Fixing Incorrect KCL Equations in Op-Amp Circuits – Circuit Analysis Tips

circuit analysiscurrentkirchhoffs-lawsmathoperational-amplifier

I have the following op-amp circuit:

schematic

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.

Best Answer

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?