use an opamp which goes up to the positive rail with its output. short r3 and r6 and take r5 out. set r2=r7 and r4=r8. Ao=r8/r7=r4/r2 Vo= i1 * r1 * Ao you can make the opamp to output current, but i thought you want to measure current and output voltage... ge At 21:25 01/07/99 +0000, Peter L. Peres wrote: >Hello, > > I have a small question. It might be stupid, but... > > I need to measure a DC current using a probe resistor in a circuit. The >resistor is in the +-branch and at the highest voltage available in the >system. This current needs to be transformed into a ground-referenced >voltage for ADC purposes. I'll try an ASCII schematic: R1 U1, I1 >-----*---/\/\/\---*-----> U1 - R1*I1, I1 | | R2 / | \ | / | \ | | | |\ *------------ ------------------+ \ | | R7 | >----*----O Vo | *---/\/\/\----*----- / | | | |/ | / | R8 | R4 \ *---/\/\/\---+ / ^ \ A | === /// GND > Assuming R2 == R4 and the differential amplifier gain is Ao = R8/R7, the >expression for the output voltage assumes the form: > > >Vo = Ao*U1*I1*R1*R5/(R3 + R5) > > >this is very nice, but I'd like to get rid of U1, for obvious reasons. U1 >is not fixed, obviously. The most direct way is, of course, to measure U1 >at the same time, and divide Vo by something proportional to U1 (in >software). (U1 IS measured at the same time by other parts of the >circuit). > >Question: Is there a cunning way to avoid this, using *simple* analog >circuits to inject a current proportional to U1 into the '-' input of the >op-amp at point A ? I know how to do this with 2 current mirrors and a R. >Any other ideas ? An op-amp voltage controlled current source referenced >to U1 ? Pros and cons ? > >Note that the expressions come off a bit of paper, and that it' 9PM here. >There may be errors. > >tia, > > Peter