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I'm trying to use a simple pi low pass filter using two capacitors and one inductor. Schematic Pi LC low pass filter

I've found the formula to calculate the value of the capacitance and of the inductance using the cut off frequency, but I would like to know where this formula comes from, I can't find any explanation on the internet. These are the formulas :

L     =     Zo / (2pi x Fc) Henries

 C     =     1 / (Zo x 2pi x Fc) Farads

 Fc     =    1 / (2pi x square root ( L x C) Hz

For example with a RC low pass filter I just use the voltage divider formula and from that, using the impedances, I find the transfer function of the filter. Then from the transfer function I'm able to find the cutoff frequency. Could be done like this also for the pi LC l.p.f? Thank you

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    \$\begingroup\$ What is important also is the schematic and the source resistance/impedance driving this filter. A small sketch would be helpful. \$\endgroup\$
    – Verbal Kint
    Commented Oct 3, 2017 at 10:49
  • \$\begingroup\$ I've posted the formula. I'm not interested in this specific case, but I would like to understand the demonstration of the formula \$\endgroup\$
    – Luca Daidone
    Commented Oct 3, 2017 at 11:26
  • \$\begingroup\$ Note: The formulas above assume the capacitors are equal. \$\endgroup\$
    – D.Deriso
    Commented Sep 2, 2021 at 20:21

1 Answer 1

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The impedance magnitude of a capacitor is:

    ZC = 1 / ωC

The impedance magnitude of a inductor is:

    ZL = ωL

The rolloff frequency of a LC filter is when these two are equal. Set the two equations above equal to each other, and solve for ω. Then remember that ω = 2Πf, where f is the frequency in Hz.

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  • \$\begingroup\$ Ok perfect, but why the rolloff frequency is when the impedances are equals? \$\endgroup\$
    – Luca Daidone
    Commented Oct 3, 2017 at 11:41
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    \$\begingroup\$ @Luca: Imagine a simple LC low pass filter. At low frequencies, the inductor is nearly a short and the capacitor nearly open. Small changes in either don't matter much. At high frequencies, the capacitor is short and inductor open. Again, small changes in either don't matter. The two interact and give a half-way result when their effects are about equal. Here, small changes in the inductance or capacitance do matter. \$\endgroup\$
    – Olin Lathrop
    Commented Oct 3, 2017 at 11:47

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