Introduction / Context:
A simple shunt zener regulator uses a series resistor R to drop excess voltage and a zener diode to clamp the load voltage at Vz. Choosing R requires considering input voltage variation and load current range so that the zener remains in regulation without reversing current (negative IZ) or dropping out at low Vin.
Given Data / Assumptions:
- Input voltage Vin: 15–20 V.
- Zener voltage Vz = 6.8 V.
- Load current IL: 5–20 mA.
- No zener minimum current specified; design to avoid IZ < 0 at Vin max and to supply IL max at Vin min.
Concept / Approach:
- Current through R: IR = (Vin − Vz) / R.
- Node equation: IR = IL + IZ, with IZ ≥ 0 for regulation.
- To avoid IZ < 0 at Vin max with IL min: (Vin_max − Vz) / R ≥ IL_min.
- To supply IL max at Vin min (just at edge of regulation with IZ ≈ 0): (Vin_min − Vz) / R ≥ IL_max.
Step-by-Step Solution:
Compute R_max from low Vin, high load: R_max = (15 − 6.8) / 0.020 = 8.2 / 0.020 = 410 Ω.Check at Vin_max and IL_min: IR = (20 − 6.8) / 410 ≈ 13.2 / 410 ≈ 32.2 mA ⇒ IZ ≈ 32.2 − 5 = 27.2 mA ≥ 0, so still regulating.Therefore, R ≈ 410 Ω satisfies both edges; nearest choice is 410 Ω.
Verification / Alternative check:
At Vin_min = 15 V and IL_max = 20 mA: IR = 8.2 / 410 = 20 mA ⇒ IZ ≈ 0 mA, right at regulation threshold, which is the limiting case.
Why Other Options Are Wrong:
- 310 Ω or 400 Ω: Work but provide unnecessary extra zener current at Vin_max (higher dissipation) and are not the computed limit.
- 550 Ω: Too large; at Vin_min cannot supply IL_max while maintaining Vz.
- None of the above: Incorrect because 410 Ω meets the design edge.
Common Pitfalls:
- Ignoring the worst-case corners of Vin and IL.
- Forgetting to verify IZ ≥ 0 at Vin_max and adequate current at Vin_min.
Final Answer:
410 Ω
Discussion & Comments