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I am trying to implement a precision current sense amplifier. I am calculating now the tolerances introduced by the resistor.

I know the maximum drop will be 20mV and I will use a 5mΩ resistor. Max drop is 18.75mV. I am almost sure that, if I understood well, on the amplifier side, the offset contribution will set the floor for the minimum detectable current. Then, on the resistor side, I can estimate the voltage (hence current) contribution due to resistor tolerance (let's say 1%). But there is the TCR. Everywhere I read that is enough to multiply the ambient temperature with it to find the resistance deviation, but what about the self heating?

The resistor I choose was this one: https://eu.mouser.com/datasheet/2/303/res_fc4l-1265349.pdf (FC4L16 serie, 5mΩ). But I see no mentioning on thermals due to self heating. I would expect a thermal resistance to allow me the temperature increase, or estimate it at least. Should I just assume that within power rating, self heating does not contribute as much as ambient temperature? The resistor will be stressed at less than 33% of its current rating (max dissipation will be 70mW, while is rated 250mW).

Hope my doubt is clear: what is the correct way to estimate errors from those resistors, if I don't get the thermal data?

EDIT: is not important how good or bad will be the performance of the resistor, I need to find out truthful error contributions, and I don't know when to make assumptions and when not, since seems I don't have full thermal data on that particular resistor - and I am not sure is normal or not.

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    \$\begingroup\$ To determine how hot the resistor will get indeed you need to know the thermal resistance from the resistor to ambient. Realize that your PCB layout will have a large influence on that thermal resistance, large copper areas means better cooling. How much is the influence of TCR? Suppose the resistor gets 100 degrees hotter, with +/- 100 ppm that means +/- 1 %. You don't mention at all what accuracy and drift you actually need. Is that 1% acceptable? \$\endgroup\$ Commented May 6, 2021 at 10:09
  • \$\begingroup\$ I don't have a requirement. I want to know with a part like that (Which does not show thermal resistance) which performance I can get out of it. My goal is to quantify it, it does not matter if is bad - so 1% or 10% are both OK as long as it is truthful. You mention 1%, based on self heating, but how can I estimate that increase? That is the specific question \$\endgroup\$
    – thexeno
    Commented May 6, 2021 at 10:13
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    \$\begingroup\$ but how can I estimate that increase? Carefully read my comment. Since I didn't have thermal resistance information I cannot estimate how hot the resistor will get. So I made an assumption, the assumption being that the resistor will get 100 degrees C hotter than ambient, that results in +/- 1% resistor value change. If that 100 C temperature increase is realistic depends on unknown factors like power dissipation in the resistor and cooling area size. \$\endgroup\$ Commented May 6, 2021 at 10:34
  • \$\begingroup\$ Thank you. But then, the Ultimate question is, what is the correct way of estimating errors from such resistors? \$\endgroup\$
    – thexeno
    Commented May 6, 2021 at 10:57
  • \$\begingroup\$ back around the circle - Bimpelrekkie gave you the answer. The unknown is how hot the resistor will get - this depends on how the resistor is mounted etc. I'd suggest most engineers would calculate the loss in Watts and actually measure it in the application and maybe in free air. \$\endgroup\$
    – Kartman
    Commented May 6, 2021 at 13:04

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Yes, you should account for self-heating using the thermal resistance.

You can estimate the thermal resistance from the derating curve. The max temperature at zero power is the max allowable substrate temperature.

enter image description here

Subtract the rated temperature at full power from the rated temperature at zero power.

125 - 70 = 55 degC

Divide by the rated power.

55 degC / 0.25 W = 220 degC/W

Now, it isn't clear whether this resistor is cooled primarily by conduction or convection. If you assume conduction, then use your local board temperature for the ambient.

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  • \$\begingroup\$ For competeness, I checked the datasheet of a MOSFET SQM120P10-10m1L. They state 375W @ Tcase = 25 °C, 125W @ Tcase = 125 °C. Which lead to 0.4°C/W, and is also explicitly stated: RthJC = 0.4°C/W. Which still makes me wonder why not put the figure explicitly, in the case of the resistor. The graph is very important, but sometimes a number is too. In other words, I verified what you said, so if there isn't an error in my findings, thanks a lot! \$\endgroup\$
    – thexeno
    Commented May 8, 2021 at 14:59

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