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[Chris Grimshaw]

Every answer given has definitely been an answer, and all correct to my understanding.  However not an answer to my question.
Maybe this will trigger what I want to know.


Identical systems in identical condition.

#1 - mic preamp overdriven to making the vocal sound fuzzy, bad, distorted, no good, output to amp and speaker where the total energy (adjusted for harmonics and power) to the tweeter is "X".

#2 Mic preamp is clean, output to amp and speaker where the total energy (adjusted for harmonics and power) to the tweeter is also the same as "X" in #1.

Is the distorted signal tweeter stressed more than the other?

If they both have the same average power as you imply, they'll be under equal thermal stress. One of the drivers will be producing more frequencies than the other, but with the level adjustments you're making, it'll level out.

Making that level adjustment would be seriously difficult, since the impedance of a speaker varies hugely with frequency. 1w of heating power might be 2.83v at one frequency, and 10v at another. Not impossible, but certainly non-trivial.

ANY amplifier driven to clipping can and will generate as much as 10x the rated output of the amplifier.

Not even close.
The power difference between a pure sine wave and a square wave of the same peak voltage is a factor of two. 3dB.
That assumes, of course, that the power supply can source the extra current required and the rails won't sag.

Chris

[Ivan Beaver]

Something else to remember.

The power rating is for a SPECIFIC TIME, with a SPECIFIC WAVEFORM.

As soon as either of these are different, the rating will be different (it could be higher or lower).

It is NOT for 24/7 with any signal you want.

[Bob Leonard]

Not even close.
The power difference between a pure sine wave and a square wave of the same peak voltage is a factor of two. 3dB.
That assumes, of course, that the power supply can source the extra current required and the rails won't sag.

Chris

I know the textbook Chris. I didn't say "does" I said can and will, which can and will be the case in many instances.

[David Allred]

I know the textbook Chris. I didn't say "does" I said can and will, which can and will be the case in many instances.

"Will" sounds pretty "does" to me.

[John Ferreira]

I think the OP wants to know if all else is equal, will one wave form cause more damage than the other. The answer is, if you add DC content, which some forms of distortion do, then YES. However, when you say "...total energy would be the same", then the same amount of heat and damage to both drivers should occur. You would have to reduce the overdriven signal, (or signal  with higher DC content) though, to have the "total energy" equal on both signals.
An audio signal with DC, or direct current, causes a higher amount of heat on drivers (tweeters, woofers, and sub-woofers).
I hope this answers your question.


Sent from my iPad using Tapatalk

[Ivan Beaver]

Not even close.
The power difference between a pure sine wave and a square wave of the same peak voltage is a factor of two. 3dB.
That assumes, of course, that the power supply can source the extra current required and the rails won't sag.

Chris

But that is not what bob was talking about.

Yes, you are correct, if the actual signal was a steady state sine wave.

But with clean musical signal, the peaks will be WELL above the average )think 10-20dB).

If you squash that (the peak will stay the same, but the average level will go up), then the actual heating effect to the voice coil could easily be 10-20dB higher than the "clean" average.

I love it when people say "but my amp never peaked", yet they were hitting 20dB of compression the whole time-------------

[David Allred]

I think the OP wants to know if all else is equal, will one wave form cause more damage than the other. The answer is, if you add DC content, which some forms of distortion do, then YES. However, when you say "...total energy would be the same", then the same amount of heat and damage to both drivers should occur. You would have to reduce the overdriven signal, (or signal  with higher DC content) though, to have the "total energy" equal on both signals.
An audio signal with DC, or direct current, causes a higher amount of heat on drivers (tweeters, woofers, and sub-woofers).
I hope this answers your question.


Sent from my iPad using Tapatalk

Thank you.  You understood the question.  It seemed that most had a preconceived question in mind, or just an answer they wanted  to give.

[Stelios Mac]

DC content is (usually) filtered out anyway. Well, it's certainly filtered out of the tweeter with a 1300Hz high-pass, it (should) be filtered out of the woofers on a properly designed active cab.

Nobody had a preconceived question in mind, it's just the fact that your question depends on a lot of factors. You first mentioned vocals, then a 5kHz sine wave. It's all VERY different.

[John Ferreira]

DC content is (usually) filtered out anyway. Well, it's certainly filtered out of the tweeter with a 1300Hz high-pass, it (should) be filtered out of the woofers on a properly designed active cab.

Nobody had a preconceived question in mind, it's just the fact that your question depends on a lot of factors. You first mentioned vocals, then a 5kHz sine wave. It's all VERY different.

DC content is not filtered out quite the way you are visualizing. You must define DC as an unchanging current regardless if it is above or below the 0Volt line. An overdriven guitar signal can have plenty of DC content, because of that unchanging current, specially in square waves, or waves with flat tops, and yet guitar solos are not filtered out by the tweeter's crossover capacitor.

You may have what is called Capacitive Reactance, or Xc, but that does not mean it "filters" the whole wave out.

[Chris Grimshaw]

But that is not what bob was talking about.

Yes, you are correct, if the actual signal was a steady state sine wave.

But with clean musical signal, the peaks will be WELL above the average )think 10-20dB).

If you squash that (the peak will stay the same, but the average level will go up), then the actual heating effect to the voice coil could easily be 10-20dB higher than the "clean" average.

I love it when people say "but my amp never peaked", yet they were hitting 20dB of compression the whole time-------------

Ivan, I'd invite you to look at the quote again.

ANY amplifier driven to clipping can and will generate as much as 10x the rated output of the amplifier.

If you take the sine wave rating of an amplifier, lets say 500w @8ohm, that's 63v RMS or 89v peak. If you push that up to full square wave clipping, the peak voltage output is the same as the RMS voltage output (by definition of a square wave), resulting in 1000w output into 8ohm.

That is the maximum possible power output from the amplifier. Ever. It's only a factor of two.

Program material and peak-to-average ratios don't come into it - the statement is patently false, and that's the bit I took issue with.

When it comes to the thermal ratings of a speaker, they're usually rated with pink noise clipped to a 6dB peak-to-average ratio, per AES standards. So, a 1000w AES rated driver (6dB p/a) would burn with 500w of sine waves (3dB p/a) or 250w of square waves (0dB p/a). Each signal has the same heating power, but different peak voltages. In theory, you could burn a 1000w rated driver with a 250w amplifier run completely into square waves. That's not the product of an amplifier putting out 4x power, but a result of the test signals used to define a speaker's thermal characteristics.


With regards to DC content, remember we're talking about speakers where each amplifier is assigned to a frequency range and drive unit. If the amplifier is allowed/made to clip, the short-term DC (if you believe in oxymorons) will be delivered to the drivers.

Chris

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