I have the following circuit.

I calculated it's transfer function, in the common way, to be: \$ \frac{Vo(s)}{E(s)} = \frac{-R_3}{s\ R_2R_4C + R_2} \$

Then i thought something else. Since there's only one current flowing in the circuit (ideal Op-Amp) from \$R_2\$, to \$R_3\$ to \$R_4\$ to \$C\$ and down to earth, if i apply Kirchoff's current Law at the \$Vo\$ node i get: \$ \frac{E(s)-V_o(s)}{R_2+R_3+R_4} = s\ C(V_o – 0) \$ which gives the transfer function: \$ \frac{Vo(s)}{E(s)} = \frac{1}{s\ C(R_2+R_3+R_4) + 1} \$

Now i'm confused with this second method, because something doesn't feel right. Also the Op-Amp introduces a phase difference so there should definitely be a minus sign in the transfer function. This second transfer function should be wrong but i don't understand why exactly. Can someone tell me your thoughts and what do you think about mine? What is wrong? Thanks in advance!

## Best Answer

There are in fact two currents flowing in this circuit. You have to consider the output of the OP-Amp as a second voltage source which sinks a second current from \$V_O\$.

With: $$V_{OP,out}=\frac{-E\cdot R_3}{R_2}$$ $$V_{OP,out}=V_O-R_4\cdot I_2$$ $$I_2 = -sCV_O$$

We get: $$\frac{-E\cdot R_3}{R_2}=V_O(1+sCR_4) \rightarrow \frac{V_O}{E}=\frac{-R_3}{sCR_2R_4+R_2}$$