Error propagation formula and components tolerance (questions) - Page 1
Forum section: electronics / general

author: NightMoth
author_rank: Contributor
timestamp: February 04, 2025, 11:23:05 am
content: Hello allI hope it is correct forum section to post my questions.I'm trying to get a rid with error propagation concept and how one can relate it to components tolerances.I found general formula for error propagation here:https://foothill.edu/psme/daley/tutorials_files/10.%20Error%20Propagation.pdfAnd now I'm trying to apply the formula for a "toy" voltage divider output problem:Vin = 10V (ideal)R1(upper) = 1k, 1% tolerance (+/-10R or min 990R ... max 1010R)R2(lower) = 1k, 1% tolerance (+/-10R or min 990R ... max 1010R)Vout = ? V, ? % toleranceWithout using error propagation formula I can estimate:Vout(ideal) = Vin*R2/(R1+R2) = 10V*1k/(1k + 1k) = 5VVout(min) = Vin*R2min/(R1max+R2min) = 10V*990R/(1010R + 990R) = 4.95VVout(max) = Vin*R2max/(R1min+R2max) = 10V*1010R/(990R + 1010R) = 5.05VSo I expect that correct answer isVout = 5V +/- 50mV or 5V with 1% tolerance.Now i'm trying to get the same result with error propagation formula and it's where my problem begins.Here is my workflow:- Step 1. First I derived error propagation formula from voltage divider formula:std_Vout^2 = Vin^2*(std_R1^2*R2^2 + std_R2^2*R1^2)/(R1+R2)^4;- Step 2. Next I need to substitute standard deviations for resistors values into the formula, but I only know tolerances of R1 and R2.So (just for calculation purposes) I made assumption that manufacturer of resistors used five-sigma interval to guarantee tolerance,hence assumed standard deviation manufacturer got when they tested their resistors was 10R/5 = 2R or 0.2% relative to 1k resistor values.- Step 3. Next I substituted all values into the formula and got standard deviation for Vout:std_Vout = 7.071 mV or 0.141% realtive to ideal 5V Vout- Step 4. Finally I calculated toleranse for Vout using five-sigma interval (because I've assumed same interval for resistor's tolerance in Step 2):Tolerance = 5*std_Vout = 5*0.007071V = 0.0354V or 0.707%,- Result. Final answer calculated with the formula is:Vout (ideal) = 5VStandard deviation = 7.071mV or 0.141%Tolerance = +/- 35.4mV or 0.707% tolerance.But expected answer is Vout = 5V with +/- 50mV or 1% tolerance.Here is my questions:1) Calculated realtive standard deviation for Vout is 0.141%, but relative standard deviations of resistors is 0.2%.As for me it seems very suspicious that uncertainty of some device's output is less than uncertainty of components parameters it made from.I always belived that some device made from many components can't be more precise that precision of any of its components.Or it depends? And actually works this way in real life?2) I realised that in order to get expected answer for tolerance I have to multiply calculated answer by sqrt(2).For example, above calulated tolerance is 0.707% and if I multiply it by sqrt(2) it will be 1% (expected answer).I tried re-calculate above problem with other values and tried some other simple problems and realised that it is always so.Where I have missed this sqrt(2) factor?Is it because tolerance is in +/- range so formula for Vout variance should be doubled?3) In general, is it ok to substitute tolerances instead of standard deviations into the error propagation formula? (it looks like in my calculations it changes nothing, just need to skip Step 2 and 4)

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2025, 11:39:23 am
content: The first calculation is for the worst case. This is different from the error propergation that calculated the standard deviation and thus more like expected limits. This allows for the chance that the errors of the 2 resistors can compensate with some properbility. It is in between the best case and worst case.It is the best estimate for the standard deviation.There is no simple factor like 3 or 5 between the standard deviation and the absolute worst case. The 3 sigma level is just an approximation / convention that works OK most if the time. With normal distributed data there is still a chance to be outside the 3 sigma and also the 5 sigma interval - it is just not very likely.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: February 04, 2025, 01:47:40 pm
content: I think an example of how lower precision parts can yield a higher precision device is the Hamon divider.