Sallen-Key Low-Pass Filter Calculator
This calculator designs and analyzes ideal second-order Sallen-Key low-pass filters using a single documented VCVS topology convention. It calculates natural frequency, Q, damping ratio, passband gain, response magnitude, phase, Butterworth gain, peaking, op-amp gain resistors, and sweep tables.
It is not a duplicate of the first-order active low-pass calculator. The first-order page sizes a simple RC stage; this page models a second-order active topology where Q depends on component ratios and non-inverting op-amp gain.
Engineering tool
Sallen-Key Low-Pass Filter Calculator
Design and analyze ideal second-order Sallen-Key low-pass filters, including natural frequency, Q, Butterworth gain, peaking, op-amp gain resistors, and frequency response.
First resistor in the adopted Sallen-Key denominator convention.
Second resistor in the adopted Sallen-Key denominator convention.
Capacitor C1 in the adopted Sallen-Key topology convention.
Capacitor C2 in the adopted Sallen-Key topology convention.
Frequency where the transfer function is evaluated.
Result console
- Natural frequency
- 1.591549kHz
- Q factor
- 0.70710656
- Damping ratio
- 0.707107
- Passband gain
- 1.585786
- Passband gain
- 4.0049dB
- Gain at frequency
- 1.12131929
- Gain at frequency
- 0.9946dB
- Normalized gain
- 0.70710631
- Normalized gain
- -3.0103dB
- Phase
- -90°
- Passband-relative -3 dB frequency
- 1.591549kHz
- Peak status
- No peaking
- Classification
- Butterworth response
This V1 model assumes an ideal op-amp with sufficient GBW, slew rate, input common-mode range, and output swing.
The natural frequency is not automatically the passband-relative -3 dB frequency unless the response is Butterworth.
Formula reference
Sallen-Key Low-Pass Formulas
Adopted topology convention: standard VCVS Sallen-Key low-pass with passband gain K = 1 + Rf/Rg. The transfer function denominator is derived from the stated R1, R2, C1, C2, and K convention.
H(s) = K / (a2s² + a1s + 1)a2 = R1R2C1C2a1 = C2(R1 + R2) + C1R1(1 - K)ω0 = 1 / √a2f0 = 1 / (2π√(R1R2C1C2))Q = √a2 / a1ζ = 1 / (2Q)Equal components: Q = 1 / (3 - K)Butterworth equal components: K = 3 - √2 ≈ 1.585786Gain dB = 20log10(|H|)Variable definitions
- R1
- First Sallen-Key resistor in the adopted denominator convention
- R2
- Second Sallen-Key resistor in the adopted denominator convention
- C1
- Capacitor C1 in the adopted topology convention
- C2
- Capacitor C2 in the adopted topology convention
- K
- Non-inverting passband gain
- f0
- Natural or pole frequency
- Q
- Quality factor
- ζ
- Damping ratio
- H(jω)
- Low-pass voltage transfer function
Worked Examples
Equal components f0
Known: R1 = R2 = 10 kΩ, C1 = C2 = 10 nF
f0 = 1/(2πRC) ≈ 1591.55 Hz.
Unity-gain equal components
Known: K = 1
Q = 1/(3 - K) = 0.5, not Butterworth.
Butterworth gain
Known: Q = 1/√2
K = 3 - 1/Q ≈ 1.58578644.
Butterworth verification
Known: K ≈ 1.58578644
General coefficient formula gives Q ≈ 0.70710678.
Butterworth frequency
Known: R = 10 kΩ, C = 10 nF
f0 ≈ 1591.55 Hz and f3dB ≈ f0 for Butterworth response.
K = 2 equal components
Known: R1 = R2, C1 = C2, K = 2
Q = 1, so response peaking is expected.
High-Q boundary
Known: K approaches 3 from below
Q becomes very large and component sensitivity increases sharply.
Invalid gain boundary
Known: K = 3 with equal components
The damping coefficient becomes zero and the normal stable design is rejected.
Solve R
Known: f0 = 1 kHz, C = 10 nF
R = 1/(2πf0C) ≈ 15.9155 kΩ.
Solve C
Known: f0 = 10 kHz, R = 10 kΩ
C = 1/(2πf0R) ≈ 1.59155 nF.
Solve Rf
Known: K ≈ 1.585786, Rg = 10 kΩ
Rf = (K - 1)Rg ≈ 5.85786 kΩ.
Low frequency
Known: f << f0
Absolute gain approaches K.
High frequency
Known: f >> f0
Gain approaches 0 for the low-pass response.
Butterworth at f0
Known: Q = 1/√2
Normalized magnitude is approximately 0.70710678.
Frequency unit equivalence
Known: 1000 Hz and 1 kHz
Both evaluate the same transfer function frequency.
Log sweep
Known: 100 Hz to 100 kHz, 50 points
The calculator generates 50 finite ordered response rows.
Unequal components
Known: R1, R2, C1, C2 not equal
f0, Q, and response are calculated from denominator coefficients.
Voltage dB
Known: Any response magnitude
Gain dB uses 20log10(|H|), not 10log10(|H|).
Sallen-Key Filter
A Sallen-Key stage is an active second-order filter using an op-amp as a VCVS gain element.
Natural Frequency
f0 depends on R1, R2, C1, and C2 through the product R1R2C1C2.
Q Factor
Q depends on component ratios and K. It is not controlled by f0 alone.
Butterworth Response
For the equal-component convention used here, Butterworth response requires K ≈ 1.585786.
Unity Gain
Equal R and C with unity gain gives Q = 0.5, so it is not Butterworth.
Peaking
Q above 1/√2 can create response peaking and ringing.
Op-Amp Limits
GBW, slew rate, output swing, common-mode range, noise, and bias current affect practical response.
Tolerance
Capacitor and resistor tolerance shift f0 and Q; high-Q designs are especially sensitive.
Common Mistakes
Support reference
FAQ
What is a Sallen-Key low-pass filter?
A Sallen-Key low-pass filter is an active second-order filter topology that uses two resistors, two capacitors, and a non-inverting op-amp gain stage.
How do I calculate its cutoff frequency?
This calculator first calculates the natural frequency f0 = 1/(2π√(R1R2C1C2)). The passband-relative -3 dB frequency is then solved from the second-order response and equals f0 only for Butterworth Q.
What is the natural frequency of a Sallen-Key filter?
For the adopted standard topology, natural frequency is f0 = 1/(2π√(R1R2C1C2)). It is also called the pole frequency.
How do I calculate Q?
The calculator uses denominator coefficients: a2 = R1R2C1C2 and a1 = C2(R1+R2)+C1R1(1-K), so Q = √a2/a1.
How does op-amp gain affect Q?
In the adopted Sallen-Key low-pass topology, increasing non-inverting gain K reduces the damping coefficient and raises Q. Too much gain can make the ideal model invalid.
What gain gives a Butterworth response?
For the equal-component case R1 = R2 and C1 = C2, Butterworth Q = 1/√2 requires K = 3 - √2, approximately 1.585786.
Why does unity gain with equal components not give Butterworth response?
With equal R and equal C, Q = 1/(3-K). If K = 1, Q = 0.5, which is more damped than the Butterworth value of 0.70710678.
What is the difference between f0 and -3 dB frequency?
f0 is the natural or pole frequency. The -3 dB frequency is defined relative to passband gain and depends on Q. They coincide for Butterworth response but not for every second-order response.
Why does a high-Q filter peak?
When Q exceeds 1/√2, the normalized second-order low-pass response can rise above the passband level near f0, creating peaking and possible ringing.
How do component tolerances affect Sallen-Key filters?
Q and f0 depend on component ratios, not only absolute values. Capacitor tolerance, resistor tolerance, and op-amp gain error can materially shift the real response.
How much op-amp bandwidth is required?
The required op-amp GBW depends on gain, Q, topology, signal amplitude, and phase-margin needs. Choose an op-amp with comfortable bandwidth margin and verify with datasheet guidance or simulation.
Can this calculator design unequal-component filters?
Yes. Analyze, natural-frequency, Q/gain, and sweep modes support unequal R1, R2, C1, and C2 values using the coefficient-based model.
What is the difference between this and an active RC low-pass calculator?
The existing active RC low-pass calculator is a first-order tool. This page is specifically for second-order Sallen-Key low-pass stages with Q, damping, peaking, and Butterworth design behavior.
Documentation
Related Engineering Resources
Guides and lookup references linked to this tool.
Engineering Guide
Active and Application-Specific Filter Topologies
Compare active, notch, PWM smoothing, anti-alias, and pi-filter implementations with loading, tolerance, and device-limit checks.
22 min · Intermediate
Engineering Guide
Butterworth Filter Design: Order, Poles, Stage Q, and Cascaded Verification
Design Butterworth filters from attenuation requirements through order, cutoff, normalized poles, stage Q, cascade implementation, tolerance, and verification.
18 min · Intermediate
Engineering Guide
Practical Filter Design: Cutoff Frequency, Q Factor, Bandwidth, and Damping
Learn cutoff frequency, -3 dB points, center frequency, bandwidth, Q factor, damping, filter order, passive and active tradeoffs, and real hardware limits.
18 min · Intermediate
Engineering References
Engineering Reference
Filter Response, Q, Bandwidth and Damping Reference
Lookup tables for filter response types, cutoff conventions, center frequency, bandwidth, Q, damping ratio, order, asymptotic slope, and approximation terminology.
14 min · Reference
Engineering Reference
Active Filter Topologies and Approximation Reference
Lookup for Sallen-Key, multiple-feedback, twin-T notch, Butterworth, Bessel, Chebyshev, elliptic, cascaded-stage, and ADC anti-alias filter terminology.
12 min · Reference
Planned Engineering Guides
Planned guide
Sallen-Key Filters Explained
Planned guide
Second-Order Low-Pass Filter Design
Planned guide
Butterworth Active Filter Design
Planned guide
Understanding Active Filter Q
Planned guide
Op-Amp Bandwidth in Active Filters
Planned guide
Component Tolerance in Sallen-Key Filters
Planned Filter Calculators
Multiple-Feedback Band-Pass Filter Calculator
Butterworth Filter Calculator
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Engineering Disclaimer
This calculator uses an ideal Sallen-Key low-pass model. It assumes ideal op-amp behavior, sufficient GBW and slew rate, no output saturation, no common-mode violation, ideal components, and no PCB parasitics. Validate critical designs with op-amp datasheets, SPICE, tolerance analysis, and bench measurement.