Question about OPT impedance in RDH4

[Alan0354]

Do you mean what I posted doesn't make sense? No surprise there then

No no, I mean the RDH4 doesn't make sense.

Everyone here talks about the primary impedance of the OT alone from loading the secondary with R₂. That is very straight forward that everyone already agreed that the impedance loading each tube is 1/4 the impedance of the anode to anode impedance. We don't need to talk about this any further.

I think the RDH4 is trying to say the loading of the opposite tube comes into play as this second only concerning with triode with low plate resistance. My scanner bite the dust, I cannot draw and write formula to post. I am looking at eqation (1), (3) and (4) in page 573 about the power series of the two current. You can see the even harmonics cancel and will not load the opposite tube. But the odd harmonics are not cancelled, so the two tubes loading each other only in the odd harmonics as shown in (5).

[Tooboob]

I am thinking about a model for this as the odd harmonics does not cancel out and even cancels. So the two tubes definitely not present a load to each other ONLY in the even harmonics. The two tubes do load each other regarding to odd harmonics. It is very confusing.

I think you need to ask yourself "Even and odd harmonics of what?"

The output stage has no idea which input frequencies are fundamentals and which are harmonics, and no sentience to track them over time. A frequency and it's harmonics pass through a low distortion output stage pretty much unmolested, whether or not they are even or odd harmonics. The power tubes can't divide by two and check for a remainder.

The cancellation of even harmonics only applies to the distortion generated in the power stage. Symmetrical distortion is generated with odd harmonics. Think of a sine wave and its third harmonic harmonic both starting at a zero crossing. The sine wave tend to square up a bit, symmetrically, and if we get the coefficients for the odd harmonics right, we get a square wave. We can get this out of a push-pull power stage if we drive it to clipping, lopping the extremes off both edges of the waveform. If things are balanced well, this is the primary distortion of a push-pull stage. Clipping and non-linearity are symmetrical, equal for the negative and positive going output = odd order. Go single-ended, and the cutoff and "saturation" distortions are different, generating assymetry, which must include even harmonics.

The push-pull power stage, by it's nature, applies the same distortion, which is generally asymmetric, with both even and odd harmonics, to both output polarities. Subtracting one from the other in the OPT forces symmetry (for a sinusoidal input), and the asymmetrical even-order harmonic distortion becomes symmetrical odd-order harmonic distortion. The Fourier analysis that would make this clear is left as an exercise for the reader, since the result is obvious by inspection, and I have no idea how to perform the analysis.

Each push-pull tube's output, when it is driving, ideally see the other tube in cutoff. The other tube isn't selectively filtering out even harmonics. It's not even on.

[Dave H]

...everyone already agreed that the impedance loading each tube is 1/4 the impedance of the anode to anode impedance.

I didn't

It's only 1/4 of the A-A impedance when the other tube is off which is most of the time in class AB but none of the time in class A. In class A or when both tubes are on in the 'class A' part of class AB the impedance loading each tube is 1/2 of the A-A impedance. That's what I think RDH4 is saying but as you say everyone(?) here is saying it's always 1/4 of the A-A impedance.

[Jazz P Bass]

My head hurts. I'm gonna go sit in the lobby .

Move over please.

[Alan0354]

I didn't

It's only 1/4 of the A-A impedance when the other tube is off which is most of the time in class AB but none of the time in class A. In class A or when both tubes are on in the 'class A' part of class AB the impedance loading each tube is 1/2 of the A-A impedance. That's what I think RDH4 is saying but as you say everyone(?) here is saying it's always 1/4 of the A-A impedance.

I meant everyone agree with that ONLY if you look at the OT alone, not taking into account of the loading of the tubes, just the OPT.

Now we are getting somewhere!!! The following is only my wild guess about the loading of the triode:

I don't think it's 1/2 of A-A impedance when the tubes are loaded. The reason is the two power tubes are driven by equal and opposite voltage like in equations (1) and (3) in page 573.

For tube 1:
(1)=>I_b1=a₀+ a₁eg¹+a₂eg²+a₂eg³+a₄eg⁴...........

For tube 2 on the opposite side:
(2)=>I_b2=a₀-a₁eg¹+a₂eg²-a₂eg³+a₄eg⁴...........

Since the two half of the primary is opposite phase, (1) coupled to the other side as -a₁eg¹-a₂eg²-a₂eg³-a₄eg⁴...........


You see on the plate of tube 2, the a₀ and the even orders are all in phase. Tube one does not have to "drive" the even harmonics. Tube 2 does not present a load to tube 1. It is like you try to move an object, but the object is moving exactly the same as your motion. It offer no resistance to you. Therefore it's not loading you down.


But if you look at the odd harmonics, they are all opposite phase. Odd harmonics of tube 2 is fighting with tube 1 and thereby presenting a load. The total load is given by (5) in page 573:
I_d=2a₁eg¹+2a₃eg³+2a₅eg⁵...........

Therefore, only the odd harmonics present load to the opposite tube. It almost looks like the push pull configuration reduce the odd harmonics distortion a little because it present a lower impedance load of 1/2 ra to the opposite tube and load it down just a little. The half is from the 2 of each term in (5).

[jazbo8]

I meant everyone agree with that ONLY if you look at the OT alone, not taking into account of the loading of the tubes, just the OPT. Now we are getting somewhere!!! The following is only my wild guess about the loading of the triode:

Now before we dig ourselves a bigger hole, for purely speculative purposes, please show us how you would draw a composite load line based NOT on RDH4's equations but on your understanding how a push-pull stage functions. Show us how YOU think the circuit behaves, since you obviously do not agree with the conventional interpretation. Given your background and insistances on the reference material's accuracy, YOU should able be to provide a valid interpretation if you want to debuke it.

We do not need to see the equations (yet) but simply how you would draw a load line, i.e., incorporating the load from the other tube of the push-pull pair, etc. If you had to do draw a load line yourself, what does it look like?

[Dave H]

Move over please.

Room for one more?

[Justin Thomas]

I was gonna post this this morning but... "It's a guitar amp, not a NASA rocket." -Enzo-

Wire it up, fire it up, crank it up, blow it up. Repeat as needed.

I hope it's a magical expanding bench...

Justin

[Dave H]

Wire it up, fire it up, crank it up, blow it up. Repeat as needed.

Priceless! I'll remember that.

[Jazz P Bass]

I hope it's a magical expanding bench...

The Flexible Love Chair fits the bill.



https://www.youtube.com/watch?featur...&v=xXutmqo_Lm8

[Alan0354]

Now before we dig ourselves a bigger hole, for purely speculative purposes, please show us how you would draw a composite load line based NOT on RDH4's equations but on your understanding how a push-pull stage functions. Show us how YOU think the circuit behaves, since you obviously do not agree with the conventional interpretation. Given your background and insistances on the reference material's accuracy, YOU should able be to provide a valid interpretation if you want to debuke it.

We do not need to see the equations (yet) but simply how you would draw a load line, i.e., incorporating the load from the other tube of the push-pull pair, etc. If you had to do draw a load line yourself, what does it look like?

I don't know, I don't have answer. I just talk and hope someone have an answer. All I see for now is the odd harmonics see the load and the even does not.

Don't take I am debunking anything, I am just looking at the logic. Don't take that I know any better. I hope you guys have more experience might know more.

[Tony Bones]

I think the book is trying to tell us that the opposite triode do present a load to the other triode. The question is how do you look at it.

This paragraph cannot be looked as simple impedance reflection to the primary from the secondary alone. It consider the impedance as a whole circuit. That's why it has 3 equations.

Exactly. If one tube is completely cut-off then its 1/2 primary is essentially open circuit. That leaves only 1/2 of the primary for the one tube that's conducting; 1/4 the PP load. That part's easy.

When both tubes are conducting they aren't loaded by each other, they're working together, in series, across the entire primary. It's true that the even order harmonics sneak out through the tranformer center tap, but that's a second order effect (no pun) that just confuses the first order understanding.

I'll be honest and admit that I have not refered to the pages in RDH4 in order to participate in this thread. The reason is that I already studied exactly those pages years ago and the only conclusion that I came to is that Crowhurst was a terrible writer. How he became such a prolific author is a mystery to me.

[Alan0354]

Ha ha, RDH4 is the best I read. That's the reason I revisit the book again. I have been going through the next few pages so far, it is fantastic!!! Remember, he wrote this book before most of us even born!!!! Electronic technology has gone through leaps and bounds. But I still like the book. It's the first I come across that explain why pp cancels even harmonics and sum odd harmonics so clearly by using mathematics. To me, it's a mile better than Morgan Jones. The way he explains the composite curves make me want to start gluing the plate curves back to back!!!!

[jazbo8]

I'll be honest and admit that I have not refered to the pages in RDH4 in order to participate in this thread. The reason is that I already studied exactly those pages years ago and the only conclusion that I came to is that Crowhurst was a terrible writer. How he became such a prolific author is a mystery to me.

Wow, you managed to put down two of the best authors on audio/tube literture in one go. IMO, very few authors even come close to Langford-Smith or Crowhurst. Like Alan, I can't find a better reference than RDH4, so if there is any discrepency or, I shudder to think, some errors in it then it's worth the time to investigate and/or clarify - at least it is for nerds like me.

[jazbo8]

I think the RDH4 is trying to say the loading of the opposite tube comes into play as this second only concerning with triode with low plate resistance. My scanner bite the dust, I cannot draw and write formula to post. I am looking at eqation (1), (3) and (4) in page 573 about the power series of the two current. You can see the even harmonics cancel and will not load the opposite tube. But the odd harmonics are not cancelled, so the two tubes loading each other only in the odd harmonics as shown in (5).

Please also read Chapter 13, sections (ii) - (v). That's why I asked you earlier to draw out the composite load line, if you have a different interpretation of the push-pull operation, then it would not look like the example given in Fig. 13.33 of RDH4.