Question

For a transistor \(\alpha\) and \(\beta\) are given as \(\alpha = \frac{I_C}{I_E}\) and \(\beta = \frac{I_C}{I_B}\). Then the correct relation between \(\alpha\) and \(\beta\) will be :

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

Hint:

Remembering the relation \( I_E = I_B + I_C \) is key. From there, you can derive any relationship between \(\alpha\) and \(\beta\). A useful way to remember the final result: \(\alpha\) (common-base gain) is always slightly less than 1, while \(\beta\) (common-emitter gain) is large. The formula \( \beta = \alpha/(1-\alpha) \) reflects this, as when \(\alpha\) is close to 1 (e.g., 0.99), \(1-\alpha\) is very small, making \(\beta\) large (e.g., 0.99/0.01 = 99).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the definitions of the current gain in common-base configuration (\(\alpha\)) and common-emitter configuration (\(\beta\)) for a transistor. We need to find the mathematical relationship between them.
Step 2: Key Formula or Approach:
The fundamental relationship between the three transistor currents (emitter current \(I_E\), base current \(I_B\), and collector current \(I_C\)) is:
\[ I_E = I_B + I_C \] We will use this equation along with the definitions of \(\alpha\) and \(\beta\) to derive the relationship.
Step 3: Detailed Explanation:
Start with the fundamental current equation:
\[ I_E = I_B + I_C \] We want to find a relationship between \(\alpha = \frac{I_C}{I_E}\) and \(\beta = \frac{I_C}{I_B}\). Let's try to express \(\beta\) in terms of \(\alpha\).
To do this, we need to eliminate \(I_E\) and \(I_B\) and have only \(I_C\) and the gains.
Divide the fundamental equation by \(I_C\):
\[ \frac{I_E}{I_C} = \frac{I_B}{I_C} + \frac{I_C}{I_C} \] From the definitions, we know:
- \( \frac{I_E}{I_C} = \frac{1}{\alpha} \)
- \( \frac{I_B}{I_C} = \frac{1}{\beta} \)
Substitute these into the divided equation:
\[ \frac{1}{\alpha} = \frac{1}{\beta} + 1 \] Now, we just need to rearrange this equation to match one of the options. Let's solve for \(\beta\).
\[ \frac{1}{\beta} = \frac{1}{\alpha} - 1 \] \[ \frac{1}{\beta} = \frac{1 - \alpha}{\alpha} \] Taking the reciprocal of both sides:
\[ \beta = \frac{\alpha}{1 - \alpha} \] This matches option (C).
Step 4: Final Answer:
The correct relation between \(\alpha\) and \(\beta\) is \( \beta = \frac{\alpha}{1 - \alpha} \).