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DIY speakers, Power handling.


Got a question, that any idiot with electrical understanding should know, but for some reason this idiot does not.

When constructing a loudspeaker, (ive read several books on this but non touched on this issue) how do you determine the loudspeakers power handling ability?

Is the power handling the same as the driver with the lowest handling ability?

Is it the sum of the total accepable power loads of all drivers combined?

Or is there an equasion used to figure this out? Like the total watt hgandling devided by number of drivers?

I am now reading some books on Solid State amplifier construction, and Randy Sloan touched on the issue that when driving a loudspeaker that the power is not evenly distributed over all of the drivers, IE, in a 2way design the woofer might use about 60-70% of the power while the tweeter uses only 30-40%. This apparently is one of the reasons there is such a sonic benefit from Bi-Amping so each driver will have access to the full amount of power it desires.

That is probably what im gonna end up doing, custom building the amplifier to match the drivers and bi-amping the little bastards.
slappy

Showing 2 responses by marakanetz

marakanetz's avatar
marakanetz
5,473 posts
 
Slap, you've got too many today!

You're right,
The woofer takes the most if it while the tweeter uses just a-little. Tweeter uses only 3...7% of the power.
marakanetz's avatar
marakanetz
5,473 posts
 
I guess that there will be soon nights dedicated to build an amp but will that worth the time spent if it realy won't sound as good as simple Nad that costs in low hundreds?

To design a good quality loudspeaker you should test and adjust a crossover at aneholic rooms or use ready to go designes of enclosures and crossovers.

Higher power implies to higher complication.

Active bi-amped system makes it simplier since you can use two lower-powered amplifiers with some effort made to build an electronic crossover. A system with active subwoofer is even more simplier in that case.
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