Multiple-Feedback (MFB) Filter Calculator
Multiple-Feedback (MFB) Filter Calculator
Second-order low-pass or band-pass in the inverting multiple-feedback topology: the three resistors and two capacitors for the cut-off or centre frequency, Q and gain you want, the nearest E-series parts, the response those real parts actually give, and the op-amp speed it needs.
Multiple-feedback section
a 1 kHz Butterworth low-pass at a gain of −2, seeded with a 10 nF capacitor and rounded to E24, with a 10 MHz op-amp
The multiple-feedback section
needs C1/C2 ≥ 4Q²(1 + |K|)
band-pass (C1 = C2 = C): H(s) = −(s/R1C) ÷ (s² + 2s/(R3C) + (1/R1 + 1/R2)/(R3C²)), Q = ω₀R3C/2, |K| = R3/2R1, needs 2Q² > |K|
- K
- the passband gain. Negative — the MFB inverts, which is the price of everything else it does well
- Q
- 0.7071 for a Butterworth low-pass, 0.5774 Bessel, 0.8637 for a 0.5 dB Chebyshev; for a band-pass it is f₀ divided by the −3 dB bandwidth and you choose it
- C1, C2
- in the low-pass, C1 is the grounded capacitor at the R-node and C2 is across the amplifier. TI’s SLOA049 numbers them the other way round; the circuit is the same
- 4Q squared
- the capacitor ratio that makes the design equations have a real solution at all. It is why a high-Q MFB low-pass is impractical
Worked example
a 1 kHz Butterworth low-pass at a gain of −2, seeded with a 10 nF capacitor and rounded to E24, with a 10 MHz op-amp
Butterworth second order is Q = 0.7071 and f₀ = fᴄ, so ω₀ = 6,283.2 rad/s
The ratio rule: C1 ≥ 4Q²(1 + K)C2 = 4 × 0.5 × 3 × 10 nF = 60 nF, so the next E12 value above a 25% margin is 82 nF
Solving (K+1)/R2² − (ω₀C1/Q)/R2 + ω₀²C1C2 = 0 gives R2 = 5.425 kΩ, R1 = R2/K = 2.712 kΩ and R3 = 1/(ω₀²C1C2R2) = 5.694 kΩ
Nearest E24 parts: R1 = 2.7 kΩ, R2 = 5.6 kΩ, R3 = 5.6 kΩ
Those give f₀ = 992.5 Hz, Q = 0.7029 and a gain of 2.074, so the real −3 dB point is 986.5 Hz — -1.35% low
The op-amp needs about 100 × gain × fᴄ = 204.6 kHz; 10 MHz clears it comfortably
What each Q and gain costs an MFB low-pass
| Case | Q | Gain | Capacitor ratio C1/C2 needed | Sallen-Key gain for the same Q | Notes |
|---|---|---|---|---|---|
| Butterworth low-pass, Q = 0.7071 | 0.7071 | 1.0 | 4.0:1 | 1.586 | The everyday case. MFB and Sallen-Key are equally good here; pick on whether you want the inversion. |
| Butterworth low-pass at a gain of 10 | 0.7071 | 10.0 | 22.0:1 | 1.586 | MFB gets the gain for free — it is R2/R1. A Sallen-Key needs a separate gain stage or the equal-component form, whose gain is not yours to choose. |
| Chebyshev 0.5 dB, Q = 0.8637 | 0.8637 | 1.0 | 6.0:1 | 1.842 | Still comfortable in either topology. |
| A Q of 3 section | 3.0000 | 1.0 | 72.0:1 | 2.667 | Here the capacitor ratio starts to bite: MFB needs C1/C2 ≥ 4Q²(1+K), which is 72 at Q = 3 and unity gain. |
| A Q of 10 section | 10.0000 | 1.0 | 800.0:1 | 2.900 | C1/C2 ≥ 800. This is where a single MFB section stops being practical and a band-pass arrangement or a state-variable filter takes over. |
When multiple feedback beats Sallen-Key
The multiple-feedback section is the other standard way to build a second-order filter round one op-amp. Where a Sallen-Key puts the amplifier in a follower and feeds signal forward into it, the MFB runs the amplifier as an inverter with two feedback paths from the output — one to the resistor node, one to the virtual ground. The consequence is the first thing to know about it: it inverts. If the sign matters and you have only one op-amp, that settles the question. If you have two, or the sign does not matter, the rest of this page is about why the MFB is often the better choice.
Three reasons to prefer it. First, the gain is free and independent: it is R2/R1 in the low-pass and R3/2R1 in the band-pass, chosen without touching Q. The equal-component Sallen-Key cannot do that — there the gain IS the Q, fixed at 3 − 1/Q, so a Butterworth section is stuck at 1.586. Second, the MFB is less sensitive to component tolerance at high Q: the Sallen-Key’s Q depends on the amplifier’s gain through 1/(3 − K), which runs away as K approaches 3 — at Q = 10 a 1% error in one gain resistor moves Q by about 20%. Third, and most practically, a Sallen-Key low-pass has a floor: above the frequency where the op-amp runs out of open-loop gain, the feedback capacitor stops being a short to a low-impedance output and the signal leaks through it, so the stop-band attenuation stops improving and eventually comes back up. The MFB’s output falls away properly, because the amplifier’s output is the only path to the load.
And the reason not to. The MFB low-pass needs its grounded capacitor to be at least 4Q²(1 + |K|) times the other one, or the design equations have no real solution at all. At Q = 0.707 and unity gain that is 4:1 and costs nothing; at Q = 3 it is 72:1; at Q = 10 it is 800:1, and you are looking for a 10 nF and an 8 µF capacitor in the same filter. So the low-pass form is for low and moderate Q, and a high-Q section wants either the band-pass arrangement, where no such constraint exists, or a state-variable topology. The band-pass has its own, milder limit — the gain at the centre must be less than 2Q², which is rarely binding.
Reading the answers. As on the Sallen-Key page, the headline is what the REAL parts give, not what you asked for: the resistors are rounded to the E-series, and the page then re-derives f₀, Q and gain from the rounded values and reports the response those produce. That is usually a per cent or two away from the target, and the honest way to close the gap is to nudge the seed capacitance until the resistors land near preferred values rather than to move to E96 blindly. The op-amp check is the same 100 × gain × frequency rule — verified here by solving the circuit with a one-pole amplifier model rather than by assertion: at exactly that gain–bandwidth product the response at the corner is within 0.7% of the ideal, and at a tenth of it the error is 8.7%.
To decide what order and corner frequency you need in the first place — which is a different question from how to build the section — see the anti-aliasing filter calculator. For a passive first-order RC or a series RLC, the RC and RLC filter calculator; for the amplifier itself, the inverting amplifier calculator, which does the closed-loop bandwidth and output-swing checks on the same stage. One assembly note: the MFB’s inverting input is a virtual-ground summing node with several components on it, so it is the highest-impedance, most noise-sensitive point in the circuit. Keep it small on the board.
Frequently asked questions
What is a multiple-feedback filter?
A second-order active filter built round one inverting op-amp with two feedback paths from the output — one to the resistor node and one to the inverting input. It gives a low-pass, band-pass or high-pass section with three resistors and two capacitors, and it inverts the signal.
MFB or Sallen-Key?
MFB when you want gain and Q chosen independently, when Q is above about 2, or when the stop-band depth matters — a Sallen-Key’s attenuation floors out where the op-amp runs out of gain. Sallen-Key when you need a non-inverting stage, unity gain with no gain resistors, or a high Q in a low-pass, where the MFB’s capacitor ratio becomes impractical.
Why does the MFB low-pass need such a big capacitor ratio?
Because the design equations only have a real solution when C1/C2 ≥ 4Q²(1 + |K|). At Q = 0.707 and unity gain that is 4:1 and harmless; at Q = 3 it is 72:1 and at Q = 10 it is 800:1. It is the one real disadvantage of the topology, and it is why high-Q sections use the band-pass form or a state-variable filter.
Does the MFB filter invert the signal?
Yes, always — the op-amp is used as an inverter, so the passband gain is negative. In a chain of two sections the inversions cancel; on its own, add an inverting buffer if the sign matters, or use a Sallen-Key.
How do I set the gain of an MFB band-pass without changing its Q?
Change R1. In the band-pass form the centre gain is R3 ÷ 2R1, the Q is ω₀R3C ÷ 2 and the centre frequency depends on R1, R2 and R3 together — so R3 sets Q, R1 sets the gain, and R2 is then used to bring the centre frequency back where you want it.
What op-amp does an MFB filter need?
The same rule of thumb as a Sallen-Key: a gain–bandwidth product of at least 100 × passband gain × the filter’s frequency. A 1 kHz section at a gain of 2 needs 200 kHz; a 100 kHz section at a gain of 10 needs 100 MHz. Below that the corner sags and the Q drifts from the design value.
Related calculators
References
- Karki J. Active Low-Pass Filter Design. Texas Instruments application report SLOA049 (rev. A, October 2000; rev. B, September 2002). The multiple-feedback section and its design procedure, including the capacitor-ratio constraint. Its transfer function is the one this page uses — the fetched text gives the denominator as (j2πf)²(R2R3C1C2) + j2πf(R3C1 + R2C1 + R2R3C1/R1) + 1 with a DC gain of −R2/R1, which is algebraically identical to the form here with the two capacitors’ labels exchanged.
- Karki J. Analysis of the Sallen-Key Architecture. Texas Instruments application report SLOA024B, and Analog Engineer’s Circuit: single-supply, second-order Sallen-Key low-pass filter, SBOA226, which states the op-amp requirement as GBW = 100 × gain × fᴄ. This page applies the same rule and verifies its consequence numerically.
- The design equations are checked in this build against a nodal solve of the actual circuit — three complex node equations with the op-amp as a one-pole gain block — at six frequencies per design across five component sets, agreeing in magnitude and phase to one part in 50,000. The band-pass’s −3 dB edges were located on that simulated response and match f₀(√(1 + 1/4Q²) ± 1/2Q) exactly.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015, chapter 6. The MFB and VCVS realisations side by side, and the general point that the MFB’s inverting summing node is the sensitive spot in the layout.