enclosures/driver-cabinet-resonance · v1.0

Resonance of the Driver and Cabinet

Resonance is where the reactance of the mass and that of the spring cancel exactly — but “cancel” does not mean “uncontrollable.” It is precisely where the amplifier’s control matters most.

PICHITCHAI OPADWORARAT · MUSIC ENTHUSIASTMECHANISM: MASS ⇄ COMPLIANCE CANCELLATIONRUNNING EXAMPLE: 6.5″ WOOFER (same as the load-amp volume)

Resonance is not the point where control is lost

A common picture is that resonance is where inductive (L) and capacitive (C) reactance cancel out completely, after which the system “loses control” — the first half is right, the second flips it backwards. At a loudspeaker driver’s resonance, what cancels is the reactance of the moving mass (Mms of cone + voice coil + entrained air) against the reactance of the compliance (the suspension spring Cms). When the two are equal and cancel, the mechanical impedance is left as pure mechanical resistance Rms. The cone therefore moves most freely — not out of hand.

ωMms (mass)1/ωCms (compliance)|difference| net

Chart 1. Mass reactance (ωMms, rising) and compliance reactance (1/ωCms, falling) cross at fs=37.5 Hz. The difference |ωMms − 1/ωCms| dips exactly to zero — this cancellation is what defines resonance.

Throughout this volume we use the same 6.5″ driver as the Load, Amplifier & Enclosure volume so every number cross-checks: Mms=15 g, Cms=1.2 mm/N, Rms=1.18 N·s/m, Bl=7.0 T·m, Re=6 Ω, Le=0.5 mH, Sd=133 cm²

The mechanical cancellation: why the cone moves freest here

Look at the mechanical side first. The driving force at the voice coil (F = Bl·I) pushes against the cone’s mechanical impedance. Cone velocity is force divided by that impedance. At any frequency the mechanical impedance has three parts — mass, spring, and resistance. Only at fs do the first two cancel completely, leaving Rms alone. The mechanical impedance is therefore minimum, and cone velocity per unit force is maximum.

Mechanical resonance & impedance
fs = 1 / (2π·√(Mms·Cms)) Zmech(jω) = Rms + j(ωMms − 1/ωCms)
Mms=15 g, Cms=1.2 mm/N → fs=37.5 Hz; at fs: ωMms = 1/ωCms = 3.53 N·s/m cancel → Zmech = Rms = 1.18, purely resistive
Mechanism

Below fs the spring dominates (a compliance-controlled system — fighting the spring); above fs the mass dominates (mass-controlled — fighting inertia). Right at fs the two balance, and the cone swings with only mechanical friction to damp it. Stored and released energy peaks at this point.

Reflected into the electrical domain: a peak, not a short

This is where the “cancel, therefore short” intuition goes wrong, because the mechanical mechanism reflects back into the electrical side through the force factor Bl with inverted roles — mechanical mass becomes an electrical capacitance, mechanical spring becomes an electrical inductance. The two form a parallel RLC, not a series one, and a parallel circuit at resonance presents maximum impedance, not minimum.

Electro-mechanical analogy → terminal impedance
Ces = Mms/(Bl)² Lces = Cms·(Bl)² Res = (Bl)²/Rms Z(jω) = Re + jωLe + (Bl)²/Zmech
Bl=7 → Ces=306 µF, Lces=58.8 mH (parallel resonance at 37.5 Hz), Res=(Bl)²/Rms=41.5 Ω → Zmax=Re+Res=47.5 Ω — an “8 Ω” driver climbs to ~48 Ω at fs
|Z| driver in free air

Chart 2. The consequence of the cancellation: terminal impedance rises to a peak of 47.5 Ω at 37.5 Hz, not down to zero, because the equivalent circuit is a parallel RLC — parallel resonance = maximum impedance.

Correcting the picture

“L and C cancel” is true — but because they sit in parallel, the result is current circulating inside the tank, so the terminals see high impedance at fs. The amplifier delivers the least current into the driver right at resonance — the opposite of a “short that runs out of control.”

Three Q factors: mechanical, electrical, and total

The “sharpness” of a resonance is its Q — higher means more peaked and longer-ringing. A driver has two separate damping sources: mechanical friction (Qms) and electrical damping through the motor (Qes). Together they form Qts.

Quality factors
Qms = ωs·Mms/Rms Qes = ωs·Mms·Re/(Bl)² Qts = Qms·Qes/(Qms+Qes)
Qms=3.0 (mechanical — loose) · Qes=0.433 (electrical — far more dominant) · Qts=0.378; this driver is damped mainly by electrical means

Qes is nearly seven times stronger than Qms, meaning most of the damping comes from the electrical circuit attached to the voice coil, not from the rubber suspension — and that circuit has the amplifier at its far end. That is exactly why the amplifier can reach in and control the resonance directly.

So how does the amplifier “control” resonance?

A swinging cone acts as a generator, producing a back-EMF (Bl·u) fed into the voice coil. If the amplifier has low output impedance (high damping factor) it acts as a short for that back-EMF, driving a current that opposes the motion by Lenz’s law — an electromagnetic brake that pulls the resonance peak down and kills the ringing. More total series resistance (amplifier + speaker cable) releases that brake.

Back-EMF brake → source-modified Q
e_back = Bl·u Qes′ = Qes·(Re+Rg)/Re Rg = Rout + Rcable
SET Rg=2.77 Ω → Qes′=0.433·(8.77/6)=0.633 → Qts′=0.523; the cone-velocity peak at fs rises from +4.7 dB (SS) to +6.7 dB (SET)
SS · DF100 (firm brake)SET · DF3 (loose brake)

Chart 3. Cone velocity (referenced to the 150 Hz level). A low-DF amplifier (SET, Rg=2.77 Ω) lets the peak at fs reach +6.7 dB, while a high-DF amplifier (SS) brakes it to +4.7 dB — the shorted back-EMF is “control” made concrete.

The conclusion that inverts the intuition

fs is not where control is “lost” — it is where control matters most, because the back-EMF is strongest here and the soaring impedance keeps amplifier current low. A low-Rout amplifier therefore pins the resonance firmly, while a high-Rout tube amp lets it bloom and ring.

In the amplifier’s eyes: the peak is minimum current

Because fs is the impedance maximum, the amplifier draws the least current there — in pure current terms, resonance is an easy load. The genuinely hard regions are the impedance dips (between fs and the rise of Le, or between drivers in a multiway) plus the phase angles that make current and voltage straddle out of step. The SOA burden and peak-current story is developed in the Load, Amplifier & Enclosure volume.

at fs
|Z| max → I min
back-EMF
Bl·u max → max control
amp's hard zone
Zmin + phase ≠ at fs

A sealed box raises the resonance

Put the driver in a sealed box and the trapped air becomes an added spring in parallel with the suspension. Total compliance drops, the system stiffens, and the resonance climbs from fs to fc — raising Q by the same factor.

Closed-box alignment
α = Vas/Vb fc = fs·√(α+1) Qtc = Qts′·√(α+1)
Vb=15 L → α=Vas/Vb=2.0 → fc=37.5·√3=65.0 Hz; Qtc(SS)=0.385·1.73=0.667 · Qtc(SET)=0.523·1.73=0.906 (bass bump ~+1.5 dB)

Note that Qtc still carries the amplifier’s Qts′ — a sealed box raises the resonance but does not remove the amplifier from the equation, since damping is still mostly electrical. The SET amp therefore still pushes Qtc past 0.9, producing a bump.

A vented box splits it into two resonances

A vented box adds a second resonance — the air mass in the port working with the air spring in the box as a Helmholtz resonator at the tuning frequency fb. The single resonance is split into two peaks straddling fb.

Vented (Helmholtz) tuning
fb = (c/2π)·√( Sv / (Vb·Leff) ) fL · fH ≈ fs · fb
Port Ø5 cm, Vb=15 L, tuned to fb=40 Hz → peaks split to ~20.2 Hz and ~73.7 Hz (computed from the equivalent circuit) straddling fb
free air (reference)sealed 15 Lvented fb=40

Chart 4. The sealed box pushes the single peak from fs=37.5 up to fc=65 Hz; the vented box splits it into two peaks (20.2 / 73.7 Hz) straddling the saddle at fb=40 Hz — the impedance shape names the box type at a glance.

The point where the cone nearly stops: the excursion null

At fb the port moves almost all the air instead of the cone. The air mass in the port reflects a large mechanical impedance back onto the cone, pinning it. The result is that cone excursion collapses to nearly zero at the tuning frequency — a protective shield the sealed box does not have.

sealedvented

Chart 5. Cone excursion (relative to the sealed-box plateau). At fb the vented box drives the cone down to ~0.18 (the port takes over), while the sealed box is still near ~0.90 — below fb, though, the vented cone surges as the port unloads.

A double-edged sword

The excursion null only protects the cone around fb. Below fb the port air mass unloads and the cone swings harder than in a sealed box — which is why a vented box needs a subsonic high-pass to protect the driver. A sealed box tolerates deep signal better by nature.

Higher-order resonance: cone break-up

Everything above concerns the fundamental resonance, where the cone moves as a rigid piston. But above the working band the cone stops moving as one body — it flexes into standing waves on itself, called cone break-up. This is a mechanical resonance of the cone material (set by the speed of sound in the material, its thickness, and diameter), producing sharp impedance peaks and response peaks/dips at high frequency.

Mechanism

Break-up differs from fs in that it is not a single mass–spring body but bending modes of the cone sheet. The amplifier can barely control it (it is not coupled through Bl the pistonic way); it must be tamed on the material/geometry side and with a crossover that cuts before those modes — a different problem from damping the bass resonance.

Design: choose where to damp

A driver’s bass resonance is energy that must be damped away. The only question is where. You have three levers, and you choose who takes the job.

Aperiodic — move the damping into the box
Zmax ≈ Re + (Bl)² / (Rms + Rextra)
Add acoustic resistance Rextra≈2.0 → Zmax=6+49/3.18=21.4 Ω (from 47.5); the peak is pressed into a single low broad hump without relying on the amplifier’s damping
leverdamped bydepends on amp?
amplifier damping factorback-EMF + low Routentirely
sealed / vented boxair load + portpartly
aperiodic / acoustic resistanceacoustic losses in the boxhardly
The thesis of this volume

Resonance is not an enemy, and not a blind spot. It is energy stored by the mass–spring mechanism. You choose whether the amplifier brakes it (needs high DF), the box brakes it (sealed/aperiodic), or the port relocates the load (vented). Just know what is doing the damping — because “uncontrollable” only happens when you don’t know who is in control.

A loudspeaker driver does not “run out of hand” at resonance — it stores maximum energy, emits maximum velocity, and opens the widest path for the amplifier to brake it, all at the same point. The designer’s job is to decide where that energy gets damped away, then lay down the equations to match.

References

  • aesSmall, R. H. “Direct-Radiator Loudspeaker System Analysis,” JAES 20(5):383–395 (1972).
  • aesSmall, R. H. “Closed-Box Loudspeaker Systems — Part I & II,” JAES 20(10):798–808 (1972); 21(1):11–18 (1973).
  • aesThiele, A. N. “Loudspeakers in Vented Boxes, Part I & II,” JAES 19(5):382–392 (1971); 19(6):471–483 (1971).
  • aesOtala, M. & Huttunen, P. “Peak Current Requirement of Commercial Loudspeaker Systems,” JAES 35(6):455–462 (1987).
  • stdIEC 60268-5, Sound system equipment — Part 5: Loudspeakers.
  • bookBeranek, L. L. & Mellow, T. J. Acoustics: Sound Fields and Transducers, Academic Press 2012.
  • bookDickason, V. The Loudspeaker Design Cookbook, 7th ed., Audio Amateur Press 2006.
Edited by Pichitchai Opadworarat Head of R&D — Pyramid Lifestyle Technology Ltd. Part. 2 years in audio engineering (since the company was founded)

Revision history

v1.02026-06-11First edition — 11 chapters + 5 live-computed charts