Re: inductance question

On Thu, 14 Jun 2007 22:06:37 -0400, John Popelish <jpopelish@xxxxxxxx>
wrote:

kell wrote:
On Jun 14, 9:49 am, D from BC <myrealaddr...@xxxxxxxxx> wrote:
On Thu, 14 Jun 2007 16:48:17 +0200, Fred Bartoli





<fred._canxxxel_this_bartoli@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Yzordderrex a écrit :
I have 50' of wire and I want to build an air core inductor and
maximize the inductance. Is 1 big turn going to give me the most
inductance?
regards,
Bob
N9NEO
Just say NEO!
NEO! It isn't just one big turn :-)
What you're after is called a Brook's coil.
The very nearly optimal coil has a square section (a x a), internal
diameter 2a (and consequently external diameter 4a ).
You'll have to solve this for your wire length and diameter.
Inductance value is L = 2.55E-9.a.N^2 (in H) with a given in mm.
Square section?? Is that why those wallwart transformers have square
bobbins?

D from BC-

square in cross section. wound round.

He'll get more inductance by tightly bundling those turns
than by winding them is a single layer solenoid of the same
diameter. The bundled form provides tighter coupling
between the turns, that most closely approaches the turns
squared effect on inductance.

The inter winding capacitance will be higher, also. I don't
know exactly what the optimum relationship between bundle
diameter and average turn diameter for optimum inductance
per wire length, but I am guessing that the bundle cross
section diameter needs to be quite a bit less than the
average turn diameter.

Just one of many web pages on the Brooks coil:

http://home.san.rr.com/nessengr/techdata/brooks/brooks.html

Making the cross section circular or hexagonal rather than square increases
the inductance by about 1%; hardly worth doing.

.

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