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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 12 January · Evening Shift · Physics · Semiconductors

In the figure, given that V _ BB supply can vary from 0 to 5.0V, V...

In the figure, given that VBB{V}_{BB} supply can vary from 00 to 5.0V,VCC=5V,βdc=200,5.0V, {V}_{CC} = 5V, {\beta}_{dc} = 200, RB=100kΩ,{R}_{B} = 100kΩ, RC=1kΩ{R}_{C} = 1kΩ and VBE=1.0V.{V}_{BE} = 1.0V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

In the figure, given that V _ BB supply can vary from 0 to 5.0V, V _ CC = 5V, β _ dc = 200, R _ B = 100kΩ, R _ C = 1kΩ

Options

  1. A25\muA25\muA and 2.8V2.8V
  2. B20\muA20\muA and 2.8V2.8V
  3. C25\muA25\muA and 3.5V3.5V
  4. D20\muA20\muA and 3.5V3.5V

Correct answer

C. 25\muA25\muA and 3.5V3.5V

Step-by-step solution

When switched on,

VCE=0{V}_{CE} = 0

VCC−RCiC=0{V}_{CC} - {R}_{C}{i}_{C} = 0

ic=VCCRC=51×103=5×10−3A{i}_{c} = \dfrac{{V}_{CC}}{{R}_{C}} = \dfrac{5}{1 \times {10}^{3}} = 5 \times {10}^{- 3}A

ic=βiB{i}_{c} = \beta{i}_{B}

iB=iCβ=5×10−3200=2.5×10−5A=25\muA{i}_{B} = \dfrac{{i}_{C}}{\beta} = \dfrac{5 \times {10}^{- 3}}{200} = 2.5 \times {10}^{- 5}A = 25\muA

using KVL at input side,

VBB−iBRB−VBE=0{V}_{BB} - {i}_{B}{R}_{B} - {V}_{BE} = 0

VBB=VBE+iBRB{V}_{BB} = {V}_{BE} + {i}_{B}{R}_{B}

=1+100×103×25×10−6= 1 + 100 \times {10}^{3} \times 25 \times {10}^{- 6}

=1+2.5= 1 + 2.5

=3.5V= 3.5V
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