Square waves and underpowered amps can't destroy speakers?

[gdougherty]

http://www.prosoundweb.com/article/unde ... dspeakers/

An interesting read and postulation from a repair center. Not sure if it's accurate or if the temporary current output of an underpowered amp swings high enough to fry a driver, but it makes some amount of sense. Seems it'd be easy enough to measure temp of a driver being driven at a given voltage with no clipping and a driver at the same voltage with clipping.

[SoundInMotionDJ]

Not sure if it's accurate

It is not accurate.

There is a very narrow place where the myth of underpowering continues to take refuge. The failure has nothing to do with the woofer "stalling" on the flat parts of the signal. The failure mode is still too much delivered power.

No amount of "clipping" will cause the earphone jack of your iPod to blow a pro sound woofer. None. The iPod can not source sufficient current to warm the voice coil to the point of failure.

Amps deliver voltage and current...not watts. Watts are what is experienced by the load as a result of the applied voltage across the impedance. A "200w" amp is more accurately described as a "40v source". A "500w" amp is more accurately described as a "64v source". In normal operation, amps are constant voltage devices, and can source sufficient current to deliver that voltage into the applied load.

The "RMS" value represents the area under the curve, and this area is proportional to the final delivered power. A pure sine wave has an RMS value that is 0.707 (one over square root of 2) of the peak value. A pure square wave has a RMS value of 1.0 (it's a square, and all the area is under the curve). The change in waveform shape from sine to square will deliver over 40% more power with the same peak voltage! This is where the myth of underpowering lives.

As the signal to the amps is increased, the average sound level is getting higher...but not the peaks. This can fool the ear into thinking that the overall noise level is lower than it is...prompting even more gain to be added to the signal.

There are few people who maintain that a "1000w speaker" can be blown with a "50w amp" even in the worst cases of clipping. In most cases the examples involve an amp that is 0.75 the rated capacity of the speakers. And now that we know that the shape of the curve can add up to 40% to the delivered power....it is easier to see that the "underpowered" amp can still deliver "too much power."

--Stan Graves

[Bill Fitzmaurice]

The author of that article is 100% wrong, and does serious damage to the credibility of the website that his post is on. This is the real story:
http://billfitzmaurice.info/forum/viewforum.php?f=10

[Radian]

A long time ago, in a galaxy far, far away.... "Hey man, your speakers sound like @$$ because you're using too small of an amp.", somehow got replaced with "Dude, you'll blow your speakers cause your amp is too small."

[Zack Brock]

It is amazing to me how these myths are started and then propogated. And then it is near impossible to clear them up. Somehow they become part of the knowledgebase of the public and just can't be refuted.

[gdougherty]

Not to mention that people see amp and speaker ratings and don't understand the whole picture. That 200W .01%THD 20Hz-20Khz @8ohm amp can deliver more than 200W. And that "1000W" speaker may only handle 250W RMS. The theory espoused sounds semi-plausible, but I'm glad to see all the clarification.

Another question on the same topic. The author assumes the clipped signal will result in a square wave output at the amp. Is that the case or will the overloaded input's THD really round off and soften up that "square wave" such that even if you could destroy a speaker with a square wave, you wouldn't be doing it with an overdriven signal?

[Ron K]

Can DC be passed to the output of a Direct coupled pro Audio amplifier when it's fully clipped and being fed an asymmetrical waveform?

If not then why?

Assuming of course the amp is not Capacitor coupled at any point other then the input.

Not a 5 watt I-pod amp but a pro audio amp capable of delivering 50watts RMS or more.

[Bill Fitzmaurice]

Can DC be passed to the output of a Direct coupled pro Audio amplifier when it's fully clipped and being fed an asymmetrical waveform?

If not then why?

No. Even the most severely asymmetric waveform is not DC, and if it was amplifiers don't amplify DC. The entire notion of DC appearing at an amp output is pure piffle, plain and simple.

[Greg Plouvier]

It can appear - along with magic smoke and cooked drivers.

[jcmbowman]

It can appear - along with magic smoke and cooked drivers.

Not if your amplifier is operating withing it's parameters. You should only get DC off the amp if there's something seriously, SERIOUSLY wrong with that amp.

[Bill Fitzmaurice]

It can appear - along with magic smoke and cooked drivers.

Not if your amplifier is operating withing it's parameters. You should only get DC off the amp if there's something seriously, SERIOUSLY wrong with that amp.

+1. Benign low voltage DC can be present as quiescent current. You only have high voltage DC when your output devices are blown.

[Greg Plouvier]

It can appear - along with magic smoke and cooked drivers.

Not if your amplifier is operating withing it's parameters. You should only get DC off the amp if there's something seriously, SERIOUSLY wrong with that amp.

+1. Benign low voltage DC can be present as quiescent current. You only have high voltage DC when your output devices are blown.

That's what I was referring to - blown outputs