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RE: MMC lead length and bends...



Original poster: "Jim Mora" <jmora@xxxxxxxxxxx>

Hi Terry et all,

I want to be sure I understand you. When you say serpentine, do you mean the
capacitors joined at opposite ends with each additional cap. Further more,
is it better to stand them up and stack the layers or lay them down with
beads for standoffs?

I built my 8" coil with very sort flat copper straps from your advice and
the archive's records. I also cut my welding cable to lentgh after the coil
was tuned. It worked great out of the gate using the Panasonic Caps and one
string of cd's .15 as it tapped out too far which gave LTR with a blown dual
1" brass RQ dowel gap.

This next coil: 12.25" 64" of 18awg 200 degree magnet wire will wear a
bonnet of 60"  x 12" which brings the coil res point near 57Khz. Is that low
enough lower the stray inductance? also I'm using a 15 degree .5" copper
tubing to push upwards toward to .18 coupling and still enough distance btw
the primary and the secondary and it is a long coil. Which I now have
learned assists in negating this issue.

What does the jury say, serpentine UP back and forth or FLAT with standoffs
back and forth and layered? I like flat as each stack will have its adjacent
cap at the same potential, but up and down could that with a little thought.

This has been a long thread but very enlightening for me.

Thanks Much,
Jim Mora

-----Original Message-----
From: Tesla list [mailto:tesla@xxxxxxxxxx]
Sent: Tuesday, November 07, 2006 4:29 PM
To: tesla@xxxxxxxxxx
Subject: Re: MMC lead length and bends...

Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx>

Hi,

As I bend and twist such things....  It seems to me that the gross
"loop area" of the primary connective wiring is far more significant
than these little capacitor inductances and such.  1uH is just a
twist of primary wiring in many cases...

I think to imagine the "area enclosed" buy the primary capacitors
wiring is pretty important.  Serpentine wiring and such really cut that
down.

Cheers,

         Terry



At 05:06 PM 11/7/2006, you wrote:
>Tesla list wrote:
>>Original poster: "Jim Mora" <jmora@xxxxxxxxxxx>
>>Antonio,
>>Did you mean to say inductors in series as the CD caps have a ~ length of
>>4.45cm, would that not look as wire length? Also does the capacitance get
>>larger or smaller with a larger diameter such as butt splices which have
an
>>outside diameter of 4mm and a length of 15mm per splice (360 of them @ 20
>>per string), (18 parallel rows that would be subtractive). I'm on the
fence
>>how to route 127cm strings! My primary and lower table will be 91.4cm's in
>>circular diameter.
>
>Everything, the leads and the body of the capacitors, count as wire length.
>The inductance is smaller with thicker wire, and so smaller too withing the
>body of the capacitors (supposing that their plates are contacted at the
>sides of the roll).
>The inductance of a straight wire can be obtained as:
>L = 2e-7*h*Ln(2*h/r-3/4) Henrys
>h=length, r=radius, in meters. Ln=natural logarithm.
>For the inductance of a capacitor string, there is a complication due to
>the different radii to be considered in the leads and in the bodies of
>the capacitors, but an average value will give reasonable precision,
>as the formula is very insensitive to the exact radius of the "wire".
>
>A more exact calculation is possible, by considering capacitors and
>leads as separate inductors in series, with the influence of the mutual
>coupling considered too. The Inca program can do the basic calculation:
>Ex: Considering a capacitor with 5 cm of length and 2 cm of diameter, in
>series with a lead with 5 cm of length and 1 mm of diameter:
>The Inca program calculates:
>Lcap=0.017 uH
>Llead=0.046 uH
>M=0.007 uH
>The total inductance is Lcap+Llead+2*M=0.077 uH
>Considering a single wire with the average of the diameters:
>h=0.1, r=0.00525 => L=0.059 uH. Not so close.
>For a string of 10 of these capacitors (5 cm capacitor, 5 cm lead),
>The total inductance would be the sum of all the terms of the
>inductance matrix. The mutual inductance between successive capacitors
>or successive leads would be Ms=0.0026 uH.
>Ignoring the mutual inductance between elements at larger distance,
>the inductance would be:
>L = 10*(Lcap+Llead)+38*M+36*Ms = 0.63+0.266+0.094=0.990 uH
>(agrees with the rule of 1 uH/meter!)
>Considering a single wire:
>h=1, r=0.00525, L=1.04 uH. Very close.
>
>Antonio Carlos M. de Queiroz
>