circuitos microelectronicos - analisis y diseño - muhammad rashid

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Chapter 1 Exercise Problems EX1.1

3 / 2 exp2

gi

En BT

kT =

GaAs: ( ) ( ) ( )( )3/ 214

6

1.42.1 10 300 exp2 86 10 300i

n =

or 6 31.8 10 in cm=

Ge: ( )( ) ( )( )3/ 213

6

0.661.66 10 300 exp2 86 10 300i

n =

or 13 32.40 10 in cm=

EX1.2 (a) majority carrier: holes, 17 310 op cm

= minority carrier: electrons, ( )2102 3 3

17

1.5 102.25 10

10i

oo

nn c

p= = = m

cm

(b) majority carrier: electrons, 15 35 10 on= minority carrier: holes,

( )2102 4 315

1.5 104.5 10

5 10i

oo

np c

n= = = m

EX1.3 For n-type, drift current density EnJ e n or ( )( )( )19 16200 1.6 10 7000 10 E= which yields

E 17.9 /V cm= EX1.4 Diffusion current density due to holes:

( )16 110 expp p

pp p

dpJ eDdx

xeDL L

= =

(a) At 0x =( )( )( )19 16 2

3

1.6 10 10 1016 /

10pJ A cm

= =

(b) At 310 x cm= 3

23

1016exp 5.89 /10p

J A cm

= =

EX1.5

( ) ( )( )( )16 17

2 26

10 10ln 0.026 ln

1.8 10a d

bi Ti

N NV Vn

= = or 1.23 biV V=

EX1.6

1/ 2

1 Rj jobi

VC CV

= +

and


( ) ( )( )( )

2

17 16

210

ln

10 100.026 ln 0.757

1.5 10

a dbi T

i

N NV V

n

V

= = =

Then ( )1/ 2 1/ 250.8 1 7.610.757jo jo

C C

= + = or

2.21 joC p= F EX1.7

exp 1DD ST

vi IV

=

so ( )3 1310 10 exp 10.026

Dv =

Solving for the diode voltage, we find ( ) 313100.026 ln 110Dv

= +

or ( ) ( )100.026 ln 10Dv

which yields 0.599 Dv V=

EX1.8

and exp DPS D D D ST

VV I R V I IV

= +

so ( ) ( )3 344 4 10 4 10DD D D VI V I = + = and

( )1210 exp0.026

DD

VI = By trial and error, we find 0.864 and 0.535 D DI mA V V EX1.9

(a) 5 0.7 1.08 4

PSD D

V VI I m

R = = = A

(b) PS PSDD

V V V VI R

R I = =

Then 8 0.7 6.79 1.075

R k= = (c)


1.251.08

0.7 2 4 6 80

Diode curve

Load lines

VD(v)

(a)

(b)

ID(mA)

EX1.10 PSpice analysis EX1.11

Quiescent diode current 10 0.7 0.465 20

PSDQ

V VI m

R = = = A

Time-varying diode current:

We find that 0.026 0.0559 0.465

Td

DQ

Vr kI

= = =

Then ( )( )

0.2sin0.0559 20

Id

d

Vv tir R k

= = + + or ( )9.97sindi t A = EX1.12

For the pn junction diode, ( ) 3151.2 10ln 0.026 ln 4 10DD T SIV VI

= or 0.6871 DV V=

The Schottky diode voltage will be smaller, so 0.6871 0.265 0.4221 DV V= = Now exp DD S

T

VI IV

or 3

101.2 10 1.07 10 0.4221exp0.026

S SI I

= = A

EX1.13

( )10 5.6 1.79 ZP I V I I mA= = = Also 10 5.6 1.79 2.46 I R k

R= = =

Test Your Understanding Exercises TYU1.1 (a) T = 400K

Si: 3/ 2 exp2

gi

En BT

kT =

( )( ) ( )( )3 / 215

6

1.15.23 10 400 exp2 86 10 400i

n =

or 12 34.76 10 in c

= m


( )( ) ( )( )Ge: 3/ 215

6

0.661.66 10 400 exp2 86 10 400i

n =

m

or 14 39.06 10 in c

= GaAs:

( )( ) ( )( )3 / 214

6

1.42.1 10 400 exp2 86 10 400i

n =

m

or 9 32.44 10 in c

= (b) T = 250 K

( )( ) ( )( )Si: 3/ 215

6

1.15.23 10 250 exp2 86 10 250i

n =

m

or 8 31.61 10 in c

=

( )( ) ( )( )Ge: 3/ 215

6

0.661.66 10 250 exp2 86 10 250i

n =

m

or 12 31.42 10 in c

=

( )( ) ( )( )GaAs: 3/ 214

6

1.42.10 10 250 exp2 86 10 250i =

m

=

(a)

( )15 3

21024 3

15

8 10

1.5 102.81 10

8 10

o d

io

o

n N cm

np cn

m

= = = = =

(b) 15 158 10 0.1 10on n n= + = + or

15 38.1 10 n c = m4

4 12.81 10 10op p p= + = +

or 14 310 p cm

TYU1.6

(a) 2lna d

bi Ti

N NV V

n =

so ( ) ( )( )( )15 17

210

10 100.026 ln 0.697

1.5 10biV V

= =

(b) ( ) ( )( )( )17 17

210

10 100.026 ln 0.817

1.5 10biV V

= =

TYU1.7

(a) exp 1DD ST

VI IV

=

14 0.510 exp0.026D

I Then, for VD = 0.5 V, 2.25 DI A= VD = 0.6 V, 0.105 DI mA= VD = 0.7 V, 4.93 DI mA= (b) 1410 D SI I A

= for both cases. TYU1.8

100T C = so 2 100 200 DV m = VVThen 0 650 0 20 0 450 DV . . .= =

TYU1.9

1 2 3 40

Diode0.450.500.55

0.0330.2251.54

Load line

VD(v)

ID(mA)

1.00.87

0.54v

VD ID

TYU1.10

( )1.05 0.7D D DP I V I= = so 1 5 DI . mA=


Now 10 0.7 6.2 1.5

PS

D

V VR R

I = = = k

TYU1.11

0.8 30.8 0.026

Dd

T

Ig mSV

= = = TYU1.12

0.026 0.0265050

Td D

D D

Vr II I

= = = or

0.52 DI mA= TYU1.13 For the pn junction diode,

4 0.7 0.825 4D

I mA= =

For the Schottky diode, 4 0.3 0.925 4D

I mA= = TYU1.14

z zo z z zo z z zV V I r V V I r= + = so ( )( )35.20 10 20 5.18 zoV V= = Then ( )( )35.18 10 10 20 5.38 z zV V= + = V


Chapter 1 Problem Solutions 1.1

/ 23/ 2 gE kTin BT e

= (a) Silicon

(i) ( )( ) ( )( )[ ]

3 / 2156

19

8 3

1.15.23 10 250 exp2 86 10 250

2.067 10 exp 25.581.61 10 cm

i

i

n

n

= = =

(ii) ( )( ) ( )( )[ ]

3 / 2156

19

11 3

1.15.23 10 350 exp2 86 10 350

3.425 10 exp 18.273.97 10 cm

i

i

n

n

= = =

(b) GaAs

(i) ( )( ) ( )( )( ) [ ]

3/ 2146

17

3 3

1.42.10 10 250 exp2 86 10 250

8.301 10 exp 32.56

6.02 10 cm

i

i

n

n

= = =

(ii) ( )( ) ( )( )( ) [ ]

3/ 2146

18

8 3

1.42.10 10 350 exp2 86 10 350

1.375 10 exp 23.26

1.09 10 cm

i

i

n

n

= = =

1.2

a. 3 / 2 exp2i

Egn BTkT

= 12 15 3 / 2

6

1.110 5.23 10 exp2(86 10 )( )

TT

=

34 3/ 2 6.40 101.91 10 expT

T =

By trial and error, 368 KT b. 9 310 cmin

=

( )( )9 15 3 / 2 61.110 5.23 10 exp

2 86 10T

T =

37 3 / 2 6.40 101.91 10 expT

T =

By trial and error, 268 KT 1.3 Silicon


(a) ( )( ) ( )( )( ) [ ]

3/ 2156

18

10 3

1.15.23 10 100 exp2 86 10 100

5.23 10 exp 63.95

8.79 10 cm

i

i

n

n

= = =

(b) ( )( ) ( )( )( ) [ ]

3 / 2156

19

10 3

1.15.23 10 300 exp2 86 10 300

2.718 10 exp 21.32

1.5 10 cm

i

i

n

n

= = =

(c) ( )( ) ( )( )( ) [ ]

3 / 2156

19

14 3

1.15.23 10 500 exp2 86 10 500

5.847 10 exp 12.79

1.63 10 cm

i

i

n

n

= = =

Germanium.

(a) ( ) ( ) ( )( ) ( ) [ ]3 / 215 18

6

3

0.661.66 10 100 exp 1.66 10 exp 38.372 86 10 100

35.9 cm

i

i

n

n

= = =

(b) ( )( ) ( )( ) ( ) [ ]3 / 215 18

6

13 3

0.661.66 10 300 exp 8.626 10 exp 12.792 86 10 300

2.40 10 cm

i

i

n

n

= = =

(c) ( )( ) ( )( ) ( ) [ ]3 / 215 19

6

15 3

0.661.66 10 500 exp 1.856 10 exp 7.6742 86 10 500

8.62 10 cm

i

i

n

n

= = =

1.4 a. 15 35 10 cm n typedN

= 15 3

0 5 10 cmdn N= =

( )2102 4 30 015

0

1.5 104.5 10 cm

5 10inp p

n= = =

b. 15 35 10 cm typedN n=

15 35 10 cmo dn N= =

( )( )( )( ) ( )

3/ 2146

3/ 214 12

6 3

1.42.10 10 300 exp2(86 10 )(300)

2.10 10 300 1.65 10

1.80 10 cm

in

= = =

( )262 4 30 015

0

1.8 106.48 10 cm

5 10inp p

n = = =

1.5


(a) n-type (b)

( )16 3

21023 3

16

5 10 cm

1.5 104.5 10 cm

5 10

o d

io

o

n N

npn

= = = = =

(c) 16 35 10 cmo dn N= =

From Problem 1.1(a)(ii) 11 33.97 10 cmin=

( )211 6 316

3.97 103.15 10 cm

5 10op

= = 1.6 a. 16 310 cm typeaN p

= 16 3

0 10 cmap N= =

( )2102 4 30 016

0

1.5 102.25 10 cm

10inn n

p= = =

b. Germanium 16 310 cm typeaN p

= 16 3

0 10 cmap N= =

( )( ) ( )( )( )( ) ( )

3/ 2156

3/ 215 6

13 3

0.661.66 10 300 exp2 86 10 300

1.66 10 300 2.79 10

2.4 10 cm

in

= = =

( )2132 10 30 016

0

2.4 105.76 10 cm

10inn n

p= = =

1.7 (a) p-type (b)

( )17 3

21023 3

17

2 10 cm

1.5 101.125 10 cm

2 10

o a

io

o

p N

nnp

= = = = =

(c) 17 32 10 cmop=

From Problem 1.1(a)(i) 8 31.61 10 cmin=

( )28 317

1.61 100.130 cm

2 10on

= = 1.8 (a)

( )15 3

21024 3

15

5 10

1.5 104.5 10

5 10

o

io o

o

n cm

np pn

cm

= = = =

(b) n-typeo on p (c) 15 35 10 o dn N cm

= 1.9 a. Add Donors

15 37 10 cmdN=


b. Want 6 3 210 cm /o i dp n N= =

So ( )( )2 6 152 3

10 7 10 7 10

exp

in

EgB TkT

= = =

21

( ) ( )( )221 15 3

6

1.17 10 5.23 10 exp86 10

TT

=

By trial and error, 324 KT 1.10

( )( ) ( )42.2 15 10 3.3 I J A EA

I I

= == = mA

1.11

1

8512

7.08 (ohm cm)

JJ EE

= = ==

1.12

( )( )( )191 1 1

1.6 10 480 0.80ap a pg N

e N e g = = 16 31.63 10 cmaN

= 1.13

( )( )( )19

15 3

0.51.6 10 1350

2.31 10 cm

n d

dn

d

e N

Ne

N

== = =

1.14 (a) For n-type, ( )( )191.6 10 8500n d de N N = For ( ) 115 19 3 410 10 1.36 1.36 10dN cm cm (b) ( ) 3 20.1 0.136 1.36 10 /J E J = = A cm 1.15

( )( ) 15 219 42

10 101.6 10 1800.5 10

576 A/cm

n n n

n

dn nJ eD eDdx x

J

= = =

=

1.16


( )( )( )( )

15

19 15

4

/

110 exp

1.6 10 15 10exp

10 10

2.4 p

p p

pp p

pp

x Lp

dpJ eDdx

xeDL L

xJL

J e

= =

= =

(a) x = 0 22.4 A/cmpJ =(b) 10 mx = 1 22.4 0.883 A/cmpJ e= =(c) 30 mx = 3 22.4 0.119 A/cmpJ e= = 1.17 a. 17 3 17 310 cm 10 cma oN p

= = ( )262 5 3

17

1.8 103.24 10 cm

10i

o oo

nn n

p = = =

b. 5 15 15 317 15 17 3

3.24 10 10 10 cm10 10 1.01 10 cm

o

o

n n n np p p p

= + = + == + = + =

1.18

(a)

( ) ( )( )( )

2

16 16

210

ln

10 100.026 ln 0.697 V

1.5 10

a dbi T

i

N NV Vn

= = =

(b) ( ) ( )( )( )18 16

210

10 100.026 ln 0.817 V

1.5 10biV

= =

(c) ( ) ( )( )( )18 18

210

10 100.026 ln 0.937 V

1.5 10biV

= =

1.19

2lna d

bi Ti

N NV V

n =

a. ( ) ( )( )16 166 210 100.026 ln 1.17 V(1.8 10 )bi biV V =

=

b. ( ) ( )( )18 166 210 100.026 ln 1.29 V(1.8 10 )bi biV V =

=

c. ( ) ( )( )18 186 210 100.026 ln 1.41 V(1.8 10 )bi biV V =

=

1.20

( ) ( )162 110ln 0.026 ln (1.5 10 )aa dbi T iNN N

V Vn

= = 0 2


For 15 310 , 0.637 a biN cm V= = V

VFor 18 310 , 0.817 a biN cm V= =

0.817

0.637

1015 1016 1017 1018 Na(cm3)

Vbi(V)

1.21

(0.026)300TkT =

T kT (T)3/2200 0.01733 2828.4 250 0.02167 3952.8 300 0.026 5196.2 350 0.03033 6547.9 400 0.03467 8000.0 450 0.0390 9545.9 500 0.04333 11,180.3

( )( ) ( )( )14 3 / 2 61.42.1 10 exp

2 86 10in T

T =

2lna d

bi Ti

N NV V

n =

T ni Vbi200 1.256 1.405 250 6.02 103 1.389 300 1.80 106 1.370 350 1.09 108 1.349 400 2.44 109 1.327 450 2.80 1010 1.302 500 2.00 1011 1.277

200

1.45

1.35

1.25

250 300 350 400 450 500

Vbi(V)

T(C) 1.22

1/ 2

1 Rj jobi

VC CV

= +


( ) ( )( )( )16 15

10 2

1.5 10 4 100.026 ln 0.684 V

1.5 10biV

= =

(a) ( )1/ 210.4 1 0.255 pF

0.684jC

= + =

(b) ( )1/ 230.4 1 0.172 pF

0.684jC

= + =

(c) ( )1/ 250.4 1 0.139 pF

0.684jC

= + = 1.23

(a) 1 2

1/

Rj jo

bi

VC CV

= +

For VR = 5 V, 1 25(0 02) 1 0 00743

0 8

/

jC . . p.

= + = F

For VR = 1.5 V, 1 21 5(0 02) 1 0 0118

0 8

/

j.C . . p.

= + = F 0 00743 0 0118( ) 0 00962

2j. .C avg . pF+= =

( ) ( ) ( ) ( )( ) t /C C C Cv t v final v initial v final e = + where

3 1( ) (47 10 )(0 00962 10 )jRC RC avg . = = = 2 or

104 52 10 . s = Then ( ) ( )1 5 0 5 0 it /Cv t . e = = +

1 /1

5 5ln1.5 1.5

re t + = = 10

1 5.44 10 t s=

(b) For VR = 0 V, Cj = Cjo = 0.02 pF

For VR = 3.5 V, ( )1/ 23.50.02 1 0.00863

0.8jC p

= + = F 0.02 0.00863( ) 0.0143

2jC avg pF+= =

( ) 106.72 10jRC avg s = = ( ) ( ) ( ) ( )( ) /tC C C Cv t v final v initial v final e = +

( )2 2/ /3.5 5 (0 5) 5 1t te e = + = so that 102 8.09 10 t s

= 1.24

( ) ( )( )( )18 15

210

10 100.026 ln 0.757 V

1.5 10biV

= =

a. 1 VRV =1/ 21(0.25) 1 0.164 pF

0.757jC

= + =


( )( )0 3 11 1

2 2 2.2 10 0.164 10f

LC = =

2

0 8.38 MHzf = b. VR = 10 V

1/ 210(0.25) 1 0.0663 pF0.757j

C = + =

( )( )0 3 11

2 2.2 10 0.0663 10f

=

2

0 13.2 MHzf = 1.25

a. exp 1 0.90 exp 1D DST T

V VI IV V

= =

exp 1 0.90 0.10DT

VV

= =

( )ln 0.10 0.0599 VD T DV V V= = b.

0.2exp 1 exp 10.026

0.2exp 1exp 1 0.026

21901

2190

F

TSF

R S R

T

F

R

VVII

I I VV

II

= =

= =

1.26 a.

11

11

11

0.5(10 )exp 2.25 mA0.0260.6(10 )exp 0.105 A

0.0260.7(10 )exp 4.93 A

0.026

I I

I I

I I

= = = = =

b. 13

13

13

0.5(10 )exp 22.5 A0.0260.6(10 )exp 1.05 mA

0.0260.7(10 )exp 49.3 mA

0.026

I I

I I

I I

= = = = =

1.27 (a) ( )/ 1D TV VSI I e=

( )/ /6 11 11150 10 10 1 10D T DV V V Ve e = T


Then 6 6

11 11

150 10 150 10ln (0.026) ln10 10D T

V V

= =

Or 0.430 DV V= (b)

6

13

150 10ln10D T

V V

=

Or 0.549 DV V= 1.28

(a) 3 0.710 exp0.026S

I = 152.03 10 ASI

= (b)

DV ( ) ( 1)DI A n = ( )( )2DI A n = 0.1 149.50 10 141.39 10 0.2 124.45 10 149.50 10 0.3 102.08 10 136.50 10 0.4 99.75 10 124.45 10 0.5 74.56 10 113.04 10 0.6 52.14 10 102.08 10 0.7 310 91.42 10 1.29 (a)

1210SI A=

VD(v) ID(A) log10ID0.10 114 68 10. 10 3. 0.20 92 19 10. 8 66. 0.30 71 03 10. 6 99. 0.40 64 80 10. 5 32. 0.50 42 25 10. 3 65. 0.60 21 05 10. 1 98. 0.70 14 93 10. 0 307. (b)

1410SI A=

VD(v) ID(A) log10ID0.10 134 68 10. 12 3. 0.20 112 19 10. 10 66. 0.30 91 03 10. 8 99. 0.40 84 80 10. 7 32. 0.50 62 25 10. 5 65. 0.60 41 05 10. 3 98. 0.70 34 93 10. 2 31. 1.30 a.

2 2 1

1

10 exp

ln (10) 59.9 mV 60 mV

D D D

D T

D T D

I V VI VV V V

= = = =


b. ( )ln 100 119.7 mV 120 mVD T DV V V = = 1.31

(a) (i) ( ) 615150 10ln 0 026 ln 10 = 0.669 V

DD t

S

D

IV V .I

V

= =

(ii) 6

15

25 100 026)ln10

= 0.622 V

D

D

V .

V

= (

(b) (i) 15 120 2(10 )exp 2 19 10 A0 026D

.I ..

= = (ii) 0DI =(iii) 1510 ADI

= (iv) 1510 ADI

= 1.32

3

14

3

12

2 10ln (0 026) ln 0 6347 V5 10

2 10(0 026) ln 0 5150 V5 10

0 5150 0 6347 V

DD t

S

D

D

IV V . .I

V . .

. V .

= = = = =

1.33

(a) exp DD St

VI IV

=

3 21.1012 10 exp 5 07 10 A0.026S S

I I . = = 1

(b) ( )21

4

1.05 07 10 exp0.026

2 56 10 A 0 256 mA

D

D

I .

I . .

= = =

1.34

(a) 23 71 010 exp 5 05 10 A0 026D

.I ..

= =

(b) 23 51 110 exp 2 37 10 A0 026D

.I ..

= =

(c) 23 31 210 exp 1 11 10 A0 026D

.I ..

= = 1.35 IS doubles for every 5C increase in temperature.

1210 SI A= at T = 300K

For 120.5 10 T 295 KSI A= =

For 1250 10 , (2) 50 5.64nSI A n= = =

Where n equals number of 5C increases. Then ( )( )5.64 5 28.2 T K = =So 295 328.2 T K


1.36 / 5( ) 2 , 155 C

( 55)TS

S

I TT

I= =

155 / 5 9(100) 2 2.147 1( 55)

S

S

II

= = 0 @100 C 373 K 0.03220T TV V = @ 55 C 216 K 0.01865T TV V =

( )(( )

)

9

9 8

13

3

0.6exp(100) 0.0322(2.147 10 )

0.6( 55) exp0.01865

2.147 10 1.237 10

9.374 10

(100) 2.83 10( 55)

D

D

D

D

II

II

= =

=

1.37 3.5 = ID (105) + VD

(a) 9 95 10 exp 0 026 ln0 026 5 10D D

D DV II V ..

= =

Trial and error. VD ID VD0.50 53 10 0.226 0.40 53 1 10. 0.227 0.250 53 25 10. 0.228 0.229 53 271 10. 0.2284 0.2285 53 2715 10. 0.2284 So

5

0 2285 V3 272 10 A

D

D

V .I .

(b)

( )( )9

9 5 4

5 10 A

5 10 10 5 10 V

3 4995 V

D S

R

D

I I

V

V .

= = = = =

1.38

( )410 2 10D DI V= + and ( ) 120.026 ln 10DD IV = Trial and error. VD(v) ID(A) VD(v) 0.50 44.75 10 0.5194 0.517 44.7415 10 0.5194 0.5194 44.740 10 0.5194

0.5194 V

0.4740 mAD

D

V

I

==

1.39


135 10 AsI=

1.2 V

R1 50 K

R2 30 K

ID

VTH

VD

VD

RTH R1 R2 18.75 K

ID

2

1 2

30(1.2) (1.2) 0.45 V80TH

RVR R

= = = +

0.45 , ln DD TH D D TS

II R V V VI

= + =

By trial and error: 2.56 A, 0.402 VD DI V= =

1.40

VI

V0

VD VD

IR

I1

I2

1 K

132 10 SI A

= 0 0.60 VV =

( )1302

1 2

0.60exp 2 10 exp0.026

2.105 mA0.6 0.60 mA1 K

2.705 mA

ST

R

R

VI IV

I

I I I

= = == == + =

31

13

0

2.705 10ln (0.026) ln2 10

0.60652 1.81 V

D TS

I D I

IV VI

V V V V

= = == + =

1.41 (a) Assume diode is conducting. Then, 0.7 DV V V= =So that 2

0 7 23 3 30R.I . A=


11 2 0 7 50

10R. .I A=

Then 1 2 50 23 3D R RI I I .= = Or 26 7 DI . A= (b) Let Diode is cutoff. 1 50 R k=

30 (1 2) 0 45 30 50D

V .= =+ . V Since , 0D DV V I< = 1.42

ID

5 V

VAVB VAVr

3 k

2 k

2 k

2 k

A&VA:

(1) 5

2 2A A

DV VI = +

A& A rV V(2) ( ) ( )5

2 2A r A r

D

V V V VI

+ =

So ( )5 5

3 2 2A r A A AV V V V V V

2r + =

Multiply by 6: ( ) (10 2 15 6 3A r A A rV V V V V + = )

A

25 2 3 11r rV V V+ + = (a) 0.6 VrV =

( )11 25 5 0.6 28 2.545 VA AV V= + = = From (1)

52.5

2 2A A

D AV V

DI V I= = Neg. 0DI =

Both (a), (b) 0DI = VA = 2.5,

2 5 2 V 0.50 V5B D

V V= = = 1.43 Minimum diode current for VPS (min)

(min) 2 , 0.7 D DI mA V V= = 2 1

2 1

0 7 5 0 7 4 3, . .I IR R

= = =1

.R

We have 1 2 DI I I= +


so (1) 1 2

4 3 0 7 2. .R R

= + Maximum diode current for VPS (max) ( )10 0 7 14 3 D D D DP I V I . I . mA= = =

1 2 DI I I= + or

(2) 1 2

9 3 0 7 14 3. . .R R

= +

Using Eq. (1), 11 1

9 3 4 3 2 14 3 0 41 . . . R . kR R

= + = Then 2 82 5 82 5R . .= 1.44 (a) Vo = 0.7 V

5 0.7 0.215 mA20

I I= =

(b) 10 0.7 0.2325 mA20 20

(20 K) 5 0.35 Vo o

I I

V I V

= =+= =

(c) 10 0.7 0.372 mA5 20

0.7 (20) 8 0.14 Vo o

I I

V I V

= =+= + = +

(d) 0(20) 5 5 Vo o

IV I V

== =

1.45

(a) ( )95 2 10 DI V= + ( ) 120.026 ln 2 10D IV = VD ID VD Vo = VD = 0.482 V 0.6 42.2 10 0.481 0.482 42.259 10 0.482 0.226 I mA=

(b) ( )410 4 10 DI V= + ( ) 120.026 ln 2 10D IV = Vo I VD 0.483 DV V= 0.5 42.375 10 0.4834 I = 0.238 mA 0.484 42.379 10 0.4834 0.24 oV V=

(c) ( )410 2.5 10 DI V= + ( ) 120.026 ln 2 10D IV = Vo I VD VD = 0.496 V 0.480 43.808 10 0.496 I = 0.380 mA 0.496 43.802 10 0.496 0.10 oV V= (d) 122 10 A

5 VS

o

I I IV

= =

1.46 (a) Diode forward biased VD = 0.7 V


5 (0.4)(4.7) 0.7 2.42 VV V= + + = (b) (0.4)(0.7) 0.28 mDP I V P= = = 1.47

(a) 2 1 1

2

02 1

1 1

0.65 0.65 mA1

2(0.65) 1.30 mA2 5 3(0.65) 1.30 2.35 K

R D D

D

I rD

I I I

IV V VI R

R R

= = = == =

= = = =

(b) 2

2 2

1 2 2

1

0.65 0.65 mA1

8 3(0.65) 3.025 mA2

3.025 0.652.375 mA

R

D D

D D R

D

I

I I

I I II

= == =

= = =

1.48

a. (0 026) 0 026 k 26

1

0 05 50 A peak-to-peak

(26)(50) A 1 30 mV peak-to-peak

Td

DQ

d DQ

d d d d

V . .I

i . Iv i v .

= = = = = == = =

b. For (0 026)0 1 mA 2600 1DQ d.I .

.= = =

0 05 5 A peak-to-peakd DQi . I = = (260)(5) V 1 30 mV peak-to-peakd d d dv i v . = = =

1.49

RS

S d

a. diode resistance d Tr V I= /

d Td S

Td SS

Td s o

T S

r V Iv vVr R RI

Vv v vV IR

/= = + + = = +

Sv

b. 260SR =


( ) ( )

0 0

0 0

0 0

0 0261 mA, 0 09090 026 (1)(0 26)

0 0260 1 mA, 0 500 026 0 1 0 26

0 0260 01 mA. 0 9090 026 (0 01)(0 26)

T

S T S S

s S

S S

v vV .I .v V IR . . v

v v.I . .v . . . v

v v.I . .v . . . v

= = = = + + = = =+= = =+

1.50

exp , lnaS a TT S

V II I V VV I

=

6

14

100 10pn junction, (0 026) ln10a

V .

=

0 599 VaV .=

Schottky diode, 6

9

100 10(0 026) ln 10a

V .

=

0 299 VaV .= 1.51

I

Schottky

pn junction

Schottky: exp aST

VI I

V

3

7

0.5 10ln (0.026) ln5 10

0.1796

a TS

IV VI

V

= = =

Then of pn junction 0.1796 0.30

0.4796aV = +

= 30 5 10

0 4796exp exp 0 026

Sa

T

I .I.V.V

= =

124 87 10 ASI .=

1.52 (a)

0.5 mA

VD

I1

I2

31 2 0 5 10I I .

+ =


8 125 10 exp 10 exp 0 5 10D DT T

V V .V V

+ = 3

8 35 0001 10 exp 0 5 10DT

V. .V

=

3

8

0 5 10(0 026) ln 0 23955 0001 10D D

.V . V ..

= =

Schottky diode, 2 0 49999 mAI .= pn junction, 1 0 00001 mAI .= (b)

VD2 VD1

I0.90 V

12 81 210 exp 5 10 expD D

T T

V VIV V

= =

1 2 0.9D DV V+ = 12 81 1

8 1

0.910 exp 5 10 exp

0.95 10 exp exp

D D

T T

D

T T

V VV V

VV V

= =

81

12

2 5 10 0 9exp exp0 02610

D

T

V .V .

=

8

1 12

5 102 ln 0 9 1 181310D T

V V . .

= + =

1 0 5907 pn junctionDV .= 2 0 3093 Schottky diodeDV .=

12 0 590710 exp 7 35 mA0 026.I I.

= = . 1.53

VPS 10 V

VZ

R 0.5 K

RL

V0

I

IZIL

0 5 6 V at 0 1 mAZ Z ZV V . I .= = =

10Zr = ( )( )0 1 10 1 mVZ ZI r .= =

VZ0 = 5.599 a. LR

10 5 599 4 401 8 63 mA0 50 0 01Z Z

. .I .R r . .= = =+ +

( )(0 5.599 0.00863 10Z Z Z ZV V I r= + = + ) 0 5 685 VZV V .= =


b. 11 5 59911 V 10 59 mA0 51PS Z

.V I ..

= = = ( )( )0 5.599 0.01059 10 5.7049 VZV V= = + =

9 5 5999 V 6 669 mA0 51PS Z

.V I ..

= = = ( )( )0 5.599 0.006669 10 5.66569 VZV V= = + =

0 05 7049 5 66569 0 0392 VV . . V . = = c. I = IZ + IL

0 0 0, , 0PS ZL ZL Z

V V V V VI I IR R r

= = =

0 010 5 5990 50 0 010 2

V V . V. . = + 0

010 5 599 1 1 1

0 50 0 010 0 50 0 010 2. V

. . . . + = + +

20.0 + 559.9 = V0 (102.5) 0 5.658 VV =

1.54

a.

( )( )9 6 8 11 mA

0 211 6 8 74 8 mW

Z Z

Z Z

.I I.

P . P .

= == =

b.

12 6 8 26 mA0 226 11 100 136%

11

Z Z.I I

.

%

= ==

( )( )26 6.8 176.8 mW176.8 74.8% 100 136%

74.8

ZP = ==

1.55

( )( )0.1 20 2 mVZ ZI r = = 0 6 8 0 002 6 798 VZV . . .= =

a. LR = 10 6 798 6 158 mA0 5 0 02Z Z

.I I .. .= =+

( )( )0 0 6.798 0.006158 20Z Z Z ZV V V I r= = + = + 0 6 921 VV .=

b. Z LI I I= + 0 010 6 798

0 50 0 020 1V V . V

. . = + 0

010 6 798 1 1 1

0 30 0 020 0 50 0 020 1. V

. . . . + = + +

359.9 = V0 (53) V0 = 6.791 V

0 6 791 6 921V . . = 0 0.13 VV =

1.56


For VD = 0, 0.1 ASCI = For ID = 0 14

0.2ln 15 10D T

V V = +

0.754 VD DCV V= =


Chapter 2 Exercise Problems EX2.1

( )( )

30 12 0.6 87 0.2

max 30 12 42

D

R

i peak mA

v V

= == + =

11

12.6sin 24.830

t =

2

By symmetry180 24.8 155.2t = =

155.2 24.8% time 100% 36.2%360= =

EX2.2 (a) 112sin 1.4 0Ov = = or 1

1.4sin 0.116612

= = which yields

1 6.7 = By symmetry, 2 180 6.7 173.3 = = Then 173.3 6.7% time 100% 46.3%

360= =

(b) 11 4sin 0 354. . = =

which yields 1 20 5. =

By symmetry, 2 180 20 5 159 5. . = = Then 159.5 20.5% time 100% 38.6%

360= =

EX2.3

( )( )( )324

2 2 60 10 0.4M

r

VCfRV

= = or 500 C F= EX2.4

M M

rr

V VV Rf RC f CV

= = or ( )( )( )675

60 50 10 4R =

Then 6.25 R k= EX2.5 10 14 , 5.6 ,PS ZV V V V = 20 100 LR

( )( )

( ) ( )( ) ( )

( ) ( )( ) ( )

5.6max 0.28 ,205.6min 0.056 100

max max min min(max)

min 0.9 0.1 max min 0.9 0.1 max

L

L

PS Z L PS Z LZ

PS Z PS PS Z PS

I A

I A

V V I V V II

V V V V V V

= =

= = =

or ( ) ( ) ( ) ( )( )( )( ) ( )( )14 5.6 280 10 5.6 56

max10 0.9 5.6 0.1 14L

I =


or ( )max 591.5 LI m= A( ) ( )( )Power(min) max 0.5915 5.6Z ZI V= =

So Power(min) = 3.31 W

Now ( )

( ) ( )max 14 5.6

max min 0.5915 0.056PS Z

iZ L

V VR

I I = =+ + or 13 iR

EX2.6

( )( ),max13.6 9For 13.6 , 0.2383 15.3 49 4 0.2383 9.9532

PS Z

L

v V I A

v V

= = =+= + =

( )( ),min11 9For 11 , 0.1036

15.3 49 4 0.1036 9.4144

PS Z

L

v V I A

v V

= = =+= + =

9.9532 9.4144Source Reg 100% 100%13.6 11

L

PS

vv = =

or Source Reg 20.7%=

( )( ),13.6 9For 0, 0.2383 15.3 49 4 0.2383 9.9532

L Z

L noload

I I A

v V

= = =+= + =

For 100 ,LI mA= ( )13.6 9 4 0.1015.3Z

Z

II

+ = which yields

( )( ), , ,

,

0.1591 9 4 0.1591 9.6363

Load Reg 100%

9.9532 9.6363 100%9.6363

Z

L full load

L noload L full load

L full load

I Av A

v vv

== + =

= =

V

or Load Reg 3.29%= EX2.7

VOR1

V2

D2

V1

D1

I

For 25 , on 5 I Ov V D V< = VThen, V2 = 4.3 V. D1 turns on when v1 = 2.5 V, Then, V1 = 1.8 V.


For 21 2

1 12.5 ,3 3

OI

I

v Rv Vv R R

> = + =

V

So that R1 = 2R2 EX2.8 For ( )0, max 2 OV v = =Now, so that 8 ,Ov V = ( )min 10 Ov V= EX2.9

10 4.44.4 , 0.5895 9.5O

v V I = = = mA Set I = ID1, then ( )( )4.4 0.6 0.5895 0.5 3.505 Iv V= = Summary: For 0 3.5 , 4.4 I Ov V v = V

D VFor turns on and when 23.5 , Iv V> 9.4 , 10 I Ov V v =

10

4.4

03.505 9.4

O(V)

I(V) EX2.10

( )1

2 2

0.6 , 00.6 10

4.27 2.2

O D

D D

V V I

I I I I

= = = = = = mA

EX2.11 At the VA node:

215

5 1A

DV I = +

5AV (1)

At the VB node: ( )2

15 0.75 1B

0B

D

V VI + + = (2)

We see that 0.7,B AV V= so Equation (2) becomes

215 0.7

5 1A A

DV VI + =

0 (2)

Solving for ID2 from Equation (1) and substituting into Equation (2), we find 15 0.7

25 15 10

A A AV V V = Then VA = 10.71 V and VB = 10.01 V

Solving for the diode currents, we obtain 1 115 10.71 0.858

5D DI I m= = A


Also 2 215 10.71 10.71 0.144

5 15D DI I m= = A

V

EX2.12 Assuming all diodes are conducting, we have 0.7 BV = and VA = 0 Summing currents, we have

15

5A

2D DV I I = + (1)

( )2 3

55

BD D

VI I

+ = (2) ( )

1

10 10 0.710 10

BD

VI

= = (3) ( )( )( )

1

2

3

From 3 , we find0.93

From 1 , we obtain0.07

From 2 , we have0.79

D

D

D

I mA

I mA

I mA

=

=

=

EX2.13 (a) phI e=

so ( ) ( ) ( )( ) ( )2

1919

6.4 100.8 1.6 10 0.52 1.6 10

I

=

or 12 8 phI . mA= (b) We have ( )( )12.8 1 12.8 .Ov V= =The diode must be reverse biased so that 12.8 PSV V> EX2.14 The equivalent circuit is

R

0.2 V

5 V

Vr 1.7 V

rf 15

I

So 5 1.7 0.2 15 f

I mAr R = =+

Or 15 1.7 0.2 3.1 0.207 15 15f

r R k + = = = Then 207 15 192 R R= =2.15

(a)

( )( )40 12 0.233 A

1200.233 12 2.8 W

ZI

P

= == =


(b) ( )( )0.233 A, 0.9 0.233 0.21 AR LI I= = =So 120.21 57.1L

L

RR

= = (c) ( )( )( )0.1 0.233 12 0.28 WP P= = Test Your Understanding Exercises TYU2.1

( )120sin 2 60 , 0.7 ,Iv t V V= = 2.5 R k and = Full-wave rectifier: Turns ratio 1:2 so that

120 0.7 119.3 119.3 100 19.3

S I

M

r

v vV VV V

== == =

So ( )( )( )3119.3

2 2 60 2.5 10 19.3M

r

VCf RV x

= = or 20.6 C F= TYU2.2

( )50sin 2 60 , 0.7 ,Iv t V V= = 10 .R k and = Full-wave rectifier ( )

( )( )( )350 1.4

2 2 60 10 10 2M

r

VCf RV

= =

or 20.3 C F= TYU2.3 Using Equation (2.10)

(a) ( )2 42

0.32775

0.327Percent time 100% 5.2%2

r

M

VtV

= = =

= =

(b) ( )2 19.32

0.569119.3

0.569Percent time 100% 18.1%

r

M

VtV

= = =

= =

(c) ( )2 22

0.28748.6

0.287Percent time 100% 9.14%

r

M

VtV

= = =

= =

TYU2.4

PS ZZ L

i

V VI I

R=

For VPS (min) and IL (max), then ( ) 11 9min 0.1 020ZI= = (Minimum Zener current is zero.)

For VPS (max) and IL (min), then ( ) ( )13.6 9max 0 max 230 20Z ZI I m= = A

The characteristic of the minimum Zener current being one-tenth of the maximum value is violated. The proper circuit operation is questionable. TYU2.5


( ) ( ) ( )minmin maxPS ZZ Li

V VI I

R=

so (10 930 max0.0153 L

I= ) which yields ( )max 35.4 LI m= A TYU2.6

4.35

2.7

2.7 6

4.7

4.7

O(V)

I(V)

TYU2.7 As vS goes negative, D turns on and .5 Ov V= + As vS goes positive, D turns off. Output is a square wave oscillating between +5 and +35 volts. TYU2.8

0.6 50

4.4

O(V)

I(V)(a)

0 3

4.4

2.4

O(V)

I(V)(b)

5


TYU2.9 (a) VO = 0.6 V for all V1.

(b) 30

3.6

0.6

VO(V)

VI (V) PSpice Exercises



R 1 K

I 0

10

0

10

I

0.6 V, 20 fV r= =

( )

( )

0

0

For 10 V, 10 0.6

1 9.41 0.02

9.22

If

Rv vR r

v

= = +

= + =

0

9.22

0.6

10

2.2

R

DI 0

D

0

0

00

ln and

ln

I D

DD T D

S

I TS

v v v

viv V iI R

vv v VI R

=

= =

=

2.3

(a) 80sin 13.33sin6s

tv t = =

Peak diode current 13.33(max)di R=


(b) (max) 13.3 VsPIV v= = (c)

( )

[ ] ( )( )

1 1( ) 13.33sin2

13.33 13.33 13.33cos 1 12 2

4.24

oT

o oo o o

o

o

v avg v t dt dtT

x

v avg V

= =

= = = =

(d) 50% 2.4

( ) ( )0 02 max max 2S Sv v V v v V = = + a. For ( ) ( ) ( )0 max 25 V max 25 2 0.7 = 26.4 VSv v= = +

1 1

2 2

160 6.0626.4

N NN N

= =

b. For ( ) ( )0 max 100 V max 101.4 VSv v= = 1 1

2 2

160 1.58101.4

N NN N

= =

From part (a) ( ) ( )2 max 2 26.4 0.7SPIV v V= = or 52.1 PIV V= or, from part (b) ( )2 101.4 0.7PIV = or 202.1 PIV V= 2.5 (a)

( )(max) 12 2(0.7) 13.4 V

13.4rms (rms) 9.48 V2

s

s s

v

v v

= + =

= =

(b)

( )( )( )

2 2 12 2222 F

2 60 0.3 150

M Mr

C r

V VV Cf R f V R

C C

= =

= =

(c)

( )

2, peak 1

2 1212 1150 0.3

, peak 2.33 A

M Md

r

d

V ViR V

i

= +

= + =

2.6 (a)

( )( ) ( ) ( )max 12 0.7 12.7

maxrms rms 8.98

2

S

SS S

v Vv

v v V

= + =

= =

(b) ( )( )( )12

60 150 0.3M M

rr

V VV CfRC fRV

= = = or 4444 C F=


(c) For the half-wave rectifier ( ), max12 121 4 1 4

2 150 2 0.3M M

Dr

V ViR V

= + = + or

, max 4.58 Di A=

2.8 (a)

( ) ( )( )peak 15 2 0.7 16.4 V

16.4rms 11.6 V2

s

s

v

v

= + =

= =

(b) ( )( )( )15 2857 F

2 2 60 125 0.35M

r

VCf RV

= = =

2.9

S

O

O

26

0.60.6

26

25.4

25.4

0

0

2.10

VB 12 V

iD

x1 x2t

R

iD

S

S

( ) 24sinSv t t=

Now ( ) ( )0

1 TD Di avg i t dtT

= We have for 1 2x t x

24sin 12.7D

xiR

=

To find x1 and x2, 124sin 12.7x =

1

2

0.558 0.558 2.584

x radx rad

=

= =

Then


( )

( )

2

1

2 2

11

1 24sin 12.722

1 24 1 12.7 6.482 4.095cos or 2 1.192 2

x

Dx

x xxx

xi avg dxR

x x RR R R R

= =

= = =

2 1 2.584 0.558Fraction of time diode is conducting 100% 100%2 2

x x

= = or Fraction = 32.2%

Power rating

( ) ( )( ) ( )

( ) ( )( )( ) ( )

2

1

2

1

2 22

111

2

2 2

0

2 22

2 2

24sin 12.72

1 24 sin 2 12.7 24 sin 12.72

1 sin 224 2 12.7 24 cos 12.72 2 4

xT

avg rms Dx

x

x

xx x

xxx

R R xP R i i dt dxT R

x x dxR

x x x xR

= = =

= + = +

For 1.19 ,R = then 17.9 avgP W= 2.11 (a)

( ) ( ) ( )15 1500.1

max max 15 0.7 or max 15.7 S o S

R

v v V v V

= =

= + = + =

Then ( ) 15.7 11.1 2S

v rms V= =

Now 1 12 2

120 10.811.1

N NN N

= =

(b) ( )( )( )15

2 2 2 60 150 0.4M M

rr

V VV CfRC fRV

= = = or 2083 C F= (c) ( ) ( )2 max 2 15.7 0.7SPIV v V= = or 30.7 PIV V= 2.12

R1

R2 RLD2

0

i

R1

R2

RL

0

i

For 0iv >

0V = Voltage across 1L iR R v+ =

Voltage Divider 01

12

Li i

L

Rv v vR R

= = +


0

20

2.13 For ( )0, 0iv V> =

R1

RLR2

i0

a.

20

2 1

2

0

||||

|| 2.2 || 6.8 1.66 k1.66 0.43

1.66 2.2

Li

L

L

i i

R Rv vR R R

R R

v v v

=

+ = =

= = +

4.3

0

b. ( ) ( ) ( )00 0maxrms rms 3.04 2v

v v V= = 2.14

( )( )

23.9 0.975 mA4

20 3.9 1.3417 mA12

1.3417 0.975 0.367 mA

0.367 3.9 1.43 mW

L

R

Z Z

T Z Z T

I I

I

I I

P I V P

= =

= =

= == = =

2.15 (a)

( )( )40 12 0.233 A

1200.233 12 2.8 W

ZI

P

= =

= =

(b) IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A

So 120.21 57.1LL

RR

= = (c) ( )( )( )0.1 0.233 12 0.28 WP P= = 2.16

RL

II

IZ IL

Ri

VZ

VI V0

6.3 V, 12 , 4.8I i ZV R V= = =


a. 6.3 4.8 125 mA

12125

I

L I Z Z

I

I I I I

= = =

25 120 mA 40 192L LI R b.

( )( )( )( )0100 4.8 480 mW

120 4.8 576 mWZ Z Z Z

L L L

P I V P

P I V P

= = == = =

2.17 a.

20 10 45.0 mA222

10 26.3 mA380

18.7 mA

I I

L L

Z I L Z

I I

I I

I I I I

= =

= == =

b.

( ) ( )( ) ( )( )

400max 400 mW max 40 mA10

min max 45 4010min 5 mA

2 k

Z Z

L I Z

LL

L

P I

I I I

IR

R

= = = = = = = =

(c) For 175iR = 57.1 mA 26.3 mA 30.8 mAI L ZI I I= = =

( ) ( )max 40 mA min 57.1 40 17.1 mA10 585

17.1

Z L

L L

I I

R R

= = == =

2.18 a. From Eq. (2-31)

( ) [ ] [ ]( )( ) ( )( )

( )( )

500 20 10 50 15 10max

15 0.9 10 0.1 205000 250

4max 1.1875 Amin 0.11875 A

Z

Z

Z

I

II

=

=

=

=

From Eq. (2-29(b)) 20 10 8.081187.5 50i i

R R= = +

b. ( )( )

( ) ( )( )01.1875 10 11.9 W

max 0.5 10 5 WZ Z

L L L

P P

P I V P

= == = =

2.19 (a) As approximation, assume ( )maxZI and ( )minZI are the same as in problem 2.18.

( ) ( ) ( )

( ) ( ) ( )0 0

0 0

max nom max10 (1.1875)(2) 12.375 V

min nom min10 (0.11875)(2) 10.2375 V

Z Z

Z Z

V V I r

V V I r

= +

= + =

= +

= + =


b. 12.375 10.2375% Reg 100% % Reg 21.4%10

= = 2.20

( ) ( )( )

( ) ( ) ( ) ( )( )( )

( ) ( ) ( )

max min% Reg 100%

max min

max min 30.05

6

L L

L

L Z z L Z z

L

Z Z

V VV nom

V nom I r V nom I rV nom

I I

=

+ +=

= =

So ( ) ( )max min 0.1 Z ZI I A = Now ( ) ( )6 6max 0.012 , min 0.006

500 1000L LI A I A= = = =

Now ( )

( ) ( )min

min maxPS Z

iZ L

V VR

I I

=

+

or ( ) ( )15 6280 min 0.020

min 0.012 ZZI A

I

= =+

Then ( )max 0.1 0.02 0.12 ZI A= + = and ( )( ) ( )max

max minPS Z

iZ L

V VR

I I

=

+

or ( ) ( )max 6280 max 41.3

0.12 0.006PS

PS

VV V

= =+

2.21 Using Figure 2.21 a. 20 25% 15 25 VPS PSV V= For ( )min :PSV

( ) ( )( )min max 5 20 25 mAmin 15 10 200

25

I Z L

PS Zi i

I

I I IV V

R RI

= + = + =

= = =

b. For ( )maxPSV ( ) ( )25 10max max 75 mAI Ii

I IR = =

For ( ) ( )min 0 max 75 mAL ZI I= = ( )( )

( ) ( )( )( ) ( )( )

0

0

0

0

10 0.025 5 9.875 Vmax 9.875 0.075 5 10.25min 9.875 0.005 5 9.90

0.35 V

Z Z Z ZV V I rVVV

= = =

= + =

= + =

=

c. ( )0

0

% Reg 100% % Reg 3.5%nomV

V

= = 2.22 From Equation (2.29(a))

( )( ) ( )

min 24 16min max 40 400PS Z

iZ L

V VR

I I

= =

+ + or 18.2iR =

Also 2 2

18.2 2 20.2

M Mr

r

i z

V VV CfRC fRV

R R r

= = + = + =

Then


( ) ( ) ( )24 9901

2 60 1 20.2C C F= = 2.23

( )( )( )

( ) ( )

0

0 0

nom 8 V8 0.1 0.5 7.95 V

max nom 12 8 1.333 A3

Z Z Z Z Z

Z Z

S Zi

i

V V I r VV VV V

IR

= + =

= + =

= = =

For 0.2 A 1.133 AL ZI I= = For 1 A 0.333 AL ZI I= =

( ) ( )( )( )

( ) ( )( )( )

( )

0

0

0

max max7.95 1.133 0.5 8.5165

min min7.95 0.333 0.5 8.11650.4 V

0.4% Reg % Reg 5.0%nom 8

2 23 0.5 3.5

L Z Z Z

L Z Z Z

L

L

M Mr

r

i z

V V I r

V V I r

VV

VV VV CfRC fRV

R R r

= +

= + =

= +

= + =

=

= = =

= == + = + =

Then ( )( ) ( )12 0.0357

2 60 3.5 0.8C C F= =

2.24 (a) For 10 0,Iv both diodes are conducting 0Ov = For 0 3,Iv Zener not in breakdown, so 1 0, 0Oi v= =

( )1

3For 3 20

3 110 1.520 2

II

Io I

vv i mA

vv v

> =

= =

At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA O(V)

I(V)10 3.0 10

43.5

(b) For 0,Iv < both diodes forward biased

10

.10

Ivi = At 110 , 1 Iv V i mA= =

For 13

3, .20I

Ivv i > = At 110 , 0.35 Iv V i mA= =


i1(mA)

I(V)10

1

3 10

0.35

2.25 (a)

V115

1 K

I 0

1 K 2 K

1 01 15 5 V for 5.7, 3 I I

V v v v= = = For 5.7 Iv V>

( )

( )

( )

( )

1 1 10 1

00 0

0 0

0 0

0

0

0.7 15 , 0.71 2 1

15 0.7 0.71 2 1

15.7 0.7 1 1 1 2.51 2 1 1 2 1

18.55 2.5 3.422.5

5.7 5.715 9.42

I

I

I

I I

I

I

v V V V v V

vv v v

v v v

v v v v

v vv v

+

+ = = +

+ =

+ + = + + = + = = += == =

0(V)

I(V)0 5.7

5.7

9.42

15 (b) 0Di = for 0 5.7Iv Then for 5.7 Iv V>

3.422.5

1 1

II

I OD

vvv vi

+

= = or

0.6 3.421

ID

vi = For vI = 15, iD = 5.58 mA


iD(mA)

I(V)5.7

5.58

15 2.26

(a) For D off, 20 (20) 10 3.33 30o

v V = = Then for 3.33 0.7 4.03 3.33 I ov V v V + = = For 4.03, 0.7;I o lv v v> = For 10, 9.3I ov v= =

O(V)

I(V)0 4.03

3.33

9.3

10 (b) For 4.03 , 0I Dv V i =

For ( )1010

4.03, 10 20

ooI D

vvv i

> + =

Which yields 3 0.60520D I

i v=

For 10, 0.895 I Dv i mA= = iD(mA)

I(V)0 4.03

0.895

10 2.27

O

I

12.510.7

10.7 3030

30

For 30 10.730 V, i 0.175 A100 10I

v = = =+

0 i(10) 10.7 12.5 Vv = + =


b. O

12.510.7

0

30

2.28

OI

5

R 6.8 K

0.6 V15sinI

Vv t

=

=

OO

4.4

19.4 2.29 a.

0 =V

3 V0

0.6 =V

2.4

0

b.

0 =V

20

5

0.6 =V

19.4

5

2.30


010

10

4.7

6.7

2.31 One possible example is shown.

Ii

Ri

Vign VRADIODZ

VZ 14 V

D L

RADIO

L will tend to block the transient signals Dz will limit the voltage to +14 V and 0.7 V. Power ratings depends on number of pulses per second and duration of pulse. 2.32

40

0

O(V)

35

50

O(V)

(a)

(b) 2.33

OI

Vx

C

a. For 0 2.7 VxV V = = b. For 0.7 V 2.0 VxV V= = 2.34

10 V

OI

C

2.35


O

I10

0

10

20

VB 0

O

O

I VB

I VB

70

1013

3

20

VB 3 V

13

0

107

3

20

VB 3 V

2.36 For Figure P2.32(a)

I

20

10

0

10

O

2.37 a.

( )1 1 2

0 1 0

10 0.6 0.94 mA 09.5 0.5

9.5 8.93 V

D D D

D

I I I

V I V

= = =+

= =

b.

( )1 1 2

0 1 0

5 0.6 0.44 mA 09.5 0.5

9.5 4.18 V

D D D

D

I I I

V I V

= = =+

= =

c. Same as (a) d.

( ) ( ) ( )( )0 0

1 2 1 2

10 0.5 0.6 9.5 0.964 mA2

9.5 9.16 V

0.482 mA2D D D D

II I

V I V

II I I I

= + + == == = = =

2.38 a. 1 2 00 10D DI I I V= = = = b.


( ) ( )( )

2 1

0 0

10 9.5 0.6 0.5 0.94 mA 0

10 9.5 1.07 VD DI I I I I

V I V

= + + = = == =

c. ( ) ( )

( )2 1

0 0

10 9.5 0.6 0.5 5 0.44 mA 0

10 9.5 5.82 VD DI I I I I

V I V

= + + + = = == =

d.

( ) ( )

( )1 2 1 2

0 0

10 9.5 0.6 0.5 0.964 mA2

0.482 mA2

10 9.5 0.842 V

D D D D

II I

II I I I

V I V

= + + =

= = = == =

2.39 a.

( )

1 2 1 2 3 0

1 2 1 2

3 1 2 3

0 , , , on 4.4 V

10 4.4 0.589 mA9.5

4.4 0.6 7.6 mA0.5

2 7.6 0.589 14.6 mA

D D D D

D D D D

V V D D D V

I I

I I I I

I I I I I

= = =

= =

= = = == + = =

b.

( ) ( )

( ) ( )( )

1 2 1 2 3

1 2 1 2

3

0 0

5 V and on, off

10 9.5 0.6 0.5 5 0.451 mA2

0.226 mA2

0

10 9.5 10 0.451 9.5 5.72 V

D D D D

D

V V D D DII I

II I I I

I

V I V

= =

= + + + =

= = = ==

= = =

c. V1 = 5 V, V2 = 0 D1 off, D2, D3 on 0 4.4 VV =

2 2

1

3 2 3

10 4.4 0.589 mA9.5

4.4 0.6 7.6 mA0.5

0

7.6 0.589 7.01 mA

D D

D

D D D

I I

I I

I

I I I I

= =

= ==

= = =

d. V1 = 5 V, V2 = 2 V D1 off, D2, D3 on 0 4.4 VV =

2 2

1

3 2 3

10 4.4 0.589 mA9.5

4.4 0.6 2 3.6 mA0.5

0

3.6 0.589 3.01 mA

D D

D

D D D

I I

I I

I

I I I I

= =

= ==

= = =

2.40 (a) D1 on, D2 off, D3 on So 2 0DI =

Now ( )

2 1 11 2

10 0.6 0.6 100.6 , 1.25 2 6D D

V V I I mAR R

= = = =+ +


( )( )( )

1 1

3

3 3 1 3

10 0.6 1.25 2 6.9

0.6 52.2

22.2 1.25 0.95

R

D R D D

V V V

I mA

I I I I mA

= =

= =

= = =

(b) D1 on, D2 on, D3 off So 3 0DI =

1 11

10 0.6 4.4 54.4 ,6D

V V IR

= = =

or 1 0.833 DI mA=

( )

( )( )

22 3

2 2 1 2

2 2 3 2

4.4 5 9.4 0.94 10

0.94 0.833 0.107

5 0.94 5 5 0.3

R

D R D D

R

I mAR R

I I I I mA

V I R V V

= = =

+

= = == = =

(c) All diodes are on 1 24.4 , 0.6 V V V V= =

( )

( )

1 11

2 22

3 33

10 0.6 4.40.5 10

4.4 0.60.5 0.5 1 5

0.6 51.5 2.93

D

R

R

I mA R kR

I mA R kR

I mA R kR

= = =

= + = = =

= = =

2.41

For vI small, both diodes off 0.5 0.0909

0.5 5O I Iv v v = = +

When 0.6,I Ov v = D1 turns on. So we have 0.0909 0.6 0.66, 0.06I I I Ov v v v = = = For D1 on

0.65 5 0.5

I O I O Ov v v v v + = which yields 2 0.612I

Ovv =

When 0.6,Ov = D2 turns on. Then 2 0.6

0.6 3.9 12I

Iv v V= =

Now for 3.9Iv > 0.6 0.65 5 0.5 0.5

I O I O O Ov v v v v v + = +

Which yields 2 5.4

; 10 1.15 22I

O I Ovv For v v V+= = =

2.42

0I

10 K

D2

D4D3

D1

10 K

10 K

10 V

10 V


For 0.Iv > when D1 and D4 turn off

( )0

10 0.7 0.465 mA2010 k 4.65 V

= =

= =

I

v I

0

I

10 4.65

4.65

4.65

4.65

10

0 for 4.65 4.65= I Iv v v

2.43 a.

10 V

10 V

V0D2

D1 R2ID1

R1

1 25 k , 10 kR R= =

D1 and D2 on 0 0V = ( )

1

1

0 1010 0.7 1.86 1.05 10

0.86 mA

D

D

I

I

= =

=

b.

( )1 2 1 2 1

0 2 0

10 k , 5 k , off, on 0

10 0.7 101.287

1510 3.57 V

DR R D D I

I

V IR V

= = =

= =

= =

2.44 If both diodes on (a)

( )

( )1

2

1 1 2 1

1

0.7 V, 1.4 V

10 0.71.07 mA

101.4 15

2.72 mA5

2.72 1.071.65 mA

A O

R

R

R D R D

D

V V

I

I

I I I II

= =

= =

= =

+ = = =

(b) D1 off, D2 on ( )

( )( )1 2

2 2

1

1

10 0.7 151.62 mA

5 1015 1.62 10 15 1.2 V

1.2 0.7 1.9 ff ,0

R R

O R O

A

D

I I

V I R VV V D oI

= = =

+= = == + = =

2.45


(a) D1 on, D2 off

110 0.7 0.93 mA

1015 V

D

O

I

V

= =

=

(b) D1 on, D2 off

110 0.7 1.86 mA

515 V

D

O

I

V

= =

=

2.46

( )0 0 0

0 0

0

0

15 0.7 0.710 20 20

15 0.7 0.7 1 1 1 4.010 10 20 10 20 20 20

6.975 V

0.349 mA20D D

V V V

V V

VVI I

+ += +

= + + = =

= =

2.47

10 K10 K

10 K 10 K VDVa

ID

V1 V2Vb

a. V1 = 15 V, V2 = 10 V Diode off

D7.5 V, 5 V V 2.5 V

0a b

D

V V

I

= = = =

b. 1 210 V, 15 VV V= = Diode on

2 1 0.610 10 10 10

15 10 1 1 1 1 1 10.610 10 10 10 10 10 10 10

42.62 6.55 V10

15 6.55 6.55 0.19 mA10 10

0.6

b b a aa b

b b

b b

D D

D

V V V V V V V V

V V

V V

I I

V V

= + + =

+ = + + + +

= =

= ==

2.48 0,Iv = D1 off, D2 on

( )( )10 2.5 0.5 mA

1510 0.5 5 7.5 V for 0 7.5 Vo o I

I

v v v

= =

= =

For 7.5 ,Iv V> Both D1 and D2 on 2.5 10

15 10 5I o o ov v v v

= + or ( )5.5 33.75I ov v= When vo = 10 V, D2 turns off

( )( )10 5.5 33.75 21.25 VIv = = For 21.25 V, 10 VI ov v> =


2.49 a. 01 02 0V V= =

b. 01 024.4 V, 3.8 VV V= =

c. 01 024.4 V, 3.8 VV V= = Logic 1 level degrades as it goes through additional logic gates. 2.50 a. 01 02 5 VV V= =

b. 01 020.6 V, 1.2 VV V= =

c. 01 020.6 V, 1.2 VV V= = Logic 0 signal degrades as it goes through additional logic gates. 2.51 ( ) ( )1 2 3 4 V AND V OR V AND V 2.52

10 1.5 0.2 12 mA 0.01210

8.310 691.70.012

681.7

IR

R

R

= = =

+

+ = =

=

2.53

( )10 1.7

80.75

10 1.7 8 0.75 2.3 V

I

I I

VI

V V

= =

= =

2.54

VR

VPS R

2 K

( )( )1 V, 0.8 mA1 0.8 22.6 V

R

PS

PS

V IVV

= =

= +

=

2.55

( )( )( )3 19 172 2

0.6 10 1 1.6 10 10

3.75 10 cm

PhI e A

A

A

=

=

=


Chapter 3 Exercise Solutions EX3.1

( )1 , 3 , 4.5 4.5 3 1 2

TN GS DS

DS DS GS TN

V V V V V VV V sat V V V

= = =

= > = = =

Transistor biased in the saturation region ( ) ( )2 2 20.8 3 1 0.2 /D n GS TN n nI K V V K K mA V= = =

(a) VGS = 2 V, VDS = 4.5 V Saturation region:

( )( )20.2 2 1 0.2 D DI I mA= = (b) VGS = 3 V, VDS = 1 V Nonsaturation region:

( ) ( )( ) ( )20.2 2 3 1 1 1 0.6 D DI I mA = = EX3.2

( )2 , 3

3 2 1 TP SG

SD SG TP

V V V VV sat V V V

= =

= + = =

(a) 0.5 NonsaturationSDV V= (b) 2 SaturationSDV V= (c) 5 SaturationSDV V= EX3.3

( ) ( )( )

2

1 2

2

160 10 3.636 160 280

0.25 3.636 2 0.669 mA

G DD GS

D

RV V V VR R

I

= = = = + + = =

( )( )( )( )

10 0.669 10 3.31 V0.669 3.31 2.21 mW

DS

D DS

VP I V

= =

= = =

EX3.4

( )( )

2

2

1

1 2 2

1.2 0.4 1.2 2.932 V

1

DQ P SG TP

SG SG

SG DD TTN DD

I K V V

V V

RV V V VR R R

= +

= =

= = +

Note K = k

( )( ) 22

11

1

12.932 200 10 682 K

682 200 283 K682

10 4 5 K1.2D

RR

R RR

R

= =

= =+

= =

EX3.5

(a) ( ) ( )21 2

4010 5 10 5 1 V40 60G

RVR R

= = = + +

( ) ( )25 = = =

SD n GS TN

S

S G GS

VI K V V

RV V V


( ) ( )( )( )

( )( )( )( )

2

2 2

2

5 1 0.5 1 2 1

0.5 3.5 0 7 2.646 V

0.5 2.646 1 1.354 mA10 1.354 3 5.937 V

GS GS GS

GS GS GS

D D

DS

V V V

V V V

I IV

= +

= = =

= == =

(b) ( )( )24 1GS n GS TNV K V V = ( )( )( )( )( )( )( )( )

(1) 1.05 0.5 0.525(2) 0.95 0.5 0.475(3) 1.05 1 1.05 V(4) 0.95 1 0.95 V

n

n

TN

TN

KKVV

= =

= =

= =

= =

(1)-(3) ( )

( )

2

2

4 0.525 2.1 1.1025

0.525 0.1025 3.421 0

0.1025 0.010506 7.1841 2.652 V2 0.525

GS GS GS

GS GS

GS

V V V

V V

V

= +

=

+= =

( )( )( )

20.525 2.652 1.05 1.348 mA10 1.348 3 5.957 V

D

DS

IV

= =

= =

(2)-(4) ( )

( )

2

2

4 0.475 1.9 0.9025

0.475 0.0975 3.5713 0

0.0975 0.00950625 6.785472 0.475

GS GS GS

GS GS

GS

V V V

V V

V

= +

+ =

+=

2.641 VGSV =

( )( )( )

20.475 2.641 0.95 1.359 mA10 1.359 3 5.924 V

D

DS

IV

= =

= =

(1)-(4) ( )( )2

2

4 0 525 1 9 0 9025

0 525 0 0025 3 5262 0GS GS GS

GS GS

V . V . V .

. V . V .

= +

+ =

( )0 0025 0 00000625 7 40502

2 0 5252 5893 V

GS. . .V

..

+=

=

( )( )( )

20 525 2 5893 0 95 1 41110 3 5 7678 V

D

DS D

I . . . .V I .

= =

= =

(2)-(3) ( )2

2

4 0.475 2.1 +1.1025

0.475 0.0025 3.4763 0

0.0025 0.00000625 6.604992(0.475)

GS GS GS

GS GS

GS

V V V

V V

V

=

+ =

+=

2

2.7027(0.475)(2.7027 1.05) 1.2973 mA10 (3) 6.108 V

GS

D

DS D

VI

V I

=

= =

= =

1.297 1.411 mA5.768 6.108 V

DQ

DS

IV

EX3.6


( )

( )( )

2

1 2

10 5

200 10 5 0 714 V350

5 5 1 2

G

S D S D

RVR R

.

V I R . I

=

+

= = = =

So ( )

( )5 1 2 0 714

4 286 1 2= =

=

SG S G D

D

V V V . I .. . I

( )24 286

1 2

=

= +

SGD

D p SG TP

. VI.

I K V V

( )( ) ( ) ( )( )( )

22

2

4 286 1 2 0 25 2 1 1

4 286 0 3 0 6 0 3

= +

= +

SG SG SG

SG SG SG

. V . . V V

. V . V . V .

( ) ( )( )( )

2

2

0 3 0 4 3 986 0

0 4 0 4 4 0 3 3 9862 0 3

SG SG

SG

. V . V .

. . . .V

.

+ =

+=

Must use sign 3 04 VSGV .+ = ( )( )

( ) ( )( )( )

20 25 3 04 1 1 04 mA

10 10 1 04 1 2 4 4 59 V

, Yes

D D

SD D S D SD

SD SD

I . . I .

V I R R . . V .

V V sat

= == + = + =>

EX3.7

( )( ) ( )2

10

10SD DQ S P

SD P SG TP S P

V I R R

V K V V R R

= +

= + +

Set SD SG TPV V V= +

( ) ( ) ( )210 0.25 5.2+ = +SG TP SG TPV V V V ( ) ( )21.3 10 0+ + + =SG TP SG TPV V V V

( ) ( )( )( )1 1 4 1.3 10

2 1.32.415 V

SG TPV V +

+ =

=

( )3.415 V 3.42 VSGV = ( )

( )( )22.415 V 2.42 V

0.25 2.415 1.46 mASD

D

V

I

=

= =

EX3.8

( ) ( )21 2

24010 5 10 5240 270

0.294 V

G

G

RVR R

V

= = + + =

( ) ( )

( )( )( )

2

2

5 5

0.084.706 4 3.9 2.4 1.442

S G GSD n GS TN

S S

GS GS GS

V V VI K V VR R

V V V

+

= = =

= +

20.624 0.4976 3.80744 0GS GSV V =


( )0.4976 0.2476 9.50337

2 0.6242.90 V

GS

GS

V

V

+=

=

( )( )( )

20.08 4 2.90 1.2 0.463 mA2

10 3.9 10 3.57 V

D D

DS D DS

I I

V I V

= = = + =

EX3.9

( )210 and SGD D p SG TPS

VI I K V VR

= = +

( )( )20.12 0.050 0.82.35 V

SG

SG

VV

=

=

10 2.35 63.75 k0.12S S

R R= = ( )

( )( ) ( )( )( )

8 208 20 0.12 63.75 0.12

20 0.12 63.75 836.25 k

0.12

= = +

=

= =

SD D S D

D

D D

V I R RR

R R

( )( )( )( )

( )( )

( )2

(1) 0.05 1.05 0.0525(2) 0.05 0.95 0.0475(3) 0.8 1.05 0.84 V(4) 0.8 0.95 0.76 V

10

P

P

TP

TP

SGD P SG TP

S

KKVV

VI K V VR

= =

= =

= =

= =

= = +

(1)-(3) ( )( ) 2

2

10 63.75 0.0525 1.68 0.7056

3.347 4.623 7.6384 0SG SG SG

SG SG

V V V

V V

= + =

( )4.623 21.372 102.263

2 3.3472.352 V 0.120 mA8.0 V

SG

SG D

SD

V

V IV

+=

=

(2)-(4) ( )( ) 2

2

10 63.75 0.0475 1.52 0.5776

3.028 3.603 8.251 0SG SG SG

SG SG

V V V

V V

= + =

( )3.603 12.9816 99.936

2 3.0282.35 V0.1208.0

SG

SG

D

SD

V

VI

V

+=

=

(1)-(4) ( )( ) 2

2

10 63.75 0.0525 1.52 0.5776

3.347 4.087 8.06685 0SG SG SG

SG SG

V V V

V V

= + =


( )4.087 16.7036 107.999

2 3.3472.279 V0.121 mA7.89 V

SG

SG

D

SD

V

VI

V

+=

=

=

=

(2)-(3) ( )( ) 2

2

10 63.75 0.0475 1.68 0.7056

3.028 4.0873 7.8634 0SG SG SG

SG SG

V V V

V V

= + =

( )4.0873 16.706 95.242

2 3.0282.422 V0.119 mA8.11 V

SG

SG

D

SD

V

VI

V

+=

=

=

=

Summary 0 119 0 121 mA7 89 8 11 V

D

SD

. I .. V .

EX3.10

( )( )( ) ( )

( ) ( )( )

( )( )

2

2 2

2

2

2

,

10 10 0.2 2

10 2 8 82 7 2 0

7 7 4 2 22 2

Use sign: 3.77 V

10 3.77 0.623 mA10

Power 0.623 3.77 Power 2.35 mW

DD GSD D n GS TN

S

GS GS GS TN TN

GS GS GS

GS GS

GS

GS DS

D D

D DS

V VI I K V VR

V V V V V

V V VV V

V

V V

I I

I V

= =

= +

= +

=

+=

+ = =

= == = =

EX3.11 (a) 4 V, IV = Driver in Non Sat.

( ) [ ]( ) ( ) ( )

22

2 22 2

2

2

5 2 4 1 5 1 4 16 8

6 38 16 0

nD I TND O O nL DD O TNL

D D D O O O

D O

K V V V V K V V V

V V V V V V

V V

= = = = +

+ =

( )38 1444 384

2 60.454 V

D

D

V

V

=

=

(b) 2 VIV = Driver: Sat

[ ] [ ][ ] [ ]

2 2

2 25 2 1 5 1nD I TND nL DD O TNL

O

K V V K V V V

V

=

=

5 4 1.76 VO OV V= = EX3.12 If the transistor is biased in the saturation region


( ) ( )( )( )

( )( )

( )

2 2

20.25 2.5 1.56 mA

10 1.56 4 3.75

3.75 2.5

D n GS TN n TN

D D

DS DD D S DS

DS GS TN TN

I K V V K V

I I

V V I R V

V V V V

= =

= == = => =

>

Yes biased in the saturation region ( )( )Power 1.56 3.75 Power 5.85 mWD DSI V= = =

EX3.13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation.

( ) ( )( )( ) ( )

22

2

2

2 5 1 0.25 0.25 4 2.06

DD DL

nD I TND O O nL TNL

nD nD

nL nL

I I

K V V V V K V

K KK K

=

=

= =

(b) ( ) ( ) 22

2 2

0.2 2

50 / and 103 /DL nL TNL nL

nL nD

I K V K

K A V K A V = = = =

EX3.14

( ) ( )( )( )

22

For

1 5 3.25 1 1.75

sat 1.75 1 0.75 V

=

= +

= + = =

= = =

N

DN DP

n GSN TN p scop TP

GSN I

o DSN o

MI I

K V V K V V

V V V

V V V

( ) ( )For : 1.75 V

sat 5 3.25 1 0.75 VSo 5 0.75 4.25 V

=

= = + = =

= =

P I

DD O SD SGP TP

ot ot

M VV V V V V

V V

EX3.15 For 10 ,DR k= VDD = 5 V, and Vo = 1 V

( )( )( ) ( )

( )( )

2

2 2

5 1 0.4 10

2

0.4 2 5 1 1 1 0.057 /

0.4 1 0.4

D

D n GS TN DS DS

D n n

D DS

I mA

I K V V V V

I K K mA V

P I V P mW

= =

= = = =

= = =

EX3.16 a. 1 2 2 25 V, 0, cutoff 0DV V M I= = =

( )( )( ) ( )

2

20 0 0

52

0.05 30 2 5 1 5

OD n I TN O O

D

VI K V V V VR

V V V

= = =


( ) ( )( )( )

20 0

2

0 0

1 1

1.5 13 5 0

13 13 4 1.5 50.40 V

2 1.55 0.40 0.153 mA

30R D R D

V V

V V

I I I I

+ =

= =

= = = =

b. 1 2 5 V= =V V

( ){ }( )( ) ( )

( ) ( )( )( )

2

20 0 0

20 0

2

0 0

1 2

52 2

5 2 0.05 30 2 5 1

3 25 5 0

25 25 4 3 50.205 V

2 35 0.205 0.160 mA

300.080 mA

On I TN O O

D

R R

D D

V K V V V VR

V V V

V V

V V

I I

I I

= =

+ =

= =

= == =

EX3.17

2 3 1 1& watched 0.4 mA = =Q REFM M I I ( )( )

23 3 2

21 1

0.4 0.3 1 2.15 V

0.4 0.6 1 1.82 VGS GS GS

GS GS

V V V

V V

= = == =

EX3.18

( ) ( )

( )

( ) ( )

2

2

2

0.040.1 15 0.62

1.177 0.040.2 1.177 0.6

2

30

0.040.2 25 0.62

1.23 V

SGC

SGC SGB

B

B

SGA

SGA

V

V V VWL

WL

V

V

=

= =

=

=

=

=

EX3.19 (a) ( ) ( )

( ) ( )

2 23 3 4 4

3 4

2 24 4

3 3

2 3 12 1 3 14

= =

= = = =

REF n GS TN n GS TN

GS GS

n n

n n

I K V V K V VV V V V

K KK K

(b) ( )

( )

22 2

2 32 2

2 2

But 2

0.1 2 1 0.1 /

=

= =

= =

Q n GS TN

GS GS

n n

I K V VV V V

K K mA V

(c) ( )( )

2 23 3

2 24 4

0.2 2 1 0.2 /

0.2 3 1 0.05 /

n n

n n

K K mA V

K K mA V

= == =

EX3.20


( )( )

2 2

22 2 2 2

22 2 2 2

55 5 0 16.7 K0.3

0.3 0.2 1.2 2.425 V 2.425 V

S S

D n GS TN

GS GS G GS S

V R

I K V V

V V V V V

= = = =

=

= = = + =

15 2.425 25.8 K

0.1DR = =

( )

( )( )

( )( ) ( )

1 2 1

1 1

21 1 1 1

21 1

1 1 1 1

21

1 2 1

2.425 5 2.575 V2.575 5

24.3 K0.1

0.1 0.5 1.2 1.647 V1.647 2.575 0.928 V

110 5 10 5

S G DSQ

S S

D n GS TN

GS GS

G GS S G

G TN

V V V

R R

I K V V

V VV V V V

RV RR R R

= = =

= ==

= == + = + =

= = +

( )( ) 11

22

2

10.928 200 10 5 491 K

491 200 337 K491

RR

R RR

= =

= =+

EX3.21

12 2

1 1 1

1 1 1

3 31 3

1 2 3

2 1 1

2 2

5 (0.25)(16) 5 1 V( ) 0.25 0.5( 0.8) 1.507 V

1.507 1 0.507 V

(5) 0.507 (5) 50.7 K500

1 2.5 1.5 V1.5 1.507

S D S

DQ n GS TN GS GS

G GS S

G

S S DS

G S GS

V I RI K V V V VV V V

R RV RR R R

V V VV V V

= = =

= = == + = =

= = =

+ + = + = + =

= + = +

2 3 2 32

1 2 3

3.007 V

(5) 3.007 (5)500G

R R R RVR R R

=

+ + = = + +

2 3

2 2

300.7300.7 50.7 250 K

R RR R

+ =

= =

1 1

2 2 2

500 250 50.7 199.3 K1.5 2.5 4 V

5 4 4 K0.25

D S DS

D D

R RV V V

R R

= == + = + =

= =

EX3.22

( ) ( ) ( )( )( )

22

sat 1.2 4.5 sat 3.3 V

1.21 12 1 6.45 mA

4.5

DS GS P DS

GSD DSS D

P

V V V V

VI I IV

= = =

= = =

EX3.23 Assume the transistor is biased in the saturation region.


( )( )( )

( )( )

2

2

1

8 18 1 1.17 V 1.173.5

15 8 0.8 8.68.6 1.17 7.43 V7.43 1.17 3.5 2.33

GSD DSS

P

GSGS S GS

D

DS

DS GS P

VI IV

V V V V

VVV V V

=

= = = =

= =

= =

= > = =

Yes, the transistor is biased in the saturation region. EX3.24

( )( )( )

( )

2

2

2

1 21 2

2

1 2

2.5 mA

1

2.5 6 1 1.42 V4

5 2.5 0.25 54.3756 6 4.375 1.625

5 1625 1.35 k2.5

202 200 k

1.42 4.375 5.795

D

GSD DSS

P

GSGS

S D S

S

DS D

D D

G GS S

G

I

VI IV

V V

V I RV

V V

R R

R RR RV V V

RVR R

=

=

= =

= =

=

= = =

= =

= + = +

= + = =

=

+ ( )

( )2 21

20 10

5.795 20 10 42.05 k 42 k200

157.95 k 158 k

R R

R

= =

=

EX3.25

2

2 2

2

0.

1

6 1 6 11 4 2 16

0.375 4 6 0

S GSS GS D

S S

GSD DSS

P

GS GS GS GS

GS GS

V VV V IR R

VI IV

V V V V

V V

= = =

=

= = +

+ =

( )( )( )

impossible

4 16 4 0.375 62 0.375

8.86 or 1.806 V

1.806 mA

GS

GS GS

GSD

S

V

V V

VIR

=

= =

= =


( )( )( )

( )0

5 1.81 0.4 5 4.2781.81 4.276 2.47 V

sat 4 1.81 2.19

D D D

SD S SD

SD P GS

V I RV V V V

V V V

= = =

= = == = =

( )So satSD SDV V> EX3.26

( )( )1 2

1 21 2

100 k

5 mA, 5 1.2 6 V12 V

in

DQ S DQ S

SDQ D S SDQ

R RR R RR R

I V I RV V V V

= = = +

= = = =

= =

&

6 12 18 V= = ( )

( )

2 2

2

1 2

18 200.4 k

5

1 5 8 14

0.838 V0.838 6 5.162

20

D D

GS GSDQ DSS

P

GS

G GS S

G

R R

V VI IV

VV V V

RVR R

= =

= = =

= + = =

=

+

( ) ( )

( ) ( )( ) ( ) ( )

11

1 22 2

1 2

2 2

15.162 100 20 387 k

100 387 100 387 100

387 100 100 387 135 k

RR

R R R RR R

R R

= =

= = ++

= =

TYU3.1 (a)

( )1.2 , 2

2 1.2 0.8 = =

= = =

TN GS

DS GS TN

V V V VV sat V V V

(i) 0.4 NonsaturationDSV = (ii) 1 SaturationDSV = (iii) 5 SaturationDSV = (b)

( ) ( )1.2 , 2

2 1.2 3.2 = =

= = =

TN GS

DS GS TN

V V V VV sat V V V

(i) 0.4 NonsaturationDSV = (ii) 1 NonsaturationDSV = (iii) 5 SaturationDSV = TYU3.2

(a)

( )( )( )( )( )

( )

148

8

82

23.9 8.85 10

7.67 10 /450 10

100 500 7.67 100.274 /

2 7

=

= = =

= =

n oxn

oxox

ox

n n

W CKL

C F cmt

K K mA V

(b) VTN = 1.2 V, VGS = 2 V


(i)

( ) ( )( ) ( )20.4 Nonsaturation

0.274 2 2 1.2 0.4 0.4 0.132

= = =

DS

D D

V V

I I mA

(ii)

( )( )21 Saturation

0.274 2 1.2 0.175

= = =

DS

D D

V V

I I mA

(iii)

( )( )25 Saturation

0.274 2 1.2 0.175

1.2 , 2

= = == =

DS

D D

TN GS

V V

I I mA

V V V V

(i)

( ) ( )( ) ( )20.4 Nonsaturation

0.274 2 2 1.2 0.4 0.4 0.658 DS

D D

V V

I I mA

= = + =

(ii)

( ) ( )( ) ( )21 Nonsaturation

0.274 2 2 1.2 1 1 1.48 DS

D D

V V

I I mA

= = + =

(iii)

( )( )25 Saturation

0.274 2 1.2 2.81

= = + =

DS

D D

V V

I I mA

TYU3.3 (a) (sat) 2 1 2 0 8 VSD SG TPV V V . .= + = =

(i) Non Sat (ii) Sat (iii) Sat (b) (sat) 2 1 2 3 2 VSDV . .= + =

(i) Non Sat (ii) Non Sat (iii) Sat TYU3.4 (a)

( )( )

14p ox

ox 8

8

8

2

C (3.9)(8.85 10 )C350 10

9.861 10

300 9.861 10(40)(2) 2

0.296 /

P

P

P

WKL Z

K

K mA V

= =

=

= =

(b) (i) 2(0.296) 2(2 1.2)(0.4) (0.4)

0.142 mA

= =

DI

(ii) [ ]2(0 296) 2 1 2 0 189 mAD DI . . I .= = (iii) ID = 0.189 mA (i) ( )( ) ( )2(0 296) 2 2 1 2 0 4 0 4

0 710 mA

= + =

DI . . . .

.

(ii) ( )( ) ( )2(0 296) 2 2 1 2 1 1=1.60 mA

= + DI . .

(iii) ( )( )20 296 2 1 23 03 mA

= +

=

DI . ..

TYU3.5


(a) ( )

( )( )20, 2.5 0.8 1.7

For 2 , 10 Saturation Region

0.1 2.5 0.8 0.289

= = == =

= =

DS

DS DS

D D

V sat VV V V V

I I mA

(b)

( ) ( )

( )( ) ( )( )

( ) ( ) ( )( )( )

1

2

2

2

0.02

1For 2

0.1 2.5 0.8 1 0.02 2 0.300

10

0.1 2.5 0.8 1 0.02 10 0.347

=

= +

=

= + = =

= + =

D n GS TN DS

DS

D D

DS

D D

V

I K V V VV V

I I mA

V V

I I mA

(c) For part (a), 0 or = = For part (b), 10.02 ,V =

( ) ( )( )( )1 12 20.02 0.1 2.5 0.8o n GS TNr K V V = = or 173 or k= TYU3.6

2 2

2 0.70 , 1 TN TNO f SB f

f TNO

V V V

V V V

= + + = =

(a) 0 , 1 SB TNV V V= = (b) ( )1 , 1 0.35 0.7 1 0.7 1.16 SB TN TNV V V V V = = + + = (c) ( )4 , 1 0.35 0.7 4 0.7 1.47 SB TN TNV V V V V = = + + = TYU3.7

( )( )

( )

2

2

2

1 2

22

1

0.4 0.25 0.8 2.06 V

2.06 7.5 68.8 k250

181.2 k

47.5 4 8.75 k

0.4

D n GS TN

GS GS

GS DD

DS DD D D

D D

I K V V

V V

RV VR R

R R

R

V V I R

R R

=

= =

= +

= =

= = =

= =

( ) , YesDS DSV V sat> TYU3.8


( )

( )( )( )

( )

2

2

5 and

5So 0.1

0.1 0.080 1.2 2.32 V5 2.32So 26.8 k

0.14.5 2.32

2.185 5 2.18 28.2 k

0.1, Yes

SD S GS

S

GSS

D n GS TN

GS GS

S S

DS D S D DS S

D

DD D

D

DS DS

VI V V

RVR

I K V V

V V

R R

V V V V V VV

VR RI

V V sat

= =

=

=

= =

= = = = + =

=

= = = >

TYU3.9

( )( )

( )( )( )

2

2

For 2.2 5 2.2 0.56

5

0.56 2.2 1

0.389 /2

389 219.4

40

DS

D D

D n GS TN

n

n oxn

V V

I I mA

I K V V

KCWK mA V

LW WL L

=

= ==

=

= =

= =

TYU3.10 (a) The transition point is

( )( )

1

1

5 1 1 1 0 05 0 01

1 0 05 0 017 236 2 236 V3 236

2 24 1 1 24

+ +=

+ /

+ + /=

+ /

= == = =

DD TNL TND nD nL

ItnD nL

It

Ot It TND Ot

V V V K / KV

K K

. .

. .. V ..

V V V . V . V

(b) We may write ( ) ( )( )2 20.05 2.236 1 76.4 D nD GSD TND DI K V V I A= = =

TYU3.11

( )( )

1

1

5 1 1 12.5

15 2.52.5 2.5 5 1.67 2.78

1.5

+ + /=

+ /

+ + /=

+ /

+ / = + / / = = / =

DD TNL TND nD nL

ItnD nL

nD nL

nD nL

nD nL nD nL nD nL nD nL

V V V K KV

K K

K K

K K

K K K K K K K K

b. For VI = 5, driver in nonsaturated region.


( ) ( )( ) [ ]( ) [ ]

( )

( ) ( )( )( )

22

22

220 0 0

220 0 0

20 0

20 0

2

0 0

2

2

2.78 2 5 1 5 1

22.24 2.78 416 8

3.78 30.24 16 0

30.24 30.24 4 3.78 160.57 V

2 3.78

DD DL

nD I TND O O nL GSL TNL

nDI TND O O DD O TNL

nL

I I

K V V V V K V V

K V V V V V V VK

V V V

V V VV V

V V

V V

=

= = =

=

= +

+ =

= =

TYU3.12 We have 1.2 1.8 DS GS TN TNV V V V V V= < = = Transistor is biased in the nonsaturation region.

( )( )( )( ) ( )

( )

( )( )

2

2

2

5 1.22 and 0.475 8

0.475 2 0 1.8 1.2 1.2

0.475 2.88 0.165

2165 2

9.4335

= = = = =

= = /=

= =

DD DSD n GS TN DS DS D D

S

n

n n

n oxn

V VI K V V V V I I mAR

K

K K mA VCWK

LW WL L

TYU3.13 (a) Transition point for the load transistor Driver is in the saturation region.

( ) ( )( )

( ) ( )

( )

2 2

2 Then 5 2 3 , 3

1

0.08 1 2 1.89 0.01

=

=

= = = == = =

=

= =

DD DL

nD GSD TND nL GSL TNL

DSL GSL TNL TNL DSL DD Ot

Ot Ot

nDIt TNL

nL

It It

I I

K V V K V VV sat V V V V V V V

V V V V

K V VK

V V V

(b) For the driver:

1.89 , 0.89 Ot It TND

It Ot

V V VV V V V

=

= =

TYU3.14

( )( ) ( )( ) ( )

2

2

2

0.050 2 10 0.7 0.35 0.35

0.319 mA10 0.35 30.3 k

0.319

D n GS TN DS DS

D

DD oD D

D

I K V V V V

IV V

R RI

= =

=

= = =

TYU3.15 (a) Transistor biased in the nonsaturation region


( )( )

2

2

2

5 1.5 12

2

12 4 2 5 0.8

4 33.6 12 0 0.374

DSD

D n GS TN DS DS

DS DS

DS DS DS

VIR

I K V V V V

V V

V V V V

= =

= =

+ = =

Then 5 1.5 0.374 261 12

R R = = TYU3.16

a. ( )( ) ( )( ) ( )

22

2 2

5 2

5 0.102 5 1 0.10 0.10 0.248 /

25

= = = =

OD n TN O O

D

n n

VI K V V V VR

K K mA V

b. ( ) ( )

( ) ( ) ( )( )

200 0

20 0 0

20 0

2

0 0

5 2 0.248 2 5 125

5 12.4 8

12.4 100.2 5 0

100.2 100.2 4 12.4 50.0502 V

2 12.4

V V V

V V V

V V

V V

= =

+ =

= =

TYU3.17

( ) ( )( )( )

( )( )

2 25 50 0.15 0.466 V

0.005 10 0.050 V 0.466 0.050 0.516 V

5 0.005 100 4.5 V4.5 0.050 4.45 V

DQ GS TN GS GS

S GG GS S GG

D D

DS D S DS

I K V V V V

V V V V V

V VV V V V

= = == = = + = + == == = =

TYU3.18

( )( )( ) ( )

2

2

2

100 2 0.7 0.2 0.1 0.1

9 A2.5 0.1 267 k

0.009

D GS TN DS DS

D

D D

I K V V V V

I

R R

= =

=

= =



210 0.08 0.333 mA/V2 1.2 2n

nkWK

L

= = = For 0.1 VDSV = Non Sat Bias Region (a) 0 0GS DV I= = (b) ( )( ) ( )21 V 0.333 2 1 0.8 0.1 0.1 0.01 mAGS DV I = = = (c) ( )( ) ( )22 V 0.333 2 2 0.8 0.1 0.1 0.0767 mAGS DV I = = = (d) ( )( ) ( )23 V 0.333 2 3 0.8 0.1 0.1 0.143 mAGS DV I = = = 3.2 All in Sat region

210 0.08 0.333 mA/V1.2 2n

K = = (a) 0DI =

(b) [ ]20.333 1 0.8 0.0133 mADI = = (c) [ ]20.333 2 0.8 0.480 mADI = = (d) [ ]20.333 3 0.8 1.61 mADI = = 3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now

( )( )( )( )

2

2

2

2

0.03 2 1.5 0.25 0.12

0.15 3 1.5 2.25 0.0666

0.39 4 1.5 6.25 0.0624

0.77 5 1.5 12.25 0.0629

n n n

n n n

n n n

n n n

K K K

K K K

K K K

K K K

= = == = =

= = =

= = =

From last three, 2(Avg) 0.0640 mA/VnK = (c) 2

2

(sat) 0.0640(3.5 1.5) (sat) 0.256 mA for 3.5 V(sat) 0.0640(4.5 1.5) (sat) 0.576 mA for 4.5 V

D D GS

D D GS

i i Vi i V

= = == = =

3.4 a.

( ) ( )0

0 2.5 2.5 GS

DS GS TN

VV sat V V V

=

= = =

i.

( ) ( )( ) ( )20.5 V Biased in nonsaturation

1.1 2 0 ( 2.5) 0.5 0.5 2.48 mADS

D D

V

I I

= = =

ii.

( ) ( )( )22.5 V Biased in saturation

1.1 0 2.5 6.88 mADS

D D

V

I I

= = =

iii. VDS = 5 V Same as (ii) 6.88 mADI = b. VGS = 2 V

( ) ( )sat 2 2.5 4.5 VDSV = = i.

( ) 20.5 V Nonsaturation

1.1 2(2 ( 2.5))(0.5) (0.5) 4.68 mADS

D D

V

I I

= = =


ii.

( ) 22.5 V Nonsaturation

1.1 2(2 ( 2.5))(2.5) (2.5) 17.9 mADS

D D

V

I I

= = =

iii.

( ) ( )( )25 V Saturation

1.1 2 2.5 22.3 mADS

D D

V

I I

= = =

3.5

( )0 2 2 DS GS TNV V V V> = = Biased in the saturation region

( )( )

2

2

20.0801.5 0 2 9.375

2

nD GS TN

k WI V VL

W WL L

=

= =

3.6

( )( )( )14 10600 3.9 8.85 10 2.071 10n oxn n ox

ox ox ox

k Ct t t

= = = =

(a) 500 A 241.4 A/Vnk = (b) 250 282.8 A/Vnk = (c) 100 2207 A/Vnk = (d) 50 2414 A/Vnk = (e) 25 2828 A/Vnk = 3.7 a.

( )( )

( )( )

148 2

80 0

8

2

3.9 8.85 107.67 10 F/cm

450 10

21 64650 7.67 102 40.399 /

ox oxox

x x

n oxn

n

Ct tC WK

L

K mA V

= = =

=

= =

b.

( ) ( )( )2 23 V Saturation

0.399 3 0.8 1.93 mAGS DS

D n GS TN D

V V

I K V V I

= = = = =

3.8

( )

( )

2

2

2 20.081.25 2.5 1.2 23.1 m

1.25 2

nD GS TN

kI V V

=

=

3.9

( ) ( )148

08 2

3.9 8.85 10

400 108.63 10 F/cm

oxox

x

Ct

= =

=


( ) ( )( )

8

5

21 600 8.63 102 2.5

1.036 10

n oxn

n

C WKL

W

K W

=

= =

( )( ) ( )

2

23 51.2 10 1.036 10 5 1 7.24 mD n GS TNI K V V

W W =

= =

3.10 Biased in the saturation region in both cases.

( )22p

D SG TP

k WI V VL

= +

(1) ( )20.0400.225 32 TP

W VL

= +

(2) ( )20.0401.40 42 TP

W VL

= +

Take ratio of (2) to (1): 2

2

(4 )1.40 6.2220.225 (3 )

46.222 2.49 2.33 3

TP

TP

TPTP

TP

VVV V VV

+= =

+

+= = =

+

Then ( )20.0400.225 3 2.33 25.12

W WL L

= =

3.11

( )5 V, 0 5 V

0.5 5 0.5 4.5 S G SG

TP SD SG TP

V V VV V V sat V V V

= = == = + = =

a.

( )20 5 V Biased in saturation

2 5 0.5 40.5 mAD SD

D D

V V

I I

= = = =

b.

( )( ) ( )22 V 3 V Nonsaturation

2 2 5 0.5 3 3 36 mAD SD

D D

V V

I I

= = = =

c.

( )( ) ( )24 V 1 V Nonsaturation

2 2 5 0.5 1 1 16 mAD SD

D D

V V

I I

= = = =

d. 5 V 0 0D SD DV V I= = = 3.12 (a) Enhancement-mode (b) From Graph VTP = + 0.5 V

( )( )( )( )

2

2

2

2

2

0.45 2 0.5 2.25 0.20

1.25 3 0.5 6.25 0.20

2.45 4 0.5 12.25 0.20

4.10 5 0.5 20.25 0.202Avg 0.20 mA/V

p p p

p p

p p

p p

p

k K K

k K

k K

k KK

= = == =

= =

= =

=

(c) 22

(sat) 0.20 (3.5 0.5) 1.8 mA(sat) 0.20 (4.5 0.5) 3.2 mA

D

D

ii

= =

= =


3.13 ( )SD SG TPV sat V V= +

(a) ( ) ( )1 2 1 SD SDV sat V sat V= + = (b) ( ) ( )0 2 2 SD SDV sat V sat V= + = (c) ( ) ( )1 2 3 SD SDV sat V sat V= + =

( ) ( ) 222 2p p

D SG TP SD

k kW WI V V V satL L

= + =

(a) ( )( )20.040 6 1 0.12 2D D

I I mA = =

(b) ( )( )20.040 6 2 0.48 2D D

I I mA = =

(c) ( )( )20.040 6 3 1.08 2D D

I I mA = = 3.14

(sat) 3 0 8 2 2 VSD SG TPV V V . .= + = =

215 0 04 0 25 mA/V1 2 2P

.K ..

= =

a) VSD = 0.2 Non Sat ( )( ) ( )20 25 2 3 0 8 0 2 0 2 0 21 mADI . . . . . = = b) VSD = 1.2 V Non Sat ( )( ) ( )20 25 2 3 0 8 1 2 1 2 0 96 mADI . . . . . = = c) VSD = 2.2 V Sat 20 25(3 0 8) 1 21 mADI . . .= = d) VSD = 3.2 V Sat ID = 1.21 mA e) VSD = 4.2 V Sat ID = 1.21 mA 3.15

( )( )( )14 110 0 0

250 3 9 8 85 10 8 629 10p oxp p ox

x x x

. . .k Ct t t

= = = =

(a) 2500 17 3 A/Vox pt k . = = (b) 2250 34 5 A/Vpk . = (c) 2100 86 3 A/Vpk . = (d) 250 173 A/Vpk = (e) 225 345 A/Vpk = 3.16

( )( )( ) ( )( )( ) ( )( )

148 2

80

8 2

8 2

3.9 8.85 106.90 10 F/cm

500 10

675 6.90 10 46.6 A/V

375 6.90 10 25.9 A/V

oxox

x

n n ox

p p ox

Ct

k C

k C

= = =

= = = =

PMOS:


( )

( )

( )

2

2

2

2

0.02590.8 5 0.6 3.192

4 m 12.8 m

0.0259 3.19 41.3 A/V2

pD SG TP

p

p p

p

p p n

k WI V VL

W WL L

L W

K K K

= +

=

circuitos microelectronicos - analisis y diseño - muhammad rashid

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