Question:
A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _______ \(\Omega\).
A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _______ \(\Omega\).
Updated On: Apr 10, 2026
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Correct Answer: 300
Solution and Explanation
Step 1: Understanding the Concept:
A Zener diode regulates voltage by operating in its breakdown region. To prevent the diode from burning out, a series resistor must limit the total current flowing through the diode so that its power dissipation does not exceed its rated maximum.
Step 2: Key Formula or Approach:
1. Maximum Zener current \( I_Z = \frac{P_{\text{max}}}{V_Z} \).
2. Series resistance \( R = \frac{V_{\text{source}} - V_Z}{I} \).
Step 3: Detailed Explanation:
Given: Zener voltage \( V_Z = 10 \text{ V} \), Max power \( P_{\text{max}} = 0.5 \text{ W} \), Source voltage \( V_S = 25 \text{ V} \).
First, calculate the maximum allowable current through the Zener:
\[ I_Z = \frac{P_{\text{max}}}{V_Z} = \frac{0.5}{10} = 0.05 \text{ A} \]
The voltage to be dropped across the series resistor \( R \) is:
\[ V_R = V_S - V_Z = 25 - 10 = 15 \text{ V} \]
To limit the current to 0.05 A, the minimum resistance required is:
\[ R_{\text{min}} = \frac{V_R}{I_Z} = \frac{15}{0.05} = 300 \text{ } \Omega \]
Step 4: Final Answer:
The minimum resistance required is 300 \(\Omega\).
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