Hi,
I'm seeking to have some of points clarified:
1. Quarter space (wall loading)
I understand that since you are only radiating into 1/4th of the total space when you put a speaker near 1 wall and a floor, the intensity of the sound will be 4 times as much (maybe a little more because of increased air impedance) as when suspended off the ground far away from any walls. I am unsure at what approximate frequency this effect stops being useful though. What is the general rule of thumb about which frequencies benefit and which don't?
2. Going from one driver to two increases the power sensitivity by 3 db
I think I have this one figure out, so I'll explain what I think and then hopefully someone can confirm or correct it.
P=V^2/R. If you connect the two drivers in parallel then the resistance will be 1/2, so at first look there is a 3db gain. However since you keep the voltage at 2.83V, that corresponds to 2 watts, so no wonder its 3db louder. If you connect them in series, the resistance will be 2x as much and it will be at -3db at 2.83V. Only 1/2 watt is being used in this case though, so of course it will be down 3db.
In other words: How you hook up the speakers does not change the power sensitivity.
So, why do people always say there is a 3db gain then? This is the tricky part that I'm not sure on. My take is that since the drivers are trying to simultaneously compress the air at the same point, this provides a better impedance match for each driver. In other words, since the air is pushing back harder, the driver is exerting more force on the air and is thus transferring more energy into it (assuming that the excursion hasn't changed much). Thus, the power sensitivity is increased. How all the variables add up to give a rather convenient 3db gain, I haven't figured out yet.
Does that seem right or am I way off?
Thanks for the clarification.
I'm seeking to have some of points clarified:
1. Quarter space (wall loading)
I understand that since you are only radiating into 1/4th of the total space when you put a speaker near 1 wall and a floor, the intensity of the sound will be 4 times as much (maybe a little more because of increased air impedance) as when suspended off the ground far away from any walls. I am unsure at what approximate frequency this effect stops being useful though. What is the general rule of thumb about which frequencies benefit and which don't?
2. Going from one driver to two increases the power sensitivity by 3 db
I think I have this one figure out, so I'll explain what I think and then hopefully someone can confirm or correct it.
P=V^2/R. If you connect the two drivers in parallel then the resistance will be 1/2, so at first look there is a 3db gain. However since you keep the voltage at 2.83V, that corresponds to 2 watts, so no wonder its 3db louder. If you connect them in series, the resistance will be 2x as much and it will be at -3db at 2.83V. Only 1/2 watt is being used in this case though, so of course it will be down 3db.
In other words: How you hook up the speakers does not change the power sensitivity.
So, why do people always say there is a 3db gain then? This is the tricky part that I'm not sure on. My take is that since the drivers are trying to simultaneously compress the air at the same point, this provides a better impedance match for each driver. In other words, since the air is pushing back harder, the driver is exerting more force on the air and is thus transferring more energy into it (assuming that the excursion hasn't changed much). Thus, the power sensitivity is increased. How all the variables add up to give a rather convenient 3db gain, I haven't figured out yet.
Does that seem right or am I way off?
Thanks for the clarification.
GestaltH said:Hi,
I'm seeking to have some of points clarified:
1. Quarter space (wall loading)
I understand that since you are only radiating into 1/4th of the total space when you put a speaker near 1 wall and a floor, the intensity of the sound will be 4 times as much (maybe a little more because of increased air impedance) as when suspended off the ground far away from any walls. I am unsure at what approximate frequency this effect stops being useful though. What is the general rule of thumb about which frequencies benefit and which don't?
Boundary wavelength to the driver and the frequencies below
the also inevitable cancellation nulls you will get. As baffle step
takes over the rear wall becomes irrelevant. FWIW a bass unit
must be mounted very close to the floor and wall over its
working range for the loading to be valid. In free spaced
loudspeakers boundary reinforcement enhances low bass.
By close spacing you lose this low bass boost.
2. Going from one driver to two increases the power sensitivity by 3 db
I think I have this one figure out, so I'll explain what I think and then hopefully someone can confirm or correct it.
P=V^2/R. If you connect the two drivers in parallel then the resistance will be 1/2, so at first look there is a 3db gain. However since you keep the voltage at 2.83V, that corresponds to 2 watts, so no wonder its 3db louder. If you connect them in series, the resistance will be 2x as much and it will be at -3db at 2.83V. Only 1/2 watt is being used in this case though, so of course it will be down 3db.
In other words: How you hook up the speakers does not change the power sensitivity.
no. parallel is +6dB and series the same for voltage sensitivity.
In terms of "efficiency" dB/W two coupled drivers are +3dB.
So, why do people always say there is a 3db gain then? This is the tricky part that I'm not sure on. My take is that since the drivers are trying to simultaneously compress the air at the same point, this provides a better impedance match for each driver. In other words, since the air is pushing back harder, the driver is exerting more force on the air and is thus transferring more energy into it (assuming that the excursion hasn't changed much). Thus, the power sensitivity is increased. How all the variables add up to give a rather convenient 3db gain, I haven't figured out yet.
They share the acoustic loading, n drivers go up 10logdB.
As a big driver can be considered to be a number of small drivers
in other words the bigger the driver the more efficient it is.
Does that seem right or am I way off?
see above
Thanks for the clarification.
/sreten.
2 speakers in parallel will add 3dB because the sound pressures are in phase, and 3dB because you've gone from 8ohm to 4ohm,
so 6dB overall output increase from the pair.
But in series, 2 speakers will add 3dB because the sound pressures are in phase, but lose 3dB because you've gone from 8ohm to 16ohm thus cutting the power in half.
Overall, then, +3dB - 3dB=0dB
so 6dB overall output increase from the pair.
But in series, 2 speakers will add 3dB because the sound pressures are in phase, but lose 3dB because you've gone from 8ohm to 16ohm thus cutting the power in half.
Overall, then, +3dB - 3dB=0dB
myhrrhleine said:
But isn't that if you assume constant voltage? If you keep that wattage the same, isn't it just 3db?
Example:
System One
Drivers: 1 (8ohm)
Watts: 1
Volts: 2.828
System Two
Drivers: 2 (4ohm total in when two x 8 ohm in parallel)
Watts: 1
Volts: 2
With both systems having the same Watts, doesn't the SPL increase by +3db from the first to second system.
Well, the bottomline is: two drivers are 3 dB more efficient than one, when they are given the same amount of total power. However, when you're looking at voltage, the change of impedance comes into play.
So two drivers in parallel have 3 dB extra sensitivity at 1 Watt (compared to one driver), but 6 dB more SPL at the same voltage because the impedance has been divided by two.
Two drivers in series have 3 dB extra sensitivity at 1 Watt (compared to one driver), but have the same SPL at the same voltage because of the impedance has multiplied by two.
So two drivers in parallel have 3 dB extra sensitivity at 1 Watt (compared to one driver), but 6 dB more SPL at the same voltage because the impedance has been divided by two.
Two drivers in series have 3 dB extra sensitivity at 1 Watt (compared to one driver), but have the same SPL at the same voltage because of the impedance has multiplied by two.
Because if your amp is stable to 4 Ohms... the 4 ohms speakers will miraculously get 2x the wattage out of the same amp then if you used a 8 Ohms speaker which equates to +3dB... it's an electrical thing.
If you cut the Resistance by half and maintain the Voltage at the same level here is what happens electrically:
If V= R x I, then V= R/2 x 2I, when you keep the same Voltage, reduce the Resistance you effectively must have double the Amps...
So for R, we have V x I = Watts
And for R/2 we have V x 2I = 2 times the Watts
If you keep the voltage at the same level, 2 speakers in parallel will SUCK 2 time more watts out of your amp.... you get a gain of +3 dB in SPL, or a gain of 1.4142 x the SPL.... Not double the SPL.
Am I off?
Re: Re: Seeking clarification of noob level questions.
Thanks for all the great replies. The question might be noob level, but apparently the answer is not.
Ok, so I understood very little of this. Could you please expand a little bit and and explain in easier to understand terms?
We appear to be saying the same thing, except that I might have misused the term "sensitivity". I purposely always said "power sensitivity" to try to clarify what I meant because I did not know the proper term. I don't think that "efficiency" is quite the right term either though. Usually that is defined to be useful power out / power in. Isn't this what Eta0 is? Or, can "efficiency" be used to describe db/w in this instance also?
When you say "share the acoustic loading," is this similar to the higher pressure effect that I described?
If not, how can this be? If we assume that 2x the input power produces 2x the output power, then if we give driver A 1/2 watt and driver B 1/2 watt and separate (isolate) them, then driver A will output 1/2 the power that a single driver at 1 watt would and the same will be true of driver B. Thus, the efficiency has not changed because it still adds up to the output power of one driver at 1 watt. Furthermore, since they are isolated and each only driven by 1/2 watt, the SPL will actually be -3db. If we bring the drivers close together, so that they are always in phase, then the pressure amplitude will be double what it was. This will result in the rms pressure being sqrt[2] higher. Because SPL is 20*LOG10[Prms/Pref] (Prms is the rms pressure and Pref is the reference pressure), this will result in a +3db gain. Thus, we are back to a +0 db change (-3db because they are driven by 1/2 watt +3db because they in phase and the pressure amplitude is 2x). Also, note that the efficiency hasn't changed.
So, none of that accounts for an increase in efficiency. In order to increase the efficiency, we need to increase the power output without increasing the power input. Thus, the amount of work done on the air by the driver needs to be increased. Since work is force*displacement, we either need a higher force or a longer displacement. What I was suggesting in my original post was that we get this increase in efficiency because of a higher force caused by increased back pressure.
Is this right? Have I missed something important?
Thanks for all the great replies. The question might be noob level, but apparently the answer is not.
sreten said:
Ok, so I understood very little of this. Could you please expand a little bit and and explain in easier to understand terms?
sreten said:
We appear to be saying the same thing, except that I might have misused the term "sensitivity". I purposely always said "power sensitivity" to try to clarify what I meant because I did not know the proper term. I don't think that "efficiency" is quite the right term either though. Usually that is defined to be useful power out / power in. Isn't this what Eta0 is? Or, can "efficiency" be used to describe db/w in this instance also?
sreten said:They share the acoustic loading, n drivers go up 10log
As a big driver can be considered to be a number of small drivers
in other words the bigger the driver the more efficient it is.
When you say "share the acoustic loading," is this similar to the higher pressure effect that I described?
If not, how can this be? If we assume that 2x the input power produces 2x the output power, then if we give driver A 1/2 watt and driver B 1/2 watt and separate (isolate) them, then driver A will output 1/2 the power that a single driver at 1 watt would and the same will be true of driver B. Thus, the efficiency has not changed because it still adds up to the output power of one driver at 1 watt. Furthermore, since they are isolated and each only driven by 1/2 watt, the SPL will actually be -3db. If we bring the drivers close together, so that they are always in phase, then the pressure amplitude will be double what it was. This will result in the rms pressure being sqrt[2] higher. Because SPL is 20*LOG10[Prms/Pref] (Prms is the rms pressure and Pref is the reference pressure), this will result in a +3db gain. Thus, we are back to a +0 db change (-3db because they are driven by 1/2 watt +3db because they in phase and the pressure amplitude is 2x). Also, note that the efficiency hasn't changed.
So, none of that accounts for an increase in efficiency. In order to increase the efficiency, we need to increase the power output without increasing the power input. Thus, the amount of work done on the air by the driver needs to be increased. Since work is force*displacement, we either need a higher force or a longer displacement. What I was suggesting in my original post was that we get this increase in efficiency because of a higher force caused by increased back pressure.
Is this right? Have I missed something important?
myhrrhleine said:
myhrrhleine said:
I was assuming that that waves would be in phase and roughly sinusoidal. This is where I came up with the value of sqrt[2] for the increase in rms pressure. If the waves don't correlate, then this effect doesn't work. Off the top of my head, I'm not sure what the rms increase in pressure would be for uncorrelated random noise being added together. My point was that even assuming perfectly in phase waves, this results in no db change because the +3 db cancels with the -3db drop from running each driver at only 1/2 watt.
myhrrhleine said:
I was also assuming constant power of 1 watt to the whole system and equally balance between the two drivers, so that no matter how they are hooked up they each have 1/2 watt.
I am interested in actual efficiency gain, not tricking the amp to supply more/less power. I didn't make this very clear in my original post. My results are for db/W not db/2.83V
myhrrhleine said:
Yes, I agree that summing identical waves gives you +3db, but that is added to an spl that is already -3db down because each wave is only powered by 1/2 watt. Does that make sense? Thus, there is no efficiency gain from this effect.
Does this make sense at all, or am I doing a poor job of trying to explain what I mean? I can try explaining what I'm getting after again
if its still unclear.
So, the only way that I can see that multiple drivers are more efficient is that there is higher back pressure for the cone to push against, similar to how the throat of a horn works.
myhrrhleine said:
Ok, I still don't think that I was being clear (or maybe I'm misunderstanding what you said):
"But 2 signals of 1/2w each coherent sines" =+0db
I'll illustrate with an example:
Wave 1: pressure=A*Sin[x], where A is the pressure amplitude corresponding to a 90db/w driver supplied with 1 watt. The SPL will be 90db.
In this case,
(1) SPL = 90db = 20*LOG10[Prms/Pref]
(2) Pref = 20 microPa (this is just what it is defined to be)
(3) Solving (1), Prms=632,456 microPa (this is just the rms value)
(4) Pressure amplitude = 2 * (rms pressure)^2. Pressure amplitude is just the A in A*Sin[x].
(5) Solving (4), A = 800,000 microPa
Thus, pressure in microPa is 800,000*Sin[x].
Wave 2: pressure=B*Sin[x], where B is the pressure amplitude corresponding to a 90db/w driver supplied with 1/2 watt. The SPL will be 86.9897db.
(1) SPL = 86.9897db = 20*LOG10[Prms/Pref]
(2) Pref = 20 microPa
(3) Prms = 447,214 microPa
(4) B = 2 * Prms^2
(5) B = 400,000 microPa
Thus pressure in microPa is 400,000 *Sin[x]
Wave 3: Identical to wave 2.
Pressure in microPa is 400,000 *Sin[x]
Wave 2 + Wave 3: Total power is 1 watt.
400,000*Sin[x]+400,000*Sin[x] = 800,000*Sin[x]
Rms pressure = 632456 microPa
SPL = 20*LOG10[632456/20] = 90db
So, you can see that running two drivers at half power produces the exact same SPL as one driver at full power if all other things are held constant. Thus, the db/W value has not changed by using two drivers. This says that the +3db gain that we know from experimenting exists has nothing to do with the adding of in phase sound waves.
sreten said:
Common error - sensitivity and efficiency are related and often confused, even by professionals.
Efficiency is measured in percent.
Sensitivity is measured in dB.
The true efficiency of a loudspeaker varies with frequency and the rated efficiency of a driver is just a relation, called "Power Available Efficiency", (PAE) based on its T/S parameters. Sensitivity is then just a PAE number, expressed in dB - either with an input of 1 Watt or with an input of 2.83V. Few manufacturers actually measure and specify this number.
Power Sensitivity is rated in dB @ 1 Watt @ 1 meter
Voltage sensitivity is rated in dB @ 2.83V @ 1 meter
The difference between these two numbers is:
diff_dB=10*log(8/Re), where Re is the DC resistance of the voice coil
Voltage sensitivity is more useful because amplifiers are constant voltage devices, for the most part.
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