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Re: [TCML] Bifilar secondary

These coils are so tightly coupled that they act as one. You could just wire them in parallel to get lower resistance and therefore higher Q. Toploads on a coil this size are probably 20-30pF, so you can just use the normal resonant equation:

                       2*pi*f = SQRT(1/(LC))

If you are going to run differential voltages up these wires, you should bypass them at each end with a big cap to make the TC think they are one wire.

You could look into how magnetron cathode heaters are connected.

--- Carl

On 3/16/2016 3:55 PM, jj@xxxxxxxxxxxxx wrote:

Hello all,
I acquired a small bifilar secondary coil with the following characteristics:
the coil form is phenolic
the coil form is 33.5" long, 4" in diameter with a wall thickness of .25"
winding length is 30.25", 13 turns per inch
wire diameter is .072" (about AWG 13)
Wire is covered with cloth insulation with a thickness of .039"
first helix has an inductance of 524 uh and the second is 528 uh

I do not have the appropriate equipment to measure its parasitic capacitance.

Would someone please help me to calculate its resonate frequency or at least point me in the right direction? I have not been able to find a Tesla coil calculator program that does bifilar coils.
Any and all help will be highly appreciated.
Thanks to everyone,
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