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Can someone explain me intuitively on why the LC filter cut off frequency should be very less than the switching frequency of the buck converter?

What is the reason and why can't it be higher than the switching frequency?

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  • \$\begingroup\$ Such a filter will start attenuating at a certain frequency, the cut-off frequency is taken at -3dB, and it attenuates more the further frequency goes up. So it should be obvious, if the cut-off frequency is higher then the switching frequency then the filter won't attenuate at all #. It needs to be low enough so that the desired amount of attenuation is reached at the switching frequeny. (# ignoring harmonics for simplicity.) \$\endgroup\$ Commented Jun 21, 2021 at 14:43

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If you want to filter out the switching frequency and pass only DC, the LC filter must have a cutoff frequency below the switching frequency. If the LC cutoff is above the switching frequency, it will pass also the switching frequency.

As the LC filter is a second order filter, it has an attenuation of 40 dB per decade. So if the PWM runs at 40 kHz, having an LC filter at cutoff of 4 kHz attenuates the PWM frequency by 40 dB, so one hundredth of the ripple is present.

If you bring the LC filter cutoff down to 400 Hz, that is two decades below PWM frequency, so attenuation is 80 dB.

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    \$\begingroup\$ And the reason the cutoff should be much less than the PWM frequency is because you usually want a very high ripple attenuation. \$\endgroup\$
    – TimWescott
    Commented Jun 21, 2021 at 14:48
  • \$\begingroup\$ Thank you for the answer. I am just trying to understand. Isn't it, 20dB/decade or 40dB/decade? I have never heard 40dB/octave. Could you tell me how one hundredth ripple will be present in the example scenario that you mentioned, please? \$\endgroup\$
    – user220456
    Commented Jun 21, 2021 at 15:40
  • \$\begingroup\$ Yes I mixed up 40dB per decade and 12 dB per octave. Why a hundredth? 20×log(100) = 40dB. \$\endgroup\$
    – Justme
    Commented Jun 21, 2021 at 16:01

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