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
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
