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Re: free standing coil

Malcolm writes:

Ed, thanks for the comments.
> 
>     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

M>I define loaded Q in the classic sense of having the coil loaded
M>(in this case by spark resistance). In a normal use of a transmission

Uhm, OK, all I meant was the Q of a coil with a terminal attached- no spark
. 
M>line it is normally by its characteristic impedance. Spark loading
M>is rather less well defined than this (depends on spark current and
M>degree of spark channel ionization).

Yeah, and since Q is usually considered to be a "steady" state quantitiy, 
it doesn't mean as much under the transient of a spark since there are many 
more frequencies present in a pulse discharge. But it still gives an idea 
of the losses in the coil...

M>L/C ratio is the definitive factor in his Q measurement. The coil 
M>former will add to R in the equation below as will spark when run-
M>ning.

R also differs due to skin and proximity effects in the coil wire itself and 
eddy currents induced in its environment.

M>(Q=1/R x SQRT(L/C)
M>Characteristic impedance approx SQRT(L/C) for high Q coil

> 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?!? 

M>It would be an interesting exercise, but since the primary will be
M>the determining factor when powered, you're probably right. System
M>(coupled) Q is an aggregate :   Qsys = SQRT(Qp x Qs)

>     A related request for the group:

M>No problem. I'll post some data directly to avoid clogging the list.
M>I'll try and get it together this weekend.
M>I once tried the same trick on usa-tesla and got no help at all :-(

Great!

> The method I use to determine Q is the 1/2 power bandwith method: 

M>Ditto