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In a \mathrm{NPN} transistor, \mathrm{108} electrons enter the emitter in \mathrm{10^{-8}s}. If \mathrm{1%} electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively

Option: 1

\mathrm{0.8\ and\ 49}


Option: 2

\mathrm{0.9\ and\ 90}


Option: 3

\mathrm{0.7\ and\ 50}


Option: 4

\mathrm{0.99\ and\ 99}


Answers (1)

best_answer

108 electrons enter the emitter in \mathrm{10^{-8}s}
\mathrm{\text { i.e. }, i_E=\frac{108 \times 1.6 \times 10^{-19}}{10^{-8}} A=172.8 \times 10^{-11} A} \\ \mathrm{\therefore 1 \%\ of\ i_E\ is\ lost\ in\ base\ i. e., i_B=\frac{i_E}{100}}
\mathrm{\Rightarrow 99 \%\ i_E\ i.e., \frac{99}{100}\ i_E\ enters\ the\ collector} \\\mathrm{\Rightarrow I_C=0.99\ i_E}
Current amplification factor
\mathrm{\beta=\frac{i_C}{i_B}=\frac{0.99 i_E}{0.01 i_E}=99}

Posted by

Gautam harsolia

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