Skip to main content

Half wave rectifier with waveforms

Short answer

A single diode passes only positive half-cycles; during negative half it blocks. DC output Vdc = 0.318 Vm, Vrms = Vm/2, ripple factor 1.21, max efficiency 40.6%, PIV = Vm. Simple but high ripple.

Draw the circuit diagram of half wave rectifier with its input output voltage waveforms and explain in brief.

A half-wave rectifier converts AC into pulsating DC by passing only one half-cycle of the input and blocking the other, using a single diode.

Circuit

        Tr          D
 AC  o-)||(--+---->|----+------+
 mains )||(  |          |      |
       )||( vs         RL     vo
       )||(  |          |      |
     o-)||(--+----------+------+

Operation

  • Positive half-cycle: the top of the secondary is positive, the diode is forward biased and conducts. Current flows through RL, and vo ≈ vs (minus 0.7 V for a Si diode).
  • Negative half-cycle: the diode is reverse biased and acts as an open switch. No current flows, so vo = 0. The diode must withstand the peak inverse voltage PIV = Vm.

Waveforms

 vs  |  _        _
     | / \      / \
   0 +/---\----/---\----> t
     |     \__/     \__/

 vo  |  _        _
     | / \      / \
   0 +/---\____/---\____> t

The output is unidirectional but pulsating: one pulse per input cycle, so the ripple frequency equals the supply frequency (50 Hz).

Important values (ideal diode)

QuantityValue
DC (average) outputV_dc = Vm/π = 0.318 Vm
RMS outputV_rms = Vm/2
Ripple factor1.21
Efficiency (max)40.6%
PIVVm

It is simple and cheap but has high ripple and low efficiency, so it is used only in low-power circuits; a filter capacitor across RL is added to smooth the output.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Electrical and Electronics Engineering chapter 4 questions →