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Re: Tank Capacitance: what is the limit
Subject: Re: Tank Capacitance: what is the limit
Date: Thu, 24 Apr 1997 16:59:59 -0700 (PDT)
From: "Edward V. Phillips" <ed-at-alumni.caltech.edu>
To: tesla-at-poodle.pupman-dot-com
CAREFUL: THIS ISN'T THE WHOLE STORY!!!
"When dealing with AC components, notably capacitors and inductors, an
interesting phenomena appears known as reactance. Reactance is a kind of
"apparent" resistance and is measured in ohms just like resistance.
Reactance, however, is frequency dependent unlike your straight forward
resistors.
If I remember correctly, it is calculated this way:
1
Xc = ---------------- Eq #1
2 * Pi * Hz * C
Where,
Pi = 3.14
Hz = frequency in Hertz
C = Capacitance in Farads
Xc = Reactance in Ohms
* means multiplication
To make sure were speaking the same language.
If you have a .01 microfarad capacitor (= 0.00000001 farads) and a
perfect, ideal 60 Hertz SINE wave hooked across the capacitor it would
give an apparent resistance of 265,393 Ohms.
Now, if either the capacitance OR the frequency were to increase, the
Reactance would decrease and vice-a-versa.
This is all important because of the relation:
V = I * R
where,
V = Volts
I = Amps
R = Resistance
Just because a transformer is rated at 10,000 volts 30 mili amps does
NOT mean that it will not or can not supply more amperage, indeed it
can!!
But it may not be DESIGNED to supply that much current. If you try to
take more current from a transformer than it was designed for you will
end up with burned up windings either primary windings or secondary
windings or both. (but of course transformers can usually be pushed
safely
past there rated values).
If we rearange the above equation we get:
V
--- = R Eq #2
I
Given a constant voltage, if we increase R the only way to make this
Equation work is to decrease the Amperage (thus decreasing power).
Conversly if we lower R the Amperage must also rise to make this
equation balance risking burning out the windings in the transformer.
With Neon transformers, if too much current is drawn from them they
start
to shunt themselves to save their own hide. They start shutting down by
lowering their voltage output resulting in lower power.
By setting Xc in Eq #1 equal to R in Eq #2 and solving for C we get.
1
C = -------------------
2 * Pi * Hz * (V/I)
and this is the standard eq for matching the transformer with the
Capacitor.
So, for a 15,000 V 30 miliamp =(.03 amp) operating at 60 hertz
(house frequency)
we get:
1
0.0000000053 = -----------------------------
2 * 3.14 * 60 * (15000/0.03)
or 0.0053 microfarads.
Hope this helps.
"
This derivation is correct. However, something needs to be
added. If you chose a capacitor of this value (or the appropriate
value for your transformer) you will have created a SERIES RESONANCE
of the secondary leakage reactance and the capacitor. [This is
only approximate, because the leakage reactance is a non-linear
function of the transformer output voltage due to magnetic circuit
saturation effects.] In a series resonant circuit the current is
limited only by the effective series resistance. If the effective
series resistance in the above example is 30,000 ohms (I would
actually expect nearer 15,000 based on scaling from 12000 volt
transformers I have here), the current will be
I=E/R=15000/30000=0.5 amperes
and the voltage across the capacitor (and transformer) will be
E=IxX=0.5x15000/0.03=250,000 volts!!!!
Obviously "something's gotta give" and in this case it will be
the transformer secondary insulation. I suspect that a large
fraction of the "blown" neon's result from series resonant
conditions coupled with too large a gap spacing.
Moral is, if you use a "matched" secondary capacitance,
or anything approaching it you had better use a safety gap
across the transformer, preferably set to just fire at the
normal UNLOADED operating voltage.
For what it's worth,
Ed