# Re: Capacitance of a coil

```Reiner,
I just posted the same request and got excellent reply's.
I've pasted the reply below for reference.

Tesla List wrote:

> Original Poster: Reinier Heeres <rwh-at-worldonline.nl>
> OK coilers,
> just for me to understand it better and get familiair with the calcs),
> but I don't know how to calculate the capacitance of a coil. I could of
> course use Medhurst's formula (is there a table for H/D ratio > 5?), but
> I wanted to know if there's another way to calc it too.
>
> Thanks, Reinier
>
> BTW: It's only going to handle metric units!

>From Antonio:
capacitance=(11.26*L+16*R+76.4*sqrt(R*R*R/L))*1e-12

From: Mike
K = 0.100976*(H/D) + 0.309630
If you want K for the entire series including (H/D) less that two, it
becomes a forth order equation.
K = 0.0005(H/D)^4 - 0.0097(H/D)^3 + 0.0648(H/D)^2 - 0.0757(H/D) + 0.4723

From: Reinhard
K = (.585-.25442*(H/D)+(.15563*(H/D)^2)-(.02777*(H/D)^3)+(.00172*(H/D)^4))

This is the equation I am using mainly because it's easy to implement in my
spreadsheet, is accurate (all are close to one another), and works well and
closest to the entire range of the B&S table. BTW, for wire sizes of 00,
000, 0000, use -1, -2, -3 etc... and 0 = 0 guage.

(for reference)
Medhurst formula: C = K * D
C = solenoid self-capacitance in picofarads
D = diameter of solenoid in centimeters
K = (originally extracted from the Medhurst table)
Cself(pF) = K * (D * 2.54) <<<converting to inches at the Cself equation

Reiner, for metric units just enter D in cm and forget about multiplying by
2.54
Bart

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