Care to post data and test setup? Your claim contradicts a LOT of published info.
It has everything to do with phase if we are to believe your objection. You asserted: "At this point the propagation delay becomes significant (easily measured with very basic test equipment), and in fact its well-known to any designer that you have to limit the bandwidth of the feedback itself, as above a certain frequency it becomes positive instead of negative feedback, causing oscillation."
It has everything to do with phase if we are to believe your objection. You asserted: "At this point the propagation delay becomes significant (easily measured with very basic test equipment), and in fact its well-known to any designer that you have to limit the bandwidth of the feedback itself, as above a certain frequency it becomes positive instead of negative feedback, causing oscillation."
Yes I think we had a semantic issue. If the peaking inside the amp changes with feedback that is because the feedback causes a peaking in the input voltage. The amp itself cannot change it's internal transfer function of course.
jan
Like I mentioned earlier, I just used a square wave as a source, drove the amp at a nominal frequency (usually around 1KHz, so I don't have to put up with a slow sweep time; I still use older analog 'scopes) and just compare input to output, overlaid. Basically you are looking at two vertical lines on the 'scope once you zoom in enough. Being analog, I only get an approximation of the time, but its not hard to see that with a fixed time difference between input and output, that the higher you go in frequency the more error the feedback will introduce.
Keep in mind that I have only looked at this on tube amps. Most tube amps have a pretty slow rise time, which might play a role (not looked into that); I've been under the impression you can see the delay time on almost any amp.
Keep in mind that I have only looked at this on tube amps. Most tube amps have a pretty slow rise time, which might play a role (not looked into that); I've been under the impression you can see the delay time on almost any amp.
The parameter linking the two is called group delay, but there are some caveats: it has to do with cutoff frequencies, etc.
In addition, it isn't flat with frequency, unlike an ideal transmission line.
Here is an example with a real amplifier:
At low frequencies, there is a large increase due to the time constants, and it looks like ~zero at higher frequencies.
It isn't in fact: it becomes much smaller and more constant, but not perfectly constant, and at the top of the bandwidth there are once again serious variations.
Note that most of the ~190ns delay are caused by the input filter: 120ns.
But do not confuse this with an actual time delay: such a first order filter reacts instantly to a stimulus, it is simply the amplitude that is affected
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It's easier to understand phase shift versus delay if you realise that in the case of phase shift, whenever the input changes, the output changes also immediately. It is ONLY in the case of a sine wave, where the derivative is ALSO a sine wave, that it LOOKS as if it is delayed!
Take the current into a cap and the voltage on that cap. If the current is a sine wave, the output looks to the inexperienced like a sine wave that is 90 degrees delayed. But, and this is the crux, at the moment the input current changes, immediately you see a change in the cap voltage. I have done a tutorial about that somewhere, see if i can find it.
Do people really think that when you switch on the current into a cap that it takes a time delay before you see a voltage on the cap? Of course not: switch a current into a cap and the voltage starts to rise immediately! There is NO delay!
If you really want to think: if the cap current goes through it's peak and starts to go down, the voltage on the cap still continues to rise because current is still going into it. BUT, because the current now starts to become smaller, the voltage rise changes from increasing in rate, going down in rate: surprise, that's a cosine! And a cosine is a sine wave - phase shifted 90 degrees.
Isn't physics beautiful - all those mathematical constructs, and reality follows it accurately. Some theory huh!
Jan
But that is not delay. If you decompose the square wave you will find that the higher component parts have more phase shift, and if you then reassemble the wave you get a square wave that seems delayed.
The rising and falling edges are composed of the highest harmonics and they are phase shifted the most (every amp is, of course, a low-pass filter).
Jan
Yes, exactly. The point is that the circuit must be designed with the use of feedback in mind to avoid such problems.
The peaking is due to the use of feedback, and does not happen open loop.
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Yes, and you will see that with higher frequency, the difference in, say, the sine zero crossing, widens.
That's what phase shift does for you.
But if you somehow could see the very start of the signal you'd see that the output appears at exactly the same moment (give of take a few nanoseconds) as the input. The output reacts to the input immediately, but in a different shape because of the phase shift.
Jan
I think the use of a sine wave here confuses matters; an unambiguous one shot pulse is really what you want to use as a test signal.
A true time delay would preserve the symmetry of the pulse. The start and stop of the edge would look alike in their curvature.
In a phase shift, the low pass characteristic would have an RC behavior, that is, a sharp (discontinuous derivative) start with an exponential (1-exp(-t/RC)) rise and asymptote. This behavior would be unmistakable as a phase shift, not a time delay.
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On an amplifier with 200KHz bandwidth- there is significant phase shift at 1KHz? I don't mean to put words in your mouth, but one of the amps tested had this sort bandwidth.
I'm always open to learning, but so far I am still seeing what looks like propagation delay, as the delay time is constant at all points in the passband.
I'm always open to learning, but so far I am still seeing what looks like propagation delay, as the delay time is constant at all points in the passband.
If you input the same signal to both scope channels, do the two waveforms track exactly?
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Ok, now if you input a square wave (symmetrical about the zero amplitude level) and compare it with the output, do the sharp corners
where the transitions begin coincide, or are they displaced in time by the same amount as what you identify as the propagation delay
around the zero amplitude level? Can you post a screen shot?
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Fully agree! I was of half a mind to put something like that in the sim but until know just didn't have a quiet moment...
Jan
This may help to frame things:
Below you see the input and output of a regular RC low pass, with a 20kHz sine wave.
RED is input from the generator, GREEN is output across the cap. Clearly, you see that the GREEN output comes with quite some delay with respect to the RED input, right?
Wrong!
See the top graphs, this is the same RC but now excited with a square wave. In the top right you see input and output; the output across the C is the basic square wave with the higher harmonics attenuated as is expected with a low-pass.
You see that the output reacts immediately to changes in the input.
The upper left is a zoom in onto the output waveform, showing even better that it reacts IMMEDIATELY to a change (in this case rising edge) of the input. So, NO delay!
The special case with sine waves is a very sneaky one because it surely LOOKS as if the output is delayed, but also here any change in input has a corresponding, non-delayed change in output.
The bummer here is that a phase shifted sine still looks like a sine and that totally misleads you.
Edit: maybe we should say that what is delayed is the SHAPE, but not the input/output causal chain. Or does that even confuse more?
Jan
Below you see the input and output of a regular RC low pass, with a 20kHz sine wave.
RED is input from the generator, GREEN is output across the cap. Clearly, you see that the GREEN output comes with quite some delay with respect to the RED input, right?
Wrong!
See the top graphs, this is the same RC but now excited with a square wave. In the top right you see input and output; the output across the C is the basic square wave with the higher harmonics attenuated as is expected with a low-pass.
You see that the output reacts immediately to changes in the input.
The upper left is a zoom in onto the output waveform, showing even better that it reacts IMMEDIATELY to a change (in this case rising edge) of the input. So, NO delay!
The special case with sine waves is a very sneaky one because it surely LOOKS as if the output is delayed, but also here any change in input has a corresponding, non-delayed change in output.
The bummer here is that a phase shifted sine still looks like a sine and that totally misleads you.
Edit: maybe we should say that what is delayed is the SHAPE, but not the input/output causal chain. Or does that even confuse more?
Jan
What you are seeing, which you misinterpret as a propagation delay, is the effect of a first-order filter low pass filter - which most amps approximate to within their passband. The 'delay' you see will be approximately equal to the time constant of the low pass filter - only it is not a delay as it can be undone with an inverse filter.atmasphere said:
Coping with filters (or integrators) is standard control/servo loop theory - see any decent textbook on the subject. Any audio designer will be familiar with this stuff.
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