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Rectifier Filter Capacitor Calculator

Size the smoothing capacitor after a rectifier for a given load current and ripple, and find the DC output voltage.

Smoothing capacitorRipple voltageBridge rectifierDC output
V rms
V
V
Minimum capacitance5,000 µF
Use standard value6,800 µF or larger
Capacitor voltage rating25 V or higher
Peak DC voltage15.57 V
Average DC voltage14.57 V
Ripple frequency100 Hz
How this result was calculated
  1. Vpeak = 12 × √2 − 2 × 0.7 = 15.57 V
  2. C = I ÷ (fripple × ΔV) = 1 A ÷ (100 × 2) = 5,000 µF
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Smoothing Capacitor Formula

C = Iload ÷ (fripple × ΔVpp)
fripple = f (half wave) or 2f (full wave)
Vpeak = √2 × Vrms − diode drops

A bridge has two diodes in the current path, a centre tap rectifier one. The capacitor rating allows 25 % above the nominal peak for mains variation and light load.

Worked example

12 V rms, 50 Hz bridge, 1 A load, 2 V ripple: C = 1 ÷ (100 × 2) = 5000 µF, so fit 6800 µF (25 V). Vpeak = 16.97 − 1.4 = 15.6 V; average about 14.6 V.

Key insight: a bigger capacitor lowers ripple but raises the peak charging current in the diodes and transformer, so do not oversize it by more than you need.
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Frequently Asked Questions

How do I calculate a smoothing capacitor?

C = I ÷ (f × ΔV), with f twice the mains frequency for a full wave rectifier.

What voltage rating should the filter capacitor have?

At least 1.25 × the peak secondary voltage; for 12 V rms that means 25 V.

Why does half wave need twice the capacitance?

The capacitor is recharged only once per cycle instead of twice.

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