All those tests seem to be done into a resistor. Where are the real world power delivery into a loudpeaker, or even a simulated losd (complex impedance)?
What an amplifier can deliver into a real world load can be significantly different (a real world example is the 1st gen Carver Cube (200W) not being able to deliver as much power as an NAD 3020 (20W) once connected to a real loudspeaker)
dave
Most speaker designs do not have a huge variance in the impedance curve, just add a zobel circuit to tame it all.
Now If you have a complex load then have fun and figure it out. In general Chucks tests are way beyond what most speaker loads represents to an amp and the thread was geared towards testing amps for subwoofer usage and those loads are very predictable.
The point is that many pro audio amps do meet their specs with a decent tolerance and we have to remember that the difference between 300Watts and 350Watts is sort of meaningless considering its < 1dB. Of course if 300 is clipping and 350Watts isnt clipping then its a world of difference.
Earl, so what is the best method of predicting multi-driver SPL gain based on real-world hardware? And real-world SPL based on real-world amp/driver variables?
..Todd
One point that has me confused is the discussion of volts instead of power. I know that my hypothetical amplifier with +-50DV rails will produce maximum 100v P/P AC signal (right?) And the power output of the amplifier is much greater than 100 watts, and the current draw is significant, and directly related to load impedance. So why is power not considered?
I have this feeling I'm too dense to understand the answer I'm going to get here.
..Todd
I have this feeling I'm too dense to understand the answer I'm going to get here.
..Todd
Power would be based on an RMS vcoltage which would be more like 35 volts. For the most part one never uses the power that a sine wave would require because the peak to RMS level of music is so much greater than for a sine wave. Its volts that are the limiting factor not watts.
So this means the following. IF you knew the volatge sensitivity of your loudspeakers in dBSPL per volt (not dB / watt) at 1 meter and you knew the MAX_VOLTAGE of the driver (not max watts) and the drievsr were all in parallel - same voltage - then as long as the the amp voltage does not exceed the speakers MAX_VOLTAGE then the SPL is simply volts times sensitivity (or add dB) plus 6 dB for each doubling of the number drivers. Then in a real room its about -3 dB for each doubling of distance back from the source specified at 1 meter. No need to worry about impedance, power disipation or any of that other stuff.
When watts are used there are so many assumptions and simplifications made that unless you know ALL of these, you don't have a good way of determining the maximum output capability of the system.
In my designs, I know that the speakers MAX_VOLTAGE is so much higher than any amps that I use that amp volts is all I need. Thus I know that the amps will clip long before the speakers are the problem. The speakers are so efficient and high impedance that power dissipation is not an issue either. It all becomes quite simple when done correctly.
Its the marketing guys that want amps and speakers in watts because this can be manipulated in so many ways that they can pretty much say whatever they want to say. It all becomes quite meaningless - and confusing to the novice. A confused consumer is a gullable consumer.
So this means the following. IF you knew the volatge sensitivity of your loudspeakers in dBSPL per volt (not dB / watt) at 1 meter and you knew the MAX_VOLTAGE of the driver (not max watts) and the drievsr were all in parallel - same voltage - then as long as the the amp voltage does not exceed the speakers MAX_VOLTAGE then the SPL is simply volts times sensitivity (or add dB) plus 6 dB for each doubling of the number drivers. Then in a real room its about -3 dB for each doubling of distance back from the source specified at 1 meter. No need to worry about impedance, power disipation or any of that other stuff.
When watts are used there are so many assumptions and simplifications made that unless you know ALL of these, you don't have a good way of determining the maximum output capability of the system.
In my designs, I know that the speakers MAX_VOLTAGE is so much higher than any amps that I use that amp volts is all I need. Thus I know that the amps will clip long before the speakers are the problem. The speakers are so efficient and high impedance that power dissipation is not an issue either. It all becomes quite simple when done correctly.
Its the marketing guys that want amps and speakers in watts because this can be manipulated in so many ways that they can pretty much say whatever they want to say. It all becomes quite meaningless - and confusing to the novice. A confused consumer is a gullable consumer.
But that is where the simple idea that doubling the watts only gains 3dB matters.
and if I want around 130dB SPL levels @ 12 feet and I have 95dB sensitivity speaker I use simple and well known formulas (many sites can do this easily) to approximate that I need probably 250Watts to make sure I do not clip.
Voltage is for those with electrical degrees. Watts works for 99.9% of the population, is it accurate? Depends on the tolerances of each individual needs.
You only think that it works. There are lots and lots of caveats imbedded in using watts as you suggest and while it may work some or most of the time, it is not going to work all of the time.
Most amps today cannot deliver the "watts" that they advertise, but then most speakers sound terrible when driven at there "Max wattage". Its all quite pointless.
Volts is harder to cheat.
Its pointless to you but Its not pointless to me. I tend to have amps that hare measured by 3rd parties. I tend to also know that when looking at amps I understand that 200Watts probably has a 20 percent tolerance.
So if Im looking at approx. 200Watts/ch I would try and buy a 250Watt amp to ensure headroom.
I 100% agree on most speakers sounding terrible driven at "Max Watts" but we are not really talking about driving them constantly that loud, we are talking about the > 20dB peaks that need to be handled by adequate power. I build pro audio designs for that reason.
Watts are extremely cheap so just buy more head room. Case in point, The new Peavy amp was just measured on AVS, 450Watts/ch into 4 ohms is sweet for < $300 AND it weighs 7 lbs...nice for testing and moving around.
IMO, I can get all the answers from Watts so its not useless to me.
Voltage is definitely a higher learning curve and audio is hard enough for most people
Voltage is definitely a higher learning curve and audio is hard enough for most people
I never accept the argument: "Its too hard to learn to do it right".
How about "It's too hard to find, comprehend and trust the information needed to learn to do it right." That's my challenge.
..Todd
Thats fine but you are not 99% of the population. Few of them have a PhD or even want to educate themselves on the topic of audio. Electronics companies know this, more or less they are never selling to a guy like you they are selling to the guy that goes into Best Buy without any electronics knowledge.
Even I care little about learning all about voltages etc since Watts are accurate enough to give a good estimation of what amp power is needed. I see little benefit to making it more confusing for the sake of just numbers. Meaning, even if Voltage was the standard I would still end up with the same power amps.
Dumbing it down is okay in the real world
It is the unwillingness of the consumer to learn right from wrong that gives them a license to rip them off.
While I agree and I complain about the ignorance daily, I can also understand that people have little time to read up and learn about voltage, etc. I doubt its an unwillingness when you consider how much time most people have any more...jobs, kids, many are single parents. Just getting bills paid and the kids educated is hard enough. Audio isnt that important to the majority and Voltage education is less important.
Watts are simple, its usually a nice big number (people like that over decimals) and we can do easy approximations to get the dB difference and that is all that matters in the end.
As for the "rip them off" I do not seem many amp companies ripping off J6P. I see high end companies ripping off audiophiles.
I have participated in a long form and then a long discussion with the FTC about ratings of power amplifiers and loudspeakers. The FTC and I concluded what most people really wanted to know what how many Volts a loudspeaker required to produce a particular SPL at a specific distortion like 10%THD, how many volts were required to produce a specific SPL, how many volts the speaker would handle, and how many volts a specific amplifier would deliver into a particular load as found in loudspeakers. This information would provide for useful direct comparisons between amplifiers and between loudspeakers. It was also the conclusion of this meeting that no big audio company would every stand for such a clear delineation of performance parameters.
Power compression sets in on small ports in woofers at pretty low power levels, much less than the driver will usually produce otherwise. Power compression on mid and high drivers is more complex to define as it can be excursion or thermal limits being reached. If power goes up linearly with input then a typical driver will have increasing THD also so the limit becomes when the driver reaches a THD which exceeds the maximum specified value for the design. In sub woofers it is not uncommon to see 30% THD at 30Hz and 100dB at 1 meter output. Please understand I am not saying all subs but many are this high.
To really know the answer to the question the definitions of THD and so on need to be known to determine when the limits have been reached. Like a simple capacitor... measured in µF, volts, current capacity, and often some form of frequency range. Then answer cannot be distilled down to a single value which describes all performance criteria.
Directly to your question, every doubling of distance increase loss by 6dB so if output is 100dB at 1 meter it is 94dB at 2 meters and 88dB at 4 meters. Having two matched coherent drivers operating in phase means that if the amplitude of motion for the single driver and the two drivers are the same time there will be an increase of 6dB in SPL. Please note at higher frequencies or other factors can effect the radiation pattern of the multiple loudspeakers and thereby affect the resultant SPL at any distance. Directivity is another subject and beyond the scope of most of the DIY persons though a few know about it like Gedlee and myself as example.
For power doubling increase SPL by 3dB so 20 watts is 3dB louder than 10 watts. For voltage doubling is 6dB (just like cone motion above) which also results in 4 times the power. If 8 volts is 8 watts then 16 volts is 32 watts.
Hope this helps some.
Power compression sets in on small ports in woofers at pretty low power levels, much less than the driver will usually produce otherwise. Power compression on mid and high drivers is more complex to define as it can be excursion or thermal limits being reached. If power goes up linearly with input then a typical driver will have increasing THD also so the limit becomes when the driver reaches a THD which exceeds the maximum specified value for the design. In sub woofers it is not uncommon to see 30% THD at 30Hz and 100dB at 1 meter output. Please understand I am not saying all subs but many are this high.
To really know the answer to the question the definitions of THD and so on need to be known to determine when the limits have been reached. Like a simple capacitor... measured in µF, volts, current capacity, and often some form of frequency range. Then answer cannot be distilled down to a single value which describes all performance criteria.
Directly to your question, every doubling of distance increase loss by 6dB so if output is 100dB at 1 meter it is 94dB at 2 meters and 88dB at 4 meters. Having two matched coherent drivers operating in phase means that if the amplitude of motion for the single driver and the two drivers are the same time there will be an increase of 6dB in SPL. Please note at higher frequencies or other factors can effect the radiation pattern of the multiple loudspeakers and thereby affect the resultant SPL at any distance. Directivity is another subject and beyond the scope of most of the DIY persons though a few know about it like Gedlee and myself as example.
For power doubling increase SPL by 3dB so 20 watts is 3dB louder than 10 watts. For voltage doubling is 6dB (just like cone motion above) which also results in 4 times the power. If 8 volts is 8 watts then 16 volts is 32 watts.
Hope this helps some.
This is correct. SPL = 20 log p, where p is the pressure.
Ignoring phase cancellations (which is the same as your assumption of coherency), then two cones oscillating in a given manner will generate twice the pressure and thus be +6dB over one cone oscillating in the *same* manner:
1 driver: SPL1 = 20 log p0
2 driver: SPL2 = 20 log (p0+p0) = SPL1 + 6dB
There is an intuitive connection between pressure and applied voltage (doubling the applied voltage doubles the pressure). So, when two drivers are connected in parallel they see the same voltage and thus each generate the same pressure, p0, as in the single driver case.
Actually, this is the limiting case as frequency goes to zero. As frequency goes to infinity, on the other hand, the gain will only be +3dB because the (noncoincident) sources will interfere. In terms of pressure, if a single cone has pressure magnitude p0, then the RMS pressure is 2*p0 (+6dB) when coherent, and sqrt(2)*p0 when incoherent (+3db).
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