Tolerance Question

Most electronic device tolerances are guaranteed limits (guaranteed worst-case error), not standard deviation.

John

My understanding of tolerance is that all, 100% of the sensor must be within the tolerance range, not some statistical grouping like 99.73%. Furthermore any that are not within the stated range may be rejected, not sold or not used in service usually with a money back guarantee. The tolerance is usually stated as a percentage of some value, often the full scale value. It is sometimes stated as the accuracy of any reading; i.e., within a percentage or number of units of true. In your example I would interpret it as meaning the reading is within +/- 2deg F of the actual temperature. If the reference is boiling water, the readings would "always" be within +/- 2 degF of 212 degF. BTW, that's a lousy tolerance for a thermocouple. Bob

A single batch of sensors may have a bimodal distribution if the better ones have been culled from the batch. Sort of a gaussian with the edges clipped and a big notch in the middle.

John

I read in sci.electronics.design that Joshua Guthrie wrote (in ) about 'Tolerance Question', on Thu, 14 Apr 2005:

You are getting off-beam answers because you are asking a statistics problem in an electronics newsgroup.

The combined distribution due to two sensors is no longer Gaussian/normal but bimodal. I think you need to ask on a math or stats newsgroup.

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I read in sci.electronics.design that Joshua Guthrie wrote (in ) about 'Tolerance Question', on Thu, 14 Apr 2005:

OK, I'm going to have another go at this, because a small dim light appeared in what I laughingly call my brain.

You may be OK to assume for this example that the tolerance band is between the 3-sigma points of the distribution, but it may not be in a practical case.

In this example, yes, but in general, there is a valley in the distribution between the two nominal values.

Standard deviations are like r.m.s. values - they add as the square root of the sum of the squares. To determine the new sigma, we can subtract

212 (the 'd.c. component') to give two 'r.m.s.' values, -(1)^2 + 0.667^2 and +1^2 + 0.667^2. The new SD or r.m.s. value is them sqrt(2*0.667^2 + 2*1^2) = 2.211.

That 2 degrees difference make a LOT of difference to the 3-sigma tolerance, going from +/- 2 degrees to +/- 6.634 degrees. But I suppose if you think about the two distribution curves added together, the result has a much broader peak than either individual curve.

Please understand that I may not have worked this out correctly. Stats are a bit out of my field. It's only the thing about sigma = r.m.s. value that tempts me to try for a solution.

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Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
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I read in sci.electronics.design that John Larkin wrote (in ) about 'Tolerance Question', on Thu, 14 Apr 2005:

Yes, the OP's question is about two sensors with a 2 degree difference when reading the same temperature.

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Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
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This is very similar to the path I was wandering down. But the thing that stumped me was this.

The more sensors I put in the water and average the results, the less confident I would get in the results. This seems counter intuitive.

I would think, as I added more sensors measuring the same phenomena, the more my tolerances in the system as a whole should tightened up. Besides, I have additional information, and know the PDF of the data points. Kinda like if I added an infinite number of sensors, the more infinitesimal my error would get.

But you all are probably right, I'm probably off topic in this group.

If the entire production lot of thermistors were skewed, adding more sensors would merely converge on the wrong point. If the mfr is selling 2% devices, why would he invest in 0.1% standards? You can buy a reel of laser-trimmed 5% resistors, for example, that are all very close to one another and are all about 2% high.

The only safe thing to do is buy a good, certified platinum RTD and use it with a good, 4-wire DVM to measure the actual temperature and calibrate your thermistor setup, or get some equivalent accurate standard.

Not at all.

John

Joshua, You aren't quite thinking about this right.

The typical tolerance on a part like this isn't gaussian, or bi-modal (at least not anymore.) Instead, they are typically uniform. When you add two sensors, what you get is a possible distribution that is more likely to be correct, but there is a slightly greater chance it is be beyond the original spec. As you add sensors, the distribution gains a peak at your specified value, and a gaussian distribution around that point occurs. As you add sensors, your average is more likely to be the correct value, and the utility of doing more rigourous statistical analysis on your measurements increases. You can then do things like, reject values that differ from the average by an amount greater than 'x', etc.

Charlie

It is counterintuitive because it is wrong.

And that is exactly right.

The PDF is not on the temperature measurement, it is on the error function of the population of available sensors from which you have selected a sample of two. The average of two fixed sensors from that population is a repeatable and deterministic function, it will not change as long as those two sensors are fixed. As you add more sensors and average, the error function approaches that of the sensor population as a whole and the STD, or root mean square error, of the sample function deviation from population mean approaches zero. If the sensor population mean error function with temperature is zero, then the tolerance will tighten up and also approach zero.

Yes- you are.

I read in sci.electronics.design that Joshua Guthrie wrote (in ) about 'Tolerance Question', on Fri, 15 Apr 2005:

Not if they are all giving you significantly different answers. In your example, the two sensors were giving results 3 sigmas apart. If they had been 0.1 sigma apart, the effect on the combined sigma would have been much less, but it would still have been larger.

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Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I think the central limit theorem may apply here.

The central limit theorem states that given a distribution with a mean m and standard deviation sd, the sampling distribution of the mean approaches a normal distribution with a mean m and a standard deviation sd/sqrt(N) as N, the sample size, increases.

So if you take the average of N readings the standard deviation of the average will be reduced by a factor of sqrt(N).

Note that the average will approach a normal distribution even if the original distribution is not normal.

As others have already pointed out, this average may not converge on the right answer, for example if all your sensors came from the same batch they could well all read a bit high or a bit low.

Gareth.

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There is an applet here which shows the central limit theorem:

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