An 18 V unregulated supply feeds a 220 Ω series resistor followed by a 9.1 V Zener diode in shunt with the load. The Zener requires a minimum of 5.0 mA to remain in breakdown, and its breakdown characteristic may be treated as ideal. The maximum DC load current the regulator can deliver at rated output voltage is most nearly:
- (A)35.5 mA
- (B)40.5 mA
- (C)45.5 mA
- (D)81.8 mA
Show worked solution
Answer: (A)
Subtracting the 5.0 mA holding current from the 40.45 mA that the series resistor delivers leaves about 35.5 mA available to the load.
Series-resistor current is set entirely by the 18 V source and the clamped 9.1 V output, regardless of how that current later divides between diode and load.
Load current is whatever the Zener does not take, so the load is maximized at the instant the diode carries only its minimum breakdown current.
FE Reference Handbook — Electronics: Diodes, Zener Diode Model and Shunt Regulation
Why the other choices appear
- (B)Omits the 5.0 mA Zener holding current and assigns the entire 40.45 mA series current to the load.
- (C)Adds the 5.0 mA minimum diode current to the series current instead of subtracting it, 40.45 + 5.0 = 45.5 mA.
- (D)Places the full source voltage across the series resistor, ignoring the 9.1 V clamp, 18/220 = 81.8 mA.