Almost every DC power supply starts with a rectifier — a circuit that turns alternating current into direct current. The full-wave bridge rectifier does it with four diodes arranged in a diamond, and it uses both halves of the AC cycle, unlike a half-wave rectifier that throws half the input away. It also needs no centre-tapped transformer, which is why it is the standard choice in practice.

Figure 1: Bridge rectifier circuit and conduction path (a), and the input, rectified and filtered waveforms (b). (Schematic, not to scale.)

How it works

On the positive half-cycle, the left AC terminal is positive: current flows through diode D1, through the load from + to −, and back through diode D4 to the right AC terminal. On the negative half-cycle, the right terminal is positive: current flows through D2, through the load in the same direction, and back through D3. In each half-cycle two diodes conduct in series, and the load current never reverses — that is full-wave rectification.

Because the current always flows the same way through the load, the output is a series of positive half-sine pulses at twice the mains frequency — 100 Hz on a 50 Hz supply. A filter capacitor then smooths these pulses into nearly steady DC.

The key formulas

• Peak secondary voltage: Vm = √2 × Vrms
• Average (DC) output, ideal diodes: Vdc = 2Vm / π = 0.636 × Vm
• With silicon diodes (two conduct at a time, 0.7 V each): Vdc = 2Vm / π − 1.4 V
• Ripple frequency: 2f = 100 Hz on a 50 Hz supply
• Ripple factor (no filter): γ = 0.48 — much lower than a half-wave rectifier's 1.21
• Peak inverse voltage per diode: PIV = Vm (half that of a centre-tapped full-wave circuit)
• Filter capacitor ripple: Vr(pk-pk) = IL / (f × C), where f is the ripple frequency (100 Hz)

Worked example — 12 V RMS secondary, 100 mA load

A bridge rectifier is fed from a 12 V RMS secondary and supplies a load drawing IL = 100 mA.

Step 1 — peak voltage: Vm = √2 × 12 = 16.97 V.

Step 2 — DC output without a filter: ideal Vdc = 2 × 16.97 / π = 10.80 V. With two silicon diodes conducting, Vdc = 10.80 − 1.4 = 9.40 V.

Step 3 — ripple factor: without a filter, γ = 0.48 (the textbook value for a full-wave rectifier).

Step 4 — add a 1000 µF filter capacitor: ripple frequency f = 100 Hz, so Vr(pk-pk) = IL / (f × C) = 0.1 / (100 × 1000 × 10⁻⁶) = 0.1 / 0.1 = 1.0 V peak-to-peak. The DC output rises to Vdc ≈ Vm − 1.4 − Vr(pk-pk)/2 = 16.97 − 1.4 − 0.5 = 15.07 V ≈ 15.1 V. The load resistance is RL = Vdc / IL = 15.07 / 0.1 = 151 Ω ≈ 150 Ω.

So the same circuit gives 9.4 V of rough pulsating DC without a capacitor, and about 15.1 V of smooth DC with a 1000 µF capacitor — the capacitor charges to near the peak each half-cycle and supplies the load between peaks.

Key points to remember

• Four diodes in a bridge: D1 + D4 conduct on one half-cycle, D2 + D3 on the other; load current always flows the same way.
• Vdc = 2Vm / π = 0.636 Vm (ideal), minus 1.4 V for two silicon diode drops.
• Ripple frequency is 100 Hz on a 50 Hz supply; ripple factor without a filter is 0.48.
• Worked example: 12 V RMS, 100 mA → Vdc = 9.40 V unfiltered; with a 1000 µF capacitor, ripple 1.0 V pk-pk and Vdc ≈ 15.1 V.
• PIV per diode is only Vm, and no centre-tapped transformer is needed — the bridge's two big practical advantages.