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NEET 2018
Exam
NEET
Year
2018
Subject
Physics
Chapter
Semiconductors

NEET 2018 · Physics · Semiconductors

In the circuit shown in the figure, the input voltage V _ i is20V,...

In the circuit shown in the figure, the input voltage Viis20V,VBE=0andVCE=0{V}_{i}\text{is}20V, {V}_{BE} = 0\text{and}{V}_{CE} = 0. The values of IB,ICandβ{I}_{B}, {I}_{C}\text{and}\beta are given by

In the circuit shown in the figure, the input voltage V _ i is20V, V _ BE = 0and V _ CE = 0 . The values of I _ B , I _

Options

  1. AIB=20μA,IC=5mA,β=250{I}_{B} = 20\mu A, {I}_{C} = 5mA, \beta = 250
  2. BIB=25μA,IC=5mA,β=200{I}_{B} = 25\mu A, {I}_{C} = 5mA, \beta = 200
  3. CIB=40μA,IC=10mA,β=250{I}_{B} = 40\mu A, {I}_{C} = 10mA, \beta = 250
  4. DIB=40μA,IC=5mA,β=125{I}_{B} = 40\mu A, {I}_{C} = 5mA, \beta = 125

Correct answer

D. IB=40μA,IC=5mA,β=125{I}_{B} = 40\mu A, {I}_{C} = 5mA, \beta = 125

Step-by-step solution

In the circuit shown in the figure, the input voltage V _ i is20V, V _ BE = 0and V _ CE = 0 . The values of I _ B , I _

Vi=IBRB+VBE{\text{V}}_{\text{i}} = {\text{I}}_{\text{B}}{\text{R}}_{\text{B}} + {\text{V}}_{\text{B}\text{E}} (By Kirchoff's Voltage Law)

20=IB×(500×103)+020 = {\text{I}}_{\text{B}} \times \left(500 \times {10}^{3}\right) + 0

IB=20500×103=40μA{\text{I}}_{\text{B}} = \dfrac{20}{500 \times {10}^{3}} = 40\mu A

VCC=ICRC+VCE{\text{V}}_{\text{C}\text{C}} = {\text{I}}_{\text{C}}{\text{R}}_{\text{C}} + {\text{V}}_{\text{C}E}(By Kirchoff's Voltage Law)

20=IC×(4×103)+020 = {\text{I}}_{\text{C}} \times \left(4 \times {10}^{3}\right) + 0

IC=5×10−3=5mA{\text{I}}_{\text{C}} = 5 \times {10}^{- 3} = 5mA

β=ICIB=5×10−340×10−6=125\beta = \dfrac{{\text{I}}_{\text{C}}}{{\text{I}}_{\text{B}}} = \dfrac{5 \times {10}^{- 3}}{40 \times {10}^{- 6}} = 125
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