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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
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What does a crossover do?

A crossover divides the audio band between drivers and also shapes phase, level and sometimes protection. Acoustic slopes matter more than component labels alone.

Give each driver a suitable job

A woofer and tweeter have different strengths. A crossover limits their operating ranges and often adjusts sensitivity or response irregularities. There is a broad overlap around the chosen crossover frequency. Within it, both drivers radiate, so level, phase, timing and spatial separation all matter. The goal is one coherent loudspeaker, not merely two operating drivers.

Passive and active arrangements

A passive network operates after the power amplifier. Its inductors, capacitors and resistors interact with the actual driver impedance and may dissipate power. An active analogue or DSP crossover divides the signal before separate power-amplifier channels. Active control makes adjustments easier, but still requires suitable drivers, correct gain structure and acoustic verification.

The electrical curve is only part of the result

A driver already has its own amplitude and phase response. The baffle and mounting geometry add further effects. A 24 dB/oct electrical filter therefore does not guarantee a 24 dB/oct acoustic slope. For development, measure each driver in the intended enclosure, preserve consistent reference conditions and use its real electrical and acoustic data.

Why two quieter branches can sum flat

For ideal aligned fourth-order Linkwitz–Riley acoustic responses, each branch is about 6 dB down at crossover. Their complex sound pressures can still add to flat magnitude. Do not add decibel values or ignore phase. Two equally strong contributions arriving in opposition can instead produce a deep cancellation. Raising one driver does not resolve the underlying alignment problem.

Listen beyond one point

Vertically separated drivers have angle-dependent path differences. A flat on-axis sum can coexist with pronounced vertical cancellations. Directivity should also match reasonably through the transition: a narrowing woofer handing over to a much wider tweeter changes the energy sent into the room. A coaxial arrangement changes the geometry but does not automatically eliminate all acoustic issues.

Choose a safe overlap region

The crossover frequency is not a universal number inferred from diameter alone. Check tweeter resonance, excursion and recommended operation, and check woofer breakup and directivity. A shallow filter may leave substantial out-of-band energy. Electrical filter order, acoustic slope and permissible operating level belong in the same decision.

Develop, measure and repeat

Start with suitable drivers and a test enclosure. Define an acoustic target, adjust filtering and levels, then measure individual branches, the combined response and several angles. For passive designs also inspect impedance. Save intermediate states and level-match listening tests. A successful crossover usually becomes unobtrusive precisely because considerable work went into it.

Explore the relationship
Interactive experiment

Two ways, one complex sum.

Start at 0 ms: two ideal LR4 branches give flat summed magnitude. Add a little delay to the high-frequency branch and watch the sum.

Model limits: Ideal filters and equal-sensitivity ideal drivers at one observation point. No baffle, directivity or real driver impedance.

A closer look · model and equation

Here LR4 uses two cascaded second-order Butterworth filters per branch. Each branch is −6.02 dB at fc. Summation is complex, including phase; extra delay multiplies the HF branch by exp(−jωt).

Workshop takeaway

Component values are the beginning. The acoustic branches and their relative phase determine whether two drivers become one coherent loudspeaker.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

06 · An ideal series inductor feeds a purely resistive 8 Ω substitute driver. Target first-order electrical low-pass at 2 kHz. Find L.

Show worked answer
  1. At cutoff, inductive reactance equals R.
  2. L = R/(2πfc).
  3. L = 0.637 mH.
Calculation
L=\frac{8}{2\pi\cdot2000}=0.637\,mH

07 · An ideal series capacitor feeds 8 Ω. Target first-order electrical high-pass at 2 kHz. Find C.

Show worked answer
  1. At cutoff, capacitive reactance magnitude equals R.
  2. C = 1/(2πfcR).
  3. C = 9.947 µF.
Calculation
C=\frac{1}{2\pi\cdot2000\cdot8}=9.947\,\mu F

09 · A speaker has 8 Ω nominal impedance. Why is task 6 not a complete real crossover design?

Show worked answer
  1. Impedance varies with frequency and is complex.
  2. The acoustic driver response also contributes.
  3. Include measured magnitude, phase and mounting geometry.

19 · Two ideal acoustic LR4 branches are each -6.02 dB at crossover and in phase. What is their sum?

Show worked answer
  1. Each branch has pressure ratio 0.5.
  2. Add in phase: 0.5 + 0.5 = 1.
  3. Sum = 0 dB.
Calculation
20\log_{10}(0.5+0.5)=0\,dB

Match drivers together →

Project connection · LNGR 2-PR-THT-PR

See how driver spacing, passive radiators, crossover design and measurements meet in this planning project. CNC geometry v1.2; not yet built or acoustically measured.

Open the project build log →

Further technical reading

Colloms & Darlington (2005): High Performance Loudspeakers, 6th ed. Chapter 6 (Systems and Crossovers), starting on printed page 263. Source information (COLLOMS-2005-6) →