How to calculate gain and draw loadline for tubes with cathode resistors?

Again, Valve Wizard comes to the rescue. This is the article I use: http://www.valvewizard.co.uk/Common_Gain_Stage.pdf

This is to find the gain and plate resistance of a triode with unbypassed cathode resistor Rk. All explained in p29 to p31. This is the detail derivation I wrote using the article to fill in the missing step so you can see step by step LOGICALLY derive the result. Using ONE SINGLE model to explain the gain and the equivalent plate resistance due to adding the unbypassed Rk.

Valve Wizard gain and Zout of triode with Rk.jpg

This is logical, easy to understand and straight forward. I can write the derivation of the Gain and plate resistance in one page using one single Thevenin equivalent from beginning to the end. I am having so much faith on Valve Wizard even though they are mainly for guitar amps. The guy know his electronics.
 
I believe I found something wrong on RDH4, I am still looking for output impedance of tubes with NFB and I went back and read page 308 to 310 of RDH4. The way RDH4 derive equation (2a) is wrong and the result is wrong:

One cannot mix open loop gain equation (1) into closed loop like that. Also A'=Eo/E'i is just simply WRONG. Then substituting (1) into A'=Eo/E'i is just wrong. You cannot mix Eo from open loop equation into closed loop equation like this. The result (2a) is wrong.

RDH4 problem 1.JPG



This is the standard way of deriving closed loop NFB equation:

NFB equation.jpg

This is supported by:

https://en.wikipedia.org/wiki/Negative-feedback_amplifier

http://www.electronics-tutorials.ws/systems/negative-feedback.html

https://www.electricaltechnology.or...-and-negative-feedback-amplifier-systems.html

http://www.talkingelectronics.com/Download eBooks/Principles of electronics/CH-13.pdf


You cannot trust any book is perfect. Always trust but verify, particularly an old book that is written in the 40s. Not just this RDH4, I challenged the famous book on Phase Lock Loop by Roland Best and he offered me a new copy!!! From my experience of reading a lot of books, there are very few books that do not have errors. Even the 6th edition of the famous power amp design book by Doug Self have numerous errors in his 6th edition. And yes, I did challenge him in DIYaudio before.

That's why I always take my time go through step by step to verify every forumla, always buy two or more books per subject to cross verify. Authors are just human, human make mistakes. Just human nature.

I am still working on equations (13), (14) and Fig.7.3 in page 310 of RDH4. Fig. 7.3b just does not look right for me the same reason as equation (2a) in p308.
 
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I think the main difference here, is Wikipedia, etc, have applied a convention where Beta is an absolute value, representing only the magnitude of negative feedback (i.e. a positive number), whereas RDH4 is using an obsolete method where the value for Beta itself is negative (look at 3a)

In your equation Eo=A(Ei-BEo), this would only be true for a positive value of Beta, but RDH4 uses a negative value in 3a,

I'm not sure if this is an error, or a demonstration of how conventions change over time,
 
I think the main difference here, is Wikipedia, etc, have applied a convention where Beta is an absolute value, representing only the magnitude of negative feedback (i.e. a positive number), whereas RDH4 is using an obsolete method where the value for Beta itself is negative (look at 3a)

In your equation Eo=A(Ei-BEo), this would only be true for a positive value of Beta, but RDH4 uses a negative value in 3a,

I'm not sure if this is an error, or a demonstration of how conventions change over time,

For resistor divider network, beta cannot be negative. Only thing that can make the term negative is if the output is INVERTED.

It is an error, the sign change from all the articles I sited. It's just that simple like I wrote how to derive it. This is a very basic equation from any books about opamp and closed loop feedback control theory. I did not invent it. And if you follow p308 step by step, you see it's just wrong to say Eo = A' X E'i. there is not other way about this. Don't believe me, just look up any book.
 
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Notice equation 3a in RDH4 states Beta is a negative quantity.
That's besides the point, putting in (1) into Eo=A' X E'i is just wrong. You can do two wrong to make it right, that does not make it right. The way I wrote is the standard way and is correct verified by many articles.

It is not just you have the right eqution, one has to know how they arrive the equation to verify. One cannot trust anything blindly. Eo=A' X E'i is just wrong. Put in some real numbers and you'll see immediately.

I am still looking at Fig. 7.3, I don't want to say anything until I go through all the equations.
 
For resistor divider network, beta cannot be negative. Only thing that can make the term negative is if the output is INVERTED.

It is an error, the sign change from all the articles I sited. It's just that simple like I wrote how to derive it. This is a very basic equation from any books about opamp and closed loop feedback control theory. I did not invent it. And if you follow p308 step by step, you see it's just wrong to say Eo = A' X E'i. there is not other way about this. Don't believe me, just look up any book.

Beta is negative there, because they derived a formula for any feedback (positive or negative), and then make beta negative to show it is inverse feedback.

The newer forumulas are already made for inverse feedback and just use a value for beta. Obviously this stuff has been developed and changed since RDH4, but I still don't think it's technically wrong.
 
I told you already, a lot of my stuff is already published, and on AK. Some in great detail, some not delivered in detail deliberately as the presentation intent was on general topology. Where are your original designs? Where are they written up? Has anybody else built them? Has anybody who built them, liked them? Answers in detail please...:)

It is indeed 'show and tell' time, and you have shown exactly nothing.

'Talking theory' requires that definitions of terms not be butchered. This is exemplified by your issues understanding RDH4 among others.
cheers,
Douglas

I'm building Alan's solid state amplifier design. No idea how it's going to sound, but it definitely looks like a technically sound design. Aspects of the design definitely tend toward overkill, but the resulting specifications seem to back up these design decisions. I'm certainly interested to hear it once it's all done.
 
Beta is negative there, because they derived a formula for any feedback (positive or negative), and then make beta negative to show it is inverse feedback.

The newer forumulas are already made for inverse feedback and just use a value for beta. Obviously this stuff has been developed and changed since RDH4, but I still don't think it's technically wrong.
Hi Max

It is the Eo=A' X E'i that is wrong. A' is defined as closed loop gain, E'i is defined as Ei- beta X Eo which is differential input voltage. For closed loop Eo = A X E'i, there is not other way to look at it. A'=Eo/Ei .

Once this is wrong, all the derivation that follows are all wrong. Then the article substitute in the open loop equation (1) Eo=A X Ei to derive (2a), you cannot do that. So it's two wrongs to get the right answer.

The one in p310 also start from Eo= A' X E'i also, it cannot be right. I am still looking for other ways to derive r'p= rp/( 1+ A X beta). I know this is correct, just how do you come up with it.

Even though this is already out of the area of this thread, but this is very relevant to what we've been talking in our PM, that's the reason I really hone in on this one. If you have another way to derive it, please let me know. I am the kind that can't let go and have to get to the bottom. I really think this is the best way to really understand the circuit and provide inspiration in the design.
 
Hi Max

It is the Eo=A' X E'i that is wrong. A' is defined as closed loop gain, E'i is defined as Ei- beta X Eo which is differential input voltage. For closed loop Eo = A X E'i, there is not other way to look at it. A'=Eo/Ei .

Once this is wrong, all the derivation that follows are all wrong. Then the article substitute in the open loop equation (1) Eo=A X Ei to derive (2a), you cannot do that. So it's two wrongs to get the right answer.

The one in p310 also start from Eo= A' X E'i also, it cannot be right. I am still looking for other ways to derive r'p= rp/( 1+ A X beta). I know this is correct, just how do you come up with it.

Even though this is already out of the area of this thread, but this is very relevant to what we've been talking in our PM, that's the reason I really hone in on this one. If you have another way to derive it, please let me know. I am the kind that can't let go and have to get to the bottom. I really think this is the best way to really understand the circuit and provide inspiration in the design.

The way it is presented makes sense, because they are normalizing everything for the same value of Eo, the idea is to find what new value of A, and new value of Ei are required to get the same Eo, for a given value of B.

Confusing? Yes, maybe. Wrong? I think not. It makes logical sense to me, for Eo to be the fixed variable, because usually you need a specific output from your circuit to achieve some goal.

And the equations do arrive at the same conclusion, once you factor out Ei and Eo, and all that remains is A, A', and B.
 
I finally work out the output impedance with feedback. I still say RDH4 is wrong in page 310 using Eo=A' X E'i. It's no point of going any further, we just agree to disagree. Here is how I do it using Eo= A X E'i where A is the open loop gain and E'i is the differential input voltage.

Feedback and output impedance.jpg

I started out with A X E'i and go through the voltage divider due to Ro. From the derivation, it arrives to the equivalent circuit with voltage source A' X Ei with output impedance R'o = Ro/(1 + A X beta)
 
I finally work out the output impedance with feedback. I still say RDH4 is wrong in page 310 using Eo=A' X E'i. It's no point of going any further, we just agree to disagree. Here is how I do it using Eo= A X E'i where A is the open loop gain and E'i is the differential input voltage.

View attachment 1056735

I started out with A X E'i and go through the voltage divider due to Ro. From the derivation, it arrives to the equivalent circuit with voltage source A' X Ei with output impedance R'o = Ro/(1 + A X beta)

It's the same conclusion RDH4 gets to, only they assume negative Beta for inverse feedback. following the same convention as for gain.

Thinking to practical consequences, one thing which interests me, is the higher output impedance of the tube with an unbypassed cathode resistor, will have a lower frequency pole compared with a stage with a bypassed cathode, when you take into account it is driving the miller capacitance of the following stage. In other words, removing the cathode bypass capacitor can affect high frequency performance as well as low frequency.

This is sure something I never thought about before.
 
It's the same conclusion RDH4 gets to, only they assume negative Beta for inverse feedback. following the same convention as for gain.

Thinking to practical consequences, one thing which interests me, is the higher output impedance of the tube with an unbypassed cathode resistor, will have a lower frequency pole compared with a stage with a bypassed cathode, when you take into account it is driving the miller capacitance of the following stage. In other words, removing the cathode bypass capacitor can affect high frequency performance as well as low frequency.

This is sure something I never thought about before.
But the pole depends on the plate load resistor. My thinking is using 6SN7, use low value plate resistor to increase the BW. Use unbypass Rk to reduce THD. Use two stages of 6SN7 to pick back up the gain.

More importantly is the local feedback of the output stage to reduce the output resistance of the 6L6GC from 16.5K to 1.5K as I PM you. That will increase the damping factor of the amp. That's why I spent so much time on this thread.

BTW, I was wrong about better than in triode mode. rp for triode mode of 6L6GC is 1.7K, I use 2 pairs of 6L6GC, so the combined rp is 1.7K/2 = 850ohm. Still better..............BUT how about using in UL mode with local feedback?.............................


Ha ha, I'm a man with a mission, not just making this thread for the fun of it.

I give up on the plate curve of the triode with unbypassed Rk, I don't think there is any other way but to plot out the graph.
 
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I drew the load line and calculated the output power driving Vg1 to +10V.

6L6GC triode 400V 50mA.jpg

Datasheet does not go that high, I have to extrapolate the curves and measure with ruler. It's 18.15W output power using +B=400, bias current of 50mA. Primary resistance of 1.9K.
 
For SE guitar amps, a 10% tap and a pentode driver( say 6EJ7 and a KT66 final ) is quite nice. KT66 g2 to the 30% U-L tap. To get these you will need a custom OPT.
cheers,
Douglas
 
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I still need to read this one, it's long. Any particular pages is important so I can skip to those?

The beginning is interesting, but the part I am suggesting to read is summarized in that graph, basically the means of applying negative feedback, it's near the end.
 
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