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