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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