Question

In an n-p-n transistor, ${10}^{10}$ electrons enter the emitter in ${10}^{-6}s$. If $2\%$ of the electrons are lost in the base, then the current transfer ratio and the current amplification factor are

• A. $0.98,49$
• B. $0.49,49$
• C. $0.98,98$
• D. $0.49,98$

Answer 1

The correct option is A $0.98,49$

Given that, in an n-p-n transistor ${10}^{10}$ electrons enter the emitter in ${10}^{-6}s$, hence the emitter current is
$I_e=\frac{\left(n_e\right)\left(e\right)}{t}$
where, $n_e$ is number of electron enters into the emitter, $e$ is charge on electron and $t$ is time.

$I_e=\frac{\left({10}^{10}\right)(1.6\times {10}^{-19})}{{10}^{-6}}$

$I_e=1.6\times {10}^{-3}=1.6mA$

$2\%$ of the electrons are lost in the base hence, the base current is

$I_b=\frac{2}{100}\times 1.6=0.032mA$

In transistors, $I_e=I_b+I_c$
$I_c=I_e-I_b=1.6-0.032=1.568mA$
Now, the current transfer ratio is
$\frac{I_c}{I_e}=\frac{1.568}{1.6}=0.98$
and the current amplification factor is
$\frac{I_c}{I_b}=\frac{1.568}{0.032}=49$