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Re: Magnifier coupling measurements, new (update) (fwd)



---------- Forwarded message ----------
Date: Sat, 11 Jul 1998 15:15:00 -0700
From: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Magnifier coupling measurements, new (update) (fwd)

John Freau wrote:

> As a final test, I configured the sec var tune cap to give a greater
> capacitance range (but lowered voltage capability).  Now an even
> longer beat could be produced, such that the 1st RF notch occured
> at 24uS, and the quench occured at that point.  This gives a k of
> 0.04, and the spark output is killed.  This is probably because so
> much energy is lost in the gap during the long transfer, that not
> enough winds up in the secondary to cause breakout.

Your interesting experiments show that it is possible to vary the
primary current beat waveforms by varying the secondary capacitance, and so
apparently control the time required for energy transfer to the secondary/
third coil system.
I believe, however, that the problem is somewhat more complex. Consider
the following:

a) A notch in the primary current beat waveform does not necessarily
correspond to a notch in the primary capacitor voltage, or in the primary
energy.
This is true even in a two-coils system (see for example the recent post by
Bert Hickman listing the particular values of the coupling coefficients 
where this occurs, in the discussions on rising the secondary). In a 
6th-order system (3 resonances), as the magnifier with significant secondary
capacitance, the situation can be much more complex. It is true that the 
longer beats have more probability of being the correct ones, but others
may exist. Or maybe none.

b) After the disconnection of the primary circuit, the energy only goes
efficiently from the secondary system (L2, C2) to the third coil system
(L3, C3) if they are in tune: L2*C2 = (L2+L3)*C3. The secondary/third
coil system behaves as a two-coils system with:
L1'=L2; C1'=C2; L2'=L2+L3; C2'=C3; k=sqrt(L2/(L2+L3))
This may be the reason for the low output when you increased C2.

c) If C2 is small these effects are not important, and the system behaves
as a two-coils system, as you have verified:
Primary: L1, C1; secondary: L2'=(L2+L3), C2'=C3, k'=k*sqrt(L2/(L2+L3))
where k is the primary-secondary coupling coefficient.
Note that the rule about the existance of optimal coupling coefficients
is valid for the effective k' too:
k'optimal= (2n-1)/(2*n^2-2n+1)
for complete energy transfer in the middle of the nth half-RF-cycle in a 
low-loss system (from Bert's post, and verified as true). Due to this 
effect, there are optimal values for L3 in an ideal magnifier with low C2:
L3=L2*((k/k'optimal)^2-1)
For correct tuning:
C3=(L1/L2)*C1*(k'optimal/k)^2

d) There are also optimal values for L3 if C2 is significant, for the
same reason:
L3=L2*(1/k'optimal-1)
For correct tuning:
C3=C2*k'optimal^2

All these equations are for lossless systems and ignore transmission line
effects, what is a good approximation for most systems. They match exactly
with simulations, and agree with what I could observe in low-power tests 
done some time ago, with two coils only.
The optimization of the effective coupling coefficient probably have only
marginal effect, but the correct tuning of the 3-coil system is important.
The question for the magnifier is how to tune the primary/secondary system 
in a way that extracts maximum energy from the primary and transfers  
maximum energy to the third coil. C2=0 with correct tuning and optimal coupling
is one solution((c) above), but not exactly realizable. With significant C2,
(b) and (d) ideally solve the problem for the secondary/third coil, but ignore
what happens when the primary is connected. The problem of exactly how is
the energy transfer from the primary to the 4th-order system L2,C2,L3,C3 is
not yet solved, even in the ideal case.

Antonio Carlos M. de Queiroz
http://www.coe.ufrj.br/~acmq