[Amps] How do I determine Class

[R.Measures]

On Sep 2, 2004, at 6:43 AM, Bill Fuqua wrote:

> I think it has to do with the close proximity and precise alignment of 
> the screen grids with the control grids so that screen grids are in 
> the electron shadow of the control grids.  I have not given it much 
> thought. In any case, any electrons hitting the screen grid will 
> produce secondary electrons provided they have enough kinetic energy. 
> At the high screen potentials that transmitting tubes use each 
> electron that hits the screen grid will produce 2 or maybe more 
> secondary electrons. The question is where will the secondary 
> electrons go? Will they  got back to the screen grid or will they go 
> to the plate?  With suppressor grids the answer is clear because the 
> electric field will be favor the secondaries going back to the screen 
> grid. But there must be something in the average tetrode that forces 
> the secondaries back to the screen grid rather than going to the 
> plate.
>
>    Here is one possibility. In a non-shadow type tube most of the 
> electron impact will be on the side of the screen grid closest to the 
> control grid. In this case the secondary electrons will be emitted 
> with a low kinetic energy ( low speed) back toward the control grid. 
> In this region the electric field in going to push the secondaries 
> back to the screen grid before they get very far from it. They will 
> only have a few electron volts of kinetic energy and easily drawn back 
> to the screen grid.
>    But in the case of the control grid shadow around the screen grid. 
> I can only think that there are situations that allow the electrons to 
> miss the front surface of the screen grid (nearest the control grid)  
> and then strike it on a surface that is nearer the plate than the 
> control grid thus the secondary electrons would be attracted to the 
> plate causing a negative net current flow to the screen grid.
>     Just some thoughts.

Bill --  This makes some sense, however, reverse screen current 
reportedly happens in tubes that show no reverse screen current region 
in the published characteristic curves.     Thus, I am using a 50k-ohm, 
100w screen bleeder in the current project.
>
> ...Many of the old designs were for maximum gain so that very low 
> power exciters could drive them. Here you can vary the drive quite a  
> bit by changing the link coupling.
>>
>> For a Bruene bridge neutralized tetrode or pentode, there is no link 
>> coupling.  The grid is driven either directly (50-ohm termination) or 
>> through a Z-step up broad band xfmr. (200 or 450 ohm termination) , 
>> and the grid XC is tuned out by a parallel roller L.
>
>    The MB-40L has a link coupling. And we are talking about a 1956 
> tetrode amplifier from the handbook.

Was the amplifier neutralized?
>
Richard L. Measures, AG6K, 805.386.3734.  www.somis.org