In an NPN transistor,10^{8} electrons enter t | vedclass

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(B) Given: Number of electrons entering the emitter $n_E = 10^{8}$,time $t = 10^{-8} \ s$.Emitter current $I_E = \frac{n_E \cdot e}{t} = \frac{10^{8}

Vedclass · Accepted Answer

$0.99$ and $99$

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(B) Given: Number of electrons entering the emitter $n_E = 10^{8}$,time $t = 10^{-8} \ s$.Emitter current $I_E = \frac{n_E \cdot e}{t} = \frac{10^{8} \cdot e}{10^{-8}} = 10^{16} \cdot e$.Since $1\%$ of electrons are lost in the base,the base current $I_B = 0.01 \cdot I_E = 0.01 \cdot 10^{16} \cdot e = 10^{14} \cdot e$.The collector current $I_C = I_E - I_B = 10^{16} \cdot e - 10^{14} \cdot e = 0.99 	imes 10^{16} \cdot e$.The fraction of current entering the collector is $\alpha = \frac{I_C}{I_E} = \frac{0.99 	imes 10^{16} \cdot e}{10^{16} \cdot e} = 0.99$.The current amplification factor $\beta = \frac{I_C}{I_B} = \frac{0.99 	imes 10^{16} \cdot e}{0.01 	imes 10^{16} \cdot e} = \frac{0.99}{0.01} = 99$.

In an $NPN$ 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 are respectively:

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