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Re: Quick questions about PFC calculations
Original poster: "Gerry Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
Hi Mike,
I think what you calculated is the C where its XC = XL of the ballast
(referenced to the primary side) and is probably an OK start
point. The one thing not taken into account is the load on the
secondary that's seen by the primary and is in series with this
ballast. The answer to this question is sometimes difficult and
depends on whether you have a static spark gap, a SRSG, or an
ARSG. If you have a kill-a-watt meter ($30 investment), you can
measure the PF, Line voltage, and Line current directly during
operation and compute the exact capacitance to use.
Gerry R.
Original poster: Mike <megavolts61@xxxxxxxxx>
Hi all,
I've never worried about power factor correction in a NST
powered coil, but I got this great lil 9/30 (Thanks again Hal) and
thought it might be cool to run it from my car via a 400W inverter
that I have. I know the inverter would not like an inductive
load. The instruction said I should probably run a resistive load
at the same time if it acted up, but I would think power factor
correction would be the way to go.
My question is: Did I calculate this cap correctly? First I
get 270VA/120V = 2.25 A, then dividing 120 by 2.25 and get 53.333 ohms.
This is the total impedance of the load presented to the line.
Now measure the resistance across the primary I got 2.2 ohm
so I assumed the impedance due to the inductance is 51.1 ohm.
Not quite. XL does not add linearly to resistance and you need both
the resistance of the primary and secondary if you want to factor the
resistance out.
R = Rp + Rs / n^2 where n is the turns ratio [9000/120 = 75].
XL = sqrt (53.3^2 - R^2) = 53.25 ohms (using your 2.2 ohms)
You can think of reactance as being at right angles to resistance
when you combine them into a total impedance (Ztotal^2 = R^2 +
X^2, Pythagorean Theorem). You can see that the resistance is a
small factor and you can probably neglect it..
XL = 53.25 ohms = XC for parallel resonance.
XC = 1/(2*pi*60*C) so C = 1/(2*pi*60*XC) which is what I think you meant.
C = 49.8 uf
So yes, you are on track. If you can measure the actual PF you will
know what the true load is counting for the effect of the secondary.
Gerry R.