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 $10\text{k}\Omega$. The common emitter current gain $\beta$ will be . (Round off to the nearest integer)

Answer 1

Given, Power gain $={10}^6$
Input resistance, $r_{in}=100$ $\Omega$
Output load resistance, $r_{out}=10\text{k}\Omega$

Now, Power gain $={\beta }^2\times \frac{r_{out}}{r_{in}}$
$\Rightarrow {10}^6={\beta }^2\times \frac{10\times {10}^3}{100}$
$\Rightarrow {\beta }^2={10}^4\Rightarrow \beta =100$