What is a Sallen-Key filter?
An active filter topology with one op-amp, two resistors and two capacitors that gives a second-order (40 dB per decade) response. It can be built as a low-pass or a high-pass filter.
CIRCUIT DIAGRAM EXAMPLE
Sallen-Key Low-Pass Filter Circuit Diagram
The Sallen-Key topology builds a second-order filter from one op-amp, two resistors and two capacitors. Its unity-gain form is one of the most common active filters, used for anti-aliasing in front of ADCs and in audio circuits.
How this circuit works
At low frequencies the capacitors are open circuits, so Vin passes through R1 and R2 to the + input, and the op-amp, wired as a voltage follower, copies it to Vout. The passband gain is exactly 1.
At high frequencies C2 shunts the + input to ground, like an RC low-pass filter. C1 is the trick: it returns to the output instead of ground. Near the cutoff the output still follows the input, so C1 carries little current and the response stays flat for longer. Far above the cutoff the output is small, C1 behaves like a second capacitor to ground, and the signal is attenuated twice.
The result is a second-order response: flat to about 1.07 kHz, then falling at 40 dB per decade, twice as steep as a single RC stage. How sharp the knee is depends on Q, which here is ½ × √(C1 / C2) ≈ 0.74. Q = 0.707 is the Butterworth response, the flattest passband without a peak.
Because the op-amp is a follower, its gain does not enter the filter formulas, which makes the unity-gain version easy to design with standard values. The op-amp still needs a gain-bandwidth well above the cutoff to behave like an ideal follower.
Formulas
- Cutoff frequency
fc = 1 / (2π × √(R1·R2·C1·C2)) = 1 / (2π × 10 kΩ × √(22 nF × 10 nF)) ≈ 1.07 kHz- Quality factor (R1 = R2)
Q = ½ × √(C1 / C2) = ½ × √2.2 ≈ 0.74- Butterworth design (R1 = R2 = R)
C1 = 2 × C2, fc = 1 / (2π × R × C2 × √2)With R = 10 kΩ, C2 = 10 nF and C1 = 20 nF this gives fc ≈ 1.13 kHz.
Parts list
| Ref | Part | Value | What it does |
|---|---|---|---|
| C1 | Capacitor | 22 nF | 22 nF from the R1–R2 junction to the output: the feedback capacitor. |
| C2 | Capacitor | 10 nF | 10 nF from the + input to ground. |
| R1 | Resistor | 10 kΩ | 10 kΩ, first resistor. |
| R2 | Resistor | 10 kΩ | 10 kΩ, second resistor. |
| U1 | Op-amp | TL071 | TL071 op-amp wired as a voltage follower. Supply pins are not drawn. |
Netlist
Recovered from the drawing itself: every pin on each net below is joined by a wire, a junction or a shared supply or ground symbol.
| Net | Connects |
|---|---|
| Vin | R1.1 |
| N1 | R1.2, R2.1, C1.1 |
| N2 | R2.2, U1.noninv, C2.1 |
| Vout | U1.inv, U1.out, C1.2 |
| GND | C2.2 |
Common mistakes
- Swapping C1 and C2. With the larger capacitor to ground, Q falls to about 0.34 and the knee becomes soft.
- Using a slow op-amp. Its gain-bandwidth must sit well above the cutoff; for audio-range filters most general-purpose op-amps are fine.
- Expecting endless attenuation. Far above the cutoff, signal leaks through C1 into the op-amp's output, whose impedance rises with frequency, so the stopband comes back up. Add a passive RC after the filter if you need deep attenuation at very high frequencies.
- Using X7R or Y5V capacitors. Their value moves with voltage and temperature; C0G (NP0) ceramic or film capacitors keep the cutoff where you put it.
Build notes
- For an exact Butterworth response (Q = 0.707) make C1 exactly twice C2, e.g. 20 nF and 10 nF.
Frequently asked questions
How do I calculate the cutoff frequency?
fc = 1 / (2π √(R1 R2 C1 C2)). With equal resistors that simplifies to 1 / (2π R √(C1 C2)).
How do I get a Butterworth response?
In the unity-gain version with equal resistors, make the feedback capacitor C1 twice the ground capacitor C2. That gives Q = 0.707.
How do I turn it into a high-pass filter?
Swap each resistor with a capacitor and each capacitor with a resistor. The cutoff formula keeps the same form.
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