Hi Paul, Chris,It might be more accurate to look at the total length of the ribbon and width to determine area and treat it as rolled capacitor separated by 1mm, using air as the dielectric, and determine C as one would with any flat plate capacitor, but in this case, as a "rolled capacitor".
C = 0.2248*k*A*(N-1)/(d*x) (in pF)
where
k = Dielectric constant (1.0006 for air)
A = Effective plate area in square inches (length x width)
N = Number of conductive plates (3 for a rolled cap)
d = individual dielectric film thickness (.03937" in this case)
x = number of stacked sheets of dielectric between plates (1 in
this case)
I think Chris is looking at about 6.4nF for the ribbon primary. So he's probably correct to ask the question given the very low tank capacitance. Primary C may be much higher than expected (a good deal of area at about 557 square inches). Should be able to verify by inserting a signal to the primary as one would with a secondary and looking in the range of 470kHz assuming inductance is about 18uH.
Regards, Bart Paul Nicholson wrote:
Chris Swinson wrote: > Does anyone know how to work out , or know of an online > calculator for crunching primary Self Capacitance ? Does JavaTC not give the Cdc of the primary? If not, use JavaTC or FANTC to work out the self capacitance of a disk. Then, divide the Cdc by two to get an approximate effective capacitance. -- Paul Nicholson -- _______________________________________________ Tesla mailing list Tesla@xxxxxxxxxxxxxx http://www.pupman.com/mailman/listinfo/tesla
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