FWIW, I found several formulas for "uncorrelated error" analysis, which goes beyond the root-sum-square stuff.

E.g., the gain of an opamp depends on Rf/Rin.

If ARf, BRin are nominal values of ohms, and
aRf, bRin are tolerances in ohms, and
Rf, Rin are what you will see in production, then
Rf/Rin = ARf/BRin +/- (ARf/BRin)x[{(aRf/ARf)^2} + {(bRf/BRf)^2}]^0.5

 

So what's your point?

__________________
Ron

 

What's an Arf? Is that like an ALF?

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Pay it forward.

 

Willbe,
This is interesting. I uses the monte carlo mehod with PSPICE to plot an envelop around the operating point.
How does this cross correlation match sensitivity and monte carlo techniques.
Thanks,
Bobmar7

 

Don't know. If "tolerance" means 95% of the values are within the tolerance limits, then these formulas should give you the interval limits.
I spent a lot of time on a spreadsheet and writing a program to do this, but in the back of a book at Border's [it might have been "Electronics for Inventors" or some such thing] I just recently noticed closed form formulas for calculating tolerances for uncorrelated errors.
These 7 formulas might be on the Web in a lot of different places. The easiest is the Root Sum Square method for adding, which means you can make a 1% resistor from 100 ea. 10% resistors in series. It also works for a series string of LEDs.
You can do it on a spreadsheet by simulating the normal dist. with 10, 20 or 50 values and then operating on each, two values at a time, but it's tedious. I never got my 100 line program to totally work.
Generally, dividing or multiplying two or more normal distributions do not result in another normal distribution, and I know this from the histograms that my spreadsheets produced, but these formulas seem to sidestep that issue. They are just looking for the extreme values, not the shape of the distribution.
The trick is to know that these are called "uncorrelated errors". Since 'formers have mutual inductance, I guess this won't work for them.

My point is, for those who want to produce many of the same analog circuit, these formulas will help pick appropriate component tolerances, depending on your final desired output level tolerance.

 

That's not true, what happens if you pick a batch of resistors who's average value is +-5% either side of the nominal value?

You're assuming a perfect Gaussian distribution around the nominal value which of course you should never assume. It might work out almost impossible for 100 10% resistors to be all 5% above or below the nominal value but in practise it's far more likely to happen if they've all been made using the same process.

I tend to assume the tolerance doesn't change for resistors in series and parallel and that errors accumulate in amplifiers - it results in higher quality designs, at a cost of course.

I've just been on a course which touched on quality, six sigma, tolerances etc. The only trouble was it was too mechanically biased but it was interesting.

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Is there a typo here? I think you meant to write
Rf/Rin = ARf/BRin +/- (ARf/BRin)x[{(aRf/ARf)^2} + {(bRin/BRin)^2}]^0.5.
How does this go beyond RSS? It looks to me like it is RSS.
If I understand this correctly, we could rewrite this as
G=Gnom*{1±√[(Tol(Rf)²+Tol(Rin)²]}
which makes it a little easier for me to understand.
Is this true?

This does not give you the extreme values of the distribution.

__________________
Ron

 

You're right on the typo, my bad.

"That's not true, what happens if you pick a batch of resistors who's average value is +-5% either side of the nominal value?"
Without correcting for non-normal distributions, there's a 2.5% chance of picking a 5% resistor of 1.05x the nominal value.
For a batch of 5 resistors the chance that they are all on the high side is .025^5 = 1 in 100 million, which is about my chance of dying within the next three hours.
Actuarial Life Table
so if you're going to pick these five, hurry up.

 

I knew you would say something like that which is perfectly true in theory.

Of course it ignores the fact that any process churning out resistors won't always produce a mean value equal to the stated component value. A batch consisting of 1 million resistors might have a mean value of 3% above the nominal value. This is because to maximise quality the process will be more consistent than the tolerance rating: the actual values kicked out might only vary by ±5%.

In other words the box of 100 resistors might say 100Ω ±10% on the label but their 'real' value might be 103R ±5%: their values will range from 97.85Ω to 108.15Ω.

After the batch of resistors has been produced a couple of boxes of resistors will go though inspection, they'll discover their value is too high and adjust the process accordingly but the resistors will all pass.

Processes often tend to drift in a certain direction which makes this kind of error more likely.

Maybe if you soruce your resistors from different suppliers then you can rely on your calculations but I wouldn't trust them. Probability and statistics can be useful but you need to be careful, especially when applying them to components made by factories you know nothing about.

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When choosing only two resistors, what are the chances of picking one that is 5% low and the other 5% high? Is it 2.5%*2.5%? I'm not challenging you here, I'm just trying to understand the implications of the equation.

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Ron

 

Yeah, what hero said =) All that math assumes a completely homogeneous sampling, and isn't based on real world manufacturing processes, and it's sampling policy.

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He's talking about probability.

If you toss a coin once there's 50% chance of heads, if you toss it twice there's only a 25% chance of getting heads each time.

In theory, it's a similar principle with resistor values. In practise it doesn't always follow since processes can be biased to either side of the stated value and will also have a tighter tolerance than specified.

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I do not answer private messages asking for help because no one else can: benefit from advice I may give or correct me if I'm wrong.

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This is why good sampling plans are so good in a manufacturing process.

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Couldn't have said it better myself. I tried, but ended up deleting it and just quoting you

 

Dear Hero999,

I wish that our parts suppliers really sent us Gaussian probability distributions so that simple random analysis would yield the square root of the sum of the squares of the tolerances. To be fair, for large quantity purchases they do. This explains why usually when I put 25 resistors in parallel or series the measured tolerance becomes 1%.

Unfortunately, this does not always happen, which is the idea that started this thread. The reason for this problem is the fact that some suppliers will not have a Gaussian distribution. Instead the distribution may be bi-modal or even uniform. These distributions tend to be correlated rather than random and although multiple parts have improved tolerance, the effect is not as pronounced.
Bobmar7.