I recently found this paper online and would like to try to simulate the circuit in Figure 6, which (for convenience) I reproduce here:

Approaching opamps for the first few times, I present some doubts:

How do I know from such a circuit what the cutoff frequency is?
If I am not mistaken, I should first calculate the transfer function, right? And from that I can calculate the cutoff frequency .
In the past I calculated it for a Sallen-Key low-pass filter ... but in this circuit here I don't know how to calculate it (I attach the few calculations in the photo..).
Does anyone know how to calculate both?

EDIT: if I am not mistaken, the gain of this configuration is the gain of a differential amplifier, right? So in this case: -R2/R1+R5/(R5+R4)*(1+R2/R1).
I kindly ask for your confirmation.

 

Since it's a differential stage, it may be easier to connect one input to ground, and then calculate the transfer function of the other side.
You then can do the same for the opposite side and then combine them together.

 

If you make C2 two 1500pF capacitors in series, then you can put the values of the top half in a filter design program.
http://sim.okawa-denshi.jp/en/OPtazyuLowkeisan.htm

 

If you make C2 two 1500pF capacitors in series, then you can put the values of the top half in a filter design program.
http://sim.okawa-denshi.jp/en/OPtazyuLowkeisan.htm

Since it's a differential stage, it may be easier to connect one input to ground, and then calculate the transfer function of the other side.
You then can do the same for the opposite side and then combine them together.

You're right! I had not thought about the "symmetry" of the circuit..
So by dividing the circuit I would get these two subcircuits (and their sub-transfer functions):

Sorry for the trivial question, these two subcircuits are connected in parallel with each other, right?
Because if so, I should add up the two transfer functions found.

 

You're right! I had not thought about the "symmetry" of the circuit..
So by dividing the circuit I would get these two subcircuits (and their sub-transfer functions):
View attachment 306924
Sorry for the trivial question, these two subcircuits are connected in parallel with each other, right?
Because if so, I should add up the two transfer functions found.

Think of it more as two different paths for the same signal.
The output from the original is the difference between the two inputs.
The output of the single-ended circuit is the difference between a single input and ground.

 

Note that since C2 would be connected to ground through the input resistor, it wouldn't be completely at ground potential if you ground one input.
But the value of the input resistors are low enough that it probably wouldn't affect the transfer function much if you assume it as at ground potential when calculating the function for half the circuit.

 

-3dB @ 2kHz = 12.56krad/s, 80 us 2nd order Butterworth.
100k*100pF= 10 us, 10k/1k = 20 dB gain, 2k*33pF = 66 ns

A different way to get 20 dB 2nd order 50 kHz below. compared to the paper.

 

More value symmetry but same gain 20 dB and 50 kHz BW

500 kHz GBW, but since the 1st cap C=220 pF has much higher Q than 1 to line up with other poles the OA GBW must be much higher than 500 kHz. It gets multiplied by the Q^2 of the pole such that OA GWB needs to be 10 MHz for the 500 kHz GBW of the filter !! That's normal for any active filter.