You are confusing time domain and frequency domain concepts.
Impedance is a frequency domain concept but you are trying to use it in the time domain which doesn't make sense. I didn't say you were wrong about the peak currents. It's just the "impedance drop" explanation that is wrong. Actually, that article you posted supports my argument that the power supply is irrelevant when considering the peak currents: "The current peaks typically last a few hundred microseconds."
The correct explanation if you want to talk about impedance is that the current contribution from each frequency component of the transient signal cause constructive interference just at that moment where the high peak current occurs. In the time domain, stored energy and induction, like you say, is a good explanation. But impedance doesn't enter into it in the way you (mis)use it and say it's the impedance that drops and causes the peak current.
Of course. But I wasn't referring to that.
No, it causes increased impedance in transformers too. It is the lack of self-induction during saturation that causes the high inrush current.
Correct - but this is a linear effect and thus included in the impedance information. You can model the cone mass and suspension springiness (with energy storage and everything) like electrical components (inductors and capacitors). There is nothing strange going on there.
Impedance is a frequency domain concept but you are trying to use it in the time domain which doesn't make sense. I didn't say you were wrong about the peak currents. It's just the "impedance drop" explanation that is wrong. Actually, that article you posted supports my argument that the power supply is irrelevant when considering the peak currents: "The current peaks typically last a few hundred microseconds."
The correct explanation if you want to talk about impedance is that the current contribution from each frequency component of the transient signal cause constructive interference just at that moment where the high peak current occurs. In the time domain, stored energy and induction, like you say, is a good explanation. But impedance doesn't enter into it in the way you (mis)use it and say it's the impedance that drops and causes the peak current.
pacificblue said:
Of course. But I wasn't referring to that.
No, it causes increased impedance in transformers too. It is the lack of self-induction during saturation that causes the high inrush current.
Correct - but this is a linear effect and thus included in the impedance information. You can model the cone mass and suspension springiness (with energy storage and everything) like electrical components (inductors and capacitors). There is nothing strange going on there.
It makes sense to me. While R and jX of the voice-coil remain constant, the impedance the amplifier sees is obviously lower than that at certain moments. If that were not the case, Ohm's law would have to be rewritten. So there must be an additional component.megajocke said:
The best I can come up with is Z=Rvc+jXvc+jXoe, where vc is voice-coil and oe is other effects. While jXoe is probably 0 most of the time, during the current peaks jXoe<0 and |jXoe|>jXvc. The question is, where jXoe comes from. In a voice-coil induction and self-induction are the only possible sources I can think of. Induction is probably too constant to explain those short peaks.
...and come from where?megajocke said:
Current flow will always lead to a voltage drop -> less ability to deliver the demanded current. Whether the short duration keeps the effect on sonic performance low depends probably on, how bad the peak really is. May be irrelevant for >90 % of all speakers, but could be relevant for some that are labeled "picky" or "tough loads".
Only for the signal that is used to measure the impedance curve. A music signal is as different from that, as it is from steady sine-waves.megajocke said:
I never said there was.megajocke said:
From what I have read of Mega's responses, in this and other threads, he is of the other half that refuse to accept that a speaker can demand more current than Re would predict.
As long as his feet are so firmly entrenched in that camp, I cannot see how the two sides are going to agree.
I think that was why he chose ridiculous examples to help explain some of the differences that might be heard or measured to put down the case for excess current demand.
I am in the camp that says the speaker can demand current well in excess, anywhere from two times to five times, that which nominal impedance would predict on steady state sinewave signals within the voltage capability of the amplifier.
I too am entrenched in this view.
As long as his feet are so firmly entrenched in that camp, I cannot see how the two sides are going to agree.
I think that was why he chose ridiculous examples to help explain some of the differences that might be heard or measured to put down the case for excess current demand.
I am in the camp that says the speaker can demand current well in excess, anywhere from two times to five times, that which nominal impedance would predict on steady state sinewave signals within the voltage capability of the amplifier.
I too am entrenched in this view.
AndrewT said:
The speaker itself won't draw more current than what Re predicts, but they system sure as hell can. I have, and am currently using a 3 way dual woofer system that uses a pair of SS 25W/8565. "Nominal" impedance 4R. It gets down to about 1.8R just above primary resonance - due to interaction with the crossover indcutor. It helps with the BSC, and is therefore desirable. But most amps get very very upset with this load so I use one of my homebrew PA amps. The high current demand in this part of the band is predicted quite well by the measured Z curve of the drivers and the model for the crossover. Nonlinear effects need not be included.
Hi Wg,
is that two 4ohm bass drivers in parallel or two 8ohm bass drivers in parallel?
What is the Re of each driver?
is that two 4ohm bass drivers in parallel or two 8ohm bass drivers in parallel?
What is the Re of each driver?
Two 8 ohm bass drivers in parallel. The combined Re is in the 3 ohm range. I experimented with a congugate network to flatten the impedance peak above fb, and in so doing , knock out all that capacitive reactance which the crossover was interacting with. That did raise the impedance where it was dipping that low, but the midbass got thin. Outdoor gated measurments at 25ms confirmed that the mild 2dB rise caused by the crossover was doing a decent job of baffle step correction so I left it that way. Using an old HK reciever was out of the question - it sounded like nails on chalkboard - very obviously limiting. In the end I even had to relieve the poor overheating Phase Linear from mains duty.
AndrewT said:
Hi,
I won't deny that the speaker can demand more current than Vcc/Re, it's pretty obivous that it can. For a single driver it is not likely to be much higher than twice this. I won't say the article is wrong either, the 6 times is with crossovers and everything.
What I doubt is that the type of signals needed occur in music at full amplitude. If you want to be on the safe side - design for it.
For very high power subwoofer amps and PA amps, it doesn't make much sense to try to do that though.
pacificblue said:
The impedance doesn't change. If you define something like instantaneous resistance, r(t)=u(t)/i(t) then you are right that it does decrease, or even becomes negative. But such a quantity isn't very useful for anything. But impedance is something else. Observe that the models used in those simulations were entirely linear, and still you get high peak currents!
The impedance of a speaker driver looks something like Z = Re + f(w) where the real part of f is always positive. w is supposed to be a small omega, angular frequency. The impedance does not change much with time for a real speaker unless the voicecoil goes out of the gap. The simulations that showed 6 times Vcc/Re were entirely linear and thus did not change impedance with time.
Not true, the impedance curve shows what happens at every frequency if a sine wave of inifinite duration is applied at that frequency. That might not seem very useful at first, but any signal can be described as a sum of sinewaves of infinite duration, where each one has a different amplitude and phase shift. (Fourier transform)
If a system is linear, superposition can be used. One calculates the current for each frequency component using the impedance curve, because this is just the kind of signal the impedance curve describes - sinewaves of inifinite duration. These currents are then added to each other.
The kind of signal used in the test contains a lot of harmonics and these are phase shifted by different amounts in such a way as to add constructively at some points in time - these points in time are when the peak currents occur.
For example, think of a resistor and capacitor in series: Z = R + 1/(jwC)
If this combination is driven by a low frequency square wave between +Vcc and -Vcc the peak current will be two times Vcc/R. This is in part because the current components caused by the harmonics of the square wave are phase shifted by different amounts.
On peak currents and power supply:
40V pk output - 4 ohm speaker load. Resisitive peak current = 10A.
Let's say the speaker can draw ten times that as peaks = 100A.
The PS has 20mF of smoothing.
100A during 100µs is a charge of 10mC.
This discharges the power supply by not more than 10mC/20mF = 0.5V
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