Design DC voltage regulator for 6 V

Given: Vz = 6 V at Iz = 20 mA, rz = 10 Ω, IZK = 2 mA, Pzmax = 500 mW. Vin = 15 V ± 30 % → Vin(min) = 10.5 V, Vin(max) = 19.5 V.

Circuit

 Vin o----[ R ]-----+------+----o Vo = 6 V
                    |      |
                  (Dz)    [RL]
                    |      |
                   GND    GND

Step 1: Zener model

The Zener is represented by Vz0 in series with rz: Vz0 = Vz − rz Iz = 6 − 10 × 0.02 = 5.8 V, so Vz = 5.8 + 10 Iz.

Step 2: Maximum Zener current

Pz = Vz0 Iz + rz Iz² = 0.5 W

10 Iz² + 5.8 Iz − 0.5 = 0 → Iz(max) = 76.2 mA, with Vz = 5.8 + 0.762 = 6.56 V.

Step 3: Choose R

The worst case for the Zener power is at Vin(max) with no load (RL open), when the entire current goes through the Zener. That current must not exceed 76.2 mA:

R ≥ (Vin(max) − Vz)/Iz(max) = (19.5 − 6.56)/76.2 mA = 169.8 Ω

Take the standard value R = 180 Ω. (Check: at Vin(max), Iz = (19.5 − 6.52)/180 = 72 mA, and Pz = 0.47 W < 0.5 W.)

Step 4: Maximum load current

The load current is largest when Vin is minimum, and the Zener is kept at the knee current IZK = 2 mA (the Zener must stay in breakdown):

Vz at 2 mA = 5.8 + 10 × 0.002 = 5.82 V

IR = (Vin(min) − Vz)/R = (10.5 − 5.82)/180 = 26.0 mA

IL(max) = IR − IZK = 26.0 − 2 = 24.0 mA

This corresponds to a minimum load resistance of about 6/24 mA = 250 Ω.

Assumption: the design is based on R fixed from the Zener power rating at no-load with maximum input.

Answer: R = 180 Ω (≥ 170 Ω); the circuit can deliver a maximum load current of about 24 mA (at Vin = 10.5 V, with Vo ≈ 5.8 V to 6.5 V).