The relation between `alpha` and `beta` of a transistor is

Options

A. `1/alpha+1/beta=1`
B. `1/alpha=beta+1/beta`
C. `1/alpha-1/beta=0`
D. `1/alpha-1/beta=1`

Answer

To find the relation between alpha (α) and beta (β) of a transistor, we can follow these steps:

### Step-by-Step Solution:

1. **Define Alpha (α) and Beta (β)**:
- Alpha (α) is defined as the ratio of the collector current (Ic) to the emitter current (Ie):
\[ \alpha = \frac{I_c}{I_e} \]
- Beta (β) is defined as the ratio of the collector current (Ic) to the base current (Ib):
\[ \beta = \frac{I_c}{I_b} \]

2. **Express Emitter Current (Ie)**:
- The emitter current can be expressed in terms of collector current and base current:
\[ I_e = I_c + I_b \]

3. **Substitute for Ie in terms of Ic and Ib**:
- From the definition of α, we can substitute for Ie:
\[ \alpha = \frac{I_c}{I_c + I_b} \]

4. **Rearranging the Equation**:
- We can rearrange this equation to express it in terms of β:
\[ \alpha (I_c + I_b) = I_c \]
- Expanding this gives:
\[ \alpha I_c + \alpha I_b = I_c \]

5. **Isolate Terms**:
- Rearranging the above equation leads to:
\[ \alpha I_b = I_c - \alpha I_c \]
- Factor out Ic:
\[ \alpha I_b = I_c (1 - \alpha) \]

6. **Express Ib in terms of Ic and α**:
- Now, we can express Ib in terms of Ic:
\[ I_b = \frac{I_c (1 - \alpha)}{\alpha} \]

7. **Substituting Ib into the Beta Equation**:
- Now substitute this expression for Ib into the definition of β:
\[ \beta = \frac{I_c}{I_b} = \frac{I_c}{\frac{I_c (1 - \alpha)}{\alpha}} = \frac{\alpha}{1 - \alpha} \]

8. **Final Relation**:
- Thus, we have the relation between α and β:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
- This can also be rearranged to give:
\[ \alpha = \frac{\beta}{\beta + 1} \]