Myth Busters: 1000W amp is only twice as loud as a 100W amp - Page 7 - diyAudio
 Myth Busters: 1000W amp is only twice as loud as a 100W amp
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 15th January 2013, 02:28 PM #61 diyAudio Member   Join Date: Sep 2004 Location: Lansing, Michigan Oh, and I think someone may have said it, but... Twice as loud is a matter of perception. Twice the sound pressure level is a measurable quantity.
 15th January 2013, 02:58 PM #62 diyAudio Member   Join Date: Apr 2008 interesting: Why Do Tweeters Blow When Amplifiers Distort?
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Join Date: Mar 2009
Location: Buenos Aires - Argentina
Quote:
 Originally Posted by Charles Darwin Ok JMFahey but lets just look at the square waves rise time for a moment: If the rise time is 50 microsec to go from zero V to 40V we must have a 20kHz component that reaches 40V. If it would be lower than 40V we would get a longer rise time because only a 20kHz sine is fast enough and all waveforms in the end are made up of sines. I may have miscalculated the actual numbers but the idea is the same: Look at the actual rise time and find a sine of frequency x that satisfies that. The faster the rise time the higher the frequency needed to achieve it.
Well, that's the point you don't seem to get.
Sorry, I don't wand to sound rough or unkind, not my style.

Quote:
 If the rise time is 50 microsec to go from zero V to 40V we must have a 20kHz component that reaches 40V.
Absolutely not. No way.

1) The waveform we are analyzing is *still* a 100Hz one, be it square, sine or whatever.
The 20KHz component you mention will be that one given by the Math posted above, there's no other, unless you invent a new Physics, Math, or both.

2) To analyze it from another point of view (which confirms my earlier statements, the beauty of Science is that it's consistent):

* Suppose the 100Hz squarewave has a leading edge which rises within 50uSec.

* Suppose it reaches 40 V.
(as you see I am so sure about what really happens that I accept all your premises)

* Peak power will be 40*40/8=200W (you must be smiling by now )

*BUT* that narrow pulse will repeat every 10000uSec (we *still* have those pesky 100Hz as base frequency )

Duty cycle will be: 10000/50=200 .

So average power dissipated by that poor voice coil will amount to 200W/200=1W.

Piece of cake.

3) To see it from a *third* point of view:

The "leading edge" of the 100Hz squarewave will reach 40V.
Fine.
But a component will be a sinewave
Which the "leading edge" is not, by definition.
The "20 KHz component" will be a sinewave (at least, that's what Fourier says).
Its amplitude will be given by the formula posted above.
Yes, this last point confirms point (1) .
Why does it not surprise me?

diyAudio Member

Join Date: Mar 2009
Location: Buenos Aires - Argentina
Sorry, typing error (although the end result is correct)=
It says:
Quote:
 Duty cycle will be: 10000/50=200 . So average power dissipated by that poor voice coil will amount to 200W/200=1W.
It should have said:
Quote:
 Duty cycle will be: 50/10000=1/200 . So average power dissipated by that poor voice coil will amount to 200W/200=1W.

 16th January 2013, 09:27 AM #65 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders input a 1kHz sine-wave and set the level to be just below clipping. Attach a passive crossover set to 3kHz. Measure the signal at the treble driver. It will not be zero. The crossover attenuates the signal, it does not eliminate the 1kHz. Now change the signal to 1kHz square wave. The amplifier will be delivering almost double the power. All of the extra power is due to the harmonics of the 1kHz fundamental. The first harmonic of the square wave series is the third, @ 3kHz. Just measure the signal at the treble driver. That measured 3rd and higher is virtually unattenuated by the crossover. You will find that the power being delivered to the treble driver will have increased to nearly the same as the amplifier was delivering when the unclipped signal was being passed. The Treble power will have increased at least one hundred fold. (+1000%) Ordinary clipping of music signals does something approaching "square waving" of the signal. The extra energy in the clipped signal is not due to higher maximum voltages but due to increased content of harmonics. Instead of 0.1% of harmonic distortion, one is getting 20%, or 50%, or in the extreme case 100% of added harmonics. The passive crossover ensures that much of the harmonics are delivered to the treble and mid drivers. Clipping destroys treble drivers. __________________ regards Andrew T. Sent from my desktop computer using a keyboard Last edited by AndrewT; 16th January 2013 at 09:29 AM.
 16th January 2013, 11:54 AM #66 diyAudio Member     Join Date: Feb 2001 Location: USA "he passive crossover ensures that much of the harmonics are delivered to the treble and mid drivers." And the clipping is largely at the lower frequencies and the harmonics feeding the tweeter are trivial (assuming a good crossover). "Clipping destroys treble drivers. " You still have not proved this, and direct examination of failures does not show damage from excess long-term average-power. The primary failure mode seen on most tweeters is mechanical failure of single-strand lead-out wires. In other words, they don't burn up, they break. Cone motion increases at a 12dB/oct rate as the frequency goes down. Program material can increase at a 6dB/oct rate going down to about 250hz or so. Unless the tweeter has an 18dB/oct crossover the lead-out wire can fracture. Most speakers having problems had either a 6dB or 12dB crossover, and a single-strand lead-out wire. I was astonished the first time I put a spectrum analyzer on a tweeter after repair. Playing program material it looked like the crossover was around 500hz, not the 4.5Khz it showed with pink noise. This was a 6dB crossover, and the single-strand lead-out wire tweeters dropped dead on a regular basis. __________________ Candidates for the Darwin Award should not read this author.
diyAudio Member

Join Date: Nov 2011
Location: UK
Quote:
 Originally Posted by Enzo Oh, and I think someone may have said it, but... Twice as loud is a matter of perception. Twice the sound pressure level is a measurable quantity.
X2 and X2 again.

Technically YES. Perceptionally DUBIOUS.

This is like the argument that comes up almost weekly - is 10W enough. Enough for WHAT ??

When I was building my first Class A it was supposed to be 40W at Class A. Was it loud enough ? Most of the time it was absolutely deafening but occasionally I wanted to drive it at high volume. It was then that the limitations of the design became apparent and the ceiling left a lot to be desired. Most of the time though it was being operated at less than 5W and it was fine. I've now got a 100W Class A and the perfermance is outstanding.

Big 1000W amplifiers are rarely used in household situations and will (nearly always) have the headroom that is required for them to operate linearly at high volume.

In club and stage situations Hi-Fi performance is often compromised but if 1000W Main Amps are being used then Sub-Sonic amps will need to be substantial in order to keep up with them.

Last edited by KatieandDad; 16th January 2013 at 12:36 PM.

 16th January 2013, 04:38 PM #68 diyAudio Member     Join Date: Oct 2012 With a single pole high pass, you will actually get roughly equal voltages from each of the squarewave's harmonics below the pole's frequency, all the way down to the fundamental. The rise in level as you go down the harmonics is pretty much exactly matched by the filter's increasing attenuation. Luckily crossovers tend to be at least double pole. Last edited by Robert Kesh; 16th January 2013 at 04:40 PM.
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Join Date: Oct 2007
Quote:
 Originally Posted by Robert Kesh Luckily crossovers tend to be at least double pole.
Not the ones found in cheap speakers. I've seen too many "3 ways" with a toy midrange and tweeter, and just a cap and an L pad on each. This kind of speaker is much more subject to abuse than something somebody spent \$2000 on. I've seen just as many failed level controls as tweeters. Luckily, they are cheap to replace, but are doomed to repeak failure without putting in a 'real' crossover. Unfortunately, putting in a stock 3-way 12dB usually sounds worse than the original two capacitor "crossover" until you spend time, effort, and money to optimize it.

 27th January 2013, 09:36 PM #70 diyAudio Member   Join Date: Jul 2007 Considering 10db is twice as loud, here is another fact 10% THD is 50% distortion, in terms of loudness assume 100 w amp, 10% THD means 10w noise out of 100w 10w is 10db less than 100w So 10% THD is 50% loudness distortion

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