George,
There will be one frequency where each driver has an impedance minimum. All you need to do is determine where they are singly and then use a group of oscillators to tickle them at the same time. Using a high value resistor in series with the device under test will essentially be a current source so measuring with a true RMS voltmeter across the loudspeaker will give the impedance. For a double check you can then substitute a resistor close to the measured value.
ES
In the context of real amplifier, it isn't actually a problem: if your stimulus signal is made up of three frequencies, the voltage amplitude of each signal needs to be 1/3rd of the maximum signal handling ability of the amp, otherwise there will always be some instant where clipping occurs.
This means that even if the instantaneous quotient of V/I might apparently be lower than the nominal impedance, the peak current will never exceed what is seen at the max out voltage on a nominal load.
Properly designed amplifiers normally have no problem driving much lower loads than nominal, provided their OP devices remain within their SOA.
What kills them is overcurrent, or exceeding the second breakdown limit.
That is valid for non-clipping conditions, and for steady-state harmonic signals, and I agree that some transient conditions may be somewhat harsher, but basically I find Vacuphile's view as the most sensible: certainly no negative resistances coming out of nowhere, because they would violate basic energy conservation laws, and perhaps some peculiar transient conditions (likely to be found somewhere, sometimes in an actual audio signal) leading to a momentary higher current than what you could expect from simple arithmetics, but no more, and it is certainly severely bound by theoretical limits(it has probably already been studied, but I have no ref to offer)
Typically a CM is placed between the OPS and the pre-driver + IPS. so its really integrated into the structure of the amp. I just cannot see how we can then ignore its contribution to reducing unwanted PSU noise on the output. It most assuredly does reduce PSU artifacts on the output so it should be considered fully as improving PSRR.
Just my two cents.
Just my two cents.
No. Three signals can each be 57% of full scale and still sum to a bit less than 100%.
And why would they be? Harmonics would but they are decreasing in amplitude. Most likely at least 20 dB down or more from driver to driver.
But you knew that. Now get back to work on your thesis...
Probably not audible artifacts, hence easy to ignore*. If there's no gain on the noise it might be seen as trivial, but that doesn't mean that it is... Subjectively I wonder what flipping a CM on and off during playback would be like.
*Audible artifacts almost don't exist, but noise does change music. Let's leave the pops and sizzles out of the discussion because an amp with those needs more than a CM.
Last edited:
Ed
OK for all you say par the mechanism by which the other two frequencies (except the Fc ) will affect the Impedance of the driver.
If I want to perform this test what will be the other two frequencies if Fc is say 45Hz?
Elvee
As far as I understand, Ed’s talk is targeting on measuring low impedance of loudspeakers.
Although this is done typically with low level signals at the speaker terminals, when a high value resistor is in series with the speaker (for achieving a current source), the voltage drop across the resistor is substantial, to the point that the amp may be driven into clipping before the voltage at speaker coil reaches an adequate level for a noiseless impedance measurement.
Having two or three frequencies driving the amp simultaneously may do amplifier clipping more risky.
George
Derf
Music isn't steady state. However clipping often sounds horrible depending on how the amplifier recovers. Some squeeg.
George
Every loudspeaker system will be different. So just sweep the first oscillator in series with a high value resistor. Then pick the frequencies that correspond to the voltage dips.
Music isn't steady state. However clipping often sounds horrible depending on how the amplifier recovers. Some squeeg.
George
Every loudspeaker system will be different. So just sweep the first oscillator in series with a high value resistor. Then pick the frequencies that correspond to the voltage dips.
With signals of more or less random phase/frequencies, there will always be an instant where all three reach their peaks simultaneously (it can be demonstrated mathematically).
The problem is similar in other fields: for DSL signals using hundreds of carriers, there has to be mitigation mechanisms to limit the extreme peaks without clipping or reducing the average power to insignificant levels.
Rogue waves are also caused by a comparable mechanism.
Yes.
These forces are strongest in a BR enclosure at resonance. But since they are not strong enough to force the cone in the opposite direction as the current flowing through the VC, they do not generate the kind of movement of the VC that would lead to a lowering of the impedance of the driver below Rdc. That was the original question: can back EMF lower the impedance below Rdc? Not in a real speaker without an outside force impacting on the cone.
Well my view is that the easiest way to explain to my grandmother is to be clear about what I mean, NOT wavering to explain things in terms that are used differently from the use by the rest of the engineering community. Audio has already too much bad rap for sloppy use of terms. What is the definition of PSRR again?
Jan
Totally agreed..
The issue is this: The voltages will be a summation as per your statement. However, the currents will also be a summation, as the amp is driving three parallel loads...
There is a problem when one considers the entities as simple real numbers. Operation of a reactive load requires the energy stored in the load be returned.
If one simply uses a real V/I ratio, one calculates negative resistance in quadrants 2 and 4. What is really going on is energy shedding by the load.
Despite any calculations, it is trivial to show that in quadrants 2 and 4, the EMF of the load is HIGHER than that of the amp, so it is trying to return power to the source.
I'm afraid that incorrect for 3 independent and uncorrelated signals.
No need, you are correct within the context of 3 uncorrelated independent signals.I understood as 3 simultaneous sinusoidal signals, so you'd hit some sort of periodic beat.
Simulations are crunching right now.
Again, the difficulty is in the definitions. When one considers the entire V/I space and the complex load, it can easily be seen that forces on the cone are contrary to your argument. Your explanation is entirely consistent with a pure resistive load, not one which stores energy. For example, scotts simple "tap the cone" demonstration showed a back emf generated by just the spider restoring the position. The forces within an enclosure with hundreds of watts of non steady state stimulus pushes the cone around far more, and the amp has to control those energies by the mechanism we consider damping factor.
The problem with the question is, it is a steady state stimulus resistive question in a transient stimulus reactive world.. If the load is resonating, trying to brake it will take significant amplifier.. Again, note the term "brake", it indicates that it is not a steady state condition.
The "outside force" is the entire energy conversion system reacting to a change in the stimulus.
John
Last edited:
Objection !
The term “brake” indicates that it is a steady state condition.
The technical term that indicates it is not a steady state condition is “quash”
George
Ya got me...Objection !
The term “brake” indicates that it is a steady state condition.
The technical term that indicates it is not a steady state condition is “quash”
George
john
Yes avoiding complex math completely will guarantee some misconceptions. Impedance and resistance are not the same. Measuring at peaks and dips in the frequency response (assuming min phase) happens to give a resistance. Maybe a better question is can you get a real resistance less than Rdc?
In my case, I was bantering back with Ed about finishing my thesis. I was in the middle of a rather intensive FEA sim (COMSOL)...thinking of which, would be a good platform to show the 4-quadrant IV behavior, and an estimate of the magnitude. (I do not have time to build such a model in the next, say, 6 months, but might put a bug in someone's brain to try it).
Think Ed is still trying to make a different point, as there seems to be two discussions at play here.
Honest question: do we really have high enough Q resonators to generate the kinds of cone forces needed to achieve quadrants 2 and 4? I understand the theoretical possibility, but don't know if they're (reasonably) achievable in normal box designs.
Take the model presented by nelson, and force the sim to view the current and voltage on an x-y plot. I assume a sim can do such. If not, export the voltage and current waveforms as excel's, use excel to graph. In that sim, try a single sine, then two, then three, with each sine targeting a specific branch of the load.
My depictions earlier in the damping thread were simple excel columns. I just made one graph with time as horizontal, and another with current as horizontal.
If you examine Nelson's load model, you see the reactances. Anytime there are reactances, there will be operation in Q2 and Q4. (unless sines are used at the exact frequencies where phase is zero.)
John
- Status
- Not open for further replies.