Driver matching

Rear chamber

Klipsch reactance annulling. Size the sealed back chamber that cancels the horn’s mass reactance above cut-off.

Sheet: 14 / 25
Units: Metric
Air: 20 °C

Fig. 1Reactances at the diaphragm

Horn mass reactanceChamber + suspensionNetHorn resistancedrag on plot to read

When the chamber is right, the net trace lies on zero everywhere above cut-off and the driver sees only the horn's resistance.

Rear chamber $V_B$: 75.1L
Total compliance $V$: 54.6L
Klipsch $2.9\,S_t\,\ell$: 54.9L
Net reactance at $f_c$: -0.00145ρc/St
Suspension share: 27.3%

Fig. 2Where the chamber sits

How it works

Just above cut-off the throat of an exponential horn is not a pure resistance. It also presents a mass-like reactance that falls with frequency:

Z_t = \frac{\rho c}{S_t}\left[\sqrt{1-\left(\frac{f_c}{f}\right)^{2}} + j\,\frac{f_c}{f}\right]

That reactance robs output in the bottom octave. Paul Klipsch's remedy, known as reactance annulling, is to put a sealed volume of air behind the driver. A closed chamber is a spring, and at the diaphragm its reactance is

X_B = -\frac{\rho c^{2}\,S_d^{2}}{\omega V}

Both terms vary as $1/f$ and have opposite sign. Referred to the diaphragm, the horn term is $\rho c\,S_d^{2} f_c/(S_t f)$ , so the two cancel at every frequency above cut-off when

V = \frac{S_t\,c}{2\pi f_c} = \frac{2 S_t}{m}

Klipsch wrote the same result as $V = 2.9\,S_t\,\ell$ , with $\ell$ the length over which the horn area doubles. The read-out shows both; they agree to within rounding.

Allowing for the suspension

The driver's own suspension is a second spring in parallel with the air spring, equivalent to a volume $V_{as}$ . The chamber only has to supply the remainder:

\frac{1}{V} = \frac{1}{V_B} + \frac{1}{V_{as}} \quad\Rightarrow\quad V_B = \frac{V\,V_{as}}{V_{as}-V}

If $V_{as}$ is smaller than $V$ the driver is already too stiff and no chamber helps. Drag the trial volume to see the net reactance move away from zero: too small a chamber leaves a stiffness (negative) residue and thins the bass; too large leaves the horn's mass reactance uncorrected.

The chamber also raises the driver's resonance to $f_s\sqrt{1+V_{as}/V_B}$ . Line it lightly; heavy stuffing increases the effective volume by up to 20 %.

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How it works

Just above cut-off the throat of an exponential horn is not a pure resistance. It also presents a mass-like reactance that falls with frequency:

Z_t = \frac{\rho c}{S_t}\left[\sqrt{1-\left(\frac{f_c}{f}\right)^{2}} + j\,\frac{f_c}{f}\right]

X_B = -\frac{\rho c^{2}\,S_d^{2}}{\omega V}

Both terms vary as $1/f$ and have opposite sign. Referred to the diaphragm, the horn term is $\rho c\,S_d^{2} f_c/(S_t f)$ , so the two cancel at every frequency above cut-off when

V = \frac{S_t\,c}{2\pi f_c} = \frac{2 S_t}{m}

Klipsch wrote the same result as $V = 2.9\,S_t\,\ell$ , with $\ell$ the length over which the horn area doubles. The read-out shows both; they agree to within rounding.

Allowing for the suspension

The driver's own suspension is a second spring in parallel with the air spring, equivalent to a volume $V_{as}$ . The chamber only has to supply the remainder:

\frac{1}{V} = \frac{1}{V_B} + \frac{1}{V_{as}} \quad\Rightarrow\quad V_B = \frac{V\,V_{as}}{V_{as}-V}

The chamber also raises the driver's resonance to $f_s\sqrt{1+V_{as}/V_B}$ . Line it lightly; heavy stuffing increases the effective volume by up to 20 %.

Rear chamber

Inputs

How it works

Allowing for the suspension

Inputs

How it works

Allowing for the suspension

§ How it works

Allowing for the suspension

§ How it works

Allowing for the suspension

How it works

How it works