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

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(C) The emitter current $I_e$ is given by $I_e = \frac{n \cdot e}{t} = \frac{10^{10} 	imes 1.6 	imes 10^{-19}}{10^{-6}} = 1.6 	imes 10^{-3} \ A = 1

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$0.98, 49$

Vedclass · Answer

(C) The emitter current $I_e$ is given by $I_e = \frac{n \cdot e}{t} = \frac{10^{10} 	imes 1.6 	imes 10^{-19}}{10^{-6}} = 1.6 	imes 10^{-3} \ A = 1.6 \ mA$.Since $2\%$ of electrons recombine in the base,the base current $I_b = 0.02 	imes I_e = 0.02 	imes 1.6 \ mA = 0.032 \ mA$.The collector current $I_c = I_e - I_b = 1.6 \ mA - 0.032 \ mA = 1.568 \ mA$.The current gain $\alpha = \frac{I_c}{I_e} = \frac{1.568}{1.6} = 0.98$.The current gain $\beta = \frac{I_c}{I_b} = \frac{1.568}{0.032} = 49$.Thus,the values are $0.98$ and $49$.

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

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