>What doesn't it really do?
The voltage at the out of the amp doesn't correspond to the position of the cone. Nor it's (and here's a good one 🙂 )
velocity ...........
The voltage at the out of the amp doesn't correspond to the position of the cone. Nor it's (and here's a good one 🙂 )
velocity ...........
hitsware said:
The author was talking about distortion measurements. Which unless you're talking about phase distortion (which he doesn't give any indication that he is), is a frequency domain measurement.
It might well be a semantics issue if he hadn't made that comment about "huge back-EMF currents" which along with his claim that voltage and current haven't any linear relationship except in purely resistive circuits leads me to believe that he doesn't know what he's talking about and that his subsequent arguments are based on erroneous assumptions.
se
hitsware said:
Not necessarily in the time domain it doesn't. But then the position of the cone in the time domain doesn't correspond to the current either.
The cone is a resonant mass/spring system so it's position isn't going to always be in phase with the applied force.
se
>The cone is a resonant mass/spring system so it's position isn't going to always be in phase with the applied force.
You're starting to get the idea.
You're starting to get the idea.
hitsware said:
What idea? The author says absolutely nothing about the position of the cone. Only that the force that moves the cone is a function of the current through the voice coil. He says we should measure the distortion of the current rather than the distortion of the voltage. But as I've been saying, as long as I = E/Z and E = I x Z, it doesn't matter whether you measure the distortion of the voltage or the current. You'll get the same result.
se
>It might well be a semantics issue if he hadn't made that comment about "huge back-EMF currents" which along with his claim that voltage and current haven't any linear relationship except in purely resistive circuits leads me to believe that he doesn't know what he's talking about and that his subsequent arguments are based on erroneous assumptions.
Ok forget the cone position part. My digression.
Would you say that an amp (what we are testing)
Will dissapate the same power putting out 10V
into an 8 Ohm resistor as it will with 10V into 8 Ohms
of reactance?
Maybe "huge back-EMF currents" is a little less than maximum eloquence. How about "considerable current"?
Obviously enough to make our voltage driven test system less than reveiling of everything going on in the system.
Ok forget the cone position part. My digression.
Would you say that an amp (what we are testing)
Will dissapate the same power putting out 10V
into an 8 Ohm resistor as it will with 10V into 8 Ohms
of reactance?
Maybe "huge back-EMF currents" is a little less than maximum eloquence. How about "considerable current"?
Obviously enough to make our voltage driven test system less than reveiling of everything going on in the system.
jam said:
Another guy who doesn't understand Ohms law. Ohms law is NOT: "an increase of the voltage produces an increase of the current" Ohms law says that V=I.Z. If you change V, and at the same time decrease Z (neg impedance) the new I will be different from what you expect if Z hadn't sneakily changed in the mean time, but still V=I.Z. In fact, the new Z is V/I BY DEFINITION.
Ohms law still holds comfortably, thank you very much.
Jan Didden
hitsware said:
Ok.
It will have the same apparent power (VA or voltamperes) but not the same average power and the power factor will be less than 1.
But so what? There will still be a linear relationship between the voltage waveform and the current waveform so if you measure distortion using either voltage or current you get the same result.
What I took issue with was the author's claim that voltage and current only have a linear relationship in purely resistive circuits and have a non-linear relationship in reactive circuits.
How many more times do I have to post Ohm's Law for resistance, reactance and impedance to get across the notion that voltage and current have a linear relationship in all three?
Here they are again. Just show me where the non-linear relationship exists between voltage and current in any of these:
Ohm's Law for resistance:
I = E/R, E = I x R, R = E/I
Ohm's Law for reactance:
I = E/X, E = I x X, X = E/I
Ohm's Law for impedance:
I = E/Z, E = I x Z, Z = E/I
Again, the LARGEST back-EMF occurs at resonance. At resonance the impedance is HIGH. So for a given voltage, at resonance, according to Ohm's Law, the current will be LOWEST. Also at resonance, current and voltage are in phase and the load behaves purely resistively. So I'm at a loss to explain his comment about these "huge back-EMF currents" being "dumped back into the output of the power amp."
What's obvious? Again, show me where the voltage and current have a non-linear relationship. Otherwise, this guy's just talking out of his rectum and a distortion measurement made of the voltage waveform will give the same result as a distortion measurement made of the current waveform.
Nothing you have brought up so far has shown otherwise.
se
I think you can be refuted mathematically, but I lack the means so let me take another tack.
BTW .... We're seldom 'at resonance' , but still driving a reactance.
Also it maybe a series rather than parrallel resonance.
Say we put a squarewave into the amp.
Comes out square into a resistor.
We change the load to a reactance.
If the amp has a low damping factor we see ringing.
This is back EMF induced.
If the amp has a high damping factor we see none.
The EMF is still there.
Why do we see no ringing?
Because the amp absorbs (swamps?) the energy.
If you look at the LM1875 datasheet they have a
paragraph about driving reactive loads that is
instructive.
The amp can go sideways due to the reactance.
Why is this?
Because the phase relationships are skewed.
But in a practicle case the amp doesn't quite
go sideways (just doesn't get the right message
back)
BTW .... We're seldom 'at resonance' , but still driving a reactance.
Also it maybe a series rather than parrallel resonance.
Say we put a squarewave into the amp.
Comes out square into a resistor.
We change the load to a reactance.
If the amp has a low damping factor we see ringing.
This is back EMF induced.
If the amp has a high damping factor we see none.
The EMF is still there.
Why do we see no ringing?
Because the amp absorbs (swamps?) the energy.
If you look at the LM1875 datasheet they have a
paragraph about driving reactive loads that is
instructive.
The amp can go sideways due to the reactance.
Why is this?
Because the phase relationships are skewed.
But in a practicle case the amp doesn't quite
go sideways (just doesn't get the right message
back)
hitsware said:
Ok.
So?
Well, since your typical dynamic loudspeaker models as a parallel RLC circuit...
Ok.
So what has any of this to do with there being a non-linear relationship between voltage and current in a reactive circuit?
No one's said that voltage and current are always in phase with each other in a reactive circuit.
Look, the author claimed that there is a non-linear relationship between current and voltage in reactive circuits and that because of this, and because it's current which ultimately produces the magnetic field which causes the speaker cone to move, we should measure amplifier's distortion by using the current waveform rather than the voltage waveform.
Well, if you change the phase relationship between voltage and current by way of reactance, you still end up with the same waveforms for both current and voltage. Therefore, your distortion measurements will give you the same result whether you're measuring current or voltage.
In other words, it doesn't matter if current and voltage are shifted relative to each other in the time domain. Distortion measurements aren't made in the time domain. They're made in the frequency domain. And except for absolute levels, you'll get the same result whether you're measuring the current waveform or the voltage waveform.
se
>Look, the author claimed that there is a non-linear relationship between current and voltage in reactive circuits
You're letting that one little (perhaps semantic) tree block your view of the forest. You owe me a beer for the (attempted) lesson.
mm 🙂
You're letting that one little (perhaps semantic) tree block your view of the forest. You owe me a beer for the (attempted) lesson.
mm 🙂
hitsware said:
But that little tree is the main premise behind the author's arguments.
Either there is a linear relationship between voltage and current or there isn't. If there is, then the author's argument that we should measure an amplifier's distortion by way of current rather than voltage flies right out the window.
se
You're bound and determined to make it a six-pack.......
>But that little tree is the main premise behind the author's arguments.
His premise is that E & I are out of phase and that I controls the cone while we measure E......Whether or not his usage of 'non linear' can be taken literally in the strictest mathematical sense.
>But that little tree is the main premise behind the author's arguments.
His premise is that E & I are out of phase and that I controls the cone while we measure E......Whether or not his usage of 'non linear' can be taken literally in the strictest mathematical sense.
hitsware said:
However you want to interpret his "linearly related" comment, what's clear is that he's implying you would get a different distortion result by measuring current versus voltage.
And if Ohm's Law holds true for resistance, reactance and impedance, you should get the same measurement either way.
se
hitsware said:
How does a non-periodic signal manage to skirt around Ohm's Law with regard to the relationship between voltage and current in a reactive circuit?
se
>How does a non-periodic signal manage to skirt around Ohm's Law with regard to the relationship between voltage and current in a reactive circuit?
It doesn't "skirt around it". It makes using it in this application unfeasible. The nature of the signal makes I & E almost
unrelated (talking in ideals and perfect cases here which
I'm sure you understand is necessary for purposes of
demonstration)
So far we agree:
1) E & I are out of phase in a reactive load (driver)
2) Cone position (movement) is governed by I
3) The error correction mechanism (feedback) is derived from E
Say we take an inverting amplifier
Hook the output to the + side of the driver
Between the - side of driver and gnd. we
insert a low value (current sensing) resistor.
AND
We have a perfect microphone to measure
system (acoustic) output
We can then monitor (and compare)
1) input
2) output (voltage) of amp
3) current through driver (and amp)
4) acoustic output of system
From our above agreement we can see that the
acoustic output will correspond to I
BUT our error correction is based on E
Perhaps with a sinewave (though I doubt it)
we would achieve 'sort' of an agreement between
E & I (though displaced in time)
BUT Since music in so complex (both in frequency and
amplitude) we have 2 (almost) unrelated events.
I.E. the corrected E will correspond to the input but
not the acoustic output....................mike
It doesn't "skirt around it". It makes using it in this application unfeasible. The nature of the signal makes I & E almost
unrelated (talking in ideals and perfect cases here which
I'm sure you understand is necessary for purposes of
demonstration)
So far we agree:
1) E & I are out of phase in a reactive load (driver)
2) Cone position (movement) is governed by I
3) The error correction mechanism (feedback) is derived from E
Say we take an inverting amplifier
Hook the output to the + side of the driver
Between the - side of driver and gnd. we
insert a low value (current sensing) resistor.
AND
We have a perfect microphone to measure
system (acoustic) output
We can then monitor (and compare)
1) input
2) output (voltage) of amp
3) current through driver (and amp)
4) acoustic output of system
From our above agreement we can see that the
acoustic output will correspond to I
BUT our error correction is based on E
Perhaps with a sinewave (though I doubt it)
we would achieve 'sort' of an agreement between
E & I (though displaced in time)
BUT Since music in so complex (both in frequency and
amplitude) we have 2 (almost) unrelated events.
I.E. the corrected E will correspond to the input but
not the acoustic output....................mike
hitsware said:
Mike, again, as long as Ohm's Law for a reactive circuit holds true, I and E must be related linearly in terms of magnitude. So I don't understand what you mean when you say they will be almost unrelated.
Except at resonance where they are in phase.
Yes, but not instantaneously.
Yes.
We're supposed to be measuring amplifier distortion. Why use a microphone? Just measure the current waveform across the current sense resistor and measure the voltage waveform across the loudspeaker terminals.
Ok.
So? Since Ohm's Law for a reactive circuit is I = E/Z and E = I x Z, any change in I must still result in a corresponding and linearly related change in E and vice versa.
It won't even correspond perfectly with the input since there will always be at least SOME delay between the input and the output, which will result in a frequency dependent phase shift. Which is one reason why negative feedback becomes less effective as frequency increases.
se
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