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Re: Primary Coil Design


>> A slug of graphite placed at a spot where you suspect you have an RF
>> heating problem would get hot long before any other object would.  Stick
>> a cheap alcohol thermometer (the kind with the red fuild)in the graphite
>> slug and you have a primitve detector.  Don't use a mercury thermometer
>> since mercury is an electrical conductor.  A half inch cube of graphite
>> would be about the right size. Drill a hole and epoxy the thermometer in
>> place.  If you leave the scale on make sure it is plastic and remove any
>> metal bits used in its construction.
>> 

>Harry,
>
>Thanks for such a great tip. I'm sure everybody thumbing through their 
>junk parts catalogs looking for graphite chunks right now! I know I will 
>be.
>
>Any ideas on the the loss of Q and the gaining of stray inductance of 
>lead distances between 12" and 20"? I've got 1500+ straight strand, 
>0000ga. copper cable for the connections between the capacitors and the 
>primary coil. Please check out "www.ddlabs-dot-com/tesla.html" to see photos 
>and renderings of this coil project.
>
>...Jeff
>


Jeff,

I am still in the planning stages for my first TC but have been working
in RF induction heating for  5+ years.  My comments and observations will
be from that perspective so  some of my thoughts may run contrary to the 
hard earned wisdom of experienced coilers out there.   

My major concern in induction heating work is to keep as much of the
RF oscillator's output doing useful work as possible.  This translates
into keeping leads carrying RF to the heating coil  as close together 
as I can without arc overs.  The smaller the area enclosed by the leads, 
the less "stray" inductance there is to "steal" power from the heating coil.
This would qualify as stray inductance in a TC application too. My suggestion
would be to keep leads as short as is practical and as close together as
is safe.  Comments anyone?

Most of the time I don't worry too much about it because I can usually
"throw another kilowatt across it".(Apologies to Richard Quick.) My machines
range from 2.5 to 25 kilowatts and produce some amazing, if unintentional,
sparks from time to time.  Usually with a shower of water, since all the
conductors are water cooled copper tubing.

As I understand (?) it, the Q of a coil is given by Q = (2 x PI x F x L)/R 
where PI=3.14159,  F = frequency in Hz , L = inductance in Henries and R is
the resistance of the coil in ohms.  In RF work the AC resistance is higher
than the DC resistance since eddy currents  exclude current from the centre
of the conductor.  The current travels at the surface of the conductor only.
The formula I remember gives the depth of current penetration into a copper
conductor as D = 6.6/SQRT(F) in centimetres. (Divide by 2.54 to get inches.)
The term SQRT(F) means take the square root of F.

Calculate the AC resistance using R = rho x l / A  where rho is the resistivity
of copper (about 1.75 x 10^-6 ohm-centimetres)and l is the length of the
conductor,
again in centimetres.

To calculate A (the area) use A = D x 2 x PI x r  where D = 6.6/SQRT(F) and r 
is the radius of the conductor, in centimeters.

So now you have the Q of the coil at some frequency F.  From the formula for
Q the easiest way to increase Q is to decrease R, that is go to a larger radius
conductor, which you have done.  Since the current only travels in the outer
layer of the conductor, the centre really serves only as thermal mass and
keeps the 
rate of temperature rise down when handling large currents at high frequencies.


Regards

Harry Adams