100nF

Active low-pass filter calculator

A 1 kHz Butterworth Sallen-Key with a 10 nF C2 needs C1 = 20 nF and R = 11.25 kΩ. The nearest E24 value, 11 kΩ, moves the corner to 1023 Hz. Q stays 0.707, because in TI SLOA049's equations only the capacitor ratio sets it. The ideal-op-amp model holds while the GBW exceeds 100·K·f0·Q, 71 kHz here.

Second-order low-pass with an op amp: pick the corner, the shape and one capacitor, and get the resistors, what the E24 values actually give you, the peaking to expect, and how fast the op amp has to be.

1.0e+2 Hz1.0 kHz10 kHz1.0e+2 kHz0 dB-20 dB-40 dB-60 dB
Fig 1 — Sallen-Key: f0 1.0 kHz, Q 0.707, −40 dB/decade above the knee.
C1 (junction to output)
20.0 nF
C2 (+ input to ground)
10.0 nF
R1 = R2, exact
11.3 kΩ → E24 11.0 kΩ
With E24: f0 / Q
1.02 kHz · 0.707
Designed f0 / Q
1.00 kHz · 0.7071
Response at fc = 1.00 kHz
-2.82 dB
Op amp GBW, at least
72.3 kHz

Flat passband, −3 dB at fc.

How this is calculated

Standard: TI SLOA049 (active low-pass filter design); TI SBOA093

ω0=1RC1⋅C2,Q=C1/C22\omega_0 = \frac{1}{R \sqrt{C_1 \cdot C_2}}, \quad Q = \frac{\sqrt{C_1 / C_2}}{2}
Sallen-Key at unity gain, with R1 = R2 = R.
C1=4Q2C2,R=12πf0C1C2C_1 = 4 Q^2 C_2, \quad R = \frac{1}{2\pi f_0 \sqrt{C_1 C_2}}
The same two relations rearranged into the parts to fit, given a corner frequency and the capacitor you already have.
ω0=1R2R3C1C2\omega_0 = \frac{1}{\sqrt{R_2 R_3 C_1 C_2}}
Multiple-feedback, with inverting gain K = R_2 / R_1.
Q=R2R3C1C2C2(R2+R3+R2R3R1)Q = \frac{\sqrt{R_2 R_3 C_1 C_2}}{C_2 \left(R_2 + R_3 + \frac{R_2 R_3}{R_1}\right)}
C1≥4Q2(1+K)C2C_1 \ge 4 Q^2 (1 + K) C_2
The multiple-feedback capacitor ratio needed for the resistors to come out real.
fpeak=f01−12Q2,∣H∣peakK=Q1−14Q2f_{peak} = f_0 \sqrt{1 - \frac{1}{2Q^2}}, \quad \frac{|H|_{peak}}{K} = \frac{Q}{\sqrt{1 - \frac{1}{4Q^2}}}
Where the response peaks and how high, for Q above 0.7071. Below that it does not peak.
GBW≥100⋅K⋅f0⋅Q\text{GBW} \ge 100 \cdot K \cdot f_0 \cdot Q
Gain-bandwidth the op amp needs so that its own roll-off does not move the corner.
∣H(f)∣=K(1−(ff0)2)2+(ff0Q)2|H(f)| = \frac{K}{\sqrt{\left(1 - \left(\frac{f}{f_0}\right)^2\right)^2 + \left(\frac{f}{f_0 Q}\right)^2}}
Second-order low-pass magnitude. Q is 0.7071 for Butterworth, 0.5773 for Bessel and 0.9565 for Chebyshev with 1 dB of ripple.

Assumptions

What sets a Sallen-Key or MFB filter's corner and Q

A second-order low-pass around one op amp, from a corner frequency, a response shape and a capacitor you already have. Sallen-Key is the unity-gain, equal-resistor form: R1 = R2 = R, C1 from the resistor junction to the op-amp output, C2 from the + input to ground. MFB (multiple feedback) is the inverting form with gain K = R2/R1, C1 to ground at the summing node and C2 in the feedback.

Both topologies, the capacitor ratio the multiple-feedback form needs for its resistors to come out real, the magnitude response, the peak a high Q puts just below f0, and the gain-bandwidth the op amp has to have are all under How this is calculated above. Where the ratio bound applies the tool fits 1.1 times it, so the arithmetic does not land on the boundary.

The shape sets Q and where the pole f0 sits relative to the corner you asked for. Butterworth: Q 0.7071, f0 = fc. Bessel: Q 0.5773, f0 = 1.2736 fc. Chebyshev 0.5 dB: Q 0.8637, f0 = 1.2313 fc. Chebyshev 1 dB: Q 0.9565, f0 = 1.05 fc. For Butterworth and Bessel fc is −3 dB; for Chebyshev it is the ripple-band edge. Equations from TI SLOA049. The tool rounds resistors to E24 and reports the f0 and Q you actually get.

Worked example: a 1 kHz Butterworth from a 10 nF capacitor

Sallen-Key, 1 kHz Butterworth, starting from a 10 nF C0G on the reel.

Q = 0.7071, f0 = 1000 Hz
C1 = 4 × 0.7071² × 10 nF = 20.0 nF
R  = 1 / (2π × 1000 × √(20n × 10n)) = 1 / (2π × 1000 × 14.14n) = 11.25 kΩ
E24: 11 kΩ  →  f0 = 1 / (2π × 11k × 14.14n) = 1023 Hz,  Q = ½ √(20/10) = 0.707 (unchanged)

Q depends only on the capacitor ratio, so resistor rounding moves the corner 2 % and leaves the shape alone. Ask for Bessel at the same 1 kHz and the tool puts the pole at 1273.6 Hz, Q 0.5773: C1 = 13.3 nF, R = 10.8 kΩ. The soft Bessel roll-off needs the pole up there for −3 dB to land on 1 kHz.

MFB, 1 kHz Butterworth, gain 2, from C2 = 1 nF.

min C1 = 4 × 0.7071² × (1 + 2) × 1 nF = 6.0 nF      tool uses 1.1× = 6.6 nF

a  = 1 / (2π × 1000)      = 1.5915e-4
p  = a² / (C1 C2)         = 2.533e-8 / 6.6e-18 = 3.838e9      (= R2 R3)
b  = a / (Q C2)           = 1.5915e-4 / 0.7071e-9 = 2.2508e5  (= R2 + R3 (1+K))
disc = √(b² − 4 (1+K) p)  = √(5.066e10 − 4.606e10) = 6.786e4
R3 = (b − disc) / (2 (1+K)) = 1.5722e5 / 6 = 26.2 kΩ
R2 = p / R3 = 3.838e9 / 26 203 = 146.5 kΩ
R1 = R2 / K = 73.2 kΩ

E24: R1 75 kΩ, R2 150 kΩ, R3 27 kΩ
  f0 = 1 / (2π √(150k × 27k × 6.6n × 1n)) = 973 Hz
  Q  = 0.708,  gain = 150/75 = 2.00
GBW wanted: 100 × 2 × 1000 × 0.707 = 141 kHz

The 6 nF minimum is the discriminant reaching zero; the 10 % back-off keeps tolerance from making the design unsolvable. A bigger C1 spreads the resistors apart, which is fine until R3 gets small enough to load the source.

Where the Sallen-Key and MFB equations stop being valid

The op amp is part of the filter. Once loop gain at f0 drops toward the section's own gain, the pole moves and Q rises. 100·K·f0·Q keeps the error near 1 %; at a tenth of that the shape is visibly off. High-Q sections at tens of kHz want megahertz op amps.

Sallen-Key comes back up. C1 dumps the stop-band signal into the op-amp output, and the output is not a short: its closed-loop impedance rises with frequency as loop gain falls. The response flattens out a decade or two above fc and then rises again; floors of 40 to 60 dB are typical. MFB's feedback capacitor works into a virtual ground and keeps falling until parasitics take over, which is why it is the better anti-alias filter when the aliasing band is far above fc.

Sensitivity. Unity-gain SK sets Q by the capacitor ratio and barely notices the resistors. Add gain (the tool does not) and the equal-component form has Q = 1/(3 − K): K of 2.9 is Q 10, K of 3 oscillates, and 1 % in the gain leg is 10 % in Q. MFB spreads the sensitivity over resistor ratios instead, so nothing blows up but every 1 % resistor contributes, and C1/C2 ratios of 10 or more pull the two capacitors from different decades.

Fourth order is not two of these. Two Butterworth sections in series give Linkwitz-Riley, −6 dB at fc. A true fourth-order Butterworth is two sections at the same f0 with Q 0.5412 and 1.3066; Bessel and Chebyshev spread f0 as well. Take the pair from the pole tables in SLOA049, run the tool once per section, and put the higher-Q section last so it does not clip on the peaking.

Common active filter mistakes

Further reading