Half vs Full Wave

Part of Vacuum Tubes

Half-wave and full-wave rectification are the two fundamental approaches to converting AC power to DC — understanding their trade-offs guides transformer design, filter sizing, and power supply selection.

Why This Matters

Every piece of tube equipment runs on DC power, but generators and mains supplies produce AC. The rectifier converts AC to pulsating DC; the filter smooths the pulsating DC to steady DC suitable for tube operation. The choice between half-wave and full-wave rectification affects the transformer size needed, the filter capacitor size required, the output voltage and regulation, and the peak stress on the rectifier tubes.

Building power supplies for radio stations, telephone exchanges, and audio amplifiers requires making this choice correctly. A half-wave supply is simpler to build with fewer components, but is limited in current output and produces more ripple. Full-wave is more efficient and better-regulated, but requires either a center-tapped transformer or four rectifier tubes. Understanding the physics behind each type allows you to choose correctly for each application.

For community communication infrastructure that must be built and maintained from salvage or locally manufactured components, the power supply is often the most demanding sub-assembly. Getting it right the first time — by choosing the correct topology — saves troublesome rework.

Half-Wave Rectification

In half-wave rectification, a single diode allows current to flow during only one half-cycle of the AC supply (the half-cycle when the diode’s plate is positive). During the opposite half-cycle, the diode blocks and no current flows.

The DC output voltage of a half-wave rectifier with a capacitor filter is approximately equal to the peak AC voltage: Vdc ≈ Vpeak = Vrms × √2. For a 250V RMS transformer secondary, the peak is 354V. After subtracting the diode forward drop (15-20V for a vacuum tube rectifier), the DC output is about 335V.

Ripple frequency equals the AC supply frequency (50 or 60 Hz). Each cycle produces one pulse of charging current for the filter capacitor, which then discharges through the load until the next pulse. With 50 Hz AC, the filter capacitor discharges for 20ms between charging pulses — a long time that requires a large capacitor to maintain acceptable ripple.

The ripple voltage for a half-wave supply with a capacitor filter is approximately:

Vripple ≈ Iload / (f × C)

For 100mA load current, 50 Hz supply, and a 200µF filter capacitor:

Vripple ≈ 0.1 / (50 × 0.0002) = 10V

A 10V ripple on a 300V supply is about 3.3% — acceptable for most applications.

Peak current through the rectifier diode can be very high. Each time the filter capacitor charges, the charging current is limited only by the transformer’s internal resistance and the diode’s forward resistance. Peak currents of 10-20 times the average DC current are typical in capacitor-input filters. This stresses the rectifier tube and can cause it to fail prematurely if the transformer impedance is too low or the capacitor too large.

Half-wave rectification is practical for low-current supplies (below 50mA), portable or simple equipment, and situations where transformer simplicity is paramount. It is not suitable for high-current amplifier power supplies where the large capacitor required and the single-diode peak current stress present problems.

Full-Wave Center-Tap Rectification

A full-wave center-tap circuit uses two diodes and a transformer with a center-tapped secondary. Each diode conducts on alternate half-cycles, so the output has twice the pulse frequency of a half-wave supply.

Circuit: the center tap of the transformer secondary connects to the DC negative output (ground). Each end of the secondary connects to one diode plate. Both diode cathodes connect together to form the DC positive output. During the first half-cycle, one end of the secondary goes positive, forward-biasing the first diode. During the second half-cycle, the other end goes positive, forward-biasing the second diode. Current flows to the load in the same direction during both half-cycles.

The DC output voltage is lower than for half-wave with the same transformer. For a center-tapped secondary, each diode uses only half the secondary winding. A transformer with 500V total secondary (250V each half) produces the same output as a 250V secondary on a half-wave circuit. To get 300V DC output from a full-wave center-tap supply, the total secondary voltage must be about 2 × (300 + 20 diode drop) = 640V.

Ripple frequency doubles to 100 Hz (for a 50 Hz supply) or 120 Hz (for 60 Hz). The filter capacitor charges twice as often, so for the same ripple voltage, you need only half the capacitance compared to a half-wave supply. From the earlier formula:

Vripple ≈ Iload / (f × C)

With f = 100 Hz instead of 50 Hz, you need only 100µF instead of 200µF for the same ripple.

Each diode must withstand the full secondary voltage as peak inverse voltage during the half-cycle when it is not conducting. For a 500V total secondary, each diode sees a PIV of 500V (the full secondary voltage, not just half). This is a disadvantage compared to the bridge circuit.

Full-Wave Bridge Rectification

The bridge rectifier uses four diodes arranged in a bridge configuration. Two diodes conduct during each half-cycle, always directing current through the load in the same direction. No center-tap is needed, so the full secondary voltage is used.

DC output: a 250V RMS secondary produces DC output near the full 250 × √2 = 354V peak, minus two diode drops (30-40V for vacuum tubes), yielding about 315V. This is higher than the center-tap circuit with the same transformer.

PIV per diode is only the peak secondary voltage — half what the center-tap circuit requires.

The disadvantage: four rectifier tubes instead of two, doubling the heater power and cost. This is why the bridge circuit, though common with cheap semiconductor diodes, was rarely used in vacuum tube equipment where two-diode rectifier tubes (both diodes in one envelope, with a single cathode) were available for center-tap service.

With vacuum tubes, center-tap rectification using a dual-diode tube (one cathode, two plates in a single envelope) is the efficient and practical choice for full-wave rectification. The 5Y3, 5U4, and 5AR4 are classic dual-diode rectifier tubes designed for this application.

Transformer and Filter Sizing

Choosing between half-wave and full-wave affects the transformer specification:

For half-wave: secondary voltage = desired DC output / 0.9 (rough rule). Only half the transformer secondary winding is used at a time, so the core must be sized for the full peak current even though average current is half.

For full-wave center-tap: secondary voltage (each half) = desired DC output / 0.9. The full core is used efficiently because both halves of the secondary conduct alternately.

For any rectifier with a capacitor-input filter, the transformer must supply not just the DC output power but also the reactive current for charging the capacitor. The ratio of peak current to average current (crest factor) means the transformer must be rated at 1.5-2× the DC power output. Design for this when salvaging or winding transformers for power supply duty.

A choke-input filter (inductor before the main filter capacitor) dramatically improves regulation and reduces peak rectifier current. With a choke-input filter, the crest factor drops to near 1 and the transformer handles current much more efficiently. The choke is harder to make or source than a capacitor but rewards the builder with better power supply performance, lower ripple, and longer rectifier tube life. Use choke-input filters for transmitter and high-current audio amplifier power supplies whenever you can obtain suitable inductors from salvage.