Sepp?

[Tim Wyatt]

I have an old amp I'm trying to figure out(no schematic). It seems the output tubes are connected to the output transformer as a push pull would be. The signal goes into the first tube and also branches off into a triode, the signal from the triode then feeds the second tube. So, the second tube is running 180 degrees out of phase from the first tube(give or take hystresis of triode path). Is this a type of single ended push pull? I have seen much easier ways to do this. Why the triode? Thanks.

 

[Sch3mat1c]

Sounds like this.

http://webpages.charter.net/dawill/tmoranwms/Circuits/Stereo_6V6.gif

Cheezy non-paraphase, just cascaded stages to invert for the other half. A voltage divider cancels the gain of the second 12AX7, turning it into a 1:1 inverter.

Tim

 

[Yvesm]

Hi Tim

Cheezy non-paraphase

Why not ?
To keep driver impedance equal on both output grids ?
If so, I suppose the unbypassed K resistor of the PI stage is "carefully" choosen to balance what the first stage sees in its own cathode, including feed back injection.
Or, was it just picked from the junk box 😀

Yves.

 

[Tim Wyatt]

Sch3mat1c, Yes very much like that. 6L6s and some different values but the same concept. This is a new one to me. So both tubes are running class A working against each other across the primary? Could this transformer still be used if converted to push-pull? Thanks for your help.

 

[Sch3mat1c]

Output tube class depends on biasing, if it uses a cathode bias resistor chances are it's class A, or mild AB (class AB self bias gives massiev compression, not the least bit hi-fi but wouldn't suprise me back in those days. I mean look at the Magnavox schem there, top 1959 tech!).

You should trace the full schematic, that's what the above schematic is (of one of these).

(Hmm, that auction is a pretty shiny one!)

Tim