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Find the power rating of the diode in the given circuit. The breakdown voltage of the diode is 5V.(a) 200 mW(b) 125 mW(c) 250 mW(d) 300 mWThe question was posed to me in class test.This interesting question is from Voltage Regulators topic in chapter Regulators of Analog Circuits

Answer»

The correct option is (C) 250 mW

To explain I would say: SOURCE current IS = 15-5/200 = 0.05A = 50 mA

The power rating of the diode is the maximum power it can dissipate which occurs when the LOAD is DISCONNECTED because then the whole current flows into the diode.

Hence maximum power rating is P = VZxIZ = 5×50 = 250 mW.



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