018 2.5x1018( )2 + 4 1010( )2

2

Evaluating this with a calculator yields n = 0, and n = ni2

p= .

No, the result is incorrect because of loss of significant digitswithin 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.

3318

6202318

i318310

7.16106

10 and 106 So

2n >> /106 and 10 re, temperaturoomAt

/cm/cmx

/cmp

nn/cmxp

cmxNN/cmn

i

DAi

======

(b)

At 200K, ni2 =1.08x1031 200( )3 exp 1.128.62x105 200( )

= 5.28x109 /cm6

ni = 7.27x104 /cm3 N A N D >> 2ni, so p = 6x1018 /cm3 and n = 5.28x109

6x1018= 8.80x1010 /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.

3317

6202317

i317310

i

333103

10 and 103 So

2n >> /103 and /10n re, temperaturoomAt

/cm/cmx

/cmnnp/cmxn

cmxNNcm

i

AD

======

(b) At 250K, ni2 =1.08x1031 250( )3 exp 1.128.62x105 250( )

= 4.53x1015 /cm6

ni = 6.73x107 /cm3 N D N A >> 2ni , so n = 3x1017 / cm3 and n = 4.53x1015

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

3318

6202318

i318310

i

716106

10 and 106 So

2n >> /106 and /10n re, temperaturoomAt (b)

/cm./cmx

/cmp

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.

3317

6202317

i317310

333103

10 and 103 So

2n >> /103 and 10 re, temperaturoomAt (b)

/cm/cmx

/cmp

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 = ni2

n= 10

20

4x1016= 2.5x103 / cm3

N D + N A = 4x1016 / cm3 | Using Fig. 2.13, n =1030 cm2

V s and p = 310cm2

V s = 1

qnn =1

1.602x1019C( )1030 cm2V s

4x1016

cm3

= 0.152 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 = ni2

p= 10

20

1018=100 / cm3

N D + N A =1018 / cm3 | Using Fig. 2.13, n = 375 cm2

V s and p =100cm2

V s = 1

q p p =1

1.602x1019C 100 cm2

V s

1018

cm3

= 0.0624 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= ni2

p= 10

20

7x1019=1.43/cm3

N D + N A = 7x1019 / cm3 | Using Fig. 2.13, n =120 cm2

V s and p = 60cm2

V s = 1

q p p =1

1.602x1019C 60 cm2

V s

7x1019

cm3

=1.49 m cm

2.34

Phosphorus is a donor : N D = 5.5x1016 / cm3 | Boron is an acceptor : N A = 4.5x1016 / cm3N D > N A : material is n - type | N D N A =1016 / cm3 >> 2ni = 2x1010 / cm3

n =1016 /cm3 | p = ni2

p= 10

20

1016=104 /cm3

N D + N A =1017 / cm3 | Using Fig. 2.13, n = 800 cm2

V s and p = 230cm2

V s = 1

qnn =1

1.602x1019C 800 cm2

V s

1016

cm3

= 0.781 cm

27

2.35

= 1

q p p | p p =1

1.602x1019C( )0.054 cm( )=1.16x1020

V cm s

An iterative solution is required. Using the equations in Fig. 2.8:

NA p p 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

= 1

q p p | p p =1

1.602x1019C( )0.75 cm( )=8.32x1018

V cm s

An iterative solution is required. Using the equations in Fig. 2.8: NA p p 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

= 1

qnn | nn nN D =1

1.602x1019C( )2 cm( )=3.12x1018

V cm s An iterative solution is required. Using the equations in Fig. 2.8:

ND n nn

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)

= 1

qnn | nn nN D =1

1.602x1019C( )0.001 cm( )=6.24x1021

V cm s

An iterative solution is required. Using the equations in Fig. 2.8:

ND n nn 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)

= 1qp p | p p pNA =

11.602x1019C( )0.001 cm( ) =

6.24x1021

V cm s

An iterative solution is required using the equations in Fig. 2.8:

NA p p 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:

= 1

q p p | p p pN A =1

1.602x1019C( )1 cm( )=6.25x1018

V cm s An iterative solution is required. Using the equations in Fig. 2.8:

NA p p 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:

= 1

q p p | p p pN A =1

1.602x1019C( )0.25 cm( )=2.5x1019

V cm s

NA p p 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 = 1.1x1017

- 1.7x1016

= 9.3 x 1016

/cm3

(b) If donors are added:

ND ND + NA n ND - NA nn 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 1016

/cm3 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 n = 1250 cm2/V-s from Fig. 2.8.

= qnn qnN D = 1.602x1019C( )1250( )1016( )= 2.00 cm Now we add acceptors until = 5.0 (-cm) -1:

= q p p | p p p N A N D( )= 5 cm( )

1

1.602x1019C= 3.12x10

19

V cm s

NA ND + NA p NA - ND p 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 p = 405 cm2/V-s from Fig. 2.8.

= q p p q pN A = 1.602x1019C( )405( )1016( )= 0.649 cm Now we add donors until = 5.5 (-cm) -1:

= qnn | nn n N D N A( )= 5.5 cm( )

1

1.602x1019C= 3.43x10

19

V cm s

ND ND + NA n ND - NA p 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 =

1.38x1023T1.602x1019

= 8.62x105T

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

= qVTn

dndx

j = 1.602x1019C( )0.025V( ) 350 cm2V s

1018 00 104

1cm4

= 14.0 kAcm2

2.46

j = qDp dpdx = 1.602x1019C( )15 cm2s

1019 / cm3

2x104cm

exp

x2x104cm

j =1.20x105 exp 5000 xcm

Acm2

I 0( )= j 0( )A = 1.20x105 Acm2

10m

2( )108cm2m2

=12.0 mA

32

2.47

jp = qp pE qDp dpdx = qp p E VT1p

dpdx

= 0 E = VT

1p

dpdx

E VT 1N AdN Adx

= 0.025 1022 exp 104 x( )

1014 +1018 exp 104 x( )E 0( )= 0.025 10221014 +1018 = 250 VcmE 5x104cm( )= 0.025 1022 exp 5( )1014 +1018 exp 5( )= 246 Vcm

2.48

jndrift = qnnE = 1.60x1019C( )350 cm2V s

1016

cm3

20

Vcm

= 11.2

Acm2

jpdrift = q p pE = 1.60x1019C( )150 cm2V s

1.01x1018

cm3

20

Vcm

= 484

Acm2

jndiff = qDn dndx = 1.60x10

19C( )350 0.025cm2s

104 10162x104cm4

= 70.0

Acm2

jpdiff = qDp dpdx = 1.60x10

19C( )150 0.025cm2s

1018 1.01x10182x104cm4

= 30.0

Acm2

jT = 11.2 484 70.0 + 30.0 = 535 Acm2

2.49 NA = 2ND E C

E A N A

HolesE

V

NA

ND

N A

ND

N A

N DE D

2.50

= hc

E= 6.626x10

34 J s( )3x108 m / s( )1.12eV( )1.602x1019 J / eV( ) =1.108 m

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 18

+ 6

12

+ 4 1()= 8 atoms

(b) V = l 3 = 0.543x109 m( )3 = 0.543x107cm( )3 =1.60x1022cm3(c) D = 8 atoms

1.60x1022cm3= 5.00x1022 atoms

cm3

(d ) m = 2.33 gcm3

1.60x10

22cm3 = 3.73x1022 g(e) From Table 2.2, silicon has a mass of 28.086 protons.

mp = 3.73x1022 g

28.082 8( )protons =1.66x1024g

proton

Yes, near the actual proton rest mass.

34

CHAPTER 3

3.1

j = VT ln NANDni2 = 0.025V( )ln1019 cm3( )1018 cm3( )

1020 cm6 = 0.979V

wdo = 2sq1

NA+ 1

ND

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11019cm3

+ 11018cm3

0.979V( )

w do = 3.73 x 106cm = 0.0373m

xn = wdo1+ ND

NA

= 0.0373m1+ 10

18cm3

1019cm3

= 0.0339 m | x p = wdo1+ NA

ND

= 0.0373m1+ 10

19cm3

1018cm3

= 3.39 x 10-3 m

E MAX = qNA xps =1.60x1019C( )1019cm3( ) 3.39x107cm( )

11.7 8.854 x1014 F /cm = 5.24 x 105 V

cm

3.2

ppo = NA = 1018

cm3 | npo = ni

2

ppo= 10

20

1018= 10

2

cm3

nno = ND = 1015

cm3 | pno = ni

2

nno= 10

20

1015= 10

5

cm3

j = VT ln NANDni2 = 0.025V( )ln1018cm3( )1015cm3( )

1020cm6= 0.748 V

wdo = 2sq1

NA+ 1

ND

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11018cm3

+ 11015cm3

0.748V( )

wdo = 98.4 x 106cm = 0.984 m

3.3

ppo = N A = 1018

cm3 | npo = ni

2

p po= 10

20

1018= 10

2

cm3

nno = N D = 1018

cm3 | pno = ni

2

nno= 10

20

1018= 10

2

cm3


1020 cm6 = 0.921V

wdo = 2sq1

NA+ 1

ND

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11018cm3

+ 11018cm3

0.921V( )

w do = 4.881x106cm = 0.0488 m

34

3.4

ppo = N A = 1018

cm3 | npo = ni

2

p po= 10

20

1018= 10

2

cm3

nno = N D = 1018

cm3 | pno = ni

2

nno= 10

20

1018= 10

2

cm3


1020 cm6 =1.04V

wdo = 2sq1

NA+ 1

ND

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11018cm3

+ 11020cm3

1.04V( )

w do = 0.0369 m

3.5

ppo = N A = 1016

cm3 | npo = ni

2

p po= 10

20

1016= 10

4

cm3

nno = N D = 1019

cm3 | pno = ni

2

nno= 10

20

1019= 10

cm3

j = VT ln N A N Dni2= 0.025V( )ln 1019 cm3( )1016 cm3( )1020 cm6 = 0.864V

wdo = 2sq1

N A+ 1

N D

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11019cm3

+ 11016cm3

0.864V( )

wdo = 0.334 m

3.6

wd = wdo 1+ VR j | (a) wd = 2wdo requires VR = 3 j = 2.55 V | wd = 0.4m 1+

50.85

= 1.05 m

3.7

wd = wdo 1+ VR j | (a) wd = 3wdo requires VR = 8 j = 4.80 V | wd =1m 1+

100.6

= 4.20 m

3.8

jn = E , = 1 =1

0.5 cm =2

cm | E =jn =

1000A cm22 cm( )1 = 500

Vcm

35

3.9

jp = E | E = jn = jn = 5000A cm

2( )2 cm( )=10.0 kVcm

3.10

j jn = qnv = 1.60x1019C( ) 4x1015cm3

107cms

= 6400

Acm2

3.11

j jp = qpv = 1.60x1019C( ) 5x1017cm3

107cms

= 800

kAcm2

3.12

jp = q p pE qDp dpdx = 0 E = Dp p

1p =

kTq

dpdx

1p

dpdx

p(x) = N o exp xL

|

1p

| E = VTL

= 0.025V104cm

= 250 Vcm

dpdx

= 1L

The exponential doping results in a constant electric field.

3.13

jp = qDn dndx = qnVT

dndx

| dndx

= 2000A / cm2

1.60x1019C( )500cm2 /V s( )0.025V( )=1.00 x 1021

cm4

3.14

10 =104 1016 exp 40VD( )1[ ]+ VD and the solver yields VD = 0.7464 V

3.15

f =10 104 ID 0.025ln ID + ISIS

| f ' = 104 0.025ID + IS

| ID' = ID ff '

Starting the iteration process with ID = 100 A and IS = 10-13A:

ID f f'

1.000E-04 8.482E+00

-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( )1.38x1023 J / K

= 290 K

3.18

VT = kTq =1.38x1023 J / K( )T

1.60x1019C= 8.63x105T

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 :

1.21.00.80.60.40.20.00

1

2

3

4

5

6

(a)

(b)

(c)

37

3.20

nVT = n kTq =1.04

1.38x1023 J / K( )300( )1.60x1019C

= 26.88 mV T = 26.88mV1.602x10-19

1.38x10-23= 312 K

3.21

iD = IS exp vDnVT

1

or

vDnVT

= ln 1+ iDIS

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

VDnVT

1

(a) VD =1.05 0.025V( )ln 1+ 7x105 A1018 A

= 0.837V | (b) VD =1.05 0.025V( )ln 1+

5x106 A1018 A

= 0.768V

(c) ID =1018 A exp 01.05 0.025V

1

= 0 A

(d) ID =1018 A exp 0.075V1.05 0.025V

1

= 0.943x10

19 A

(e) ID =1018 A exp 5V1.05 0.025V

1

= 1.00x10

18 A

3.23

VD = nVT ln 1+ IDIS

| ID = IS exp

VDnVT

1

(a) VD = 0.025V ln 1+ 104 A

1017 A

= 0.748V | (b) VD = 0.025V ln 1+

105 A1017 A

= 0.691V

38

(c) ID =1017 A exp 00.025V

1

= 0 A

(d) ID =1017 A exp 0.06V0.025V

1

= 0.909x10

17 A

(e) ID =1017 A exp 4V0.025V

1

= 1.00x10

17 A

3.24

ID = IS exp VDVT

1

=10

17 A exp 0.6750.025

1

= 5.32x10

6 A = 5.32 A

VD = VT ln IDIS+1

= 0.025V( )ln 15.9x10

6 A1017 A

+1 = 0.703 V

3.25

VD = nVT ln 1+ IDIS

= 2 0.025V( )ln 1+ 40A1010 A

=1.34 V

VD = 2 0.025V( )ln 1+ 100A1010 A

=1.38 V

3.26

(a) IS = IDexp VD

nVT

1

= 2mAexp 0.82

0.025

1

=1.14x1017 A

(b) ID =1.14x1017 A exp 50.025

1

= 1.14x10

17 A

3.27

(a) IS = IDexp VD

nVT

1

= 300Aexp 0.75

0.025

1

= 2.81x1017 A

(b) ID = 2.81x1017 A exp 30.025

1

= 2.81x10

17 A

39

3.28

VD = nVT ln 1+ IDIS

| 10

-14 IS 1012 | VD = 0.025V( )ln 1+ 103 A1012 A

= 0.518 V

VD = 0.025V( )ln 1+ 103 A1014 A

= 0.633 V | So, 0.518 V VD 0.633 V

3.29

VT =

1.38x1023 307( )1.60x1019

= 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 =1.38x1023 J / K( )T

1.60x1019C= 8.63x105T

For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV

3.31

kTq

= 1.38x1023 303( )

1.60x1019= 26.1 mV | VD = 0.0261V( )ln 1+ 1032.5x1016

= 0.757 V

V = 1.8mV / K( )20K( )= 36.0 mV | VD = 0.757 0.036 = 0.721 V

40

3.32

kTq

= 1.38x1023 298( )

1.602x1019= 25.67 mV | (a) VD = 0.02567V( )ln 1+ 1041015

= 0.650 V

V = 2.0mV / K( )25K( )= 50.0 mV(b) VD = 0.650 0.050 = 0.600 V

3.33

kTq

= 1.38x1023 298( )

1.602x1019= 25.67 mV | (a) VD = 0.02567V( )ln 1+ 2.5x1041014

= 0.615 V

b( ) V = 1.8mV / K( )60K( )= 50.0 mV VD = 0.615 0.108 = 0.507 V(c) V = 1.8mV / K( )80K( )= +144 mV VD = 0.615+ 0.144 = 0.758 V

3.34

dvDdT

= vD VG 3VTT

= 0.7 1.21 3 0.0259( )300

= 1.96 mVK

41

3.35

IS2IS1

= T2T1

3

exp EGk

1T2

1T1

=

T2T1

3

exp EGkT1

1

T1T2

f x( )= x( )3 exp EGkT1

1

1x

x =

T2T1

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+ VR j | (a) wd =1m 1+

50.8

= 2.69 m (b) wd =1m 1+ 100.8 = 3.67 m

3.37

j = VT ln N A N Dni2= 0.025V( )ln 1016cm3( )1015cm3( )1020cm6 = 0.633 V

wdo = 2sq1

N A+ 1

N D

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11016cm3

+ 11015cm3

0.633V( )

wdo = 0.949 m | wd = wdo 1+ VR jwd = 0.949m 1+ 10V0.633V = 3.89 m | wd = 0.949m 1+

100V0.633V

= 12.0 m

42

3.38

j = VT ln N A N Dni2 = 0.025V( )ln1018cm3( )1020cm3( )

1020cm6=1.04 V

wdo = 2sq1

N A+ 1

N D

j =

2 11.7 8.854x1014 F cm1( )1.602x1019C

11018cm3

+ 11020cm3

1.04V( )

wdo = 0.0368 m | wd = wdo 1+ VR jwd = 0.0368m 1+ 51.04 = 0.0887 m | wd = 0.0368m 1+

251.04

= 0.184 m

3.39

Emax =2 j + VR( )

wd= 2 j + VR( )

wdo 1+ VR j= 2 j

wdo1+ VR j

3x105 Vcm

= 2 0.6V( )104cm

1+ VR0.6

VR = 374 V

3.40

E = 2 jwdo

= 2 0.748V( )0.984x104cm

=15.2 kVcm

| j + VR = Emax2wdo j

=3x105 V

cm0.984x104cm( )

2 0.748V

VR = 291.3 0.748 = 291 V

3.41

VZ = 4 V; RZ = 0 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 >> j, so j can be neglected.

Emax = qN A x pS =qN Awd

S =qN AS

2Sq

VRN A

N A = Emax2 S

2qVR= 3x10

5( )2 11.7( )8.854x1014( )2 1.602x1019( )1000 = 2.91 x 1014 / cm3

43

3.43

j = VT ln N AN Dni2 = 0.025ln10151020

1020= 0.864V

wdo = 2Sq1

N A+ 1

N D

j =

2 11.7( )8.854x1014( )1.602x1019

11015

+ 11020

0.864 =1.057x10

4cm

Cjo" = S

wdo=

11.7 8.854x1014( )1.057x104

= 9.80x10-9 F / cm2 | Cj = Cjo" A

1+ VR j= 9.80x10

-9 0.05( )1+ 5

0.864

=188 pF

3.44

j = VT ln N AN Dni2 = 0.025ln10181015

1020= 0.748V

wdo = 2Sq1

N A+ 1

N D

j =

2 11.7( )8.854x1014( )1.602x1019

11018

+ 11015

0.748 = 0.984x10

4cm

Cjo" = S

wdo=

11.7 8.854x1014( )0.984x104

= 10.5x10-9 F / cm2 | Cj = Cjo" A

1+ VR j= 10.5x10

-9 0.02( )1+ 10

0.748

= 55.4 pF

3.45

(a) CD = IDTVT =104 A 1010 s( )

0.025V= 400 fF (b) Q = IDT =104 A 1010 s( )=10 fC

(c) CD =25x103 A 1010 s( )

0.025V=100 pF | Q = IDT = 5x103 A 1010 s( )= 0.50 pC

3.46

(a) CD = IDTVT =1A 108 s( )0.025V

= 0.400 F (b) Q = IDT =1A 108 s( )=10.0 nC(c) CD =

100mA 108 s( )0.025V

= 0.04 F | Q = IDT =100mA 108 s( )=1.00 nC

44

3.47

j = VT ln N AN Dni2 = 0.025ln10191017

1020= 0.921V

wdo = 2Sq1

N A+ 1

N D

j =

2 11.7( )8.854x1014( )1.602x1019

11019

+ 11017

0.921 = 0.110 m

Cjo = S Awdo =11.7 8.854x1014( )104( )

0.110x104= 9.42 pF / cm2 | Cj = Cjo

1+ VR j= 9.42 pF

1+ 50.921

= 3.72 pF

3.48

j = VT ln N AN Dni2 = 0.025ln10191016

1020= 0.864V

wdo = 2Sq1

N A+ 1

N D

j =

2 11.7( )8.854x1014( )1.602x1019

11019

+ 11016

0.864 = 0.334m

Cjo = S Awdo =11.7 8.854x1014( )0.25cm2( )

0.334x104= 7750 pF | Cj = Cjo

1+ VR j= 7750 pF

1+ 30.864

= 3670 pF

3.49

VDC

RFC

C10 H 10 H

C

L =

C = Cjo1+ VR j

(a) C = 39 pF1+ 1V

0.75V

= 25.5 pF | fo = 12 LC =

1

2 105 H( )25.5 pF = 9.97MHz

(b) C = 39 pF1+ 10V

0.75V

=10.3pF | fo = 12 LC =

1

2 105 H( )10.3pF =15.7 MHz

3.50

(a) VD = 0.025V( )ln 1+ 50A107 A

= 0.501 V | (b) VD = 0.025V( )ln 1+ 50A1015 A

= 0.961 V

45

3.51

(a) VD = 0.025V( )ln 1+ 4x103 A1011 A

= 0.495 V | (b) VD = 0.025V( )ln 1+

4x103 A1014 A

= 0.668 V

3.52

RS = Rp + Rn Rp = p LpAp = 1 cm( )0.025cm0.01cm2

= 2.5

Rn = n LnAn = 0.01 cm( )0.025cm0.01cm2

= 0.025 RS = 2.53

3.53

(a) VD' = VT ln 1+ IDIS

= 0.025V( )ln 1+ 10

3

5x1016

= 0.708V

VD = VD' + ID RS = 0.708V +103 A 10( )= 0.718 V(b) VD = VD' + ID RS = 0.708V +103 A 100( )= 0.808 V

3.54

c =10 m2 Ac =1m2 RC = cAc =10 m2

1m2 =10 / contact5 anode contacts and 14 cathode contacts

Resistance of anode contacts = 105

= 2

Resistance of cathode contacts = 1014

= 0.71

3.55

(a) From Fig. 3.21a, the diode is approximately 10.5 m long x 8 m wide. Area = 84 m2. (b) Area = (10.5x0.13 m) x (8x0.13m) = 1.42 m2.

46

3.56

(a) 5 =104 ID + VD | VD = 0 ID = 0.500mA | ID = 0 VD = 5VForward biased - VD = 0.5 V ID = 4.5V104 = 0.450 mA(b) 6 = 3000ID + VD | VD = 0 ID = 2.00mA | ID = 0 VD = 6VIn reverse breakdown - VD = 4 V ID = 2V3k = 0.667 mA(c) 3 = 3000ID + VD | VD = 0 ID = 1.00mA | ID = 0 VD = 3VReverse 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 mAForward biased - VD = 0.5V ID = 9.5V5k =1.90 mA(b) -10 = 5000ID + VD | VD = 0 ID = 2.00 mA | VD = 5 V ID = 1.00 mAIn reverse breakdown - VD = 4V ID = 6V5k = 1.20 mA(c) 2 = 2000ID + VD | VD = 0 ID = 1.00 mA | ID = 0 VD = 2 VReverse 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 A .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 mAIn reverse breakdown - VD = 4 V ID = 10 (4)V10k = 0.600 mA(b) 10 =104 ID + VD | VD = 0 ID =1.00 mA | VD = 5 V ID = 0.500 mAForward biased - VD = 0.5 V ID = 10 0.5V10k = 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 k: For vD = 0, iD = 6V/4 k = 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 k: For vD = 0, iD = -6V/3 k = -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 k: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) (d) V = +12 V and R = 8 k: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) (e) V = -25 V and R = 10 k: 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 0.025ID + 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 =104 ID + VD | ID = 0 VD =1V | VD = 0 ID = 0.1mA 50 A, 0.5 V( )

Mathematical model: f =1109 exp 40VD( )1[ ]+ VD 49.9 A, 0.501 V( )

Ideal diode model: ID = 1V/10k = 100 A; (100 A, 0 V) Constant voltage drop model: ID = (1-0.6)V/10k = 40.0A; (40.0 A, 0.6 V)

3.66 Using Thvenin equivalent circuits yields and then combining the sources

+-

+-+-

VI 1 k 1.2 k

1.5 V1.2 V

0.3 V

+-

+-

VI 2.2 k

(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 =

2.2k0.3V = 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

CVD:

0.3 V+-

0.6 VI

2.2 k +-

onV

(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 =

2.2k0.3V 0.6V = 136 A.

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 - +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 =1015 A exp 0.3V0.025V

1

=163 pA, a very small current.

3.67 The nominal values are:

VA = 3V R2R1 + R2

= 3V

2k2k + 3k

=1.20V and RTHA =

R1R2R1 + R2

= 2k 3k( )2k + 3k =1.20k

VC = 3V R4R3 + R4

= 3V

2k2k + 2k

=1.50V and RTHC =

R3R4R3 + R4

= 2k 2k( )2k + 2k =1.00k

IDnom = 1.50 1.20

1.20 +1.00

Vk =136 A

For maximum current, we make the Thvenin equivalent voltage at the diode anode as large as possible and that at the cathode as small as possible.

VA = 3V1+ R1

R2

= 3V1+ 2k 0.9( )

2k 1.1( )=1.65V and RTHA = R1R2R1 + R2

= 2k 0.9( )2k 1.1( )2k 0.9( )+ 2k 1.1( )= 0.990k

VC = 3V1+ R3

R4

= 3V1+ 3k 1.1( )

2k 0.9( )=1.06V and RTHC = R3R4R3 + R4

= 3k 1.1( )2k 0.9( )3k 1.1( )+ 2k 0.9( )=1.17k

IDmax = 1.651.06

0.990 +1.17

Vk = 274 A

For minimum current, we make the Thvenin equivalent voltage at the diode anode as small as possible and that at the cathode as large as possible.

VA = 3V1+ R1

R2

= 3V1+ 2k 1.1( )

2k 0.9( )=1.350V and RTHA = R1R2R1 + R2

= 2k 1.1( )2k 0.9( )2k 1.1( )+ 2k 0.9( )= 0.990k

VC = 3V1+ R3

R4

= 3V1+ 3k 0.9( )

2k 1.1( )=1.347V and RTHC = R3R4R3 + R4

= 3k 0.9( )2k 1.1( )3k 0.9( )+ 2k 1.1( )=1.21k

IDmin = 1.350 1.347

0.990 +1.21

Vk =1.39 A 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( )16k = 0.625 mA

(b) Diode is forward biased :V = 5 + 0 = 5 V | I = 5 5( )16k = 0.625 mA

(c) Diode is reverse biased : I = 0 | V = 5+16k I( )= 5 V | VD = 10 V(d ) Diode is reverse biased : I = 0 | V = 7 16k I( )= 7 V | VD = 10 V (b)

(a) Diode is forward biased :V = 3 0.7 = 2.3 V | I = 2.3 7( )16k = 0.581 mA

(b) Diode is forward biased :V = 5 + 0.7 = 4.3 V | I = 5 4.3( )16k = 0.581 mA

(c) Diode is reverse biased : I = 0 | V = 5+16k I( )= 5 V | VD = 10 V(d ) Diode is reverse biased : I = 0 | V = 7 16k I( )= 7 V | VD = 10 V

3.70 (a)

(a) Diode is forward biased :V = 3 0 = 3 V | I = 3 7( )100k =100 A

(b) Diode is forward biased :V = 5 + 0 = 5 V | I = 5 5( )100k =100 A

(c) Diode is reverse biased : I = 0 A | V = 5+100k I( )= 5 V | VD = 10 V(d ) Diode is reverse biased : I = 0 A | V = 7 100k I( )= 7 V | VD = 10 V (b)

54

(a) Diode is forward biased :V = 3 0.6 = 2.4 V | I = 2.4 7( )100k = 94.0 A

(b) Diode is forward biased :V = 5 + 0.6 = 4.4 V | I = 5 4.4( )100k = 94.0 A

(c) Diode is reverse biased : I = 0 | V = 5+100k I( )= 5 V | VD = 10 V(d ) Diode is reverse biased : I = 0 | V = 7 100k I( )= 7 V | VD = 10 V

3.71 (a)

(a) D1 on, D2 on : ID2 =0 9( )22k = 409A | ID1 = 409A

6 043k = 270A

D1 : 409 A, 0 V( ) D2 : 270 A, 0 V( )(b) D1 on, D2 off : ID2 = 0 | ID1 = 6 043k =140 A | VD2 = 9 0 = 9V D1 : 140 A, 0 V( ) D2 : 0 A, 9 V( )

(c) D1 off, D2 on : ID1 = 0 | ID2 =6 9( )65k = 230 A | VD1 = 6 43x10

3 ID2 = 3.92 V D1 : 0 A,3.92 V( ) D2 : 230 A,0 V( )(d ) D1 on, D2 on : ID2 =

0 6( )43k =140 A | ID1 =

9 022k 140 A = 270 A

D1 : 140 A,0 V( ) D2 : 270 A,0 V( ) (b)

(a) D1 on, D2 on :

ID2 =-0.75 0.75 9( )

22k = 341A | ID1 = 341A 6 0.75( )

43k =184AD1 : 184 A, 0.75 V( ) D2 : 341 A, 0.75 V( )(b) D1 on, D2 off :

ID2 = 0 | ID1 = 6 0.7543k =122A | VD2 = 9 0.75 = 9.75VD1 : 122 A, 0.75 V( ) D2 : 0 A, 9.75 V( )

55

(c) D1 off, D2 on :

ID1 = 0 | ID2 =6 0.75 9( )

65k = 219A | VD1 = 6 43x103 ID2 = 3.43V

D1 : 0 A, 3.43 V( ) D2 : 219 A, 0.75 V( )(d) D1 on, D2 on :

ID2 =0.75 0.75 6( )

43k =140A | ID1 =9 0.7522k 400A = 235A

D1 : 235 A, 0.75 V( ) D2 : 140 A, 0.75 V( )

3.72 (a)

(a) D1 and D2 forward biased

ID2 =0 9( )

15Vk = 600A ID1 = ID2

6 0( )15

Vk = 200A

D1 : 0 V, 200 A( ) D2 : 0 V, 600 A( )

(b) D1 forward biased, D2 reverse biased

ID1 = 6 015Vk = 400A VD2 = 9 0 = 9 V

D1 : 0 V, 400 A( ) D2 : -9 V, 0 A( )

(c) D1 reverse biased, D2 forward biased

ID2 =6V 9V( )

30k = 500A VD1 = 6 15000ID2 = 1.50VD1 : 1.50 V, 0 A( ) D2 : 0 V, 500 A( )

(d) D1 and D2 forward biased

ID2 =0 6( )

15Vk = 400A ID1 =

9 0( )15

Vk ID2 = 200A

D1 : 0 V, 200 A( ) D2 : 0 V, 400 A( )

(b)

(a) D1 on, D2 on :

ID2 =-0.75 0.75 9( )

15k = 500A | ID1 = 500A 6 0.75( )

15k = 50.0AD1 : 50.0 A, 0.75 V( ) D2 : 500 A, 0.75 V( )

56

(b) D1 on, D2 off :

ID2 = 0 | ID1 = 6 0.7515k = 350A | VD2 = 9 0.75 = 9.75VD1 : 350 A, 0.75 V( ) D2 : 0 A, 9.75 V( )

(c) D1 off, D2 on :

ID1 = 0 | ID2 =6 0.75 9( )

30k = 475A | VD1 = 6 15x103 ID2 = 1.13V

D1 : 0 A, 1.13 V( ) D2 : 475 A, 0.75 V( )

(d) D1 on, D2 on :

ID2 =0.75 0.75 6( )

15k = 400A | ID1 =9 0.7515k 400A =150A

D1 : 150 A, 0.75 V( ) D2 : 400 A, 0.75 V( )

3.73 Diodes are labeled from left to right

(a) D1 on, D2 off, D3 on : ID2 = 0 | ID1 = 10 03.3k + 6.8k = 0.990mA

ID3 + 0.990mA =0 5( )2.4k 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 = 0ID1 =

10 0( )V8.2k +12k = 0.495mA | VD2 = 5 10 8200ID1( )= 0.941V

ID3 =0 5V( )

10k ID1 = 0.005mAD1 : 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( )

8.2k V =1.22mA > 0 | I12K =0 2( )12k V = 0.167mA | ID2 = ID1 + I12K =1.05mA > 0

I10K =2 5( )10k 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 = 0ID3 =

12 5( )4.7 + 4.7 + 4.7

Vk =1.21mA > 0 | VD1 = 0 5+ 4700ID3( )= 0.667V < 0

VD2 = 5 12 4700ID3( )= 1.33V < 0D1 : 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( )3.3k + 6.8k = 0.990mA

ID3 + 0.990mA =0.6 5( )

2.4k ID3 = 0.843mA | VD2 = 5 10 0.6 3300ID1( )= 1.13VD1 : 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 = 0ID1 =

10 0.6 5( )8.2k +12k +10kV = 0.477mA | VD2 = 5 10 0.6 8200ID1( )= 0.490V

VD3 = 0 5+10000ID1( )= +0.230V < 0.6V so the diode is offD1 : 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( )

8.2Vk =1.07mA > 0 | I12K =

0.6 1.4( )12

Vk = 0.167mA

ID2 = ID1 + I12K = 0.906mA > 0 | I10K =1.4 5( )

10Vk = 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 = 0ID3 =

11.4 5( )4.7 + 4.7 + 4.7

Vk =1.16mA > 0 | VD1 = 0 5+ 4700ID3( )= 0.452V < 0

VD2 = 5 11.4 4700ID3( )= 0.948V < 0D1 : 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( )

10k +10k =1.50mA | ID2 = 0

ID3 =0 10( )

10k =1.00mA | VD2 =10 104 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 3.9k3.9k +11k = 6.28V | RTH =11k 3.9k = 2.88k

IZ = 6.28 42.88k = 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

iD

-6 -5 -4 -3 -2 -1

Q-Point: (-0.8 mA, -4 V)

3.80

IS = 27 915k =1.20mA IL

9V1.2mA

= 7.50 k

3.81

IS = 27 915k =1.20mA | P = 9V( )1.20mA( )=10.8 mW

3.82

IZ = VS VZRS VZ

RL= VS

RSVZ 1RS

+ 1RL

| PZ = VZ IZ

IZnom = 30V

15k 9V1

15k +1

10k

= 0.500 mA | PZ

nom = 9V 0.500mA( )= 4.5 mWIZ

max = 30V 1.05( )15k 0.95( ) 9V 0.95( )

115k 0.95( )+

110k(1.05)

= 0.796 mA

PZmax = 9V .95( )0.796mA( )= 6.81 mW

IZmin = 30V 0.95( )

15k 1.05( ) 9V 1.05( )1

15k 1.05( )+1

10k(0.95)

= 0.215 mA

PZmin = 9V 1.05( )0.215mA( )= 2.03 mW

3.83

(a) VTH = 60V 100150 +100 = 24.0V | RTH =150 100 = 60 | 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

RSVZ 1RS

+ 1RL

| PZ = VZ IZ

IZnom = 60 15( )V

150 15V

100 =150 mA | PZnom =15V 150mA( )= 2.25 W

IZmax = 60V 1.1( )

150 0.90( )15V 0.90( )1

150 0.90( )+1

100(1.1)

= 266 mA

PZmax =15V 0.90( )266mA( )= 3.59 W

IZmin = 60V 0.90( )

150 1.1( ) 15V 1.1( )1

150 1.1( )+1

100(0.9)

= 43.9 mA

PZmin =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.

T = 1120

2VrVP

| Vr = ITC =5

0.1 60( )= 0.8333VT = 1

1202 0.8333( )

6 = 1.40 ms

3.86

VD = nVT ln 1+ IDIS

= 2 0.025V( )ln 1+ 48.6A109 A

=1.230 V

62

3.87

Von = nVT ln 1+ IDIS

| VD = Von + ID RS

VD =1.6 0.025V( )ln 1+ 100A108 A

+100A 0.01( )=1.92 V

Pjunction VonIDC = Von IPT2T = 0.92V100A

2

1ms16.7ms

= 2.75 W

PR 43T

T

IDC

2 RS = 4316.7ms

1ms

3A( )

20.01 = 2.00 W

Ptotal = 4.76 W

3.88

VDC = 1T v t( )dt0T = 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 iD2 t( )RS dt

0

T = 1T IP2 1 tT

2

RS dt0

TPD = IP

2 RST

1 2tT +t 2

T 2

2

dt0

T = IP2 RST t t2

T +t 3

3T 2

0

T

PD = IP2 RST

T T + T3

=

13

IP2 RS

TT

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 = CVP = 2 60( )1.05( )6.3 2( )= 3530 A(e) T = 1

2VrVP

= 12 60( )

2 .25( )6.3 2

= 0.628ms | IP = Idc 2TT =7.91.5

260

1.628ms

= 841A

3.92

VOnom = VP Von( )= 6.3 2 1( )= 7.91V

VOmax = VPmax Von( )= 6.3 1.1( ) 2 1[ ]= 8.80V

VOmin = 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 = 9.00V0.25 160s 10.5F =1.20VISC = CVP = 2 60( )0.5( )10( )=1890A | T = 1 2VrVP =

12 60( )

2 1.2( )10

=1.30ms

IP = Idc 2TT =9

0.252

601

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 = CVP = 2 400( )0.158( )6.3 2( )= 3540 A(e) T = 1

2VrVP

= 12 400( )

2 .25( )6.3 2

= 94.3s | IP = Idc 2TT =7.91.5

2400

194.3s = 839A

3.95

(a) Vdc = VP Von( )= 6.3 2 1( )= 7.91V (b) C = I TVr = 7.910.25 10.5 1105 = 633F(c) PIV 2VP = 2 6.3 2 =17.8V (d ) Isurge = CVP = 2 105( )633F( )6.3 2( )= 3540A(e) T = 1

2VrVP

= 12 105( )

2 .25( )6.3 2

= 0.377s | IP = Idc 2TT =7.91.5

2105

10.377s = 839A

3.96

65

(a) C = I TVr

= 13000 0.01( )

160

= 556 F (b) PIV 2VP = 2 3000 = 6000V

(c) Vrms = 30002

= 2120 V (d ) T = 12VrVP

= 12 60( ) 2 0.01( )= 0.375ms

IP = Idc 2TT =12

60

10.375ms

= 88.9A (e) Isurge = CVP = 2 60( )556F( )3000( )= 629A

3.97 Assuming Von = 1 V:

C = VP VonVr

T 1R

= 10.025

160

303.3

= 6.06 F | PIV = 2VP = 2 3.3 +1( )V = 8.6 V | Vrms = 3.3 +12 = 3.04 V

T = 12TRC

VP VonVP

= 12 60( )

20.110 6.06F( )

160

s

3.3V4.3V

= 0.520 ms

IP = Idc 2TT = 30260

s

10.520ms

=1920 A | Isurge = CVP = 2 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 = 20.2V0.5 10.25V 1120s =1.35 F(c) PIV 2VP = 2 15 2 = 42.4 V (d ) Isurge = CVP = 2 60( )1.35( )15 2( )=10800 A(e) T = 1

2VrVP

= 12 60( )

2 .25( )15 2

= 0.407 ms | IP = Idc TT =20.2V0.5

160

s

10.407ms

=1650 A

3.101

(a) Vdc = VP Von( )= 9 2 1( )= 11.7 V (b) C = I Vr T2 = 11.7V0.5 10.25V 1120s = 0.780 F(c) PIV 2VP = 2 9 2 = 25.5 V (d ) Isurge = CVP = 2 60( )0.780( )9 2( )= 3740 A(e) T = 1

2VrVP

= 12 60( )

2 .25( )9 2

= 0.526 ms | IP = Idc TT =11.7V0.5

160

s

10.407ms

= 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 = 13.431

8001

22000F = 0.254 V

T = 1120

2VrVP

= 1120

2 0.254( )14.1

= 0.504 ms

IP = Idc TT =13.4V

31

60s 1

0.504 ms=150 A

Simulation with RS = 0.02 .

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 VonVr

T 1R

= 10.025

1s120

30A3.3V

= 3.03 F (b) PIV = 2VP = 2 3.3+1( )V = 8.6 V

(c) Vrms = 3.3+12

= 3.04 V (d) T = 12VrVP

= 12 60( )

2 0.025( )3.3( )4.3

= 0.520 ms

(e) IP = Idc TT = 30A1

60s

10.520ms

= 962 A | Isurge = CVP = 2 60 / s( )3.03F( )4.3V( )= 4910 A

3.104

(a) C = IVr

T2

= 13000 0.01( )

12 120 =139 F (b) PIV 2VP = 6000 V

(c) VS = 30002

= 2120 V (d ) T = 1 2VrVP

= 12 60( ) 2 0.01( )= 0.375 ms

IP = Idc TT =1160

s

10.375ms

= 44.4 A (e) Isurge = CVP = 2 60 / s( )139F( )3000V( )=157 A

3.105 The circuit is behaving like a half-wave rectifier. The capacitor should charge during the first 1/2 cycle, but it is not. Therefore, diode D1 is not functioning properly. It behaves as an open circuit.

3.106

(a) Vdc = VP 2Von( )= 15 2 2( )= 19.2 V (b) C = I Vr T2 = 19.2V0.5 10.25V 1120 s =1.28 F(c) PIV VP =15 2 = 21.2 V (d ) Isurge = CVP = 2 60 / s( )1.28F( )15 2( )=10200 A(e) T = 1

2VrVP

= 12 60( )

2 .25( )15 2

= 0.407 ms | IP = Idc TT =19.2V0.5

1s60

10.407ms

=1570 A

3.107

(a) C = I Vr

T2

=

1A3000V 0.01( )

1120

s

= 278 F (b) PIV VP = 3000 V

(c) VS = 30002

= 2120 V (d ) T = 12VrVP

= 12 60( ) 2 0.01( )= 0.375 ms

IP = Idc TT =1A160

s

10.375ms

= 44.4 A (e) Isurge = CVP = 2 60( )278F( )3000( )= 314 A

69

3.108

(a) C = I Vr

T2

=

30A0.025( )3.3V( )

1120

s

= 3.03 F (b) PIV Vdc + 2Von = 3.3+ 2( )= 5.3 V

(c) Vrms = 5.32

= 3.75 V (d ) T = 12VrVP

= 12 60( ) 2

0.025 3.3( )5.3

= 0.468 ms

IP = Idc TT = 30A160

s

10.468ms

=1070 A (e) Isurge = CVP = 2 60 / s( )3.03F( )3.3V( )= 3770 A

3.109 V1 = VP - Von = 49.3 V and V2 = -(VP -Von) = -49.3V.

3.110 *Problem 3.110 VS1 1 0 DC 0 AC 0 SIN(0 35 60) VS2 0 2 DC 0 AC 0 SIN(0 35 60) D1 1 3 DIODE D4 2 3 DIODE D2 4 1 DIODE D3 4 2 DIODE C1 3 0 0.1 C2 4 0 0.1 R1 3 0 500 R2 4 0 500 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 10US 50MS .PRINT TRAN V(3) V(4) .PROBE V(3) V(4) .END

0s 10ms 20ms 30ms 40ms 50ms

Time

40V

20V

0V

-20V

-40V

v1

v2

3.111

(a) Vdc = VP 2Von( )= 15 2 2( )= 19.2 V (b) C = I Vr T2 = 19.2V0.5 10.25 1120 =1.28 F(c) PIV VP =15 2 = 21.2 V (d ) Isurge = CVP = 2 60 / s( )1.28F( )15V 2( )=10200 A(e) T = 1

2VrVP

= 12 60( )

2 .25( )15 2

= 0.407 ms | IP = Idc TT =19.2V0.5

160

s

10.407ms

=1570 A

70

3.112 3.3-V, 15-A power supply with Vr 10 mV. Assume Von = 1 V.

Rectifier Type Half Wave Full Wave Full Wave Bridge

Peak Current 533 A 266 A 266 A

PIV 8.6 V 8. 6 V 5.3 V

Filter Capacitor

25 F 12.5 F 12.5 F

(i) The large value of C suggests we avoid the half-wave rectifier. This will reduce the cost and size of the circuit. (ii) The PIV ratings are all low and do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 15 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components.

3.113 200-V, 3-A power supply with Vr 4 V. Assume Von = 1 V.


Peak Current 189 A 94.3 A 94.3 A

PIV 402 V 402 V 202 V

Filter Capacitor

12,500 F 6250 F 6250 F

(i) The the half-wave rectifier requires a larger value of C which may lead to more cost. (ii) The PIV ratings are all low enough that they do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 3 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components.

71

3.114 3000-V, 1-A power supply with Vr 120 V. Assume Von = 1 V.


Peak Current 133 A 66.6 A 66.6 A

PIV 6000 V 6000 V 3000 V

Filter Capacitor

41.7 F 20.8 F 20.8 F

(i) A series string of multiple capacitors will normally be required to achieve the voltage rating. (ii) The PIV ratings are high, and the bridge circuit offers an advantage here. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers but neither is prohibitively large. (iv) We must choose between use of a center-tapped transformer (full wave) or extra diodes (bridge). With a PIV of 3000 or 6000 volts, multiple diodes may be required to achieve the require PIV rating.

3.115

iD 0+( )= 5V1k = 5 mA | IF = 5 VD1k = 5 0.61k = 4.4 mA

Ir = 3 0.61k = 3.6 mA | S = 7ns( ) ln 1 4.4mA3.6mA

= 5.59 ns

3.116 *Problem 3.143 - Diode Switching Delay V1 1 0 PWL(0 0 0.01N 5 10N 5 10.02N -3 20N -3) R1 1 2 1K D1 2 0 DIODE .TRAN .01NS 20NS .MODEL DIODE D TT=7NS IS=1E-15 .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END

0s 5ns 10ns 15ns 20ns

Time

10

5

0

-5

-10

v1

vD

Simulation results give S = 4.4 ns.

72

3.117

iD 0+( )= 5V5 =1 A | IF = 5Von5 = 5 0.61 = 0.880 A

IR = 3 0.65 = 0.720 A | S = 250ns( ) ln 1 0.880 A0.720A

= 200 ns

3.118 *Problem 3.145(a) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 0 .01N 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 0s 5ns 10ns 15ns 20ns 25ns

Time

2.0

1.0

0

-1.0

-2.0

v1

vD

For this case, simulation yields S = 3 ns.

*Problem 3.145(b) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END

0s 10ns 20ns 30ns 40ns

Time

2.0

1.0

0

-1.0

-2.0

v 1

vD

For this case, simulation yields S = 15.5 ns.

73

In case (a), the charge in the diode does not have time to reach the steady-state value given by Q = (1mA)(50ns) = 50 pC. At most, only 1mA(7.5ns) = 7.5 pC can be stored in the diode. Thus is turns off more rapidly than predicted by the storage time formula. It should turn off in approximately t = 7.5pC/3mA = 2.5 ns which agrees with the simulation results. In (b), the diode charge has had time to reach its steady-state value. Eq. (3.103) gives: (50 ns) ln (1-1mA/(-3mA)) = 14.4 ns which is close to the simulation result.

3.119

IC =11015 exp 40VC( )1[ ] A | For VC = 0, ISC =1AVOC = 140 ln 1+

11015

= 0.864 V

P = VC IC = VC 11015 exp 40VC( )1[ ] [ ]dPdVC

=11015 exp 40VC( )1[ ] 40x1015VC exp 40VC( )= 0

Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts

3.120 (a) For VOC, each of the three diode teminal currents must be zero, and

VOC = VC1 + VC 2 + VC 3 = 140V ln 1.05x1015( )+ ln 1.00x1015( )+ ln 0.95x1015( )[ ]= 2.59 V

(b) For ISC, the external currents cannot exceed the smallest of the short circuit currentof the individual diodes. Thus, ISC = min 1.05A,1.00A,0.95A[ ]= 0.95 ANote that diode three will be reversed biased in part (b).

Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts

3.121

= hcE

(a) = 6.625x1034 J s 3x108 m / s( )

1.12eV 1.602x1019 j / eV( ) =1.11 m - far infrared(b) = 6.625x10

34 J s 3x108 m / s( )1.42eV 1.602x1019 j / eV( ) = 0.875 m - near infrared

74

CHAPTER 4

4.1 (a) VG > VTN corresponds to the inversion region (b) VG

4.6

a( ) vn = nE = 500 cm2V s

2000

Vcm

= 1.00x10

6 cms

a( ) vn = nE = 400 cm2V s

4000

Vcm

= 1.60x10

6 cms

4.7 The carrier velocity must increase as the carriers travel down the channel to compensate for the decrease in carrier density

4.8

a( ) 0 < 0.8V ID = 0(b) VGS - VTN = 0.2V , VDS = 0.25V Saturation regionID = Kn

'

2WL

VGS VTN VDS2

VDS =

2002

AV 2

5m0.5m

1 0.8( )

2 = 40.0 A(c) VGS -VTN = 1.2V , VDS = 0.25V triode regionID = Kn' WL VGS VTN

VDS2

VDS = 200

AV 2

5m0.5m

2 0.8

0.252

0.25( )= 538 A

(d ) VGS - VTN = 2.2V , VDS = 0.25V triode regionID = Kn' WL VGS VTN

VDS2

VDS = 200

AV 2

5m0.5m

3 0.8

0.252

0.25( )=1.04 mA

e( ) Kn = Kn' WL = 200 AV 2

5m0.5m

= 2.00

mAV 2

4.9

a( ) Kn = Kn' WL = 200 AV 2 60m3m

= 4.00

mAV 2

b( ) Kn = 200 AV 2 3m0.15m

= 4.00

mAV 2

| c( ) Kn = 200 AV 2 10m0.25m

= 8.00

mAV 2

4.10

a( ) 0

(d) VGS - VTN = 2V , VDS = 0.1V triode regionID = Kn' WL VGS VTN

VDS2

VDS = 250

AV 2

10m1m

31

0.12

0.1( )= +488 A

e( ) Kn = Kn' WL = 250 AV 2

10m1m

= 2.50

mAV 2

4.11 Identify the source, drain, gate and bulk terminals and find the current I in the transistors in Fig. P-4.3.

-0.2 V

I

+5

+0.2 V

I

+5

(a) (b)

WL

= 101

WL

= 101

D

G

S

B

D

G

S

B

+

-V

GS

+

-V

GS

+

-

VDS

+

-

VDS

(a)

VGS = VG VS = 5V VDS = VD VS = 0.2V Triode region operationI = ID = Kn' WL VGS VTN

VDS2

VDS = 100

AV 2

101

5 0.70

0.22

0.2 = +840 A

(b)

VGS = VG VS = 5 0.2( )= 5.2V VDS = VD VS = 0 0.2( )= 0.2V I = IS = 100 AV 2

101

5.2 0.70

0.22

0.2 = 880 A

4.12

(a) I = 100 AV 2

101

5 0.75

0.52

0.5 = 2.00 mA

(b) I = 100 AV 2

101

3 0.75

0.22

0.2 = 430 A

4.13

a( ) VGS = VG VS = 5 0.5( )= 5.5 V VDS = VD VS = 0 0.5( )= 0.5 V I = IS = 100 AV 2

101

5.5 0.75

0.52

0.5 = 2.25 mA

b( ) VGS = VG VS = 3 1( )= 4 V VDS = VD VS = 0 1( )=1 V I = IS = 100 AV 2

101

4 0.75

12

1= 2.75 mA

77

4.14

Kn = Kn' WL a( ) W = KnKn' L =

4mA /V 2

100A /V 2 0.5m( )= 20 m b( ) W = 800A /V2

100A /V 2 0.5m( )= 4 m

4.15

a( ) Ron = 1Kn

' WL

VGS VTN( ) = 1

100x106 1001

5 0.65( )

= 23.0

b( ) Ron = 1100x106 100

1

3.3 0.50( )

= 35.7

4.16

Ron = 1Kn

' WL

VGS VTN( ) or W

L= 1

Kn' VGS VTN( )Ron

(a) WL

= 1100x106 5 0.75( )500( )=

4.711

| (b) WL

= 1100x106 3.3 0.75( )500( )=

7.841

4.17

Ron 0.1V5A = 0.020 = 20m | Kn =ID

VGS VTN 0.5VDS( )VDS =5

5 2 0.5 0.1( )( )0.1( )=17.0A

V 2

Note that this will require WL

= KnKn

' = 17.00.1 =170

1

4.18 Picking two values in saturation :

395A = Kn2

4 VTN( )2 and 140A = Kn2 3 VTN( )2

Taking the ratio of these two equations :

395140

= 4 VTN( )23 VTN( )2 VTN =1.5 V Kn =125

AV 2

WL

=125 A

V 2

100 AV 2

= 1.251

From the graph, VTN is somewhat less than 2 V. VTN > 0 enhancement - mode transistor

78

4.19 Using the parameter values from problem 4.22:

0V 1.0V 2.0V 3.0V 4.0V 5.0V 6.0V

800uA

600uA

400uA

200uA

0A

Drain-Source Voltage (V)

Dra

in C

urre

nt (A

)

VGS = 5 V

VGS = 4.5 V

VGS = 4 V

VGS = 3.5 V

VGS = 3 V

VGS = 2.5 V

VGS = 2 V

4.20a( ) For VGS = 0, VGS VTN and ID = 0 b( ) For VGS =1 V , VGS = VTN and ID = 0c( ) VGS VTN = 2 -1 =1V and VDS = 3.3 | VDS > VGS VTN( ) so the saturation region is correct

ID = 3752AV 2

5m0.5um

2 1( )

2V 2 =1.88 mA | Kn = Kn' WL = 375AV 2

5m0.5um

= 3.75

mAV 2

d( ) VGS VTN = 3-1 = 2V and VDS = 3.3 | VDS > VGS VTN( ) so the saturation region is correct ID = 3752

AV 2

5m0.5m

31( )

2V 2 = 7.50 mA

4.21

a( ) For VGS = 0, VGS < VTN and ID = 0 b( ) For VGS =1 V , VGS < VTN and ID = 0c( ) VGS VTN = 2 -1.5 = 0.5V and VDS = 4 | VDS > VGS VTN( ) so the saturation region is correct

ID = 2002AV 2

10m1um

2 1.5( )

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Solutions Manual Microelectronic Circuit Design 4th Ed