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In an $n-p-n$ transistor (in common emitter mode) $10^{10}$ electrons enter the emitter in $10^{-}$ ${ }^6$ s. Only $2 \%$ of the electrons are lost in the base. Calculate the current amplification factor. (Charge on electron is $1.6 \times 10^{-19} \mathrm{C}$ ).
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Verified Answer
The correct answer is:
$49$
$i_e=\frac{q}{t}=\frac{10^{10} \times 1.6 \times 10^{-19}}{10^{-6}}=1.6 \mathrm{~mA}$
Base current,
$i_b=2 \%$ of $i_e=\left(\frac{2}{100}\right) \times 1.6 \mathrm{~mA}=0.032 \mathrm{~mA}$
$\therefore \quad i_c=i_e-i_b=1.568 \mathrm{~mA}$
Therefore current amplification factor,
$\beta=\frac{i_c}{i_b}=\frac{1.568}{0.032}=49$
Base current,
$i_b=2 \%$ of $i_e=\left(\frac{2}{100}\right) \times 1.6 \mathrm{~mA}=0.032 \mathrm{~mA}$
$\therefore \quad i_c=i_e-i_b=1.568 \mathrm{~mA}$
Therefore current amplification factor,
$\beta=\frac{i_c}{i_b}=\frac{1.568}{0.032}=49$
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