I was going through a few buck converter questions in this forum to understand deeply about the circuit and I found this question. However, I am not focused too much on the ripple right now. But when I read the question, the question says the cut off frequency of the LC filter at the buck converter should be lower than the switching frequency.

Can someone explain me why should it be lower than the switching frequency? Would like to understand it more intuitively.

And whether the LC network contributes 90 degree phase shift to the control loop of the converter or 180 degrees? If LC network contributes 180 degrees phase shift, the internal op-amp of the IC would also contribute 180 degrees phase shift and hence we would have positive feedback of more than 360 degrees. So, I guess the LC network would contribute only 90 degree phase shift. Am I correct?


1 Answer 1


A buck converter can be conveniently modeled as a low-impedance square-wave generator feeding a \$LC\$ filter as shown in the below sketch:

enter image description here

Because you want a clean dc output voltage without too much ripple, you understand that you need to filter out the square wave. For that purpose, you need to set the corner frequency of the filter low enough to remove all the harmonics as this is the average value that you want. I have shown in my book on switching converters that you can choose the output ripple \$\Delta V\$ (solely based on the capacitance contribution, not the ESR) by applying the following formula in which \$f_0\$ represents the cut-off frequency of the \$LC\$ filter: \$\frac{\Delta V}{V_{out}}=\frac{\pi^2}{2}(\frac{f_0}{F_{SW}})^2(1-D)\$

Now, the \$LC\$ filter together with the resistive load is modeled by a 2nd-order transfer function. As such, in high frequencies, the phase responses approaches -180°. Should you now add the equivalent series resistance of the capacitor, the ESR, it creates a zero which brings the phase back to -90° after the zero.

enter image description here

See the above Bode plots where the ESR brings the phase back to -90° at higher frequencies. The ESR helps building a better phase margin while compensating the converter but having a highly-variable ESR also brings drawbacks as the compensation strategy must ensure stability in all cases. There are also cases where the output capacitors exhibit a very small ESR and, in this case, there is no phase boost in the crossover region.

Regarding compensation, you indeed cannot close the loop with such a phase lag and you need to tailor the compensator to create a so-called phase boost in the crossover region. See the below example for a voltage-mode-controlled buck converter:

enter image description here

You can see the compensation elements are calculated so that the phase goes up in the region of crossover to purposely compensate the phase deficiency brought by the \$LC\$ filter. Another option is to resort to current-mode control and, in this case, the low-frequency response of the buck becomes that of a 1st order. This example is part of the free 60+ SIMPLIS templates I released to illustrate my last book on transfer functions of switching converters.

  • \$\begingroup\$ Thank you for the answer. I was trying to absorb and understand the information that you have said. But I have few questions. 1. Could you explain me a little more on, "For that purpose, you need to set the corner frequency of the filter low enough to remove all the harmonics as this is the average value that you want." this statement of yours. I am trying to understand it more intuitively. And so, the LC filter adds 180deg phase lag, right? \$\endgroup\$
    – user220456
    Jun 21, 2021 at 14:24
  • \$\begingroup\$ Well, if you want to get rid of a switching frequency that is let's say at 100 kHz, you can set the corner frequency well below the frequency you want to remove otherwise you filter nothing. Does it make sense? You want to feed your load with a flat dc voltage, not a voltage affected by a lot of switching noise, right? Regarding the filter, yes, it adds a 180° lag but as the ESR of the cap. comes in, it may mitigate the response in the end (see my plots). \$\endgroup\$ Jun 21, 2021 at 15:08
  • \$\begingroup\$ thank you very much for the clarification. For the other question which I asked, since LC adds 180deg and the internal op-amp also adds 180deg, wouldnt there be +ve feedback in the control loop & cause oscillations? \$\endgroup\$
    – user220456
    Jun 21, 2021 at 18:31
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    \$\begingroup\$ Again, please read my explanation: the compensator adds phase lead at crossover - this is the phase boost - so that the total phase is away from the 360° limit. Please check my APEC 2019 dedicated to the buck converter and you will plenty of useful information on modeling. \$\endgroup\$ Jun 21, 2021 at 18:55

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