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.
Skip to the animationA 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-wave | Full-wave bridge | Centre-tapped | |
|---|---|---|---|
| Diodes | 1 | 4 | 2 |
| Average output | 0.318 V_peak | 0.637 V_peak | 0.637 V_peak |
| Ripple frequency | 50 Hz | 100 Hz | 100 Hz |
| Forward drops in the path | 1 | 2 | 1 |
| Transformer | Ordinary | Ordinary | Needs 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
Check yourself
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