Anode loaded 6080 giving less than unity gain, Why?? - Page 2 - diyAudio
 Anode loaded 6080 giving less than unity gain, Why??
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Join Date: Oct 2002
Location: Chicagoland
Blog Entries: 2
Quote:
 For a push-pull output transformer, the each tube sees half the plate to plate impedence.
Not 1/4?
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 23rd May 2006, 12:16 AM #12 diyAudio Member     Join Date: Nov 2005 Impedance looking into the primary of the OPT is the square of the turns ratio. 1/2 the turns (one leg) means 1/4 the impedance... all other things being equal.
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Join Date: Jun 2003
Location: Dallas,TX
Quote:
 Impedance looking into the primary of the OPT is the square of the turns ratio. 1/2 the turns (one leg) means 1/4 the impedance... all other things being equal.
Not for a push-pull scheme. It is a more complicated matter than what appears to be two single-ended OPT's in series. As a matter of fact, the load each tube sees varies along the load-line (each tube sees approximately one half the plate to plate impedence at the quiescent point. As the grid for one tube swings to its most positive value [while the other tube's grid swings the other way] the impedence it sees approaches the full a-a value. Conversely as it swings toward its most negative value [to 0 volts], the impedence it sees approaches [a-a]/4), but to keep it simple, we'll just call it 1/2 the plate to plate impedence.

John

 23rd May 2006, 12:50 AM #14 diyAudio Member     Join Date: Nov 2005
 23rd May 2006, 01:36 AM #15 diyAudio Member     Join Date: Nov 2005 Sorry... just being a goof. The transformer impedance is what it is. Voltages and currents are transformed by the turns ratio and impedances by the square. The Zin is simply a function of the load resistance and the turns ratio (perfect transformer). You are talking about a method for PP analysis though, and I would like to learn more. I'm having trouble following why the impedance changes. You are talking about the effects of primaries driven simultaneously and a non superposition way of looking at the circuit? Are you talking about transistions in operating class? Can you elaborate?
diyAudio Member

Join Date: Jun 2003
Location: Dallas,TX
Quote:
 Can you elaborate?
The analysis is rather complicated, but with two tubes "running in opposite directions" on the same primary of a transformer, let's just say there is an autoformer effect going on.

Quote:
 Are you talking about transistions in operating class?
This is a class A situation. Other modes of operation get even weirder.

I suppose I could go down to my garage and dig out an old textbook and quote verbatim a longwinded explanation complete with mathematical equations, but if you don't already accept the rule of thumb of each tube seeing 1/2 the plate to plate impedence of an OPT, maybe you ought to study the problem yourself.

John

 23rd May 2006, 03:06 AM #17 diyAudio Member     Join Date: Nov 2005 I'm OK with that... and I will dig into RDH4 as well. I was, however, under the impression that the rule of thumb was (a-a)/4. The autoformer effect seems clear to me; but that would be based on whether currents are flowing through the "other" primary, and, the Zin, or effective Zin, would be a function of this. Hence my question re operating class. Curiouser and curiouser...
 23rd May 2006, 04:04 AM #18 diyAudio Member     Join Date: Jun 2003 Location: Dallas,TX "Please see RDH, pages 199-203." I swiped this from another thread. John
 23rd May 2006, 06:12 AM #19 diyAudio Member     Join Date: Aug 2002 Location: Eire "gain=mu RL / ( rp + RL + Rk ( mu + 1 ) ) i just try : 80/110=.72 2x600/(280+600+280(2+1))=.6977 .72 ~= .6977 seem ok to me" That looks to be about exactly the situation I have, so I have nothing to worry about. Thanks everyone, I understand a little bit better now. Shoog
 23rd May 2006, 08:36 AM #20 diyAudio Member   Join Date: May 2005 Location: Wisconsin! poobah, it's (a-a)/4 when you're in Class B mode. Either actually operating that way, or when you hit a peak in Class AB. When one of the tubes goes into cutoff, the other one sees only half the winding, the other half is open-circuited by the cutoff tube. Half the winding open circuit halves the rated winding ratio so the other tube sees 1/4 the rated impedance. I *think* this is the right way to look at class A operation. If you look at each tube individually, you can see it. Basically each tube sees the full primary TWICE, once directly, and again through the coupling of the two halves of the primary together. So it sees the full primary (the a-a) value, in parallel with the full primary again. I'm having trouble visualizing the voltage analysis that shows it directly but I'm pretty sure that's the mechanism, rather than each tube somehow seeing 1.414 of its winding.

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