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Fluorescent Lamps

Back to basics. This is the way I thought they worked. The ballast provides a high voltage to get the mercury to ionize and then the voltage drops and the filaments on each side maintain the glow. However, my magnifying lamp wasn't working so I delved into what the problem might be. I made a schematic but now my concepts seem to be wrong. I'm confused.
Turns out the tube wasn't getting properly seated.
 

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Yes, that's the way it works afaik too. A tube is made to operate on say around 170mA, and has a voltage drop according to length/wattage (some 40-70V).
The ballast provides a filament current while the momentary is closed, an inductive kick when it's released, and a ~ constant current in normal operation.
If one of the filaments are out you won't get preheating and hence no inductive kick either. Shorting the broken filament may enable re-lighting the tube.
 

davenn

Moderator
i dont get it !

If that circuit is correct, and that upper switch is truely momentary, ie. doesnt stay closed when the lower switch is closed, then there is gap in the circuit and no current can flow sustaining the light ??

Dave
 
Yes, the lower switch is permanent and the upper momentary. The current flow is dependent on the ionized gas in the tube conducting from one end to the other.
 
Yea, that's what I was thinking. It's for sure a momentary pushbutton. When pushed you can see the filaments glow and when released the entire tube fires up so to speak.
 
The ballast limits the current flow when the lamp is operating. The vapor is highly conductive, the bulb would fracture if it wasn't there. A ballast is an inductor, it will maintain the voltage at a level to maintain ionization but limit the current to milliamps
 

davenn

Moderator
Yes, the lower switch is permanent and the upper momentary. The current flow is dependent on the ionized gas in the tube conducting from one end to the other.

Doh of course, what was I thinking ??!! just having a "blonde" moment

apologies to all the blonds out there ;)

Dave
 
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