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So, I'm starting on the front-end for my digital project, figured out the ratios (100kΩ, 900kΩ, 9MΩ for ÷10 and ÷100), tolerances (0.05% - 0.1%) and temperature coefficients (< 50ppm/°C might work) I wanted and went to buy parts and was surprised to see that there are barely any 0.1% low tempco resistors in the megaohms range. Why is that? Even switching to 9.09M to stay within E96 and E192 still only gives a handful of results...

I did find the Caddock 1776-C67 and 1776-C671 which are better because of matching relative ratio tolerancs and temperature coefficiencients; however, I can't seem to find more options.

There's nothing wrong with the 1776 series; in fact, it's perfect, but it makes me think that I'm doing something wrong if this is the only option (or at least there are very few options.) Am I not supposed to be using a resistive voltage divider or at least not in these decades?

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  • \$\begingroup\$ If you open up a precision multimeter you will find laser trimmed resistors on ceramic carriers... \$\endgroup\$ – PlasmaHH Jan 16 '18 at 15:36
  • \$\begingroup\$ Do you need 0.1%? I was using some 10Meg metal film, 1% (but I hand selected some to be closer to 10Meg.) I found that soldering them into place increased the resistance by ~0.1% or so. \$\endgroup\$ – George Herold Jan 16 '18 at 15:52
  • \$\begingroup\$ I'm not really sure you have asked a meaningful EE question. \$\endgroup\$ – Andy aka Jan 16 '18 at 16:01
  • \$\begingroup\$ Why's that, Andy? I'm asking whether the design is incorrect (which implies the question of whether there's another way to do it based on the error margin which you can infer from what I was seeking) and the question furthermore hints at asking how other meters do it. Joren seemed to have understood all of that and answered perfectly. \$\endgroup\$ – Anthony Jan 16 '18 at 17:21
  • \$\begingroup\$ @GeorgeHerold, for tolerance, no, only the tempco matters. I was already considering calibration like in Joren's answer but I was just surprised I couldn't really do it that way even if I wanted to. Thanks for adding your experience though, that's helpful. \$\endgroup\$ – Anthony Jan 16 '18 at 21:39
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This somewhat depends on what multimeter class you are dealing with. Rather surprising, the top-of-the-line bench multimeters (6.5 digit and up) don't rely on high-accuracy resistors at their input. Those fancy laser-trimmed resistors are rather for the ratios between different reference voltages and such in the actual multislope A/D converter.

What they need is a resistor that won't drift - both over time and temperature. They could very well be 5% tolerance parts, as long as they do not move away from that. Why? They can just calibrate any error out. If you know the error, in the divider, and you just apply the opposite transform to your measured data.

The reason they do this is because it is just impossible to get a resistor that is that accurate, and even if you did, just hooking it up to a circuit will blow that rating away. For illustration, let us consider a 0.001% resistor, which is the best I have ever seen on the market (and we are talking >100 EUR/USD per resistor here). That is still only 5.5 digits. It is far easier to just get a resistor that doesn't drift, and compensate for it after the fact.

Also, most of these high range multimeters actually have a much higher input impedance at low ranges. Of the top of my head, the Keithley 2000 has a rated input impedance of >10 GOhms for all but the two highest ranges (so up to 10 volts full scale). The range changes the gain of the amplifier, and doesn't use a input divider to scale. (ofcourse, at 100 V and 1000 V range this is not really possible since you won't find a JFET that can handle anywhere near that voltage. They use JFETs for the combination of input impedance and 1/f noise).

In lower-end multimeters (hand held ones) you might indeed find precision matched networks on ceramic carriers to do this input division. Alternatively, you can also use calibration, though that might not be desired as it is an extra processing step.

TLDR; the reason you can't really find them is because in the applications you want to use them, people don't need that accuracy - they use calibration instead.

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  • \$\begingroup\$ Thanks. This was exactly what I was looking for. Already accepted. It's funny you mentioned the Keithley because I almost added that as a second part to the question... I was thinking to myself... I know they must have much higher input impedances and I'm pretty sure they don't just hook up buffer and in-amps directly to the input! Thanks a lot. \$\endgroup\$ – Anthony Jan 16 '18 at 17:23
  • \$\begingroup\$ I mentioned the Keithley mainly because that is the meter I use by far the most in my lab, and hence am also most familiar with. Make no mistake - there is still a lot of protection circuitry in front of that buffer (after all, you need to be able to just plug in 1 kV and it has to take it)! That is the advantage of having a buffer with many, many gigaohm input impedance (likely limited by leakage current and not the buffer itself) - You can put a lot in series with the signal without introducing errors because of it. \$\endgroup\$ – Joren Vaes Jan 16 '18 at 17:46
  • \$\begingroup\$ I was looking through its manual and repair manual. Really impressive. Unfortunately, somewhat obfusciated circuit and parts. Also, I think you meant "for all but the two highest ranges", not lowest, right? (or maybe lowest resolution?) I just wanted to get one clarification (this is getting off topic), do you mean that the input to the ADC frontend is ~10V? Thus for the 10V range, it's just a unity gain (JFET) buffer? \$\endgroup\$ – Anthony Jan 16 '18 at 21:22
  • \$\begingroup\$ I think that is the case. This is quite common in multimeters, I repaired at least 2 others meters that had the same. \$\endgroup\$ – Joren Vaes Jan 16 '18 at 21:25
  • \$\begingroup\$ Alright, that makes sense. Thanks, very interesting and helpful. \$\endgroup\$ – Anthony Jan 16 '18 at 21:35

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