Question

A common emitter amplifier is designed with NPN transistor $(\alpha =0.99)$. The input impedance is $1$ K$\Omega$ and load is $10$ K$\Omega$. The voltage gain will be?

• A. $9.9$
• B. $99$
• C. $990$
• D. $9900$

Answer 1

Answer: (C)

$990$

Given input impedance $=1K\Omega$ and load$=10K\Omega$

Thus resistance gain $=\frac{load}{inputimpedance}$

$R_{gain}=\frac{10\times {10}^3}{1\times {10}^3}=10$

We know voltage gain$V_{gain}=\beta \times R_{gain}$

$\beta =\frac{\alpha }{\alpha -1}$

$\beta =\frac{0.99}{1-0.99}$

$\beta =99$

Thus$V_{gain}=99\times 10=990$