 
|
Audio Asylum Thread Printer Get a view of an entire thread on one page |
For Sale Ads |
|
|
Audio Asylum Thread Printer Get a view of an entire thread on one page |
For Sale Ads |
205.188.116.7
In Reply to: keep going...sooner or later, you'll get it. posted by Sector-7G on November 3, 2004 at 19:35:57:
well... either a whole bunch of folks like Norman Crowhurst, Frederick Termin, RCA engineers, Steve Bench, RDH contributors, John Broskie, etc, et al... all got this wrong...Or you need to go back to and have an honest fresh look at their material and try to understand and master the methods which they have outlined and demonstrated.
I can't and won't do it for you.
I put up the information (and offered no opinions of my own) as references for people to explore and evaluate the articles on their own given the incessant harping and baiting that has accompanied the exploration of this technical issue by primarily one other person.
And I have no interest in arguing or debating these issues with you or anyone else... the sources speak for themselves and the trueness or falseness of their respective thesis is strictly independent of whether I am able to or care to defend (or oppose) the several theses that they have offered. I have adopted this position since this technical debate has been transformed by some into a lithmus test and\or a religious crusade and has been used as a battle axe.
However, I did find it interesting (notice... I did not say I agree or disagree) that the RCA engineers state explicitly that each triode in a pure class A PP amp sees one quarter of raa as it's reflected load impedance... and I stumbled across this when I was just killing some time waiting to pick up my girls from school and was reading the front of the RCA manual which I referenced.Now... if they are wrong and *IF* I did agree with them or have a tendency to agree with their stated views... then, none-the-less, that would not be in keeping with bad company would it?
So... why the near religious dimension to this debate?
MSL
Follow Ups:
and if Einstein had said to himself...Maxwell got it all, and everybody says he is right, we would not get his theory of relativity, and his recomendation of my grandfather for an academic post would have been worthless, comming from a patent clerk.It would have saved my grandfather a lot of work translating said relativity work into english( and correcting some of the math errors ).
mindless adherance to some idea, however well referenced is not good for anybody.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
We learned some of the Einstein`s math in "Mechanics of fluids", and I remembered blank looks through the class :-). But, theory sometimes work in another world besides practical "engineering" math methods...
from MSL> > > well... either a whole bunch of folks like Norman Crowhurst, Frederick Termin, RCA engineers, Steve Bench, RDH contributors, John Broskie, etc, et al... all got this wrong...that is often the case. And even more frightening is the certainty with which they are followed.
Also NC was not *ALL* wrong, and you have yet to respond to this instance where his own text contradicts your interpretation of the whole. So please quit saying that NC says a-a/4, he actually showed something else, and that which he showed also matched the eq'n on p.287 in RDH3.
Don't lump all those RCA Engineers into this. I have met a few of them and they certainly did not embrace this dogmatic approach of yours, or its conclusions either.
get out and do wome of your own work. some analytical thinking. not the research you have been using( which has led you astray ). Like I said, you ought to know better.
It does not matter, you have again stuck your nose into this, and all you have contributed is more dogmatic confusion. No wonder it has a religious tone to you.
A first indication that a lot of the lit. is wrong should be:
none of the a-a/4 examples obey conservation of energy for class A operating conditions. *** except as Dave Cigna showed: by accident***that it does not make any common sensee ought to be a second.
back to the question: in this PP stage, what happens to the load on a tube if you pull the other? same, higher or lower load? or work backwards, with one valve,working into half the primary( under an uncontested a-a/4, what happens to the load of you put in a valve on the other side of the primary?
Since you have once more indicated aninterest in discussing this topic it wold be interesting to see what your analysis of the last question is. Please remain calm and continue the discussion.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
Doug wrote:::::Also NC was not *ALL* wrong, and you have yet to respond to this instance where his own text contradicts your interpretation of the whole. So please quit saying that NC says a-a/4,...::::
Well, you have to do all the math correctly first... and know what the values on the graphs mean and if they represent the values of one tube or two tubes and etc...
Norman apparently anticipated your confusion, and stated;
"All this may seem a little confusing at first, but the important point to realize is that the practical circuit makes the two loads appear to be in series., because the widings of the transformer which combine the output are in series. But the magnetizing effect in the transformer is differential, so you subtract the smaller current from the larger."
"This means that the current change represented on the curves is double that which occurs in each half-winding, and the voltage difference on the curves is what occurs across one half of the winding (either half). Hence, the effective impedance considered on the load line is 1/4 of the impedance from plate to plate."
To get the correct answers you must use the right math and the right values for each of the parameters..
Now, if you would be kind enough to show me where (in his text) Norman states explicitly (please quote him directly) that the load seen by each tube is raa/2 I would be most appreciative. Your pulling raw numbers off of the graphs without understanding the formulas, relationships, and values of the parameters. And this, as Norman predicted, will lead to confusion on your part.
next... your cry about
"....conservation of energy for class A operating conditions."So you think that Steve Bench, John Broskie, RCA engineers, Termin, Crowhurst, etc, et al... all have violated this dictum?
but... first we would need to know how power is developed in each half of the primary and then ascertain how that power is then "combined" in the output (see Norman's text above)...
if the two primary half windings are in series then surely the value of each half cannot be raa/2. If you have a 5K reflected impedance on either side of ct... then the combined impedance from end to end will be 20K not 10K.
Plus we know that if the reflected impedance of the whole primary is (n1/n2)^2 times the load on the secondary... then... each half primary will have just one quarter the imepedance of the whole which follows mathmetically from ((n1/2)\n2)^2) times the load on the secondary.
If you double the turns you quadruple the impedance. And this axiom is not and cannot be violated.
Transformed series impedances do not add like resistors. Just as two eight ohm windings in series with each other combines or adds up to 32 ohms not 16 ohms.
so your thesis (raa/2) would also violate the "conservation of energy" theorem.... you would have less power than is generated by the tubes by half when two 5K series connected windings are shown to have an end to end impedance of 20K and not the 10K as you have proposed.
MSL
from MSL:so your thesis (raa/2) would also violate the "conservation of energy" theorem.... you would have less power than is generated by the tubes by half when two 5K series connected windings are shown to have an end to end impedance of 20K and not the 10K as you have proposed.The half winding *BY ITSELF* is a-a/4. The half winding, operating *WITH* the other one( PP Class A ) is a-a/2. the load applied to each of the valves is half the total.
why do you think it is less than half? where does the other, remaining fraction go? it cannot appear and dissapear to suit your analysis. If the whole is a-a, and in Class A PP, it *IS* by definition. with no cut off conditions occuring, then it cannot load the two ends with anything less than a sum equaling a-a.
You continue to analyze the half winding as an independant entity, which is incorrect. When taken as a whole, the conservation of energy still holds. your claim that it does not, is false.
I suspect that it is true, you have painted yourself into a corner and cannot excape anymore with what you think is your dignity intact. It is my opinion that just the opposite it true, there will be nothing left of your so-called dignity if you continue to stick you your false conclusion.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
first of all, I am not crying about anything. and you continue to dodge my question.from MSL:Now, if you would be kind enough to show me where (in his text) Norman states explicitly (please quote him directly) that the load seen by each tube is raa/2 I would be most appreciative.
read the second paragraph, and measure the slope of the dashed line he refers to. a few folks have done just that. I will leave the execise to the student. What is the slope of that line?
and further:
so your thesis (raa/2) would also violate the "conservation of energy" theorem.... you would have less power than is generated by the tubes by half when two 5K series connected windings are shown to have an end to end impedance of 20K and not the 10K as you have proposed.The conservation theory is only violated when you work it as you just describe. When you take the end-to-end impedance, power in = power out. it is when you decide that half of this winding equals a quarter that it breaks down. When operating together they cannot be considered independently as you have proposed.
What is the voltage across the winding on a turn by turn basis? is there a discontinuity anywhere? do any of the turns carry a different voltage as you progress fron one end of the winding to the other?
regards,.
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
Doug wrote:::::from MSL:Now, if you would be kind enough to show me where (in his text) Norman states explicitly (please quote him directly) that the load seen by each tube is raa/2 I would be most appreciative.::::
:::read the second paragraph,::::
on which page and in which coloumn... or better yet quote the text you are referring to. Show me in Norm's words where he sez that the loadline seen by each tube is raa/2.
:::...and measure the slope of the dashed line he refers to. a few folks have done just that. I will leave the execise to the student. What is the slope of that line?::::the slope of the line in degrees? That's all you can measure directly... is what is the slope of the line referenced to the horizontal axis in degrees....
otherwise... you need to account for the voltages and currents as prescribed by the author and whose text re: these parameters I have already quoted. But you cannot read voltage divided by current directly off of these curves without introducing errors.
you need to go through Norman Crowhursts' examples step by step... and when you encounter something different than what he sez... figure that your probably making an error of some kind. Prudence might suggest that Crowhurst (author of 600 plus books and articles) probably had proof read and was satisfied with his work before publishing it.
Funny... how you guys cannot provide one published example of a PP loadline having raa/2 as it's effective or operating impedance for each half primary.
On the other hand, I have provided mutlitple published sources which state explicitly that each tube in a class A PP amp will see raa/4 as it's reflected load impedance.
If brains bigger and better and smarter than my own cannot satisfy you then I doubt that I will be able to explain it any better than say Steve Bench, Broskie, Crowhurst, RCA engineers, etc, et al.I'll just wait for the Dougie text on electronics to be published.
msl
We provided you many references, but you never commented them, just keep going Raa/4. We gave you many examples, why you never calculate this example of yours (4 Ohms /16 Ohms /2 x 8 Ohms)? You just use some sentences from various sources that saying about Raa/4.
No problem for me, any more. For the end, just the reference from the RDH4, I wrote it 100 times...Page 571 & 572, "Fundamental principles of push-pull operation":
"The load resistance R2 is connected across the secondary, and the reflected resistance across the whole primary is:
Rl=4R2(N1^2/N2^2)
and the reflected resistance across half the primary is
Rl``= 1/2 Rl = 2R2(N1^2/N2^2)
where N1=turns in half primary winding
and N2=turns in secondary windingIf valve V2 in fig. 13.31 were removed from its socket, the load resistance effective on V1 would then be
Rl` = 1/4 Rl = R2(N1^2/N2^2)
which is half the load resistance on V1 under push-pull conditions. This is the condition which occurs when one of the valves reaches plate current cut-off."
Well, this is it - it can`t be expressed more clearly.
It's from Eastman, "Fundamentals of Vacuum Tubes". Dave Slage deserves credit for digging it up. I just found it's link on another website. Also look at Damir's explanation below...the one I said I like. The URL is http://131.109.59.51/images/tubes/ppamp.jpg
Russ
Hi Russ:
I've seen the article you make reference to before. Note that it looks at not how power is distributed (or combined) across the two series connected loads (i.e., porimary halves) by the output transformer.. meaning it does not show you the plate curves and the loads across the plate curves for each tube... nor how the tubes develop power into a load... rather it describes the condition which occurs *after* each half pri has developed power across a load and the transfomer has combined these two series connected loads... look toward Crowhurst (I mentioned this earlier in a different response) for this distinction.And speaking of references... an earlier reference listed on Ken Gilbert's page is to an MIT article... the url of which is
http://131.109.59.51/images/pdf/Push_Pull_Theory_MIT.pdf
if you go to page 19 in this pdf... you will see a PP loadline for the 45 tubes operating in pure CLASS A1. And you will note that the loadline from end to end is four times greater than the loadline that each tube works into....
so.... we can all call on our experts... but I would say the chief difference is that the article you refer to does not demonstrate the loadline or magnitude of the loadline nor how power is developed across this loadline but rather how it is or can be viewed once the two halves of the primary have been combined by the transformer....
remember... the tube can only develop power if it operates into a load. Leave the secondary open circuited and no power can be developed across the primary. The primary can only be loaded via a load being placed on the secondary. This is the point Broskie makes very well about a tranney does not have an intrinsic impedance... it is the load placed on the secondary mulitplied by the impedance ratio ot the primary which establishes or creates a load on the primary that a generator or a tube (or tubes) can then work into...
the tubes see the impedances presented by the transformer... they do not "make" the impedances... take all the load off the secondary and you will have essentially an open circuit and the tubes cannot develop any power at all... so the tubes must have an impedance presented to them in order for them to function.
the MIT illustration on p19 demonstrates once again that the load that each tube works into (in this case a pure Class A1 circuit) is raa/4. Which also follows analytically from the now fimiliar turns ratio (or impedance ratio) formulas.
I.E., double the turns quadruple the impedance. Or alternately, halve the turns and quarter the impedance.
Otherwise, like I said, this is quickly becoming a rehash of the same arguments that were bantied about six weeks ago...
I thought I had something "new" to add... in the sense that I had found what I would have considered reliable authoritative articles written by the RCA engineering staff which addressed this issue... and posted up to simply make people aware of other literature and technical references out there.
also... in the meantime I had purchased John Broskies excellent PP tubecad program which also sez that each tube in a PP application sees as it's load impedance raa/4 because each tube only sees half the primrary winding and hence will be loaded by only one quarter of it's end to end impedance.
cheers,
MSL
"I've seen the article you make reference to before. Note that it looks at not how power is distributed (or combined) across the two series connected loads (i.e., porimary halves) by the output transformer.. meaning it does not show you the plate curves and the loads across the plate curves for each tube... nor how the tubes develop power into a load... rather it describes the condition which occurs *after* each half pri has developed power across a load and the transfomer has combined these two series connected loads... "The tube would not give the hair off a possum's tail to know what the transformer or anything else is up to.
It sees a load z = v/i . In this case v is 1/2 the v across the entire primary.
Anything inconsistent with this is either double-talk or ignorant nonsense... or ignorant double-talk.
some from column A, some from column B.regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
Never said it was a comprehensive or convincing one. I also figured you had already seen it. I had also seen the article from Pearl. Up till then I only knew them from tube coolers and never knew they made OPT's.This whole discussion reminds me of one we had here about a concertina phase splitters balance. I jumped right out there and said that since there was an impedance difference between the signal taken off the plate and the one taken off the cathode it couldn't be truly balanced. One can find numerous respected references that say about the same thing. But in the end they are wrong because they all fail to look at the whole situation and instead tried to look at each signal by itself as if the other one wasn't happening to the same tube at the same time.
I don't think we can get a proper understanding by trying to look at one tube at a time in a push pull circuit where they work together. So to me the point is moot. Frankly I'll just look in a book and see what plate to plate impedance is advised for a given tube and run with it. Usually going with twice what is advised for a SE(tube) configuration ends up being about the same. I guess I am easy and just go with what has worked for countless folks in the past.
Oh just saw this, Lundahl's data sheet states, "In class A each tube sees 1/2 and in class B sees 1/4 of transformers primary impedance". So I guess we might call that a second published source...of sorts. But really who cares:) I can see valid points on both sides. Certainly the majority of the literature says Raa/4. And anyone who wants to say Raa/2 has to also say "only under strict and ideal class A operation". However I do agree that when you pull one tube in the pair the remaining tube simply has to work into a different impedance. I have seen AB1 amps run okay with one dead tube up to a certain volume level. I mean you could certainly hear something wasn't right but they did function.
In the end all that matters to me is the ability to pick a suitable transformer and I don't think that's really all that much of a problem. Triodes do well with quite a wide range. Frankly it has always seemed to me that pentodes would actually do better with a much higher impedance than the norm but I reckon trying to make a 100 watt 20K transformer just isn't feasible.
And hey, again, please accept my apology regarding the post about Damir's moniker. All I can say in my behalf is that I was mistaken and should have looked more fully into the matter before accusing you (or anyone for that matter) of such a thing.
One of these days I need to ask you some questions about grid chokes when fixed bias is used. I assume your forum is okay for that?
Russ
of folks who have gotten this issue wrong.Take any example you want, none of them maintain a balance of power if you load each tube with a-a/4.
Start anywhere you want, establishing turns ratios, secondary load, and input ( to the primary ) voltages and the answer is only right for a-a/2.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
the part of interest from NC:Then, on the composite curves ( which are drawn straight ), a composite load line is dreawn which represents a plate to plate value of 6k8 Ohms. From this load line the dottd curves are drawn, using the original curves to represnet the load line applied to each tube by this composite load. This is the practical working load for each tube.
Earth to Mike! what is the slope of the line called the 'practical working load for each tube? ( I know it changes, but from each pair of points, you can calculate a slope to answer this question ).
hint: it is *NOT* a-a/4
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
I said 'second paragraph'. That means the one right after the first. Can you find the first paragraph? it is at the beginning...The part of interest is its last sentence, perhaps the next to last as well. You can read, I needn't post it again.
Slope of the dashed line in ohms please. volts per ampere yeilds ohms.
by inspection it is clearly not as steep as the a-a/4 line which is solid...so, what is its slope?
Just because there were other folks who got this wrong is no excuse for you ( or anybody else for that matter ) to do the same. None what so ever.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
why not try working out Steve Bench's example as provided on his website? Wherein the anode to anode impedance is 8800 ohms while each tube works into a primary reflected impedance of 2200 ohms.Maybe Steve's examples and methods will be less confusing to you.
Mike
what do you know of confusing?I may go and poke a proper hole in Mr. Bench. Do you claim it is impossible?
You ignore all from NC which indicates you are wrong, why would this example prove any different? Does not look worth doing. I believe you will keep your head in the sand for as long as you wish, and not a second less. Just let us know when the time is up, and we'll talk.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
Doug proclaims;"I may go and poke a proper hole in Mr. Bench. Do you claim it is impossible?"
never shy nor ambivalent about his own greatness....that's what I find so dinstinctive about Doug.
the compliment is accepted.As far as discussing this with either Mr.Bench or Mr.Broskie, I am certain that Iff they are indeed wrong, they will be big enough to see that.
It would not take a very extensive measurement set-up to look at what is going on with a PP 2A3 amp. Measure the individual tube currents and voltages, along with output power and the result will quickly become quite clear. Each valve is working into a-a/2. you will not need to apply a big signal, and plate z will not change much, and perhaps change the sharing of load as is predicted in RDH3.
It should be noted that the load line NC called attention to as being the practical working load for each valve is a-a/2 and that when compared to the plate z at its operating pionts follows the RDH3 prediction fairly closely. There is room for some error, the curves are not a detailed plot. It is detailed enough to see that it is *NOT* a-a/4, no matter how many times you, or NC say it is.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
As Norman points out you cannot compute or read the values directly off of the graph... since the two windings are series aiding opposition...some of the voltages must be added and subtracted... same goes for the plate currents and etc...
Norman explains this in the third and succeeding paragraphs....
and given your use of his 6800 ohm example.... Norman states that
"...the overall plate load is 6800 ohms... but the load for each tube will be only 1/4 of 6800 ohms, or 1700 ohms, when drawn against the composite loadline.
And in the fifth paragraph (from the beginning) Norman talks about where the "actual load" in connected... and walks you through the turns ratios and impedance ratios to illustrate that 1/2 of the pp primary will have only 1/4 of the impedance of the whole primary.
msl
the explanation is quite clear, and with the figure, shows you are mistaken. There is no issue with the idea of half the primary, taken independantly from the other having a quarter of the a-a load.It is interesting to note that you cannot answer the question raised as to the slope of the line NC drew in dashes and referred to as working load for each valve.
It is the situation( as in a class A circuit ) where one side is *NOT* operating independantly from the other that has confused you( and a few others ) so thoroughly.
that your analysis does not maintain conservation of energy *OUGHT* to be enough of a red flag to indicate that you have made a mistke somewhere. For a good Engineer it would be, not that you need to be an engineer to understand this stuff. Just careful and thorough.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
We went trough some literature citates before, and I gave some references, too. I didn`t mention, for example "Valve amplifiers 3", pages 461-463.
Why is confusion? Composite load-lines are somewhat complicated, and second - difference between loading of the both half of the primary vs. loading only one half of the primary.
Your example with CT secondary (or two secondary windings) is the best for understanding this - I`ll repeat this:
- let`s say we have 300 turns primary and two secondary of 10 turns. Turns ratio between primary and one or another sec. is 30:1.
- when we connect one 4-Ohms speaker on one or another secondary (only one, other stays "open"), our tube at the primary "see"
(30^2)*4=3600 Ohms reflected load. Let`s say that it "produce" 100Vrms at the primary, and now our speaker "see" 3,33Vrms, and dissipate (3,33^2)/4 = 2,78W.
- we can connect our secondaries in series, and connect 16 Ohms speaker. Turns ratio is now 300/20=15, but reflected impedance is the same like in the first example (15^2)*16=3600 Ohms. Power is the same - 100/15=6,67Vrms, P=6,67^2/16=2,78W.
- but, we can connect TWO speakers of 8 Ohms, one across each secondary, every speaker now gets P/2=2,78/2=1,39W, AND our reflected primary resistance is the same, or 3600 Ohms. Every speaker "see" 3,33Vrms, and P1=P2=P/2=(3,33^2)/8=1,39W.
- in another words - full CT secondary is 16 Ohms, half is 4 Ohms or 1/4 and two halves 2 x 8 Ohms, or if the both halves are loaded, each "see" 1/2.
- this is it, very simple. No one proved that this is wrong. I get just lot of trolling and blablah.
.
This post is made possible by the generous support of people like you and our sponsors:
Top of Page ]
[ Contact Us ]
[ Support/Wish List ]
[ Copyright © 2025, Audio Asylum, all rights reserved ]
