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Question about testing a CRT in unknown condition:
If a CRT has gone to air, but nobody has powered the filament since before the vacuum was lost, could such a CRT be restored to operation by re-evacuating the tube, without have to do a total rebuild (assuming any leak would be extremely slow or could be found and closed)? I say this, because I wonder if a test of an unknown CRT should start with a very low voltage continuity test of the filament, then, if there is any conductivity, doing some type of test for any presence of air in the tube. Rather than just hooking up the tester, which would destroy the gun of the tube if in fact the tube had gone to air? |
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#2
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I'm sure someone who really knows the answer to this will chime in.
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#3
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Testing an unknown CRT is certainly a common scenario. Perhaps one of the sages could tell us all what the potential risks might be.
Phil Nelson |
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Phil Nelson |
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Last edited by andy; 12-07-2021 at 01:39 PM. |
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Phil |
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Last edited by andy; 12-07-2021 at 01:38 PM. |
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#8
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However, once the heater is burned out, a rebuild would be the ONLY option. No possibility of a rework (however slim) would exist. I guess that I would prefer to leave as many options open for as long as possible. Perhaps an attempt to seal the leaks followed by a getter re-flash and cathode re-activation could still rescue the tube, as long as the heaters remain intact. However, this is very unlikely and perhaps a waste of time and money, but at least the option remains open. If I had access to an RF generator and a getter "wand", I would try a getter re-flash in a New York Nanosecond... not much to loose! jr Last edited by jr_tech; 11-10-2009 at 01:27 PM. |
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![]() I was fortunate, the heaters appeared to power up normally. Had they not, I really had no game plan, and I doubt that the tube could have been saved anyway... Perhaps long term low voltage on the heaters would "pump" the tube somewhat... perhaps the getters could be "re-flashed"... but at least the heaters would be intact for further experiments. jr Last edited by jr_tech; 11-10-2009 at 03:34 PM. Reason: add info |
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#10
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During the activation of the cathode BaCO3 is dissociated to BaO. Subsequent heating at a higher temperature then reduces some of the BaO to Ba with the help of a reducing agent like Si. When the tube goes up to air the Ba oxidizes back to BaO or BaO2. Mostly BaO I believe. There are other alkali's involved too, but Barium is the major one.
You would need to reactivate it, but the reducing agent should have been mostly used up during the first activation. There would a significant expense in trying it, with no way of knowing the status of the cathode before hand. I don't think it would be worth it. If I spent that much I'd want to know that I had a fresh gun in the end. John |
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