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I have the following circuit.

enter image description here

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!

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  • \$\begingroup\$ Your assumption (...only one current...from R2...) is not correct. You have TWO voltage sources - and you must apply the superposition rule in case you want to start from the beginning (without starting with the opamps gain). \$\endgroup\$ – LvW Sep 12 '15 at 15:04
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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\$. Current paths

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

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  • \$\begingroup\$ That's very clear. Thanks. You have forgotten a sign in \$V_o/E\$. Just something i noticed about current \$I_2\$. In my opinion the direction of this current should be towards ground since current flows from higher to lower potential(conventional current flow), therefore i don't think it would go back to the op-amp where it certainly has higher potential. Although it should't make any difference to the calculations. \$\endgroup\$ – Nikos Sep 12 '15 at 15:44
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    \$\begingroup\$ I added the sign in the last equation. Regarding the current \$I_2\$ you have to think about how OP-Amps work. As the positive input is connected to 0V, the OP-Amp tries to force the negative input to 0V as well (it always tries to equal out the voltage differential of both inputs). If we input a positive voltage, the output of the OP-Amp will go below 0V to force the negative input to the same potential as the positive one. Therefore the current \$I_2\$ will go into the OP-Amp (as well as \$I_1\$) \$\endgroup\$ – Martin Sep 12 '15 at 16:06
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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.

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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

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

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  • \$\begingroup\$ Yes, this is the way i initially did it. Thanks for answering. \$\endgroup\$ – Nikos Sep 12 '15 at 15:49

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