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Iam trying to do nodal analysis of this op-amp circuit, but iam not quite sure if I have done it correctly or if I have missed something. The op-amps are considered being ideal so there is no current flowing in or out. I hope the picture is clear enough. enter image description here

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    \$\begingroup\$ If they are ideal opamps, then their inputs do not accept or inject current. Without current, R3 and R6 cannot drop a voltage. So the output of their respective opamps must appear at their input. So, I'm not at all sure where you got your "0" values at the opamp inputs. Must be an error in your perspectives (or a unique value for Vin.) \$\endgroup\$ – jonk Nov 8 '20 at 21:15
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Your node A equation is incorrect because the voltage across R2 is \$V_A\$, not \$V_B\$. (It will turn out in this circuit that these voltages are the same, but for the purpose of doing nodal analysis, you can't assume this a priori)

Your node B equation is incorrect because you didn't include current coming from the op-amp output pin.

Your equation set is incorrect because you didn't include an equation for the output node.

The fact that it's difficult to write the equation for node B without knowing the internals of the op-amp, is why for hand calculation we don't usually use the nodal analysis. Instead we use equations derived from the knowledge that in negative feedback the two input terminals of the op-amp will be driven to the same voltage.

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  • \$\begingroup\$ Alright, because Iam thinking that the equation for node B is maybe not necessary because of the negative feedback of the op-amp. \$\endgroup\$ – Håkan Mjölk Nov 8 '20 at 18:16
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Assuming ideal op-amp for analysing.

At Node A: $$\frac{(V_A-V_{in})}{R_1}+\frac{V_A}{R_2}=0$$

\$V_A\$ is the voltage at non-inverting input. Voltage at inverting and non-inverting inputs should be equal for negative feedback. Therefore:

At Node B: $$\frac{(V_B-V_{A})}{R_3}+\frac{(V_B -V_C)}{R_4}-i_0=0$$

\$i_0\$ is assumed to be the current flowing out of op-amp output to node B.

Current flowing into inverting input should be zero, means \$\frac{(V_B-V_{A})}{R_3}=0\$, so \$V_B=V_A\$.

At Node C: $$\frac{(V_C-V_{B})}{R_4}+\frac{V_C}{R_5}=0$$

and finally \$V_{out} = V_C\$ just like we saw in the case of \$V_B = V_A\$.

If \$V_{in}\$ and all resistor values are known then every equations can be solved.

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    \$\begingroup\$ On Node C, do you mean VC/R5 ? \$\endgroup\$ – Håkan Mjölk Nov 8 '20 at 18:23
  • \$\begingroup\$ Oh yes. Thanks for correcting! \$\endgroup\$ – Meenie Leis Nov 8 '20 at 18:25
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Following Meenie's explanation, since ideal opamp has no input current in either inverting/non-inverting inputs the current throughout \$R_3\$ and \$R_6\$ can be considered as zero so \$V_b=V_a\$ and \$V_{out}=V_c\$, then reaching to $$V_{out}=V_{in}\frac{R_2}{R_1+R_2}\frac{R_5}{R_4+R_5}$$ \$R_3\$ and \$R_6\$ are not necessary in the ideal case. You always can simulate the circuit with this fantastic tool provided in here, I made it for you so you can confirm by simulating the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

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