Chapter 2 Problem Solutions 2.1

R 1 K

I 0

10

0

10

I

0.6 V, 20 fV r= = Ωγ

( )

( )

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 tω ω= =

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

π

π

π

π π π

= = ×

= − = − − − =⎡ ⎤⎣ ⎦

=

∫ ∫

(d) 50% 2.4

( ) ( )0 02 max max 2S Sv v V v v Vγ γ= − ⇒ = +

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 Vγ= − = −

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 V13.4rms (rms) 9.48 V

2

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 μ

= ⇒ =

= ⇒ =

(c)

( )

2, peak 1

2 1212 1150 0.3

, peak 2.33 A

M Md

r

d

V ViR V

i

π

π

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦

=

2.6 (a)

( )

( ) ( ) ( )

max 12 0.7 12.7 max

rms rms 8.98 2

S

SS S

v Vv

v v V

= + =

= ⇒ =

(b) ( )( )( )

1260 150 0.3

M Mr

r

V VV C

fRC fRV= ⇒ = = or 4444 C Fμ=

(c) For the half-wave rectifier ( ), max

12 121 4 1 42 150 2 0.3

M MD

r

V Vi

R Vπ π

⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or

, max 4.58 Di A=

2.8 (a)

( ) ( )

( )

peak 15 2 0.7 16.4 V16.4rms 11.6 V

2

s

s

v

v

= + =

= =

(b) ( )( )( )

15 2857 F2 2 60 125 0.35

M

r

VCf RV

μ= = =

2.9

S

O

O

26

0.60.6

26

25.4

25.4

0

0

2.10

VB 12 V

iD

x1 x2

t

R

iD

S

S

( ) 24sinSv t tω=

Now ( ) ( )0

1 T

D Di avg i t dtT

= ∫

We have for 1 2x t xω≤ ≤ 24sin 12.7D

xiR

−=

To find x1 and x2, 124sin 12.7x =

1

2

0.558 0.558 2.584

x radx radπ

== − =

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

−⎡ ⎤= = ⎢ ⎥⎣ ⎦

⎛ ⎞ ⎛ ⎞= − − ⋅ = − ⇒ = Ω⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫π

π π

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

x xπ π− −= × = × 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

π

π

π

−⎡ ⎤= ⋅ = = ⎢ ⎥⎣ ⎦

⎡ ⎤= − +⎣ ⎦

⎡ ⎤⎡ ⎤= − − − +⎢ ⎥ ⎦⎣ ⎣ ⎦

∫ ∫

∫

For 1.19 ,R = Ω 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 Vγ

= = Ω

= + = + =

Then ( ) 15.7 11.1 2Sv rms V= =

Now 1 1

2 2

120 10.811.1

N NN N

= ⇒ =

(b) ( )( )( )

152 2 2 60 150 0.4

M Mr

r

V VV C

fRC fRV= ⇒ = = or 2083 C Fμ=

(c) ( ) ( )2 max 2 15.7 0.7SPIV v Vγ= − = − or 30.7 PIV V= 2.12

R1

R2 RL

D2

0

i

R1

R2

RL

0

i

For 0iv >

0Vγ = Voltage across 1L iR R v+ =

Voltage Divider 01

12

Li i

L

Rv v v

R R⎛ ⎞

⇒ = =⎜ ⎟+⎝ ⎠

0

20

2.13 For ( )0, 0iv Vγ> =

R1

RLR2

i0

a.

20

2 1

2

0

||||

|| 2.2 || 6.8 1.66 kΩ1.66 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 0

maxrms 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 A120

0.233 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

= ⇒ = Ω

(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= = Ω =

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≤ ≤ ⇒ ≤ ≤ Ω b.

( )( )( )( )0

100 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 kΩ

Z Z

L I Z

LL

L

P I

I I I

IR

R

= ⇒ = =

⇒ = − = −

⇒ = =

⇒ =

(c) For 175iR = Ω 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

= ⇒ = − =

= ⇒ = Ω

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 iR R−= ⇒ = Ω

+

b. ( )( )

( ) ( )( )0

1.1875 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 Z

Z

I AI

−= ⇒ =+

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

( ) ( )max

max minPS Z

iZ L

V VR

I I−

=+

or ( ) ( )max 6

280 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

= + = + =− −= = ⇒ = Ω

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 rVV

V

= − = − == + == + =

Δ =

c. ( )

0

0

% Reg 100% % Reg 3.5%nomV

VΔ

= × ⇒ =

2.22 From Equation (2.29(a))

( )( ) ( )

min 24 16min max 40 400PS Z

iZ L

V VR

I I− −= =

+ + or 18.2iR = Ω

Also 2 2

18.2 2 20.2

M Mr

r

i z

V VV C

fRC fRVR R r

= ⇒ =

≅ + = + = Ω

Then

( ) ( ) ( )24 9901

2 60 1 20.2C C Fμ= ⇒ =

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

= += + == += + =

Δ =Δ

= = ⇒ =

= ⇒ =

= + = + = Ω

Then ( )( ) ( )

12 0.0357 2 60 3.5 0.8

C 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= =

( )

13For 3

203 110 1.5

20 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

Iv

v 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 IV 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 v

i

⎛ ⎞− +⎜ ⎟− ⎝ ⎠= = or 0.6 3.42

1I

Dv

i−

= 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 30ov 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 Ii 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 30

30

30

For 30 10.730 V, i 0.175 A100 10Iv −= = =

+

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

γ

ω==

O

O

4.4

19.4 2.29 a.

0γ =V

3 V

0

0.6γ =V

2.4

0

b.

0γ =V

20

5

0.6γ =V

19.4

5

2.30

0

10

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 Vγ = ⇒ = b. For 0.7 V 2.0 VxV V= ⇒ =γ 2.34

10 V

OI

C

2.35

O

I10

0

10

20

VB 0

O

O

I VB

I VB

7

0

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 DV V I I mA

R 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 ,6DV V I

R− −= = =

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

− −= = ⇒ = Ω

− −= + = = ⇒ = Ω

− − −= = ⇒ = Ω

2.41

For vI small, both diodes off 0.5 0.09090.5 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 0.612

IO

vv

−=

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

0.6 3.9 12

II

vv 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 22

IO I O

vv 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

−= =

= Ω =

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

V0

D2

D1 R2ID1

R1

1 25 k , 10 kR R= Ω = Ω

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

= Ω = Ω =

− − −= =

= − ⇒ = −

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 V2

Vb

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 mA15

10 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

− −= = =+

+ = = Ω

= Ω

2.53

( )

10 1.78

0.7510 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 17

2 2

0.6 10 1 1.6 10 10

3.75 10 cm

PhI e A

A

A

− −

−

= Φ

× = ×

= ×

η