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Question: Consider an NPN transistor amplifier in common-emitter configuration. The current gain of the transi...

Consider an NPN 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

A

1.1 mA

B

1.01 mA

C

0.01 mA

D

10 mA

Answer

1.01 mA

Explanation

Solution

Current gain β=ΔicΔibΔib=1×103100=105A\beta = \frac{\Delta i_{c}}{\Delta i_{b}} \Rightarrow \Delta i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5}A

= 0.01mA.

By using Δie=Δib+ΔicΔie\Delta i_{e} = \Delta i_{b} + \Delta i_{c} \Rightarrow \Delta i_{e}= 1.01 + 1 = 1.01mA.