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Exploring Fuse Blowing with Non-Branded 220V Bulbs vs. Branded Ones

-Obywatel- 20046 17
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Treść została przetłumaczona polish » english Zobacz oryginalną wersję tematu
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  • #1 15548106
    -Obywatel-
    Level 17  
    For some time now I have had a problem with blowing fuses - including the pre-meter protection - when a 220V bulb burns out.

    This was not a problem when normal branded bulbs were available in stores. It has appeared since I was forced to switch to cheap "special purpose" bulbs. How is it possible that this problem did not occur with branded bulbs?

    The problem is that I have a larger stock of these bulbs and I don`t feel like throwing them into the trash. Would adding a smaller fuse on the lamp cable solve the problem? Are there no strong ones for an electric arc and will blow the "traffic jams" anyway?
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  • #2 15548673
    jdubowski
    Tube devices specialist  
    -Obywatel- wrote:
    For some time now I have had a problem with blowing fuses - including the pre-meter protection - when a 220V bulb burns out.


    How have you done your security grading?

    -Obywatel- wrote:
    This was not a problem when normal branded bulbs were available in stores. It has appeared since I was forced to switch to cheap "special purpose" bulbs. How is it possible that this problem did not occur with branded bulbs?


    The problem may be caused by the bulbs themselves, but not only. It may be caused by changed protections or by stiffening the network - short-circuit currents are higher today.

    -Obywatel- wrote:
    Would adding a smaller fuse on the lamp cable solve the problem? Are there no strong ones for an electric arc and will blow the "traffic jams" anyway?


    Theoretically, you have a fuse in the bulb - one of the wires in the leg is thinner. An additional fuse would solve the problem if it had adequate breaking power - no glass fuses without sand backfill are out of the question.

    Fogy wrote:
    The rule is to use a type B fuse, which will trip at a lower short-circuit current (within 10 ms at a current of 5 x Iz) before a type C fuse (which will trip at the same time at a current of 10 x Iz).


    Ordinary "ES" devices do not have switching selectivity, using B/C will support the selectivity of operation, but will not guarantee it.
  • #3 15548742
    Fogy
    Level 11  
    Quote:
    Ordinary "ES" devices do not have switching selectivity, using B/C will support the selectivity of operation, but will not guarantee it.


    Can you explain this?


    In addition, there remains the issue of the residual current circuit breaker tripping. During an electric arc that may occur when a bulb burns out, this switch will certainly operate, cutting off the power.
    Do I understand this correctly "krzychol66"?
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  • #4 15548854
    398216 Usunięty
    Level 43  
    -Obywatel- wrote:
    How is it possible that this problem did not occur with branded bulbs?
    Some of them had their own fuse built inside the bulb. Theoretically, when an arc occurs when the bulb burns out, it should be the first to work. Theoretically, because not always...
    In addition, the bulb filament itself is also important - if it is long, it is arranged on supporting springs in the shape of a circle - the risk of an arc occurring between the voltage supplies to the filament is small. Bulbs that have short filaments are more susceptible to such an arc. Halogen bulbs in particular are built this way, but since they are usually powered by a transformer, it "takes over" the current shock caused by the arc and "softens" the effect of a sudden increase in current "seen" by the power supply network. Of course, it is also important what type of fuse it is - with what delay it interrupts the circuit.
  • #5 15550013
    -Obywatel-
    Level 17  
    The socket circuit is protected with a B16 fuse. B40 pre-meter protection.

    Does it have any significance in this situation? :?:
  • #6 15550422
    Adam-T
    Level 41  
    -Obywatel- wrote:
    The socket circuit is protected with a B16 fuse. B40 pre-meter protection.

    And the lighting circuit?
    Fogy wrote:
    In addition, there remains the issue of the residual current circuit breaker tripping. During an electric arc that may occur when a bulb burns out, this switch will certainly operate, cutting off the power.

    If the RCD is without an overcurrent element, it will not work.
    jdubowski wrote:
    Theoretically, you have a fuse in the bulb - one of the wires in the leg is thinner.

    At work I once had such a batch (30 pieces) of light bulbs that in 90% of cases the fuse burned out and the filament remained intact.
    Fogy wrote:
    What if we used a C circuit breaker instead of B40? Or instead of the previous B16, use A16 (if available)?

    I have not encountered characteristic A. However, a better solution would be to replace the B40 with 40A fuse links with gG time-lag characteristics, provided that the short-circuit loop impedance measurements allow it.
  • #7 15550476
    -Obywatel-
    Level 17  
    Adam-T wrote:
    -Obywatel- wrote:
    The socket circuit is protected with a B16 fuse. B40 pre-meter protection.

    And the lighting circuit?


    B10 lights, but this is a lamp that plugs into a socket. The bulb is Ukrainian Iskra E14 40W.

    But your question gave me some food for thought - in "upper" lighting I also have bulbs shooting - sometimes even the bubble will fly away :D , but they don`t blow fuses. There, in turn, there is E27 100W placed horizontally. In the ill-fated lamp, the bulb is positioned vertically with the bulb down.
  • #8 15550546
    Adam-T
    Level 41  
    -Obywatel- wrote:
    The bulb is Ukrainian Iskra E14 40W

    Well, I guess the only explanation left is "this guy is like that".

    I deleted the rest of the post. It referred to a deleted post, and secondly, it was incorrect. [retrofood]
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  • #9 15550841
    jdubowski
    Tube devices specialist  
    Fogy wrote:
    Quote:
    Ordinary "ES" devices do not have switching selectivity, using B/C will support the selectivity of operation, but will not guarantee it.


    Can you explain this?


    In a classic fuse, the function is simple - there is only one element responsible for both short-circuit and overload operation (wire).
    In an "esa" or other "automatic fuse" we have two elements - a thermal release - a bimetal and an electromagnetic short-circuit release. 10A from 20A differs mainly in the bimetal trigger.

    Fogy wrote:
    In addition, there remains the issue of the residual current circuit breaker tripping. During an electric arc that may occur when a bulb burns out, this switch will certainly operate, cutting off the power.


    Why will it work? If it works, it`s only "by accident".

    I removed some of the content. [retrofood]

    Mariopi wrote:
    It used to be possible to repair a burnt-out light bulb - current designs no longer allow this, because the leads burn out - hence such sensations as an electric arc flash and the failure of the protection devices.


    On the contrary - the arc is the cause after which the leads "evaporate".
    Arc causes:
    - use of carbonated bulbs (nitrogen + argon) to increase effectiveness
    - use of double-stranded cables
    - power supply from a network with low short-circuit loop resistance.
  • #10 15557820
    rafbid
    Level 33  
    luke666 wrote:
    I give all my points for a rational explanation as to why a residual current circuit breaker would trip when a light bulb burns out.
    When the filament burns out, a small electric arc is created in the bulb. In the area of the electric arc, the gas is strongly ionized and constitutes plasma. Ions are created when an atom or molecule loses or gains charge. As a result of applying appropriate energy, the molecule loses an electron and becomes a positive ion (predominance of positive charges over negative ones), or gains an electron and becomes a negative ion. The key feature of ions is that they are electrically unstable and seek out other molecules to bind to. Based on this, it can be concluded that current leakage occurs.
  • #11 15562055
    retrofood
    VIP Meritorious for electroda.pl  
    jdubowski wrote:

    Why will it work? If it works, it`s only "by accident".


    NO. Due to exceeding the appropriate value of short-circuit current.

    rafbid wrote:
    When the filament burns out, a small electric arc is created in the bulb.

    It`s rising miscellaneous electric arc, sometimes very large.

    Added after 1 [minute]:

    -Obywatel- wrote:
    I apologize in advance for digging up such an old topic, but I couldn`t find a fresher one.


    Yeah? The last one was in March. And at the latest in February.
  • #12 15562771
    CYRUS2
    Level 43  
    Fogy wrote:
    The rule is to use a type B fuse, which will trip at a lower short-circuit current (within 10 ms at a current of 5 x Iz) before a type C fuse (which will trip at the same time at a current of 10 x Iz).
    There is no such rule.
    jdubowski wrote:
    Ordinary "ES" devices do not have switching selectivity, using B/C will support the selectivity of operation, but will not guarantee it.
    There is no support. Fuses are zero-one devices, or "digital".
    rafbid wrote:
    When the filament burns out, a small electric arc is created in the bulb.
    Nonsense, there is no arc. A flash of light is not an arc.
    rafbid wrote:
    or it gains an electron and becomes a negative ion. The key feature of ions is that they are electrically unstable and seek out other molecules to bind to. Based on this, it can be concluded that current leakage occurs.
    You`ve read too many books and you fantasize.
  • #13 15563035
    jdubowski
    Tube devices specialist  
    CYRUS2 wrote:
    rafbid wrote:
    When the filament burns out, a small electric arc is created in the bulb.
    Nonsense, there is no arc. A flash of light is not an arc.


    There is a discharge in a gas with a pressure close to atmospheric - if not an arc, what other term do you propose?
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  • #14 15563729
    CYRUS2
    Level 43  
    A flash of light.
    The tungsten evaporated from the fiber fragment, it became thinner and therefore overheated = temperature increase and increased light.
    Notice that the halogen bulb is much brighter than the classic 60W bulb. Due to higher temperature. Not an arc at all.
  • #15 15563751
    retrofood
    VIP Meritorious for electroda.pl  
    CYRUS2 wrote:
    A flash of light.
    The tungsten evaporated from the fiber fragment, it became thinner and therefore overheated = temperature increase and increased light.
    Notice that the halogen bulb is much brighter than the classic 60W bulb. Due to higher temperature. Not an arc at all.


    What is interesting is that the flashes of light cause the evaporation and spraying of not only the filament, but even the wires inside the bulb, into the bulb. Flashes of light can sometimes even scatter bubbled glasses all over the room.

    The occurrence of an electric arc when the filament burns out in some light bulbs is obvious and more and more common. The current is then forced to be many times higher than the rated current, which causes the activation of all protections in series, and even RCD protection.
    There is no way to prevent this, although this phenomenon occurs less frequently in bulbs from reputable manufacturers.
  • #16 15563806
    rafbid
    Level 33  
    CYRUS2 wrote:
    The tungsten evaporated from the fiber fragment, it became thinner and therefore overheated = temperature increase and increased light.
    Tungsten evaporates from the filament over the life of the bulb and therefore has a limited lifespan. Over time, their light parameters change (for the worse), so the resistance should increase until failure occurs.
  • #17 15564176
    zbich70
    Level 43  
    rafbid wrote:
    Tungsten evaporates from the filament over the life of the bulb and therefore has a limited lifespan. Over time, their light parameters change (for the worse), so the resistance should increase until failure occurs.

    The only problem is that tungsten does not evaporate equally along the entire length of the filament.
    And where it evaporates faster, the cross-section decreases, the resistance increases and therefore the heat release increases.
    And we know the consequences...
  • #18 15564503
    Akrzy74
    Rest in Peace  
    Search engine disgust. I`m closing.
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Topic summary

The discussion revolves around the issue of fuses blowing when non-branded 220V bulbs burn out, a problem not encountered with branded bulbs. Participants suggest that the issue may stem from the bulbs themselves, changes in electrical protections, or increased short-circuit currents. It is noted that some bulbs have built-in fuses that should ideally prevent such occurrences, but this is not always effective. The type of fuse used in the circuit is also critical, with recommendations for using time-lag fuses to better handle the current surges caused by electric arcs during bulb failure. The positioning of bulbs (vertical vs. horizontal) and filament design are discussed as factors influencing the likelihood of arcing. The conversation highlights the importance of understanding the electrical characteristics of both the bulbs and the circuit protections in place.
Summary generated by the language model.
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