AD8219 output

Hello. We tried to calculate the output voltage and simulate it to determine the expected output. Given the configuration, we can say that the application is a current sense amplifier wherein there is a small voltage drop expected due to the current flowing to the shunt resistor, may we clarify this? Also, the voltage drop would then be converted and amplified by the current sense amplifier into a small output voltage.

Based on our calculation, there would be a voltage drop of 0.3939 mV, so V2 would be 6.4996V. These values were also acquired through simulation. Regarding the output, the calculated output voltage is approximately 24 mV, and the simulated output is about 29 mV (There may have been some additional offsets that we didn’t initially consider on the computation).

For reference, this is the simulated circuit showing the V2 and Vout since V1 is 6.5 since it is direct from the source.

Can  you measure the voltages at V1 and V2 shown on the image below because if you acquired output voltage of 0V, maybe because there is no voltage drop or voltage difference across the Rshunt?

Hi mArms,

Thanks for your replay, I removed the 100Ohms resistors and connected the shunt resistor directly to the IN+ and IN- pins, and now it works properly.

Dor.