Author Topic: SELA Abadie 6002 DC Blacklight Glow  (Read 537 times)
rjluna2
Member
*****
Offline

Gender: Male
View Posts
View Gallery

Robert


GoL
SELA Abadie 6002 DC Blacklight Glow « on: September 30, 2021, 01:56:24 PM » Author: rjluna2
I just wanted to comment about James' SELA Abadie 6002 DC Blacklight Glow.  I'm wondering if this would work with 40 Watts incandescent bulb in series in 120 Volts circuit?  I also wondered how would this starting method if this was in the DC circuit as described at that site ???
Logged

Pretty, please no more Chinese failure.

Rommie
Administrator
Member
*****
Offline

Gender: Female
View Posts
View Gallery

Andromeda Ascendant


Re: SELA Abadie 6002 DC Blacklight Glow « Reply #1 on: September 30, 2021, 02:24:20 PM » Author: Rommie
I very much doubt it, the lamp is designed for a DC supply, and the control circuit doesn't look all that complicated to build, so I wouldn't really want to try it and risk frying such a rare lamp  :poof:
Logged

Ria (aka Rommie) in Aberdeen

"There is no shame in not knowing; the shame lies in not finding out." (Russian proverb)
Administrator, UK and European time zones. Any questions or problems, please feel free to get in touch :love:

Medved
Member
*****
Offline

Gender: Male
View Posts
View Gallery

Re: SELA Abadie 6002 DC Blacklight Glow « Reply #2 on: October 01, 2021, 03:08:57 AM » Author: Medved
I'm wondering if this would work with 40 Watts incandescent bulb in series in 120 Volts circuit?

I would be careful here. The lamp is designed to operate from a 28V supply and have 12V drop with a resistive ballast. That is the operating point, where the real power varies only very little even when the arc voltage starts to drift due to any reason (temperature, so mercury pressure; these are saturated vapor lamps after all...), so it will inherently suppress virtually any thermal instabilities. Constant current drive (that the 120V feed via a resistor is) will make the real power proportional to the actual arc voltage, so if the lamp will have higher than normal drop, the power will get higher, heating the bulb further, releasing more mercury, increasing the drop even further. And because there is a filament parallel to the discharge, this will get overheated and fail soon in the process.

Plus not sure, if this lamp will be able to handle first both polarities (so whether the cathode emission mix is really on both electrode end sections of the filament). The description indicates the polarity (negative on cap), so it is likely the filament is formed into the cathode (thicker, coated with emission mix,...) only on one of its ends.
Second the lamp may increase its pressure so arc voltage as it warms up. That may mean the lamp won't be able to reignite after the zero cross anymore, so may start to cycle. Or the reignition voltage would be so high, it would cause the filament still connected in parallel to the arc to overheat and so fail soon. The DC supply it is designed for keeps there the constant 12V and that is what the filament is designed for to handle long time.


I also wondered how would this starting method if this was in the DC circuit as described at that site ???

Initially it is just an incandescent, where the heat from the filament will warm up the "electrode" sections (the thicker end sections, usually coated with an emission mix as well) of the filament to the thermionic emission. Once electrons get released, the discharge ignites.
« Last Edit: October 01, 2021, 03:12:55 AM by Medved » Logged

No more selfballasted c***

Print 
© 2005-2023 Lighting-Gallery.net | SMF 2.0.19 | SMF © 2021, Simple Machines | Terms and Policies