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Re: Here is an old published example

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



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