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