Enter values and press Calculate.
After a rectifier, a large reservoir (smoothing) capacitor charges to the peak voltage and then supplies the load while the rectified waveform dips, discharging a little each cycle. That discharge is the peak-to-peak ripple voltage. Assuming the capacitor discharges at roughly constant current between peaks, the ripple is Vr = Iload/(n·fline·C), where n = 1 for a half-wave rectifier (one charge pulse per cycle) and n = 2 for a full-wave rectifier (two pulses per cycle).
| Quantity | Formula |
|---|---|
| Required capacitor | C = Iload / (n × fline × Vr) |
| Ripple voltage | Vr = Iload / (n × fline × C) |
| Ripple pulses / cycle (n) | 1 = half-wave, 2 = full-wave |
| Approx DC output | Vdc ≈ Vpeak − Vr/2 |
Because full-wave rectification recharges the capacitor twice as often, it needs only half the capacitance for the same ripple. This constant-current approximation is accurate for the small-ripple designs used in real supplies (ripple under ~10% of the peak).
Use C = Iload/(n·f·Vr), where Iload is the DC load current, f is the line frequency, Vr is the allowed peak-to-peak ripple, and n is 1 for half-wave or 2 for full-wave rectifiers.
A full-wave rectifier charges the capacitor twice per cycle instead of once, so the capacitor only has to hold the load for half as long between peaks. That halves the required capacitance for the same ripple.
A common rule is to keep ripple below about 10% of the DC voltage before a regulator, and much lower (tens of millivolts) if there is no regulator. Lower ripple simply needs a larger capacitor.
Yes. The capacitor charges to the peak voltage, so rate it at least 1.3–1.5× the peak (Vm=√2×Vrms) to allow for mains surges and tolerance.
Use your local mains frequency. The formula already accounts for full-wave doubling through the factor n, so you enter the line frequency (50 or 60 Hz), not the ripple frequency.
It is a constant-current approximation that is accurate when the ripple is small (under ~10% of the peak), which covers almost all practical designs. For very large ripple the true value is slightly different.
It raises the average output close to the peak voltage. The DC output is approximately Vpeak minus half the ripple, so a bigger capacitor gives a slightly higher and steadier DC level.
The ripple voltage rises, the DC output sags under load, and the rectifier draws higher peak currents in short bursts, which stresses the diodes and transformer and can cause hum in audio circuits.
Ripple falls, but the inrush current at power-on and the repetitive peak charging current rise, stressing the diodes and possibly the transformer. Very large capacitors may need an inrush limiter.
For clean, fixed output voltages, yes. The capacitor removes most ripple and a linear or switching regulator removes the rest and holds the voltage constant as the load or mains varies.
Yes. Paralleling capacitors adds their capacitance and lowers total ESR, which improves ripple performance and current handling. It is common in high-current supplies.
Yes, in practice. Besides the capacitance-based ripple, the capacitor ESR adds a ripple component equal to the peak charging current times the ESR. Use low-ESR capacitors for high-current or switching supplies.
Rectifier Average & RMS • Ripple & Form Factor • SMPS Output Ripple • All Calculators