MCQ
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
  • A
    $6\,mA$ and leaving the base
  • B
    $3\,mA$ and leaving the base
  • $6\,mA$ and entering the base
  • D
    $3\,mA$ and entering the base

Answer

Correct option: C.
$6\,mA$ and entering the base
c
$I _C=24\,mA$

$\text { and } I _{ E }=\frac{ I _{ C }}{\alpha}$

$I _{ E }=\frac{24 mA }{0.8}=30\,mA$

$\therefore I _{ B }= I _{ E }- I _C$

$=6\,mA \text { (into the base) }$

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