EE331W11E2solutions

EE331W11E2solutions - EE-331 Devices and Circuits 1 Prof R...

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(radians) t on θ cond 2 π f := t on 0.197 ms = Jaeger approximation: [C] Capacitor size: T 1 2 π f 2V rip V p := t decay πθ cond 2 π f := t decay 1.05 ms = T 0.195 ms = C t decay Rln V DC V DC V rip := C 78.9 µ F = [D] Peak diode current during the first half cycle: t 1 1 2 π f asin V on V p := t 1 0.016 ms = I Cpeak1 2 π f V p C cos 2 π f t 1 () := I Cpeak1 4.95 amp = (t 1 is small enough to be ignored.) Choose a 5 A rated diode to achieve this. Note that diodes with higher current ratings cost more, so the more cost effective design will choose a diode whose current rating is comfortably higher than the maximum calculated current, but no more higher than this, since that would be spending money unnecessarily. Over-rating a part is just as bad as under-rating a part from an economic standpoint. EE-331 Devices and Circuits 1 Prof. R. B. Darling Exam # 2 Solutions Problem 1: Design of a Capacitively Filtered Rectifier.
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This note was uploaded on 09/10/2011 for the course EE 331 taught by Professor Taicheng during the Winter '08 term at University of Washington.

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EE331W11E2solutions - EE-331 Devices and Circuits 1 Prof R...

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