Question

A Zener diode has a voltage rating of 10 volts and the maximum power drop across the Zener diode is 0.5 watt. What resistance (in Ohm) should be connected in series with the Zener diode so that it can be operated safely by a battery of 25 volts?

• A. 300 \(\Omega\)
• B. 200 \(\Omega\)
• C. 30 \(\Omega\)
• D. 20 \(\Omega\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A Zener diode acts as a voltage regulator. To operate safely, the current flowing through it must not exceed its maximum rated current, which is derived from its power rating. The series resistor drops the excess voltage from the source.

Step 2: Key Formula or Approach:
1. Maximum current \( I_{max} = \frac{P_{max}}{V_Z} \).
2. Voltage across series resistor \( V_R = V_s - V_Z \).
3. Series resistance \( R = \frac{V_R}{I_{max}} \).

Step 3: Detailed Explanation:
Given: \( V_Z = 10 \, \text{V} \), \( P_{max} = 0.5 \, \text{W} \), \( V_s = 25 \, \text{V} \). First, find the maximum safe current through the Zener: \[ I_{max} = \frac{0.5}{10} = 0.05 \, \text{A} \] Next, find the voltage that the series resistor must drop: \[ V_R = 25 - 10 = 15 \, \text{V} \] To ensure the current does not exceed 0.05 A, the resistance must be: \[ R = \frac{15}{0.05} = \frac{1500}{5} = 300 \, \Omega \]

Step 4: Final Answer:
The required series resistance is 300 \(\Omega\).