An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100$\Omega$ and the output load resistance is 10 K$\Omega$. The common emitter current gain '$\beta$' will be ________. (Round off to the Nearest Integer).
Answer (integer)
100
Solution
Power gain = 10<sup>6</sup><br><br>Input resistance = 100$\Omega$<br><br>Output load resistance = 10K$\Omega$<br><br>Power gain = ${\beta^2} \times {{{r_{out}}} \over {{R_{in}}}}$<br><br>$\Rightarrow$ ${10^6} = {\beta ^2} \times {{10 \times {{10}^3}} \over {100}}$<br><br>$\Rightarrow$ $\beta$<sup>2</sup> = 10<sup>4</sup><br><br>$\Rightarrow$ $\beta$ = 100
About this question
Subject: Physics · Chapter: Electronic Devices · Topic: Transistors
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