As the title suggests, I have done all the necessary calculations to implement a gain into my second order stage of my 6th order high pass filter but for some odd reason, it would change the amplitude response of the circuit. I have recalculated the resistance that affect the filter by implementing K into their equations and I cross checked it with analog.com which proved that it was spot on so Im not really sure on why the response changes hence changing the cut-off frequency.
The difference in the cut-off frequency is huge, the original design is at 100Hz whilst the gain version reduces it to 180Hz.
The schematic without the gain:
The schematic with a gain:
And also if someone could explain why resistance of 150kOhms and above only work with the gain of the amplifier that would be appreciated!
Best Answer
In all Sallen-Key filters, the opamp does not only provide gain to the input signal. It is the main purpose of the active block to allow positive feedback. Only in this case, the circuit is able to produce a conjugate-complex pole pair - which is the precondition for allowing pole-Q values (Qp) larger than 0.5 (Remember: Qp=0.7071 for Butterworth response and Qp=0.9565 for Chebyshev with 1dB ripple).
That means: Larger gain (all other elements remain unchanged) makes more positive feedback (equal to higher Qp-values). And - there is a direct relationship between Qp and the form of the transfer function in the pole frequency region (amplitude peaking).
(Comment: The shown second circuit does NOT provide a gain value above unity).