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Re: Math Doodling
Hi Dave, all,
My comments below. Sorry for the slight delay on a very
interesting subject.
> Original Poster: David Sharpe <sccr4us-at-erols-dot-com>
> Here is a simple math analysis situation that blew Richard Hull
> and Alex Tajnsek away. Based on equations in the Heise paper and
> assuming lossless transfer of power:
>
> Vo = Vin * sqrt ( Ls/Lp )
>Where Vo = max Vout from resonator
> Vin = Vin applied to tank circ.
> Ls = Inductance of resonator
> Lp = Inductance of tank pri.
>
> If the following equation is assumed to be correct in the time domain:
>
> Vin = Iin * sqrt ( Lp/Cp )
> Where Vin = Vin applied to tank circ.
> Iin = peak tank current
> Lp = Inductance of tank pri.
> Cp = Capacitance of tank C
>
> AUTHORS NOTE: This is RMS tank current times Surge Impedance equals
> applied voltage to tank circuit.
>
> Then substituting equation 2 into 1 and simplifying results in:
>
> Vo = Iin * sqrt ( Ls/Cp ) Variables as listed above
>
> This suggests that Cp should be made a small as possible, and
> to maximize Vo, as high a Vin as possible should be employed. This
> makes sense because Iin will go up with higher Vin, and bang energy is
> .5*C*V^2.
>
Looking at your last equation (Vo = Iin*sqrt(Ls/Cp) would suggest
something different to me. It tells me, the more current I have
flowing in the primary, the better my Vo will be. (However, it is
quite different than seen at first glance. Read on...).
For a given FresS (LsCs) I obviously need a fitting primary
FresP (LpCp). So I can either go for a:
1.) A big cap and a small primary inductance.
2.) Large primary inductance and a small cap.
Let us look at the true effects of each with an example:
1.)
Xformer: 7500Vrms
Cap: 67nF
Input Joules = 0.5*7500V*sqrt(2)*(67*10^-9) = 3.77J
Pri induc = 23.44µH
Pri current = V*SQRT(C/L) = 10606*SQRT(67*10^-9/23.44*10^-6) = 567A
LpCp= 1570.48
Ls = 36.5mH
Using the equation Vo = Iin * sqrt ( Ls/Cp ), for this setup
Vo= 567A* sqrt(36.5*10^-3/67*10^-9) = 418.5kV
====> Vo (low voltage case) = 418.5kV
2.)
Xformer: 15kVrms
Cap (for equal Joule input this is 1/4 of the cap in example #1)
->67/4 = 16.75nF
Test: 0.5*15000^2*sqrt(2)*(16.75*10^-9) = 3.77J
For equal Fres Lp*Cp must be equal to case 1,
so our new primary must be: 1570.48/16.75 = 93.76µH
Pri current = 21213.2*SQRT(16.75*10^-9/93.76*10^-6) = 283.53A
Ls = 36.5mH
Using the equation Vo = Iin * sqrt ( Ls/Cp ), this means for this setup
Vo= 283.53A* sqrt(36.5*10^-3/16.75*10^-9) = 418.5kV
====> Vo (high voltage case) = 418.5kV
By no coincedence, the primary current in case two is exactly one half
of the primary current in case one. As the square law (V vs. Ws)
flows into the equation Vo = Iin * sqrt (Ls/Cp), it makes no difference,
if I use a big primary and a small cap or a small primary and a big
cap. The output voltage is the same (as I expected). However,
(empirically) for case one, I will probably need a larger toroid, which
means I MIGHT be able to get longer sparks from this setup. In the
real world both coil #1 and coil #2 will give similar spark lengths.
Lets look at the equation Vo = Iin * sqrt ( Ls/Cp) a little differently:
sqrt (Ls)
Vo = Iin * -----------
sqrt (Cp)
For a given secondary coil, sqrt (Ls) is a constant and can be
ignored for the moment, so I get: =>> Vo= Iin/sqrt(Cp). As
Cp gets larger, so will Iin, and thus Vo remains constant.
If you are in for the outmost Vo, then you will need to build a
secondary coil with a very high inductance, get a small Cp
and a high primary current. However, for a small Cp you will
need a large primary (µH wise). The current equation (I=Vo*
sqrt (Cp/Lp) "works against" you. So it will be difficult to find
such a combination (large Ls, small Cp and large Ipri) and I
am not sure if it is worth the trouble. Luckly for us coilers,
the length of the spark output is NOT soley dependant on
Vo. In almost every case (sure, there are limits) going for a
larger toroid (resulting in less Vo) will get you longer sparks.
My thoughts behind this: The initial arc length depends on
voltage, but the rate of growth and the "end length" of the
arc depends soley upon the current the coil can supply. The
bigger your topload, the higher the current reserve is, that
the sparks can feed on (sort of like using LTRs on the
secondary side ;o) ) There will be a point of diminishing
returns, where Ctop drops the voltage so far that it will
result in shorter arcs. So, now the question is, what is
better: a low or high inductance in the secondary to
feed a large Ctop?
I had Richie do some simulations for me and my planned setup
(200bps rotary, etc) to see what cap size would be optimum.
One thing we (Richie did the work) found out is that my (low
voltage) supply is extremely stiff. The secondary leakage
inductance of my xformer setup is very low. This might be of
some interest, esp. in higher break rate and / or DC driven
setups. The stiffer the supply, the better (faster) it can
recharge a certain capacitor in time X.
I purposely chose the values in the example above the way I did,
in order to show it makes no difference what input voltage one
choses to drive the coil and to clean up with the myth that low
voltage coils are not capable of making long sparks. ;o). A high
voltage driven coil WILL be less finicky about tank connections,
etc. This means the low voltage setup will require more thought
on construction and need slightly "heavier" equipment, because
of the high currents it runs with. There is, of course, a limit either
way. For a low voltage setup, you will HAVE to use an xformer
that will fire across a reasonable gap and for a high voltage
setup you will be limited to a certain voltage, if you want to
save yourself insulation headaches (corona, etc).
> Also, if C is made smaller, dielectric losses maybe REDUCED, with a
> given capacitor (since dielectric area and volume are reduced).
> This is the first time that in doodling with the equations, a
> possible mathematical validation of what has been touted by the TCBOR
> all along is derived, make tank capacitors small, and leverage energy
> by the use of very high voltages.
While this sounds true (and I will believe it) our main losses occur
within the spark gap. I think losses in Cpri, the coil former, etc are
neglectably small. Of course, when I make this statement, I mean if
properly used and / or constructed. E.g: A badly built SW cap will
have high losses, that WILL make a difference in spark length output.
Afternote:
I realize most of this has been posted, but I thought an example
would clarify this subject. John Freau´s post on low vs. high
input voltage empirically proved my thoughts. Thanks John!! and
yes I really am pleased (;o)) + my low voltage coil is doing pretty
good !!!
Coiler greets from germany,
Reinhard