Re: Coupling coeff. vs Voltage gain (was Re: Who needs a quenching gap ?)

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

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






"Tesla list" <tesla-at-pupman-dot-com> on 15.12.2000 01:35:21

To:   tesla-at-pupman-dot-com
cc:    (bcc: Marco Denicolai/MARTIS)
Subject:  Re: Coupling coeff. vs Voltage gain (was 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:
>Vc1 initially reverts polarity, reaching -5.6 kV and climbs back to
>zero in 4 uS, touching zero only.
>Vc2 goes to 180 kV, and then reverts to the peak of -316 kV, at 4 uS.
>Il1 goes to 108 A, reverts to -52 A, and croses 0 at 4 uS.
>Il2 goes to -1.7 A, reverts to 3.4 A, and crosses 0 at 4 uS.

Hi Antonio.

I tried the values above and got the same values than you. Now I see what you
mean: primary voltage rises to zero (from -5.6kV) and then starts becoming
negative again. You have no polarity reversal THERE (-at- 4 us) but you have it
BEFORE and AFTER this instant: I mean, Vc1 is oscillating all the time and
getting negative and positive values.

I see also that with k=0.6 it is very clear (after 1 primary cycle) the instant
when Il1=Il2=Vc2=0. Just as you said. The problem was I didn't realize your
tabuled values were good only for f1=f2 (it was not mentioned there).

>> So, if I understood well, the instant when both CURRENTS are zero is the
>total
>> energy transfer instant, correct?
>
>Yes, but this is not enough. With zero current in both inductors the
>voltages on both capacitors are at peaks (dV/dt=0). But for complete
>energy transfer the primary voltage must go to zero too. In the case
>mentioned in that paper, it doesn't go to zero.

I see now that condition can be reached indeed.

>I see that it says this. It's a mistaken interpretation. The problem
>is that the true maximum gain rule is not this, but sqrt(C1/C2), as
>the initial energy was in C1, and ends all in C2. This rule is
>independent of L1, L2, and k, and is only reached in the usual
>conditions of f1=f2 and k in the families that I mentioned.

Makes sense to me that a better maximum gain is sqrt(C1/C2), as C1 is actually
the initial energy storage element. I tried now in several cases and, yes, you
are correct: you can get even a higher gain when f1=f2. Trying to optimize on
sqrt(L2/L1) won't really give much insight.
The Phung paper really tackles the problem from a wrong point of view: at least
the maths looks correct and interesting (at least to me).

So would this "optimizing" procedure be right?

1. make f1=f2 by tuning/measuring the primary and secondary circuits
(uncoupled,
as much as you can)
2. increase k as much as your assembly, spark gap, etc, allows to
3. aim to the next lower k value from your table, to get a perfect (?) total
energy transfer

Of course, practical issues, losses, uncertainty, etc. will makes an actual
tuning different from the above mentioned :)


>The paper is readable. Do you have at hand the reference [5]?
>
>Antonio Carlos M. de Queiroz

Sadly not. If you are able to get it, please, let me have a copy.

Thanks for the educational discussion.

Regards