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diyAudio Member

Join Date: Nov 2006
Location: Munich, Bavaria
Just read your post and um...I don't know what exactly you want to point out
Maybe I just try to comment:

Quote:
 You have a voltage, and a current, divide V / R as per ohms law, and you get a resistance.
(I think you ment V/I, not V/R. I write Ua/I)
And that's almost everything. This just tells you how much voltage you loose at specific current. Just static data, not any information on dynamic operation. You must take your look on ∂Ua/∂I to get any information about operating when a changing signal is applied. And not just that, ideally you want to have constant parameters (like the inner resistance ∂V/∂I), so you also must watch (∂^2)Ua/∂(I^2) (second derivation) which should be ideally constant !
And that's the reason why those simple Ua - Ia graphs are just the beginning of the development.
You have to look at the others, too:

See how our parameters change ! (ri=∂Ua/∂I ; gm=∂I/∂Ug ; D=∂Ug/∂Ua ; D=1/µ ; D is called "punch through" and tells how much you would have to change grid voltage to get the original current when Anode current changes.) Yes, when you look at those parameters you see that ri*gm*D=1 since the derivates simply cancel out, but these are all functions of Ia and Ua ! So ri[Ia,Ua]*gm[Ia,Ua]*D[Ia,Ua]=1. So these parameters aren't static. It is more like a huge ugly partial differential equation.
Just with your eyes you can't see this from the simple Ua - Ia graph.

Quote:
 If you have say, 100 volts plate voltage, and just say for eg. 4ma. for 6sn7
I know, this is an example, but 4mA would be a pretty bad OP. Just see my picture above. µ and ri are heavily varying at that current. That's why a 6SN7 should be operated at over 10mA. It much more quiet in that region as you see, even gm varies less.

Quote:
 There are generally 2 contradictory load resistances we can design by, 1/ is for maximum voltage swing, this will yield a flattish load line, with its attending hi value load resistor. 2/ is for maximum linearity, of lack of distortion, which will yield a steep loadline, the reasons will become clear.
I don't really share your opinion there. What you have to separate is a Power stage and a simple voltage amplification stage. I don't want to point out the mathematical way, but you can make an Ansatz with "totally consuming power" and "power which the tubes deliver to the next stage/speakers". Derivating this Ansatz and equalizing to zero leads you to some maxima and thus you know how to squeze out the max. power from a tube. That's basical the way how tube manufacturers calculate the optimum Ra and Raa values in their datasheets. I think that is not what you want. Simple voltage amplifier stages are calculated different, they're just optimised for low distortion respecting max. input signal. Just like shown above with different graphs.

Quote:
 1/ is for maximum voltage swing, this will yield a flattish load line, with its attending hi value load resistor. 2/ is for maximum linearity, of lack of distortion, which will yield a steep loadline, the reasons will become clear.
I don't think this is even right. Think....a µ-follower ideally has a flat loadline and has very low distortion. So this completely disagrees with what you said. And when you look at the my posted graph again, you see that what I said is right. A constant current source as Anode resistance forces constant current und thus we have just a vertical line at a specific current where all parameters are (at the first approach) constant.

I could write for hours on this topic, but I have plenty other stuff to do in university
Regards, simon

diyAudio Member

Join Date: May 2005
Location: USA
Blog Entries: 6
Quote:
 Originally Posted by lt cdr data For those baffled by load lines, I have made some startling discoveries. Please correct me if I am wrong, which I never am
First time for everything.

Quote:
 There are generally 2 contradictory load resistances we can design by, 1/ is for maximum voltage swing, this will yield a flattish load line, with its attending hi value load resistor. 2/ is for maximum linearity, of lack of distortion, which will yield a steep loadline, the reasons will become clear.
This is just not right. Steep loadlines (low Rp's) give the worst distortion performance, especially where triodes are employed. This also applies to pentodes as well.

Getting the distortion down by making Rp as large as possible to flatten the loadline also applies to pentodes. Unless you have some special requirements for wideband performance (like a video amp) then make that plate load as large as possible. Small signal pents often use plate loads of 100K and more, and operate at some very small plate currents and Vpk's. (Consistent, of course, with drive capability and voltage gain.)

Even with power stages, you'd rather operate into a larger load resistance and sacrifice some output in favor of improved distortion performance.
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 29th June 2011, 11:34 AM #4 diyAudio Member   Join Date: May 2007 The OP may be confusing static resistance (Va/Ia) with dynamic resistance (dVa/dIa). As it happens, for many triodes at typical bias settings, these two values are not too different in numerical value but they are two quite different concepts. You can't just say I have 2.82V peak, so pick 3V bias. Its not as simple as that. True, 3V bias is the minimum (for a grounded cathode stage) but it might not be the optimum. For some higher mu valves 3V bias will be at or beyond cutoff! Its good that you have started to understand load lines, and grid bias. You still have further to go.

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