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How to figure output from efficiency in db ?

I always understood that doubling power in watts added 3 db to the speakers output in db. And that 10 times more watts caused a 10db increase. Thus, if your speakers have an efficiency of, say, 90 db/watt, a 2 watt signal produces a 93db output, a 10 watt input would provide a 100db output, and a 100 watt input would produce a 110db output (all other factors being equal).
Lately I have seen, on the Musical Fidelity website and many places elsewhere, the claim that, for example, a 100 watt increase causes only a 13db increase. How can this be?
Isn't the formula still the one I have been using?
(This has nothing to do with allowances for distance, room conditions, etc. This is entirely a matter of their using some other formula.)
Can anyone clear this up for me?
rpfef

Showing 1 response by gs5556

gs5556
1,209 posts
 
I also think they are taking listening distance into account. A -7dB to -12dB attenuation translates into 7 to 13 feet (2.2 to 4.0 meters) from the source.

The attenuation by distance:

dB = 20 LOG (D/Do) where Do is the reference distance of one meter and D is the listening distance. For every doubling of distance the intensity drops by 6dB. To get to 13dB for a 100 Watts, 7dB has to be subtracted from 20dB, so the attenuation distance:

At -7dB: D/Do = 10^(7/20) = 2.2

D = 2.2 * Do = 2.2 * 1.0 = 2.2 meters (7'-4")

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