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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 is accepted by the collector, then the base current is numerically:
Detailed Solution
$I_C=0.8I_E \Rightarrow I_E=\frac{24}{0.8}=30$ mA. $I_B=I_E-I_C=30-24=6$ mA, and for an n-p-n transistor the conventional base current enters the base.