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Old 30th December 2003, 08:43 PM   (permalink)
Default cut-off frequency

I am build a full-bridger single phase inverter that uses PWM to switch the transistors at 18kHz. I am using the following equation to come up with an LC filter:

fc = 1/(2*pi*sqrt(LC))

Does anyone know a range of cut-off frequencies that would eliminate most of the harmonics and still keep the 60 HZ fundamental frequency with minimal attenuation?
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Old 31st December 2003, 03:57 AM   (permalink)
Default

I don't think a resonant circuit is the way to go. I would use a series inductor and shunt capacitor. This is an example of the design process I would use:
Assume you have a 100 watt load and can tolerate 5% voltage drop in the filter, then Rload = E^2/P = 12100/100 = 121 ohms. 5% of that is 6 ohms = the reactance of the inductor. Now, L=Xl/2/pi/F = 6/6.28/60 = 16mH. The reactance of 16mH at 18kHz is: 6.28*18K*.016 = 1800 ohms. If you want to attenuate the 18kHz by a factor of 100, then the shunt capacitor reactance is: 18 ohms and C=1/2/pi/F/Xc = 1/6.28/18k/18 = .5uF.
The output voltage will rise 5% with no load, but you can't get something for nothing.
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