Root Sum Squared Error Analysis - Page 1
Forum section: electronics / projects

author: edba
author_rank: Contributor
timestamp: May 01, 2025, 04:32:53 pm
content: HiIf you have two resistors in series that are say 100 ohm and 10 ohm with tol 1% then RSS analysis suggests it will have a resistance of 110 ohm +/- SQRT(1%^2 + 1%^2) = 110 ohm +/- 1.414%. This makes sense as it is less than worst case analysis of +/-2%. What if the resistors are in parallel? What form does the RSS calculations take?Edba

author: moffy
author_rank: Super Contributor
timestamp: May 01, 2025, 04:43:55 pm
content: If 100 is +1% high = 101if 10 is +1% high = 10.1result: (101 + 10.1)/110 = 111.1/110 = 1.01 or a +1% error

author: Tation
author_rank: Regular Contributor
timestamp: May 01, 2025, 05:15:32 pm
content: Revise the calculations because 100 + 10 (1 % tol.) gives 110 with 0.913 % tol. Already discussedhere.Also discussed, in the same thread (and post), the parallel case.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 01, 2025, 09:33:25 pm
content: I'm really struggling with this. Assuming the OP's 100 ohm and 10 ohm 1% resistors, in series, then it is completely obvious that the maximum possible value of resistance when both resistors are 1% high is 111.1 ohms, which is 110 ohms + 1%.So it is surely the case that putting 1% resistors in series MUST give a tolerance of 1% on the summed value.It seems to me that RSS analysis gives aprobability curveof some kind. That is, RSS saysthe probability of both resistors being a full 1% high is lower than the probability of an individual resistor being 1% high. That feels intuitively true.However, it is surely wrong to claim that a pair of 1% resistors has a tolerance of less than 1%.The definition of tolerance is the worst case situation:1% tolerance means the worst case error is 1% high (or low). The worst case error for a pair of 1% resistors is - must be - 1% high or low.So I think it is misleading to say that - thanks to RSS - the tolerance of a pair of 1% resistors is less than 1%. It is still 1%, but the probability curve of the combined values is narrower - that is, the combined value is more likely to be close to the centre of the tolerance band.Am I wrong?I also suspect another shortcoming of RSS: it doesn't (can't) take into account the probability curve of the individual resistors' tolerance band, because it doesn't know it.

author: bdunham7
author_rank: Super Contributor
timestamp: May 01, 2025, 09:46:28 pm
content: Quote from: SteveThackery on May 01, 2025, 09:33:25 pmNot necessarily, you'd have to actually define the term for any of this discussion to make any sense. If the resistor values are a standard distribution and the tolerance is defined as x std devs, then the RMS calculation will give you a new tolerance also as x std devs. If the distribution is rectangular, not random or even completely unknown then the math changes. In any case, when it comes to these matters many assumptions are often made and not all of those have solid support. If the spec is absolutely guaranteed--say by selection--and you want the result to also be absolutely guaranteed then you just add the tolerances as your intuition suggests.

author: Tation
author_rank: Regular Contributor
timestamp: May 02, 2025, 08:18:20 am
content: If:the resistance of the resistors distributes normallyand the tolerance is defined as being proportional to the standard deviationand the tolerance is smalland the resistances of different resistors are not correlatedthen the series or parallel combination of 2 or more resistors also distributes normally and its tolerance (defined as above) is less than that of the original resistors.Under other sets of assumptions the result may be other, of course.

author: edba
author_rank: Contributor
timestamp: May 02, 2025, 08:36:32 am
content: Hi I think I have worked it out:For two resistors in series each 1 % tolerance:Worst case gives:100 Ω + 10 Ω +/- (1 Ω + 0.1 Ω) =110 Ω +/-1.1 Ωor110 Ω +/- 1 %RSS gives:100 Ω + 10 Ω +/- √((1 Ω)^2 + (0.1 Ω)^2) =110 Ω +/-1.005 Ωor110 Ω +/- 0.91 %Now two resistors in parallel each 1 % tolerance:First of all convert to admittances:1/100 Ω = 0.01 S +/- 0.0001 S, 1/10 Ω = 0.1 S +/- 0.001 SNow use the same formula:Worst case gives:0.01 S + 0.1 S +/- (0.0001 S + 0.001 S) = 0.11 S +/-0.0011 S= 9.09 Ω +/- (0.0011 S/0.11 S)=9.09 Ω +/- 1%RSS gives:0.01 S + 0.1 S +/- √0.0001 S^2 + 0.001 S^2 = 0.11 S +/-0.000100499 S= 9.09 Ω +/- (0.000100499 S/0.11 S)=9.09 Ω +/- 0.92 %With the formulas you can have resistors of different values and tolerances,BrgdsEdba

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 10:29:18 am
content: Quote from: bdunham7 on May 01, 2025, 09:46:28 pmAbsolutely right.But that is not how resistor tolerances are given.When you buy a 100 ohm 1% resistor:1/ All you know is that it is somewhere in the range 99 to 101 ohms.2/ You are given no information on the shape of the distribution curve.3/ If the resistors are assigned to tolerance bins after manufacture, you can be virtually certain that the values donotfollow a normal distribution curve, or anything like it.This is the problem with mathematics - the hidden assumptions built into it may be far away from reality. I'm guessing the RSS method assumes a normal distribution curve around the nominal value?

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 10:44:32 am
content: Quote from: edba on May 02, 2025, 08:36:32 amWhat is the point of the RSS calculation when it is wrong? Or rather, the assumptions it makes (normal distribution?) are untested.If you are designing a circuit in which resistors arenotindividually hand-picked for their value, then you must design the circuit to work over the full range of possible values: from -1% to +1%. If you design for the RSS value then some of your circuits won't work. Why would you do that?(I'm assuming for the sake of argument that the resistor values are critical to circuit function; I realise that is rare.)

author: gf
author_rank: Super Contributor
timestamp: May 02, 2025, 10:48:55 am
content: Quote from: SteveThackery on May 02, 2025, 10:29:18 amMy understanding is that variance(a) + variance(b) = variance(a + b) holds for random variables a and b with arbitrary probability distributions (even for different distributions for a and b), but a and b must beuncorrelated.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: May 02, 2025, 12:40:22 pm
content: I've seen RMS used for many things because people crave some reasonable sounding information. I think we use it because it's easy and we don't have enough information to use anything better. Does it actually predict reality? I suspect it does poorly at best.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 02, 2025, 01:29:32 pm
content: Just remember that resistor value errors definitely can be correlated! Assumptions of errors averaging out to better accuracy are wrong because manufacturers sometimes do binning of parts: they sell parts that fail the tighter tolerance as lower tolerance parts: ten 1% resistors canallbe off by 0.5% to 1% all in the same direction!Besides, even if they were completely randomly distributed, statistical analysis makes no sense at N=2 or N=3 or so. The chances that two resistors are both +1%, is just too high for any practical use.So I fail to see what is the practical purpose of such analysis. Just do the good old worst-case analysis instead.

author: bdunham7
author_rank: Super Contributor
timestamp: May 02, 2025, 02:04:19 pm
content: Quote from: SteveThackery on May 02, 2025, 10:29:18 am1/ You don't really even know that for sure2/ Most of the time, but some precision lines do give you additional information.3/ Binning is one issue and IDK how common it is nowadays. I think modern process control methods are more likely to give you batches of components that are all very close in value to each other but only within the tolerance of the nominal value.Yes, a standard distribution around the nominal value is assumed. I don't know that this assumption isalwayswrong, but I suspect it usually is. But I also think it is a mistake to replace it with another unsupported assumption such as the specified tolerance indicating the "worst case". If your needs are critical then you should investigate further or characterize the parts on your own.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 02:30:26 pm
content: Quote from: bdunham7 on May 02, 2025, 02:04:19 pmWhat? Of course you do! 1% tolerance means plus or minus 1% of the nominal value. What else can it mean?

author: eugene
author_rank: Frequent Contributor
timestamp: May 02, 2025, 02:38:44 pm
content: The problem I'm having with this discussion is the assumption that the actual values of components in a batch will fall in a normal distribution around the nominal value. If that's truly the case then RSS does make sense and you could put four 1% resistors in series/parallel and end up with the same nominal resistance with a tolerance better than 1%. In this case tolerance really is a matter of probabilities because the claim is that any random sample will fall within, for example, 6 sigma of the entire distribution.I would guessthat a more likely situation with modern manufacturing is that the components in a single batch fall in a normal distribution around themean valueof the batch. That mean value might not line up with the nominal value. However, presumably some minimum percentage of the batch will still fall within the tolerance of the nominal value. But all of this is just a guess and is not based on any actual knowledge of the industry. (I did once work at a company that made thick film resistors, but there every resistor was laser trimmed by ablating any of a finite number of 'fuses.' I don't know what the final distribution looked like.)Going back many decades to the origin of E6, E12, E24, etc series', the story I heard was that they would simply make a whole bunch if carbon comp resistors with essentially random values, then bin them into 6, 12, 24, etc equal size bins. Obviously, if this story is true then the distribution of values around nominal would not be Gaussian, but uniform. This also debunks what I used to hear a lot (either when those resistors were still in use, or by people that still remembered using them) that if you bought a bag of 100k, 10% resistors, thennoneof them would be within 5% of 100k because all of those went into the 5% bin. In any case, that particular fear couldn't be justified once you realize that they also needed 110k 5% resistors. Where did they come from if all of the 109k resistors went into the 100k 10% bin?My final conclusion: there are too many plausible possibilities to make any predictions unless the manufacturer provides information about the probability distribution of those particular components.

author: eugene
author_rank: Frequent Contributor
timestamp: May 02, 2025, 02:43:01 pm
content: Quote from: SteveThackery on May 02, 2025, 02:30:26 pmIt could mean that some minimum portion of the distribution of values falls within +/- of the tolerance. If they're manufacturing to 6 sigma then there is still a 0.0000002% chance that any random sample will fall outside of tolerance. If they're manufacturing to 3 sigma then as many as 0.3% could be outside tolerance. I don't know anything about any particular series by any particular manufacturer, but it's certainly possible.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 02:43:13 pm
content: Quote from: bdunham7 on May 02, 2025, 02:04:19 pm1% means the resistor will be within +/- 1% of the nominal value. We agree that does not say anything about the distribution curve, so any assumptions about that are just that - assumptions.As I see it, 1%mustbe the worst case. If a resistor is outside those limits then it isdefective. If it is at or inside those limits, then it isin specification. The worst case is at the tolerance limits. Where else could it be?Assuming the resistor will not exceed +/- 1% is not an "unsupported assumption" - it is a published specification by the manufacturer.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 02:46:57 pm
content: Quote from: eugene on May 02, 2025, 02:43:01 pmIn which case it isdefective. Do you think they don't test the value of each resistor? I think they are bound to, in order to assign a tolerance band to it. And that test could and should exclude even that 0.0000002%.If you are arguing that resistors are not individually tested, but the manufacturer relies on the accuracy of the manufacturing process, then I totally agree: we should expect to see something close to a normal distribution with some outside the claimed 1% (say).But I would point out that this appears to make the RSS calculation even less reliable, because it assumes none of the resistors exceed their tolerance. Doesn't it?

author: eugene
author_rank: Frequent Contributor
timestamp: May 02, 2025, 02:55:43 pm
content: Quote from: SteveThackery on May 02, 2025, 02:46:57 pmWe like to make that assumption, and it might be true, but I am not prepared to assume that every manufacturer tests every every resistor. Even though it's a comparatively easy thing to do. many of these resistors cost a small fraction of a penny each. I cannot make the assumption that there aren't manufacturers that get their line going so that all but 0.0000002% are within tolerance and ship without testing each part. It's economics. And lets face it, the vast majority of my designs will continue to work just fine even if one of the nominally 1% resistors happens to be 1.00000001% outside of tolerance. If I really need that kind of tolerance then I go with 0.1% in the first place.Again, I get that we want to make the assumption, and most of us have, but in this context I can't.

author: bdunham7
author_rank: Super Contributor
timestamp: May 02, 2025, 03:25:24 pm
content: Quote from: SteveThackery on May 02, 2025, 02:46:57 pmUnless the manufacturer supplies a confidence interval or gives you a 1000% money-back guarantee that absolutely none of their parts will ever exceed tolerances, you don't know the expected defect rate. If you declare that any resistor outside of tolerance is defective, then by your definition all non-defective resistors will meet their specified tolerance. But that is circular reasoning and the lack of any defectsby your definitionwas not included in the original statement about buying a 1% resistor.QuoteI think you'll more likely see a small distribution of some kind centered around a mean value that is within the tolerance specification of the nominal value. Dave Jones did a video on this and that was what he found with his samples. YMMV (your manufacturer may vary) and I wouldn't make this a universal assumption. The easiest way to meet a 1% tolerance is to use a manufacturing process that gives you 0.1% or less variations. But there's always a chance that something goes wrong.QuoteNo, a standard distribution, even if the specification is many std devs like "six sigma", allows for a mathematically defined number of out-of-tolerance results. So if your component or device has a given standard deviation and you set the specified tolerance at 3X that amount (k=3) then there is a 99.5% confidence level--but still a 0.5% chance of a particular unit being out of tolerance. Six sigma (k=6) reduces that to 0.0004%, but not zero.

author: iMo
author_rank: Super Contributor
timestamp: May 02, 2025, 03:32:39 pm
content: The distribution of the resistor's values is uniform, my bet. Otherwise the manufacturers will not use the wording "1k 5% tolerance", but "1k x_sigma" instead. Also they do not guarantee the "batches with certain distribution", afaik. Moreover, a guy who once did the binning of a larger "batch" of resistors told me the distribution he saw was uniform (any value between +/- x% from the nominal value has got the same probability)..

author: Siwastaja
author_rank: Super Contributor
timestamp: May 02, 2025, 04:12:32 pm
content: Quote from: bdunham7 on May 02, 2025, 03:25:24 pmThis. Specifications of low-cost commodity components are rarely absolute guarantees. Some better datasheets even specify confidence numbers for min/max and tolerances.One can try asking money back if out-of-tolerance resistor is found... the whole 0.1 cents.Anything of extraordinary accuracy and reliability requirements need working with manufacturers and buying special grade parts, and that is going to be at least 100 times more expensive than working with normal components off Digikey.And if I buy 100 000 pcs of 1% resistors 0.1 cents each and one of them is 1.05% off nominal value, I will let it be. Then again, if it's a systematic and significant failure of keeping within spec, I will definitely complain and name the manufacturer to others.

author: Victorman222
author_rank: Regular Contributor
timestamp: May 02, 2025, 04:49:32 pm
content: For illustration here are some real resistor value distibution "graphs" i made over the yearsPlot 1 is of Tesla TR191 5% resistors, i was surprised how almost all are within 1%Plot 2 is of S2-29V 1% precision thin film resistors, 1% is the loosest spec for them, can see that all resistors within 0.1% or even 0.2% of value were taken out. Right when i needed a 0.1% one...Plot 3 is of S5-25 1% precision wirewounds, worth noting this isn't one batch but 6x 10pcs boxesPlot 4 is of S2-29V 0.05% PTF resistors, interestingly here all resistors are in range 0...+0.05%Plot 5 is of S5-50 0.05% PWW resistors

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: May 02, 2025, 07:15:12 pm
content: Let me offer an example of something I can't prove, but very strongly suspect. I suspect electrolytic capacitor manufacturers can now control their process to a much higher precision than their standard tolerance. Since they can do that, they save a lot of money on etched foil by building all their capacitors on the low side of nominal. You very rarely, if ever, find one at or above the nominal value. At least this has been my experience with higher values for the last decade or so.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 02, 2025, 09:41:55 pm
content: Quote from: bdunham7 on May 02, 2025, 03:25:24 pmYou've been describing the maths around devices with a normal distribution, and the idea of standard distributions and confidence levels.But you are not describing RSS. Please read the very first post. The input to the RSS calculation is the nominal value and the specified tolerance limit (1%, in our case) for each resistor. The calculation produces an answer that istightertolerance than the worst case.No distribution curve is input to the calculation.So please answer these questions:1/ What distribution curve does the RSS assume for the resistor values? What justification is there for that answer?2/ Does the RSS assume some resistors will be outside the specified tolerance? If so, how many?3/ The RSS calculation produces one number as its answer. What grounds are there for assuming any particular distribution curve around that value?When you can answer those fundamental questions we'll be getting somewhere. Otherwise we are vigorously debating something we have no data for: the distribution curve of resistor values. Probably they differ from one manufacturer to another, and probably they differ over time as new manufacturing processes are adopted.

author: moffy
author_rank: Super Contributor
timestamp: May 02, 2025, 10:37:34 pm
content: Quote from: Conrad Hoffman on May 02, 2025, 07:15:12 pmI've noticed that also, some can be slightly higher than nominal but only marginally.

author: bdunham7
author_rank: Super Contributor
timestamp: May 02, 2025, 11:56:11 pm
content: Quote from: SteveThackery on May 02, 2025, 09:41:55 pmThe idea is that if you have twouncorrelatedsources of error and both have a normal distribution, then when you combine them (say put two resistors in series) you can take the root of the sum of the squares (RSS) of the two standard deviations and that will be the standard deviation of the error of the combined unit. Note that these errors should be expressed in absolute terms, not percentages. You need set up the problem logically, so for example if you are using parallel resistors, you'd convert to Siemens or mhos first. Also, if you use a number that is a multiple of the std devs, say k=5, as long as you're consistent this multiple will fall out of the math and you'll get a k=5 result. So assuming this is what we mean:1/ a normal distribution and there's no justification given for the assumption other than RSS is a math operation that works on std devs.2/ yes, as explained in a previous post it depends on how many std devs the tolerance stated represents but a normal distribution always implies the possiblity of some samples being beyond any fixed limit3/ if the inputs are normal distributions, the output will be also--that's just math, but for a mathematical proof you'll have to look it up somewhere. If you apply RSS to other distributions I don't think you get a useful result. If you take two rectangular (uniform) distributions and combine them you should get a triangular result, but IDK how you'd even assign a single number to that let alone calculate it.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 03, 2025, 11:59:41 am
content: Thanks for a great reply, @bdunham7.I think I can reasonably claim the following:Unless you know that the resistor values follow a normal distribution curve, the output from the RSS calculation is not meaningful.I don't think we can always assume the distribution curve is "normal". If resistors are manufactured to a specific value and target tolerance, and then not tested, then of course it is absolutely reasonable to assume their values should follow a normal distribution curve around the nominal value.However, I seriously doubt* that resistors could be manufactured to (say) 0.1% tolerance such that no testing is required. I think a test-and-select process would be necessary. Perhaps that is true for wider tolerances - including 1% resistors? I don't know. It is very difficult to find reliable information.*I know that "seriously doubt" doesn't make a good argument, but in support of that position I can describe how resistors used to be made. Back in the day, all resistors were tested after manufacture and assigned to tolerance bins. That meant two things:1/ With the exception of the tightest tolerance band,no other tolerance band had a normal distribution. In fact they were anything but: the distribution curve had a big hole in the middle because those resistors had already been assigned to a tighter tolerance bin.2/ It really was the case thatthe assigned tolerance band acted as a "brick wall"(that is, no tails outside the band, like a normal distribution), because those resistors would be assigned to the next wider tolerance bin.If any element of that tolerance binning is still in use (which I dont know, but strongly suspect is the case), then the assumptions inherent in RSS (a normal distribution) are wrong and RSS does not produce a meaningful result.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 03, 2025, 05:40:47 pm
content: Quote from: SteveThackery on May 03, 2025, 11:59:41 amYes. And more specifically, we know from practical experience (e.g. see above by Victorman222, similar datasets have been posted by many others too) that the opposite is true: that resistor values almost never follow a normal distribution curve, not even close, and even if the shape is that of normal distribution, it would not be centered at the nominal resistance but some unknown mean instead.

author: gf
author_rank: Super Contributor
timestamp: May 03, 2025, 06:50:17 pm
content: Quote from: SteveThackery on May 03, 2025, 11:59:41 amThe RSS calculation applies toanyprobability distribution, but it tells how thestandard deviationsof two random variables add up (provided that the samples are uncorrelated):stdev(a+b) = sqrt( stdev(a)^2 + stdev(b)^2 )or equivalently, variance(a+b) = variance(a) + variance(b)It does not tell you how "tolerances" add up if "tolerance" is not defined as the standard deviation, but in some other way.In the special case where the distributions of a and b happen to be normal, the calculation also applies to quantiles.

author: Tation
author_rank: Regular Contributor
timestamp: May 04, 2025, 10:17:07 am
content: Quote from: SteveThackery on May 02, 2025, 09:41:55 pmThe application of the RSS method to other industries (e. g.: mechanics) predates its eventual application to electronics, so such questions were answered many years ago.The main point of RSS is not assuming that the components distribute normally (although it is a common assumption with theoretical support, more on this later), but that the values of different components are uncorrelated.E. g.: suppose that 10 pieces of some assembly are mounted, in the final product, stacked. And then they must fit inside some cavity. Say there is a 1 mm tolerance to fit the stack. A worst-case analysis gives 1/10 mm tolerance for each piece. But it is extremely improbable for 10 different pieces, of different shapes and manufactured by different machines, to be selected in a way that all 10 are in the upper edge of their tolerance (here is where no-correlation/independence is expected and assumed). Thus, maybe, we can work with more ample tolerances for the pieces and still fit the cavity, in most cases.With numbers, working with a tolerance of 3-sigmas (99.73 % of manufactured pieces are inside stated tolerance of 0.1 mm, more on this later), then the probability of all 10 pieces being outside their upper tolerance limit is ~2e-27 %, truly negligible and, surely, all assembled stacks will fit the cavity.But, if the tolerances of each piece are wider,0.368 mm instead of 0.1 mm, the analysis (assuming independence and normality) tells us that, still, 99.73 % of the assemblies (3-sigmas) will fit the cavity. This yield may be enough and there is an obvious advantage in manufacturing the pieces with a x3.7 wider tolerance.But in electronics, scenarios like this are not common. Can anybody show a realistic situation when some components are added and the final combination must fit a given error bound?Why normality?Because it is known that the sum of many random, uncorrelated, variables, no matter their distributions, tends to converge to a normal distribution. In the previous example, it is not needed for the dimensions of each piece to distribute normally, but the dimension of the 10-pieces stack will do. In case the dimensions of each piece distribute uniformly, the dimension of the 10-pieces stack will distribute according to the Irwin-Hall distribution with n=10. The figure below compares this distribution with a normal one with same mean and variance (not having a PC here, so asked ChatGPT to generate the plot. Hope it has not screwed it up...). As seen, it is hardly distinguishable from a Gaussian. Thus, the result of RSS, if there is a sufficient number of combined components, can be assumed to be normal. The mathematical foundation for this is the central limit theorem.Thus, if some industrial process is influenced by many random variables that, in some way, combine additively, the result of such process should resemble a normal distribution. If there is no evidence against this, assuming normality is usually safe. This way, the machine producing resistors will surely produce resistors whose resistance distributes normally. But if, afterwards, measures them, bins them, and also rejects those out of tolerance, then they will no longer distribute normally.RSS can still be used to compute the tolerance of the combination of various pieces even when they do not distribute normally (but are, still, non-correlated) by means of the so-called "inflation coefficients". Such coefficients multiply each component tolerance (prior to squaring) and depend on how such component value distributes. This coefficient for the case of normal distributions equals 1, and that for uniform distributions is sqrt(3).Tolerance and sigmaGiven than the output of the process uses to be normal, there's a non-zero probability of some manufactured pieces to be out of tolerance. The fraction of products out of tolerance may be expressed in sigmas, meaning how many standard deviations fit inside the tolerance limits. The fractions of out of tolerance components are {31.7, 4.5, 0.27, 6.8e-4, 2e-7} % for {1, 2, 3, 4.5, 6} sigmas. Such out of tolerance components may be rejected by the manufacturer, sold as lower grade parts or sent undetected (or detected) to clients.

author: bson
author_rank: Unknown
timestamp: May 05, 2025, 03:26:31 am
content: Quote from: gf on May 03, 2025, 06:50:17 pmThis is only true if they have the same mean.Otherwise, measure 100x 100 ohm 1% resistors and 100x 10 ohm 1% resistors, and try to add their standard deviations. Clearly that's not meaningful. Combined as a single distribution they would range from ~10 to ~100, which a HUGE standard deviation, and a mean around ~55. The only way you can combine their std devs is by first recalculating the combined mean, then recalculating their std devs around that mean. (Which I'm not entirely sure is meaningful either, but I could possibly be convinced oherwise... I mean, meaningful, as in combined mean. Pun not intended.)

author: ledtester
author_rank: Super Contributor
timestamp: May 05, 2025, 04:50:23 am
content: Quote from: bson on May 05, 2025, 03:26:31 amWhat you seem to be describing is a sample space consisting of only 200 values: the 100 "100R 1% resistors" together with the 100 "10R 1% resistors", and if you take the average value of those 200 values you get something around 55R.However, we're talking about the space of all possible pairs of resistors, one being a 100R and the other being a 10R for a total of 10,000 possible outcomes. If you take the average value of all of those pairs you get something around 110R.QuoteThis is indeed more like what is going on.

author: bdunham7
author_rank: Super Contributor
timestamp: May 05, 2025, 04:56:22 am
content: Quote from: bson on May 05, 2025, 03:26:31 amNo, you wouldn't even have to know what the actual values or the mean was to do that calculation. And the two values of resistors would be separate, they would be your A and B groups. The (A+B) refers to a unit consisting of one A and one B, in this case in series.

author: youngda9
author_rank: Regular Contributor
timestamp: May 05, 2025, 11:31:23 am
content: For component WCA, you need to take into account four factors:- BOL Tolerance- Temperature Variation- Age Variation- Radiation Variation - only needed for space-bound components.So you have three error sources.- BOL Tolerance is your 1% that you've chosen.- Temperature Variation is defined by the Temperature Coefficient of the resistor (100ppm/DegC, 12ppm/DegC, etc.). You need to calculate the operational temperature range of the resistor in order to factor this into the equaiton.- Age Variation, you'll have to look this up to find the correct equations...it's been a while and I'm a little rusty with this.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 05, 2025, 11:50:08 am
content: Quote from: youngda9 on May 05, 2025, 11:31:23 amPlease define rare/special acronyms you use. I have been doing this for 25 years and never heard the terms WCA and BOL Tolerance. Even Googling for "BOL Tolerance" does not result a single useful result. WCA is ungooglable as well. This isn't some secret club, write out the words you want to convey.

author: MisterHeadache
author_rank: Regular Contributor
timestamp: May 05, 2025, 04:30:56 pm
content: I'm pretty sure that the RSS calculation is inappropriate for use on a parallel combination of resistances because of another assumption - that the normal effect of the tolerances is just simple arithmetic addition. In all the literature that I am familiar with (and I do teach introductory tolerance stackups in my company including statistical modeling techniques like RSS), the fundamental mathematics behind how the RSS formula is derived is implicitly based on that.We also occasionally do Monte Carlo statistical stackups, which are appropriate to use on this problem and might shed some insight. I ran several 10,000 run simulations of a 100 ohm and a 10 ohm resistor in parallel, each with a tolerance of 1% and each having a centered normal distribution with a standard deviation of 1/3 of their tolerance (those are assumptions behind the scenes of a traditional RSS calcuation by the way). The results spit out consistently the correct result for the resistance combination (9.09... ohms) and a standard deviation that ranged from 0.0272 to 0.0280. FWIW...

author: Tation
author_rank: Regular Contributor
timestamp: May 06, 2025, 06:33:01 am
content: Quote from: MisterHeadache on May 05, 2025, 04:30:56 pmCertainly RSS applies to sums of random variables, thus to series of resistors.But, if a random variable distributes normally and its standard deviation is much smaller than its mean (that is: its percent tolerance is small), then the inverse of such random variable does also distribute as gaussian, with mean equal to the inverse of the mean of the original variable and maintaining the same ratio of deviation to mean (thus maintaining the percent tolerance).This makes RSS extensible to parallel combinations of resistors by working with percent tolerances, as the parallel is the inverse of the sum of inverses.With 100 & 10 Ohm 1 % resistors, the tolerance of their combination (series or parallel) is 0.913 %. In parallel this gives +-83 mOhm and, thus, a deviation of 0.02768 Ohm, agreeing with your Monte Carlo analysis.

author: edba
author_rank: Contributor
timestamp: May 06, 2025, 08:32:47 am
content: Hi everyone,Have a read of this article it is very interesting on the subject of tolerance analysis:https://www.edn.com/for-error-analysis-its-hip-to-be-squared/BrgdsEdba

author: SteveThackery
author_rank: Super Contributor
timestamp: May 06, 2025, 09:11:01 am
content: Quote from: edba on May 06, 2025, 08:32:47 amGood arricle - thank you.

author: edba
author_rank: Contributor
timestamp: May 06, 2025, 10:50:32 am
content: It covers the case where components are selected through sorting into bins. A 1% resistor would then have a uniform distribution i.e. equal probability of selecting any value between +/-1%. If you use the RSS technique you multiply the tolerance by 1.73, so the resistor tolerance converts to +/-1.73% resistor. I have done a comparison of normal and uniform distribution below:Normal100 Ω + 10 Ω ±√(1 Ω^2) + (0.1 Ω^2) =110 Ω ±1.0050 ΩUniform100 Ω + 10 Ω ±√(1.73 Ω^2) + (0.173 Ω^2) =110 Ω ±1.5806 ΩBrgdsEdba

author: SteveThackery
author_rank: Super Contributor
timestamp: May 06, 2025, 07:56:03 pm
content: Quote from: edba on May 06, 2025, 10:50:32 amI'm afraid that's a faulty assumption, too. When resistors are binned you get a big hole in the middle of the distribution because those resistors have already been assigned to the next higher tolerance bin.Assuming your tightest tolerance is 0.1%, that bin will have a normal distribution, but chopped off at the ends so none of them exceed 0.1% away from the nominal value. The next bin - say 0.2% - will have a hole in the range from -0.1% to +0.1%, so it will look nothing remotely like a normal distribution. The 0.5% bin will have a hole from -0.2% to +0.2%, and so on down the tolerances.It's not looking good for RSS.

author: eugene
author_rank: Frequent Contributor
timestamp: May 07, 2025, 02:10:01 pm
content: Quote from: SteveThackery on May 06, 2025, 07:56:03 pmThat's a faulty assumption.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 07, 2025, 07:06:09 pm
content: Quote from: eugene on May 07, 2025, 02:10:01 pmIt's not an assumption. It's how binning works (but see later): once a resistor is in the 0.1% bin it isn't available to go in the 0.2% bin. If all the 0.1% resistors are taken out of the population requiring sorting, then you know for sure that no 0.1% resistors will be in the 0.2% bin. Hence the hole in the middle of the 0.2% distribution.Two points: years ago all resistors were sorted into tolerance bins after production. This isnt controversial - it was widely known amongst the EE community. I was taught it in college. However, I don't know know if that is true these days. I think it almost certainly will be for the very tight tolerance parts, in which case you'll inevitably get the two-humped distribution in the bins below the tightest tolerance. I suspect for wider-tolerance components, modern-day manufacturing is probably accurate enough to not require test-and-bin sorting, but simply to rely on three-sigma (or better) accuracy in production. In that case binning isn't required, and you'll get normal distributions.Second point: if binning is used, but more resistors end up in the 0.1% bin than are required, then instead of discarding the surplus it would make sense to throw them in with the 0.2% resistors. And so on down the train. In that case you won't get a hole in the 0.2% distribution. It might not end up as a "normal" distribution, but somewhere in between. Or perhaps even a large peak, depending on how many resistors get "demoted". I've no idea if this is actually done - it's a theoretical possibility.The problem we face is that it seems to be extremely difficult to get a reliable source of information on whether binning is actually used, and if so how widely. Apart from a few instances reported earlier, the manufacturers don't seem to publish much about this topic.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 08, 2025, 12:16:23 am
content: Very interesting video - thanks for posting it.We have one limitation with Dave's demonstration: we cannot know if the Gaussian curve would apply to the next tolerance band out (2%). That's because when resistors are binned, the top bin (tightest tolerance) does, indeed, have a normal distribution. It is only the lower bins that have a hole in the middle.We don't know what the top level tolerance is on this particular family of resistors. It could be the 1% resistors Dave measured, or maybe it's tighter.If binning is not used, and the tolerance is achieved purely by tightly controlled manufacturing processes, then we could reasonably expect all the tolerance bands to have a normal distribution.

author: bdunham7
author_rank: Super Contributor
timestamp: May 08, 2025, 04:51:41 am
content: Quote from: SteveThackery on May 08, 2025, 12:16:23 amThat depends on the binning process and probably other factors. Other people have assumed that binning would result in a more uniform (rectangular) distribution because you are selecting from the center of the peak. But I don't think that's correct either. Suppose you take a group of resistors with a more or less uniform distribution over at least the range of +/- 1% and with fat tails beyond 1%--say an std dev of 5% or something. Your binning selector machine itself has a random error of 0.5% so you tighten up the spec and reject any resistor that is more then 0.5% from nominal. What does the distribution of the output look like?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: May 08, 2025, 10:13:25 am
content: As an aside, I was always impressed that Linear Technology would publish statistical distributions of, I think, noise, maybe other things, in their data sheets.

author: eugene
author_rank: Frequent Contributor
timestamp: May 08, 2025, 02:34:19 pm
content: Quote from: SteveThackery on May 07, 2025, 07:06:09 pmUnless you have specific knowledge about the process, it is an assumption. Some might even call it a wild guess.Knowledge is often not absolute; we think some things are probably true, we're pretty sure of other things and we'realmostcompletely sure of other things. Recognizing and being cognizant of this is essential to any scientist or engineer. We need to keep straight in our own mind what we know to be true and what we guess to be true.Some people willfully choose to believe whateverfeelsright. Let's try to avoid that. It leads to misunderstanding at best and conspiracy theories at worst.

author: MisterHeadache
author_rank: Regular Contributor
timestamp: May 08, 2025, 04:08:56 pm
content: I ran two more Monte Carlo 10k run simulations. With the original, here are the parameters:Simulation #1: Each resistor has a tolerance of +/-1% and each has a centered normal distribution with a standard deviation of 1/3 of their tolerance.Simulation #2: Each resistor has a tolerance of +/-1% and each has a uniform distribution of their values (meaning any value between the lower and upper tolerance limits has an equal probability of occurring)Simulation #3. Same as simulation #2 except they have been ‘binned’ such that all values between +/-0.5% have been removed from the population. So you are left with a uniform distribution with a ‘hole’ in the middle.Results are as follows:Simulation #1: As I stated before, you get a gaussian distribution with a standard deviation of approximately between 0.0272 to 0.0280. See attached image ‘gaussian’ as an example of one 10k trial run.Simulation #2: You get a ‘mostly uniform’ distribution with a standard deviation of approximately 0.048. However, I don’t think the standard deviation is a good measure here due to the ‘mostly uniform’ distribution resulting from the combo of just two resistors. As you increase the number of resistors in the stack, the resulting distribution shifts to be Gaussian (a direct effect of the Central Limit Theorem…) Anyway, see attached image ‘uniform’ as an example of one 10k trial run.Simulation #3: This one took a bit of adjustments to my software, but I was able to set it such that for each random trial, there was an equal/uniform chance that a value of 99-99.5 or 100.5-101 would be chosen. For the 10 ohm resistor, that is 9.9-9.95 or 10.05-10.1. The standard deviation is approximately 0.063, but you end up with a distinct bimodal distribution. No surprise, there has to be a hole in the middle, but I was surprised (based on simulation #2) that even with just two resistors we see a gaussian shape in them rather than two mostly uniform shapes.Anyway, I have definitelyat this point. And let me paraphrase George Box: “ALL MODELS ARE WRONG, BUT SOME ARE USEFUL.” So hopefully there is something useful here to the OP question.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 08, 2025, 07:38:19 pm
content: Quote from: eugene on May 08, 2025, 02:34:19 pm1/ Perfectly good point. As I have mentioned earlier, I was actually taught all the stuff I've been saying at college. I've also said it was a long time ago (late '80s to early' 90s). Back then, binning worked as I've described, with the unusual effects on tolerance distributions.2/ I've also said that I don't know if binning is used today, although I suspect (but don't know) it must be for the really tight tolerances (0.1%, etc).3/ Finally, if binning is used, I don't see how it could be significantly different in implementation or effect from how it used to be forty years ago. So far nobody has offered a convincing alternative.My first paragraph is simply a true account of my past, and as such isn't really open to argument. My second paragraph is simply a statement of my opinion, and as such is no better or worse than any other. My third paragraph is where some new insights might be gained.If anyone can think of a binning regime that is substantially different in terms of results from the system I've described, that would be a great thing to discuss. I am totally open to it.In closing, I was interested to see that Dave (in the video) also said how incredibly difficult it is to find any hard information on how present day resistors are classified in terms of their tolerance, and what the distribution curves look like.

author: Tation
author_rank: Regular Contributor
timestamp: May 08, 2025, 08:36:50 pm
content: Quote from: MisterHeadache on May 08, 2025, 04:08:56 pmThe PDF of the sum of 2 random variables is the convolution of their PDFs, so:1) the convolution of 2 gaussians is another gaussian => OK2) the convolution of 2 rectangles is a trapezoid. If the rectangles are equal width, degenerates into a triangle. If are much dissimilar, degenerates into the wider one with not vertical sides. => OK3) each tail may resemble a triangle. Convoluting 2 triangles gives a bell shaped, 3 order (may vary) spline, starts to look gaussian => OK

author: bdunham7
author_rank: Super Contributor
timestamp: May 08, 2025, 08:43:39 pm
content: Quote from: SteveThackery on May 08, 2025, 07:38:19 pmI mentioned it a few posts back. Suppose you want to select for +/-1.0% but your binning method or device has a 0.5% maximum error (unspecified distribution, you can add that in later). If the goal is to end up with all resistors within the 1.0%, you are much more concerned with a false accept than a false reject. If you have a resistor at say -1.5% but the machine made an error of +0.5%, then you'd have a false accept and an out of tolerance product. So you might set the threshold for +/- 0.5%, this would mean that anything beyond +/- 1% would surely be rejected, but also some resistors that are in tolerance will also be rejected. The probability of a false reject depends on the exact value of the resistor. One with -0.9% would very likely be rejected, one with +0.1% would very likely be accepted. Resistors at exactly 0.0% would always be accepted, those at the extremes would almost always be rejected. What sort of distribution would the resulting batch of resistors have? Now you may see that the answer to that question depends on the earlier question about the distribution of the binning machine error.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 08, 2025, 09:38:04 pm
content: Just a quick point: surely in reality the accuracy of the binning machine wouldsignificantlyexceed that of the resistors it is binning. I mean, like an order of magnitude, maybe?Even so, your suggestion looks good. I'll have to take some time to think about it.

author: bdunham7
author_rank: Super Contributor
timestamp: May 08, 2025, 10:41:01 pm
content: Quote from: SteveThackery on May 08, 2025, 09:38:04 pmIt depends on the numbers. If you make 1M 1% current sense resistors and you need 1000 selected for 0.1%, it may be easier to make a machine that measures to 0.05% than 0.01%, especially if you consider throughput. Think about what it takes to measure resistance that accurately with a high-end DMM. The binning process will probably use a good DMM or close derivative of one plus an automated feed fixture.

author: eugene
author_rank: Frequent Contributor
timestamp: May 08, 2025, 11:06:08 pm
content: I have no actual knowledge of how binning is or ever was performed, but the way that I alwaysimaginedbinning of, say, the E series resistors of yore was that a collection of components were sorted into bins corresponding tooneof the series (E6, E12, E24, ...) Note that the requirement for binning here is that there is one and only one bin for every possible value. In other words no (significant) overlap and no gaps between bins. With E6 there are 6 bins per decade, etc.So it makes no sense to try to bin two series at the same time, or even sequentially. If you bin for E24 then all of the resistors will go into one bin or another. There will not be any left over to bin E12. So the popular claim that 100 ohm parts from the E12 series (10%) will not have any within 5% of 100 ohms (because those parts went into the E24 5% bin) makes no sense to me. It implies that the the 82 and 110 ohm E24 bins are empty (all of those components went into the 100 ohm E6 bin.) Wait, what? This is confusing, sure. Like I said, the only thing that makes sense to me is that you bin a collection of parts for one tolerance only. If you want a different tolerance, then you start with a different collection of parts. Consequently, there will be no "holes" in the middle of the distribution.

author: edba
author_rank: Contributor
timestamp: May 09, 2025, 07:14:23 am
content: Even though I started this post I am now thinking there are so many assumptions being made about distribution shape, binning etc. when all we really know is what the manufacturer is stating in their datasheet; that the value can be nominal or nominal - tolerance or nominal + tolerance. To say it another way if we don't really know how the values are distributed so must assume worst case. In engineering there are only yes, no or I don't know answers. If you don't know then go and find out and if you can't find out then make an educated guess. So maybe it is the end of the road for RSS?

author: iMo
author_rank: Super Contributor
timestamp: May 09, 2025, 09:52:42 am
content: Quote from: edba on May 09, 2025, 07:14:23 amI have here a P&P wheel with X thousands of 0805 0R 5% resistors. It is statistically relevant number so perhaps after binning them we may get how the distribution is like..

author: SteveThackery
author_rank: Super Contributor
timestamp: May 09, 2025, 09:30:08 pm
content: Quote from: eugene on May 08, 2025, 11:06:08 pmI think this is a great suggestion and it does make sense. I have a question: are E12 resistors (for example) always the same tolerance? And do E24 resistors always have a twice-as-tight tolerance than E12? I ask, because that is implied the system you describe. In that system, no resistor is rejected because the bins in any series all join up without gaps.

author: TimFox
author_rank: Super Contributor
timestamp: May 09, 2025, 09:58:20 pm
content: Quote from: SteveThackery on May 09, 2025, 09:30:08 pmE12 and E24 are two sequences of component values, where E12 is convenient for 10% tolerance and E24 is convenient for 5% tolerance components.However, it is common (not always done) to see components in a given series with tighter tolerances (e.g., 1%) available as standard products in E24 values.

author: eugene
author_rank: Frequent Contributor
timestamp: May 09, 2025, 10:37:37 pm
content: Quote from: SteveThackery on May 09, 2025, 09:30:08 pmThe E series system goes way back to the days where carbon comp was used for most applications. Wire wound resistors were available, but were mostly used wjere higher performance was required and much higher cost could be justified. Anyway, E6 was typically considered 20% (OMG!), E12 10%, and E24 5%. Those were the common options available. The system may have been developed for the benefit of manufacturers but it was mainly advertised as a way for engineers to choose a value. So if the engineer calculated a desired value and decided that 10% was good enough, then they would find the closest E12 bin and write that value on the (hand drawn) schematic. The idea, from the engineer's perspective, was that no matter what value they calculated to be ideal, it always fall within tolerance of some existing value in any series. In those days, most of the resistors were 10% (E12) with more critical parts being 5% (E24). Only special applications required wire wound precision resistors. (Check out the schematics of the 500 series Tektronix scopes. Those guys knew what they were doing!)

author: SteveThackery
author_rank: Super Contributor
timestamp: May 10, 2025, 12:51:13 pm
content: Quote from: TimFox on May 09, 2025, 09:58:20 pmAh, dammit, so that puts us back to close to where we were. I think @eugene's proposal is ingenious, but the use of contiguous bins does require that the tolerance is immutably fixed to the particular E series. For example, if we are making E24 resistors the bins would need to be +/- 5% in order to be contiguous.Suppose we want to make a set of E24 resistors with 1% tolerance, as @TimFox describes. This would require the bins to be non-contiguous. Therefore resistors will fall into the gaps. What to do with those? One thing would be to create a new range of resistors: an E24 2% range, even though that wasn't the original plan. Then some of the leftover resistors will drop into those bins. There are still gaps between the bins, so we now need to create a third range: an E24 5% range. At last, the bins are contiguous and the remaining resistors will all find an appropriate bin.So, we set out to make an E24 1% range, but in order to not waste resistors we also had to create an E24 2% range and an E24 5% range in order to use up all the resistors. And guess what, we're back to having holes in the distributions for the 2% and 5% resistors.Another approach making use of the spare resistors is to "promote" them to an E48 1% range, because that has twice as many bins (although still not contiguous) so will catch many of these spare resistors. Unfortunately, every alternate bin will be empty because those values have already ended up in the E24 1% bins. E48 1% bins are still not contiguous, so we can promote the few remaining resistors to E96 1%, where the bins are contiguous and there will be no unassigned resistors.So, we set out to make an E24 1% range, but this time we had to create an E48 1% range and an E96 1% range to use up those spare resistors. Unfortunately there will be empty E48 and E96 bins because the relevant resistors have already been "stolen" by the E24 binning machine.Nevertheless, at first glance it looks like the resistors in this promotion-based binning system will have a normal distribution. I think. I need to work on that.In summary, the original system I described (for each value, widen out the tolerances until all the resistors are assigned) results in non-Gaussian distributions in all the bins apart from the top level bin. The promotion-based system described above, based on @eugene's suggestion and in which resistors are promoted to a higher E series, results in some empty bins in those higher E-series bins, but the occupied bins have a normal distribution.Phew! This is much more complicated than it first appears.

author: TimFox
author_rank: Super Contributor
timestamp: May 10, 2025, 02:14:50 pm
content: Film resistors for 1% tolerance can be “helixed” during production to within the band, leaving negligible numbers within the dead bands. Metal foil resistors are usually laser trimmed, with appropriate patterns on the original part to facilitate the trimming. Similarly, wirewound resistors can be made to tolerance by careful automatic machines.I once had a conversation with a factory engineer about film capacitors, who said they could adjust the winding machines to achieve the tolerance, without further binning.Such parts probably are close to a normal (rather than uniform) distribution around their nominal value.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 10, 2025, 03:21:16 pm
content: Quote from: TimFox on May 10, 2025, 02:14:50 pmThat's great to know. We've been wondering about whether resistors aremadeto a certain accuracy, orselectedto a certain accuracy.I've been arguing that if they are tested-and-binned we would usually see non-Gaussian distributions in the bins. If the resistors aremadeto the required tolerance, then we should expect to see a normal distribution around the nominal value.Dave's video showed a Gaussian distribution, albeit not reaching out to the tolerance limits. Your information, @TimFox, seems to confirm that resistors are not normally binned. These are the only two bits of information we've got so far, I believe, which just shows how difficult it is to find relevant hard data. Tim, if you'd said this at the beginning think how much server space we would have saved! 😂🤣I feel I have flogged this thread to bits, so I'm inclined to stop here (hooray!) and depart knowing that the (limited) evidence so far suggests that accurate resistors are made, not selected. If that turns out to be true, then we can all use RSS with pretty good confidence in the results. I also depart with atinysuspicion that the most accurate resistors might be really hard to make using the methods @TimFox describes, so are test-and-selected instead. It would seem to make sense. As far as I can see, thiscouldbreak RSS. I have no evidence to support that suspicion, though.

author: bdunham7
author_rank: Super Contributor
timestamp: May 10, 2025, 03:54:16 pm
content: Quote from: SteveThackery on May 10, 2025, 03:21:16 pmI would think that given the vast numbers and wide variety of resistors available in the past century or so that almost every method you could imagine has been tried. Making a 1% film or 2% WW power resistor is likely a much different art than making a selection of carbon composite resistors. I think it's clear just from what has been presented in that there's ample evidence of this.The most accurate resistors are test-and-adjusted, usually by laser trimming asTimFoxstated. Some of the most ultra-precision resistor networks appear to be hermetically sealed with a clear side so that they can be laser trimmed after they are packaged.https://xdevs.com/article/res_f8846a/

author: TimFox
author_rank: Super Contributor
timestamp: May 10, 2025, 03:54:52 pm
content: Among the most accurate resistors are Vishay’s bulk foil planar resistors.The basic geometry is serpentine on the plane, to reduce inductance, but they can contain regions with specific trim points, so that individual bits can be “blown” in a laser trim after a high resolution resistance measurement in an automated setup.Vishay Intertechnology published a useful book on resistorsResistor Theory and Technology(2001). Appendix A discusses these patterns for trimmable foil resistors. One example (p. 262) shows a pattern that can be trimmed to obtain values between 5 and 1000 Ohms, with 0.5 Ohm steps, by cutting individual lines in a set of coarse and fine path adjustment possibilities.During my career, I bought semi-custom resistors from Texas Components Corp., who adjust Vishay foil resistor chips to order before packaging them.

author: TimFox
author_rank: Super Contributor
timestamp: May 10, 2025, 04:04:55 pm
content: “RSS” is always true for standard deviations of uncorrelated (statistically independent) random distributions, but “tolerance” vs. “standard deviation” varies depending on the type of distribution (normal, uniform, bimodal, etc.).

author: iMo
author_rank: Super Contributor
timestamp: May 10, 2025, 04:46:45 pm
content: Out of my curiosity I've watched several YT videos on the smd resistors manufacturing. From what I've seen they produce the resistors on aprox 10x10cm large ceramic wafers in many steps (pretty complex stuff with several high temperature steps).They laser trim entire wafer by putting it into a laser machine which cuts an L uniformly into all resistors (for example) with pre-calculated lengths .My current understanding is they do not measure each resistor during the laser trimming as it would take time.In one video I saw similar trimming "to given value" where it took aprox 1sec per item with measuring (special precision parts).There is say 2000 - 10000 resistors (based on size) on the wafer, thus trimming "to given value" while measuring each resistor may take hours.So my current understanding is they, based on some reference measurement for given resistive paste composition and a test on couple of resistors, trim all resistors with laser without measuring during the trimming, then process them (at least one hot step after the trimming) and (perhaps) they do final binning or just some random sampling.Would be interesting to know more details about it, indeed..PS: for example the manufacturing process in steps:

author: SteveThackery
author_rank: Super Contributor
timestamp: May 11, 2025, 09:37:24 am
content: Quote from: TimFox on May 10, 2025, 04:04:55 pmOK, so do you think it will produce a meaningful result with the distribution shown below? That is the shape you get from the old-fashioned binning system.

author: gf
author_rank: Super Contributor
timestamp: May 11, 2025, 11:00:48 am
content: Quote from: SteveThackery on May 11, 2025, 09:37:24 amYes, it applies to any distribution if you measure uncertainty in terms of standard deviation.But if you measure uncertainty as Maximum Absolute Deviation (MAD), then RSS does not apply at all, but MAD(A+B) = MAD(A) + MAD(B). MAD is, however, meaningess for unbounded distributions (like Gaussian and others), since MAD of these distributions is always infinite.

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2025, 02:31:04 pm
content: Choose the statistic that is appropriate for your application, to get a statistical estimate that is measurable and meaningful.

author: bdunham7
author_rank: Super Contributor
timestamp: May 11, 2025, 02:59:47 pm
content: Quote from: SteveThackery on May 11, 2025, 09:37:24 amLet's say your graph is approximately represented by a set of values (5,5,5,5,5,6,6,6,6,7,7,7,8,8) and their corresponding negative values. The std dev of that would be 6.35. The RSS calculation for two same-value resistors in series would give you std dev of about 9.0, or 4.5% if you convert it back into a percentage of the combined resistance. The combined distribution would fill in the center and would look like a blend of a gaussian curve and a triangle, I'm not sure what you'd call it. I'm not ambitious enough to calculate it out and graph it. The 'maximum absolute deviation' would be still be 8% for the combination, although this would occur much less often than in the original batch of single resistors. Thus combinations of 2S2P, 3S3P and so on would yield combinations that are much more likely to be nearer the nominal value but the 'MAD' would always be 8%.Is that meaningful?Edit: I misinterpreted your graph even though it really is pretty clear. The numbers I gave assumed that you'd taken out everything up to 5% and the remaining were intended to be 10% tolerance with the extreme values of 9 and 10 also removed. In this case (2,2,2,2,2,3,3,3,3,4,4,4,5,5) plus the corresponding negative values would be a reasonable representation of your picture. That gives you a std dev of 3.4 and 2 resistors in series would have a std dev of 4.8 or 2.4% of the combined value, again with a 'MAD' of 10 or 5% of the combined value. So in the case of one resistor the likelihood of being close to the MAD is high, with two in series it is much more probable the value will be near nominal.

author: SteveThackery
author_rank: Super Contributor
timestamp: May 11, 2025, 06:02:24 pm
content: Quote from: bdunham7 on May 11, 2025, 02:59:47 pmThank you - I think it is. 😊I do need to think it through carefully, though.