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UCC28742: Filter inductor value in design tool

Brian Antheunisse
Expert 5305 points
Part Number: UCC28742


Hey guys,

Customer is using the design tool to design their UCC28742 schematic, but they aren't sure what to size this inductor:

since the CB1 and CB2 calculations on the excel spreadsheet are present, why is there nothing mentioned about the unnamed inductor?

Thanks,

Brian

  • 0
    TI__Guru* 86286 points
    Hello Brian,

    That is a differential mode inductor and will attenuate the high frequency ripple current from CB2 from going back on the line. I am not sure why the applications engineer who developed this tool did not include it.

    This is just a double power filter fp=1/((2*3.14*fsw*10)^2*(L*CB1)^0.5. I would recommend setting L to put a pole at 1/10th the switching frequency. L=1/((2*3.14*(fsw/10))^2*CB2).

    The current going through this inductor should be the low frequency RMS current that should be calculated by the excel tool. If not let me know because I know there are several application notes that show how to calculate this current.

    Regards,

    Mike
  • 0 in reply to Mike O'
    TI__Expert 5305 points
    Can you post the app notes you were thinking of?

    Thanks!
    Brian
  • 0 in reply to Brian Antheunisse
    TI__Guru* 86286 points
    Hello Brian,

    The following link will get you to an application note discussing how to select the input capacitance for flyback. It goes through calculations for the low frequency and high frequency RMS currents. www.ti.com/.../slua908

    Regards,

    Mike
  • 0 in reply to Mike O'
    TI__Expert 5305 points

    The input capacitance actually was calculated on the spreadsheet calculator tool.

    with regards to your equation above, Is the above calculation sufficient for CB1 and CB2?  Do I still need to do the calculations on the low freq and high freq currents?

    Thanks,

    Brian 

  • 0 in reply to Brian Antheunisse
    TI__Expert 5305 points
    I never got an answer here, let me elaborate on my concerns:

    Why is the Fsw divided by factor 10 to determine the Inductor value?
    When I plug the capacitor and inductor calculated values in, I get a resonant frequency in the MHz range.
    Shouldn't the resonant frequency of the LC combination be within the operating switching range of the UCC chip?


    For the UCC28742 the max switching frequency is 84.755kHz.
    So is this the value you are specifying to divide by 10 ?

    thanks,
    brian
  • 0 in reply to Brian Antheunisse
    TI__Guru* 86286 points
    Hello Brian,

    Sorry this was missed I will look into now for you.

    CB1 will have half of the low frequency and no high frequency current.

    CB2 will have all of the high frequency current and half of the low frequency current.

    The following application not will discuss how to calculate the current in these capacitors and how to select their value as well. www.ti.com/.../slua908.pdf

    Setting the pole of the PI filter to 1/10th of the maximum frequency was done to block the high frequency ripple voltage from the line seen at VBULK. This is a double pole filter so you if put a pole at 1/10th the maximum switching frequency you will have a 40 dB reduction in ripple voltage seem at CB1.

    Regards,

    Mike