Question

The depletion region width of a p-n junction is \( 4 \times 10^{-6} \, \text{m} \) and the potential barrier is \( 0.8 \, \text{V} \). The electric field intensity in the depletion region is:

• A. \( 1 \times 10^5 \, \text{V/m} \)
• B. \( 2 \times 10^5 \, \text{V/m} \)
• C. \( 4 \times 10^5 \, \text{V/m} \)
• D. \( 8 \times 10^5 \, \text{V/m} \)

Hint:

Always double-check the power of ten in units. Writing \( 0.8 / 4 \) as \( 0.2 \) and then shifting the decimal to get \( 2 \times 10^5 \) prevents simple calculation errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the electric field intensity in the depletion region of a p-n junction when its width and the barrier potential are given.

Step 2: Key Formula or Approach:
The average electric field intensity \( E \) in the depletion region is related to the potential barrier \( V \) and the width \( d \) by the relation:
\[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:
Given:
- Potential barrier \( V = 0.8 \, \text{V} \)
- Depletion region width \( d = 4 \times 10^{-6} \, \text{m} \)
Substitute these values into the formula:
\[ E = \frac{0.8}{4 \times 10^{-6}} \] \[ E = 0.2 \times 10^6 \, \text{V/m} \] \[ E = 2 \times 10^5 \, \text{V/m} \] Thus, the electric field intensity in the depletion region is \( 2 \times 10^5 \, \text{V/m} \).

Step 4: Final Answer:
(B) \( 2 \times 10^5 \, \text{V/m} \)