Zener Regulator With Load
A 15 V unregulated supply feeds a 10 V Zener diode through a 250 ohm series resistor, and a 1000 ohm load resistor is connected in parallel with the Zener. What current actually flows through the Zener diode in this regulated circuit?
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- zener regulator with load
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
10 mA
In a loaded Zener regulator the series current splits at the diode node between the load branch and the Zener itself, so the Zener current is found by Kirchhoff's current law: (I_Z = I_S - I_L). The regulator works by letting the Zener automatically absorb whatever current the load does not draw, holding the output fixed at the breakdown voltage even as the load demand changes. First, the series resistor carries the total current set by the voltage across it, which is the input minus the regulated Zener voltage: (I_S = (V_{in} - V_Z)/R_S = (15 - 10)/250 = 0.02,\text{A} = 20,\text{mA}). Next, the load sees the constant Zener voltage of 10 V, so (I_L = V_Z / R_L = 10 / 1000 = 0.01,\text{A} = 10,\text{mA}). The Zener therefore carries (I_Z = 20 - 10 = 10,\text{mA}). The value 20 mA forgets to subtract the load current. The value 30 mA wrongly adds load and series currents. The value 5 mA misuses the input voltage for the load branch. As a plausibility check, the Zener current is positive, which is essential: a regulator only works while (I_Z > 0), since the diode must stay in breakdown to hold its voltage steady.
This hard difficulty physics question is from the chapter semiconductor electronics, covering the topic of zener regulator with load. It appeared in the 2025 exam.
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