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Design zener shunt regulator (5.6 V, 15 V supply)

Short answer

With Iz(min)=5mA (5×Izk) and IL(max)=15mA, R = (15V–5.6V)/20mA = 470Ω. Maximum Zener power at no load is Pz(max) = 5.6V × 20mA = 112mW.

A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. Izk is specified to be 1mA. The zener is to be used in the design of a shunt regulator fed from a 15V supply. The load current varies over the range of 0 mA to 15 mA. Find a suitable value for the resistor R. What is the maximum power dissipation of the zener diode?

Given: Vz = 5.6 V (constant for Iz ≥ 5 Izk), Izk = 1 mA, Vs = 15 V, IL varies from 0 to 15 mA.

Circuit

 15 V o----[ R ]-----+-------+---- Vo = 5.6 V
                     |       |
                   (Dz)     [RL]
                     |       |
                    GND     GND

Step 1: Minimum Zener current

The Zener voltage is constant only for Iz ≥ 5 Izk = 5 × 1 mA = 5 mA. So the minimum Zener current must be Iz(min) = 5 mA.

Step 2: Find R

The worst case is the largest load current, IL = 15 mA. At this condition Iz should be at its minimum, 5 mA:

IR = IL(max) + Iz(min) = 15 + 5 = 20 mA

R = (Vs − Vz)/IR = (15 − 5.6)/20 mA = 9.4 V / 20 mA = 470 Ω

(470 Ω is a standard value; the supply is assumed to be fixed at 15 V.)

Step 3: Maximum Zener power

The Zener current is largest with no load (IL = 0), when the whole of IR flows in it:

Iz(max) = IR = 20 mA

Pz(max) = Vz Iz(max) = 5.6 V × 20 mA = 112 mW

Hence a Zener rated at 250 mW (or 400 mW) is suitable.

Answer: R = 470 Ω; maximum Zener dissipation = 112 mW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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