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JEE Main 202127 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
27 August
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2021 · 27 August · Evening Shift · Physics · Semiconductors

For a transistor α and β are given as α = I _ C I _ E and β = I _ C...

For a transistor α\alpha and β\beta are given as α=ICIE\alpha = \dfrac{{I}_{C}}{{I}_{E}} and β=ICIB\beta = \dfrac{{I}_{C}}{{I}_{B}}. Then the correct relation between α\alpha and β\beta will be:

Options

  1. Aα=1−ββ\alpha = \dfrac{1 - \beta}{\beta}
  2. Bαβ=1\alpha\beta = 1
  3. Cβ=α1−α\beta = \dfrac{\alpha}{1 - \alpha}
  4. Dα=β1−β\alpha = \dfrac{\beta}{1 - \beta}

Correct answer

C. β=α1−α\beta = \dfrac{\alpha}{1 - \alpha}

Step-by-step solution

In the case of the transistor the relation between emitter current, base current and, the collector current is, iE=iB+iC{i}_{\text{E}} = {i}_{\text{B}} + {\text{i}}_{\text{C}}, ⇒iEiC=iBiC+1⇒ \dfrac{{i}_{E}}{{i}_{C}} = \dfrac{{i}_{B}}{{i}_{C}} + 1,

⇒1α=1β+1⇒ \dfrac{1}{\alpha} = \dfrac{1}{\beta} + 1

⇒β=α1−α⇒ \beta = \dfrac{\alpha}{1 - \alpha}

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