Re: Chaotic Resonance(Solid State Coilers)

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "harvey norris by way of Terry Fritz <twftesla-at-uswest-dot-net>" <harvich-at-yahoo-dot-com>



> On Mon, 8 Jan 2001, Tesla list wrote:
> 
> > Original poster: "Peter Lawrence by way of Terry
> Fritz
> <twftesla-at-uswest-dot-net>" <Peter.Lawrence-at-Eng.Sun-dot-com>
> > 
> > I find this interesting because my
> back-of-the-envelope calculation using
> > the traditional resonance-capacitor formula
> yeilded a different value.
> > 
> > 	1 / (2 * PI * 8,300/0.020 * 50) = 7.6nF
Duh, I have done that math backwards predicated on the
4.7 nf and came up and also wondered if that was a
correct capaicity to resonate? This is how things work
with my math!
>I recently did a series of measurements and
calculations on a Boiler
>Ignition Transformer  (8.3kV/20mA)  to determine,
amongst other things,
>what value capacitor will resonate with the secondary
winding. Because 240V
>mains and 8.3kV are not friendly, I fed 18V to the
primary winding instead
>and measured 630V output, a predictable result. I
then applied various loads
>to the transformer and measured and recorded primary
and secondary voltages
>and currents.
>
>Under these conditions, the transformer secondary
took a fancy to a
> 0.0047uF (4.7nF) capacitor. That's where it
resonated at 50Hz.{Must be Europe,HDN} The output
>voltage sky rocketed to 4000V. 
{HDN comments}
By calculation the acting inductance of that secondary
should then be 2158 Henry as determined by resonance
formula,ASSUMING that this was the cited actual best
resonant value to be used. So here we see the small
amount of voltage rise:only 4000/630 = 6.35 as an
acting Q factor. In contrast an air core coil of only
56 henry (of 23 gauge wire) in comparison(still a vast
quantity) will easily deliver a 15 fold voltage rise
at 60 hz resonance,in comparison to the ferromagnetic
example. Now the measured resistance of my 15,000 volt
30 ma Magnetek NST current limited secondary is
Magnetek NST is ~20,000 ohms. Let us calculate by
impedance considerations the maximum amperage that
could occur if the outputs were shorted. In this
situation the 15,000 volts would permit 15,000/20,000
Amps or 750 ma if the secondary were to approach
resonance. Thus the theoretical Q would be 750/30 or
25. The REAL Q values are never approached as the
mathemathical IDEAL values in ferromagnetic resonance,
which shows its disadvantages in resonant
applications. Now let us assume that the 30 ma output
that is rated for short conduction value is the actual
value that would occur if that were the value
established by impedance. In this way an ohmic value
for the impedance can be assigned, thereby a L value
also obtained. The impedance Z is known to be that
ohmic value to allow 30 ma conduction -at- 15,000 volts
or Z=15,000/.03=500,000 ohms. Thus the actual
resistance of 20,000 ohms appears as 500,000 ohms at
60 hz. The impedance is actually the inductive
reactance squared + the resistance squared with this
sum itself taken as the square root or Z= sq
rt{X(L)^2+R^2} Thus we now know that of this 500,000
ohms Z quantity of impedance, most of the ohmic
resistance is due to the inductive reactance or X(L).
X(L) needs to be determined to ascertain the acting L,
since this L quantity is beyond typical LCR meter
measurements. The quantity in parentheses{} is then
500,000^2= 25*10^10. Subtracting the 4*10^8 or 20,000
ohms resistance squared leaves 24.96*10^10 as the
inductive reactance [X(L)]squared figure. The square
rooot of this quantity yeilds the actual inductive
reactance of ~499,600 ohms. Thus of the actual assumed
impedance made by 30 ma current -at- short of secondary,
the supposedly vast resistance of the 20,000 ohms of
the secondary only contributes 400/500,000 or .08%
rather than 20,000/500,000 or 4% of the assumed
impedance dictated by a 30 ma conduction -at- 15,000
volts output.

Now... after some of these mathematical shenanigans
that book taught learning gives us, we have deduced
the inductive reactance, which can further lead us to
an acting inductance L figure for the secondary. This
is related by the equation
X(L)=2[pi]{freq}(L)=6.28*60*L where L then
=499,600/6.28(60)=1325.9 Henry.

Now the only thing left to calculate is what capacity
would be used to resonate this secondary. Again this
is made by the resonant formula R(f)=1/2[pi]sq rt{LC}
where C is the unknown quantity.

60 = 1/6.28{1326 C}^1/2 or LC=[1/6.28*60]^2 or
C = [1/376.8]^2/1326 = 5.31*10^-9 F

10^-9 F = 1 nanofarad or for this NST transformer
example we might expect a 5.3 nf capacity to resonate
at 60 hz.

Sincerely mathematical; HDN


> > I wonder whats wrong here. Did you try a capacitor
> value near 7.6?
> > 
> > -Pete Lawrence.
> > 
> 
> It's true that the capacitance value you calculated
> will draw roughly 20mA
> from the transformer when it is the only thing
> connected to the
> transformer, but that has absolutely nothing to do
> with resonance.
> Resonance is when the internal inductance is
> cancelled by the external
> capacitance producing a purely resistive load and a
> large voltage across
> the capacitor. This capacitance is independent of
> the current capacity of
> the transformer and the voltage you drive it with,
> it only depends on the
> internal inductance which should be roughly fixed.
> 
> Darren Freeman
> 
oops, wrongway 
> 


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