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Constant Current Neon?
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I changed the title so people would know what the thread is about!
Fr. Tom McGahee
----------
> From: Tesla List <tesla-at-poodle.pupman-dot-com>
> To: Tesla-list-subscribers-at-poodle.pupman-dot-com
> Subject: Re: Top Toroid
> Date: Friday, February 07, 1997 2:25 AM
>
> Subscriber: Esondrmn-at-aol-dot-com Fri Feb 7 00:09:43 1997
> Date: Wed, 5 Feb 1997 16:00:08 -0500 (EST)
> From: Esondrmn-at-aol-dot-com
> To: tesla-at-pupman-dot-com
> Subject: Re: Top Toroid
>
> In a message dated 97-02-05 03:38:52 EST, you write:
>
> << John and all
>
> - snip-
>
> I use only neons. The charge current is absolutely limited by the neon
> to either 60 or 120ma(depends on neon being used). It would be
> interesting if someone out there can calculate the time it actually
> takes to charge a cap from a current limited neon. I have been unable to
> come up with a formula which satisfies me. The formula would have to
> assume a rated neon voltage and current and incorporate those parameters
> in a formula which includes the cap value. With the timing determined ,
> it would be easy to determine if multiple firings are even possible.
>
> By the way , in some testing I have done, it appears that the neon will
> put out its max current only when the input voltage is at maximum. At
> say 60 volts input you can only get 30 ma from a 60ma neon. This must be
> taken into account in any formula and has been a stumbling point for me.
>
> Help anyone!!!!
>
> Skip
First I'll answer one of Skip's questions about the constant current being
less when the input voltage is less. Skip, this makes perfect sense when
you think of how the neon uses magnetic shunts to accomplish this magic.
When you lower the primary voltage you are bound to be lowering the
magnetic flux density from the primary, no? OK, so you have less magnetic
flux from the primary, but you still have these magnetic shunt thingies
that are going to siphon off some of the magnetic force. How much? Well,
they operate sort of like this. The more the secondary draws current, the
more the primary responds by increasing the primary *current* and therefore
the flux intensity. Primary voltage is kept fairly constant. The increase
in flux intensity is countered by the magnetic shunts which have gaps such
that only after the magnetic field reaches a certain value do they kick in
and create a good magnetic coupling. When they *do* kick in, this makes the
magnetic flux left over for the secondary pretty much the same as it was
before. So the current in the secondary tends to remain constant. When one
shunt becomes saturated it lets some of the extra magnetic flux from the
primary get past itself... where it is promptly gobbled up by the next
shunt in line. The point at which the magnetic shunts kick in is a factor
of the primary *voltage* and their gap setting. Gap size remains constant,
but primary voltage can be changed with a variac. With a lower primary
voltage the magnetic shunts cut in more weakly, reducing the current they
will shunt away... but there is already less current to begin with, because
the primary voltage is lower. Net result: lower primary voltage, lower
constant current in secondary. I do not think that you will find that the
relationship is perfectly linear, however. Also, at very low voltages the
resistance of the secondary wires will predominate, and percent of magnetic
loss due to the core characteristics will swamp all other factors.
> >>
> Skip,
>
> It takes 5 time constants (if I remember correctly) to get to full
charge. I
> don't remember the specs on your system. My new 3" coil uses two 3,000
ohm
> series resistors to help protect the 15 kv, 30 ma transformer. Using
6,000
> ohms and .0047 mfd, one TC = 2.82 x 10 -5 seconds. 5 TC = .14 ms. One
half
> cycle is 8.3 ms.
>
> For a current limited transformer, I think we really need to use the
> effective impedance of the transformer for R. In this case it would be
> 500,000 ohms. So one TC = 2.3 ms and 5 TC = 11.75 ms. If one half cycle
at
> 60 cycles is 8.3 ms, then my cap will never get fully charged. I think I
am
> missing something here.
>
> Fr. McGahee, what's the deal here?
>
> Ed Sonderman
Ed,
how can I refuse when you asked for me by name? You are right that when
using a simple RC series configuration the DC charge rate is T=R*C where T
is time in seconds to reach about 63% of the DC level to which we are
charging. Because the charge curve in such a circuit is not linear but a
curve that is asymptotic, it will take 5 of these T times to reach about
95% of full charge.
Ah, but Ed, my friend, we is not dealing with the friendly DC here. We are
trying to charge up to a voltage that is constantly *changing*. Oh Joy! Mr.
AC he is so wonderful! Let's skip all the dreadful math and simply say that
the net result of all this is that the capacitor is always charging
*towards* the changing AC level. It isn't a matter of whether the AC wave
is going up or going down, but whether it is above or below the capacitor
voltage at any given instant. *That* is what determines the *direction* of
the capacitor voltage.
If you were to be so silly as to plot out what this looks like on a piece
of graph paper, then you would find that after a few cycles the capacitor
voltage waveform will settle down to being a sine wave that is shifted by
up to 90 degrees from the AC sine wave that is charging it. The amplitude
of the waveform at the capacitor is a function of R, C, the AC voltage
charging the capacitor, and the frequency. Assume that R=1 ohm and C=.001
mfd... then in this case the capacitor voltage would track the original AC
signal almost exactly as to both phase and amplitude. Increasing either R
or C would increase the phase difference (to a maximum of -90 degrees),
and decrease the amplitude of the voltage waveform across the capacitor to
something approaching zero. Increasing frequency will cause the maximum
voltage amplitude to decrease also. (The capacitor has less time to
charge). And none of this stuff is linear! No wonder my students prefer DC
theory to AC theory any day of the week!
So that's one reason why your math fails to give you a reasonable answer,
Ed. But have no fear!!! It is not important at all that your silly
capacitor never gets charged to *the* maximum (if by maximum you mean the
maximum voltage from the transformer). What is important is that there is a
maximum charge reached by your capacitor, and *that* is the voltage you
want your spark gap to fire at. It will be a -90 degrees phase shifted sine
wave for all practical purposes.
*** Some stuff on Rotary Gaps related to the Peak Firing Point
In a rotary gap if you are going for Big Bang, you are trying to match the
rotary position of the gap elements to the actual peak of the capacitor
voltage such that the capacitor is at its peak value when the gaps are the
exact distance apart to initiate breakdown. Where exactly that is depends
on the presenting distance of the gap and its phase angle. In a synchronous
rotary spark gap you adjust the phase angle by adjusting the rotational
relationship between the motor housing and the rotating elements mounted to
the motor. If the presenting distance is too close, then the gap may fire
again before the rotating element gets far enough away from the stationary
element to prevent a second discharge. Both the size of the gap elements
and their distance come into play. When run in Big Bang mode there would be
120 pulses per second. The speed of the motor and the arrangement of rotor
elements determines the number of Breaks Per Second. Anything less that 120
Breaks Per Second (at 60 Hz) and you are wasting stored energy. Faster than
120 Breaks Per Second and you will sometimes be firing early and therefore
not reaching peak voltage before the gap fires. You will, however have more
firings per second, and although the energy per Bang is less, the results
may be more to your liking, as the character of the sparks is different at
different break rates.
An asynchronous rotary spark gap is one in which the speed of the motor can
be varied. These type gaps are easier to build than the synchronous ones,
and allow for more experimentation, but are difficult to get to synchronize
if desired.
By the way, if we remove the energy at any other time than the peak, we
*may* still get the same total energy out, but we will have lost our
wonderful *Peaks*.
There are also timing scenarios with a rotary gap that cause unnecessary
losses.
If you had capacitors that could take it, and you
moved your spark gap opening out to let's say some ridiculous distance like
a foot, then the gap would never fire, and *all* the energy would be wasted
every single cycle. Talk about wasteful! By the way, the wasted energy
would show up primarily as heating of the transformer as it alternately
pumped current into the capacitor and then worked to pump it out again.
*** Matching Transformers to Capacitors: Constant Current Approach
On a slightly different note, matching the transformer and tank capacitor
helps us to maximize the ratio of the useable power out to the available
power in. This has very much to do with the rate at which the transformer
can charge the capacitor.
There is a way to calculate the relationship of transformer and capacitor
such that we can maximize the efficiency with which we deliver the *charge*
that we store in the capacitor. Once we have this maximized stored charge,
we want to get the most *use* out of it. That happens when we dump the
energy through the spark gap at the instant the capacitor voltage is at its
maximum voltage level.
I want to say something about the neon transformer as a constant current
device. A perfect constant current source will charge a capacitor linearly.
Thus a perfect constant current AC source would charge and discharge a
capacitor in what would be a Triangle Wave instead of a sine wave. But the
Neon transformer is *not* a perfect constant current device. So I would
imagine that if you were to look at the charge/discharge cycle of a neon
transformer and capacitor (no spark gap or any other load), then you would
see a waveform that was a cross between a triangle and a sine wave. The top
of the waveform would be rounded like a sine wave, but the sides would be a
bit straighter than a sine wave.
As far as charging time goes, it is very easy to determine for a true
constant current source. V=I*T/C
That is: Voltage at time T is equal to constant current in amps times Time
in seconds divided by capacitance in farads
Rearranging in terms of T we get T=V*C/I
Please note that in the above equation you must qualify the voltage as
being the voltage at time T. Thus the formula says that the Time in seconds
for the Voltage to reach a particular value is equal to the Capacitance in
farads multiplied by the voltage *at that time* divided by the constant
current in amps.
Now the "*at that time*" is important. Imagine 0 to 180 degrees of a sine
wave. The formula implies that if the sine wave represents the maximum
available voltage that the constant current source can produce *at any
instant in time* then the output voltage can be represented as a straight
line drawn from the 0 degree point to the time point desired along the sine
wave. For example, if its a 60 Hz sine wave with a peak value of 12KV, then
90 degrees would be (90/360)*1/60=.004166 of a second. That would be the T.
Rearrange the equation in terms of C for a 30 ma transformer and you get
C=T*I/V C=.004166*.030/12000 C=.010415 microfarads
So a .01 mfd capacitor will charge to a Peak value of 12KV in 1/4 of a
cycle if the constant current is 30 ma. But please remember that the neon
transformer does *not* come near this ideal! The constant current is more
like a maximum constant current. It falls short of the ideal both near the
low end (0 volts) and the high end (12KV) of its cycle.
*** Matching Transformers to Capacitors: Capacitive Reactance Approach
For those who are interested in the derivation of the formula most commonly
used to calulate the proper matching capacitor for a transformer (neon or
pole pig):
It is based on the formula for Capacitive Reactance:
Xc=1/(2*pi*f*C) where Xc is capacitive reactance in ohms, f is frequency
in hertz, and C is capacitance in Farads.
Now, since matching is actually based on the fact that at resonance Xc=Xl,
and since the Xl of a transformer is equal to its effective resistance, and
since R=V/I people substitute V/I for Xc.
So now (V/I)=1/(2*pi*f*C) rearranging for C we get
C=1/(2*pi*f*(V/I))
Substituting values for a 12KV 30 ma neon we get
C=1/(2*3.14*60*12000/.030) so C=.00634 mfd
*** Reality Check
Now it should really be mentioned here that the above formula makes the
assumption that Inductive Reactance=V/I for a transformer. No distinction
is made in the formula for the differences that exist between a pole pig
and a neon. This disturbs my sense of reality a bit. I mean, if you short
out a neon rated at 12KV and 30 ma the thing current limits. Take a 14.4KV
-at- 1 amp pole pig and short the two leads together, and I sure hope you
connected an arc welder in the primary circuit someplace, because otherwise
it is meltdown time! In a pole pig *you* are responsible for supplying the
current limiting.
Also, the fact that the values used on known operating Tesla coils match
closely to the values given in the formula *could* very well simply mean
that someone used that formula to arrive at their value. *That* says
*nothing* to me about the validity of the formula per se, because that is
an example of circular reasoning. I am somewhat content with the fact that
people like Richard Hull are satisfied with the values this formula gives.
According to them, these are good working values. Richard Hull *also* says
that some people use bigger values with some success. Perhaps these
*bigger* values correspond to the values we get using the formula derived
from the constant current scenario. Maybe what we should be telling people
is that the value of capacitor to be used for a 12KV 30 ma rated
transformer (of any type, assuming it has some form of current limiting or
ballast) is between .00634 and .010415 mfd. That at least would jibe with
the fact that these transformers are performing as something less than a
pure constant current generator, but better than a purely resistive type
source. So the value we should use is *somewhere* in there, *between* those
two values. I can sleep at night with reasoning like that.
And so, I will go off to bed now for a good night's sleep. Comments
welcome.
Fr. Tom McGahee