Question

In a common emitter configuration of a transistor, the voltage drop across a $500\Omega$ resistor in the collector circuit is $0.5\text{V}$. When the collector supply voltage is $0.96\text{V}$ and $\alpha =0.96$, then the base current is

• A. $\frac{1}{20}\mu \text{A}$
• B. $\frac{1}{24}\text{mA}$
• C. $\frac{1}{5}\mu \text{A}$
• D. $\frac{1}{20}\text{mA}$

Answer 1

The correct option is B $\frac{1}{24}\text{mA}$

Given that$\alpha =0.96$

We know that,

$\beta =\frac{\alpha }{1-\alpha }=\frac{0.96}{1-0.96}=24$

Collector current is
$I_C=\frac{V_C}{R}=\frac{0.5}{500}={10}^{-3}\text{A}$

And $I_B=\frac{I_C}{\beta }=\frac{{10}^{-3}}{24}=\frac{1}{24}\text{mA}$.

Hence, $\left(D\right)$ is the correct answer.