Question

In a npn transistor, $10$${}^{10}$ electrons enter the emitter in $10$${}^{-6}s$. $4\%$ of the electrons are lost in the base. The current transfer ratio will be

• A. $0.98$
• B. $0.97$
• C. $0.96$
• D. $0.94$

Answer 1

The correct option is C $0.96$

No. of electrons reaching the collector,
$n_c=\frac{96}{100}\times {10}^{10}=0.96\times {10}^{10}$

Emitter current, $I_E=\frac{n_E\times e}{t}$

Collector current, $I_c=\frac{n_C\times e}{t}$

$\therefore$ Current transfer ratio,
$\alpha =\frac{I_c}{I_E}=\frac{n_C}{n_E}=\frac{0.96\times {10}^{10}}{{10}^{10}}=0.96$