Question

What are '\(\alpha\)' and '\(\beta\)' parameters of transistor? What is their relation?

Hint:

Remember the typical values: \(\alpha\) is always close to, but less than, 1. \(\beta\) is always much greater than 1. This can help you check if your derived relation is reasonable. If \(\alpha=0.99\), then \(\beta = \frac{0.99}{1-0.99} = \frac{0.99}{0.01} = 99\).

Answer 1

Step 1: Understanding the Concept:
In a transistor, the flow of a small current in one part of the device controls a much larger current in another part. The parameters \(\alpha\) and \(\beta\) are "current gains" that quantify this control for different circuit configurations.
Step 2: Key Formula or Approach:
The fundamental relationship between the currents in a transistor is:
Emitter Current (\(I_E\)) = Base Current (\(I_B\)) + Collector Current (\(I_C\))
\[ I_E = I_B + I_C \] Definitions:
- Common-Base Current Gain (\(\alpha\)): \(\alpha = \frac{\text{Output Current}}{\text{Input Current}} = \frac{I_C}{I_E}\)
- Common-Emitter Current Gain (\(\beta\)): \(\beta = \frac{\text{Output Current}}{\text{Input Current}} = \frac{I_C}{I_B}\)
Step 3: Detailed Explanation and Derivation of Relation:
Definition of \(\alpha\):
Alpha is the DC current gain in the common-base configuration. It represents the fraction of emitter current that reaches the collector. Since \(I_C\) is always slightly less than \(I_E\) (because a small portion becomes base current), the value of \(\alpha\) is always slightly less than 1 (typically 0.95 to 0.99).
Definition of \(\beta\):
Beta is the DC current gain in the common-emitter configuration. It represents how much the collector current is amplified compared to the base current. Since \(I_B\) is very small compared to \(I_C\), the value of \(\beta\) is large (typically 50 to 300).
Relation between \(\alpha\) and \(\beta\):
We start with the fundamental current equation:
\[ I_E = I_B + I_C \] Divide the entire equation by \(I_C\):
\[ \frac{I_E}{I_C} = \frac{I_B}{I_C} + \frac{I_C}{I_C} \] Recognizing the reciprocals of our definitions (\(\frac{1}{\alpha} = \frac{I_E}{I_C}\) and \(\frac{1}{\beta} = \frac{I_B}{I_C}\)):
\[ \frac{1}{\alpha} = \frac{1}{\beta} + 1 \] Now, we rearrange to solve for \(\beta\):
\[ \frac{1}{\beta} = \frac{1}{\alpha} - 1 \] \[ \frac{1}{\beta} = \frac{1 - \alpha}{\alpha} \] Taking the reciprocal of both sides gives the relation:
\[ \beta = \frac{\alpha}{1-\alpha} \] Step 4: Final Answer:
\(\alpha\) and \(\beta\) are the current gains for common-base and common-emitter transistor configurations, respectively, and are related by the formula \(\beta = \frac{\alpha}{1-\alpha}\).