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  • For a transistor amplifier in common emitter configuration for load impedance of <img alt="\mathrm{1\ k\Omega }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B1%5C%20k%5COmega%20

For a transistor amplifier in common emitter configuration for load impedance of \mathrm{1\ k\Omega } \mathrm{(h_{fe} = 50\ and\ h_{oe} =25 \mu\ AV^{-1})}, the current gain is

Option: 1

-5.2


Option: 2

-15.7


Option: 3

-24.8


Option: 4

-48.78


Answers (1)

For a transistor amplifier in common emitter configuration, current gain
\mathrm{A_i=\frac{h_{f e}}{1+h_{o e} R_L}}
where \mathrm{h_{fe}} and \mathrm{h_{oe}} are hybrid parameters of a transistor.
\mathrm{\therefore A_i=\frac{50}{1+25 \times 10^{-6} \times 1 \times 10^3}=-48.78}

Posted by

Ramraj Saini

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