Sallen-Key 2nd-Order Low-Pass Filter Calculator

Understanding the Sallen-Key low-pass filter — concept and principle

The Sallen-Key topology builds a second-order active low-pass filter from two resistors, two capacitors and one op-amp wired as a voltage follower. It rolls off at −40 dB per decade, twice as steep as a single RC stage, and is widely used for anti-aliasing before an ADC, audio crossovers and sensor smoothing.

C1 pushes the output back into the middle node; the C1 : C2 ratio sets Q = 0.7416, close to the flat Butterworth 0.707.

How it is calculated

Natural frequency: f₀ = 1 / (2π√(R1·R2·C1·C2)), with R in ohms and C in farads (inputs are in nF, so the calculator multiplies by 10⁻⁹).
Quality factor (unity gain): Q = √(R1·R2·C1·C2) / (C2·(R1 + R2)).

Here C1 is the capacitor from the junction of R1 and R2 back to the op-amp output, and C2 is the capacitor from the op-amp's non-inverting input to ground. With R1 = R2 the Q formula simplifies to Q = ½·√(C1/C2).

The principle

The two RC sections set the frequency scale, which is why f₀ depends on the geometric mean of the four parts. The trick is C1: it feeds the output back to the middle node. Near f₀ this positive feedback lifts the response and sharpens the corner; Q measures how strong that lift is. Q = 0.5 or lower gives a soft, critically damped corner; Q = 1/√2 ≈ 0.707 gives the Butterworth response, the flattest possible passband; higher Q produces a resonant peak. Above f₀ both capacitors short the signal path, giving the second-order −40 dB/decade slope.

Worked example

Inputs: R1 = R2 = 10,000 Ω, C1 = 22 nF, C2 = 10 nF.
R1·R2·C1·C2 = 10⁴ × 10⁴ × 22×10⁻⁹ × 10×10⁻⁹ = 2.2 × 10⁻⁸, and its square root is 1.4832 × 10⁻⁴ s.
f₀ = 1 / (2π × 1.4832 × 10⁻⁴) = 1,073.022407 Hz.
Q = 1.4832 × 10⁻⁴ / (10×10⁻⁹ × 20,000) = 1.4832 × 10⁻⁴ / 2 × 10⁻⁴ = 0.7416198487.
Check with the shortcut: ½·√(22/10) = 0.7416, slightly above Butterworth.

Each point is the calculator re-run with only “R1 (Ω)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

R1 (Ω)	Cut-off frequency (Hz)
2,500	2,146.044815
5,000	1,517.482841
7,500	1,239.019552
10,000	1,073.022407
12,500	959.7404178
15,000	876.1191269
20,000	758.7414207

How to read the result

At f₀ the gain of a second-order low-pass equals Q, so with Q = 0.7416 the response is about −2.6 dB at 1,073 Hz rather than exactly −3 dB. The −3 dB point coincides with f₀ only when Q = 0.707. A Q this close to Butterworth gives a passband peak of well under 0.1 dB. For pure Butterworth with equal resistors, choose C1 = 2·C2.

Common mistakes

Swapping C1 and C2, which changes Q (here it would drop to about 0.34) while f₀ stays the same.
Assuming f₀ is the −3 dB frequency for every Q.
Using an op-amp whose gain-bandwidth is too low; aim for well over 100 × f₀ × Q as a rough rule.

Keep in mind: The formulas assume an ideal unity-gain op-amp and exact component values. Capacitor tolerance and op-amp limits shift f₀ and Q, so check critical filters by simulation or measurement.

Frequently asked questions

How do I design a Butterworth Sallen-Key quickly?

Pick equal resistors R, set C2 = C and C1 = 2C; then f₀ = 1 / (2πRC√2) and Q = 0.707.

Can I make it high-pass?

Yes, swap the positions of resistors and capacitors. The f₀ formula stays the same form, but the Q expression changes.

How do I get a steeper filter?

Cascade stages with specific Q values from a filter table; for example a 4th-order Butterworth uses Q ≈ 0.541 and 1.307.

Sources: R. P. Sallen and E. L. Key, "A Practical Method of Designing RC Active Filters", IRE Transactions on Circuit Theory, 1955 · Texas Instruments, Analysis of the Sallen-Key Architecture, Application Report SLOA024

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

Your numbers