High side current monitor (INA193) solution?

I need to find a solution to have a high side current monitor for high common mode voltage inputs and at high temperatures (stable up to 125C).

Currently, I have tried using the INA193 shunt monitor IC with a 7.5 ohm shunt and common mode voltage that can reach up to 58VDC (which the IC claims to be able to handle). The supply is 3.6V and GND. Load current varies between 10mA and 30mA.

At ambient, the circuit works fine, but if I put any heat applied to the chip the gain appears to drift downward. It is specified to be 20V/V but seems to drift down to almost 10V/V at temperature. Letting it cool restores the gain to workability.

Is there a solution to this issue or another high side current monitor solution I could use better? Thank you.

The datasheet suggests the error should be much smaller than you describe over the full temperature range. Can you post your detailed measurements of Vout vs Is at different temperatures? How are you measuring Is and Vout during the test?

The biggest temperature error I can see in the TI data sheet is that the output current changes a lot with temperature. In their circuit it is very small, 10s of micro amps? You have got plenty of headroom? i.e. Vo/Vcc is < .8.
Frank

Hi,

I have used the INA193 many times.
I never recognised a drift this much.
So either you have a defective INA, or the drift is produced else where.

As an example let's use 20mA.
20mA x 7.5 ohms = 150mV.

150mV x 20 = 3000mV = 3V

From the datasheet I see that 150mV is on the top limit of Vsense. So measuring 30mA of load current with a 7.5 Ohm shunt is out of range.

--> use a shunt of about 4 Ohms, this gives you a little headroom.

Try and see if it works with this setup.

Klaus

If it goes into an ADC as a differential voltage/a voltage across a resistor, maybe it's okay to dispense with the monitor and feed the voltage across the shunt directly into the converter hi and lo pins. 0.03A (max?) x 6 ohm resistor = 180mV, that would only be 5mW across the resistor. A 60 ohm resistor would give 1.8V for 30mA, and be about 54mW.