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Zener regulator design: Rs and IZM

Short answer

Load current IL = 20 mA. Rs = (30−20)/20 mA = 500 Ω. At max input, current through Rs = (50−20)/500 = 60 mA, so IZM = 60−20 = 40 mA.

Design a Zener diode voltage regulator that will maintain an output voltage of 20 V across 1 kΩ load with an input that will vary between 30 V and 50 V. Determine the proper value of Rs and the maximum Zener current IZM.

A Zener regulator uses a series resistor Rs and a Zener diode in parallel with the load. The Zener keeps the load voltage at Vz; Rs drops the extra input voltage.

 Vi o--[ Rs ]--+-------+
     30-50 V   |       |
             Zener    RL = 1 kΩ
             Vz=20V    |
 o-------------+-------+

Given: Vz = VL = 20 V, RL = 1 kΩ, Vi = 30 V to 50 V. Assume an ideal Zener (Iz(min) ≈ 0, rz = 0), as in the usual textbook design (Boylestad).

Load current (constant)

IL = VL/RL = 20/1000 = 20 mA

Choosing Rs (worst case: minimum input)

At Vi(min) = 30 V the Zener must still be in breakdown, so the current through Rs must at least supply the load (Iz ≥ 0):

Rs = (Vi(min) − Vz)/IL
   = (30 − 20)/20 mA
   = 500 Ω

Maximum Zener current (maximum input)

I_Rs(max) = (Vi(max) − Vz)/Rs
          = (50 − 20)/500 = 60 mA
IZM = I_Rs(max) − IL = 60 − 20 = 40 mA

Answer: Rs = 500 Ω, IZM = 40 mA.

Zener power rating needed: PZM = Vz × IZM = 20 × 40 mA = 0.8 W, so select a 20 V Zener of at least 1 W.

Rs dissipates (50 − 20)²/500 = 1.8 W at maximum input, so use a 2 W (or higher) resistor. In practice Rs is chosen a little below 500 Ω (e.g. 470 Ω standard value) so a small minimum Zener current Iz(min) flows at 30 V input and regulation is kept.

✓ Answer checked by Claude Opus 5.5 (high effort)

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