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When a fluorescent light fixture goes out, it often flickers and then after a while stops shining altogether. When this happens and when it finally dies is the circuit broken and does it still use electricity?

I don't understand too much what exactly is spent or breaks in a fluorescent light bulb when it is at the end of its life and I would appreciate any explanation. Here's a video about what I'm talking about:

https://www.youtube.com/watch?v=NDnKEOeFJn0

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  • \$\begingroup\$ Electronic circuits can do whatever the designer has failed to design them not to do - so can draw power when the tube is dead if the designer did not think things through well enough. And some do. Even some older school fluorescent drive circuits can dissipate power when the tube is dead. A few days ago I disconnected a fluro fitting with "iron ballast" where the fitting around the ballast area was discernibly above ambient temperature but the bulb had long ago ceased trying to light up the world. I do not know circuit details in that case. \$\endgroup\$ – Russell McMahon Oct 22 '14 at 0:05
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    \$\begingroup\$ Don't know the science, but I changed 2 bad ballast in my office of 12 lights when a necessary light went out and the electric bill dropped by half. Wished I had done it a long time ago. \$\endgroup\$ – user77644 May 30 '15 at 11:48
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I'm going to go out on a limb and say this question is valuable from the point of view of electronic design, as it pertains to some fundamental understanding on how fluorescent lights work.

Fluorescent lights work by accelerating electrons from the cathode to the anode in an almost-vacuum environment. In this vacuum is mercury vapour, and when the electron hits a mercury atom, that Hg atom goes into an excited state and outputs one or more photons of UV light upon decay. These UV photons then hit the phosphor-based coating on the inside of the glass tube, which converts these UV photons to visible white light.

So, in order to function, it is vitally important for these lights to have a lot of 'free' electrons available to shoot at the mercury. One way to make electrons more mobile and likely to shoot off the cathode is to heat it up, and this is what a so-called 'starter' circuit does: it is essentially nothing more than a high voltage generator and a heating coil. The heating coil heats up the electrode to mobilize the electrons and the high voltage generator (usually just a resonant LC pump) creates enough voltage for the initial 'spark' to ignite the bulb. Once electrons start flowing and the lamp is 'on', the gas inside the lamp looks more like a plasma and is very conductive, so neither the high voltage nor the addition of heat is necessary to keep it working. Hence, it's just a starter, once the bulb is on, it is shut down.

Old-style starters would keep trying to fire the bulb even when the electrodes were entirely spent. This means that that heating coil would be running until its filament would burn out. In a lot of cases this would mean the bulb has a higher power consumption after it's died.

Modern electronic starters 'give up' after a few tries when they detect that the bulb won't start. After that they use up no or almost no energy until power is cycled to the starter.

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    \$\begingroup\$ Excellent answer, but what causes the light to flicker (lower frequency flicker, not the refresh rate) when the light is near the end of it's life? \$\endgroup\$ – Jarrod Christman Oct 16 '14 at 18:05
  • \$\begingroup\$ Back in the olden days, fluorescent light fixtures had a ballast and a starter. The starter was replaceable; its job was to shut off the filaments once the lamp had heated up. There's no longer a separate starter in the old sense, but I think the correct term for that heavy thing that controls the lamp is still ballast, even though it does the jobs of both of the older components. \$\endgroup\$ – Pete Becker Oct 16 '14 at 18:06
  • \$\begingroup\$ On a vaguely related note, it should perhaps be noted that CFLs can also use more power after failing, making them hotter than expected. (Something to watch out for when changing a dead bulb!) I'm guessing that CFLs are designed to be as inexpensive as possible so they don't have the same cut-out logic as modern fluorescent ballasts. \$\endgroup\$ – Harry Johnston Oct 16 '14 at 21:40
  • \$\begingroup\$ @HarryJohnston I did some research on that too and found that CFLs usually end up burning out the starter or the electronics that control the starter so they actually end up being an open circuit. It is true that they are cheaper and don't have the protections in the modern fluorescent ballasts. \$\endgroup\$ – PressingOnAlways Oct 17 '14 at 3:31
  • \$\begingroup\$ "neither the high voltage nor the addition of heat is necessary to keep it from working" Is there a spare "from" in there, in case it turns out to be needed later? :) \$\endgroup\$ – a CVn Oct 17 '14 at 7:41
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A couple of months ago I changed out 8 old-school fluorescent tubes in two fixtures in my basement ceiling. I hate changing the tunes out,so I waited until absolutely necessary - we're talking YEARS here...I don't remember exactly how many were flickering, but in at least one of the fixtures, two bulbs were completely dead and the other two were flickering gently to each other. I'm not sure how the circuit is setup, but there is some sort of interdependency between the tubes. To make a long story short, even with the addition of an electric car two months ago, my electric bill has been cut in half. I'm not ready to blame the bulbs yet, it could be a coincidentally bad meter read, but it totally makes sense that a flickering bulb would draw a heck of a lot more current.

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I inserted an amp meter inline with a 2 tube fluorescent fixture. With both tubes working, the current draw was .74 amps. With both lamps removed, the current dropped to .45 amps. I was amazed it drew that much current. This is an old magnetic type ballast. I wonder if the electronic ballasts draw less when unloaded.

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    \$\begingroup\$ Did the ballast have power factor correction capacitors? If so the majority of the current could be at leading power factor and not charged for, or causing any dissipation. \$\endgroup\$ – Kevin White Sep 21 '15 at 21:14
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The answer is "yes". Ballasts will still draw power even without fluorescent tubes in place, or with blown tubes. Two ways to show this: 1) the ballast still gets warm (although less so than with a tube), and 2) if you unscrew the line leads to the ballast with the current on (not recommended for safety reasons), you will see a little spark, indicating current flow.

Forget fluorescent altogether, anyway. LEDs have finally reached the level of quality consistency and price point to be a viable alternative. If you have questions, let me know.

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