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Re: free standing coil

CC:	
Subj:	RE: Sparks thru the core


Malcolm says:
M>the coils have been capped at the top so it has not been possible
M>to check for inside/outside preference. Perhaps Ed could risk his
M>free-standing coil ? (I shouldn't be so mean :-) 

Ah, why not?! :-)
I would be happy to destroy my free standing coil, however, it was 
only a small test coil so it would not really be suitable for
real electrical breakdown tests (it's too squat). If you supply me 
with a former and some wire, I'll make you one, though ;)


M> I also have
M>plans to build a free-standing coil so have more than a passing
M>interest in ideas expressed on this topic.
M>     My two cents worth is this : considering a single secondary
M>turn; it is shielded from seeing vertically beyond its immediate
M>neighbouring (above and below) turns. But it is not so shielded
M>from seeing turns on the opposite side of the coil (consider looking
M>down the coil from top to bottom).
M>     Please keep the ideas flowing. There are things to be explained
M>and whenever that occurs, NEW methods can emerge.


	On the practical side of the Q issue, I asked Richard Hull about
the Q's they get for their Magnifiers. He estimated that the loaded (
meaning with terminal connected ) coil Q was about 80. This is probably
since they used a fairly thick (1/4"?)1ft diameter PVC form, but they still 
get 100+ inch arcs at 7kW of input power. So, perhaps it is not worth
the investment to obtain the highest Q's?!? 

	A related request for the group:
In my numerical investigations of secondary/extra coils, I don't have 
enough experimental data to compare to. It would be a great help
If all of you could send your Q data to me. For those not exactly 
familair with the physical meaning of the term, it is:
Q=6.282 x energy stored in the resonator/energy loss per cycle
The 6.282 is just an agreed upon convention.

The method I use to determine Q is the 1/2 power bandwith method: 
1) find the resonant frequency f0 of your coil with/without terminal
using a sine wave signal generator and note either the peak current into
the coil (or the peak voltage at the top of the coil with a very high
impedance probe)

2) Find the two frequencies, f1,f2, at which the power into the coil is 
half that of the resonant value. In terms of current or voltage amplitude 
this is .707 of the peak values at resonance

3) Then Q=f0/(f1-f2)

I'd like to see the effects on Q of:
coil geometry      
coatings
former material and thickness
wire types and insultaion
terminal capacitances
environment (distance to floor and floor material etc...)

Probaly too much to ask, but I'll take anything you all have....


-Ed Harris