Full-Wave Rectification, Smoothing and Zener Regulation

From Wall Socket to Steady d.c.

The mains supply is 230 V alternating current, but phones, laptops, radios and most electronic circuits need a low, steady direct voltage such as 5 V or 12 V. A power supply converts one into the other in four stages: a transformer reduces the voltage, a rectifier makes the current flow one way, a capacitor smooths out the bumps, and a regulator holds the output steady. This tutorial explains each stage, including half-wave and full-wave rectification, the bridge rectifier, smoothing capacitors and ripple, and Zener diode voltage regulation, with worked design examples. It builds on the tutorial on semiconductors and diodes.

Stage 1: The Step-Down Transformer

A transformer lowers the 230 V mains to a convenient a.c. voltage, such as 12 V r.m.s., using the turns ratio Vs/Vp = Ns/Np. It also isolates the low-voltage circuit from the mains, which improves safety.

Remember that the r.m.s. value is not the peak. A 12 V r.m.s. supply has a peak of 12 × √2 ≈ 17.0 V. After rectification and smoothing, the d.c. output is close to the peak, not the r.m.s. value.

Stage 2: Rectification

A diode conducts when forward biased and blocks current when reverse biased. A silicon diode needs about 0.7 V across it before it conducts (its forward voltage drop). Rectifiers use this one-way behaviour.

Half-Wave and Full-Wave Rectification

Half-wave rectification. A single diode in series with the load lets through only the positive half-cycles. During negative half-cycles the diode is reverse biased and no current flows. The output is a series of positive pulses with gaps between them.

Advantages: very simple and cheap.
Disadvantages: half the input waveform is wasted, the output has large gaps, and it is hard to smooth. The ripple frequency equals the supply frequency, 50 Hz.

Full-wave bridge rectification. Four diodes are arranged in a bridge. During one half-cycle, two diodes on opposite sides conduct; during the other half-cycle, the other two conduct. In both half-cycles, current flows through the load in the same direction, so the negative half-cycles are flipped upwards.

Advantages: uses both halves of the a.c., gives twice as many pulses (a ripple frequency of 100 Hz from a 50 Hz supply), and is much easier to smooth.
The current passes through two diodes at a time, so the peak output is about 1.4 V less than the input peak.

Example 1. A 12 V r.m.s. transformer feeds a silicon bridge rectifier. Find the peak output voltage.

Input peak = 17.0 V. Two diodes drop 2 × 0.7 = 1.4 V, so the output peak ≈ 15.6 V. With a single-diode half-wave rectifier it would be 17.0 − 0.7 = 16.3 V.

An older full-wave design uses a centre-tapped transformer and only two diodes, each conducting on alternate half-cycles. It drops only one diode voltage but needs a transformer with twice the secondary winding.

Stage 3: Smoothing with a Capacitor

A large capacitor (a reservoir capacitor, typically hundreds or thousands of microfarads) is connected across the output of the rectifier.

As the rectified voltage rises towards its peak, the capacitor charges up to the peak voltage.
As the voltage from the rectifier falls, the diodes switch off and the capacitor supplies the load, discharging slowly.
When the next pulse arrives and rises above the capacitor's voltage, it recharges the capacitor.

The output is now a nearly steady d.c. voltage with a small sawtooth variation called ripple. The capacitor discharges for nearly a whole ripple period T = 1/f, at a current I, losing a charge of about IT. Since ΔV = ΔQ/C, the peak-to-peak ripple is approximately

ΔV ≈ I / (fC),

where f is the ripple frequency (100 Hz for full-wave, 50 Hz for half-wave from a 50 Hz supply).

Example 2. A full-wave rectifier with a 2,200 μF reservoir capacitor supplies 0.50 A. Estimate the ripple.

ΔV ≈ 0.50 / (100 × 2200 × 10⁻⁶) ≈ 2.3 V. With half-wave rectification the ripple would double to about 4.5 V, showing why full-wave is preferred.

Example 3. What capacitance gives a ripple of only 1.0 V at 0.50 A with full-wave rectification?

C = I/(fΔV) = 0.50/(100 × 1.0) = 0.005 F = 5,000 μF (a standard 4,700 μF capacitor would give just over 1 V).

Larger capacitors and smaller load currents give less ripple. The capacitor must be rated for at least the peak voltage, and electrolytic capacitors must be connected the right way round.

Stage 4: Regulation with a Zener Diode

Even after smoothing, the output still has some ripple and it changes when the mains voltage or the load current changes. A regulator holds the output steady.

A Zener diode is designed to be used in reverse bias. When the reverse voltage reaches its breakdown voltage Vz (for example 5.1 V), it conducts, and the voltage across it stays almost constant at Vz over a wide range of currents. Placed in parallel with the load and fed through a series resistor R, it acts as a simple voltage regulator:

if the input voltage rises, more current flows through R and the extra is taken by the Zener, so the output stays at Vz;
if the load draws more current, the Zener takes less, again keeping the output at Vz.

The series resistor drops the difference between the input voltage and Vz, and limits the current.

Example 4. Design a Zener regulator to supply 5.1 V to a 100 Ω load from a 12 V smoothed supply. Allow at least 5 mA through the Zener.

Load current = 5.1/100 = 51 mA.
Total current through R = 51 + 5 = 56 mA.
R = (12 − 5.1)/0.056 ≈ 123 Ω; choose the standard value 120 Ω, giving 6.9/120 = 57.5 mA, so the Zener carries 6.5 mA.
Worst case for the Zener: if the load is disconnected, all 57.5 mA flows through it, dissipating 5.1 × 0.0575 ≈ 0.29 W. A 0.5 W Zener diode is therefore suitable.

The resistor dissipates 6.9 × 0.0575 ≈ 0.4 W, so it should be rated at 0.5 W or more. Zener regulators are simple but wasteful, so they are used for small currents.

Integrated Circuit Regulators and Switch-Mode Supplies

Most linear power supplies now use three-terminal regulator chips such as the 7805 (5 V) or 7812 (12 V), which contain a voltage reference, an amplifier and a power transistor. They give very steady outputs with built-in protection against overheating and short circuits. They need an input a couple of volts above their output.

Phone chargers and laptop adapters use switch-mode power supplies instead. These rectify the mains directly, switch it on and off tens of thousands of times a second, and use a small high-frequency transformer. They are much smaller, lighter and more efficient than transformer-based supplies, which is why a modern phone charger is tiny compared with old transformer "power packs".

Summary

A linear power supply has four stages: step-down transformer, rectifier, smoothing capacitor and regulator.
Half-wave rectification uses one diode and wastes half the waveform; full-wave rectification with a bridge uses four diodes and both halves.
A reservoir capacitor charges to the peak voltage; ripple ΔV ≈ I/(fC), with f = 100 Hz for full-wave from 50 Hz mains.
A Zener diode in reverse breakdown holds a constant voltage; a series resistor sets the current.
Regulator ICs and switch-mode supplies are the modern versions of these ideas.