How combining resistors affects the overall tolerance of the total resistance.
Combining Multiple Resistors to Improve Tolerance
Series and parallel resistor combinations are one of the first things you learn when studying introductory electronics theory (see Figure 1). The resulting total resistance formulas are simple and well known.
Figure 1 - Resistor Combinations
A much more interesting, and as we will see more difficult problem, is the following:
"What is the Tolerance of the series and parallel combinations given the Tolerance of the individual resistors?"
This is a question I wondered about and looked for answers online. As is typical on the web, I found many opinions and "half-baked" formulations. I also found some more rigorous analysis, but nowhere (that I found) was the topic clearly summarized and, perhaps more importantly, tied back to a practical measurement. In this article I tried to summarize what I learned and also provide a practical experiment to support the theory.
Why Combine Multiple Resistors?
As we will see, combining multiple resistors can (under certain assumptions) improve the overall precision of the resulting total Resistance. This may be your main motivation for some circuit design requiring precision components. Another benefit is that you can potentially increase the power handling capability as the power is 'distributed' among multiple resistors. There are also disadvantages of course, the most obvious being an increase in space and cost and perhaps a reliability penalty as you must assemble more components in the board.
First the Conclusion
I find this type of problem very interesting, but admittedly this is more of a statistics problem than an Electrical Engineering one. Therefore, for those readers that prefer to just get the results and are not so interested in 'how' they were obtained, here's the conclusion:
Starting with N identical resistors of mean resistance Ri = μ and tolerance Ti, the tolerance Tt of the series and parallel combination RT can be approximately estimated with the following formula:
There are some caveats to this formula however. If you want to understand this better, I encourage you to read on or, as a minimum, read also the "Conclusions" section. For those of you that, like myself, enjoy mathematics, this is actually a very interesting problem. I'll start with the simpler, perhaps less interesting case of the series combination and then later analyze the parallel case.
Series Resistor Combination
We start with N resistors of mean value E(Ri) = μ and tolerance Ti. We first must define what we mean by "Tolerance" more precisely. We all know that tolerance is usually specified as a percentage value. So, as an example, for a 1KΩ resistor (Ri = 1KΩ) with 1% tolerance (Ti = 1%), we expect the average value to be 1KΩ and, for any given random resistor Ri:
To analyze this problem statistically though, we will formulate the "Tolerance" Ti in terms of the "standard deviation" of each Resistor (which we denote as σ):
In other words, the tolerance Ti is a ratio of a measure (kσ) of the resistor variance around the mean divided by the average value. This is of course a simplification. In reality, I can imagine the resistor manufacturer might 'screen' resistors that are above a certain tolerance and through away the rest. However, it is a reasonable and intuitive model for Tolerance that is also mathematically convenient.
For the series combination of Ri resistors we can write:
Let's assume that each resistor 'Ri' follows a normal distribution with standard deviation σ and mean μ. We can write for each resistor Ri:
This again is a simplification. As we will see under "Experimental Results" the actual distribution isn't usually normal. However, this is a good starting point and still provides a pretty good estimate. We can now calculate the expected (mean) value of RT using known statistical relations:
So as expected, the mean value of the resulting resistor RT is N times each resistor value Ri. The variance result is more interesting though. Writing it in terms of standard deviation σ we get:
This means that as we increase the total number of resistors N, we get an increase in standard deviation by a factor of sqrt(N) which was not obvious at first. What does this do to the Tolerance you ask? Even though the total variance increased by sqrt(N), we must recognize that this variance is over a resistor RT that is N times larger! So let's see how the RT resistor tolerance compares with the original tolerance. Using our definition of "Tolerance" above we get:
So in actuality, the tolerance of the resulting resistor RT, is better than each invidual Ri resistor by a factor of sqrt(N). It's an interesting result. Intuitively, one expects the combination of resistors to result in a resistor with better tolerance as some of the variance will "cancel" out. The expression above gives a more precise formulation to this intuition.
Parallel Resistor Combination
The parallel combination case is more difficult to analyze and also more interesting. For a combination of N, Ri resistors in parallel, the well know relation (a) applies to the total resistance RT:
Let's assume again that each resistor 'Ri' follows a normal distribution with standard deviation σ and mean μ. We can write for each resistor Ri:
Each of the 1/Ri terms in equation (a) above is therefore in the form:
Simple and innocent as this formula (b) might look, it actually presents a considerable difficulty. We need to compute the inverse of a random variable and extract the resulting distribution characteristics such as its mean and variance. Unfortunately, when dealing with random variables, the mean and variance of the inverse distribution is generally *not* the inverse of the original mean and variance. Therefore, in order to calculate the (1/Ri) mean and variance, some simplification is needed. We can apply a so called "linear" approximation by using the following Taylor Series expansion:
To convert expression (b) into something in the form (c) we simply divide numerator and denominator by μ:
For decent enough resistors we can expect μ >> σ , i.e. the average value of the resistor is much larger than the variance from sample to sample. If not, you should really be looking for another supplier! With that assumption, we get σ/μ << 1 and thus the second, third and higher order terms in equation (d) become negligible and we can discard them. This yields:
Equation (e) is much easier to handle mathematically and we can determine the Expected value (mean) and the Variance using know properties for these statistical functions. Therefore:
In summary, we can state that:
So we found with equations (f) a good approximation for the mean and variance of the inverse resistance. Now we need to determine the actual mean and variance for the total parallel combination... Feeding (f) back into (a) we obtain:
So we have computed the mean and variance of 1/RT. Now we need the mean and variance of RT. This may be a bit of recursive thinking, but if we look at expressions (f) and think of them as generic approximations for the mean and variance of the inverse of a random variable (1/X), we can apply the same approximation again to (g), now thinking of RT as the random variable. In other words, for a generic random variable X where the mean is much larger than its variance we have the following approximations:
In summary, starting with N resistors of mean μ and variance σ, the mean and variance of the parallel combination RT can be estimated with the following formulas:
So after all this math, we conclude that the distribution for the parallel resistor has a mean of μ/N (fully expected for parallel resistors) and, more interestingly, a variance that decreases with the cube on the number of resistors. This is not the end of the story however, because this new variance is over a smaller resistor. Using the same definition of "Tolerance" that we used above for the "series" case, we can write:
Interestingly we arrive at the same formula for the tolerance of the combined parallel resistor as we did for the series resistor combination! Isn't math great? Now, for the more statistically inclined among you, you may see intuitively that this result sounds a lot like the famous 'central limit theorem'. In fact, when you sum many IID random variables, the central limit theorem tells us that the resulting random variable is approximately normal with standard deviation equal to the individual standard deviation divided by sqrt(N). In a sense, both the series and parallel combinations are sums of many independent random variables so this result makes intuitive sense...
Math is a beautiful thing, but does it really apply to practice? In other words, can we confirm the formulations above using real-world measurements and experiments? As Sir Arthur Eddington, who provided experimental tests of Einstein's theory of gravity, famously said:
"Do not trust a theory until it has been supported by experiment; and do not trust experimental results until they have been supported by theory"
So, let's see how far can we trust our "Theory". The ideal way to prove this would be to measure thousands of resistors, calculate their parallel combination for different "N" and see the variance from batch to batch. Of course this is an impractical task for most hobbyists like me. In fact, only a resistor manufacturing facility would be able to conduct such a study. So instead, I decided to do the following:
1) Measure a total of 50, 1KΩ +/- 1% Resistors from my parts bin (See Figure 2)
2) Use the statistical technique of "Bootstrap Re-sampling" to infer the variance for the parallel combination of these resistors
Figure 2 - Resistors Used in this Experiemnt
Item 2) above (Bootstrap Re-sampling) is an interesting topic on its own right, and explaining it in a rigorous way is beyond the scope of this article. It's however a well accepted statistical method to derive a variance of (almost) any statistic when we have a limited sample of the population. We will use the technique here to estimate what would be the variance of the parallel combination of *any* sample of 50 resistors from this manufacturing batch given my 50 resistor 'sample'.
I measured each of the resistors using my Fluke 87 multimeter in "High-resolution" mode. The Fluke 87 is quite accurate (+/-0.2% in this resistance range) so should be adequate for the task. The histogram in Figure 3 summarizes the results. Note that the average value for this set of 50, 1KΩ resistors came actually a little below 1KΩ though well within the 1% specification. A more interesting conclusion from looking at Figure 3 is that the actual distribution of values does not look very 'normal'. In one of the forums where this topic was discussed, someone mentioned that it's expected to be "bi-modal". I'm not sure if this data supports that assertion or not, but I can say that it does not look exactly normal, and it does seem to have two distinct "bumps" in the histogram.
Figure 3 - Resistor Measurement Histogram and Statistics
At this stage, you might reasonably think that the "normality" assumption I made in the theoretical analysis above is wrong and therefore the conclusions are not valid. As we shall see though, the formulas derived are still a pretty good approximation of reality even assuming this type of empirical distribution.
Figure 4 shows the results of Bootstrapping the parallel resistors values (corresponding to step 2 above). One way to look at this data is as the result of "simulating" a total of 10000 batches of 50 resistors each from this batch of resistors, calculating the parallel value for each, and plotting the resulting histogram. Note that the histogram looks 'normal' and is centered around the mean value of 997/50 ~= 20 as expected.
Figure 4 - Bootstrap (simulation) Histogram for Parallel of 50, 1K Resistors
The really interesting data that we can extract here though is the standard error for the estimate, or the 'variance' for these 10000 resistor parallel combinations which is listed as "Standard Error of Bootstraps" in Figure 4. This was calculated using a small program I wrote for Excel VBA. Compare that against the predicted value from equation (i):
The error is remarkably small (about 0.5%) showing that our original formula is still a good approximation even for this "not-so-normal" resistance distribution!
Even though the results above show good correlation between theory and practice, a few words of advice are in order. First is that I fully expect different manufacturers to have quite different quality standards, and that might affect the results you get. Even in keeping with the same manufacturer, there is no guarantee that there will be differences from batch to batch. If your design is critical, allow some margin on top of the calculations above. However, the results clearly show that there are some statistical gains to be had by combining multiple resistors and it's a handy technique to keep in mind if you ever need it. I love this type of "deceptively simple" problem that straddle the border between Engineering and Mathematics. I hope you enjoyed it too.
UPDATE: As noted by some readers in the eevblog, some manufacturers seem to introduce a sort of 'biasing" into their selection where the average value is slightly below the nominal. It seems to be the case in my measurements where the average came-up at 997.25 Ohm and not 1000 Ohm. Presumably this accounts for temperature coefficients and aging tending to *increase* the resistor value. Keep in mind that the way I defined "Tolerance" above was relative to this "biased" average value and not relative to the nominal value. If you know the average (biased value) for a given set of resistors, then the tolerance gain relative to this average can be estimated by dividing by sqrt(N) as shown above. This analysis does not imply that the tolerance relative to the nominal value will improve by combining resistors!
Comments, questions, suggestions? You can reach me at: contact (at sign) paulorenato (dot) com