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MultiWay Conventional loudspeakers with crossovers 

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4th September 2003, 02:59 PM  #1 
diyAudio Member
Join Date: Jul 2002
Location: England

Speaker impedance not what it seems?
I have a question about speaker impedances. This is a fairly simple question, but it has me puzzled.
Often manufacturers rate a "characteristic impedance" which as we all know does not tell us much about the actual load the amplifier "sees" during operation. Impedance curves give a better view of the problem as it can be seen that the impedance dips and peaks at various mechanical and electrical resonances, etc. Manufacturers often also rate a minimum impedance  the lowest dip on the impedance graph, but my querie is that this may not be the actual lowest impedance the amp "sees". Say a loadspeaker has an essentially flat impedance graph but with two dips. The manufacturer will quote the lower of the two as the minimum impedance, but what happens when the amplifier outputs a signal that is the combination of two sinusoidal signals at the frequencies that relate to the impedance dips? Does the impedance of the speaker fall to the parallel combination of the two dip impedances?
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4th September 2003, 04:29 PM  #2 
diyAudio Member

No, it does not.
The impedance of the speaker will be the lowest trough for the one wave, and the secondlowest for the other. (Think about it: when you put an 8ohm woofer and 8ohm tweeter in a crossover together, the DCR stays 8ohms.) The crucial thing the amp "sees" is current draw, and current draw is simply voltage^2 / impedance for the given freqency. So, it's important (with solidstate amps) that the impedance curve doesn't drop to a point that makes the amp clip, because maybe, just maybe, the voltage at that frequency might go too high and the current draw might injure the amp / speakers. Assuming white noise, the max current draw will be the voltage^2 of the white noise over the lowest impedance on the graph. Music more closely resembles pink noise, however, so the impedance graph doesn't tell the whole story. In short, don't worry about it except for very low frequencies; they're the only ones with enough power to do serious damage, and they'll probably bottom out your woofers before they'll burn out your amp. 
6th September 2003, 07:25 PM  #3 
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Join Date: Jul 2002
Location: England

In your reply you stated:
"The impedance of the speaker will be the lowest trough for the one wave, and the secondlowest for the other." I'm not sure you read the question correctly  the impedance graph shows the measured impedance for a sinusoid at a single frequency. My query was with regards to a signal containing TWO sinusoids at frequencies that coincide with BOTH impedance dips.
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6th September 2003, 08:12 PM  #4  
frugalphile(tm)
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Re: Speaker impedance not what it seems?
Quote:
The attached (from one of Richard Heyser's papers) is not an impedance curve, but will hopefully get the idea across... let the projection of the spiral in 1 direction be the impedance magnitude and in the perpendicular direction the phase... a 3rd projection perpendicular to the other 2 would give the Nyquist impedance plot. dave
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6th September 2003, 09:07 PM  #5  
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Location: Calgary, Alberta

Quote:
Is that purely for the test sinusoid condition only? What is the effect/artifact when an actual music source is the condition?
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6th September 2003, 09:31 PM  #6  
frugalphile(tm)
diyAudio Moderator

Quote:
dave
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6th September 2003, 09:44 PM  #7 
diyAudio Moderator Emeritus
Join Date: Jan 2002
Location: Michigan

That’s a cool graph Dave.
Another way to look at it is to realize that the impedance is frequency dependent. What this means is that each frequency being reproduced will see a different impedance at the driver based on the drivers impedance curve. The driver differentiates each frequency by the rate of change of the current coming from the amp. The driver sees the rate of change of the current and presents the impedance that corresponds to that frequency. The phase also plays a roll in what impedance is presented, but that will complicate the discussion. Let’s stick will current rate of change for now. When two or more frequencies are combined they form a composite wave form. The rate of current change will vary at different points in this wave form. The impedance presented by the driver will also vary with the wave form as I described above.
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Rodd Yamashita 
6th September 2003, 09:58 PM  #8 
diyAudio Member

Hi Rodd,
Annex666's initial question is fascinating. I have no idea (yet) what the implications are. One the one hand one could see the combination of two tones as an integral waveform. OTOH, we are always comfortable with FFT etc to dissect a waveform in it's constituent frequencies, and take these in account separately. Lets say the amp gives a 8VRMS 1kHz tone plus a 8VRMS 2kHz tone in an 8 Ohms speaker (just for the discussion). Can we say that this causes 1 Amp 1kHz PLUS 1Amp 2 kHz? I guess so. What is then the peak current, that then depends on the phase relation, right? If the peaks of the two waves coincide, we have 2 Amps peak. I can also imagine a situation where the two tones are opposite in phase and the peak current (at that moment) is zero. So, how do I translate that to the original question of the compound impedance the amp sees? Interesting... Jan Didden 
6th September 2003, 11:06 PM  #9 
diyAudio Moderator Emeritus
Join Date: Jan 2002
Location: Michigan

Hi Jan,
Yes, we would have to go back to Dave’s graph and take phase into account. We would have to determine the phase relationship between applied voltage (the original signal from the amp) and the resultant current through the driver (the resultant driving signal). The phase of the signal from the amp at a given point in time is also required information. Then we have to take the electrical equivalent circuit model to the driver and the resultant frequency dependent impedance response. This impedance will have a purely resistive component (for which Ohm’s lay will hold) and a reactive component (represented by phase). When we look at any instantaneous point (in time) on the “wave form” that represents the voltage signal, we will find it to have amplitude and frequency (corresponding to the instantaneous rate of change of the signal determined by its phase). The driver will present an impedance to this signal at that instant in time. This impedance will determine the resultant current. The resistive component of the impedance will determine the amplitude of the current and the reactive (capacitive or inductive) component will determine the phase (the instantaneous rate of change) for the resultant current at that instant in time. So going back to your example, we would have to know the phase relationship between the to signal frequencies. With that we would calculate the voltage amplitude and phase (rate of change) at any point in time = t of the composite wave form. Then we would need the impedance response (with frequency) of the driver. With those two pieces of information, we could than calculate the resultant voltage/current signal at any point in time = t which will include amplitude and phase.
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8th September 2003, 01:13 AM  #10 
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Join Date: Jun 2003
Location: USA

> what happens when the amplifier outputs a signal that is the combination of two sinusoidal signals at the frequencies that relate to the impedance dips? Does the impedance of the speaker fall to the parallel combination of the two dip impedances?
No. Say we have two impedance dips, say 4Ω at 400Hz and 5Ω at 999Hz. Say we play a signal that is 400Hz and 999Hz, both at the same amplitude. Say that the amplifier's peak output voltage is 40 Volts. Say that the amp is rated for a 4Ω load; it therefore can deliver 40V/4Ω = 10 Amps peak. I picked 400 and 999 because we can be sure (if we play for more than a short burst) that the phases will pass each other and the peaks will sit on top of each other. The highest level we can play without voltage clipping is 20Vpk at 400 and 20Vpk at 999, because 20V+20V= 40V, the amp's voltageclip point, and we can't go higher than that. A 20V peak on the 400Hz wave in the 4Ω impedance is 5A peak. A 20V peak on the 999Hz wave in the 5Ω impedance is 4A peak. The amplifier must deliver 5A+4A= 9A peak. The amplifier "sees" a 40V/9A= 4.44Ω load. In general the amp will never see less than the "minimum impedance". Here is another factoid. For a "simple speaker" (no crossover), the lowest impedance and the highest current under ANY condition will be limited by the DC resistance. A "4Ω" speaker is typically a 3Ω DC resistance. While the complex impedance can be very, uh, complex, the current will never rise above Vpk/3Ω However, certain resonant crossovers can "invert impedance" over narrow frequency bands, and you can "see" an effective 2Ω and corresponding high current in a a nominal 4Ω woofer. In fact this seems inevitable using a classic 12dB/oct crossover below 800Hz with a heavy woofer. 
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