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Discussion Starter · #1 ·
My wife gave me the green light for a SI 18" driver. I'm thinking 4 ohm per coil so it could be driven by an 8 ohm amp. I have such an amp but in bridged mode it's only 100 W. Eventually I'd like to add something like an inuke but in the mean time, are there any dangers in under-powering a driver?
Rubus
 

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The problem with under driving a speaker is that you are more likely to damage it from clipping the amp. You can also damage a speaker from too much power but in general as long as you keep the amp from clipping, and do not go over the max power suggested by the speaker manufacturer you will be ok.
 

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With non-damaging levels of clipping you will also hear a lot more distortion from the underpowered amp. Upgrading to a proper amp will be a big noticeable change.
 

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Discussion Starter · #4 ·
Thank-you ellisr63 and fusseli,
I ordered the driver yesterday. It may very well be that this is going to be a moot issue. It may take just as long to save up for a inuke as it will to design and build the cabinet. In case I finish the cabinet first, I do have 100 W I could connect to it. It wouldn't be a permanent arrangement. I'm not sure I understood you ellisr63. Actually I'm sure I didn't understand you because I don't know what clipping is or which device might get damaged. How can I tell if I am clipping?
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You won't damage a driver by supplying too little power. If that were the case then everything would blow up when you turned it down. Also clipped waves themselves will not damage your driver but you do need to remember that a clipped 100W waveform could actually be supplying double the power that a clean 100W sine wave.

Even though your amp will supply a lot less than the max rating for the driver you can still easily damage the speaker by delivering a frequency too low for the system. So a proper high pass filter would still be in order even with the small amp.
 

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Thank-you ellisr63 and fusseli,
I ordered the driver yesterday. It may very well be that this is going to be a moot issue. It may take just as long to save up for a inuke as it will to design and build the cabinet. In case I finish the cabinet first, I do have 100 W I could connect to it. It wouldn't be a permanent arrangement. I'm not sure I understood you ellisr63. Actually I'm sure I didn't understand you because I don't know what clipping is or which device might get damaged. How can I tell if I am clipping?
Rubus
I guess I didn't explain what I meant properly... Maybe this will explain it better. :T
http://en.wikipedia.org/wiki/Clipping_(audio)

When a amplifier is driven into clipping I have found that it will not sound as clear as when the amplifier is playing at a lower level. I have also found it is most noticeable on the higher frequencies as they will sound more jumbled and sound distorted. If you are listening to a known source and it starts to sound distorted... I turn it down as it is a very good chance that it is clipping the amp.

One thing to note is that if you go with a bigger amp then suggested and drive it to clipping you will most likely blow a speaker faster than using a smaller amp that is clipping. I believe it has to do with the driver heating up and or the driver is exceeding it's max throw. When a amplifier is clipping it is losing control of the speaker... What can happen is that when the driver pushes outward the amp that is clipping may not be able to stop the piston from coming out of the cylinder. If the piston comes out of the cylinder the speaker is blown. This is what I was told years ago... If I am wrong please correct me.

I have always turned the music down when it started to behave like the amp was clipping, and I have never blown a speaker. I have heard others playing to levels where the sound was distorted, and told them that the amp was clipping, and they didn't turn it down... The next thing you knew they had blown a speaker.
 

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Discussion Starter · #7 ·
Thank you dboomere and Ron,
The wiki link was pretty interesting. I'm a bit naive on these issues and it seems counter-intuitive. How does a wave that is clipped have more area than an unclipped wave? I need to study this some more. Having just spent $200 on a driver I won't be able to replace I want to be sure I don't ruin it. I appreciate the generosity of all who share their expertise.
Paul
 

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This of the sound wave as just that, a gentle wave that will flow from a flat horizontal plane such as a calm pond to a larger height back to calm and then down below calm as the water is moved, sort of like dropping a rock in the pond.
So you have seen the wave form, smooth up, middle and then down, that must then be measured against a square wave which represents clipping. The square if measured mathematically will cover more area then the gentle curve. One has to measure straight up, across then straight down, whereas the gentle wave is more like a short cut to the next note.

In addition, the speaker can move in a fashion much more similar to the gentle ebb and flow that is the clear sound wave, however, no speaker can do what a square wave asks. Try turning the front wheels of a car
in a square, it just does not work.
 

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Clipping in a way is like running the driver into a brick wall it will damage it over time because the signal is no longer a smooth wave rather a hard stop and start.
 

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A clipped wave itself causes no more damage or stress than any other waveform to your drivers. That is simply a myth. There are plenty of clipped waveforms that occur in the recorded music that you play through your system.

OTOH overdriving a speaker with too much power can happen regardless of waveform and that is a problem.


..and herein lies the fallacy kind of. If the clipping is used for a reason, guitars for example it is very controlled and the amplifier will be able to cope. If the clipping is due to an amplifier hitting its maximum range of power delivered, it will try to give more and this will then cause the amplifier to send out clipped and at times, square wave signals. WIKI : The transition between minimum to maximum is instantaneous for an ideal square wave; this is not realisable in physical systems.
Not realizable in physical systems, in other words at some point the speaker will be pinned out or in or both, and this causes the coil to overheat and melt the teeny copper wire. Yes the same can happen with high powered systems but usually not because of clipping, but rather because the speaker was not designed for that. A speaker can usually handle more power of the good kind against its rating than less power of the bad.
So for a speaker of 100 Watts, 100+ non clipped good watts may be doable where as 25 clipped watts could burn it up.
 

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Discussion Starter · #11 ·
I am really enjoying this discussion. Thank you. I think I'm starting to get over my confusion. What I envisioned was comparing a smooth waveform to what you get when you take scissors and clip off the tops of the waves. In my mind it seemed obvious that the clipped wave would have less area by the amount you clipped off. Smoothness I get. I get that a clipped wave would be tough for a driver to follow. I get that if you ask a driver to do a lot of tough work, then it will quickly fail. I'm just not sure what area has to do with it. Smoothness is described mathematically as a derivative. A wave that is approximately square has large derivatives. The Fundamental Theorem of Calculus tells us that determining smoothness and area are inverse operations. Can you feel my pain? One way I see it has to do with what you can do with that pond after it's been in a polar vortex. Ice skaters interact with each other with grace and smoothness, much the way a driver interacts with a smooth wave. Hockey players are different. They are more like a driver meeting up with a square wave. They call you for clipping in football, do they in hockey?
I am pleased to be able to celebrate Chinese New Year tomorrow knowing Fed Ex has picked up my 600 W Stereo Integrity HT 18 D4 driver. I'm really hoping it's not the year of the hoarse.
 

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I am really enjoying this discussion. Thank you. I think I'm starting to get over my confusion. What I envisioned was comparing a smooth waveform to what you get when you take scissors and clip off the tops of the waves. In my mind it seemed obvious that the clipped wave would have less area by the amount you clipped off.
But you left out that you then push up that clipped waveform to the ceiling pulling up everything else with it. So the "duty cycle" is longer than a sine wave of the same peak voltage.
 

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The only drivers that I blew because the amp was probably driven into clipping is the tweeter.
When your amp is driven too hard the output will looks like a square wave, a square wave has the original wave plus multiple waves that are a multiple of the original frequency. (i.e. a note at 60HZ, will also have content at 120Hz, 180Hz, 240Hz, etc). and tweeter cannot take as much power.

but if your amp is only driving your sub, you will never be able to blow your driver.
If I remember my math correctly a square wave a 1.41 time the power of a sinus (hey don't flame me on this, I've been out of college for 40 years:eek:lddude:).
 

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I was speaking metaphorically. With a pure sine wave the peak voltage would only touch the top rail of the power supply at one point. With a perfect square wave it would touch it for the entire period of the wave ( a long line).
 

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The only drivers that I blew because the amp was probably driven into clipping is the tweeter.
When your amp is driven too hard the output will looks like a square wave, a square wave has the original wave plus multiple waves that are a multiple of the original frequency. (i.e. a note at 60HZ, will also have content at 120Hz, 180Hz, 240Hz, etc). and tweeter cannot take as much power.

but if your amp is only driving your sub, you will never be able to blow your driver.
If I remember my math correctly a square wave a 1.41 time the power of a sinus (hey don't flame me on this, I've been out of college for 40 years:eek:lddude:).
Actually it doubles.
 

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Discussion Starter · #17 ·
The sine curve has an area of 2 over the first pi radians. (The integral of sin(x) is -cos(x), evaluated at the boundaries gives -cos(pi) + cos(0) = 2.) A square (rectangular) wave has an area of pi. Pi/2 is about 1.57. A square wave has about 1.57 times the area of a sine wave. As far as power is concerned I have no idea. The square root of two is about 1.41. That might show up in a power comparison calculation but I don't get the modeling.
I heard from Nick at Stereo Integrity today. "You can use a 100 watt amplifier on the woofer. Too little power never killed any speaker. :)"
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So a sine wave has a 3dB crest factor and a true square wave has none. It's just full on during the entire period. So when you look at it that way the sine wave has 1/2 the power. Of course with music as your source it's almost impossible to actually get perfect square waves although you will get clipping.

Speaker power ratings are usually based on shaped power over some period of time (usually at least 2 hours). So a few clips even at double power probably won't destroy them right away. Besides it probably will sound so awful that you just won't go there.
 
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