To simplify my question lets assume:

I need to convert +/-7.07A RMS AC current into +/-707mV RMS voltage with high acurracy (0.1% or better).

It may be also higher voltage, up to +/-2.5Vpk.

For now I just measure voltage on Ohmite 15FR100E resistor:

  • 100mOhm
  • 90PPM/°C
  • 35°C/W
  • 5W

When I'm trying to measure 5A RMS current It probably heats up by about 87.5°C (calculated from thermal resistance and power) and that gives about 0.788% resistance change (calculated: 87.5°C * 90PPM/°C).

I want to improve this and I'm thinking about two options:

OPTION 1. Use expensive resistor: Powertron FPR 4-T221 0R100 S 1% Q

  • 100mOhm
  • 25PPM/°C
  • 4.8°C/W
  • 3°C/W heatsink
  • 15W
  • 4-terminal

And this should give me 0.024% resistance change (from 5A RMS).

OPTION 2. Use INA250 current sense amplifier with fixed gain and integrated 2mOhm shunt.

INA250 op amp features from datasheet:

enter image description here

I think, that I can assume

  • I can forgot about thermal problems because shunt R is very small
  • I can compensate 0.3% gain error
  • If I pick 200mV/A version - I'll get 2x higher voltage vs my shunt and this is fine

But I'm not sure about (QUESTIONS):

  • Can I just ignore 50mA offset error (this is ~0,7% of my 7.07A peak current) when I need to measure sine RMS acurrately?
  • should I expect some extra sources of errors when I replace shunt with INA250?
  • is this that simple? really? $4 op-amp replaces big expensive shunt just like that?


Cheap Ohmite resistor which I'm using now:

enter image description here

Expensive Powertron resistor:

enter image description here

If someone asks why I'm not using resistor like 10mOhm to reduce heat - the answer is because I tried to avoid 10x gain op-amp in circuit, that would be extra source of errors. Maybe I was wrong. Actually this is not about measurements, I'm building current source similar to this:


  • \$\begingroup\$ "...with high acurracy (0.1% or better)." - why? \$\endgroup\$ May 26, 2020 at 6:57
  • \$\begingroup\$ Because this circuit is supposed to be used as calibrator for some product (measurement device) which is capable to archive 0.1% acurracy. With 0.1% acurracy calibrator I will be able to reach about 0.2% product acurracy. This is good enough for product purpose, but if it is possible to get 0.05% calibrator for few bucks (lke 20-50USD) more - why not. And maybe I will be able to measure smaller currents more accurately on single range (one resistor) instead of making range switch. \$\endgroup\$
    – Kamil
    May 26, 2020 at 11:42

1 Answer 1


If you want o build a calibration tool, you tool should be more accurate than the product you are attempting to calibrate. You seem to be making a meal out of sensing with relatively high value series sensing resistors. You should be trying to minimize the sense R.

You could start here or here for pointers on how to measure accurately (ignore the fact that the links are about measuring low currents). These will give you a much better understanding of the accuracy and performance you might achieve.

One great article to ponder is this, on the uCurrent Gold. enter image description here The basic design could scale to many amps with ease, and with high accuracy and very low burden power dissipation.

  • \$\begingroup\$ "You should be trying to minimize the sense R." I'm worried about noises and offsets from op-amp input at low voltages/resistances. I shouldn't? How low can I go? \$\endgroup\$
    – Kamil
    Jun 1, 2020 at 0:59
  • \$\begingroup\$ Well, I know a bit about uCurrent and I have parts to build it, but I'm not sure if I can just put MAX4239 with 10-20x gain inside feedback loop of OPA541. \$\endgroup\$
    – Kamil
    Jun 1, 2020 at 1:02
  • \$\begingroup\$ Low sense R means low noise....obviously. If the 'noise' you describe is in the external current path then it is not 'noise' Tha amplifier is already a low noise part, so noise from the amp should not be a concern. \$\endgroup\$ Jun 1, 2020 at 14:35
  • \$\begingroup\$ I'm more concerned about noise and offset of signal source that feeds OPA. If there is 1mV of noise from signal source - half resistance will double noise to signal ratio. My first prototype has no signal source and I'm using Analog Discovery 2 generator and it looks like there is more than 1mV RMS of noise. This is not a big deal if I'm generating few amps, but when I need pretty 50mA RMS sinewave - noise and offset are huge relatively to signal. Thats why I picked 100mOhm resistor (circuit operates on higher input voltages). \$\endgroup\$
    – Kamil
    Jun 1, 2020 at 22:51

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