Question about OPT impedance in RDH4

[Alan0354]

I have old books from the days I was in college in the 70s. Then I just restudied calculus all the way to partial differential equations. The new books are totally different as I have both from the old days and the new. Also I have old EM books and the newer ones. New ones like "Field and Waves Electromagetics" by David K Cheng, Introduction to Electrodynamics by David griffith, and books by Kraus, Balanis, Ulaby etc. Those are so much more in detail and descriptive compare to the old dry EM books from the older days. People do change in their style of writing. The older books are very dry, but it does not mean they are bad. They are just harder to read, and have to sweat it out like what I (we)are doing here, chew the chapter line by line and hope to get more out of it.

But this is very normal, this is advanced book, not for the weak of heart. My experience is if you get through 3 to 5 pages a day, it's a good day. A lot of equations, I had to refer back to the calculus and ODE books. The equations on post #35 are all in the second semester of calculus using power series representation of the plate current.

I am not young anymore and I don't go to school. Took me over 3 years to really studied through Cheng's book on EM because I had to actually stopped and studied over one semester of PDE to appreciate the boundary condition problem. This time, I study for myself, so I go into a lot more detail than in school. But that's how it is. More often, I have to refer to other EM books like JD Jackson, D Griffiths on some of the descriptions.

A lot of the old books do not even have problem sets and exercise. The new ones have examples, problem sets at the end of the chapter AND the most important of it all, they even have the solution manual. I worked out most of the problems in Chengs and Ulaby and over half of Griffiths to learn EM.

Back to the thread, I really need to get through (iii) (iv) and (v) on all the things regarding to Class A triode first.

[jazbo8]

With so many cross-postings in this thread, I will respond to the first post, sorry if it looks out of sequence...

If you are referring to RDH4 then the answer is out of thin air.

As I posted earlier, the text clearly stated - and the reflected resistance across half of the primary is R_L'' = 1/2 R_L - which simply follows from the equation above for the resistance across the whole primary. If you refuse to accept that's the right interpretation, then simply treat R_L'' as a definition. Either way, it means the same thing.

Not true, If you read p 576, he used R'_L =R_L/4 for load line all the way for Class A where both sides are on.

Well, I can see how some confusion could arise from the text on p.576 when looking at Fig. 13.33, let me try to explain and hopefully without mucking up the matter further. Fig. 13.33 shows a load line with a slope of 750R, or RL=3k. But with -60V bias, the 2A3 is not operating in Class A but rather Class AB, since the plate currents reach cutoff, i.e., the load line is for Class B. But if we re-bias the tubes for Class A while retaining the same load (note the bias is not really optimized in any way but simply used to meet the Class A condition), we find that the load line is way too steep and the tubes' Pdiss are greatly exceeded as shown below:

Which is a long way of saying that R'_L =R_L/4 may not be the ideal load for Class A push-pull output... So what is? Could you work it out?

[Alan0354]

Thanks for joining in. Actually the current is not bad even if you raise the quiescent current to 100mA to make it Class A and keep the max current at 200 mA as shown in your graph at E_pi=150V at the 750ohm load line intersecting the E_g1=0V. That's the very reason I posted in another thread http://music-electronics-forum.com/t38400/. I think you use RMS current rather than peak current. In this case, for quiescent current of 100mA and peak at 200mA. The RMS current is still 100mA as the negative half cycle bottom to 0mA.

I have a question, in Fig 13.33, we know the quiescent bias is E_c =-60V for both. Why do you need the other composite characteristic line for E_c1=-50 with E_c2=-70, E_c1=-40 with E_c2=-80 etc. In another word, why do we need a whole bunch of composite characteristics lines?

I don't know why they want to construct the load line curve for V1 in Fig. 13.34. Seems like one composite characteristics line for the quiescent point is good enough to tell about the distortion already.

[jazbo8]

Thanks for joining in. Actually the current is not bad even if you raise the quiescent current to 100mA to make it Class A and keep the max current at 200 mA as shown in your graph at E_pi=150V at the 750ohm load line intersecting the E_g1=0V.

The 2A3 has a maximum plate dissipation of 15W, if Ip0 = 100mA then Pda = 300 x 0.1 = 30W! So please try again, I think if you work through the numbers, it will help you with the visualization.

That's the very reason I posted in another thread http://music-electronics-forum.com/t38400/. I think you use RMS current rather than peak current. In this case, for quiescent current of 100mA and peak at 200mA. The RMS current is still 100mA as the negative half cycle bottom to 0mA.

As mentioned in the other thread, the maximum plate current is neither peak nor RMS but the average current.

I have a question, in Fig 13.33, we know the quiescent bias is E_c =-60V for both. Why do you need the other composite characteristic line for E_c1=-50 with E_c2=-70, E_c1=-40 with E_c2=-80 etc. In another word, why do we need a whole bunch of composite characteristics lines? I don't know why they want to construct the load line curve for V1 in Fig. 13.34. Seems like one composite characteristics line for the quiescent point is good enough to tell about the distortion already.

You are right, the composite characteristic is not really needed and RDH4 said the same thing, Fig. 13.33 & 34 were just shown for completeness I suppose. It is usually sufficient to use just the Ep-Ip characteristic for one tube to work out the output power and distortion, the methods were also included in RDH4.

[Dave H]

the text clearly stated - and the reflected resistance across half of the primary is R_L'' = 1/2 R_L - which simply follows from the equation above for the resistance across the whole primary. If you refuse to accept that's the right interpretation, then simply treat R_L'' as a definition. Either way, it means the same thing

Where's the 'exasperated' smiley when you need it?

Of course I accept it. Haven't I posted to that effect about five times already in this thread? It's clearly stated but I think with insufficient explanation which is why Alan is still asking where it came from when the thread is 48 posts long.

[jazbo8]

Where's the 'exasperated' smiley when you need it?

Of course I accept it. Haven't I posted to that effect about five times already in this thread? It's clearly stated but I think with insufficient explanation which is why Alan is still asking where it came from when the thread is 48 posts long.

Roger that, I will shut up now.

[Alan0354]

With so many cross-postings in this thread, I will respond to the first post, sorry if it looks out of sequence...

As I posted earlier, the text clearly stated - and the reflected resistance across half of the primary is R_L'' = 1/2 R_L - which simply follows from the equation above for the resistance across the whole primary. If you refuse to accept that's the right interpretation, then simply treat R_L'' as a definition. Either way, it means the same thing.

No the text is not clear. If you follow the resistance across the whole primary from anode to anode, it should be 1/4R_L for one side of the primary. The author clearly said with the other tube in and not in cutoff, then the only side of the primary is R"_L = 1/2 R_L.

Notice the author never used R"_L = 1/2 R_L in (ii) and (iii). He used only R'_L = 1/4 R_L.

Well, I can see how some confusion could arise from the text on p.576 when looking at Fig. 13.33, let me try to explain and hopefully without mucking up the matter further. Fig. 13.33 shows a load line with a slope of 750R, or RL=3k. But with -60V bias, the 2A3 is not operating in Class A but rather Class AB, since the plate currents reach cutoff, i.e., the load line is for Class B. But if we re-bias the tubes for Class A while retaining the same load (note the bias is not really optimized in any way but simply used to meet the Class A condition), we find that the load line is way too steep and the tubes' Pdiss are greatly exceeded as shown below:

Which is a long way of saying that R'_L =R_L/4 may not be the ideal load for Class A push-pull output... So what is? Could you work it out?

I looked at your diagram vs fig 13.33, I notice your composite characteristic lines cross the plate curve at the point of one tube cutoff and going into Class B. The composite characteristic lines in Fig.13.33 never cross the plate curve, it just asymptotes to the curve, but never even touch it. Seems like you are drawing the straightest portion of the composite characteristic in ONLY the Class A portion. The book actually draw the straight line (line FBE in Fig.13.32 in page 574)closest to the composite characteristic curve ( it's not really straight).

Thanks

[Alan0354]

These are new questions, I kind of give up on the original question as the author never even use R"_L' = 1/2 R_L any more, he used only R'_L' = 1/4 R_L for everything else.

My new questions are:

1) In page 578 equation (20). P₀=0.2E_bb*I_bm for 2 tubes. I don't understand that. From my understanding, Fig.13.36 show ONLY half the E_bb swing and half of the I_b swing . So those are only the peak values, NOT peak to peak values and that's for one tube, not two. So the RMS value is 0.707 of those value. That should be the same as shown in equation (22) stated P₀ = 1/2 I_bm( E_bb-E_min) = (0.707 I_bm)*(0.707*0.4 E_bb). The total power of two tubes should double that.


2) In page 578 Fig 13.37. What is curve A which is called Dynamic Characteristic?It said it's from mutual characteristics, but I don't know nor can I find on web!!!

Thanks

[Tony Bones]

this is the most detail book I've seen that explains when a lot of other books regurgitate the result but never border to explain why. Like why for triode, the optimal primary impedance should be 2ra of the tube, like why in pp, even harmonics cancelled and odd harmonics added etc.

I believe in order to deeply understand and design amps, this will give more insight than just remember the guide lines. I think it's worth chewing on this slowly. Just because I asked, read and post on this thread since yesterday, I feel I learn a lot already.

I do get that, and I completely agree that it's worth taking the time to understand completely.

FWIW, I have no general problem with RDH4. It certainly wins the prize for being the most comprehensive. I realize now that the pages refered to at the start of this thread were actually written by Langford-Smith, not Crowhurst. Years ago I struggled with this topic and recall becoming especially frustrated with the treatment in RDH4 as well as some well regarded papers by Crowhurst. I struggled with exactly the same thing as this thread, namely reconciling a few equations that shouldn't be so hard to understand. I continue to feel that way (and continue to feel that Crowhurst is over rated ) but I don't hate RDH4 as a whole.

It's almost always helpful to get information from multiple sources. For me it has always been easiest to start with the concepts and end with the math rather than the other way around, but I might be weird.

[Alan0354]

Actually (iv) on distortion skip a lot of steps and not very clear. Anyone has suggestion another book I can buy?

[Dave H]

For me it has always been easiest to start with the concepts and end with the math rather than the other way around, but I might be weird.

If it's good enough for Einstein...

[jazbo8]

No the text is not clear. If you follow the resistance across the whole primary from anode to anode, it should be 1/4R_L for one side of the primary. The author clearly said with the other tube in and not in cutoff, then the only side of the primary is R"_L = 1/2 R_L. Notice the author never used R"_L = 1/2 R_L in (ii) and (iii). He used only R'_L = 1/4 R_L.

Agree to disagree, moving on...

I looked at your diagram vs fig 13.33, I notice your composite characteristic lines cross the plate curve at the point of one tube cutoff and going into Class B. The composite characteristic lines in Fig.13.33 never cross the plate curve, it just asymptotes to the curve, but never even touch it. Seems like you are drawing the straightest portion of the composite characteristic in ONLY the Class A portion. The book actually draw the straight line (line FBE in Fig.13.32 in page 574)closest to the composite characteristic curve ( it's not really straight).

Kindly follow the direction given in RDH4 (or elsewhere) to draw the composite plate curves yourself if you think the curves in my figure were not drawn correctly. Looking forward to see what you come up with.

[Alan0354]

Agree to disagree, moving on...


Kindly follow the direction given in RDH4 (or elsewhere) to draw the composite plate curves yourself if you think the curves in my figure were not drawn correctly. Looking forward to see what you come up with.

No, I think yours is correct for the class A. If you look at RDH4, it takes the average, but if you look at only the straight linear portion with both tubes conducting, it should cross the plate curve like what you drawn.

[jazbo8]

My new questions are:1) In page 578 equation (20). P₀=0.2E_bb*I_bm for 2 tubes. I don't understand that. From my understanding, Fig.13.36 show ONLY half the E_bb swing and half of the I_b swing . So those are only the peak values, NOT peak to peak values and that's for one tube, not two.

Don't forget that half of the output power is dissipated by the tubes (ignoring other losses in the circuit), i.e., some power are wasted and not transferred to the load.

2) In page 578 Fig 13.37. What is curve A which is called Dynamic Characteristic?It said it's from mutual characteristics, but I don't know nor can I find on web!!!

You don't really need the web for this one, RDH4 has detailed descriptions of both, basically the mutual characteristics (aka transfer characteristic) is Ip/Ig2 vs. Eg and can be found in many datasheets; and the dynamic characteristic is the mutual characteristic with load.

No, I think yours is correct for the class A. If you look at RDH4, it takes the average, but if you look at only the straight linear portion with both tubes conducting, it should cross the plate curve like what you drawn.

I'm glad that we are on the same page... Fig. 13.33 shows the class AB/B composite characteristic, while my figure show the class A composite characteristic, although as I mentioned it isn't really usable for a real circuit since the tubes would be under too much stress.

[Alan0354]

You don't really need the web for this one, RDH4 has detailed descriptions of both, basically the mutual characteristics (aka transfer characteristic) is Ip/Ig2 vs. Eg and can be found in many datasheets; and the dynamic characteristic is the mutual characteristic with load.


I'm glad that we are on the same page... Fig. 13.33 shows the class AB/B composite characteristic, while my figure show the class A composite characteristic, although as I mentioned it isn't really usable for a real circuit since the tubes would be under too much stress.

I was questioning the way RDH4 drew the composite characteristic, why don't they want to follow the straightest part in Class A region instead of the two points where the amp cross into Class B. That's why I want to confirm with you about that. Is the slope of the line suppose to be the plate resistance of the "composite tube" which is about 1/2 of the plate resistance of the individual tube?

But really come to think about the big picture, these are all approximate only. In real life, unless you wind your own OPT, what you see on the market is what you get. Not to mention the impedance of the speaker varies with frequency, you never get an exact match anyway.