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Let's say \$ V_s \$ is some time-varying voltage source. Is it necessarily true that the current going through \$ C_1 \$ is zero? I would think that the answer is yes, because the current flowing through R_1 must be equal to the current flowing through \$ R_2 \$ (as these are the current out and in of the voltage source). I think the op-amp should be in negative feedback and shouldn't contribute any current through its terminals.

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    \$\begingroup\$ If no current was flowing through \$C_1\$ then it would not be influencing the circuit. But \$C_1\$ does influence the circuit, look up the transfer function of a Sallen-Key filter. The opamp is in negative feedback mode, - input and output are shorted so it behaves as a voltage buffer. Note how one side of \$C_1\$ is connected directly to the output. The output of the opamp does carry a current, the current through \$C_1\$ + any current the load on the output needs. \$\endgroup\$ – Bimpelrekkie Dec 13 '18 at 9:03
  • \$\begingroup\$ Thanks for pointing me towards the Sallen-Key filter! I am still a bit confused though -- let's say there isn't any additional load on the output. Isn't it true that the current flowing out of the voltage source must be equal to the current flowing into the voltage source, which implies that the current flowing through $R_1$ is equal to the current flowing through $R_2$? \$\endgroup\$ – user1825464 Dec 13 '18 at 9:19
  • \$\begingroup\$ the current flowing out of the voltage source must be equal to the current flowing into the voltage source That is true but that says nothing about the currents through R1 and R2. If the I(R1) = I(R2) was true then I(C1) would need to be zero. Also I could then replace R1 and R2 by a single resistor. Again, that would change the behavior of the circuit. So I(R1) ≠ I(R2). \$\endgroup\$ – Bimpelrekkie Dec 13 '18 at 9:27
  • \$\begingroup\$ According to Kirchoff \$ I_{R1} + I_{R2) + I_{C1} = 0 \$. That's the current into node 'A'. Now assume there is no current in either op-amp input and both inputs are at the same voltage you should be able to calculate its response from here. As pointed out by others this is a Sallen-Key filter. \$\endgroup\$ – Warren Hill Dec 13 '18 at 11:35
  • \$\begingroup\$ Your assumption about current flow is wrong. There's a path for the current to return to the source ground via the output of the op-amp (i.e. through its low output impedance). \$\endgroup\$ – Chu Dec 13 '18 at 12:05
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Consider this picture: -

enter image description here

The voltage at the non-inverting input is \$A\times\dfrac{XC_2}{R_2+XC_2}\$ and, given that the op-amp is configured as a unity gain buffer (gain = 1) the output voltage is the same.

This means that the voltage across C1 cannot be zero and hence current will flow through C1 if that voltage has an AC signal content produced by Vs. If the voltage produced by Vs is purely DC then no current will flow once the circuit has stabilized.

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  • \$\begingroup\$ The derivative of the voltage difference needs to be nonzero for current to flow \$\endgroup\$ – Scott Seidman Dec 13 '18 at 12:02
  • \$\begingroup\$ ... or billions of bypass caps just got very very hot \$\endgroup\$ – Scott Seidman Dec 13 '18 at 12:05
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    \$\begingroup\$ @ScottSeidman I shall amend to mention that this applies to AC voltages! \$\endgroup\$ – Andy aka Dec 13 '18 at 12:06
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This is a low pass filter. When the input frequency is high the capacitors impedance will be low and of course there will be some current flowing through C1 and then goes to the op amp and returns to it's source. The output voltage will be damped due to C2 low impedance which somehow grands the output. In low frequencies however there is a very low current flowing through all the circuit and of course C1.

You can use simulation tools to see for yourself.

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  • \$\begingroup\$ It's not a high pass filter, as DC bias passes on from input to output. \$\endgroup\$ – Justme Dec 13 '18 at 11:37
  • \$\begingroup\$ @Justme Thanks. it was just a silly mistake. \$\endgroup\$ – MoHaMaD InSoMnIaC Dec 13 '18 at 12:16

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