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Re: Who needs a quenching gap ?



Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-uswest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
> 
> Original poster: "Marco Denicolai by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <Marco.Denicolai-at-tellabs.fi>

> Thanks for your explanation. I have been plotting (simulating) now also
primary
> and secondary currents and understand a little bit more what you mean. But:
> 
> 1. primary current and voltage are always 90 degrees shifted (of course):
> so how
> can you get all zeroes besides secondary voltage? I mean, should I look for a
> zero in both currents only?

The currents cross zero. The primary voltage touches zero and increases
again, without a polarity reversal.
 
> 2. therefore the total energy transfer time instant is only influenced by K,
> having those magic K values you listed minimizing the energy left in the
> primary.

Correct, assuming that the system is tuned.

> But in the literature the K "magic" values are typically found according to
> another criteria, namely maximum voltage developed at the secondary. 

This is equivalent. The energy in the system is limited, and so if the
secondary voltage is maximum, all the energy is in the secondary 
capacitance, with no energy in the secondary inductor or in the primary.

> The
> procedure relates tuning ratio (uncoupled primary/secondary frequency
> ratio) and
> K to the maximum voltage gain, resulting in a set of [tuning_ratio,K] for
> obtaining maximum gain. The maximum gain is 1.18*sqrt(L2/L1) when K=0.546 and
> f1/f2=0.735 (paper from B.T.Phung).

Something strange in this. The maximum gain can't exceed sqrt(L2/L1),
with the two untuned frequencies identical.
In this condition and with one of the ks that I listed all the
energy goes to the secondary capacitance after a few cycles. 
There is no energy in the system to produce a larger secondary voltage,
for fixed primary and secondary capacitances.
I tried a simulation with the numbers that you give, and obtained the
following:

Test system (just 2 capacitors and 2 coupled inductors, with initial
charge in the primary capacitor):
L1=0.1 mH;  C1=10 nF
L2=100 mH;  C2=10 pF
k=0.6
Results in voltage gain of 31.6 (sqrt(L2/L1)=sqrt(1000)=31.6).

Other system:
L1=0.1 mH * 1/0.735^2 = 0.185 mH; C1=10 nF
L2=100 mH;                        C2=10 pF
k=0.546
This corresponds to f1/f2=sqrt(l2*c2/(l1*c1))=0.735
A simulation shows that the maximum gain reaches only 27.4.
Note that I retained the values of C1 and C2. If I increase C1
instead of L1, the output voltage really increases, but I am 
putting more initial energy in the system, and can obtain even 
higher gain if I retune L1. 

Would the subject of the paper be to show that for fixed L1, L2, 
and C2 there is an optimum combination of C1 and k that is not the 
usual one? My simulations appear to show this. If I increase
C1 instead of L1 to have f1/f2=0.735, and set k=0.546, the voltage 
gain climbs to 37.2 (1.18*sqrt(L2/L1), as you say), and is always
smaller for values of C1 and k close to these values.
This is really interesting!

But the gain could go to sqrt(C1/C2)=sqrt(10e-9*1.85/10e-12)=43.0 
with L1 retuned.
 
> Have you been investigating how these two families of K values compare with
> each
> other?

I don't know the paper that you mention. Can you put a copy somewhere
in the web?

Antonio Carlos M. de Queiroz