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Question: Answered & Verified by Expert
In a CE transistor amplifier, the audio signal voltage across the collector resistance of $2 \mathrm{k} \Omega$ is $2 \mathrm{~V}$. If the base resistance is $1 \mathrm{k} \Omega$ and the current amplification of the transistor is 100 , the input signal voltage is
PhysicsSemiconductorsNEETNEET 2012 (Screening)
Options:
  • A $0.1 \mathrm{~V}$
  • B $1.0 \mathrm{~V}$
  • C $1 \mathrm{mV}$
  • D $10 \mathrm{mV}$
Solution:
2905 Upvotes Verified Answer
The correct answer is: $10 \mathrm{mV}$
Current amplification factor
$\beta=\frac{\Delta I_C}{\Delta I_B}$
Collector resistance
$\Delta I_C=\frac{2 \mathrm{~V}}{2 \times 10^3 \Omega}=1 \times 10^{-3} \mathrm{~A}$
Base current
$\Delta I_B=\frac{V_B}{R_B}=\frac{V_B}{1 \times 10^3}=V_B \times 10^{-3}$
Given, $\beta=100$
Now, $100=\frac{10^{-3}}{V_B \times 10^{-3}}$
$V_B=\frac{1}{100} \mathrm{~V}=10 \mathrm{mV}$

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