Transistor

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Question

In an n-p-n transistor, $10^{8}$ electrons enter the emitter in $10^{- 8} \sec .$ If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively

Moderate

Solution

$\begin{array}{l} I_{E} = \frac{10^{8} \times 1 .6 \times 10^{- 19}}{{10^{-}}^{8}} A = 1 .6 mA . \\ I_{B} = \frac{1}{100} I_{E} = 0.01 I_{E} and I_{C} = 99 % of I_{E} = 0 .99 I_{E} . \\ ∴ β = \frac{I_{c}}{I_{B}} = \frac{0 .99 I_{E}}{0 .01 I_{E}} = 99 . \end{array}$

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