Question about OPT impedance in RDH4

[Alan0354]

I am reading Chapter 13.5 page 571 and 572 of RDH4 regarding to reflected impedance to the primary by load R₂.

From my understanding, If I define turns from anode of a pp primary to the CT is N₁, turns of secondary is N₂, then the reflected impedance of half primary ( anode to CT) is

R'_L = R₂[(N₁²)/N₂²]

Total impedance across the whole primary ( anode to anode) is

R_L= R₂[(2N₁²)/N₂²]=4R'_L


But in the top of page 572, It said reflected resistance across half the primary is R"_L=1/2 R_L=2R₂[(N₁²)/N₂²]

And only if the opposite tube is cut off, then R'_L = R₂[(N₁²)/N₂²]

I don't know what that mean whether the opposite tube is cut off or not.

[J M Fahey]

Turns ratio squared is impedance ratio, always.

If you find 2 statements somewhere which look contradicting, please post the relevant page here, in full, so we don't need to dig the book and read it for you.

I'm certain there must be some explanation about that, maybe one tube "seeing" the other through the transformer, but don't want to guess something which is probably explained there.

EDIT: by the way, you committed a typo there:

N1^2)/N1^2

is always =1 , by definition

[Tony Bones]

You can view the two PP tubes (class A) as being in series. Remember that there is only DC current through the CT. That means that all of the AC signal current must go between the two ends of the primary, through both tubes in series.

So suppose the OT presents a load of R ohms from one end of the primary to the other. Q: How do the two tubes (in series) share the load? A: 1/2 R each, assuming class A, matched tubes, blah blah blah, all of which is assumed when we assume there's only DC through the center tap.

There's at least three ways to prove this that I can think of, but don't make me do all the work. (Hint: conservation of energy.)

[Alan0354]

Turns ratio squared is impedance ratio, always.

If you find 2 statements somewhere which look contradicting, please post the relevant page here, in full, so we don't need to dig the book and read it for you.

I'm certain there must be some explanation about that, maybe one tube "seeing" the other through the transformer, but don't want to guess something which is probably explained there.

EDIT: by the way, you committed a typo there:

is always =1 , by definition

this is the link to RDH4 http://www.tubebooks.org/Books/RDH4.pdf

You can see at the bottom of p 571 to top of p572.

Thanks

[Dave H]

I think they are saying that with only one tube conducting the conducting one will 'see' half the total (p-p) primary turns and therefore 1/4 the total impedance but if both tubes are conducting the contribution from the other tube means that each tube will 'see' 1/2 the total impedance not 1/4.

[J M Fahey]

The 3 formulas say exactly the same, just are written in slightly different ways.

What all 3 say is that the impedance ratio between half and full primary is 4:1

To give you a similar example, imagine: 4 Yen=1 Dollar

then one formula might say:
> 4 Yen=1 Dollar

another might say:
> 8 Yen=2Dollar

and yet another
> 1 Yen=1/4 Dollar

all look different but are exactly the same.

What may confuse you on RDH4 is that instead of directly comparing 1/2 primary to full primary they compare it to secondary, so they look different ... they are not.

To make things more confusing, they always use the letter R but either alone, or accompanied by one " ' " ot by two " ' ' " meaning different sections .

Reread it 1 or 50 times, in a given moment it will click in your mind.

Rl=4R2 (N1^2/N2^2)

Rl ' ' = 1/2Rl (N1^2/N2^2)= 2R2 (N1^2/N2^2)<-- see that here you divide both sides of the equation by 2 ; the actual ratio is still 4:1 (shown as 2:1/2) ... see it?

Rl ' = 1/4 R2 (N1^2/N2^2) (4:1 is the same as 1:1/4)

Rl is not Rl ' is not RL ' '
There is a verbal description of each in those pages.

Don't overthink it ... if you think they are contradicting themselves ... read it again.

The actual Bible contradicts itself in some points ... RDH4 does not .

[Dave H]

The 3 formulas say exactly the same, just are written in slightly different ways.

What all 3 say is that the impedance ratio between half and full primary is 4:1

That last statement is true but the three formulas are not the same. The first formula says 'the reflective resistance across the whole primary is 4n'. The second formula says 'the reflected resistance across half the primary is 2n' (NB not n) and the third formula says 'if V2 is removed the load resistance effective on V1 is n which is half the load resistance on V1 under push pull conditions'. (n=R2*V1^2/V2^2)

RDH4 may not contradict itself but the writer of that section must have had a masters degree in obfuscation.

[J M Fahey]

Yes, my point is that it's matematically correct but poorly written.

Instead of basically the same formula written 3 times (which in stead of clearing the idea muds it) they should have used 2 different but complementary ones:

1 ) end to end impedance to secondary impedance ratio is yada yada yada .

2) half primary to full primary impedance ratio is 4X

no matter how you read them, they never seem to contradict

And any particular case can be derived from these two.

[Alan0354]

The 3 formulas say exactly the same, just are written in slightly different ways.

What all 3 say is that the impedance ratio between half and full primary is 4:1



all look different but are exactly the same.

What may confuse you on RDH4 is that instead of directly comparing 1/2 primary to full primary they compare it to secondary, so they look different ... they are not.

To make things more confusing, they always use the letter R but either alone, or accompanied by one " ' " ot by two " ' ' " meaning different sections .

Reread it 1 or 50 times, in a given moment it will click in your mind.

Rl=4R2 (N1^2/N2^2)

Rl ' ' = 1/2Rl (N1^2/N2^2)= 2R2 (N1^2/N2^2)<-- see that here you divide both sides of the equation by 2 ; the actual ratio is still 4:1 (shown as 2:1/2) ... see it?

Rl ' = 1/4 R2 (N1^2/N2^2) (4:1 is the same as 1:1/4)

Rl is not Rl ' is not RL ' '
There is a verbal description of each in those pages.

Don't overthink it ... if you think they are contradicting themselves ... read it again.

The actual Bible contradicts itself in some points ... RDH4 does not .

No, it specified that R'_L is only for IF one tube is removed from the socket. Read it again. There are R''_L and R'_L depends on whether one tube is removed or at cut off.

[Alan0354]

Also in this same line of thoughts, ra of the tube in triode mode is quite low, in 300 to 600 ohm range. So one tube will looks like a load to the other tube from the two half of the primary. So as long as both tubes are on, one tube affect the loading of the other. How do you look at that?

Thanks

[jazbo8]

Deja vu... Have you considered re-writing (a portion of) RDH4 and publishing it? Since you managed to find so many in-consistencies and mistakes in it, in the meantime, just use it as a doorstop.

[Alan0354]

Actually reading this make me question about the load impedance seen by each of the triode in the push pull. This is different from pentode where output impedance is very high and the opposite tube does not present load to the first tube. As I described in post #10, the low ra of the second tube might present a load to the first tube. Maybe this is what RDH4 trying to say.

RDH4 is about the best on tubes I found so far. I don't take anything for granted and assume it's a typo or mistake.

[Dave H]

Also in this same line of thoughts, ra of the tube in triode mode is quite low, in 300 to 600 ohm range. So one tube will looks like a load to the other tube from the two half of the primary. So as long as both tubes are on, one tube affect the loading of the other. How do you look at that?

Thanks

I don't think it matters if it's triode or pentode. When both tubes are on each one has the same RMS current and contributes half the total power into the load so the impedance each tube is driving looks like 1/2 the total p-p impedance but when one tube cuts off the other one suddenly finds itself having to drive 1/4 of the p-p impedance and all the power. I'm guessing the sudden change in impedance as one tube cuts off is one cause of the crossover distortion kink you can see on a scope.

[jazbo8]

RDH4 may not contradict itself but the writer of that section must have had a masters degree in obfuscation.

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.

[J M Fahey]

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.