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:

Correct answer: C. 6 ~mA and entering the base As per question, we have, I_ c =24 ~mA , and α=0.8 Now, I _ E = I _ c α =(24)/(0.8)=30 ~mA As, I _ E = I _ B + I _ C Hence, I _ B = I _ E - I _ C =6 ~mA (where, the symbols have their usual meanings)

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Semiconductors

NEET 2022 · Physics · Semiconductors

The collector current in a common base amplifier using n - p -n...

The collector current in a common base amplifier using

n

p

-n transistor is

24 \mathrm{~mA}

. If

80 \%

of the electrons released by the emitter is accepted by the collector, then the base current is numerically:

Options

A $6 \mathrm{~mA}$ and leaving the base
B $3 \mathrm{~mA}$ and leaving the base
C $6 \mathrm{~mA}$ and entering the base
D $3 \mathrm{~mA}$ and entering the base

Correct answer

C. $6 \mathrm{~mA}$ and entering the base

Step-by-step solution

As per question, we have,

\mathrm{I}_{\mathrm{c}}=24 \mathrm{~mA}

, and

\alpha=0.8

Now,

\quad

\mathrm{I}

_{\mathrm{E}}=\frac{

\mathrm{I}

_{\mathrm{c}}}{\alpha}=\frac{24}{0.8}=30 \mathrm{~mA}

As,

\quad

\mathrm{I}

_{\mathrm{E}}=

\mathrm{I}

_{\mathrm{B}}+

\mathrm{I}

_{\mathrm{C}}

Hence,

\quad

\mathrm{I}

_{\mathrm{B}}=

\mathrm{I}

_{\mathrm{E}}-

\mathrm{I}

_{\mathrm{C}}

=6 \mathrm{~mA}

(where, the symbols have their usual meanings)

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