Hand-matching resistors: high-end DMM vs Wheatstone bridge

Question

I wrote an article dealing with the question of hand-matching resistors to high tolerances using a DMM. The driving purpose behind that article is to show that it's harder to do that than one might naïvely think, rather than to come up with the best possible way to achieve the end goal. So, for the purposes of this question, let us say that my article is unimpeachably correct within that limited scope.

What I want to do here is re-ask a question I got via email: would it be better to use a Wheatstone bridge instead?

From Wikipedia:

The idea behind the question is that a few precision resistors plus an eBay'd galvanometer (V_G) would be cheaper than a DMM good enough to achieve the match.

It seems to me there's a serious problem with this idea, which is that to get the benefit from the low-current measuring ability of the galvanometer, you can't make R2 adjustable, as you would in setting up a Wheatstone bridge to measure an unknown resistor. All that does is buy you a new measuring problem, either:

1. a resistance measurement, the now-unknown value of R2 once you've finished nulling the differential current across V_G; or

2. an angle measurement, if you've made R2 a rotary pot

I initially thought problem #2 would be easier to tackle. You could use a precision pot and an indicator dial knob on it large enough to get the angle measurement accurate to the degree level. For the 1000 ppm sort of measurement I discuss in my article, a 10 turn pot takes care of 1/10 of that, allowing the angle measurement to be accurate to within a few degrees. It seems suitable pots for this range from about $15 to $150, no doubt being a function of repeatability of measurement and such.

The problem is that the pot itself isn't perfect, so all you've done is turn it back into measuring problem #1. You merely get the choice to fully characterize the pot up front or instead treat it as an unknown from the start and measure its value at the end. You therefore still need a high-precision ohmmeter somewhere. I guess it could live at a cal lab somewhere so you don't have to pay for it, but you do still have to "rent" it via the calibration service.

Therefore, instead of making R2 adjustable, I think it would be better to buy three precision resistors instead of two: R1 = R2 = R3. The precision of the resistors has to be at least twice the accuracy you need for the match, so you're talking about at least $20 in parts here, and possibly much higher. On top of that, you now have to build a separate bridge for each resistance value you want to measure.

Yes, I understand that R2 doesn't have to equal R_x, but if they aren't close, you've wasted the potential (ahem) of using a galvanometer, haven't you? Isn't the idea to turn the resistance measurement into a high-precision near-zero current measurement?

Answer 1

The whole point is to match 2 different resistor values to each other, correct?

In that case, get 2 high quality fairly closely matched resistors R1, R3.

Then irregardless of the actual values of R2 and Rx, as long as they are closely matched the voltage divider circuits R1/R2 and R3/Rx will produce a near-zero voltage Vg. The smaller the voltage, the closer the match (assuming very good R1/R3 matching). You can now use R2/Rx in your circuits and they will be pretty well matched to each other.

There is another issue with resistor matching, though: temperature coefficients. These are often specified as max values and there's no telling if two given resistors will drift at the same rate or even in the same direction.

Answer 2

As Warren's original article referred to balancing stereo amps (CMoy) by matching the gain setting resistors; I also suggested that perhaps this could be done without the precision resistors by putting the left and right feedback resistors in the top of the bridge (R1 & R3) and the matching L&R ground resistors (non-inverting op amp) in the bottom (R2 & Rx). If I'm right, this would allow the Left and Right gain setting pairs to be matched pretty accurately using just a Galvanometer (or other suitable meter). Personally I would probably sort the resistors using the first method, before using the second method to pair them up.

Answer 3

It is possible, even with crude technologies such as existed 100 years ago, to make a potentiometer which may be modeled as an ideal pot which is which has a very small (less than 0.01%) bit of mechanical slop between knob placement and wiper position, and a not-so-small series resistor on the wiper whose value may change by orders of magnitude as the pot is moved around.

If one tries to replace a single resistor in the bridge with the pot (using the wiper and one end of the end terminals), accuracy and precision will be very poor because of the essentially random wiper resistance. On the other hand, if one replaces both R1 and R2 with the sides of the pot, accuracy and precision may be excellent. If one maps out the highest and pot setting where the meter visibly reads "negative", and the lowest setting where it reads "positive", the correct setting will be between them. The lower the wiper resistance, the closer those two settings will be (and thus the greater the available precision), but even monstrous variations in wiper resistance won't cause erroneous readouts; they'll simply increase the amount of uncertainty that's known to exist [e.g. with an "ideal" pot, a particular meter might be sensitive enough allow a resistor to be measured as being between 991.2 and 991.7 ohms; a pot with annoying variations in wiper resistance may only allow it to be discerned as being between 989.3 and 996.3--less precise, but no less correct].

Prior to the invention of calibrated meter movements, potentiometers were the key to making precise electrical measurements (thus the term "meter" in the name).

Answer 4

The original question is two years old, but I'm going to have a go at it anyway. In one sense the potentiometer can be replaced by a fixed resistor, from the same batch that supplied R1 and R3. then replace Vg with a DMM that can measure mili-volts or better. What that creates is a binning mechanism, whereby you can plug in various examples of Rx, and get a relative measurement of where they exist within the spectrum of values from the original batch. Examples which exhibit very close relative readings, are then candidates for closer examination. If you have 4 (or more) candidates within a given bin, you can then plug them into the wheatstone and see if they really give a 0.0 reading across the center. If they do, then swap opposite corners and see what happens. A continued 0.0 suggests that you have a obtained matched set of 4. You do not know the specific resistance, only that they match.

Answer 5

No point in answering a very old question, unless you can bring something new to the party, which is not mentioned in any of the existing answers.

What a Wheatstone bridge arrangement does is make it possible to measure a resistance to high resolution, without an expensive high resolution DMM. Essentially what the bridge does is to subtract a large constant voltage from the voltage across the unknown resistor, leaving only the LSBs to be measured at the bridge output by a cheap DMM.

To this end, the bridge does not need any expensive high quality resistors, only what are in the original nominally same resistance batch being sorted. Leave the three resistors in the bridge, and measure the voltage across the bridge, to mV, when the 4th resistor is changed.

However, this is not a pre-screening a la Cosmicray to see what resistors are suitable for a closer look. This is a measurement of resistance, less a large, but consistent, constant. As you cycle each resistor in your batch through the 4th position, write down the mV measurement against it. When you have 4 resistors with the same reading, you have found your 4 matching resistances.