Electronic – Are two (or N) resistors in series more precise than one big resistor


Let's say I have one 2 kΩ resistor with 5% tolerance. If I replace it with two 1 kΩ resistors with 5% tolerance, will resulting tolerance go up, down, or remain unchanged?

I'm bad with probabilities, and I'm not sure what exactly tolerance says about resistance and its distribution.

I am aware that in the 'worst case' it will be the same; I'm more interested in what will happen on average. Will the chance of getting a more precise value increase if I use a series of resistors (because deviations will cancel each other out)?

On 'intuitive level' I think that it will, but I have no idea how to do the math with probabilities and find out if I'm actually right.

Best Answer

The worst case won't get any better. The result of your example is still 2 kΩ ±5%.

The probability that the result is closer to the middle gets better with multiple resistors, but only if each resistor is random within its range, which includes that it is independent of the others. This is not the case if they are from the same reel, or possibly even from the same manufacturer within some time window.

The manufacturer's selection process may also make the error non-random. For example, if they make resistors with a wide variance, then pick the ones that fall within 1% and sell them as 1% parts, then sell the remaining ones as 5% parts, the 5% parts will have a double-hump distribution with no values being within 1%.

Because you can't know the error distribution within the worst case error window, and because even if you did, the worst case stays the same, doing what you are suggesting is not useful to electronic design. If you specify 5% resistors, then the design must work correctly with any resistance within the ±5% range. If not, then you need to specify the resistance requirement more tightly.