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On 7/21/06 at 9:28 AM Tom W8JI wrote:
>> I could see this if the tube was operating and the
>> electrons were flowing between the cathode and grid then
>> back to the cathode via ground like it supposed to. Once
>> the fuse/resistor opens, the grid could be less some
>> electrons. Since the cathode is still emitting electrons
>> that are being drawn to the anode across the grid, the
>> grid would accept enough electrons to come into
>> equilibrium. I can't see it though taking more charge
>> after it's equalized, and just keep charging up to some
>> higher potential which is what I was talking about earlier
>> in other posts. That's just not possible, or I cant see it
>> would be. Equilibrium yes, but other no.
>
>Will,
>
>What you imagine happens runs contrary to how the systems
>actually behave in real life.
>
>1.) The rated dissipation is determined by the point of
>secondary emission by heat. So if we trip the grid off from
>excess current in operation is tripping when the grid is
>acting like a cathode.
Let me ask a question here. Lets say we took away the current from a
cathode/heater and let it cool off completely. Then we grounded both ends of
the cathode/heater where no current would flow through it to heat it up. After
this we applied plate voltage. Would any current flow from the cold cathode to
the anode?
>
>2.) During an arc or anode to grid fault, the tube has
>plasma or significant leakage path inside. Opening the grid
>does not instantly make that plasma or path open. This is
>especially true when a very poor fuse like a small resistor
>is used to interrupt a few kV of voltage.
Here's an experiment for you. Take a 1/4 watt resistor the size and type Rich
mentiones, I believe he said 30 ohm 1/4 watt, and apply say 2500-3000 Vdc
across it. Let me know what happens and how long it took to obliterate it.
>
>Surely you can see all that?
>
>73,
>Tom
>
>
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Best,
Will
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