The collector current in a common base amplif | vedclass

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Question

(C) Given collector current $I_C = 24\,mA$.The current gain $\alpha$ is the ratio of collector current to emitter current,which is given as $80\% = 0.

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$6\,mA$ and entering the base

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(C) Given collector current $I_C = 24\,mA$.The current gain $\alpha$ is the ratio of collector current to emitter current,which is given as $80\% = 0.8$.We know that $I_C = \alpha I_E$,so $I_E = \frac{I_C}{\alpha}$.Substituting the values: $I_E = \frac{24\,mA}{0.8} = 30\,mA$.Using Kirchhoff's current law for a transistor,$I_E = I_B + I_C$,so $I_B = I_E - I_C$.$I_B = 30\,mA - 24\,mA = 6\,mA$.In an $n-p-n$ transistor,the emitter current enters the transistor,while the collector and base currents leave the transistor. However,the question asks for the base current magnitude and direction relative to the base terminal. Since $I_E = I_B + I_C$,the base current $I_B$ flows into the base region to compensate for the recombination of charge carriers. Thus,the current is $6\,mA$ and entering the base.

The collector current in a common base amplifier using $n-p-n$ transistor is $24\; mA$. If $80\%$ of the electrons released by the emitter are accepted by the collector,then the base current is numerically

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