On Jul 21, 2006, at 4:33 AM, Tom W8JI wrote:
>>> 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.
When did we get arcing?
>
>> 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.
In a good vacuum?
> 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.
But this is apparently a real special kind of arc because it
apparently leaves no trace behind inside the tube.
>
> 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.
>
Quicksand
R L MEASURES, AG6K. 805-386-3734
r@somis.org
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