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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 9 April · Morning Shift · Physics · Semiconductors

An NPN transistor is used in common emitter configuration as an...

An NPNNPN transistor is used in common emitter configuration as an amplifier with 1kΩ1kΩ load resistance. Signal voltage of 10mV10mV is applied across the base-emitter. This produces a 3mA3mA change in the collector current and 15\muA15\muA change in the base current of the amplifier. The input resistance and voltage gain are:

Options

  1. A0.33kΩ,1.50.33kΩ, 1.5
  2. B0.67kΩ,3000.67kΩ, 300
  3. C0.67kΩ,2000.67kΩ, 200
  4. D0.33 k Ω,300Ω, 300

Correct answer

B. 0.67kΩ,3000.67kΩ, 300

Step-by-step solution

loadresistace=outputresistaceRL=1kΩ=1000Ωloadresistace = outputresistace{R}_{L} = 1kΩ = 1000Ω
inputvoltage=ΔVi=10mV=10×10−3Vinputvoltage = \Delta{V}_{i} = 10mV = 10 \times {10}^{- 3}V
inputcurrent=basecurrent=ΔiB=15\muA=15×10−6Ainputcurrent = basecurrent = \Delta{i}_{B} = 15\muA = 15 \times {10}^{- 6}A
outputcurrent=collectorcurrent=ΔiC=3mA=3×10−3Aoutputcurrent = collectorcurrent = \Delta{i}_{C} = 3mA = 3 \times {10}^{- 3}A
ΔiBRi=ΔVi=0.67kΩ\Delta{i}_{B}{R}_{i} = \Delta{V}_{i} = 0.67kΩ
Ri=10×10−315×10−6=0.67kΩ{R}_{i} = \dfrac{10 \times {10}^{- 3}}{15 \times {10}^{- 6}} = 0.67kΩ
Voltage gain =ΔICRLΔVi=3×10−3×100010×10−3=300= \dfrac{\Delta{I}_{C}{R}_{L}}{\Delta{V}_{i}} = \dfrac{3 \times {10}^{- 3} \times 1000}{10 \times {10}^{- 3}} = 300

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