Sallen Key Filter Calculator: Formulas, Worked Examples & Component Sizing

If you need a second-order (-40 dB/decade) low-pass or high-pass roll-off without resorting to bulky, expensive inductors, the unity-gain Sallen-Key topology is your default active filter choice. A Sallen Key filter calculator relies on two core equations to determine the cutoff frequency and the Quality factor (Q). By manipulating these formulas, you can size standard E24 resistors and E12 capacitors to hit exact audio crossover or ADC anti-aliasing targets.

The Core Sallen-Key Low-Pass Formulas & Symbol Definitions

The standard unity-gain Sallen-Key low-pass filter uses two resistors, two capacitors, and an operational amplifier configured as a voltage follower (gain = 1). The governing equations for the cutoff frequency (fc) and the Quality factor (Q) are:

Cutoff Frequency:
fc = 1 / (2π √(R1 · R2 · C1 · C2))

Quality Factor:
Q = √(R1 · R2 · C1 · C2) / (R1C1 + R2C1)

Symbol Definition Table

When This Formula Applies (and Its Assumptions)

These exact formulas assume a unity-gain configuration (the op-amp's inverting input is tied directly to its output). If you are designing a non-unity gain Sallen-Key filter (where gain > 1 via a voltage divider on the inverting input), the denominator of the Q equation changes to include the gain term (3 - K). Furthermore, the math assumes an ideal op-amp with infinite Gain-Bandwidth Product (GBP) and zero output impedance. For high-frequency designs (>100 kHz), you must select an op-amp with a GBP at least 50 to 100 times higher than your target fc to prevent phase shift errors from altering your actual Q.

Standard Component Sizing for Audio Crossovers

Before running raw math, it helps to know what a realistic answer magnitude looks like. For audio-frequency Sallen-Key filters (20 Hz to 20 kHz), capacitor values typically fall between 1 nF and 470 nF, while resistors sit between 1 kΩ and 100 kΩ. If your calculator spits out 0.05 Ω or 4 Farads, you have made a unit conversion error.

Below is a data-dense reference table mapping common audio crossover targets to standard E12 capacitor and E24 resistor values. These assume a Butterworth response (Q ≈ 0.707) where possible, achieved by scaling the capacitor ratio.

Rearranged Forms & The Unit-Tracking Trap

When using a Sallen Key filter calculator in reverse—starting with a desired frequency and working backward to find components—you need the algebraically isolated forms. Assuming you fix three components to solve for the fourth, here are the rearranged equations based on the fc formula:

• Solve for R₁: R1 = 1 / ( (2π · fc)2 · R2 · C1 · C2 )
• Solve for R₂: R2 = 1 / ( (2π · fc)2 · R1 · C1 · C2 )
• Solve for C₁: C1 = 1 / ( (2π · fc)2 · R1 · R2 · C2 )
• Solve for C₂: C2 = 1 / ( (2π · fc)2 · R1 · R2 · C1 )

Which Unit Mistakes Break the Math?

The most common bench failure when calculating Sallen-Key values is the micro vs. nano trap. Capacitors in the audio range are usually labeled in nanofarads (nF) or microfarads (μF). If you enter 100 into a calculator assuming it means 100 nF, but the calculator expects base Farads, your output resistance will be off by a factor of 107. Always convert to base SI units before calculating:

• 1 nF = 1 × 10-9 F
• 1 μF = 1 × 10-6 F
• 1 kΩ = 1 × 103 Ω

For deeper topological variations and high-Q limitations, refer to the Electronics Tutorials Sallen-Key Guide or All About Circuits' active filter design breakdown.

Worked Example 1: Designing a 1 kHz Anti-Aliasing Filter

Scenario: You are feeding an audio signal into a 12-bit ADC sampling at 4 kHz. You need a 2nd-order low-pass anti-aliasing filter with a cutoff of fc = 1000 Hz. A Butterworth response isn't strictly necessary here; a simple equal-component design is acceptable to minimize BOM count.

Step 1: Choose Capacitors
Let's use equal capacitors for simplicity: C1 = C2 = 10 nF.
Convert to Farads: 10 nF = 10 × 10-9 F = 10-8 F.

Step 2: Choose Resistors
Let R1 = R2 = R. The cutoff formula simplifies to:
fc = 1 / (2π · R · C)

Step 3: Solve for R with Unit Tracking
R = 1 / (2π · fc · C)
R = 1 / (2π · 1000 Hz · 10-8 F)
R = 1 / (6283.18 · 10-8)
R = 1 / 0.0000628318
R = 15,915 Ω

Step 4: Select E24 Standard Value and Verify
The nearest E24 resistor is 16 kΩ (16,000 Ω).
Recalculate actual fc:
fc = 1 / (2π · 16000 · 10-8) = 994.7 Hz.
This is a 0.5% error, perfectly acceptable for an anti-aliasing stage. Note that because R and C are equal, Q = 0.5, yielding a critically damped, slightly rounded roll-off rather than a flat Butterworth passband.

Worked Example 2: Sizing a 150 Hz Butterworth Subwoofer Crossover

Scenario: You need a 150 Hz low-pass filter for a subwoofer amplifier. To ensure a maximally flat passband with no peaking at the cutoff, you must design for a Butterworth response, which requires Q = 0.707 (or 1/√2).

Step 1: Establish the Capacitor Ratio for Q = 0.707
To keep the math clean, let's set R1 = R2 = R. The Q formula simplifies to:
Q = 0.5 · √(C2 / C1)
Set Q = 0.707:
0.707 = 0.5 · √(C2 / C1)
1.414 = √(C2 / C1)
Square both sides: C2 / C1 = 2.
Therefore, C₂ must be exactly twice the value of C₁.

Step 2: Select Standard Capacitors
Let's choose C1 = 100 nF (100 × 10-9 F).
Then C2 = 200 nF (200 × 10-9 F).
*(Note: If 200 nF is unavailable, parallel two 100 nF C0G/NP0 film capacitors on the bench).*

Step 3: Solve for R with Unit Tracking
Using the full cutoff formula rearranged for R (where R1 = R2 = R):
R = 1 / (2π · fc · √(C1 · C2))
R = 1 / (2π · 150 · √(100 × 10-9 · 200 × 10-9))
R = 1 / (942.47 · √(2 × 10-14))
R = 1 / (942.47 · 1.414 × 10-7)
R = 1 / 0.0001332
R = 7,507 Ω

Step 4: Select E24 Standard Value
Fortunately, 7.5 kΩ is a standard E24 value. Using 7.5 kΩ for both R₁ and R₂ yields an actual cutoff of 149.8 Hz and a true Butterworth Q of 0.707.