John Curl's Blowtorch preamplifier part III

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Joe, DF is a figure of merit which allows one to calculate the Zout of an amplifier if so desired. This Zout can be used to calculate effective Qes and this new Qes can be used to calculate the resulting response shape. No magic.

I don't quite understand your underlined question about seeing and telling amplifiers, but the words like 'output Z', ' load Z' and 'impedance' all have well defined meanings that are readily understood and which can be used to predict both the behaviour of the amplifier as that of the loudspeakers.

Oh dear, the 'claim' word again, "march that man to the scaffold!"

Gotta point this out: It seems that you have deconstructed the DF as a way of calculating the output impedance? Why would you do that?

In what universe would you do that calculation since you would already know what it is?

You have to know the output impedance before you can calculate the DF, don't you? Yikes! My head is spinning...what convoluted logic - please make sense. This makes no EE sense.

But I note you avoided this question earlier:

"My turn to ask you a question, can the amplifier tell where the divide is between its own output Z and the load Z? In other words, can the amplifier in fact see more than single impedance at any one frequency?"

Could you please addresses that. Seems reasonable to ask.

And I want an EE answer to that, OK?

[Edited the question for clarity's sake.]
 
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I might as well ask everybody else here the same question:

Can the amplifier tell where the divide is between its own output Z and the load Z? In other words, can the amplifier in fact see more than single impedance at any one frequency?"

Yes... or no?

Let's take a vote on this, should be interesting.

[Edited the question for clarity's sake.]
 
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Gotta point this out: It seems that you have deconstructed the DF as a way of calculating the output impedance? Why would you do that?
He didn't. This is just a marketing phrase from half a century ago. Taking offense now is too late.

If your real point is that a loudspeaker's motor model must include the driving amplifier's impedance, yes, and nobody disagrees.

We're a modeling species. Even our perception of reality is only a model. We don't see in rectilinear vision. There is no such thing as "red". Hearing has a zillion layers before it becomes listening. Taste and smell are so primitive we'll probably never understand the mechanics. Our processing latency at best is just slightly less than a quarter second. We exist behind the times because our model isn't real-time.

Only the model is discussable. How does your model differ from the conventional accepted one?

All good fortune,
Chris
 
This crossover looks setup to do impedance compensation and be a easy load.
I am missing something on this?
You have that part correct, the point is that optimal damping is supplied passively and locally.

This means the amplifier has no return energy to deal with.
For the best of amplifiers probably of no great consequence, for lesser amplifiers much improvement.


Dan.
 
Joe Rasmussen said:
Those who say that DF is for real has yet to come up with a satisfactory explanation.
You are falling into the classic trap of equating 'no explanation which I am willing to understand and accept' to 'no explanation'. The explanation is clear, it is available in any textbook and it has been repeated here by several people.

Then you are falling into the trap of accusing those who are trying to teach you the truth about this of having failed to 'understand' the equation. The reality is that we understand the equation; you do not.

Then you say 'my expert friend agrees with me'. Sorry, that cuts no ice at all. Two reasons:
1. we don't know your friend; he may be a genius, or he may be an (electrical) idiot.
2. we were not present at the conversation, so we don't know whether you misunderstood him or he misunderstood you.
Much better to claim 'this well-known textbook agrees with me'. Then we can investigate, although even then it is possible that you have deeply misunderstood the book.

I know of somebody who could end up writing a paper on this for the AES.
Someone we don't know might or might not write a paper which might or might not pass peer review and might or might not contain the truth. Not a clear technical argument, is it?
 
I might as well ask everybody else here the same question:

Can the amplifier tell where the divide is between its own output Z and the load Z? In other words, can the amplifier in fact see more than single impedance at any one frequency?"

Yes... or no?

Let's take a vote on this, should be interesting.

[Edited the question for clarity's sake.]
Set a single circuit with an interrupter and a coil in your AC plug.
Set a second one with a equivalent resistance at 50/60 Hz.
Compare the sparks when you open the two circuits.

I was tempted to ask you, if you have an oscilloscope, to compare the phases of the currents.
And then I thought, if you had an oscilloscope, you probably would not have asked the question ;-)
 
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Well, at least the terminals are.
The drivers, no.

Jn

@Destroyer Yes I forgot to mention adding shunt networks at the speaker terminals does not change the impedance seen by the drivers themselves with a voltage out PA. If the shunt networks are added to an existing properly aligned multi-way speaker there would be no change in driver induced distortion, certainly driven from an ideal voltage amplifier.
 
Not sure why, it worked when I posted them.

I snipped a portion of the original URLs and got these to load for me. The + sign might break it, so you may need to copy and paste:

http://pmacura.cz/13+14kHz_voltagedrive.jpg

http://pmacura.cz/13+14kHz_currentdrive.jpg

ETA: They work (for me) from clicking in this posted message.

The current drive graph definitely shows lower spurs around the two tones, but the 1k spur on the left appears slightly HIGHER in the current drive. Can anyone give a technical explanation for any of these differences? Especially since the speaker has a near-flat impedance response, it seems the same signal from either amp would give the same signal at the output of the amplifier and the same signals at the driver terminals.
 
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You have that part correct, the point is that optimal damping is supplied passively and locally.

This means the amplifier has no return energy to deal with.
For the best of amplifiers probably of no great consequence, for lesser amplifiers much improvement.


Dan.

It's also worth noting, there are a lot of people out there who run low / zero FB SET amps and these speaker's benign load would be an advantage.

T
 
It's also worth noting, there are a lot of people out there who run low / zero FB SET amps and these speaker's benign load would be an advantage.
Yes, that is the bottom line.....amplifier output resistance/impedance becomes 'irrelevant' when the loudspeaker presents as a resistive load......no voltage divider frequency response dependencies and no return of stored energy dependencies, win/win.


Dan.
 
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Joined 2003
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Yes, if you make impedance "equalization" to resistor-like shape, which in fact means you decrease impedance at some frequency intervals, then the crossover distortion becomes higher than if no impedance equalization was used.

Makes sense because the load impedance is lowered, so the amplifier current demands increase. But, a modern amp will not be phased by dynamic load and you can see this in a typical EF3 with moderate amounts of feedback: THD stays in low double digit PPM for any load > 3 Ohms and wide phase angles.

But, I get that the Elsinore is a nice easy load and this would be a clear advantage when using a tube or ZGFB amplifier - both have their places and I love my low power class A amps for acoustic and jazz.
 
Yes, if you make impedance "equalization" to resistor-like shape, which in fact means you decrease impedance at some frequency intervals, then the crossover distortion becomes higher than if no impedance equalization was used.
What is more natural to hear ? A distortion that varies depending of the frequency, or a monotonic one, a little higher at some frequency ?
Anyway, with the performance of modern amplifiers, the distortion is so low that I hardly believe somebody can hear-it.
 
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