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:

Question

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 (C) 6 mA and entering the base (D) 3 mA and entering the base

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e90b4a54521c9f75d22888db8c6ff13a-.png" /> I C =24 mA and I E =( I C /α) I E =(24 mA /0.8)=30 mA ∴ I B = I E - I C =6 mA (into the base)

Q. The collector current in a common base amplifier using $n - p - n$ transistor is $24 m A$ . If $80%$ of the electrons released by the emitter is accepted by the collector, then the base current is numerically:

$6 m A$ and leaving the base

$3 m A$ and leaving the base

$6 m A$ and entering the base

$3 m A$ and entering the base

$I_{C} = 24 m A$
and $I_{E} = \frac{I _{C}}{α}$
$I_{E} = \frac{24 m A}{0.8} = 30 m A$
$∴ I_{B} = I_{E} - I_{C}$
$= 6 m A$ (into the base)

Q. The collector current in a common base amplifier using n−p−n transistor is 24mA. If 80% of the electrons released by the emitter is accepted by the collector, then the base current is numerically:

6mA and leaving the base

3mA and leaving the base

6mA and entering the base

3mA and entering the base