>>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 is already arcing. That is what makes the resistor fail.
>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).
You can doubt it, but if you start an arc with 50 volts the
lead can pull back 1/4 inch and still maintain the arc
through plasma. With 3000 volts it is quite easy to sustain
an arc for a few inches. I suppose people don't think about
this much, but that is why high voltage fuses are filled
with "sand" and have an element several inches long. Anyone
who thinks a 1/4 watt resistor will quench an arc needs to
spend some time doing HV experiments.
> 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.
You are mixing examples. I clearly separated the two forms
of grid current. Please reread carefully. I never said the
arc heated the entire grid, although it will cause plasma
inside the tube and that certainly will instantly heat the
area where the arc is.
What I actually said is there are TWO distinct sources of
excessive grid current. One a tube HV fault to the grid. The
other an electron bombardment (kinetic energy) problem from
excessive grid current.
This isn't as complex as it is being made.
73 Tom
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