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LC Filter Voltage Drop Considerations

  1. Aug 24, 2013 #1
    Hi, I've gotten a little confused and would like some guidance and confirmation,

    I've designed an H-Bridge inverter which should output 100A at 20Vrms onto a 200mΩ load. The current target output peak voltage (with max duty cycle of 98% considered) is 28.86V. The LC filter is made up of a 1500μF Cap and a 60μH L with a Fc of 531Hz, the fundamental is 50Hz, switching is 16KHz.

    I'm trying to calculate what compensations (if any) will I need to make to the output peak voltage from the H-Bridge for losses in the filter to ensure 20Vrms is outputted from the filter.

    I've found the following calculation but I'm a little confused from its results,

    Xc=1/2*pi()*f*C= 1/2*pi()*50*1.5E-3 = 2.4485Ω
    Xl=2*pi()*f*L = 2*pi()*50*60E-6 = 53.0516Ω
    Z=sqrt(R^2+(Xl-Xc)^2) = sqrt(0.2^2+(53-2.4)^2) = 50.6Ω
    I=20Vrms/50.6 = 0.395A
    Vlc=I*(Xl-Xc) = 0.395*(53-2.4) = 19.99V

    1) Should the frequency in the reactance calcs be the fundamental or the switching?
    2) Is I in this case the current through the LC network?
    3) Would this result indicate that no or little alteration would be needed to the peak voltage for switching as the voltage across the network is the output voltage?

    Many Thanks in advance,

  2. jcsd
  3. Aug 24, 2013 #2
    Xc = 1/(2*pi()*f*C) - Don't forget the parentheses. I get 2.122.
    Xl = 2*pi()*f*L - I get 0.01885, how did you get 53.0516?
    It looks like you plan to put both components in series with the load. That's fine but ideally you would want both reactances to be equal in magnitude but opposite in sign (conjugates) at 50 Hz and each value should be at least 10 times the value of the load. I think you will need a much, much bigger inductor and at 100 amps that may be a problem.
  4. Aug 24, 2013 #3
    Xc = 1/(2*pi()*f*C) - Don't forget the parenthesis. The value I get is 2.122066.
    Xl = 2*pi()*f*L - The value I get is 0.018850. Are you sure you meant 60 microHenries?

    It looks like you intend to put the capacitor and inductor in series with the load. I suggest you pick values that are equal in value but opposite in sign (conjugates) at 50 Hz with each component having a reactance at least 10 times the value of the load.
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