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Sometimes when encountering an opamp or comparator input I see an RC network topology similar to the following one: enter image description here

And in this case let's say we know everything but the value of C2, and we want to set C2 to a value to obtain the right low-pass filter effect. But to calculate the cut-off frequency one needs to know the equivalent resistor value of all the network before C2 (only at the left side of C2 the equivalent resistance matters since the signal is coming from left?).

Only after knowing this approximate equivalent resistance one can set C2 to obtain a desired low-pass filter.

First of all for simplicity the unknown output impedance of the input signal is neglected(even-though it might have a big impact if it is big). Here are the steps I followed to thevenize this circuit.

To thevenize the circuit, Vcc is shorted and C3(DC blocking capacitor) is shorted as follows(R2 and R4 becomes in parallel): enter image description here

Thevenizing R7 and R9 the circuit becomes(Thevenin voltage becomes Input/2): enter image description here

Now second thevenizing yileds(Thevenin voltage becomes 5*Input/32): enter image description here

So the final RC circuit simplifies to: enter image description here

And now I want to compare the frequency response of the original and thevenized circuits:

Here is the freq. response of the original circuit: enter image description here

And here is the freq. response of the thevenized circuit: enter image description here

My questions are:

The cut-off frequency definition is where the ouput is 3dB less than the output at DC. Following this definition yileds very similar cut off frequencies. The cut-off freq. in both are very close. One is around 7.8Hz the other is around 9.3Hz. I think that's the effect of neglecting C3 at the beginning.

1-) The resistors in the original network drops the input voltage at each stage until the output. But in the Thevenin circuit we have no similar loss. The final Thevenin voltage is 5/32 of the original one. So I would expect a power ratio (5/32)^2 which is around 32dB. But the difference at DC is around 18dB. What could be the reason?

2-) Is my way of using Thevenin here correct for practical rough estimate? And similarly can I use the Thevenin with R7 R9 C3 (ignoring the circuit after C3) to see the high-pass filter effects roughly?

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    \$\begingroup\$ (1) Sanity check : 5/32 is about 1/6, between 1/4 (-12dB) and 1/8 (-18dB) so should give around 16dB attenuation. If you're calculating power ratios, remember that 20dB is 10:1 in voltage ratio but 100:1 in power, so 32dB is clearly wrong. (2) Try it. \$\endgroup\$ – Brian Drummond Oct 11 '16 at 18:47
  • \$\begingroup\$ I am looking at your original response and your thevinized response. They should be identical and the are not. The difference is not much but it might be due to C3. Did you check that? In the meanwhile i work also on the rest. Saw that you were already thinking of C3. \$\endgroup\$ – Decapod Oct 11 '16 at 19:01
  • \$\begingroup\$ @BrianDrummond: I am trying to learn from this also. It is more the 50 years ago since I made these calculations. \$\endgroup\$ – Decapod Oct 11 '16 at 19:08
  • \$\begingroup\$ I would simply re-simulate the original with C3 shorted, to test that hypothesis. \$\endgroup\$ – Brian Drummond Oct 11 '16 at 19:11
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    \$\begingroup\$ Just looking at the circuit casually, with C2 and C3 being so close in value, I would be very hesitant to draw conclusions from results derived by shorting out one of them. \$\endgroup\$ – rioraxe Oct 11 '16 at 23:31

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