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Rectifiers and Smoothing

Watch the negative half disappear and the peaks flatten — then find out why the capacitor that smoothed it made the current spiky.

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A rectifier uses the diode's one-way conduction to turn alternating voltage into a pulsating unidirectional one, and a reservoir capacitor smooths it — at the cost of making the input current a series of narrow, tall spikes.

Why the negative half is a problem

Mains supplies alternating voltage and almost everything downstream wants a steady one. The negative half is not merely useless — it reverse-biases the circuit it feeds. The diode's one-way behaviour is exactly the tool for removing it.

Half-wave and full-wave

Half-waveFull-wave bridgeCentre-tapped
Diodes142
Average output0.318 V_peak0.637 V_peak0.637 V_peak
Ripple frequency50 Hz100 Hz100 Hz
Forward drops in the path121
TransformerOrdinaryOrdinaryNeeds a centre tap

Doubling the ripple frequency halves the smoothing capacitance needed for the same ripple, so full-wave wins twice over — a higher average and an easier filter.

Smoothing

A reservoir capacitor charges to the peak and then discharges into the load until the next peak arrives. The ripple is approximately V_r ≈ I_load/(f·C) — worse with more load current, better with more capacitance, better with a higher ripple frequency.

Design to the ripple trough, not the average. If the trough falls below a downstream regulator's dropout voltage, the output dips every cycle and the result is audible hum.

The cost of the capacitor

The diode conducts only while the input exceeds the capacitor's voltage — a narrow window near each peak. All the charge has to be delivered in that window, so the peak current is many times the average.

  • It sizes the diodes for a surge far above the DC output current.
  • It heats the transformer, because RMS current is much higher than the average.
  • It injects harmonics into the mains, which is why power-factor correction is mandatory above modest power levels.
  • Inrush at switch-on is worse still, since the capacitor starts empty — hence NTC thermistors and soft-start circuits.

Where the diode drops matter

A bridge puts two forward drops in the current path. At 12 V that is a 12% loss; at 5 V it is 28%. Schottky diodes roughly halve it, and synchronous rectification — MOSFETs turned on in place of diodes — nearly eliminates it, which is standard in modern low-voltage supplies.

A switch-mode supply avoids most of this by rectifying the mains first and then switching at 100 kHz, where the transformer and capacitors shrink by orders of magnitude.

The numbers you will be asked for

Half-wave average

V_avg = V_peak / π = 0.318 V_peak

Full-wave average

V_avg = 2V_peak / π = 0.637 V_peak

Ripple voltage

V_r ≈ I_load / (f · C)

Peak inverse voltage, bridge

PIV = V_peak

Peak inverse voltage, centre-tapped

PIV = 2 V_peak

Watch it work

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Why is a full-wave bridge so much easier to smooth than a half-wave rectifier?
Adding a smoothing capacitor makes the transformer run hotter. Why?
A 5 V supply is built with a bridge rectifier. What is the objection?
You have specified a smoothing capacitor from the average output voltage. What is likely to go wrong?

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4 still unanswered — the dots above jump straight to them.