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I am building an audio crossover with time alignment based on Douglas Self's design. I desided to make an algorithm to calculate best resistor pair. The first part is easy:

  1. Make a list of all resistors in E24 series.
  2. Calculate parallel resistance for all combinations of two resistors.
  3. Compare it to the target value and get error = calculated / target.
  4. Return value with smallest error.

Now here's the part that I can't figure out. I use resistors with 1% tolerance. If I use two in parallel the combined deviation supposedly goes down. How do I calculate the exact value by how much the deviation goes down and how do I combine it with error from previous part to get the best pair?

enter image description here

enter image description here

UPDATE:

On @jonk 's recommendation:

r1_deviation_facror = 1 ( 1 + R1/R2)

r2_deviation_facror = 1 ( 1 + R2/R1)

combined_deviation = sqrt(r1_deviation_facror^2 + r2_deviation_facror^2)

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    \$\begingroup\$ No matter how you combine the resistors, the tolerance remains the same. For example consider a 100Ohm resistors with tolerance 1%. Then the maximum value of parallel connection would be (101*101)/202=50.5, that is 50 plus 1% of 50. \$\endgroup\$
    – Eugene Sh.
    Apr 23, 2021 at 19:49
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    \$\begingroup\$ The statistical distribution is unknown. It is possible that all 4.7k resistors are manufactured on the same line. Those within 0.1% are marked and sold as 0.1%, and thus removed from the population. Those meeting 1% are marked and sold as 1% and are thus removed from the population. Then 5%. At this point, the resistors which remain in the population will be clustered at the high and low ranges. This is not a normal distribution. Note: I am not saying this is how they do it. But it could be. \$\endgroup\$
    – user57037
    Apr 23, 2021 at 20:02
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    \$\begingroup\$ @Viktor I've only bothered to measure a batch of resistor values once in order to get its distribution. (We knew we had a problem, so the time was spent to see.) In almost no cases did we get resistors near the specified value. The population was bimodal (with strong kurtosis), instead. Obviously, selection was taking place and the manufacturer was selling tighter tolerance parts for more money, selling us the remainders. I've no idea today how that may apply and, if so, where it is more or less likely with respect to rated % tolerances. \$\endgroup\$
    – jonk
    Apr 23, 2021 at 20:09
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    \$\begingroup\$ Without getting into the statistics and guessing about distributions, if you are only making one piece I don't see a lot of downside to measuring a part chosen to be slightly high and paralleling with with another to trim it in. My experience is that modern resistors are trimmed in and tend to be within about 1/3 of rated tolerance and often the average is biased to one particular side of nominal. \$\endgroup\$ Apr 23, 2021 at 22:06
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    \$\begingroup\$ Aside from mathematical elegance, do you have any reason to believe that you can hear such tiny variations in audio amplitude? \$\endgroup\$ Apr 23, 2021 at 22:42

5 Answers 5

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If you are just looking for the mathematics, then it's just a sensitivity analysis. Suppose:

$$\begin{align*} R_{parallel} &= \frac{R_1\,R_2}{R_1+R_2} & R_{series} &= R_1+R_2 \end{align*}$$

Then, for parallel:

$$\begin{align*} \% R_{parallel} &= \frac{\text{d}\,R_{parallel}}{R_{parallel}}\\\\ &= \frac1{R_{parallel}}\left[\frac{R_2^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_1 + \frac{R_1^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_2\right]\\\\ &=\frac{R_1+R_2}{R_1\,R_2}\left[\frac{R_2^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_1 + \frac{R_1^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_2\right]\\\\ &=\frac{R_2}{R_1+R_2}\,\frac{\text{d}R_1}{R_1} + \frac{R_1}{R_1+R_2}\,\frac{\text{d}R_2}{R_2}\\\\ &=\frac{\%R_1}{1+\frac{R_1}{R_2}}+\frac{\%R_2}{1+\frac{R_2}{R_1}} \end{align*}$$

It's pretty easy to see from the above that the smaller resistor's %-deviation dominates the %-deviation for parallel combinations. And that just makes sense, of course.

And, for series:

$$\begin{align*} \% R_{series} &= \frac{\text{d}\,R_{series}}{R_{series}}\\\\ &= \frac1{R_{series}}\bigg[\text{d}R_1 + \text{d}R_2\bigg]\\\\ &=\frac1{R_1+R_2}\bigg[\text{d}R_1 + \text{d}R_2\bigg]\\\\ &=\frac{R_1}{R_1+R_2}\,\frac{\text{d}R_1}{R_1} + \frac{R_2}{R_1+R_2}\,\frac{\text{d}R_2}{R_2}\\\\ &=\frac{\%R_1}{1+\frac{R_2}{R_1}}+\frac{\%R_2}{1+\frac{R_1}{R_2}} \end{align*}$$

It's pretty easy to see from this case above that the larger resistor's %-deviation dominates the %-deviation for series combinations. And that just makes sense, too, of course.

That's the mathematics, sans Beyes and probabilities. And I've no idea if that's where you wanted to go. So I'm just putting it out there for your consideration.

Notes

I left the above with the following two statements:

$$\begin{align*} \% R_{parallel} &=\frac{\%R_1}{1+\frac{R_1}{R_2}}+\frac{\%R_2}{1+\frac{R_2}{R_1}}\\\\ \% R_{series} &= \frac{\%R_1}{1+\frac{R_2}{R_1}}+\frac{\%R_2}{1+\frac{R_1}{R_2}} \end{align*}$$

Let's set \$k_1^{'}=\frac1{1+\frac{R_2}{R_1}}\$ and \$k_2^{'}=\frac1{1+\frac{R_1}{R_2}}\$. But note also that \$1 = k_1^{'}+k_2^{'}\$! So let's use \$k=k_1^{'}\$ and therefore \$\left(1-k\right)=k_2^{'}\$:

Then:

$$\begin{align*} \% R_{parallel} &= k_2^{'}\cdot \%R_1+k_1^{'}\cdot \%R_2\\\\&= \left(1-k\right)\cdot \%R_1+k\cdot \%R_2\\\\ \% R_{series} &= k_1^{'}\cdot \%R_1+k_2^{'}\cdot \%R_2\\\\&= k\cdot \%R_1+\left(1-k\right)\cdot \%R_2 \end{align*}$$

The above is just algebraic manipulation. But there's a deeper meaning.

You were asking about resistor values that are near to each other versus resistor values that are far from each other. Your tables show entries for \$199:\$1 and \$39:1\$ ratios, for example. How does that apply here?

A ratio expressed as \$a:b\$ is no different from saying \$\frac{a}{a+b}:\frac{b}{a+b}\$ and as it turns out \$1=\frac{a}{a+b}+\frac{b}{a+b}\$. But recall that \$1 = k_1^{'}+k_2^{'}\$. So we can just as well write: \$ k_1^{'}:k_2^{'}\$ or \$k:1{-k}\$, in equivalent form.

If we choose things so that \$R_2\ge R_1\$ (and we can always do that), then it follows that \$\frac{R_2}{R_1}:1\$ is the equivalent relationship. Set \$r=\frac{R_2}{R_1}\$, where \$R_2\ge R_1\$ and \$\therefore r\ge 1\$, so that \$r:1\$ applies. (As \$r\to \infty\$, \$k\to 0\$ and as \$r\to 0\$, \$k\to 1\$.) Then:

$$\begin{align*} \% R_{parallel} &= \frac1{r+1}\bigg[r\cdot \%R_1+ \%R_2\bigg]\\\\ \% R_{series} &= \frac1{r+1}\bigg[ \%R_1+ r\cdot\%R_2\bigg] \end{align*}$$

Since \$r\ge 1\$, and keeping in mind \$R_2\ge R_1\$, the above makes it pretty clear in each case whether the larger or smaller valued resistor is dominant.

Now, if the % specification is considered Gaussian (integral of Poisson events) then you can use that fact in your computations:

$$\begin{align*} \% R_{parallel} &= \frac1{r+1}\sqrt{r^{\,2}\cdot \%R_1^{\:2}+ \%R_2^{\:2}}\\\\ \% R_{series} &= \frac1{r+1}\sqrt{ \%R_1^{\:2}+ r^{\,2}\cdot\%R_2^{\:2}} \end{align*}$$

Let's take the case of a \$39:1\$ and where both are 1% resistors placed in parallel or series. Then we'd compute \$\frac{1}{39+1}\sqrt{\left[ 1\%\right]^2+\left[39\cdot 1\%\right]^2}\approx 0.97532\%\$. (Doesn't matter whether we use parallel or series.) Not much of an improvement.

To relate this to your table, where the author proposes the sum of two \$10\:\Omega\$ resistor with 1% tolerance to make a single \$20\:\Omega\$ resistor, you'd now expect \$20\:\Omega\cdot \pm 0.97532\%\approx \pm 0.1951 \:\Omega\$, which is what you'll see in your table for the case of \$39:1\$.

(However, Self calls this value the "standard deviation," which absent any clarification implies to me \$1\,\sigma\$. If so, I disagree. More on that in the Summary below.)

For \$4:1\$, find \$20\:\Omega\cdot \frac{1}{4+1}\sqrt{\left[1\%\right]^2+\left[4\cdot 1\%\right]^2}\approx \pm 0.1649\:\Omega\$ and this also matches the table entry you have in your question.

Summary

The only problem I have with Self's discussion is that he appears to assume that a 1% resistor has a single \$\sigma\$ of 1%. I frankly doubt this and instead earlier believed that it is a \$3\,\sigma\$ specification. But I'm no expert on the subject. Perhaps someone will add a reference to an official document from a resistor manufacturer on this topic.

However, that said, the above shows that the blind application of statistical rules (absent any real empirical knowledge of the statistics of modern resistors made by modern manufacturers and sold "unbinned") combines nicely together with common sensitivity analysis on parallel and series equations to achieve similar results to the use of Excel and hundreds of monte carlo runs in a grand numerical exercise by Douglas Self.

Douglas Self didn't really need to resort to numerical, monte carlo methods to get to the same place. It could have been similarly achieved merely by adding sensitivity analysis. That analysis can also be applied to far more complex combinations and more interesting questions, allowing wide-ranging exploration, if needed.

Best wishes!

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  • \$\begingroup\$ That's more in the direction of what i need. Let me dwell on it for a couple days. \$\endgroup\$
    – Viktor
    Apr 23, 2021 at 21:55
  • \$\begingroup\$ I added the picture above, which I am trying to understand. I see that the values of standart deviation are a square root of a sum of squares of resistor values devided by a 10 to the power of something, but where that "`10 to the power of something" came from I can't understand. And how I can measure this values against the error from the first part of the algorithm. \$\endgroup\$
    – Viktor
    Apr 23, 2021 at 22:00
  • \$\begingroup\$ I've tried your formula for parallel resistance. You can see the result for 97.7 Ohm in the picture above. It shows that with two equal resistors the deviation is 1%, and it grows from that with increasing resistros ratio. But according to Self it should decrease for two, and the closer their values are, the more it decreases. Unless I grossly misunderstood the concept. \$\endgroup\$
    – Viktor
    Apr 23, 2021 at 22:18
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    \$\begingroup\$ @Viktor It's not that complicated. If you have a parallel/series combination of two resistors where there is a 199:1 relationship in their magnitudes, then the statistics of the smaller/larger resistor is all that will matter (almost all.) That's obvious, since one of the resistors dominates. Self shows this by saying that you "improve from 0.200 to 0.199" in this case (not much, at all.) But when neither dominates, in 1:1, then you "improve from 0.200 to 0.141", which is a sqrt(2) improvement, as expected. But that's about how statistics combine. \$\endgroup\$
    – jonk
    Apr 24, 2021 at 0:39
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    \$\begingroup\$ @Viktor But that assumes Self's point that resistors fall into nice Gaussian curves. Which they very well may do, if the manufacturer doesn't "bin" parts before selling them. If the manufacturer does bin parts, then the statistical assumption is false and therefore the Excel approach of monte carlo runs has far less value. \$\endgroup\$
    – jonk
    Apr 24, 2021 at 0:41
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Your assumption that placing resistors in parallel will always reduce their tolerance is incorrect, in general. This would only be true if you could ensure that both resistors came from a population where the mean value was exactly equal to the specified value. You can not assume that this is the case unless you have measured the resistors yourself.

If you purchase a batch of resistors their mean value is unlikely to be exactly the specified (marked) value. So, if you put two of these resistors in parallel, their combined value should be closer to their mean value, but not necessarily closer to the marked value.

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    \$\begingroup\$ Many years ago I measured a case where the distribution was bimodal, with resistors very near the specified value almost entirely absent. Obviously the result of a selection process. I don't know if manufacturers still bother doing that. But it has happened at least once. \$\endgroup\$
    – jonk
    Apr 23, 2021 at 20:02
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    \$\begingroup\$ @jonk I've seen that behavior a lot with resistors of older vintage (one high cluster and/or one low cluster) but when I tested various recently-manufactured (last year or so) resistors the readings I got were all clustered closely around the nominal value. e.g. 1% nominal reading as 0.3% actual. So YMMV, test to be sure. \$\endgroup\$ Apr 24, 2021 at 5:19
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    \$\begingroup\$ @AlexHajnal Thanks! That experience of mine was two decades back. It does seem reasonable to me that circumstances have changed over time. I very much appreciate the updated info!! You've also confirmed something else. That is that a 1% spec is probably a \$3\sigma\$ spec. That's nice to hear. \$\endgroup\$
    – jonk
    Apr 24, 2021 at 5:24
  • \$\begingroup\$ And then there's also the problem that whatever measuring equipment people have at home usually isn't the most accurate either. With calibrated Agilent/Fluke equipment you know how much the maximum deviation is, with the average, non-calibrated DIY multimeters from the local hardware store you don't. \$\endgroup\$
    – Mast
    Apr 24, 2021 at 19:28
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Since the probability distribution will not be known, I'd like to suggest a different approach.

Assuming resistors in a (terrible) uniform distribution between -1% and +1%, this is distribution of two 1 kOhm resistors in parallel:

enter image description here

And this is for 1 kOhm in parallel with 10 kOhm:

enter image description here

This clearly confirms the intuition (also detailed in jonk's math) that shows why you get larger standard deviation, which you will not be able to determine without knowing the original distributions.

Since you are writing code to find the best value by exhaustive search, my suggestion is: 1) calculate the maximum and minimum value for each association, and choose based on this worst case range 2) since you are using 1% resistors, don't combine resistors with a ratio of 1:100 (as you mentioned in the comments) or worse.

It is pointless to base your decision on unknown probability distributions, which change the resulting standard deviation.

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Combining gaussian distributions example: If you have a resistor population with mean 100 ohms and standard deviation 5 ohms then a series combination of two of them will have a mean 200 ohms standard deviation of 7.07 ohms, the square root of the sum of the squares of the individual deviations. If 200 ohm resistors have the same ratio of standard deviation to mean, then their standard deviation would be 10 ohms. The deviation does go down. But the objections in the other answers and comments are correct. The distribution won't be gaussian for starters. @jonk's experience for some resistors was expectable for many: 10% resistors were the ones that failed 5%.

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If I wanted to create a resistance to match a specific (probably non-standard) value, I would:

  1. Pick the smallest standard value that was guaranteed to be higher;
  2. Measure it
  3. Add the smallest parallel resistance that is guaranteed to produce a result higher than the target value;
  4. Go back to step 2 until within my target tolerance

The value of the parallel resistances you add will increase quickly.

Of course this only works when you're building just one or two.

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