How do I work out the effect of ppm/°C and tolerance on current sense resistor?

[David_]

Hello.

I know that ppm is parts per million, but I am choosing between using two parallel 2mΩ resistors each rated with 200ppm/°C, 0,5% tolerance, 2W or one single 1mΩ resistor that's rated with 30ppm/°C, 0,5% tolerance, 3W but the single 1mΩ resistor costs twice as much as the two 2mΩ resistors but I don't know how to evaluate the ppm/°C value for both of these options.

First of all, what does the two parallel 200ppm/°C resistors amount to in terms of ppm/°C?
I know that two parallel 0,5% tolerance resistors is probably more accurate than a single 0,5% tolerance resistor, but I can't explain to my self why that is, so I don't know if ppm/°C could be treated in the same way.

Is the answer to the above that since each resistors value are decreased, each 2mΩ resistor amounts to 0,5mΩ(total amount of 1mΩ, there are two resistors hence 0,5mΩ), so could it be said that the tolerance is 0,5% out of 0,5mΩ which will be smaller than 0,5% out of 1mΩ?

How and why does the two parallel 0,5mΩ(2mΩ) resistors each with 0,5% tolerance combine into the final tolerance value, is the value of the combination still 0,5%?

I know that the 2Ω resistors aren't 0,5Ω but are still 2mΩ, or are they?, I guess I am trying to think of this in a way to develop a more intuitive feel for the results.

As for ppm/°C, so each °C brings about a change in resistance of x ppm, can I simply divide the resistor value by 1 000 000 and multiply the resulting value with in these cases 200 and 30?

I feel as I should have done more google searching but I have pretty severe ADD and I have simply not manage to get that task done while I have been able to open this thread.

Regards

[Cerebus]

Hello.

I know that ppm is parts per million, but I am choosing between using two parallel 2mΩ resistors each rated with 200ppm/°C, 0,5% tolerance, 2W or one single 1mΩ resistor that's rated with 30ppm/°C, 0,5% tolerance, 3W but the single 1mΩ resistor costs twice as much as the two 2mΩ resistors but I don't know how to evaluate the ppm/°C value for both of these options.

First off, unless you see an explicit indication to the contrary, you should assume that temperature coefficients are ± values. That is 200ppm/C implies ±200ppm/C. You will sometimes see specifications that call out asymmetric limits such as -50/+100ppm/C. If it helps, convert the ppm to percentages first if you're more comfortable with percentages (1ppm = 0.0001%).

First of all, what does the two parallel 200ppm/°C resistors amount to in terms of ppm/°C?
I know that two parallel 0,5% tolerance resistors is probably more accurate than a single 0,5% tolerance resistor, but I can't explain to my self why that is, so I don't know if ppm/°C could be treated in the same way.

Parallel resistors (of equal tempco) will exhibit the same tempco limits as the individual resistors. That is, two 2mΩ ±200ppm/C resistors paralleled will produce a 1mΩ ±200ppm/C resistor. However, because of statistical effects, there will be a tendency for paralleled resistors to average out their tolerances and tempcos, so you will, on average, get better results in tolerance terms from paralleled resistors. The improvement is on the order of the square root of the number of resistors combined but it is statistical so it's not a guaranteed improvement.

Is the answer to the above that since each resistors value are decreased, each 2mΩ resistor amounts to 0,5mΩ(total amount of 1mΩ, there are two resistors hence 0,5mΩ), so could it be said that the tolerance is 0,5% out of 0,5mΩ which will be smaller than 0,5% out of 1mΩ?

No. All the operations involved (multiplication, division) are commutative so you get the same result whichever way you cut it.  If the individual resistors have 0.5% tolerance then a parallel combination has the same tolerance. (Subject, as before, to a slight statistical improvement, that you can't rely on. So you might as well use the worst case figures straight and count any statistical improvement as a welcome bonus.)

How and why does the two parallel 0,5mΩ(2mΩ) resistors each with 0,5% tolerance combine into the final tolerance value, is the value of the combination still 0,5%?

I know that the 2Ω resistors aren't 0,5Ω but are still 2mΩ, or are they?, I guess I am trying to think of this in a way to develop a more intuitive feel for the results.

To get an intuitive grip, think about what the current is doing, see the circuit from the perspective of the current flowing through it - "be the current" if you will. In this case, the current splits in half to pass through two equal valued resistors. Each seperate stream of current 'sees' the tolerances of the components that it passes through, and then the currents recombine on the other side (altered by the tolerances) to add together to make a whole again.

As for ppm/°C, so each °C brings about a change in resistance of x ppm, can I simply divide the resistor value by 1 000 000 and multiply the resulting value with in these cases 200 and 30?

Yup. Don't forget that it's 'delta T' (∂T), that is, the temperature difference from nominal. So if nominal temperature is the usual 25C, and the actual working temperature is 40C, then ∂T = 15C and for a ±200ppm/C tempo that's a difference of ±3000ppm, or ±0.3%.

For what it's worth I'd regard 200ppm/C as rather poor for a current shunt (remember it will suffer I²R heating) and 30ppm/C as a much more desirable figure.

[Andreas]

Parallel resistors (of equal tempco) will exhibit the same tempco limits as the individual resistors. That is, two 2mΩ ±200ppm/C resistors paralleled will produce a 1mΩ ±200ppm/C resistor.

You forget the wiring in between the resistors. (which might be relevant when having shunts in 1 mOhm range).
Copper has around 3800 ppm/K tempco. This may spoil the 200 ppm/K value.

And how do you do the kelvin connections on 2 parallel shunts?
I think paralleling low ohmic resistors is not a good idea.

with best regards

Andreas

[Cerebus]

Parallel resistors (of equal tempco) will exhibit the same tempco limits as the individual resistors. That is, two 2mΩ ±200ppm/C resistors paralleled will produce a 1mΩ ±200ppm/C resistor.

You forget the wiring in between the resistors. (which might be relevant when having shunts in 1 mOhm range).
Copper has around 3800 ppm/K tempco. This may spoil the 200 ppm/K value.

And how do you do the kelvin connections on 2 parallel shunts?
I think paralleling low ohmic resistors is not a good idea.

with best regards

Andreas

When a man says he has severe ADD you try to take account of that and try to answer his questions as directly and as simply as you can. Throwing in too much extra, tangentially, all at once, gets very difficult to follow for someone with ADD. One step at a time, my friend, one step at a time. Pile too much on this chap's plate in one go (who has been brave enough and helpful enough to reference his ADD) and you risk driving him off. A little patience is called for.

We'll get to the nitty gritty when David_ is comfortable with ppms and combining them tolerances, paralleling and all that.