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I'm having trouble figuring out how to set up the KCL equations for this op amp. Everything I've tried seems to lead to a dead end. I've tried setting up three KCL equations at nodes v_n(-), v_p(+), and V_out, and using just node -v_d and v_out. Anyone have any ideas?

This is what I've tried using \$-v_d\$ as a node: \$\frac{-v_d - v_2}{R} + \frac{-v_d - v_{out}}{R}+\frac{-v_d}{R_i}=0\$ and \$\frac{v_d-v_1}{R}+\frac{v_d}{R}+\frac{v_d}{R_i} = 0\$ and \$\frac{V_{out}+v_d}{R}+\frac{V_{out}-Av_d}{R_o} = 0\$ but this seems to be trivial as the first and second equation reduce to something that doesn't involve other parts of the circuit.

  • \$\begingroup\$ Show us your equations, and we'll point out where you've gone wrong. \$\endgroup\$ Dec 21, 2016 at 13:43
  • \$\begingroup\$ @ScottSeidman Edited to include what I've done. \$\endgroup\$
    – Oliver G
    Dec 21, 2016 at 13:56
  • \$\begingroup\$ \$v_d\$ is not ground referenced. This makes things tough. I suggest representing this without that variable. \$v_d\$ is then simply the current through \$R_i\$ times \$R_i\$ \$\endgroup\$ Dec 21, 2016 at 14:01
  • \$\begingroup\$ .. then \$v_-\$ is just \$v_+\$ plus the voltage across \$R_i\$ \$\endgroup\$ Dec 21, 2016 at 14:03
  • \$\begingroup\$ @ScottSeidman So you're saying: \$\frac{v_{-} - v_2}{R} + \frac{v_{-} - v_{out}}{R}+\frac{v_{-}-v_{+}}{R_i}=0\$ and \$\frac{v_{+}-v_1}{R}+\frac{v_{+}}{R}+\frac{v_{+}-v_{-}}{R_i} = 0\$ and \$\frac{V_{out}-v_{-}}{R}+\frac{v_{out}-A(v_{+}-v_{-})}{R_o} = 0\$, where \$v_d = v_{+} - v_{-}\$? \$\endgroup\$
    – Oliver G
    Dec 21, 2016 at 14:16

1 Answer 1


There are only three nodes where the voltages are not already determined. So, three equations are needed.

Non-inverting input: $$ \frac{\text{Vp}-\text{V1}}{R}+\frac{\text{Vp}}{R}+\frac{\text{Vp}-\text{Vn}}{\text{Ri}}=0 $$ Inverting input: $$ \frac{\text{Vn}-\text{V2}}{R}+\frac{\text{Vn}-\text{Vout}}{R}+\frac{\text{Vn}-\text{Vp}}{\text{Ri}}=0 $$ And the output: $$ \frac{\text{Vout}-A (\text{Vp}-\text{Vn})}{\text{Ro}}+\frac{\text{Vout}-\text{Vn}}{R}+\frac{\text{Vout}}{\text{RL}}=0 $$ Using a computer algebra program to solve this system yields $$ \text{Vout} = \frac{\text{RL} (2 A R \text{Ri} (\text{V1}-\text{V2})+\text{Ro} (R (\text{V1}+\text{V2})+2 \text{Ri} \text{V2}))}{R (2 \text{Ri} ((A+2) \text{RL}+2 \text{Ro})+3 \text{RL} \text{Ro})+4 R^2 (\text{RL}+\text{Ro})+2 \text{Ri} \text{RL} \text{Ro}} $$

This reduces to Vout = $V1-V2$ for \$Ro = 0\$ and \$Ri = \infty\$


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