Question

If \(\alpha\) and \(\beta\) are the current gain in the CB and CE configurations respectively of the transistor circuit, then \(\dfrac{\beta - \alpha}{\alpha\beta}\) is

• A. infinite
• B. 1
• C. 2
• D. 0.5

Hint:

Key relations: \(\alpha = \dfrac{\beta}{1+\beta}\) and \(\beta = \dfrac{\alpha}{1-\alpha}\). Also: \(\dfrac{1}{\alpha} - \dfrac{1}{\beta} = 1\) always holds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Relation between \(\alpha\) (CB gain) and \(\beta\) (CE gain): \(\beta = \dfrac{\alpha}{1-\alpha}\), so \(\alpha = \dfrac{\beta}{1+\beta}\).
Step 2: Detailed Explanation:
\[ \frac{\beta - \alpha}{\alpha\beta} = \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1+\beta}{\beta} - \frac{1}{\beta} = \frac{1+\beta - 1}{\beta} = 1 \]
Step 3: Final Answer:
\(\dfrac{\beta - \alpha}{\alpha\beta} = 1\).