How it works
In a bridge rectifier, during the positive half-cycle diodes D1 and D3 conduct; during the negative half-cycle D2 and D4 conduct. Current always flows in the same direction through the load RL. DC output is Vdc = 2Vm/π, roughly 0.637Vm, and ripple factor drops to 0.482 — far better than the half wave's 1.21. Each diode must withstand a PIV of Vm in the bridge configuration. The centre-tap rectifier uses only two diodes but needs a centre-tapped transformer; here each diode sees a PIV of 2Vm because the full secondary voltage appears across the reverse-biased diode. Efficiency of both full wave types is 81.2%, double that of half wave.
Key points to remember
Critical values: Vdc = 2Vm/π, ripple factor = 0.482, efficiency = 81.2%, and ripple frequency = 100 Hz (for 50 Hz supply). Bridge rectifier PIV = Vm per diode; centre-tap rectifier PIV = 2Vm per diode — this distinction is a guaranteed exam question. The filter capacitor value C = Idc/(2f × Vr) for full wave, where Vr is allowable ripple voltage; notice the factor of 2f because ripple frequency is doubled. TUF = 0.812 for bridge, making it more efficient use of the transformer than the centre-tap design. Both diode voltage drops (2 × 0.7 V) in the bridge reduce effective output compared to centre-tap's single 0.7 V drop.
Exam tip
The examiner always asks you to compare PIV of bridge versus centre-tap full wave rectifiers — bridge PIV = Vm and centre-tap PIV = 2Vm, and the reason (full secondary appears across reverse diode in centre-tap) must be stated for full marks.