Question

In CE NPN transistor ${10}^{10}$ electrons enter the emitter in ${10}^{-6}$s when it is connected to battery. About $5$% electrons recombine with holes in the base. The current gain of the transistor is______
$(e=1.6\times {10}^{-19}C)$

• A. $0.98$
• B. $19$
• C. $49$
• D. $0.95$

Answer 1

Answer: (B)

$19$

Given:

The number of electrons entering the emitter is ${10}^{10}electrons$.

The time taken by the electrons is ${10}^{-6}s$.

Current gain common emitter

$\beta =\frac{I_C}{I_B}$

The emitter current is given as:

$I_E=\frac{q}{t}$

$\Rightarrow \frac{{10}^{10}\times 1.6\times {10}^{-19}}{{10}^{-6}}$

$I_E=1.6mA$

Now, $5\%$ of the electrons recombine in the base region. So the base current will be:

$I_B=5\%\times 1.6mA$

$I_B=0.08mA$

We know that the emitter current is the sum of the base current and collector current.

$I_E=I_B+I_C$

So, $I_C=1.6-0.08=1.52mA$

$⟹\beta =\frac{1.52}{0.08}$
$=19$