I want to measure very small resistances (0.5 to 1 Ohm) using a high side current sense amplifier. The device I have in mind is MAX9611. The typical application circuit from datasheet is like:

enter image description here

My idea is to replace the Rsense with the unknown resistor which should be measured. Then, I would connect the Rsense to a precision 1A current source. Is this idea going to work? I do not know how the current source would work, so I am not sure if it shows me the voltage it is using to force the 1A through that Rsense resistor or not (I will propably use a bench current source) to simply use the R=V/I formula.

One more thing is, I do not know if I would need the LOAD in the current path or not (effectively shorting the current source to ground through my small Rsense unknown resitor).

I have really no idea if this idea is going to work or not. I would appriciate if you can shed some light over this idea!

  • \$\begingroup\$ What is your big picture goal? Instruments exist for this, and you can even use a regular volt meter instead of the current sense amplifier. The basic idea is OK, but will be limited by the precision of the current sense amplifier. The precision could possibly be improved by using a square wave current source. Then measure the square wave voltage amplitude. This would or could sort of automatically cancel out voltage offset. \$\endgroup\$
    – user57037
    May 15, 2016 at 18:10
  • \$\begingroup\$ Why not just run 1 A to ground through the resistor under test and measure the voltage across it? \$ R = \frac {V}{I} = \frac {V}{1} = V~\Omega \$. Too easy? \$\endgroup\$
    – Transistor
    May 15, 2016 at 22:27
  • \$\begingroup\$ @transistor I am not sure if that chip measures the current or votage...I mean wouldn't it be a equation with 2 unknowns? \$\endgroup\$
    – Dumbo
    May 16, 2016 at 0:40
  • \$\begingroup\$ @mkeith I want to build my own low resistance measurement \$\endgroup\$
    – Dumbo
    May 16, 2016 at 0:41

1 Answer 1


Yes, this will probably work. High side current sense amplifiers are just differential voltage amplifiers, usually with fixed gain, or gain that is select-able over several discrete options. After glancing at the datasheet, it seems this one has three gain settings.

The current sense amplifier does not "force" current through anything. It just measures the voltage drop on the shunt resistor, rsense. Where the datasheet shows a load, you would connect your precision current sink.

This part does have an output, and in the datasheet, they show how to use the output to create an inrush current limiter. The set pin and voltage divider would be utilized for this.

The basic technique for low Ohm measurement is to force a known current through the part in question, and measure the voltage developed. I usually just use a lab supply and a volt meter. If you want to be really accurate, you can measure the current as well as voltage (if you have two meters). But I usually just assume the lab supply display is accurate enough.

Final note is that the offset voltage specification of the current sense amp is not that great (considering the small full-scale input voltage). I recommend that you pulse your current sink on and off. This will cause the current to turn on and off. On the microprocessor side, you will get one voltage reading from the current sense amplifier when the current sink is on. Call that Vhigh. Then you will get another reading when the current sink is off. Call that Vlow. The difference is the true voltage you are trying to read. This will basically calibrate out the offset error.

  • \$\begingroup\$ Thanks for the details, so you mean this chip measures voltage? I was thinking it measures current! So if I use a precision 1A current then through the I2C I can read the voltage drop of Rsense then simple R = the voltage I read? \$\endgroup\$
    – Dumbo
    May 16, 2016 at 15:54
  • \$\begingroup\$ @Sean87, yes, the current flows through the shunt, not the amplifier. \$\endgroup\$
    – user57037
    May 16, 2016 at 21:02
  • \$\begingroup\$ Can you please have a look here electronics.stackexchange.com/questions/248424/… \$\endgroup\$
    – Dumbo
    Jul 28, 2016 at 11:05

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