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

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(A) The emitter current $I_e$ is given by $I_e = \frac{n_e 	imes e}{t}$.Since $2 \%$ of electrons recombine in the base,the number of electrons reach

Vedclass · Accepted Answer

$0.98, 49$

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(A) The emitter current $I_e$ is given by $I_e = \frac{n_e 	imes e}{t}$.Since $2 \%$ of electrons recombine in the base,the number of electrons reaching the collector is $98 \%$ of the emitter electrons.Thus,the collector current $I_c = 0.98 \ I_e$.The current gain $\alpha$ is defined as $\alpha = \frac{I_c}{I_e} = \frac{0.98 \ I_e}{I_e} = 0.98$.The current gain $\beta$ is defined as $\beta = \frac{\alpha}{1 - \alpha}$.Substituting the value of $\alpha$: $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.Therefore,the values are $\alpha = 0.98$ and $\beta = 49$.

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter in $10^{-6} \ s$ and $2 \%$ of electrons recombine with holes in the base. The current ratios $\alpha$ and $\beta$ of the transistor are respectively (nearly):

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