Power Factor Correction Capacitor

Sep 15, 2006 7 Replies

Hello- I am running an AC refrigeration compressor on a sailboat using an inverter. By determining the DC watts and the AC VA i figure a power factor of about .5 . I would like to correct this power factor, and have calculated that I need a 150 uF capacitor to do so. My question is, what kind of capacitor should I get with that capacitance that will handle 120 V AC? I appreciate the help. John



You need an AC rated capacitor, oil filled or polymer film capacitor or bank of paralleled capacitors.

But be warned that your inverter may not like having a power factor capacitor connected across it. Example of oil filled type:

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Example of film capacitors:
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As has been pointed out it would serve no useful purpose and may cause you additional problems. By the way, how did you arrive at the 150uF figure?

Thank you all for your help. First of all the reason I wanted to correct the problem was because I have both a DC Ammeter and an AC ammeter on the electrical panel, as well as an AmpHour battery monitor, and its just nice for them all to correspond if at all possible. But if there's a chance that it will hurt things to add a capacitor I definitely won't do it. The only thing I really care about is the DC Ammeter and the DC battery monitor. My question is are these DC readings still accurate as far as true amount of amps being drawn from the battery banks even when the inverter is putting out way more power AC? Right now to run the compressor I get about 45 amps DC and 9 amps AC... so should the 45 amps DC be accurate? Thanks again for your help.

The problem is that the inverter probably outputs a waveform with some fast rising and falling edges, rather than a smooth sine wave. Even if it did produce a perfect sine wave, there is a fast edge every time you turn it on, or connect the capacitor. The current through a capacitor is proportional to the rate of change of voltage across it. Fast edges imply large currents.

I have seen power factor capacitors added by having small inductances in series, to limit the peak current, but it is tricky to keep the LC resonance from being excited by any harmonic in the system (which would also produce large current at that harmonic frequency).

The DC meter should represent an accurate measure of load on the battery. AC power factor current represents energy borrowed and returned each half cycle, so it heats the inverter up a bit, as it sloshes back and forth through it, adding to its losses, but it doesn't change the average DC load much. It just adds ripple to it. But a DC meter should average that ripple and give an accurate average answer.

It is an accurate measure of battery drain. It might drop a few percent if you could eliminate the power factor current, because the inverter would be wasting a little less power because the borrowed and returned energy would be sloshing back and forth between motor and capacitor, instead of between motor and battery, through the inverter.

Old rule: "If it ain't broke, don't `fix' it". It's good advice.

Did you measure it ? There is tool called "cos fi" meter. But it does measure correct only sinusoidal shapes (as your AC voltmeters and ammeters) There is a coefficient of 4.44 to 4 (shape coefficient) between sinusoidal and rectangular waves. About power factor, if you have a regular energimeter in your house, and such a small cos fi, there is no compensation required because there will be no effect in the money you'll pay for your energy...

If you gave an active energimeter and also a reactive energimeter in your house (I have doubts you do) then it worth the trouble of mounting capacitors...

greetings, Vasile

I would like to correct this power factor,

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