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I am measuring the current from a battery (12v) using a shunt resistor. But if i can i want to use the normal rail as supply for the op-amp:

differential current measurement circuit

Now my thinking is that because of the two voltage-dividers (r311/r314 and r312/r315), my non-inverting input will be closer to 1.8v than to 0, and my inverting input will 'start off' at 0v and then be pulled to ~1.8v by the op-amp.

Is this possible, or am i forgetting something? I didn't find any reading on this anywhere.

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  • \$\begingroup\$ If you are unsure get yourself a free sim tool like LTSpice. \$\endgroup\$ – Andy aka Jun 30 '17 at 14:33
  • \$\begingroup\$ I did that, and it seems to work, but i did have my gain-setting back to front. so it seems to be that pulling the lower voltage level up, goes at the cost of the gain \$\endgroup\$ – David van rijn Jun 30 '17 at 15:11

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