Getting the wrong KCL equations in op-amp circuit

Question

I have the following op-amp circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When I use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{K}_1:\text{I}_4=\text{I}_1+\text{I}_7\\ \\ \text{K}_2:\text{I}_7=\text{I}_2+\text{I}_3\\ \\ \text{K}_3:0=\text{I}_1+\text{I}_2+\text{I}_8\\ \\ \text{K}_4:\text{I}_9=\text{I}_5+\text{I}_{10}\\ \\ \text{K}_5:0=\text{I}_3+\text{I}_6+\text{I}_{10}\\ \\ \text{K}_6:\text{I}_5=\text{I}_4+\text{I}_6+\text{I}_8+\text{I}_9 \end{cases}\tag1 $$

In the system of equations, \$\text{K}_6\$ gives the KCL of the incoming and outgoing current from the sources to the ground.

Question: there is a mistake in my KCL equations, but where is it? Because I get the wrong answers.

Answer 1

Jan, I won't bother with much (because you can pound out equations as well as the next one) except to say that I wrote the following:

var('r1 r2 r3 r4 r5 r6 io1 io2 vm1 vp2 vm2 vo2 vn i4')
e1 = Eq( 0/r1 + 0/r2 + 0/r3, vp2/r1 + vp2/r2 + vo2/r3 + i4 )   # opa1 (-) input is virtual ground
e2 = Eq( vp2/r1 + vp2/r2, io1 )                                # opa1 output is op2 (+) input
e3 = Eq( vm2/r4 + vm2/r5, vo2/r5 )                             # opa2 (-) nodal
e4 = Eq( vo2/r3 + vo2/r5 + vo2/r6, io2 + vm2/r5 + vn/r6 )      # opa2 output nodal
e5 = Eq( vm2, vp2 )                                            # opa2 (-) = opa2 (+)
ans = solve( [e1, e2, e3, e4, e5], [vm2, vp2, vo2, io1, io2] )

So, for example, I find the impedance seen by \$V_n\$ as:

$$R=\frac{V_n}{I_4}+\frac{R_4+R_5}{\left[\frac{R_4}{R_1\mid\mid R_2}+\frac{R_4+R_5}{R_3}\right]}$$

KCL just works so far as I can tell. I guess I'm not sure what the question might be?

Answer 2

K6 gives the KCL of the incoming and outgoing current from the sources to the ground.

Output currents in the op. amps. should not be assumed as coming from the ground. In fact, the power rails are not even considered in the circuit (with ideal op. amps.):

Summary: don't use KCL at op. amp. outputs. Instead, use voltages. The equations should be created considering only the voltages at \$K_3\$ and \$K_4\$. Relying on \$I_8\$ and \$I_9\$ is a problem since they depend on circuits which are not part of the schematic you are analyzing, with ideal op. amps.

Now you can built equations, using Ohm's law, for \$i_1\$, \$i_2\$ and \$i_5\$. E.g.:

$$i_5 = \frac{V_4 - V_3}{R_5}$$

$$i_5 = \frac{V_3}{R_4}$$

[...]

and so on.

Another issue: the circuit forces the two inputs of the first op. amp. to different voltages. This breaks the ideal op. amp. model. Maybe there is a resistor missing or misplaced in the schematic?

Edit: disregard the last paragraph. The new schematic you posted, with an ideal current source, solves the problem as it only forces the total current to the three input resistors. Now you can consider \$V_1 = 0\$ (@ \$K_1\$) for your new equations.