How do the tolerances combine when you're using two resistors? I'm pretty sure they'd add together if in series (so two 5%'s become 10%), but I'm having trouble easily intuiting what happens if in parallel. Do they combine in the same way that resistances combine when in parallel?
edit: Actually, I'm not so sure anymore that the tolerances would add up in series... I should probably just look this stuff up, since I'm not awake enough to intuit correctly, I think.
Values (for resistors) add in series and sort of divide-average in parallel.
In either case though, the tolerance divides.
The combined tolerance becomes more accurate the more resistors there are in total, whether parallel or serial. The highs and lows, and the chances of high or low, cancel each other out and you get a final actual value that is closer to the nominal statistical center of the bell curve the more individual parts there are. (same goes for other components, just resistors are simpler to talk about because their behavior is simple.)
In series, a single 10K might really be 9K or 11K, but if you chain 10 10Ks in series, you don't get a "maybe 90K maybe 110K". That is technically possible but statistics means that what what you actually get is if there was N% chance that a given 10K is 9K or 11K, the there is 1/10th of N% chance (or less, I bet the actual equation is more complicated) that the chain of 10 is 90K or 110K. If the individual 10Ks were 10%, then you get 100K with something like 1% tolerance.
(except also in reality, there is such a thing as batches, where all the parts in a given batch are all high or low the same way, because the process was drifting a little high or low while it was cranking out thousands of them that hour. So Ideally your 10 individuals need to come from 10 different batches or even 10 different manufacturers if that were practical or in a pure math world.)
In parallel, the statistical division is the same though the value centers on the value/N rather than value*N. 10 10% 10Ks in parallel = 1 1% 1k
I’m a bit tired otherwise I’d write something more rigorous. There are different ways the central limit theorem is expressed and proved here—there are more powerful ways to state it that require more complicated proofs, and there are simpler versions that are simple to prove.
A simple version will suffice here. Treat the resistors as iid variables with finite variance σ². When you average them, the variance of the average is σ²/n. More or less… this means that if your resistors are ±10%, and you have 16 of them, you get something with (fuzzy math) ±10%/√16 = ±2.5%.
There’s a lot of unstated assumptions in what I just wrote. But you’ll see the “grows proportional to √n” a lot in stats.
If the resistor values are distributed as a Gaussian where the tolerance is some confidence interval, then the total resistance of resisters in series would be distributed as the sum of those Gaussians whose tolerance would be the root-sum-square of the individual tolerances: sqrt(tol1^2 + tol2^2 + ... tolN^2), or if all the tolerances are the same then sqrt(N)*tol.
edit: Actually, I'm not so sure anymore that the tolerances would add up in series... I should probably just look this stuff up, since I'm not awake enough to intuit correctly, I think.