Zener regulator Vo, load effect, minimum RL

Given: Vz = 6.8 V at Iz = 5 mA, rz = 20 Ω, Izk = 0.2 mA, V+ = 10 V ± 1 V, R = 0.5 kΩ.

Zener model

Vz0 = Vz − rz Iz = 6.8 − 20 × 0.005 = 6.7 V. So the Zener is 6.7 V in series with 20 Ω.

(a) No load, V+ = 10 V

I = (V+ − Vz0)/(R + rz) = (10 − 6.7)/(500 + 20) = 6.346 mA

Vo = Vz0 + rz I = 6.7 + 20 × 6.346 mA = 6.827 V

(b) Change in Vo when a load draws IL = 1 mA

The load current is taken from the Zener (the current in R is almost unchanged), so the Zener current decreases by about 1 mA:

ΔVo = −rz (R/(R + rz)) IL = −(20 ∥ 500) × 1 mA = −19.2 Ω × 1 mA

ΔVo = −19.2 mV (new Vo ≈ 6.808 V). Load regulation = −19.2 mV/mA.

Exact check: Vo = (V+/R + Vz0/rz − IL)/(1/R + 1/rz) = (0.02 + 0.335 − 0.001)/0.052 = 6.808 V ✓.

(c) Minimum RL for the diode to stay in breakdown

The Zener stays in breakdown while Iz ≥ Izk = 0.2 mA. At this limit:

Vz = 6.7 + 20 × 0.0002 = 6.704 V

Current in R = (10 − 6.704)/500 = 6.592 mA (at nominal V+)

IL(max) = 6.592 − 0.2 = 6.392 mA

RLmin = 6.704 / 6.392 mA ≈ 1.05 kΩ (at V+ = 10 V)

For the worst case V+ = 9 V: IR = (9 − 6.704)/500 = 4.592 mA, IL = 4.392 mA, so RLmin = 6.704/4.392 = 1.53 kΩ. If the supply can drop to 9 V, RL must be larger than about 1.5 kΩ.

Answer: Vo = 6.83 V; ΔVo = −19.2 mV; RLmin ≈ 1.05 kΩ (nominal supply), 1.53 kΩ (V+ = 9 V).