Zener Voltage Regulator
An unregulated 10 V source is connected through a 100 ohm series resistor to a 5 V Zener diode operating in breakdown, with no load resistor attached across the diode. What current flows through the series resistor?
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- zener voltage regulator
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
50 mA
A Zener diode operated in reverse breakdown holds an almost constant voltage across itself, here its rated 5 V, even as the current through it varies. In this unloaded regulator, the entire current supplied by the source passes through the series resistor and then through the Zener, since there is no parallel load branch to divert any of it. The series resistor is essential: it absorbs the surplus voltage and limits the current so the Zener is not destroyed by excessive dissipation. The voltage dropped across the series resistor equals the difference between the input voltage and the Zener voltage: (V_R = V_{in} - V_Z = 10 - 5 = 5,\text{V}). Applying Ohm's law to the series resistor gives the current (I = V_R / R = 5 / 100 = 0.05,\text{A} = 50,\text{mA}). The value 100 mA wrongly divides the full 10 V input by the resistor, ignoring the Zener voltage. The value 150 mA adds the two voltages instead of subtracting. The value 5 mA misplaces a factor of ten. As a final check, the resistor drop of (0.05 \times 100 = 5,\text{V}) plus the 5 V held by the Zener restores the 10 V input, confirming Kirchhoff's voltage law is satisfied.
This medium difficulty physics question is from the chapter semiconductor electronics, covering the topic of zener voltage regulator. It appeared in the 2025 exam.
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