Passive Crossover Design Calculator

C1 equals 0.159 divided by R sub H times f; L1 equals R sub L divided by 6.28 times f
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Solution

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1st Order Butterworth Crossover

The simplest crossover design with a 6 dB/octave roll-off slope. Minimal phase shift but a wide driver overlap band.

C₁ = 0.159 / (R_H × f), L₁ = R_L / (6.28 × f)

2nd Order Crossover

12 dB/octave slope with multiple alignment choices: Butterworth, Linkwitz-Riley, Bessel, and Chebychev.

C = k_C / (R × f), L = k_L × R / f

3rd Order Butterworth Crossover

18 dB/octave roll-off for tighter frequency separation. Good for protecting tweeters from low-frequency power.

C₁ = 0.1061 / (R_H × f), L₁ = 0.1194 × R_H / f

4th Order Crossover

Steepest 24 dB/octave roll-off with excellent driver isolation. Available in Linkwitz-Riley, Bessel, Butterworth, Legendre, Gaussian, and Linear-Phase alignments.

C = k_C / (R × f), L = k_L × R / f

How It Works

A passive crossover network splits an audio signal into separate frequency bands for different speaker drivers. The crossover uses capacitors and inductors to create high-pass and low-pass filters. The filter order determines the roll-off slope: 1st = 6 dB/octave, 2nd = 12 dB/octave, 3rd = 18 dB/octave, 4th = 24 dB/octave.

Example Problem

Design a 1st-order Butterworth crossover at 3,000 Hz with R_H = 8 Ω (tweeter) and R_L = 4 Ω (woofer).

  1. Identify the knowns. Tweeter (high-pass) impedance R_H = 8 Ω, woofer (low-pass) impedance R_L = 4 Ω, target crossover frequency f = 3,000 Hz.
  2. Identify what we're solving for. We need the capacitor C₁ that feeds the tweeter and the inductor L₁ that feeds the woofer for a 1st-order (6 dB/octave) Butterworth network.
  3. Write the 1st-order Butterworth formulas: C₁ = 0.159 / (R_H × f) and L₁ = R_L / (6.28 × f). The 0.159 and 6.28 constants come from 1 / (2π) and 2π at the −3 dB point.
  4. Substitute the known values: C₁ = 0.159 / (8 × 3,000) and L₁ = 4 / (6.28 × 3,000).
  5. Simplify the arithmetic. C₁ = 0.159 / 24,000 = 6.625 × 10⁻⁶ F. L₁ = 4 / 18,840 = 2.1231 × 10⁻⁴ H.
  6. State the final results with units. **C₁ = 6.625 μF** and **L₁ = 0.212 mH**. Use the closest standard non-polarized film capacitor and air-core inductor values; tolerances within 5% will hold the designed response curve.

Higher-order networks produce more components with different coefficients.

Key Concepts

Passive crossover networks use frequency-dependent impedance of capacitors and inductors to split an audio signal into bands. The filter order (1st through 4th) determines roll-off steepness: 6, 12, 18, or 24 dB/octave. Higher orders provide better driver isolation but add complexity and component count. Different alignment types (Butterworth, Linkwitz-Riley, Bessel, Chebychev) trade off between flat amplitude response, phase behavior, and transient response.

Applications

  • Home audio: designing two-way and three-way passive speaker crossover networks
  • Car audio: building custom crossovers for component speaker systems in vehicles
  • Studio monitors: matching crossover design to driver specifications for accurate sound reproduction
  • PA systems: calculating component values for high-power passive crossovers in live sound equipment

Common Mistakes

  • Using nominal impedance instead of actual driver impedance at the crossover frequency — impedance varies with frequency and can differ significantly from the rated value
  • Choosing a crossover frequency outside the drivers' overlap range — both drivers must operate comfortably at the crossover point
  • Ignoring component tolerances — capacitor and inductor values in audio applications should be within 5% to maintain the designed response curve
  • Mixing crossover types between high-pass and low-pass sections — both filters must use the same alignment (e.g., both Linkwitz-Riley) for proper summation

Frequently Asked Questions

What is a passive crossover network?

A passive crossover network uses non-powered components (capacitors and inductors) to divide an audio signal by frequency. Unlike active crossovers, passive crossovers sit between the amplifier and the speaker drivers, requiring no external power supply.

What is the difference between Butterworth and Linkwitz-Riley crossovers?

A Butterworth crossover is −3 dB at the crossover frequency, meaning each filter passes half power at that point. A Linkwitz-Riley crossover is −6 dB at crossover, so when the outputs are summed acoustically, the result is perfectly flat.

How do I choose the right crossover frequency?

The crossover frequency should fall within the operating range of both drivers. Tweeter crossover frequencies are typically between 2,000 Hz and 5,000 Hz for two-way systems.

Why do the results show values in farads and henrys?

Farads (F) measure capacitance and henrys (H) measure inductance. In audio crossovers, typical values are in the microfarad (μF) range for capacitors and millihenry (mH) range for inductors.

Can I use this calculator for three-way speaker systems?

This calculator computes component values for a two-way crossover network. For a three-way system, design two separate two-way crossovers: one between the woofer and midrange, and another between the midrange and tweeter.

Why do steeper crossover slopes need more components?

Each filter order adds one reactive component per leg. A 1st-order high-pass uses one capacitor; a 4th-order uses two capacitors and two inductors in series-parallel arrangements. The extra components shape the response curve into the steeper roll-off.

Do I need to flip the tweeter polarity with an even-order crossover?

Yes — 2nd and 4th order Linkwitz-Riley alignments introduce a 180° phase shift between the high-pass and low-pass outputs at the crossover frequency. Reversing the tweeter wiring restores acoustic in-phase summation. Butterworth and 1st/3rd order networks have a 90° offset and typically do not require polarity reversal.

Can I scale the component values for a different impedance?

Capacitor values scale inversely with impedance and inductor values scale directly. Doubling impedance from 4 Ω to 8 Ω halves the required capacitance and doubles the required inductance at the same crossover frequency. Always recalculate using the actual driver impedance at the crossover point rather than the nominal rating.

Reference: Dickason, Vance. 1991. The Loudspeaker Design Cookbook. Audio Amateur Press. 4th ed.

Worked Examples

Two-Way Bookshelf — 1st Order Butterworth

What capacitor and inductor do I need for a 2.5 kHz tweeter-to-woofer split on 8-ohm drivers?

A simple two-way bookshelf monitor uses an 8-ohm dome tweeter and an 8-ohm 6.5-inch mid-woofer crossing at 2.5 kHz. A 1st-order Butterworth (6 dB/octave) network is the cheapest and least phase-disruptive option when the drivers blend well acoustically. Compute the series capacitor for the tweeter and the series inductor for the woofer.

  • Knowns: Rh = 8 Ω (tweeter), Rl = 8 Ω (woofer), f = 2500 Hz
  • C1 = 0.159 / (Rh × f) = 0.159 / (8 × 2500) = 0.159 / 20,000
  • C1 ≈ 7.95 × 10⁻⁶ F = 7.95 μF (high-pass cap in series with tweeter)
  • L1 = Rl / (6.28 × f) = 8 / (6.28 × 2500) = 8 / 15,700
  • L1 ≈ 5.10 × 10⁻⁴ H = 0.51 mH (low-pass coil in series with woofer)

Use 7.5 μF film cap + 0.50 mH air-core coil — closest standard values.

Round up the cap to 8.2 μF if the tweeter is sensitive in the 2–3 kHz range; rounding down protects the tweeter more aggressively but raises the acoustic crossover frequency.

Three-Way Mid-Tweeter — 2nd Order Linkwitz-Riley

Which Linkwitz-Riley LR2 values cross a mid driver into a tweeter at 3.5 kHz?

A three-way studio monitor uses a 4-inch midrange and a 1-inch tweeter, both 8 Ω, blending at 3.5 kHz. An LR2 (12 dB/octave) network gives a flat acoustic sum at the crossover frequency when both drivers are in-phase polarity flipped relative to each other. Compute the four passive components.

  • Knowns: Rh = 8 Ω, Rl = 8 Ω, f = 3500 Hz
  • Coefficients (Linkwitz-Riley 2nd-order): kC = 0.0796, kL = 0.3183
  • C1 = kC / (Rh × f) = 0.0796 / (8 × 3500) ≈ 2.84 μF (tweeter HP cap)
  • C2 = kC / (Rl × f) = 0.0796 / (8 × 3500) ≈ 2.84 μF (woofer LP cap)
  • L1 = kL × Rh / f = 0.3183 × 8 / 3500 ≈ 0.728 mH (tweeter HP coil to ground)
  • L2 = kL × Rl / f = 0.3183 × 8 / 3500 ≈ 0.728 mH (woofer LP coil)

Use 2.7 μF caps + 0.75 mH coils.

LR2 networks need the tweeter wired in reverse polarity; otherwise you'll get a 6 dB notch at the crossover frequency from the 180° phase relationship between Butterworth-aligned drivers.

Pro-Audio Subwoofer-to-Mid — 4th Order LR

What 4th-order LR4 components cross a 4-ohm subwoofer into a mid-bass at 120 Hz?

A pro-audio top cabinet hands off to a sub-bass system at 120 Hz. Both sub and mid-bass drivers are 4-ohm units; the 4th-order LR (24 dB/octave) crossover gives the steepest practical roll-off so the sub stays out of vocal range. Compute the four caps and four inductors per side.

  • Knowns: Rh = 4 Ω, Rl = 4 Ω, f = 120 Hz
  • LR4 cap coefficients: C1 = 0.0844, C2 = 0.1688 (on Rh side); C3 = 0.2533, C4 = 0.0563 (on Rl side)
  • LR4 ind. coefficients: L1 = 0.1000, L2 = 0.4501 (Rh side); L3 = 0.3000, L4 = 0.1500 (Rl side)
  • C1 = 0.0844 / (4 × 120) ≈ 175.8 μF
  • L4 = 0.1500 × 4 / 120 = 5.0 mH

Sub-mid LR4 at 120 Hz uses 4 caps (~117–528 μF) + 4 coils (3.3–15 mH) per side.

Passive LR4 networks at 4 Ω need very large coil values (15+ mH air-core costs serious money). Most pro-audio rigs do this crossover actively with DSP and bi-amp the cabinets — but the passive-LR4 math still drives the textbook design.

Passive Crossover Formulas

Each filter order has its own component-value formulas. The high-pass leg feeds the tweeter through a capacitor; the low-pass leg feeds the woofer through an inductor. Higher orders add cascaded LC stages with alignment-specific coefficients.

C₁ = 0.159 / (R_H × f)1st-order Butterworth — high-pass capacitor
L₁ = R_L / (6.28 × f)1st-order Butterworth — low-pass inductor
C = k_C / (R × f),   L = k_L × R / f2nd- and 4th-order — k_C, k_L depend on alignment (Butterworth, Linkwitz-Riley, Bessel, Chebyshev, Legendre, Gaussian, Linear-Phase)
C₁ = 0.1061 / (R_H × f),   L₁ = 0.1194 × R_H / f3rd-order Butterworth — first stage (additional C₂, C₃, L₂, L₃ stages follow)

Where:

  • C — series or shunt capacitance in farads (F); audio values usually appear in microfarads (μF)
  • L — series or shunt inductance in henries (H); audio values usually appear in millihenries (mH)
  • R_H — high-frequency driver impedance in ohms (tweeter, typically 4 or 8 Ω)
  • R_L — low-frequency driver impedance in ohms (woofer, typically 4 or 8 Ω)
  • f — crossover frequency in hertz (the −3 dB or −6 dB intercept depending on alignment)
  • k_C, k_L — alignment-specific dimensionless coefficients tabulated in loudspeaker design references

The 0.159 and 6.28 constants come from 1 / (2π) and 2π evaluated at the −3 dB intercept of a Butterworth response. The 2nd- and 4th-order coefficients shape the cutoff into Butterworth (maximally flat), Linkwitz-Riley (flat acoustic sum), Bessel (linear phase), or Chebyshev (ripple-trade-off) responses depending on the alignment.

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