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In Reply to: Here is an old published example posted by Russ57 on December 1, 2004 at 10:03:15:
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
Follow Ups:
"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!!!
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