Question

For a transistor, $\alpha_{dc}$ and $\beta_{dc}$ are the current ratios, then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc}\times\beta_{dc}}$ is}

• A. 2.5
• B. 2
• C. 1.5
• D. 1

Hint:

Always remember $\beta$ is usually much larger than $\alpha$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for a simplified value of an expression involving common-base current gain (\( \alpha \)) and common-emitter current gain (\( \beta \)).

Step 2: Key Formula or Approach:
The fundamental relationship between \( \alpha \) and \( \beta \) is:
\[ \beta = \frac{\alpha}{1 - \alpha} \quad \text{or} \quad \alpha = \frac{\beta}{1 + \beta} \]

Step 3: Detailed Explanation:
Starting with \( \beta = \frac{\alpha}{1 - \alpha} \):
\[ \beta(1 - \alpha) = \alpha \] \[ \beta - \alpha\beta = \alpha \] \[ \beta - \alpha = \alpha\beta \] Divide both sides by \( \alpha \times \beta \):
\[ \frac{\beta - \alpha}{\alpha \times \beta} = 1 \]

Step 4: Final Answer:
The value of the expression is 1.