Question

The correct relation between \(\alpha\) (ratio of \(I_C\) to \(I_E\)) and \(\beta\) (ratio of \(I_C\) to \(I_B\)) of a transistor is:

• A. \(\beta = \frac{\alpha}{1 + \alpha}\)
• B. \(\alpha = \frac{\beta}{1 - \alpha}\)
• C. \(\beta = \frac{1}{1 - \alpha}\)
• D. \(\alpha = \frac{\beta}{1 + \beta}\)

Hint:

\(\alpha\) is always slightly less than 1 (common base gain), while \(\beta\) is much greater than 1 (common emitter gain).

Answer 1

The Correct Option is D

Step 1: Use the fundamental transistor current equation: \(I_E = I_B + I_C\).
Step 2: Divide the entire equation by \(I_C\): \[\frac{I_E}{I_C} = \frac{I_B}{I_C} + \frac{I_C}{I_C} \implies \frac{1}{\alpha} = \frac{1}{\beta} + 1\]
Step 3: Solve for \(\alpha\): \[\frac{1}{\alpha} = \frac{1 + \beta}{\beta} \implies \alpha = \frac{\beta}{1 + \beta}\]