# Measure the resistance at the ends of capacitor LTSpice

#### kalemaxon89

Joined Oct 12, 2022
231
I have the following circuit and have calculated the gain and cutoff frequency by hand using the open circuit time constants method.

The difficult part of this method is to find the resistances seen at the ends of the capacitors, so I would like to know if they can be found using LTSpice while avoiding hand calculations.

In the circuit below we analyze the case of C1:

I need Vx and Ix oh C1, these are the calculations:

Is there any way in simulation to find Vx/Ix = 83581.58 ohm without performing the calculations?
I don't mean finding ESR etc. of the capacitor ..
.. but finding the resistance seen at the ends of the capacitor (which in this case is treated as an open circuit) .. kind of like you do with Thevenin

I attach the simulation.

#### Attachments

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#### Papabravo

Joined Feb 24, 2006
20,994
It is not entirely clear what you are asking. In LTspice, the model of a capacitor includes three resistance parameters according to the following diagram.

Is this what you are referring to?

#### kalemaxon89

Joined Oct 12, 2022
231
It is not entirely clear what you are asking. In LTspice, the model of a capacitor includes three resistance parameters according to the following diagram.
View attachment 308347

Is this what you are referring to?
I don't mean finding ESR etc. of the capacitor ..
.. but finding the resistance seen at the ends of the capacitor (which in this case is treated as an open circuit) .. kind of like you do with Thevenin
In this case I should find Vx/Ix

#### BobTPH

Joined Jun 5, 2013
8,661
That is called reactance and it is dependent on frequency.

R = 1/(2 pi f C)

#### kalemaxon89

Joined Oct 12, 2022
231
That is called reactance and it is dependent on frequency.

R = 1/(2 pi f C)
Maybe I explained myself wrong in the main post.
Does the simulation on LTSpice allow me to find Vx/Ix = 83581.58 ohms ?

@Papabravo

#### ericgibbs

Joined Jan 29, 2010
18,646
Hi k89,
If I follow you correctly, you are asking for the impedance seen by the Cap, when its looking into the output of an OPA, in parallel with that 19k
How can that be ~83K?
E

#### kalemaxon89

Joined Oct 12, 2022
231
Hi k89,
If I follow you correctly, you are asking for the impedance seen by the Cap, when its looking into the output of an OPA, in parallel with that 19k
How can that be ~83K?
E
In the time constants method "open circuit", the cutoff frequency is: fc = 1/2*pi * 1/Cj*Rj (with j=1,2 .. number of capacitors).
In this case j = 3 because there are C1, C2 and C3.
By hand, I calculated each Rj and, applying the formula, found the cutoff frequency of the circuit.

Let's take the case of C1 (the picture attached in the main post):
Cj1 with the calculations is ~83k = R2 * [1 + ..]

If I want to avoid doing all the calculations by hand as I have shown in the picture, can I use LTSpice to find 83k?

If you are not convinced 83k, I kindly ask you to tell me (in the calculations I have shown) where, if any, I made a mistake.

#### ericgibbs

Joined Jan 29, 2010
18,646
hi k,
This LTS shows the impedance at that Node.
E

#### ericgibbs

Joined Jan 29, 2010
18,646
hi k,
This plot shows the Current in that line.
E

#### kalemaxon89

Joined Oct 12, 2022
231
hi k,
This LTS shows the impedance at that Node.
E
View attachment 308354
I think I understand that what I am asking for cannot be accomplished by simulation.

The method of open-loop time constants (for example in the case of C1) requires analyzing such a configuration:

(the other two capacitors are open circuits and the voltage source is a shortcircuit)

But in this configuration (in my opinion) Vx/Ix on LTSpice cannot be measured!
How can I measure the ratio of Vx to Ix without an input volatge? I think it's impossibile :/

In other words, I'm asking to the simulator to solve all those calculations that I solved by hand ... that is, I'm asking it to get this formula:

Correct me if I'm wrong but it can't be done, right?

@Papabravo @BobTPH

#### Papabravo

Joined Feb 24, 2006
20,994
I think I understand that what I am asking for cannot be accomplished by simulation.

The method of open-loop time constants (for example in the case of C1) requires analyzing such a configuration:
View attachment 308360
(the other two capacitors are open circuits and the voltage source is a shortcircuit)

But in this configuration (in my opinion) Vx/Ix on LTSpice cannot be measured!
How can I measure the ratio of Vx to Ix without an input volatge? I think it's impossibile :/

In other words, I'm asking to the simulator to solve all those calculations that I solved by hand ... that is, I'm asking it to get this formula:
View attachment 308365

Correct me if I'm wrong but it can't be done, right?

@Papabravo @BobTPH
Part of the problem here is that you are using imprecise terminology. Capacitors do not have resistance per se. An ideal capacitor will have a reactance which is frequency dependent. A non-ideal capacitor will have an impedance with a real part and an imaginary part. Therefore the ratio of Vx(s)/Ix(s) cannot be the result of any combination of real valued resistors because it ignores the fact that both Vx(s) and Ix(s) are frequency dependent.

I've never seen anybody do the type of analysis you are doing, and I don't think it has much of a foundation or relevance.

#### BobTPH

Joined Jun 5, 2013
8,661
Why do you say you cannot measure it? LTSPICE can measure the current into any component and the voltage between any nodes.

#### kalemaxon89

Joined Oct 12, 2022
231
Part of the problem here is that you are using imprecise terminology. Capacitors do not have resistance per se. An ideal capacitor will have a reactance which is frequency dependent. A non-ideal capacitor will have an impedance with a real part and an imaginary part. Therefore the ratio of Vx(s)/Ix(s) cannot be the result of any combination of real valued resistors because it ignores the fact that both Vx(s) and Ix(s) are frequency dependent.

I've never seen anybody do the type of analysis you are doing, and I don't think it has much of a foundation or relevance.
Part of the problem here is that you are using imprecise terminology. Capacitors do not have resistance per se. An ideal capacitor will have a reactance which is frequency dependent. A non-ideal capacitor will have an impedance with a real part and an imaginary part. Therefore the ratio of Vx(s)/Ix(s) cannot be the result of any combination of real valued resistors because it ignores the fact that both Vx(s) and Ix(s) are frequency dependent.

I've never seen anybody do the type of analysis you are doing, and I don't think it has much of a foundation or relevance.
Not knowing english well I probably made a mistake with the terms, however, I seem to have written "the resistance seen at the ends of the capacitor" and not "the impedance of the capacitor." Perhaps they are the same thing grammatically?

I thank you for your fair and accurate observation.

Time constants method (up.pt)

#### Alec_t

Joined Sep 17, 2013
14,214
Assume a and b are the nodes at the ends of a cap C. You could use a behavioral voltage source to generate an output V=v(a,b)/i(C). Is that what you want?

#### kalemaxon89

Joined Oct 12, 2022
231
Assume a and b are the nodes at the ends of a cap C. You could use a behavioral voltage source to generate an output V=v(a,b)/i(C). Is that what you want?
I should have found a way.

For each capacitor (they are 3) I apply the method of time constants (opencircuit) by finding the ratio Vx/Ix:

Resistances match those found in the calculations

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#### kalemaxon89

Joined Oct 12, 2022
231
Why do you say you cannot measure it? LTSPICE can measure the current into any component and the voltage between any nodes.
You are right! I ended up using QUCS, but the concept is the same