In the common-base configuration,a transistor | vedclass

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(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.We know that the relationship between $\alpha$ and $\beta$ (curren

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$38$

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(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.We know that the relationship between $\alpha$ and $\beta$ (current amplification factor in common-emitter configuration) is given by $\beta = \frac{\alpha}{1 - \alpha}$.Substituting the value of $\alpha$:$\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} = 19$.In common-emitter configuration,the current gain $\beta$ is defined as the ratio of change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$:$\beta = \frac{\Delta I_C}{\Delta I_B}$.Given $\Delta I_B = 2 \mu A$,we can calculate $\Delta I_C$:$\Delta I_C = \beta 	imes \Delta I_B = 19 	imes 2 \mu A = 38 \mu A$.

In the common-base configuration,a transistor has a current amplification factor of $0.95$. If the transistor is used in a common-emitter configuration and the base current changes by $2 \mu A$,then the change in the collector current is: (in $\mu A$)

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