# Amplifier circuit: Voltage divider x KCL

I was checking the solution to one question when I found out I couldn't use voltage divider rule to easy the formulas, the amplofier circuit (infinite gain): My first attempt: $$V_{x}=\frac{R_4}{R_1+R_2}.V_{out}$$ $$V_y=\frac{R_2}{R_2+R_3}.V_x$$ Then, after few manipulations: $$V_{out}/V_y =(1+\frac{R_3}{R_2}).(1+\frac{R_1}{R_4})$$

But usising Kirchhoff' current law: $$\frac{V_{y}}{R_2} = \frac{V_x-V_y}{R_3}= \frac{V_{out}-V_x}{R_1}-\frac{V_x}{R_4}$$

Again, after few manipulations we get: $$V_x=R_3.(\frac{1}{R_2}+\frac{1}{R_3}).V_y \\$$ $$\frac{V_y}{R_2} = \frac{V_{out}}{R_1}-(\frac{1}{R_1}+\frac{1}{R_4}).R_{3}.(\frac{1}{R_2}+\frac{1}{R_3}).V_y$$ Which reduces to: $$V_{out}/V_y = \frac{R_1}{R_2} + (1+\frac{R_3}{R_2}).(1+\frac{R_1}{R_4})$$

The latter agrees with the official answer, but why is the first invalid? How did I lost the R1/R2 factor using it?

• I don't understand your very first equation. It seems wrong to me, though there may be some explanation to it that I don't see. R4 is in parallel to R2+R3, when forming a divider with R1 to make Vx. Why isn't that fact showing up in the first equation? Commented Nov 5, 2023 at 1:03
• Yup, now it seems obvious to me as I remove the line connecting to the negative input of the amplifier, it really shows $$(R_2+R_3) || R_4$$Thanks everyone.
– Duca
Commented Nov 6, 2023 at 1:32

$$\ V_x= {{R_4||(R_2+R_3)}\over{R_1+R_4||(R_2+R_3)}}\times V_{out}\$$ ....R4 is in parallel with (R2+R3).
The second equation for $$\V_y\$$ is correct.