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RE: Bart's Coil



Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


Malcolm -

The 68.493 KHz (Fo) is with the secondary without a toroid and not shorted.
The 114.56 KHz (Fr) is also with the secondary without the toroid. In other
words the secondary coil is the same for both frequencies but the
frequencies are different.

John Couture

----------------------------


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Wednesday, November 28, 2001 6:38 AM
To: tesla-at-pupman-dot-com
Subject: RE: Bart's Coil


Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<m.j.watts-at-massey.ac.nz>

John,
      You've got my interest:

On 27 Nov 2001, at 21:07, Tesla list wrote:

> Original poster: "John H. Couture by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>
>
>
> Bart -
>
> Thank you for making the K Factor Test. It looks like the K of your coil
is
> definitely 0.165 and the 653.52 is the correct mutual inductance. This is
> proof that these two tests can certainly be used to find these two
> parameters accurately when tests are properly made. I admit the tests are
> not easy to perform and you did a good job.
>
> The reason I wasn't sure that you could make the test is because when the
K
> becomes smaller (looser coupling) the two frequencies get closer together
> and it is harder to read the frequencies. A frequency counter would make
it
> easier to read the frequencies than an oscilloscope. Note that it is a
> coincidence that the operating frequency is near these two frequencies.
> Changing the toroid size would change the operating frequency but not the
K
> factor frequencies.
>
> Several years ago I made a rough graph of the K factor vs Fo/Fs ratios. It
> is obvious from the graph that for K's lower than 0.2 the two frequencies
> get very close together. As an example when the K = 0.165 the ratio of
Fo/Fs
> is 0.986 (7.2/7.3) and when K = 0.10 the ratio is 0.995, a difference that
> will require careful measuring. For the graph click on
>
>        www.mgte-dot-com/graph10.pdf
>
> You will need Acrobat 4 or 5 (free) to review and print the graph. I lived
> in California when I made the graph but now live in Jacksonville Florida.
>
> Here is an unsolved problem for advanced coilers. What is the relationship
> between the resonant frequency of the unloaded secondary coil (114.56 KHz)
> and the open secondary (unloaded) test frequency Fo (68.493 KHz)? Is there
a
> theoretical equation that relates these parameters?
<snip>

Would you mind re-stating the last paragraph please? I'm not sure
exactly what it is you're asking. Is one frequency quoted with the
secondary shorted end-end?

Thanks,
Malcolm