Current measurement circuit

connect that pin of to uP to op-amp whose current you want to measure ,,

connect it to the non-inverting terminal as shown in the diagram

sorry I meant inverting terminal ,,
just make sure that the current you want to measure flows throw op-amp ,,via inverting terminal ...as in your diagram ,,butt in opposite direction

please exuse my ignorence. Look at this diagramm, I think this the way to measure the current flowing through a resistor which s supplied by 3V. Is this true?

Now I want to achieve 2 things: connect the power-supply-pin of the uP with the voltage source, and then measure its current. In the diagramm shown here, the resistor has 2 pins, one goes to power supply and one to -ive input of the Amp... so how about the uP? which pin goes to the Amp

sorry I think the image has not been attached

heyy ,,first let me be clear ,,you want to measure current ...ryt ??

then you dont need a resistr at input side ,, just a feedback resistr is enough .

and I dont know about single power supply op amp ,,
SI used dual power supply op-amp
with +5 and -5V powr sply

Yes rumman, I want to measure the current and the diagramm above shows the circuit, used to measure the current through 'Last' resistor. Yes we don't need it, and it does not belong the measurement circuit. its gonna be replaced by the uP.

Now which pin exactly do I have to connect to the -tive pin if the amp?

and I dont know about single power supply op amp ,,
SI used dual power supply op-amp
with +5 and -5V powr sply

rumman89

Ok forget about the op amp first, the thing I don't understand is which pin of the uP do I have to connect to the IN- of the amp!

Thank you for your patience :-)

see its like this,, you need to connect that pin from which current to be measured should flow throw the op-amp ,,as the op-amp have infinite input res ideally all this currenl wil flow throw the feedback path ..

Sorry Rumman, I did not really figured out what do you mean, it would be really awesome if you can post a diagram and mention exactly how does your circuit looks like... I was talking to an anlog designer and he told me this never would be possible!