Question

A certain p-n junction, having a depletion region of width 20mm was found to have a breakdown voltage of 100V. If the width of the depletion region is reduced to 1 mm during its production, then it can be used as a Zener diode for voltage regulation of

• A. 15V
• B. 5V
• C. 7.5V
• D. 2000V

Hint:

Remember that for Zener diodes, the breakdown voltage is inversely proportional to the width of the depletion region. Reducing the width increases the breakdown voltage.

Answer 1

The Correct Option is B

Step 1: Understanding the relationship.
In a Zener diode, the breakdown voltage is inversely proportional to the width of the depletion region. The breakdown voltage \( V \) can be related to the width of the depletion region \( W \) by the equation: \[ V_1 W_1 = V_2 W_2 \] Where: - \( V_1 = 100V \) and \( W_1 = 20 \, \text{mm} \) - \( W_2 = 1 \, \text{mm} \) (new width)
Step 2: Apply the relation.
Substitute the known values into the equation to solve for \( V_2 \): \[ 100 \times 20 = V_2 \times 1 \] \[ V_2 = 100 \times 20 = 5V \]
Step 3: Conclusion.
The Zener diode will have a breakdown voltage of 5V when the depletion region width is reduced to 1 mm. Therefore, the correct answer is (2) 5V.

Answer 2

We can reach the same result by reasoning from the physical mechanism of Zener breakdown rather than a direct ratio equation. In a Zener diode, avalanche/Zener breakdown occurs once the electric field across the depletion region reaches a fixed critical value \( E_c \) that depends only on the semiconductor material, not on the diode's geometry. Since the field across a depletion region of width \( W \) under a reverse voltage \( V \) is \( E = \dfrac{V}{W} \), breakdown happens exactly when \[ \dfrac{V}{W} = E_c \ (\text{constant}) \] which means \( V \) must be directly proportional to \( W \).

Using the first (original) condition to fix this constant: \[ E_c = \dfrac{V_1}{W_1} = \dfrac{100}{20} = 5 \ \text{V per unit width} \] Applying this same critical field to the new, narrower depletion region of width \( W_2 = 1 \): \[ V_2 = E_c \times W_2 = 5 \times 1 = 5\text{V} \]

1. 15V: This does not correspond to the field-strength ratio computed above.
2. 5V: This matches exactly the value obtained from \( V_2 = E_c \times W_2 \), confirming it is correct.
3. 7.5V: Not consistent with the constant critical-field relationship derived from the given data.
4. 2000V: This would instead result from treating \( V \) and \( W \) as inversely related, which is the opposite (and incorrect) relationship for this physical situation, since narrowing the depletion region lowers, not raises, the breakdown voltage.

Therefore, the correct answer is 5V.