Rectifier Filter Capacitor Calculator (Ripple Voltage)

How it works

C = I ÷ (k × f × V_ripple) with k = 2 for full-wave, 1 for half-wave · V_peak = V_rms × √2 − diode drops (2 for a bridge) · V_dc(avg) ≈ V_peak − V_ripple ÷ 2 · ripple RMS ≈ V_ripple ÷ (2√3)

After rectification the reservoir capacitor charges to the peak voltage and then supplies the load alone until the next peak, discharging by ΔV = I × Δt ÷ C. With a full-wave rectifier the peaks come twice per mains cycle (100 or 120 Hz), so half the capacitance gives the same ripple as a half-wave circuit. The formula is a slightly conservative approximation that treats the discharge as lasting the whole period. Choose a capacitor with a voltage rating at least 25% above the peak and a ripple-current rating above the load current, and remember electrolytics are often −20%/+80% tolerance.

Worked example

A 12 V RMS transformer feeding a bridge rectifier on 50 Hz mains with a 1 A load and 1 V of allowed ripple: C = 1 ÷ (2 × 50 × 1) = 0.01 F = 10,000 µF. The peak DC voltage is 12 × 1.414 − 1.4 = 15.57 V, the average output is 15.07 V and the capacitor should be rated 25 V. Ripple RMS is about 0.289 V (1.92% ripple factor).

Sources: Wikipedia — Ripple (electrical): capacitor-input filter · Wikipedia — Rectifier (smoothing)

Frequently asked questions