POWER ELETRONICS CIRCUITS, DEVICES, AND APPLICATIONS SOLUTIONS MANUAL [3, THIRD ed.]

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SOLUTIONS MANUAL POWER ELECTRONICS CIRCUITS, DEVICES, AND APPLICATIONS THIRD EDITION

MUHAMMAD H. RASHID

PEARSON Prentice Hall

Upper Saddle River, New Jersey 07458

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CHAPTER 2 POWER SEMICONDUCTOR DIODES AND CIRCUITS

Problem 2-1 ^tm~ 5 us and di/dt = 80 A/MS (a) From Eq. (2-10), QRR = 0.5 (di/dt) trr2 = 0.5 x 80 x 52 x 10"5 = 1000 |JC (b) From Eq. (2-11), — = V2x 1000x80 = 400 A dt Problem 2-2 VT = 25.8 mV, VDi = 1.0 V at IDi = 50 A, and VD2 = 1.5 V at ID2 = 600 A

Taking natural (base e) logarithm on both sides of Eq. (2-3), l-u,

T

—

TV,

J

_L

D

which, after simplification, gives the diode voltage VD as /•

\

vD=TjVTIn\D If IDI is the diode current corresponding to diode voltage VD1, we get /

Vm=fjVTIn\D\,

-^ L

if VD2

is the diode voltage corresponding to the diode current ID2/

we get '/,

V ' D2 -nV — / T Jn\ \

Therefore, the difference in diode voltages can be expressed by

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Y

-V

-nV

Jn\ D1

VD\~rlVTm\D2

L

(D\)

(a) For VD2 = 1.5 V, VDi = 1.0 V, ID2 = 600 A, and ID1 = 50 A, 1.5-1.0 = 77x0.0258x/«| — ], which give ^ = 7.799 (b) For VDI = 1.0 V, IDi = 50 A, and t\ 7.999

1.0 = 7.799 x 0.0258 In

50

, which gives Is = 0.347 A.

Problem 2-3 VDI = VD2 = 2000 V, Ri = 100 kfi (a) From Fig. P2-3, the leakage current are: Isi = 17 mA and Is2 = 25 mA IRI = VDI/RI = 2000/100000 = 20 mA (b) From Eq. (2-12), ISi + IRI = IS2 + IR2 or 17 + 20 = 25 + IR2, or IR2 = 12 mA R2 = 2000/12 mA = 166.67 kQ

Problem 2-4 For VD = 1.5 V, Fig. P2-3 gives IDi = 140 A and ID2 = 50 A

Problem 2-5

IT = 200 A, v = 2.5 Ij = i2i i x = iT/2 = 200/2 = 100 A For Ii = 100 A, Fig. P2-3 yields VDi = 1.1 V and VD2 = 1.95 V v = VDI + Ii Ri Or 2.5 = 1.1 + 100 RI or R x = 14 mQ v = VD2 + I2 R2 Or 2.5 = 1.95 + 100 R2 or R2 = 5.5 mQ

Problem 2-6

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R! = R2 = 10 kft, Vs = 5 kV, Isi = 25 mA, Is2 = 40 mA From Eq. (2-12), ISi + IRI = Is2 + IK*

or Isi + VDI /Ri = IS2 + VD2 /R2 25 x 10"3 + Voi/10000 = 40 x 10"3 + VD2/10000 VDI + VD2 = Vs = 5000 Solving for VDi and VD2 gives VDi = 2575 V and VD2 = 2425 V Problem 2-7

ti= 100 MS, t2 = 300 MS, t3 = 500 MS, f = 250 Hz, fs = 250 Hz, Im = 500 A and Ia = 200 A (a) The average current is Iav = 2Im fti/n - Ia (t3 - t2)f = 7.96 - 10 = - 2.04

A. (b) For sine wave, / . =/ JftJ2 = 55.9 A and for a rectangular negative m\' i

ri

wave,

•

-f 2 )= 44.72 A

The rms current is Irms =V55.922 +44J222 = 71.59 A (c) The peak current varies from 500 A to -200 A. Problem 2-8

ti = 100 MS, t2 = 200 MS, t3 = 400 MS, t4 = 800 MS, f = 250 Hz, Ia = 150 A, Ib = 100 A and Ip = 300 A

(a) The average current is lav = la fts + Ib f(t5 - tO + 2(IP - Ia) f(t2 - ti)/7i = 15 + 5 + 2.387 = 22.387 A.

(b) / , = ! / a-/ v ' r\ p /

0=7

r2

2

= 16.77 A,

[ft- = 47.43 A and I =l,Jf (t-tA) = 22.36 A aV 3 r3 6V 5 4 y

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The rms current is 7rm =^/(16.7722 + 47A322 + 22.3622) = 55.05 A Problem 2-9 R = 22 ft, C = 10 |JF, V0 = 220 V 0 = VR +VC = VR+~\

With initial condition: v c (t=0) = -V0/ the current is

V -tlRC W.-L. The capacitor voltage is vc(t) = -Ri = -V0e

-t/RC

= -220e

-fx!0 6 /220

(b) The energy dissipated is

W = 0.5 C V02 = 0.5 x 10 x 10-6 x 220 x 220 = 0.242 J. Problem 2-10

R = 10 Q, L = 5 mH, Vs = 220 V, Ii = 10 A The switch current is described by

, di „ . Vr =L — + Ri t> dt T7

With initial condition: i(t=0) = Ii, Vs

R

-tRIL

-tRIL

-2000? = 22-12e

A

Problem 2-11

With initial condition: i(t=0) = I0 and vc(t=0) = 0, we get

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I- • L

= 1 cos(a>

o

,0

The capacitor voltage is vc(0 = ^J/*=/j£si \~s

\^

where « = i/Jci o Problem 2-12 Fig. o2-12a: (a) L di/dt or i(t) = Vs t/L (b) di/dt = Vs/L; (d) di/dt (at t= 0) = Vs/L. Fig. p2-12b: (a)

V-V

i ±li

^

(d)

-t/RC

R2C

At t = 0, di/dt = (Vs - V0)/(R2 C)

Fig. p2-12c:

(a)

L

di

~tRIL

+

vs (d)

At t = 0, di/dt = VS/L

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Fig. p2-12d: f \/ di' , 1 7 (a) V T =L-j-+

o

at

C

With initial condition: i(t=0) = 0 and v c (t=0) /— /) = / p ' T sin(0, it is an under-damped case and the solution is of the for z(0 -e

[A cos(a> t) + A^ sin(a> t)]

At t= 0, i(t=0) = 0 and this gives Ai = 0. —n t

0

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— -o) cos(a> dt

r

a

r

-at

_ = 0) .~=-± dt /=o or

A2 = Vs/(cor L) = 220 x 1000/(3893 x 5) = 11.3

The final expression for current i(t) is

i(0 = 11.3xsin(3893/)« (b) The conduction time is corti = TI or ti = Tt/3893 = 807 us (c) The sketch for i(t) is shown. / ^s ___ Problem 2-14

V s = 200 V, Lm = 150 [iH, N! = 10, N2 = 200 and ti = 100 us The turns ratio is a = N2/Ni = 200/10 = 20 (a) From Eq. (2-52) the reverse voltage of diode, V D = 200 x (1 + 20) = 4620 V (b) From Eq. (2-55) the peak value of primary current, I0 = 220 x 100/150 = 146.7 A (c) The peak value of secondary current lo1 = lo/a = 146.7/20 = 7.3 A (d) From Eq. (2-58) the conduction time of diode, t2 = 20 x 100 = 2000 MS. (e) The energy supplied by the source

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W=Pvidt=(l m

__ From Eq. (2-55), W = 0.5 Lm I022 = 0.5 x 150 x 10-6 x 146.72 = 1.614 J

Problem 2-15 (a) ic = id + Im

With initial condition: ict=0) = Im and vc(t=0) = - Vs/

Ic

*'(/) = r 0 . — s m ( a > n + I cos(fo t) ^\L o m o where « =i/VZc o

vc (0 cvy

i f

IT

= — I ic (0 dt =1 .1— sin(a> t) - Fc C* w\ oy 5

1C (/ (0 = vs --

(b) For id (t = ti) = 0 \C —

or

- /„ = 0

cos(a)sin( 0, the foad current is continuous. (iii) By numerical integration of iL2 between the limits cot = 7i/3 to 2:1/3, we get the rms diode current as Ir = 30.2 A. (iv) The rms output current Irms = V3 Ir = V3 x 30.2 = 52.31 A Problem 3-14 RF = 5%, R = 200 Q and f = 60 Hz

(a) Solving for Ce in Eq. (3-62),

C =- 1 e 4x60x200

1

V2x0.05.

= 315.46

(b) From Eq. (3-61), the average load voltage Vdc is 169.7

4x60x200x415.46xlO'

= 169.7 - 11.21 = 158.49V

Problem 3-15 RF = 5%, R = 200 Q, and f = 60 Hz

(a) For a half-wave rectifier, the frequency of output ripple voltage is the same as the supply frequency. Thus, the constant 4 in Eq. (3-62) should be changed to 2. Solving for Ce in Eq. (3-62),

1 2x60x200

1 V2x0.05

= 630.92

(b) From Eq. (3-61), the average load voltage Vdc is

16 www.elsolucionario.net

I

Vd =169.7 rfc

- i = 169.7-22.42= 147.28V 2x60x200x415.46xlO- 6

Problem 3-16 co = 2 7t x 60 = 377 rad/s, Vdc = 48 V, Vs = 120 V, Vm = A/2 x 120 = 169.7 V (a) Voltage ratio x = Vdc/Vm = 48/169.7 = 28.28 % a = sin'1 (x) = 16.43° Solving Eq. (3-70) for 6 gives: 6 = 117.43° Equation (3-105) gives the current ratio Wlpk = 13.425 % Thus, Ipk = WO.13425 = 186.22 A The required value of inductance is U = Vm/(cQ Ipk) = 169.7/(377 x 186.22) = 2.42 mH. Equation (3-106) gives the current ratio IrmS/Ipk = 22.59 % Thus Irms = 0.2259 x Ipk = 0.2259 x 186.22 = 42.07 A (b) Idc = 15 A, U = 6.5 mH, Ipk = Vm/(co U) = 169.77(377 x 6.5 mH) = 69.25 A y = Wlpk = 15/69.25 = 21.66% Using linear interpolation, we get X = X n - (X n +i - X n ) (y n - Y)/(Yn+l - Yn)

= 10 - (15 - 10) (25.5 - 21.66)7(21.5 - 25.5) = 14.8%

Vdc = x Vm = 0.148 X 169.7 = 25.12 V a = an - (ctn+i - an) (yn - y)/(yn+i - yn) = 5.74 - (8.63 - 5.74) (25.5 - 21.66)7(21.5 - 25.5) = 8.51° . 13 = Bn - (Bn+i - Bn) (yn " Y)/(Yn+l ' Yn)

= 139.74 - (131.88 - 139.74) (25.5 - 21.66)7(21.5 - 25.5) = 132.19°. Z = Irms/Ipk = Zn - (Zn+l • Xn) (Yn ' Y)/(Yn+l ' Yn)

= 37.06 - (32.58 - 37.06) (25.5 - 21.66)7(21.5 - 25.5) = 32.76% Thus Irms = 0.3276 x Ipk = 0.3276 x 69.25 = 22.69 A

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(a) f := 60

Vdc := 48

Idc := 25 co := 2-71 -f

Vs := 120

Vm := >/2-Vs

Vdc Vm

a := asin(x)

x :=

co = 376.99

180 — = 16.43 71

100-k = 69.49

Idc

Ipk:= Lcr :=

Ipk = 35.97

Vm

1000 -Lcr= 12.51

mH

oo-Ipk

1

/•a +7i

|_(cos(a) ~ cos^ (())) - x-((j) - a) J dra I sin1cot 27

sin 3

nn

a