Rectifier Filter Capacitor Calculator
Size the smoothing capacitor after a rectifier for a given load current and ripple, and find the DC output voltage.
- Vpeak = 12 × √2 − 2 × 0.7 = 15.57 V
- C = I ÷ (fripple × ΔV) = 1 A ÷ (100 × 2) = 5,000 µF
Smoothing Capacitor Formula
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.
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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