Question

What is the relationship between current gain $$\alpha$$ in common-base mode and current gain $$\beta$$ in common-emitter mode?

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

Answer 1

The correct option is B

$\beta =\frac{\alpha }{1-\alpha }$

Step 1: Given

The current gain in common-base mode is given by $\alpha$

The current gain in common-base mode is given by $\beta$

Step 2: To find

The relation between $\alpha$ and $\beta$

Step 3: Calculation

We know, the current gain in common-base mode is given by,

$$\begin{gathered} \alpha =\frac{I_c}{I_e} \\ \Rightarrow \frac{1}{\alpha }=\frac{I_e}{I_c}----------------\left(1\right) \end{gathered}$$

where $I_c$ is the collector current and $I_e$ is the emitter current.

The current gain in common-base mode is given by,

$$\begin{gathered} \beta =\frac{I_c}{I_b} \\ \Rightarrow \frac{1}{\beta }=\frac{I_b}{I_c}----------------\left(2\right) \end{gathered}$$

where $I_c$ is the collector current and $I_b$ is the emitter current.

We also know that in a transistor, sum of collector current and base current is equal to the emitter current.

$\Rightarrow I_c+I_b=I_e$

Dividing the equation by $I_c$

$$\begin{gathered} \Rightarrow \frac{I_c}{I_c}+\frac{I_b}{I_c}=\frac{I_e}{I_c} \\ \Rightarrow 1+\frac{1}{\beta }=\frac{1}{\alpha }\left(From\left(1\right)and\left(2\right)\right) \\ \Rightarrow \frac{1}{\beta }=\frac{1}{\alpha }-1 \\ \Rightarrow \frac{1}{\beta }=\frac{1-\alpha }{\alpha } \\ \Rightarrow \beta =\frac{\alpha }{1-\alpha } \end{gathered}$$

So, the relationship between current gain $$\alpha$$ in common-base mode and current gain $$\beta$$ in common-emitter mode is $\beta =\frac{\alpha }{1-\alpha }$.

Therefore, option B is correct.