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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 10 April · Morning Shift · Physics · Semiconductors

An NPN transistor operates as a common emitter amplifier, with a...

An NPN\text{NPN} transistor operates as a common emitter amplifier, with a power gain of 60dB60dB . The input circuit resistance is 100Ω100Ω and the output load resistance is 10kΩ10kΩ . The common emitter current gain β\beta is:

Options

  1. A6×1026 \times {10}^{2}
  2. B102{10}^{2}
  3. C104{10}^{4}
  4. D6060

Correct answer

B. 102{10}^{2}

Step-by-step solution

Av×β=Pgain{A}_{v} \times \beta = {P}_{gain}

60=10log⁡10(PP0)60 = 10\log_{10}\left(\dfrac{P}{{P}_{0}}\right)
P=106=β2×RoutRin=β2×104100P = {10}^{6} = {\beta}^{2} \times \dfrac{{R}_{out}}{{R}_{in}} = {\beta}^{2} \times \dfrac{{10}^{4}}{100}
β2=104{\beta}^{2} = {10}^{4}
β=100=102\beta = 100 = {10}^{2}

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