9 min readPathway sheetSection A objective (2 marks, two parts); Section B short answer (3 marks) on how the Zener regulator works; Section B long answer (5 marks) with a design numerical, often with a choice

Zener diode as a voltage regulator

A reverse-biased Zener across the load, fed through a series resistor Rs, keeps the output at Vz. Rs absorbs input changes; the Zener current absorbs load changes.

Series currentIs = (Vi − Vz)/Rs

Current splitIs = Iz + IL, so Iz = Is − IL; Iz must stay above a minimum, about 5 to 20 mA

Power in the ZenerPd = Vz × Iz; it is largest when the load current is zero

Condition to workThe input voltage Vi must be greater than Vz

Why we need a regulator

A rectifier with a filter gives nearly steady DC, but its voltage still changes. If the load draws more current, the filter capacitor discharges more and the output drops. If the mains voltage changes, the output changes too. A changing supply can spoil the working of an amplifier or a charger. A Zener diode fixes this. A supply that uses one is a regulated power supply.

How to draw the regulator circuit

Draw the unregulated DC input Vi, with + at the top.
From +, draw a series resistor Rs, then a junction point.
From that point to the − line, draw the Zener diode with its cathode (bar) at the top, so it is reverse biased.
Draw the load RL in parallel with the Zener. Label the output Vo = Vz.
Mark the currents: Is through Rs, Iz down through the Zener and IL through the load.

How it keeps the output steady

1
Vi is more than Vz, so the Zener is in breakdown. Its voltage stays at Vz over a wide range of current, so Vo = Vz.
2
The current through Rs is Is = (Vi − Vz)/Rs. It splits: Is = Iz + IL.
3
Load changes: if the load draws more current IL, the Zener draws less Iz by the same amount. Is and Vo stay the same.
4
Input changes: if Vi rises, Is rises. The extra voltage drops across Rs, and the extra current goes through the Zener. Vo stays at Vz.
5★
So the output is held at Vz against changes in both load current and input voltage, as long as Iz stays above its minimum.

★ key step

What happens when the input voltage rises

The output Vo = Vz does not change.

Worked example: designing a 9 V regulator

Problem

Design a Zener regulator for a steady 9 V output. The load current varies from 0 to 50 mA, and the input varies from 15 V to 18 V. Take the minimum Zener current as 10 mA. Find Rs and the power rating of the Zener.

Working

Choose a 9 V Zener.
At the lowest input with the largest load, Rs must still pass Is = 50 + 10 = 60 mA = 0.060 A.
Rs = (Vi min − Vz)/Is = (15 − 9) V ÷ 0.060 A = 100 Ω.
The Zener current is largest at the highest input with no load: Iz = (18 − 9) V ÷ 100 Ω = 0.090 A.
Power Pd = Vz × Iz = 9 V × 0.090 A = 0.81 W. Choose a rating above this, such as 1 W.

Answer

Rs = 100 Ω, with a 9 V, 1 W Zener diode

Worked example: currents in a regulator

Problem

A 6 V Zener regulator has Vi = 12 V and Rs = 200 Ω. The load is 600 Ω. Find Is, IL, Iz and the power in the Zener.

Working

Is = (12 − 6) V ÷ 200 Ω = 0.030 A = 30 mA.
IL = Vz/RL = 6 V ÷ 600 Ω = 0.010 A = 10 mA.
Iz = Is − IL = 30 − 10 = 20 mA.
Pd = Vz × Iz = 6 V × 0.020 A = 0.12 W.

Answer

Is = 30 mA, IL = 10 mA, Iz = 20 mA and Pd = 0.12 W

In real life

A household voltage stabiliser for a fridge or TV has the same goal: a steady output while the mains voltage swings. Inside chargers and set-top boxes, small Zener circuits hold sensitive chips at a fixed voltage.

Exam angle

3 marks: “With a circuit diagram, explain how a Zener diode keeps the output voltage constant when (a) the load current changes and (b) the input voltage changes.” Rs in series; Zener reverse biased across RL. Is = (Vi − Vz)/Rs = Iz + IL. (a) If IL rises, Iz falls equally, so Vo = Vz. (b) If Vi rises, Is rises; the extra voltage drops across Rs and the extra current flows through the Zener, so Vo = Vz.

5 marks: “A Zener regulator must give 5 V. The input varies from 10 V to 12 V and the load current from 0 to 40 mA. Taking the minimum Zener current as 10 mA, find Rs and the largest power in the Zener.” Is = 40 + 10 = 50 mA; Rs = (10 − 5) V ÷ 0.050 A = 100 Ω. Largest Iz = (12 − 5) V ÷ 100 Ω = 70 mA, at no load. Pd = 5 V × 0.070 A = 0.35 W, so use a 0.5 W Zener. Draw the circuit.

2 marks: “In a Zener regulator, why does the Zener diode dissipate the most power when the load current is zero?” Is is fixed by Vi and Rs. With IL = 0, all of Is flows through the Zener, so Iz, and with it Pd = Vz × Iz, is largest.

Remember it

Rs soaks up the extra volts, the Zener soaks up the extra amps: Is = Iz + IL and Vo = Vz.