Zener regulator (no-load): An unloaded zener regulator has source voltage 20 V, series resistor 330 Ω, and zener voltage 12 V. What is the zener current under these conditions?

Introduction / Context:
A basic shunt zener regulator uses a series resistor to drop excess voltage so the zener diode can clamp the output near its zener voltage. With no load attached, nearly all current through the series resistor flows through the zener. Computing this current is a straightforward application of Ohm’s law.

Given Data / Assumptions:

• Source voltage V_S = 20 V.
• Zener voltage V_Z = 12 V (assume ideal regulation at this value).
• Series resistor R_S = 330 Ω.
• No-load (unloaded) condition: I_S ≈ I_Z.

Concept / Approach:
Series resistor current I_S = (V_S − V_Z) / R_S. With no load, the same current becomes the zener current I_Z. This ensures the zener remains in regulation as long as I_Z stays within its allowable range.

Step-by-Step Solution:
Compute voltage across R_S: V_RS = V_S − V_Z = 20 − 12 = 8 V.Compute I_S: I_S = V_RS / R_S = 8 / 330 A.I_S ≈ 0.02424 A = 24.24 mA.Therefore I_Z ≈ 24.2 mA (no-load).

Verification / Alternative check:
Sanity check: If R_S were 400 Ω, I_Z would be 20 mA; with 330 Ω, a bit higher than 20 mA is expected, consistent with 24.2 mA.

Why Other Options Are Wrong:
(a) and (c) overstate the current for the given resistor. (d) is far too high and would require a much smaller resistor.

Common Pitfalls:
Forgetting to subtract the zener voltage from the source; mixing mA and A; ignoring load conditions which, if present, would reduce I_Z accordingly.

Final Answer:
24.2 mA