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Semiconductor Electronics: Materials, Devices And Simple Circuits

Question
CBSEENPH12039741

For a transistor amplifier in common emitter configuration having load impedance of 1 kΩ (hfe = 50 and hoe = 25) the current gain is

  • -5.2

  • -15.7

  • -24.8

  • -48.78

Solution

D.

-48.78

For a transistor amplifier in common emitter configuration, current again
straight A subscript straight i space equals space minus space fraction numerator straight h subscript straight f subscript straight e end subscript over denominator 1 plus straight h subscript oe straight R subscript straight L end fraction
therefore space straight A subscript straight i space space equals space fraction numerator 50 over denominator 1 plus space 25 space straight x space 10 to the power of negative 6 end exponent space straight x space 1 space straight x space 10 cubed end fraction space equals space minus space 48.78
where  hfe and hoe are hybrid parameters of a transistor