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Why don't fluorescent bulbs blow up when electricity is connected?

  1. Apr 15, 2013 #1
    Why don't fluorescent bulbs blow up when electricity is connected? My understanding is that gas inside heats up to plasma to make light. Wouldn't the heat of plasma expand the gas, add extra pressure and the pressure would break the glass? Maybe super low pressure inside the bulb to start with?
     
  2. jcsd
  3. Apr 15, 2013 #2
    I'm a noob myself, so I'm not sure about your question. However, I got this off Wikipedia:

    "A fluorescent lamp tube is filled with a gas containing low pressure mercury vapor and argon, xenon, neon, or krypton. The pressure inside the lamp is around 0.3% of atmospheric pressure." http://en.wikipedia.org/wiki/Fluorescent_lamp
     
  4. Apr 15, 2013 #3
    Yea I found that after my post too. .3% is really nothing, this means most fluorescent bulbs are basically empty vacuum with tiny tiny bit of gas.

    Aaaand they all use ballasts, which basically controls how much electricity goes through the bulb. So I'm guessing here since the electricity is limited so is the temp/pressure inside the bulb. In other words, ballast is controlling pressure inside the bulb indirectly. Which keeps the pressure from exceeding the pressure at which the bulb glass would blow up.

    Learn something new everyday
     
  5. Apr 16, 2013 #4

    sophiecentaur

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    Without a ballast, the fuse would blow or the glass envelope would melt before the gas pressure became 'explosive'. Try working out the temperature you need to raise gas that starts at 0.3% AP before it would reach, say 2AP. (Assume the gas laws apply)
     
  6. Apr 16, 2013 #5

    HallsofIvy

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    Not everything "blows up" when you add pressure. The neon light bulb is strong enough to withstand the slight increase in air pressure inside it.
     
  7. Apr 16, 2013 #6
    Yes but tuning something into plasma isn't a slight increase in temp, its huge so the pressure would be huge too.


    But the use of a ballast to limit current and use of super low pressure in tubes explains why no booom!
     
  8. Apr 16, 2013 #7

    sophiecentaur

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    I don't think the limit would be due to temperature-related pressure. It would be blown fuses long before that happened. You'd be limited, briefly, to an arc power of around 1.5kW in any domestic lighting circuit. That's not many Joules in the short time a fuse takes to operate.

    But, of course, as you say, that's what the ballast is there for.
     
  9. Apr 16, 2013 #8
    This is too much of a simplification. The emission isn't due to the temperature of the gas, it's due to the transfer of energy from the current in the ionised gas (which does not have to be particularly hot) to the electronic orbitals of the mercury vapour and the fluorescent material. These electrons then fall back to lower energy orbitals and emit in the visible light.
     
  10. Apr 17, 2013 #9

    sophiecentaur

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    Modern fluorescent lights produce UV which excites phosphors on the tube surface to produce broad band white(ish) light.

    I have a feeling that there would be a limit to the current capacity of the column of gas - due to the number of available charge carriers - and hence the possible maximun power dissipation. I will try a back of a fag packet calculation to establish a ball park figure. Unless someone can do it first.
     
  11. Apr 17, 2013 #10
    I think sophiecentaur is right you have a given amount of mercury vapor which by the way is the current carrier in a gas discharge lamp not the gas itself because the voltages in a typical gas discharge lamp are not so high for them to make an arc that long over a gas like xenon and others.
    So after the mercury vapor "burns out" the lamp can't start anymore or is failing after several seconds of operation.
    Also using one without the ballast will in most cases result in a current overload which will burn out the mercury vapor and make the lamp useless again.
    But even in these situations no explosion occurs a gas discharge lamp is not a grenade guys :)
     
  12. Apr 17, 2013 #11

    SteamKing

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    Is it OK to say 'fag packet' these days?
     
  13. Apr 17, 2013 #12
    I don't think anything I said contradicts this.

    Strictly speaking though, mercury has emission bands in the visible range.
     
  14. Apr 17, 2013 #13

    sophiecentaur

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    I don't think I was contradicting you, either.
    It is true that mercury vapour produces visible emissions. However, the light that is produced 'directly' is pretty unpleasant as a light source and plain old mercury lamps would only be used for high efficiency industrial lighting (and, I believe, the UV lamps that used to be available in Boot's the Chemist). It is the UV component that makes them suitable for domestic lighting because of the phosphors that are included these days. Iirc, the high power mercury lamps are of a fairly different construction to the domestic type - more along the lines of a mercury arc rectifier than a simple medium voltage fluo tube.
     
  15. Apr 17, 2013 #14

    sophiecentaur

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    BTW, Did a 'cigarette' packet calculation and I reckon that, even with the ions moving at c, with the density of (singly ionised) ions available, over a cross section of 1cm^2, you would limit at about 1800A. If you take into account the mean free path at that pressure and the application of only 200V total, I reckon the limiting current would be not many tens of amps because of the limited speed that the electrons can achieve in such a congested place.
    That calculation was pretty rough and ready and I could be way out, of course. I may re-visit it when the mood takes me - or someone else could show just how wrong I could be. (Challenge)
    I agree about the non-grenadelike nature of the process though!
     
  16. Apr 17, 2013 #15
    calculations are good ofcourse but this is one of those question to which you know the answer when you have atleast a basic understanding in physics and electricity.
     
  17. Apr 17, 2013 #16

    sophiecentaur

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    umm. Not sure about that. The numbers really count in cases like this one, I think. The actual value of supply volts is highly relevant, for instance.
     
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