What's in self shunting Xmas light bulbs?

I understand there's also a "tester" which forces shunts to close when they didn't when the bulb failed. Anyone know about where to obtain and how well it works? (I have a 12' pre-wired tree with one section out.)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Jim Thompson
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I happened to be reading the "How Stuff Works" article on Xmas lights and was intrigued by the description of the self shunting feature of today's series string bulbs as described on this page:

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I'm assuming that the little shunt pictured might be just several wraps of a fine wire with a coating on it which somehow changes state or "punches through" when subjected to full line voltage.

Can someone tell us what that material is and what actually occurs?

Thanks guys,

Jeff

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Jeffry Wisnia

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Jeff Wisnia

It's an anodized aluminum foil wire wrapped around the filament support pins. The very thin (something like 1 micron) oxide coating typically breaks down (along with any oxide on the Dumet wires) when subjected to full AC line voltage, but does not break down at the few volts normally across the bulb (if it does, the bulb goes out just as if the filament failed). It's similar to a low-voltage electrolytic capacitor sans electrolyte. Heat is not involved except at the tiny spots where the insulation breaks down and a little weld forms.

A small capacitor charged to a high voltage (and connected only to the string with the bad bulb shunt) should work. Maybe 0.1J @500V or 1kV? Removing single bulbs in other paralleled strings should isolate the bad string. If the bulb has actually fallen out of its holder, then none of this will work, of course.

Best regards, Spehro Pefhany

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Spehro Pefhany

I find it interesting that you are assuming voltage punch-through when the howstuffworks.com web page claims heat punch-through, especially considering that you are right and howstuffworks.com is wrong!

"If you look closely at a bulb, you can see the shunt wire wrapped around the two posts inside the bulb. The shunt wire contains a coating that gives it fairly high resistance until the filament fails. At that point, heat caused by current flowing through the shunt burns off the coating and reduces the shunt's resistance. (A typical bulb has a resistance of 7 to 8 ohms through the filament and 2 to 3 ohms through the shunt once the coating burns off.)" -howstuffworks.com

This is, of course, easy to test; break open a couple of bulbs and measure the resistance with the filament broken, with the shunt wire removed, and with both intact. Then measure the resistance of a burnt-out bulb. Unless the design has changed since the last time I designed a christmas bulb making machine, you will find that the shunt starts with infinite resistance and becomes a low resistance when it is doing its shunting job.

I don't have any mini christmas bulbs, so I would appreciate it if someone would do the experiment I just described and post the resistance readings. (If you want to do advanced research, measure the voltage and current in circuit while the string is operating; the resistance of the filament gets lower when it is hot.)

The shunt is nothing more than aluminum wire. The christmas bulb making machine heats it (around 400 or 500 degrees Celsius if I remember correctly) in a steam chamber. This creates a thin layer of aluminum oxide, which is a good insulator but has a fairly low breakdown voltage if it is very thin. Some christmas bulb making machines use aluminum foil instead of wire, but I wouldn't sign off such a design - too hard to control the tension of the shunt across the filament supports.

When the lamp fails, the shunt sees the full line voltage, which punches through the oxide and welds the shunt to the filament support wires. Older series streetlights have the same basic setup, except the shunt is in the base and consists of metal plates with thin silk between them. An open streetlight lamp causes thousands of volts across the silk dielectric, which punches through and welds the plates together. Look at

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and search for the word "cutout" for details.

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Guy Macon
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Guy Macon

High voltage across the entire string, current limited so that when the shunt closes it doesn't supply enough current to burn out the bulbs.

Reply to
Guy Macon

I did some searching and found that someone already has:

From Jon Titus, Editor in Chief, Test & Measurement World:

"I ran a few experiments on series-string lights ... The shunt wire may "spot weld" itself to form a shunt across a bad bulb. Here's what I did:

"I carefully broke the glass on three replacement bulbs so as to maintain the integrity of the filament. Prior to breaking the glass, I checked each bulb to ensure it worked properly in the string.

"1. First bulb inserted in the string with intact but exposed filament. The string works fine. I expected the filament to go "poof," and then I would observe the results. A closer look showed that the filament had heated, but it was still intact. Small traces of white (oxidation?), but mainly blue discoloration in the middle section of the filament.

"2. Second bulb inserted with intact but exposed filament. The string works fine again. The filament looks intact, but discolored as above. Used a toothpick to break the filament in the hot circuit while I observed any action on the shunt wire. As I broke the filament, I saw small sparks at the small shunt wire. Increased current through the shunt may have caused more "welding" across resistive connections.

"3. Third bulb inserted with a purposely severed filament. As soon as I inserted this bulb in the string I observed the small sparks at the shunt wire. The bulb completed the circuit, and the bulbs on the string all lit.

"So, I'd say the small wire does act like a shunt, getting welded due to the initial high potential. That's neat. It must take some interesting QC to ensure a uniform coating on the wire. Or perhaps the surface is just oxidized slightly.

"An interesting effect was noticed in bulb #1. Even though exposed to air, the filament didn't heat up and break. Perhaps the filament started to overheat and oxidize, thus raising its resistance. As the resistance increases, the potential across the bulb increased enough to cause the shunt to weld itself in place. Thus the filament fails but does not necessarily break. In years past, I have looked at several series of dead lights with a magnifying glass to try to find a broken filament. In most cases, I could not find such a bulb. This effect may be the reason. As the bulb starts to fail, the shunt kicks in so quickly that the filament never has to break. Some clever engineering in a $3 string of cheap lights."

Reply to
Guy Macon

There can be one impressive side effect of these self shunting bulbs. If you don't replace the bad bulbs the whole string can go out in a successive failure as the voltage increases on the remaining bulbs.

This was demonstrated once while I was at my parents place for Christmas a few years back. I had noticed one string of lights on the tree was a bit brighter than the others, and that it had several bulbs out.

Later, another bulb or two went out right when I was looking at the tree. Then a few seconds later several more went, and the remaining bulbs got much brighter. But that didn't last long. Just as I was thinking it was time to go unplug the tree, the remaining bulbs went off like old fashioned camera flash bulbs in a series of flashes.

Now I kind of want to go get a string of lights, and short out a few bulbs and see if I can make it happen again. :-)

Reply to
Carl D. Smith

My guess: a) almost constant current through the rest of the filaments b) this open air filament is cooled, stays low-ohmic and low-voltage.

Ever tried to start a halogen bulb on a constant voltage / constant current source where the current limit is just above the normal operating current?

I did. It doesn't.

Regards, Arie de Muynck.

Reply to
Arie de Muynck

"Arie de Muynck" ...

current

That was of course one with a current foldback, not constant current or it would have worked. I did change the supply to constant current. Typically R_cold = 0.1 R_hot

Arie.

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Arie de Muynck

I read in sci.electronics.design that Guy Macon wrote (in ) about 'What's in self shunting Xmas light bulbs?', on Tue, 7 Dec 2004:

That'll be the day.

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John Woodgate

Sounds clever. I haven't seen that, but this one works 100% of the time. Home Depot sells them.

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It's dead simple - wand it over the dark bulbs to find the last one with live AC. That's the dead one.

Pre-wired trees are a little harder to trace the string, but it still works great.

Cheers, Richard

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Richard H.

"Carl D. Smith" wrote

Great product design: If the customer is too cheap to buy a box of replacement bulbs he gets to buy a whole new string ...

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Nicholas O. Lindan, Cleveland, Ohio
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Nicholas O. Lindan

Except... a whole new string is cheaper than replacement bulbs! :-)

Reply to
Richard H.

A Variac should do the trick.

Reply to
Guy Macon

Note to self: next time, smoke the crack *after* posting...

Seriously, I had trouble sleeping last night so I got up and read a few newsgroups, and I manages to get several things exactly backwards. D'oh! The resistance of the filament gets higher when it is hot. The current through the filament gets lower when it is hot.

I blame Bush for any errors I may have made.

Reply to
Guy Macon

Or by underestimating american taste.

See this page I couldn't resist putting up:

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Happy Holidays and thanks for answering my question guys,

Jeff

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Jeffry Wisnia

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Jeff Wisnia

I wouldn't know, but it looks to me like a mother-daughter pair.

But, I just noticed that the buildings behind them have *functional* shutters, not like the ubiquitous fake shutters I see most everywhere nowadays which are usually the wrong width to "fit" if they *could* hinge closed, and are invariably mounted with the louvers pointing the opposite of where they should.

Jeff

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Jeffry Wisnia

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Jeff Wisnia

You know that, I know that, but "No one has ever gone broke by underestimating man's intelligence".

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Nicholas O. Lindan, Cleveland, Ohio
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Nicholas O. Lindan

"Jeff Wisnia" wrote

America has no special lock on bad taste.

Would this be from EuroDisney?

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Nicholas O. Lindan, Cleveland, Ohio
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Nicholas O. Lindan

Aha, an uncropped version:

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Bourbon Street, New Orleans.

Heck, that's close enough to being French in my book.

-- Nicholas O. L>

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Nicholas O. Lindan

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