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?
The idea behind the question is that a few precision resistors plus an eBay'd galvanometer (VG) 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:
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a resistance measurement, the now-unknown value of R2 once you've finished nulling the differential current across VG; or
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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 Rx, 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?
Best Answer
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.