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Ch02p

Ch02p
Ch02p

Chapter 2

Exercise Problems

EX2.1

()()30120.687 0.2

max 301242 D R i peak mA v V

??===+= 1112.6sin 24.830t ???=??

???

ω°° 2By symmetry 18024.8155.2t =?=ω 155.224.8% time 100%36.2%360

?=×=

EX2.2

(a) 112sin 1.40O v =?=θ or 1 1.4sin 0.116612

θ== which yields

1 6.7θ=°

By symmetry, 2180 6.7173.3θ=?=° Then 173.3 6.7% time 100%46.3%360

?=

×= (b) 114sin 0354

..θ== which yields

1205.θ=° By symmetry, 21802051595..θ=?=° Then 159.520.5% time 100%38.6%360

?=×=

EX2.3

()()()

3242260100.4M r V C fRV ==? or 500 C F μ=

EX2.4

M M r r V V V R f RC f CV =?= or ()()()

6756050104R ?=× Then

6.25 R k =Ω

EX2.5

1014 , 5.6 ,PS Z V V V V ≤≤=20100 L R ≤≤Ω

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

5.6max 0.28 ,20

5.6min 0.056 100

max max min min (max)min 0.90.1max min 0.90.1max L L PS Z L PS Z L Z PS Z PS PS Z PS I A I A V V I V V I I V V V V V V ====?????????

???=????? or ()()()()()()()()()

14 5.628010 5.656max 100.9 5.60.114L I ???=??

or

()max 591.5 L I m =A ()()()Power(min)max 0.5915 5.6Z Z I V =?=

So Power(min) = 3.31 W

Now ()()()max 14 5.6max min 0.59150.056

PS Z i Z L V V R I I ??==++ or 13 i R ?Ω

EX2.6

()(),max 13.69For 13.6 , 0.2383 15.34

940.23839.9532 PS Z L v V I A v V

?===+=+= ()(),min 119For 11 , 0.1036 15.34

940.10369.4144 PS Z L v V I A v V

?===+=+= 9.95329.4144Source Reg 100%100%13.611

L PS v v Δ?=

×=×Δ? or Source Reg 20.7%=()(),13.69For 0, 0.2383 15.34

940.23839.9532 L Z L noload

I I A v V ?===+=+= For 100 ,L I mA = ()13.6940.1015.3Z Z I I ?+????

=?

which yields

()(), ,, , 0.1591 940.15919.6363 Load Reg 100%9.95329.6363100%9.6363

Z L full load L noload L full load L full load

I A

v A v v v ==+=?=×?=

×V

or

Load Reg 3.29%=

EX2.7

O

2

?I

2I O Then, V 2 = 4.3 V.

D 1 turns on when v 1 = 2.5 V ,

Then, V 1 = 1.8 V.

For 212112.5 ,33

O I I v R v V v R R Δ>=?Δ+=V ? So that R 1 = 2R 2

EX2.8

For ()0, max 2

O V v γ==Now, so that

8 ,O v V Δ=()min 10 O v V =?

EX2.9

10 4.44.4 , 0.5895 9.5

O v V I ?==

=mA Set I = I D 1, then ()()4.40.60.58950.5 3.505 I v V =??= Summary: For

0 3.5

, 4.4I O v V v ≤≤=V D V For turns on and when

23.5 , I v V >9.4 , 10 I O v V v ≥=?O (V)

I (V)

EX2.10

()

1220.6 , 0

0.610 4.27 2.2O D D D V V I I I I I =?=???==?==mA

EX2.11

At the V A node:

21551A D V I ?=+5A V (1) At the V B node:

()2150.751B 0

B D V V I ?++= (2) We see that

0.7,B A V V =? so Equation (2) becomes

2150.751A A

D V V I ??+=0

(2’) Solving for I D 2 from Equation (1) and substituting into Equation (2’), we find

150.72515

10A A A V V V ?????=???? Then V A = 10.71 V and V B = 10.01 V

Solving for the diode currents, we obtain 111510.710.858 5

D D I I m ?=?=A

Also 221510.7110.710.144 515

D D I I m ?=??=A V

EX2.12

Assuming all diodes are conducting, we have 0.7 B V =? and V A = 0 Summing currents, we have

155

A 2D D V I I ?=+ (1) ()2355

B D D V I I ??+= (2) ()110100.71010

B D V I ???== (3) ()()()123From 3, we find

0.93From 1, we obtain 0.07From 2, we have

0.79D D D I mA

I mA

I mA

===

EX2.13

(a) Φph I e η=

so ()()()()()219196.4100.8 1.610

0.52 1.610I ?????×??=××????

or 128 ph I .mA = (b) We have

()()12.8112.8 .O v V ==The diode must be reverse biased so that

12.8 PS V V >

EX2.14

The equivalent circuit is

0.2 V

r ? 1.7 V ? 15 ?

So 5 1.70.215 f I mA r R ??==+ Or 15 1.70.2 3.10.207 ?1515

f r R k ??+=

== Then

20715192 ?R R =??=2.15

(a) ()()40120.233 A 120

0.23312 2.8 W Z I P ?====

(b)

()()0.233 A,0.90.2330.21 A R L I I ===So 120.2157.1L L R R =

?=Ω (c) ()()()0.10.233120.28 W P P =?=

Test Your Understanding Exercises

TYU2.1

()120sin 260, 0.7 ,I v t V V γπ== 2.5 ?R k and =

Full-wave rectifier: Turns ratio 1:2 so that

1200.7119.3 119.310019.3 S I

M r v v V V V V

==?==?= So ()()()

3119.32 260 2.51019.3M r V C f RV x == or 20.6 C F μ=

TYU2.2

()50sin 260, 0.7 ,I v t V V γπ==10 ?.R k and = Full-wave rectifier

()()()()

350 1.42 26010102M r V C f RV ?==× or 20.3 C F μ=

TYU2.3

Using Equation (2.10)

(a)

Δ0.3270.327Percent time 100% 5.2%2t ωπ===??=×=???

?

(b) Δ0.5690.569Percent time 100%18.1%t ωπ===??=×=????

(c)

Δ0.2870.287Percent time 100%9.14%t ωπ===??=×=????

TYU2.4

PS Z Z L i

V V I I R ?=? For V PS (min) and I L (max), then ()119min 0.1020

Z I ?=?= (Minimum Zener current is zero.) For V PS (max) and I L (min), then ()()13.69max 0max 230 20

Z Z I 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 max PS Z Z L i V V I I R ?=? so (10930max 0.0153

L I ?=?) which yields ()max 35.4 L I m =A

TYU2.7

As v S goes negative, D turns on and .5 O v V =

+ As v S goes positive, D turns off. Output is a square wave oscillating between +5 and +35 volts.

TYU2.8

?

O (V)

I (V)(a)?O (V)I (b)

TYU2.9

(a)

(b)

V O (V)

V I (V) PSpice Exercises

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