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In an experiment I am conducting, I am trying to measure low currents by wiring a radio device in series with a small-resistance resistor setup, composed of 9 of 1 ohm +/- 5% resistors in series and parallel as shown below, resulting in a total of 1 ohm.

By my understanding, resistors are manufactured such that there is a distribution of values centered by the marked value with most of them lying in the +/- 5% range, but in theory, there will always be resistors outside such range.

While in theory (by error analysis formulas) my series-parallel setup would have the same error as each one of the resistors, intuitively, doing so is in essence taking the average of the 9 resistors and thus tightening up that distribution curve, such that the error %age is lowered. This was my intent in doing those 9 resistors as opposed to only 1 resistor.

So my question is: What is the error of the 9 resistor setup? Theoretically, I calculate 5% but by intuition, it is less. If it's the former, I would like to know why, and if it's the latter, I would like to know how to find the final percentage error in this situation. I hope it is the latter so that I can qualify my experimental technique (oh and speaking of which, please assume that the voltage drop of the resistor setup does not drop the current through the circuit).

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  • \$\begingroup\$ dangerousprototypes.com/2010/07/01/… \$\endgroup\$
    – user3109679
    Feb 9, 2015 at 0:50
  • \$\begingroup\$ In my experience, modern carbon- or metal-film resistors will be very close to their marked values. I once measured 50 10K 5% resistors of both metal and carbon film, and probably of different production batches, and the total spread of values was about 1% (spread, low to high, not +/- 1%). The average was just under 10 K, so I concluded my meter read slightly low. It is likely that the older carbon composition resistors had a wider spread, but I don't know if the distribution implied in the dangerousprototypes link was common. \$\endgroup\$
    – Peter Bennett
    Feb 9, 2015 at 1:22
  • \$\begingroup\$ For a randomly distributed error, the error goes to zero for series and parallel as the number of resistors increases. For a biased deviation from the expected value, it converges to a constant. physics.stackexchange.com/questions/160764/… \$\endgroup\$
    – C. Towne Springer
    Feb 9, 2015 at 6:31
  • \$\begingroup\$ For many enough series resistors, the residual distribution will approach the normal distribution, thus the tolerance standard deviation will decrease by a factor \$1/\sqrt{N}\$. Although the absolute tolerance stays at the initial value, the probability of coming close to that value (for many enough resistors) eventually becomes so low that it won't happen any time, anywhere in the universe. \$\endgroup\$
    – HKOB
    Feb 9, 2015 at 12:22

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You can't improve the expected accuracy by combining resistors with the same accuracy. But you will improve the standard deviation of the result. What this means is that if you had 10 individual 1 ohm, 5% resistors, the standard deviation of the 10 resistors would be close to the standard deviation of the lot they were manufactured from. However, if you combine 9 of them as you did to form a 1 ohm resistor, and did this 10 times with different resistors from the same lot, the standard deviation of these ten 1 ohm resistor networks would be smaller than the first set by a factor of 3 (the square root of 9). What this means is that you have improved the probability of being closer to 1 ohm by using a network of 9 resistors because you have narrowed the width of the probability distribution around 1 ohm.

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  • \$\begingroup\$ I think this page gives the proof of what you said. \$\endgroup\$
    – nidhin
    Feb 12, 2015 at 15:46
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The x% tolerance rating of your resistors don't give you any information about the distribution of the errors, like the article in dangerous prototypes linked by user3109679, and this experiment by Dave Jones have found out, it is usually U shaped, or worse, might be heavily biased towards one side of the mean depending on the batch/process.

Without knowing the actual distribution of values you really can't say where the average will lay. The spread may be lowered but it may not converge to the nominal value as you add resistors if the distribution is lopsided.

Also, do not forget that carbon resistors have a non-trivial thermal coefficient (datasheet link) which for 1Ω resistors of that brand is 350ppm/°C, which for a 20°C variation (say, for ambient temperatures between 10 and 30°C) would mean an aditional variation of 0.7% in the resistance value!

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  • \$\begingroup\$ For certain classes of resistors it's common for the resistances for a given marking to be evenly distributed within the tolerance range, since the resistors are made, then measured and marked. \$\endgroup\$
    – Hot Licks
    Feb 9, 2015 at 2:06

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