Why can't you buy a 912-digit multimeter?

Isn't there any need for it? Is a 812-digit multimeter the latest high end you can buy? I have tried Keysight, Keithley, and Fluke, but there is nothing higher than 812 digits.

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    \$\begingroup\$ What is your goal? \$\endgroup\$ Jul 27, 2019 at 22:04
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    \$\begingroup\$ You'd need a 10.5 digit meter to calibrate it. \$\endgroup\$
    – Transistor
    Jul 27, 2019 at 22:09
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    \$\begingroup\$ What's the true accuracy of even that 8.5 digit DMM? Probably not 1 part in 100 million... \$\endgroup\$ Jul 27, 2019 at 23:03
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    \$\begingroup\$ Check NIST. I think you'll find that accuracy of their best Josephson Junction devices is on the order of a some parts in \$10^{10}\$. So, roughly speaking, about 9.5 digits. And that's under optimal conditions. The equipment used to provide such a standard costs about $300k each and will probably require a PhD to operate well. There's a recent 2018 paper on the topic, "Impact of the latest generation of Josephson voltage standards in ac and dc electric metrology" by Rüfenacht, et al. DOI: 10.1088/1681-7575/aad41a. You can buy a voltage standard that is 9.5 digits. Not a multimeter, though. \$\endgroup\$
    – jonk
    Jul 28, 2019 at 0:46
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    \$\begingroup\$ Every potential customer must answer for themselves the following questions: What are you trying to measure? Why do you require or want an accuracy of around a part in 1000 million? How would you calibrate it? What is your budget? \$\endgroup\$
    – Russell McMahon
    Jul 28, 2019 at 2:54

5 Answers 5


Four reasons:

  • Because modern meters have an autoranging function.
  • Because the dynamic range of the analog system would not support 912 digits, with a range of 1 V the noise floor would be in the nanovolts (you can't get lower than nanovolts because of thermal noise, without significant cooling of what your measuring, the cables and the meter to reduce the thermal noise temperature), and all of the digits below the 9th would be noisy.
  • ADCs usually have a 5 V range, and even with a 24-bit ADC, you'll have roughly 60 nV per bit which restricts the resolution of the last digits.
  • On a 6.5 digit meters that are commonly used, for most measurements around a normal lab have noise in the uV range. And the last digits are commonly noisy on a 6.5 digit meter. One more digit might be nice for some applications, 3 more digits would be frivolous.

Even nanovolt meters don't have 912 digits.

For most measurements 6 (or so) digits will suffice because great care must be taken to lower the noise floor below 1 μV.

Here is a cool scale that also illustrates the point:

enter image description here

Source: Understanding and Applying Voltage References

It's difficult to obtain gains larger than 140dB with an analog subsystem, and about that point your also limited on resolution. Gaining doesn't help because of the noise inherent in all analog electronics, you gain up the signal, you also gain up the noise.

The marketing departments can ask for more digits, but it's not going to help engineers.

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    \$\begingroup\$ Autoranging with relays can play havoc with sensitive experiments, so in the physics lab I work in we normally turn it off. Thus for one experiment we need a 6.5 digit multimeter to get 3.5 digits at the start and not saturate the photodiode at the end. \$\endgroup\$
    – Chris H
    Jul 28, 2019 at 18:20
  • \$\begingroup\$ You can take your signal, filter it, then amplify it with a gain of 2 or more, and measure three digits there (then divide by your gain for your measurement). This is usually what is done when you need more precision with a 10-bits ADC (like the one you find on most microcontrollers) for pretty much a couple of cents in parts. \$\endgroup\$ Jul 30, 2019 at 2:58
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    \$\begingroup\$ @DrunkenCodeMonkey Yes, this works for 10-bits, 10bits is equivalent to 3.5digits. There is no way to increase the SNR for the frequency's you are measuring. Filtering will reduce the overall pk-pk noise, but will do nothing for sensor noise. If the noise per given frequency is 10nv/Hz at the input then there is no amount of gaining or filtering to get better SNR. The other problem is every time you add a filter or analog stage, you increase the noise. \$\endgroup\$
    – Voltage Spike
    Jul 30, 2019 at 3:05
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    \$\begingroup\$ Seeing the dB scale there really makes it sink in just how precise an instrument an 8½ digit multimeter is... \$\endgroup\$
    – Hearth
    Dec 20, 2022 at 3:18

Putting aside the signal-processing challenges, let’s examine some noise floors.

A 62 ohm resistor produces 1 nanovolt/rtHz RMS noise at 290 Kelvin, and ignoring various crystal-flaw contributors, some of which are current-level dependent and may boost that nanovolt by orders of magnitude.

So we have a 1 nanovolt random noise floor, in a 1 volt input full scale range. If you constrain the effective-noise-bandwidth to 1 cycle-per-second.

This gives us 9 decimal digits, or 30 bits (or with sign, 31 bits).

How much input signal power must we have?

Using Vnoise_cap = sqrt(K * T / C) for a switched-capacitor filter, we learn a 10 pF capacitor at 290 degrees Kelvin will produce 20 microvolts RMS random noise. This noise comes from the SWITCH (e.g. a FET, as the FET turned off).

We need to reduce the noise floor by a factor of 20,000.

This requires a capacitor of size 10 pF * 20,000 * 20,000 = 4,000 * 1,000 * 1,000 pF.

Or 4 millifarads.

What sensor energy does this require?

Power = frequency * capacitance * voltage^2

Sensor power = 1 * 0.004 farad * 1 volt^2

Sensor power = 0.004 watts

What sensors produce 4 milliwatts? A moving-coil phono-cartridge with 10 ohms (resistance of the coil) may produce 200 microVoltsRMS output; using Power = Vrms^2/Resistance, we find Power = 4e-8/10 = 4e-9 = 4 nanoWatts; thus we should not expect 30 bit music from vinyl records, even for severely filtered tones.

Now, for fun, guess what is the effective-noise bandwidth of 62 ohms and 0.004 Farads? The -3dB corner is about 4 radians per second. Integrating from DC to infinity, you get 6.28 radians per second.

Ain't nature fun?


Aside from the matter of need and accuracy from what I understand, there are two other issues: Leakage and noise.

If you go to high voltages (e.g., measuring 100 volts to 9.5 digits) you run into leakage issues: the voltage causes tiny currents to flow between lots of different points (e.g., between the positive and negative terminal cables in a coaxial cable, inside the switches of the meter, etc), which makes your last digit not that useful compared to an 8.5 digit meter already out there.

But when you go to lower voltages, say 1 volt, you run into noise and thermal offset issues. The last digit on 1 volt would be 1 nanovolt. Given the input impedance you would want (as even the smallest loading will have effect at 9.5 digits), you need incredibly long measurement times to get rid of thermal noise. At that point, 1/f noise really comes into the picture and makes everything even worse. And as if it wasn't even enough: thermal voltages (voltage generated between two metals when there is a temperature gradient across them) can be on the order of microvolts!

So all of these things require incredible control to get around, beyond what is realistically possible in a lab (In fact, to get the true performance out of a 6.5 digit meter at the lower ranges you already need to take things like thermal EMF and leakage into account), unless you are doing extreme calibration. And in those cases, the absolute reference labs will usually use custom Josephson-junction based references, where cryogenic temperatures and quantum-physics are used to turn a measurement of time (frequency, really) into a measurement of voltage. These can cost many hunderds of thousands of dollars and require a lot of expertise to operate.


In a previous project I worked on, we built, tested and used precision voltage sources for Penning trap experiments. We needed \$100\,\text{V}\$ sources to be stable (i.e. precise, not accurate) in the sub-\$\mu\text{V}\$ range.

One problem with 8.5 digit multimeters and measurements at that level is that you have to deal with thermal potentials and contact potentials, which severely degrade your accuracy. Also, both effects are usually temperature dependent, which degrades your precision, unless you have good thermal stability of the test setup. If you had a 9.5 digit multimeter, you would have to have even better control over the measurement environment.

If you really, really need a 9.5 digit multimeter, current ADC technology is not sufficient. I suppose you could set up cryogenic Penning trap for that purpose. It would have to be custom-built, cost a few hundred thousand dollars and one to two PhD students. But it can be done! Calibration would be the most tricky part, but can be done against a Josephson junction array (primary standard).


Probably, there is a need for it, but not a big need. Not many people need that much accuracy, only some high end companies who probably make machines having also that much accuracy (for the parts that need to be measured with a 9.5 digit DMM). However, I can imagine there is a 'need' for it, or at least a wish.

The reason why there are none, is that it is probably very expensive to make one with that accuracy; if it is possible at all, it is too costly and nobody will buy it.

An analogy is a well known wafer stepper company which makes machines on nm accuracy. These machines are heavily depending on the quality of optical lenses. There are very few companies on this world who can make good lenses, and this wafer stepper company would like to have better lenses, but just at the cost that they can earn it back from customers.

  • \$\begingroup\$ Wafer stepper overlay accuracy must be below nanometers by now, no? Whether it's 10's or 100's of picometers, though, I'm not sure. \$\endgroup\$
    – The Photon
    Jul 28, 2019 at 0:10
  • \$\begingroup\$ @ThePhoton True indeed, overlay (which is the smallest accuracy for placing layers is in 0.5 nm now afaik, maybe even smaller), however internally for some steps picometers might be used. \$\endgroup\$ Jul 28, 2019 at 0:20
  • \$\begingroup\$ Also, if you have to measure or encode such precise values in an analog fashion, you'll try very hard to take them to the time instead of voltage domain ASAP. \$\endgroup\$ Jul 29, 2019 at 3:11
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    \$\begingroup\$ @MSalters Not of the new layer needs to be on top of the existing layer. \$\endgroup\$ Jul 29, 2019 at 11:20
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    \$\begingroup\$ that's what he said. \$\endgroup\$ Jul 26, 2020 at 12:12

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