# Why do I get two different transfer functions for the same gain?

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.

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.

• Applying KCL at O isn't productive because you don't know either the output current of the opamp or the load current. Commented Jun 12, 2023 at 18:08
• @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 $O$ ? Commented Jun 12, 2023 at 19:07
• Your missing relationship is $I_{in} = \dfrac{V_{in}-V_{x}}{R_{3}}$. Then the load on $V_{in}$ is $\dfrac{V_{in}}{I_{in}} = \dfrac{V_{in}}{V_{in}-V_x} R_3$ Commented Jun 12, 2023 at 19:13
• @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. Commented Jun 12, 2023 at 21:27
• @BenVoigt Oh, right! Thanks! Commented Jun 13, 2023 at 17:20

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