Simple Theory question about Circuit Analysis

Question

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!

Answer 1

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

Answer 2

The assumption of an ideal op-amp does lead to the conclusion that the current in R3 is equal to the current in R2. However the current in R4 is the sum of the current in R3 and the output current of the op-amp. Thus your second method is incorrect. That is why the result does not agree with the first method.

Answer 3

As pointed out in the other answer the ideal amplifier simplification assumes the current into, or out of, the inverting and non-inverting inputs is zero. The output of the amplifier is capable of supplying or sinking current.

We can therefore assume that the current in \$R_3\$ equals the current in \$R_2\$ the output we source or sink the current required to make this true.

So the transfer function from the input to the output of the op-amp is:

$$\dfrac{-R_3}{R_2}$$

From the output of the op-amp we have a simple low pass filter the transfer function from the output of the op-amp to the output of the circuit is

$$\dfrac{\dfrac{1}{s \cdot C}}{R_4 + \dfrac{1}{s \cdot C}} = \dfrac{1}{1 + s \cdot C \cdot R_4}$$

Making the transfer function of the entire circuit

$$- \dfrac{R_3}{R_2} \cdot \dfrac{1}{1 + s \cdot C \cdot R_4} $$