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Old October 6th 13, 01:44 AM posted to rec.radio.amateur.antenna
Jeff Liebermann[_2_] Jeff Liebermann[_2_] is offline
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First recorded activity by RadioBanter: Jun 2007
Posts: 1,336
Default Homebrew Coil Form Factor

On Sat, 05 Oct 2013 15:24:01 -0500, John S
wrote:

The microwave oven test is not very useful. The microwave will test
whatever at about 2450MHz. I doubt he is trying to make a loading coil
for anything near that frequency.


Actually, it's very useful. In my admittedly limited experience,
anything that survives the microwave oven test, will usually work
nicely at HF frequencies. Those that get warm in the oven, might
work ok. Those that melt, burn or explode, will probably not work
very well at HF. While testing at 1000 times the frequency, and 30
times the power, might be considered overkill, it does yield some
useful information.

Incidentally, the loss tangent changes over frequency on many
materials. See table at:
http://www.kayelaby.npl.co.uk/general_physics/2_6/2_6_5.html

There are other resources for that information. Please search.


Well, there's the loss tangent tables for various materials.
http://www.rfcafe.com/references/electrical/dielectric-constants-strengths.htm
Is that what you mean? The problem here is that the ABS or PVC coil
form is hollow. There's only a little plastic and plenty of air. If
the PVC tube were thin enough, it might be considered air core.

You are correct. But, the losses (percentage-wise) will be the same
barring heating effects.


The losses are fairly small, but the effects on the material at
different power levels quite different. For example, a 1dB power loss
at the 5 watt level will produce about 1 watt of heat. That same 1dB
loss at 150 watts will produce 30 watts of heat, which will probably
melt the coil form.

It wouldn't be air core considering that it would be over
wood (dry and waterproofed) or PVC pipe.


For a hollow PVC or ABS pipe, it's mostly air core. (For a solid wood
coil form, it's all wood with some absorbed water.) I'm not sure how
to calculate the effective dielectric constant of a hollow plastic
pipe. Probably by measuring the inductance of a surrounding coil with
and without the pipe, and calculating what dielectric constant was
necessary to produce the change. Unfortunately, that doesn't help
calculate the combined losses or loss tangent.

Most posters here do not state all of the facts initially. That's
because most of their ideas are still brewing in the brain and some of
the details are not formalized. It is the reason we ask for other
people's inputs. Cut him some slack and continue to ask how you can help.


Of course. However, I post far too many "answers" in far too many
newsgroups and mailing lists to remember everyones situation. If I'm
expected to spend my time answering a question or solving a problem, I
expect to be provided with:
1. What problem are you trying to solve? A one line description is
sufficient.
2. What do you have to work with? (Equipment, site limitations,
available test equipment, numbers, numbers, numbers).
3. What have you done so far and what happened?

--
Jeff Liebermann
150 Felker St #D
http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558