# Self resonant frequency of buck converter

I know that one can calculate the self-resonant frequency for an LC filter of a simple buck converter(having one inductor and one capacitor at the output) by the formula 0.5*pi*sqrt(LC). My question is if we add another inductor at the output of the converter following the capacitor, forming T type LC filter, will the self resonant frequency for this circuit remain the same ? Or will it change ? What i mean to ask is that the self resonant frequency of circuit (a) and circuit (b) would be the same ? The reason for asking this question is that i am supposed to design a current controller for my buck converter. And the controller should be designed in such a way that the controller bandwidth is higher than the circuit's self-oscillation frequency. So initially i had no T type filter, just a simple circuit like (a), and i designed a current controller whose bandwidth is greater than 0.5*pi*sqrt(LC), but later i added another inductor to filter out the current more (which is being supplied to the load). So now i am not sure if my circuit filters self resonating frequency would be same or not (which means my old controller design would work fine or not)

P.S my question may be very basic or stupid, but i dont have detail idea of power electronic components, so i posted the question as i couldnt find anything regarding this on the internet.

Thankyou!

• A schematic would be helpful here. – AaronD Nov 25 '15 at 18:05
• i just added a rough sketch – yiipmann Nov 25 '15 at 18:12
• i added explanation to why i am asking this question, may be now it is more clear – yiipmann Nov 25 '15 at 18:29
• The LC resonant frequency is of little consequence for the buck converter since it is actively modulated and the controls adds poles/zeros to the overall transfer functions. Open-loop modeling as you have depicted it is not widely used. – SunnyBoyNY Nov 25 '15 at 19:04
• "bandwidth is greater than 0.5*pi*sqrt(LC)" - what does this mean? – Andy aka Nov 25 '15 at 22:16 