Active Filter Calculator

Filter Type
Response
2
Hz
Topology
V/V
Stage Count1
Deviation at Cutoff0.00 dB

Bode Plot (Magnitude)

142545791583166311k2k4k7k14k32k100k-80 dB-60 dB-40 dB-20 dB0 dB

Component Values per Stage

StageR1 (exact)R1 (E96)R2 (exact)R2 (E96)C1C2
12.25 kΩ2.26 kΩ1.13 kΩ1.13 kΩ100 nF100 nF

Active filters use op-amps with resistors and capacitors to shape a frequency response without bulky inductors. This designer takes a target response (Butterworth, Chebyshev, or Bessel), an order, and a cutoff frequency, computes the pole locations, and synthesizes each second-order stage as a Sallen-Key or multiple-feedback (MFB) circuit, snapping the resistor values to E96 and E24 standard parts.

Formula

R2 = Q / (2π·f·C), R1 = 1 / (2π·f·C·Q·K) (Sallen-Key, C1 = C2 = C)

f
Stage natural frequency = normalized pole frequency × cutoff
Q
Stage quality factor from the pole location
C
Chosen stage capacitor (1 nF to 1 µF depending on cutoff)
K
Passband gain of the stage
R1, R2
Computed resistors, then snapped to E96/E24 standard values

How it works

  1. Choose the response type, filter order, cutoff frequency, topology (Sallen-Key or MFB), passband gain, and—for Chebyshev—the allowable ripple in dB.
  2. The engine places the normalized poles (evenly spaced on the unit circle for Butterworth, on an ellipse for Chebyshev, from a lookup table for Bessel) and groups them into one stage per conjugate pole pair, each with its own natural frequency and quality factor Q.
  3. For each stage it picks a convenient capacitor for the frequency band, solves for R1 and R2 from the Q and stage frequency, snaps them to the nearest E96 and E24 values, and generates Bode magnitude/phase data across four decades around the cutoff.

Worked example

A 2nd-order Butterworth low-pass filter at 1 kHz, Sallen-Key topology, unity gain.

  1. A 2nd-order filter has one conjugate pole pair, so it is realized in a single stage with Q ≈ 0.707.
  2. For a 1 kHz cutoff the engine selects C = 100 nF (the 100 Hz–1 kHz band).
  3. R2 = Q / (2π·f·C) = 0.707 / (2π·1000·100e-9) ≈ 1125 Ω; R1 = R2 / Q² relation gives ≈ 2251 Ω.
  4. Snapping to E96: R1 ≈ 2.26 kΩ, R2 ≈ 1.13 kΩ.

One stage: R1 ≈ 2251 Ω (E96 2.26 kΩ), R2 ≈ 1125 Ω (E96 1.13 kΩ), C1 = C2 = 100 nF, gain 1, with a flat Butterworth −3 dB point at 1 kHz.

Frequently asked questions

What is the difference between Butterworth, Chebyshev, and Bessel responses?
Butterworth gives the flattest passband with no ripple. Chebyshev trades passband ripple for a steeper roll-off near cutoff. Bessel sacrifices roll-off sharpness to preserve a linear phase (constant group delay), which best maintains the shape of pulse and audio waveforms.
When should I use Sallen-Key versus MFB topology?
Sallen-Key is a non-inverting stage that is simple and stable for low to moderate Q and unity or modest gain. MFB (multiple feedback) inverts the signal and generally offers better high-Q performance and lower sensitivity to component tolerances, at the cost of an inverted output.
Why are component values snapped to E96 and E24?
Resistors are sold in standardized decade series. E24 (5% parts, 24 values per decade) and E96 (1% parts, 96 values per decade) are the most common, so the tool rounds the ideal calculated resistors to the nearest stock value you can actually buy.
Why does a higher-order filter use multiple stages?
Each Sallen-Key or MFB stage realizes one second-order (or first-order) section. An order-4 filter therefore needs two cascaded stages, an order-6 needs three, and so on, with each stage tuned to its own pole-pair frequency and Q.