I'm trying to derive the expression for \$ A_X = \frac{V_X}{V_{IN}}\$ but I'm getting two different ones relatively to the \$KCL\$ I use to calculate it.

enter image description here

With \$ KCL \; A)\; -\frac{V_{IN}}{R_S} = \frac{V_{IN}-V_O}{R_F}\$
therefore with \$V_O = \left(1+\frac{R_F}{R_S} \right)V_{IN}\$ , if I use:

  • \$KCL \; X)\; \frac{V_{IN}-V_X}{R_3}+sC\left(V_O-V_X\right) = \frac{V_X}{R_4}\$ ,
    then I get \$ R_4V_{IN}-R_4V_X + sCR_3 R_4 \left(1+\frac{R_F}{R_S} \right)V_{IN} - sCR_3 R_4V_X=R_3 V_X\$ ,
    therefore \$ A_X = \frac{V_X}{V_{IN}} = \frac{R_SR_4+sCR_3R_4\left( R_S+R_F\right)}{R_S\left(R_3+R_4 \right)+sCR_3R_4R_S} \$ .
  • \$ KCL\;O)\; \frac{V_{IN}-V_O}{R_F}=sC\left(V_O-V_X\right)\$ ,
    then I get \$ V_{IN}-\left(1+\frac{R_F}{R_S} \right)V_{IN} = sCR_F\left(1+\frac{R_F}{R_S} \right)V_{IN}-sCR_FV_X\$ ,
    therefore \$ A_X = \frac{V_X}{V_{IN}} = \frac{1+sC\left(R_S+R_F\right)}{sCR_S}\$ .

I was thinking that maybe not considering the current that goes into the output of the op-amp when applying Kirchhoff's current law at the \$ O\$ node is an error thus resulting in the second expression being wrong, but I'm not sure.

  • \$\begingroup\$ Applying KCL at O isn't productive because you don't know either the output current of the opamp or the load current. \$\endgroup\$
    – Ben Voigt
    Jun 12, 2023 at 18:08
  • \$\begingroup\$ @BenVoigt Thanks! That's what I though. But now, if I wanted to calculate the impedance felt by the voltage source, how could I derive a third relationship between \$V_X\$ and \$ V_{IN}\$ without relying on the \$KCL\$ at \$O\$ ? \$\endgroup\$
    – RitterTree
    Jun 12, 2023 at 19:07
  • \$\begingroup\$ Your missing relationship is \$I_{in} = \dfrac{V_{in}-V_{x}}{R_{3}}\$. Then the load on \$V_{in}\$ is \$\dfrac{V_{in}}{I_{in}} = \dfrac{V_{in}}{V_{in}-V_x} R_3\$ \$\endgroup\$
    – Ben Voigt
    Jun 12, 2023 at 19:13
  • \$\begingroup\$ @RitterTree Actually, you get the same result if you also include the output node in your KCL. If you set up va/rs+va/rf = vout/rf and vx/r4+vx/r3+vx*s*c = vout*s*c+vin/r3 and vout/rf+vout*s*c = iout+vx*s*c+va/rf and va = vin and solve for va, vx, iout, and vout then you will get vout/vin just as you got. It works fine to include the output amp's output current. \$\endgroup\$ Jun 12, 2023 at 21:27
  • \$\begingroup\$ @BenVoigt Oh, right! Thanks! \$\endgroup\$
    – RitterTree
    Jun 13, 2023 at 17:20

1 Answer 1


Your application of KCL at O accounts for currents in Rf and C, but not for op-amp output current, and so it isn't correct.

You can treat the op-amp output as a ground-referenced voltage source, with potential \$V_O\$, which allows you to draw the system of R3, R4 and C like this:


simulate this circuit – Schematic created using CircuitLab

KCL applied at X gets you:

$$ \frac{V_{IN}-V_X}{R_3} + (V_O-V_X)sC = \frac{V_X}{R_4} $$

As you stated, \$V_O = \left(1+\frac{R_F}{R_S}\right)V_{IN}\$, and you can substitute \$V_O\$ above with this expression for \$V_O\$ to obtain a potential \$V_X\$ in terms of only \$V_{IN}\$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.