OT: Shortening Christmas Light Strand

Could you draw a circuit diagram?

I believe it is two parrallel circuits sharing a common return

(ignore the period characters, which are used as placeholders to prevent trimming during the post)

circuit 1 ---+---+---+---+---+---+ .............O...O...O...O...O...O common ------+---+---+---+---+---+ .............O...O...O...O...O...O circuit 2 ---+---+---+---+---+---+

each circuit has its own 'flash' component

so the current flows along the circuit 1 "rail", through a bulb on that rail, and finally to the common return

so the current flows along the circuit 2 "rail", through a bulb on THAT rail, and finally to the common return

the "dead end" wire he observed is the end of the rail ending prior to the last bulb

- iz

Darren J L> >

Be careful about this. Some strands are built to have a certain number of bulbs (read "load") in order to be cool enough to be safe, and also meet the wire rating specs.

I know this personally as I have melted a strand due to shortening one.

~ Duane Phillips.

More likely two series strings woven together.

Plug in the string. Assume all the bulbs light. Remove one bulb from it's socket. I suspect that half the lamps will go out.

Shortening a string is not really to practical, as that would require changing all the bulbs to a new voltage as the number in each series string is reduced.

if the circuits are indeed series, which can be tested as you described

- iz

DaveL wrote:

Could be 1 series with a 110v plug on the end too (feed thru)

Option 1- Buy shorter light set. Option 2- Cover the unneeded bulbs with black "electrical" tape to hide them. Option 3- Double the wire back on itself between each bulb the same distance and secure it with black (or color-matched) tape. Option 4- Double full length of wire back on itself near the midpoint placing one half of the bulbs between the other half for even spacing (reduced lenght by half).

Cutting won't work since the bulbs are in series and rely on having most of them operational to prevent exceeding load/voltage maximums for each bulb. -Scott

Norm Dziedzic wrote:

they are not necessarily in series. Unplug a single bulb and see if half the string goes out

I have seen parrallel-wired light strings (they're alot easier to find bad bulbs in)

- iz

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

C4 and C9 bulb strings are parallel, but mini lights with a 3-wire cord are generally series. If you look at the rating of each bulb, they run in the 2 to 6 volt range so series is used to divide up the 120v amongst them. To give you added length, they will divide the set into 2 or 3 sections, hence the extra wire(s). A common set size has 100 lights divided into 2 sections of 50 with each bulb rated for 2.5v to give you the voltage drop on each bulb that 'uses up' the available 120v. If you have self-shunting bulbs, each time one burns out, the set gets a little brighter... at least up to a point. Self-shunting bulbs are becoming very common and could be mistaken for a parallel setup since the whole string doesn't go out when a single bulb fails. If you remove the bulb, the string will go out though. The real problem with these mini light sets is when every manufacturer uses different sized bulb holders and different voltage ranges. You can never find the right replacement bulbs again. :| -Scott

Ismaeel Abdur-Rasheed wrote:

Yup, that's my analysis from last year trying to revive a couple of old light strings (not very cost effective, but educational). 120 VAC with 35 bulbs in series gives you about 3.5 volts across each bulb. My strings had 3 wires down to two between the middle bulbs.

So who can explain the design of the bulb that lets it (usually) still conduct when the filament burns out? It can be pretty hard to track down the bad bulb when this mechanism doesn't work unless you have an EMF detector since the whole 1/2 string goes dark.

Brad Hitch

oic :)

thanks for the explanation

gee, you sure know alot about christmas lights! :)

- iz

Scott Alecks> C4 and C9 bulb strings are parallel, but mini lights with a 3-wire cord

Its probably not a bulb design, its a circuit design.

Most incandescent bulbs provide a current path by way of the filament; open the filament (burn it out), and no current flows through that bulb.

Its exactly like igniters. Cluster igniters are wired in parallel with each one across the battery leads. If one burns out, it doesn't affect the current to the other ones (the other lights stay on). If the igniters were wired in series, as soon as one burned through, it would stop the current flow to all the others (the entire strand would go out).

But everyone has probably seen a single unlit bulb in a series strand. Typically, this is caused (its not a "feature") by a glob (or broken piece) of filament shorting out the filament leads inside the bulb (or the external leads or lamp socket itself are shorted). Like an igniter, the filament is a thin strand of coiled resistance wire across two leads. If the filament overheats, it can melt, glob together, and fall across the two leads inside the bulb. The glob can conduct current without heating to incandescense. Its like shorting the clips to an igniter; you'll have continuity, but the igniter won't fire. If you pull such an unlit bulb out of it's socket, the rest of the lights in a series strand will go out.

A parallel circuit light string is the only way to go, IMHO, though it is hard to find them with small lamps.

It is a bulb design. Each bulb has a shunt that conducts enough to keep the circuit working, if the filament burns out. But, it increases the voltage on the other bulbs, and if you have enough go out, the higher voltage will cause the life to get shorter and shorter for the remaining bulbs.

I have observed the same thing with my experimentation. Now for a question that I have never been able to figure out: how does the shunt work???

I've done many experiments, and not been able to figure it out.

I took a bulb and broke the glass, and physically broke the filament. Now it measures infinite resistance. I can see the thin wire shunt wrapped several times between the leads that hold the filament.

I tried raising the voltage on a bulb until it burns out (usually around

7.5-8.0 volts). The shunt does not seem to do anything. The bulb measures infinite resistance, just like the case where I broke the filament.

So - how does the shunt enter the circuit? Does it somehow "melt" or something when the 120V open-circuit voltage is across the shunt? That is the only explaination I can think of, since my experiments were with lower voltages when the filament goes out.

I've never seen a string like that, but if that's how they really are wired, then it would seem to me that the string could be shortened anywhere between "bulb pairs".

Old Christmas lights were my main source of split-bulb BP initiators in the old days. They were pretty reliable in operation, but I did break a few filaments during construction. I could identify the broken ones cause they measured open with an ohm meter.

I've eyed many a lamp internals and never spotted a shunted bulb like you describe. Interesting idea, although pretty wasteful of power.

this might violate a Fire Code to shorten ?

All the miniature bulbs have shunts now. That's how they achieve series wiring (for low-voltage operation) but still allow the series to light if one bulb goes out.

But, during normal operation, the shunt doesn't appear to be a parallel resistance to the filament. That's what I can't figure out!

The shunt has higher resistance so it doesn't get the current until the filament breaks?

steve