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In Reply to: I have done the same, posted by j rodney on November 29, 2004 at 12:19:55:
"I have no idea where Joel is coming from as the RC coupling is pretty wide range in the amps I have seen, and, at least by 1940, the bandwidths are in the 40-10kHz range, -3db (fram manufacture specs). "That's an interesting point, though I admit I only have a partial grasp of the idea Joel mentions. But I would also point out that we haven't even talked about the tolerance windows that parts of the day exhibited. 20% tolerances and higher on big electrolytics. Did anyone expect anything like precision timing from RC networks built parts with such high potential deviations? Maybe I'm missing something here....
dh
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Follow Ups:
> > Maybe I'm missing something here.... > >
I still giggle about the idea that all caps sound the same. I brought one of my first projects to a friend with a well practiced ear. It was a linestage. We pluged in and listened and after about 30 seconds, he turned to me and accused me of putting in Solen metalized poly caps in as coupler caps. He was right of course, as I had done just that.He took pity on me, handed me some motor runs and said, 'solder these in, and listen again'. I did just that, and smiled to myself, wishing it had been me to put those oilers in in the first place.
I think the old 'lytic tolerancing was something like -20/+50%. With worst case resistor values, this represents about 2:1 diffeence in Tau value.
As far as proof goes, just because most posters believe it does make a diff., it is not absolute proof of the effect. There are folks who don't believe that the load applied to each of the valves in a PP Class A output stage is half the a-a load. That one was fairly easy to set out the reasoning and mathematical basis of. And then to step further into what corrections need be applied to improve the model. Did you really think you were going to get unanimity on such an issue?
let's try an easier one first: which is best, PP or SE?
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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"As far as proof goes, just because most posters believe it does make a diff., it is not absolute proof of the effect."I understand that. Nobody's ears are NIST certified.
"There are folks who don't believe that the load applied to each of the valves in a PP Class A output stage is half the a-a load."
I was really hoping we had seen the last of this one. Oh well....
"Did you really think you were going to get unanimity on such an issue?"
No, but I'm not sure that's ever the point.
Thanks.
Hey-hey!!!,
It was just an example of an unresolved issue. Just because I don't feel in any way unresolved about the proof I have measured on the subject does not require others to have the same feelings.It is a public place, it somebody wishes to present an oposing POV, they should plan to defend the idea that was put forth. It is one thing to start by saying I believe...but to claim as fact, and then get upset because the proof is a pile of BS and it is being contested in an enthusiastic manner is just plain foolish.
Compared to past discussions, it went fairly smoothly. It remains to be seen if I will get banned from posting in his forum again.
regards,
Douglas
!
Sector7g wrote:::::There are folks who don't believe that the load applied to each of the valves in a PP Class A output stage is half the a-a load. That one was fairly easy to set out the reasoning and mathematical basis of.;;;;
The following text is quoted from the following source;RCA Receiving Tube Manual
RC-17 (1954) page 20
"The method for determining the proper load resistance for triodes in push-pull is as follows: Draw a load line through Imax on the zero-bias curve and through the Eo point on the zero-current axis. Four times the resistance represented by this load line is the plate-to-plate load for two triodes in a CLASS A push-pull amplifier.""Expressed as a formula,
plate to plate load (rpp) = 4 X (Eo - 0.6Eo)/Imax
where Eo us expressed in volts, Imax in amperes, and Rpp in ohms."
If you have the earlier RCA Receiving Tube Manual published in 1940 the same text is found as quoted above on page 17.
Another reference for those considering wether each tube (or pri half) sees raa/2 or raa/4 is the following abbreviated text from Broskie's PP Tubecad program."Transformers are usually specified by their plate-to-plate impedance. This is actually their REFLECTED imepedance, as in itself a transformer has no intrinsic impedance, only a turns ratio btwn primary and secondary windings (just as a magnifying glass has a magnification ratio, but no intrinsic magnified image)."
"The full plate-to-plate impedance is not the impedance used in drawing a load line on the composite plate curves, as the composite tube only sees ONE QUARTER of this impedance. This makes sense, as each output tube only sees half the primary, wich means that it sees an effective winding ratio only half that of the plate-to-plate ratio to the secondary."
and, of course, *if* each tube (each pri half) sees only half the turns of the full primary then it's (each half's) reflected impedance will be only one quarter of the full primary's plate to plate reflected impedance.
further references on this issue can be found at the following urls;
http://members.aol.com/sbench102/composit.html (which is an excellent tutorial on PP loadlines) which states in part that;"The same rules we used in the single ended case apply, with one exception. The impedance you plot is the impedance AS SEEN BY A SINGLE TUBE. Thus it is 1/4 the plate to plate load impedance. Let's use a 8800 ohm plate to plate load. In this case, each tube sees 2200 ohm load..."
http://www.ax84.com/media/ax84_m225.pdf
http://www.audioexpress.com/resource/audioclass/ga400ac.pdf
and RDH on pages 575 and 576.
The RCA analysis is nothing more then handy little approximation of composite load line method. You can see that - they use load line through 250V/0 mA and another point - where vertical line through 0,6*Eo = 0,6*250=150V "cut" the zero bias curve, and this is Imax=0,2A. In their example, "optimum" load line Raa=4*(250-150)/0,2 = 2000 Ohms (!?). Everyone can see that this formulas and analise is made with maximum power in mind, they claim 10W from 2A3 PP pair, operated at 250V. This is different then usual SE OP (250V/60mA/Ra=2k5, P=3,5W). With Raa where each tube "see" 2k5 you can expect 7W, and this is Raa=5k.
Composite load line method is complicated, and it is easier (without much error) to just find "optimum" SE load line, and for PP use double value of our SE Ra, or PP Raa=2*Ra.
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Hi Damir:you wrote:
:::The RCA analysis is nothing more then handy little approximation of composite load line method.:::
wow. This seems to be a value laden statement if I have ever seen one. What the method does is to use the plate curves from a 2A3 and then give you step by step instructions on how to create a PP loadline for this tube (using the actual plate curves).If you don't like their example... then you could alter or modify the illustrated example... and the RCA tube manual has complete instructions on how one could modify or change the example for varying plate voltages, plate currents, grid bias, and primary impedance.
As opposed to being difficult... it appears to be relatively straightforward. And actually working through several examples would probably give most folks enough ease with the method to find it both practical and useful.
Steve Bench's method (and he also states that each tube works into raa\4) is a bit more complex. But again, best way to evaluate these models is to run through them and see "how they work".
Simpler yet would be to purchase the PP tubecad program which does most of the work for you. converting your input (your specify primary impedance from plate to plate) and, in the background, the program does all the calcs and etc considering that the load that each tube works into is raa/4. For $29 this is a deal and a half.
:::In their example, "optimum" load line Raa=4*(250-150)/0,2 = 2000 Ohms (!?). Everyone can see that this formulas and analise is made with maximum power in mind,;;;;and so we should throw out the whole method because you don't approve of the specific illustration or example that they used?
Does the method they use only work with the singular example they provide? That if you change any parameter you must throw out the method\model? Or does the RCA tube manual also show you how to modify\alter the the relevant parameters (plate voltages, currents, grid bias, load impedance, and etc)?
:::With Raa where each tube "see" 2k5 you can expect 7W, and this is Raa=5k.::::
the reflected primary impedance of each half primary will be 1/4 of the whole primary impedance considered end to end.
follows from ((n1/2)/n2)^2 times the load on the secondary will be only one fourth of (n1/n2)^2 times the load on the secondary.
:::Composite load line method is complicated, and it is easier (without much error) to just find "optimum" SE load line, and for PP use double value of our SE Ra, or PP Raa=2*Ra.:::no. if you want each tube to see a reflected half primary impedance equal to the value chosen for SE operation then the transformer's impedance from end to end must be four times the value you would have chosen for SE operation.
But.. the "beauty" of PP is that you do not need to use four times the value of load impedance as you would in a SE application for each tube... and that in PP you can get greater output power with lower distortion than in SE operation with a lower primary impedance for each tube individually. That is the classic argument for PP... note... again... I am presenting the classic argument in favor of PP operation NOT that I am necessarily endorsing said argument.
MSL
MSL:"The "beauty" of PP is that you not need to use four times the value of load impedance as you would in SE application for each tube...and taht in PP you can get greater output power with lower distortion then in SE operation with a lower primary impedance for each tube individually."How? This mean that both tubes "somehow" work together, no? And that their "work together" is somewhat different then one tube through the only half of the primary? That`s I`m "preachin`" about... :-).
I KNOW that you don`t advise your customers that they actually need Raa=20k OPT if they want 5k load for each 300B in PP pair... Raa=10k will somehow do...:-). (Note, this is a little fun, with :-) signs all over, don`t start... :-)).And about "PP Tubecad" - I don`t know the whole text and never use one. But "realistic" SPICE simulators you can download for free. Using one can be real "eye opener". Why? You can "construct" PP amp with, say 2 x 2A3 models and "see" all the DC and AC voltages and currents on every point in the diagram. You can place "oscilloscope" too and see phases and amplitudes of your waveforms. You can change Raa, see Bode-plots, distortion, etc. Without this pictures, many people have difficulties to mentally imagine PP operation. And you can use only one tube and half the primary...etc.
But, this is "dangerous" device... maybe after using one, somebody can start thinking that maybe I actually know what I`m talking about...huh...:-).
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from your previous:the reflected primary impedance of each half primary will be 1/4 of the whole primary impedance considered end to end.follows from ((n1/2)/n2)^2 times the load on the secondary will be only one fourth of (n1/n2)^2 times the load on the secondary.
This only works when considering the two halves independantly of each other.
This is not what is happening with a PP, class A stage.
Have you calculated the slope of the dashed line in the NC paper yet?
If you have, care to share the results? ( you can use the search f'n, Grover did just that and posted the results on DIY ).
Dougie wrote;:::This only works when considering the two halves independantly of each other.:::
::::This is not what is happening with a PP, class A stage.::::
accdg to who? Certainly not the RCA engineers who addressed this issue head on and state clearly that in PP CLASS A each tube works into raa\4 as it's reflected load impedance.Nor Steve Bench. Pick apart Steve's thesis. Show us where he steered you and everyone else wrong.
Nor Norman Crowhurst's article.
Nor John Broskie's text.
Funny... with seventy to eighty years (at least) of research and publishing you guys cannot find one published source who explicitly states and shows a PP loadline as being constructed or comprised (for each half) of a slope (i.e., loadline) equal to raa/2 (for each tube) as you propose.
Every bit of research that I have been able to explicate to date... shows and illustrates that each tube in a PP arrangement works into a reflected impedance of raa\4.
next, it will be suggested that we burn the textbooks.
msl
first of all, Mike, you asked for a quote, I gave it to you, and you ignore it. Do the footwork and see how wrong yo are about including NC in this gang who claims the single valve is seeing a-a/4 when run as part of a Class A pair.Nobody has suggested burning anything, only that they be recognized as wrong if indeed they are. I am not going to go through yor ref's and see which is which. It is of no intrest to me.
Your claim that one valve is operating independently of the other in a PP class A pair is interesting, and I think goes far in showing the depth of your misunderstanding of the issue at hand.
Russ also posted an article which shows what is going on. Their math contains no obvious errors, and plainly states that when run as a *PAIR* in PP clas A, the load is half that of a single valve( or when one valve is pulled from its socket so to speak ). The single valve is shown to be operating into a-a/4, and to refresh your mamory, nobody has said otherwise.
Answer the question you requested the quote for.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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God, this almost sounds like one camp is looking at a potential instantaneous value of a parameter, and the other camp is looking at the same parameter with a "long sample average" measurement.
Hey Dave, I don't think that is the case. The load seen by each triode certainly does vary a bit. In the RDH3 an equation is offered which relates this change to plate resistance(for Triodes ). I am not sure I buy the whole thing. The plate z does change across a swing, and varies by a factor of ~2.5it appears to be completely alien that the multi-grid valves would be operated at anything else but max power conditions. Plae z certainly varies a bit with pentodes from g1=0 to near-cut off at te other end. I have not seen many data sheets with PP class A pentode op points thoroughly docuymented very often.
The a-a/2 looks like a place to start correcting in order to arrive at a more comprehensive explanation. From first principles, like conservation of energy, it looks to be a reasonable place to start.
regards,
Douglas
!
OK, just a faint flashback to a moment I had learning about RF signals. Apparently not an appropriate parallel. Clearly this is a complex topic, probably with more than one way of arriving at a reasonably satisfactory observation. Thanks for the reply.
Doug wrote:::::The load seen by each triode certainly does vary a bit. In the RDH3 an equation is offered which relates this change to plate resistance(for Triodes ). I am not sure I buy the whole thing.::::
As John Broskie points out... there is no intrinsic pri impedance. Pri impedance is determined solely by the load placed on the secondary and the turns ratio btwn the primary and secondary windings.
impedance ratio equals (n1\n2)^2
in pp each half pri's impedance ratio would be ((n1/2)\n2)^2
thus if the load on the secondary is 600 ohms and we have a 4:1 turns ratio (a 16:1 impedance ratio) from full pri to sec then the primary impedance will be 9600 ohms.
each half primary will have turns ratio of 2:1. Thus an impedance ratio of 4:1. Multiply this impedance ratio by the load on the secondary and we get an impedance of 2400 ohms for each primary half.
A tube or tubes, a generator or generators each work into the primary impedance they don't "make" or "define" the primary impedance which is determined solely by the load on the secondary and the turns ratios btwn the primary (or any sub-portion) and the secondary.
Thus, this is why, we can "ratio" transformers and change the primary impedance... simply by changing the load on the secondary winding.
The generator or tube may be more or less happy working into a given primary impedance... it may load the generator more or less than an alternative value of primary impedance... and the generator may develop more or less distortion operating into a given primary impedance.... but the generator or tube "sees" this impedance as it's load and works into this impedance... the generator or tube does not "define" or "make" the primary impedance.
as regards the behaviour of the tube and it's plate resistance and the "movement" of it's plate resistance... Steve Bench addresses this issue in his article on PP loadlines. I put a hot link to his text below.
msl
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Yes, the reflected impedance for each tube is AA/4
And with only one tube in circuit, that's the only load for that tube to see.Do you think that each tube is seeing ONLY the reflected impedance of it's half of the primary and the other tube makes no difference in the load as seen by each tube?
So the load for one tube is the same with or without the other tube in circuit(in a PP pair)?
That would be impossible. But would have to be the case. Otherwise the answer HAS it be something other than AA/4.
From RDH3 Load for each tube is AA/4 X (1+ Rp1/Rp2)
Thanks....Tre'
He gets it....and She gets it...and They get it...why don't You get it?some things are worth explaining more than once. No student left behind and all that.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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Doug wrote:::::He gets it....and She gets it...and They get it...why don't You get it?::::
::::some things are worth explaining more than once. No student left behind and all that.::::
have you straightened out Steve Bench yet? Has Broskie been converted to your religious crusades yet?
Too late to do any good for Crowhurst... unless, of course, he's stuck in purgatory till you are able to make him see the light....
and there is a whole brigade of RCA engineers... who had the utter audacity to say that each tube in PP class A sees raa\4....
ya got your mission cut out for ya Doug.
Frederick Terman?
Gordon Rankin?
MSL?
In a Class A PP amplifier with only one of the output tubes in place, What is the load for that tube?Thanks....Tre'
Gang,First off... You cannot look at 1/2 the solution to find the entire argument. You must concentrate on the entire system and how it works.
This is not a new argument, this was done by the entire engineering community more than 60 years ago.
Why now do you think that you have stumbled onto something new?
Douglas dragged me into this and I made this point in an email I just sent him.
~~~~~~
Douglas,First you cannot look at PP as 1/2 of the equation then add the other. You have to look at it as a total system. You could do it your way if you did dual SET amplifiers with common B+ and parallel secondaries.
BUT.... in that relationship the tubes are not running on 1/4 of the impedance because the primaries are not in parallel, they are totally seperate.
There is just a ton of writing on this Douglas. I don't think you going to change the minds of 1000 of engineers who wrote about this for years.
This is the way it works, live with it.
~~~~~~~
Well I got work to do...
(GR)First off... You cannot look at 1/2 the solution to find the entire argument. You must concentrate on the entire system and how it works.exactly, which is why i am confused about all of this discussion about 1/2 the primary. for the class A ac circuit the CT is not there electrically, so the half of the winding that gives you 1/4 a-a is not there either.
dave
Well, I mathematicaly described 30:1 turns ratio SE OPT with CT secondary, see below. In short, you can connect one 4 Ohms speaker on the half of the sec., one 16 Ohms speaker across the whole sec., or two 8 Ohms speakers across both halves. In all cases, our tube on the primary will "see" the same reflected impedance and "gave" the same power. Well, let`s reverse the thing:1.) We have, say 300 turns CT primary (150+150 turns), and 10 turns secondary. Turns ratio from whole primary to secondary is 30:1, or from halves of the primary 15:1.
2.) Let`s connect the speaker of 8 Ohms across the secondary. Then our reflected impedance to the whole primary is (30^2)*8 = 7200 Ohms.
Reflected impedance across the one or another half of the primary is (15^2)*8=1800 Ohms, or 1/4 of the whole primary.3.) Let`s connect the tube, or better - for simplicity AC generator that gives 100Vrms no matter of load. We can connect the generator through the whole primary, Raa=7200 Ohms. Our 100 Vrms will give the power Ppr`=U^2/Raa=100^2/7200=1,3889W. That power is "transferred" to the secondary load. Usec is Upr/turns ratio, or Usec`=Upr/30=3,333Vrms, and power is Psec=Usec`^2/Rsp=3,33^2/8=1,3889W.
We can connect our generator across only the half primary, our generator "see" Raa/4=1800 Ohms, and gives power Ppr``=100^2/1800=5,555W. On the speaker - Usp``=100/15=6,667Vrms, Psp``=6,667^2/8=5,555W.
4.) Now, we feel like goin` PP, and connect two identical generators with 100Vrms output, but in antiphase. Connection - one wire from the each generator on the one side of the primary, then remaining two wires connect together and grounding (the same like with tubes). We can now say that we have two generators in series, in antiphase, connected on the primary, and each one gives 100Vrms.
When we measured AC voltage across our primary, we measured 200Vrms!
Now, Ppr```=200^2/7200=5,555W, and our Usec```=200/30=6,667Vrms, and Psp```=6,667^2/8=5,555W.5.) The question is - what load each tube/generator in PP A class (both tubes working) "see"?
The Law of power conservation - Ppr=Psec. Every tube/generator gives half of the total voltage and half of the total power on the primary (and transferred to the secondary load).Usec```^2/Rsp = Ugen1^2/Rload1 + Ugen2^2/Rload2
We know that Ugen1=Ugen2=100Vrms, we know Usec```^2/Rsp=5,555W, and we know that Rload1 = Rload2 = Rload1,2 = unknown resistances of the one and another halves of the primary in PP case.
Then we have: 5,555 = 2*(100^2)/Rload1,2 and
Rload1,2 = 20000/5,5555 = 3600 Ohms (half of the total Raa=7200 Ohms).
We can express this with turn ratios, the result is the same :
1/Rsec = 1/(Rload1*(Nsec/Nhalf pr1.)^2)+1/(Rload2*(Nsec/Nhalf pr2)^2)
1/8=1/(Rload1,2*(10/150)^2) + 1/(Rload1,2*(10/150)^2)
and from that Rload1,2 = 3600 Ohms.
6.) Summary and conclusion:
Reflected secondary load across the whole primary is the square of the turns ratio times Rsec, or in this case Raa=7200 Ohms. This load (7200 Ohms) "see" the tube connected across the whole primary.
When we connect the tube across the only half of primary, our tube "see" Raa/4, or in this case, 1800 Ohms.
When we connect two tubes in PP A class, each tube "see" half of the primary resistance, Raa/2, or in this case 3600 Ohms.
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since Mike can't answer this, maybe you can. Iff those Bell Labs Engineers taught you anything of use...Take a Class A, PP power stage. It is working into some load a-a, reflected by the secondary. Each of the tubes is working into some load. Pull one valve, what happens to the load on the remaining one?
up, down, or remains the same will be a sufficient answer.
core saturation is not an issue for this case.
Let's see if you have any better grip on this that your Iron supplier does.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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well your push pull amp ain't going to work very well with just one tube...with the requisite number of tubes in the PP amp... in the proper arrangement. then;
each half primary will have an reflected impedance of (n1/2)/n2)^2 times the load on the secondary.
the whole primary from end to end will have a reflected impedance of (n1/n2)^2 times the load on the secondary.
As Mr. Broskie so aptly pointed out... transformers don't come with instrinsic impedances. The primary reflected impedance is simply the product of the turns ratio squared times the load on the secondary. For each half.. it is the turns ratio squared (from half pri to sec) times the load on the secondary.No load on secondary... no reflected impedance on primary. Place a load on the secondary and the impedance ratio (times the load on sec)is solely determinative of the magnitude of the reflected primary impedance.
A tube or tubes, a generator or generators, will then work into the primary (which the tube or generator will see as a load)...
the tube or generator might be happy with the load or it might not be happy with the load presented to it. It may under certain conditions based on the load and it's own electrical capabilities experience current or voltage clipping. Or it might distort like a banshee.... with the "wrong" primary impedance...
but... the primary imedance is ONLY determined by the load on the secondary multiplied by the applicable impedance ratio.
Tubes work into this reflected primary load impedance... they don't create a load impedance (assuming an ideal model) on their own idividually or in concert with other tubes... if there is no load impedance reflected to the primary then the tube (or tubes) will not deliver any power whatsoever...
i.e., the tube can only work into the load that is reflected onto the primary or each primary half winding by the secondary as the case may bee. And that is the magnitude of the load that each tube will see and work into.
yes I would put a proper load on the secondary. If only one tube is in place then the only load that is there for the tube is the reflected load from the secondary on the one half of the primary equaling 1/4 of the plate to plate imp. Right?Thanks...Tre'
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Have Fun and Enjoy the Music
"Still working the problem"
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Hey-Hey!!!,
well, if the load applied to one valve in a class A PP pair is a-a/4...which valve is it? is it the case for both? at the same time? are both valves loaded by a-a/4? can't possibly be right, since the load from end to end is a-a and two valves are doing it...probably magical transformer action that allows each valve in a PP pair to be loaded by a-a/4...I have seen a 3 YO argue better than this, and about comlex thigs like bedtime.
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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where the Church made a very good job of controlling the information.philosophy: questions w/o answers
religion: answers that may never be questioned
Fortunatley, the Church is no longer in charge and it is possible to question and challenge answers provided in the Old Texts.
that the power delivered to the speaker from each tube will be in accordance with AA/4.
but at what distortion will that power be delivered? A distortion relative to the load line that is represented by AA/4 (the load presented to the tube only by the reflected imp. of one half of the secondary (aa/4) with no accounting for the load that is in series as presented by the other half of the primary (which is seen by the tube as well)? And would lower the distortion by raising the imp. seen by the tube giving a more horizontal load line?You may or may not be able to follow the above. And if you can't, I don't blame you. But if the load with only one tube in place is AA/4 then the load with both tubes in place HAS it be something other than AA/4
Tre'
Have Fun and Enjoy the Music
"Still working the problem"
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it seems an impossible task. Why not let GR weigh in on this himself. I doubt he will...you removed me from your customer list for less. With the result I had to get my own stuff made.So back to the NC article? what is the slope of that line Mike? the one NC says is the working load on each valve? the one which is clearly *NOT* a-a/4? regardless of how confused the rest of the article got, that part is clear.
You have misread NC once agian it seems...
regards,
Douglas
!
mmmmmhhhh...Forbidden Donut!!!
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:::Why not let GR weigh in on this himself.::::He has, in the last roundabout on this topical matter GR stated that each tubes in PP sees raa/4. Look it up.
:::I doubt he will...you removed me from your customer list for less. With the result I had to get my own stuff made.::::
Total BS. When did I remove you from our customer list? In fact, I continue to treat and "service" you well... providing information on authentic Peerless products which you come over to the MQ board and ask for details on. Within the last week you were asking for details on a Peerless output trans on our board and I took the time to research the answers and get back to you. It's right on our MQ forum for all to see.
And while I take the time to provide answers to you as a service. All the while you were soliciting sales for hot rodded copies of tranneys that I freely gave you information on. Shame on you.
You really do have a warped sense of reality. In fact, I provided you much of the information needed to design and build a 4 or 5 hundred watt output transformer... that Norman Crowhurst did for me before he passed away in the very early nineties. And at no charge to you whatsoever. I barely got a thank-you.
Your not barred or "removed from our customer lists".... not yet anyway :=)..,
and at this point.. since you seem to be solely focused now on making personal attacks I will ignore you (i.e, do not feed trolls) as I have sucessfully done in the past.
bye dougie
I do feel the need to take some of your nasty accusations apart bit by bit.**All the while you were soliciting sales for hot rodded copies of tranneys that I freely gave you information on. ***
I am not soliciting sales of anything. I had a Peerless OPT unwound so I could have a pair wound for my own experimentation. They turned out to be remarkably inexpensive when wound with a fast NC machine.
You or one of your goons accused me of stealing intellectual property from you. I stole nothing. Then you stated that it could not possibly be done well. Chop shop was the wording you used. Scam was another. I don't care if you *NEVER* respond to one of my posts again. Your behaviour is open to anybody who cares to click on one of these links( until you twist the arm of those with the delete button again ).
Yes you did send me instructions for winding an opt, allegedly authored by NC...I have my doubts. I have not broken any agreement I made regarding it until now when I confirm your sending me something fitting that description.
I have destroyed it w/o copying it.
regards,
Douglas
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Huh, Mike, without you we`ll discuss only about tube rolling, etc. This is real fun today, after many boring weeks. Only disadvantage is that you are made of such inflammable material, and party`s ending:-).
OT - know someone who can return my moniker? :-)
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He is one of the really good guys and your fight was never with him now was it? You have made your point. I do see your side of the equation and can understand your feelings but again Damir wasn't the reasonable party. Enough is enough. And Damir, if you get your moniker back do consider registering it. The asylum is worth the support.
Russ:I am disappointed that you would have taken such an ill founded, unwarranted, and inuspportable position.
But, as Bob Dylan said... "it ain't me babe your looking for"
and of course, Damir_the real one could (to protect himself) register his handle... but it is his choice... so far, even after being burned twice, he'd apparently rather cry after the fact than be sensible and register his namesake as thousands of other people have done.
just sorry to see you (of all people) join in the juvenile gamesmanship that plagues Tube DIY.
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But, as Bob Dylan said... "it ain't me babe your looking for">Then please forgive me for stating so. No point in me screwing up an apology by saying something more than I am truly sorry.
I agree and thought I said as much above:)
Again you will have to forgive me as my intent was quite the opposite...but alas I lack socail skill and grace and will forever suffer from athletes tounge.
Russ.Your apology means a lot to me. So I thank-you.
same difference...and it is not a large leap to assume you had a hand in it. Regardless of your denials both private and public, I remain convinced as well.
regards,
Douglas
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mmmmmhhhh...Forbidden Donut!!!
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Haha, Mike I didn`t cried in much more serious situations then your identity games. I just make a little fun of it, without malice - believe it or not.
And little numbers in the top of the message mean something. The same as with numbers in YOUR 4/16/2x8 Ohms example.
But, let`s end it here, I don`t want that technical discussion and fun outgrow in something else...
P.S. Thanks, Russ - you have a good intention.
Bye, it`s 22.40h here...
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why don't you look at the question I put to you a bit ago. think for yourself, and examine what is put before you.so what happens ot the PP stage when you pull one valve?
or what happens when you put the missing one in?
If it makes any difference, I'd rather you look at the latter. The single valve working into half the turns, is seeing a load of a-a/4. what happens when we put the second valve in?
none of this has anything to do with the magnitude of the load applied to the secondary, and no difference of opinion has been put forth on that aspect. It is a fairly direct question you have been dodging with great regularity, clouding the waters with partial truths and literature ref's.
probably 'later'.so why does the law of conservation of energy not apply to your claim of a-a/4?
The measurements were posted, taken from a real OPT. no matter what sort of literature you dig up to confirm your opinion, it is not going to change what is actually happening. I see that you have not made ention of NC, whose article actually showed that he contradicted himself a few times, in addition to confusing a number of folks who should know better( yourself included ). NC made a one sentence notice of the actual load, and provided a plot of said load.
It was *NOT* a-a/4.
If you're going to stick your neck out, why not look around first. RDH3 on one of its pages gave a description, and equation and refrence to something other than a-a/4, and that equation matches the actual data in the NC article in the Glass Audio page. I have not looked after their reference yet.
regards,
Douglas
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mmmmmhhhh...Forbidden Donut!!!
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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?
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
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mmmmmhhhh...Forbidden Donut!!!
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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...
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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
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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
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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
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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.
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."
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
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.
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some from column A, some from column B.
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?
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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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
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
!
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.
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
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mmmmmhhhh...Forbidden Donut!!!
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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
!
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.
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!!!
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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.
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