13 ms·

tolerance should actually go down since the errors help cancel each other out. reference: https://people.umass.edu/phys286/Propagating_uncertainty.pdf https://

tolerance should actually go down since the errors help cancel each other out. reference: https://people.umass.edu/phys286/Propagating_uncertainty.pdf https://people.umass.edu/phys286/Propagating_uncertainty.pdf disclaimer: it will be a relatively small effect for just two resitors aleph's comment is also correct. the bounds they quote are a "wost-case" bound that is useful enough for real world applications. typically, you won't be connecting a sufficiently large number of resistors in series for this technicality to be useful enough for the additional work it causes.

rexer 2y ago
Note that tolerance and uncertainty are different. Tolerance is a contract provided by the seller that a given resistor is within a specific range. Uncertainty is due to your imprecise measuring device (as they all are in practice). You could take a 33k Ohm resister with 5% tolerance, and measure it at 33,100 +/- 200 Ohm. At that point, the tolerance provides no further value to you.
- MobiusHorizons 2y ago
  It’s not nearly that simple:) Component values change with environmental factors like temperature and humidity. Resistors that have a 1% rating don’t change as much over a range of temperatures as 5% or 10% components do. This is typically accomplished by making the 1% resistors using different materials and construction techniques than the lower tolerance parts. Just taking a single measurement is not enough.
immibis 2y ago
If values are normally distributed, random errors accumulate with the square root of the number of components. Four components in series have 2x the uncertainty over all, etc, but if you divide that double uncertainty by four times the resistance, it's half the percentage uncertainty as before. (I avoid using the word "tolerance" because someone will argue whether it really works this way) In reality, some manufacturers may measure some components, and the ones within 1% get labeled as 1%, then it may be that when you're buying 5% components that all of them are at least 1% off, and the math goes out the window since it isn't a normal distribution.
- afiori 2y ago
  I wonder about the effect of different wiring patterns. For example you can can combine N^2 resistors in N parallel strips of N resistors in serie. I expect that in this case the uncertainty would decrease
  - afiori 2y ago
    Iterating either of f(x) = 3/(1/x + 1/110 + 1/90) g(x) = 1/(1/(3x) + 1/(3110) + 1/(3*90)) Seems to show that 100 is a stable attractor. So I will postulate without much evidence that if you link N^2 resistors with average resistance h in a way that would theoretically give you a resistor with resistance h you get an error that is O(1/N)
  - RetroTechie 2y ago
    In the article's example, I'd prefer 2 resistors in parallel. That way result is less dramatic if 1 resistor were to be knocked off the board / fail. Eg. 1 resistor slightly above desired value, and a much higher value in parallel to fine-tune the combination. Or ~210% and ~190% of desired value in parallel. That said: it's been a long time since I used a 10% tolerance resistor. Or where a 1% tolerance part didn't suffice. And 1% tolerance SMT resistors cost almost nothing these days.
    - bombela 2y ago
      This might be why pretty much all LED lightbulbs/fixtures have two resistors in parallel. Used for the driver chip control pin, that sets the current to deliver via some specific resistance value. It's always a small and a large resistor. The higher this control resistance, and the lower the driving current. Cut off the high value resistor to increase the resistance a bit. In my experience this often almost halves the driving current, and up to 30% of the light output (yes, I measured). Not only most modern lights are too brights to start with anyways, this fixes the intentional overdriving of the LEDs for planned obsolescence. The light will last pretty much forever now.
nomel 2y ago
> tolerance should actually go down since the errors help cancel each other out. Complete nonsense. The tolerance doesn't go down, it's now +/- 2x, because component tolerance is the allowed variability, by definition, worst case, not some distribution you have to rely on luck for. Why do they use allowed variability? Because determinism is the whole point of engineering, and no EE will rely on luck for their design to work or not. They'll understand that, during a production run, they will see the combinations of the worst case value, and they will make sure their design can tolerate it, regardless. Statistically you're correct, but statistics don't come into play for individual devices, which need to work, or they cost more to debug than produce.
- tempestn 2y ago
  The total tolerance is not +/- 2x, because the denominator of the calculation also increases. You can add as many 5% resistors in series as you want and the worst case tolerance will remain 5%. (Though the likely result will improve due to errors canceling.) For example, say you're adding two 10k resistors in series to get 20k, and both are in fact 5% over, so 10,500 each. The sum is then 21000, which is 5% over 20k.
- deleted 2y ago
  [deleted]
- jjmarr 2y ago
  > Statistically you're correct, The Central Limit Theorem (which says if we add a bunch of random numbers together they'll converge on a bell curve) only guarantees that you'll get a normal distribution. It doesn't say where the mean of the distribution will be. Correct me if I'm wrong, but if your resistor factory has a constant skew making all the resistances higher than their nominal value, a bunch of 6.8K + 6.8K resistors will not on average approximate a 13.6K resistor. It will start converging on something much higher than that. Tolerances don't guarantee any properties of the statistical distribution of parts. As others have said, oftentimes it can even be a bimodal distribution because of product binning; one production line can be made to make different tolerances of resistors. An exactly 6.8K resistor gets sold as 1% tolerance while a 7K gets sold as 5%.

rexer 2y ago

Note that tolerance and uncertainty are different. Tolerance is a contract provided by the seller that a given resistor is within a specific range. Uncertainty is due to your imprecise measuring device (as they all are in practice). You could take a 33k Ohm resister with 5% tolerance, and measure it at 33,100 +/- 200 Ohm. At that point, the tolerance provides no further value to you.

MobiusHorizons 2y ago
It’s not nearly that simple:) Component values change with environmental factors like temperature and humidity. Resistors that have a 1% rating don’t change as much over a range of temperatures as 5% or 10% components do. This is typically accomplished by making the 1% resistors using different materials and construction techniques than the lower tolerance parts. Just taking a single measurement is not enough.

immibis 2y ago

If values are normally distributed, random errors accumulate with the square root of the number of components. Four components in series have 2x the uncertainty over all, etc, but if you divide that double uncertainty by four times the resistance, it's half the percentage uncertainty as before. (I avoid using the word "tolerance" because someone will argue whether it really works this way) In reality, some manufacturers may measure some components, and the ones within 1% get labeled as 1%, then it may be that when you're buying 5% components that all of them are at least 1% off, and the math goes out the window since it isn't a normal distribution.

afiori 2y ago
I wonder about the effect of different wiring patterns. For example you can can combine N^2 resistors in N parallel strips of N resistors in serie. I expect that in this case the uncertainty would decrease
- afiori 2y ago
  Iterating either of f(x) = 3/(1/x + 1/110 + 1/90) g(x) = 1/(1/(3x) + 1/(3110) + 1/(3*90)) Seems to show that 100 is a stable attractor. So I will postulate without much evidence that if you link N^2 resistors with average resistance h in a way that would theoretically give you a resistor with resistance h you get an error that is O(1/N)
- RetroTechie 2y ago
  In the article's example, I'd prefer 2 resistors in parallel. That way result is less dramatic if 1 resistor were to be knocked off the board / fail. Eg. 1 resistor slightly above desired value, and a much higher value in parallel to fine-tune the combination. Or ~210% and ~190% of desired value in parallel. That said: it's been a long time since I used a 10% tolerance resistor. Or where a 1% tolerance part didn't suffice. And 1% tolerance SMT resistors cost almost nothing these days.
  - bombela 2y ago
    This might be why pretty much all LED lightbulbs/fixtures have two resistors in parallel. Used for the driver chip control pin, that sets the current to deliver via some specific resistance value. It's always a small and a large resistor. The higher this control resistance, and the lower the driving current. Cut off the high value resistor to increase the resistance a bit. In my experience this often almost halves the driving current, and up to 30% of the light output (yes, I measured). Not only most modern lights are too brights to start with anyways, this fixes the intentional overdriving of the LEDs for planned obsolescence. The light will last pretty much forever now.

nomel 2y ago

> tolerance should actually go down since the errors help cancel each other out. Complete nonsense. The tolerance doesn't go down, it's now +/- 2x, because component tolerance is the allowed variability, by definition, worst case, not some distribution you have to rely on luck for. Why do they use allowed variability? Because determinism is the whole point of engineering, and no EE will rely on luck for their design to work or not. They'll understand that, during a production run, they will see the combinations of the worst case value, and they will make sure their design can tolerate it, regardless. Statistically you're correct, but statistics don't come into play for individual devices, which need to work, or they cost more to debug than produce.

tempestn 2y ago
The total tolerance is not +/- 2x, because the denominator of the calculation also increases. You can add as many 5% resistors in series as you want and the worst case tolerance will remain 5%. (Though the likely result will improve due to errors canceling.) For example, say you're adding two 10k resistors in series to get 20k, and both are in fact 5% over, so 10,500 each. The sum is then 21000, which is 5% over 20k.
deleted 2y ago
[deleted]
jjmarr 2y ago
> Statistically you're correct, The Central Limit Theorem (which says if we add a bunch of random numbers together they'll converge on a bell curve) only guarantees that you'll get a normal distribution. It doesn't say where the mean of the distribution will be. Correct me if I'm wrong, but if your resistor factory has a constant skew making all the resistances higher than their nominal value, a bunch of 6.8K + 6.8K resistors will not on average approximate a 13.6K resistor. It will start converging on something much higher than that. Tolerances don't guarantee any properties of the statistical distribution of parts. As others have said, oftentimes it can even be a bimodal distribution because of product binning; one production line can be made to make different tolerances of resistors. An exactly 6.8K resistor gets sold as 1% tolerance while a 7K gets sold as 5%.