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Question regarding esistor tolerances

Started by MaxMaxfield 4 months ago15 replieslatest reply 4 months ago84 views

When I was starting out in engineering ~40 years ago, I was told that if a resistor manufacturer was making a bunch of 10K resistors, for example, then their actual values would range arouns a standard "bell curve" distribution.

I was also told that the manufacturer would measure the values of all the resistors -- the ones that were within 1% of 1K were marked as 1% tolerance and removed from the batch. Out of the resistors that remained, the ones that were within 5% of 1K were marked as 5% tolerance and removed from the batch.Similarly, of the resistors that remained, the ones that were within 10% of 1K were marked as 10% tolerance and removed from the batch.

The result of all this is that the 1% tolerance resistors would range between 990 and 1010 ohms. Meanwhile, the 5% resistors -- which you would think would range between 950 and 1050 ohms would actually have values between 950 and 990 or 1010 and 1050 ohms, because the 1% ones had been removed. As a result, you could guarantee that the 5% resistors would be at least 1% off the nominal value.

Similarly, instead of ranging between 900 and 1100 ohms, the 10% resistors would actually range between 900 and 950 or 1050 and 1100 because the 5% and 1% resistors had been removed. As a result, you could guarantee that the 10% resistors would be at least 5% off the nominal value.

Does anyone know (a) if this was indeed  the way things worked and (b) is it still this way today?

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Reply by DilbertoSeptember 19, 2020

HI, MaxMaxfield!

Glad to read from you again! I always learn something new with your questions!

I worked in an electronics components manufacturing facility in the second half of the eighties that manufactured capacitors and semiconductors, not resistors ( a firm of the Siemens group ).

But the sorting method of capacitors was exactly as you described and I think that with resistors should not be different. I know that because my job was repairing the machines that tested and sorted the capacitors.

I don't know if this sorting method holds nowadays, because maybe the manufacturing precision available today may have precluded the existence of the "non 1% pieces".

Your conclusions about tolerance are correct with respect to the capacitors.  A commercially 5% tolerance capacitor really has its capacitance between -5% and -1% or between +1% and +5% of the rated value.

Cheers!

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Reply by jms_nhSeptember 20, 2020

I trust your recollection, but I find it hard to believe that this is a viable strategy. (See my comment elsewhere on this page.) How could the capacitor company tune their yields up/down to meet demand for 1% or 5% components?

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Reply by DilbertoSeptember 20, 2020

Hi, jms_nh!

I've read the excerpt from the article "Tolerance Analysis" that you attached in your answer to CustomSarge.

Well, first of all, things used to happen exactly as I said in that manufacturing facility, I'm sure.

The machines had a set of bins in which pieces fell off according to the test results:

First bin was for catching short-circuited parts.

Second, for parts with excess leak current.

Third, for parts with excess loss ( tangent of delta ).

Then came the bins for grading:

Fourth, for 1% grade parts.

Fifth, for 5% grade parts.

Sixth, for 10% grade parts.

Seventh, for 20% grade parts.

And, finally, a bin for the remaining parts.

I don't know the answer to your question about how could the capacitor company tune their yields up/down to meet the demand for 1% or 5% components, but I suspect that that wasn't the case. Simply the process had that yielding and the company had to cope with that.

Besides that,

1. I'm the one who worked in the passive components industry.

2. Reported facts date from the eighties.

3. You must add the "marketing factor" to the analysis.

4. And there is the Chinese market ... :-)

Cheers!

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Reply by jms_nhSeptember 20, 2020

Oh -- I completely believe you. Just curious, what kind of ratios did bins 4/5/6/7 have?

I suppose that as long as the 1% parts that failed tolerance spec had equal or better specs than the specs of the 5%/10%/20% capacitors then they could re-use the failed parts for lower grades.

Still doesn't seem like a good business strategy though to depend on it.

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Reply by DilbertoSeptember 21, 2020

I don't remember the ratios, but I think the majority of parts fell off in the 1% bin.

This, alongside with the testing method ( see below ) made the reclassification of parts unnecessary.

The testing machine was based on an analog selection circuit.

There was a measuring point in which a capacitance bridge measured the part and produced a pair of voltage signals, one proportional to the deviation of capacitance and another proportional to tg( d ).

Then a set of analog comparators gave the verdict of what bin should receive the part. Of course, the circuit was complemented by a shift register in order to fire the opening solenoid of the correct bin, as the bins were aligned with the testing conveyor length.

So, to reclassify the 'non 1%' parts, as the Germans use to say, it would be "viel arbeit, wenig geld".

Be happy!

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Reply by MaxMaxfieldSeptember 21, 2020

One way would be to vary the price -- if you make 1% resistors 100X more expensive, then a lot of designers will decide that 5% is "good enough" LOL

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Reply by CustomSargeSeptember 19, 2020

Ok, at the risk of inferring naivete: all mfgrs grade their products by yield.

Logic chips used to be graded on speed, but they're fast enough now to be irrelevant.

Complex sequential circuits (microprocessors, microcontrollers, etc) are still sorted by speed by yield. You'd have to be an idiot to not do this -  wafers aren't dead uniform.

Check "overclocking" with respect to "pushing" performance across the years - dubious at best.

So, in wafer output testing, you get to characterize performance. Faster is better and $$$.

I'll entertain differing opinions...  L8R  <<<)))

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Reply by jms_nhSeptember 20, 2020

I was also told that the manufacturer would measure the values of all the resistors -- the ones that were within 1% of 1K were marked as 1% tolerance and removed from the batch. Out of the resistors that remained, the ones that were within 5% of 1K were marked as 5% tolerance and removed from the batch.Similarly, of the resistors that remained, the ones that were within 10% of 1K were marked as 10% tolerance and removed from the batch.

The result of all this is that the 1% tolerance resistors would range between 990 and 1010 ohms. Meanwhile, the 5% resistors -- which you would think would range between 950 and 1050 ohms would actually have values between 950 and 990 or 1010 and 1050 ohms, because the 1% ones had been removed. As a result, you could guarantee that the 5% resistors would be at least 1% off the nominal value.

I'd heard essentially the same thing, but I believe the story is apocryphal, or perhaps dates to vacuum tube days when human labor was cheaper.

See the commentary toward the end of my article "Tolerance Analysis". I've attached an excerpt below, in case you're impatient. This is my argument against use of grading nowadays, although you'd have to ask someone who's worked in the passive components industry to know for sure.

Grading may still be done for some electronic components (perhaps voltage references or op-amps), but it’s not a great manufacturing strategy. The demand for different grades may fluctuate with time, and is unlikely to match up perfectly with the yields from various grades. Suppose that you are the manufacturing VP of Danalog Vices, Inc., which produces the DV123 op-amp in two grades:

  • an “A” grade op-amp with input offset voltage less than 1mV
  • a “B” grade op-amp with 1mV - 5mV input offset.

Suppose also that the manufacturing process ends up with 40.7% in the “A” grade, 57.2% in the “B” grade, and 2.1% as yield failures.

Maybe in 2019, there were orders for 650,000 DV123A op-amps and 800,000 DV123B op-amps. To meet this demand, Danalog Vices fabricated wafers with enough dice for 1.8 million parts: 732,600 DV123A, 1,029,600 DV123B, and 37,800 yield failures, meeting demand and a little extra. At the end of the year, there are 82,600 excess DV123A in inventory and 229,600 excess DV123B in inventory.

Now in 2020, the forecasted orders are 900,000 DV123A and 720,000 DV123B op-amps. (Some major customer decided they needed higher precision.) You don’t have many options here… making 2.2 million dice would produce 895,400 DV123A op-amps and 1,258,400 DV123B op-amps. Combined with the previous year’s inventory, this would be enough to meet demand plus 78,000 extra DV123A op-amps and 768,000 DV123B. Tons of excess B grade op-amps.

That’s not going to work very well. If the fraction of customer orders of A grade parts is much higher than the natural yield of A grade parts, then there will be an excess of B grade parts. If we had too many A grade op-amps, Danalog Vices could package and sell them as B grade op-amps, but an excess of B grade op-amps will end up as scrapped inventory.

There’s no realistic way to shift the manufacturing process to make more A grade op-amps through grading alone. We could add a laser-trimming step on the manufacturing line to improve B-grade dice until they meet A-grade specs, which adds some cost.

To jump out of this grading quagmire and back to the overall point I am trying to make: you cannot be sure of the error distribution of components. Can’t can’t can’t. The best you might be able to do is get characterization data from the manufacturer, but this would be for one sample batch and may not be representative of the manufacturing process through the product’s full life cycle.

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Reply by MaxMaxfieldSeptember 21, 2020
"...perhaps dates to vacuum tube days when human labor was cheaper..." I wasn't trying to suggest that the grading was done by hand -- just that the 1% parts were taken out of the mix, then the 5% parts, then the 10% parts, and so on -- but the testing cold be performed by machine as in Dilberto's response.
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Reply by antedeluvianSeptember 20, 2020

Max

Here’s another twist. You may have noticed that 5% resistors have a higher wattage rating than the 1% resistors in the same package. I have been told that they are the same device. It’s just that the temperature rise at the greater wattage causes the resistance value to stray outside the 1% but remains inside the 5%

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Reply by jms_nhSeptember 21, 2020

Cite source for this information?

That's not how it works. 1% and 5% tolerance apply at 25 degrees C. The manufacturers have other specs for temperature coefficient.

For SMT resistors one major issue that dictates power-dissipating capability is whether the parts are thick-film or thin-film; thick film components can handle more power for the same package because of their construction and their thermal characteristics.

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Reply by antedeluvianSeptember 21, 2020

Jason

Cite source for this information?

I got this information from someone who I regard with the same degree of authority as I ascribe to you. But it could certainly be apocryphal and derived from the days of through hole resistors. So I am now torn.

1% and 5% tolerance apply at 25 degrees C.

Intuitively I think I would agree, but I went to look at a few data sheets. I could not find this stated in the 3 or 4 that I looked at. In a Vishay data sheet (https://www.vishay.com/docs/28773/crcwce3.pdf ) it would seem that if it is at a particular temperature (and that is by no means clear to me) it is at 70 deg C. (See STANDARD ELECTRICAL SPECIFICATIONS at the top.) (Edit added). 70 degrees is where the power derate curve begins its decline.

If you look at the operational temperature range of 210degC (-55 to 155) at 100ppM this results in a range of 210*100 ppM which by my calculation is 2% (+/- 1 %). For a 200ppM this would be +/-2%, and not +/-5%, so the +/-1% for the 1% resistor is probably a coincidence.

Do you have any source for 25 degC measurement?

For SMT resistors one major issue that dictates power-dissipating capability is whether the parts are thick-film or thin-film;

I was only looking at thick film...

thick film components can handle more power for the same package because of their construction and their thermal characteristics.

I think the cause and effect are the other way around. I am not sure that they differ that much in their construction, and certainly the thermal characteristics play a large role, but try an find a 50 or 20ppM thick film resistor, or a 0.1% one. They can handle more power because the thermal characteristics are such that they cannot improve on 1% 100ppm


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Reply by MaxMaxfieldSeptember 21, 2020
I wish one of us knew someone who works in a resistor manufacturing facility today!!!
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Reply by MaxMaxfieldSeptember 21, 2020

OMG -- that is another twist!!!   I would never have thought of this one -- Max

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Reply by antedeluvianSeptember 23, 2020

It appears the whole question is even more complicated because "tolerance" is apparently very loosely defined. Read "Reading Between the Lines in Resistor Datasheets" ( http://www.vishaypg.com/docs/63517/FACTS-121.pdf ) to make everything clear as mud.