Question about OPT impedance in RDH4

[Dave H]

May be I need better glasses, but I just don't see the obfuscation - start with the full primary, then half of the primary, and finally with one tube out of the circuit.

My glasses are fine too I understand it OK but it has caused confusion in this thread. The obfuscation is the unnecessarily confusing notation used (RL,RL',RL'' etc). Your descriptions above are better.

EDIT
The point I was trying to make is that RL''=1/2RL (not 1/4RL) so it's not the same formula written three different ways. The tube is driving half the primary turns but when the other tube is conducting it 'sees' 1/2 the primary impedance not 1/4.

[Jazz P Bass]

Impossible?

Improbable.

[jazbo8]

The point I was trying to make is that RL''=1/2RL (not 1/4RL) so it's not the same formula written three different ways.

I guess you could read it that way, but for me, equation 2 simply means what the text said "the reflected resistance across half of the primary...", it made no reference to the output tubes. WYSIWYG...

[Dave H]

Impossible?

Improbable.

What is? I'm not making it up. It's straight out of RDH4.

[Dave H]

I guess you could read it that way, but for me, equation 2 simply means what the text said "the reflected resistance across half of the primary...", it made no reference to the output tubes. WYSIWYG...

You're getting me now It says 'the reflected resistance across half the primary RL''=1/2RL" which is what I wrote above isn't it? A little lower down it does make reference to the output tubes. It says "RL'=1/4RL which is half the load resistance on V1 under push-pull conditions. This is the condition which occurs when one of the valves reaches plate current cut-off"

[Jazz P Bass]

I was redefining the term, that is all.
Impossible is a pretty heavy word.

[jazbo8]

You're getting me now It says 'the reflected resistance across half the primary RL''=1/2RL" which is what I wrote above isn't it? A little lower down it does make reference to the output tubes. It says "RL'=1/4RL which is half the load resistance on V1 under push-pull conditions. This is the condition which occurs when one of the valves reaches plate current cut-off"

Sorry about that, we were saying the same thing but in different ways, which just goes to show how easy it is to obfuscate without even trying...

[Alan0354]

Personally when I find something I don't understand, don't consider it a typo but simply that I don't understand.

It's impossible that a classic book written over 60 years ago and read by Millions (literally) can have an uncorrected typo.

I don't think it's a typo, I think the book is trying to say you have to consider the opposite triode present a load on the other triode because the plate resistance is low for triode. That you cannot assume it's high impedance as in the pentode case. That's the reason I post this to clarify. As I said, this is the best book I've seen, The 3 equations are specified one for anode to anode, one for one side with the other side loaded by the other tube, then one for either the other tube is removed or cutoff.

[Alan0354]

You're getting me now It says 'the reflected resistance across half the primary RL''=1/2RL" which is what I wrote above isn't it? A little lower down it does make reference to the output tubes. It says "RL'=1/4RL which is half the load resistance on V1 under push-pull conditions. This is the condition which occurs when one of the valves reaches plate current cut-off"

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.

[jazbo8]

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.

I'm lost, so what was your question?

[Alan0354]

I'm lost, so what was your question?

Question is how the book come up with the formula when both tubes are in and not in cut off. If you read through p574, you'll see how the two current sum and cancel. 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.

[g1]

My head hurts. I'm gonna go sit in the lobby .
Wed-love-to-chat-but-we-gotta-sit-in-the-lobby-and-wait-for-the-limo.mp3

[Dave H]

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.

If the opposite valve presented a load on the other valve wouldn't the load resistance on the remaining valve go up when a valve is removed? It doesn't, the load resistance is halved when a valve is removed.

With only one valve it is easy to see that it is driving 1/4 of the p-p primary impedance. When both valves are driving however they are each providing half the current at the the same load voltage so the load impedance looks like it has doubled to 1/2 of the p-p primary impedance.

[Alan0354]

That's why I post. That doesn't make sense either.

Because of the sign of the power series, there is a difference in loading of the odd and even harmonics. I don't think the opposite triode presents loading in even harmonics, only loading in the odd harmonics.

[Dave H]

That doesn't make sense either.

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