Extremely Disappointed

[NDMstang65 403]

Ok, I posted that before I realized there was another page.

You took voltage away from the square wave...to make it "equal" using the assumption of .707 is the RMS figure (the 3.535dx from x=0 to pi), the sine wave you assigned 5 volts to, the square wave where rms=peak you took power away from it to make it "equal" by RMS figures.

Exactly as that is how you describe power. Again, average is implied. Has to be unless you are referring to instantaneous but then it wouldn't be a stress test showing heat as you can't measure instantaneous heat changes accurately.

I definitely see what you are saying...which begs the question, does voltage actually drop with sine vs. square with same wattage? But at that point how could the same 100 watts for example be the same? If voltage is down on the square wave, and current is definitely down due to a higher resistance of the coil due to induced heat...you are not getting anywhere near the 100 watts that you were with the sine wave, you are getting far less..and there is way more heat present?

You should think about your calculation of watts. By definition it is average power. You are taking a shortcut of course and using a simple measurement device that only measures instantaneous which is confusing this as it is the wrong unit of measurement for the task at hand. You need to account for the delta T otherwise the whole discussion of heat and cooling makes no sense.

Still..

If there is less power by definition using RMS figures..

Why is there more heat present?

clipping.

[///M5 2,833]

Ok, I posted that before I realized there was another page.

You took voltage away from the square wave...to make it "equal" using the assumption of .707 is the RMS figure (the 3.535dx from x=0 to pi), the sine wave you assigned 5 volts to, the square wave where rms=peak you took power away from it to make it "equal" by RMS figures.

Exactly as that is how you describe power. Again, average is implied. Has to be unless you are referring to instantaneous but then it wouldn't be a stress test showing heat as you can't measure instantaneous heat changes accurately.

I definitely see what you are saying...which begs the question, does voltage actually drop with sine vs. square with same wattage? But at that point how could the same 100 watts for example be the same? If voltage is down on the square wave, and current is definitely down due to a higher resistance of the coil due to induced heat...you are not getting anywhere near the 100 watts that you were with the sine wave, you are getting far less..and there is way more heat present?

You should think about your calculation of watts. By definition it is average power. You are taking a shortcut of course and using a simple measurement device that only measures instantaneous which is confusing this as it is the wrong unit of measurement for the task at hand. You need to account for the delta T otherwise the whole discussion of heat and cooling makes no sense.

Still..

If there is less power by definition using RMS figures..

Why is there more heat present?

clipping.

If I understood what you did for a test right then you are comparing MORE power on the square wave to less on the sine. More power = more heat.

[NDMstang65 403]

No..can't.

If amplifier is running wide open at 2000 watts...sine wave, it is making 200 volts and 10 amps. That is all it will make...that is all the power supply will support, that is all the gates on the fets will switch.

Flip the generator over to square wave, by definition in this thread it should have less power...according to RMS calculations done. It was already making maximum power before with the sine wave.

There is FAR more heat present...we stuck the fluke probe right to the coil. It was around 130 degrees hotter if memory serves me correctly right after turning the amp off..that is absolute fact. It was hotter. Period.

If 2000 watts is 2000 watts...and there is no difference according to a speaker with a square wave form vs a sine wave form..why was there 130 degrees more heat? The amp didn't suddenly make 4000 watts to get it the 130 degrees hotter.

The only thing that changed was the wave form, and the time period that it was on...100%. Instead of .707 RMS of a sine wave..because RMS and Peak is the same within a square wave form..

[///M5 2,833]

I am still confused as to your point. You either have an instantaneous voltage reading or are talking about average power. If you have the same output in volts at the peak of the wave the average power is higher for a square wave than a sine wave.

If it helps you think about it, do the same test at 100w on an amplifier that is capable of 1000w.

If the time period changes, the power changes. Delta T is part of the power equation and cannot be removed.

[NDMstang65 403]

My point:

If clipping does not break speakers.

An amplifier is making maximum power with a sine wave, it is then driven with a square wave.

The Coil is 130 degrees hotter..without more power being present...because the amplifier can not make any more power it was already making maximum power with the sine wave previously.

The only thing that is changed is the wave form itself, and the time that the coil is "on".

[///M5 2,833]

My point:

If clipping does not break speakers.

An amplifier is making maximum power with a sine wave, it is then driven with a square wave.

The Coil is 130 degrees hotter..without more power being present...because the amplifier can not make any more power it was already making maximum power with the sine wave previously.

The only thing that is changed is the wave form itself, and the time that the coil is "on".

You showed the same instantaneous voltage in your example though which by default is more average power. And yes more average power = more heat.

[NDMstang65 403]

Thinking more..

We are not talking about taking a potential 1000 watt amp and making it run at 100 watts sine wave and then square waving it and getting 400 watts...that is a no brainer why you would have more heat. It is more power.

We're talking about a 1000 watt amp, putting out maximum power with a sine wave at 1000 watts...it is then driven with a square wave, more heat is present than with a sine wave with the same 1000 watts. It can only make 1000 watts..that's all it can do. 100 volts and 10 amps for sake of the discussion...square wave..sine wave..pink noise..white noise..what have you. 1000 watts.

But we really can't talk about watts at this point (because of the rms value here), what you need to look at is voltage. The rails are capable of doing 100 volts peak to peak..that is what it is capable of doing, that is what the square wave does. We are talking about a peaked out rail here...+/- 50 volts the sine wave just touches it. Your square wave hits it 100% of that energy for that period of time.

Why is the coil 130 degrees hotter with the square wave form? If..'clipping' does not blow speakers or burn parts up.

It has to be the amount of time that the speaker is 'on' 100% at the extreme north and south end where it literally sits there, does not move, and burns. This is not even looking into the MASSIVE acceleration factors and forces going on here either..

Edit:

At the two extreme points the coil is 'stopped' or 'dead' and doing nothing but taking the grunt of the amp for the time value of the operating frequency of the output transistors...it's not doing anything (Back to the rev limiter analogy here)

It is just like pushing the clutch in and holding the gas pedal to the floor bouncing the engine off of the rev limiter and jerking the gear shift into reverse. Still holding the accelerator to the floor and slamming it back in first..the time that the speaker is stopped is the period of time in which it takes you to physically switch the gear from reverse to forwards without using the clutch....which shanks gears in transmissions, snaps spiders in subs..

Which is due to acceleration factor and more energy per time division..which needs to be tested by an accelerometer..which i'll do when I have time.

There is no clutch, there is no braking or slow down. You are 100% forward, and instantly stopping and 100% rearword...and burning in-between.

Hope this helps.

[///M5 2,833]

I was referencing the other amp situation as it is the only way to test the heat and your statement. From the test scenario you described you can't draw that conclusion but if you did it at a level that you could easily confirm they have exactly the same average power then fine.

[Dutchie. 10]

Hey guys I just wanna thank you for having this detailed conversation. I've been following it and it has helped me really understand what can damage a speaker.

[Duran 387]

Nick, make me a woofa that I can plug into my 480 volt, 30 amp wall sockets.

Please and thank you!

[///M5 2,833]

If the power is the same, the heating is the same, otherwise Joule was wrong as he even used heat going through a wire to define power. In other words the effect of the amplifier on the coil wire will be the same. if you are measuring something different it is because of a different factor. The time is already dealt with in the power equation so that is not the case. If you wanted to blame this on some other aspect of the design fine, but not when you are applying equal power.

[kirill007 52]

DIRECT QUOTE from the link honda95 gave : "A speaker that fails at 100 watts fails at 100 watts. 100 watts of a sine, 100 watts of a square, 100 watts of music averaged..... This is a fact..."

He tries to tell square wave=sine wave , that this is a fact, which it isn't.(he only concluded it from his tests.)

You don't even understand the test. And now you are just making shit up, lol... If you had a clue, you would understand that I never said all waveforms were the same, quite the opposite really.

This is why you, and most other people who arm chair quater back this are full of BS... You just make crap up out of gut feelings.

But anyway, this is truly a dead horse... I am done beating it...

I don't know what this means, my knowledge of english doesn't goes that far.

But i do understand, only problem is all of you're graphs are down.

But i'm done discussing since my language makes up for a lot of mistakes and NDMSTANG tells exactly what i want to tell.

[Beandip 3]

Wow this is a great read! Its really nice to see these types of threads develop since you can learn so much.

sticky worthy for sure

Edited February 9, 2011 by beandip

[Impious 1,476]

But we really can't talk about watts at this point (because of the rms value here), what you need to look at is voltage. The rails are capable of doing 100 volts peak to peak..that is what it is capable of doing, that is what the square wave does. We are talking about a peaked out rail here...+/- 50 volts the sine wave just touches it. Your square wave hits it 100% of that energy for that period of time.

If you have a signal that's 50V peak, run a sine wave and then a square wave at that same peak voltage into a 1ohm load you'll have average power of (50*.707)^2/1 = 1250w with the sine wave but 50^2/1 = 2500w with the square wave. Both signals will be within the rail voltage of the amplifier. The peak voltage of the signal hasn't changed, but the average power of the signal has increased significantly with the square wave.

I guess my point is.....even if you take the same amplifier and run the same peak signal voltage for a sine wave and square wave through it.....the square wave will have significantly more average power than the sine wave even though the peak voltage of the signal hasn't changed (and hence both signals are within the rails of the amp). That 1000w amplifier is a 1000w amplifier with a sine wave, not a square wave. Average power will increase with the square wave, which leads to the build up of heat. Granted you may hit current or power supply limitations which will not allow the amplifier to output a full 2500w continuously.....but you'll be at the limits of the amplifier, which will result in an increase in average power with a square wave compared to a sine wave with the same amplifier, same peak voltage, within the rail voltages of the amplifier.

If you are going to do a normalized average power comparison, which is what we are trying to compare, you have to reduce the peak voltage of the square wave so that both the sine wave and square wave have the same RMS voltage, which would give both signals equivalent average power. So a 50V peak sine wave would have an RMS voltage of 35.35V. The square wave would need to be reduced to a level of 35.35V for an equal average power comparison. So a square wave wouldn't hit the peak 50V and stay there, because it would peak at 35.35V. Only the sine wave would peak at 50V.

This is what Sean, 95Honda and myself are arguing. With the same average power (i.e. the same RMS voltage, not peak voltage), there won't be a significant difference in heat and/or failure time within the driver. We can't just talk in terms of voltage alone, we have to differentiate between peak and RMS signal voltage, because that dramatically affects the average power of the signal. And you aren't going to make this comparison by only flipping the signal generator between sine and square wave, as the peak and RMS voltages of the signal would need to be adjusted to properly compare average power.

I was thinking about something else on my morning drive......the square wave assumes essentially an instantaneous change in voltage level. If the signal is 5V peak to peak, the signal has a 10V swing in level with virtually zero change along the y-axis of time. Loudspeakers have inductance and energy storage......I would imagine there has to be some sort of "lag" present in the driver's ability to respond to an instantaneous change in voltage like that. The driver isn't going to be able to respond instantaneously and go from a position of +5V to -5V in almost literally no time. That's almost exactly what inductance describes...the ability of the driver to respond to a change in voltage. So I wonder how much "rounding" of the square wave there is in the actual response of the loudspeaker to the signal, as this would affect (reduce) the amount of time the driver was actually at the plateau of square wave, which would reduce the amount of heat build up you are suggesting occurs. By the time the speaker finally gets up there to the plateau of the square wave, the signal is turning around heading back down in the other direction, so the speaker turns around and follows it back down (obviously overly simplified). I wonder if it would almost "track" the same path as a sine wave due to the inductance induced lag.

You're the speaker engineer Nick What do you think?

[ssh 924]

But we really can't talk about watts at this point (because of the rms value here), what you need to look at is voltage. The rails are capable of doing 100 volts peak to peak..that is what it is capable of doing, that is what the square wave does. We are talking about a peaked out rail here...+/- 50 volts the sine wave just touches it. Your square wave hits it 100% of that energy for that period of time.

If you have a signal that's 50V peak, run a sine wave and then a square wave at that same peak voltage into a 1ohm load you'll have average power of (50*.707)^2/1 = 1250w with the sine wave but 50^2/1 = 2500w with the square wave. Both signals will be within the rail voltage of the amplifier. The peak voltage of the signal hasn't changed, but the average power of the signal has increased significantly with the square wave.

I guess my point is.....even if you take the same amplifier and run the same peak signal voltage for a sine wave and square wave through it.....the square wave will have significantly more average power than the sine wave even though the peak voltage of the signal hasn't changed (and hence both signals are within the rails of the amp). That 1000w amplifier is a 1000w amplifier with a sine wave, not a square wave. Average power will increase with the square wave, which leads to the build up of heat. Granted you may hit current or power supply limitations which will not allow the amplifier to output a full 2500w continuously.....but you'll be at the limits of the amplifier, which will result in an increase in average power with a square wave compared to a sine wave with the same amplifier, same peak voltage, within the rail voltages of the amplifier.

If you are going to do a normalized average power comparison, which is what we are trying to compare, you have to reduce the peak voltage of the square wave so that both the sine wave and square wave have the same RMS voltage, which would give both signals equivalent average power. So a 50V peak sine wave would have an RMS voltage of 35.35V. The square wave would need to be reduced to a level of 35.35V for an equal average power comparison. So a square wave wouldn't hit the peak 50V and stay there, because it would peak at 35.35V. Only the sine wave would peak at 50V.

This is what Sean, 95Honda and myself are arguing. With the same average power (i.e. the same RMS voltage, not peak voltage), there won't be a significant difference in heat and/or failure time within the driver. We can't just talk in terms of voltage alone, we have to differentiate between peak and RMS signal voltage, because that dramatically affects the average power of the signal. And you aren't going to make this comparison by only flipping the signal generator between sine and square wave, as the peak and RMS voltages of the signal would need to be adjusted to properly compare average power.

I was thinking about something else on my morning drive......the square wave assumes essentially an instantaneous change in voltage level. If the signal is 5V peak to peak, the signal has a 10V swing in level with virtually zero change along the y-axis of time. Loudspeakers have inductance and energy storage......I would imagine there has to be some sort of "lag" present in the driver's ability to respond to an instantaneous change in voltage like that. The driver isn't going to be able to respond instantaneously and go from a position of +5V to -5V in almost literally no time. That's almost exactly what inductance describes...the ability of the driver to respond to a change in voltage. So I wonder how much "rounding" of the square wave there is in the actual response of the loudspeaker to the signal, as this would affect (reduce) the amount of time the driver was actually at the plateau of square wave, which would reduce the amount of heat build up you are suggesting occurs. By the time the speaker finally gets up there to the plateau of the square wave, the signal is turning around heading back down in the other direction, so the speaker turns around and follows it back down (obviously overly simplified). I wonder if it would almost "track" the same path as a sine wave due to the inductance induced lag.

You're the speaker engineer Nick What do you think?

Wow, great post. I could somewhat understand both sides, but after reading that, I fully understand both arguments. It still seems even if the drivers response to a square wave could have some lag and round out the edges, that this cycle would be happening to so many times, so fastly (sp?) that it would all still add up and cook the coil much quicker then a sine wave.

[///M5 2,833]

Wow, great post. I could somewhat understand both sides, but after reading that, I fully understand both arguments. It still seems even if the drivers response to a square wave could have some lag and round out the edges, that this cycle would be happening to so many times, so fastly (sp?) that it would all still add up and cook the coil much quicker then a sine wave.

No. Basically: Square wave = more power at the same voltage. Turn down the voltage so it has the same power as the sine wave and the heating of the coil will be the same. Brad's lag comment actually says the contrary to what you think you understood that being the speaker cannot follow the path directly but lags and rolls that it will then come closer to following the sine wave and it would then make less heat not more. Either way, again motion and everything else is confusing this. It is really very simple. Average power through a wire creates heating. Same average power = same heating. There is nothing else to discuss on the heating side. If the topic switched to cooling and some other response because of waveform shape fine, but in the realm of heating via the Physics of power there is no difference in waveforms of the same average power. I will add that anyone that discusses instantaneous power as a heat source doesn't understand basic Physics as ignoring the time in the equation is not possible.

Sorry that I am not as eloquent in my descriptions as Brad is, but hopefully my points have been clear.

Don't stop asking questions or clarifying though, we don't always realize that you all don't follow

[NDMstang65 403]

Go take a 120 volt lightbulb..put it on a sine wave

Put a 120 volt light bulb now on a square wave.

Tell me if the square wave one is not brighter.

Rail voltage on an amplifier does not necessarily scale back...and it is not normalized. There is absolutely no point in normalizing things.

[///M5 2,833]

Go take a 120 volt lightbulb..put it on a sine wave

Put a 120 volt light bulb now on a square wave.

Tell me if the square wave one is not brighter.

Of course it will be as it is MORE average power.

Rail voltage on an amplifier does not necessarily scale back...and it is not normalized. There is absolutely no point in normalizing things.

Yes there is, power kills speakers. Or more correctly stated, there are two ways to kill a driver with a signal:

1) Overexcursion - cause by too much instantaneous power

2) A thermal death - caused by too much average power

When describing a signal that is detrimental to the driver it has to be for one of the two reasons. And the only way heat can be increased is to increase the average power.

[Fi Car Audio 64]

Im sorry, I simply dont have time to read everything and make a point by point post... but when we tested amps and waveforms it was done with amps that were capable of making full voltage into both a square or a sine wave. At that, it is what I feel a real world clipping situation is. While it is pretty damn close to impossible to get a perfect square wave (and even more so into a reactive load, ie a speaker) we could tweak and get it fairly close. Points are, that most amps are capable of producing a pretty dang "square" square wave and that you can, within a percent or two, approximate Vrail as Vrms for a sqaure wave. While Vrms for a sine is .707Vrail. This equates to a situation where 40% more power is applied to the speaker in a MASSIVELY clipped scenario. But, we took it several steps further to use a laser based measurement system , an accelerometer, and two different thermal measurements (thermocouple as well as infrared) to measure things. At that we found the obvious, things heat up more, and often significantly more. This is due (based off of measurements of heat and excursion) the increased power and reduced excursion (for like power in a square vs sine wave the square wave has less excursion.... and at higher power and excursion levels where our cooling system comes in to play, was even more drastic). Now, the interesting side of things is even when normalized at higher power levels, we found that the square wave does induce more heat. Watts is Watts, so the reduced excursion simply means that our cooling system does work.

Of note, but not particular to what has been discussed is the force being applied to the speaker in a square wave vs sine wave. A square wave is significantly more brutal (an order of magnitude more) on things mechanically. That is not fun for a speaker. Not too mention that getting smacked like that can lead to a already thermally stressed coil to come undone in a situation that similar power from a sine wave would hold together long enough to cool back down (less force/stress and higher excursion/cooling).

All in all it doesnt equate to anything beneficial. Sounds like poop and is rough on speakers (and Im sure amps, but not as much my concern). Keep it clean and things stay happier.

Thanks,

Scott

[Adrian_D 520]

Here's a video which i think is useful in the discussion....

Amp rail voltage (blue) vs output voltage (yellow):

When the rail voltage has the same value as the output voltage, the amp starts to clip. When it starts to clip, peak voltage will stay the same, nothing is normalized => extra RMS power (or extra peak power applied for a longer period of time).

Obviously it wasn't in an ugly clipping situation, just the tops of the sine were affected. The point is, beyond the limit shown in the video, the signal becomes closer and closer to a square signal. You can also hear some noise in the amp just when it starts to clip.

[SubSam 21]

The Square wave form does not see more power. It sees 5 volts for a longer period of time.

[...]

It sees over 50% more energy for that given period of time, that one cycle. That builds up over an extended period of time more..and more..and more. It is all about time, not the power itself.

Energy over time is the definition of power.

There is one thing left about this statement that has yet to be explored, and it is the most obvious and explanatory fact that we can learn here: there's two kinds of power at hand here. Electrical, in mathematical terms P=iV (current x voltage) and mechanical, W=Force*d/t. Obviously at the flat end of the clipped curve the woofer just burns (no freq), as Nick was stating above. The electrical energy doesn't translate to mechanical energy ( force x distance ), but increases dissipation energy, as you stated above, iV/t, in electrical terms. Now if the voltage is capped, but the average power has increased (square wave integral vs sinal), where does that power come from? Taking V out of the energy dissipation, where V=IR, E=(I^2)R/t. At clip, the t remains constant over each cycle, but the current increases which dramatically effects heat dissipation which in turn raises R which also directly relates to energy dissipation. Point in case, clipping causes more heat. Period. Over long periods, ur effed.

[///M5 2,833]

The Square wave form does not see more power. It sees 5 volts for a longer period of time.

[...]

It sees over 50% more energy for that given period of time, that one cycle. That builds up over an extended period of time more..and more..and more. It is all about time, not the power itself.

Energy over time is the definition of power.

There is one thing left about this statement that has yet to be explored, and it is the most obvious and explanatory fact that we can learn here: there's two kinds of power at hand here. Electrical, in mathematical terms P=iV (current x voltage) and mechanical, W=Force*d/t. Obviously at the flat end of the clipped curve the woofer just burns (no freq), as Nick was stating above. The electrical energy doesn't translate to mechanical energy ( force x distance ), but increases dissipation energy, as you stated above, iV/t, in electrical terms. Now if the voltage is capped, but the average power has increased (square wave integral vs sinal), where does that power come from? Taking V out of the energy dissipation, where V=IR, E=(I^2)R/t. At clip, the t remains constant over each cycle, but the current increases which dramatically effects heat dissipation which in turn raises R which also directly relates to energy dissipation. Point in case, clipping causes more heat. Period. Over long periods, ur effed.

Two kinds of electrical power?? No. In AC it is just average power. Sine or square the wave is described appropriately by it.

[jesseadkins187 5]

Id like to follow up and let everyone know that the amplifier was at fault. I appreciate you pointing out the amplifier nick, I hooked everything up last night and sub was getting stinky at extremely low volume. I checked the output and was receiving a steady 51 volts dc. I'm hoping I haven't burnt up my other sub now.

Also note fi was very helpful in providing a recone and I really apreciate their assistance.

sure glad they helped you out with one even you admitted it was the amplifiers fault..... must be nice

[jesseadkins187 5]

Nick, make me a woofa that I can plug into my 480 volt, 30 amp wall sockets.

Please and thank you!

you have 480 volt wallsockets LMAO

[holley2346 6]

Id like to follow up and let everyone know that the amplifier was at fault. I appreciate you pointing out the amplifier nick, I hooked everything up last night and sub was getting stinky at extremely low volume. I checked the output and was receiving a steady 51 volts dc. I'm hoping I haven't burnt up my other sub now.

Also note fi was very helpful in providing a recone and I really apreciate their assistance.

sure glad they helped you out with one even you admitted it was the amplifiers fault..... must be nice

The whole objective of the thread was to get a recone whether they "helped" me or not. Things happen, amplifiers fail. This issue has been long resolved so an attempt at trying to be an ass is quite pointless.

Fi has been extremely helpful with the whole situation over the past couple of weeks and after several calls with nick I feel more confident than ever running fi audio.

Also - there is some extremely valuable information on this thread, thanks for all the info guys.

Edited February 16, 2011 by holley2346