Skip to content
LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
LNGR / Knowledge

Sealed, bass reflex or passive radiator?

Three common enclosure concepts solve the low-frequency problem in different ways. The right choice depends on the driver, target bandwidth, available volume, excursion and practical construction.

Why the cone needs an enclosure

Front and rear radiation from a bare cone tend to oppose each other at low frequencies. A cabinet separates the paths and gives the rear radiation a defined acoustic load. The trapped air also acts like a spring. The cabinet is therefore part of the system, not a decorative box fitted after the driver has been chosen.

Sealed: an additional air spring

A sealed cabinet adds stiffness to the driver suspension. Changing volume alters resonance and damping. It is mechanically straightforward, but not limitless: deep output still demands excursion and amplifier capability. Electronic extension is possible only within those reserves. A higher-Q alignment can create a response peak; a lower-Q response is more gradual but is not automatically preferable in every installation.

Vented: use a tuned air mass

A vent couples its air mass to the enclosure compliance. Near tuning it can contribute substantial output while active-cone movement is reduced. Below the useful resonant region, acoustic output falls and cone excursion can become problematic. The design must therefore consider tuning, port airflow and any protective filtering together.

A bigger port needs more space

For a given volume flow, a larger port area reduces average air velocity. At unchanged cabinet volume and low tuning, it also generally requires a longer effective port. That length occupies space and can introduce construction and resonance problems. A slot port follows the same principles; narrow gaps, bends, sharp edges and obstructed ends are not free solutions.

A passive radiator replaces the air column

A passive diaphragm responds to pressure inside the cabinet and supplies a moving mass without an open port. Its own suspension compliance and losses make it more than a literal substitute for an ideal tube. It can solve an impractically long port, but adds cost, mounting area and another excursion limit. Effective area and usable stroke both matter.

Is sealed automatically faster?

Descriptions such as tight or slow often mix frequency balance, room decay, crossover alignment and system limits. Resonant alignments differ in time and phase behaviour, but one group-delay figure does not explain every listening impression. Compare complete designs at similar level and response, and check whether unwanted noises or compression appear at the intended operating level.

Compare the complete compromises

Set usable size, low-frequency target and required level before choosing the alignment. Deduct drivers, bracing, port and electronics from internal volume. Model response together with active excursion and port velocity or passive-radiator stroke. Build a simple test cabinet, verify airtightness and tuning, then assess the system in its intended room. The best option is the one whose compromises fit the job.

Explore the relationship
Interactive experiment

How strongly does the air spring act?

A sealed cabinet adds an air spring behind the cone. Reduce its volume: resonance frequency and Q rise in the model. The driver itself stays unchanged.

Model limits: Ideal sealed cabinet: no leakage, room effects, excursion or thermal limits. This curve does not simulate a vented or passive-radiator cabinet.

A closer look · model and equation

fc = Fs·√(1 + Vas/Vb); Qtc = Qts·√(1 + Vas/Vb). The curve is a normalised sealed small-signal model. The −3 dB point follows from the complete response and is not always equal to fc.

Workshop takeaway

Set the target and driver first, then choose the cabinet type. Volume and tuning belong together; an attractive response curve does not replace checking excursion, port and construction.

Work it out · answers unfold

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

09 · A measurement gives 2 V input and 1 V output. Find voltage gain in dB.

Show worked answer
  1. Ratio = 0.5.
  2. 20 log10(0.5).
  3. Gain = -6.02 dB.
Calculation
G=20\log_{10}\frac{1}{2}=-6.02\,dB

10 · A time window is 5 ms long. Estimate characteristic frequency resolution using 1/T.

Show worked answer
  1. T = 0.005 s.
  2. Take reciprocal.
  3. About 200 Hz; not a universal hard lower accuracy limit.
Calculation
\Delta f\approx\frac{1}{T}=200\,Hz

11 · What window duration gives 50 Hz resolution by the same approximation?

Show worked answer
  1. Duration = 1/resolution.
  2. 1/50 = 0.02 s.
  3. 20 ms.
Calculation
T\approx\frac{1}{50}=20\,ms

20 · Create a short measurement record for an RC low-pass with at least six documented fields.

Show worked answer
  1. Record circuit and component values including tolerances.
  2. Note source, instruments, probe/load and frequency points.
  3. Document input/output voltage, phase reference and model deviations.

Match drivers together →

Further technical reading

Colloms & Darlington (2005): High Performance Loudspeakers, 6th ed. Sections 4.3–4.5 (closed-box, reflex and band-pass), printed pages 146–169. Source information (COLLOMS-2005-6) →