Sallen-Key low-pass filter design, without the guesswork

Pick a response shape, read its Q, and get the two capacitors and one resistor. Then the four things that decide whether the built filter matches the design.

At unity gain the whole design is three lines. Choose the response shape and read its frequency scaling factor and quality factor from a table; choose the capacitor you want to use; then:

$C_1 = 4Q^2 C_2, \qquad R = \frac{1}{2\pi\,\text{FSF}\cdot f_c \sqrt{C_1 C_2}}$

For a 1 kHz Butterworth starting from a 10 nF C0G part, that is $C_1 = 20$ nF, $C_2 = 10$ nF and $R = 11.25\ \text{k}\Omega$ — fit 11.3 kΩ and the corner lands at 996 Hz. The active filter calculator does exactly this, in both directions.

What makes the topology worth using is the ordering. $Q$ is a ratio of two capacitors and the corner is set by a resistor, so the two decisions do not interact. Nothing about choosing 11.3 kΩ instead of 11.25 kΩ moves the shape of the response; it moves only where the shape sits. That property is not free — it is a consequence of choosing unity gain and equal resistors, and three of the four simplifications TI’s SLOA049 (Active Low-Pass Filter Design) offers give it up.

The unity-gain Sallen-Key low-pass: two equal resistors in series from the input to the non-inverting op-amp input, a capacitor from that input to ground, and a second capacitor from the junction between the resistors back to the output, with the output tied to the inverting input. — Fig 1 — The unity-gain Sallen-Key, and the only two decisions in it. The ratio of the two capacitors sets Q; their geometric mean with R sets the corner. Note which capacitor is which: SLOA049 numbers them the other way round, so its Q expression looks inverted next to the calculator's.

Which capacitor is C1

Before anything else, a labelling trap that costs an afternoon. SLOA049’s transfer function for the Sallen-Key is

$H(f) = \frac{K}{(j2\pi f)^2 R_1 R_2 C_1 C_2 + j2\pi f\left(R_1 C_1 + R_2 C_1 + R_1 C_2 (1-K)\right) + 1}$

and the $R_1 C_1 + R_2 C_1$ in the middle term identifies $C_1$ as the capacitor from the non-inverting input to ground — only a shunt element multiplies both resistances. In that numbering, at unity gain and with equal resistors, $Q = \tfrac{1}{2}\sqrt{C_2/C_1}$ .

The calculator on this site numbers them the other way: $C_1$ is the feedback capacitor, from the resistor junction to the output, and so $Q = \tfrac{1}{2}\sqrt{C_1/C_2}$ . Both are the same circuit and the same maths. The physical statement, which no numbering can confuse, is that the capacitor going back to the output raises Q and the capacitor going to ground lowers it — because the feedback capacitor is what feeds energy back into the network, and Q is a measure of how much.

The standard form, and what FSF and Q are

Every second-order low-pass, whatever the topology, is the same function. SLOA049 writes it as

$H_{LP}(f) = \frac{K}{-\left(\dfrac{f}{\text{FSF}\cdot f_c}\right)^2 + \dfrac{1}{Q}\cdot\dfrac{jf}{\text{FSF}\cdot f_c} + 1}$

with three regimes it spells out: well below the corner the gain is $K$ ; at $f = \text{FSF}\cdot f_c$ the response is $jKQ$ , so the signal is shifted 90° and multiplied by Q; well above it the gain falls as the square of the frequency ratio.

FSF and Q are not properties of the circuit — they come from the filter polynomial, and SLOA049 gives the conversion from its pole locations explicitly:

$\text{FSF} = \sqrt{\text{Re}^2 + \text{Im}^2}, \qquad Q = \frac{\sqrt{\text{Re}^2 + \text{Im}^2}}{2\,\text{Re}}$

So the whole “filter type” question reduces to two numbers per stage, and the circuit design that follows is the same for all of them:

Second-order shape	FSF	Q	overshoot to a step
Bessel	1.2736	0.5773	0.4 %
Butterworth	1.0000	0.7071	4.3 %
1 dB Chebyshev	1.0500	0.9565	14.6 %
3 dB Chebyshev	0.8414	1.3049	27.2 %

FSF and Q are SLOA049’s tables 9-1 to 9-4; the overshoot column is the closed-form step response of each, computed rather than quoted.

Magnitude responses of four second-order low-pass shapes on log-frequency axes, all scaled to the same one kilohertz corner. The Bessel rolls off earliest and most gently, the Chebyshev peaks before its corner and falls fastest, and the Butterworth sits between them. — Fig 2 — The four shapes, all evaluated at the same stated corner from SLOA049's FSF and Q pairs. They differ only in where the pole sits and how sharp it is; every one of them is 40 dB per decade eventually, and the choice is entirely about the first octave either side of the corner.

Choose the shape from the signal, not from the roll-off

The magnitude plot makes the Chebyshev look strictly better — it falls fastest past the corner. The step response is where that is paid for, and SLOA049’s own summary table is blunt about the trade:

Bessel — Constant group delay – no overshoot with pulse input. Slow rate of attenuation above fc

3-dB Chebyshev — Fast rate of attenuation above fc. Large overshoot and ringing in response to pulse input

Between the Bessel and the 3 dB Chebyshev that is the difference between 0.4 % and 27 % overshoot on a step, from filters with the same corner and the same part count. The decision rule that follows:

Anti-aliasing ahead of an ADC — the signal is a spectrum and the requirement is attenuation at the Nyquist frequency. Butterworth, or Chebyshev if the sample rate is tight.
Conditioning a pulse, a PWM signal or a sensor’s step — the signal is a shape in time and overshoot is distortion. Bessel.
Not sure — Butterworth. SLOA049 calls it “the best all-around filter response”, and 4 % overshoot rarely breaks anything.

For a single-pole requirement none of this applies and the RC filter calculator is the right tool; a second-order active stage earns its op amp only when the roll-off has to be steeper than 20 dB per decade or the source cannot drive the load.

Step responses of the same four filter shapes. The Bessel rises smoothly with no overshoot, the Butterworth overshoots slightly, and both Chebyshev responses overshoot substantially and ring for several cycles. — Fig 3 — The same four filters answering a step, computed from the closed-form second-order response. This is the trade the frequency plot hides: SLOA049 gives the Bessel "constant group delay — no overshoot with pulse input" and the 3 dB Chebyshev "large overshoot and ringing", and here that is a difference of about 27 % of full scale.

The design procedure, in the order the decisions bind

The reason to start with Q rather than with the corner is that Q is fixed by a ratio, and the ratio has to be buildable before anything else matters.

A four-step design sequence drawn as boxes: choose the response shape and read its FSF and Q, choose a capacitor, compute the second capacitor from Q, then compute the resistor from the corner frequency. — Fig 12 — The whole procedure at unity gain, in the order the decisions actually constrain one another. Q is fixed before either component value, because it is a ratio; the corner is set last, because it is the only thing left to set.

SLOA049’s component guidance is worth following literally, because both halves of it have a failure mode:

Capacitors — Avoid values less than 10 pF; Use C0G (NP0) dielectrics; Use 1%-tolerance components

Resistors — Values in the range of a few hundred ohms to a few thousand ohms are best

Resistors too large make the capacitors small enough that stray capacitance — the op amp’s input capacitance, the pad-to-plane capacitance under the node — becomes a design component. Resistors too small make the op amp drive them. And the dielectric is not a detail: an X7R capacitor’s capacitance loss under DC bias is tens of per cent, which moves both the corner and Q by far more than any resistor tolerance in this article.

Why the capacitor ratio is the real constraint

$C_1 = 4Q^2C_2$ grows quadratically, and that decides which stages this topology can build at unity gain:

Q = 0.5773  (Bessel)                       ratio  1.33 : 1
Q = 0.7071  (Butterworth)                  ratio  2.00 : 1
Q = 1.3049  (3 dB Chebyshev, 2nd order)    ratio  6.81 : 1
Q = 1.9320  (6th-order Butterworth, last)  ratio 14.9  : 1
Q = 12.79   (6th-order 3 dB Cheb, last)    ratio  654  : 1

A 2:1 or 7:1 pair is two stock C0G values. A 654:1 pair is 1 nF against 654 nF, which is not one dielectric — the large part would be X7R or bigger, and its tolerance and bias behaviour would then set Q. That is the point at which a unity-gain Sallen-Key stops being the right circuit, and either the gain-setting version or the multiple-feedback topology takes over.

The required ratio of the two capacitors plotted against Q on log axes, rising as the square of Q. Marked points show that ordinary filter shapes need ratios near two, while high-Q Chebyshev stages need ratios in the hundreds or thousands. — Fig 4 — Why high-Q stages are hard in this topology at unity gain. The capacitor ratio goes as 4Q², so a Butterworth needs 2:1 and the last stage of a sixth-order 3 dB Chebyshev needs 654:1 — a pairing you cannot buy in one dielectric, which is the practical reason such stages go to MFB or to a gain-setting network.

The simplification that looks easiest and is worst

SLOA049 lists four Sallen-Key simplifications, “ordered from harder to easier”, with the warning that “the easier the design becomes, the more the design freedom is limited”. The last one sets all four passive components equal:

$\text{FSF}\cdot f_c = \frac{1}{2\pi RC}, \qquad Q = \frac{1}{3-K}$

This is genuinely attractive — one resistor value, one capacitor value, and the corner and Q are now independent. The catch is in that denominator. $Q$ has a pole at $K = 3$ , and differentiating gives the sensitivity:

$\frac{dQ/Q}{dK/K} = QK$

For a Butterworth, $K = 1.586$ and the sensitivity is 1.12 — a 1 % gain error is a 1.1 % Q error, which is nothing. For a $Q$ of 10 the gain is 2.9 and the sensitivity is 29: a 1 % error in a resistor divider becomes a 29 % error in Q, and the filter has a peak where the design had none.

The unity-gain version has no such term at all, which is SLOA049’s own observation about the architecture: “the unity-gain Sallen-Key inherently has the best gain accuracy because the gain is not dependent on component values.” Default to it, and reach for gain elsewhere in the chain.

Q plotted against passband gain for the equal-component Sallen-Key simplification. The curve rises without limit as the gain approaches three, so a small gain error near that point produces a large Q error. — Fig 5 — The simplification that looks easiest and is the least robust. With all four passive parts equal, Q depends only on the gain: Q = 1/(3 − K), which has a pole at K = 3. The sensitivity of Q to the gain is Q·K, so a Butterworth tolerates a gain error and a Q of 10 multiplies it by 29.

Where the tolerance actually lands

Differentiate the two unity-gain design equations and the error budget falls straight out. With $f_0 = 1/(2\pi R\sqrt{C_1C_2})$ and $Q = \tfrac{1}{2}\sqrt{C_1/C_2}$ :

$\frac{\Delta f_0}{f_0} = -\frac{\Delta R}{R} - \frac{1}{2}\left(\frac{\Delta C_1}{C_1} + \frac{\Delta C_2}{C_2}\right), \qquad \frac{\Delta Q}{Q} = \frac{1}{2}\left(\frac{\Delta C_1}{C_1} - \frac{\Delta C_2}{C_2}\right)$

Two things follow. The resistors do not appear in the Q expression — with $R_1 = R_2$ they cancel, so resistor tolerance moves the corner and leaves the shape alone. And the capacitors enter Q as a difference, so two capacitors from the same reel, which track, give a better Q than their individual tolerances suggest.

With SLOA049’s simulation tolerances — 1 % resistors, 2 % capacitors — the worst case is 3 % on the corner and 2 % on Q, and the root-sum-square figures are 1.73 % and 1.41 %. Both are smaller than the shift a class 2 dielectric would contribute on its own.

Bars showing how each component tolerance contributes to the error in corner frequency and in Q, with worst-case and root-sum-square totals for one per cent resistors and two per cent capacitors. — Fig 6 — Where the error actually comes from, differentiating the two design equations. The corner frequency collects the resistor error and half of each capacitor error; Q collects only half the difference of the capacitor errors, and no resistor error at all, because the resistors are equal and cancel.

Higher orders are different stages, not repeated ones

An $n$ th-order filter needs $n/2$ second-order stages, and they are not copies of each other. A sixth-order Butterworth uses stages of $Q$ 0.5177, 0.7071 and 1.9320 — all at the same $f_0$ , all with different capacitor ratios. Cascading three Butterworth filters instead gives a response that is 9 dB down at the corner rather than 3, and is not a Butterworth of any order.

SLOA049 also fixes the order:

Theoretically, the order of the stages makes no difference, but to help avoid saturation, the stages are normally arranged with the lowest Q near the input and the highest Q near the output.

The reason is visible in the plot: the $Q = 1.932$ stage peaks 6.0 dB above its own passband. Put it first and a full-scale input at the corner clips before the later stages have attenuated anything. Put it last and the earlier stages have already taken 6 dB out at that frequency.

Magnitude responses of second, fourth and sixth-order Butterworth filters built by cascading stages, with the individual stage responses of the sixth-order case drawn faintly beneath the composite. One stage peaks above zero decibels on its own. — Fig 7 — A sixth-order Butterworth is three different second-order stages, not three identical ones. SLOA049's table gives them Q of 0.5177, 0.7071 and 1.9320; the last peaks 6.0 dB above the passband on its own, which is why the note puts the lowest Q first and the highest Q last.

The op amp has to be much faster than the filter

Every equation above assumes an ideal amplifier. The bound the calculator applies is

$\text{GBW} \ge 100 \cdot K \cdot f_0 \cdot Q$

and it is not a conservative rule of thumb: SLOA049’s own simulated Sallen-Key responses, built around a TLV9062, “are almost identical from 10 Hz to about 80 kHz” — with a 1 kHz corner. Departure at 80× the corner is what a good modern part does.

A 1 kHz Butterworth wants 71 kHz of gain-bandwidth, which is nearly any op amp. A 100 kHz Butterworth wants 7.1 MHz, which is a decision. The last stage of a 100 kHz sixth-order Chebyshev, at $Q = 12.8$ , wants 128 MHz — and that stage is also the one asking for a 654:1 capacitor ratio. High-Q stages are expensive twice over.

Required op-amp gain-bandwidth product plotted against filter corner frequency on log axes, with one line for each of several stage Q values. The requirement is a hundred times the corner frequency multiplied by the gain and the Q. — Fig 8 — What the amplifier has to be, before any of the passive design matters. The bound the calculator uses is GBW ≥ 100·K·f0·Q, and it is not conservative: SLOA049's own Sallen-Key simulations depart from the ideal response above roughly 60–90 kHz with a 1 kHz corner.

The stop-band comes back up

The Sallen-Key has one behaviour that surprises people measuring their first one: past some frequency the attenuation stops improving and then reverses. SLOA049 explains the mechanism precisely, and it is structural rather than a component defect:

The assumption made here is that C1 and C2 are effective shorts when compared to the impedance of R1 and R2, so the input of the amplifier is at AC ground. In response, the amplifier generates an AC ground at the output, limited only by the closed-loop output impedance ZOUT.

In other words, at high frequency the two capacitors short out the amplifier’s role entirely. The feedback capacitor ties the $R_1$ – $R_2$ junction to the output and the shunt capacitor grounds the non-inverting input, so what is left — and what SLOA049’s Figure 11-1 draws — is $R_1$ from the input to the output node, with $R_2$ and the op amp’s closed-loop output impedance $Z_{OUT}$ both shunting that node to ground. The leak is the divider $R_2 \parallel Z_{OUT}$ against $R_1$ , which is near enough $Z_{OUT}/R_1$ because $Z_{OUT}$ is ohms and $R_1$ is kilohms. A real op amp’s closed-loop output impedance rises with frequency as its loop gain runs out, so the leak grows.

The remedy SLOA049 applies is a passive pole after the amplifier — “a 100-Ω resistor is placed in series with the output and a 47-nF capacitor is connected from the output to ground”, which computes to a pole at 33.9 kHz. It costs two parts and some output impedance, and it puts the stop-band back.

The MFB topology has no such path, because there is no route from input to output that survives the capacitors becoming shorts; SLOA049’s high-frequency model of it (Figure 11-3) is left with only the parasitic capacitance from input to output, which is a layout matter rather than a circuit one. That, rather than component sensitivity, is usually the reason to pick it.

The ideal second-order roll-off compared with the real Sallen-Key response, which stops falling and comes back up at high frequency as the amplifier output impedance rises, and with the same filter after a small resistor and capacitor are added at the output. — Fig 9 — The Sallen-Key stop-band does not keep going down. Above cutoff both capacitors are effectively shorts: the feedback capacitor ties the resistor junction to the output and the shunt capacitor grounds the amplifier input, so what is left is R1 from the input to the output node, with R2 and the amplifier's closed-loop output impedance both shunting that node to ground. The leak is the divider R2 ∥ ZOUT against R1, and ZOUT rises with frequency. SLOA049's fix is a 100 Ω and 47 nF at the output; that pole computes to 33.9 kHz.

When to use MFB instead

SLOA049’s own summary is short enough to quote whole:

MFB — Less sensitive to component variations and excellent high-frequency response. Less simplifications available to ease design

The MFB inverts, its gain is a resistor ratio rather than a fixed unity, and its capacitor constraint is stiffer: $C_1 \ge 4Q^2(1+K)C_2$ before the resistors come out as real numbers at all, with $C_1$ the shunt capacitor at the summing node as the calculator numbers it. The same labelling trap as the Sallen-Key applies here: SLOA049’s Figure 7-1 and its Equation 22 carry $C_1$ in the middle term of the denominator, and in an MFB only the feedback capacitor multiplies every resistor there, so the note’s $C_1$ is the feedback capacitor and its $C_2$ the shunt one. Read the constraint against the note with the subscripts swapped. In exchange the MFB gives a stop-band that keeps falling and lower sensitivity to every component.

Use MFB when the stop-band depth matters — an anti-aliasing filter whose whole job is what happens above the corner — or when the stage needs gain anyway. Use unity-gain Sallen-Key when the passband accuracy matters, when the signal must not be inverted, or when the design has to be obvious to whoever reads the schematic next.

The multiple-feedback low-pass topology: an inverting stage in which the input resistor, the feedback resistor and a capacitor to ground all meet at one node, a further resistor carries that node into the inverting input, and a second capacitor closes the loop from the inverting input to the output. — Fig 10 — The multiple-feedback alternative. It inverts, its gain is a resistor ratio, and no path survives both capacitors becoming shorts — which is why its stop-band keeps falling. The cost is a capacitor ratio that must satisfy C1 ≥ 4Q²(1 + K)C2 before the resistors come out as real numbers, with C1 the shunt capacitor as the calculator numbers it. SLOA049 numbers the MFB capacitors the other way round as well: its C1 is the feedback capacitor.

Checking the arithmetic against the source

SLOA049 publishes three worked Sallen-Key designs at a 1 kHz corner with every part rounded to a stock value. Putting those published component values back through the note’s own transfer function is the cheapest available check that the equations on this page are the equations in the document:

                R1, R2          C1, C2       target f0   built f0   target Q   built Q
Butterworth     4.22k, 18.4k    10n, 33n      1000 Hz     994 Hz     0.7071    0.7077
Bessel          7.23k, 14.5k    10n, 15n      1274 Hz    1269 Hz     0.5773    0.5771
3 dB Chebyshev  7.32k, 7.32k    10n, 68n       841 Hz     834 Hz     1.3049    1.3038

Every corner lands within 1 % and every Q within 0.1 %. That asymmetry is worth noticing: rounding to stock values costs frequency accuracy and almost no shape accuracy, because Q is a ratio and the ratio survives rounding far better than the absolute value does.

A table comparing the corner frequency and Q that SLOA049 intended for its three example circuits with the values recomputed from the standard component values it actually specifies, showing small errors from the rounding to stock parts. — Fig 11 — SLOA049's own Table 10-1 put back through the design equations. The note rounds every part to a stock value, and the resulting corner frequencies land within 1 % of target while every Q lands within 0.1 % — the rounding costs frequency accuracy, not shape, because Q is a ratio and ratios survive rounding better than absolutes.

The short version

Read FSF and Q for the shape you want; nothing else about the filter type reaches the circuit.
Pick the shape from the step response if the signal is a pulse, and from the magnitude response if it is a spectrum.
At unity gain, $C_1 = 4Q^2C_2$ sets the shape and $R$ sets the corner, and the two do not interact.
Keep the capacitor ratio under about 100:1 and both capacitors C0G.
Give the op amp at least $100\,K f_0 Q$ of gain-bandwidth.
Expect the stop-band to come back up, and add the output RC if it matters.
Higher orders are different stages, lowest Q first.

Work the values out in the active filter calculator, which carries the same equations, reports the gain-bandwidth the stage is asking for, and refuses the MFB capacitor combinations that have no real solution.