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In Reply to: Re: MOSFETs work fine too... posted by Robert Karl Stonjek. on December 11, 2001 at 03:30:36:
Impedance rises at the resonant frequency in a sealed enclosure. In a ported enclosure the impedance is lowest at the resonant frequency. Likewise, the cone motion is greatest at the resonant frequency for a sealed enclosure, and slows almost to a halt at the resonant frequency in ported enclosures.Yes, it's low at the resonant Helmholz frequency of the enclosure/port. But that's also a point at which the amplifier has greatest control over cone motion as back-EMF from the voice coil is pretty much zilch.
Further, I was not referring to sine wave sound but transient spikes.
Then I think you're going to have to better define what you mean by "transient spikes." A transient spike is simply a signal of short duration with a relatively high amplitude. A single cycle of a full-swing sine wave is a transient spike. In fact, more of a transient spike than you'll ever encounter in any music signal. So I don't understand why you're differentiating sine waves from transient spikes as if they were mutually exclusive.
Say the speaker is asked to produce a nice big low frequency sine wave.
Ok.
Half way through the duty cycle the speaker is asked to produce a transient thump.
Ok.
The cone is moving into the speaker and passing its normal rest position when it must suddenly push out in the opposite direction with significant force. (ie the transient ‘thump’ has the opposite phase to the cones motion).
Ok. I don't know what this has to do with your claim of reduced impedance during transients, but ok.
If the amplifier were disconnected right when the cone is passing its normal rest position (while reproducing a low frequency sine wave) then it would continue on and produce a voltage - opposite in polarity to the voltage that was pushing it. This is not back EMF. This reverse voltage is seen by the amplifier as a very low impedance (amplifier not disconnected).
You're going off into apples and oranges land here. You describe something which might occur if the amplifier is disconnected, and then go on to say that this is what the amplifier sees if it's not disconnected. What?
If you want to know what the amplifier sees when it's not disconnected, then you describe what it sees when it's not disconnected. And if you don't disconnect it, the amplifier sees back-EMF of the same polarity as the voltage being output by the amplifier. Which means that it sees the same impedance it sees any other time at whatever frequency your sine wave happens to be.
Compare this with a speaker that is not producing a low frequency sine wave at the moment it is asked to produce the transient thump. The amplifier sees a higher impedance and an easier load.
But that's essentially what you described earlier when you were referring to some transient occurring at 180 degrees into some low frequency sine wave. Since back-EMF is always in phase with the driving voltage, at 180 degrees, the driving voltage will be zero and back-EMF will be zero. And the energy that was stored during the first 90 degrees of that sine wave were returned during the second 90 degrees. So your transient is effectively starting out as if there were no sine wave.
Back EMF is the voltage produce by a charged inductor (coil) when the charging voltage is removed. People forget that coils are much like capacitors in that they can be charged and discharged.
Ok. Don't know what this has to do with transients and low impedances though.
I’ll give an analogy to show my point. Lets say you lift a 1 kg weight one meter up in the air. The resistance to being lifted can be thought of as the impedance. Now consider the impedance if the weight were falling at the time you begin to lift it. The impedance is much much greater.
Well if you want to use this analogy, then you're defeating your own argument. Your original claim was that transients were of a LOWER impedance. Your analogy here is saying that they are of a HIGHER impedance.
Now consider a cone that is moving at some frequency at some amplitude when you apply a ‘lifting’ transient. Obviously the impedance at that moment will not correspond to any sine wave sweep or resonant frequency.
And according to your analogy, it would have to have a HIGHER impedance, which is contrary to your claim of LOWER imedance during transients.
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.
Huh? If you're talking about typical voltage feedback in voltage source amplifiers, the amplifier doesn't push any harder regardless. How hard the amplifier "pushes" is reflected in its output voltage which is just an amplified version of its input voltage. Feedback makes sure that the output voltage doesn't divert from the input voltage save for linear voltage gain.
The overshooting speaker cone produces back-EMF which is of the same polarity as the driving voltage and manifests itself as a higher impedance.
Passive crossovers may cause problems with amplifier damping (and feedback), but in well designed systems the amplifier with feedback can count the speaker cone as part of its own output circuit and correct for any errors.
Not in any typical voltage feedback voltage source amplifier it can't. Now if you mounted an accelerometer or somesuch to the cone and tied its output into the feedback network you could do something along those lines. But nothing like that is happening in a typical amplifier/speaker arrangement.
se
Follow Ups:
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. 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.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.
You do realise that back EMF is usually described in relation to stationary inductors, not speakers.
"Then I think you're going to have to better define what you mean by "transient spikes." A transient spike is simply a signal of short duration with a relatively high amplitude. A single cycle of a full-swing sine wave is a transient spike. In fact, more of a transient spike than you'll ever encounter in any music signal. So I don't understand why you're differentiating sine waves from transient spikes as if they were mutually exclusive."
Square wave and triangular waves are not sine waves. By sine wave I’m referring to "a signal of short duration with a relatively high amplitude". 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).
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.
"Ok. I don't know what this has to do with your claim of reduced impedance during transients, but ok."
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.
But the driving voltage will meet a voltage that has the opposite polarity. Lets say we place a resistor in series with the positive line and run this experiment twice. We are looking at the voltage drop across the resistor at the instant when the ‘out’ voltage is applied.
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).
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).
If you disagree with this then it is time to get some test equipment and try the experiment for your self. 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.
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.
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).
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). In a MOSFET amplifier, the higher impedance/lower load causes a change in the output voltage which the feedback circuit immediately corrects for (No need for an accelerometer). Without feedback this doesn’t happen.
Kind Regards,
Robert Karl Stonjek.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)
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
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 impedance is at the frequency/frequencies involved with the transient.
Of course the speaker responds, but not in the same way that it would if it weren’t moving in the opposite direction to the transient voltage in the first place. Before it can move in the opposite direction it must first stop.
Loudspeakers with instant response have not been invented yet. If the speaker is not moving in the first place then it does not first have to stop. The delay in changing directions is exactly what I am talking about. In that instant, if there were a resistor in series with the speaker the voltage drop would be greater for the cone moving in the opposite direction to the driving voltage than a cone at rest when that driving voltage was applied. That translates to a lower than usual impedance which a MOSFET amplifier with feedback responds to (see demonstration below).
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 impedance 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.
Yes, but you are including the voltage produced by an overshooting speaker as back EMF. It isn’t. Charge a coil and then take away the charging voltage then a voltage will be produced by the coil as the field collapses (lookup Lenz’s Law). If the applied voltage is AC then we can see this on a trace as a phase change. Obviously speaker coils are inductors as well. But the overshoot voltage is not back EMF (not caused by the collapsing of a field) unless the term is used more loosely than I had realised.
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.
Your obviously from the pre-digital era. Fourier transform reduces
to sine waves, Wigner transform reduces to square waves. Look it up.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.
Before it can move in the opposite direction it must stop. Before it can stop its momentum must be overcome. This momentum does not have to be overcome if the speaker is not moving in the first place. Before the cone stops moving, it will produce a voltage that you call Back EMF that is the opposite in polarity to the voltage that is trying to send the cone off in the opposite direction. This is the same as a transient low impedance that a MOSFET amp with feedback can correct for (though not perfectly).
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.
Not so. For a MOSFET amp (of the Hitachi MOSFET variety), the output changes in response to the load as I’ve maintained all along. Feedback is required to maintain a constant voltage so when the load increases, the output pushes harder, that is, the current rises. If the speaker produces a voltage, either from the back EMF caused by a collapsing field or voltage produced by the overshooting speaker, the amplifier corrects for those as well.
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 controlling 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.
Maybe for a dozy old low frequency sine wave. For transients, maybe its down to listening. I’m not convinced. If I repair an amplifier that has no feedback and it ends up being a bit dull on the transients then it is doing what it supposed to do. A transistor amplifier restored to its gritty worst - the customer will be happy. Properly designed MOSFET - now that’s much better.
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 .
No, your overplaying the back EMF card. The coil will have a back EMF as any coil will, I never said it wouldn’t. But if you push on a speaker cone, it also produces a voltage. If the cone has some momentum then it will overshoot at the end of its cycle (moving in or out too far) and produces a voltage with the same polarity as you mention. This voltage makes the impedance appear to rise (effectively it does rise).
We agree on the fundamentals of these points. Consider a resistor in series with the driver. We see that the voltage drop over the resistor is small when the speaker overshoots because driving voltage (from the amplifier) and the voltage produced by the speaker have the same polarity.
Now what happens if, as the driver moves inward, the amplifier sends a signal with the opposite polarity? Lets call it a transient - the leading edge of a drum sound. The voltage drop over the resistor is going to be higher than if the driver were stationary when the thump signal came through. This only occurs for a fraction of a second after which the speaker is moving in the direction the amplifier is pushing it.
But that fraction of a second makes the difference between a dull thud and a crisp faithful reproduction. The momentary lower impedance causes MOSFET output to drop which is immediately corrected for by the feedback circuit.
I will perform a simple test right now. I will hook up a woofer to one of my amps and push down on the cone. This simulates a cone that momentarily does not move where it is supposed to, say, for whatever reason, it was moving in the opposite direction to the driving voltage. That motion must be overcome before it moves as it should (this only takes a fraction of a second).
(10 minutes later) I ran the test. I connected the scopemeter to the gate of the MOSFETs of the power amplifier and a small woofer to the output. I shorted the input of the amplifier. Now I pressed down on the cone. 100mV spike shows up in the gate (using Min/Max/average trace on the scopemeter - so that any spikes remain on the screen).
Now I connect the scopemeter only to the speaker. Pushing the cone in the same way I get up to 400mV.
Now I connect the speaker to the amplifier and check the voltage in the same way. When I push down on the speaker as hard as I can, no voltage shows up at all - the amplifier has corrected for the speaker’s voltage so that none shows up on the scopemeter. This would be typical of an op amp circuit with feedback - very hard to measure the correcting voltage as it all happens so fast - but a voltage does show up in the amplifier’s feedback circuit.
Just thought of a final test - with the amplifier off, do I still get no voltage at the speaker (when the speaker is connected)?
2 Minutes later. The Voltage is about half what it is for an open driver. Switched the amp back on and the driver voltage does not show no matter how hard I push on it.
This is proof positive - a very simple experiment to perform (anyone with a MOSFET amplifier and scope and a couple of minutes can perform it) and it shows that the amplifier is correcting for the voltages produced by the speaker.
So if it overshoots when reproducing a simple sine wave, say at its resonant frequency, the amplifiers driving voltage drops so that the voltage that speaker produces itself and the voltage of the output devices does not exceed the correct voltage (input multiplied by gain). Without feedback the voltage would rise.
The voltage does rise (for MOSFET amps of the Hitachi variety)- I measured it and reported the rise in my last post.
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.
A ported enclosure with an 8 ohm speaker typically has an impedance ranging from 8 to 16 ohms, a sealed enclosure may have an even higher impedance peak. As illustrated above, without feedback the amplifiers output would change with frequency as the impedance of the speaker varies with frequency.
Electrostatic loudspeakers typically have an impedance dip at high frequency, typically at 10kHz, and at very low impedance, typically of 2 ohms or less. An amplifier without feedback would have an output varying by over 3dB.
Note that Hitachi MOSFETs (a typical MOSFET) varies in output according to load. The change in load is caused by the changing impedance of the driver. If the amplifier changes its output voltage in response to a change in the speakers impedance, then it is sensing the speaker and correcting for changes in the speakers impedance.
A low amplifier output impedance would achieve much the same thing, as you mention, but I’m not convinced that a MOSFET amp will perform properly without feedback and I’m also not convinced that corrections performed by the feedback circuit of a MOSFET amp does not control the loudspeaker to a degree (My systems are all active only, no passive X components).
But perhaps any finer points must be listened to (in a listening test, I mean). Or performed in controlled conditions (Lab conditions) which I’m not prepared to do even though I have the equipment to do a good approximation.
Kind Regards,
Robert Karl Stonjek.
Sorry, busy day today and this thing's become something of a monster so I didn't have time to get a reply banged out. I'll set aside the time tomorrow.se
Understood. Half the disagreements on newsgroups and message boards start when one or both parties do not have enough time to outline their case clearly enough.Reiterating:- the debate began on the subject of feedback and whether or not it is really necessary. I pointed out that MOSFET amps need feedback to maintain a constant output with frequency when the load changes with frequency.
Someone pointed out that there are several varieties of MOSFETs and not all perform in the manner I outlined. I specified Hitachi MOSFETs in later discussions.
You said that feedback was not necessary for MOSFETs either but later said that it is necessary to correct for "inherent limitations" of some designs, and also maintained that there are other (non-feedback) methods of achieving the same thing.
I also mentioned that MOSFET amps can correct for the errant voltage output of a driver, such as might be generated by an overshooting speaker.
I did perform an additional experiment. I build a subwoofer that uses a compound woofer - basically the compound woofer is two woofers glued together face to face.
So I have some of these lying around. In the experiment I powered one of the woofers. The other was passively driven, simulating an overshoot condition.
The input voltage was 6VRMS at 20Hz. The voltage generated by the passively driven woofer was 2.3V with no load.
I then connected various resistors to act as loads/damping and checked the voltage output of the passively driven woofer:-
No Load.2.3V
10r.....1.2V
1r......220mV
0.033R...12mV
Amp.......9mV (passive woofer connected to power amp; amp input shorted)
Short.....9mVNote that the amp gave the same result as the short. The error bars are two big to say more than that. Accuracy of around ±2mV and 1.9mV measured noise.
Also, the leads I was using were just regular hookup cables with alligator clips on the ends. To short the woofer, I connected them together, but when a voltage still appeared, I tried shortening the lead and found the voltage dropped (as expected). But this does say something about cables and damping - the two leads were only 300mm in length.
Notably, the amplifier’s feedback circuit showed a 0.5V error voltage passed to the gates of the MOSFETs. This seems to show that the amp was ‘pushing against the voltage’ produced by the speaker.
If I had a couple days spare to set up the test properly I might find something interesting, but is it really worth it? My original point was that feedback is required to keep some (ie Hitatchi) MOSFET amps form varying output voltage with load. You say there is another approach to achieve the same thing.
Anyway, if a person was really serious about sound they would dispense with the passive crossover and drive each speaker with a separate power amplifier. I think there is something to be said for specialist subbass, bass-mid and treble amps, though that is one further step I don’t think I’ll bother taking!! :)
Kind Regards,
Robert Karl Stonjek.
Understood. Half the disagreements on newsgroups and message boards start when one or both parties do not have enough time to outline their case clearly enough.Yes. And becuase one or both parties do not have enough time to actually read what's actually being said by the other party.
Reiterating:- the debate began on the subject of feedback and whether or not it is really necessary.
It did? Granted, my memory's not what it used to be, but I could have sworn that it began on the subject of a loudspeaker's impedance dropping during transients. Oh well. Let's move on.
I pointed out that MOSFET amps need feedback to maintain a constant output with frequency when the load changes with frequency.
Yes. And as I beleive I pointed out, such behavior is the practical consequence of an amplifier's non-zero output impedance. And as I also believe I pointed out, an amplifier's output imepedance can be lowered verus what it would be otherwise by the application of feedback.
Someone pointed out that there are several varieties of MOSFETs and not all perform in the manner I outlined. I specified Hitachi MOSFETs in later discussions.
Any device with a non-zero output impedance will behave in that manner to one degree or another. And since every device that I'm aware of has a non-zero output impedance, it really boils down to a matter of degree and what degree the designer feels is acceptable.
You said that feedback was not necessary for MOSFETs either but later said that it is necessary to correct for "inherent limitations" of some designs, and also maintained that there are other (non-feedback) methods of achieving the same thing.
First, I didn't say it wasn't necessary. Whether it's necessary is ultimately a decision for the designer to make. What I did was point out that, in the context of controlling the motion of a loudspeaker cone, particularly at its fundamental resonant point, the amplifier's output impedance doesn't have quite the substantial effect that many people believe. That even a thousand fold difference in output impedances results in a very marginal difference in the amplifier's control over the loudspeaker.
Second, I didn't say that feedback was necessary to correct for inherent limitations. I simply said that that's what the feedback is doing in this particular context. If a designer feels it's necessary to apply feedback in order to reduce the amplifier's output impedance, then obviously the amplifier's inherent output impedance (i.e. without feedback) is a limitation. I made no mention of its being inherently necessary as again that's a call for the designer to make. Other designers might be willing to live with the inherent limitation.
Thirdly, I believe I said that you can design an amplifier with low output impedance without having to resort to feedback, but I didn't intend this to apply to your specific topology/parts compliment.
I also mentioned that MOSFET amps can correct for the errant voltage output of a driver, such as might be generated by an overshooting speaker.
And I mentioned that it does this by way of its output impedance. Doesn't matter if it's a MOSFET amp, a BJT amp, a tube amp, or what have you. With feedback or without feedback. Any "errant voltage" applied across the amplifier's output will see the amplifier's output impedance and behave accordingly.
I think now might be a good time to address this whole "overshooting speaker" thing as I think you may be looking at this issue from some erroneous assumptions.
For example your assumption that any time the speaker is moving in one direction, then before it can move in the opposite direction, its momentum must first be overcome, which you say cannot occur instantaneously.
While it's true that any mass which is moving (in this context in response to an applied force) has momentum seeing as momentum is mass times velocity, not every mass which is moving in response to an applied force GAINS momentum. And it's only momentum GAINED which would need to be overcome before you could change its direction.
Take a hockey puck for example. Set it down on a smooth sheet of ice and with your finger apply a little force in one direction for a certain distance and then stop applying the force. The hockey puck will continue to move a bit beyond the point at which you stopped applying the force. That distance represents the momentum that was GAINED by the hockey puck. And if you were to apply a force in the opposite direction at the same moment you stopped applying the initial force, you would first have to overcome this gained momentum before you could get it moving in the opposite direction.
Now do the same thing only this time with the hockey puck sitting on a sheet of rough sandpaper. This time the hockey puck doesn't move past the point at which you stopped applying the force. That's because the much higher frictional losses of the sandpaper versus the ice prevented the hockey puck from gaining momentum.
Loudspeakers of course have losses as well, in all three domains; electrical, mechanical and acoustic. With respect to raw drivers, the relationships between losses and reactances are described in the various Q parameters, Q ES , Q MS and Q TS . What they describe is the damping effect the losses have on the reactive elements.
Q ES describes the damping of the electrical portion of the driver, Q MS of the mechanical portion, and Q TS is the product/sum of the two others, i.e. (Q ES x Q MS ) / (Q ES + Q MS ).
Anyway, the point here is that these figures (well, Q TS really) are useful in determining a driver's propensity to overshoot. A Q TS of 0.707 will have a maximally flat response with no overshoot. Above this is considered "underdamped" and you will begin to get overshoot as the driver's internal losses are overcome by the reactive elements. Below this is considered "overdamped" and internal losses dominate.
Q TS for typical drivers ranges from around 0.2 to 0.6.
If it's not obvious yet, another way of putting this is that for any Q TS at or below 0.0707, the driver's internal losses are such that they prevent the cone from GAINING momentum. And if the cone doesn't gain any momentum, then when your transient comes along that wants to move the cone in the opposite direction, then as far as the transient is concerned, it sees nothing different than if the cone were otherwise at rest. It's like the hockey puck on the sandpaper.
Of course raw drivers are typically used in enclosures which in combination with the driver's Q TS will result in a different system Q (Q TC , which will always be at least as high or higher than the driver's Q TS ). And depending on the designer, one may decide on a Q TC greater than 0.707.
Anyway, this is just a side note to address any notion that simply because the cone is moving it must have gained momentum.
And now back to our regularly scheduled program...
I did perform an additional experiment. I build a subwoofer that uses a compound woofer - basically the compound woofer is two woofers glued together face to face.
So I have some of these lying around. In the experiment I powered one of the woofers. The other was passively driven, simulating an overshoot condition.
The input voltage was 6VRMS at 20Hz. The voltage generated by the passively driven woofer was 2.3V with no load.
I then connected various resistors to act as loads/damping and checked the voltage output of the passively driven woofer:-
No Load.2.3V
10r.....1.2V
1r......220mV
0.033R...12mV
Amp.......9mV (passive woofer connected to power amp; amp input shorted)
Short.....9mVNote that the amp gave the same result as the short. The error bars are two big to say more than that. Accuracy of around ±2mV and 1.9mV measured noise.
Ok. Basically this is a good illustration of the effects of an amplifier's output impedance. As I said previously, the reason your amplifiers don't have a constant voltage into a changing load is because of the amplifier's output impedance.
Let's say your amplifier, which isn't using any feedback at this time, has an inherent output impedance of 1 ohm. And into a 16 ohm load you're drawing 1 amp of current. Ohm's Law says that you've got 16 volts across the load.
And since the amplifier's output impedance is in series with the laod, all of the current flowing through the load is also flowing through the amplifier's output impedance. And Ohm's Law says that you must be dropping 1 volt across the amplifier's output impedance.
So basically you've got an amplifier that's ouputting 17 volts (since the total voltage is the sum of all the voltage drops) but because of its output impedance, it's only delivering 16 volts to the load.
Keeping all else equal, drop the load impedance down to 4 ohms. 17 volts into 5 ohms (4 ohms for the load, 1 ohm for the amplifier's output impedance) gives you 3.4 amps. Now you've got 13.6 volts across the load and 3.4 volts dropped across the amplifier's output impedance.
The effect of this is a 2.4 volt (1.4 dB) difference between the voltages applied across a 16 ohm and 4 ohm load.
Now compare the difference if the amplifier's output impedance is reduced two orders of magnitude, from 1 ohm to 10 milliohms. In the case of the 16 ohm load, you've got 16.99 ohms across the load and 0.012 ohms dropped across the amplifier's output impedance. In the case of the 4 ohm load, these numbers become 16.96 volts and 0.042 ohms respectively. So you've gone from a 2.4 volt (1.4 dB) difference to a 0.03 volt (0.015 dB) difference.
When you apply feedback, what you're doing is reducing voltage gain, which also has the effect of lowering output impedance. How much you lower output impedance (and reduce gain) depends on how much feedback you apply with respect to the amplifier's open loop voltage gain.
And this also goes toward your claim that the amplifier is "pushing harder" when you apply feedback. Since the feedback lowers voltage gain, the amplifier's output voltage will be lower for a given input voltage. So even though you've reduced the amplifier's output impedance, you've reduced its voltage gain by the same amount. Which means that the amplifier is pushing just as hard into the same load with feedback as it was without it.
Notably, the amplifier’s feedback circuit showed a 0.5V error voltage passed to the gates of the MOSFETs. This seems to show that the amp was ‘pushing against the voltage’ produced by the speaker.
If the amplifier was indeed "pushing against the voltage" produced by the speaker, then it would be producing a voltage of the same polarity as the voltage from the speaker in which case you'd be measuring something more like the 2.3 volts you got from the non-loaded speaker. And if that were the case, your feedback would be positive rather than negative.
If I had a couple days spare to set up the test properly I might find something interesting, but is it really worth it? My original point was that feedback is required to keep some (ie Hitatchi) MOSFET amps form varying output voltage with load. You say there is another approach to achieve the same thing.
What you're really saying though is that the MOSFET amps you're familiar with have inherently high output impedances and you feel that feedback is required to lower output impedance to an acceptable level.
If you want lower output impedance without feedback, then use devices in the output stage which have an inherently lower output impedance. I don't know that you'll be able to reach the level you'd like using MOSFETs and it appears that MOSFETs are the only devices you'd consider.
Anyway, if a person was really serious about sound they would dispense with the passive crossover and drive each speaker with a separate power amplifier. I think there is something to be said for specialist subbass, bass-mid and treble amps, though that is one further step I don’t think I’ll bother taking!! :)
No argument there. One of my friend's pet peeves are all the outrageously expensive, five figure loudspeakers out there which utilize passive crossovers. I'm inclined to agree. :)
se
Regarding ‘gained momentum’.There is no such a term mentioned in any of my undergrad physics text books. The momentum of a mass is given by its velocity and mass alone. There are no other variables to add. The force required to bring a mass that is travelling at some velocity to a halt is the same as the force required to accelerate that mass in the first place.
Introducing friction complicates things. In particular, the environment in which a speaker cone moves can be thought of as a mass, a force and a spring. The mass is made up of the speaker’s cone, the coil and any other moving parts, and the air mass that is moving with the cone.
The spring can be thought of as having two components - compression and extension. The compression spring can be thought of as all the forces that impact on the cone and prevent it moving. The compressibility of the air in front of the cone would be the main component (for a speaker moving outwardly).
Extension springs can be thought of as pulling against the speaker cone and so hindering its motion. The speakers spider and the surround at the top of the cone, electrical damping, and the damping that the box may provide can all be added together with the compression spring to give one figure for the spring (resistance to the cones motion).
This is not the same as friction. Friction prevents motion. Your hockey puck does not spring back. Speaker cones do. If a voltage were applied to a speaker from an infinite impedance source in one polarity only, say zero to 9 volts, then the driver returns to its rest position of its own accord. All piston-cone drivers will overshoot (decaying oscillations).
As it returns, the speaker produces a voltage.
If there is a low impedance source then the driver will return to its original position slower and will overshoot less.
Regardless of the tension of the spring, a driver will still have to slow down before it can move off in the other direction. That is physics. Whether or not the period in which this slowdown occurs is significant concerns Audio Engineers.
You claim that the ‘friction’ nulls the overshoot condition is erroneous. Leave aside the overshoot and just look at the conditions at the time the reverse voltage occurs (for the low frequency sine wave and drum beat with opposite phase to the sine wave at the instant the drum is struck). If it were a friction effect, like the hocky puck analogy, and ignoring any overshoot then you are 100% correct.
But if we replace friction with spring then at the moment the drum beat arrives there is a spring effect. This spring is going to pull the speaker in the same direction the amplifier was wanting to push it anyway. So the impedance will be higher that it would be if the speaker was stationary (for the overshoot condition, it would be momentarily lower then higher once it started moving in the right direction).
In an extreme case, the voltage produced by the speaker would be significant. A MOSFET amp (with feedback)can take the voltage produced by the speaker into account when calculating the correct voltage over the load. It does this by sensing the voltage at the output and comparing that to the input voltage.
It is because of the intrinsically higher impedance of the MOSFET amp that they are able to do this. An intrinsically lower impedance amp effectively blinds the feedback from any activity beyond the output devices.
And since the amplifier's output impedance is in series with the load, all of the current flowing through the load is also flowing through the amplifier's output impedance. And Ohm's Law says that you must be dropping 1 volt across the amplifier's output impedance.
No, the amplifiers output is in parallel with the load. The only series circuit is between the amplifier’s output and the supply rails via the output devices.
The amplifiers impedance is in parallel with the speaker. If the amplifier’s output impedance was zero ohms then this would be equivalent to a dead short across the speaker (assuming no loss on the leads).
The voltage drop across the amplifiers output is always the same as the voltage drop across the speaker (assuming no lead loss).
Your calculations are erroneous. When calculating the current drawn by the circuit you only need the voltage output of the amplifier and the impedance of the speaker at the test frequency. I’ve done this many times with an accurate current probe and it works out exactly as expected. The amplifier’s output impedance never enters the calculation. (I have done this with high impedance non-feedback MOSFET amps as well - no difference).
Output impedance disappears when the amplifier is switched off. A push-pull transistor amplifier has each pair of devices biased so that they push against each other (electrically). The output resists being changed from the value set by the bias (when the amplifier is idling).
The reason why amplifiers using Hitachi MOSFETs and a feedback circuit have effectively low output impedance is that any voltage appearing at the output, whether generated by the amplifier itself or the load, is compared with the input. An error voltage is generated to correct for any voltages that don’t match.
Apart from output impedance, the feedback corrects for non-linearities in the voltage amplifier stage. All the voltage amplification in most MOSFET amps is done by a transistor voltage amplifier stage.
The speed at which this can be done is roughly equivalent to a 3 megahertz frequency (depending on the stability of the amplifier and the placement and value of various capacitors in the voltage amplifier and feedback circuits).
Keeping all else equal, drop the load impedance down to 4 ohms. 17 volts into 5 ohms (4 ohms for the load, 1 ohm for the amplifier's output impedance) gives you 3.4 amps. Now you've got 13.6 volts across the load and 3.4 volts dropped across the amplifier's output impedance.
No, if the load is 4 ohms and the voltage is 17VRMS then the current drawn is 4.25ARMS. Note that this is not the case if you are measuring the amps voltage at the gates rather than at the output.
Just let me reiterate: the amplifier’s output impedance is measured between the amplifier’s two output terminals (usually load and Earth/zero volt). This can be done by using a constant current sine wave voltage generator to pass a voltage to the amplifier’s output via a resistor. By measuring the voltage across the resistor you can use Ohm’s law to calculate the amplifier’s output impedance. If the voltage is higher, then the impedance is lower etc.
If you want lower output impedance without feedback, then use devices in the output stage which have an inherently lower output impedance. I don't know that you'll be able to reach the level you'd like using MOSFETs and it appears that MOSFETs are the only devices you'd consider.
I took one of my amplifiers around to a certain advocate of Plinius amps and substituted one of his giant class A amps for mine. Compared to the transistor monster, my amp had a clearer midrange (the Plinius was ‘thicker’) and sweeter highs (driving 10’ high electrostatics).
He bought the amp. :)
Kind Regards,
Robert Karl Stonjek.
I suspect the "open loop" output impedance also depends on the type of MOSFET you're using as well. It is well known that the lateral geometry devices (the ubiquitos Hitachi 2SK135/2SJ50 and similar) are not low impedance devices in the same way that the IRFP240 (for example) is. I suspect that's partof the reason that the IR/Harris devices show up much more (than the lateral devices)in single ended and other class-a designs.I beleive blanket generalizations about MOSFET amps is less than optimally useful.
Quite right. I've always referred to them as depletion and enhancement type MOSFETs. Hitachi are enhancement types. Are the IRFP240 depletion types?Depletion type MOSFETs are routinely used in switchmode power supplies (I think).
I note that complementary pairs are only linear when used in pairs and so a singled ended amplifier would not be linear if only half of the pair were present.
Counterpoint and Forte (among others) used what I suspect are depletion devices. They had to be in matched sets and these amplifiers had no feedback. They also destroy themselves in a cloud of smoke - I know because I got to rebuild around 20 or so amps. I used Hitachi MOSFETs in a redesigned output and the comments from owners was unanimous - the amps never sounded so good, not even when they were new!!
That was a surprise as the first one I rebuilt was for a HiFi shop manager who just wanted it going - we didn’t expect it to sound better than new though I designed the new output stage and apart from having no feedback it was a fairly regular design and similar to the original.
But the upshot is that I have to agree with you regarding MOSFET devices - there are at least two classes of MOSFETs and it is not good science to generalise about them. Toshiba MOSFETs are very different to the Hitatchi , for instance.
Kind Regards,
Robert Karl Stonjek.
There are no depletion mode high power devices that I'm aware of, however I remember reading about a company that was planning to do some medium power depletion mode devices. I've heard of lateral, vertical and HEXFETs. The IR devices are HEXFETs.
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