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Q.

Consider an n-p-n transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current?

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a

1.01 mA

b

1.1 mA

c

0.01 mA

d

10 mA

answer is A.

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Detailed Solution

$\begin{array}{l} Current gain = \frac{Change in collector current}{Change in base current} \\ β = \frac{{Δi}_{C}}{{Δi}_{B}} \\ Also i_{E} = i_{B} + i_{C} \Rightarrow {Δi}_{E} = {Δi}_{B} + {Δi}_{C} \\ β = \frac{{Δi}_{C}}{{Δi}_{E} - {Δi}_{C}} \\ Given, β = 100, {Δi}_{C} = 1 mA \\ ∴ 100 = \frac{1}{{Δi}_{E} - 1} {Δi}_{E} - 1 = \frac{1}{100} = 0 .01 \\ {Δi}_{E} = 1 + 0 .01 = 1 .01 mA \end{array}$

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