Electronic – Sallen Key Cut-off Frequency

filter

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

Best Answer

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.

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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: -

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