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Question

Physics Question on Transistors

The transistor parameters, namely α\alpha and β\beta of a transistor are related as

A

1α+1β=1\frac{1}{\alpha} +\frac{1}{\beta} = 1

B

1α=β+1β\frac{1}{\alpha} =\beta+ \frac{1}{\beta}

C

1α−1β=0\frac{1}{\alpha} - \frac{1}{\beta} = 0

D

1α−1β=1\frac{1}{\alpha} - \frac{1}{\beta} = 1

Answer

1α−1β=1\frac{1}{\alpha} - \frac{1}{\beta} = 1

Explanation

Solution

For common-emitter transistor β=α1−α\beta=\frac{\alpha}{1-\alpha} ⇒1−αα=1β\Rightarrow \frac{1-\alpha}{\alpha}=\frac{1}{\beta} or 1α−1β=1\frac{1}{\alpha}-\frac{1}{\beta}=1