JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2019
- Hard
- MCQ
An \(npn\) transistor operates as a common emitter amplifier, with a power gain of \(60\, dB\). The input circuit resistance Is \(100\,\Omega \) and the output load resistance is \(10 \,k\,\Omega \). the common emitter current gain \(\beta \) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(10^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}_{\mathrm{v}} \times \beta=\mathrm{P}_{\text {gain }}\)
\(60=10 \log _{10}\left(\frac{P}{P_{0}}\right)\)
\(P=10^{6} =\beta^{2} \times \frac{R_{\text {out }}}{R_{\text {in }}}\)
\(=\beta^{2} \times \frac{10^{4}}{100}\)
\(\beta^{2}=10^{4} ; \quad \beta=100\)
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