Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.
Sign up to join this community
I designed a unity gain low pass Sallen key filter with equal resistor values of 10k and capacitors of 8nF. The cut-off frequency to my knowledge at the -3dB point is defined as:
\$f=\frac{1}{2\pi\sqrt{R1R2C1C2}}\$
Upon simulation the cut-off designed for, which is 2kHz is at the -6dB point. Is this formula defined for the -6dB point since this is a second order system? I'm confused.
Circuit:
Frequency Response:
Cut-off is at -6dB.
A sallen key filter like this (or any 2nd order low pass filter) has gain-magnitude of "Q" at the cut-off frequency. Given that your values produce a Q of 0.5 then the amplitude at the cut-off frequency is 6 dB down.
If you instead made the feedback capacitor \$\sqrt2\$ higher at 11.3 nF and the grounding capacitor lower by the same amount (5.66 nF) then Q would be about 0.707 and your filter would have a cut-off frequency of about 1990 Hz and an amplitude of -3 dB down at this frequency.
This would be the ideal (so-called) butterworth response.
You can try it for yourself at THIS website.
Proof that Q is the value of the T.F. at the natural cut-off frequency: -
Bonavia - the formula as shown by you does NOT give the 3-dB-cut-off frequency fc but the pole frequency fp. In your simplified case (unity gain with equal components), the poile quality factor Qp is only Qp=0.5. This value can be achieved with a passive lowpass also.
The classical design procedure is to select the values for fc and Qp and find the component values as a last step.
Example: Qp=0.7071 for a Butterworth response.
This can be realized with two equal resistors but a capacitor ratio C1/C2=2.
In this case, the cut-off frequency will be fc=fp=0.7071/(R1C2)
