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Author Topic: Why is there a real bright light before a light bulb burns out?  (Read 10582 times)

paul.fr

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another_someone

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You are talking here about incandescent lights.

Incandescent lights have a thin filament wire that heats up when current passes through it.  The amount of heat and light it produces is proportional to the resistance of the wire, which is proportional to the thickness of the wire.  As the bulb ages, the wire gets thinner, so its resistance rises, so its light output rises.  Initially this is a fairly gradual process, but as the wire gets ever thinner, it begins to become a runaway effect, where as the wire gets thinner, it gets hotter, and as it gets hotter so the rate at which it gets thinner increases, which then makes the wire hotter yet; and towards the end this happens so rapidly as to become obviously noticeable.
 

Offline lightarrow

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You are talking here about incandescent lights.

Incandescent lights have a thin filament wire that heats up when current passes through it.  The amount of heat and light it produces is proportional to the resistance of the wire, which is proportional to the thickness of the wire.  As the bulb ages, the wire gets thinner, so its resistance rises, so its light output rises.  Initially this is a fairly gradual process, but as the wire gets ever thinner, it begins to become a runaway effect, where as the wire gets thinner, it gets hotter, and as it gets hotter so the rate at which it gets thinner increases, which then makes the wire hotter yet; and towards the end this happens so rapidly as to become obviously noticeable.
Yes, but I think it's not totally correct: the amount of heat produced doesn't increase with the resistance, since it's voltage constant, not current. It's true that the filament's temperature increases when it gets thinner, but it's not clear to me why; we also have to consider that the effect of impedences is reduced when resistance increases, so the current doesn't decrease proportially to resistance, but it decreases less.
 

Offline syhprum

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As the lamp ages eventually a break occurs in the filament and an arc occurs, now the resistance of the arc is somewhat higher than the normal filament but the overall resistance of the lamp and the consequent current flowing thru it is only is little affected.
Due to the higher resistance of the arc carrying much the same current the power dissipated in this section is higher hence is has a higher temperature and emits More light.
This is a very unstable state and the filament either collapses or the lamp fails to light next time due to a gap in the filament
 

Offline chris

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(syhprum came in with the answer as I was typing this, hence the double response!)

Yes, lightarrow is correct - the temperature of the filament - i.e. the brightness - is proportional to the current flowing through it. The higher the current the hotter the filament and the greater the light that is emitted. But as the filament becomes hotter its resistance rises, which is what limits the current and determines the ultimate temperature of the filament.

When a bulb undergoes the Edisonian-equivalent of a supernova (for want of a better phrase), the current through the filament suddenly becomes much higher. This heats the tungsten wire beyond its operating temperature, melting it and then cutting the circuit. But in the fraction of a second before the filament burns out it glows white hot, and that's the flash you see.

Why does it happen? This occurs due to a "tungsten arc". The filament breaks and develops an arc between the broken ends. The arc melts some tungsten, vapourising it and in the process effectively shortening the length of the remaining filament. This reduces the resistance, allowing a higher current to flow across the arc and in the remaining intact filament, which glows even more brightly than usual.

Hey presto, the big bang in a bulb!
 

Offline syhprum

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In low voltage lamps with a robust filament a break and an arc can persist for a long time, this used to be a problem on the 50W 12V lamps that were the source of illumination on the HELL scanners on which I used to work.
When this occurred the scanning spot became distorted and the colour balance was upset.
 

another_someone

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Yes, but I think it's not totally correct: the amount of heat produced doesn't increase with the resistance, since it's voltage constant, not current. It's true that the filament's temperature increases when it gets thinner, but it's not clear to me why; we also have to consider that the effect of impedences is reduced when resistance increases, so the current doesn't decrease proportially to resistance, but it decreases less.

If we have no capacitive or inductive load, then is not impedance merely a synonym for resistance (unless you are talking about source impedance, in which case if that is significant, then we are not talking about constant voltage).
 

Offline lightarrow

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Yes, but I think it's not totally correct: the amount of heat produced doesn't increase with the resistance, since it's voltage constant, not current. It's true that the filament's temperature increases when it gets thinner, but it's not clear to me why; we also have to consider that the effect of impedences is reduced when resistance increases, so the current doesn't decrease proportially to resistance, but it decreases less.
If we have no capacitive or inductive load, then is not impedance merely a synonym for resistance (unless you are talking about source impedance, in which case if that is significant, then we are not talking about constant voltage).
Wires connecting the lamp have impedence too (mostly inductive).
« Last Edit: 31/07/2007 23:33:50 by lightarrow »
 

another_someone

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Wires connecting the lamp have impedence too (mostly inductive).

But an inductance in an AC circuit will cause a drop in voltage over the inductor (albeit with a phase shift), and thus you cannot refer to a constant voltage on the filament wire. You have an series LR circuit, and an increase in R will increase in voltage drop over the resistor, and the expense of reduced voltage drop over the inductor.

That having been said, I cannot imagine the inductance of the wires to be that great to have a very significant impact on the overall circuit.
 

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