In a PNP transistor working as a common-base | vedclass

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(C) Given: Current gain $\alpha = 0.96$ and emitter current $I_e = 7.2 \, mA$.In a common-base configuration,the current gain $\alpha$ is defined as $

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$0.29$

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(C) Given: Current gain $\alpha = 0.96$ and emitter current $I_e = 7.2 \, mA$.In a common-base configuration,the current gain $\alpha$ is defined as $\alpha = \frac{I_c}{I_e}$.Therefore,the collector current $I_c = \alpha 	imes I_e = 0.96 	imes 7.2 \, mA = 6.912 \, mA$.According to Kirchhoff's current law for a transistor,$I_e = I_c + I_b$.Thus,the base current $I_b = I_e - I_c$.$I_b = 7.2 \, mA - 6.912 \, mA = 0.288 \, mA$.Rounding to two decimal places,$I_b \approx 0.29 \, mA$.

In a $PNP$ transistor working as a common-base amplifier,current gain is $0.96$ and emitter current is $7.2 \, mA$. The base current is.....$mA$

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