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Re: Capacitor oil / Capacitor question / Coil length formula


From: 	Jim Monte[SMTP:JDM95003-at-UCONNVM.UCONN.EDU]
Sent: 	Wednesday, January 07, 1998 7:32 PM
To: 	tesla-at-pupman-dot-com
Subject: 	Re: Capacitor oil / Capacitor question / Coil length formula


>> What are we doing here creating new physics? The capacitance may increase
>> with higher dielectric oil but only because of the space being displaced
>> between the foil and the poly. This may make a higher capacitance but it has
>> to lower the breakdown voltage by stressing the poly more. You just don't
>> get something for nothing. I remember seeing the posts about doubling
>> capacitance, but I don't believe it.
>
>
>Dave, it's not in violation of physics! There is no lowering of the
>breakdown voltage either since the dielectric strength of the Sunsil oil is
>the same as that of mineral oil which doesn't differ much from that of the
>polyethylene.  If one makes a sandwich out of several dielectrics with the
>same strength but differing dielectric constants, the resultant capacitors
>value seems to be determined more by the higher dielectric constant than by
>anything else.

The overall value is dependent on both dielectrics, and replacing oil
with another type of higher dielectric constant will stress the poly
more than it was previously stressed.  Below are the details.

Consider a parallel plate cap with two dielectrics.  Let the dielectrics
be parallel to the conducting plates and have dielectic constants e1 and
e2.  Assuming a plate separation that is small relative to the plate
dimensions so that the non-uniform areas near the boundaries of the plate
can be neglected, the electric flux is directed perpendicular to the
plates from positive to negative.  If the positive plate has a surface
charge density ps and the negative one a corresponding charge density
-ps, the flux density will be D=ps.  Since the electric field is related
to D via D=eE, (e is my attempt to represent the permittivity of the
dielectric), the E in dielectric 1 is E1=D/e1 and in dielectric 2 is
E2=D/e2.

Now C = Q/V = SurfaceIntegral(ps)dS / LineIntegral(E).dL.  For this
problem, ps is uniform and within each dielectric, so is E.  Letting
the dielectrics have thickness d1 and d2 and letting the plate area be S,

  the voltages across the dielectrics are
      V1= ps*d1/e1 and V2=ps*d2/e2 and

                ps*S                  S
  C =  --------------------- = ---------------
        ps*d1/e1 + ps*d2/e2     d1/e1 + d2/e2

Clearly the value of C is influenced by both dielectrics.
Since VT=V1+V2=constant, decreasing V1 by increasing e1 will increase
V2.  So if region 1 corresponds to oil and region 2 to poly, the poly
would be stressed more.  No free lunch here.



While I'm posting, I would like some feedback regarding a cap I am
considering making.  The capacitor will essentially be a cylindrical
capacitor with the outer cylinder being 3/4" copper tubing and the inner
one being 1/2" copper tubing.  This will be implemented as shown below:


                     | + |
                     |   |
   -- ------ ------ ------ ------ ------ ------ --
    | |    | |    | |    | |    | |    | |    | |
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   || ||  || ||  || ||  || ||  || ||  || ||  || ||   stage 1
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   |   |  |   |  |   |  |   |  |   |  |   |  |   |
   -- ------ ------ ------ ------ ------ ------ --
              .
                .                                    intermediate
                  .                                     stages
   -- ------ ------ ------ ------ ------ ------ --
    | |    | |    | |    | |    | |    | |    | |
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   || ||  || ||  || ||  || ||  || ||  || ||  || ||
   || ||  || ||  || ||  || ||  || ||  || ||  || ||    stage n
   |   |  |   |  |   |  |   |  |   |  |   |  |   |
   -- ------ ------ ------ ------ ------ ------ --
                     |   |
                     | - |

The tubing would be TIG welded to copper plate and the whole assembly
would be submerged in oil.  Each stage would look something like a
bed of nails, with the inner stages having nails on both sides.  The
holes in the copper plate would be to allow the oil to pass.  Insulating
spacers would prevent each stage from shorting.

This design seems to have an advantage over using a solid dielectric in
that if you did manage to break down the oil, an oil change is all that
is needed to bring the cap back to good-as-new condition.  Being copper
with fairly short conducting paths, it should be very well-suited for
pulse applications, and since copper tubing is mass produced, it should
be reasonably cheap.  Disadvantages I see are mainly in the time it will
take to weld everything together with sufficient precision.  Any others?

I have seen a lot regarding oil dielectric constants and breakdown
voltages, but I do not recall seeing anything about losses of various
oils.  Can someone supply information on loss tangents for different
oils?

With oil having a breakdown voltage of approx 200 kV/inch, using 1/2"
and 3/4" tubing would give a breakdown voltage of about 25 kV/stage,
but I've also read that asking a single cap stage for more than 10 kV
is asking for trouble.  Any suggestions on what would be a reasonable
rating for this cap per stage?

Incidentally, a cylindrical cap has
  C= 2*pi*dielectric_permittivity*length / ln(outer_radius/inner_radius).



Finally, here is the the info on coil length:
A few weeks ago Adam Smith gave a formula for the length of wire
required to wind a pancake coil.  The result is always an underestimate
but is accurate to within less than 1 percent unless the number of turns
is near 0 or there are few turns and the first turn begins at a small
radius.  Below is a formula which only neglects any stretching or
shrinking that the wire or tubing undergoes while being bent.  It is
applicable to any conical, pancake, or solenoidal coil with constant
spacing between turns.  Derivation was via an arc length integral,
starting with
 x(t)=((R0+k*t)*cos(t),(R0+k*t)*sin(t),A*t) in Cartesian coordinates.

WireLength = (R1*f(R1) - R0*f(R0)) / (2*k)     +
             ((k*k+A*A)/(2*k)) * ln((R1+f(R1))/(R0+f(R0))) R0<R1, N>0
                             OR
             2*pi*N*sqrt(R0*R0+A*A) R0=R1, N>0 (solenoidal)
                             OR
             sqrt((R1-R0)*(R1-R0) + H*H) N=0 (straight piece of wire)

f(x)=sqrt(x*x + k*k + A*A)
k=(R1-R0)/(2*pi*N)
A=H/(2*pi*N)
R0= inner radius of coil, measured from centerline to center of the wire
R1= outer radius of coil, measured from centerline to center of the wire
    R0=R1 for a solenoidal coil.
N= number of turns
H= height of coil measured at centerline.  This is NOT the diagonal
   length for a conical.  H=0 for pancake.
      \            |           /
       \           |          /
        \          |         /
         \         H        /
          \        |       /
           \       |      /
            \      |     /
             \     |    /

Jim Monte