Re: [TCML] Primary selfC

[FIFTYGUY@xxxxxxx]

In a message dated 7/24/08 9:45:44 P.M. Eastern Daylight Time,  
bartb@xxxxxxxxxxxxxxxx writes:

>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".



If it was just one turn, the capacitance would  be very obviously only 
isotropic, no?
    If you look at it from the standpoint of *many*  turns, the turn-turn 
voltage is so small as to make the charge between  individual turns also small. A
nd the capacitance from each turn to all the  others quickly becomes pretty 
darn small. But overall, the capacitance is  determined by the shape and size of 
the coil and is pretty much  isotropic.
    So I think the interesting case to examine would  be a 2 turn ribbon 
primary. The voltage at 1 turn is 1/2 the voltage  applied to the whole primary. 
At all points along the primary the voltage  to the adjacent turn is 1/2 Vpri. 
Now draw out some distributed  capacitances: from the outermost end of the 
primary ribbon to the midpoint  (first turn finishes/second/inner turn starts) is 
a capacitor  with voltage 1/2 Vpri on it. But the same node at the midpoint 
also has a  capacitance to the innermost end of the primary (zero volts). So 
the total  charge at that discrete point is effectively dependent on two 
capacitances in  series to ground, each at 1/2 Vpri. So the total overall capacitance 
at  that point would be treated as 1/4 of what the parallel plate formula 
would  give. This is intuitive to me as if you had two closed ribbons, 1/2 Vpri 
on  one of them and zero V on the other.
 
>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)


So if the effective capacitance is as 1/4 of that  of a single 
turn-to-single-turn coil, using the above I calculate around  200pF. Still I bet in real 
life it's much less than that!

>I think Chris is looking at about 6.4nF for the ribbon primary.  



If we could get that much capacitance out of  reasonable ribbon primaries, we 
wouldn't bother with primary tank caps!
 
-Phil LaBudde
Center for the Advanced Study of Ballistic  Improbabilities



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