Re: Oil for Caps (fwd)

To: tesla-at-pupman-dot-com
Subject: Re: Oil for Caps (fwd)
From: Tesla List <mod1-at-pupman-dot-com>
Date: Mon, 12 Jan 1998 19:27:09 -0700 (MST)
Approved: mod1-at-pupman-dot-com


---------- Forwarded message ----------
Date: Mon, 12 Jan 98 20:56:22 EST
From: Jim Monte <JDM95003-at-UCONNVM.UCONN.EDU>
To: tesla-at-pupman-dot-com
Subject: Re: Oil for Caps


Hi,

Here's another angle to the oil selection question.

  <snip of Fr. Tom agreeing that the cap value will not dramatically
   change by using a different oil for caps with thick poly>

>HOWEVER, I believe that the looked-for increase probably WOULD show up
>if the cap was made using 6 mil poly!!! The reason here is that the
>thickness of the oil interface will now be much larger in proportion
>to the poly thickness. Note that this would imply TWO things at least:
>Higher dielectric constant oil COULD be used to increase the capacitance
>of caps made using THIN poly, AND such use of high dielectric constant
>oil WOULD put greater electric stress on the poly, because the
>proportion of thickness would be quite different with thin poly.
<snip>

I've been thinking about the electric stress on the poly again, and
maybe MORE stress on it via a higher-k oil would actually be a good
thing, regardless of any cap value improvement.  Why?  Because more
stress on the poly = less stress on the oil.  If the oil has voltage
breakdown, the poly is doomed.  So it may be better to balance the
load via k so that each dielectric is operating at nearly the same
percentage of its breakdown voltage, instead of poly at 50% and oil
at 90%, for example.

For a parallel plate cap, the electric flux is nearly constant between
the plates, so if Dp= flux density in poly and Do= flux density in oil,
  Dp = Do (1),
Now flux density is proportional to the electric field via the
permittivity of the dielectric -> D = e*E.  Using the same subscript
notation and substituting into (1)
  ep*Ep = eo*Eo (2)
Let the system operate at some fraction f of the breakdown voltage for
each dielectric, and let ExBD= breakdown voltage (per unit length) of
dielectric x.  Substituting into (2),
  ep*f*EpBD = eo*f*EoBD (3)
Solving (3) for eo,
       EpBD                  EpBD
  eo = ---- * ep   or   ko = ---- * kp
       EoBD                  EoBD

Since a piece of poly has a breakdown voltage/unit length several times
higher than oil, to equalize the stress, the dielectric constant of the
oil should be proportionally higher.

BTW, the oil and poly thicknesses or even relative thicknesses did not
show up anywhere.  It makes no difference if the cap is loosely taped
together or squeezed in a press and then tightly clamped.  These two
cases would produce caps with different properties, but to equalize
voltage stress on each dielectric, the same k ratios should be used.

It seems reasonable to want to balance the stress since failure of
either dielectric will kill the cap.  Also, for given thicknesses and
breakdown voltages, this will produce the cap with the highest overall
breakdown voltage.  Other ideas?

Jim Monte