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
_______________________________________________
Tesla mailing list
Tesla@xxxxxxxxxxxxxx
http://www.pupman.com/mailman/listinfo/tesla