Sallen-Key Low-Pass Filter

A second-order active low-pass built from one op-amp and four passives. It is the default building block for anti-aliasing filters in front of an ADC.

Transfer function

For the unity-gain configuration (op-amp wired as a buffer):

$$H(s)=\frac{{\omega }_0^2}{s^2+\frac{{\omega }_0}{Q}s+{\omega }_0^2}$$

where

$${\omega }_0=\frac{1}{\sqrt{R_1{R}_2{C}_1{C}_2}}Q=\frac{\sqrt{R_1{R}_2{C}_1{C}_2}}{C_2(R_1+R_2)}$$

Two knobs: ${\omega }_0$ moves the corner, $Q$ sets what happens at the corner.

Choosing Q

$Q$	Response	Use it for
$0.5$	Critically damped	Fastest settling, no overshoot
$0.577$	Bessel	Flat group delay — pulses keep their shape
$0.707$	Butterworth	Maximally flat magnitude — the usual default
$>1$	Chebyshev-ish	Steeper rolloff, bought with peaking and ringing

Q peaks the noise too

A $Q$ of 2 does not just peak the signal near ${\omega }_0$ — it peaks whatever noise lives there by the same 6 dB. On an anti-aliasing filter in front of a 16-bit ADC that is a real loss of ENOB, not a cosmetic one.

A worked 1 kHz Butterworth

Fix the capacitors first — E12 capacitor values are coarser than resistor values, so you want the resistors to absorb the rounding.

Pick $C_1=22\mathrm{nF}$, $C_2=10\mathrm{nF}$. For $Q=0.707$ and $f_0=1\mathrm{kHz}$, solving the pair above gives

$R_1\approx 8.2\mathrm{k}\Omega R_2\approx 14\mathrm{k}\Omega$

Both are stock E96 values, so the built filter lands within about 2 % of the target corner — comfortably inside the tolerance of the capacitors themselves.

Check the op-amp before you build it

From Op-Amp Basics, gain–bandwidth is finite. The rule of thumb is that the op-amp needs

$\text{GBW}\gtrsim 100\cdot Q\cdot f_0$

For this filter that is 70 kHz — trivially met by anything modern. Push the same topology to a 100 kHz corner at $Q=2$ and you suddenly need 20 MHz, which rules out most of the jellybean parts in your drawer.

See also

• Op-Amp Basics — where the GBW constraint comes from
• Back to the vault