Zener regulator design: with Vin ranging 15–20 V, Vz = 6.8 V, and load current 5–20 mA, choose the series resistor value R for regulation.

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:

Verification / Alternative check:

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: