A Zener diode is specified having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle.

Options

A. 40 mA
B. 60 mA
C. 50 mA
D. 45 mA

Explanation

To find the maximum current that a Zener diode can handle given its breakdown voltage and maximum power dissipation, we can use the relationship between power, voltage, and current. The formula we will use is:

\[ P = V \times I \]

Where:
- \( P \) is the power (in watts),
- \( V \) is the voltage (in volts),
- \( I \) is the current (in amperes).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Breakdown Voltage (\( V_{Z} \)) = 9.1 V
- Maximum Power Dissipation (\( P_{max} \)) = 364 mW = 364 × 10⁻³ W

2. **Convert Power to Watts**:
- Since the power is given in milliwatts, convert it to watts:
\[ P_{max} = 364 \text{ mW} = 364 \times 10^{-3} \text{ W} \]

3. **Use the Power Formula**:
- We know that the maximum power can also be expressed as:
\[ P_{max} = V_{Z} \times I_{max} \]
- Rearranging this formula to find \( I_{max} \):
\[ I_{max} = \frac{P_{max}}{V_{Z}} \]

4. **Substitute the Values**:
- Substitute the known values into the equation:
\[ I_{max} = \frac{364 \times 10^{-3}}{9.1} \]

5. **Calculate \( I_{max} \)**:
- Performing the division:
\[ I_{max} = \frac{364}{9.1} \times 10^{-3} \text{ A} \]
- Calculate \( \frac{364}{9.1} \):
\[ I_{max} \approx 40 \times 10^{-3} \text{ A} \]

6. **Convert to Milliamperes**:
- Convert the current to milliamperes:
\[ I_{max} = 40 \text{ mA} \]

### Final Answer:
The maximum current that the Zener diode can handle is **40 mA**.