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The first figure shows the layout of a metal wire and current flowing in the metal from left to right.
Assume that that the sheet resistance of the metal is Rs (Ω/sq) then the resistance of the square 3W x 3W with such current flowing is just same as Rs, the sheet resistance of the metal.

enter image description here

Now consider the case where there is a zero-ohm contanct in the center of the square. The current flows from left side to the square and entering the contact from all directions and then though the contact to the bottom layer.
So I'm trying to calculate the resistance of the blue part if current flowing in all directions to the contact as shown in the image.
The result is 0.65*Rs but I would like to know if there is a general approach to prove this. I heard that it can be done by using integral but I haven't figured out how to do that.

enter image description here

emnha
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    In the real world, the contact resistance will be dominated by the resistance at the exact boundary between the two contacts. Conditions at the interface between the two contacts are what matter most. Material type, 3D contact geometry, surface smoothness, mating force, etc. Using material resistance only will greatly underestimate the contact resistance (to the point of folly). – user57037 Sep 21 '21 at 19:10
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    You need to model this with finite elements. Back in the day we instead built a model resistor from carbon-filled paper. Contacts were made with electrolytic silver. – Janka Sep 21 '21 at 19:10
  • @Janka any more detail how to model that? – emnha Sep 22 '21 at 04:53
  • By creating a resistor network with a finite number of resistors that have values that roughly resemble the distances between their endpoints. Start with the centers of each of the eight squares around the pin. That gives you ten nodes and 17 resistors. – Janka Sep 22 '21 at 12:53
  • @Janka I still don't quite get it. How close am I? https://ibb.co/9pxj5TD – emnha Sep 22 '21 at 19:25
  • Not very close. I write you an answer. – Janka Sep 22 '21 at 20:27
  • One of the things that throws me with this question is where are you wanting to calculate the resistance from and to? Normally resistance is from a single point to another point. But it almost sounds like you want to measure the resistance from an outer equipotential outline to an inner equipotential square. In any event, I think the question would be more clear if you somehow specify exactly what resistance it is that you want. – user57037 Sep 23 '21 at 03:57
  • @mkeith would it be clear now? – emnha Sep 23 '21 at 06:23
  • If that is really the problem, I don't think there is going to be a simple solution. If you imagine that the perimeter of the 3W square is all at the exact same voltage, and the current flows toward the small (WxW) square, then I expect you could set up a simple integral and solve it. But if the current enters the big square from the trace, well, it is going to be more complicated like everyone is saying. Not a simple integral. – user57037 Sep 23 '21 at 06:52
  • @mkeith yes, it's from the trace to the square. – emnha Sep 23 '21 at 07:09

1 Answers1

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You need to model this with finite elements. That means you create a resistor network with a finite number of resistors that have values that roughly resemble the distances between their endpoints.

For example, start with the centers of each of the eight squares around the pin. That gives you ten nodes and 17 resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, it gets quite tedious to solve this by hand. That's why there's computer programs that do it.

Back in the day we instead built a model resistor from carbon-filled paper. Contacts were made with electrolytic silver.

Janka
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  • thanks, i see the idea. Is it possible to do it exact? I heard that it is possible using integral but i haven't figured out how. – emnha Sep 22 '21 at 20:55
  • The integral method is the carbon paper method. – Janka Sep 22 '21 at 20:56
  • For very simple geometries as for example concentric circles, you can solve this by making the finite elements circular resistors and indeed find a nice formula by pure mathematical considerations. This is useful for calculating electrical fields and radial currents in coaxial configurations for example. A seasoned engineer looks up the geometry in his handbook and applies the formula. Solving integrals is for mathematicians. – Janka Sep 22 '21 at 21:04
  • I would like to know a general method to calculate it exact so I can apply for other shape as well. – emnha Sep 23 '21 at 17:04
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    The general method is called *finite element method* or FEM. Look it up. There is no way to calculate arbitrary shapes exactly. – Janka Sep 23 '21 at 19:13
  • I checked some previous posts on this topic and they all calculated the resistance between two nodes but in my case it's not two nodes so can I apply that? – emnha Sep 24 '21 at 10:17
  • This is one post: https://electronics.stackexchange.com/questions/342340/how-do-you-determine-the-effective-resistance-of-a-finite-grid-of-resistors?zarsrc=30&utm_source=zalo&utm_medium=zalo&utm_campaign=zalo – emnha Sep 24 '21 at 10:18
  • You have two nodes. One is the edge of the square where it touches the track to the left. The other is the pin. I chose that particular network topology above because it gives you a rough figure about the result resistance on first look. It must be somewhat around 0.75Ω (per unit). You can chose any other topology as well as long as you change the resistor values accordingly. – Janka Sep 24 '21 at 15:18
  • The contact is of zero ohm resistance so it's an equipotential node. However, how do you know that all points on the edge where the current enters on the left are of same potential? – emnha Sep 24 '21 at 17:57
  • I simulated your model and got the equivalent resistance of the whole circuit is 1.5945 ohms and the resistance of the square part is 0.5945 ohms. The result is good but I still want to prove the exact value. – emnha Sep 24 '21 at 20:52
  • It's only a very simple model. Don't expect models to be more than 10% accurate. Reality usually has effects you haven't even thought about, for example non-neglible unlineareties. – Janka Sep 24 '21 at 22:15
  • How do you know that all points on the edge where the current enters on the left are of same potential? – emnha Sep 25 '21 at 06:56
  • It's what you have to assume from the way that problem is formulated. All the measures it gives are about the square. The result is about the square. The track width on the left is also meant as an information about the square. – Janka Sep 25 '21 at 12:40
  • So with FEM you cannot calculate the exact resistance value? From what I read the result approaches the exact one when the number of elements increases. – emnha Sep 25 '21 at 12:58
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    There is no exact solution for this kind of problem. There are a lot of problems for which there is no exact solution. For a very simple one, lookup the circumference of an ellipsis. There should be an exact solution, no? Just like for a circle. There isn't. – Janka Sep 25 '21 at 13:05
  • Is it possible to make a model that you can easily extend the result for a bigger number of elements? Like you first make 17 nodes and you got the result 0.5 ohms. Now if you want to 100 nodes then you just have to change the number of nodes from 17 to 100 in your program and run it again to get the more accurate result. – emnha Sep 25 '21 at 13:50
  • Yes, there are FEM programs out there that do exactly this sort of thing. – Janka Sep 25 '21 at 21:39
  • Can you recommend some of them? – emnha Sep 26 '21 at 03:53
  • can you recommend some software to do the model as you said? – emnha Sep 27 '21 at 08:01
  • For that simple kind of problem without any nonlinearities? No. Any FEM software will do. – Janka Sep 27 '21 at 08:24