In an n-p-n transistor,10^{8} electrons enter | vedclass

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(A) Given: Number of electrons $N = 10^{8}$,time $t = 10^{-8} \,s$.Emitter current $I_{E} = \frac{N 	imes e}{t} = \frac{10^{8} 	imes 1.6 	imes 10^{

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$0.99$ and $99$

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(A) Given: Number of electrons $N = 10^{8}$,time $t = 10^{-8} \,s$.Emitter current $I_{E} = \frac{N 	imes e}{t} = \frac{10^{8} 	imes 1.6 	imes 10^{-19}}{10^{-8}} = 1.6 	imes 10^{-3} \,A$.Since $1\%$ of electrons are lost in the base,the collector current $I_{C}$ is $99\%$ of $I_{E}$.Fraction of current entering the collector = $\frac{I_{C}}{I_{E}} = 0.99$.Base current $I_{B} = 1\%$ of $I_{E} = 0.01 	imes I_{E}$.Current amplification factor $\beta = \frac{I_{C}}{I_{B}} = \frac{0.99 	imes I_{E}}{0.01 	imes I_{E}} = \frac{0.99}{0.01} = 99$.Thus,the values are $0.99$ and $99$.

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

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