Question

If the breakdown voltage of a Zener diode is 7.2 V and the maximum power dissipated is 216 mW, then the maximum current through the Zener diode is

• A. 30 mA
• B. 4.16 mA
• C. 3.33 mA
• D. 2.4 mA

Hint:

$P = VI$ is the fundamental power equation. Always check units (mW vs W).

Answer 1

The Correct Option is A

Step 1: Formula for Power:
Power dissipated in a Zener diode is given by $P = V_Z \cdot I_Z$. So, maximum current $I_{max} = \frac{P_{max}}{V_Z}$.
Step 2: Substitution and Calculation:
Given: $P_{max} = 216 \, \text{mW} = 216 \times 10^{-3} \, \text{W}$. $V_Z = 7.2 \, \text{V}$. \[ I_{max} = \frac{216 \times 10^{-3}}{7.2} \] \[ I_{max} = \frac{216}{7.2} \text{ mA} \] \[ I_{max} = \frac{2160}{72} \text{ mA} \] Since $72 \times 3 = 216$, $2160/72 = 30$. \[ I_{max} = 30 \, \text{mA} \]
Step 3: Final Answer:
The maximum current is 30 mA.