[Amps] How do I determine Class

[Bill Fuqua]

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.

At 06:18 PM 9/1/2004 -0700, R.Measures wrote:

>On Sep 1, 2004, at 2:19 PM, Bill Fuqua wrote:
>
>>At 09:59 AM 9/1/2004 -0700, R.Measures wrote:
>>
>>>On Sep 1, 2004, at 8:22 AM, Jim Isbell wrote:
>>>
>>>>  This is EXACTLY the answer I needed.  The amp uses a pair of 4-125As 
>>>> and the screens are at "350V - 400V"   Plates at "2000V - 3000V" (but 
>>>> no terminating resistor) so I guess that makes it a class AB1 linear.
>>>
>>>Jim --  Class AB1 amplifiers always have a grid-terminating R because 
>>>the grid draws zero current.  .
>>
>>Not always. You don't  have to have a grid-terminating resistor for AB1 
>>there is always a parallel resistance due to the loss in the grid tank 
>>circuit.  Q= R(parallel)/ X. The tank circuit is loaded by the drive 
>>impedance also, so it is not totally unloaded Q.
>
>Bill -- What SWR would the transceiver see if the grid-coil R was the only 
>load?   My guess is somewhat less than wonderful.

    Not many transceivers available in 1956 to most hams.


>>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.



>>>>The screen grids have several bypass caps to ground but no resistor.
>>>
>>>A shunt resistor is needed to sink reverse screen-I.  Typically, about 
>>>25% of normal screen-I is shunted by this resistor.
>>
>>The 4- series does not have problems with the negative screen current 
>>like the 4CX series. I just checked the data sheets.
>
>Interesting observation.  Any theories on why this is the case?
>
>Richard L. Measures, AG6K, 805.386.3734.  www.somis.org
>
>_______________________________________________
>Amps mailing list
>Amps at contesting.com
>http://lists.contesting.com/mailman/listinfo/amps