JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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
- A \(6\times10^2\)
- B \(10^2\)
- C \(60\)
- D \(10^4\)
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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