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Re: MOSFETs work fine too...

I concede an error - in the analogy I should have said 'Load' not 'impedance'. The load is greater if the weight you are about to lift is falling.

Ok.

Likewise, if the cone is moving in the opposite direction (to a transient thump, say a drum sound) because of a lower frequency tone and assuming it has some momentum, then the load on the amplifier is greater as the impedance will be lower at that point.

You seem to be assuming here that the loudspeaker isn't responding to the transient. If it's not responding to the transient, then it's not reproducing it. And if it's not reproducing it, we can't hear it. So what's the point?

Now if you assume a loudspeaker that is responding to the transient, then the impedance will be whatever the loudspeaker's impednace is at the frequency/frequencies involved with the transient.

Further, I repeat that back EMF is the voltage produced by a charged coil. What you are referring to as back EMF is just the speaker producing a voltage due to its motion (the coil moving past a magnet). If we were to run with that non-scientific definition then all the voltage, thus the signal, produced by a microphone would be due to ‘back EMF’. It is not.

Er, "non-sicentific definition"? My definition of back-EMF is any reactive voltage which is in opposition to an applied voltage. Since microphones do not have any voltage applied to them (other than polarizing DC voltages on condenser mics), then my definition would not include a microphone.

You do realise that back EMF is usually described in relation to stationary inductors, not speakers.

I don't know what circles you move in, but where I come from, back-EMF is rather routinely described in relation to loudspeakers. In fact, you can't discuss a loudspeaker's electrical impednace without implicitly discussing back-EMF. It's precisely back-EMF that makes your loudspeaker a reactive load as opposed to a nice static, purely resistive load.

Square wave and triangular waves are not sine waves.

No, they're not. I don't recall saying or implying that they were. Though I might mention that square waves and triangular waves can be described and produced as sums of multiple sine waves.

By sine wave I’m referring to "a signal of short duration with a relatively high amplitude".

Huh? In reference to a sine wave or a transient? Or did you mean to say you're "not referring to"?

Sine wave usually refers to a complete and continuous signal, like a rolling organ note. But if we are talking physics then that’s another thing (most of the people at this message board are not trained in physics or engineering).

The issue is transients. And a transient can be of a wholly sinusoidal nature. Transient simply refers to the amplitude envelope of an event. So if you want to talk transients, I don't see why you need to exclude transients simply because they're sinusoidal in nature. I mean, it doesn't change your fundamental argument and a sinusoidal transient makes the basic analysis easier.

By transient spike read single non recurrent signal that would look more like a square wave for a drum beat, triangular wave for other spikes.

Why must it be that? Are you saying that a sinusoidal transient wouldn't exhibit the lower impedance you claimed? What difference would it make?

Consider just a long coil and magnet such that the coil passes through the magnet’s gap - like a speaker. The coil has some mass. If the coil moves in one direction, say ‘in’, then it must have some momentum. If the driving voltage is reversed, the coil must move in the opposite direction.

Ok.

But the driving voltage will meet a voltage that has the opposite polarity.

Not if the cone (or more specifically the voice coil) moves in the opposite direction as you state above. If the voice coil moves in the opposite direction due to the driving voltage being reversed, then it will see an opposing voltage. I.e. back-EMF.

It is fairly clear that if the coil is moving ‘in’ at the instant the ‘out’ voltage is applied, then the voltage drop will be much higher than if the coil were initially stationary. This is because the voltage the coil produces has the opposite polarity to the driving voltage (that wants to push the coil in the opposite direction).

In which case your speakers aren't reproducing the transient they're supposed to. If they were, then the voice coil would be moving in the opposite direction as it's supposed to and you wouldn't be seeing that reverse voltage.

Now, a simple method of testing impedance is to place a resistor in series with the positive lead and check the voltage drop across the resistor for a given frequency. The higher the voltage drop the lower the impedance, the lower the voltage drop the higher the impedance (and the smaller the load).

That would give you an indication of impedance provided the driving voltage remained constant. It won't provide you with any useful information under transient conditions however.

If you disagree with this then it is time to get some test equipment and try the experiment for your self.

I don't agree that this simple method would prove useful for testing transient conditions.

It is noteworthy that the load of the cabinet shows up on impedance plots. If you were to physically touch the cone while an impedance test were in progress then that shows up as well. If the cone were moving of its own accord (say it had been producing a low frequency sine wave) then that will show up as well.

And how long would it be moving of its own accord once the test signal has been applied in order to measure the impedance?

As for feedback, MOSFET devices sag as the load increases (output voltage drops if there is no feedback to correct for it). The maximum gate-drain voltage is limited to 15 volts for Hitachi MOSFETs. Any more and they die. You must place zener diodes across the gate-drain to limit this voltage - usually a 12 volt zener does the trick.

As you know the gate voltage is always higher than the drain voltage when the MOSFET amplifier has a load. For a given gate voltage, the drain voltage will vary according to the load (ie, impedance of speaker).

With feedback, the drain voltage will remain the same for a given input voltage regardless of the load until the zener diodes start to clip the input (to the gate) and so the output (via the drain).

When the impedance of a load suddenly changes, as clearly outlined above, the output voltage changes and so, via the feedback circuit, the gate voltage is raised - the amplifier pushes harder.

When the impedance of a load suddenly changes, as clearly outlined above, the output voltage changes and so, via the feedback circuit, the gate voltage is raised - the amplifier pushes harder.

What your feedback is doing here is trying to amoeliorate inherent limitations in the amplifier itself. Let's go back to what you said orignially:

Without feedback, the amplifier produces a voltage that would move the speaker cone in the desired direction if the cone were stationary in the first place. With feedback the amplifier pushes hard enough to overcome the overshooting speaker cone.

What the amplifier is trying to do, whether it uses feedback or not, is to maintain a voltage across its output identical to that across its input except higher in amplitude (assuming the amplifier has voltage gain greater than 1). Period. In order for it to "push harder" it would have to increase the voltage across its output to greater than it would be otherwise (i.e. input signal times voltage gain). And if it's increasing the voltage across its output beyond this point, i.e. "pushing harder" then it's doing something it shouldn't be doing.

When the impedance of a load suddenly changes, as clearly outlined above, the output voltage changes and so, via the feedback circuit, the gate voltage is raised - the amplifier pushes harder.

In conclusion, if the driver was reproducing a low frequency sine wave at the same time a short high amplitude signal representing the leading edge of a drum sound or other percussive sound, then the amplifier without feedback will fail to control the driver properly - the change in impedance will be ignored. The amplifier with feedback will correct for some of the speaker overshoot (no amplifier/speaker is perfect) and produce a more faithful sound (in this case, a drum sound).

The amplifier controls the driver ultimately by virtue of its low output impedance. While feedback can help reduce output impedance, you can still achieve low output impedances without it. And even relatively high output impedances don't have a huge effect on controling the cone's motion. That's largely taken care of by the voice coil's DC resistance.

An amplifier with an output impedance of 8 ohms (a damping factor of 1 into a nominal 8 ohm speaker) only has marginally less control over cone motion than an amplifier with an output impedance thousands of times lower.

BTW for a driver that is producing a single sine wave, and assuming the driver is liable to overshoot, then as the voltage delivered by the amplifier changes (toward the maximum positive end of the cycle, for instance), then the drivers impedance drops (it produces a voltage of its own).

This is incorrect. The point at which the driver has the MOST overshoot is at its fundamental resonance. And this is a point of MAXIMUM impedance. That "voltage of its own" you're talking about here is back-EMF. Which because it's in phase with the driving voltage manifests as a HIGHER impedance.

The fundamental resonance of a typical dynamic loudspeaker models as a parallel RLC resonant circuit across the amplifier's output. Do some modeling with that and perhaps you'll gain a bit better insight as to what's going on.

PS I just ran a quick test on a MOSFET amp I just finished (sounds sweet).
With 100mV input, I tested the gate voltage for various loads. If there were no feedback, the output would vary rather than the gate voltage. My dummy loads are all 10R 200W, so when wired in parallel I get 10, 5, and 3.3 ohms.
100mV in, 4.38V out (for all loads)
4.40 (gate voltage, no load)
4.99 10R load
5.42 5R load
5.80 3.3R load (2.4dB rise over no load voltage)

Yes. Again this illustrates that what your feedback network is doing is trying to compensate for the inherent limitations in the amplifier, not correcting problems with the speaker.

se




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