SOLUTIONS MANUAL
POWER
ELECTRONICS
CIRCUITS, DEVICES,
AND APPLICATIONS
THIRD EDITION
MUHAMMAD H. RASHID
PEARSON
Prentice
Hall
Upper Saddle River, New Jersey 07458
, 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
/ -^
L
Vm=fjVTIn\D\, 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
, 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
50
1.0 = 7.799 x 0.0258 In , 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
, 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 •
ri m\' i
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-/ = 16.77 A,
v ' r\ p 2
/ 0=7 [ft- = 47.43 A and I =l,Jf (t-tA) = 22.36 A
r2 aV 3 r3 6V 5 4 y
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller a_plus_work. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $9.49. You're not tied to anything after your purchase.
