Full wave bridge rectifier operation

A full wave bridge rectifier converts the complete ac cycle into pulsating dc using four diodes connected as a bridge. No centre-tapped transformer is needed.

Circuit

                    A
                    o
                  /   \
       D4 (N -> A)     D1 (A -> P)
                /       \
   N (-) o-----+         +-----o P (+)
                \       /
       D3 (N -> B)     D2 (B -> P)
                  \   /
                    o
                    B

 ac input (transformer secondary) across A and B
 RL connected between P (+) and N (-)
 (X -> Y means anode at X, cathode at Y)

The ac input is applied at nodes A and B. D1 and D3 conduct when A is positive; D2 and D4 conduct when B is positive. The current through RL always flows in the same direction (from P to N).

Working

• Positive half cycle (A positive): D1 and D3 are forward biased, D2 and D4 are reverse biased. Current flows A → D1 → RL → D3 → B.
• Negative half cycle (B positive): D2 and D4 conduct, D1 and D3 are off. Current flows B → D2 → RL → D4 → A, in the same direction through RL.

So both half cycles appear across RL as positive pulses.

Waveforms

 Input vs      Output vo
  /\  /\        /\  /\  /\  /\
 /  \/  \      /  \/  \/  \/  \
 ----------    -------------------

Expressions (vs = Vm sin ωt, diodes ideal)

• Peak output: Vm (Vm − 1.4 V with real silicon diodes, as two diodes are in the path)
• Average (dc) value: Vdc = 2Vm/π = 0.636 Vm; Idc = 2Im/π
• RMS value: Vrms = Vm/√2 = 0.707 Vm; Irms = Im/√2
• Ripple factor: r = √((Vrms/Vdc)² − 1) = 0.482
• Form factor = Vrms/Vdc = 1.11
• Efficiency η = Pdc/Pac = 81.2% (max)
• Peak inverse voltage (PIV) of each diode = Vm
• Output ripple frequency = 2f (100 Hz for a 50 Hz supply)
• Transformer utilisation factor ≈ 0.812

Advantages: high efficiency, low PIV per diode (Vm), no centre tap. Disadvantages: four diodes are needed, and there are two diode drops in the path (about 1.4 V lost), so it is poor for low voltages.