Solution Manual of Power Electronics 3rd Edition by M H Rashid

 

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 = 600A 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=TjVTInD If IDI is the diode current corresponding to diode voltage VD1, we get / -^ L Vm=fjVTInD, if VD2 is the diode voltage corresponding to the diode current ID2/ we get '/, V -nV Jn' D2 — / T

 

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