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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2023 · 8 April · Evening Shift · Physics · Semiconductors

For a given transistor amplifier circuit in CE configuration V _ CC...

For a given transistor amplifier circuit in CE configuration VCC=1V,RC=1kΩ,Rb=100kΩ{V}_{CC} = 1V, {R}_{C} = 1kΩ, {R}_{b} = 100kΩ and β=100\beta = 100. Value of base current Ib{I}_{b}is

For a given transistor amplifier circuit in CE configuration V _ CC = 1V, R _ C = 1kΩ, R _ b = 100kΩ and β = 100 .

Options

  1. AIb=1.0μA{I}_{b} = 1 . 0\mu A
  2. BIb=0.1μA{I}_{b} = 0 . 1\mu A
  3. CIb=100μA{I}_{b} = 100\mu A
  4. DIb=10μA{I}_{b} = 10\mu A

Correct answer

D. Ib=10μA{I}_{b} = 10\mu A

Step-by-step solution

Considering the transistor in saturation mode, VCE=0{V}_{CE} = 0

Using Kirchhoff's voltage rule,

−ICRC+VCC=0⇒IC=VCCRC⇒IC=1V103Ω=10−3A- {I}_{C}{R}_{C} + {V}_{CC} = 0 ⇒ {I}_{C} = \dfrac{{V}_{CC}}{{R}_{C}} ⇒ {I}_{C} = \dfrac{1V}{{10}^{3}Ω} = {10}^{- 3}A

So,

β=ICIb=10−3AIb⇒Ib=10−3102=10−5A=10μA\beta = \dfrac{{I}_{C}}{{I}_{b}} = \dfrac{{10}^{- 3}A}{{I}_{b}} ⇒ {I}_{b} = \dfrac{{10}^{- 3}}{{10}^{2}} = {10}^{- 5}A = 10\mu A

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