RF toolkit capacitance calculation

The exact expression is indeed the acosh expression used in Matlab, and we would expect no less. The log expression is an approximation, and it's a bit of a holdover from the times when scientific software was not readily available. acosh was not so easy to calculate, but log tables were common. Leaving out the factor of pi*epsilon, the approximation goes as follows. Using the variable Cinv for 1/C, then

Cinv = acosh(D/2a)
so
cosh(Cinv) = D/2a
(exp(Cinv) + exp(-Cinv))/2 = D/2a

Now if D/2a is large, then Cinv is large and you can drop the exp(-Cinv) term compared to the exp(Cinv) term. Then

exp(Cinv) = D/a
Cinv = log(D/a)

The natural question is, how large is large?

Da = 2:.001:20;    % D over a
Cinv1 = acosh(Da/2);
Cinv2 = log(Da);
plot(Da,Cinv1,Da,Cinv2)
legend('acosh(d/2a)','log(D/a)','location','northwest')

The plot shows a pretty good approximation for D/a greater than about 5.

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Derek Neal 2020-11-12

David-

Thank you for your reply. I wondered if I might ask you about the ac resistance formula on the rfckt.twowire page. R=1/πaσδ, where "a" is the RADIUS of the conductor. Most other texts that I've been studying use the diameter. Can you help me understand? Also, is this formula for a solid conductor? Is the skin effect negligible if the conductor is stranded?

David Goodmanson 2020-11-12

编辑：David Goodmanson 2020-11-17

Hi Derek,

It's good to check these things out as you are doing. Looks like they are using radius instead of diameter because that doubles the resistance, and in this situation you have two wires carrying equal current, not just one wire.

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