Well - my speakers caught fire.

[DirtyDuke]

No, I hate those cables. I roll my own. I also have a kickass bass and killer tube preamp running into a 3000w amp.

Nice!!!

 

[RoadRanger]

So 20Hz^2=400 watts?

Yah, that's why you should never run any midrange - 200Hz ^ 2 is 40,000 watts <eek>.

 

[Vince Klortho]

Your math and electrical terminology is off, but I agree that a pedal or other processing can indeed compress the dynamic range such that if you really crank the amp you don't see dangerous peaks but the average power is really high. Possibly high enough to overheat and burn out a loudspeaker driver, particularly if is weakened in some way.

What part of the math is wrong? Here is my take on it : for a sine wave peak voltage is RMS voltage times the square root of 2.0 and for a square wave of 50% duty cycle the RMS value is the same as the peak value. Also, P = V²/Z so V² = PZ

400W RMS into 4Ω means it supplies 40V RMS and 56.6V peak for a sine wave. A 56.6V peak square wave is 56.6 V RMS and RMS power into 4Ω would be 800W (56.6² / 4).

The math looks correct to me.

(edited to correct terminology: voltages are not peak to peak values - they are peak values symmetric around zero.)

 

[BurningSkies]

How much predicted headroom do you spend ?

It probably depends on the gig and the situation. I generally run my power amp bridged into a 4 ohm load. I've been doing outdoor shows recently without any PA support for bass and have enough volume to not miss a full FOH and still fill the space. Indoor gigs (like last night's) I'm not cranking but have more than enough wiggle room to get louder than needed without any loss of headroom.

The real idea is for me to remove the concept of 'headroom limits' from as many stages of my setup as possible. The only limitations I have are the speakers themselves and if they start to complain (and I've actually heard them do it once or twice) I back of just enough to clean it up. That's loud, given I'm using 3015LFs.

 

[bdplaid]

By the way, I'm a German guy thus don't feel you have to apologise for nothing. I'm used to direct words.
To the opposite I myself should write in more civilized words sometimes.

Actually I'm of German heritage myself and speak very directly. which has gotten me into no end of trouble this week, hence the frustration. My fellow Americans have gotten quite touchy. vielen dank!

 

[bdplaid]

So 20Hz^2=400 watts?

Sarcasm of course

uh, no...

 

[bdplaid]

Second consideration what also may happened to your loudspeaker,
Heavy distorted signal with very low crest. But speakers need sufficiant movement for cooling the VC.
Two things, lower going frequency cycles reduce air ventialtion around the VC (easy to understand).
Compressed signals reduce ventialtion due to reduced movement of the VC (also easy to understand). But compressed signals give way to squeeze more power out of an amplifier without clipping.

So both in combination, excessive compressed signal in combination of very low sub hatmonic content is a very good bet to wreck woofers.

You bring up an interesting situation -
I was thinking that the worst thing you can do to a speaker is to send it a square wave - basically going from full excursion to full excursion - for any length of time. I've lived by that mantra and thus have overpowered nearly every cab I've every had to keep things clean, with good results.

But I see the danger in a highly compressed, loud signal - it's going to keep voltage near peak going through the voice coils. Sounds like burnout time, because, as you say, there is limited movement of the coil to allow it to cool. That sounds exactly my friend's situation - running a large PA with heavy amounts of compression on the bass leg of the crossover (we used to run a comp after each leg of the crossover in a very large three-way system.)

 

[Doner Designs]

Sounds like this might be the best cab for metal.

 

[Mark Reccord]

You bring up an interesting situation -
I was thinking that the worst thing you can do to a speaker is to send it a square wave - basically going from full excursion to full excursion - for any length of time. I've lived by that mantra and thus have overpowered nearly every cab I've every had to keep things clean, with good results.

Except it doesn't actually work like that. Even with a pure square wave the driver will only move to the excursion points dictated by the voltage and frequency of the waveform. So if a sine wave with a 10V amplitude and a frequency of 100Hz moves a driver +/-2mm maximum excursion, a square wave with a 10V amplitude and a fundamental frequency of 100Hz will result in approximately the same excursion. The square wave will dissipate twice as much power in the voice coil, though. It won't automatically drive the speaker to its excursion limits just because it's a square wave. Even if we go to the extreme of DC, the excursion of the cone is proportional to the voltage. In other words the point where the force provided by the motor is equal to the force exerted by the suspension in the opposite direction is what determines how far the cone moves.

 

[ThisBass]

What part of the math is wrong? Here is my take on it : for a sine wave peak voltage is RMS voltage times the square root of 2.0 and for a square wave of 50% duty cycle the RMS value is the same as the peak to peak value. Also, P = V²/Z so V² = PZ

400W RMS into 4Ω means it supplies 40V RMS and 56.6V peak to peak for a sine wave. A 56.6V p-to-p square wave is 56.6 V RMS and RMS power into 4Ω would be 800W (56.6² / 4).

The math looks correct to me.

As far as I learned that stuff correctly. In electrical terminology RMS voltage is of most interest. The number of Crest determines relationship between RMS and Peak. Crest is synonymical to Peak To RMS Ratio which is expressed in dB.
Peak to Peak numbers are fairly uncommon (at least in Europe). A pure Peak to Peak number tells nothing about RMS whereas RMS is of interest for any serious power draw.
(By the way Peak to Peak for 40V RMS sine results in 113 Volt)

Once a supply voltage or rail voltage is known than it's possible to predict RMS for a distinct Crest number.
For example 50 Volt supply and a signal with 9dB Peak To RMS Ratio (or Crest).
RMS: 18 Volt

Sine:
crest 1.41
Peak To RMS Ratio 3dB
RMS 35 Volt

Square
crest 1
Peak To RMS Ratio 0dB
RMS 50 Volt

AES-1986 Pink Noise (testing of loudspeaker power handling)
Crest 2
Peak To RMS Ratio 6dB
RMS 25 Volt

Bass guitar signal (none distorted and none compressed)
Crest VERY large
Peak To RMS Ratio VERY lots of dB
Assumption it was around 12dB
RMS 13 Volt

So if a bass guitar signal was beheaded by 1/2 due to clipping (or compression) the remaining bottom stays (for now) far away from beeing squared.
RMS 25 Volt

For the record, the examples don't demand for precise correctness.
IMO it's a good idea to show the way how different Crest numbers impact gainable RMS at a given system.

 

[Bob Lee (QSC)]

and for a square wave of 50% duty cycle the RMS value is the same as the peak to peak value.

No, the rms voltage is then equal to the peak voltage. The peak-to-peak voltage is 2× the peak voltage.

 

[Bob Lee (QSC)]

send it a square wave - basically going from full excursion to full excursion

A square wave is not necessarily "full excursion to full excursion." In fact, it rarely is.

 

[BurningSkies]

A square wave is not necessarily "full excursion to full excursion." In fact, it rarely is.

I've never understood how it could be...because the wave shape and voltage would be dependent upon an amp rather than the cab, and since cabs come in a whole variety of power handling...what would be 'full excursion' on one cab with a square wave wouldn't be on another.

 

[Bob Lee (QSC)]

Except it doesn't actually work like that. Even with a pure square wave the driver will only move to the excursion points dictated by the voltage and frequency of the waveform. So if a sine wave with a 10V amplitude and a frequency of 100Hz moves a driver +/-2mm maximum excursion, a square wave with a 10V amplitude and a fundamental frequency of 100Hz will result in approximately the same excursion. The square wave will dissipate twice as much power in the voice coil, though. It won't automatically drive the speaker to its excursion limits just because it's a square wave. Even if we go to the extreme of DC, the excursion of the cone is proportional to the voltage. In other words the point where the force provided by the motor is equal to the force exerted by the suspension in the opposite direction is what determines how far the cone moves.

If you have a 10V rms sine wave at 100 Hz driving a loudspeaker and then change it to a 10V rms square wave of 100 Hz, it is true that the excursion will not change much. The amount of power dissipated in the voice coil will also stay about the same.

The thing to remember is that at frequencies above resonance, voltage applied to a loudspeaker driver produces an accelerative force, and the suspension acts mainly as a centering spring force. Below resonance (including DC), voltage produces a positional force because it is acting directly against the force of the suspension.

 

[Mark Reccord]

If you have a 10V rms sine wave at 100 Hz driving a loudspeaker and then change it to a 10V rms square wave of 100 Hz, it is true that the excursion will not change much. The amount of power dissipated in the voice coil will also stay about the same.

The thing to remember is that at frequencies above resonance, voltage applied to a loudspeaker driver produces an accelerative force, and the suspension acts mainly as a centering spring force. Below resonance (including DC), voltage produces a positional force because it is acting directly against the force of the suspension.

Haha! Yeah, I should have written sine wave with 10V peak amplitude as that's what I meant and it also makes my power dissipation math correct.

I think what I said about excursion holds as well?

 

[Vince Klortho]

No, the rms voltage is then equal to the peak voltage. The peak-to-peak voltage is 2× the peak voltage.

Right - my mistake. So that means it's a terminology error - not a mathematical error.

I will correct my post.

 

[Bob Lee (QSC)]

Haha! Yeah, I should have written sine wave with 10V peak amplitude as that's what I meant and it also makes my power dissipation math correct.

I think what I said about excursion holds as well?

A 10V peak sine wave at 100 Hz will produce lower excursion than a 10V peak square wave at the same frequency. The fundamental alone of the square wave would be about 1.28 times the sine wave.

 

[ThisBass]

Fourier Analysis
Sometimes imposing what's going on at frequency domain.

At time domain this means the peak value will grow if the square is low pass filtered by a LPF.
Seems to be magical but it's absolutely true.

 

[Mark Reccord]

A 10V peak sine wave at 100 Hz will produce lower excursion than a 10V peak square wave at the same frequency. The fundamental alone of the square wave would be about 1.28 times the sine wave.

Yes, of course. Even though I know better I sometimes oversimplify and think of a square wave as a single frequency. At any rate, my original point stands even though my math is a bit off. Cheers!

 

[Vince Klortho]

FWIW, here's a page that shows the coefficients of the harmonics (odd) that comprise a square wave : Invalid Link Removed and an image :

The 1.28 (1.2732 actually) value Bob mentioned comes from the 4/Π coefficient of the first harmonic or fundamental frequency.