Question

In a $CE$ transistor amplifier, the audio signal voltage across the collector resistance of $2k\Omega$ is $2V$ , if the base resistance is $1k\Omega$ and the current amplification of the transistor is $100$ then the input signal voltage is

• A. $10mV$
• B. $1mV$
• C. $0.1V$
• D. $1V$

Answer 1

Answer: (A)

Current amplification factor,
$\beta =\frac{\Delta I_C}{\Delta I_B}$
Collector current,
$\Delta I_C=\frac{2V}{2\times 10^3\Omega }=1\times 10^{-3}A$
Base current,
$\Delta I_B=\frac{V_B}{R_B}=\frac{V_B}{1\times 10^3}=V_B\times 10^{-3}$
Given, $\beta =100$
Now, $100=\frac{10^{-3}}{V_B\times 10^{-3}}$
$\Rightarrow V_B=\frac{1}{100}V=10mV$