You run a small signal stability sweep (across frequency) in the presence of another large signal excitation. Easiest way is an admittance probe - inject a small current, read the voltage that develops. A negative real part with a zero crossing in phase indicates instability. Rigorously, you could extract the poles and zeros of the admittance function and judge it that way. If there are real poles in the wrong half plane you can find them that way - you won’t with the frequency sweep.
It’s only valid for that one set of conditions. Drive at another frequency or amplitude, with the load reactance opposite and you get different result. To be rigorous, it is VERY time consuming.
It’s only valid for that one set of conditions. Drive at another frequency or amplitude, with the load reactance opposite and you get different result. To be rigorous, it is VERY time consuming.
Looking back on this, with 20/20 hindsight, I can see that my hypothesis that the recursive summing from a feedback loop somehow causes the FFBP effect is fundamentally wrong. B. Putzeys, in Jan's post #173, described the actual mechanism, but my math skills weren't up to the task (still aren't).
This will be no surprise to the various contributors who tried to gently guide me, but if someone reads this in the future, please don't be lead down a similarly wrong path.
"In the classical case of a perfect polynomial nonlinearity you get new harmonics after adding feedback because the solution is not a polynomial. Its Taylor expansion goes on forever." (B. Putzeys) That classical case is independent of delays and summed calculation, but requires half century old memories of obscure and acedemic maths. Much thanks to all who contributed and helped me with my sloooow, still very incomplete, understanding.
All good fortune,
Chris
This will be no surprise to the various contributors who tried to gently guide me, but if someone reads this in the future, please don't be lead down a similarly wrong path.
"In the classical case of a perfect polynomial nonlinearity you get new harmonics after adding feedback because the solution is not a polynomial. Its Taylor expansion goes on forever." (B. Putzeys) That classical case is independent of delays and summed calculation, but requires half century old memories of obscure and acedemic maths. Much thanks to all who contributed and helped me with my sloooow, still very incomplete, understanding.
All good fortune,
Chris
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@Chris Hornbeck this is one of the best posts I’ve ever read.
I’ve never read a similar post on tech forums, where someone works on others’ opinions and after a long time admit others were right and helps all users saying that others’ opinion was right.
You really made my day!
Thank you!
I’ve never read a similar post on tech forums, where someone works on others’ opinions and after a long time admit others were right and helps all users saying that others’ opinion was right.
You really made my day!
Thank you!
An interesting experiment maybe to listen to a feedback amplifier isolated from the reactive load.
I.e. each channel would have two identical amplifiers, inputs driven in mono:
Amp 1. Driving a resistive load
Amp 2. Driving the speaker, but the feedback taken from the output of the resistive load amplifier
I.e. One could even have a switch for A/B comparism.
And probably speaker protection, in case it all goes horrible wrong.. 🙂
Perhaps this has already been done?
I.e. each channel would have two identical amplifiers, inputs driven in mono:
Amp 1. Driving a resistive load
Amp 2. Driving the speaker, but the feedback taken from the output of the resistive load amplifier
I.e. One could even have a switch for A/B comparism.
And probably speaker protection, in case it all goes horrible wrong.. 🙂
Perhaps this has already been done?
How would you set that up? How would you 'listen' to the resistive load?
You can't take feedback from one amp and use that in the other, conditions are never fully identical I think.
But give it a try!
Jan
You can't take feedback from one amp and use that in the other, conditions are never fully identical I think.
But give it a try!
Jan
The idea is that the resistor amplifier output is the 'correct' feedback for a low distortin signal. I.e. the feedback makes the amplifier perform as intended, but ignore the speaker - because there is no speaker, just a resistor.
When this feedback is used for the amplifier driving the speaker, the amplifier basically ignores the speaker, because it's using the feedback look from the resistor driving amplifier.
So the amplifier is being corrected for it's flaws, but allowing the speaker to do what it likes.
The equivalent circuit would for example be a power MOSFET amplifier, (where we could ignore the extra load of two additional power MOSFETs upon the driver section), using one pair to drive a resistor to close the feedback look, and the 2nd pair hooking up the drains to a different PSU, and connect the sources to the speaker..
I.e. to decouple the speaker from the GNFB loop. I'm curious as to the effect on the sound 🙂
If there was no active feedback on the MOSFETS driving a real speaker, then they would have zero(ish) impedance into an 8 ohm load, (because the gate signal is corrected by feedback from the MOSFETS driving an 8 ohm resistor, with zeroish impedance),
... but the reactive component would be ignored, yes, because there's no feedback path for that.
So i was curious about how this would affect the sound.
Stability would be unaffected by load, no Zobel required, and into 8 ohm, it's to the design spec.
I think I know what the outcome is:
Because - as you say - #2 doesn't know anything about it's load, it cannot react to any changes there. Be it resistive or reactive.
That is equivalent to no fb at all. (except gain reduction).
To prove my point I did a quick sim.
Amps were pentode input, splitter, pp triode output, fb to cathode from opt.
Open loop output impedance 5.7 ohm, closed loop 0.6 ohm.
Varying load on #2 between 6 and 10 ohm (constant 8 ohm on #1) allows calculation of #2's output impedance.
Result: 5.7 ohm. The same as open loop.
This is confirmed by looking at the frequency response with a speaker sim model on #2, which showed the same resonance peaks as with the same amp w/o fb.
qed.
An expensive way to have fb reduce gain but not output impedance.
Because - as you say - #2 doesn't know anything about it's load, it cannot react to any changes there. Be it resistive or reactive.
That is equivalent to no fb at all. (except gain reduction).
To prove my point I did a quick sim.
Amps were pentode input, splitter, pp triode output, fb to cathode from opt.
Open loop output impedance 5.7 ohm, closed loop 0.6 ohm.
Varying load on #2 between 6 and 10 ohm (constant 8 ohm on #1) allows calculation of #2's output impedance.
Result: 5.7 ohm. The same as open loop.
This is confirmed by looking at the frequency response with a speaker sim model on #2, which showed the same resonance peaks as with the same amp w/o fb.
qed.
An expensive way to have fb reduce gain but not output impedance.
What you need is an active load compensator. Some complementary Mosfets across the output acting as an auxiliary load, would be controlled so as to draw a total amplifier current proportional to the output V.
Amplifier output current is monitored and the load controller makes sure that total amplifier current drawn is proportional to Vout. So the speaker can do what it wants and the amplifier can do what it is expected into a constant load. The compensator does nothing when an ideal constant R speaker is attached. The load compensator is a high impedance current output amplifier controlled to make current loading on the V amplifier proportional to Vout.
This might make badly behaving speakers sound worse, but it's the speaker's bad design showing thru.
Back EMFs from the speaker get cancelled and load impedance variations get nulled out.
Amplifier output current is monitored and the load controller makes sure that total amplifier current drawn is proportional to Vout. So the speaker can do what it wants and the amplifier can do what it is expected into a constant load. The compensator does nothing when an ideal constant R speaker is attached. The load compensator is a high impedance current output amplifier controlled to make current loading on the V amplifier proportional to Vout.
This might make badly behaving speakers sound worse, but it's the speaker's bad design showing thru.
Back EMFs from the speaker get cancelled and load impedance variations get nulled out.
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Audio mythology is full of circular logic because the author doesn't actually know how to achieve their ideas. A "load compensator" would be another amplifier that has to deal with the load instead of the original amplifier. So why bother with the first amp? Just make an amp that can handle the load in the first place. I posted a spread sheet for mixed feedback to set the output impedance of an amplifier, ie without simply adding a build-out equal to that impedance. https://www.diyaudio.com/community/threads/bridge-amplifier-with-current-sensor.404596/#post-7490801
"circular logic"
It's called a composite V & I amplifier. The literature is plentiful on such parallel devices. The V amplifier concentrates on accurate voltage output and the I amplifier concentrates on reducing the current loading variation. Sure, a single zero ohm output V amplifier can do the job too, but that's not easy to achieve with a tube amp with an OT. The separate SS I amplifier keeps that high bandwidth function out of the OT.
It's called a composite V & I amplifier. The literature is plentiful on such parallel devices. The V amplifier concentrates on accurate voltage output and the I amplifier concentrates on reducing the current loading variation. Sure, a single zero ohm output V amplifier can do the job too, but that's not easy to achieve with a tube amp with an OT. The separate SS I amplifier keeps that high bandwidth function out of the OT.
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The load compensator approach can also be added onto any existing tube amplifier without tearing into it. A far more attractive product.
But the feedback doesn;'t know what the load is. All it does is (try to) make the output a faithful copy of tghe input, whatever the load. If any.
Jan
Yes, exactly - that's what I think would be interesting to listen to.
The speakers are driven with a signal that is 'perfect' (as perfect as possible) designed for an 8 ohm resistor, which as we know enjoy the best music HiFi can give.
But the speaker is not interfered with, it just runs with what it's given. With a separate PSU and output MOSFETs the speaker's reactions are then entirely decoupled from the amplifier.
So it's a new type of feedback, a GNFB amplifier driving a speaker without feedback. Each part getting on with what it's best at 🙂. At least, that's the philosophy of the idea. And if it works well, of course smaller mosfets and bigger resistances can be selected to mimic the curves of the power MOSFETs - to provide the correct correction.
I did simulate the Maplin MOSFET amplifier by taking the feedback back from the gate driver stage, it did pretty well but the distortion of the power MOSFETs was uncorrected, and probably not prepared for the 8ish ohm load of the speaker.
... so the idea of having a sibling output pair or amplifier drive the 8 ohm resistor is that the output of the 'feedbackless' MOSFETs for the speaker would already be pre-corrected for the nominal load.
I'm sorry, but I really, really don't understand the goal here. Why should an amplifier's output not just be a replica of its input? Voltage in, voltage out. Why predistort it to cater to speaker impedances?
All good fortune,
Chris
All good fortune,
Chris
I'm not sure it will be pre-distorted, TBH, as the resistor is a linear...
It's really to make sure the output of the devices driving the speaker is linear at the predicted load.
The load on the dummy (GNFB) devices can be experimented, even in the simulator, I may have a go at that when I get the chance, and find a suitably horrible loudspeaker subcircuit 🙂