SPL for more than 1 speaker

[jimboyogi]

Hi All,

I've run numerous searches both here and on google, so sorry if this has been covered before, but I still haven't found a definitive answer. Actually I have found three conflicting answers, so I wanted to ask those that I trust the most ie. you good people.

Simple question. I have 2 speakers with the same efficiency, eg. 100dB/W/m. I then connect them together, either in series or parallel. What SPL do I get now? (ignoring impedance changes, cabs etc.)

So far in my searches, I have read 0dB/W/m gain, 3dB/W/m gain, or somewhere in between

If there is a good thread about this that someone knows please point me in the right direction. If not, what to believe?

Cheers

[Chuck H]

AFAIK 100dB is 100dB no matter how you slice it. A three dB difference from one speaker to the next is the same as as double the wattage on the more efficient speaker as far as volume. So when pairng speakers it's usually best to know the strengths of the speakers and what you want. Otherwise it can be unpredictable. With two speakers of the same efficiency you can look at the plots and figure an even balance of the strength of either speaker. It's a best case scenario.

I can see where some guy's might think there is a volume advantage. One speaker demonstrates it's efficiency in the upper mids and the other in the bass so overall there is an increase in efficiency... That might be true if you were comparing two of one speaker next to two of the other. But taken individually the fequencies balance. Your efficiency is still 100dB because there is no doubling of any advantage.

Chuck

[redelephant]

+1. But the max spl, or the spl where the speakers start to break up/compress is 3dB louder. But for same watts in, the spl is the same. One watt is one watt, and 100dB per watt is just that.

[Steve Conner]

Many, many threads on this in the archives, some of them quite ill-tempered. Use the search function unless you fancy having the same argument all over again.

My personal answer: If the total power delivered to the speaker system remains the same, then the total radiated sound power (SWL) remains the same, because conservation of energy demands it.

But the SPL measured on axis goes up by 3dB, because of directivity gain. (Each speaker gets -3dB power because it's split between them, but the wavefronts from them sum coherently on-axis, giving twice the sound pressure: +6dB, for a total of +3dB.)

Since we don't have gigantic ears that can envelop a speaker system completely, then you can argue that on-axis SPL is the more relevant measurement, and claim that the efficiency did indeed go up by 3dB. But this just means that more sound gets thrown towards the audience, and less gets wasted in useless directions.

[jimboyogi]

Thanks people.

Steve, I tried the search function, trying a bunch of different key words/phrases, and came up with zip on this topic. I did learn a bunch of other interesting things though as I read through a heap of threads looking

I think that I must be using the search incorrectly, I figured that this MUST have been covered before at length.

But anyway, I think that you three have covered it off beautifully, so thanks again.

[J M Fahey]

Saying the same but in another way.
1) In fact *there is* a slight efficiency increase, because, to put it in layman language, when one piston starts compressing air in front of it, that air is already somewhat compressed by the other piston, and viceversa.
The problem is, to double efficiency you should have air with double density, meaning 2 Atmospheres pressure.
You will have nothing like that, by several orders of magnitude, so the increase although physically possible, is unmeasurable.
That effect *is* used in compression drivers (wonder why they call them that way [] where they achieve it by letting the piston radiate into a very small tight chamber ... but on free space, it gets unmeasurable.
2) What you will have, is an increase in directivity, which will produce an increase in SPL *on axis* , don't know if up to the theoretical +3dB and definitely only at certain frequencies.
All of this is accurately predicted by the dipole radiator theory, as compared to point source radiation.
As stated above, you get this extra SPL on axis, at the expense of lower SPL off axis, with the added problem of very bumpy response, phase cancellations, you name it.
EDIT: these "Line Array" PA guys do it to good effect, but they are *very* careful about phase delay, getting a coherent wavefront across many octaves, etc.
They "manufacture" a cylindrical wave (instead of the regular soherical one) and really get to the back of the Stadium with quality sound and high SPL, but the system construction is very critical.

[m-fine]

You do get the 3 db gain. If you have a 1 watt amp and a 100 db/w speaker, you get 100 db of spl. If you add a second amp and speaker in phase with the first, you will get 106 db spl (convert db to joules to add). If you simply double the power in the first amp, you will not get 106 db, only 103 db. By doubling the number of drivers you are getting an additional 3 db of spl. This is not magic, it is simply a matter of the spl output being non linear (two drivers with lower excursion being more efficient than a single driver with higher excursion). The key assumption is that they are in phase. As long as the drivers are close together they will be in phase a lower frequencies, but at higher frequencies you can get zones of gain and zones of cancelation depending on the spacing and the measurement/listening location. This is called comb filtering.

This concept is common knowledge in the hifi world and an important aspect of speaker design where multiple woofers or mid drivers are matched with a single more efficient tweeter. You can often see this by looking at a line of speakers that use the same drivers. The larger multiple driver models will typically have a higher sensitivity than the single woofer bookshelf model.

[R.G.]

Saying the same but in another way.
...The problem is, to double efficiency you should have air with double density, meaning 2 Atmospheres pressure.
You will have nothing like that, by several orders of magnitude, so the increase although physically possible, is unmeasurable.
That effect *is* used in compression drivers (wonder why they call them that way [] where they achieve it by letting the piston radiate into a very small tight chamber ... but on free space, it gets unmeasurable.

Horn enclosures are acoustic transformers. The problem with loudspeakers in general is that there is a big impedance mismatch between speakers (high density, solid) and air (low density, gaseous). The loudspeaker can't get a good "grip" on the air. Hence the bigger the loudspeaker and the lighter the radiating surface, the better "matched" its impedance is to air.

One analogy which is easy to grasp is the difference between walking and swimming. When you're walking, you interact with a solid surface with your feet. When you swim, the water is much less dense and solid than you feet (and hands, and so on) so power transfer to the water is much poorer. To do better, you can put on fins which have a bigger surface and match your muscular effort to the water's resistance better. Speakers have a much harder time, as air is 800 times less dense than water.

I guess I could have used the "flap your arms to fly" analogy but it seemed further out. 8-)

In a horn enclosure, the speaker does not radiate into open air, but into the entry of a tube whose cross sectional area provides a better impedance match to the speaker radiating area. The speaker can drive the air just in the horn throat with much better power transfer. The slowly-expanding throat of the horn matches the high pressure waves at the throat to the low pressure, but wide extent waves at the mouth.

m-fine
You do get the 3 db gain. If you have

Yes. What he said and how he explained it.

[J M Fahey]

The problem with loudspeakers in general is that there is a big impedance mismatch between speakers (high density, solid) and air (low density, gaseous).

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").

[Enzo]

And when you get these conflicting answers you found, remember there are separate questions being lumped together.


You can;t just say put two speakers of equal whatever together, unless you also plan to separately power them and what all. As soon as you connect two speakers to the same amp, they are now in parallel which, them being equal, drops the impedance of the load in half. (Or they are in series which doubles the impedance and halves the power.) Depending upon the amp type, it may double the power from the amp or may not. This cannot be ignored, but is a separate issue entirely from acoustic coupling, and beaming of sound and accoustic efficiency.


And then there is that whole thing with your subwoofers coupling to the floor, and also to the wall, and even to a corner.

[J M Fahey]

(Or they are in series which doubles the impedance and halves the power.)

Enzo, you gave me an excellent idea.
Actual testing beats everything else.
I'm finishing a 2x12" 100W Bass amplifier, to be picked tomorrow by a customer.
If I have 1 "free" hour, I'll make the following experiment:
1) I'll feed pink noise into the power amp, to get, say, 4 volts output.
2) I'll wire both speakers in series, getting 16 ohms total, they'll receive 1W RMS nominal.
Measure SPL with my RION NL-05 precision meter.
3) changing nothing, short one of the speakers.
Impedance halves, power doubles (+3dB), total efficiency *should* halve according to theory (-3dB), SPL measurement should be the same as before, *if* adding an extra speaker adds +3dB.
If you want me to write the same but going the other way:
same conditions as above, but I start with *one* of the speakers shorted, the other one receiving nominal 2 W (4 volts into 8 ohms), I record SPL; then open the switch , power goes to 1W (4 V into 16 ohms), losing 3 dB; if SPL stays the same, the system is twice as efficient as before (+ 3dB), theory is proved; if not, it's disproved.
Any hints?
You are all welcome, tomorrow I'll post the actual results.
PS: obviously I'll put the microphone element centered, at the same distance from both cone centers.
Both speakers will be in the same enclosure.

[m-fine]

If you have a head with multiple output taps, you can test while maintaining the impedance match.

[redelephant]

You do get the 3 db gain. If you have a 1 watt amp and a 100 db/w speaker, you get 100 db of spl. If you add a second amp and speaker in phase with the first, you will get 106 db spl (convert db to joules to add). If you simply double the power in the first amp, you will not get 106 db, only 103 db. By doubling the number of drivers you are getting an additional 3 db of spl. This is not magic, it is simply a matter of the spl output being non linear (two drivers with lower excursion being more efficient than a single driver with higher excursion). The key assumption is that they are in phase. As long as the drivers are close together they will be in phase a lower frequencies, but at higher frequencies you can get zones of gain and zones of cancelation depending on the spacing and the measurement/listening location. This is called comb filtering.

This concept is common knowledge in the hifi world and an important aspect of speaker design where multiple woofers or mid drivers are matched with a single more efficient tweeter. You can often see this by looking at a line of speakers that use the same drivers. The larger multiple driver models will typically have a higher sensitivity than the single woofer bookshelf model.

We are talking about adding one driver to the mix and still using the same ONE amplifier. So if u have one watt into a 100dB/W driver, my guess is u get 100dB SPL at one meter from that single driver. Now add one identical driver either in series or parallel, and keep the power at one watt from that same ONE amp, each driver sees 0.5watts b/c they share exactly the whole watt, and u end up with no more SPL than u started with. Watt for watt.

My pennies in this pot is that u get potentially higher SPL at the point where the speakers begin to compress. In other words, if max SPL of those drivers is 110dB, then two of those will give you 113dB. But that requires an amp that can drive them to their max, and now we are not talking about the same as was the question.

[m-fine]

Not correct.

If each driver got 1 watt, the SPL WOULD BE 106, but since each only gets half a watt, the SPL is 103 db. This can be easily confirmed in empiracle testing or through fairly basic mathematics. You need to convert db to joules to be able to add and subtract the energy transfered, but when you convert back to db you will see how it works.

Simply put (assuming you are in phase and within certain
parameters)...

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.


So, take a 1 watt amp and add a second driver while maintaining an impedance match, and you gain 3 db. Keep the single driver and go to 2 watts and you get the same 3 db. Add a second 1 watt amp or double power and the number of speakers and you add 6 db.

[m-fine]

Here us another way to look at it. 1 watt into 1 100 db/w speaker. You have 100 db at 1m. Now cut the power to 0.5 watts and you get 97 db. Anyone dissagree with this (I hope not). Now add a second driver also getting 0.5 watts. No one should argue that that driver is also producing 97 db. Finally add two 97 db signals that are in phase, and the answer is 103 db.

Google "db SPL" for the formulas to convert db spl to joules and back to db. I am posting from a phone so I can not paste it in here at the moment.