1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Modems and routers Calculator Camcorder Carbon monoxide detector Cash register CD and DVD players Ceiling fan (remote) Cellular phones Coffee maker Compass Copy machine Cordless phone Depth finder Digital Camera Digital watch Digital voice recorder Digital scale Digital thermometer Electronic dart board Electric guitar Electronic door bell Electronic gas pump Elevator Exercise machine Fax machine Fish finder Garage door opener GPS Hearing aid Invisible dog fences Laser pointer LCD projector Light dimmer Keyboard synthesizer Keyless entry system Laboratory instruments Metal detector Microwave oven

Model airplanes MP3 player Musical greeting cards Musical tuner Pagers Personal computer Personal planner/organizer (PDA) Radar detector Broadcast Radio (AM/FM/Shortwave) Razor Satellite radio receiver Security systems Sewing machine Smoke detector Sprinkler system Stereo system Amplifier CD/DVD player Receiver Tape player Stud sensor Talking toys Telephone Telescope controller Thermostats Toy robots Traffic light controller TV receiver & remote control Variable speed appliances Blender Drill Mixer Food processor Fan Vending machines Video game controllers Wireless headphones & speakers Wireless thermometer Workstations Electromechanical Appliances* Air conditioning and heating systems Clothes washer and dryer Dish washer Electrical timer Iron, vacuum cleaner, toaster Oven, refrigerator, stove, etc. *These appliances are historically based only upon on-off (bang-bang) control. However, many of the high end versions of these appliances have now added sophisticated electronic control.

1-1

R. C. Jaeger & T. N. Blalock 6/9/06

1.2B = 19.97 x 100.1997(20201960) = 14.5 x 1012 = 14.5 Tb/chip

1.3 (a) 0.1977(Y2 1960) B2 19.97 x10 0.1977(Y2 Y1 ) 0.1977(Y2 Y1 ) = = 10 so 2 = 10 0.1977(Y1 1960) B1 19.97 x10Y2 Y1 = log2 = 1.52 years 0.1977

(b)

Y2 Y1 =

log10 = 5.06 years 0.1977

1.4

N = 1610 x101.5

0.1548(20201970)

= 8.85 x 1010 transistors/P(2 ) N 2 1610 x10 0.1548(Y2 Y1 ) = = 10 0.1548(Y1 1970) N1 1610 x10 log2 ( a ) Y2 Y1 = = 1.95 years 0.1548 log10 ( b ) Y2 Y1 = = 6.46 years 0.15480.1548 Y 1970

1.6

F = 8.00 x10

0.05806(20201970)

m = 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/P will be 8.85 x 1010. in 2020. Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors.

1-2

R. C. Jaeger & T. N. Blalock 6/9/06

1.8

1.5W tube)= 113 MW! P = 75 x106 tubes (1.9 1.10

(

)

I=

1.13 x 108W = 511 kA! 220V

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

10.24V 10.24V 10.24V = = 2.500 mV VMSB = = 5.120V 12 2 2 bits 4096 bits 1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500 mV )= 5.855 V VLSB =1.11VLSB = 5V mV 5V = = 19.53 bit 2 bits 256 bits8

and

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

2.77V = 142 LSB mV 19.53 bit

1.12VLSB = 2.5V 2.5V mV = = 2.44 bit 2 bits 1024 bits10

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

(

)

10

= 36510

2.5V VO = 365 = 0.891 V 1024

1.13

mV 6.83V 14 10V = 0.6104 and 2 bits = 11191 bits 14 10V bit 2 bits 1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1) 10 VLSB = 1119110 = 1010111011011121.14

(

)

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.1.15VLSB = 5.12V mV V 5.12V = 1.25 and VO = ( 1011101110112 ) VLSB LSB = 12 bit 2 2 bits 4096 bits 11 9 8 7 5 4 3 VO = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV 0.0625V

(

)

10

VO = 3.754 0.000625 or 3.753V VO 3.755V

1-3 6/9/06

1.16

IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A1.17

VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000 t Volts1.18

vCE = [5 + 2 cos (5000t)] V1.19

vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V1.20

V = 10 V, R1 = 22 k, R2= 47 k and R3 = 180 k.+V1

I2 R2

R V

1

+V

I

3

2

R

3

-

V1 = 10V V2 = 10V

22 k + 47 k 180 k

(

22 k

)

= 10V

22 k = 3.71 V 22 k + 37.3k

37.3k = 6.29 V Checking : 6.29 + 3.71 = 10.0 V 22 k + 37.3k 180 k 180 k 10V I2 = I1 = = 134 A 47 k + 180 k 22 k + 37.3k 47 k + 180k 47k 47k 10V I3 = I1 = = 34.9 A 47 k + 180k 22 k + 37.3k 47 k + 180k 10V = 169A and I1 = I2 + I3 22 k + 37.3k

Checking : I1 =

1-4

R. C. Jaeger & T. N. Blalock 6/9/06

1.21

V = 18 V, R1 = 56 k, R2= 33 k and R3 = 11 k.+V1

I2

R V

1

+V

I3 R

R

2

2

3

-

V1 = 18V

56 k

56 k + 33k 11k 18V

(

)

= 15.7 V

V2 = 18V

33k 11k

56k + 33k 11k

(

)

= 2.31 V

Checking : V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct. I1 = 56k + 33k 11k

(

)

= 280 A

I2 = I1

11k 11k = (280 A) = 70.0 A 33k + 11k 33k + 11k Checking : I2 + I3 = 280 A

I3 = I1

33k 33k = (280 A) = 210 A 33k + 11k 33k + 11k

1.22

I1 = 5mA

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

I2 = 5mA

2.4 k = 1.03 mA 9.2 k + 2.4 k

V3 = 5mA 2.4 k 9.2 k

(

k = 3.72V )5.6k3.6 + 3.6 k and

Checking : I1 + I2 = 5.00 mA1.23

I2 R2 = 1.03mA(3.6 k)= 3.71 V

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

(

)

1-5 6/9/06

1.24+v v R1

s

+ g vm

v

th

-

Summing currents at the output node yields: v + .002v = 0 so v = 0 and v th = vs v = vs 5x10 4+v R1

g vm

ix

vx

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

Thvenin equivalent circuit:

495 vs

1-6

R. C. Jaeger & T. N. Blalock 6/9/06

1.25 The Thvenin equivalent resistance is found using the same approach as Problem 1.24, and 1 1 Rth = + .025 = 39.6 4 k +v vs R1

g vm

in

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

0.0253v

s

39.6

1-7 6/9/06

1.26 (a)i vs R1

+ R2 v th -

i

Vth = Voc = i R2

but

i =

vs R1

andix

Vth = vs

R2 39 k = 120 vs = 46.8 vs R1 100 k

i R1

R2 i

Rth

v

x

Rth =

vx ; ix

ix =

vx + i R2

but

i = 0 since VR 1 = 0.

Rth = R2 = 39 k.

Thvenin equivalent circuit:

39 k 58.5v

s

(b)i is

+ R2 v th -

R

1

i

i Vth = Voc = i R2 where i + bi + is = 0 and Vth = s R2 = 38700 is + 1

1-8

R. C. Jaeger & T. N. Blalock 6/9/06

i R1

R2 i

Rth

v

x

Rth =

vx ; ix

ix =

vx + i but R2

i + i = 0 so i = 0

and Rth = R2 = 39 k

Thvenin equivalent circuit:

39 k

38700i s1.27

i vs R1

R2 i

in

in = i but i =

vs R1

and in =

R1

vs =

100 vs = 1.33 x 103 vs 75k Norton equivalent circuit:

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

0.00133v

s

56 k

1-9 6/9/06

1.28is vs

i R1

R2 i

is =

vs v v +1 i = s + s = vs R1 R1 R1 R1

R=

vs R 100k = 1 = = 1.24 is + 1 81

1.29 The open circuit voltage is vth = g mv R2 and v = +is R1.

vth = g m R1 R2is = (0.0025) 105 106 i s= 2.5 x 108 is For is = 0, v = 0 and Rth = R2 = 1 M

( )( )

1.30

5V 3V f (Hz) 0 01.31

500

1000

2V f (kHz) 0 9 10 11

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

v = 4sin (20000t )sin (2000t )=

[

]

236 o A = 5 0 = 2 x105 36 o 10 0

A = 2 x105

A = 36o

1-10

R. C. Jaeger & T. N. Blalock 6/9/06

1.33

(a) A =1.34

101 12 o 102 45o o A = = 5 45 = 100 12 o (b) 2 x103 0 o 103 0 o R2 620 k 180 k = = 44.3 (b) Av = = 10.0 14k R1 18 k 62 k = 38.8 1.6 k

(a) Av = 1.35vo (t ) =

(c) Av =

R2 v s (t )= (90.1 sin 750t ) mV R1

IS =

VS 0.01V = = 11.0A and R1 910

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 vo v + i + = 0 R2 R1

or

vs v o vs + =0 R2 R1

and A v =

vo R = 1+ 2 vs R1

1.38 Writing a nodal equation at the inverting input terminal of the op amp givesv v v1 v v2 v + = i + o R1 R2 R3 vo =

but v- = v+ = 0

and

i- = 0

R3 R v1 3 v2 = 0.255sin 3770t 0.255sin10000t volts R1 R2

1-11 6/9/06

1.39

b b b 0 1 1 1 0 0 vO = VREF 1 + 2 + 3 (a) vO = 5 + + = 1.875V (b) vO = 5 + + = 2.500V 2 4 8 2 4 8 2 4 8

b1b2b3

vO (V) 0 -0.625 -1.250 -1.875 -2.500 -3.125 -3.750 -4.375

000 001 010 011 100 101 110 111

1.40 Low-pass amplifier

Amplitude 10

f 6 kHz

1-12

R. C. Jaeger & T. N. Blalock 6/9/06

1.41 Band-pass amplifier

Amplitude 20

f 1 kHz 5 kHz

1.42

High-pass amplifier

Amplitude 16

f 10 kHz1.4315000t ) vO (t ) = 10 x 5sin (2000t )+ 10 x 3cos(8000t )+ 0 x 3cos( vO (t ) = 50sin(2000t )+ 30cos(8000t ) volts

[

]

1.4412000t ) vO (t ) = 20 x 0.5sin (2500t )+ 20 x 0.75cos(8000t )+ 0 x 0.6cos( vO (t ) = 10.0sin (2500t )+ 15.0cos(8000t ) volts

[

]

1.45 The gain is zero at each frequency:

vo(t) = 0.

1-13 6/9/06

1.46

t=linspace(0,.005,1000); w=2*pi*1000; v=(4/pi)*(sin(w*t)+sin(3*w*t)/3+sin(5*w*t)/5); v1=5*v; v2=5*(4/pi)*sin(w*t); v3=(4/pi)*(5*sin(w*t)+3*sin(3*w*t)/3+sin(5*w*t)/5); plot(t,v) plot(t,v1) plot(t,v2) plot(t,v3)2 1 0 -1 -2 0

1

2

3

4

(a)10 5 0 -5 -10 0

5 x10-3

1

2

3

4

(b)

5 x10-3

1-14

R. C. Jaeger & T. N. Blalock 6/9/06

10 5 0 -5 -10 0

1

2

3

4

(c)

5 x10-3

10 5 0 -5 -10 0(d)1.47

1

2

3

4

5 x10-3

( c ) 3000( 1 .10) R 3000( 1 + .10) or 2700 R 3300 Vnom = 2.5V V 0.05V 0.05 = 0.0200 or 2.00% 2.50

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

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

1.48

T=

1.49 1.50

1 .5) C 20000F ( 1 + .2) or 10000F R 24000F 20000F ( 8200( 1 0.1) R 8200( 1 + 0.1) or 7380 R 9020 The resistor is within the allowable range of values.

1-15 6/9/06

1.51

(a) 5V ( 1 .05) V 5V ( 1 + .05) or 5.75V V 5.25V V = 5.30 V exceeds the maximum range, so it is out of the specification limits. (b) If the meter is reading 1.5% high, then the actual voltage would be 5.30 = 5.22V which is within specifications limits. Vmeter = 1.015Vact or Vact = 1.0151.52

TCR = R nom

R 6562 6066 = = 4.96 o T 100 0 C = R o + TCR (T)= 6066 + 4.96(27)= 62000 C

1-16

R. C. Jaeger & T. N. Blalock 6/9/06

1.53+R V1

V1

I2 R2

+V2

I3 R3

-

Let RX = R2 R3

then V1 = V

R1 = R1 + RX

R min X = V1max =

47k(0.9)+ 180 k(0.9) 10( 1.05) 33.5k 1+ 22 k( 1.1) = 4.40V

180 k)(0.9) 47 k(0.9)(

V1 R 1+ X R1

= 33.5k R max = X V1min =

10(0.95)

47 k( 1.1)+ 180 k( 1.1) = 3.09V

180 k)( 1.1) 47 k( 1.1)(

= 41.0 k

41.0 k 1+ 22 k(0.9) V R1 R2 R3

I1 =

V R1 + RX

and I2 = I1

R3 = R2 + R3

R1 + R2 +

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)+ R2 = R2 + R3 V R1 + R3 + R1 R3 R2

1.1) 22000( 1.1)(47000)( 180000(0.9)

= 114 A

I3 = I1

I3max =

10( 1.05)

22000(0.9)+ 180000(0.9)+

180000)(0.9) 22000(0.9)( 47000( 1.1)

= 43.1 A

I3min =

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

180000)( 1.1) 22000( 1.1)( 47000(0.9)

= 28.3 A

1-17 6/9/06

1.54I1 = I

R2 + R3 =I R1 + R2 + R3

1 1+ R1 R2 + R3

and similarly I2 = I

I1max = 1+ I2max = 1+

5600( 1.05)+ 3600( 1.05) 5600(0.95)+ 3600(0.95) 2400( 1.05) I 1 1 R + + 2 R1 R3 R1 R3 5( 1.02) mA = 1.14 mA I2min =

2400(0.95)

5( 1.02)

1 R + R3 1+ 2 R1

mA = 4.12 mA

I1min = 1+

5600(0.95)+ 3600(0.95) 1.05) 5600( 1.05)+ 3600( 2400(0.95) 5(0.98) mA = 0.936 mA

2400( 1.05)

5(0.98)

mA = 3.80 mA

1+

V3 = I2 R3 =

V3max =

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

5( 1.02)

= 4.18 V

V3min =

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

= 3.30 V

1.55From Prob. 1.24 : Rth = 1 gm + 1 R1 = 619 min Rth =

max Rth =

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

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

= 412

1-18

R. C. Jaeger & T. N. Blalock 6/9/06

1.56

For one set of 200 cases using the equations in Prob. 1.53. R1 = 22000 * (0.9 + 0.2 * RAND())

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

R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND()) V1 Min Max Average 3.23 V 3.71 V 3.71 V I2 116 A 151 A 133 A I3 29.9 A 40.9 A 35.1 A

1.57

For one set of 200 cases using the Equations in Prob. 1.54: R1 = 2400 * (0.95 + 0.1* RAND())

R1 = 5600 * (0.95 + 0.1* RAND()) R3 = 3600 * (0.95 + 0.1* RAND()) I1 Min Max Average 3.82 mA 4.09 mA 3.97 mA I2 0.96 mA 1.12 mA 1.04 mA V3 3.46 V 4.08 V 3.73 V

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

1.58 1.59

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

1-19 6/9/06

CHAPTER 22.1 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 I max 2.4 108 cm2 7 A = 10 1m) = 500 mA (5m)( 2 cm 2 m

EG ni = BT 3 exp 5 8.62 x10 T 31 For silicon, B = 1.08 x 10 and EG = 1.12 eV:ni = 2.01 x10 /cm30 3 -10 3

6.73 x10 /cm 2.27 x10 /cm13 3

9

3

8.36 x 10 /cm . 8.04 x 10 /cm .15 3

13

3

For germanium, B = 2.31 x 10 and EG = 0.66 eV: ni = 35.9/cm 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 /cm 2.614 3

for T = 506 K

ni = 10 /cm3 for T = 739 K

16

EG ni = BT 3 exp 5 8.62 x10 T T = 100 K: ni = 6.03 x 10-19/cm3

with

B = 1.27x1029 K 3cm66 3 11 3

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

T = 500 K: ni = 2.79 x10 /cm

20

2.7

cm2 V 6 cm vn = n E = 700 2500 = 1.75 x10 V s cm s V cm2 5 cm v p = + p E = +250 2500 = +6.25 x10 cm V s s 1 cm 4 A jn = qnvn = 1.60 x1019 C 1017 3 1.75 x106 = 2.80 x10 s cm cm2 1 cm 10 A j p = qnv p = 1.60 x1019 C 103 3 6.25 x105 = 1.00 x10 s cm 2 cm

(

)

(

)

2.8

E ni2 = BT 3 exp G kT 10 2 31

B = 1.08 x1031

(10 ) = 1.08 x10

1.12 T 3 exp 8.62 x105 T 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 K2.9

v=

j 1000 A / cm 2 cm = = 105 2 Q s 0.01C / cm

2.10C cm MA 6 A j = Qv = 0.4 3 10 7 =4 2 = 4 x10 2 cm sec cm cm

21

2.11

V cm2 6 cm vn = n E = 1000 2000 = +2.00 x10 V s cm s V cm 2 5 cm v p = + p E = +400 2000 = 8.00 x10 V s cm s 1 cm 10 A jn = qnvn = 1.60 x1019 C 103 3 +2.00 x106 = 3.20 x10 s cm2 cm 1 cm 4 A j p = qnv p = 1.60 x1019 C 1017 3 8.00 x105 = 1.28 x10 s cm cm2

(

)

(

)

2.12

(a )2.13

E=

V 5V = 5000 4 cm 10 x10 cm

(b )

V 4 V = 105 10 x10 cm = 100 V cm

(

)

1019 cm 7 A j p = qpv p = 1.60 x1019 C 3 10 7 = 1.60 x10 s cm2 cm A i p = j p A = 1.60 x10 7 2 1x104 cm 25 x104 cm = 4.00 A cm

(

)

(

)(

)

2.14 For intrinsic silicon, = q (n ni + p ni )= qni (n + p )

1000( cm) for a conductor1

ni

cm 2 1.602 x1019 C ( 100 + 50) v sec 39 1.73 x10 E n2 = BT 3 exp G with i = 6 cm kT

q (n + p )

=

1000( cm)

1

=

4.16 x1019 cm3

B = 1.08 x1031 K 3cm6 , k = 8.62x10-5 eV/K and EG = 1.12eV

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

22

2.15 For intrinsic silicon, = q (n ni + p ni )= qni (n + p )

105 ( cm) for an insulator1

ni

cm 2 1.602 x10 C (2000 + 750) v sec E 5.152 x1020 n2 = = BT 3 exp G with i 6 cm kT

q (n + p )

=

105 ( cm)

1

(

19

)

=

2.270 x1010 cm 3

B = 1.08 x1031 K 3cm6 , k = 8.62x10-5 eV/K and EG = 1.12eV

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

Si

Si

Si

Donor electron fills acceptor vacancy

P

B

Si

Si

Si

Si

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. 2.20E= A V = j = 10000 2 (0.02 cm ) = 200 , a small electric field. cm cm j

2.21

C cm A jndrift = qnn E = qnv n = 1.602 x1019 1016 3 10 7 = 16000 2 s cm cm

(

)( )

2.22 1015 atoms 104 cm 3 N = 10m)(0.5m ) 1m)( ( = 5,000 atoms 3 cm m

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

If we assume N A N D >> 2 ni = 1014 / cm 3 : p = N A N D = 9 x1014 /cm3 | n = If we use Eq. 2.12 : p =12 3

ni2 251026 = = 2.78 x1012 /cm3 p 9 x1014

9 x1014

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

(9 x10 ) + 4(5x10 ) = 9.03x1014 2 13 2

14

2.24 N A > N D: N A N D = 5 x1016 1016 = 4 x1016 /cm 3 >> 2ni = 2 x1011 /cm 3p = N A N D = 4 x1014 /cm 3 | n = ni2 10 22 = = 2.50 x10 5 /cm 3 p 4 x1016

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

2ni = 2 x1017 /cm3 ; Need to use Eq. (2.11) n= p= 1017 2 i

( ) ( ) = 1.62 x101017 + 4 1017 234 2 2

17

/cm3

n 10 = = 6.18 x1016 /cm3 n 1.62 x1017

24

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

(2.5x10 ) + 4(10 )18 2 10

2

2

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

ni2 = . 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. At room temperature, ni = 1010 /cm3 and N A N D = 6 x1018 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 6 x10 /cm and n = = = 16.7 /cm3 18 3 p 6 x10 /cm (b) 3 1.12 9 6 At 200K, ni2 = 1.08 x1031 (200) exp 8.62 x105 (200) = 5.28 x10 /cm 18 3

ni = 7.27 x10 4 /cm 3

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

5.28 x109 = 8.80 x1010 /cm 3 18 6 x10

2.28

(a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The material is n-type. At room temperature, n i = 1010 / cm 3 and N D N A = 3 x1017 / cm 3 >> 2n iSo n = 3 x1017 /cm 3 and p = 10 20 /cm 6 ni2 = = 333 /cm 3 17 3 n 3x10 /cm 3 1.12 15 6 (b) At 250K, ni2 = 1.08 x1031 (250) exp 8.62 x105 (250) = 4.53 x10 /cm ni = 6.73 x10 7 /cm 3 N D N A >> 2 ni , so n = 3 x1017 / cm 3 and n = 4.53x1015 = 0.0151/ cm 3 17 3x10

2.29

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

25

(b) At room temperature, n i = 1010 / cm3 and N A N D = 6 x1018 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 6 x10 /cm and n = = = 16.7 /cm3 18 3 p 6 x10 /cm18 3

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. (b) At room temperature, ni = 1010 /cm3 and N A N D = 3x1017 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 3x10 /cm and n = = = 333 /cm3 17 3 p 3x10 /cm17 3

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 = 4 x1016 / cm3 >> 2 ni = 2 x1010 / cm 3 n = 4 x1016 / cm3 | p = n2 1020 i = = 2.5 x103 / cm 3 16 n 4 x10 cm2 cm 2 and p = 310 V s V s

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

=

1 qn n

=

(

cm 2 4 x1016 1.602 x10 C 1030 V s cm 3 19

1

)

= 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 >> 2 ni = 2 x1010 / cm 3 p = 1018 / cm318

|

n=3

n2 1020 i = 18 = 100 / cm 3 p 10

cm2 cm 2 N D + N A = 10 / cm | Using Fig. 2.13, n = 375 and p = 100 V s V s 1 1 = = = 0.0624 cm q p p cm 2 1018 19 1.602 x10 C100 V s cm 3

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 = 7 x1019 /cm3 >> 2 ni = 2 x1010 /cm3 p = 7 x1019 /cm3 |19

n=3

ni2 1020 = = 1.43 /cm3 19 p 7 x10

cm2 cm 2 N D + N A = 7 x10 / cm | Using Fig. 2.13, n = 120 and p = 60 V s V s 1 1 = = = 1.49 m cm q p p cm 2 7 x1019 19 1.602 x10 C 60 3 V s cm

2.34 Phosphorus is a donor : N D = 5.5 x1016 / cm 3 | Boron is an acceptor : N A = 4.5 x1016 / cm 3N D > N A : material is n - type16 3

|

N D N A = 1016 / cm3 >> 2 ni = 2 x1010 / cm 3

ni2 1020 n = 10 /cm | p = = 16 = 10 4 /cm3 p 10 cm2 cm 2 N D + N A = 10 / cm | Using Fig. 2.13, n = 800 and p = 230 V s V s 1 1 = = = 0.781 cm q n n cm 2 1016 19 1.602 x10 C 800 V s cm 3 17 3

27

2.35

=

1 q p p

| p p =

(1.602 x10 C)(0.054 cm)19

1

=

1.16 x1020 V cm s

An iterative solution is required. Using the equations in Fig. 2.8:NA 1018 1.1 x1018 1.2 x 1017 1.3 x 1019 p 96.7 93.7 91.0 88.7 p p 9.67 x 1020 1.03 x 1020 1.09 x 1020 1.15 x 1020

2.36

8.32 x1018 = | p p = = q p p 1.602 x1019 C (0.75 cm) V cm s 1

(

1

)

An iterative solution is required. Using the equations in Fig. 2.8:NA 1016 2 x 1016 3 x 1016 2.4 x 1016 p 406 p p 4.06 x 1018 363 7.26 x 1018 333 1.00 x 1019 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 qn n

| n n n N D =

(1.602 x10 C)(2 cm)19

1

=

3.12 x1018 V cm s

An iterative solution is required. Using the equations in Fig. 2.8:ND 1015 2 x 1015 2.5 x 1015 n 1350 1330 1330 nn 1.35 x 1018 2.67 x 1018 3.32 x 1018

28

2.3 x 1015

1330

3.06 x 1018

29

2.39 (a) 1 1 6.24 x1021 = | n n n N D = = qn n 1.602 x1019 C (0.001 cm) V cm s

(

)

An iterative solution is required. Using the equations in Fig. 2.8:ND 1019 7 x 1019 6.5 x 1019 n 116 96.1 96.4 nn 1.16 x 1021 6.73 x 1021 6.3 x 1021

(b)

=

1 q p p

| p p p N A =

1 6.24 x10 21 = (1.602 x1019 C)(0.001 cm) V cm s

An iterative solution is required using the equations in Fig. 2.8:NA 1.3 x 1020 p 49.3 p p 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 1 6.25x1018 = | p p pN A = = q p p 1.602 x1019 C ( 1 cm) V cm s

(

)

An iterative solution is required. Using the equations in Fig. 2.8:NA 1016 1.5 x 1016 1.7 x 1016 p 406 383 374 p p 4.1 x 1018 5.7 x 1018 6.4 x 1019

30

To change the resistivity to 0.25 ohm-cm: 1 1 2.5 x1019 = | p p pN A = = q p p 1.602 x1019 C (0.25 cm) V cm s

(

)

NA 6 x 1016 8 x 1016 1.1 x 1017

p 276 233 225

p p 1.7 x 1019 2.3 x 1019 2.5 x 101917 16 16 3

Additional acceptor concentration = 1.1x10 - 1.7x10 = 9.3 x 10 /cm (b) If donors are added:ND 2 x 1016 1 x 1017 8 x 1016 4.1 x 1016 ND + NA 3.7 x 1016 1.2 x 1017 9.7 x 1016 5.8 x 101616 3

n 1060 757 811 950

ND - NA 3 x 1015 8.3 x 1016 6.3 x 1016 2.4 x 1016

nn 3.2 x 1018 6.3 x 1019 5.1 x 1019 2.3 x 1019

So ND = 4.1 x 10 /cm 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. 2.00 = qn n qn N D = 1.602 x1019 C ( 1250) 1016 = cm -1 Now we add acceptors until = 5.0 (-cm) :

(

)

( )

= q p pNA 1 x 1017 2 x 1017 1.8 x 1017

|

3.12 x1019 p p p (N A N D )= = 1.602 x1019 C V cm sND + NA 1.1 x 1017 2.1 x 1017 1.9 x 1017 p 250 176 183 NA - ND 9 x 1016 1.9 x 1017 1.7 x 1016 p p 2.3 x 1019 3.3 x 1019 3.1 x 1019

5( cm)

1

31

2.43

Boron is an acceptor: NA = 1016/cm3 and p = 405 cm2/V-s from Fig. 2.8. 0.649 = q p p q p N A = 1.602 x1019 C (405) 1016 = cm -1 Now we add donors until = 5.5 (-cm) :

(

)

( )

= q n nND 8 x 1016 6 x 1016 4.5 x 1016

|

n n n (N D N A )=ND + NA 9 x 1016 7 x 1016 5.5 x 1016 n

5.5( cm)

1

1.602 x1019 C

=

3.43 x1019 V cm sp p 5.8 x 1019 4.5 x 1019 3.4 x 1019

ND - NA 7 x 1016 5 x 1016 3.5 x 1016

832 901 964

2.44

VT =

kT 1.38 x1023 T = = 8.62 x105 T q 1.602 x1019 50 4.31 75 6.46 100 8.61 150 12.9 200 17.2 250 21.5 300 25.8 350 30.1 400 34.5

T (K) VT (mV)2.45

dn dn j = qDn = qVT n dx dx

cm2 1018 0 1 kA j = 1.602 x1019 C (0.025V ) = 14.0 2 350 4 4 V s 0 10 cm cm

(

)

2.46

cm2 1019 / cm 3 dp x = 1.602 x1019 C 15 exp 4 4 s 2 x10 cm 2 x10 cm dx x A j = 1.20 x105 exp5000 2 cm cm 108 cm2 A I (0)= j (0)A = 1.20 x105 2 10m2 = 12.0 mA 2 cm m j = qD p

(

)

(

)

32

2.47

j p = q p pE qD p

dp 1 dp 1 dp = q p p E VT = 0 E = VT dx p dx p dx

1022 exp 10 4 x 1 dN A E VT = 0.025 14 N A dx 10 + 1018 exp 10 4 x

(

(

)

)

E (0)= 0.025

10 V = 250 18 cm 10 + 10 22 10 exp(5) V E 5x104 cm = 0.025 14 = 246 18 cm 10 + 10 exp(5)14

22

(

)

2.48

V cm2 1016 A jndrift = qn nE = 1.60 x1019 C 350 3 20 = 11.2 2 cm V s cm cm V cm2 1.01x1018 A drift 19 20 j p = q p pE = 1.60 x10 C 150 = 484 2 3 cm V s cm cm A cm2 10 4 1016 dn = 70.0 2 jndiff = qDn = 1.60 x1019 C 350 0.025 4 4 s 2 x10 cm dx cm A cm2 1018 1.01x1018 dp diff jp = qD p = 1.60 x1019 C 150 0.025 = 30.0 2 4 4 s 2 x10 cm dx cm A jT = 11.2 484 70.0 + 30.0 = 535 2 cm

(

)

(

)

(

)

(

)

2.49EC ED

NA = 2NDND

ND

ND

EA EV

NA

NA

NA

NA

Holes

2.50

34 8 hc 6.626 x10 J s 3x10 m / s = = = 1.108 m E (1.12eV ) 1.602 x1019 J / eV

(

(

)(

) )

33

2.51Al - Anode Al - Cathode Si02 n-type silicon

p-type silicon

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.Mask Photoresist Si02 p-type silicon n+ p-type silicon Ion implantation

n-type silicon

n-type silicon

Structure after exposure and development of photoresist layer

Structure following ion implantation of n-type impurity

Mask for ion implantation Side view

Top View

2.53

1 1 1 = 8 atoms ( a ) N = 8 + 6 + 4() 8 2 ( b ) V = l 3 = 0.543x109 m = 0.543x107 cm = 1.60 x1022 cm3 8 atoms atoms = 5.00 x1022 1.60 x1022 cm3 cm3 g ( d ) m = 2.33 3 1.60 x1022 cm3 = 3.73 x1022 g cm (c ) D = ( e ) From Table 2.2, silicon has a mass of 28.086 protons. mp = g 3.73x1022 g = 1.66 x1024 proton 28.082(8)protons

(

) (3

)

3

Yes, near the actual proton rest mass.

34

CHAPTER 33.1

1019 cm3 )( 1018 cm 3 ) ( NA ND j = VT ln 2 = (0.025V )ln = 0.979V ni 10 20 cm 6 2( 11.7 8.854 x1014 F cm1 ) 2s 1 1 1 1 w do = + 19 3 + 18 3 (0.979V) j = 19 10 cm q NA ND 1.602 x10 C 10 cm w do = 3.73 x 106 cm = 0.0373m w do 0.0373m w do 0.0373m -3 xn = = = m 18 3 = 0.0339 m | x p = 19 3 = 3.39 x 10 ND N 10 10 cm cm A 1+ 1+ 1 + 19 3 1 + 18 3 NA ND 10 cm 10 cmE MAX = qN A x p

s1018 cm 3

(1.60 x10 =| n po =

19

C) (1019 cm3 )(3.39 x107 cm)14

11.7 8.854 x10

F / cm

= 5.24 x 10 5

V cm

3.2 p po = N A = n i2 10 20 10 2 = = p po 1018 cm 3

1015 cm 3 ) 1018 cm3 )( ( NA ND j = VT ln 2 = (0.025V )ln = 0.748 V ni 10 20 cm6 w do = 2( 11.7 8.854 x1014 F cm1 ) 2s 1 1 1 1 + = 18 3 + 15 3 (0.748V) j 19 10 cm 10 cm q NA ND 1.602 x10 C

1015 n no = N D = cm 3

n i2 10 20 10 5 | p no = = = n no 1015 cm 3

w do = 98.4 x 106 cm = 0.984 m3.3

p po = N A =

1018 ni2 1020 102 | n = = = po p po 1018 cm3 cm3

1018 ni2 1020 102 nno = N D = 3 | p no = = = nno 1018 cm3 cm 1018 cm 3 ) 1018 cm3 )( ( NA ND j = VT ln 2 = (0.025V )ln = 0.921V ni 10 20 cm 6 w do = 2( 11.7 8.854 x1014 F cm1 ) 2s 1 1 1 1 + = 18 3 + 18 3 (0.921V) j 19 10 cm 10 cm q NA ND 1.602 x10 C

w do = 4.881x106 cm = 0.0488 m

34

3.4

p po = N A = nno = N D =

1018 ni2 1020 102 | n = = = po p po 1018 cm3 cm3

1018 ni2 1020 102 | p = = = no nno 1018 cm3 cm3 (1018 cm3 )(1020 cm3 ) = 1.04V N N j = VT ln A 2 D = (0.025V )ln ni 10 20 cm 6 2( 11.7 8.854 x1014 F cm1 ) 2s 1 1 1 1 w do = + 18 3 + 20 3 (1.04V) j = 19 10 cm 10 cm q NA ND 1.602 x10 C w do = 0.0369 m3.5

p po = N A = nno = N D =

1016 ni2 1020 10 4 | n = = = po p po 1016 cm3 cm3

1019 ni2 1020 10 | p = = = no nno 1019 cm3 cm3 1019 cm3 1016 cm 3 N AND j = VT ln = (0.025V )ln = 0.864V ni2 1020 cm 6

(

)(

)

2 11.7 8.854 x1014 F cm1 2s 1 1 1 1 wdo = + j = 19 3 + 16 3 (0.864V) 19 q N A ND 10 cm 1.602 x10 C 10 cm wdo = 0.334 m3.6wd = wdo 1 + VR

(

)

j

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

5 = 1.05 m 0.85

3.7wd = wdo 1 + VR

j

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

10 = 4.20 m 0.6

3.8

jn = E , =

1

=

1 2 j 1000 A cm 2 V = | E= n = = 500 1 0.5 cm cm cm 2( cm)

35

3.9

j p = E3.10

|

E=

jn

= jn = 5000 A cm 2 (2 cm)= 10.0

(

)

kV cm

4 x1015 10 7 cm A j jn = qnv = 1.60 x1019 C = 6400 2 3 cm cm s

(

)

3.11

5 x1017 10 7 cm kA j j p = qpv = 1.60 x1019 C = 800 2 3 cm cm s

(

)

3.12

j p = q p pE qD p

D 1 dp kT 1 dp dp p = 0 E = = dx q p dx p p dx 1 dp 1 V V 0.025V | E = T = 4 = = 250 L p dx L cm 10 cm

x p ( x ) = N o exp | L

The exponential doping results in a constant electric field.3.13

j p = qDn

dn dn dn 2000 A / cm 2 1.00 x 1021 = qnVT | = = dx dx dx 1.60 x1019 C 500cm2 / V s (0.025V ) cm 4

(

)(

)

3.14 10 = 10 4 1016 exp(40VD ) 1 + VD

[

]

and the solver yields VD = 0.7464 V

3.15

f = 10 10 4 I D 0.025ln

ID + IS IS

| f ' = 10 4

0.025 ID + IS

| I'D = I D -13

f f'

Starting the iteration process with ID = 100 A and IS = 10 A:ID 1.000E-04 9.275E-04 9.426E-04 9.426E-04 f 8.482E+0 0 1.512E01 3.268E06 9.992E16 f' -1.025E+04 -1.003E+04 -1.003E+04 -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 -15 (b) Changing IS to 10 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

19 qVT 1.60 x10 C (0.025V ) T= = = 290 K k 1.38 x1023 J / K

3.18

23 kT 1.38 x10 J / K T VT = = = 8.63 x105 T 19 q 1.60 x10 C For T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV

(

)

3.19

40V D Graphing I D = I S exp 1 yields : n 6 5

(b)

4 3

(a)

2

1(c)

0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

37

3.20

1.38 x1023 J / K (300) kT = 1.04 = 26.88 mV nVT = n q 1.60 x1019 C

(

)

T = 26.88mV

1.602x10-19 = 312 K 1.38x10-23

3.21

v i vD = ln1 + D iD = I S exp D 1 or nVT IS nVT i 1 v For i D >> I S , D ln D or ln (I D )= v D + ln (IS ) nVT IS nVT -4

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 A for vD = 0.60 V and iD = 10 A for vD = 0.20 V. Then there are two equations in two unknowns: 40 8 ln 10-9 = .20 + ln (IS ) or 9.21 = + ln (IS ) n n 40 24 ln 10-4 = .60 + ln (IS ) or 20.72 = + ln (IS ) n n -12 Solving for n and IS yields n = 1.39 and IS = 3.17 x 10 A = 3.17 pA.-9

( ) ( )

3.22

V I VD = nVT ln1 + D | I D = I S exp D 1 IS nVT 7 x105 A 5 x106 A = 1.05 0.025 V ln (a) VD = 1.05(0.025V )ln1 + = 0.837 V | (b) V 1 + = 0.768V ( ) D 1018 A 1018 A 0 (c) I D = 1018 Aexp 1 = 0 A 1.05 0.025V 0.075V 19 (d) I D = 1018 Aexp 1 = 0.943x10 A 1.05 0.025V 5V 18 (e) I D = 1018 Aexp 1 = 1.00 x10 A 1.05 0.025V V I VD = nVT ln1 + D | I D = I S exp D 1 IS nVT 104 A 105 A (a) VD = 0.025V ln1 + 17 = 0.748V | (b) VD = 0.025V ln1 + 17 = 0.691V 10 A 10 A

3.23

38

0 (c) ID = 1017 Aexp 1 = 0 A 0.025V 0.06V 17 (d) ID = 1017 Aexp 1 = 0.909 x10 A 0.025V 4V 17 (e) ID = 1017 Aexp 1 = 1.00 x10 A 0.025V 3.24

V 0.675 6 I D = I S exp D 1 = 1017 A exp 1 = 5.32 x10 A = 5.32 A 0.025 VT 15.9 x106 A I VD = VT ln D + 1 = (0.025V )ln + 1 = 0.703 V 17 IS 10 A

3.25

I 40 A VD = nVT ln1 + D = 2(0.025V )ln1 + 10 = 1.34 V 10 A IS 100 A VD = 2(0.025V )ln1 + 10 = 1.38 V 10 A

3.26ID 2mA = = 1.14 x1017 A V 0.82 exp D 1 exp 1 nVT 0.025 5 17 ( b ) I D = 1.14 x1017 A exp 1 = 1.14 x10 A 0.025 (a ) IS =

3.27

ID 300A = = 2.81x1017 A V 0.75 exp D 1 exp 1 nVT 0.025 3 17 ( b ) I D = 2.81x1017 A exp 1 = 2.81x10 A 0.025 (a ) IS =

39

3.28

I VD = nVT ln1 + D | 10-14 I S 1012 IS 103 A VD = (0.025V )ln1 + 14 = 0.633 V 10 A 1.38 x1023 (307)

| |

103 A VD = (0.025V )ln1 + 12 = 0.518 V 10 A So, 0.518 V VD 0.633 V

3.29

V D = 0.0264 V | I = I exp 1 D S 1.60 x1019 0.0264 n Varying n and IS by trial-and-error with a spreadsheet: VT =n 1.039 VD 0.500 0.550 0.600 0.650 0.675 0.700 0.725 0.750 0.775 IS 7.606E-15 ID-Measured 6.591E-07 3.647E-06 2.158E-05 1.780E-04 3.601E-04 8.963E-04 2.335E-03 6.035E-03 1.316E-02 ID-Calculated 6.276E-07 3.885E-06 2.404E-05 1.488E-04 3.702E-04 9.211E-04 2.292E-03 5.701E-03 1.418E-02 Error Squared 9.9198E-16 5.6422E-14 6.0672E-12 8.518E-10 1.0261E-10 6.1409E-10 1.8902E-09 1.1156E-07 1.0471E-06

Total Squared Error

1.1622E-06

3.30

23 kT 1.38 x10 J / K T = = 8.63 x105 T VT = q 1.60 x1019 C For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV

(

)

3.31 23 103 kT 1.38 x10 (303) = = 26.1 mV | V = 0.0261 V ln 1 + = 0.757 V ( ) D 16 q 1.60 x1019 2.5 x10

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

40

3.32

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

23 104 kT 1.38 x10 (298) = = 25.67 mV | ( a ) VD = (0.02567V )ln1 + 15 = 0.650 V q 1.602 x1019 10

3.33

(b) V = (1.8mV / K )(60 K )= 50.0 mV ( c ) V = (1.8 mV / K )(80 K )= +144 mV3.34

23 2.5 x104 kT 1.38 x10 (298) = = 25.67 mV | ( a ) V = 0.02567 V ln ( ) 1 + 1014 = 0.615 V D q 1.602 x1019

VD = 0.615 0.108 = 0.507 V VD = 0.615 + 0.144 = 0.758 V

mV dvD vD VG 3VT 0.7 1.21 3(0.0259) = = = 1.96 dT T 300 K

41

3.353 E 1 1 T 3 E T I S 2 T2 = exp G = 2 exp G 1 1 I S1 T1 k T2 T1 T1 kT1 T2 E 1 3 T x= 2 f (x ) = (x ) exp G 1 T1 kT1 x 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 1.00000 1.00500 1.01000 1.01500 1.01400 1.01422 1.01922 1.02422 1.02922 1.03422 1.03922 1.04422 1.04922 1.04880 1.10000 1.10239

f(x) 1.00000 1.27888 1.63167 2.07694 1.97945 2.00051 2.54151 3.22151 4.07433 5.14160 6.47438 8.13522 10.20058 10.00936 90.67434 100.0012

Delta T 0.00000 1.50000 3.00000 4.50000 4.20000 4.26600 5.76600 7.26600 8.76600 10.26600 11.76600 13.26600 14.76600 14.64000 30.00000 30.71610

3.36wd = wdo 1 + VR

j

| (a) wd = 1m 1 +

5 10 = 2.69 m (b) wd = 1m 1 + = 3.67 m 0.8 0.8

3.37

1016 cm 3 1015 cm 3 N AND j = VT ln = (0.025V )ln = 0.633 V ni2 1020 cm6 2 11.7 8.854 x1014 F cm1 2s 1 1 1 1 + wdo = j = 16 3 + 15 3 (0.633V) 19 q N A ND 10 cm 1.602 x10 C 10 cm wdo = 0.949 m wd = 0.949m 1 + | wd = wdo 1 + VR

(

)(

)

(

)

j

10V = 3.89 m 0.633V

|

wd = 0.949m 1 +

100V = 12.0 m 0.633V

42

3.38

1018 cm 3 1020 cm 3 N AND j = VT ln = (0.025V )ln = 1.04 V ni2 1020 cm6 2 11.7 8.854 x1014 F cm1 2s 1 1 wdo = + j = q N A ND 1.602 x1019 C wdo = 0.0368 m wd = 0.0368m 1 +3.39

(

)(

)

(

)

1 1 1.04V) + ( 18 3 1020 cm3 10 cm

|

wd = wdo 1 +

VR

j

5 = 0.0887 m 1.04 2( j + VR ) wdo 1 + VR

|

wd = 0.0368m 1 +

25 = 0.184 m 1.04

Emax =

2( j + VR ) wd

=

=

2 j V 1+ R wdo j

j

3 x1053.40

2(0.6V ) V V = 4 1 + R VR = 374 V cm 10 cm 0.6

2(0.748V ) 2 kV = 15.2 | E= j = 4 wdo 0.984 x10 cm cm VR = 291.3 0.748 = 291 V3.41

E w j + VR = max do = 2 j

3 x105

V 0.984 x104 cm cm 2 0.748V

(

)

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 p

S

=

qN A wd

S

=

qN A

S

2S VR q NA

2 3 x105 ( 11.7) 8.854 x1014 Emax S NA = = = 2.91 x 1014 / cm3 19 2qVR 2 1.602 x10 1000 2

(

) (

(

)

)

43

3.43

j = VT ln

N AND 10151020 = 0.025ln = 0.864V ni2 1020

2( 11.7) 8.854 x1014 1 2S 1 1 1 4 wdo = + + j = 0.864 = 1.057 x10 cm q N A ND 1.602 x1019 1015 1020 C =" jo

(

)

Swdo

=

11.7 8.854 x1014 1.057 x104

(

) = 9.80x10

-9

F / cm

2

| Cj =

C" jo A 1+ VR

=

9.80x10-9 (0.05) 5 1+ 0.864

= 188 pF

j

3.44

j = VT ln

N AND 10181015 = 0.025ln = 0.748V ni2 1020

2( 11.7) 8.854 x1014 2S 1 1 wdo = + j = q N A ND 1.602 x1019 C =" jo

(

)

1 1 0.748 = 0.984 x104 cm + 18 15 10 10 | Cj = C" jo A 1+ VR = 10.5x10-9 (0.02) 10 1+ 0.748 = 55.4 pF

Swdo

=

11.7 8.854 x1014 0.984 x104

(

) = 10.5x10

-9

F / cm

2

j

3.45

4 10 I D T 10 A 10 s (a) CD = = = 400 fF VT 0.025V

(

)

(b) Q = I D T = 104 A 1010 s = 10 fC

(

)

(c) CD =3.46

25x103 A 1010 s 0.025V

(

) = 100 pF

| Q = I D T = 5 x103 A 1010 s = 0.50 pC

(

)

8 I D T 1A 10 s (a) CD = = = 0.400 F VT 0.025V

(

)

(b) Q = I D T = 1A 108 s = 10.0 nC

(

)

(c) CD =

100mA 108 s 0.025V

(

) = 0.04 F

| Q = I D T = 100 mA 108 s = 1.00 nC

(

)

44

3.47

j = VT ln

N AND 10191017 = 0.025ln = 0.921V ni2 1020

2( 11.7) 8.854 x1014 1 2S 1 1 1 wdo = + + j = 0.921 = 0.110 m q N A ND 1.602 x1019 1019 1017 C jo =

(

)

S Awdo

=

11.7 8.854 x1014 104 0.110 x104

(

)( ) = 9.42 pF / cm

2

| Cj =

C jo 1+ VR

=

9.42 pF 5 1+ 0.921

= 3.72 pF

j

3.48

j = VT ln

N AND 10191016 = 0.025ln = 0.864V ni2 1020

2( 11.7) 8.854 x1014 2S 1 1 wdo = + j = q N A ND 1.602 x1019 C jo =

(

)

1 1 + 0.864 = 0.334m 1019 1016 | Cj = C jo 1+ VR = 7750 pF 3 1+ 0.864 = 3670 pF

S Awdo

=

11.7 8.854 x1014 0.25cm2 0.334 x104

(

)(

) = 7750 pF

j

3.49L=

RFC

CVDC C 10 H 10 H

C=

C jo 1+ VR

(a) C =

39 pF 1V 1+ 0.75V

= 25.5 pF | f o =

1 2 LC 1

=

j

2

(

1

105 H 25.5 pF = 15.7 MHz

)

= 9.97 MHz

(b) C =

39 pF 10V 1+ 0.75V

= 10.3 pF | f o =

1 2 LC

=

2

(

105 H 10.3 pF

)

3.50

50 A 50 A (a) VD = (0.025V )ln1 + 7 = 0.501 V | (b) VD = (0.025V )ln1 + 15 = 0.961 V 10 A 10 A

45

3.51

4 x103 A 4 x103 A (a) VD = (0.025V )ln1 + = 0.025 V ln = 0.495 V | (b) V ( ) 1 + 1014 A = 0.668 V D 1011 A

3.52

RS = R p + Rn Rn = n3.53

Rp = p

Lp 0.025cm =( 1 cm) = 2.5 Ap 0.01cm 2 RS = 2.53

Ln 0.025cm = (0.01 cm) = 0.025 An 0.01cm 2

' VD = VD + I D RS = 0.708V + 103 A( 10)= 0.718 V

I 103 ' (a) VD = VT ln1 + D = (0.025V )ln1 + = 0.708V 16 IS 5x10

' (b) VD = VD + I D RS = 0.708V + 103 A( 100)= 0.808 V

3.54

10 m2 c = 10 m Ac = 1m RC = = = 10 / contact Ac 1m2 5 anode contacts and 14 cathode contacts2 2

c

10 = 2 5 10 = 0.71 Resistance of cathode contacts = 14 Resistance of anode contacts =

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 = 10 4 I D + VD | VD = 0 I D = 0.500 mA | I D = 0 VD = 5V Forward biased - VD = 0.5 V I D = (b) 6 = 3000 I D + VD 4.5V = 0.450 mA 10 4 | VD = 0 I D = 2.00 mA | I D = 0 VD = 6V 2V = 0.667 mA 3k | VD = 0 I D = 1.00 mA | I D = 0 VD = 3ViD 2 mA

In reverse breakdown - VD = 4 V I D = (c) 3 = 3000 I D + VD Reverse biased - VD = 3 V I D = 0

1 mA (c) Q-point -6 -5 -4 -3 -2 -1 1 (b) Q-point -1 mA 2 3 4 5 6 (a) Q-point vD

-2 mA

3.57

(a) 10 = 5000 I D + VD | VD = 0 I D = 2.00 mA | VD = 5 V I D = 1.00 mA 9.5V = 1.90 mA 5k | VD = 0 I D = 2.00 mA | VD = 5 V I D = 1.00 mA

Forward biased - VD = 0.5V I D = (b) -10 = 5000 I D + VD

In reverse breakdown - VD = 4V I D = (c) 2 = 2000 I D + VD Reverse biased - VD = 2 V I D = 0iD 2 mA (a) Q-point

6V = 1.20 mA 5k | VD = 0 I D = 1.00 mA | I D = 0 VD = 2 V

1 mA (c) Q-point -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 vD

-1 mA (b) Q-point

-2 mA

47

3.58

*Problem 3.58 - Diode Circuit V 1 0 DC 5 R 1 2 10K D1 2 0 DIODE1 .OP .MODEL DIODE1 D IS=1E-15 .END

SPICE Results VD = 0.693 V ID = 0.431 A

3.59 (a) 10 = 10 4 I D + VD | VD = 0 I D = 1.00 mA | VD = 5 V

I D = 0.500 mA

In reverse breakdown - VD = 4 V I D = (b) 10 = 10 4 I D + VD

10 ( 4)V = 0.600 mA 10 k | VD = 0 I D = 1.00 mA | VD = 5 V I D = 0.500 mA 10 0.5V = 0.950 mA 10 k | VD = 0 I D = 2.00 mA | I D = 0iD 2 mA

Forward biased - VD = 0.5 V I D = (c) 4 = 2000 I D + VD

VD = 4 V

Reverse biased - VD = 4 V I D = 0

1 mA (c) Q-point -6 -5 -4 -3 -2 -1

(b) Q-point

vD 1 2 3 4 5 6

(a) Q-point -1 mA

-2 mA

48

iD (A)

0.002

0.001

-7

-6

-5

-4

-3

-2

-1

v D (V)

1

2

3

4

5

6

7

-0.001

-0.002

49

3.60R i V +D

-

+ v

D

-

The load line equation: V = iD R + vD

We need two points to plot the load line.

(a) V = 6 V and R = 4k: 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 = 3k: 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 = 3k: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) (d) V = +12 V and R = 8k: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) (e) V = -25 V and R = 10k: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA)i (A)D

.002

Q-Point (0.5V,1.45 mA) (d)

Q-Point (-3V,0 mA) -7 -6 -5 -4 -3 -2

.001 Q-Point (0.5V,1.4 mA) -1 1 2 3 Load line for (a) 4 5 6 7 v (V)D

Q-Point (-4V,-0.67 mA)

(c) -.001

Load line for (b) -.002 Q-Point (-4V,-2.1 mA) (e)

50

3.61 -9 Using the equations from Table 3.1, (f = 10-10 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), i V = iD R + VT ln D IS

or 10 = 10 4 iD + 0.025ln 1013 iD .

(

)

4 13 We want to find the zero of the function f = 10 10 iD 0.025ln 10 iD

(

)

iD .001 .0001 .0009 .00094

f -0.576 8.48 0.427 0.0259 - converged

3.63

I 0.025 f = 10 10 4 ID 0.025ln1 + D | f ' = 10 4 ID + IS IS x 1.0000E+00 9.2766E-04 9.4258E-04 9.4258E-04 9.4258E-04 f(x) -9.991E+03 1.496E-01 3.199E-06 9.992E-16 9.992E-16 f'(x) -1.000E+04 -1.003E+04 -1.003E+04 -1.003E+04 -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 I D + VD | I D = 0 VD = 1V | VD = 0 I D = 0.1mA (50 A, 0.5 V )

9 Mathematical model: f = 1 10 exp(40VD ) 1 + VD (49.9 A, 0.501 V )

[

]

Ideal diode model: ID = 1V/10k = 100A; (100A, 0 V) Constant voltage drop model: ID = (1-0.6)V/10k = 40.0A; (40.0A, 0.6 V)3.66 Using Thvenin equivalent circuits yields and then combining the sources

1.2 k 1.2 V +

I V +

1k + 1.5 V

I V +

2.2 k + 0.3 V

-

-

-

(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 0.3V I= = 0.136 mA . This current is greater than zero, which is consistent with the diode 2.2 k being "on". Thus the Q-pt is (0 V, +0.136 mA).I V+ I 0.6 V 2.2 k + 0.3 V Von

+

2.2 k + 0.3 V

-

-

Ideal Diode:

CVD:

(b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it 0.3V 0.6V = 136 A . is "on". Substituting the CVD model with Von = 0.6 V yields I = 2.2 k 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).V + I=0

2.2 k

+ 0.3 V

52

(c) The second estimate is more realistic. 0.3 V is not sufficient to forward bias the diode into -15 significant conduction. For example, let us assume that IS = 10 A, and assume that the full 0.3 V appears across the diode. Then 0.3V iD = 1015 Aexp 1 = 163 pA , a very small current. 0.025V 3.67 The nominal values are: R2 2 k VA = 3V = 3V = 1.20V 2 k + 3k R1 + R2

and RTHA =

R4 2 k 2 k(2 k) R3 R4 VC = 3V = = 1.00 k = 3V = 1.50V and RTHC = R3 + R4 2 k + 2 k 2 k + 2 k R3 + R4 1.50 1.20 V nom ID = = 136 A 1.20 + 1.00 k 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 = 3V 3V = = 1.65V and R1 2 k(0.9) 1+ R2 1 + 2k( 1.1) 3V 3V = = 1.06V and R3 3k( 1.1) 1+ R4 1 + 2k(0.9) RTHA = 2 k(0.9)2 k( 1.1) R1 R2 = = 0.990 k R1 + R2 2 k(0.9)+ 2 k( 1.1) 3k( 1.1)2 k(0.9) R3 R4 = = 1.17 k R3 + R4 3k( 1.1)+ 2 k(0.9)

2 k(3k) R1 R2 = = 1.20 k R1 + R2 2 k + 3k

VC =

RTHC =

1.65 1.06 V max ID = = 274 A 0.990 + 1.17 k 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 =

3V 3V = = 1.350V and R1 2k( 1.1) 1+ R2 1 + 2 k(0.9) 3V 3V = = 1.347V and R3 3k(0.9) 1+ R4 1 + 2 k( 1.1)

RTHA =

2 k( 1.1)2 k(0.9) R1 R2 = = 0.990k R1 + R2 2 k( 1.1)+ 2 k(0.9) 3k(0.9)2 k( 1.1) R3 R4 = = 1.21k R3 + R4 3k(0.9)+ 2 k( 1.1)

VC =

RTHC =

1.350 1.347 V min = = 1.39 A 0 ID 0.990 + 1.21 k

53

3.68 SPICE Input *Problem 3.68 V1 1 0 DC 4 R1 1 2 2K R2 2 0 2K R3 1 3 3K R4 3 0 2K D1 2 3 DIODE .MODEL DIODE D IS=1E-15 RS=0 .OP .END Results NAME D1 MODEL ID VD

DIODE 1.09E-10 3.00E-01

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 =

( 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)

= 0.625 mA 16k 5 (5) = 0.625 mA ( b ) Diode is forward biased : V = 5 + 0 = 5 V | I = 16k ( c ) Diode is reverse biased : I = 0 | V = 5 + 16 k(I )= 5 V | VD = 10 V

3 (7)

( d ) Diode is reverse biased : I = 0 | V = 7 16k(I ) = 7 V | VD = 10 V3.70 (a)

= 0.581 mA 16k 5 (4.3) ( b ) Diode is forward biased : V = 5 + 0.7 = 4.3 V | I = = 0.581 mA 16k ( c ) Diode is reverse biased : I = 0 | V = 5 + 16 k(I )= 5 V | VD = 10 V

( a ) Diode is forward biased : V = 3 0 = 3 V | I =

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

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

3 (7)

54

= 94.0 A 100 k 5 (4.4) ( b ) Diode is forward biased : V = 5 + 0.6 = 4.4 V | I = = 94.0 A 100k ( c ) Diode is reverse biased : I = 0 | V = 5 + 100 k(I ) = 5 V | VD = 10 V ( a ) Diode is forward biased : V = 3 0.6 = 2.4 V | I = ( d ) Diode is reverse biased : I = 0 | V = 7 100k(I ) = 7 V | VD = 10 V0 (9)

2.4 (7)

3.71 (a)( a ) D1 on , D2 on : I D2 = D1 : (409 A, 0 V )

22 k D2 : (270 A, 0 V )

= 409A | I D1 = 409A

60 = 270A 43k

( b ) D1 on , D2 off : I D 2 = 0 | I D1 = 140 A, 0 V ) D1 : (

60 = 140 A | VD2 = 9 0 = 9V 43k D2 : (0 A, 9 V )

( c ) D1 off, D2 on : I D1 = 0 | I D2 = D1 : (0 A, 3.92 V ) 0 (6)

6 (9)

65k D2 : (230 A,0 V )

= 230 A | VD1 = 6 43 x103 I D2 = 3.92 V

( d ) D1 on, D2 on : I D 2 = 140 A,0 V ) D1 : ( (b) (a) D1 on, D2 on : I D2 =

43k D2 : (270 A,0 V )

= 140 A | I D1 =

90 140 A = 270 A 22k

-0.75 0.75 (9)

22 k 184 A, 0.75 V) D2 : (341 A, 0.75 V) D1 : ( (b) D1 on, D2 off :

= 341A | I D1 = 341A

6 (0.75) 43k

= 184A

6 0.75 = 122A | VD 2 = 9 0.75 = 9.75V 43k 122 A, 0.75 V) D2 : (0 A, 9.75 V) D1 : ( I D2 = 0 | I D1 =

55

( c ) D1 off, D2 on : I D1 = 0 | I D2 =

6 0.75 (9)

65k D1 : (0 A, 3.43 V ) D2 : (219 A, 0.75 V ) (d) D1 on, D2 on : I D2 =

= 219A | VD1 = 6 43 x103 I D2 = 3.43V

0.75 0.75 (6)

43k D1 : (235 A, 0.75 V)

9 0.75 400A = 235A 22 k D2 : ( 140 A, 0.75 V) = 140A | I D1 =

3.72 (a) (a) D1 and D2 forward biased15 k D1 : (0 V, 200 A) I D2 = 0 (9) V = 600A I D1 = I D2

D2 : (0 V, 600 A)

6 (0) V = 200A 15 k

(b) D1 forward biased, D2 reverse biased 60 V = 400A VD2 = 9 0 = 9 V 15 k D1 : (0 V, 400 A) D2 : (-9 V, 0 A ) I D1 = (c) D1 reverse biased, D2 forward biased = 500A VD1 = 6 15000 I D2 = 1.50V 30 k D1 : (1.50 V, 0 A) D2 : (0 V, 500 A) I D2 = 6V (9V )

(d) D1 and D2 forward biased 15 k D1 : (0 V, 200 A) I D2 = 0 (6) V = 400A I D1 =

9 (0) V

15 k D2 : (0 V, 400 A)

I D 2 = 200A

(b)(a) D1 on, D2 on : I D2 = -0.75 0.75 (9) 6 (0.75) 15k

15k D1 : (50.0 A, 0.75 V) D2 : (500 A, 0.75 V)

= 500A | I D1 = 500A

= 50.0A

56

(b) D1 on, D2 off : 6 0.75 = 350A | VD 2 = 9 0.75 = 9.75V 15k D1 : (350 A, 0.75 V) D2 : (0 A, 9.75 V) I D2 = 0 | I D1 = ( c ) D1 off, D2 on : I D1 = 0 | I D2 = 6 0.75 (9)

= 475A | VD1 = 6 15 x103 I D2 = 1.13V 30 k D1 : (0 A, 1.13 V ) D2 : (475 A, 0.75 V ) (d) D1 on, D2 on : I D2 = 0.75 0.75 (6)

15k D1 : ( 150 A, 0.75 V)

9 0.75 400A = 150A 15k D2 : (400 A, 0.75 V) = 400A | I D1 =

3.73 Diodes are labeled from left to right

(a) D1 on, D2 off, D3 on : I D 2 = 0 | I D1 = I D3 + 0.990 mA = 0 (5)

10 0 = 0.990mA 3.3k + 6.8 k

I D 3 = 1.09 mA | VD 2 = 5 ( 10 3300 I D1 ) = 1.73V 2.4 k 1.09 mA, 0 V) D1 : (0.990 mA, 0 V) D2 : (0 mA, 1.73 V) D3 : (( b ) D1 on, D2 off, D3 on : I D2 = 0 | I D3 = 0 = 0.495mA | VD 2 = 5 ( 10 8200 I D1 )= 0.941V 8.2 k + 12 k 0 (5V ) I D1 = 0.005mA I D3 = 10k D1 : (0.495 mA, 0 V ) D2 : (0 A, 0.941 V ) D3 : (0.005 mA, 0 V ) I D1 =

(10 0)V

57

(c) D1 on, D2 on, D3 on I D1 =

0 (10) 0 (2) V = 1.22mA > 0 | I12 K = V = 0.167 mA | I D 2 = I D1 + I12 K = 1.05mA > 0 8.2 k 12 k 2 (5) V = 0.700mA | I D3 = I10 K I12 K = 0.533mA > 0 I10 K = 10k 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 : I D1 = 0, I D 2 = 0 V = 1.21mA > 0 | VD1 = 0 (5 + 4700 I D3 ) = 0.667V < 0 4.7 + 4.7 + 4.7 k 12 4700 I D 3 ) = 1.33V < 0 VD2 = 5 ( I D3 = 1.21 mA, 0 V ) D1 : (0 A, 0.667 V ) D2 : (0 A, 1.33 V ) D3 : ( 12 (5)

3.74 Diodes are labeled from left to right

(a) D1 on, D2 off, D3 on : I D 2 = 0 | I D1 = I D3 + 0.990 mA = 0.6 (5)

10 0.6 (0.6) 3.3k + 6.8 k

= 0.990 mA

I D3 = 0.843mA | VD2 = 5 ( 10 0.6 3300 I D1 )= 1.13V 2.4 k 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 : I D 2 = 0 | I D 3 = 0 V = 0.477 mA | VD 2 = 5 ( 10 0.6 8200 I D1 )= 0.490V 8.2 k + 12 k + 10 k VD3 = 0 (5 + 10000 I D1 )= +0.230V < 0.6V so the diode is off I D1 = D1 : (0.477 mA, 0.600 V) D2 : (0 A, 0.490 V) D3 : (0 A, 0.230 V)0.6 (9.4) V 8.2 0.6 ( 1.4) V 12 k

10 0.6 (5)

( c ) D1 on, D2 on, D3 on I D1 = k = 1.07mA > 0 | I12 K =

= 0.167 mA

I D2 = I D1 + I12 K = 0.906 mA > 0 | I10 K = D1 : ( 1.07 mA, 0.600 V)

1.4 (5) V = 0.640 mA | I D 3 = I10 K I12 K = 0.807 mA > 0 k 10 D2 : (0.906 mA, 0.600 V) D3 : (0.807 mA, 0.600 V)

58

11.4 (5) V = 1.16mA > 0 | VD1 = 0 (5 + 4700 I D3 ) = 0.452V < 0 4.7 + 4.7 + 4.7 k VD2 = 5 ( 11.4 4700 I D 3 ) = 0.948V < 0 I D3 = D1 : (0 A, 0.452 V ) D2 : (0 A, 0.948 V ) D3 : ( 1.16 mA, 0.600 V )

(d) D1 off, D2 off, D3 on : I D1 = 0, I D 2 = 0

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 MODEL DIODE DIODE ID 4.74E-04 -4.22E-13 VD 6.95E-01 -4.21E-01 NAME D1 MODEL DIODE ID 8.79E-03 VD 7.11E-01 D3 DIODE 2.67E-11 2.63E-01

D2 D3 DIODE DIODE 1.05E-03 7.96E-04 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

= 1.50 mA | I D 2 = 0 10 k + 10 k 0 (10) I D3 = = 1.00mA | VD2 = 10 10 4 I D1 0 = 5.00V 10k 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

I D1 =

10 (20)

The simulation results are very close to those given in Ex. 3.8.3.783.9 k = 6.28V | RTH = 11k 3.9 k = 2.88 k 3.9 k + 11k 6.28 4 IZ = = 0.792 mA > 0 | (I Z , VZ )= (0.792 mA,4 V ) 2.88 k

VTH = 24V

60

3.79

6.28 = 2880 I D + VD | I D = 0, VD = 6.28V | VD = 0, I D =-6 -5 -4 -3 -2 -1 vD

6.28 = 2.18 mA 2880

Q-point

-1 mA

-2 mA

i

D

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

3.80

IS =3.81

27 9 9V = 1.20mA I L < 1.20 mA | RL > = 7.50 k 15k 1.2mA 27 9 = 1.20mA | P = (9V )( 1.20 mA) = 10.8 mW 15k

IS =3.82

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

1 VS VZ VZ VS 1 = VZ + | PZ = VZ I Z RS RL RS RS RL 1 30V 1 nom nom IZ = 9V + = 0.500 mA | PZ = 9V (0.500 mA)= 4.5 mW 15k 15 k 10 k 30V ( 1.05) 1 1 max IZ = 9V (0.95) + 15k 0.95 10k(1.05) = 0.796 mA 15k(0.95) ( ) IZ = 1 1 = 9V ( 1.05) 15k 1.05 + 10 k(0.95) = 0.215 mA 15k( 1.05) ( ) 30V (0.95)

PZmin = 9V ( 1.05)(0.215mA)= 2.03 mW3.83

( a ) VTH = 60V

100 24 15 = 24.0V | RTH = 150 100 = 60 | I Z = = 150 mA 150 + 100 60 60 15 = 300 mA | P = 15 I Z = 4.50 W P = 15 I Z = 2.25 W | ( b ) I Z = 150

61

3.84

IZ =

1 VS VZ VZ VS 1 = VZ + RS RL RS RS RL

|

PZ = VZ I Z

nom IZ =

max IZ

PZmax = 15V (0.90)(266mA)= 3.59 Wmin = IZ

= 150 mA | PZnom = 15V ( 150 mA) = 2.25 W 150 100 60V ( 1.1) 1 1 = 15V (0.90) + 150 0.90 100(1.1) = 266 mA 150(0.90) ( ) 60V (0.90) 1 1 15V ( 1.1) + 150 1.1 100(0.9) = 43.9 mA 150( 1.1) ( )

(60 15)V 15V

1.1)(43.9mA)= 0.724 W PZmin = 15V (

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 = T =3.86

1 120 1 120

IT 5 2Vr | Vr = = = 0.8333V VP C 0.1(60) 2(0.8333) 6 = 1.40 ms

I 48.6 A VD = nVT ln1 + D = 2(0.025V )ln1 + 9 = 1.230 V 10 A IS

62

3.87

I Von = nVT ln1 + D | VD = Von + I D RS IS 100 A VD = 1.6(0.025V )ln1 + 8 + 100 A(0.01) = 1.92 V 10 A 100 A 1ms I T = 0.92V Pjunction Von I DC = Von P = 2.75 W 2T 2 16.7 ms 2 4 T 2 4 16.7 ms PR I DC RS = (3 A) 0.01 = 2.00 W 3 T 3 1ms Ptotal = 4.76 W

3.88

VDC =

1 T

T

v (t )dt = T (V 0

1

P

Von )T

VDC = 0.975( 18V )= 17.6 V3.89

0.05( VP Von ) TVr = 0.975( VP Von ) VP Von ) = ( 2 2

1 PD = T

T

2 1 T 2 t 1 RS dt i (t )RS dt = T I P T 0 0 2 DT T

2 2 2t t2 IP t2 t3 RS 1 + dt = t + T T 2 T T 3T 2 0 0 2 I R T 1 2 T PD = P S T T + = I P RS T 3 3 T

I2R PD = P S T

3.90 Using SPICE with VP = 10 V.15V Voltage 10V

5V

0V

-5V

-10V

t -15V 0s 10ms 20ms 30ms 40ms 50ms

63

3.91

VP Von ) = 6.3 2 1 = 7.91V (a) Vdc = ( (c) PIV 2VP = 2 6.3 2 = 17.8V (e) T = 1

(

)

(b) C =

I T 7.91 1 1 = = 1.05F Vr 0.55 0.5 60

(d) I surge = CVP = 2 (60)( 1.05) 6.3 2 = 3530 A

(

)

2( .25) 2Vr 2T 7.91 2 1 1 = = 0.628 ms | I P = I dc = = 841 A T .5 60 .628 ms VP 2 (60) 6.3 2

3.92

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

1.1) 2 1 = 8.80V VOmax = VPmax Von = 6.3( VOminmin P

( = ( V

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

Circuit3_93b-Transient-8 +0.000e+000 +10.000m +20.000m +30.000m +40.000m +50.000m +60.000m +70.000m

Time (s)

+10.000

+5.000

+0.000e+000

-5.000

-10.000

V(2)

*REAL(Rectifier)*

SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A I T 9.00V 1 1 Vdc = ( VP Von ) = ( 10 1)= 9.00V | Vr = = = 1.20V C 0.25 60 s 0.5F

10)= 1890 A | T = I SC = CVP = 2 (60)(0.5)( I P = I dc 9 2 1 2T = = 923 A T 0.25 60 1.3ms

2( 1.2) 1 2Vr = = 1.30ms VP 2 (60) 10 1

64

Circuit3_93b-Transient-11 (Amp) +0.000e+000 +20.000m +40.000m +60.000m +80.000m +100.000m +120.000m

Time (s) +140.000m

+10.000

+5.000

+0.000e+000

-5.000

-10.000

V(1)

V(2)

*REAL(Rectifier)*

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

VP Von )= 6.3 2 1 = 7.91V ( a ) Vdc = ( ( c ) PIV 2VP = 2 6.3 2 = 17.8V ( e ) T = 1

(

)

( b) C =

I T 7.91 1 1 = = 0.158 F Vr 0.25 0.5 400

( d ) I surge = CVP = 2 (400)(0.158) 6.3 2 = 3540 A

(

)

2( .25) 2Vr 2T 7.91 2 1 1 = = 94.3s | I P = I dc = = 839 A T VP 2 (400) 6.3 2 .5 400 94.3s

3.95

( a ) Vdc = ( VP Von )= 6.3 2 1 = 7.91V ( c ) PIV 2VP = 2 6.3 2 = 17.8V ( e ) T =3.96

(

)

( b) C =

I T 7.91 1 1 = = 633F Vr 0.25 0.5 105

( d ) I surge = CVP = 2 105 (633F ) 6.3 2 = 3540 A = 0.377s | I P = I dc 2T 7.91 2 1 = = 839 A 5 .5 10 0.377s T

( )

(

)

1 2Vr = VP 2 105 1

( )

2( .25) 6.3 2

65

(a) C =

1 IT 1 = = 556 F Vr 3000(0.01) 60 3000 = 2120 V ( d ) T = 1

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

( c ) Vrms =

2 2T 2 1 = 1 I P = I dc = 88.9 A T 60 0.375ms

1 2Vr = 2(0.01) = 0.375ms VP 2 (60) ( e ) I surge = CVP = 2 (60)(556F )(3000)= 629 A

3.97 Assuming Von = 1 V: V Von 1 1 1 30 3.3 + 1 C= P T = V = 8.6 V | Vrms = = 3.04 V = 6.06 F | PIV = 2VP = 2(3.3 + 1) Vr R 0.025 60 3.3 2

T =

1

1 3.3V 1 2T VP Von 2 = s = 0.520 ms RC VP 2 (60) 0.110(6.06 F ) 60 4.3V 2 1 2T = 30 s = 1920 A | I surge = CVP = 2 (60 / s)(6.06 F )(4.3V ) = 9820 A T 60 0.520 ms

I P = I dc3.9840V

vO

20V

v1

0V

vS Time-20V 0s 5ms 10ms 15ms 20ms 25ms 30ms

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) .END4.0kV

3.0kV

vO

2.0kV

1.0kV

vS0V

-1.0kV

Time -2.0kV 0s 20ms 40ms 60ms 80ms 100ms

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 = ( c ) PIV 2VP = 2 15 2 = 42.4 V ( e ) T =3.101

(

)

I T 20.2V 1 1 = = 1.35 F Vr 2 0.5 0.25V 120 s

( d ) I surge = CVP = 2 (60)( 1.35) 15 2 = 10800 A

( )

2( .25) 1 20.2V 1 1 2Vr T = = 0.407 ms | I P = I dc = = 1650 A s VP 2 (60) 15 2 T 0.5 60 0.407 ms 1

( a ) Vdc = ( VP Von )= 9 2 1 = 11.7 V ( b ) C = ( c ) PIV 2VP = 2 9 2 = 25.5 V ( e ) T = 1

(

)

I T 11.7V 1 1 = = 0.780 F Vr 2 0.5 0.25V 120 s

( d ) I surge = CVP = 2 (60)(0.780) 9 2 = 3740 A

( )

2( .25) 1 1 2Vr T 11.7V 1 = = 0.526 ms | I P = I dc = s = 958 A VP 2 (60) 9 2 T 0.5 60 0.407 ms

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 R303 .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

20V

10V

vS

0V

-10V

vO

Time -20V 0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms

Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 AVDC = VP Von = 10 2 0.7 = 13.4 V | Vr = T = 1 120 2Vr 1 = VP 120 2(0.254) 14.1 1 13.4 1 = 0.254 V 3 800 22000F

= 0.504 ms

I P = I dc

1 T 13.4V 1 = s = 150 A 3 60 0.504 ms TCircuit3_102-Transient-15 Time (s) +10.000m +12.000m +14.000m

Simulation with RS = 0.02 .+0.000e+000 +2.000m +4.000m +6.000m +8.000m

+15.000

+10.000

+5.000

+0.000e+000

-5.000

-10.000

-15.000

V(1)

V(2)

*REAL(Rectifier)*

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 =

1 1s 30 A VP Von 1 T = V = 8.6 V = 3.03 F (b) PIV = 2VP = 2(3.3 + 1) R 0.025 120 3.3V Vr 3.3 + 1 2 = 3.04 V (d) T = 2(0.025)(3.3) 2Vr 1 = = 0.520 ms VP 2 (60) 4.3 1

( c ) Vrms =

( e ) I P = I dc

1 T 1 = 30 A s = 962 A | I surge = CVP = 2 (60 / s)(3.03F )(4.3V )= 4910 A T 60 0.520 ms

3.104

(a) C = ( c ) VS = I P = I dc

I T 1 1 = = 139 F Vr 2 3000(0.01) 2 120 3000 2 = 2120 V ( d ) T = 1

( b ) PIV 2VP = 6000 V

2

Vr 1 = 2(0.01) = 0.375 ms VP 2 (60) 139F )(3000V )= 157 A ( e ) I surge = CVP = 2 (60 / s)(

1 1 T = 1 s = 44.4 A T 60 0.375ms

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 = ( c ) PIV VP = 15 2 = 21.2 V ( e ) T =3.107

(

)

I T 19.2V 1 1 s = 1.28 F = Vr 2 0.5 0.25V 120

( d ) I surge = CVP = 2 (60 / s)( 1.28 F ) 15 2 = 10200 A

( )

2( .25) 1 2Vr T 19.2V 1s 1 = = 0.407 ms | I P = I dc = = 1570 A VP 2 (60) 15 2 T 0.5 60 0.407 ms 1 1 I T 1A s = 278 F = Vr 2 3000V (0.01) 120 3000 2 = 2120 V ( d ) T = 1

(a) C = ( c ) VS = I P = I dc

( b ) PIV VP = 3000 V

1 2Vr = 2(0.01) = 0.375 ms VP 2 (60) ( e ) I surge = CVP = 2 (60)(278F )(3000) = 314 A

1 1 T = 1 A s = 44.4 A T 60 0.375ms

69

3.108

(a) C =

1 I T 30 A s = 3.03 F = Vr 2 (0.025)(3.3V ) 120 5.3 2 = 3.75 V ( d ) T = 1

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

( c ) Vrms = I P = I dc

0.025(3.3) 1 2Vr = 0.468 ms = 2 VP 2 (60) 5.3 ( e ) I surge = CVP = 2 (60 / s)(3.03F )(3.3V )= 3770 A

1 T 1 = 30 A s = 1070 A T 60 0.468 ms

3.109 V1 = VP - Von = 49.3 V

and

V2 = -(VP -Von) = -49.3V.40V

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 3.111

v1

20V

0V

-20V

v2Time -40V 0s 10ms 20ms 30ms 40ms 50ms

VP 2Von )= 15 2 2 = 19.2 V ( a ) Vdc = ( ( c ) PIV VP = 15 2 = 21.2 V ( e ) T = 1

(

)

( b) C =

I T 19.2V 1 1 = = 1.28 F Vr 2 0.5 0.25120

( d ) I surge = CVP = 2 (60 / s)( 1.28 F ) 15V 2 = 10200 A

(

)

2( .25) 1 2Vr T 19.2V 1 1 = = 0.407 ms | I P = I dc = s = 1570 A T 0.5 60 0.407 ms VP 2 (60) 15 2

70

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

Rectifier Type Peak Current PIV Filter Capacitor

Half Wave 533 A 8.6 V 25 F

Full Wave 266 A 8. 6 V 12.5 F

Full Wave Bridge 266 A 5.3 V 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.

Rectifier Type Peak Current PIV Filter Capacitor

Half Wave 189 A 402 V 12,500 F

Full Wave 94.3 A 402 V 6250 F

Full Wave Bridge 94.3 A 202 V 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.

Rectifier Type Peak Current PIV Filter Capacitor

Half Wave 133 A 6000 V 41.7 F

Full Wave 66.6 A 6000 V 20.8 F

Full Wave Bridge 66.6 A 3000 V 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.3.115iD 0 + = 5V 5 VD 5 0.6 = 5 mA | I F = = = 4.4 mA 1k 1k 1k 3 0.6 4.4 mA Ir = = 3.6 mA | S = (7 ns) ln1 = 5.59 ns 1k 3.6 mA

( )

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

10

5

v1

vD0

-5

Time -10 0s 5ns 10ns 15ns 20ns

Simulation results give

S = 4.4 ns.

72

3.117iD 0 + = 5V 5 Von 5 0.6 = 1 A | IF = = = 0.880 A 5 1 5 3 0.6 0.880 A = 0.720 A | S = (250 ns) ln1 IR = = 200 ns 5 0.720 A

( )

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

2.0

v1

1.0

vD

0

-1.0

Time -2.0 0s 5ns 10ns 15ns 20ns 25ns

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 .END2.0

v1

1.0

vD

0

-1.0

Time -2.0 0s 10ns 20ns 30ns 40ns

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 = 1 1015 exp(40VC ) 1 A | For VC = 0, I SC = 1 A VOC = 1 1 ln1 + 15 = 0.864 V 40 10

[

]

P = VC IC = VC 1 1015 exp(40VC ) 1

[

[

]]

dP = 1 1015 exp(40VC ) 1 40 x1015 VC exp(40VC ) = 0 dVC Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts3.120 ( a) For VOC, each of the three diode teminal currents must be zero, and 1 VOC = VC1 + VC 2 + VC 3 = V ln( 1.05 x1015 )+ ln( 1.00 x1015 )+ ln(0.95 x1015 ) = 2.59 V 40 (b) For ISC, the external currents cannot exceed the smallest of the short circuit current

[

]

[

]

of the individual diodes. Thus, ISC = min[1.05 A,1.00 A,0.95 A] = 0.95 A Note 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 Watts3.121 hc = E

) = 1.11 m - far infrared 1.12eV ( 1.602 x10 j / eV ) 3 x10 m / s) 6.625 x10 J s( = 0.875 m - near infrared ( b) = 1.42eV ( 1.602 x10 j / eV )(a) = 6.625 x1034 J s 3 x108 m / s19 34 8 19

(

74

CHAPTER 44.1 (a) VG > VTN corresponds to the inversion region (b) VG 0 enhancement - mode transistor395 (4 VTN ) A = 2 VTN = 1.5 V K n = 125 140 (3 VTN ) V22

78

4.19 Using the parameter values from problem 4.22:800uA

VGS = 5 V

600uA

VGS = 4.5 V

Drain Current (A)

400uA

VGS = 4 V

VGS = 3.5 V200uA

VGS = 3 V VGS = 2.5 V VGS = 2 V

0 A 0V

1.0V

2.0V

3.0V

4.0V

5.0V

6.0V

Drain-Source Voltage (V)

4.20 (a) For VGS = 0, VGS VTN and ID = 0 (b) For VGS = 1 V , VGS = VTN and ID = 0

(c ) VGS VTNID =

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

mA 375 A 5m A 5m 2 2 ' W = 375 2 = 3.75 2 (2 1) V = 1.88 mA | K n = K n 2 V 2 V 0.5um L V 0.5um = 3 -1 = 2V and VDS = 3.3 | VDS > (VGS VTN ) so the saturation region is correct 375 A 5m 2 2 (3 1) V = 7.50 mA 2 2 V 0.5m

(d) VGS VTNID =

4.21 (a) For VGS = 0, VGS < VTN and ID = 0 (b) For VGS = 1 V , VGS < VTN and ID = 0

(c ) VGS VTNID =

= 2 -1.5 = 0.5V and VDS = 4 | VDS > (VGS VTN ) so the saturation region is correct

200 A 10m A 10m mA 2 2 ' W V = 250 A | K = K = 200 2 1.5 = 2.00 2 ( ) n n 2 2 2 V 1um L V 1um V = 3 -1.5 =1.5V and VDS = 4 | VDS > (VGS VTN ) so the saturation region is correct 200 A 10m 2 2 (3 1.5) V = 2.25 mA 2 2 V 1m

(d) VGS VTNID =

79

4.22 (a) VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 0.2 V. VDS < VGS - VTN so the transistor is operating in the triode region.

V 0.2 W A 10 VGS VTN DS VDS = 200 2 2 0.75 0.2 = 460 A 2 2 L V 1 (b) VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 2.5 V. VDS > VGS - VTN so the transistor is operating in the saturation region. 200 A 10 K' W 2 2 ID = n (2 0.75) = 1.56 mA (VGS VTN ) = 2 2 V 1 2 L (c) VGS < VTN so the transistor is cutoff with ID = 0. 300 ' ID K n so (a) ID = (d) 460A = 690A (b) ID = 2.34 mA (c ) ID = 0 200 ' ID = K n

4.23 (a) VGS - VTN = 4 V, VDS = 6 V. VDS > VGS - VTN --> Saturation region

(b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.5 V, VDS = 0.5 V. VDS = VGS - VTN --> Boundary between triode and saturation regions (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 - 1 = 1.5 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 3 - (-6) = 9 V, VGS - VTN = 9 - 1 = 8.0 V, and VDS = 6 V --> triode regionD --> 'S' G + 2V (e) + S --> 'D' VG'S' = +2.5 V VD'S' = +0.5 V G 0.5 V + 3V (f) + S --> 'D' V G'S' = +9.0 V V D'S' = +6.0 V D --> 'S' -

6.0 V

80

4.24 (a) VGS - VTN = 2.6 V, VDS = 3.3 V. VDS > VGS - VTN --> Saturation region

(b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1.3 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.8 V, VDS = 0.5 V. VDS < VGS - VTN --> triode region (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 0.7 = 1.8 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 3 - (-3) = 6 V, VGS - VTN = 6 0.7 = 5.3 V, and VDS = 3 V --> triode region4.25

R

R2

4

D G B S R R

V

DD

+ -

1

3

4.26+VDD

D G

I B

S

+VDD

DD G S D B G S B

I B

G S

D B G S

(a)4.27

(b)

81

VDS = 3.3V, VGS VTN = 1.3 V; VDS > VGS - VTN so the transistor is saturated.

(a) gm = K n (VGS VTN ) = 250

A 20m (2 0.7) = 6.50 mS V 2 1m A 20m (3.3 0.7) = 13.0 mS V 2 1m

(b) gm = K n (VGS VTN ) = 2504.28

(a) gm =

iD 760 140 A = = 310 S | As a check, we can use the results from Problem 4.22. vGS 53 V

gm = K n (VGS VTN ) = 125

(b) gm =

iD vGS

(4 1.5)V = 313 S V2 A 390 15 A = = 188 S | Checking : gm = 125 2 (3 1.5)V = 188 S V 42 V

A

4.29 VDS > VGS - VTN so the transistor is saturated.

Kn 250 A 2 2 5 0.75) ( 1 + 0.025(6)) = 2.60 mA (VGS VTN ) (1 + VDS ) = 2 ( 2 2 V K 250 A 2 2 (b) ID = n (VGS VTN ) = 5 0.75) = 2.26 mA 2 ( 2 2 V ( a) ID =4.30 VDS > VGS - VTN so the transistor is saturated.

Kn 500 A 2 2 4 1) ( 1 + 0.02(5)) = 2.48 mA (VGS VTN ) (1 + VDS ) = 2 ( 2 2 V K 500 A 2 2 (b) ID = n (VGS VTN ) = 4 1) = 2.25 mA 2 ( 2 2 V ( a) ID =4.31 (a) The transistor is saturated by connection.ID = 12V VGS 100 x106 10 A 2 = 2 (VGS 0.75V ) 5 1 V 10 2

2 12.5VGS 17.8VGS 4.97 = 0 VGS = 0.266V , 1.214V VGS = 1.214 V since it must exceed 0.75V

ID =

12 1.214 = 108 A 10 5

Checking :

100 x106 10 A 2 2 (1.214 0.75V ) = 108 A 1 V 2

(b) ID =

12V VGS 1000 x106 A 2 = V 0.75V ) (1 + 0.025VGS ) 5 2 ( GS 10 2 V

82

Starting with the solution from part (a) and solving iteratively yields VGS = 1.20772 V and ID = 108 A, essentially no change. (c) 12V VGS 100 x106 25 A 2 ID = = 2 (VGS 0.75V ) 5 1 V 10 22 62.5VGS 91.75VGS + 11.16 = 0 VGS = 0.446V , 1.046V VGS = 1.046 V since VGS must exceed the threshold voltage.

12 1.046 ID = = 110 A 10 5

100 x106 25 A 2 Checking : ID = 2 (1.046 0.75V ) = 110 A 1 V 2

4.32 (a) The transistor is saturated by connection. 12V VGS 100 x106 10 A 2 ID = = 2 (VGS 0.75V ) 4 1 V 5 x10 22 31.25VGS 45.88VGS + 5.58 = 0 VGS = 0.0588V , 1.401V VGS = 1.401 V since VGS must exceed the threshold voltage.

12 1.401 100 x106 10 A 2 = 212 A Checking : I = 2 (1.401 0.75V ) = 212 A D 4 1 V 5 x10 2 6 12V VGS 1000 x10 A 2 = V 0.75V ) (1 + 0.02VGS ) (b) ID = 4 2 ( GS 5 x10 2 V ID =

Starting with the solution from part (a) and solving iteratively yields VGS = 1.3925 V and IDS = 212 A, essentially no change4.33 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated. ' ' W W Kn Kn 2 2 I = V V and I = ( GS1 TN ) (VGS 2 VTN ) . Therefore D1 D2 2 L 2 L From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' Kn Kn W W 2 2 I = ID1 = ID 2 or (VGS1 VTN ) = (VGS 2 VTN ) 2 L 2 L which requires VGS1 = VGS2. Using KVL:

VDD = VDS1 + VDS 2 = VGS1 + VGS 2 = 2VGS 2 V VGS1 = VGS 2 = DD = 5V 2 ' K W 100 A 10 2 2 I= n (VGS1 VTN ) = (5 0.75) V 2 = 9.03 mA 2 2 V 1 2 L (b) The current simply scales by a factor of two (see last equation above), and ID = 18.1 mA. (c) For this case,

83

' ' Kn Kn W W 2 2 V V 1 + V and I = ( GS1 TN ) ( (VGS 2 VTN ) (1 + VDS2 ). DS1 ) D2 2 L 2 L Since VGS = VDS for both transistors ' ' Kn Kn W W 2 2 ID1 = (VGS1 VTN ) (1 + VGS1 ) and ID 2 = (VGS 2 VTN ) (1 + VGS2 ) 2 L 2 L and ID1 = ID2 = I ' K' W Kn W 2 2 (VGS1 VTN ) (1 + VGS1 ) = n (VGS 2 VTN ) (1 + VGS2 ) 2 L 2 L which again requires VGS1 = VGS2 = VDD/2 = 5V. K' W 100 A 10 2 2 I= n (VGS1 VTN ) (1 + VDS ) = (5 0.75) V 2 (1 + (.04 )5) = 10.8 mA 2 2 L 2 V 1

ID1 =

4.34 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (by' ' W W Kn Kn 2 2 I = V V and I = (VGS 2 VTN ) . ( ) D 1 GS 1 TN D 2 Therefore 2 L 1 2 L 2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' 10 40 Kn Kn 2 2 V V = I = ID1 = ID 2 or ( GS1 TN ) (VGS 2 VTN ) 2 1 2 1 which requires VGS1 = 2VGS2 - VTN. Using KVL:

connection).

VDD = VDS1 + VDS 2 = VGS 2 + VGS1 = 3VGS 2 VTN V + VTN 10 + 0.75 VGS 2 = DD = = 3.583V VGS1 = 6.417 3 3 100 A 10 K' W 2 2 I= n (VGS1 VTN ) = (6.417 0.75) V 2 = 16.1 mA 2 2 V 1 2 L 100 A 40 2 Checking : I = (3.583 0.75) V 2 = 16.1 mA which agrees. 2 2 V 1 (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' W K' W 2 2 ID1 = n (VGS1 VTN ) (1 + VGS1 ) and ID 2 = n (VGS 2 VTN ) (1 + VGS2 ) 2 L 1 2 L 2

where VGS2 = VDD VGS1. Therefore, ' 10 K ' 40 Kn 2 2 1 + (10 VGS1 )) (VGS1 VTN ) (1 + VGS1 ) = n (10 VGS1 VTN ) ( 2 1 2 1 VGS1 = 6.3163, VGS2 = 3.6837, ID1 = 20.4 mA, Checking: ID2 = 20.4 mA which agrees.

84

4.35 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (by' ' W W Kn Kn 2 2 V V and I = (VGS 2 VTN ) . ( ) GS1 TN D2 Therefore 2 L 1 2 L 2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' 25 12.5 Kn Kn 2 2 I = ID1 = ID 2 or (VGS1 VTN ) = (VGS 2 VTN ) 2 1 2 1 Solving for VGS2 yields: VGS 2 = 2VGS1 2 1 VTN

connection).

ID1 =

(

)

Also, VDD = VDS1 + VDS 2 or VGS1 = 10 VGS 2 VGS1 = I= 10 +

( 2 1)V1+ 2

TN

= 4.271V VGS 2 = 5.729V

' 100 A 25 Kn W 2 2 2 (4.271 0.75) V = 15.5 mA (VGS1 VTN ) = 2 2 V 1 2 L 100 A 12.5 2 2 Checking : I = (5.729 0.75) V = 15.5 mA - agrees. 2 2 V 1 (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' W K' W 2 2 ID1 = n (VGS1 VTN ) (1 + VGS1 ) and ID 2 = n (VGS 2 VTN ) (1 + VGS2 ) 2 L 1 2 L 2

where VGS2 = VDD VGS1. Therefore, ' ' 10 40 Kn Kn 2 2 1 + (10 VGS1 )) (VGS1 VTN ) (1 + VGS1 ) = (10 VGS1 VTN ) ( 2 1 2 1 VGS1 = 4.3265 V, VGS2 = 5.6735 V, ID1 = 19.4 mA, Checking: ID2 = 19.4 mA both agree4.36 VGS - VTN = 5 - (-2) = 7 V > VDS = 6 V so the transistor is operating in the triode region. 6 6 (a) ID = 250 x10 5 (2) 6 = 6.00 mA 2 (b) Our triode region model is independent of , so ID = 6.00 mA. 4.37 Since VDS = VGS, and VTN < 0 for an NMOS depletion mode device, VGS - VTN will be greater than VDS and the transistor will be operating in the triode region.

85

4.38 ( a) VDS = 6V | VGS VTN = 0 (3) = 3V so the transistor is saturated2 Kn 250 A 2 0 (3V )] = 1.13 mA (VGS1 VTN ) = 2 [ 2 V 2 2 250 A 0 (3V )] ( 1 + 0.025(6)) = 1.29 mA (b) ID = 2 [ 2 V

ID =

4.39+10 V 100 k D G S (a) (b) I DS W = 10 L 1 G D -10 V 100 k S I DS W = 10 L 1

(a) If the transistor were saturated, then 100 x106 10 2 ID = (2) = 2.00 mA 1 2 but this would require a power supply of greater than 200 V (2 mA x 100 k). Thus the transistor must be operating in the triode region.10V VDS VDS 3 = 10 0 2 VDS ( ) 10 5 2 10 VDS = 50VDS (4 VDS ) and VDS = 0.0504V using the quadratic equation. 0.0504 10V VDS ID = 103 2 = 99.5 A 0.0504 = 99.5 A Checking : 2 10 5 (b) For R = 50 k and W/L = 20/1, 10V VDS V = 2 x103 0 (2) DS VDS 4 5 x10 2 10 VDS = 50VDS (4 VDS ), the same as part (a). 0.0504 10V VDS ID = 2 x103 2 = 199 A 0.0504 = 199 A Checking : 5 x10 4 2 (c) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region.

86

10V VDS V = 1000 x106 VDS (2) DS VDS 5 10 2 10 VDS = 50VDS (4 + VDS ) and VDS = 0.04915V using the quadratic equation. 0.04915 10V VDS ID = 103 0.04915 (2) = 99.5 A 0.04915 = 99.5 A Checking : 10 5 2 (d) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 10V VDS V = 2000 x106VDS (2) DS VDS 4 5 x10 2 10 VDS = 50VDS (4 + VDS ) Same as part (c). VDS = 0.04915V using the quadratic equation. 0.04915 10V VDS ID = 103 0.04915 (2) = 99.5 A 0.04915 = 99.5 A Checking : 10 5 2 4.40 See figures in previous problem but use W/L = 20/1. 25 x106 20 2 I = (1) = 250 A but this would require (a) If the transistor were saturated, then D 2 1 a power supply of greater than 25 V. Thus the transistor must be operating in the triode region.

20 10V VDS V = 100 x106 0 (1) DS VDS 5 1 10 2 10 VDS = 100VDS (2 VDS ) and VDS = 0.05105V using the quadratic equation. 0.05105 10 0.0510 ID = 2.00 x103 1 V = 99.5 A 0.05105 = 99.5 A Checking : 2 10 5 (b) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 20 V VDS = 10 ( 10 5 )( 100 x106 ) VDS (1) DS VDS 1 2 V VDS = 10 200VDS 1 + DS and VDS = 0.04858V using the quadratic equation. 2 0.04858 10 - 0.04858 V = 99.5 A ID = 2000 x106 1 + 0.04858 = 99.5 A Checking : 2 10 5

87

4.41 ( a) VTN = 0.75 + 0.75 1.5 + 0.6 0.6 = 1.26V

(

)

VGS VTN = 2 1.26 = 0.74V > VDS = 0.2V Triode region 10 0.2 ID = 200 x106 2 1.26 0.2 = 256 A (compared to 460 A) 1 2 (b) VGS VTN = 2 1.26 = 0.74V < VDS = 2.5V Saturation region 200 x106 10 2 ID = (2 1.26) = 548 A (compared to 1.56 mA) 1 2 (c) VGS < VTN s