Voltage at non inverting pin of op amp

alison_941

Joined Dec 21, 2022
18
Could anyone here confirm what will be the voltage at pin 3 of op amp for below schematic in terms of mathematical equation. I want to proof read my equation if its correct or not. my final equation is 0.6V+{(R7/R5)*I1*R4} ignoring offset voltage of opamp.

MrChips

Joined Oct 2, 2009
28,473
If the opamp is working correctly with negative feedback applied, the voltage on the non-inverting input will be the same as the voltage on the inverting input, otherwise the output of the opamp will reach one of the supply rails.

crutschow

Joined Mar 14, 2008
32,026
With negative feedback, the high open-loop gain of the op amp will always try to adjust the output voltage so as to make the voltage difference between the two inputs essentially 0V.

Alec_t

Joined Sep 17, 2013
13,424
Neither input is ground-referenced, so I don't see how the voltage at either input can be calculated.

crutschow

Joined Mar 14, 2008
32,026
Neither input is ground-referenced, so I don't see how the voltage at either input can be calculated.
The input reference is the 0.6V source connected to R5.

WBahn

Joined Mar 31, 2012
28,467
Could anyone here confirm what will be the voltage at pin 3 of op amp for below schematic in terms of mathematical equation. I want to proof read my equation if its correct or not. my final equation is 0.6V+{(R7/R5)*I1*R4} ignoring offset voltage of opamp.
View attachment 289212
I'm assuming that Voffset1 is modeling the offset voltage of the opamp (that you are ignoring)?

Your result indicates that Vx (what I'm calling the voltage at the opamp inputs) depends on the value of R7 but does not depend on the value of R8.

Does that make sense?

Isn't the voltage you are looking for (Vx), taking Voffset1 to be 0 V, just going to be the 0.6 V plus the voltage drop across R5 (taking the current to be flowing left-to-right in R5)?

If so, then does it make sense for a larger value of R5 to result in a smaller difference between the 0.6 V input and Vx?

Draw an equivalent DC circuit with I1 and R4 converted to the Thevenin equivalent. I think you will see what needs to be done from there.