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Can't blow a speaker by under powering it.......
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Imagine a sine wave that is 2 volts peak to peak. If we take its average level, it will be 1.414 volts.
Imagine a square wave that is 2 volts peak to peak. If we take its average level it will still be 2 volts.
As you clip a signal, it becomes less sinudoidal and more square. The power formula is P=V^2/R. Into a 4-ohm load, a sine will put out half of a watt and a square will put out one watt.
Imagine a square wave that is 2 volts peak to peak. If we take its average level it will still be 2 volts.
As you clip a signal, it becomes less sinudoidal and more square. The power formula is P=V^2/R. Into a 4-ohm load, a sine will put out half of a watt and a square will put out one watt.
As simply as you can put it, you ain't going to overfill a 400ml glass with 200ml of water.
As others have pointed out, the myth tends to come from people who are pushing their amps too hard, so not really because they are underpowering the cab, but because they are trying to get too much from their amp.
As others have pointed out, the myth tends to come from people who are pushing their amps too hard, so not really because they are underpowering the cab, but because they are trying to get too much from their amp.
Show me an amp that puts out DC to the speakers and I'll show you an amp that is broken.
I'm not referring to amp failure due to shorted output transformer, output transistor or some failure in the driver circuit that would cause the power section to go partial DC.
Beg pardon but I've had thousands of amps in perfect working order on a bench that when driven hard enough to hit the power supply rail will put out a wave form that will be a flat top sine wave. There will be some duration to the rise but the waveform at it's max amplitude will be flat enough to be DC for all practical purposes although there will be a very small slope that depends on the power supply filtering and regulation. The duration of the flat portion will be dependent of the frequency of the input. The lower the freq, the longer the duration and that will be 90% or over of the frequency, depending on the amp. Any amp that won't do this has either a compressor or limiter of some type like Peavey's DDT circuit.
Whether or not a speaker system will or will not be damaged by a square wave input will depend on the total power generated by the amp and whether or not the speaker system can handle the power and heat generated.
mech
Even with a pure square wave at very low frequencies, the cone is still moving "in a curve" so to speak. It is still subject to the laws of physics. It will oscillate (and hence accelerate) in one direction, and then near the peak of the wave, it will slow down, stop at the peak like a ball thrown straight up in the air, and then accelerate in the other direction to start the process again.
So it's not like the cone is at its outermost position, and then magically jumps to its inner most position 4-10mm or so away, without traveling through the intermediary positions in some amount of time, albeit in a matter of milliseconds.
So it's not like the cone is at its outermost position, and then magically jumps to its inner most position 4-10mm or so away, without traveling through the intermediary positions in some amount of time, albeit in a matter of milliseconds.
How long does DC have to be DC before it's DC?
What you are saying is that the speaker gets DC, until the signal alternates to the opposite polarity and then gets DC again, until it alternates again.
By that logic, there is no such thing as AC at all, because at whatever + or - voltage you are at, it could be said to be "DC".
But it isn't.
What you are saying is that the speaker gets DC, until the signal alternates to the opposite polarity and then gets DC again, until it alternates again.
By that logic, there is no such thing as AC at all, because at whatever + or - voltage you are at, it could be said to be "DC".
But it isn't.
Edit: Whoa, beaten to the point by others by the time I posted:
A square wave is still AC. Technically, I guess you could say you're getting very short amounts of DC at the top and bottom of each clipped wave. Still, very different from connecting a constant (battery) DC voltage to a voice coil.
Now if you connected the battery for a fraction of a second, changed polarity, and reconnected it the next fraction of a second, then you would have a square wave. This would still cause the speaker to move in and out, not stay in one place.
I'm no guru, but that's how I understand it.
CL400Peavey said:
No, it wouldn't. It'd be overpowering it. It's the reason a 300 watt amp can kill a 350 watt speaker.
I can also see why people seem to think that clipping is the same as DC. A 2 volt square wave delivers the same power as 2 VDC. They're wrong about it being DC, but it delivers the same power.
That's the way a power amp works. The output stage varies the amount of a DC supply that is allowed to be seen by the speaker.
A square wave is not DC but has a LARGE DC component. A squared off sine wave has a smaller but still significant DC component.
If there were a speaker which would perfectly recreate the input waveform and it's input were a perfect square wave, it would move back and forth between the positive and negative amplitudes in zero seconds and be stopped during the duration of the DC component of the square wave.
Can anyone show me that there is no DC component to the two waveforms in this pic or that a speaker will act other than described? This is a decent 100Hz square wave on the top with over 99% DC duration and a clipped 100Hz sine wave with about 50% DC duration on the bottom.
mech
OK, so what effect does reducing the gain going to the power amp ultimately have on the output terminals?
And if it is the "same amount of power" (energy), how does it somehow result in less movement by the drivers (work)?
Aside from the semantic gymnastics of any gain control/output discussion... turning off the power switch definitely reduces the power level.
Mech the point is, what you have pictured is not what the speaker is really doing.There is some level of acceleration where the driver has to slow down and move the other direction. Its not really stationary.
It reduces the sensitivity of the input of the power amp. A higher input level is required to drive it to X amount of output power.
If you lower the input sensitivity of a power amp and drive it with a higher level input the result is normally a better signal to noise ratio. Not really meaningful for bass amps. Especially if your bass has single coils.
mech
Do the Fourier analysis: multiply your square wave function by cos(0) and integrate over one period. The result will be proportional to the amplitude of the DC component.
Show your work.
Hint: the answer is exactly zero.
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