Mastering the Butterworth Filter Equation: Sizing, Derivations, and Worked Examples

The Butterworth filter equation defines a "maximally flat" magnitude response with zero passband ripple, making it the default choice for audio crossovers, ADC anti-aliasing, and sensor signal conditioning. For an n-th order low-pass filter, the general magnitude response is governed by a single polynomial. However, in practical 2nd-order active Sallen-Key designs, this mathematical requirement forces the quality factor (Q) to exactly 0.7071 (or $1/\sqrt{2}$), which directly dictates your resistor and capacitor ratios.

Below is the complete derivation, symbol mapping, and unit-tracked calculation framework you need to move from a target cutoff frequency to a physical bill of materials.

The Core Butterworth Filter Equation and Symbol Definitions

The general magnitude response for an n-th order Butterworth low-pass filter is:

|H(jω)| = H₀ / √[1 + (ω / ωc)^(2n)]

When translating this to a physical 2nd-order active circuit (the Sallen-Key topology), we use the component-level equations for cutoff frequency (fc) and Quality Factor (Q):

fc = 1 / (2π √(R1·R2·C1·C2))
Q = √(R1·R2·C1·C2) / [R1·C1 + R2·C1 + R1·C2·(1 - K)]

For a true Butterworth response, Q must equal 0.7071. To simplify component sourcing, designers typically use a unity-gain configuration (K = 1) and set R1 = R2 = R. Substituting these into the Q equation reveals that C1 must equal 2 × C2.

Rearranged Forms for Component Sizing

Assuming the highly practical unity-gain, equal-resistor Sallen-Key topology (K=1, R1=R2=R, C1=2C2), let C2 = C. The cutoff equation simplifies to:

fc = 1 / (2π · R · C · √2)

Here are the rearranged forms to solve for any missing variable. Keep these on your bench:

• Solving for R: R = 1 / (2π · fc · C · √2)
• Solving for C (which is C2): C = 1 / (2π · fc · R · √2)
• Solving for fc: fc = 1 / (2π · R · C · √2)
• Finding C1: C1 = 2 · C

Worked Examples with Strict Unit Tracking

Abstract formulas fail on the bench when unit prefixes are ignored. Here are two real-world sizing problems with explicit unit tracking.

Problem 1: Audio Subwoofer Crossover at 80 Hz

Goal: Design a 2nd-order Butterworth low-pass filter with fc = 80 Hz. You have a kit of 100 nF (0.1 µF) C0G capacitors. Find R, C1, and C2.

1. Assign knowns: fc = 80 Hz. Let C2 (which is C) = 100 nF = 100 × 10⁻⁹ F.
2. Select formula: R = 1 / (2π · fc · C · √2)
3. Substitute with base SI units:
   R = 1 / (2 · 3.14159 · 80 Hz · 100 × 10⁻⁹ F · 1.4142)
4. Calculate denominator:
   2 · 3.14159 · 80 · 1.0 × 10⁻⁷ · 1.4142 = 0.00007108
5. Solve for R:
   R = 1 / 0.00007108 = 14,068 Ω
6. Select standard part: Choose the nearest 1% E96 resistor value: 14.0 kΩ.
7. Calculate C1: C1 = 2 · C2 = 2 · 100 nF = 200 nF (Use two 100 nF caps in parallel).

Problem 2: ESP32 ADC Anti-Aliasing at 15 kHz

Goal: The ESP32 SAR ADC samples at roughly 40 kHz. To prevent aliasing, we need a Butterworth filter at fc = 15 kHz. To limit thermal noise, we select R = 4.7 kΩ. Find C2 and C1.

1. Assign knowns: fc = 15,000 Hz. R = 4,700 Ω.
2. Select formula: C = 1 / (2π · fc · R · √2)
3. Substitute with base SI units:
   C = 1 / (2 · 3.14159 · 15000 · 4700 · 1.4142)
4. Calculate denominator:
   2 · 3.14159 · 15000 · 4700 · 1.4142 = 626,389
5. Solve for C (C2):
   C = 1 / 626,389 = 1.596 × 10⁻⁹ F = 1.596 nF
6. Select standard part: Choose nearest standard C0G cap: 1.5 nF (or parallel 1.0nF + 560pF for exactness).
7. Calculate C1: C1 = 2 · 1.596 nF = 3.19 nF (Select standard 3.3 nF).

When the Formula Applies (and When It Breaks)

The Butterworth filter equation assumes an ideal, linear, time-invariant system. In physical hardware, it applies perfectly only under these conditions:

• Op-Amp Gain Bandwidth Product (GBP): The op-amp's GBP must be at least 100 × fc × Q. For a 20 kHz filter, you need an op-amp with a GBP > 1.4 MHz. If the GBP is too low, the op-amp's internal phase shift destroys the Butterworth Q, causing peaking or early roll-off.
• Component Linearity: The capacitors must not exhibit voltage coefficient or microphonics.
• Parasitics: The formula ignores PCB trace capacitance and op-amp input capacitance. This is negligible at audio frequencies, but above 500 kHz, parasitic capacitance (often 2-5 pF) will shift your fc downward.

When it breaks: If you attempt to use this exact equation for a 5 MHz RF filter using standard through-hole components, lead inductance and stray capacitance will render the calculated values useless. For RF, you must use S-parameters and microwave simulation tools like Keysight ADS.

Common Unit Mistakes and Magnitude Sanity Checks

The most common reason a simulated filter fails on the breadboard is a unit prefix error. Watch for these traps:

Magnitude Sanity Check: What should your answer look like? For Audio (20 Hz - 20 kHz), realistic resistor values are 1 kΩ to 100 kΩ, and capacitors are 1 nF to 1 µF. For RF / High-Speed (> 1 MHz), resistors drop to 50 Ω - 600 Ω, and capacitors drop to 10 pF - 500 pF. If your audio filter calculation spits out a 2 Ω resistor and a 4,000 µF capacitor, you dropped a decimal or forgot a micro- prefix.

Decision Path: Selecting Your Filter Topology and Components

Use this decision tree to lock in your physical component choices based on your target frequency and application constraints.

For automated verification of these manual calculations, cross-check your final component values using the Analog Devices Filter Wizard or consult the foundational theory at Electronics Tutorials to ensure your specific op-amp's input capacitance isn't skewing your high-frequency response.