0
\$\begingroup\$

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 ?enter image description here

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.

Your helpful suggestions and comments would be appreciated.

Thankyou!

\$\endgroup\$
  • \$\begingroup\$ A schematic would be helpful here. \$\endgroup\$ – AaronD Nov 25 '15 at 18:05
  • \$\begingroup\$ i just added a rough sketch \$\endgroup\$ – yiipmann Nov 25 '15 at 18:12
  • \$\begingroup\$ i added explanation to why i am asking this question, may be now it is more clear \$\endgroup\$ – yiipmann Nov 25 '15 at 18:29
  • \$\begingroup\$ 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. \$\endgroup\$ – SunnyBoyNY Nov 25 '15 at 19:04
  • \$\begingroup\$ "bandwidth is greater than 0.5*pi*sqrt(LC)" - what does this mean? \$\endgroup\$ – Andy aka Nov 25 '15 at 22:16
1
\$\begingroup\$

the controller should be designed in such a way that the controller bandwidth is higher than the circuit's self-oscillation frequency.

Well, for a 1 MHz switching converter (as an example), the LC circuit will have a resonant frequency that is probably no more than one-tenth of the switching frequency and quite possibly this could be one hundredth i.e. 10kHz.

Here's an example: -

enter image description here

I know it's not a straightforward buck regulator but it's still using a 6.8uH inductor and a 22uF output cap. Resonant frequency is 13 kHz and ripple should be pretty good.

So, to answer your question directly, I think you are getting muddled up!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.