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In \mathrm{NPN} transistor, \mathrm{10^{10}} electrons enter in emitter region in \mathrm{10^{-6}s}. If \mathrm{2%} electrons are lost in base region then collector current and current amplification factor \mathrm{(\beta )} respectively are

Option: 1

\mathrm{1.57\ mA, 49}


Option: 2

\mathrm{1.92\ mA, 70}


Option: 3

\mathrm{2\ mA, 25}


Option: 4

\mathrm{2.25\ mA, 100}


Answers (1)

best_answer

\mathrm{I_e=10^{10} \times 1.6 \times 10^{-19} \times \frac{1}{10^{-6}}=1.6 \mathrm{~mA}\left[\because I=\frac{Q}{t}\right]}
Since \mathrm{2%} electrons are absorbed by base, hence \mathrm{98%} electrons reach the collector,
\mathrm{\begin{aligned} & \text { i.e., } \alpha=0.98 \\ &\mathrm{ \Rightarrow I_c=\omega I_e=0.98 \times 1.6=1.568 \mathrm{~mA} \approx 1.57 \mathrm{~mA}} \end{aligned}}
Also current amplification factor \mathrm{\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{0.02}=49}

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