In a p-n-p transistor working as a common bas | vedclass

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(C) In a common base amplifier for a $p-n-p$ transistor, the current gain is given by $\alpha = 0.96$. Given emitter current, $I_E = 7.2 \,mA$. We kno

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$0.29$

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(C) In a common base amplifier for a $p-n-p$ transistor, the current gain is given by $\alpha = 0.96$. Given emitter current, $I_E = 7.2 \,mA$. We know the relation for current gain: $\alpha = \frac{I_C}{I_E}$. Substituting the values: $0.96 = \frac{I_C}{7.2}$. Calculating collector current: $I_C = 0.96 	imes 7.2 = 6.912 \,mA$. We also know that the emitter current is the sum of collector current and base current: $I_E = I_C + I_B$. Therefore, base current $I_B = I_E - I_C$. $I_B = 7.2 \,mA - 6.912 \,mA = 0.288 \,mA$. Rounding to two decimal places, we get $I_B \simeq 0.29 \,mA$.

In a $p-n-p$ transistor working as a common base amplifier, when the current gain is $0.96$ and emitter current is $7.2 \,mA$, the base current is (in $\,mA$)

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