You’re using the wrong formula - equating electrical power to sound power. The db formula for sound power is :
I was using the right formula as I was referring to amplifier power, not sound pressure and was pretty clear about that. When you halve the force and double the area, you wind up with roughly
the same sound power that you began with. But improved acoustical
coupling with increased area improves net efficiency (electrical power
in versus sound power out). Again, halving the **voltage** reduces the force to 1/4th. |
" Again, halving the **voltage** reduces the force to 1/4th. " - atmasphere
Really? Are you serious? What do you think voltage is, my friend? Voltage is "electro motive FORCE". Halving the voltage does not reduce the force to 1/4. It reduces power to 1/4 since the formula for power expressed simply with voltage and resistance is:
V squared divided by resistance.
Look, I didn’t come here to give lectures on Electrical Theory Basics. This is getting embarrassing. Please review your textbooks. I can’t continue to carry on a technical conversation with you if you fail to grasp the basics.
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When is a force not a force? When it’s a voltage. Electromotive force is not (rpt not) a force, as fate would have it. Otherwise, Volts would have units of pounds or kilos or grams or whatever. It’s not the voltage that throws you across the room. It’s the amps, baby,!
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Really? Are you serious? What do you think voltage is, my friend?
Voltage is "electro motive FORCE". Halving the voltage does not reduce
the force to 1/4. It reduces power to 1/4 since the formula for power
expressed simply with voltage and resistance is: The problem here is that voltage does not exist without current. The two together are power. Even though we refer to speakers as 'voltage driven' that is a bit of a charged term (referring to the fact that many amplifiers are designed to act as 'voltage sources' and many speakers are designed to expect a voltage source to be driving it); but in fact power is actually making the speaker move. If you can somehow make a speaker move with voltage but *without* current, you will have a new branch of physics :) A 3db reduction is halving amplifier power, which is what drives a loudspeaker. Its the 'force' that makes a speaker move and because of that if you reduce the voltage by half, the power to move the speaker is reduced to 1/4 of previous. |
I believe that the references to force in this discussion are unnecessary and are contributing to confusion, and I believe that Atmasphere is correct. A short while ago I posted as follows in the other thread I referenced above, in relation to this matter: Almarg 4-20-2018
Assuming that a speaker is operating in a reasonably linear manner, meaning for example that it is not being over-driven to the point that thermal compression becomes significant, it seems to me that the relation between acoustic power out and electrical power in will remain constant to a close approximation. And electrical power in will be proportional to the square of the applied voltage.
Therefore it would seem to me (and I believe also to Erik, Atmasphere, and Kijanki) that since a 50% reduction in applied voltage will result in a 75% reduction in electrical power in, which corresponds to a 6 db reduction in electrical power in, the result will be a 6 db reduction in acoustic power out. My comment was seconded in that thread by Erik_Squires, who is particularly knowledgeable and experienced in speaker design. An excerpt from his post: Erik_Squires 4-20-2018
... the SPL at a reference distance, measured in dB, changes
in proportion to the power OR voltage when either is expressed as dB
assuming there is no compression in the driver.
Said another way, for a single driver:
Delta V dB = Delta W dB = Delta SPL dB
That's what's so cool about dBs!
Regards, -- Al |
