The SI ampere is still defined via the force between two conductors. But that is impractical to realize, so standard labs derive it via Ohm's law. There is some work going on to redefine the SI ampere based on the charge of a proton. But the BIPM members are discussing that for years now, and they don't seem to be in a hurry to do the change.
The resistors will have something like a "bell curve" normal distribution. If they've got a standard normal distribution, the chance of getting all four of them at the very bottom or very top of the range is very slim. The very worst case would be that the percentage error in the combined resistance would be as bad as the percentage error on one of them. But the average case is much better than that, even if you're not specially selecting the resistors in matched sets.You can't assume a normal distribution (although the Fluke engineers might have had additional information), especially not one centered around the nominal value, so worst case is that you should assume a uniform distribution within the stated tolerance. Still, the standard deviation (and any confidence interval you care to calculate) of the parallel (or series) combination will be smaller than the single resistor. I believe the freely available book 'Analog SEEKrets' by Leslie Green contains a decent discussion of this topic.
Dave said something that interested me. There were four 40 ohm resistors in parallel to make 10 ohms. Each resistor was at 0.05% tolerance and he said having four of them makes that better than 0.05%. Why would that be? Obviously later in the video he discovers they are all specially matched which could make this true but he said this before knowing that detail.
So each 40 ohm resistor could have a resistance between 39.98 and 40.02 ohms and still be within tolerance. If each resistor just happened to be 39.98, you would have a total resistance of 9.995. That is still 10 ohms at 0.05%.
The resistors will have something like a "bell curve" normal distribution. If they've got a standard normal distribution, the chance of getting all four of them at the very bottom or very top of the range is very slim. The very worst case would be that the percentage error in the combined resistance would be as bad as the percentage error on one of them. But the average case is much better than that, even if you're not specially selecting the resistors in matched sets.
If you're randomly selecting the resistors, you can only put a statistical confidence level on the likelihood of the worst case not happening. But if you individually select the resistors, combining high ones with low ones and medium ones with each other, you can guarantee that the worst possible case doesn't happen.
When fluke describes the resistors as matched sets, they're doing something far better than relying on reducing standard deviations. As mentioned in the video, they'll be binned, that is, very accurately measured and grouped ("binned") according to measured value. Take a resistor from the 50.004 ohm bin and pair it with a resistor from the 49.996 bin, and hey presto, you're doing much, much better than the root-4 improvement you can expect to get from random chance.
I'm asking for help understanding the math on that aspect because I can see how if you select a random set of four, on average the resistance should be on a normal distribution but each resistor is probably on a normal distribution too. Only a few would be on the extreme value but that's why there's a normal curve instead of them all being an exact value. However, can it be stated mathematically that if you combine four items on a normal curve, the final product will have a non-normal curve but will be more narrow? How does that math work?
I believe the freely available book 'Analog SEEKrets' by Leslie Green contains a decent discussion of this topic.
See chapter 3. Google should give you a link to the (legal) PDF version within a few seconds, hosted on this very site. It explains when and when not to use statistical tolerances, and gives some rules of thumb.
But I'll explain the basic case with the assumption that the resistors follow a normal distribution, which is not necessarily a fair assumption (see the book for reasons why and how to handle that). To keep it simple I'll first explain the series case. Assume the resistance of the 40 Ohm resistors is normally distributed with a mean of 40 Ohm and a standard deviation (or any multiple of the standard deviation, like a 95% confidence interval) of 0.05% * 40 Ohm, or 0.02 Ohm. The probability distribution of the total resistance of the four resistors in series is the sum of four normal distribution, which is another normal distribution with mean = sum of the means and variance (which is the standard deviation squared) = sum of the variances. If the total variance is the sum of the variances, then the total standard deviation is the root of the sum of squared standard deviations. So in this case the mean total resistance would be 160 Ohm with a standard deviation of sqrt(4*0.02^2) = 0.04 Ohm. This is 0.025% of the mean, i.e. half that the tolerance of the original resistors. A quicker way to calculate this for identical standard deviations is sd / sqrt(n), where sd represents the standard deviation of the individual resistor and n the number of resistors.
For the parallel case you can show that for normal distributions which are well above zero, the distribution of the reciprocal value is another normal distribution with mean 1/40 Ohm and standard deviation 0.05% * 1/40 Ohm. You can then add conductances to get the conductance of the four resistors in parallel, and again get a factor sqrt(4) improvement in standard deviation.
Note that this is about what the majority (eg. 99%) of the circuits is doing, as you note the worst case does not improve.
If you were to design a resistance reference box like this today, wouldn't it be better to use solid state relays instead? You wouldn't have to worry about contact resistance issues and the leakage is very low like 0.006 pA at 80 degrees C. And only 0.003 pA at 40 degrees C. And if you're concerned about that small leakage, putting a couple in series couldn't hurt because they are only around $1 each.
Without doing much research about the best ones to use, I picked a random one and looked at the specs and they are pretty good compared to a conventional relay. http://www.clare.com/home/pdfs.nsf/www/CPC1016N.pdf
With conventional relays, the contact resistance can be low in the beginning but increase in time as the contacts become worn and seems that can mess with the calibration as you use it. With the solid state relays, I would think the resistance would be consistent throughout the life of the product. The internal resistance of the IC could be compensated with different value resistors if that is important.
Without doing much research about the best ones to use, I picked a random one and looked at the specs and they are pretty good compared to a conventional relay. http://www.clare.com/home/pdfs.nsf/www/CPC1016N.pdf
Where are these 0.006pA leakage current solid state relays? The one you linked has 1uA leakage, which is certainly unacceptable. I'd be pleasantly surprised if you could find a reasonably priced solid state relay that had a reasonably low on-state resistance (all the ones on digikey are basically immediately too high for 2-wire measurement; as I mentioned in an earlier post, though, on-state resistance is irrelevant in four-wire mode unless the relays are burning out or loading down the current source) and low leakage current (beware the distinction between control-to-relay leakage, and off-state leakage).
One point that comes to mind, at least for two-wire measurement, is that the on-state resistance would be a function of relay drive voltage. If you want the on-state resistance to be very repeatable (a far greater concern than absolute value), this is a very bad trait. At least a relay is very much either on or off.
It has an on-resistance of max. 16 ohm. A mechanical replay can get easily 16 milliohm. That's 3 orders of magnitude better. And the resistance of the solid state relay changes with temperature. That's unacceptable for a calibrator.
It says it in the graph titled "Typical Leakage vs. Temperature Measured Across Pins 3&4" The pA was too small on the pdf so I had to copy and paste it into a text editor to confirm it was pA and not uA or mA.
It says it in the graph titled "Typical Leakage vs. Temperature Measured Across Pins 3&4" The pA was too small on the pdf so I had to copy and paste it into a text editor to confirm it was pA and not uA or mA.
On that graph I read :
- 0.006µA @ 80°C
- 0.002µA @ 20°C
@cengland0: I see uA too
graph attached.
The SI ampere is still defined via the force between two conductors. But that is impractical to realize, so standard labs derive it via Ohm's law. There is some work going on to redefine the SI ampere based on the charge of a proton. But the BIPM members are discussing that for years now, and they don't seem to be in a hurry to do the change.
Naa, that's not correct. The Proton experiment was too unprecise so it's not followed up.
... then defining fixed (exact) values of e, ...
But that is the charge of a proton, isn't it? I thought the idea was fixing e (one value in Coulomb = As), and then "simply" deriving the Ampere from the charge.
Great video!! @ 27min I'm guessing the low resistor are wound on mica card.