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Multi-Way Conventional loudspeakers with crossovers

Cone displacement vs power input?
Cone displacement vs power input?
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Old 13th January 2003, 02:12 AM   #1
Circlotron is offline Circlotron  Australia
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Default Cone displacement vs power input?

Lets say you drive your speaker with a 5 volt rms low frequency tone and it vibrates 1mm p/p. If you double the drive to 10 volts rms then the current should double as well (ignoring voice coil heating, and airgap flux and suspension nonlinearities. So if the current doubles the magnetic force and therefore the cone movement would double as well. Now the power has quadrupled as well because we have doube voltage *and* double current. So it appears the cone movement goes up proportionally as the drive voltage increase, and as the square root of the power increase? Does this sound correct?

Putting it another way, if it needs 1 volt and say 1 watt (1 ohm speaker) to drive a cone 1mm p/p then it would it take 10 volts and therefore 100 watts to drive it to 10mm p/p?
Best-ever T/S parameter spreadsheet.
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Old 13th January 2003, 04:08 AM   #2
kelticwizard is offline kelticwizard  United States
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Cone displacement vs power input?

Yes, you have it right.

Another way to look at it is to say that 4 times the power yields an extra 6 dB of SPL.

If you are planning to use this as a formula to see how much cone excursion is necessary to produce what SPL at a given tone, please check out the charts in the following thread. These are for sealed systems only, where the cone provides 100% of the bass output. for ported systems, you can multiply the speaker Volume of Displacement, Vd, by four. the first chart is for normal measurement, the second is for Metric.


PS: Hitting F11 makes reading the charts easier.
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