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Old 24th June 2003, 01:29 AM   #1
future is offline future  Brazil
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Default Speaker sensibility

If I have a speaker that have 90dB SPL and I put two in a box, will my system have 93dB SPL? and if I keep doubling the number of speakers will I have an exponencial increase in sensibility?
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Old 24th June 2003, 01:40 AM   #2
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Theoretically yes, but practically no. The drivers have to be in such proximity to each other that their outputs add at the point in space where you are making the measurement. As you can imagine, as the number of drivers start to increase, the smaller the chance that the drivers outputs are going all sum at a single measurement point.
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Old 24th June 2003, 01:56 AM   #3
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I asked that because I saw some loudspeakers bigger than I am, with loads of speakers and wonder why they are built that way... they have to be conected in series/parallel in a way that it acts like a single 8ohms speaker and uses a single crossover...
I have a pair of JBL N24II 86dB and a acoustic research 108PSB subwoofer, I like its sound but it doesnt play too loud.
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Old 24th June 2003, 02:49 AM   #4
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Rodd:

My understanding is that 2 10" woofers placed next to each other will raise the SPL for any given drive voltage level 6 dB. Isn't that true? Of course, adding one more speaker in parallel converts a single 8 ohm speaker into a 4 ohm pair, which will draw twice the current at any given voltage level, which means that the pair draws more watts from the amp. Still, there is 3 dB advantage for the pair over the single driver, I thought.
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Old 24th June 2003, 05:00 AM   #5
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Hi Kelticwizard,

The advantage from two drivers over one should be qualified bexause it is both acoustical and electrical. If we keep the delivered power constant and double only the drivers, the gain is 3db. If we, as you suggest, maintain a constant voltage, and wire the two drivers in parallel, the current through the pair will double therefore doubling the delivered power. The gain would then be both acoustcal and electrical for a total gain of 6db.
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Old 24th June 2003, 08:23 AM   #6
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Default Dear Aunty DIY, I have a problem with this..

Dear Aunty DIY

I have a problem and would like your help.

Connecting two drivers in parallel halves the impedance and increases the sensitivity by 3dB. This we can be sure of and is immediately obvious when we have to turn the volume control down for the same SPL (loudness).

Regarding the increase in efficiency, I'm not at all convinced.

If we take another two drivers and connect all four in series/parallel so the impedance is the same as a single driver then we should get a 6dB increase in efficiency relative to the single driver. Now (and here's where I get a bit hazy) shouldn't this give us a 6dB rise in SPL for the same volume control setting? Please help me here Roddyama - it's still a bit early in the day for me.

Anyway, when I compare one drive unit with four in series/parallel, the four are NOT 6dB louder than the one. Nothing like. Maybe 2-3dB. Perhaps Roddyama's "proximity" thing comes into play here but, as you all know, I use 2" drivers and they're quite close together.

I think that something's not right with the conventional wisdom here. I've thought so for a while actually but have been too ashamed to mention it. I could be ostrasized from the entire audio community.

Please advise
"Worried" of Surrey
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Old 24th June 2003, 12:15 PM   #7
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Dear "Worried":

This has been covered before in this forum. I will try to dig up the thread.

My understanding is that the 6 dB in sensitivity-those who deal with it actually say it is "near 6 dB"-only occurs if the the 4 speakers are put into a square arrangement:

OO
OO


When you put them into vertical arrangement,

O
O
O
O

the increase in sensitivity becomes considerably less as the "mutual loading" concept lessens.

I'll try to locate the thread. I do remember that P. Lacombe mentioned that a pair had only an increase of 1.3 dB, and therefore a vertical arrangement of four speakers would have the 2 or 3 dB increase you mentioned.

Edit: Here it is:
System Efficiency when using multiple drivers
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Old 24th June 2003, 12:24 PM   #8
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Default Speaker Array Sensitivity

While ultimately the SPL of the speaker array would best be measured, in the development process you can compute the system sensitivity (or efficiency) impact of an array. First, assume that an individual driver in the line has a known SPL value. Next assume that the drivers in the array have overlapping acoustical radiation patterns and are spaced within a wavelength center-to-center from each other. Thus, the acoustic improvement in efficiency gain at 1 watts, 1 meter is given by:

Efficiency Gain = 10*log (Number of Drivers Driven)

while the sensitivity gain or loss at 2.83v, 1 meter is:

Sensitivity Gain/Loss = 10*log (Nominal Driver Impedance/Nominal Array Impedance)

If the nominal array impedance is less than the nominal driver impedance, the sensitivity increases or is a gain. If the array impedance is greater than the individual driver impedance, then the sensitivity decreases or becomes a loss.

Hence, for the overall system

System Efficiency Gain = SPL + Efficiency Gain

System Sensitivity = SPL + System Efficiency Gain + Sensitivity Gain/Loss

As an example, consider a case wherein we have 4 drivers connected in 2 parallel groups of 2 in series. Each individual driver is 8 ohms impedance and has an SPL of 85 dB. Hence, the efficiency gain, total impedance, and sensitivity gain are:

Efficiency Gain = 10 log 4 = 6.02 dB

Total Impedance of the Combination = 1 / (1/8 + 1/8) = 8 ohms

Sensitivity Gain = 10 log (8/8) = 0 dB

Hence, the sensitivity increase of the array is:

System Sensitivity = 85 + 6.02 + 0 = 91.02 dB.

Botom line though is that I find that measurements are preferred as the theoretical system sensitivity calculation tends to overstate the actual sensitivity for a large array.
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Old 24th June 2003, 12:34 PM   #9
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Default Ooops on Impedance Calculation for Example

For the example in my previous I screwed up the impedance calculation. Two 8 ohms drivers in series would be 16 ohms and then if you parallel two groups of these 2 in series, you get 8 ohms. The calculation is thus,

Z (overall) = 1/(1/16 + 1/16) = 8 ohms
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Old 24th June 2003, 03:11 PM   #10
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Being math challenged and simple minded, and since 'sound is round' as hitware on the FR forum likes to say, I try to envision the impact (or lack thereof) of two or more relatively transparent balls having a 'coming together' at some point in space. When they just barely touch, this ~represents pi*WL, the frequency where particle density summing begins and according to (proven) theory will be +3dB when they ~completely overlap (some much lower frequency), becoming as one. This implies that beyond pi*WL, any gain is due to reflections and varying rates of decay over distance of the surrounding BW so will vary depending on where in the polar response it's measured if all drivers are on the same plane.

Not very scientific, but has worked well for me in 'calculating' a real world approximation.

GM
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