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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:

Low Pass Sallen Key

Frequency Response:

Cut-off is at -6dB.

Frequency Response

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  • \$\begingroup\$ For this special case (R = R1 = R2 and C = C1 = C2) the -3dB cut-off frequency is: $$fc=\frac{\sqrt{\sqrt{2}-1}}{2 \pi R C}$$ \$\endgroup\$ – G36 Mar 19 '18 at 17:20
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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.

enter image description here

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: -

enter image description here

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  • \$\begingroup\$ I accept your answer however why is the highest order polynomial negative for the denominator of H(s)? I derived the same transfer function in terms of s and looks something like this.[link]ee.surrey.ac.uk/Projects/CAL/control/img/equ_70.gif \$\endgroup\$ – Bonavia Mar 19 '18 at 21:53
  • \$\begingroup\$ 1/j =-j. Is that what you are questioning? \$\endgroup\$ – Andy aka Mar 19 '18 at 22:02
  • \$\begingroup\$ Or is it that j squared = -1? \$\endgroup\$ – Andy aka Mar 19 '18 at 22:05
  • \$\begingroup\$ Just at the very bottom of your answer, the standard frequency equation, I was wondering why one of the denominator term is negative. I thought it was suppose to be positive. Other than that you basically answered my question perfectly, thank you for that. \$\endgroup\$ – Bonavia Mar 20 '18 at 0:01
  • \$\begingroup\$ s squared produces j squared which equals minus 1 \$\endgroup\$ – Andy aka Mar 20 '18 at 0:29
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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)

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