If I have a lot of woofers with the following parameters:
8 ohms (flat)
100dB@1W/1m.
Then I connect two i paralell, and I get 106 dB@2,83 volts.
Then I connect anoter pair in series with the two. Then i get back to 8 ohms, but with still 106dB@2,83 volts.
Then I connect another four woofers in paralell, and then another 8 woofers in series we should have 112dB@2,83 volts and still 8 ohms.
If we go on with another 16 woofers in paralell and another 32 in series we are at 118 dB@2,83 volts. And on and on and on.
But if we had 1,2677E+30 drivers (a relatively large and expensive speaker) and connect them the same way. We should have about 400dB@1W/1m.
This does not make sense to me, but wat is actually wrong in the theory?
8 ohms (flat)
100dB@1W/1m.
Then I connect two i paralell, and I get 106 dB@2,83 volts.
Then I connect anoter pair in series with the two. Then i get back to 8 ohms, but with still 106dB@2,83 volts.
Then I connect another four woofers in paralell, and then another 8 woofers in series we should have 112dB@2,83 volts and still 8 ohms.
If we go on with another 16 woofers in paralell and another 32 in series we are at 118 dB@2,83 volts. And on and on and on.
But if we had 1,2677E+30 drivers (a relatively large and expensive speaker) and connect them the same way. We should have about 400dB@1W/1m.
This does not make sense to me, but wat is actually wrong in the theory?
I think you'd hit some point of diminishing returns where the meager power you're putting into the system is dissipated as low level heat due to resistance in the thousands of miles of wire it's going through. 
Also 1/(1,2677E+30) watt in any one driver probably won't be enough to overcome the internal friction involved in moving the spider/suspension out of it's resting position. Starting friction is always greater than moving friction, they say, and I would assume this would apply to moving the cone of a driver from rest (stopped).
All of these factors would eat away at effiency of a Very Large Array speaker like this when the number of drivers passes a certain point... This seems like a really fun thing to figure out with limit theory, but I don't think I'm up to the math.
Also 1/(1,2677E+30) watt in any one driver probably won't be enough to overcome the internal friction involved in moving the spider/suspension out of it's resting position. Starting friction is always greater than moving friction, they say, and I would assume this would apply to moving the cone of a driver from rest (stopped).
All of these factors would eat away at effiency of a Very Large Array speaker like this when the number of drivers passes a certain point... This seems like a really fun thing to figure out with limit theory, but I don't think I'm up to the math.
I was thinking about this not so long ago, and it occured to me that when you get such a large number of drivers, the wavelength of the sound would start to get in the way of using them all together. So, it would depend on their configuration and the desired bandpass.
I read here that an acoustic watt is equivalent to an SPL of 109 db at one meter, radiating spherically; therefore it's impossible to have something that produces more than 109 db from one watt in an omnidirectional pattern. Of course, some horns are over 109 db/watt/meter- but their radiation pattern is far from spherical.
Perhaps somebody with more knowledge about the equations used can let us know what the true relationship is between radiating surface area and efficiency. It's a log scale, and 3 db/doubling is really only a rule of thumb that applies over a certain range. Don't forget that when you're talking about efficiency, 100% is 100%
I read here that an acoustic watt is equivalent to an SPL of 109 db at one meter, radiating spherically; therefore it's impossible to have something that produces more than 109 db from one watt in an omnidirectional pattern. Of course, some horns are over 109 db/watt/meter- but their radiation pattern is far from spherical.
Perhaps somebody with more knowledge about the equations used can let us know what the true relationship is between radiating surface area and efficiency. It's a log scale, and 3 db/doubling is really only a rule of thumb that applies over a certain range. Don't forget that when you're talking about efficiency, 100% is 100%
True, though I think a super-array like this would have a quasi-planar radiation, since with any reasonable sized driver the array would be larger than the longest audible wavelength. You'd get something like "high-frequency" beaming from the array at all audible wavelengths. I would think the acoustic dropoff with distance would be different, in this case.
Snickers-is said:
Hi, Snickers-is. A fun idea.
I think the above sentence is the flaw. You get back to 8 ohms, and also back to 100dB.
By adding the series speaker you are potentially adding 6dB in total, but now each speaker is receiving half power, so EACH speaker is producing 6dB less. So the effect is 'plus 6dB' and 'minus 12dB', for a net loss of 6dB, back down from 106dB to 100dB.
Regards ... /Dave
Once you put two speakers in series and apply 2.83V, this voltage will be divided into two: 2.83/2 = something smaller.
Power = V^2/R (assuming ohmic resistance for simplicity)
These two points above is in principle enough to figure out what is going on with 1,2677E+30 drivers and resolve the paradox. Cheers,
Murat
Power = V^2/R (assuming ohmic resistance for simplicity)
These two points above is in principle enough to figure out what is going on with 1,2677E+30 drivers and resolve the paradox. Cheers,
Murat
Snickers-is said:
The error is that the formula assumes that the motion of the cone is mass-controlled. If you add a sufficient number of drivers, the radiation resistance finally dominates the mechanical impedance, and the efficiency flattens out towards something less than 100%.
Another way of looking at it, is that the model assumes that the cone motion of the individual drivers remain the same as you add more and more speakers. In the real world, given a sufficient number of drivers, the sum of their sound pressures reduce the movement of the cones. Again, this limits the efficiency to something less than 100%.
Another problem in this formula is that you can't add too much drivers or you lose the efficiency advantage.
Let say we want to reproduce a 10 kHz sinewave, we have a wavelength of about 3,4 cm. Even with 2.5 cm tweeters, using the pythagore formula and standing about 4 meters from the center of the baffle, let me do a triangle... If we know that at 4,017 meters from the side of the baffle, the tweeter will be out of phase. We have less than 37 centimeters maximum on each side before going out of phase with the listener. That means to get absolutely no cancellation, we need to stay under 18 cm per side so a baffle of 36 cm wide total.
Let say with something bigger than a 14x14 array of 2,5 cm tweeters, I guess you'll get cancellation with drivers being out of phase according to the listening position, so decreasing total efficiency.
But, a 14x14 array of tweeters should give at least a 110 dB / 1W / 1m ?
Let say we want to reproduce a 10 kHz sinewave, we have a wavelength of about 3,4 cm. Even with 2.5 cm tweeters, using the pythagore formula and standing about 4 meters from the center of the baffle, let me do a triangle... If we know that at 4,017 meters from the side of the baffle, the tweeter will be out of phase. We have less than 37 centimeters maximum on each side before going out of phase with the listener. That means to get absolutely no cancellation, we need to stay under 18 cm per side so a baffle of 36 cm wide total.
Let say with something bigger than a 14x14 array of 2,5 cm tweeters, I guess you'll get cancellation with drivers being out of phase according to the listening position, so decreasing total efficiency.
But, a 14x14 array of tweeters should give at least a 110 dB / 1W / 1m ?
First of all: Two similar drivers (or driver systems) connected in series will obtain their voltage sensitivity.
I am not sure if anyone really has got into the real center of the problem I believe we have some kond of loss factor that is controlled by an exponent, making the factor very low when we are dealing with reasonable numbers, but in my example I believe it at one point reaches a limit where this factor limits the output.
I am not sure if anyone really has got into the real center of the problem I believe we have some kond of loss factor that is controlled by an exponent, making the factor very low when we are dealing with reasonable numbers, but in my example I believe it at one point reaches a limit where this factor limits the output.
Dave's result is correct but I think
"By adding the series speaker you are potentially adding 6dB in total..." part is not. Whatever you add in series the total sensitivity should decrease. Here goes the argument:
1) "Then I connect two in paralell, and I get 106 dB@2,83 volts."
This is Ok.
2) "Then I connect anoter pair in series with the two. Then i get back to 8 ohms, but with still 106dB@2,83 volts."
No. You get back to 8 Ohms but not 106dB@2,83 volts. Think as follows: Let P denote the power of one driver,
P = V^2 / R = 2.83^2 / 8 = something
at 2.83 V. (Again we assume "lossy" DC resistance for simplicity but same arguments will apply for more realistic but much complicated AC driven inductive+mechanical load)
Now for the case 2) You will have V/2 at each driver. So, the power of EACH unit will be
P' = (V/2)^2 / R = something / 4
and the total power = 4 x something / 4 = something = P again.
Sensitivity can be defined in terms of power of the sound pressure; so you will get back to 8 ohms, 100dB. Cheers,
Murat
"By adding the series speaker you are potentially adding 6dB in total..." part is not. Whatever you add in series the total sensitivity should decrease. Here goes the argument:
1) "Then I connect two in paralell, and I get 106 dB@2,83 volts."
This is Ok.
2) "Then I connect anoter pair in series with the two. Then i get back to 8 ohms, but with still 106dB@2,83 volts."
No. You get back to 8 Ohms but not 106dB@2,83 volts. Think as follows: Let P denote the power of one driver,
P = V^2 / R = 2.83^2 / 8 = something
at 2.83 V. (Again we assume "lossy" DC resistance for simplicity but same arguments will apply for more realistic but much complicated AC driven inductive+mechanical load)
Now for the case 2) You will have V/2 at each driver. So, the power of EACH unit will be
P' = (V/2)^2 / R = something / 4
and the total power = 4 x something / 4 = something = P again.
Sensitivity can be defined in terms of power of the sound pressure; so you will get back to 8 ohms, 100dB. Cheers,
Murat
You gain 3 dB in total efficiency at each time you double the number of drivers frame-to-frame.
Let say the speaker is 87 dB / 1W / 1m and 87 dB @ 2,83V
2 = 90 dB / 1W / 1m total system efficiency
parallel = 93 dB @ 2,83V total system sensivity
series = 87 dB @ 2,83V total system sensivity
4 = 93 dB / 1W / 1m total system efficiency
parallel-parallel = 99 dB @ 2,83V total system sensivity
series-series = 87 dB @ 2,83V total system sensivity
series-parallel = 93 dB @ 2,83V total system sensivity
8 = 96 dB / 1W / 1m total system efficiency
parallel-parallel-parallel = 105 dB @ 2,83V total system sensivity
series-series-series = 87 dB @ 2,83V total system sensivity
series-parallel-series-parallel = 96 dB @ 2,83V total system sensivity
...
256 = 111 dB / 1 W / 1m total system efficiency
series-series-series-series-series-series-series-series = 87 dB @ 2,83V total system sensivity
parallel-parallel-parallel-parallel-parallel-parallel-parallel-parallel = 135 dB @ 2,83V total system sensivity
series-parallel-series-parallel-series-parallel-series-parallel = 111 dB @ 2,83V total system sensivity
...
Let say the speaker is 87 dB / 1W / 1m and 87 dB @ 2,83V
2 = 90 dB / 1W / 1m total system efficiency
parallel = 93 dB @ 2,83V total system sensivity
series = 87 dB @ 2,83V total system sensivity
4 = 93 dB / 1W / 1m total system efficiency
parallel-parallel = 99 dB @ 2,83V total system sensivity
series-series = 87 dB @ 2,83V total system sensivity
series-parallel = 93 dB @ 2,83V total system sensivity
8 = 96 dB / 1W / 1m total system efficiency
parallel-parallel-parallel = 105 dB @ 2,83V total system sensivity
series-series-series = 87 dB @ 2,83V total system sensivity
series-parallel-series-parallel = 96 dB @ 2,83V total system sensivity
...
256 = 111 dB / 1 W / 1m total system efficiency
series-series-series-series-series-series-series-series = 87 dB @ 2,83V total system sensivity
parallel-parallel-parallel-parallel-parallel-parallel-parallel-parallel = 135 dB @ 2,83V total system sensivity
series-parallel-series-parallel-series-parallel-series-parallel = 111 dB @ 2,83V total system sensivity
...
Use of the @2.83V makes things very confusing...
2.83 V into 8 ohms is 1 W.
2.83V into 4 ohms is 2 W
(assuming the load is fairly resistive and the amp can deliver the current require)
So if you put 2 8 ohm speakers in parallel you are comparing the output of 1 driver with 1 W and 2 drivers with 2 W.
Every time you double the number of drivers you get 3 DB.
Some amps put out more power into a lower impedance, some don't...
dave
2.83 V into 8 ohms is 1 W.
2.83V into 4 ohms is 2 W
(assuming the load is fairly resistive and the amp can deliver the current require)
So if you put 2 8 ohm speakers in parallel you are comparing the output of 1 driver with 1 W and 2 drivers with 2 W.
Every time you double the number of drivers you get 3 DB.
Some amps put out more power into a lower impedance, some don't...
dave
Re: Re: Sensitivity of multiple drivers?
Snickers is correct and the explanations from other supports this.
Adding an identical driver in series reducec power drawn from a voltage source with 1/2 which is 3dB. Since the Bl factor goes up due to two magnets and VC´s we get an increase in 3dB efficiency while we stay at the same voltage SENSITIVTIY. Adding a second series pair in paralell reduces impedance with 1/2 and we draw again 3dB more power from a voltage source while we again have doubled the magnets and VC´s for another 3dB increase of efficiency.
One 8ohm driver 90db/2.83V
Two in series, 16ohm, 90dB/2.83V
Four in series paralell, 8ohm, 96dB/2.83V
Of course this is only valid for low freqencies where the radiation is wide enough so the drivers can act as a virtual surface or "mirror" for eachother. Also it´s only true for voltage sources with low impedance... think Krell or ICE, not son of zen or some 300B-SE.
The number of drivers you can use to increase efficiency and sensitivity while keeping the impedance identical as with one driver is; 4, 9, 16, 25, 36, 49... and so on.
/Peter
DavidH said:
Snickers is correct and the explanations from other supports this.
Adding an identical driver in series reducec power drawn from a voltage source with 1/2 which is 3dB. Since the Bl factor goes up due to two magnets and VC´s we get an increase in 3dB efficiency while we stay at the same voltage SENSITIVTIY. Adding a second series pair in paralell reduces impedance with 1/2 and we draw again 3dB more power from a voltage source while we again have doubled the magnets and VC´s for another 3dB increase of efficiency.
One 8ohm driver 90db/2.83V
Two in series, 16ohm, 90dB/2.83V
Four in series paralell, 8ohm, 96dB/2.83V
Of course this is only valid for low freqencies where the radiation is wide enough so the drivers can act as a virtual surface or "mirror" for eachother. Also it´s only true for voltage sources with low impedance... think Krell or ICE, not son of zen or some 300B-SE.
The number of drivers you can use to increase efficiency and sensitivity while keeping the impedance identical as with one driver is; 4, 9, 16, 25, 36, 49... and so on.
/Peter
I am also quite sure about the simple theory here. When you have a powered subwofer an adds another one at same power level you gain 6 dB. When reducing the power to 1/4 of the original on each driver you are back at +0dB. That is the same as with the drivers in series.
Drivers in parallell adds 6 dB.
But one very interesting thing: When will some strange things hammen makeing the theory collapse, and what particukar thing would it be?
Drivers in parallell adds 6 dB.
But one very interesting thing: When will some strange things hammen makeing the theory collapse, and what particukar thing would it be?
Snickers-is said:
Not quite... if the original power is 1 W, and you add another woofer and keep the total power the same, ie 1W, you only get 3 dB. If you double the total power (ie you add another amp so each driver gets 1 W or your amp doubles power into half the impedance) you get another 3 dB.
At the same total power adding a driver in series or in parallel gives 3 dB extra level.
dave
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