Electrical & Electronics
Active Filter Calculator
Design low-pass, high-pass and band-pass filters with component selection and response plots.
Your inputs
Your results
- Pole / center frequency
- —
- Quality factor
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- Voltage gain
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- Cutoff frequency
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- Voltage gain
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- Phase shift
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- fL · −3 dB
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- fH · −3 dB
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- Resonance bandwidth
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—Distinguish filter frequency, cutoff and response
Analyse the displayed Sallen-Key low-pass/high-pass or multiple-feedback band-pass network. Compare natural frequency, Q and response at a chosen frequency. The model uses ideal amplifiers and nominal components; it does not include device bandwidth or tolerance.
Step by step
- Select the filter topology and analyse or design mode. Match resistor and capacitor labels to the circuit diagram.
- Enter components and gain, or target frequency and Q with preferred component series. Apply a design before inspecting its rounded values.
- Review the response plot, magnitude and phase. Compare natural frequency with cutoff or band edges rather than treating them as identical.
Settings and limits
- Frequency and Q
- Natural frequency locates the model's poles; Q describes damping. Low-pass/high-pass cutoff is the −3 dB point relative to passband gain and depends on Q.
- Topology and gain
- Low-pass/high-pass use the displayed Sallen-Key network. Band-pass uses an inverting multiple-feedback network; its component relationships and gain constraints differ.
- Rounding and validity
- Preferred values alter frequency and Q. Inspect achieved errors after applying them. Invalid gain/component combinations may imply non-positive damping and are rejected.
Worked example
Use the equal-component low-pass example below. At natural frequency, the response is below the passband by more than the cutoff definition.
Example input
LP; R1 = R2 = 10 kΩ; C1 = C2 = 10 nF
K = 1; f = f0| f0 / Q / fc / |G| / φ | ≈ |
|---|---|
| f0 | 1591.549431 Hz |
| Q | 0.5 |
| fc (−3 dB) | 1024.312067 Hz |
| |G(f0)| | 0.5 V/V |
| G(f0) | −6.020600 dB |
| φ(f0) | −90° |
The calculated cutoff is lower than natural frequency. This demonstrates why a component formula alone does not identify the −3 dB corner for every Q.
Questions and troubleshooting
Why is there a peak?
A sufficiently high Q can produce resonance. Compare the full response and passband gain; a peak is not a flat-response guarantee.
Why did the design miss my target?
Resistors and capacitors are rounded independently to preferred values. Check both achieved frequency and Q, and compare another supported series.
Does this predict the built filter?
It excludes amplifier gain-bandwidth, slew rate, supply limits and component tolerances. Check device suitability and verify the assembled circuit.