Resistance of a trapezoidal geometry

[acme]

Hi Guys,

I am a little bit confused about this problem. I couldn't find any other post that could help me, so I am posting here for the first time.

Consider a layer (sheet) of resistive material, in which it's shape resembles a trapezium. What is the resistance between two electrodes placed on the "tilt" sides? The resistance of the electrodes is assumed to be very low compared to the other material

Well, for a squared-shape geometry with two electrodes in two opposite sides, I know that

dR = Rsheet * dx/b

- Rsheet is the sheet resistance of the material
- dR is the infinitesimal resistance
- x is the horizontal axis
- y is the vertical axis
- a is a value along the x axis
- b is a value along the y axis
The result is well-known, R = Rhseet * (a/b), where (a/b) is a geometrical factor.

By symmetry, for this squared shape geometry, no current flows along the "y" direction so turning the problem into a 1D. Not very complicated.

Now, returning to the trapezoidal geometry, I am not so sure if the problem is still going to be 1D. I am guessing there will be some current flowing in the "y" direction, namely "Iy". Also, not sure how to deal with the two extra triangles on each side of the contacts.

I appreciate any contributions or directions.

Thank you

 

[TheDestroyer]

Hello acme, welcome to PF.

Under equilibrium, the resistance of both the rectangular and trapezium is the same if they have the same area and the same material. I mean by equilibrium, after the current becomes stable (which is a matter of microseconds if not less).
It's totally meaningless to say that you have a sheet that you know its resistance and then divide it again to differential part and then integrate again, why would you do that, simply say "this is the resistance"!. This would be useful if you have the density of the new trapezium or rectangular as functions of x and y coordinates, like what we do in calculating the moment of inertia, if we have constant density and simple single-valued position radius, we simply apply the law I=mr^2. Here you apply the law V=IR, and it's over!
What do those geometric factors mean? please give a little explanation or simple graph for the situation.

Hope this helps :)

bye

 

[acme]

TheDestroyer, thank you so much for your reply.

I am not sure I understood what you said, and vice-versa...maybe I was not clear.

Please, refer to this link for a brief explanation about the problem

http://en.wikipedia.org/wiki/Sheet_resistance

Now, compare my equation for a squared shape geometry, R = Rhseet * (a/b), to the equation shown in this Wikipedia, R = Rs * (L/W). It's the same, agree?

Basically, that means that the resistance will change according to the shape of the layer. For a narrow layer, or let's call "stripe", the resistance will be very different if you measure along the "long" axis, or along the short axis.

My problem is to do the same thing for a trapezium. Actually, trapezium, circle, and a very complicated shape that looks like a star (with 5 vertices). But I am stucked with the trapezium right now.

 

[TheDestroyer]

Sorry to break it to you bud! but seems this guy in Wikipedia doesn't know what he's writing.

When we write the resistivity, we define it in terms of length, meaning the unit is Ohm times meter, so what does it mean physically to have a surface resistance then multiply it by a unit-less ratio! actually for me as a physicist I would say it's meaningless.

And if we consider this right anyhow. Let's try to calculate the resistance of this trapezium, I'm going to slice it to stripes, and calculate the resistance, this has given me a divergence. If you can solve it, then you got it ;) see the attachment ;)

Any ideas? am I wrong?

Good luck :)

 

[TheDestroyer]

If you have any idea say! this is weird! people asking then disappearing!

 

[acme]

I didn't have a chance to look at the your file...for whatever reason it was blocked. Of course I'll get back to you. Just Hold on...

About the sheet resistance problem, the equation that you saw at Wikipedia is 100% correct. it's very well-known and it is also shown on many textbooks.

Thanks

 

[acme]

TheDestroyer, thanks for your help. I really appreciate.

Few questions about the way you solve the problem.

i) I can't see why you have used the term "y^2" in the dR=Rs * x(y)/y^2 dy

ii) You are considering a trapezoidal geometry, good! That's what we want. But during the functions's definition, you have just defined the functions for two intervals corresponding to the two triangles (is it right?) I think we have to consider three intervals:
(based on your drawing)

0 < x < L1
L1 < x < L2
L2 < x < L

iii) If the final solutions is indeed correct and not integrable, we might want to solve for the power dissipation by Joule heating loss in the resistive layer

dP = I^2 * dR

Please, when you get a chance, see the attached file. It's the solution of the problem for a squared shape geometry. Thanks

 

[acme]

TheDestroyer, I apologize...I've just realized that the attachment wasn't loaded.

Now, let me know you if you can see the file and what do you think. And please, could you check my doubts about your solution?

Thanks