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I'm working on a small project which involves current measurement.

I'm using R10 to measure Voltage drop (100 mV) when 10 mA current goes through it.

Next I'm amplifing that signal around 37 times using an op-amp (LM358N) which is powered using 0-5 VDC and works as differential amplifier. There should be 3.73 V on the output BUT...

My problem is:

The op-amp is completely ignoring the inputs and it still outputs a value around VCC minus 1.2 V.

Making a symmetrical power supply does not help. I checked everything about 100 times and can't find any solution... If anybody of you know how to help me I'd be grateful.

Pic.1 Schematic

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    \$\begingroup\$ LM358 doesn't have rail-to-rail inputs. Either give it more than 7 V as its positive supply when measuring the current of the 5 V supply, or do low-side sensing, or choose a different op-amp. \$\endgroup\$ – The Photon Mar 30 at 20:56
  • \$\begingroup\$ Tried suppying it to 9VDC but its output was 8V so it still ignores input. But thanks! I will need to change op-amp I think... \$\endgroup\$ – Michał Rybczyński Mar 30 at 21:07
  • \$\begingroup\$ The circuit seems to be performing correctly to me. You have two similar potential dividers on the IN+ and IN- but the IN- divider is fed by a voltage 0.1 V lower than the IN+ divider. The negative feedback forces the op-amp output to move to a point where IN- = IN+. With a gain of 37 this will happen when the output voltage is 37 x 0.1 V above the voltage at the bottom of the IN+ divider. Since the bottom of the IN+ divider is at 0 V then the op-amp output will drive to +3.7 V. Perfect behavior! \$\endgroup\$ – Transistor Mar 30 at 22:18
  • \$\begingroup\$ Sure but if you put VCC to 8V it outputs 6,3V That is not perfect behavior in my opinion. \$\endgroup\$ – Michał Rybczyński Mar 31 at 23:36