I'm not against. For 30 dB, If you want to listen to the average power 10W, I have to use at least 10kW amplifier! Then I consider the technical possibility
Large peaks that some music have, or the large peaks that an amp can cleanly produce, or how loud the speaker can play is limited, in my example, to the max excursion that is reached by the speaker at 10 watts crossed at 500 Hz. Using Unibox again for this example, this I beleive translates into 101db for 1 speaker. If the music has 10 db peaks, that therefore means that my average listening level must be 91 db, which again using Unibox for this example is equivalent to 1 watt.
In the example above there any advantage to running a 100 watt amp versus a 10 watt amp of the same topography? So maybe the question becomes if both amps are asked to produce a 10 watt peaks, will these to peaks scope out the same?
What is so facinating to me is that I just realized that my midrange is only going to be using 1 watt running at 91 db.
In the example above there any advantage to running a 100 watt amp versus a 10 watt amp of the same topography? So maybe the question becomes if both amps are asked to produce a 10 watt peaks, will these to peaks scope out the same?
What is so facinating to me is that I just realized that my midrange is only going to be using 1 watt running at 91 db.
This was the first out of the bag
Loudspeaker Impedance
Your question is to simplistic... the load of a speaker is nothing like a pure resistance.
Look at this way, your 10watt amp (and I will define that as 10 watts RMS into 8 ohm) means it can put a voltage across the 8 ohm resistor of 8.91 volts RMS which is 12.6volts peak or 25.2 peak to peak.
An amp running on -/+16 volt rails would achieve that.
How do you define the transient capability of that amp... if it does indeed run on those supplies then the peak output is as I describe.
Within those limits the output from a 100 watt amp will be identical.
Suppose the amp has an agressive protection scheme... that may limit the peaks to less than that, if say the load draws more current than our pure test resistor. The speaker may be only 4 or 5 ohms at some particular frequency and the current and voltage may not be in phase... a reactive load opposed to a pure resistance, which may limit the max power if the protection cuts in. The "power" as you keep calling it may be less than "10 watts" then.
In this case the 10 watt will distort while the 100 watt wouldn't.
Suppose it's a nominal 10 watt amp running on higher supplies, supplies that sag a bit under load and supposing the amp hasn't a limiter. The peak output may be considerably higher now, maybe more like a 20 or 30 watt amp.
Loudspeaker Impedance
Your question is to simplistic... the load of a speaker is nothing like a pure resistance.
Look at this way, your 10watt amp (and I will define that as 10 watts RMS into 8 ohm) means it can put a voltage across the 8 ohm resistor of 8.91 volts RMS which is 12.6volts peak or 25.2 peak to peak.
An amp running on -/+16 volt rails would achieve that.
How do you define the transient capability of that amp... if it does indeed run on those supplies then the peak output is as I describe.
Within those limits the output from a 100 watt amp will be identical.
Suppose the amp has an agressive protection scheme... that may limit the peaks to less than that, if say the load draws more current than our pure test resistor. The speaker may be only 4 or 5 ohms at some particular frequency and the current and voltage may not be in phase... a reactive load opposed to a pure resistance, which may limit the max power if the protection cuts in. The "power" as you keep calling it may be less than "10 watts" then.
In this case the 10 watt will distort while the 100 watt wouldn't.
Suppose it's a nominal 10 watt amp running on higher supplies, supplies that sag a bit under load and supposing the amp hasn't a limiter. The peak output may be considerably higher now, maybe more like a 20 or 30 watt amp.
Yes, the topic is not simple at all. The link between amplifier and speaker (or even better driver) is complex (there aren't just watts) and should be judged on a one-basis.
For the reasons stated by Federmann, I have ruled out mid- and low-efficient speakers altogether. I start building the chain from the speakers, and there is where one should start if the room is not an option as well. Then, trying to avoid bottlenecks, I dig into what would be the best amplifier for such speakers (for each driver, or group thereof, in my case of active amplification).
1W, maybe 2W is the power that should be used for the average SPL. Less than that, and the availability of drivers and configurations are too low (read only horns).
It is nice to know, however, that just 10dB of headroom are sufficient to cover all but the highest peaks.
Also, the peak power is usually higher than the rated one. The same applies for the drivers. This shouldnt be forgotten.
It would be interesting to dig this further in a more technical level, i.e. which kind of circuits allow for an "easy" and "undistorted" peak power several times higher than their rated equivalent.
I haven't read about current in this thread, but I suspect that it plays a significant role in the distortion measured at transient peaks.
For the reasons stated by Federmann, I have ruled out mid- and low-efficient speakers altogether. I start building the chain from the speakers, and there is where one should start if the room is not an option as well. Then, trying to avoid bottlenecks, I dig into what would be the best amplifier for such speakers (for each driver, or group thereof, in my case of active amplification).
1W, maybe 2W is the power that should be used for the average SPL. Less than that, and the availability of drivers and configurations are too low (read only horns).
It is nice to know, however, that just 10dB of headroom are sufficient to cover all but the highest peaks.
Also, the peak power is usually higher than the rated one. The same applies for the drivers. This shouldnt be forgotten.
It would be interesting to dig this further in a more technical level, i.e. which kind of circuits allow for an "easy" and "undistorted" peak power several times higher than their rated equivalent.
I haven't read about current in this thread, but I suspect that it plays a significant role in the distortion measured at transient peaks.
Another thing to remember is that nearly all speakers are reactive loads except for planars. If passive crossovers are used with planars then they will represent reactive loads as well.
At the crossover frequency depending on your design 6,12, 18 or 24db per octave the power to the drive will be much reduced at the crossover frequency. A simple 6db per octave crossover at the crossover frequency will have an impedance equal to the speaker if the speaker is 8ohms. The resulting voltage will be evenly divided across the speaker and the crossover component. The power across the driver will be reduced by 1/4.ex 8 watts cross the speaker terminal = 8volts x 1 amp current, at the freq of interest 8 volts across 16 ohms ( 8+8) = .5 amps, .5 amps x 4 volts= 2 watts. This calculation does not take into account the reactive loads, where current and voltage are not in phase nor resonant frequency where the driver impedance rises etc.. ie capacitors do not dissipate any power as the voltage and current are out of phase, same with pure inductors. You can take any line voltage rated non polarized capacitor and plug it across the line voltage. It should never get hot. Capacitors are comonly used as voltage dropping devices in almost every ceiling fan and room cooling fans. So getting back to your original problem your 10 watt driver will be actually looking at around 40 watts from the amplifier when it is at the 10 watt max excursion level.
don't worry be happy
At the crossover frequency depending on your design 6,12, 18 or 24db per octave the power to the drive will be much reduced at the crossover frequency. A simple 6db per octave crossover at the crossover frequency will have an impedance equal to the speaker if the speaker is 8ohms. The resulting voltage will be evenly divided across the speaker and the crossover component. The power across the driver will be reduced by 1/4.ex 8 watts cross the speaker terminal = 8volts x 1 amp current, at the freq of interest 8 volts across 16 ohms ( 8+8) = .5 amps, .5 amps x 4 volts= 2 watts. This calculation does not take into account the reactive loads, where current and voltage are not in phase nor resonant frequency where the driver impedance rises etc.. ie capacitors do not dissipate any power as the voltage and current are out of phase, same with pure inductors. You can take any line voltage rated non polarized capacitor and plug it across the line voltage. It should never get hot. Capacitors are comonly used as voltage dropping devices in almost every ceiling fan and room cooling fans. So getting back to your original problem your 10 watt driver will be actually looking at around 40 watts from the amplifier when it is at the 10 watt max excursion level.
don't worry be happy
The question about headroom is the same as the question how many watts do I need and that appears regularly here ;-)
While you may end up with huge numbers for required power if one calculates peak currents into 2R reactive loads, there's a surprising number of people happily listening to lowish Watt (3-20W) amps.
The reason for this discrepancy is usually that the character of music is usually neglected as it's much easier to calculate and measure sine-waves.
Music is simply put, not very demanding with transients of 3-4 muV/s (Nelson mentions this number in one of his articles and suggests a slew rate of 20 muV/s to be plenty).
So you have to look at music: digitalize a few songs you like and play often. Use a sound editor to determine dynamic range: if you listen to todays pop for example the difference between average SPL and peak is tiny, a few single dB. Well recorded classic pieces are usually much more demanding.
Then, listen to music at your prefered SPL and measure the voltage at the speaker terminals. Regularly people are surprised how little watts they actually need (1W).
Add dynamic headroom from your music and be done ;-)
If you oversize, just because you can be aware that the quality of your watts may suffer. Big amps have to make more compromises than small amps.
Have fun, Hannes
While you may end up with huge numbers for required power if one calculates peak currents into 2R reactive loads, there's a surprising number of people happily listening to lowish Watt (3-20W) amps.
The reason for this discrepancy is usually that the character of music is usually neglected as it's much easier to calculate and measure sine-waves.
Music is simply put, not very demanding with transients of 3-4 muV/s (Nelson mentions this number in one of his articles and suggests a slew rate of 20 muV/s to be plenty).
So you have to look at music: digitalize a few songs you like and play often. Use a sound editor to determine dynamic range: if you listen to todays pop for example the difference between average SPL and peak is tiny, a few single dB. Well recorded classic pieces are usually much more demanding.
Then, listen to music at your prefered SPL and measure the voltage at the speaker terminals. Regularly people are surprised how little watts they actually need (1W).
Add dynamic headroom from your music and be done ;-)
If you oversize, just because you can be aware that the quality of your watts may suffer. Big amps have to make more compromises than small amps.
Have fun, Hannes
I assume the Dayton reaches full excursion @ 600Hz.
The cross over will already be 3db down at that frequency, or 6dB down if a linkwitz reilly has been implemented.
That means you need to feed in 20W or 40W @ 600Hz to reach full excursion with that driver.
However, full excursion at the minimum recommended frequency of the driver is not how you should be assessing necessary amplifier power.
Most treble and mid drivers would have an amplifier of approximately double the maximum continuous power recommended by the manufacturer.
If the Dayton can handle 60W of continuous power then the amplifier should be around 120W.
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As stated before, power and headroom are related but not the same. The type of power needs to be specified; avereage power=RMS (sine) or peak instantanoues power. Its the peak/average ratio that = headroom (its not how loud the loudest sections of music are compared to quiet parts but how "peaky" the loudest parts are . For sines its about 2db for drum tracks probably around 30db. So it depends totally on the source. With the loudness wars going on most pop music only needs a couple db of headroom, older pop maybee 10 db (tape is an automatic peak limiter and can also be used as compressor). Classical music recorded digitaly can go up to 20db but this is very rare. (most digital recorders are set up so 0VU is about -20dbfs (20 db below clipping) or 20db of headroom).
The peakiest recordings in your collection will be the CDs that at there loudest will sound quieter than any other CD. They are always mastered so the peak voltage on the entire CD is just below clipping. (PS short peaks, even 20db over VU can be inaudable but still take up headroom(decrease the overall level of the recording), thats why the are limited).
The peakiest recordings in your collection will be the CDs that at there loudest will sound quieter than any other CD. They are always mastered so the peak voltage on the entire CD is just below clipping. (PS short peaks, even 20db over VU can be inaudable but still take up headroom(decrease the overall level of the recording), thats why the are limited).
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