If the forward voltage in a semiconductor is doubled, the width of depletion layer will

Options

A. become half
B. become one fourth
C. remain unchanged
D. become double

Answer

To solve the problem of how the width of the depletion layer in a semiconductor changes when the forward voltage is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Depletion Layer**:
- In a PN junction semiconductor, the depletion layer is the region where mobile charge carriers (electrons and holes) are depleted. This region forms due to the diffusion of carriers across the junction, leading to the formation of an electric field.

2. **Initial Conditions**:
- Let the initial forward voltage applied across the PN junction be \( V \).
- The initial width of the depletion layer is denoted as \( W_d \).

3. **Effect of Forward Voltage**:
- When a forward voltage is applied, it reduces the potential barrier at the junction, allowing more charge carriers to recombine and reducing the width of the depletion layer.
- The relationship between the width of the depletion layer \( W_d \) and the applied voltage \( V \) can be expressed as:
\[ W_d \propto \frac{1}{\sqrt{V}} \]
- This means that the width of the depletion layer is inversely proportional to the square root of the applied voltage.

4. **Doubling the Voltage**:
- If the forward voltage is doubled, i.e., \( V \) becomes \( 2V \), we can substitute this into our relationship:
\[ W_d' \propto \frac{1}{\sqrt{2V}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{V}} \]
- This indicates that the new width \( W_d' \) of the depletion layer is:
\[ W_d' = \frac{W_d}{\sqrt{2}} \]

5. **Conclusion**:
- Since \( \sqrt{2} \) is approximately 1.414, the width of the depletion layer decreases when the voltage is doubled. Specifically, it becomes:
\[ W_d' = \frac{W_d}{\sqrt{2}} \approx 0.707 \cdot W_d \]
- Thus, the width of the depletion layer is reduced to about 70.7% of its original width, which is not exactly half, but indicates a significant reduction.

### Final Answer:
The width of the depletion layer will decrease when the forward voltage is doubled, becoming approximately \( \frac{W_d}{\sqrt{2}} \).