RE: Secondary size - the Why

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

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?


Luke Galyan
Bluu-at-cox-dot-net

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Monday, January 19, 2004 9:57 PM
To: tesla-at-pupman-dot-com
Subject: Re: Secondary size - the Why

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

Tesla list wrote:

  > Original poster: Mddeming-at-aol-dot-com

  >      The problem thus becomes: "For a given length and diameter of
wire, is
  > there a particular diameter and length that minimizes C self?" After
  > running a number of coils with 1667 ft, 2000 ft, and 6000 ft. of #24
wire
  > through an Etesla6 analysis, it turns out that in each case I tried
with
  > secondaries from 2in. to 20 in.diameters, the Cself min. point
occured when
  > the H/D ratio was between 4:1 and 5:1.

Trying to verify:

I will start with a secondary coil that I have that has 318.5 m of #31
wire, 1152 turns, 32 x 8.8 cm, and will try to keep the wire length
while changing the aspect ratio. I will change only the number of
turns for this, keeping the same winding pitch. Calculations with
my Teslasim program.

Diameter (cm) Length (cm) Turns Inductance (mH) C-self (pF) A. ratio

8.8           32          1152  28.2            5.55        3.6
8             35.2        1267  26.1            5.63        4.4
7             40.2        1448  23.4            5.88        5.7

The minimum is not in this direction.

9             31.7        1126  28.7            5.55        3.5
10            28.2        1014  31.1            5.58        2.8
11            25.6         922  33.2            5.71        2.3

Cself increases in this direction too. So, it appears that my coil is
close to optimal in this sense, at 3.6 aspect ratio.
Interesting.

By fixing the wire length you are actually fixing the area of the
cylinder where the coil is wound, because the coiled wire covers
a fixed area. The question then is: "What is the aspect ratio of
a cylinder with fixed area that results in the smallest Medhurst
capacitance for a coil wound over it?". If my calculation above
is correct, the answer would be always around 3.6.

Some manipulation of the formula for the Medhurst capacitance:
C=r*(11.26*(h/r)+16+76.4*sqrt(r/h)) pF,
where r is the coil radius and h its height (meters), leads
to C minimized for a constant cylinder area when:

h/(2*r)=3.33697.

The Medhurst capacitance is about one half of the free-space
capacitance of an open cylinder. Starting with a cylinder with
1 m of diameter and 3.6 m of length and varying the aspect ratio
while keeping the same area:

0.9x4:    116.0 pF
1x3.6:    114.8 pF
1.1x3.27: 114.2 pF
1.2x3:    114.2 pF
1.3x2.77: 114.5 pF

The minimum in this case is between 3:1 and 2.5:1, but 3.6:1 gives
practically the same result.

Another reason for the range of suitable aspect ratios is coupling to
the primary coil. But the range of aspect ratios that result in about
the same coupling coefficient is quite large.

Antonio Carlos M. de Queiroz