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I have an unknown system that can be modelled as a high-order RC circuit:

circuit

I performed the impedance spectroscopy measurement of it and fitted the equivalent components. Now I should calculate the time constants of the circuit, three in this case.

The \$\tau_i= R_iC_i\$ approximation doesn’t work for me.

I found some great suggestions online and on this forum but I can't answer to the problem yet.

I thought to use the parallel/series formula: $$Z=R_0+\left(R_1 \parallel \frac{1}{C_1s}\right)+\left(R_2 \parallel \frac{1}{C_2s}\right)+\left(R_3 \parallel \frac{1}{C_3s}\right)$$

Obtaining:

$$Z=R_0+\frac{R_1}{1+R_1C_1s}+\frac{R_2}{1+R_2C_2s}+\frac{R_3}{1+R_3C_3s}$$

But now I'm stuck. Should I find the roots of \$Z\$ (with MATLAB)? Are the \$1/s\$ values the time constants?

With the help of the microcap software I simulated the circuit using the values used by Verbal Kint (blue, red, green). I calculated the time required for each capacitor to reach 63.2% of the maximum voltage or V at infinite time. I also calculated the time constants considering the Thevenin circuit by eliminating the capacitors (cyan, magenta, light green).

Circuits VvsT

I also noticed that the function that I wrote above has zeros for s = -610.5, -89403 and -281440. Calculating formula I obtain very similar values to the other methods, but I don't know why of this result.

Function

Finally I compared al these "time constants" with the values of a and b obtained by Verbal Kint:

enter image description here

At this point I am wondering which one of these results is the most correct but I guess the definition of time constant it is a bit forced. What do you think? Which one would you pick?

I repeated the same considerations made above but with a circuit having very similar components. In this case it is clear visually that the time constants should be around 0.25 ms. The results are shown in the figure. You can see that in reality with no method you get all three correct values, except the graphical method which, however, is inconvenient.

circuit2

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    \$\begingroup\$ Take a big step back and explain what you are trying to achieve. As shown, the time constants are dependent on source impedance and loading impedance and you haven't shown them so, what are you really trying to achieve here? In other words, explain this: I have to solve this problem for my research \$\endgroup\$
    – Andy aka
    Commented Sep 16, 2022 at 12:47
  • \$\begingroup\$ What are your data? How did you "fit" the equivalent components? \$\endgroup\$
    – Antonio51
    Commented Sep 16, 2022 at 16:19
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    \$\begingroup\$ Theiden, you really do need to clarify (answer points made in comments.) If you respond, you may get a useful answer. And no, 1/s is not a time constant. \$\endgroup\$
    – jonk
    Commented Sep 16, 2022 at 20:52
  • \$\begingroup\$ Theiden: As presented the circuit is incomplete. The load and source resistances will influence the result. Yes there will be three found as the poles of a 3rd order differential equation or transfer function. \$\endgroup\$
    – RussellH
    Commented Sep 17, 2022 at 3:02
  • \$\begingroup\$ By definition, a system must have at least one input, and at least one output. You need to specify these. \$\endgroup\$
    – Chu
    Commented Sep 17, 2022 at 10:34

4 Answers 4

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It is almost impossible to obtain this type of 3rd-order transfer function in a meaningful expression without resorting to the fast analytical techniques or FACTs as described in my book on the subject. By going through a series of sketches without writing a single line of algebra as in this case, you can obtain the transfer function you want.

You start in dc, with \$s=0\$ in which you open all caps. It is easy here and you sum all resistors. Then you have to determine the time constants in various conditions once the stimulus is turned off. In an impedance determination, the stimulus is a test generator \$I_T\$. When turned it to 0 A, it becomes open-circuited and you are left with a simple circuit as shown below:

enter image description here

For the zeroes, an impedance represents an easy case because you have to null the response across the current source to determine the time constants in this mode. A nulled current source (0 V across its terminals) can be conveniently replaced by a short circuit. It is a so-called degenerate case. The new circuit is shown below and what you have to do is determine the resistance \$R\$ in these various conditions:

enter image description here

Finally, you assemble all these time constants to form a low-entropy expression as shown in the below sheet:

enter image description here

And if you plot all expressions, you have the general shape:

enter image description here

The difficulty here is to reorganize the expressions to unveil poles and zeroes by approximating the response. You need to check the time constants and see which ones dominate or can be grouped together. Good luck with this exercise!

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  • \$\begingroup\$ I remember that, in my "youngest" time, I "used" a Fortran procedure which could make a similar "thing" (identify the terms of a sum of three exponentials like this problem). SSP IBM subroutines ... EE&O \$\endgroup\$
    – Antonio51
    Commented Sep 17, 2022 at 13:06
  • \$\begingroup\$ Thank you for the in-depth analysis of the problem. I didn't fully understand how to get the expessions b1, b2 and b3 and the difference between b and a but in general I think I understand the procedure. You have combined the time constants of all circuit combinations considering the capacitors open or short circuits. \$\endgroup\$
    – Theiden
    Commented Sep 26, 2022 at 14:24
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    \$\begingroup\$ Yes, this is the idea behind the FACTs, determining the time constants of the circuit in different conditions. Have a look at my APEC 2016 seminar in which I offered a smooth introduction to the subject. The \$a\$ are the numerator coefficients while the \$b\$ are the denominator coefficients. The \$b\$ are determined with a zeroed stimulus while the \$a\$ are obtained with a nulled response. A bit weird in the beginning but becoming clearer as you acquire the skill : ) \$\endgroup\$ Commented Sep 26, 2022 at 20:14
  • \$\begingroup\$ Thanks for sharing your presentation, I will read it carefully! However, a doubt immediately arises, why did you exclude some situations in your discussion? You have reported t1N, t2N, t3N, t(1)2N, t(1)3N, t(2)3N and t(12)3N, but shouldn't be considered also t(2)1N, t(3)1N, t(3)2N, t(13)2N and t(23)1N? The numbers in brackets is where are present the short circuits. \$\endgroup\$
    – Theiden
    Commented Sep 27, 2022 at 14:19
  • \$\begingroup\$ Doing the math I understood that some t are redundant! But I don't understand yet if I have to consider a or b as the time constants. In the original post I posted an "update 2", as you can see by using components with very similar values, the times obtained are not very good with any method tried for now. \$\endgroup\$
    – Theiden
    Commented Sep 27, 2022 at 15:38
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It reminded me of a distributed RC delay, where the Elmore Delay approximation seen also here, might be useful for you. The first link provides a tutorial and discusses some limitations and constraints of the approximation.
I tested it on a completely different physical problem (hydraulic transient analysis for flexible pipes submitted to step-pressure response) and could find “reasonable” approximations, even when considered non-linearities of the hydraulic model.
If a quick and dirty solution (+/- 50%) is desired as an order of magnitude estimation, you can even assume certain simplifications. If empirical data is used to fine-tune the parameters of the model, the approximation was even better: +/- 10~20%.

Surely you would need to check if your case is within the restriction of such approximation.

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  • \$\begingroup\$ What a good read .What uni has this wonderful course? raedy actually axplains stuff which never happened in the 1980s \$\endgroup\$
    – Autistic
    Commented Sep 17, 2022 at 6:49
  • \$\begingroup\$ Ufsc = University Federal of Santa Catarina - in Brazil (confirmed by its suffix Br). Main campus is in the State’s capital, Florianópolis - SC. \$\endgroup\$
    – EJE
    Commented Sep 18, 2022 at 12:01
  • \$\begingroup\$ Maybe I could go the UNI in Brazil .I would enjoy such a course .I could relive my youth .I wonder what the power electronic course would be like ,What would electrical machines be like .I am sure that I could learn heaps and teach heaps , \$\endgroup\$
    – Autistic
    Commented Sep 20, 2022 at 9:47
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A simple RC circuit is described by a differential equation.When you have multiple capacitor you need to solve a system of differential equations.The differential equations of the system come from doing nodal analysis to each node of the circuit.The resulting system is in most cases not solvable because there exist solutions for only a specific type of system of differential equations.

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  • \$\begingroup\$ I tried to write the system. I verified that it was correct for an R+C and R0+R1//C circuits, in these cases I found solutions. Increasing the complexity I found some problems: I tried to solve R0+R1//C1+R2//C2. But in this case I can't get a solution. Not because it is impossible but because I get identities, maybe I'm wrong some equation? I used 5 equations: (1) V=R0*i0+R1*i1+R2*i2; (2) V=R0*i0+Q1/C1+R2*i2; (3) V=R0*i0+R1*i1+Q2/C2; (4) V=R0*i0+Q1/C1/+Q2/C2; (5) i0=i1+i3; also i0=i2+i4 is true but it is redundant . i3 and i4 are the C1 and C2 currents. \$\endgroup\$
    – Theiden
    Commented Sep 28, 2022 at 13:19
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I guess that you measured "impedance" versus "frequency".
The same simulation can be made if measurements are "impedance" versus "time".

Here is a graphical view of a system where "constants" are very different from each other (simplified case).
The center capacitor varies from the value of the left (/10) to the value of the right (x10).
The resistors are all same values.
One can easily view all choices of components and "determine" graphically all "values" of the model.
Note that, with microcap v12, one can "identify" component values interactively.

enter image description here

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  • \$\begingroup\$ I didn't know this program, thank you! It's making my work easier. \$\endgroup\$
    – Theiden
    Commented Sep 26, 2022 at 14:24

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