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Re: Mode Questions ??? (fwd)
---------- Forwarded message ----------
Date: Thu, 08 Nov 2007 22:46:07 +0100
From: Finn Hammer <f-h@xxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: Mode Questions ??? (fwd)
Dave,
This is best answered by Antonio himself, but in an archive search, here
is something for you:
.....................................................................................
To: tesla-at-pupman-dot-com
Subject: Re: Dr. Antonio's Papers - Questions regarding mode of
operation in Tesla Coils
From: "Tesla list" <tesla-at-pupman-dot-com>
Date: Sat, 13 Dec 2003 20:54:35 -0700
Resent-Date: Sat, 13 Dec 2003 20:57:38 -0700
Resent-From: tesla-at-pupman-dot-com
Resent-Message-ID: <LBGS-B.A.1SE.s-92_-at-poodle>
Resent-Sender: tesla-request-at-pupman-dot-com
--------------------------------------------------------------------------------
Original poster: "Antonio Carlos M. de Queiroz"
<acmq-at-compuland-dot-com.br>
dhmccauley-at-spacecatlighting-dot-com wrote:
> I've recently been looking at a lot of Dr. Antonio's papers
regarding tesla
> coils, multiple resonance networks, and so forth.
>
> I had some questions regarding the operational modes of these systems.
>
> I assume that a conventional tesla coil has two resonant networks
> magnetically coupled and has a mode of operation of k:l where k and l
> are the two resonant frequencies of the primary and secondary circuit
> respectively.
Not exactly. l/k is the ratio of the two resonance frequencies of the
combined primary-secondary system.
> I was wondering if you could go into a little more detail of how k
and l are
> chosen, how they relate to the coupling coefficient, and energy transfer
> characteristics, and how they truly relate to a resonant network.
All the considerations refer to a lossless system, but apply quite
well for any reasonably efficient system.
Everything refers to the energy transfer process between the primary
and the secondary systems, just after the primary spark gap starts
to conduct, and while it is still conducting.
All this happens only when two resonant systems are coupled. With
just one, there is nothing of this. With more than two, something
similar happens too.
I will use a:b instead of k:l, to avoid confusions between "l" and
"one".
When two resonant circuits are coupled, in any way (not necessarily
through magnetic coupling), their resonance frequencies change, and
both start to show the two new resonances.
In a Tesla transformer designed for maximum energy transfer, the primary
and the secondary tanks are initially tuned to the same frequency.
When they are coupled, two different resonance frequencies appear,
one above and other below the original resonance frequency.
After a quite lengthy calculation, it's possible to show that when the
coupling coefficient is k12=(b^2-a^2)/(b^2+a^2), the two resonances
frequencies are in the ratio b/a.
It happens also that complete energy transfer is only possible when
b and a are integers with odd difference, as 1:2, 2:3, 1:4, etc.
This causes the existence of a set of values of the coupling coefficient
that result in maximum energy transfer:
Mode 1:2, energy transfer in 1.0 cycles: k= 0.6000000000
Mode 2:3, energy transfer in 1.5 cycles: k= 0.3846153846
Mode 3:4, energy transfer in 2.0 cycles: k= 0.2800000000
Mode 4:5, energy transfer in 2.5 cycles: k= 0.2195121951
Mode 5:6, energy transfer in 3.0 cycles: k= 0.1803278689
Mode 6:7, energy transfer in 3.5 cycles: k= 0.1529411765
Mode 7:8, energy transfer in 4.0 cycles: k= 0.1327433628
Mode 8:9, energy transfer in 4.5 cycles: k= 0.1172413793
Mode 9:10, energy transfer in 5.0 cycles: k= 0.1049723757
Mode 10:11, energy transfer in 5.5 cycles: k= 0.0950226244
Mode 11:12, energy transfer in 6.0 cycles: k= 0.0867924528
Mode 12:13, energy transfer in 6.5 cycles: k= 0.0798722045
Mode 13:14, energy transfer in 7.0 cycles: k= 0.0739726027
Mode 14:15, energy transfer in 7.5 cycles: k= 0.0688836105
Mode 15:16, energy transfer in 8.0 cycles: k= 0.0644490644
The use of these exact values is only important when a and b are low,
as 1:2 or 2:3, maybe 3:4. Above this, there is little difference.
Most Tesla coils designed to produce sparks operate with a:b above
7:8, and so exact coupling coefficients are not important, but it
doesn't cost to design aiming at a certain optimal mode.
To chose a particular a:b mode, consider that the energy transfer
takes b/2 full cycles of the oscillation at the primary.
Use the lowest possible value, to minimize the number of oscillations,
since with long energy transfer times most of the energy is
dissipated in the primary circuit before being transferred.
The value of a is always b-1 in practical systems. If a=b-3, there is
an incomplete first notch before the true first notch of the primary
voltage, and anyway it's difficult to force operation on these modes
(k12 becomes too high). With a=b-5 there are two incomplete notches,
and so on.
If a=b-2, or the ratio reduces to a ratio of odd integers, with
a "premature complete first notch" at b/4 cycles, or the result is a
mode where energy transfer is never completed.
It's difficult to build a coil operating in a mode below, maybe, 5:6.
The coupling coefficient becomes too high, forcing small distance
between primary and secondary, what causes all sorts of insulation
problems.
To get a better feeling about what happens, try my programs mrn4
and teslasim, that you can download from:
http://www.coe.ufrj.br/~acmq/programs
Antonio Carlos M. de Queiroz
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also:
......................................................................................
> Original poster: "Finn Hammer" <f-hammer@xxxxxxxxxxxxx>
> I believe it is possible to determine the coupling from this trace, how
> is that done?
To have an idea, I list below where is the first notch of the primary
voltage for the first optimum coupling coefficients. Look at the
comments
at the end, facts that I have just observed:
First series:
These are the most usual modes, with total energy transfer at the 1st
envelope notch.
mode k cycles (primary)
1,2 3/5 = 0.600 1.0
2,3 5/13 = 0.385 1.5
3,4 7/25 = 0.280 2.0
4,5 9/41 = 0.220 2.5
5,6 11/61 = 0.180 3.0
6,7 13/85 = 0.153 3.5
7,8 15/113 = 0.133 4.0
8,9 17/145 = 0.117 4.5
9,10 19/181 = 0.105 5.0
10,11 21/221 = 0.095 5.5
11,12 23/265 = 0.087 6.0
12,13 25/313 = 0.080 6.5
13,14 27/365 = 0.074 7.0
14,15 29/421 = 0.069 7.5
15,16 31/481 = 0.064 8.0
16,17 33/545 = 0.061 8.5
17,18 35/613 = 0.057 9.0
18,19 37/685 = 0.054 9.5
19,20 39/761 = 0.051 10.0
20,21 41/841 = 0.049 10.5
Second series:
There modes result in total transfer at the -second- envelope notch.
I don't list the modes equivalent to the 1st series.
mode k cycles (primary)
1,4 15/17 = 0.882 2.0
2,5 21/29 = 0.724 2.5
4,7 33/65 = 0.508 3.5
5,8 39/89 = 0.438 4.0
7,10 51/149 = 0.342 5.0
8,11 57/185 = 0.308 5.5
10,13 69/269 = 0.257 6.5
11,14 75/317 = 0.237 7.0
13,16 87/425 = 0.205 8.0
14,17 93/485 = 0.192 8.5
16,19 105/617 = 0.170 9.5
17,20 111/689 = 0.161 10.0
19,22 123/845 = 0.146 11.0
20,23 129/929 = 0.139 11.5
22,25 141/1109= 0.127 12.5
23,26 147/1205= 0.122 13.0
25,28 159/1409= 0.113 14.0
26,29 165/1517= 0.109 14.5
28,31 177/1745= 0.101 15.5
29,32 183/1865= 0.098 16.0
31,34 195/2117= 0.092 17.0
32,35 201/2249= 0.089 17.5
34,37 213/2525= 0.084 18.5
35,38 219/2669= 0.082 19.0
37,40 231/2969= 0.078 20.0
38,41 237/3125= 0.076 20.5
40,43 249/3449= 0.072 21.5
In general: a=integer, b=a+odd integer:
mode=a,b; k=(b^2-a^2)/(b^2+a^2); full primary cycles=b/2
Or k~=1/(2*cycles), as mentioned in other posts.
Note the curious fact that it's possible to have total energy transfer
at the 1st notch (modes a,a+1), at the second notch (modes a,a+3), or
at the nth notch (modes a,a+2*n-1).
Close to each optimum k for total transfer at the 1st notch there are
two values of k the result in total transfer at the second notch.
Close to these ks there are other values that result in total transfer
at the 3rd noth, and so on. There is also a set of high optimum
couplings
corresponding to modes 1,1+odd integer, and the families that appear
around them.
I don't believe, however that a real spark gap is sensitive enough
to the primary energy to quench precisely where the primary energy
disappears, because the differences among the primary energies at all
the
envelope notches is small. But there is a tendency for this.
Antonio Carlos M. de Queiroz
.....................................................................................
Hope this helps.
Cheers, Finn hammer