I'm trying to measure DC 25A across a .005 ohm shunt resistor. From my research I've found the TI INA169 which is a current shunt monitor ic. However the datasheet says it has a total output error of ~1%. At 25A, this would be .25A. I'm trying to measure to 0.01A accuracy.

With 25A through the resistor, that is 125mV. If I had a differential ADC like the ADS1115 (16-bit, 1 for sign and 15 for value), then if I had it measuring the voltage differential between the shunt would that not be a more accurate solution?

With the INA169, what is the real accuracy I could expect if I measured the output with the ADS1115 in a single ended configuration? Would it realistically be more than 1%?

  • \$\begingroup\$ One chip is an ADC, the other produces a current proportional to the input voltage differential. Two totally different applications, not sure why you try to compare them. \$\endgroup\$
    – PlasmaHH
    Jul 4, 2016 at 10:12
  • \$\begingroup\$ With the current sense amplifier chip, I would measure the output with an ADC. However with the differential ADC, I can just measure the voltage differential of the shunt resistor without the chip. I'm wondering if that is actually possible. \$\endgroup\$
    – Peter
    Jul 4, 2016 at 10:14
  • 2
    \$\begingroup\$ Instead of "throwing" a more accurate (is it really more accurate ??) component in your design, you need to think about the tolerances in the complete system. The INA169 is a completely different component than the ADs1115 so how can those be compared ? You should learn about tolerances and accuracies first before you can have a meaningful discussion about these. You can build a better than 1% setup with the INA169 if you calibrate it. \$\endgroup\$ Jul 4, 2016 at 10:17
  • \$\begingroup\$ 0.01A accuracy over 25A is 0.04%. So, to ensure this, the first thing that comes to my mind is the tolerance of the shunt resistor. Did you actually find a 5mOhm resistor with 0.04% tolerance? I'm not sure what you want to achieve is realistic. \$\endgroup\$
    – dim
    Jul 4, 2016 at 10:18
  • 1
    \$\begingroup\$ You do not need overall 0.04% intrinsic tolerance/accuracy as long as you can calibrate. Also component drift can be calibrated out but when using components with a high drift (they change properties over time) you will have to calibrate more often. \$\endgroup\$ Jul 4, 2016 at 10:20

1 Answer 1


The advantage to the INA169 is that it can measure current on the high side of the voltage, 30 or 40 volts above your logic level supply. You could make a more accurate measurement with the ADC directly, but you must have a voltage that is both proportional to the current and in the 5-volt range of the ADC inputs, so you will need additional circuitry to perform this task. It is a challenge to build this type of circuit, since all of the components will have to be at the precision you are looking for. Remember that the shunt resistor you have chosen will dissipate 3.125 W so I agree with @dim that the sense resistor will probably not remain in this accuracy range as it heats up, and even board traces will have an effect on the measurement at this low resistance.

  • 1
    \$\begingroup\$ Also, the INA169 has bandwidth up to 100's of kHz. The ADC runs at 860 Hz in 16-bit mode or 3300 Hz in 12-bit mode. By the way, I came up with 3.125W. 25*25*0.005 = 3.125, no? \$\endgroup\$
    – mkeith
    Jul 4, 2016 at 16:39
  • \$\begingroup\$ @mkeith you are correct! 3.125W \$\endgroup\$ Jul 5, 2016 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.