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*To*: tesla@xxxxxxxxxx*Subject*: Ratio of Med C or equivalent C to intrinsic C was Re: LC III*From*: "Tesla list" <tesla@xxxxxxxxxx>*Date*: Fri, 08 Apr 2005 12:00:06 -0600*Delivered-to*: chip@pupman.com*Delivered-to*: tesla@pupman.com*Old-return-path*: <teslalist@twfpowerelectronics.com>*Resent-date*: Fri, 8 Apr 2005 12:01:56 -0600 (MDT)*Resent-from*: tesla@xxxxxxxxxx*Resent-message-id*: <8inIQD.A.XnD.2bsVCB@poodle>*Resent-sender*: tesla-request@xxxxxxxxxx

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

> Original poster: "Antonio Carlos M. de Queiroz" <acmdq@xxxxxxxxxx> > > > For the coils that I have measured, or calculated, the > Medhurst capacitance is close to one half of the capacitance > of a hollow cylinder to "infinity". > And there is a good reason for this, because if the distributed > capacitance of the whole coil is split in two, one at each end > of the coil, Each of them is at the position of the Medhurst > capacitance when the other end is grounded. Of course, this > assumes that the grounding at one end doesn't affect much the > distributed capacitance at the free end of the coil, what > appears to be true. > > Examples (calculated): > A coil with 1 m of height and 20 cm of diameter: > Medhurst capacitance: 15.28 pF > Distributed capacitance of a cylinder with this shape: 27.57 pF > A coil with 1 m of height and 40 cm of diameter: > Medhurst capacitance: 21.29 pF > Distributed capacitance of a cylinder with this shape: 38.07 pF

For a secondary with uniformly distributed L and C with no top load, the lumped equation gives fr = 1/(2*pie*(Ld*Cl)^1/2)) and the transmission line equation gives fr = 1/(4*(Ld*Cd)^1/2) (this assumes a long coil compared to its diameter and that the wave length is long compared to the mutually inductance function) Where Ld is the inductance, Cd is the intrinsic capacitance and Cl is the equivalent lumped C that produces the correct resonate frequency in the lumped equation with Ld with no top load. Hence Cd/C/ = (Pie^2)/4 ie 2.467.

When the secondary is resonated with a large C the secondary voltage linearly increases along its length. If the intrinsic C is assumed to be uniform then 1/2 the current would flow in the intrinsic C. ie uniformly distributed capacitance when referred to the end is halved. Hence Medhurst C is half the intrinsic C ie Cd/Cm = 2, (only valid when resonating with a large top load) Actually the intrinsic C is distributed more toward the ends of the coil with the greatest at the grounded end (its closer to the ground connection). The C at the top end has a greater effect but is small value than the bottom C. The bottom C is larger but has less of an effect. So the effect of the uneven distribution biased towards the bottom end tends to cancel so I would still expect the ration to be near 2. (with large top load).

Note Medhurst test conditions differ from a typical secondary that operates over a much closer ground plain and with a much smaller top load.

Robert (R. A.) Jones A1 Accounting, Inc., Fl 407 649 6400

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