Re: Secondary size - the Why

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: "Luke" <Bluu-at-cox-dot-net>
 >
 > May I ask why you want to fix the wire size?
 > So you ever design a coil with a wire length in mind?
 > I thought you aimed for a total number of turns.
 >
 > Why is it important for this calculation to fix the wire length?

Really, this would be significant only if you have a limited amount of
wire to use.

Something more realistic would be to ask:
"For a given inductance to be realized, what is the aspect ratio that
results in the minimum self-capacitance?"
This problem can be solved too, using Wheeler's formula for the
inductance in the form:
L=u0*Pi*N^2*r/((h/r)+0.9) H
and considering that the number or turns, N=h/d, where d is the wire
diameter (or winding pitch).
 >From these relations, r can be obtained as a function of (h/r) and
several constants, and substituted in the formula for the self-
capacitance:
C=r*(11.26*(h/r)+16+76.4*sqrt(r/h)) pF
Finding the minimum (easier graphically in this case), I obtain:

h/(2*r)=2.5781

This minimum does not depend on the wire used, the winding pitch,
or on the inductance (as long as we assume that the formulas are valid).
It is a broad minimum, with little degeneration for larger ratios.

For something more realistic, the effect of the top load would have to
be taken into account, as it reduces the charge density at the top of
the coil, reducing its effective self-capacitance.

Antonio Carlos M. de Queiroz