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Old 23rd May 2008, 09:59 AM   #1
graaf is offline graaf  Poland
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Default "The Dynamical Loudspeaker Impedance"?

Hi everybody!

I mean the following problem:
http://www.audiograph.se/subpages/te...damplifier.htm

Can anybody explain this please?
Is it common that an amp "sees" an impedance 4 times less than the nominal impedance of the loudspeaker? Or can it be 2 times less or 2 times more and so on?
What are the deciding factors?

Is it a part of a larger and overlooked "amplifier-loudspeaker inteface" problem?

best,
graaf
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Old 23rd May 2008, 10:18 AM   #2
AndrewT is offline AndrewT  Scotland
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Hi,
my reading has revealed that it is possible for the speaker to demand much larger peak currents than a simple Vpk/Re or Vk/Rload would predict.

I have accepted the argument that states that peak transient current approaches three times the current that using nominal impedance would predict and design the output stages to meet that demand.
But, I have no proof of how accurate these guessimates/findings are.
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Old 23rd May 2008, 11:35 AM   #3
forr is offline forr  France
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Hi,
Douglas Self addressed this problem in this article :
"Loudpseakers undercurrents"
Electronics World, February 1998, p98-99.
More current demand than the voice coil DC resistance would demand can happen with asymetrical signals. Self thinks that a previous paper by Ottala in 1983 overestimated the problem.
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Old 23rd May 2008, 11:41 AM   #4
sreten is offline sreten  United Kingdom
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Hi,

I notice they use a speaker with a spectacularly bad impedance curve.

The square wave impedance curve is a joke and is simply not impedance.
Once you start fudging about you can show almost anything you like.

I would not argue with AT's guesstimate for the current limiting
of a nominal imnpedance load for a load tolerant amplifier.

/sreten.
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Old 23rd May 2008, 11:43 AM   #5
graaf is offline graaf  Poland
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Quote:
Originally posted by forr
Hi,
Douglas Self addressed this problem in this article :
"Loudpseakers undercurrents"
Electronics World, February 1998, p98-99.
More current demand than the voice coil DC resistance would demand can happen with asymetrical signals. Self thinks that a previous paper by Ottala in 1983 overestimated the problem.
and what about the measurement the graph of which I have attached above?
this is an actual measurement of an actual loudspeaker

the upper red corresponds to a conventional measurement the loudspeaker impedance with sine wave signal
the lower blue graph corresponds to a measurement using a square wave signal

Quote:
The reason for this is that square waves consist of a large amount of sine waves, as does music. The square wave is of course not an equivalent of music, but for this test it was an easy way of showing that a complex signal (not just a simple sine wave signal) may make the load, from the amplifier point of view, very low.
best,
graaf
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Old 23rd May 2008, 11:48 AM   #6
graaf is offline graaf  Poland
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Quote:
Originally posted by sreten
Hi,
I notice they use a speaker with a spectacularly bad impedance curve.
true, perhaps to illustrate the problem better?

Quote:
Originally posted by sreten

The square wave impedance curve is a joke and is simply not impedance.
a joke? what do You mean?
what is impedance? is there "using of a sine wave signal" in the definition of impedance?
I am just asking

best,
graaf
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Old 23rd May 2008, 11:59 AM   #7
Mooly is online now Mooly  United Kingdom
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You can plot the impedance of your own speaker easily with a meter that covers the required frequency range. Doug Self's method works well, connect a 600 ohm resistor in series with the speaker and drive this from 6 volts RMS. You can read in millivolts across the speaker the equivalent impedance at an frequency. So 4 millivolts is 4 ohms etc. Think I have remembered the values correctly.
And yes it is crazy- My B&W 703 an 8 ohm speaker has a 3 ohm minima !
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Old 23rd May 2008, 12:01 PM   #8
graaf is offline graaf  Poland
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Quote:
Originally posted by Mooly
You can plot the impedance of your own speaker easily with a meter that covers the required frequency range. Doug Self's method works well, connect a 600 ohm resistor in series with the speaker and drive this from 6 volts RMS. You can read in millivolts across the speaker the equivalent impedance at an frequency. So 4 millivolts is 4 ohms etc. Think I have remembered the values correctly.
And yes it is crazy- My B&W 703 an 8 ohm speaker has a 3 ohm minima !
would this measurement be equivalent to a measurement that uses square wave signal?

best,
graaf
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Old 23rd May 2008, 12:10 PM   #9
Mooly is online now Mooly  United Kingdom
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Good question, and I don't really know the answer to that one. I suspect not, certainly in the way you mean. Hard to explain this, the effect of the 600 ohms means the speaker is undamped, it's not like feeding it from a low impedance source. Dunno is the answer, try it
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Old 23rd May 2008, 12:17 PM   #10
Mooly is online now Mooly  United Kingdom
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I will add to that, squarewave testing while incredibly useful in amplifier measurements etc, bears no relation whatsoever to anything you will see off a disc. The C.D. format cannot even reproduce a 1Khz squarewave that would be considered even "acceptable" in anything else. The 44.1 Khz sample rate ensures that. Music certainly contains nothing like this anyway.
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