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Is there a standard formula to determine the total capacitance of:

schematic

simulate this circuit – Schematic created using CircuitLab Where C1 and C are different capacitances and for n number of C1 and C. Maybe one can create an equivalent circuit design, but I can't quite see it.

Thanks!

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  • \$\begingroup\$ In my particular case n = 20, on the off chance that is relevant. \$\endgroup\$ – Q.P. Jan 25 '17 at 19:51
  • \$\begingroup\$ I don't see an easy closed form formula for that (at least I can't derive it..). Is it an exercise presuming there should be one? \$\endgroup\$ – Eugene Sh. Jan 25 '17 at 20:05
  • \$\begingroup\$ Take one section of C1 and C2, and get an expression for the capacitance Cin looking into C1, when C2 is connected by Cout to ground. Now solve for Cin = Cout. This is the iterative capacitance, which the input capcitance will tend to for large n, equal for n=infinity, and equal if a finite n ladder is terminated in Cout. \$\endgroup\$ – Neil_UK Jan 25 '17 at 20:07
  • \$\begingroup\$ @Neil_UK Thanks for the suggestion. So C1,2 = 1/(1/C1 +1/C2), and then 1/(1/C1 + 1/C1,2)? I'm not sure I can follow exactly, are you able to write a formula in an answer - I can credit you properly then as well. \$\endgroup\$ – Q.P. Jan 25 '17 at 20:23
  • \$\begingroup\$ @EugeneSh. No. It's just a circuit I am using to try and model something 'easily' I know it's some kind of iteration procedure but I can't quite see it. \$\endgroup\$ – Q.P. Jan 25 '17 at 20:24
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For each segment, we have two capacitors in series which are then in parallel with a third, so we have a total capacitance of:

$$C_{leg} = C_1 + \frac{1}{\frac{1}{C_2} + \frac{1}{C_1}}$$

We can generalise this into an iterative formula for each leg if we assume that the leg consists of only the two capacitors, and the final leg consists of just \$C_1\$:

$$C_{leg}(m+1) = C_1 + \frac{1}{\frac{1}{C_2} + \frac{1}{C_{leg}(m)}} \quad\quad\quad\mathrm{where}\quad C_{leg}(0)=C_1$$

You can then simply iterate the formula \$n\$ times until you have the total capacitance - i.e. \$C_{total} = C_{leg}(n)\$.


Just for the fun of it, you will essentially you end up calculating iteratively something like this (example for n=4):

$$C_{total} = C_1 + \cfrac{1}{\frac{1}{C_2} + \cfrac{1}{C_1 + \cfrac{1}{\frac{1}{C_2} + \cfrac{1}{C_1 + \cfrac{1}{\frac{1}{C_2} + \cfrac{1}{C_1 + \cfrac{1}{\frac{1}{C_2} + \frac{1}{C_1}}}}}}}}$$

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  • \$\begingroup\$ Nice. Thanks for the answer - I was hoping for a nice compact expression but this will do nicely!! \$\endgroup\$ – Q.P. Jan 25 '17 at 20:32
  • \$\begingroup\$ A version of the same iterative formula for the golden ratio I believe. \$\endgroup\$ – Andy aka Jan 25 '17 at 22:08
  • \$\begingroup\$ @Andyaka pretty much. Make \$C_1 = C_2 = 1\mathrm{F}\$ and with enough legs you have a \$\phi\mathrm{F}\$ capacitor. \$\endgroup\$ – Tom Carpenter Jan 25 '17 at 22:19
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As a supplement to the answer by Tom Carpenter, if you have an infinite chain and the series converges, then $$ C_{leg}(m + 1) \rightarrow C_{leg}(m) $$ $$C_{leg} = C_1 + \frac{1}{\frac{1}{C_2} + \frac{1}{C_{leg}}}$$ $$ \Rightarrow C_{leg} = \frac{1}{2}(\sqrt{4 C_1 C_2 + C_1^2} - C_1) $$

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  • \$\begingroup\$ And then I can simply say n * Cleg and in my case n = 20? \$\endgroup\$ – Q.P. Jan 25 '17 at 21:17
  • \$\begingroup\$ No. The value given above is for n approaches infinity. But when n is 20, the series is long enough that the result is probably very close to the value as given (definitely do not multiply with n). Someone good in math might be able come up with an error estimate. \$\endgroup\$ – rioraxe Jan 26 '17 at 1:07
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There is probably a compact form based on z-transform generating functions. See the SE question and answer on a related resistor ladder network: Closed form expression for a resistor ladder network.

Also see the catalogue of generating functions at : http://www.lacim.uqam.ca/~plouffe/articles/MasterThesis.pdf

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If we write $$\begin{pmatrix}U_1\\I_1\end{pmatrix}=\begin{pmatrix}1&Z\\Y&1\end{pmatrix}^n\begin{pmatrix}U_2\\I_2\end{pmatrix}$$

with $$Z = \frac{1}{2 \pi jf C_2}$$ and $$Y = 2 \pi jf C_1$$ then $$\begin{pmatrix}1&Z\\Y&1\end{pmatrix}^n=\frac{1}{2\sqrt{YZ}}\begin{pmatrix}\sqrt{Z}&-\sqrt{Z}\\\sqrt{Y}&\sqrt{Y}\end{pmatrix}\begin{pmatrix}(1+\sqrt{YZ})^n&0\\0&(1-\sqrt{YZ})^n\end{pmatrix}\begin{pmatrix}\sqrt{Y}&\sqrt{Z}\\-\sqrt{Y}&\sqrt{Z}\end{pmatrix}$$ and now I have a headache from the diagonalization and somebody else should take it from there.

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    \$\begingroup\$ Do not flag as low quality unless it meets these criteria: its not an answer, its spam or its a comment. Help the user by explaining that they could better answer the OP's question by applying these equations to capacitors. OR edit the answer. OR if you really don't like it downvote it. \$\endgroup\$ – Voltage Spike Jan 25 '17 at 23:09
  • \$\begingroup\$ If you could apply these equations to capacitors, you'd actually start to make this a workable answer -- as of right now, it's sitting in a limbo between "answer" and "half-baked un-answer" \$\endgroup\$ – ThreePhaseEel Jan 26 '17 at 3:47

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