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Registered Member #223
Joined: Mon Feb 20 2006, 06:42PM
Location: Ottawa, Ontario, Canada
Posts: 125
I recently bought two image intensifier tubes off of ebay for $30. however I am having difficulty reading the data sheet because it is in dutch(i think). I have tried babble fish but the translations are rather sketchy. It is a Phillips XX1080 image intensifier tube. I figured out the pin out okay. The only problem is with the focus potential. Apparently it is 450V max. but in respect to what? The anode or the cathode?
The anode and focuser connectors have ceramic insulators around the connectors. So I'm guessing the focuser is still at HV but only 450v less then the anode. This is the impression I got off the translation. Or is the focuser 450 above ground? a voltage divider could do either. I'm just not sure which one. There really isn't much information this type of tube. I really don't want to ruin a perfectly good tube. I have attached the data sheet. The sheet seems to talk about different tubes but the XX1080 is near the bottom. Any help would be appreciated. ]1164090336_223_FT0_xx1080.pdf[/file]
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
I don't know Dutch but the technical words are pretty much the same as German. :P
wrote ... You can get a higher gain, around 1000x, with the XX1080 (see figs 8 and 9) but here you need a voltage divider for the focusing electrode, which needs to be about 450V positive with respect to the cathode potential. A 1000 Mohm resistor between screen and focusing electrode, and one of 30Mohm between this electrode and cathode, gives about the right ratio with a supply voltage of 14kV.
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Image intensifier tube
