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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
herman
2,255 posts
 
.

It appears they do take the room into account

1. Convert amplifier power in watts to dB watts.
2. Add the amplifier power in dB watts to the loudspeaker sensitivity in dB.
3. Deduct the listening position attenuation (between 7dB -12dB).
4. You now have your system’s peak level ability at your listening position.

you got +20 and it looks like they did too but subtracted the 7

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Post removed 
gs5556
1,234 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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