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 − ⎛ ⎞= ⇒⎜ ⎟⎝ ⎠

ω °

°

2

By symmetry180 24.8 155.2t = − =ω

155.2 24.8% time 100% 36.2%360−

= × =

EX2.2 (a) 112sin 1.4 0Ov = − =θ

or 11.4sin 0.116612

θ = =

which yields 1 6.7θ = °

By symmetry, 2 180 6.7 173.3θ = − = °

Then 173.3 6.7% time 100% 46.3%360−

= × =

(b) 11 4sin 0 354. .θ = =

which yields 1 20 5.θ = °

By symmetry, 2 180 20 5 159 5. .θ = − = °

Then 159.5 20.5% time 100% 38.6%360−

= × =

EX2.3

( )( )( )3

242 2 60 10 0.4

M

r

VC

fRV= = ⇒ or 500 C Fμ=

EX2.4

M M

rr

V VV R

f RC f CV= ⇒ = or

( )( )( )6

7560 50 10 4

R−

=×

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

( )

( )

( ) ( )( ) ( )

( ) ( )( ) ( )

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 56max

10 0.9 5.6 0.1 14LI− − −

=− −

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 ≅ Ω

EX2.6

( )( ),max

13.6 9For 13.6 , 0.2383 15.3 49 4 0.2383 9.9532

PS Z

L

v V I A

v V

−= = =

+= + =

( )( ),min

11 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

vvΔ −

= × = ×Δ −

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

0.1015.3

ZZ

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

VO

R1

V2

D2

V1

D1

I

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

For 2

1 2

1 12.5 ,3 3

OI

I

v Rv V

v R RΔ

> = ⇒Δ +

=

V−

So that R1 = 2R2 EX2.8 For ( )0, max 2 OV vγ = =

Now, so that 8 ,Ov VΔ = ( )min 10 Ov V= − EX2.9

10 4.44.4 , 0.5895 9.5Ov 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 V

I− −

+ =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) ΦphI eη=

so ( ) ( )( )( ) ( )

219

19

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 Ω15 15fr R k− −

+ = = =

Then 207 15 192 ΩR R= − ⇒ =2.15

(a)

( )( )

40 12 0.233 A120

0.233 12 2.8 W

ZI

P

−= =

= =

(b) ( )( )0.233 A, 0.9 0.233 0.21 AR LI I= = =

So 120.21 57.1LL

RR

= ⇒ = Ω

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

( )120sin 2 60 , 0.7 ,Iv t V Vγπ= = 2.5 ΩR 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 ( )( )( )3

119.32 2 60 2.5 10 19.3

M

r

VCf RV x

= = or 20.6 C Fμ=

TYU2.2

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

( )( )( )( )3

50 1.42 2 60 10 10 2

M

r

VC

f RV−

= =×

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

(a) ( )2 42Δ 0.32775

0.327Percent time 100% 5.2%2

r

M

VtV

ω

π

= = =

⎛ ⎞= × =⎜ ⎟⎝ ⎠

(b) ( )2 19.32Δ 0.569

119.3

0.569Percent time 100% 18.1%

r

M

VtV

ω

π

= = =

⎛ ⎞= × =⎜ ⎟⎝ ⎠

(c) ( )2 22Δ 0.287

48.6

0.287Percent time 100% 9.14%

r

M

VtV

ω

π

= = =

⎛ ⎞= × =⎜ ⎟⎝ ⎠

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

( ) ( ) ( )min

min maxPS ZZ L

i

V VI I

R−

= −

so (10 930 max0.0153 LI−

= − ) 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