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Re: 7200 V too low? (reactance question)

Hi Nathan

This post is in response to your question about reactance and pigs.
Don't know what happened to the original, went to snip and the whole
thing just dissapeared!

 anyway, the thing is you are not going to have any reactance to match
a cap to with a pig. They are close enough to an ideal transformer to
treat as such for practical perposes. You will have to supply some
external current limiting to the pig to keep it under control. You can
them "match" _this_ current limiting to the tank cap, if you so
desire.

OK. You ask "just what is reactance anyway?" or something like that.
The reactance is just the impedance of a capacitor or an inductor at
some frequency expressed in Ohms. The reactance is a vector quantity,
and will be at 90deg to the resistance. In a cap the current leads
voltage, in an inductor current lags. When the capacitive reactance is
= to the inductive reactance at some particular frequency a condition
exists we call resonance.

You want to make the pig look like a NST to the spark gap. This can be
done by adding some resistance or inductance to the input to the pig.
The only parameter you will need from the pig for the calculation is
the turns ratio. If you input 240VAC and get out 7200VAC, the turns
ratio is 1:30. (7200/240=30) If you put in say 6 Amps to the 240 volt
winding, you will get (6/30= 200mA) at the 7200 volt winding.

So let us use a resistor for ballast and look at what happens. Let us
assume we wish a short circuit current of 300mA on the high voltage
windings. 30 X .3 = 9 Amps current draw by the 240V winding. 240V / 9
Amps = 26.66 Ohms.  So putting a 27 Ohm resistor in series with the
240V winding will limit the output current to about 300mA.

Now let us use an inductor for the ballast. Let us say we want the
short circuit current to be 150mA. 150mA X 30 = 4500mA or 4.5 Amps max
pri. current. So we need to find an inductor that will limit the
primary current to this value.
240V/4.5Amps = 53.33 Ohms. Inductive reactance is found by the formula
Xl = 2*pi*f*L. Solving for L, we get L = Xl/(2*pi*f). Plugging in
values. L = 53.33/(2*3.1416*60) = 141mH

Now let us match a ballast inductor to a tank cap for resonant
charging. Let us also pick a very small tank cap for a pig, lets say
10nf. Xc at 60Hz will be:
Xc = 1/(2*pi*f*C); Xc = 1/(2*3.1416*60*.00000001) = 265258 Ohms. Xl
needs to be equal to this value for resonance, but the inductor is on
the other side of the transformer, so we need to find Xl over there.
To find voltage or current on the other side of a transformer we
simply multiply or divide by the turns ratio. Impedance transforms as
the square of the turns ratio, however. So to find the required
reactance we divide 265258 Ohms by 30^2. 265258/900 = 295 Ohms. So L =
295/(2*3.1415*60) = 783 mH.
Now let us determine the short circuit current of the transformer with
295 Ohms of inductive reactance as a ballast. 240/295 = .814 Amps / 30
= 27mA
If we now plug in 7200V and .027 Amps in the trans cap match box in
Wintesla, we get a cap size of .0099uf, pretty darn close to the
.010uf we started with.

I would suggest using resistive ballast for your pig to start out
with. Get a heat element out of a furnace or water heater and measure
the resistance. A 5KW heat strip out of a typical furnace for 240V
operation is about 12.5 Ohms and will limit input current to 20 Amps
with the 7200V winding shorted. Maximum power will be delivered to the
load at 10 Amps input, 2.5KW. (This power limiting is only true with
resistive ballast, an inductor stores energy then gives it back, so
you can get more power out.) A water heater element is usually 4.5KW,
though they can be had in 3.5KW or 5.5KW as standard sizes. This would
make a nice jacobs ladder, and give you a feel for what is going on
with the ballast.

I have an old furnace that I use for resistive ballast. It has 4 heat
strips in it. I can hook it up in any way from all four in series to
all four in paralell for a range of resistive values from 50 Ohms to 3
Ohms. I also have a welder with an adjustable shunt and a 1.5 Ohm
resistor made from a single furnace heat strip cut and doubled so it
can take 40 Amps without getting too hot. (I don't need to have a fan
on it to keep it cool) I always start out with resistive ballast till
I get everything working well, then switch to the welder for the
higher power levels. Of course if you use a sync RSG, the phase angle
will have to be ajusted differently for resistive ballast than it is
for inductive ballast as the inductor changes the phase of the
current, the resistor doesn't.

Please remember that pigs can kill, no questions ask, so be carefull
when playing with them.
 Hope this helps some,
later
deano