Re: Ratio of Med C or equivalent C to intrinsic C was Re: LC III

["Tesla list" <tesla@xxxxxxxxxx>]

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

Correction: I think in a previous post I suggested that the difference
between Medhurst C and the value of C that gives the correct Fr for a no top
load was about 40%. From numerical simulation some time ago I had remembered
18% and incorrectly assumed that was in frequency hence the "approximately
40%" figure I gave for C.
Apparently from my post below the variation was in C not fr ie approximately
9% in fr which is a figure compatible with the observed errors between
Medhurst fr and measurements.

Robert (R. A.) Jones
A1 Accounting, Inc., Fl
407 649 6400
----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
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Sent: Friday, April 08, 2005 11:00 AM
Subject: Ratio of Med C or equivalent C to intrinsic C was Re: LC III


> 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
>
>