The question I have is......Why? Why would you want to lower the PF?
On Sat, Dec 3, 2016 at 5:51 PM, Gary Lau <glau1024@xxxxxxxxx> wrote:
> Hi Ron,
>
> I remember trying to meter the mains current with my coil running and try
> various PFC cap values, looking for the lowest drawn current.
> Unfortunately, particularly when using a static spark gap, the mains
> current varies wildly from moment to moment. I was utterly unable to
> experimentally determine if one value was better than another, although it
> was clear that current with a PFC cap was lower than without. I think the
> takeaway is that the optimal value is by no means critical, as if you were
> trying to hit some kind of resonance.
>
> Regards, Gary Lau
> MA, USA
>
> On Sat, Dec 3, 2016 at 5:32 PM, Ronald Reeland <ronreeland@xxxxxxx> wrote:
>
> > Hi:
> >
> > I have been researching adding an A.C. power factor correction capacitor
> > across the 120 volt primary of a 15,000 volt, 30 milliamp secondary neon
> > sign transformer for a spark gap Tesla coil application.
> >
> >
> > It appears that the power factor of an ordinary transformer without a
> > built-in correction capacitor is 40 to 50 %. That means that the input
> > volt/amps is around twice that of a "perfect" power factor of 100%. Of
> > course we can never achieve a perfect 100%.
> >
> >
> > But it appears a person can attain a 90% or greater power factor thus
> > reducing the primary input current draw by calculating an approximate
> value
> > of correction capacitor.
> >
> >
> > Here is a power factor correction capacitor formula that I found on the
> > web and in Brent Turner's book; "The Tesla Coil Book", how they work &
> how
> > they are built" :
> >
> > PF Capacitor= "corrected kVA" x (10^9 divided by 2TT x Frequency x
> > primary volts^2) or in a more compact form : C=kVA (10^9/2TT f e^2).
> >
> >
> > There is a partial chart in Brent's book titled" PF-Corrected Transformer
> > ratings" and complete charts on the web for various transformer output
> > volts and current. It appears that the "corrected" output kVA is one half
> > in some instances and 55% or so in other cases.So if I have computed the
> > formula properly, a 15,000 volt, .030 amp transformer requires a 46 ufd
> > A.C. capacitor across the transformer primary. This is based on the
> charts
> > "corrected kVA" of .250. (The un-corrected kVA would be 15,000 x .030 or
> > .450 kVA.)
> >
> >
> > 1. I am asking for confirmation that all of above is true and that I
> > manipulated the formula properly.
> >
> > 2. I assume that since the formula deals with secondary output, the
> > lowered kVA is reflected in the 120 volt primary input and the current
> draw
> > there is reduced in proportion.
> >
> > 3. Also, is the formula a starting point for adjusting the capacitor
> value
> > higher or lower with an A.C. ammeter in the primary until the lowest
> input
> > amperage is achieved?
> >
> > 4. I also assume the secondary must be under load while fine tuning the
> > power factor capacitor value.
> >
> > 5. I have some 56 ufd, 250 volts A.C. motor-run capacitors (EPCOS brand)
> > that I would like to try.
> >
> >
> > Any help or corrections to my assumptions would be greatly appreciated.
> >
> >
> > Thanks,
> >
> > Ron Reeland
> >
> >
> >
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> >
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--
Chris Boden
President
The Geek Group National Science Center
www.thegeekgroup.org
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