Relating SPL to Force?
I'm currently doing some work on optimising speakers for use with class-d amplifiers. So far, I've got the power down to about 0.2W for an 80dB SPL at 50cm. Now I'm looking for an equation to relate the force on the coil (and thus acceleration) to the SPL output, in order to predict the effect of increased gap field on the SPL output. Can anyone help?
I would suggest you get:
"Acoustics" by L.L. Beranek
"Theory and Design of Loudspeaker Enclosures" by J.E. Benson
JAES articles from Thiele or Small in the early 1970's
Force on the coil is B*l*i
B= Gap field strength
l = length of wire perpendicular to B
i = VC current
Trouble is, current is affected by cone motion, (Motional impedance - Lenz's Law)
-Sound pressure is proportional to cone acceleration, assuming that the cone moves like a rigid piston, and that it is small compared to wavelength.
-Cone *velocity*, ie time integral of acceleration, is proportional to the force divided by the mechanical impedance.
-Mechanical impedance is jw*Mms+Rms+1/(jwCms), or simply jw*Mms (a bit) above fs.
-Force is proportional to I*B*L, where I is the current, B is the magnetic flux density, and L is the length of the voice coil wire in that magnetic field.
-Current is proportional to voltage divided by electrical impedance.
-Electrical impedance is Re+jw*Le+((1/(jw*Mms/(BL)^2)) // (jw*(BL)^2*Cms) // ((BL)^2/Rms)) ("a//b" means a and b in parallel, or ab/(a+b)=1/(1/a+1/b)).
The above holds for a driver mounted in an infinite baffle, or a cloesed box, if Cms is taken to include box stiffness and Rms includes box losses.
I hope I got it right. :dead:
SPL ~ F^2
From the FAQ section of www.linkwitzlab.com, the 3rd paragraph of A21, I infer that velocity of a piston is proportional to current througn a voice coil driving the piston:
Vp ~ I
Driving force of the coil F is proportional to I (as posted above by Ron E and Svante):
F ~ I
Vp ~ F
Linkwitz in his A21 states that acoustic power Ap is proportional to piston velocity squared:
Pa ~ (Vp)^2
Pa ~ F^2
Assuming that SPL is proportional to Pa, then
SPL ~ F^2
I think the above implies the following interesting result. As frequency drops from two to three times the resonant frequency of a driver to the resonant frequency, electrical power to the driver decreases at the inverse of the ratio of increased impedance, while the driving force decreases to the square of the inverse of that ratio. (This assumes that the driving amplifier is a voltage source.)
correction to my post #4
At the bottom, instead of driving force F, I meant to say that acoustic power Pa decreases to the square of the inverse of increased impedance. That is, the counterintuitive result seems to be that if impedance at the resonant frequency Fc is three times impedance at two to three times Fc, then electrical power is reduced by one-third while acoustic power is reduced by one-ninth.
|All times are GMT. The time now is 11:32 PM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2015 DragonByte Technologies Ltd.
Copyright ©1999-2015 diyAudio