Coupling capacitor for DC filtering

[bonzer]

Hello everyone. Please help me to understand a thing.
I try to eliminate DC component of an input triangular wave. Average value 7.5V and peak to peak value 5V (so from 5 to 10V). Frequency varies from 200Hz to 40 kHhz. I also want to amplify it (about 2 but anyway).
I use an op amp (opa277) configured as a non inverting amplifier with gain 2.
What should I worry about with this configuration? (see figure).
I think even for low frequency it should work, common mode input resistance of my op amp is something like 250 GOhm, therefore even at 200Hz (1/sC= 800kOhm) I shouldn't have attenuation in the forming voltage divider. The other components are all odd multiplies of 200Hz so I don't have anything below this frequency right? I'm afraid of loosing amplitude because of that, or to ruin the quality of my waveform.
Unfortunately my simulator suffers because of that long needed time constants till it gets alternating and so its computationally difficult for me to check it. I don't have any way to check it practically (still closed laboratory because of covid). I hope your theoretical knowledge would help me.

[Manul]

To put it simply, for AC coupling capacitor to remove DC bias, it needs to first charge to that bias voltage. And to charge it needs to have a path for that. Op amp input is very high DC resistance and it may take forever. So you need to have some resistance after capacitor which creates a path for fast charging to this DC bias voltage and settle. Notice, that it is now a high pass filter. So you need to calculate your components for required low frequency cut off.

High pass filtering is another perspective how you can look at AC coupling. Notice, that if you have infinite resistance after capacitor, this filter will pass 0 Hz (DC), because it has infinite low cut off (no cut off at all) frequency. To block DC you must create high pass filter, there is no other way here. So you choose some frequency, lets say 100Hz, as a cut off frequency. Now all lower frequencies, including DC (0 Hz) will be attenuated (and for DC it will be infinite attenuation, it means complete blocking) and this is exactly whats is AC coupling.

[Manul]

Just for fun, if input resistance is 250 GOhm, and capacitor 1nF, like in your circuit, it is of course also a high pass filter, but cut off is 0.0006369427 Hz.

Another perspective - RC time constant. It is 250 seconds for these values. So if we say that we need 10 x RC time constants to fully settle, it means that it will settle in 41 minutes. Not practical at all. This is of course purely theoretical, because in practice capacitor will have some leakage, and capacitor will never reach DC bias voltage, so it will pass some DC.

[Siwastaja]

Also remember that the opamp inputs can't be left floating; there has to be a path for DC currents. This is because the amplifier internally "leaks" a bit, called bias current. If this current can't go anywhere, the amplifier inputs slowly drift towards positive or negative supply (whichever happens to be in the direction of the bias current).

The added 1Mohm resistor provides path for the bias current, as well.

Because you can't use arbitrarily high resistor values (to avoid problems with the opamp bias current and offset current), the C often ends up being quite a large value (even tens of microfarads), if you need to pass frequencies of a few dozen Hz or so (like in audio).