ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(B) \(10^2\)

Step-by-step Solution

Detailed explanation

\(\mathrm{A}_{\mathrm{v}} \times \beta=\mathrm{P}_{\text {gain }}\)

\(60=10 \log _{10}\left(\frac{P}{P_{0}}\right)\)

\(P=10^{6} =\beta^{2} \times \frac{R_{\text {out }}}{R_{\text {in }}}\)

\(=\beta^{2} \times \frac{10^{4}}{100}\)

\(\beta^{2}=10^{4} ; \quad \beta=100\)

Same subject
Explore more questions on app