In an NPN transistor,10^{10} electrons enter | vedclass

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(A) The emitter current $I_e$ is given by the rate of flow of charge: $I_e = \frac{n \cdot e}{t} = \frac{10^{10} 	imes 1.6 	imes 10^{-19} \, C}{10^{

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$1.57 \, mA, 49$

Vedclass · Answer

(A) The emitter current $I_e$ is given by the rate of flow of charge: $I_e = \frac{n \cdot e}{t} = \frac{10^{10} 	imes 1.6 	imes 10^{-19} \, C}{10^{-6} \, s} = 1.6 	imes 10^{-3} \, A = 1.6 \, mA$.Since $2\%$ of electrons are lost in the base,the fraction of electrons reaching the collector is $\alpha = 1 - 0.02 = 0.98$.The collector current is $I_c = \alpha I_e = 0.98 	imes 1.6 \, mA = 1.568 \, mA \approx 1.57 \, mA$.The current amplification factor $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{0.02} = 49$.

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

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