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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.

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  • \$\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

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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:

schematic

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}\$.

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