PSU Ripple Calculator — reservoir capacitor sizing

Smoothing capacitor size for a target ripple voltage after rectification.

Example: 4.7 mF at 1 A gives about 2.128 V of ripple at 100 Hz.

Inputs

Load current(A)

Supply frequency(Hz)

Rectifier

Reservoir capacitance(F)

4700u is 4700 µF. Leave blank to size it from a target ripple instead.

Target ripple (optional)(V)

Used only when capacitance is left blank.

Rectified peak voltage(V)

The transformer secondary RMS × 1.414, minus diode drops.

Result

Ripple voltage

2.128 V

Capacitancegiven: 4.7 mF
Ripple frequency: 100 Hz
Peak diode current: 6.213 A
Capacitor RMS current: 2.283 A
Ripple as % of peak: 12.52%
Minimum DC voltage: 14.87 V

Diodes conduct for only a fraction of each cycle, so the peak charging current is around 6.213 A — far above the 1 A average. Rectifier surge rating and transformer heating both depend on that peak.
The capacitor carries about 2.283 A RMS of ripple current continuously. Electrolytic lifetime is dominated by ripple-current heating — check the part's ripple rating, not just its capacitance and voltage.

Ripple against capacitance at 1 A

Capacitance	Ripple	% of peak
1 mF	10 V	58.8%
2.2 mF	4.545 V	26.7%
4.7 mF	2.128 V	12.5%
10 mF	1 V	5.9%
22 mF	454.5 mV	2.7%

Formula

V_ripple ≈ I_load / (f_ripple × C)

Worked example

4.7 mF at 1 A gives about 2.128 V of ripple at 100 Hz.

f_ripple = f_supply × pulses per cycle
50 Hz × 2
100 Hz
V_ripple ≈ I_load / (f_ripple × C)
1 A / (100 Hz × 4.7 mF)
2.128 V
Assumes the capacitor discharges linearly for the whole cycle — slightly pessimistic, which is the safe direction.

Frequently asked questions

What size smoothing capacitor do I need?

C = I_load / (f_ripple × V_ripple). For 1 A with 1 V of ripple on a 50 Hz full-wave supply, f_ripple is 100 Hz, so you need 10,000 µF. Reservoir capacitors are large for exactly this reason.

Why is ripple frequency twice the mains frequency?

A full-wave rectifier produces two charging pulses per mains cycle — one from each half. That doubles the ripple frequency and halves the time the capacitor has to discharge, which is why full-wave needs half the capacitance of half-wave.

Why do the rectifier diodes see such high current?

Because they only conduct near the peak of each cycle, while the capacitor is being topped up. All the charge the load drew over the whole cycle has to be replaced in that short window, so the peak is many times the average. Rectifier surge ratings exist for this.

Why do reservoir capacitors fail?

Ripple-current heating. The capacitor passes substantial RMS current continuously, and that heats it from the inside, drying out the electrolyte. A capacitor within its voltage and capacitance ratings but over its ripple-current rating will still fail early.

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