Question about component tolerance and probability

[rtarbell]

More of a probability question than electronics, but it can be applied to many electronic components:

Let's say we have two 100 ohm resistors, 5% tolerance each. If we put them in parallel, the equivalent resistance is 50 ohms, but what is the equivalent tolerance? How can we compute this?

Just to make things more difficult, a follow-up question:

Say we have a 50 ohm resistor, 5% tolerance, and a 200 ohm resistor, 1% tolerance. Their equivalent parallel resistance is 40 ohms, but what is the equivalent tolerance?

[richie00boy]

Purely theoretically, it's 1/((1/5) + (1/5)) with equal weighting to the resistance, like in your first case.

However this takes the assumption that tolerance spread is distributed evenly and thus some cancellation occurs. I personally would not use it as a way to get closer tolerance, I would just use 1% parts from the start.

Your second case I'm not sure but maybe the tolerance of the higher resistor is weighted in proportion to your lower one, e.g. it contributes a 1/4 of the overall value. Multiply the tolerance of this part by 4 using the above equation.

[PigletsDad]

In reality, if you use two 100 Ohm resistors from the same production batch, you will typically find they have very similar values, so the tolerance is not altered.

[AudioFreak]

Worst case would be that one of the resistors is 5% high and the other resistor is 5% low giving a worst case tolerance of 10%. As you add more and more in parallel, statistics would generally dictate a normal distribution which should have enough of the resistors close to the nominal value that the tolerance of the whole network is seen to improve. In practice, you'd be relatively unlucky to pick two 5% resistors from the same batch that are at either end of the allowable tolerance.

[richie00boy]

No that's wrong, your example would have a tolerance of 0% - think about it...

[Geoff]

Originally posted by AudioFreak
Worst case would be that one of the resistors is 5% high and the other resistor is 5% low giving a worst case tolerance of 10%.

Originally posted by richie00boy
No that's wrong, your example would have a tolerance of 0% - think about it...

Sorry, you are both wrong. If you parallel a 105R and a 95R resistor (ie 100R +/-5%) the end result is 49R875 which is 50R -0.25%

Paralleling two equal resistors results in a halving of both the value and the tolerance, so two 100R 5% resistors become 50R +/-2.5%.

[Geoff]

Originally posted by rtarbell


Just to make things more difficult, a follow-up question:

Say we have a 50 ohm resistor, 5% tolerance, and a 200 ohm resistor, 1% tolerance. Their equivalent parallel resistance is 40 ohms, but what is the equivalent tolerance?

The 'worst case' situation (both resistors maximally high or low) results in a deviation from 40R of 3.23%.

[Boris_The_Blade]

Originally posted by Geoff





Sorry, you are both wrong. If you parallel a 105R and a 95R resistor (ie 100R +/-5%) the end result is 49R875 which is 50R -0.25%

Paralleling two equal resistors results in a halving of both the value and the tolerance, so two 100R 5% resistors become 50R +/-2.5%.

not true...

paralleling two 100R resistors with 5% tolerance, that say are 105R each will give 52R5. which is 50R +/- 5%, tolerance stays the same...

[Geoff]

Originally posted by Boris_The_Blade



not true...

paralleling two 100R resistors with 5% tolerance, that say are 105R each will give 52R5. which is 50R +/- 5%, tolerance stays the same...

I stand corrected. You are absolutely right. I must have been having a senior moment when I edited my post to add the second paragraph.

However, this means that richie00boy's equation in Post #2 is also incorrect.