Hi... I was here trying to solve an OpAmp problem, but I'm stuck. I can't find a way of writing an equation for v_R1 in function of va and vb, which I know and confirmed to be equal to v1 (measured with the voltmeter). I found v_R1 = (R1*vo)/(R1 + R2) and that vo = ((R1 + R2)*v_R1)/R1 By the OpAmp thumbs-up rules, I know that the OpAmp will try everything it cans to keep v+ and v- at the same potential, so v1 = v_R1. Now, I need to get rid of v1 from the vo equation, but I can't find the way! Any help would be appreciated!
Oh, my mistake... But now, if I try to turn it upside down on LTSpice and try to run the simulation again, LTSpice keeps simulaitong for ever, and if I cancel the simulation it returns this error: Singular matrix:check nodes u1:11 and v1 Iteraction Nº. 1 I don't know what this is!
Can you write v1 in terms of va and vb? Can you write v_R1 in terms of vo? Now set v1 equal to v_R1 and, viola, both v1 and v_R1 are gone.
I could only write v_R1 in terms of vo... I was asking help for the other one. v1 in terms of va and vb!
Forget about the opamp part of the circuit completely. The opamp has no effect on v1. You have two voltage sources, va and vb, connected by two resistors. What is the voltage at the junction of the two resistors? How much current is flowing in the two resistors? What is the voltage drop across one of the resistors? What is the voltage at the center node give the voltage drop across one of the resistors and the voltage on the other side of the resistor?
It will be v_R1, because the OpAmp will always try to set the same voltage in both it's inputs so that the voltage drop between them is approximately 0 V. v_R1 = vo *R1(R1 + R2). Here I'm not sure if I can simply divide va/Ra and vb/Rb to find those 2 currents. By inspection I can tell the ia = -ib, because at v1 cannot be any current flowing, so ia and ib must cancel each other so that no current gets out of that junction into the non-inverting input of OpAmp. I don't know how to answer this with a formula. I can only say that as these 2 branches are in a parallel setup, the voltage drop of each branc must be equal! I'm not sure but I think I didn't understood the question!
I tried this: v_R1 = R1*vo/(R1+R2) Then for the non-inverting input I wrote: va - vb + Rb*I1 + Ra*I1 = 0 I1 = (vb - va)/(Ra + Rb) v1 = va + Ra*I1 or v1 = vb - Rb*I1 replacing I1 in the above equation by (vb - va)/(Ra + Rb), I get: But is this the easiest way? Jesus, I struggled a lot!
It was a big struggle for me, not for the math, but for the time I took to understand how to take v1 in terms of va and vb... And of course that I need to come up with some mnemonic to shorten the time I take in the analysis itself! I can't take this long on an exam! I can yet, give it another look:
But now, my question is: Could I write this equation only from knowing the basic non-inverting amplifier gain equation, which is:
Well, yes. All you have to do is to solve for V1 in terms of Va and Vb. V1 = Va - I1*Ra and I1 : I1 = (Va - Vb)/(Ra + Rb) So we have V1 = Va - (Va - Vb)/(Ra + Rb) *Ra = Va - (Va - Vb)*Ra/(Ra + Rb) Vo = (1 + R2/R1) * [Va - (Va - Vb)*Ra/(Ra + Rb)] And notice that my solution is slightly different than yours. Or we can use a superposition and solve for Vo directly.
Oh yes, but that is roughly what I did, right? You just took out the minus sign inside the vb-va parenthesis and therefore it comes -(va-vb). I'm asking this because I always hear our teacher just throwing out the vo expressions with just quick analysis by inspection and I have the idea that I'm really missing something here!