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Speaker sensitivity and fs in multiples.
How does using speakers in arrays affect their sensitivity and
total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. Or a pair, or trio, either in parallel or series -- is there a formula to determine the sensitivity change in a group of similar speakers units when you know what it is for an individual driver? Also, when grouping woofers, say I have four 8 inch woofers in a box, each has an fs of 55 hz. How does the res. freq. change with the multiplication of woofers, and therefore the calculation of the port if it's not a sealed box? TIA ________________ Marc Stager |
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#2
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Speaker sensitivity and fs in multiples.
On Wed, 26 Nov 2003 01:43:24 GMT, Stager
wrote: How does using speakers in arrays affect their sensitivity and total resonant frequency? specific questions snipped I don't know the answers to your questions, but they can all be found in Vance Dicakason's book Loudspeaker Design Cookbook. If you don't get any answers before I come home tonight (which I am sure you'll get) I will look them up in the book for you. Perhaps you would need the book yourself? It is full of facts and procedures and easy to understand. Per. |
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#3
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Speaker sensitivity and fs in multiples.
On Wed, 26 Nov 2003 01:43:24 GMT, Stager
wrote: How does using speakers in arrays affect their sensitivity and total resonant frequency? specific questions snipped I don't know the answers to your questions, but they can all be found in Vance Dicakason's book Loudspeaker Design Cookbook. If you don't get any answers before I come home tonight (which I am sure you'll get) I will look them up in the book for you. Perhaps you would need the book yourself? It is full of facts and procedures and easy to understand. Per. |
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#4
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Speaker sensitivity and fs in multiples.
Stager wrote in message ...
How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Or a pair, or trio, either in parallel or series -- is there a formula to determine the sensitivity change in a group of similar speakers units when you know what it is for an individual driver? Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedanmce of the array is, and then examine how it is divided amongst the drivers. Three drivers in series, for example, will in total dissipate 1/3 the power of a single driver, since its total impedance is 3 times higher, with each driver talking 1/3 of that total. Those same three drivers in parallel will take 3 times the power, since the impedance is 1/3 of the original. So, in general, you can say that n drivers in parallel will have n times the sensitivity of one, while n in series will have 1/n the sensivitity. Also, when grouping woofers, say I have four 8 inch woofers in a box, each has an fs of 55 hz. How does the res. freq. change with the multiplication of woofers, and therefore the calculation of the port if it's not a sealed box? Putting n drivers in a box makes the box look n times smaller than for a single driver. In other words, if you put 4 drivers in a single box with a volume of 100 liters, it's the same (at low frequencies) as putting each driver in a 25 liter box. In general, the effect is the same whether the box is vented or not. |
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#5
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Speaker sensitivity and fs in multiples.
Stager wrote in message ...
How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Or a pair, or trio, either in parallel or series -- is there a formula to determine the sensitivity change in a group of similar speakers units when you know what it is for an individual driver? Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedanmce of the array is, and then examine how it is divided amongst the drivers. Three drivers in series, for example, will in total dissipate 1/3 the power of a single driver, since its total impedance is 3 times higher, with each driver talking 1/3 of that total. Those same three drivers in parallel will take 3 times the power, since the impedance is 1/3 of the original. So, in general, you can say that n drivers in parallel will have n times the sensitivity of one, while n in series will have 1/n the sensivitity. Also, when grouping woofers, say I have four 8 inch woofers in a box, each has an fs of 55 hz. How does the res. freq. change with the multiplication of woofers, and therefore the calculation of the port if it's not a sealed box? Putting n drivers in a box makes the box look n times smaller than for a single driver. In other words, if you put 4 drivers in a single box with a volume of 100 liters, it's the same (at low frequencies) as putting each driver in a 25 liter box. In general, the effect is the same whether the box is vented or not. |
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#7
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Speaker sensitivity and fs in multiples.
(Dick Pierce) wrote:
In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Dick, Is there a point (or range) in the overall frequency response that increases, due to mutual coupling from using multiple drivers? Harvey Gerst Indian Trail Recording Studio http://www.ITRstudio.com/ |
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Speaker sensitivity and fs in multiples.
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Speaker sensitivity and fs in multiples.
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Speaker sensitivity and fs in multiples.
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Speaker sensitivity and fs in multiples.
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Speaker sensitivity and fs in multiples.
I'm still not too clear on the answer. I'm not actually
looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. I use several such multi-speaker arrays. There are four Audax mids (each is 100 dB @ 1W) in each mid-range box, paired at each stack, and four 8 ohm JBL 8 ohm 2405's in each HF cluster (each about 104 dB) in these stacks: http://www.naumburgconcerts.org/gallery/orc2.htm http://www.naumburgconcerts.org/gallery/sarah-reh-1.htm ( At Lincoln Center and Central Park ) I have smaller mid-high speakers with 3 Audax mids and 2 2405's in each. All are triamped. http://www.marc.stager.com/sss/A_Shankar.jpg ( At Met Mseum of Art ) They are all very efficient, but I was curious, exactly HOW efficient. ________________ Marc Stager Stager Sound Systems NYC |
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Speaker sensitivity and fs in multiples.
I'm still not too clear on the answer. I'm not actually
looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. I use several such multi-speaker arrays. There are four Audax mids (each is 100 dB @ 1W) in each mid-range box, paired at each stack, and four 8 ohm JBL 8 ohm 2405's in each HF cluster (each about 104 dB) in these stacks: http://www.naumburgconcerts.org/gallery/orc2.htm http://www.naumburgconcerts.org/gallery/sarah-reh-1.htm ( At Lincoln Center and Central Park ) I have smaller mid-high speakers with 3 Audax mids and 2 2405's in each. All are triamped. http://www.marc.stager.com/sss/A_Shankar.jpg ( At Met Mseum of Art ) They are all very efficient, but I was curious, exactly HOW efficient. ________________ Marc Stager Stager Sound Systems NYC |
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Speaker sensitivity and fs in multiples.
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#17
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Speaker sensitivity and fs in multiples.
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#18
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Speaker sensitivity and fs in multiples.
Stager wrote in message ...
I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. |
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#19
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Speaker sensitivity and fs in multiples.
Stager wrote in message ...
I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. |
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#20
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Speaker sensitivity and fs in multiples.
On 29 Nov 2003 10:45:12 -0800, (Svante)
wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. d _____________________________ http://www.pearce.uk.com |
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#21
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Speaker sensitivity and fs in multiples.
On 29 Nov 2003 10:45:12 -0800, (Svante)
wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. d _____________________________ http://www.pearce.uk.com |
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#22
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 18:49:46 +0000, Don Pearce
wrote: On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Wrong, if the speakers are close together in terms of wavelength. Although in the case being discussed 4 v 1 driver, the total input power is the same (4 * 1/4) the total displacement is 4 * 1./2. The drive voltage is 1./2 ( sqrt (1/4)). If the speakers are close together the individual displacements simply add. |
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#23
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 18:49:46 +0000, Don Pearce
wrote: On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Wrong, if the speakers are close together in terms of wavelength. Although in the case being discussed 4 v 1 driver, the total input power is the same (4 * 1/4) the total displacement is 4 * 1./2. The drive voltage is 1./2 ( sqrt (1/4)). If the speakers are close together the individual displacements simply add. |
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#24
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Speaker sensitivity and fs in multiples.
"Svante" wrote in message
om... (Dick Pierce) wrote in message . com... Stager wrote in message ... How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. |
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#25
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Speaker sensitivity and fs in multiples.
"Svante" wrote in message
om... (Dick Pierce) wrote in message . com... Stager wrote in message ... How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. |
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#26
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Speaker sensitivity and fs in multiples.
"Goofball_star_dot_etal" wrote in
message ... In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. The question was about sensitivity which includes such things as efficiency and directivity. Well, I thought this was about combining drivers of known sensitivity. Dick said drivers in a series/parallel combination that results in the same impedance will have the same sensitivity. Someone posted Dick was wrong, but in fact he is correct. In general n drivers in parallel have n times the sensitivity of one, while n in series will have a sensitivity of 1/n. This can be extended to series/parallel combinations: Take identical drivers each with a sensitivity of n: Put two in parallel you get 2n. Put two in series you get n/2. Put two in parallel and series with another two in parallel: You get (2n*2n)/(2n+2n)=4n^2/4n=n (i.e. the same as a single driver). Dick was right. |
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#27
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Speaker sensitivity and fs in multiples.
"Goofball_star_dot_etal" wrote in
message ... In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. The question was about sensitivity which includes such things as efficiency and directivity. Well, I thought this was about combining drivers of known sensitivity. Dick said drivers in a series/parallel combination that results in the same impedance will have the same sensitivity. Someone posted Dick was wrong, but in fact he is correct. In general n drivers in parallel have n times the sensitivity of one, while n in series will have a sensitivity of 1/n. This can be extended to series/parallel combinations: Take identical drivers each with a sensitivity of n: Put two in parallel you get 2n. Put two in series you get n/2. Put two in parallel and series with another two in parallel: You get (2n*2n)/(2n+2n)=4n^2/4n=n (i.e. the same as a single driver). Dick was right. |
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#28
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Speaker sensitivity and fs in multiples.
In article , "Rusty Boudreaux" wrote:
"Svante" wrote in message . com... (Dick Pierce) wrote in message .com... Stager wrote in message ... How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. Somewhere there is confusion here. Let me add to the confusion and see if I'm right. I always work with these things on paper to add things up. I,m going to simplify. If you take two series 8 ohm drivers, they will play at the same volume with half the input power. Now add another paralleled set. They will now absorb twice the input power and the output volume will increse 6 dB. You have doubled the sound field. So I say the efficiency incresed by 3 dB and you have a 93db sensitivity. If you had 4 series 8 ohm drivers, it would play at the same level with 1/4 the power of if one driver. If you took 4 paralled sets of these, it would absorb the power of one, and play at 12 dB louder. But I don't know for sure. I think approaching 10 dB efficiency is a hazy area, where its really difficult to maintain phase coherence. One interestin fact solid state amps and 8 ohm drivers. Two in parallel, and you can quadruple the amplifier rated watts of one driver rated power. Four in series, and you can almost forget about blowing them out. This is only true when we are speaking of 8 ohm power rating of an amplifier. When making arrays, the overall system response tends to go downward of one driver, because of the tendancy of the higher frequencies not being as phase coherant. In reality the lower end does not change, its the upper end that changes. greg |
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#29
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Speaker sensitivity and fs in multiples.
In article , "Rusty Boudreaux" wrote:
"Svante" wrote in message . com... (Dick Pierce) wrote in message .com... Stager wrote in message ... How does using speakers in arrays affect their sensitivity and total resonant frequency? For example, say I have four 8 ohm mids, each individually rated at 90 dB sensitivity (at 1 meter, 1 watt). If I do a series-parallel array, I know I'll still be at 8 ohms, but how will the sensitivity change from the original individual 90 dB. In a series parallel array of the kind you're talking about, for a given voltqge across the array, the total power in the array is the same, since the impedance is the same, but each driver dissipates 1/4 of the total power, leading to the same sensitivity as a single driver. Sorry, Dick, I think you are wrong here. The efficiency is actually Nope, Mr. Pierce is quite correct and you are wrong. Since the impedance is the same and the drive level is the same the total output is the same. Please explain how four drivers at one quater 'X' watts can be anything other than the same as a single driver driven at 'X' watts. Somewhere there is confusion here. Let me add to the confusion and see if I'm right. I always work with these things on paper to add things up. I,m going to simplify. If you take two series 8 ohm drivers, they will play at the same volume with half the input power. Now add another paralleled set. They will now absorb twice the input power and the output volume will increse 6 dB. You have doubled the sound field. So I say the efficiency incresed by 3 dB and you have a 93db sensitivity. If you had 4 series 8 ohm drivers, it would play at the same level with 1/4 the power of if one driver. If you took 4 paralled sets of these, it would absorb the power of one, and play at 12 dB louder. But I don't know for sure. I think approaching 10 dB efficiency is a hazy area, where its really difficult to maintain phase coherence. One interestin fact solid state amps and 8 ohm drivers. Two in parallel, and you can quadruple the amplifier rated watts of one driver rated power. Four in series, and you can almost forget about blowing them out. This is only true when we are speaking of 8 ohm power rating of an amplifier. When making arrays, the overall system response tends to go downward of one driver, because of the tendancy of the higher frequencies not being as phase coherant. In reality the lower end does not change, its the upper end that changes. greg |
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#30
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 20:54:50 -0600, "Rusty Boudreaux"
wrote: Well, I thought this was about combining drivers of known sensitivity. Yes Dick said drivers in a series/parallel combination that results in the same impedance will have the same sensitivity. Someone posted Dick was wrong, but in fact he is correct. Dick is always right. He said "Ignoring acoustic effects" and then went out for lunch, it seems, without elaborating on the magnitude of what he was ignoring. In general n drivers in parallel have n times the sensitivity of one, while n in series will have a sensitivity of 1/n. This can be extended to series/parallel combinations: Take identical drivers each with a sensitivity of n: Put two in parallel you get 2n. Put two in series you get n/2. Put two in parallel and series with another two in parallel: You get (2n*2n)/(2n+2n)=4n^2/4n=n (i.e. the same as a single driver). Dick was right. Ok, lets try again. In this four driver series-parallel arrangement the applied voltage and current to each driver is halved so the power to each (V*I) is one quarter. Well above resonance where mass dominates Half the current results in half the acceleraton (F=ma). Half the acceleration results in half the linear displacement, half the velocity, half the volume displacement, half the volume velocity, half the acoustic prresure etc. ( all at constant frequency). The same proportionality applies over the whole operating range. Mecanical work is force times distance or to get the same units pressure times volume. Power is the rate of doing work so pressure times volume velocity. Each of the drivers ion its own if supplied with quarter of the power will produce half the pressure and half the volume velocity and quatrer of the accoustic power. However in close proximity to other drivers the pressures will add up 1/2 +1/2+1/2+1/2 =2 and the volume displacements will add up 1/2+1/2+1/2+1/2 = 2 and 2 * 2 = 4 I think you must agree. Dick has written Specifically, the efficiency is determined by the following relationship: 2 2 2 p0 B l Sd n0 = ------ * ---- * ------- 2 pi c Re Mms^2 where n0 is the reference efficiency of the driver p0 is the densiy of air = 1.18 kilograms per cubic meter pi is 3.1415926535... c is the velocity of sound in air =~ 342 meters per second B is the flux density of the magnetic field in the voice coil gap in Tesla (1 T = 10,000 Gauss) l is the length of the voice coil wire in the magnet field B in meters Re is the DC resistance of the voice coil in Ohms Sd is the area of the cone in square meters Mms is the moving mass of the cone in kilograms Phew! End quote It is not obvious how this relates to multiple drivers but it is clear that the acoustic output power is proportional to the volume displacement squared. |
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#31
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 20:54:50 -0600, "Rusty Boudreaux"
wrote: Well, I thought this was about combining drivers of known sensitivity. Yes Dick said drivers in a series/parallel combination that results in the same impedance will have the same sensitivity. Someone posted Dick was wrong, but in fact he is correct. Dick is always right. He said "Ignoring acoustic effects" and then went out for lunch, it seems, without elaborating on the magnitude of what he was ignoring. In general n drivers in parallel have n times the sensitivity of one, while n in series will have a sensitivity of 1/n. This can be extended to series/parallel combinations: Take identical drivers each with a sensitivity of n: Put two in parallel you get 2n. Put two in series you get n/2. Put two in parallel and series with another two in parallel: You get (2n*2n)/(2n+2n)=4n^2/4n=n (i.e. the same as a single driver). Dick was right. Ok, lets try again. In this four driver series-parallel arrangement the applied voltage and current to each driver is halved so the power to each (V*I) is one quarter. Well above resonance where mass dominates Half the current results in half the acceleraton (F=ma). Half the acceleration results in half the linear displacement, half the velocity, half the volume displacement, half the volume velocity, half the acoustic prresure etc. ( all at constant frequency). The same proportionality applies over the whole operating range. Mecanical work is force times distance or to get the same units pressure times volume. Power is the rate of doing work so pressure times volume velocity. Each of the drivers ion its own if supplied with quarter of the power will produce half the pressure and half the volume velocity and quatrer of the accoustic power. However in close proximity to other drivers the pressures will add up 1/2 +1/2+1/2+1/2 =2 and the volume displacements will add up 1/2+1/2+1/2+1/2 = 2 and 2 * 2 = 4 I think you must agree. Dick has written Specifically, the efficiency is determined by the following relationship: 2 2 2 p0 B l Sd n0 = ------ * ---- * ------- 2 pi c Re Mms^2 where n0 is the reference efficiency of the driver p0 is the densiy of air = 1.18 kilograms per cubic meter pi is 3.1415926535... c is the velocity of sound in air =~ 342 meters per second B is the flux density of the magnetic field in the voice coil gap in Tesla (1 T = 10,000 Gauss) l is the length of the voice coil wire in the magnet field B in meters Re is the DC resistance of the voice coil in Ohms Sd is the area of the cone in square meters Mms is the moving mass of the cone in kilograms Phew! End quote It is not obvious how this relates to multiple drivers but it is clear that the acoustic output power is proportional to the volume displacement squared. |
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Speaker sensitivity and fs in multiples.
"The same proportionality applies over the whole operating range" Should read "At any given frequency the displacement is proportional to applied voltage or current" |
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#33
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Speaker sensitivity and fs in multiples.
"The same proportionality applies over the whole operating range" Should read "At any given frequency the displacement is proportional to applied voltage or current" |
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 19:19:30 GMT, ow
(Goofball_star_dot_etal) wrote: On Sat, 29 Nov 2003 18:49:46 +0000, Don Pearce wrote: On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Wrong, if the speakers are close together in terms of wavelength. Although in the case being discussed 4 v 1 driver, the total input power is the same (4 * 1/4) the total displacement is 4 * 1./2. The drive voltage is 1./2 ( sqrt (1/4)). If the speakers are close together the individual displacements simply add. No, you haven't understood. d _____________________________ http://www.pearce.uk.com |
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Speaker sensitivity and fs in multiples.
On Sat, 29 Nov 2003 19:19:30 GMT, ow
(Goofball_star_dot_etal) wrote: On Sat, 29 Nov 2003 18:49:46 +0000, Don Pearce wrote: On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Wrong, if the speakers are close together in terms of wavelength. Although in the case being discussed 4 v 1 driver, the total input power is the same (4 * 1/4) the total displacement is 4 * 1./2. The drive voltage is 1./2 ( sqrt (1/4)). If the speakers are close together the individual displacements simply add. No, you haven't understood. d _____________________________ http://www.pearce.uk.com |
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Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Sorry, you're wrong. For HIGH frequencies (averaged over directions and frequencies), yes, but for low frequencies the decreased level off axis that you describe will never occur. The contributions from the speakers will be in phase, and thus add up to a pressure corresponding to n*p1 (n = number of speakers, p1 being the sound pressure from one speaker). Input power will be multiplied by n, but output power by n^2, since acoustic power is proportional to sound pressure squared. Thus efficiency increases linearly with the number of speakers, (n^2)/n. |
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Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Sorry, you're wrong. For HIGH frequencies (averaged over directions and frequencies), yes, but for low frequencies the decreased level off axis that you describe will never occur. The contributions from the speakers will be in phase, and thus add up to a pressure corresponding to n*p1 (n = number of speakers, p1 being the sound pressure from one speaker). Input power will be multiplied by n, but output power by n^2, since acoustic power is proportional to sound pressure squared. Thus efficiency increases linearly with the number of speakers, (n^2)/n. |
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Speaker sensitivity and fs in multiples.
On 30 Nov 2003 02:06:41 -0800, (Svante)
wrote: Don Pearce wrote in message . .. On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Sorry, you're wrong. For HIGH frequencies (averaged over directions and frequencies), yes, but for low frequencies the decreased level off axis that you describe will never occur. The contributions from the speakers will be in phase, and thus add up to a pressure corresponding to n*p1 (n = number of speakers, p1 being the sound pressure from one speaker). Input power will be multiplied by n, but output power by n^2, since acoustic power is proportional to sound pressure squared. Thus efficiency increases linearly with the number of speakers, (n^2)/n. Please read my comment about the far field. That covers everything you have said here. But I will repeat for clarity "EFFICIENCY IS NOT INCREASED BY ADDING DRIVERS". d _____________________________ http://www.pearce.uk.com |
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Speaker sensitivity and fs in multiples.
On 30 Nov 2003 02:06:41 -0800, (Svante)
wrote: Don Pearce wrote in message . .. On 29 Nov 2003 10:45:12 -0800, (Svante) wrote: Stager wrote in message ... I'm still not too clear on the answer. I'm not actually looking for a simple answer but a way to calculate the efficiency given the individual sensitivity and the number of drivers in a cluster. OK, the answer is really simple; If the efficiency is measured on linear scale (%), the efficiency is proportional to the number of drivers. So, if a single speaker has an efficiency of 0.1% a system made of two such speakers would have an efficiency of 0.2%, three 0.3%, four 0.4% etc. On a log scale (dB) each doubling of the number of speakers would give you 3 dB a increase in efficiency. So, if a single speaker gives 90 dB @ 1W a system made of two such speakers would give you 93 dB, four 96 dB, eight 99 dB etc. The above holds for low frequencies (such that the dimensions of the system is smaller than the wavelength) and for a reasonable number of drivers. Obviously, the efficiency can never go above 100%. None of this changes the efficiency, these apparent gains are in fact increases in directivity. Move off axis and the result of multiple speakers is actually reduced, not increased volume. Seen this way, it is clear that there is actually no theoretical limit to the amount of "gain" available from multiple speakers. Of course the increased level is purely a far-field phenomenon, and the distance from the speakers you need to reach the far field goes up with each added driver. Sorry, you're wrong. For HIGH frequencies (averaged over directions and frequencies), yes, but for low frequencies the decreased level off axis that you describe will never occur. The contributions from the speakers will be in phase, and thus add up to a pressure corresponding to n*p1 (n = number of speakers, p1 being the sound pressure from one speaker). Input power will be multiplied by n, but output power by n^2, since acoustic power is proportional to sound pressure squared. Thus efficiency increases linearly with the number of speakers, (n^2)/n. Please read my comment about the far field. That covers everything you have said here. But I will repeat for clarity "EFFICIENCY IS NOT INCREASED BY ADDING DRIVERS". d _____________________________ http://www.pearce.uk.com |
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Speaker sensitivity and fs in multiples.
(Bob-Stanton) wrote in message . com...
(Svante) wrote in message Sorry, Dick, I think you are wrong here. The efficiency is actually quadrupled in this case leading to a 10*log(4)=6dB increase in efficiency. I know this seems terribly wrong, but is true, at least for low frequencies. It is easiest to understand like this; 4 sources gives 4 times the sound pressure, which leads to a 20*log(4)=12dB increase of the SPL compared to the single driver. However, this example assumes that each of the drivers received the same amount of power as the single driver, which is 10*log(4)=6 dB more. The net gain is 6 dB, which is an efficiency increase. Another way of looking at it is to calculate the power produced by the motion of the cone, which should be the F*x/t, F being the force from the air as sensed by the cone, x being the displacement and t the time. If there is a sound pressure from another driver, this will increase the output power of the driver, and thus increase the efficiency. This reasoning holds for low frequencies, ie when the drivers are mounted close to each other compared to the wavelength. For higher frequencies it holds straight in front of the speaker (anechoic conditions) but to the sides, interference will decrease the sound pressure. So for higher frequencies, on average (over frequencies and directions) your statement ends up correct (ie the efficiency/sensitivity is the same) I think what you are saying is: the efficiency of a quad array is higher, (at low frequencies) because the Zma (radiation impedance) is higher. Ok, that's another way of looking at it. For pratical reasons, four separate drivers must have some physical distance between them. Maybe 3 cm, edge to edge? That separation would make the Zma a little lower. Yes, but if we assume that the frequency is low the difference is neglectible. It is never the distance in itself that is important, but the distance compared to the wavelength. Have you made a measurement of the actual efficiency of single driver and of an array of those drivers? I must confess, no. But it is basic acoustics. |
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