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Re: RCA and other stories... :-)

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


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