tolerance

[Willbe]

If you put n (e.g., 5) +/- 10% resistors in series the tolerance of the composite resistor will be less than +/-10%.
It seems to me the formula may have involved dividing by the square root of n.
Does anyone have this formula?

 

[Hero999]

I don't know what you're talking about, the tolerance will be exactly the same, 10%.

Think about it, suppose you're got two 10Ω 10% tolerance resistors.Let's look at what the tolerance would be if we connected them in series to form a 20Ω resistor.

R1 = 10 - 10% = 9Ω
R2 = 10 - 10% = 9Ω
R1 + R2 = 18Ω

R1 = 10 - 10% = 9Ω
R2 = 10 + 10% = 11Ω
R1 + R2 = 20Ω

R1 = 11Ω
R2 = 11Ω
R1 + R2 = 22Ω

20 + 10% = 22Ω
20 - 10% = 18Ω

I hope you can see that it doesn't matter.

The only time you have to worry is if you're connecting two resistors with different tolerances in series, but even so the over all tolerance will be tighter than the resistor with the widest tolerance.

 

[Willbe]

I don't know what you're talking about, the tolerance will be exactly the same, 10%.

If you assume that 9, 10 and 11Ω values are equally likely. In a batch of 10Ω 10% resistors you would hardly ever find an 11Ω value.

Since I posted I happened onto a math forum. They've told me that the tolerance for two 10% resistors in series would be 10 +/- 1.4Ω = +/- 14%.
I tried this on a spreadsheet and it does seem that the sum of two distributions of values is more spread out than each individual distribution, but the shape is retained.

For more complex stuff, like the gain of an inverting opamp depending on the values of two resistors with specified tolerances I think I need to write a computer program to calc. upper and lower limits on the expected gain.
As in "Monte Carlo Simulation."

I also tried this on a spreadsheet and it does seem that dividing two distributions of values results in a weird-looking distribution of values.



One application would be figuring out the limits on the current through an LED that is not driven by a current source.

 

[3v0]

I have to agree with hero999.

Given equal tolerances: The tolerance of the total will be within the tolerance of any given part.

Say you have 10 10% 10 Ohm resistors
The worst case in one direction would be 10 11 Ohm resistors.
10*11 Ohm in series is 110 Ohm which would be the limit for a 10% 100 Ohm resistor.

Could it be that you were not clear about the problem on the math forum.

When you start talking "likely values" you are getting into statistics. Tolerance does not say anything about likely or average values.

Tolerance simply says that the values fall between a max and a min. If all the resistors in a shipment were at the high limit of tolerance, but still within, that would be acceptable.

3v0

 

[HarveyH42]

People still use 10% tolerance? Haven't seen them since Radio Shack as a kid...

 

[3v0]

People still use 10% tolerance? Haven't seen them since Radio Shack as a kid...

Much of 1/4 and 1/8 watt resistors I purchased are sold as 5% but could quite possibly spec at at 1% or 2%. I have not used or measured them all but they seem consistent.

 

[Pommie]

Since I posted I happened onto a math forum. They've told me that the tolerance for two 10% resistors in series would be 10 +/- 1.4Ω = +/- 14%.

Yes, but that would be a 1.4Ω tolerance on a 20Ω resistor which is 7%.

One thing to bear in mind when talking about tolerances is how the resistors are made. I thought that after manufacture the resistors were sorting into the tolerance band that they best fit. The ones within 1% would go into one bit, 2% into another etc. If this is the case then 10% resistors would never be within 5% of the actual value.

Mike.

 

[OutToLunch]

Tolerance is process and yield dependent. A 1% resistor would go through a different process than a 10% resistor. Then comes the pass/fail test. Measurements of thousands of parts will yield a distribution. Test limits are then set so that a maximum number of parts will pass.

I don't know this for sure, but a resistor manufacture may be able to pass most resistors and get close to 100% yield. If a resistor is expected to be within a given tolerance band and it is outside of it, then it may actually fall within the tolerance band of the standard resistor value that is adjacent to it. This would allow them to sell it, but just use different markings. Who knows, though - a resistor that falls out of the tolerance band could just be defective and it would behoove the mfg to just scrap it.

 

[MrAl]

Hi,

If you have a 10 ohm resistor that is 10 percent high that makes it 11 ohms.
If you have a 100 ohm resistor that is 10 percent high that makes it 110 ohms.
If you connect these two in series you get a resistor that is 121 ohms, or 10
percent higher than a 110 ohm resistor. Thus, it's still a 10 percent resistor.
This is worst case however.

If you look at it in a probabilistic way you get a different answer but you
also then have to assume that the two resistors were picked at random
from a set of resistors that do not have a particular relationship to each
other, or if they do then you know what that relationship is and can
work it into the probability equation. In either
of these two cases we can then compute the probable value which is
going to come out statistically lower than the direct sum of the two.
The problem is though, this doesnt mean that EVERY two choices will
yield a better value, as the max will still be the max, it will just occur
less often over the course of a large number of experiments.

Thus, if your design requires a tolerance of 10 percent then with two
10 percent resistors you can be sure you will get 10 percent, not 14,
but you can not assume you will get LOWER than that, like 5 percent,
because there will be a percentage of experiments that yield over
5 percent. 2x 10 ohms case, you can not expect to get a 21.4 ohm
resistor all the time, but that 22 ohms will come up now and then.
In the case of the two 10 ohm resistors, i would think that that 1.4
ohms comes after adding the two to get 20 ohms, so the max value
(statistically) would come out to 21.4 rather than 22 ohms.

If the final value is going to be that important, perhaps a quick
ohms measurement to ensure the proper gain spread really does
occur. For that matter, perhaps a complete gain measurement
to ensure the whole section is working.

As a side note, not too long ago i found a 15k resistor that was
color banded as 5 percent but measured 19.9k in an ohms test.
The resistor (1/4 watt) was some 20 years old and was never used.

 

[Hero999]

That's not true because the resistance value between either end of the tollerance isn't always entirely random, for example one batch of 10k 10% resistors might be nearer 11k and another batch might be nearer 9k.

When used in more complicated circuits like op-amps, the tollerance effectively gets worse, for example a non-inverting amplifier with a gain of 10 will have a tollerance of 20% if the feedback resistors were 10%.

 

[Willbe]

Yes, but that would be a 1.4Ω tolerance on a 20Ω resistor which is 7%.
Yes, 7%, so it does reduce the composite resistor standard deviation.

One thing to bear in mind when talking about tolerances is how the resistors are made. I thought that after manufacture the resistors were sorting into the tolerance band that they best fit. The ones within 1% would go into one bit, 2% into another etc. If this is the case then 10% resistors would never be within 5% of the actual value.

Mike.

That is a problem; the normal distribution has notches in it where the resistors that fell in the middle were removed.
The problem with using a 5% resistor as a 1% unit is probably stability of the value over time and temp and moisture exposure.

 

[Willbe]

Thus, if your design requires a tolerance of 10 percent then with two
10 percent resistors you can be sure you will get 10 percent, not 14,
but you can not assume you will get LOWER than that, like 5 percent,
because there will be a percentage of experiments that yield over
5 percent.

I intend to test that.

Using 20 "boxes" whose total area is 0.96, arranged like so

x
xx
xxx
xxxx
xxxx
xxx
xx
x

from -2 to 0 to +2 standard deviations, I can simulate a normal distribution on a spreadsheet.
For two resistors in series, each value must be added to each other value, 400 additions in all. Doing this on a spreadsheet might be as quick as debugging two nested loops.
Then sort and find the new 1σ and 2σ values from the histogram.

For multiplying or dividing I don't believe the distribution is normal but it is handled just as easily.

 

[3v0]

I intend to test that.

Where are you going with this ?
What are you attempting to prove ?
Seems pointless.

 

[Willbe]

Where are you going with this ?
What are you attempting to prove ?
Seems pointless.

It may tell me something about what I can expect if I put many LEDs in series. It also should predict production yield for building several of the same circuit. And, the need for Monte Carlo simulation is gone; all values are checked.

I used to work on hi-rel stuff and I guess the stat part stayed with me. And, along with probability and Game Theory I think I can debunk a lot of what I hear and read.

Along with these tools and this book
https://www.amazon.com/unSpun-Findi...bs_sr_1?ie=UTF8&s=books&qid=1222995096&sr=8-1
I should be in good shape.
Or not. . .

 

[MrAl]

Hi Willbe,


Lets take those thoughts one step further and calculate the
distribution using groups of 1 percent each, and 10 million experiments.
I think this might tell you what you want to know, but feel free
to expound a bit further on this if you like.

Using the example of two 10 ohm, 10 percent resistors connected
in series...

The distribution i found looks like this:
{0.19,0.17,0.15,0.13,0.11,0.09,0.07,0.05,0.03,0.01,0}

and in terms of approximate percent of the total population:

0.19 came in between 0 and 1 percent
0.17 came in between 1 and 2 percent
0.15 came in between 2 and 3 percent
0.13 came in between 3 and 4 percent
0.11 came in between 4 and 5 percent
0.09 came in between 5 and 6 percent
0.07 came in between 6 and 7 percent
0.05 came in between 7 and 8 percent
0.03 came in between 8 and 9 percent
0.01 came in between 9 and 10 percent
0.00 came in between 10 and 11 percent
0.00 came in over that (ie none).

and in terms of the 10 million experiments performed:

1903814 came in between 0 and 1 percent
1700131 came in between 1 and 2 percent
1500144 came in between 2 and 3 percent
1296440 came in between 3 and 4 percent
1098226 came in between 4 and 5 percent
899554 came in between 5 and 6 percent
700402 came in between 6 and 7 percent
500856 came in between 7 and 8 percent
300607 came in between 8 and 9 percent
99826 came in between 9 and 10 percent
0 came in between 10 and 11 percent
0 came in over that (ie none).

This tells me that on average the combined tolerance will be better
than the two, but there are still some that will hit the 10 percent
mark, and that none will ever be over 10 percent.

Of course there are some things that would invalidate the above
results, one being if the manufacturer sometimes produces resistors
that are already out of tolerance (such as 11 percent instead of 10),
and another being if the manufacturer produces resistors with
some sort of non uniform trend such as (on average):

1 resistor 10 ohms
2 resistors 10.1 ohms
1 resistor 9.9 ohms

If this is really that important to you perhaps the best thing
to do is contact the manufacturer and see what they have
to say about the error tolerance distribution they expect
to see with their resistors. They may actually modify the
+/- 10 percent to maybe +9 and -11 or something like that
which would change the above numbers.

 

[Willbe]

The distribution i found looks like this:
{0.19,0.17,0.15,0.13,0.11,0.09,0.07,0.05,0.03,0.01,0}

Hi, Mr. Al:

What dist. is that?

BTW, I ran a spreadsheet for a V of 10v +/- 1v divided by a resistor of 7 ohms +/- 1 ohm; there isn't hardly a simpler and more useful calculation than V/R. The normal distributions for V and R were approximated by 9 values each.

The calc'd average is 10/7 = 1.43 amps.
The first column of the histogram below is the "amps bin" and the second is how many values were in that bin.

1.1.. 3
1.2.. 12
1.3.. 17
1.4.. 19
1.5.. 18
1.6.. 9
1.7.. 2
1.8.. 1
______________
...... 81 total values

The distribution has half the expected values falling between 1.2 and 1.5 amps, a lopsided dist. shaped kind of like an ocean wave.

If you build this circuit and you're OK with this 50-50 gamble on what current you get, then we're done. Otherwise it's time to tighten/loosen the tolerances.

Here's another for 10/5 = 2 amps. 1 ohm in 5 is now a 20% tolerance on the resistor.
1.5.. 8
1.7.. 16
1.9.. 23
2.1.. 20
2.3.. 8
2.5.. 5
2.7.. 1
with half the values between 1.7 and 2.1, and yet a different shaped, more lopsided distribution.

 

[MrAl]

Hi Willbe,


That is the distribution you get when you combine lots of 10 ohm
resistors to form 20 ohm resistors and sort the resulting true tolerance
of the 20 ohm resistors into bins of 1 percent each, starting with
1 percent and going to 11 percent and above, but paying more
attention to 1 to 11 percent.

I didnt want to add another level of abstraction to this problem so
i went with the actual resistor values as produced by the series
connection of pairs of 10 ohm resistors each 10 percent tolerance.
The distribution of errors in the individual resistors is such that
all resistors have an equal chance of being anywhere between 0
and 10 percent off.

In other words, you take 10 million pairs of resistors (each 10
ohm, 10 percent), connect them in series, then measure the
resistance and compute the tolerance from a perfect 20 ohms.
You then bin this pair into bins of 1 percent each:
Bin1=0 to 1 percent off
Bin2=1 to 2 percent off
Bin3=2 to 3 percent off
etc.
Bin9=8 to 9 percent off
Bin10=9 to 10 percent off
Bin11=10 to 11 percent off
Bin12=anything over 11 percent off

Doing this tells us how many come in better, how many come in
just the same (10 percent), and how many (if any) come in
over 10 percent.

The experiments had shown that out of all the experiments performed

19 percent came in between 0 and 1 percent tolerance,
17 percent came in between 1 and 2 percent tolerance,
15 percent came in between 2 and 3 percent tolerance,
etc., and
1 percent came in between 9 and 10 percent tolerance,
0 percent came in between 10 and 11 percent tolerance,
and none came in above that.

That is what this line shows:
{0.19,0.17,0.15,0.13,0.11,0.09,0.07,0.05,0.03,0.01 ,0}

Also, adding the percentages from 0 to 5 shows that
75 percent of the resulting resistors came in at
5 percent or less, and 25 percent came in between
5 and 10 percent, and none came in above 10 percent.

This also tells me that on average, the resulting resistor
value will have a closer tolerance than either resistor,
except in 1 percent of the cases, where the tolerance
will be the same (bad luck).

Is this clearer now?

I'll repeat the actual results here for reference:

1903814 came in between 0 and 1 percent tolerance,
1700131 came in between 1 and 2 percent,
1500144 came in between 2 and 3 percent,
1296440 came in between 3 and 4 percent,
1098226 came in between 4 and 5 percent,
899554 came in between 5 and 6 percent,
700402 came in between 6 and 7 percent,
500856 came in between 7 and 8 percent,
300607 came in between 8 and 9 percent,
99826 came in between 9 and 10 percent,
none came in over 10 percent.

 

[Hero999]

Here's a design problem where using resistor tollereances can help you save money.

You require an accurate as possible (preferably <0.2% tollereance) 63k resistor.

Your supplier sells 0.1% tollerance precision resistors but only does them in E192 preferred values. From the table below the nearest values are 62k6 and 63k4 which are +/-0.6349% out so it's no good.

A good way of solving this problem is to use the expensive 62k6 0.1% resistor and a cheap 5% tollereance 390R (a standard E24 value, see bottom table) resistor in series which would make 62k99, an error of only -0.000159% (ignoring the reistor tollereances).

Now lets see what the over all tollerance is when we take into account the resistor tollerances.

390R + 5% = 409R5
62k6 + 0.1% = 62k6626
62k6626 + 409R5 = 63k0721 an error of just +0.114%.

390R - 5% = 370R5
62k6 - 0.1% = 62k5374
62k5374 + 370R5 = 62k9079 an error of just -0.146%

This explains why manufacturers don't bother making resistors in E384 preferred values - it's so easy just to select the nearest E192 value and stick an E24 value in series to get the desired value.

E192 (0.5%)
100 101 102 104 105 106 107 109
110 111 113 114 115 117 118 120
121 123 124 126 127 129 130 132
133 135 137 138 140 142 143 145
147 149 150 152 154 156 158 160
162 164 165 167 169 172 174 176
178 180 182 184 187 189 191 193
196 198 200 203 205 208 210 213
215 218 221 223 226 229 232 234
237 240 243 246 249 252 255 258
261 264 267 271 274 277 280 284
287 291 294 298 301 305 309 312
316 320 324 328 332 336 340 344
348 352 357 361 365 370 374 379
383 388 392 397 402 407 412 417
422 427 432 437 442 448 453 459
464 470 475 481 487 493 499 505
511 517 523 530 536 542 549 556
562 569 576 583 590 597 604 612
619 626 634 642 649 657 665 673
681 690 698 706 715 723 732 741
750 759 768 777 787 796 806 816
825 835 845 856 866 876 887 898
909 920 931 942 953 965 976 988

E24 (5%)
10 11 12 13 15 16 18 20 22 24 27 30
33 36 39 43 47 51 56 62 68 75 82 91

Tables from Wikipedia.
https://en.wikipedia.org/wiki/Preferred_number#Capacitors_and_resistors

 

[Willbe]

I have to digest these last two posts.
In the meantime, one more LED circuit!

A 3v supply {V} in series with 2v LED {Vz} and a 50.000Ω resistor gives If = 20 mA.
The tolerance on the 3v is +/- 0.3v, so 96% of the time V will be between 2.7v and 3.3v
and the tolerance on the 2v LED is +/- 0.4v.

Since the subtraction of two normal distributions is another normal distribution,
2σ = sqrt([.3^2] + [.4^2]) = 0.49, so 96% of the time If will be between [(3-2) + .49]/50 and [(3-2) - .49]/50, that is between 10 and 30 mA.
This brightness difference is probably noticeable.

+/- 0.7σ gives 50% of the cases, so half the time If will be between (1+0.14)/50 and (1-0.14)/50 = 23 and 17 mA.
Noticeable. . .? Dunno'!

If it weren't for the post about putting 156 LEDs in series, not in my whole career would I have stumbled onto this, this putting two and two together, so to speak.

For resistor tolerances other than zero I need my spreadsheet setup because multiplying/dividing normal distributions does not result in a normal dist..

Between this root-sum-square formula and a spreadsheet approximation I think the overall tolerance of almost all circuits can be determined.

And I think some of the handheld calculators today can do spreadsheets, so in principle you could predict production yield on the back of the same envelope on which you just designed your circuit, over lunch.
That should keep a boss happy, even that pointy-haired jerk from the Dilbert cartoons.

Ha!!!

 

[Willbe]

If you put a +/- 1Ω tolerance on the 50Ω resistor, 50% of the time you will get between 16 to 29 mA with this LED circuit, with 10 to 29* mA for 96% of the time. The curve is definitely lopsided.

bin #
.....in bin
10 32
13 79
16 85
19 83
22 59
25 39
29 20

*It should be slightly above 30 mA since the post above shows 30 with a perfect resistor. I have to assume this is a rounding error for using 20 boxes to represent the area under the curve of a normal distribution for each component.

With this spreadsheet working, for components being add'd/sub'd/mult'd/div'd, the tolerance for any arbitrary distribution, including notched, can be modelled.