CBSE · Class 12 · Physics · Semiconductor ElectronicsThe relation between current amplification factors $\alpha$ and $\beta$ of a transistor is:
Step-by-Step Solution
The current amplification factor in Common Base configuration is $\alpha = I_c / I_e$ and in Common Emitter configuration is $\beta = I_c / I_b$. Since $I_e = I_b + I_c$, algebraic manipulation yields $\beta = \alpha / (1 - \alpha)$ or $\alpha = \beta / (1 + \beta)$.
Detailed Options Breakdown
Option : $\beta = \frac{\alpha}{1-\alpha}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $\alpha = \frac{\beta}{1-\beta}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Semiconductor Electronics.
Option 2: $\beta = 1 - \alpha$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Semiconductor Electronics.
Option 3: $\alpha = 1 + \beta$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Semiconductor Electronics.
💡 Study Guide: This question tests core syllabus concepts from Semiconductor Electronics. For formulas, key summaries, and mock exam reference guides, read the full Semiconductor Electronics Revision Notes.