LC Filter Voltage Drop Considerations

In summary: In this case, you would need a much, much bigger inductor and it may be difficult to get 100 amps out of it. At 20 volts rms, the filter will dissipate 0.395 amps.
  • #1
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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,

Dale
 
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  • #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.
 
  • #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.
 

1. What is an LC filter?

An LC filter is an electronic circuit that uses inductors (L) and capacitors (C) to filter out specific frequencies from an input signal. It is used to smooth out fluctuations in voltage and reduce noise in a circuit.

2. How does an LC filter work?

An LC filter works by creating a low impedance path for certain frequencies and a high impedance path for others. The inductor and capacitor work together to create a resonant circuit that can filter out unwanted frequencies.

3. Why is voltage drop a consideration for LC filters?

Voltage drop is a consideration for LC filters because the inductor and capacitor can cause a drop in voltage across the filter. This can affect the overall performance of the circuit and may need to be taken into account when designing the filter.

4. How can I calculate the voltage drop in an LC filter?

The voltage drop in an LC filter can be calculated using the formula Vr = Vm * √(1 - (f/f0)^2), where Vr is the voltage drop, Vm is the maximum input voltage, f is the frequency of the input signal, and f0 is the resonant frequency of the filter.

5. What are some ways to minimize voltage drop in an LC filter?

There are a few ways to minimize voltage drop in an LC filter, such as using larger inductors and capacitors with lower resistance, using multiple stages of filtering, and choosing a lower resonant frequency. It is also important to ensure proper placement and grounding of the filter components.

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