5 ms·

I also thought that was interesting. Also, wouldn't the tolerance be doubled when you add them in series? Or does it still average out to +/- 5%?

aleph_minus_one 2y ago
> Also, wouldn't the tolerance be doubled when you add them in series? Or does it still average out to +/- 5%? Neither. Let R_{1, ideal}, R_{2, ideal} be the "ideal" resistances; both with the same tolerance t (in your example t = 0.05). This means that the real resistances R_{1, real}, R_{2, real} satisfy (1-t) R_{1, ideal} ≤ R_{1, real} ≤ (1+t) R_{1, ideal} (1-t) R_{2, ideal} ≤ R_{2, real} ≤ (1+t) R_{2, ideal} Adding these inequalities yields (1-t) (R_{1, ideal} + R_{2, ideal}) ≤ R_{1, real} + R_{2, real} ≤ (1+t) (R_{1, ideal} + R_{2, ideal}) So connecting two resistors with identical tolerance in series simply keeps the tolerance identical.
riedel 2y ago
Fun fact is that afaik component values are often distributed in a bi-modal way because actually +-5% often means that they sorted out already the +-1% to sell as a different more expensive batch. At least it used to be that way. Wonder if it is still worth doing this in production. So I guess one could also measure to average things out otherwise the errors will stay the same relatively.
- bluGill 2y ago
  I'm not sure where the line is, but at some point things like temperature a matter and so a low % resister cannot be high % that passes tests.
- dmurray 2y ago
  If you can measure them with that precision, would it make sense to sell them with that accuracy too? So if you tried to manufacture a resistor at 68kΩ +/- 20%, and it actually ended up at 66kΩ +/- 1%, couldn't you now sell it as an E192 product which according to TFA are more expensive? Selling with different tolerances only makes sense to me if the product can't be reliably measured to have a tighter tolerance, perhaps if the low- quality ones are expected to vary over their life or if it's too expensive to test each one individually and you have to rely on sampling the manufacturing process to guess what the tolerances in each batch should be.
  - retrac 2y ago
    Resistors with worse tolerances may be made out of cheaper, less refined wire, which will vary resistance more by temperature. The tolerance and resistance is good over a temperature range. For more reading looking up "constantan".
    - RetroTechie 2y ago
      Most resistors don't use wire, but some film of carbon (cheaper, usually the E12 / 5% tolerance parts) or metal (E24, or 1% and tighter tolerances) onto a non-conducting body. Wires mean winding into a coil, which means increased inductance. I suspect in most cases the tolerances are a direct result from the fabrication process. That is: process X, within such & such parameters, produces parts with Y tolerance. But there could be some trimming involved (like a laser burning off material until component has correct value). Or the parts are measured & then binned / marked accordingly. Actual wire is used for power resistors, like rated for 5W+ dissipation. Inductance rarely matters for their applications.
      - Gibbon1 2y ago
        Accuracy depends on the technology used. Carbon comp tends have less accuracy then carbon film. And it's not true that higher accuracy is always better. Some accurate resisters are essentially wound coils and have high inductance and will also induce and pick up magnetic interference. Stuff like that matters often a lot.
    - projektfu 2y ago
      Thanks, always good to remember that the tolerance of a resistor is not just a manufacturing number but also defined over the specified temperature range.
  - dylan604 2y ago
    Depends on where in the production line they are being tested. If they are tested after they've had their color bands applied, then you wouldn't be able to sell it as a 66kH since the markings would for a 68kH
- projektfu 2y ago
  Unless the components are expensive, that proposition seems dubious. It's much more economical to take a process that produces everything within 12% centered on the desired value and sell it as ±20%. 100% inspection is generally to be avoided in mass production, except in cases where the process cannot reach that capability, chip manufacturing being the classic example. For parts that cost a fraction of a penny, nobody is inspecting to find the jewels in the rough.
  - riedel 2y ago
    Actually it seems to be really the case that multimodal distribution are rather the result of batches not having a mean. So it is rather the effect of systematic error [1]. I guess it is really a myth (we did low cost RF designs back in 2005 and had some real issues with frequencies not aligning die to component spread and I really remember that bi modality problem, but I guess okhams razor should have told me that it makes no economical sense) [1] https://www.eevblog.com/2011/11/14/eevblog-216-gaussian-resistor-redux/ https://www.eevblog.com/2011/11/14/eevblog-216-gaussian-resi...
    - projektfu 2y ago
      Yup, forever the reason for the trim pot.
thornewolf 2y ago
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.
Sohcahtoa82 2y ago
Nope, still averages to +/- 5%. To give an example, let's say you've got two resistors of 100 Ohm +/- 5%. That means each is actually 95-105 Ohm. Two of them is 190-210 Ohm. Still only a 5% variance from 200 Ohm.
- sram1337 2y ago
  Can you assume that +/-5% isn't linearly distributed? If so, the tolerance in practice may likely end up even smaller.
  - sophacles 2y ago
    There's a fundamental misunderstanding here. Tolerance is a specification/contractual value - it's the "maximum allowable error". It's not the error of a specific part, it's the "good enough" value. If you need 100 +/- 5%, any value between 95 and 105 is good enough. Using two components to maybe cancel out the error as you describe. On average, most of the widgets you make by using 2 resistors instead of one may be closer to nominal, but any total value between 95 and 105 would still be acceptable, since the tolerance is specified at 5%. To change the tolerance you need to have the engineer(s) change the spec.

aleph_minus_one 2y ago

> Also, wouldn't the tolerance be doubled when you add them in series? Or does it still average out to +/- 5%? Neither. Let R_{1, ideal}, R_{2, ideal} be the "ideal" resistances; both with the same tolerance t (in your example t = 0.05). This means that the real resistances R_{1, real}, R_{2, real} satisfy (1-t) R_{1, ideal} ≤ R_{1, real} ≤ (1+t) R_{1, ideal} (1-t) R_{2, ideal} ≤ R_{2, real} ≤ (1+t) R_{2, ideal} Adding these inequalities yields (1-t) (R_{1, ideal} + R_{2, ideal}) ≤ R_{1, real} + R_{2, real} ≤ (1+t) (R_{1, ideal} + R_{2, ideal}) So connecting two resistors with identical tolerance in series simply keeps the tolerance identical.

riedel 2y ago

Fun fact is that afaik component values are often distributed in a bi-modal way because actually +-5% often means that they sorted out already the +-1% to sell as a different more expensive batch. At least it used to be that way. Wonder if it is still worth doing this in production. So I guess one could also measure to average things out otherwise the errors will stay the same relatively.

bluGill 2y ago
I'm not sure where the line is, but at some point things like temperature a matter and so a low % resister cannot be high % that passes tests.
dmurray 2y ago
If you can measure them with that precision, would it make sense to sell them with that accuracy too? So if you tried to manufacture a resistor at 68kΩ +/- 20%, and it actually ended up at 66kΩ +/- 1%, couldn't you now sell it as an E192 product which according to TFA are more expensive? Selling with different tolerances only makes sense to me if the product can't be reliably measured to have a tighter tolerance, perhaps if the low- quality ones are expected to vary over their life or if it's too expensive to test each one individually and you have to rely on sampling the manufacturing process to guess what the tolerances in each batch should be.
- retrac 2y ago
  Resistors with worse tolerances may be made out of cheaper, less refined wire, which will vary resistance more by temperature. The tolerance and resistance is good over a temperature range. For more reading looking up "constantan".
  - RetroTechie 2y ago
    Most resistors don't use wire, but some film of carbon (cheaper, usually the E12 / 5% tolerance parts) or metal (E24, or 1% and tighter tolerances) onto a non-conducting body. Wires mean winding into a coil, which means increased inductance. I suspect in most cases the tolerances are a direct result from the fabrication process. That is: process X, within such & such parameters, produces parts with Y tolerance. But there could be some trimming involved (like a laser burning off material until component has correct value). Or the parts are measured & then binned / marked accordingly. Actual wire is used for power resistors, like rated for 5W+ dissipation. Inductance rarely matters for their applications.
    - Gibbon1 2y ago
      Accuracy depends on the technology used. Carbon comp tends have less accuracy then carbon film. And it's not true that higher accuracy is always better. Some accurate resisters are essentially wound coils and have high inductance and will also induce and pick up magnetic interference. Stuff like that matters often a lot.
  - projektfu 2y ago
    Thanks, always good to remember that the tolerance of a resistor is not just a manufacturing number but also defined over the specified temperature range.
- dylan604 2y ago
  Depends on where in the production line they are being tested. If they are tested after they've had their color bands applied, then you wouldn't be able to sell it as a 66kH since the markings would for a 68kH
projektfu 2y ago
Unless the components are expensive, that proposition seems dubious. It's much more economical to take a process that produces everything within 12% centered on the desired value and sell it as ±20%. 100% inspection is generally to be avoided in mass production, except in cases where the process cannot reach that capability, chip manufacturing being the classic example. For parts that cost a fraction of a penny, nobody is inspecting to find the jewels in the rough.
- riedel 2y ago
  Actually it seems to be really the case that multimodal distribution are rather the result of batches not having a mean. So it is rather the effect of systematic error [1]. I guess it is really a myth (we did low cost RF designs back in 2005 and had some real issues with frequencies not aligning die to component spread and I really remember that bi modality problem, but I guess okhams razor should have told me that it makes no economical sense) [1] https://www.eevblog.com/2011/11/14/eevblog-216-gaussian-resistor-redux/ https://www.eevblog.com/2011/11/14/eevblog-216-gaussian-resi...
  - projektfu 2y ago
    Yup, forever the reason for the trim pot.

thornewolf 2y ago

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.

Sohcahtoa82 2y ago

Nope, still averages to +/- 5%. To give an example, let's say you've got two resistors of 100 Ohm +/- 5%. That means each is actually 95-105 Ohm. Two of them is 190-210 Ohm. Still only a 5% variance from 200 Ohm.

sram1337 2y ago
Can you assume that +/-5% isn't linearly distributed? If so, the tolerance in practice may likely end up even smaller.
- sophacles 2y ago
  There's a fundamental misunderstanding here. Tolerance is a specification/contractual value - it's the "maximum allowable error". It's not the error of a specific part, it's the "good enough" value. If you need 100 +/- 5%, any value between 95 and 105 is good enough. Using two components to maybe cancel out the error as you describe. On average, most of the widgets you make by using 2 resistors instead of one may be closer to nominal, but any total value between 95 and 105 would still be acceptable, since the tolerance is specified at 5%. To change the tolerance you need to have the engineer(s) change the spec.