Ripple voltage and the smoothing capacitor: sizing the reservoir after a rectifier
A rectifier turns AC into a string of humps; the reservoir (smoothing) capacitor fills in the gaps. It charges to the crest of each hump and the load drains it until the next, so the output is the peak less a sawtooth of ripple: Vpp ≈ I / (fripple × C), with fripple twice the mains frequency for a bridge or centre-tap and equal to it for half-wave. This calculator finds the peak after the diode drops, the ripple, the DC and the minimum for your load and capacitor, designs the capacitor for a target ripple, and works out what few people check: the short, heavy bursts in which the diodes recharge the capacitor, the ripple current the capacitor must carry, the step and heat its ESR adds, and the voltage ratings the parts need.
How to use the ripple calculator
- Type the transformer's secondary voltage (RMS, as printed on it), the mains frequency and the load current. Pick the rectifier in the menu row: bridge, centre-tap or half-wave.
- Type the capacitor to check it, or type the ripple you want and leave the capacitor blank: it is designed and rounded up to the next standard value.
- Add the capacitor's ESR for the step and the heat, and change the diode drop for Schottky parts. The drawing shows the capacitor's voltage riding the rectified humps with the peak, DC and minimum marked; the Capacitor sizes tab compares the standard values.
The peak
Vpeak = VRMS × √2 − n × Vdiode
A 12 V secondary peaks at 17 V; through a bridge (two diodes in the path, about 0.7–1 V each at the charging peaks) the capacitor reaches about 15.6 V. The transformer's voltage is its full-load figure; unloaded it is 5–10 % higher, which is the voltage the capacitor and the regulator see at switch-on. So "12 V" becomes a 15 V DC supply, and a "9 V" transformer with a bridge and capacitor gives about 11.5 V, which is why a 7805 is happy on it.
The ripple formula
Vpp ≈ I / (fripple × C) C = I / (fripple × Vpp)
Between crests the capacitor supplies the load alone and its voltage falls by I × t / C; taking t as the whole ripple period (10 ms at 50 Hz full-wave, 8.3 ms at 60 Hz, double for half-wave) gives the peak-to-peak ripple, slightly pessimistically because the recharge takes part of the period. The DC is about the peak less half the ripple, the minimum the peak less all of it, and the sawtooth's RMS is Vpp / 2√3. Turned round, the formula sizes the capacitor: 500 mA at 100 Hz with 1 V of ripple needs 5000 µF. A rule of thumb for a supply feeding a regulator is 2000 µF per amp for about 10 % ripple at 50 Hz.
Conduction bursts and ripple current
cos θ = 1 − Vpp/Vpeak Icharge ≈ I / share IC,RMS ≈ I × √(1/share − 1)
The diodes conduct only while the secondary is above the capacitor's voltage, a short angle θ either side of each crest, typically 10–20 % of the period. All the charge the load drew over the whole period has to be replaced in that window, so the charging current is five to ten times the load current: short, heavy pulses that the diodes, the transformer and the capacitor all carry. The capacitor sees them as ripple current, whose RMS is the figure on its data sheet that limits its life; a bigger capacitor makes the ripple smaller but the bursts sharper, so the ripple current barely falls.
ESR
An electrolytic's equivalent series resistance, tens of milliohms for a few thousand microfarads, does two things. Each charging burst drops Icharge × ESR across it, a fast step on top of the sawtooth that a bigger capacitor does not remove; and the ripple current heats it by IC,RMS² × ESR, inside a sealed can, which is what dries the electrolyte and ends the capacitor's life. Low-ESR, 105 °C parts, or two capacitors in parallel, are the answer where the ripple current is high.
Ratings
The capacitor's voltage rating must exceed the no-load peak with a margin: 25 V for a "12 V" transformer, not 16 V. Its ripple-current rating must cover the figure above at the ripple frequency. The diodes need an average current rating above the load, a surge rating for the switch-on inrush (the empty capacitor looks like a short), and a reverse voltage above the secondary peak, twice that for a centre-tap, with a margin: 1N4007 or a 1.5 A bridge for small supplies, Schottky where the drops matter.
Your supply, step by step
- Peak: 12 V × √2 − 2 × 700 mV = 15.6 V.
- Ripple: 500 mA / (100 Hz × 2200 µF) = 2.27 V (16%); DC 14.4 V, minimum 13.3 V.
- Bursts: the diodes conduct 1.74 ms per pulse (17%) at about 2.87 A; ripple current 1.09 A RMS.
- Ratings: capacitor ≥ 21.2 V; diodes 700 mW, PIV ≥ 25.5 V.
Worked example: 12 V secondary, bridge, 500 mA, 2200 µF
The secondary peaks at 12 × √2 = 17 V; after the bridge's two drops the capacitor reaches 15.6 V. At 50 Hz a bridge gives 100 Hz ripple, so with 500 mA and 2200 µF the ripple is 0.5 / (100 × 0.0022) = 2.27 V peak-to-peak, about 16%: the DC is 14.4 V and the trough 13.3 V, enough headroom for a 7812 (which needs about 14.5 V) only just, and plenty for a 7809. The diodes conduct for 1.74 ms of each 10 ms pulse (17%), in bursts of about 2.87 A, and the capacitor carries 1.09 A RMS of ripple current: it should be a 25 V part (no-load peak 17 V plus margin) rated for at least that current. To bring the ripple down to 1 V the capacitor would need to be 0.5 / (100 × 1) = 5000 µF, so 6800 µF.
Questions
Why is my "12 V" supply 16 V?
Because the capacitor charges to the peak, not the RMS: 12 × √2 = 17 V less the diodes, and more at light load when the transformer runs high.
Does a bigger capacitor always help?
It lowers the ripple voltage, but the charging bursts get shorter and sharper: the ripple current, the inrush and the diode stress rise. Past a few thousand µF per amp, use a regulator for a clean output.
Why 100 Hz ripple on a 50 Hz supply?
A bridge or centre-tap rectifier uses both halves of the cycle, so there are two charging pulses per cycle. Half-wave uses one, and its ripple is at 50 Hz and twice as large.
What ripple is acceptable?
For a linear regulator, anything that keeps the trough above its dropout: 10–20 % is common. For a bare supply feeding audio or sensors, a few percent, or add a regulator or a capacitance multiplier.