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The correct relation between the emitter current Ie and the base current Ib is given by(a) Ib = (1 + α) Ie(b) Ib = (α – 1) Ie(c) Ie = (1 – ß) Ib(d) Ie = (1 + ß) IbI had been asked this question by my college director while I was bunking the class.This is a very interesting question from BJTs Current-Voltage Characteristics in section Bipolar Junction Triodes (BJTs) of Electronic Devices & Circuits

Answer»

Correct CHOICE is (d) Ie = (1 + ß) Ib

Easy EXPLANATION: The correct MATHEMATICAL EXPRESSION are Ie = (1 – ß) Ib and Ib = (1 – α) Ie respectively.



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