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Florescent lighting?

Discussion in 'Lighting' started by Tim, Mar 31, 2005.

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  1. Tim

    Tim Guest

    Is it feasable, or doable to dim an 18 watt Tri phosphor white tube ( 2
    foot)?
    Looking for about 50%intensity range.

    I think they need a voltage across them to maintain ionization, but perhaps
    current control?
    Any pointers?
     
  2. You haven't given us the exact lamp type but almost all fluorescent
    lamps are dimmable. The only exceptions I can think of are lamps that
    have an integral glow switch starter, and almost all of these are
    compact fluorescent lamps. (The reason dimming is a problem when the
    lamp has a glow switch attached, is that the lamp voltage rises when
    the current is decreased and this can trigger the glow switch for some
    lamp types.)

    Instant start fluorescent lamps, those with only one pin on each end
    or those that have two pins on each end that are shorter together, are
    dimmable, but the dimming range is limited since the electrodes cannot
    be heated from an external source.
    All discharge lamps, including fluorescent lamps, must be operated on
    a ballast that controls the lamp current while providing the voltage
    that the lamp wants to run at. To repeat - you choose the current. The
    lamp chooses the voltage. Just the reverse of the more common voltage
    source.
    Google is your friend. Also, International Rectifier, www.irf.com, has
    a number of application notes on ballast design, including circuit
    diagrams. Note that all their circuits are designed to use their
    ballast control ICs, which are not necessary in other designs.

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  3. Tim

    Tim Guest

    Thanks for the feedback, I would be happy with being able to dimm from
    100% brightness, to about 50% level.
    The tube type is Sylvania, 18W/860 daylight with a diameter of about an
    inch, and two pins at either end. I was wondering if I can add any other
    components to a standard ballast, to vary/limit the current. I would like
    to be able to vary the brightness to see what output is right for a custom
    made light fitting, and then set the level at that setting pemamnently. I
    guess I may need a variable current source, and some understanding of the
    operation of the existing ballast, or is there a better, simpler way of
    achieving what I need??
     
  4. Whet kind of ballast? If you are using a line frequency EM ballast
    then you can dim the lamp to some extent by adding additional
    impedance between the lamp and ballast - this was done in the early
    days of the energy crunch in the 1970's, before low energy lamps and
    ballasts were available. However, if the lamp is rapid start you still
    need a way to heat the lamp electrodes. Those early low wattage lamps
    with built-in series impedance also had a small 1:1 transformer added
    to carry the electrode heating current around the additional series
    impedance.

    If you have an electronic ballast, the various designs are too
    different to suggest an external solution. If you understand
    electronics, you could go into the ballast and change the operating
    frequency, which will change the lamp current. However, this requires
    a substantial bit of knowledge. Messing with an electronic ballast is
    quite difficult and risky, for both you and the ballast.

    Have you considered purchasing a dimming ballast made for this lamp?
    Or a ballast made for a lamp that runs at about 1/2 the current of
    your lamp but the same voltage?

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  5. You should be able to buy an electronic diming ballast with a
    0-10V control voltage. When using just a single ballast, with
    some of these you can just connect a small potentiometer to the
    dimming input as they contain a low current internal 10V supply
    just for this purpose.
     
  6. JM

    JM Guest

    Get rid of the starter, and use a low voltage transformer to supply the
    cathode heaters. Now just dim the lamp, while keeping the cathode heaters at
    full power. Or, otherwise get a regular dimming ballast. I'm sure they are
    available as the 18w lamps are "their" equivalent to "our" 17w lamps.
     
  7. Using a capacitor at line frequency is a bad idea unless the
    calculations are done carefully. If the capacitive reactance is
    smaller than the inductive reactance, the capacitive reactance will
    always subtract from the inductive reactance, reducing the 50 Hz or 60
    Hz impedance, instead of increasing it, and increasing lamp current
    instead of decreasing it. Once the capacitive reactance is greater
    than the inductive reactance, the capacitor will dominate, but the
    inductive reactance will subtract from the capacitive reactance.
    Better to use an inductor, which will always decrease current for a
    simple series inductive ballast.

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  8. Also tends to produce terrible lamp CCF at line frequency, though it
    might not if placed in series with an inductor with sufficient
    inductance to ballast the lamp by itself.

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  9. No, a C only ballast very quickly wrecks a lamp on 240V 50Hz -- I've
    tried it.

    CL is fine -- Thorn Lighting had a patent for many years for their
    lead-lag twin tube fittings getting near unity power factor.
     
  10. We're getting repetitive. Typical CL ballasts on 120-volt mains have
    terrible CCF. They are used in very cheap "shoplight" fixtures and
    lamp life is very short. As stated earlier, if the L in a CL ballast
    is a large as it would need to be in an L-alone ballast, then I
    suspect the CCF would be OK, but have not done any calculations or
    measurements. I can't comment on 240-volt circuits.

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  11.  
  12. TKM

    TKM Guest

    If you don't need dynamic dimming, just sleeve the lamp with a neutral
    density filter (theatrical gel works). It's not energy efficient, of
    course.

    Terry McGowan
     
  13. Gee Terry - Just like you to think of a simple solution that
    works instead of the Rube Goldberg solutions the rest of us are
    suggesting :)

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  14. TKM

    TKM Guest

    Well, I was going to suggest how we really do it when fluorescent lamps are
    too bright -- say behind a piece of diffuse plastic that is part of a
    transilluminated sign. We wrap the lamps with black electrical tape. But
    that's pretty crude, the tape gets messy with the heat and it's probably
    unsafe; so I don't recommend it.

    Terry McGowan
     
  15.  
  16. Not really - see below.
    They are actually very comparable as far as the negative incremental
    impedance problem is concerned, which is the issue here. Why do you
    think that are not comparable?
    The current runaway in fluorescent lamps is not related to lamp
    temperature, it is a result of increasing electron density. And, the
    electron density can increase quite rapidly. If you were to connect
    your 60-volt lamp to the 240-volt power line the current would rise to
    very high values in less than 1 msec, probably less than 100 usec.
    Other than the free electrons, which have so little mass that they
    don't add to the "temperature" of the lamp as we conventionally think
    of temperature, nothing would get hot except for lamp electrodes which
    would be vaporized by the high current.
    It is not "evened out." The current varies as a distorted sine wave,
    while the voltage is almost a square wave, so the conductance is
    changing constantly over a 50 Hz half-cycle. And, the only reason why
    the current follows the shape of the power input is because a proper
    ballast is being used.
    Near the zero crossings the conductivity has decreased quite a bit due
    to electron-ion recombination. This can be seen from the fact that the
    lamp voltage is still quite high while the current is almost zero. At
    the zero crossing itself the lamp voltage finally collapses. Even on
    an inductor ballast there is a "reignition" spike on the lamp voltage.
    One advantage of an inductive ballast is the voltage kick that is
    generated at the zero crossing of the lamp current that helps to
    restart the lamp.
    If you had voltage waveforms you would see that the lamp voltage is
    much higher than ~ 60 volts at the start of each cycle when you are
    operating from a C ballast. And, the longer you keep the lamp off, the
    higher the reignition voltage spike will be. The reason your lamp
    restarts without intervention from you (such as reheating the
    electrodes) is because your peak operating voltage is more than 5.5
    times the operating voltage of the lamp, and the electrodes have a
    reasonable thermal mass.
    There is a classic paper by PC Drop and J Polman titled "Calculations
    on the Effect of Supply Frequency on the Positive Column of a
    Low-Pressure Hg-Ar AC discharge," J. Phys. D: Appl Phys., Vol. 5,
    1972. Figure 2 of that paper shows the various quantities in the
    fluorescent lamp discharge as a function of time over a 50 Hz
    half-cycle. The electron density generally follows the current
    waveshape. At the start and end of the 50 Hz half-cycle the electron
    density is about 25% of the peak value. There is a clear "reignition"
    peak in the electric field at the start of each half-cycle which is
    about 1.4 times the mean value of the electric field.

    There is no significant amount of ionized Ar in the positive column.
    Any ionized Hg must be balanced by free electrons since the plasma
    must be neutral, so the Hg ion density varies as the electron density.
    I don't understand your comment that the xenon lamp dumps its load.
    The energy is stored in the capacitor, not the lamp. The energy is
    "dumped" from the capacitor into the lamp. The flash is intense and of
    short duration because of the nature of a capacitor and the high
    conductance of the xenon discharge.
    The lamp current waveform will follow the mains waveform only with a
    proper current limiting ballast.
    See discussion above about "thermal" effects. Thermal inertia as you
    normally think of it does not control current runaway in a low
    pressure discharge.
    I don't understand. Using the data from Drop and Polman, if you had an
    inductive ballast the voltage at the start of each cycle would be
    about 1.4 times the average operating voltage. If the discharge is off
    for longer than the short period of time than is typical of an
    inductive ballast, the voltage spike at the start of each cycle would
    be higher.
    As discussed above, this is not about the thermal inertia of a
    tungsten filament.
    But you still haven't said that you ever looked at the lamp current
    waveform. Certainly, with a low enough lamp voltage the current will
    be the same shape as the line voltage. When a capacitor is placed
    directly across an AC power line, the current through it matches the
    shape of the voltage waveform. So if we imagine that we have a lamp
    whose operating voltage is, say, 1/1000 of the line voltage, then the
    current distortion caused by this very low voltage non-linear load
    placed in series with the capacitor would be small and the current
    would still be substantially sinusoidal. As the voltage of the
    non-linear lamp becomes a larger fraction of the line voltage, the
    amount of current distortion will increase. I believe that your case
    will have significant current distortion, but until I model it or
    measure it I cannot tell you how much distortion.
    I am assuming a perfect capacitor, that is zero ESR. If you want to
    add a partial resistive ballast later that is fine, and it will
    improve the waveform. And, yes a 60-volt lamps on 240-volt mains will
    have lower CCF than a 60-volt lamp on 120-volt mains. A 100-volt lamp
    will not operate on 120-volt mains with a capacitor ballast. In fact a
    100-volt lamp will not run properly on 120-volt mains with an inductor
    ballast, but that doesn't explain how those cheap LC ballasts manage
    to operate 100-volt lamps from 120-volt power lines. .

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  17. The reason that high lamp CCF damages electrodes is that for short
    periods of time each cycle the lamp current rises to values higher
    than those that can be provided by the electrode via thermionic
    emission. When thermionic emission is not sufficient to provide the
    required number of electrons, the cathode fall voltage rises, which
    increases the ion impact energy. The impact of high velocity ions on
    the surface of the electrode releases the additional electrons
    required by the discharge, but these impacts also sputter the electron
    emissive coating off the electrode, shortening its life.

    You are correct that the thermal inertia of the electrode keep it from
    physically melting during these cyclic short-term high current events,
    but it is that same thermal inertia that makes it impossible for the
    electrode temperature to rise quickly enough to provide sufficient
    thermionic emission during the high current portions of a high CCF
    waveform - and this is when the real electrode damage takes place.
    It's not about temperature. See above. As for how fast the energy from
    the capacitor is dissipated in the lamp, I suggest you either measure
    or model the current waveform.
    It is not about temperature.
    I suggest you read good text on low pressure discharges. The role of
    the ballast is to control lamp current, and the current can rise to
    destructive values in far less time than it takes any part of the
    "lamp" to heat up. The free electrons heat quickly to about 11, 000 K,
    but their total mass is so small that if you placed a thermometer in
    the lamp it would measure the temperature of the Hg and rare gas
    atoms, which are only slightly above room temperature, not the
    temperature of the electrons.

    I can understand your "final purpose" comment in that the final result
    of current runaway is often the physical destruction of parts of the
    lamp - usually the lamp electrodes - and the time constant for the
    electrodes to melt is limited by their thermal inertia. But, as was
    discussed above, the electrodes are damaged by short-term high current
    events long before their temperature rises.

    In the end a ballast must be designed to control lamp current on the
    time scale of electron creation and destruction process, not the
    thermal time constants of the electrodes or any other part of the
    lamp. It the ballast cannot control current on the time scale of the
    electron creation/destruction process, the lamp discharge will run
    away.
    The electron emissive coating is damaged by a high CCF lamp waveform.
    Can you post a picture of this lamp? Are the ends clean or dark? It is
    possible that your electrodes have been sputtered down to the bare
    tungsten but you lamp continues to run since your supply voltage is so
    high compared to the normal lamp operating voltage that you can
    support the higher cathode fall voltage of bare tungsten compared to
    barium coated tungsten. This sometimes happens in the US when lamps
    operating on instant start ballasts lose their electron emissive
    coating. In fact, lamps have been known to run on IS ballasts after
    the electrode has been broken or even after the electrode has fallen
    off the support wires - in that case the support wires become the
    electrodes. This is a dangerous mode of operation since the high
    cathode fall voltage heats the glass near the electrodes to high
    temperatures. This can cause the glass to break which can, in turn,
    allow the lamp to fall out of the socket.
    The thermal mass only keeps the electrodes from melting. It does not
    prevent damage to the electron emissive coating.
    As I said, the 1.4 factor was for an inductive ballast that had
    minimum off time between half-cycles. For a capacitor ballast that
    produces longer off time, the voltage at the start of each half-cycle
    will be significantly higher.
    It is not about temperature.

    [snip]
    Yes, a capacitor does have finite energy each cycle and does provide
    "energy control", but it also allows very high peak currents which
    will exceed the thermionic emission capability of the electrode and
    thus damage the electrode electron emissive coating as described at
    the top.


    [snip]
    I suggest you do look at the waveform.

    [snip]
    The small advantage to the grid is a big disadvantage to the lamp.
    Not necessarily. Fluorescent lamps are less efficient when operated at
    high peak currents with long off times.

    This will be my last reply to your messages in this thread. I have
    explained lamp current runaway and its consequences to the best of my
    ability.

    I did dig out my PSPICE lamp model last night and reran old
    simulations for both resistive and inductive ballasts at 60 Hz just to
    make sure it ran on this relatively new computer. If I have time I
    will run a C ballast simulation and post the waveforms to the group.
    I might even make some real measurements :)

    --
    Vic Roberts
    http://www.RobertsResearchInc.com
    To reply via e-mail:
    replace xxx with vdr in the Reply to: address
    or use e-mail address listed at the Web site.

    This information is provided for educational purposes only.
    It may not be used in any publication or posted on any Web
    site without written permission.
     
  18. R.Lewis

    R.Lewis Guest

    <snip>

    There seems to be a fair amount of 'mumbo jumbo' coming out here.
    for example:-
    What has the charge (in coulomb, not joule) got to do with the price of
    cheese?
    Provided we know how fast this charge is passed through the 'tube we may
    well be able to glean something about the (electron) energy density of the
    discharge as this time.

    That, per se, is not a major failure mechanism of commercial fluo tubes.
    Pulling metal atoms off the cathodes is.
     
  19. Nope, none of the text you quoted was from me.
     
  20. I expect any electron cloud around a fluorescent lamp filament to supply
    high peak currents for only a small fraction of a microsecond, probably
    less than a nanosecond. How much capacitance does that electrode cloud
    have - a few picofarads? The excessive peak current that results from a C
    ballast lasts several microseconds.
    Was that lamp run on a purely C ballast or one with both C and L in
    series? If the ballast was purely C, can you tell us the capacitance
    value, capacitor type, and lamp type so that one can attempt to duplicate
    these results?

    - Don Klipstein ()
     
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