# High-side current sensing with a rail-bound instrumentation amplifier?

I know that current shunt monitor ICs and high common-mode voltage differential amplifiers like the INA117 and AD629 exist but I'd like to know if it's possible to use a regular instrumentation amplifier where its input range is bounded by its power supply (e.g. AD623, INA177) and pre-attenuate the common-mode voltage using discrete or semidiscrete components.

I thought that using a voltage divider on the input and then compensating the gain factor proportionally would work if the in-amp has a high enough CMRR for the desired input range. I vaguely suspect that it's not that simple. If so, are there other (better) options to do this? I read something about using a current mirror but I don't really understand how that applies in this case.

• If you pre-attenuate the inputs, the precision of the resistors of the voltage dividers is very important. Are you aware of this ? Commented Dec 27, 2017 at 19:56
• You can buy .1% tolerance resistors cheap now days, or .025% tolerance for $20 USD to$75 USD each (going to extremes). Then use 100 ohm 25 turn trim pots to null out errors. At this point your current sense resistor becomes the biggest issue. Personally I would choose the higher quality Analog Devices parts over others.
– user105652
Commented Dec 27, 2017 at 20:41

A better way is to use a rail-to-rail input precision op-amp and a transistor.

simulate this circuit – Schematic created using CircuitLab

Do an error budget calculation on your proposed homemade differential amplifier and I think you will find that 'Oh, that way madness lies'. The circuit above requires no matched parts. Voltage across R1 is 10mV/A and across R3 it is 1V/A. A MOSFET or Darlington can be used instead of Q1 for a bit better ideal accuracy. The main limits on accuracy are the tolerance of the resistors and the offset voltage of the op-amp. A 1% error in any resistor results in a 1% error in the output. A 10uV offset voltage represents a 1mA error in the output.

Suppose we divide down the 12V to 6V on each input. We still have the same sensitivity to R1, but a 1% error in one of the four resistors will cause an error of 30mV at the input to the in-amp. That's equivalent to a 60mV error at R1, which is an error of 6A!

Yes, you could trim the resistors to get rid of most of the error for a while at one temperature but they would quickly drift out of spec. Even \$25 0.01% resistors would not yield as good and stable a result as the 0.1 cent 1% resistors. That.. is engineering.

• I didn't mean a homemade differential amplifier, rather I just meant using the in-amps gain directly after protecting it from common-mode voltage. Maybe I didn't phrase the question correctly. Commented Dec 27, 2017 at 22:41
• Same thing. You are dividing it down. Commented Dec 27, 2017 at 22:48
• Okay. I'm having some trouble understanding your answer. In the first paragraph, you're suggesting that the precision op-amp and transistor is a better method than the dividng down. Then in the last two paragraphs, you're elaborating on how dividng down is nonsense. Then, I don't quite understand how your circuit works but as far as error, it only depends on the Vos of OA1 and the resisters. Is this pretty much it and I should just study the circuit now? :) Commented Dec 27, 2017 at 23:04
• @Trevor OPA192 is a good one. 36V and 5uV Vos. Commented Dec 28, 2017 at 4:12
• Another trick is to create a supply relative to the high voltage supply (by a regulator, Zener+resistor, DC-DC converter etc.) which alllows the use of a cheaper 5V op-amp. Only the transistor sees the high voltage. Commented Dec 28, 2017 at 14:58

Yes of course you can.

That is basically all those fancy chips are.... They have simply packed up the high precision resistors inside the package for you to make it a more integrated and more reliable solution.

But, you can still do it with the discrete parts provided you can chose parts that meet your accuracy requirements. The higher "high-side" is the more troublesome that becomes though.

BTW: There are various other methods to measure current. Current mirrors though are not typically a good choice since they basically double your load current and as such are better used to measure small currents.