[SOLVED] Differential filter cutoff frequency

Hy, I am analyzing a signal conditioning circuit that is connected to a ICP accelerometer. The accelerometer is directly connected to the IN net

The circuit has a input filter connected to a programmable gain amplifier and an output filter conncted to the input of a ADC.

Concerning the input filter I am in doubt how to calculate the cutoff frequency for the presence of the resistor R1 and the capacitor C2( I guess that it has only a function of bypass and it does not have influence on the input frequency filter).

Concerning the output filter I am in doubt how to calculate the cutoff frequency for the presence of the resistor R6. (Without the resistor R6 the cutoff frequency should be 1/(2*pi*R4*C3) = 3120Hz )


I attach the schematic of this session

Thanks in advance

If we assume that C2 is too large for the Operating Frequency, the circuit is simplified.This node can be considered as short circuit to the GND
R1C1 pair creates a Pole and a Zero, similarly R4C3R6 and R5C3R6 triplet creates identical 2 Zeros and 2 Poles
All you have to do is to divide the circuit fictionally by 2 then calculate the Frequency Response..

Input filter: C2 is a bypass cap I think, and then fc_in=1/(2*pi*(R1||Rs))*C1), where Rs is the input voltage source's internal resistance, and you have to consider the input capacitor of the OPAmp is negligible.
Output filter: one of the input of your OPAmp is connected to AC ground, the output signal depends on the amplifier internal structure, I am not sure both of your OPAmp output will change differentially. If the output signal is not differential fc_out=1/(2*pi*(R6||R4)*C3), else it is fc_out=1/(2*pi*((0.5*R6)||R4)*(2*C3)). I also neglected the input impedance of your AGC here.

Hi,

It doesn't matter whether the OPAMP outputs act differentiel or not....
There may be one output active while the other stays fixed, or opposite, or both differential or both independent...

For calculating the output fc you need to see that R4 and R5 are in series...resulting in 102 ohms.
And R6 additionally lowers the overall impedance like it was in parallel to the 102 Ohms --> 96.53 Ohms.
Fc = 1 / (2 × Pi × 96.53 ohms × 1uF) = 1649 Hz

Klaus