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In Reply to: RE: OTL vs SET posted by Caucasian Blackplate on January 13, 2015 at 18:36:02
OTL amps have fewer high dollar parts, but in general the circuit design is considerably more complicated than a SET amplifier. There is also significantly more diversity in OTL amplifier circuits than single ended circuit. Though feedback isn't 100% required for operation, it is often times used to drop the output impedance without having to add an insane amount of extra output devices (so we could say it's 99% required).
First, we have to pop the bubble about feedback. It does not reduce output impedance at all. If it did, a basic law of electricity, Kirchoff's Law, would be violated. The only way you can reduce output impedance is with more tubes, bigger output transformers, etc. Of course, what is meant by the term 'output impedance' must then be under question, but put simply you can't drive a lower impedance simply by adding feedback. Do do that requires the greater number of tubes etc. I just mentioned.
We don't use a whole lot of feedback- the M-60 employs 2 db (and some people have the amps built with zero feedback) and the S-30 uses 4 (and also works fine without it). Otherwise our amps have no feedback. The question is whether we have an 'insane' number of output devices... in terms of cost, probably not, as often it costs more to retube an SET than it does our amps.
Finally there is the issue of complexity. Our amps have one gain stage. An SET usually has two or three. Which is more complex? As far as the signal is concerned, probably the SET. Certainly our amps have more parts. So its my opinion of course but I just don't see this statement as reality.
Follow Ups:
The OP was asking about "averages", so I answered considering averages.
Bringing up feedback was not targeted at you at all, you have made your case clearly in the past about how little feedback you use in your designs, but again, it's averages.
(BTW - how is the M-60 a 2 stage amp? I see 3...)
At this point it might be that we have changed the averages when it comes to OTLs, being the oldest and largest manufacturer. You had used the 99% value, which I have to assume is just not right.
My girlfriend complains that I am very literal...
Oh, I wasn't intending averages by manufacturing, more like out of 50 OTL designs reviewed, I would expect nearly all of them to use more than 10dB of feedback.
I still stand by my assertion that properly designing on OTL amp is tougher than designing a good SET (but designing a good SE output transformer is harder than both put together) .
I think we might have more reviews that other OTL manufacturers too. I know there are more than are shown on our website.
But I do agree- especially that part about SET OPTs!
"First, we have to pop the bubble about feedback. It does not reduce output impedance at all. If it did, a basic law of electricity, Kirchoff's Law, would be violated. The only way you can reduce output impedance is with more tubes, bigger output transformers, etc. Of course, what is meant by the term 'output impedance' must then be under question, but put simply you can't drive a lower impedance simply by adding feedback..."
It seems to me you are conflating two rather different issues here. The output impedance of a device can be defined in various different but related ways. Perhaps the cleanest would be to say that we take the two output terminals, drive an externally-sourced current (Delta I) through them, and measure the resulting voltage change (Delta V) across the terminals. The output impedance is then defined as
Z = (Delta V)/(Delta I).
The output impedance so defined will certainly be lowered if negative feedback is used. And this lowering of the output impedance will have all the usual concomitant features, such as increasing the damping factor for a loudspeaker connected to the output, and so on. This is an absolutely standard piece of basic circuit theory, which applies to OTLs, operational amplifiers, or whatever.
There is a caveat that must be included in the above. Namely, the discussion is applicable only provided that one is never calling on the device to supply, or source, more current than it is capable of handling. This, it seems to me, is the point you are really trying to get at; the amplifier will not be able to drive any more current through the loudspeaker simply because one has added feedback. To get it to handle more current, and thus produce more power into a given load, one would need to add more output tubes.
This same caveat would apply to any other example too. Thus the low output impedance of an op amp with lots of feedback would only be valid provided one did not try to get it to pass more current than it could handle.
But provided one never asks the amplifier to pass more output current than it is capable of, the output impedance, as defined in any standard way, is reduced by the use of feedback.
Chris
We're actually in agreement on this but the term 'output impedance' as you are using it (and for that matter, most of the audio industry) is not actually the output impedance of the amp. This is why I stated
Of course, what is meant by the term 'output impedance' must then be under question
-because in most cases its really a measure of the servo gain of the feedback loop, and not an actual measure of the impedance of the output section. However, I tend to work with the Power Paradigm rules (which predate the Voltage Paradigm rules, see link below) which measure the impedance of the output section rather than the 'output impedance'. A lot of people are confused by this topic, especially if the Voltage Paradigm is their only source of the terms.
My contention is simply that the term as it is usually seen is misleading, because so many people use the concept to suggest that if you add feedback to an amplifier, its output impedance goes down therefore it can drive lower impedance loads. As you correctly point out, there is a current limit involved which was what I was referring to with Kirchoff's Law. We just have different ways of expressing it is all :)
"the term 'output impedance' as you are using it (and for that matter, most of the audio industry) is not actually the output impedance of the amp."Well, we could probably debate that endlessly. In my view it is better to stick with conventional usage of well-defined concepts such as impedance = (Delta V)/(Delta I), since it is meaningful and is understood in almost all of electronic circuit theory.
But as you say, we are probably in agreement about the actual underlying facts, and it is only the words used to describe it that differ.
Chris
Edits: 01/15/15
That difference has a lot to do with the difference between the Voltage and Power paradigms.
I use the word 'paradigm' as such is defined as a platform of thought, outside of which nothing exists or else is blasphemy. IOW one schooled in the Voltage Paradigm will have trouble seeing how 'output impedance' can be expressed in any other terms- I have run into this a lot.
I like to use the black box approach similar to how one measures the impedance of a speaker. In this case, the amp has an unknown impedance but we can easily find out what it is by simply connecting it to a variable load. We then find out at what impedance it makes its maximum power and then continue to reduce the load until that value is cut in half. At that point, half of the power is being dissipated in the output circuit itself (causing it to heat up) and the other half in the load. At this point the load must be equal to that of the output circuit.
The interesting thing about this approach is that the value is predicted by formula and corresponds in practice.
Under the Voltage Paradigm, the output impedance is that where the circuit makes maximum power, which corresponds to the (Delta V)/(Delta I) method.
The Power technique is measuring the impedance of the output circuit, the Voltage method is measuring the dynamic response of the circuit. In the former, the measurement does not change regardless of feedback, in the latter, it does.
IOW, at this point we are simply talking semantics. I agree that most of the industry is using the Voltage Paradigm, and in lo- and mid-fi applications this is perfectly justified. In high end, it isn't so much on account of higher performance loudspeakers like ESLs, horns and others that don't conform to Voltage Paradigm rules.
In case you think I am making this up (some have in the past), Google the Fisher A-55 amplifier and one of the first hits you will see is a YouTube image of the damping control of the amp. It is marked 'Constant Voltage' at one extreme, 'Constant Current' at the other, and where they cancel (zero feedback) it reads 'Constant Power'.
"I like to use the black box approach similar to how one measures the impedance of a speaker. In this case, the amp has an unknown impedance but we can easily find out what it is by simply connecting it to a variable load. We then find out at what impedance it makes its maximum power and then continue to reduce the load until that value is cut in half. At that point, half of the power is being dissipated in the output circuit itself (causing it to heat up) and the other half in the load. At this point the load must be equal to that of the output circuit."The thing that worries me about your definition of output impedance is that it doesn't give a sensible answer if we test it on the classic example of an idealised device that is modelled as a perfect voltage source (i.e. zero impedance and voltage V) in series with a resistance R. This device unambiguously has an output impedance R.
If we now drive this into an external impedance Z, the power dissipated in the external load is P= V^2 Z/(R+Z)^2, and this is, of course, maximised at Z = R. Let's call this Z1, so we have Z1 = R. We also have Pmax = V^2/(4R). If we now follow your rule, and ask for the smaller of the two values of Z for which P = Pmax/2, we get Z = (3 - 2* sqrt2) R. Calling this value Z2, we have approximately Z2 = 0.17 R.
Thus in this case of an idealised output device, for which all conventional ways of calculating the output impedance will give the answer Zout = R, your way of defining the output impedance will give the answer Zout = 0.17 R, which is about one sixth of the correct answer.
And, by the way, the value of Z for which the power dissipated in the device equals the power dissipated in the external load is also given by Z = R. When Z = 0.17 R, which is your definition of the output impedance when applied to this idealised case, the power dissipated in the external load is about one sixth of the power dissipated in the device itself.
I suppose your argument must be that the idealisation of the output device as a perfect zero-impedance voltage source in series with a resistor R is not an accurate model of a real amplifier output stage. But it does seem to be troubling that your definition gives a result that is so much at odds with the actual answer, if applied to the usual idealisation.
Chris
Edits: 01/17/15
Thus in this case of an idealised output device, for which all conventional ways of calculating the output impedance will give the answer Zout = R, your way of defining the output impedance will give the answer Zout = 0.17 R, which is about one sixth of the correct answer.
In practice this is not borne out. I stated earlier what the differences are. However I should point out that its also impractical to assume that we are going for a perfect voltage source. In fact we are not, in our case we are going more for a power source. All amplifiers were power sources in the old days- SETs still are, as are many tube amps without loop feedback.
Obviously a perfect voltage source will have infinate power at zero ohms. It does not exist.
Its a different way of looking at it. What I recommend is that you play with the equipment and take measurements as they will reveal more than the math seen so far. Or you could trust that the numbers I suggest are right- I've done these measurements many times. Did you read the article at the link? There is a reason why someone would want to design an amplifier that has a higher output impedance like this- oddly, its because you can get less coloration- quite the opposite of the 'accepted wisdom'. This is because of how the human brain processes sound, which ultimately is the final arbiter.
"Did you read the article at the link? There is a reason why someone would want to design an amplifier that has a higher output impedance like this- oddly, its because you can get less coloration- quite the opposite of the 'accepted wisdom'. This is because of how the human brain processes sound, which ultimately is the final arbiter."Yes, I read the article you linked. It seems to me there are several distinct questions here. Some of them, such as the voltage paradigm vs the power paradigm, can be discussed even within the conventional model of an amplifier output stage as a perfect voltage source in series with an internal impedance R.
What is certainly true is that if the output impedance R is tiny, such as with a typical solid state amplifier, then for a given signal level the power delivered into a typical loudspeaker load impedance Z falls approximately inversely with Z. On the other hand, if the output impedance R is comparable with the typical loudspeaker impedance Z, then one is "sitting on the top of the curve" in the plot of power dissipated into the load as a function of Z. Thus, to a good approximation, moderate variations in Z around the value Z=R will then not cause the power into the load to change by too much. This is illustrated in the first plot here, of power into the load Z, as a function of Z, for the case of output impedance R=8 Ohms. (x-axis is Z, y-axis is power into load Z.)
By contrast, in the case of a very low output impedance amplifier, the power into Z=16 Ohms would be one quarter of the power into Z=4 Ohms, as can be seen in this second plot (normalised to the same power at Z=8 ohms):
In other words, you don't need to invoke any non-standard definition of "output impedance" in order to make your point about the power paradigm vs the voltage paradigm. It is already visible, just with the usual definition, and the usual model of an output device. If the output impedance is tiny, then for a load impedance Z in the range between 4 and 16 ohms, the power lies in a large range, with Pmin being 25% of Pmax. On the other hand, if the output impedance is 8 ohms then for the same range of load impedances the power lies in a relatively small range, with Pmin almost 89% of Pmax.
Another rather different question is whether the usual model of an amplifier output stage as a perfect voltage source in series with an internal resistance R is a good one or not. I tried various experiments in the past with my OTL amplifiers, and it seemed to me that the model worked reasonably well. Now, in all of my amplifiers there is quite a lot of negative feedback. I am interested now to experiment a bit more, in particular with how well the idealised model works if I turn off the negative feedback. I suppose my expectation would be that the model would still be reasonable, but now the value of R would be a lot larger. But I will keep an open mind until I have done the experiments.
Concerning the non-achievability of the "perfect" zero impedance voltage source, yes, of course I agree with you it cannot literally be realised in practice. But that doesn't mean it isn't a useful concept in circuit analysis. As long as its "actual" output impedance is very small compared to that of the internal resistance R that one adds in the model of the output stage, then the imperfection of the "perfect" source is insignificant. I still feel that if a definition of "output impedance" gives a radically wrong answer when tested on the conventional model of an output device, then there is reason to question the validity of that definition. But then, as I said above, I don't think you even need your non-standard definition of output impedance in order to make your point about the power vs voltage paradigms.
Anyway, I shall try to carry out some further experiments soon.
Chris
Edits: 01/21/15 01/21/15 01/21/15 01/21/15 01/21/15
The non-standard output impedance method is actually based on the formula for calculating the output impedance of a cathode follower. But as you point out, the graphs really make the point.
The problem is that quite often the application of loop negative feedback contributes to higher ordered harmonic content (re.: Norman Crowhurst). Since the ear uses such harmonics to calculate sound pressure it is very sensitive to these harmonics; so much so that even trace amounts that are hard to measure are readily audible, which is why transistor amps tend to sound bright. The brightness is not measurable as a frequency response variation, in just the same way that the 'warmth' of a tube amp is not.
Quite literally the application of feedback causes a violation of one of the most fundamental rules of human hearing perception. The Power Paradigm is a means of getting around that problem. The idea is that while frequency response is likely to not be as accurate, colorations due to distortion (which are interpreted by the ear as tonality) will be reduced. Since no speaker is really flat with any source, the actual real-world result is that you can get what sounds like very uncolored sound.
"In this case, the amp has an unknown impedance but we can easily find out what it is by simply connecting it to a variable load. We then find out at what impedance it makes its maximum power and then continue to reduce the load until that value is cut in half. At that point, half of the power is being dissipated in the output circuit itself (causing it to heat up) and the other half in the load. At this point the load must be equal to that of the output circuit."
OK, I need to think about this a bit. May I check that I have your procedure correct? You first adjust the load impedance until the maximum power dissipation in the load is achieved. Let's say that the load impedance for this is Z1. Then, you reduce the load impedance further until you reach a point where the power dissipated in the load is one half of what it was at the maximum. This second, lower, load impedance is, say, Z2.
And you then define Z2 as the output impedance of the amplifier?
Thanks,
Chris
No, that's not quite it. You have to start out at a significantly higher value (i usually start at 100 ohms for an unknown box); then decrease it until the maximum power transfer point is found, then decrease from there to 1/2 power transfer.
"No, that's not quite it. You have to start out at a significantly higher value (i usually start at 100 ohms for an unknown box); then decrease it until the maximum power transfer point is found, then decrease from there to 1/2 power transfer."
I think that's what I was saying; it's what I meant to say at any rate. First find Z1 load impedance at which maximum power into load is reached, and then reduce further until the power into the load drops to one half that, and this defines your Z2.
Chris
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