# Op-amp voltage divider

I have troubles understanding the way the circuit below is solved. I figured out that the solution is: $$V_+=\frac{R_2}{R_1+R_2}V_1 \\ V_a=\frac{R_3+R_4}{R_4}V_+$$ But what is actually the reason that R2 is not considered when calculating V_a? After all V_a=V_3+V_2 forms a mesh right? So the solution would be: $$V_a=\frac{R_3+R_4||R_2}{R_4||R_2}V_+$$ I know that is wrong, but I can't grasp why. As V+=V- isn't R4 in parallel to R2? Probably I am not understanding something basic. Can someone help me out?

simulate this circuit – Schematic created using CircuitLab

• If you substitute V+ in eq2 for V+ from eq1, you get Va as a function of V1. Try that. May 12 '16 at 19:27
• Yes, I know, I wrote it that way to make it clearer what I mean, because what is bugging me is the 2nd part of calculating Va.
– Daiz
May 12 '16 at 19:29
• It's not really in parallel- no current flows between the inputs. Va is driven by the op-amp to an unknown voltage such that V+ == V-. May 12 '16 at 19:43
• In short, Eq 3 is wrong because it's constructing parallel resistor arrangements that aren't there in the given circuit. Eq 1 and Eq 2 are correct. And when you make the substitution you get the correct results. In order to discuss some mesh with a V_2 and V_3, perhaps you could post a drawing of a mesh with a V_2 and V_3 on it. May 16 '16 at 17:01

V+ is a voltage divider of V1 (V+=V1/2) because the OpAmp draws (virtually) no current on its input pins. And no, R4 and R2 are not in parallel. Even if the voltage across them is the same (thanks to the OpAmp enforcing a virtual short), they are not sharing the same nodes!

Once you know V+, the rest is easy:

1. If you are used to these circuits, you may see the circuit as a non-inverting amplifier of V+.
2. You can compute the current through R4, notice that the current through R3 is the same and compute the output Va accordingly -rebuilding the non-inverting amplifier theory.