Skip to main content

Center tapped full wave rectifier

Short answer

A centre-tapped full-wave rectifier uses a transformer with a grounded centre tap and two diodes. On each half cycle one diode conducts, directing current through the load in the same direction. This yields Vdc = 2Vm/π, ripple factor 0.482, efficiency 81.2%, ripple frequency 2f, and each diode must block a PIV of 2Vm.

Draw and explain the operation of the full-wave rectifier circuit using center tapped transformer.

A centre-tapped full wave rectifier uses a transformer with a centre-tapped secondary and two diodes. It rectifies both half cycles of the input.

Circuit

  secondary of a        D1
  centre-tapped    X o-----|>|-----+
  transformer        )             |
                     ) upper half  |
                     )             +--------o K (+)
                CT   o----o GND    |        |
                     )             |       [RL]  vo
                     ) lower half  |        |
                     )             |       GND
                   Y o-----|>|-----+
                        D2
  (both GND points are the same node: the centre tap)

The centre tap (CT) is the common (ground) point. The load RL is connected between the junction of the two diode cathodes (K) and the centre tap. Each half of the secondary gives a voltage vs = Vm sin ωt, with the two halves 180° out of phase.

Operation

  • Positive half cycle: the top end X of the secondary is positive. D1 is forward biased and conducts; D2 is reverse biased. Current flows through D1 and RL (towards the centre tap).
  • Negative half cycle: the bottom end Y becomes positive. D2 conducts; D1 is off. Current flows through D2 and RL in the same direction.

So the current through RL is unidirectional in both half cycles.

 vo  /\  /\  /\  /\
    /  \/  \/  \/  \

Main expressions

  • Vdc = 2Vm/π = 0.636 Vm
  • Vrms = Vm/√2
  • Ripple factor = 0.482; efficiency = 81.2%
  • Ripple frequency = 2f
  • PIV = 2Vm

Advantages: only two diodes and one diode drop in the path. Disadvantages: it needs a centre-tapped transformer, and each diode must withstand a PIV of 2Vm.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Basic Electronics Engineering chapter 2 questions →