/diy/ - Do It Yourself

Anonymous Fri Sep 6 03:03:00 2019 No.1677988 [View]
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I'm testing this simple current to voltage converter for measuring the current through a load (the potentiometer in the diagram). V_out is measured with reference to the positive input of the op-amp. I_out is then computed by dividing V_out with the known resistor value in the feedback loop. This doesn't seem to work well when the load drops below 1K. I get a pretty large error compared to measuring the current with a multimeter. Any ideas? The op-amp is an AD824.

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/diy/ - Do It Yourself

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