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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Physics
Chapter
Semiconductors

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Semiconductors

In a common emitter configuration with suitable bias, it is given...

In a common emitter configuration with suitable bias, it is given than RL\mathrm{R}_{\mathrm{L}} is the load resistance and RBER_{B E} is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: [ β\beta is current gain, IB,IC,IEI_B, I_C, I_E are respectively base, collector and emitter currents]

Options

  1. AβRLRBE,ΔIEΔIB,β2RLRBE\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_E}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}
  2. Bβ2RLRBE,ΔICΔIB,βRLRBE\beta^2 \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta \frac{R_L}{R_{B E}}
  3. Cβ2RLRBE,ΔICΔIE,β2RLRBE\beta^2 \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_E}, \beta^2 \frac{R_L}{R_{B E}}
  4. DβRLRBE,ΔICΔIB,β2RLRBE\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}

Correct answer

D. βRLRBE,ΔICΔIB,β2RLRBE\beta \frac{R_L}{R_{B E}}, \frac{\Delta I_C}{\Delta I_B}, \beta^2 \frac{R_L}{R_{B E}}

Step-by-step solution

Current gain β=ΔICIB\beta=\frac{\Delta I_C}{I_B} Voltage gain Av=\mathrm{A}_{\mathrm{v}}= Current gain ×\timesResistance gain =βRLRBE=\beta \frac{R_L}{R_{B E}} Power gain Ap=( Current gain )2×A_p=(\text{ Current gain })^2 \times Resistance gain =\beta^2 RLRBE\frac{R_L}{R_{B E}}
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