Question

In common emitter configuration of a transistor, if the change in collector current is \(99.5\%\) of the change in emitter current, then the common emitter current amplification factor is:

• A. \(99.5\)
• B. \(199\)
• C. \(99\)
• D. \(49.5\)

Hint:

Remember the important relation: \[ \beta=\frac{\alpha}{1-\alpha} \] A value of \(\alpha\) very close to unity gives a very large value of \(\beta\).

Answer 1

The Correct Option is B

Concept: The common-base current gain is \[ \alpha=\frac{\Delta I_C}{\Delta I_E} \] and the common-emitter current gain is \[ \beta=\frac{\alpha}{1-\alpha} \] These two transistor current gains are related through the above expression.

Step 1: Determine the value of \(\alpha\). Given that the change in collector current is \(99.5\%\) of the change in emitter current. Therefore, \[ \alpha=\frac{99.5}{100} =0.995 \]

Step 2: Use the relation between \(\alpha\) and \(\beta\). \[ \beta = \frac{\alpha}{1-\alpha} \] Substituting \(\alpha=0.995\), \[ \beta = \frac{0.995}{1-0.995} \] \[ = \frac{0.995}{0.005} \] \[ =199 \] Hence, \[ \boxed{\beta=199} \]