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Re: 8 kHz Tesla Coil



Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Bart,

Would it be a reasonable estimate to say that proximity effects cause the conducting area to be halved thus increasing the Rac by a factor of two. If this is close, the Q would drop in two (but still over 300 for my coil). No, I haven't measured the Q yet.

Gerry R


Original poster: "Barton B. Anderson" <bartb@xxxxxxxxxxxxxxxx>

Hi Gerry,

The plotting might be of value if there were no proximity losses, current flowing in dielectrics, etc... Unfortunately, all that is real and likely far more important to R.

Have you measured the Q on the 82 kHz coil and compared?

Take care,
Bart


Tesla list wrote:

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Malcolm,

I found a set of formulas on the web for skin depth (SD) of round solid conductors that are said to be conservative and overpredict the AC effect (with no proximity effect). The author compared the results to measured Rac to verify his claim. The following equations were given:

for SD < r ;
SD = 2837 / sqrt(f)      (Skin Depth - mils)

Rac = Rdc * (r^2) / (r^2 - (r-S)^2) [based on effective area reduction due to SD]


My coil uses 1440 turns of 23awg where Rdc is 61 ohms. At 82KHz, the skin
depth is 9.9mils. The copper diameter is 22.6mils (~2 SD's) so Rac is 2% higher than
Rdc (or 62 ohms). Seems like the resistance is hardly affected by skin
effect for this case. The Q for a series RLC circuit is:


Q = sqrt(L/C) / R

R = 62 ohms
L = 83mH  (Les)
C = 46pf    (Ces)

Q = 685

I just got my 8x36 coil with the larger toroid (8x32) working, tuned, and optimized for spark length. The wall power is 2100 watts. My longest power arc currently is 86 inches. Granted, proximity effects have not been included, but this coil, using wire that is only 2 SD's thick, performs with a Freau factor of 1.9

I think by using this formula, one could easily make a table for a given coil size and plot the Q and L as a function of AWG.

Gerry R.

Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

HI Gerry,

On 23 Sep 2005, at 10:53, Tesla list wrote:

> Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>
> Hi Malcolm,
>
> No comment is a waste of time in the persuit of knowledge and
> truth.  I think I understand what you were getting at.   If one uses a
> wire size where the current density is low in the center (that I
> called a waste of copper), one might opt to fix that by using a
> smaller guage and this will result in an increase of AC resistance.
>
> One question unresolved in my mind is how important it is to have the
> optimum unloaded Q considering that once breakout starts, the coil
> becomes loaded and Q drops to a low number.  Is the issue just one of
> power dissapated in the wire or are there other issues???
>
> Gerry R.

A lot of people place a strong emphasis on the low loaded Q of the
system and claim that this makes secondary Q relatively unimportant.
However, it is demonstrably true (and a waste of money) that even an
unloaded Q as low as 40 (c.f. a loaded Q of about 10 e.g.) results in
an abysmal performance. I have always found that high unloaded Q
secondaries perform well. I put that down to two things: maximizing
output voltage on first ringup and maximizing energy retention with
light loading (e.g. air streamers) for a second bite of the cherry
within a single primary shot. At some stage I'd like to see (or
produce) a formula which produces an optimal secondary (nominated
topload included) for a given size coilform. By optimal I mean with
the highest inductance before or about the point where Q starts
dropping.

Malcolm