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.
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.