> From: Peter Chadwick <Peter.Chadwick@gpsemi.com>
> To: amps <amps@contesting.com>
> Subject: [AMPS] failure
To: <amps@contesting.com>
> Date: Tue, 17 Jun 97 14:04:00 +0000
> I think the 8122 has a cathode rather than a filament, so my guess is
> that it drew a lot of current while the cathode cooled down, which takes
> some time. I suppose its possible that it could draw enough current to
> heat the cathode with no heater volts because of dissipation in the
> resistance of the cathode, but I doubt it. However, I wouldn't bet more
> than a beer or two on that without more data!
Hi Pete,
Don't bet more than a beer on that!
The main heating of tube elements (other than the filament) is by
kinetic energy from electrons. Since the electrons are leaving the
cathode, they are removing energy with their departure.
The I^R heating is negligible in all tube elements except the
filament. Otherwise it's the speed of the electrons (related to
voltage difference between the source element and the target) and the
number of electrons striking it (related to current).
Isn't it simple? E times I gives the dissipation, where E is the
potential difference between the electron source and the target
(ie the cathode and anode) and I is the target current (ie anode
current).
That's why we calculate the anode or grid dissipation as E*I
integrated over time, and we don't have to worry about it heating
the cathode!
Turning off the filament and bias and leaving the anode on will
most likely ruin a metal oxide cathode tube. If the cathode
isn't stripped by the attempt to draw beyond safe emission limits
(a thoriated tungsten filament emitter does NOT have this problem,
it will simply saturate at lower currents as temperature falls) the
anode and screen will be ruined from heat produced by the many
electrons smashing into them.
73, Tom W8JI
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