A zener diode has a contact potential of 1 V in the absence of biasing. It undergoes Zener breakdown for an electric field of `10^(6)` V/m at the depletion region of p-n junction. If the width of the depletion region is 2.5 `mu`m, what should be the reverse biased potential for the Zener breakdown to occur ?

Options

A. (a)3.5 V
B. (b)2. 5 V
C. (c)1.5 V
D. (d)0.5 V

Answer

To solve the problem, we need to find the reverse biased potential required for Zener breakdown to occur in the given Zener diode. Here’s the step-by-step solution:

### Step 1: Understand the given parameters
- Contact potential (V_contact) = 1 V
- Electric field (E) = \(10^6 \, \text{V/m}\)
- Width of the depletion region (D) = 2.5 µm = \(2.5 \times 10^{-6} \, \text{m}\)

### Step 2: Calculate the potential required for breakdown
The potential (V) across the depletion region can be calculated using the formula:
\[ V = E \times D \]
Substituting the values:
\[ V = (10^6 \, \text{V/m}) \times (2.5 \times 10^{-6} \, \text{m}) \]

### Step 3: Perform the multiplication
Calculating the above expression:
\[ V = 10^6 \times 2.5 \times 10^{-6} = 2.5 \, \text{V} \]

### Step 4: Calculate the reverse biased potential
The reverse biased potential (V_reverse) required for Zener breakdown is given by:
\[ V_{\text{reverse}} = V - V_{\text{contact}} \]
Substituting the values:
\[ V_{\text{reverse}} = 2.5 \, \text{V} - 1 \, \text{V} = 1.5 \, \text{V} \]

### Final Answer
The reverse biased potential required for Zener breakdown to occur is **1.5 V**.

---