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

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0.4 rnA

0.2 rnA

0.29 rnA

0.35 rnA

answer is C.

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Detailed Solution

$α = \frac{i_{c}}{i_{e}} = 0.96 and i_{e} = 7.2 mA i_{c} = 0.96 \times i_{e} = 0.96 \times 7.2 = 6.91 mA ∴ i_{e} = i_{c} + i_{b} \Rightarrow 7.2 = 6.91 + i_{b} \Rightarrow i_{b} = 0.29 mA$

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