Effective Electrical Resistance

On my opinion, there are three resistances in series:

R1+R2+R1

R1=RHO*t/(a*b) ; R2=RHO*(L-2*a)/(b*t)

Rtot=2*R1+R2

I am not sure to have explained clearly the problem. May I add :

- the contact between pad and resistive sheet is perfect almost no resistance in comparison with the sheet own resistivity

- the sheet is thin as indicated t/b<<1.

- frequency is low could be considered as DC

I consider it as the case of a resistive conductor which has at each end a zone where the potential is the same all over.

Why I have a problem is that if one considers 2 narrow slices of the contact zone at each end one and symmetrical the distance in between is variable thus also the resistance but the voltage difference is the same which means that every slice brings a different current element. The whole current should be obtained by integrating all partial values over the contact area.

Any comment?

Thanks to all of you.

Tha approximation Leq= L-2a is valid only if L/a >>1.

I wanted a confirmation which I got, the equation is :

Leq/L=2* (a/L)/ln(1+2*(a/L)/(1-2*(a/L)))

If a/L becomes small then ln ( ...) → 2*(a/L)/(1-2*(a/L)) and Leq/L= 1-2*(a/L) corresponding to sketch.

As I mentioned I am confronted with ratios a/L=3....4 so that a rough estimation was not acceptable.

There are some difficulties related to the dimensions of the resistor which are in the range of a couple of mm, this cannot unfortunately allow much measurements. The need was to dimension from the design phase the geometry in such a way that the possible measurements will be only a confirmation and less a check for a "cut and try"approach which could be in this case very expensive.

One sees that resistance along the contact zone decreases and if a/L important the variation is also not neglectable, thus my concern.

The difference between approximation and equation results is quite important for the range in which I work.

Thanks again it helped me a lot.

Hold the presses! I don't see how your formula:

Leq/L = 2 * (a/L)/ln(1 + 2 * (a/L)/(1 - 2 * (a/L)))

can possibly be correct. It's purely geometric, and does not include the thickness of the conductive sheet or the relative conductivities of the pad and the sheet.

The ratio a/L is almost entirely irrelevant to Leq under any conditions in which one would refer to the conductive sheet as a "sheet". It starts to become relevant if the thickness, t, is comparable to or larger than 'a'. In that case you'd be dealing with a block of conductive material, not a sheet, and the 3D distribution of current would be important to the solution. But as long as it is a conductive sheet and the conductivity of the pad is >> the conductivity of the sheet, then both the pad material and the sheet material that are further than 't' from the inside edges of the pads simply do not participate in the current flow. The value of 'a' is therefore irrelevant.

You speak of the resistance along the contact zone decreasing, and show plots for R(x)/R(0) as a function of x/a. If the sheet is isolated from the pads, and you are simply measuring the resistances between pairs of points on the sheet, then the plots look reasonable. However, as soon as the sheet and the pads are in contact, the isolated resistances between points away from the inner edges of the contact pads cease to matter. No current flows through the sheet away from those inner edges.

I think the model on which you've based your solution is a little different. You seem to be assuming that there is a highly resistive layer *between* the conductive pad and the resistive sheet. That's the only way you could sustain a voltage gradient across the resistive sheet above the pads and have a current flow through that material.

I agree that the ratio a/L is essentially irrelevant, and that adjustments to the length (L-2a) are largely unnecessary unless the sheet is very thick, in which case the adjustment would be based on sheet thickness, not a/L.

In this diagram, (a top view) which has a little smoother gradient that the last, "blueness" indicates voltage. We'll say the darkest blue is 5 volts and white is 4 volts. The voltage drop across the resistor is 1 volt. The entire left pad is 5 volts; the entire right pad is 4 volts. 1/20 of the way across the effective length from the left, the voltage is 4.95... etc. Changing dimension A has no effect at all on the effective length regardless of weather the ratio of L to A is 2 or 10... or 20.

As you say, if one were going for a real first or second order approximation, then sheet thickness and relative conductivities become more important, as does the connection effectiveness between the materials.

Yet another good answer, by the way. I try to keep my ratio of good answers to drivel to 10%. Your ratio is already much better than that.

AS I explained the ratio a/L was relevant in the concept of independent "current tube". It is irrelevant if the pad is not totally leading current to the resistor sheet.

As the computer simulation shows the current flows from the pad to the resistor over a limited zone so that in fact the voltage on the pad-resistor are is not constant all over.

As you have noticed I am not stubborn, as soon as I started to have doubts about the model I considered I tried to find a second way for the solution this is the reason I made the download and simulated the current in the sheet. Simulation which shows that the pads are not on all their surface flooded.

My error consisted in assuming the "current tube" as independent and for this erronated assumption the formula was correct from all points of view.

When I noticed that more said that the model was wrong I thought it over and decided to find the thinking flaw. It was this concept of independent "current tube".

I brought the proof that I was wrong and that I know why.

Shooting further is of no value since a wrong recognized concept is dead and does not deserve too much effort to be demonstrated as wrong.

I continue to thank to all since this helped me to see the flaws in my concept and this is of value.

My answer concerns your remark and the one of Rixter. It is true that within a pad there is no current between 2 points of SAME pad. But every point of a pad is via the resistive sheet (very thin) connected to the other pad and this leads to a current since there is a potential difference between the 2 pads.

In fact I should have done a double integral but I tried to simplify (not over a limit as you see the "L-2*a" approach is too much for big a/L ratios) by considering 2 very narrow slices on each pad -symmetrical- and the in between length of resistor as a "current tube" and made a simple integral instead of a double.

Even if, may be, the result is not perfect it is nearer to the exact one than the L-2*a approximation which is correct for low values a/L.

Never the less I thank for comments since it helped me to consolidate my model.

I'd think that if the current is high frequency, the effective length would be very close to the unsupported length, (L-2a) due to skin effect. In other words at high frequency, most of the conductivity is near the skin and the mounting area would have little effect.

At DC, the effective length would depend partly on the ratio of conductivity between the resistive sheet and the mounts. But you could assume, I think, that current is equally distributed over the cross section of the resistive sheet. If the mounting interface were less that this cross-sectional area, resistance of the assembly would be increased beyond that dictated by L-2a. So, the contact area should be at least equal to the sheet cross section area, and any additional contact area is superfluous, electrically. In this case, effective length would be extended at each side by 1/2 the sheet thickness, so Leqivalent = L-t.

I suppose that you could put a couple of these together, and test them. More obviously, you could model them in FEA... and perhaps find that I am completely wrong in my assumptions.

Oh, that skin effect of current flux density. When you mentioned skin effect I immediately thought of the skin effect from AC flowing through any conductor that caused the development of Litz wire.

Directly related to your astute observation (a GA from me by the way) is the relative smoothness of the resistive material and pad overlap area and any oxidation that happens between these materials. This will magnify the current flux density effect here.

Assuming that the pads have a much lower resistance than the resistive material, then almost no current will flow underneath the pads... the reason being that the voltage potential difference across the pad is approximately zero. So you should figure the effective length as L-2*a.

It is the case, the pads have a very low resistivity in comparison with the resistive sheet. This means that the resistive sheet is between 2 zones with a voltage difference in between. There is no current between 2 point on same pad as you very well notice but only between the 2 pads.

In fact one could say that every point on pad "right", for instance, gets current from all points from pad "left" and this has to be integrated on all pad "right". For every point to point current line there is a different R value.

This brings the difficulty, the distributed effect.

I can tell you that in the microelectronics field, R is calculated using L-2a as the length of the resistive element. Of course, these calculations assume that the resisivity ratio of the two materials is large, which is apparently true in your case.

Does the measured value corresponds to the assumed ? I had some surprises this was the reason for this attempt to find a better approximation.

The measured value normally corresponds to the assumed value. However, there are some things that can go wrong. For example, the two materials can react with each other forming an alloy or an intermetallic layer of unexpectedly high resistivity. This will complicate the situation.

Gentlemen,

I downloaded a soft with which I could do a numerical simulation. Here are the results. They are in between this shows that my hypothesis was not correct but that the distance limited at the free resistor between the pads is slightly too small.

The high density points are at pads end (red) but it is also a zone behind which s flooded at the density in the "free" resistor zone. The simulation is fora ration a/L=0.25.

The assumption that the "free" distance can be in 1st approximation accepted as resistor length is OK.

I tried to find a correct explanation for my error and now I know where it was.

Thanks again for all support.

I wanted to show the correct result since it is not important if I was right or not, important is what one of us says no compromise with truth.