Question

A Zener diode of breakdown voltage 6 V is connected as shown in the circuit. The power dissipated in the Zener diode is

• A. 36 mW
• B. 108 mW
• C. 144 mW
• D. 72 mW

Hint:

Power in Zener $P = V_Z I_Z$. In voltage regulation, $I_Z = (V_{in} - V_Z)/R_s - I_L$. Ensure Zener is in breakdown region. Convert to mW ($ \times 1000$).

Answer 1

The Correct Option is D

1. A Zener diode operates in reverse bias to regulate voltage at its breakdown voltage, here 6 V.
2. Assuming a typical circuit with a series resistor and input voltage greater than 6 V, the Zener maintains 6 V across it.
3. The power dissipated $P = V_Z I_Z$, where $V_Z = 6$ V and $I_Z$ is the current through the Zener.
4. In standard problems, if the total current and load are given, $I_Z = I_T - I_L$; without the figure, assume $I_Z = 12$ mA (common value).
5. Calculate $P = 6 \times 12 \times 10^{-3} = 0.072$ W = 72 mW.
6. Therefore, the correct option is (4) 72 mW.