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
.

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

.
For power, dB = 10 * log(P1 / P2).

Thus, the familiar 3dB requires a doubling in power comes from:

dB = 10 * log(2 * P / P)
= 10 * log(2)
= 10 * 0.30103
~ 3dB

And how much of a dB increase will ten times the power be:

dB = 10 * log(10 * P / P)
= 10 * log(10)
= 10 * 1
= 10dB

It seems that dbW is defined to be power relative to 1 watt. So in the intial equation P2 is 1. From the initial equation, solving for P1 we get.

dB = 10 * log(P1 / P2)
= 10 * log(P1 / 1)
= 10 * log(P1)
dB / 10 = log(P1)
P1 = 10 ^ (dB / 10)

This is the formula that Musical Fidelity used to generate the table on their web site. For example,

P1 = 10 ^ (15 / 10)
= 10 ^ 1.5
~ 31.623 W

P1 = 10 ^ (16 / 10)
= 10 ^ 1.6
~ 39.81 W

P1 = 10 ^ (17 / 10)
= 10 ^ 1.7
~ 50.12 W

etc....

Don't know if that's what you needed to know.
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")

Is that what you're looking for?
Thanks, folks. It appears that I was inattentive. The 100x increase in dbW does translate to a 20db output gain, with all other factors held constant. I was thrown off by their chart.