See below;
*********** REPLY SEPARATOR ***********
On 7/20/06 at 9:57 PM Tom W8JI wrote:
>> The problem is, you stated all this can occur after the
>> fuse or resistor opens. It can not, only before it does.
>
>So the grid instantly cools, all leakage currents or arcs to
>the anode vanish, and the grid remains at zero volts at the
>instant of grid opening.
>
>Is that what you are proposing?.
If the grid to ground circuit is open (fuse or resistor blown open), it is
impossible for the anode to arc to the grid. It's impossible for any open
circuit to carry current unless it was to arc across the break. In this case,
it would have to arc
between the blown fuse caps or resistor ends to ground (I doubt there would be
much left to arc). I also highly doubt the
whole grid getting real hot from the initial arc that opens the fuse since it's
a very fast occurance. It would probably
get hot where the arc hit the grid, maybe enough to cut into the grid wire, but
that's so fast it couldn't heat the whole grid up. There would have to be a
continuous arc for some time to cause excessive heating through the whole grid.
Also, it's impossible for the grid to take a charge with the circuit open as it
would need a path for the electron flow. If it's not connected to ground (open
from the fuse blowing) where does the electron flow come from? In order for a
charge to
take place, the circuit has to be complete while one element looses electrons,
and another gains them. The electrons
can not move from one to another in an open circuit.
>
>>
>> "Now when the tube faults and if the resitor opens, the
>> grid can rise to full anode voltage"
>>
>> If the grid doesn't make a connection to anything, it
>> can't conduct and hurt one thing, it's just there in
>> space.
>
>Never mind, I probably did a poor job of explaining
>something very simple and we will have to agree to disagree.
>Repeating it five more times won't make any difference to
>anyone. People either get it, or they don't.
>
>73 Tom
Best,
Will
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