Brad,From what I understand you can achieve an extent of pf correction. The LC tank acts as a bandstop network. Select the inductive Henry value so it creates resonance (at 60 Hz) with the capacitive load. This solution works the same way as in the conventional case of an inductive load (motor, etc.) The result is a bandstop filter in either case.
Simulation below shows raw capacitive load causes Ampere waveform out of alignment with Voltage waveform. Parallel inductor pulls Ampere waveform back to coincide with Voltage waveform.
The lissajous scope figures at bottom make this clear. The left-hand shows extreme phase difference. The right-hand shows phase alignment.
View attachment 192764
Not seeing a problem is some way from there not being a problem, if you had worked in installing similar PFC - you would know there are issues.I don't see a resonance problem, LC resonator is shorted by grid impedance.
Well said Easy Peasy ha ha, if you have experience, any insight into my latest response? I would appreciate it, this is my first go at power factor correction, like most things it is not as easy in practice.Not seeing a problem is some way from there not being a problem, if you had worked in installing similar PFC - you would know there are issues.
Capacitive impedance is governed by the formula:Brad,
Thank you for your incredibly thorough explanation, it reassures me that my grasp of the concept isn't as far off as I feared. Not only do I agree with your calculations, but I also actually ordered a 13 (12.8) mH inductor and installed it in my circuit, the total current was 20 amps and since I had taps on the inductor I was able to observe the total current rise if I increased or decreased the inductance by 1 mH. This is where my confusion lies, is a total current lower than 20 amps (your simulated 24.4) possible? The intention was to try to have around 6 amps total current as 60 amps is present without the correction and no load, this is the current draw we are trying to greatly reduce.
Example:I thought the shunt inductor, creating a tank circuit, could reduce the total current to something lower than 20 amps
- that's because the reasons have not been given here - but just to assist you - imagine the L & C have very low R and sit at a resonance freq of 60Hz - what will be the current in each ? - what will be the external current to the parallel pair ?I didn't yet hear substantial considerations why a parallel PFC inductor needs extra damping due to resonance effect.
What is confusing me FvM, is I built the circuit, with a 12.8 mH inductor, in parallel with a capacitor bank of 550 uF, and I am measuring 60 amps on the cap bank line, 60 amps on the inductor line, and 20 amps on the input current before the node.The thread isn't about cap current. 550 uF filter caps are already there and the reason why the OP is looking for compensation, review post #1. I didn't say the caps won't possibly cause problems, I stated there's no particularly resonance problem when adding a parallel PFC inductor.
That's the inexplicable thing about power factor. I had to dig into detailed articles before getting any grasp of its nature. It's not that the power company's generators need to work any harder or burn an abnormal amount of coal. However they need to (or would need to) install thicker wiring everywhere in order to carry the extra Amperes.my confusion is why it still reads 20 amps. 60 amps across the capacitor, 60 amps across the inductor, but 20 amps on the input and ground, total current.
do you understand how vector arithmetics or vector drawing works?What is confusing me FvM, is I built the circuit, with a 12.8 mH inductor, in parallel with a capacitor bank of 550 uF, and I am measuring 60 amps on the cap bank line, 60 amps on the inductor line, and 20 amps on the input current before the node.
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