Kelvin (4 wire) resistance measurement - i source and V sense leads not interchangeable?

Factual Questions
1

When doing a Kelvin or 4 wire resistance measurement, under what circumstances do you have to treat the current leads differently from the voltage leads? The context is accurate measurements in the 100 to 250 ohm range, and less accurate measurements in the 1 to 1000 milliohm and the 10 to 10,000 megohm range for the purpose of troubleshooting platinum resistance temperature measurements and measurement systems.

I ask because of resistance standards that have labels for “current source” and “voltage sense”, and not just red and black connections.

In the case of shunts for measuring large currents, which are a special case of Kelvin resistance measurement but are usually not called that, the current source leads and terminals are huge and the voltage sense leads are tiny, because huge leads are expensive. Note that special case and set it aside.

My nice Tinsley 100 ohm has a bigger and a smaller terminal on each side. My “air resistors” from IET have separate labels. Yet, my nice Hart readout devices all treat them as interchangeable.

While thinking about how to construct a zero ohm resistor I picture an ideal copper sphere with four equidistant holes into which 4 wires are soldered. The idea is that there is no pair of wires that should sense any voltage given that a large current flows through the other pair. I picture a FEM diffusion model of potential for this sphere and then try to extend it to situations like Vishay foil resistors with two leads coming out of each end, but have not really settled the whole thing.

2

In that range of resistance, I agree that the resistance of the leads will be negligible, so it shouldn’t make a difference if you swap the I-source and V-sense leads.

However, I have seen some measurement setups that have resistors in the V-sense path. Kelvin low-current meters (4156A/B/C) have this built into the meter so that you can operate it without the V-sense leads attached if you want - the resistor ties the V-sense internally. On the other hand, even here it’s built into the meter so the leads shouldn’t matter.

3

CurtC, do you mean there could be a problem with the current burden imposed by the voltage measurement (which is usually quite small) and the lead resistance on the sense circuit causing a drop in the voltage that remains to be measured? I guess this would be in cases of relatively low meter input impedance and relatively high lead resistance. You’d make this better by making the v sense leads fat - meaning the fat terminals are NOT for the current connections.

Or do you mean that in some meters like (I guess?) the Agilent/HP 4156A semiconductor tester there are internal resistors that weakly tie each v sense line to its corresponding current supply line, so that if the lead resistance is much larger than the internal resistor (as it would be if the lead wasn’t even present), the meter still gives a resistance reading? This sounds like a surprising compromise to do something that could have been done without error by a switch or even some kind of routine that recognizes when some leads aren’t present.

4

If a standard resistor specifies which terminals to use for I and which to use for V, yet they appear interchangeable, it’s probably because they want you to use the same terminals that they used when they calibrated it. There could be some subtle differences between the terminals that are electrically detectable. Why? V[sup]+[/sup] and I[sup]+[/sup] are connected (usually soldered) at a point (sphere) with finite dimensions, and V[sup]-[/sup] and I[sup]-[/sup] are connected at a point with finite dimensions. The electric current flux and field lines will probably be slightly different depending on which leads get the current. In other words, the effective electrical resistance may be slightly different depending on which leads are connected to the current source due to the (differing) locations of the electrical sense lines in each case.

Another thing to think about is thermoelectric voltages showing up on the voltage sense lines. This is a function of the alloy used to make the voltage sense wires. If you are not averaging out the thermoelectric voltages (using a variety of methods), then there could definitely be a difference between the different arrangements.

5

With modern meters, a voltage drop along the sense lines usually isn’t an issue for most measurements. A much bigger issue is thermoelectric voltages on the sense lines. In any case, I never use large gage wire for the sense lines, and never use overly-large wires for the current lines. Why? Our resistors are in an oil bath. The bath temperature is not going to be the exact same temperature as the surrounding air. A large gage wire acts as an efficient heat path between the resistor and surrounding air. If the room was fairly cold, for example, heat would flow from the resistor (which is kept at 25 C in the oil bath) to the outside air via the wire. This will cool the resistor, and hence is undesirable.

6

Crafter_Man, thanks for coming along; I was hoping you would.

Your point about thermoelectric voltages is important and I forgot to mention that the meters I am using do DC measurements one way and then the other, changing the polarity and averaging the two to cancel out these emfs. I would guess that if this wasn’t the case, thermal emfs would be the biggest error in this system.

Your point about lead wires soldered to a sphere with finite dimensions is the problem I was thinking of with my copper sphere, or more accurately a relative of it for one end of a resistor rather than for a zero ohm resistor. In fact yesterday afternoon I tried making myself a zero ohm resistor by joining four copper wires in a big wad of finer copper wire, trying to make a sphere with four equidistant connection points on its surface, like the carbon in a molecular model of methane. I got about an 8 mm wad that was kind of spherical and kind of equidistantly sprouting the leads, filled with solder. Later today I will try it.

I do a lot of finite element modeling and have been trying to picture this problem in terms of it, wondering if some basic principle says that any four wire nonreactive resistor will measure the same way as long as you consider these two leads as one side and those two as the other, or trying to envision an exception to that proposed rule. Also I am trying to picture a resistor network with terminals A, B, C and D to the outside world such that 4 wire measurements with A and B as one side and C and D as the other side would give different results. I haven’t come down clearly on either side of the question, and may go wire up a dozen random resistors between 1 k and 10 k to see if a 4 wire measurement of them is independent of lead swapping. That would be a pretty convincing test, I think.

7

By the way, how much do you worry about parasitic conductances in your wiring? That is, how often is the insulation leaky enough that it changes your measurements? Would you happily dive in to setting up automated test setups with switches and reed relays, or would you instinctively veer away from all those possible leakage paths between contacts and the like?

I’ve addressed this a little with leakage specifications (which of course are for products as made new) and calculation. I have on order a tester for high resistance, a low voltage one that is not trying to test breakdown. When I have it I am going to pull rolls of multiconductor cable off the shelf and find old lengths of 4 wire RTD cable of FEP or woven glass hiding in drawers, and look at cable runs whose ends I can temporarily disconnect in operating equipment, and try sensor-to-sheath measurements on RTD assemblies, and otherwise build up some experience.

8

I think the following experiment strongly supports the view that interchanging the sense and current leads does not change a Kelvin resistance measurement.

I made a triangular network of resistors. There are three apex nodes constructed with several twisted bare leads and saturated with solder to construct a short fat stub to clip onto. There are 15 resistors, all with values of at least 1 ohm and less than 2 ohms, 5%, 1/4 W. I forget which, but they’re either metal film (liklier) or perhaps carbon. The resistors meet at 8 different nodes including the 3 apex nodes. I tried to construct this in a random way, avoiding symmetry with respect to any of the three apex nodes.

Then I measured a Vishay precision foil resistor plus this network. On one end of the precision resistor were one set of sense and current leads. The other end of the precision resistor connected to one of the three apex nodes of the network, at a binding post. Then I put the other set of sense and current leads on the remaining two network nodes. The question becomes: if I interchange these sense and current leads between the two remaining network nodes, does the measurement change?

I can try this experiment 3 times, because I can pick any of the 3 apex nodes to tie to the precision resistor at the binding post. Of course, some of the network, especially the resistors terminating at the binding post apex node, is part of the resistance being measured, so trials using different nodes at the binding post should produce different measurements. It only matters whether swapping the sense and current between the other two apex nodes changes the reading within one trial.

I used a Hart Scientific 1502A “Tweener” readout to measure the resistance. Among other refinements, this does DC measurements in opposite directions and averages out any thermal emfs or similar effects.

On trial 1, I got 100.351 and 100.351 ohms. On trial 2, I got 100.239 and 100.239 ohms. On trial 3, I got 100.359 and 100.359 ohms. I conclude interchanging the sense and current leads did not change the measurement. This conclusion requires assuming there is a cancellation of thermal emfs and similar effects, as it is obvious that a small DC voltage impressed on a part of the network between the two nodes with leads on them would have to change the reading if there were no cancellation.

I also tried measuring the resistances between apex nodes of my network. If these are identical (unlikely considering how I constructed it), my above conclusion is moot. I got 0.6261, 0.7461 and 0.6335 ohms, so these resistances are different at the 0.1 ohm level, while my earlier results didn’t detect any difference at the 0.001 ohm level.

All this suggests there is some way to prove interchanging the leads won’t matter, and further suggests that assuming as much is a practical tactic in further work, though it isn’t really proof by itself. Again, this is contingent on there being no impressed emfs or a cancellation of them by reversing polarity during the measurement. However, unless it occurs to me how to prove the matter, or somebody else posts something convincing either way, I’m going to proceed on the provisional basis that it is true; my job is actually to use such measurements for other purposes, not to establish a proof.

9

Leakage may - or may not - be a significant factor. You have to do the math and determine how much leakage would be a problem.

Our ASL F18 bridge is about 20 feet from our resistor bath. I wanted to use twisted pair shielded cabling between the resistor and the bridge. (Two cables, each cable containing two conductors plus shield.) But then I had a quandary: should one twisted pair cable be for the sense voltage, and the other for the current? Or should one twisted pair cable be for V[sup]+[/sup] and I[sup]+[/sup], and the other for V[sup]-[/sup] and I[sup]-[/sup]?

My initial thought was to use one twisted pair cable for the sense voltage (V[sup]+[/sup] and V[sup]-[/sup]), and the other for the current (I[sup]+[/sup] and I[sup]-[/sup]). I liked this arrangement because it minimized loop area (and hence magnetically-induced noise), and it helped minimize noise on the voltage signal. But… it also meant some of the current would not be flowing through the resistor - some of it would be flowing through the insulation between the two current wires. So I tried the other arrangement (one twisted pair cable for V[sup]+[/sup] and I[sup]+[/sup], and the other for V[sup]-[/sup] and I[sup]-[/sup]). There didn’t seem to be a noise problem, and I didn’t have to worry if some of the current was flowing through the current wires insulation. We are still using that arrangement and it appears to work fine.

Note also that, the lower the value of the STD resistor, the less of an issue this is, obviously.

10

You could always use shielded cables for the leads, with the shields tied to the center conductors at the meter end for the current leads and tied to each other at the meter end for the sense leads. That way the leakage current would be between shields, like a guard ring on a circuit board or a conductivity tester. Note I am using the shields here as leakage guards and not EMI shields as intended.