Resonant Frequency of LC Circuit in Switching Converter

Question

Why do we set resonant frequency of LC circuit in a buck converter to be less than 10% of Fsw? For example, if Fsw is 300 kHz, why should resonant frequency be around 30KHZ. I am guessing it is due to voltage spikes at resonance, but can't it be just twice, if that's the case?

Answer 1

Let's consider a synchronous buck converter. It produces a square wave of (usually) fixed frequency and variable duty cycle to compensate against changes of supply voltage and changes in load current.

On it's output is a series inductor feeding a capacitor tied to ground. These components form a low pass filter. Let's say the inductor is 10 uH and the capacitor is 22 uF. Let's also say there is no-load (or just a light-load) on the output of the switcher.

If you did a bode plot of the filter you would see this: -

What if the switching mechanism where aligned to 10 kHz (instead of say 300 kHz)? Answer you'd get a massive sine wave superimposed on top of the DC output that the switcher (and load circuit) is not equipped to cope with. You'd also get substantial dc current taken by the switcher because, at resonance the LC network acts as a short with only the R part offering a non-zero impedance of 0.1 ohm on this example.

If the load current was larger you'd find that the Q of the LC low pass filter would substantially reduce and this could be livable-with but there is no significant benefit designing the filter frequency to match the switching frequency.

Here is the site that the LCR calculator was on

Answer 2

The buck converter generates a spectrum of sidebands and harmonics (perturbation frequency, switching frequency plus and minus perturbation frequency, etc.) If the controller 'sees' those higher harmonics, you'll get aliasing in the control output - the controller will be influenced by the higher harmonic content and you can see things like a strong low-frequency response to a high frequency perturbation.

The low-pass characteristic of the output LC filter will strongly attenuate those higher frequency components. The 10:1 rule will mean that the filter will effectively squish those higher harmonics. It may be possible to push the two closer together, but you need to be aware of the sideband issue and its ramifications.