[Elecraft] Optimal Coil Dimensions - now building it

Fri Jan 30 09:53:16 2004

I cannot believe this.  I have been researching this very thing on the net for the last week and playing with the equations on an Excel spreadsheet.  I would like to do some work with tuners, but the dog-on coils are to expensive.  I want to make the coil and the capacitor from scratch.  So, I have been playing around with the equations for the last week.  I've also research liturature about making homebrew coils and have found some very interesting articles.  I am also researching homebrew electroplating which seems to be be quite easy to do.  I want to try some tin plating and some silver plating.  I am most facinated with the fabrication process.

Now one would ask why and I doing this?

Well, since I built the K2/100 ... I have gotten back into homebrew.  I had forgot the fun of it all.

Not bad for a guy with a humanities degree that squeaked through college algebra and has a hard time balancing the checkbook.

Lee - K0WA

"David A. Belsley" <[email protected]> wrote:
The article cited earlier on this reflector 
dealing with optimal coil 
dimensions presents an interesting problem that, in fact, has a closed 
and simple solution. Wheeler's formula is used

L = (n^2 r^2)/(9r + 10h),

where L is inductance in uH, n is number of turns, r is radius (in.), h 
is length (in.). In this form, the relation to the variables of 
interest, such as the total length of the wire and the ratio of r to h 
is not clear. However, with a change in variables

n = len/(2Pi*r), where len is the total length of the wire
h = a*r, where a is the ratio of the length of the coil to its radius
and
t = n/h = len/(2Pi*a*r^2), the turns per inch,

one gets

len = 2Pi ((L(9+10a))^2/a*t)^(1/3).

If one wishes to find the ratio of r to h (a) that gives shortest wire 
(len) capable of providing a given inductance (L) with a given turns 
per inch (t), one differentiates this expression with respect to 'a', 
sets it equal to zero, and solves. I expected this to provide a rather 
nasty expression in a, t, and L, and the derivative indeed is. 
However, the numerator of this derivative has a factor

[(20(9+10a)L^2)/at - ((9+10a)^2*L^2)/(ta^2)],

which, when set equal to zero, implies a = .9, regardless of L or t.

In other words, a coil of a given inductance can be created using the 
shortest piece of wire when the coil has a length (h) that is 90% of 
its radius (r). This is assuming, of course, that the wire gauge is 
such that the given length of wire (len) can indeed be wound within the 
length (h) of the coil.

Consistent with this, the author of the web article cited above did 
some calculations for a special set of assumptions and obtained some 
empirical evidence that a = 1 is somewhat better than a = 2. Now we 
can see that, regardless of any special assumptions, a = .9 is better 
yet -- indeed best.

best wishes,

dave belsley, w1euy
-------------------------------------
david a. belsley
professor of economics

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