MHT CET · Physics · Semiconductors
For a common-emitter amplifier, the voltage gain is 40 . Its input and output impedances are \(100 \Omega\) and \(400 \Omega\), respectively. The power gain of the \(\mathrm{CE}\) amplifier will be
- A 450
- B 400
- C 300
- D 500
Answer & Solution
Correct Answer
(B) 400
Step-by-step Solution
Detailed explanation
Power gain \(=(\) voltage gain \() \times(\) current gain \()\)
\(
\begin{aligned}
& \text {Current gain }=\frac{\text {output current }}{\text {Input current }}=\frac{\frac{\mathrm{V}_0}{\mathrm{z}_0}}{\frac{\mathrm{V}_{\mathrm{i}}}{\mathrm{z}_{\mathrm{i}}}} \\
& =\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}} \cdot \frac{\mathrm{z}_{\mathrm{i}}}{\mathrm{z}_0}=\frac{40 \times 100}{400}=10
\end{aligned}
\)
Power gain \(=40 \times 10=400\)
\(
\begin{aligned}
& \text {Current gain }=\frac{\text {output current }}{\text {Input current }}=\frac{\frac{\mathrm{V}_0}{\mathrm{z}_0}}{\frac{\mathrm{V}_{\mathrm{i}}}{\mathrm{z}_{\mathrm{i}}}} \\
& =\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}} \cdot \frac{\mathrm{z}_{\mathrm{i}}}{\mathrm{z}_0}=\frac{40 \times 100}{400}=10
\end{aligned}
\)
Power gain \(=40 \times 10=400\)
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