OT: Shortening Christmas Light Strand

Usually the parallel-wired strings are the ones with the larger screw-base bulbs (which operate at 120V) and the series-wired strings have the small push-in low voltage bulbs.

-dave w

The shunt has lower resistance once the filament blows, and is inactive before. Exactly why these puzzled me so much. I wasn't able to see the mechanism by comparing new, blown, and inoperable bulbs even under significant (ca. x10) magnification. A new bulb indicates the normal filament resistance and the shunt is inactive. The low resistance shunt can't be active and get the high(er) resistance filament to light at the same time. A blown bulb with a broken filament and an operating shunt will indicate a near short. The shunt is apparently activated by the act of blowing the normal filament - and if it doesn't work right it blacks out the whole strand. Using an EMF detector quickly allows you to see where the AC supply stops along the strand, otherwise you have to work down the string with a good bulb (and hope its only one that is defective) or take a couple of bulb bases and make probes for your VOM and start dividing the string in halves to zero in. Which is why I bought a cheap EMF detector, though you could also make one using a high input impedance op amp with a floating input.

Brad Hitch

Yup, I remember getting to this point now and keying in on the "wire" wrapped around the base of the filament leads. When the filament blows, the voltage across the shunt goes up to the line supply voltage (ca. 165 volts P-P for 120 V rms), causing it to arc over and establish the shunt connection. I couldn't see the coating on the wire shunt, however, so I still wasn't really sure how it was implemented. That link to HowStuffWorks (above) fills the gap, though in this case it's more like how stuff doesn't work.

It seems a little sporty to me to intentially arc a part with line voltage in the case of a malfunction, especially on a circuit that is in intimate contact with an easily combustible object. You are essentially leaving the fuse as the last line of defense.

Brad Hitch

120V (or 165v P-P) will not arc across that 1/4" gap where the filament is. It would take a much higher voltage to arc that far. I think the mechanism is more that there is a high resistance coating on the shunt, which exposed to a 120V differential conducts enough to heat up and melt, joining the shunt wire directly to the leads.

I suspect that the shunt is somehow designed to remain an open circuit until the filament opens and full line voltage is applied across the bulb. Perhaps the little wrapping of wire around the base of the filament support is coated with solder covered with some resistive varnish that conducts very little current at low voltage (compared to the filament itself), but if that filament opens and there is 120 volts across the shunt, it conducts enough current to heat up, melt through the varnish, and fuse itself into a solid metallic connection across the terminals. (I'm just guessing here, since I don't know the details...)

-dave w

Well, exactly, of course it won't arc across the filament gap. I meant that 0.1 mil layer of dielectric coating on the shunt wire. Probably a pretty safe design, actually, but I'm not used to seeing circuits that intentionally arc anywhere unless it's an ignitor. Of course there aren't any MOSFET's in a string of Christmas lights.

Have we strung thus out enough yet? he says, brightening at the prospect of ending this thread!

Brad Hitch

Besides, the remaining lights are still in series, so there shouldn't be any extreme current surges (beyond the increase of current when line voltage is now applied to the remaining N-1 bulbs) when the shunt bypasses a failed one.

-dave w

Yup. All of this talk makes me want to measure it. I think I'll enlist my ten-year-old to dig out an old string and replace N-1 bulbs with shunted ones and then we'll hook up the DSO to capture the transient across a current-sensing resistor when we plug in that last good bulb.

I must not have enough to do.

Brad Hitch

I think long before you get to N-1 shunts and 1 good one, you'll probably hit a cascading failure mode where the bulbs start popping filaments at a faster and faster rate - and at some point one of the shunts is going to open like a little fuse and shut the whole thing down.

-dave w

Why not measure the impedence of the loop, shorten, measure again, and add the approriate amount of resistance to compensate?

Philip

Scott Alecks> Cutting won't work since the bulbs are in series and rely on having most

That should work. But it prompts me to want to issue a reminder: Making changes like this is what often results in a fire.

For starters, the resistance measured with an ohmmeter may read much lower than the actual resistance under operating conditions - ie, the filaments will heat up and exhibit more resistance at operating levels. Hence, the resistor selected based on the ohmmeter reading could be too low, and result in too high of current.

If a resistor is used to replace the bulbs, it should be overrated. If you think a 1/4 watt is suffiicient, use a 1/2 watt resistor for extra safety.

Lastly, the resistor should be soldered in place, and wrapper thoroughly in electrical tape.

No sloppy work allowed here.

My 2 cents. FWIW.

Doug

Heat shrink tubing would be much cleaner...

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

There's a 2A fuse in line with each series; that should blow long before you hit that scenario.

And - just to drag the thread out further - it was my understanding that the coating was high resistance and easily melted, so it heats up and melts when the potential across it rises from 2.5V to 120V. I don't think it requires any kind of arc, even across the coating.

You guys are too much. I'll never get the last word.... Come on, Not even a Little Arc? RF ringing is an arc signature, isn't it? We can see if it's there.

Brad Hitch