SPL for more than 1 speaker

[pdf64]

When this was last topical here in February, I sent a query regarding it to Celestion's DrDecibel

'ENQUIRY : Hi Dr dB, I've long been of the understanding that doubling
the number of speakers increases the sensitivity by 3dB, ie 1 greenback
97dB/watt, 2 greenbacks 100dB/watt, 4 103dB/watt. However, I can't
remember where I read it, and can't google anything useful to confirm
it. Can you advise whether this is correct, and point me at a textbook
where I could try to understand the theory?'

'Hi Pete
This is correct. I looked on line for a decent reference but
couldn't find one. Something like Martin Colloms 'High Performance
Loudspeakers' would give you some background and your local library
might have a copy.
Regards
The Doc'

Unfortunately my library doesn't have a copy. Pete.

[J M Fahey]

Dear M-Fine

Doubling power to the same driver adds 3db.
Doubling drivers with the same power adds 3db (note: impedance mismatch may mean you are losing power so you have to watch this in a test)
Doubling power and drivers adds 6db.

That's exactly what I'll test tomorrow.
Not simulation or calculation, actual testing.
I have no preference for one result or the other , I'll post what I see.
Dear PDF: Dr Decibel should run for President, no matter which party, no matter which Country, he'll win anyway.
He´s the King of answering ... without answering.

[redelephant]

I admit I find the concept of sound pressure and sound level confusing. But
1. if doubling the power increases spl by 3dB, so 100dB/watt goes to 103dB with 2W. OR:
2. doubling the number of speakers that are each fed one watt also increases the spl with 3dB.
3. I agree if BOTH is done, as in doubling the speaker drivers AND doubling the power (one watt each) then the increase is 6dB.
BUT
4. we are halfing the power into double the speakers, so half the watts gives -3dB, but doubling the speakers gives +3dB, so we end up with -3dB + 3dB = 0dB increase in spl.
???
Really, I do find this confusing, and you are most likely correct, but my math can't grasp it.

[Steve Conner]

Well, I asked Dr. Decibel how to get the magnet bell off an Alnico Gold, and he gave me instructions which I followed, and it worked.

If you want to see a real political answer, ask him for the Thiele-Small parameters of their guitar speakers.

Redelephant: The reason for the +3dB gain with "double the speakers, same total power" is that an array of 2 speakers is more directional than a single speaker. It beams more sound pressure towards the measuring mic because of constructive interference between the two speakers.

But if you add up all of the acoustic power that passes through a spherical surface around the array (which is the definition of sound power level) you see no gain, because the increased SPL at the mic location comes at the expense of decreased SPL elsewhere.

Because loudspeaker efficiency is defined as the SPL at 1 meter on axis produced by 1 watt, then it looks as if the efficiency increases when speakers are placed in an array. But in actual fact, they are no more efficient at turning electrical power into acoustic power than they were when used singly. They're just better at directing it to where it's wanted.

This is my understanding at least.

[m-fine]

I admit I find the concept of sound pressure and sound level confusing. But
1. if doubling the power increases spl by 3dB, so 100dB/watt goes to 103dB with 2W. OR:
2. doubling the number of speakers that are each fed one watt also increases the spl with 3dB.
3. I agree if BOTH is done, as in doubling the speaker drivers AND doubling the power (one watt each) then the increase is 6dB.
BUT
4. we are halfing the power into double the speakers, so half the watts gives -3dB, but doubling the speakers gives +3dB, so we end up with -3dB + 3dB = 0dB increase in spl.
???
Really, I do find this confusing, and you are most likely correct, but my math can't grasp it.

You are close, 1/2 watt into one speaker is -3 db from the 1 watt level, but you are also adding another 1/2 watt into the second speaker (double drivers, double watts = +6 db) for a net of +3 db (-3 + 6 = +3).

There are a lot of ways to conceptualize what is a fairly abstract concept since our brains dont tend to think well on db scales even though our senses seem to perceive things this way.

[m-fine]

Redelephant: The reason for the +3dB gain with "double the speakers, same total power" is that an array of 2 speakers is more directional than a single speaker. It beams more sound pressure towards the measuring mic because of constructive interference between the two speakers.

But if you add up all of the acoustic power that passes through a spherical surface around the array (which is the definition of sound power level) you see no gain, because the increased SPL at the mic location comes at the expense of decreased SPL elsewhere.

Because loudspeaker efficiency is defined as the SPL at 1 meter on axis produced by 1 watt, then it looks as if the efficiency increases when speakers are placed in an array. But in actual fact, they are no more efficient at turning electrical power into acoustic power than they were when used singly. They're just better at directing it to where it's wanted.

This is my understanding at least.

This is not correct for most frequencies we are concerned about until the array gets much longer. What happens when the array is long enough is also a different boost in efficiency. As you move back from a normal speaker which is effectively a point source, the sound expands in a sphere (3 dimensions) and SPL drops by 6 db every doubling of distance. If you have 100 db at 1 m, you get 94 at 2m and 88 at 4m etc. The line array acts like a line source and the sound expands in a cylinder with a length equal to the length of the array ( only growing in 2 dimensions) and SPL decreases by 3 db with each doubling of distance. 100 db at 1 m becomes 97 at 2m, 94 at 4m, 91 at 8m and 88 at 16m etc. You can see the obvious advantages for filling a large stadium since you not only have more sound at the further distances, the SPL levels are more consistent between different locations.

Now for our guitar speaker. The motor in the speaker needs to move the cone and attached mass plus the air. If you apply a sine wave, the motor needs to accelerate the cone forward, then at the correct point (space and time) it needs to apply opposite acceleration to slow it down, stop it, and then reverse the direction to pull the driver rearwards. To move twice as much air, the cone not only needs to move further, it also has to do it faster. The time between the forward peak and rear peak is set by the frequency of the signal, so as the distance (amplitude or volume) goes up, the acceleration increases meaning more force AND the force over a longer distance. You need 4 times the power to move twice as much air, or in other words, the louder the speaker plays, the less efficient it is at turning electrical energy into moving air. See the comments from an earlier poster about the impedance mismatch between driver and air. Basically more and more of the energy goes into moving the cone weight.

The flip side is that less cone movement allows a speaker to be more efficient. That is why multiple speakers moving up and back a small amount are more efficient than a single driver with a higher excursion but moving the same amount of air. This also is seen in cone size and weight. Bigger cone area is more efficient, higher weight less efficient. It is real hard to make an 6" speaker as efficient as a 12" even though to 12" cone needs to be a little heavier to remain stiff enough. The additional area usually more than offsets the additional weight.

In summary, the reason for the extra 3 db from adding another speaker is the 2 speakers are both moving their cone mass less and are operating more efficiently than a single speaker moving twice as much.

[Diablo]

Amen brother !!!
That's the single worst point.
I make my own speakers, and the best I can get (in regular 12" speakers) is no less than 30/32 gram "brute" moving mass (cone+coil+suspension+dome+wires+adhesives) for 2 to 3 gram "useful" air, the air displaced by such a piston.
Useful payload less than 10% of what my motor has to push back and forth.
Thinnest usable cones I can get are around 14 grams; any thinner they crack in a bad way, and before that they "cry" with horrible harmonics, so it's not a easy task.
Anyway paper (cellulose) still rules in our trade; Polypropilene is too heavy (and very difficult to glue), Kevlar cones even worse (way too heavy and "dead").

Thanks for the info on cone weight vs cone cry. I have a pair of speakers that have bad cone cry even at low volume. They use lightweight cones and no surround doping. It was recommended to add dope, but that experiment didn't work. All that did was to dull the high frequency. The cone cry remained - it generates a ghost note an octave below the fundamental on the high E string of a guitar above the 12th fret. I assume the reason is because the cone lacks the required stiffness?

[Steve Conner]

M-Fine, I don't really buy that explanation. The speaker is supposedly a linear device, so it should be as efficient at 1 watt input as it is at 2 watts.

If anything like you describe does really happen, then maybe it's "coupling gain". If a speaker is surrounded by a bunch of similar cones all pushing air in unison, then the air that it pushes can only go forward. If it tried to go sideways, it would be pushed back by an equal and opposite compression from the neighbouring speakers. With one speaker standing on its own, the compression wave can go in all directions.

Maybe this focusing of the compression wave makes a better acoustic impedance match to the speakers, just like a horn does by physically boxing in the compression wave with pieces of wood. And that would make them more efficient.

The same explanation could also work for larger cones being more efficient than smaller ones.

But to me, the above is just the same as directivity gain by constructive interference, I don't see the difference.

[m-fine]

Steve that is definitely not the case. You can do it with two speakers 10+ feet apart or two 4x12 cabs some distance apart. It also holds for a pair of hifi speakers either drivers or full multi driver speakers. They need to be in phase at the mic, which usually means equal distance to the mic. You can also get room interference so an indoor test in a smaller room might give funky results.

[Steve Conner]

I still don't buy it. If they "need to be in phase at the mic" then my original explanation, directivity gain, surely applies.

[m-fine]

The in phase at the mic is required because if you combine two sine waves perfectly out of phase they cancel instead of adding. The room reflections won't cancel so you will hear it, but low frequencies will be mostly gone.

If you have a head with multiple impedance taps and two equal speakers, test for yourself. No need to believe me or not believe me.

[m-fine]

Steve, your linear comment actually gave me another way to describe what is happening, using ohms law. I had to log into my laptop to type it all though!

Speakers are linear with respect to voltage so they can follow the input signal without compressing it. (within a certain range of course) So, lets look at the power required in a few hypothetical calcs. For simplicity I will use volts for an AC sine wave treat the impedance as steady (say 8 ohms).

Ohms law, V = IR
power = VI

1) 8 volts = 1A * 8 Ohms
Power = 8 watts

2) half the voltage since we are now going to split the air movement to 2 speakers, so each one gets;

4 volts = 1/2A * 8 ohms
power = 2 watts

since we have two of them, the total power to keep the same air movement or SPL is 4 watts. That is half of what we needed with one speaker, or 3 db more efficient.

3) Or we can use two speakers keeping the power the same, so each one gets 4 watts to sum to the 8 watts in example 1.

5.66 volts = .707A * 8 ohms
Power = 4 watts

With 4 watts to each speaker or a total of 8 watts you are moving more air than the single speaker in example 1. With two speakers getting 5.66 volts each, you are moving the same air as a single speaker fed 11.3 volts

11.3 volts = 1.414A * 8ohms
Power = 16 watts

So, to get 3 db in SPL we can double the power to one speaker from 8 watts to 16 watts or we can double the speakers and feed them 4 watts each.

4) double the voltage to double the excursion (move twice as much air as example 1)
16 volts = 2A * 8 ohms
Power = 32 watts

So double the output from one speaker takes four times the power since both volts and current double. Double the air movement gives you +6db, which agrees with the 3db gain for every doubling of the power. Again, you can see this by adding the SPL in joules and converting to DB.

[Steve Conner]

I still don't buy it. You just state that two speakers, each with half the voltage, move the same air as one speaker with 1x voltage, but you don't offer any explanation of why this happens.

Are you just saying that doubling the cone area doubles the efficiency, because it "moves twice as much air" or something?

[m-fine]

The excursion of a speaker is linear with voltage. Air displaced is cone area (fixed) tines excursion. Double the voltage swing and you double the cone excursion and double the air moved which is measured as 6db higher SPL.

If you could double the cone area of a speaker while keeping everything else, including the moving mass, the same, you would add 3 db to sensitivity.

[Steve Conner]

I agree with you that the 3dB efficiency increase exists, and that larger speakers are more efficient by nature. I've just never quite understood the physical mechanism behind it.

I agree with the above statements except for two things. First, connecting two speakers together doubles the moving mass as well as the cone area, so where does that 3dB come from after all?

Second: Double air movement only means +6dB SPL if the wavefronts from the two drivers interfere constructively. In other places, they can cancel each other out and make the SPL lower, as you hinted in one of your earlier posts.

Maybe at bass frequencies the waves are so big relative to the array of drivers that they interfere constructively everywhere, and the naive "air movement" explanation holds.