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| Definition of terms: |
| f-3 |
-3dB half-power frequency. |
| fs |
resonance frequency of driver |
| fc |
resonance frequency of the closed box system |
| Q |
ratio of reactance to resistance (series circuit) or resistance to reactance (parallel circuit) |
| Qts |
total Q of driver (woofer) at fs, considering all driver resistance. |
| Qtc |
total Q of speaker system at fc, including all system resistance. |
| Vas |
volume of air having the same acoustic compliance as the driver suspension. |
| Vab |
volume of air having the same acoustic compliance as the enclosure. |
| Xmax |
peak linear displacement of driver cone. |
| Sd |
effective surface area of driver cone. |
| Vd |
peak displacement volume of driver cone. |
| Vb |
net internal volume of enclosure. |
| a |
compliance ratio (alpha). |
| no |
reference efficiency |
| Cas |
acoustic compliance of the driver suspension. |
| Cab |
acoustic compliance of the air in the enclosure. |
| Driver "Q" and Enclosure response |
| The point of fitting a woofer to a box is to control the response characteristics. |
| The objective method is to measure and align de Q-factor. Q is a composite term, used to describe resonant magnification in speaker boxes. |
| It represents the degree to which the electrical, mechanical and pneumatic circuits of the woofer/box combination interact to control resonance. |
| Certain values of Qtc have specific response characteristics, which can be categorized as follows : |
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| Qtc = 0,5 critically damped -- transient perfect |
| Qtc = 0,707 Butterworth-response(B2)-- max falt amplitude response with minimum cutoff |
Qtc > 0,707 Chebychev-response -- equal Ripple response - nax power handling and max efficiency
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| Qtc = 0,5 is usually regarded as excessivly taut and overdamped. |
| Some authorities still consider this value (0.5 - 0.6) to be optimal. |
| Peak(db) = 20 * Log (Qtc4 / Qtc2 - 0,25)0,5 |
| fgmax = 1 / (1 - 1 / 2 * Qtc2)0,5 |
| The frequency of this peak (fgmax) is given as a ratio with the box resonance (fc). |
| Qtc, Qts, A(lpha), Vas, fs |
Qtc=0,707 , Qts=0,37 , Vas = 293 Ltr , fs = 20Hz |
| A = (Qtc/Qts)2 -1 |
(0,707/0,37)2 -1 = 2,65 |
| Vab = Vas/A |
293/2,65 = 110,56 Liter |
| Vb = Vab/1.2 |
110,56/1,2 = 92 Liter |
| fc = fs*(Qtc/Qts) |
20*(0,707/0,37) = 38,2 Hz |
Firstly, calculate the Efficiency Bandwidth Product: EBP = fs/Qe.
- 50 or less = best used in a sealed enclosure.
- 50 - 90 = flexible enclosure options.
- 90 or greater = best used in ported enclosure.
If you have Qe and Qm, you can take external resistance into account when calculating Qts.
Qts = 1/((1/Qm) + Rs/((R+Rs)Qe)) where R is the resistance of the wiring, typically 0.5 ohm.
Alternatively, use Qts as specified in the TS parameters. |
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Source: The Loudspeaker Design Cookbook, Vance Dickason
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