17
\$\begingroup\$

I use light bulbs (common incandescent) in the room and they burn out periodically, ca. one bulb in 4 months burns out. Very often, exactly at the moment when it happens, I see an arc (blueish light flash) in the switch, so the bulb always burns when being switched on. I don't see an arc anytime the bulb is normally switched, but only when the bulb burns out.

Question: does the bulb burn after the arc appears, or is it vice versa - when the filament burns it causes the arc in the switch?
My thought is that the bulb burns simply because of worn out filament, but why is there almost always an arc in the switch at this moment?

\$\endgroup\$
  • 5
    \$\begingroup\$ you should only turn on the switch at zero crossings to avoid blown bulbs... \$\endgroup\$ – dandavis Oct 8 '17 at 6:43
  • \$\begingroup\$ Inrush surge. Consider a switch type dimmer, as you turn it on the power will rise from zero at a slower rate & the same for shutdown. \$\endgroup\$ – Optionparty Oct 9 '17 at 1:34
  • \$\begingroup\$ In Germany it's common to separate wall switch from lamp by a relay. I mean there is a dedicated relay box in almost every home. This reduces bouncing. Also prevents fitality from bad wall switches \$\endgroup\$ – Sean87 Oct 10 '17 at 18:04
  • 1
    \$\begingroup\$ @Sean87 I'm in Germany, so i think my home is not a common one :) \$\endgroup\$ – Mikhail V Oct 10 '17 at 23:45
  • \$\begingroup\$ @MikhailV I assume you are not in a DDR territory :D \$\endgroup\$ – Sean87 Oct 11 '17 at 12:02
13
\$\begingroup\$

When the lamps fails, it is often the case that the delicate filament collapes on itself, causing a short circuit. This causes a momentary peak of current. So much current that the short almost immediately blows itself open circuit again, due to a teeny explosion.

The lamp usually fails when switched on because the resistance of most materials, including the lamp filament, rises with temperature. When cold, it draws far more current than when hot- 10 or 15 times as much. So that is when the filament is under the most stress as it very rapidly heats up, so by far the most likely time for it to break. Additionally, it is because the AC wave is partway through rather than passing through zero, causing a sudden buildup of the electromagnetic field, causing a physical shock to affect the filament too.

Incandescent filament lamps last much longer if you never switch them off and on, for these reasons.

So to answer the question, the flash in the switch is caused by the failure.

\$\endgroup\$
  • \$\begingroup\$ This answer explains why the failure of the lamp happens at the particular time of it being turned on, but not the actual question: how does that create an arc in the switch? The large current you describe doesn't start until you close the switch, right? And once the switch is closed, the current ought to go through the closed switch rather than arc there. \$\endgroup\$ – Henning Makholm Oct 8 '17 at 8:43
  • \$\begingroup\$ @HenningMakholm: Contact bounce, as suggested by Tony Stewart, might explain it. (Or, as an alternative explanation, maybe the lamp filament could've already failed short when it was last switched off, so that the switch will arc as soon as the contacts get close enough for a spark to jump the gap. Then, presumably, the shorted filament acts as a fuse and quickly fails open before the actual fuse or circuit breaker has time to trigger. But that's really just a wild guess, and I've no idea if it even makes sense.) \$\endgroup\$ – Ilmari Karonen Oct 8 '17 at 11:53
  • 2
    \$\begingroup\$ Okay, well, the current starts flowing very slightly before the contacts make due to a little arc through the air as they make. This is one reason switch and relay contacts deteriorate over time. (Paradoxically, very very low currents through contacts also cause deteroriation as there is no tiny explosion from that arc to clean crud of the contact area). So your arc has already started when the contacts make. So you've got a very high current flowing as the contacts make as the filament simultaneously fails, enough of an arc to cause a visible flash. \$\endgroup\$ – Ian Bland Oct 8 '17 at 14:45
  • 1
    \$\begingroup\$ With the relatively crude contacts on a domestic electrical switch, high current loads may show a visible flash in a darkened environment every time you switch them, not just when there is a failure. Of course you can simulate this any time by (taking safety precautions) tapping a live, loaded wire onto a brass contact. Don't try this at home if you don't know what you're doing, kidz. \$\endgroup\$ – Ian Bland Oct 8 '17 at 14:47
  • 1
    \$\begingroup\$ "when the filament is under the most stress as it very rapidly heats up, so by far the most likely time for it to break." Would this mean that installing "dimmers" that slowly ramp up & ramp down the current so that the lights 'fade' on and off instead of a hard switch would significantly increase the lifespans of the bulbs? \$\endgroup\$ – Robotnik Oct 8 '17 at 23:15
7
\$\begingroup\$

Contact bounce at 10x rated current with inductive lines = Arc + surge current forces to separate the filament. Since breaking an inductive current causes an arc, the switch is now at the filament, so the arc occurs at the switch during contact bounce during a cold filament turn on.

This is the short explanation for the switch arc and failure during turn on. (POOF).

The best way to extend the life is allow convection cooling and never keep in sealed enclosures, ( leave a gap for air vent). The next best way is a threaded NTC insert which allows a soft start but still has the accelerated aging from elevated temperatures. ( but these are pretty rare now )

A cool lamp using a ZCS with no contact bounce will last up to 10x longer but then you need a fan with good air velocity behind the lamp. It's called the Arrhenius effect. but bad for bulb sales.

\$\endgroup\$
6
\$\begingroup\$

There are several things that cause an arc in the switch.

One of them is worn out contacts. This can be seen on repeatedly/frequently used switches.

Another one is excessive load. Incandescent lamps' filaments have very low resistance when they are cold. When you switch it on, it will draw high current (e.g. 5 times the nominal current) for a moment (i.e. until the filament gets hot). And if the filament is highly worn out (because of frequent use) then it can draw extremely high currents right before breaking off. So, this short-time excessive load can cause an arc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.