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Pi Filter Attenuation Calculator

Enter inductance, capacitance and ripple frequency to estimate attenuation.

Result

15.1 dB
50.3 Hz
2.38

How it is calculated

The tool treats the filter as an ideal second-order LC low-pass. Cutoff: fc = 1 / (2π·√(L·C)), with L in henries and C in farads (not mH or µF). Attenuation above the cutoff is A = 40·log10(f / fc) dB, which is the 40 dB per decade slope of two poles. If the ripple frequency f is at or below fc, the tool reports 0.0 dB rather than a negative value. It also shows the ratio f / fc. All three inputs must be above zero. The result is an asymptotic estimate: it ignores resonance peaking, component resistance (ESR) and load.

Worked example

A supply filter uses L = 0.01 H (10 mH) and C = 0.001 F (1000 µF) to clean up 120 Hz ripple. First, √(L·C) = √0.00001 = 0.003162, so fc = 1 / (2π × 0.003162) = 50.3 Hz. The ratio is 120 / 50.3 = 2.38. The attenuation is 40 × log10(2.38) = 40 × 0.377 = 15.1 dB. In other words, the ripple is reduced to roughly one sixth of its voltage.

Try this example →

Common mistakes

  • Typing millihenries or microfarads as if they were henries and farads. 10 mH must be 0.01 and 1000 µF must be 0.001, otherwise the cutoff is off by orders of magnitude.
  • Entering the mains frequency (50 or 60 Hz) instead of the ripple frequency. A full-wave rectifier produces ripple at twice the mains frequency, 100 or 120 Hz.
  • Choosing a cutoff above the ripple frequency and expecting some reduction. At or below the cutoff the tool shows 0.0 dB, so fc must sit well below the ripple.
  • Treating the dB figure as the exact result of a real circuit. It is the ideal rolloff; ESR, the load and resonance near fc can make the actual attenuation worse.
  • Reading 15 dB as a 15 times reduction. Voltage ratio is 10dB/20, so 15.1 dB is about 5.7 times.

About this calculator

A pi filter (named for its schematic resemblance to the Greek letter π) is an LC low-pass filter — typically capacitor, inductor, capacitor — used to smooth ripple out of a rectified power supply or to reject unwanted frequencies in RF circuits. This tool treats it as one inductor and one capacitor, a second-order (two-pole) filter; a full three-element section falls off faster. Beyond its cutoff frequency f_c = 1/(2π√(LC)) its attenuation rolls off at a standard 40 dB per decade (equivalently 12 dB per octave) — twice as steep as a single-pole RC filter.

This calculator takes the inductance, capacitance, and the ripple frequency you need to suppress, and estimates how many decibels of attenuation the pi filter provides at that frequency using the two-pole rolloff. Power-supply designers use exactly this kind of estimate to size L and C so that mains-frequency ripple (100/120 Hz after rectification) is knocked down to an acceptable residual level, and RF engineers use the same topology to filter unwanted harmonics from a signal path.

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Comments (1)

  • Editorial team

    Be sure of the ripple frequency before judging the result. With 0.01 H and 0.001 F, a 120 Hz ripple is cut by 15.1 dB, but 100 Hz, as from a full-wave rectifier on 50 Hz mains, gets only 11.9 dB. The tool models an ideal two-pole section, so treat the figure as a best case. What mains frequency and rectifier type are you designing around?

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