About this calculator
Rectifying AC gives you a series of humps, not DC. A large capacitor across the output fills in the gaps between them, holding the voltage up while the rectifier is not conducting. How large it needs to be depends on how much current you draw and how much ripple you can tolerate.
This calculator sizes that capacitor, or works backwards to the ripple a capacitor you already have will produce. It also shows the peak DC voltage, the required voltage rating, and the surprisingly large currents the diodes have to survive.
How it works
Start with the peak. A transformer's secondary is quoted in RMS volts, and the capacitor charges to the peak, which is √2 times higher — less the diode drops. A bridge rectifier puts two diodes in the path, so subtract about 1.4 V.
Between peaks the load discharges the capacitor. Treating the load as constant current gives a linear ramp down, and the amount it falls is the ripple: V_ripple = I / (f_ripple × C). Rearranged, that is the capacitor you need.
Note f_ripple, not the line frequency. A full-wave rectifier produces a hump for each half-cycle, so the gaps are half as long and the ripple frequency is twice the line frequency — 120 Hz on a 60 Hz supply. A half-wave rectifier gets no such help and needs roughly double the capacitance.
The consequence people miss is diode current. The capacitor only draws current near the peaks, in short tall spikes — the conduction angle might be 20% of each half-cycle, so the diodes carry several times the average DC current. That is also why a transformer feeding a capacitor-input filter needs to be rated well above the DC watts you are drawing.
Worked example
A 12 V RMS transformer, bridge rectifier, 60 Hz mains, feeding 500 mA. You want ripple under 1 V so a 7805 downstream stays out of dropout.
- V_peak = 12 × 1.414 − 1.4 = 15.57 V
- f_ripple = 60 × 2 = 120 Hz
- C = 0.5 / (120 × 1) = 4167 µF → use 4700 µF
- Actual ripple with 4700 µF: 0.5 / (120 × 4700e-6) = 0.89 V
- Valley voltage: 15.57 − 0.89 = 14.68 V — plenty above the 7 V a 7805 needs
- Capacitor rating: 1.5 × 15.57 ≈ 23 V → use a 25 V part
4700 µF at 25 V gives 0.89 V of ripple on a 15.57 V peak. The regulator downstream has abundant headroom — arguably too much, since every extra volt becomes heat in the 7805.
Practical notes
- Rate the capacitor for at least 1.5× the peak voltage. The peak appears at switch-on before any load is drawing, and mains voltage itself varies.
- Check the capacitor's ripple current rating, not just its voltage and capacitance. Ripple current heats the capacitor internally and is the usual cause of early electrolytic failure.
- Inrush at power-on can be tens of amps as an empty capacitor charges. Use diodes with adequate surge ratings, and consider an NTC inrush limiter on large supplies.
- More capacitance is not automatically better. It raises peak diode current and worsens inrush, for diminishing returns on ripple.
- The linear-discharge approximation used here is accurate to a few percent for ripple below about 20% of the peak. Past that it becomes optimistic.
- Mains wiring is dangerous. If you are not confident working with it, build from a pre-made supply module instead.
Frequently asked questions
How do I calculate a smoothing capacitor?
Use C = I / (f_ripple × V_ripple), where f_ripple is twice the line frequency for a full-wave rectifier. For 500 mA and 1 V of ripple at 120 Hz you need about 4200 µF.
Why is my DC voltage higher than the transformer rating?
Because the capacitor charges to the peak, not the RMS value. A 12 V RMS secondary peaks at about 17 V, and after two diode drops you still see around 15.6 V of DC.
What ripple voltage is acceptable?
If a linear regulator follows, whatever still keeps the ripple valley above the regulator's dropout — often 1–2 V is fine. For unregulated audio or analogue supplies, aim for well under 1% of the DC level.
Does a bigger capacitor always help?
Up to a point. Doubling the capacitance halves the ripple, but it also narrows the conduction angle, raising the peak current through the diodes and transformer. Past a few thousand microfarads you are usually better off regulating.
Why do the diodes get hot when average current is low?
They conduct only near the voltage peaks, for perhaps 20% of each half-cycle. To deliver 500 mA on average they must pass 2–3 A during those brief windows, and the losses scale with the square of the current.