ISEGORIA / MATH ENCYCLOPEDIA
Loudspeakers: the moving-coil driver
A coil in a magnetic gap, glued to a cone on a spring: the moving-coil driver is a mechanical oscillator welded to an electrical circuit by the force law F = BlI. Its impedance curve reads out the mechanics; its box turns a second-order system into a fourth-order one and trades size for bass; and at high frequencies the cone stops radiating in all directions and starts to beam, which is why speakers use more than one driver.
Before you begin: Damped oscillators, AC circuits and Bessel functions
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Impedance: the mechanics seen through the coil
The voice coil sits in a radial magnetic field B with a length l of wire in it, so a current I pushes it with force BlI and its velocity u generates a back voltage Blu. The cone, coil and air load form a mass on a spring with friction, and through that coupling the mechanics appears in the electrical impedance as a parallel resonant circuit: the mass as a capacitance, the suspension compliance as an inductance, the mechanical damping as a resistance. At the resonance fₛ the cone moves most and its back voltage nearly cancels the drive, so the impedance peaks at many times the coil resistance. Drag along the impedance curve to choose a frequency and watch the cone; change Bl, the moving mass and the compliance and see how the Thiele–Small parameters, the language every driver datasheet is written in, follow.
Worked example. A 6.5-inch woofer with \(R_e=6\ \Omega\), \(Bl=7.5\ \mathrm{T\,m}\), \(M_{ms}=15\ \mathrm g\), \(C_{ms}=1.0\ \mathrm{mm/N}\), \(R_{ms}=1.0\ \mathrm{kg/s}\) and cone area \(S_d=133\ \mathrm{cm^2}\) resonates at \(f_s=41.1\ \mathrm{Hz}\) with \(Q_{ms}=3.87\), \(Q_{es}=0.41\), \(Q_{ts}=0.37\). The impedance peak is \(R_e+(Bl)^2/R_{ms}=62\ \Omega\). The suspension is as stiff as \(V_{as}=24.6\ \mathrm L\) of air, and the reference efficiency \(\eta_0=\rho(Bl)^2S_d^2/(2\pi cR_eM_{ms}^2)=0.40\%\), 88 dB at 1 W and 1 m.
Watch out. The model is the lumped-parameter one: it holds while the cone moves as a rigid piston, below its first break-up mode, typically a few kilohertz for a woofer. The voice-coil inductance is treated as a pure inductor; real coils behave more like a lossy semi-inductance because of eddy currents in the pole piece, so the rising impedance at high frequency is gentler than drawn. \(M_{ms}\) includes the air mass the cone carries with it, which is why it depends on the mounting. The cone excursion is drawn for 2.83 V, the voltage that puts 1 W into 8 Ω.
Why does the mass of the cone appear as a capacitance in the electrical impedance?
The coupling is a gyrator: force is \(Bl\) times current, and voltage is \(Bl\) times velocity. Seen from the terminals, a mechanical impedance \(Z_m\) (force over velocity) becomes an electrical impedance \((Bl)^2/Z_m\), the inverse. A mass has \(Z_m=j\omega M\), growing with frequency; its inverse \((Bl)^2/(j\omega M)\) falls with frequency like a capacitor of value \(M/(Bl)^2\). By the same inversion the spring, \(Z_m=1/(j\omega C_{ms})\), appears as an inductance \((Bl)^2C_{ms}\), and series mechanical elements become parallel electrical ones.
2. The box: sealed or vented
A bare driver cancels its own bass: the back of the cone pushes air round to the front in antiphase. A box stops that, but the trapped air is a second spring. In a sealed box that spring stiffens the suspension by the compliance ratio α = V_as/V_b, raising the resonance and its Q together: the result is a second-order high-pass filter, falling 12 dB per octave. A vented box adds a port, a slug of air in a tube that resonates with the box volume as a Helmholtz resonator at f_b. Near f_b the port does the radiating and the cone barely moves: the response goes lower, then falls at 24 dB per octave, a fourth-order filter. Choose the box, drag the gold point to change the tuning or the volume, and compare the bass and the cone’s excursion.
Worked example. For the woofer of the previous experiment (\(f_s=41.1\ \mathrm{Hz}\), \(Q_{ts}=0.37\), \(V_{as}=24.6\ \mathrm L\)), a sealed box of 9.5 L gives \(\alpha=2.59\), \(f_c=77.8\ \mathrm{Hz}\) and \(Q_{tc}=0.707\), the maximally flat Butterworth alignment, \(-3\ \mathrm{dB}\) at 78 Hz. A vented box of 24.6 L tuned to 41 Hz reaches \(-3\ \mathrm{dB}\) at 41 Hz: nearly an octave deeper, from a box 2.6 times larger. At \(f_b\) the vented cone’s excursion falls to a small fraction of its value a few hertz either side.
Watch out. The curves are Small’s lumped models with a box-leakage \(Q_L=7\) for the vented box and a lossless sealed box; real boxes add absorption losses and port turbulence at high levels. Both responses are normalised to their passband level, so the comparison is of shape, not loudness; the efficiency is set by the driver. Below \(f_b\) the vented cone is unloaded and its excursion rises steeply, which is why vented systems need a high-pass filter against subsonic signals. Excursion is shown relative to the driver’s static displacement in free air.
Why does the cone almost stop moving at the port’s tuning frequency?
At \(f_b\) the port air and the box air form a resonator whose impedance, seen from the back of the cone, becomes very large: the box pressure swings strongly and pushes back on the cone exactly in opposition to the drive. The cone is held nearly still while the port’s air slug moves with large amplitude and does almost all the radiating. Energy is fed through the small cone motion into the high-Q Helmholtz resonance, which is also why port output lags the cone and why the total acoustic output drops so fast below \(f_b\), where port and cone move in antiphase.
3. Beaming: when a cone stops radiating sideways
A cone much smaller than the wavelength it radiates sends sound equally in all directions in front of it. As the frequency rises the wavelength shrinks, and waves from opposite edges of the cone start to arrive at an off-axis listener out of step: the sound narrows into a beam around the axis. For a rigid circular piston in a wall the pattern is exactly 2J₁(x)/x with x = ka sin θ, the same Bessel function that shaped the positional error of a watch balance. The transition happens around ka = 1 to 2, where the circumference is one or two wavelengths. Drag along the frequency axis on the right and watch the polar pattern on the left close in, then sprout side lobes; that is why a woofer hands over to a small tweeter at a few kilohertz.
Worked example. A 6.5-inch woofer has an effective radius of about 6.5 cm. At 1 kHz, \(ka=2\pi\times1000\times0.065/343=1.19\) and 60° off axis the level is down only \(20\log_{10}(2J_1(1.03)/1.03)=-1.2\ \mathrm{dB}\). At 4 kHz, \(ka=4.76\), and 60° off axis lies just past the first null, where \(ka\sin\theta=3.83\), 25 dB down: the woofer has become a spotlight. A 25 mm dome tweeter, \(a\approx1.3\ \mathrm{cm}\), reaches the same \(ka\) only at 20 kHz.
Watch out. The formula is for a flat rigid piston set in an infinite baffle, radiating into a half-space. A real cone is not flat, flexes at high frequencies, and sits in a finite box whose edges diffract; at low frequencies a box speaker radiates into the full sphere, not a half-space, so its directivity index tends to 0 dB rather than 3 dB. The shape of the beaming transition and its dependence on \(ka\) carry over well, and are what crossover design uses. Coaxial drivers such as a tweeter mounted at the centre of a woofer cone use the woofer’s cone as a waveguide to match the two patterns near the crossover.
Why does the pattern have nulls and side lobes at high frequency?
Off axis, each strip of the piston contributes a wave delayed in proportion to its distance across the face, \(x\sin\theta\). Summing them is a Fourier transform of the piston’s shape. When the path difference across the face is large enough, the contributions from some strips cancel those from others exactly: the first null is at \(ka\sin\theta=3.832\), the first zero of \(J_1\). Between nulls the cancellation is incomplete and a smaller lobe survives. A uniform piston is the circular analogue of a uniformly lit slit, whose diffraction pattern has the same nulls and lobes.
In practice
Where this mathematics and physics is at work, in explainers that take the real thing apart.