|Multi-Way Conventional loudspeakers with crossovers|
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|7th May 2010, 10:06 AM||#1|
Join Date: Jun 2005
Please Check Some Basic Calculations of T/S
I'm trying to find out how much power (or separate V & I) to drive a naked driver to its Xmax.
The main woofer(s) I'm currently using is Eminence SigmaPro18, some of its parameters are:
Re = 6.3 Ohm
BL = 22.1 TM
Cms = 0.24 mm/N
Xmax = 6 mm
Let's assume Xmax happens at very low frequencies. And the driver is naked (poorly loaded), so I can probably throw away air load related parameters etc...
For 6 mm displacement, I'd need 6/0.24 = 25 N of force.
And the motor strength is 22.1, so I'd need 25/22.1 = 1.13 A to have that force.
If at DC, then it's only 1.13 * 6.3 = 7.1 V (or 8 W) to drive it into Xmax.
If at fs, at which point the measured impedance is 190 Ohm, than it's 1.13 * 190 = 214.7 V (or 242.6 W) to drive it into Xmax.
Hmmm... the last one doesn't look right to me.
Besides, an amp has an output voltage swing of 214.7V should be rated @ 5762W into 8 Ohm load !! WTF, something must be wrong here.
In my own experiences, some poor little amps have the ability to drive the woofer into pretty large excursion already. I can't imagine what a fullblown 5~6kW amp can do. Shoot me some cones at a distance?
I also remember Mr. SL mentioned in his web pages that only very little power is needed to drive an ordinary woofer to its Xmax at fs (20~30W IIRC).
So my previous calculation at fs must be wrong. But how? Some other parameters should join in? Is it possible to get the right calculation by just these T/S parameters? Or something else is needed? "Simple" harmonic motions?
|7th May 2010, 12:01 PM||#2|
Join Date: Feb 2006
I don't know how you can calculate max. W if it varies with frequency. Anyway if you use WinISD you can simulate W for max. cone excursion@frequency wanted.
(1) Green is 200W (Closed box/1200L)
(2) Grey is 300W (Vented/258L/46Hz)
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Last edited by Inductor; 7th May 2010 at 12:04 PM.
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