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I came across the following self-study problem, taken from The Analysis and Design of Linear Circuits (8th edition). enter image description here

I was not intending to solve the problem, since I am just looking over all problems in the book. Something caught my attention, however. I was looking at node c, the one labeled with \$v_c\$ in the Figure. If I were to apply Kirchhoff's Current Law to then perform node-voltage analysis, I would not know what currents to write down. I know the current flowing into the positive input of the second OpAmp is zero, so that means there is either current going into/coming out of the output of the first OpAmp, and current through the 150K resistor; or else, no current goes through the 150K or out of/into the first OpAmp output. Then, what is the point of cascading two OpAmps?

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    \$\begingroup\$ As far as \$v_b\ne v_c\$, there will be a current flowing through 150k resistor and hence to/from op-amp 1 output. Think. \$\endgroup\$
    – nidhin
    Commented Oct 18, 2018 at 15:45
  • \$\begingroup\$ Thank you. I actually wrote the equations down, solved them, and got results that I later verified with LT Spice. My intuition just failed me, I guess. \$\endgroup\$
    – Bee
    Commented Oct 18, 2018 at 15:59

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Yes, current can flow into and/or out off the output of an op-amp.

However, an op-amp provides a voltage output. It is the circuit that surrounds the op-amp that dictates what current will flow into/out of the op-amp's output.

To apply hand analysis to your circuit you would assume:

  • \$V_b = 0\$ (due to op-amp action)
  • \$V_e = V_c\$ (due to op-amp action)

Apply KCL at the \$V_b\$ and \$V_o\$ nodes.

You now have 2 equations with 2 unknowns (\$V_o\$ and \$V_c\$).

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  • \$\begingroup\$ Thank you. I did that and got to a plausible solution. I would suggest that you don't need to apply KCL at \$V_0\$, since you get one equation from applying KCL at \$V_b\$, and the other from using the voltage divider equation to get a relation between \$V_0\$ and \$V_e\$. \$\endgroup\$
    – Bee
    Commented Oct 18, 2018 at 16:12
  • \$\begingroup\$ @Bee Absolutely use whatever is most comfortable to you. You could also solve \$V_c\$ as a function of \$V_o\$ (non-inverting amplifier) and apply KCL once at \$V_b\$. \$\endgroup\$
    – sstobbe
    Commented Oct 18, 2018 at 16:22

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