Operating points theory...

[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

You will need to navigate to this page, and print out some graphs if you wish to follow me.

The 6SN7 Tube

What you will read here has to my mind never really been approached, I wish to share it to hopefully stimulate some discussion and further crystallise my theories, with partial thanks to paul joppa for "the standard"

the standard, btw, is paul's reference.

Pick an operating point, that is, a plate current and a plate voltage.

this is for small signal triodes, I have no idea if it works with power tubes, and I can't do pentodes as they have a screen current which complicates things.

Now our reference operating point, here, you read off the plate voltage, and plate current.

You have a voltage, and a current, divide V / R as per ohms law, and you get a resistance.

This resistance, paul labels the operating resistance.

If we wish to design a preamp, I would suggest using grid volts 3v bias, because cd players put out 2 volts RMS, which translates to 2.82 volts peak, so 3v should give us a bit of margin.

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 hope you have printed out your graphs of small signal triodes, try 12au7, 12ax7, 6dj8 and 6sn/sl7, they all appear to work.

Now some startling relations are apparent.

If you have say, 100 volts plate voltage, and just say for eg. 4ma. for 6sn7

If you draw your loadline to double the volts at 200v, and you can magically double the current to get 8ma.

Your load resistance is paul joppa's operating resistance, the instantaneos plate voltage divided by instantaneous plate current.

that is my double relation, now the loadline you can also calculate the load resistance by 200 volts divided by 8ma, also, this is the only value that this works as.

The load resistance works out approx 4x the plate resistance of the tube.

Now we can do the same for a triple relation, ie triple the plate volts to 300v, and the current works out to 3/2, or 1.5 times the steady state, idling, or quiescent value.

your plate resistance will be about 8 times the plate resistance, you cannot really go up much more, perhaps to...

4 times the plate volts, ie 400volts, this works out to around 12 times the plate resistance and is roughly as far as you would wish to go, and co-relates to the notion that triodes can be loaded up to roughly 10x the plate resistance.

that is the reason, your plate current will not go up by much, and you have a pretty flat loadline,

at this point, you have the maximum voltage swing, measured by the length of the loadline and with the plate volts on the bottom of the graph.

this hi load gives more distortion, too, which is why for hi loads you don't get owt for nowt and there is that payback.

now going the other way, we can similarly steepen the back to our double relation, or twice the quiescent/steady state/idle where we started from.

then we can go steeper and aim for more linearity, you should use a ruler and see the loadline getting steeper as you gradually rotate it through your operating point, which is fixed and the ruler rotates around it.

you can get nice round figures and after doubling it, go for triple the operating current, say to 3ma, the voltage on the plate will intersect it at 3/2 times the idle voltage, magic figures, because the tube is running with more current, it heats up....current makes heat, so its a hot operating condition

if we have this triple relation ie trebling the steady state current, the load is twice the plate resistance.

so that is approx our range for loading a triode, twice the plate resistance, to around 12x it, with some magic numbers in for fun.

I am only just crystallizing this theory, its incomplete and am working on the related numbers.

I don't know if its to do with childs law, but if anyone wishes to further help me develop it, please do.

I hope you will print out graphs and try playing with the loadlines and magic number relations and hope I haven't lost anyone, but you must do it with graphs, I have linked to them at the top, I can'at really print them here and show what I mean, so it will mean work for some ppl.

the "operating resistance " or instantaneous vols / instant current, is a nice load to go for, around 4x the plate resistance, or perhaps an average...mean, median, or mode, don't know which yet, which for a 6sn7 will give around 33k ohms, around 6 times the plate resistance for the tube its all up in the air yet and my theory is incomplete...

hope its helpful somewhat, play around and have fun....you wil have to work at it to understand this....

best wishes

[the_manta]

Just read your post and um...I don't know what exactly you want to point out
Maybe I just try to comment:

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.

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.

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.

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

[Miles Prower]

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.

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.

[DF96]

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.