Relating SPL to Force?
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 Multi-Way Conventional loudspeakers with crossovers

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 13th March 2004, 10:12 AM #1 Kaoss   diyAudio Member   Join Date: Mar 2004 Location: Bristol 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?
 13th March 2004, 02:42 PM #2 Ron E   diyAudio Member     Join Date: Jun 2002 Location: USA, MN I would suggest you get: "Acoustics" by L.L. Beranek or "Theory and Design of Loudspeaker Enclosures" by J.E. Benson or 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) __________________ Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Armaments, universal debt, and planned obsolescence--those are the three pillars of Western prosperity. Aldous Huxley
 13th March 2004, 03:12 PM #3 Svante   diyAudio Member     Join Date: Feb 2004 Location: Stockholm -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. __________________ Simulate loudspeakers: Basta! Simulate the baffle step: The Edge
 12th August 2008, 12:19 AM #4 MCPete   diyAudio Member   Join Date: Nov 2005 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 Thus Vp ~ F Linkwitz in his A21 states that acoustic power Ap is proportional to piston velocity squared: Pa ~ (Vp)^2 Thus 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.) Regards, Pete
 13th August 2008, 04:17 AM #5 MCPete   diyAudio Member   Join Date: Nov 2005 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. -Pete

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