Hi,
As per the output voltage formula in the data sheet , the output voltage should be zero volts when Rs=open & Is =0 Amps, for the same scenario when the model is simulated in TINA TI, output is saturated to supply voltage of the IC. Please find the below attached simulated file. Could you please help me to know why its not matching with theoretical value.
Regards,
Kishore
Hello Kishore,
Thank you for your question! I think I need to clarify the operation of the INA193. It is designed to amplify a small differential voltage across its input pins, usually tens of mV generated by load current through a small sense resistor.
With your circuit as drawn, you have actually applied a 12V differential input to the device, so the output would be 12V * 20V/V = 240V. Of course, the amplifier cannot output 240V with a 5V supply and saturates somewhere below 5V.
I've made a few modifications to your circuit to better meet our general recommendations, shown below on the right (original on the left for comparison). The input resistors have been reduced to 10 ohms, and they can be removed unless you plan to add a differential input capacitor for filtering. A shunt resistor Rs = 10 mohms has been added as well as a load generator IG2 to model changing load current.
In this case you can see that the differential input voltage is small, caused by the input offset voltage of the INA193 as well as its input bias current flowing through the input resistor network. The output ideally would be 90uV * 20V/V = 1.8mV, however the swing to GND limit of the device dominates so Vout = 300mV.
You can modify the circuit on the right to meet your particular design requirements. The TINA schematic is attached below.
Best regards,
Ian Williams
Applications Manager
Current Sensing
