| Home |
| Search |
| Today's Posts |
|
#42
|
|||
|
|||
Speaker sensitivity and fs in multiples.
"Rusty Boudreaux" wrote in message ...
"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. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#43
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
"Rusty Boudreaux" wrote in message ...
"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. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#44
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
ow (Goofball_star_dot_etal) wrote in message ...
On Sat, 29 Nov 2003 13:27:44 -0600, "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. The question was about sensitivity which includes such things as efficiency and directivity. Errhh... Yes, you are right, but I don't see how this helps. I think the easy way of understanding it is to think about low frequencies, where directivity is zero, and sensitivity and efficiency is the same (given that you measure sensitivity in dB at X watts (rather than dB at Y volts)). Maybe it did help... |
|
#45
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
ow (Goofball_star_dot_etal) wrote in message ...
On Sat, 29 Nov 2003 13:27:44 -0600, "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. The question was about sensitivity which includes such things as efficiency and directivity. Errhh... Yes, you are right, but I don't see how this helps. I think the easy way of understanding it is to think about low frequencies, where directivity is zero, and sensitivity and efficiency is the same (given that you measure sensitivity in dB at X watts (rather than dB at Y volts)). Maybe it did help... |
|
#46
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Sun, 30 Nov 2003 09:47:42 +0000, Don Pearce
wrote: No, you haven't understood. Fine, I will wait until the mechanic returns from lunch. There is no point in us oily rags arguing amongst ourselves.. |
|
#47
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Sun, 30 Nov 2003 09:47:42 +0000, Don Pearce
wrote: No, you haven't understood. Fine, I will wait until the mechanic returns from lunch. There is no point in us oily rags arguing amongst ourselves.. |
|
#48
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#49
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#51
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#52
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
"Rusty Boudreaux" wrote in message ...
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). This is exactly what is wrong. It does not take into account what happens on the acoustic side of the speaker. If we go back to the (two-in-series) in parallel with (two-in-series)ie four speakers connected such that the impedance is the same as for one speaker the following will happen: -Each speaker will receive 1/4 of the power. -The system will receive the same power. -Each speaker will receive 1/2 of the voltage. -Each speaker will displace 1/2 the amount of air (displacement is proportional to the voltage at a given frequency, assuming linearity and that the pressure from the other speakers does not affect the motion (both are reasonable assumptions)) -The system will displace twice the amount of air. -The acoustic output power will quadruple (acoustic power is proportional to the volume displacement squared) Net: Input power is the same, output power is quadrupled, efficiency has quadrupled (for low frequencies). So has sensitivity, since the electrical impedance is the same and the directivity pattern is spherical (for low frequencies). Dick was right. Nope. I wish Dick could think this through again and tell us if he stands by his original statement. Dick? |
|
#53
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
"Rusty Boudreaux" wrote in message ...
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). This is exactly what is wrong. It does not take into account what happens on the acoustic side of the speaker. If we go back to the (two-in-series) in parallel with (two-in-series)ie four speakers connected such that the impedance is the same as for one speaker the following will happen: -Each speaker will receive 1/4 of the power. -The system will receive the same power. -Each speaker will receive 1/2 of the voltage. -Each speaker will displace 1/2 the amount of air (displacement is proportional to the voltage at a given frequency, assuming linearity and that the pressure from the other speakers does not affect the motion (both are reasonable assumptions)) -The system will displace twice the amount of air. -The acoustic output power will quadruple (acoustic power is proportional to the volume displacement squared) Net: Input power is the same, output power is quadrupled, efficiency has quadrupled (for low frequencies). So has sensitivity, since the electrical impedance is the same and the directivity pattern is spherical (for low frequencies). Dick was right. Nope. I wish Dick could think this through again and tell us if he stands by his original statement. Dick? |
|
#54
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
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". I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". |
|
#55
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
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". I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". |
|
#56
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#57
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#58
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#59
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#60
|
|||
|
|||
|
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. Hi: In an array of four, each driver will see 1/2 the current of a single driver. The force a voice coil exerts is directly proportional to the current in the voice coil. The sound pressure caused by the driver is directly proportional to the force exerted by the voice-coil/diaphram. For 1/2 the voice-coil current, the sound pressure will be 1/2. If each driver exerts 1/2 the force of a single driver than: P = 20*log(0.5) = -6.02 dB. (Power output for each driver in array.) 4 drivers * -6.02 = 0 dB = No change in power output (or efficiency). You cannot sum the powers. You have to sum the sound pressures. The sum of four 1/2 pressures is 2. So, sound pressure will double (+6.02 dB) for the same input power. 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. P = F * x/t If current in each driver is 1/2, than F is 1/2. F/2 means the displacement (for time T) will be x/2. P = (F/2) * (x/2) = -6.02 dB 4 drivers * -6.02 = 0 dB = No increase in power output (or efficiency). No, again you cannot sum the individual powers. For the system, total displacement will be x*2. This is acheived with the the same input power. This corresponds to an increased efficiency of +6.02 dB. 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) The power output of an array (relative to a single driver) does not change with frequency. At high frequencies, energy beams out in many lobes. The sound pressure, on axis will change, but that is not the same thing as efficiency changing. No. Given an ideal piston driven to an acceleration corresponding to the signal, the level EXACTLY straight in front of and far away from the piston does NOT vary with frequency. To the sides it does. To the sides the level will vary towards higher frequencies due to interference from the different parts of the piston. This will cause on-axis level to be constant (vs frequency), but efficiency to drop towards higher frequencies. Thus, the efficiency gain I have described in this thread is only acheived in the low-frequency region. |
|
#61
|
|||
|
|||
|
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. Hi: In an array of four, each driver will see 1/2 the current of a single driver. The force a voice coil exerts is directly proportional to the current in the voice coil. The sound pressure caused by the driver is directly proportional to the force exerted by the voice-coil/diaphram. For 1/2 the voice-coil current, the sound pressure will be 1/2. If each driver exerts 1/2 the force of a single driver than: P = 20*log(0.5) = -6.02 dB. (Power output for each driver in array.) 4 drivers * -6.02 = 0 dB = No change in power output (or efficiency). You cannot sum the powers. You have to sum the sound pressures. The sum of four 1/2 pressures is 2. So, sound pressure will double (+6.02 dB) for the same input power. 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. P = F * x/t If current in each driver is 1/2, than F is 1/2. F/2 means the displacement (for time T) will be x/2. P = (F/2) * (x/2) = -6.02 dB 4 drivers * -6.02 = 0 dB = No increase in power output (or efficiency). No, again you cannot sum the individual powers. For the system, total displacement will be x*2. This is acheived with the the same input power. This corresponds to an increased efficiency of +6.02 dB. 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) The power output of an array (relative to a single driver) does not change with frequency. At high frequencies, energy beams out in many lobes. The sound pressure, on axis will change, but that is not the same thing as efficiency changing. No. Given an ideal piston driven to an acceleration corresponding to the signal, the level EXACTLY straight in front of and far away from the piston does NOT vary with frequency. To the sides it does. To the sides the level will vary towards higher frequencies due to interference from the different parts of the piston. This will cause on-axis level to be constant (vs frequency), but efficiency to drop towards higher frequencies. Thus, the efficiency gain I have described in this thread is only acheived in the low-frequency region. |
|
#62
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Svante) wrote in message ...
"Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#63
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Svante) wrote in message ...
"Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#64
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(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. Sorry, Svante, U KNOW you are wrong here, because you did not read my post well. I said: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." |
|
#65
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(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. Sorry, Svante, U KNOW you are wrong here, because you did not read my post well. I said: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." |
|
#66
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
In article , (Svante) wrote:
Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg |
|
#67
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
In article , (Svante) wrote:
Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg |
|
#68
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Mon, 01 Dec 2003 02:41:17 GMT, (gregs)
wrote: In article , (Svante) wrote: Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg You've just given us a perfect description of an experiment demonstrating directivity - and you draw the conclusion that it isn't so, and efficiency has been increased. I'm stunned. d _____________________________ http://www.pearce.uk.com |
|
#69
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Mon, 01 Dec 2003 02:41:17 GMT, (gregs)
wrote: In article , (Svante) wrote: Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg You've just given us a perfect description of an experiment demonstrating directivity - and you draw the conclusion that it isn't so, and efficiency has been increased. I'm stunned. d _____________________________ http://www.pearce.uk.com |
|
#70
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
On Mon, 01 Dec 2003 02:41:17 GMT, (gregs) wrote: In article , (Svante) wrote: Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg You've just given us a perfect description of an experiment demonstrating directivity - and you draw the conclusion that it isn't so, and efficiency has been increased. I'm stunned. OK, in my terminology, 40Hz is a HIGH frequency if the drivers are placed 16 feet (~4.8 m) apart. The wavelength at 40 Hz is 8.6 m which in the same order of magnitude as the driver distance. In this case there will be directivity, and the efficiency, averaged over directions would not increase as much or at all. If the speakers would have been close together (and f=40Hz still (which is now LOW frequency)) the directivity would be almost zero, but the level straight in front of the speakers would still be +6dB. ...and in all other directions, since there is no directivity. So the efficiency has increased, since input power is only +3dB. I agree that the experiment does not demonstrate increased efficiency, but it demonstrates increased sensitivity, straight in front of the speaker. If the speakers would have been moved close together, it would have demonstrated an increased efficiency as well. |
|
#71
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
Don Pearce wrote in message . ..
On Mon, 01 Dec 2003 02:41:17 GMT, (gregs) wrote: In article , (Svante) wrote: Don Pearce wrote in message . .. On 30 Nov 2003 02:06:41 -0800, (Svante) wrote: I understand what you mean, and it is all based on that there is directivity. At low frequencies there is no directivity. The increased level will occur at all directions. I will also repeat for clarity: "EFFICIENCY IS INCREASED BY ADDING DRIVERS". Forget directivity. At low frequencies there is plenty phase shifting. It doesn't take much to vary the phases. Many repeat having the need to closly space drivers. Well this help control the phasing, especially if multiple drivers are directed into a channel or common waveguide. I was trying to remember the term, but EV used a common chamber for some of their LF PA boxes. I tried an experiment in the backyard afew years ago. To speakers about 16 feet apart. At the exact center, the sound was 6 dBhigher over one speaker. Even at my low test frequency of perhaps 40 Hz, it only took inches either direction with the SLM toget significant change in SPL. So its basically true, if the wavefronts are combined at the source, it helps maintain high constant SPL when you move around. Yes, efficiency is increased by adding drivers. greg You've just given us a perfect description of an experiment demonstrating directivity - and you draw the conclusion that it isn't so, and efficiency has been increased. I'm stunned. OK, in my terminology, 40Hz is a HIGH frequency if the drivers are placed 16 feet (~4.8 m) apart. The wavelength at 40 Hz is 8.6 m which in the same order of magnitude as the driver distance. In this case there will be directivity, and the efficiency, averaged over directions would not increase as much or at all. If the speakers would have been close together (and f=40Hz still (which is now LOW frequency)) the directivity would be almost zero, but the level straight in front of the speakers would still be +6dB. ...and in all other directions, since there is no directivity. So the efficiency has increased, since input power is only +3dB. I agree that the experiment does not demonstrate increased efficiency, but it demonstrates increased sensitivity, straight in front of the speaker. If the speakers would have been moved close together, it would have demonstrated an increased efficiency as well. |
|
#72
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Sun, 30 Nov 2003 09:47:42 +0000, Don Pearce
wrote: No, you haven't understood. I think I see the problem now. With your background you probably have in the back of your mind a radio antenna where you can not have gain in all directions and it is a matter of "squeezing the balloon" to get gain. Now here is the crunch: A properly matched antenna is 100% efficient but what if the antenna is a bit of wet string and 99% of the power is reflected. Can this "bit of wet string" be improved upon? There is no need to read what Svante and I have written if it must violate your antenna law, right? |
|
#73
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
On Sun, 30 Nov 2003 09:47:42 +0000, Don Pearce
wrote: No, you haven't understood. I think I see the problem now. With your background you probably have in the back of your mind a radio antenna where you can not have gain in all directions and it is a matter of "squeezing the balloon" to get gain. Now here is the crunch: A properly matched antenna is 100% efficient but what if the antenna is a bit of wet string and 99% of the power is reflected. Can this "bit of wet string" be improved upon? There is no need to read what Svante and I have written if it must violate your antenna law, right? |
|
#74
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Dick Pierce) wrote in message . com...
(Svante) wrote in message ... "Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? I think your error is to assume that ignoring acoustic effects, you can still calculate sensitivity. For some reason you to not include the next paragraph where you wrote: ----quote 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. ----End quote Given that you ignore acoustical effects, this is correct, but misleading since acoustical effects are so important. Could you please enlighten us regarding your view on the sensitivity for low frequencies for the 2x2 array, and for higher frequencies straight in front of the same 2x2 array? Including acoustic effects and assuming free field conditions. From what I have seen in this group you have provided many good insights, and I beleive that you would say the same thing as I do, regarding this. I think your input could help this thread to "settle". The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#75
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Dick Pierce) wrote in message . com...
(Svante) wrote in message ... "Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? I think your error is to assume that ignoring acoustic effects, you can still calculate sensitivity. For some reason you to not include the next paragraph where you wrote: ----quote 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. ----End quote Given that you ignore acoustical effects, this is correct, but misleading since acoustical effects are so important. Could you please enlighten us regarding your view on the sensitivity for low frequencies for the 2x2 array, and for higher frequencies straight in front of the same 2x2 array? Including acoustic effects and assuming free field conditions. From what I have seen in this group you have provided many good insights, and I beleive that you would say the same thing as I do, regarding this. I think your input could help this thread to "settle". The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#76
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Dick Pierce) wrote in message . com...
(Svante) wrote in message ... "Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? I think your error is to assume that ignoring acoustic effects, you can still calculate sensitivity. For some reason you to not include the next paragraph where you wrote: ----quote 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. ----End quote Given that you ignore acoustical effects, this is correct, but misleading since acoustical effects are so important. Could you please enlighten us regarding your view on the sensitivity for low frequencies for the 2x2 array, and for higher frequencies straight in front of the same 2x2 array? Including acoustic effects and assuming free field conditions. From what I have seen in this group you have provided many good insights, and I beleive that you would say the same thing as I do, regarding this. I think your input could help this thread to "settle". The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#77
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
(Dick Pierce) wrote in message . com...
(Svante) wrote in message ... "Rusty Boudreaux" wrote in message ... "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 Sorry, Mr. Svante, you can't read what I wrote and you chose to omit the crucial piece of text that I EXPLICITLY wrote. Nope, Mr. Pierce is quite correct and you are wrong. Nopenope... :-) Since the impedance is the same and the drive level is the same the total output is the same. Nope 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. There is an efficiency gain on the acoustic side, due to the interaction between the drivers. You apparently did not read or chose to ignore what I wrote in my original replay. I said the following, QUITE clearly: "Ignoring acoustic effects, and just concentrating on the electrical properties, you essentially examine what the total impedance of the array is, and then examine how it is divided amongst the drivers." Do you see how I EXPLICITLY chose to NOT talk about the acoustical effects? The reason being is precisely BECAUSE of the complex depednecny on frequency. Now, within the context that I had writtenm as I have just reiterated, would you care to show where my error was? I think your error is to assume that ignoring acoustic effects, you can still calculate sensitivity. For some reason you to not include the next paragraph where you wrote: ----quote 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. ----End quote Given that you ignore acoustical effects, this is correct, but misleading since acoustical effects are so important. Could you please enlighten us regarding your view on the sensitivity for low frequencies for the 2x2 array, and for higher frequencies straight in front of the same 2x2 array? Including acoustic effects and assuming free field conditions. From what I have seen in this group you have provided many good insights, and I beleive that you would say the same thing as I do, regarding this. I think your input could help this thread to "settle". The total electric input power to the speakers will be the same in the 2x2 series/parallel connection, but the acoustic output power will be quadrupled, due to this acoustic coupling, see my other posts in this thread. I know, it seems terribly wrong, but is not. It is acoustics... :-) |
|
#78
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#79
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
|
#80
|
|||
|
|||
|
Speaker sensitivity and fs in multiples.
|
| Reply |
| Thread Tools | |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| rec.audio.car FAQ (Part 4/5) | Car Audio | |||
| rec.audio.car FAQ (Part 3/5) | Car Audio | |||
| rec.audio.car FAQ (Part 2/5) | Car Audio | |||
| rec.audio.car FAQ (Part 1/5) | Car Audio | |||
| Question re. Speaker Sensitvity | High End Audio | |||
Linear Mode
