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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 12 April · Morning Shift · Physics · Semiconductors

The transfer characteristic curve of a transistor, having input and...

The transfer characteristic curve of a transistor, having input and output resistance 100Ω100Ω and 100kΩ100kΩ respectively, is shown in the figure. The voltage and power gain, are respectively:

The transfer characteristic curve of a transistor, having input and output resistance 100Ω and 100kΩ respectively, is

Options

  1. A5×104,2.5×1065 \times {10}^{4}, 2.5 \times {10}^{6}
  2. B2.5×104,2.5×1062.5 \times {10}^{4}, 2.5 \times {10}^{6}
  3. C5×104,5×1065 \times {10}^{4}, 5 \times {10}^{6}
  4. D5×104,5×1055 \times {10}^{4}, 5 \times {10}^{5}

Correct answer

A. 5×104,2.5×1065 \times {10}^{4}, 2.5 \times {10}^{6}

Step-by-step solution

Current gain (β)=ΔICΔIB=5×10−3100×10−6=50\left(\beta\right) = \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}} = \dfrac{5 \times {10}^{- 3}}{100 \times {10}^{- 6}} = 50
Voltage gain (AV)=β(RoutRin)=50×100×103100=5×104( {\text{A}}_{\text{V}} ) = \beta\left(\dfrac{{R}_{out}}{{R}_{in}}\right) = 50 \times \dfrac{100 \times {10}^{3}}{100} = 5 \times {10}^{4}
Power gain =AV×β=50×5×104=2.5×106= {A}_{V} \times \beta = 50 \times 5 \times {10}^{4} = 2.5 \times {10}^{6}
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