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Physics Question on Semiconductor electronics: materials, devices and simple circuits

In a common emitter configuration, a transistor has β=50\beta = 50 and input resistance 1 kΩ1\, k\Omega. If the peak value of a.c. input is 0.01 V0.01\, V then the peak value of collector current is

A

0.01 μA0.01\, \mu A

B

0.25 μA0.25\, \mu A

C

100 μA100\, \mu A

D

500 μA500\, \mu A

Answer

500 μA500\, \mu A

Explanation

Solution

Given that the current gain of the transistor is β=50\beta=50
Input resistance =1kΩ=1 k \Omega, input voltage =0.01 V=0.01\, V
Hence, 50=ΔiCΔiB50=\frac{\Delta i_{C}}{\Delta i_{B}}
⇒ΔiC=50ΔiB\Rightarrow \Delta i_{C}=50 \Delta i_{B}
Also the change in base current is
ΔiB= input voltage  input resistance =0.011×103=10−5 A\Delta i_{B}=\frac{\text { input voltage }}{\text { input resistance }}=\frac{0.01}{1 \times 10^{3}}=10^{-5}\, A
so, ΔiC=50×10−5=500 μA\Delta i_{C}=50 \times 10^{-5}=500 \,\mu A