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Re: secondary frequency problem (fwd)



---------- Forwarded message ----------
Date: Wed, 19 Sep 2007 07:21:32 +0100
From: Chris Swinson <list@xxxxxxxxxxxxxxxxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: secondary frequency problem (fwd)

Hi Bart,

>
> 2.7MHz is closer to 3 than 2. If you build a coil that with a
> fundamental frequency at 2.7MHz (calc), then you must have some means of
> driving it with 2.7MHz if using a signal generator and scope setup. If
> you can't get past 2MHz on your generator, how can you say that it
> doesn't ring at 2.7MHz? This must be an really small coil. Do me a

Its going by other coils I have tested. normally just after the Fres the 
voltage just keeps falling. I would knock up a 555 circuit though I don't 
think they go past 2mhz. Saying that, If I was driving it on 800khz as a 
harmonic, then my voltage gain experiments will not be valid. My x17 would 
be based on the 800khz harmonic.... that's a good question, *is* 800khz a 
harmonic of 2.7mhz ?



> favour, send me a Javatc output so I can see the details. Your probably
> just reading lower harmonics at the moment.

http://www.future-technologies.co.uk/temp/javatc.pdf

It looks like you fixed the DC resistance in the update too, I think last 
time it came out as 3ohms.



>
> A resonant transformer isn't a typical turns ratio transformer such as
> an NST or other iron core transformer. This is a resonant transformer
> where the output potential is a function of the resonant condition which
> exist at the fundamental frequency. If you want to calc output, use the
> capacitances or inductances as follows:
>
> Vout = Vin x sqrt(Cpri/Csec)

Thanks for that. I know about the calculations, I am just doing some low 
voltage experiments just to collect some bits of data.


> Typically, coil potentials range from 200kV to 500kV. Imagine if the
> number of turns (pri and sec) were used. Say for example an average 1000
> turn coil with 350kV output. This means the primary would have to have
> 350 turns! Can you imagine that? So, how do we get to 350kV with only
> say 10 turns? The answer is resonance (or that point where the inductive
> and capacitive reactances are minimum). At that point, we then get a
> high current characteristic due to reduced overall impedance which in
> turn provides the high output potential, so think of it as a function of
> the coils reactances.


I have in part been trying to get my head about it. I think of it as when 
the tank dumps its energy, it takes time to discharge. It creates a field 
which pumps the secondary. The longer the field is there more "time" the 
secondary is being pumped. However in order to pump efficiently they have to 
be in tune. Probably a few technicalities more to it, but its how I look at 
it.




>
> As far as the Q output? Way off for your coil I expect (and is why I
> would really like to see your detail). Q is calc'd from the Fraga AC
> resistance and it's this AC resistance which is probably off. The value
> is usually good for coils greater than 3:1 h/d as stated in the help
> output of Javatc. But, your coil I'm sure is small and there are no
> actual Q measurements for a coil that small. If I were you, I would
> ignore those two outputs and simply measure it yourself. Just click on
> the Q text in front of the output box. I describe in the help file popup
> how to measure Q. It is a difficult measurement as far as accuracy and
> takes some practice. In my opinion, the greatest difficulty is equipment
> resolution.


Q seems a bit of a problem to me. I wrote my own calculator to crunch the 
numbers, though I used DC resistance for the Q measurement. I am not sure 
but I am sure Tesla used DC resistance for Q. He had factors in 10,000 or 
more. My coils is actually pretty large as you will see from the data.

btw, as a suggestion, could the blue background be changed back to what it 
was before the update, I think it was a light grey. The blue is a bit harsh 
to the eyes and the red text turns into a bit of a blur. Sorry to be picky 
:P

Cheers,
Chris






>
> Take care,
> Bart