Question

An NPN transistor operates as a common emitter amplifier with a power gain of ${10}^6$. The input circuit resistance is $100\Omega$ and the output load resistance is $10K\Omega$. The common-emitter current gain ‘$\beta$’ will be ______. (Round off to the nearest integer)

Answer 1

Given data:

Power gain, $P_{gain}={10}^6$

Input resistance, $R_{in}=100\Omega$

Input resistance, $R_{out}=10K\Omega =10000\Omega$

Finding common-emitter current gain ‘$\beta$’.

Formula used:

$P_{gain}={\beta }^2\times \frac{R_{out}}{R_{in}}$

$\Rightarrow$${\beta }^2=P_{gain}\times \frac{R_{in}}{R_{out}}$

$\Rightarrow$${\beta }^2={10}^6\times \frac{100}{10000}$

$\Rightarrow$$\beta =\sqrt{{10}^4}$

$\Rightarrow$$\beta =100$

Hence, The common-emitter current gain will be $100$