I can't tell you how many times I've been asked why with mini-lights, when one burns out, how do the rest keep working?  Well thanks to Marc Kummel and his webpage at http://www.rain.org/~mkummel/stumpers/07jan00a.html, here's a detailed answer.

I thought I understood Christmas tree lights. If the bulbs are wired one after the other in series, they all go out when any one does. If they are wired in parallel like rungs on a ladder, the other bulbs stay bright, but they tend to run hot. The new lights on our tree have three wires, not two. If a bulb burns out, the others stay lit. Then I noticed a bigger mystery as I played with them. If I gently twist a bulb in its socket, it goes out and the others stay on. But if I twist too much or remove the bulb, they all go out. How do Christmas lights work?


Above is a circuit wired in "parallel"



Above is a circuit wired in "series"

Notes:

Dunn Middle School student ART sent this email:

I am still thinking about your Christmas light stumper. I've noticed that most people do not usually think about how or why everyday things work. I knew that our Christmas tree lights were in series circuits because last year, my mom and I spent two days going through three strings of lights to find the one light that did not work. It was anything but fun.

When I was a kid in the 50s, checking the Christmas tree lights was the first sign that Christmas was near. My Mom and I would test each string of lights. If it didn't light, I had to test the bulbs one by one in a string of good lights until I found the bad one. (You had to have a good string.) Tedious, but Christmas was near! Most Christmas lights have gotten easier.

 

One discovery I made while researching this stumper is that Christmas lights are incredibly cheap after Christmas. I bought several strings of 100 lights for about $1.50 each. Radio Shack wants nearly a buck for just three of these low-voltage bulbs with leads attached. Teachers with no budget take note!

 

Twisting a bulb in its socket twists the fine wire leads at the base of the bulb together and creates a shortcut past the bulb. So that bulb goes out while the others stay on. If you twist too much, the wires will break from the bulb, which removes the shunt from the circuit. Sometimes the wires stay twisted together and the lights work, but usually they come apart and they all go out.

 

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I should mention that the instructions on the box say explicitly NOT to twist the bulbs. I see why. The series bulbs divide the house voltage so each bulb only gets 120/100 = 1.2 or 120/50 = 2.4 volts. Bypassing bulbs decreases the denominator of these fractions and increases the voltage to the rest of the bulbs. At some point they'll blow.

 

Here's where I'm still confused, and I hope some reader can help. The box warns that "it is important that burned out bulbs be quickly replaced to prolong the life of the remaining bulbs." But the usual rule with parallel resistances suggests otherwise:

 

The effective resistance of a parallel circuit is always less
than the smallest resistance contained in any one branch.

If two resistors R1 and R2 are wired in parallel, then the total resistance Rt is given by the formula:

                  1               1           R1 x R2
      Rt  =  -----------  =   ---------  =   ---------
               1     1         R1 + R2        R1 + R2
              --- + ---        ------- 
               R1    R2        R1 x R2

For example, keeping R1 fixed at 10 ohms, and varying R2:

	R1	R2	Rt
	10	0.1	.099
	10	1	.9
	10	10	5
	10	100	9
	10	1000	9.9

 

As R2 gets very small, the total resistance approaches zero. That's a short circuit. And as R2 gets very large, the total resistance approaches R1, as though R2 weren't there at all. Rt is always less than R2. So burning out a bulb takes R2 out of the circuit leaving a higher total resistance than when the bulb was on. This larger resistance should drop more voltage in series with the other bulbs, so the rest of the bulbs should run cooler.

 

It's tricky to actually measure the resistance of a light bulb with a voltmeter because the resistance changes as the filament warms up. I tried it, and it seemed that both good and burned out bulbs had a resistance of about 10 ohms, though my digital voltmeter kept jumping around between 7 and 30. When I tried measuring the actual current through the bulbs connected to a 3 volt battery, the burned out bulb pulled over 230 ma while the good bulb only pulled about 40 ma. Ohm's Law is not working here! Does the resistance of the shunt wire change when the filament burns out? Is it heat or current sensitive?

 

Darn, I still don't understand Christmas lights!

 

I didn't find much help on this stumper, though I found thousands of Web sites on series and parallel circuits.

• Bill's Antique Christmas Light Site traces the history of Christmas lights with lots of images. I'm not that old!
• Bill Drennon's How do Christmas Lights Work? really doesn't have much to do with Christmas lights. It addresses a different stumper. With DC current, there are actual electrons moving through my lights, but with AC current the electrons just vibrate back and forth. How does that work?

Update!

 

I just received this interesting explanation of how shunt devices work from Rick Delair:

I have your answer to why the resistance doesn't do what you expect, and why burned out lamps shorten the life of the remaining lamps!
 

First, you should understand how the shunt device works in the lamps. It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets; 100 sets ALWAYS use two 50 light circuits, unless 5-way flashers, whence 6-volt lamps are used; 3.5 volts for 35 lights; 6-7 volts for 18-20 light sets; and 12 volts for 10-12 light sets), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights. The reason for the 230+ Ma reading on a bad lamp is the shorted shunt, and your batteries' short-circuit current limit. A good lamp reads the actual cold resistance of the filament.

 

This very same principal was applied to the old series incandescent street lights. The open circuit volts here though was typically either 2400 volts, or 4160 volts, and the shunt was in the socket, between two prongs. It was called a "FILM-DISC CUTOUT" and was a pair of metal discs about the size of a dime, or a little smaller, separated by a thin piece of silk cloth. When a lamp burned out, the high voltage would puncture through the silk, shorting the lamp socket out, and completing the circuit. A voltage/current regulator compensated for blown lamps, so the others would not get more voltage! The mini-lights have no regulator, so as they fail and short, the remaining ones get more voltage.

 

The new Noma "Stay-Lit" sets are still in series, but a semiconductor shunt is used in each socket, making the lamp shunt unnecessary. If a lamp fails, or is broken, removed, etc, the semiconductor "senses" the higher open-circuit voltage mentioned above, and completes the circuit. And, the natural resistance of the device matches the lamp's resistance, so the remaining lamps still get proper voltage. This is similar to a regulator in series street lights. A very clever idea, indeed! (But worthless if a wire breaks, though.) It may interest you to know that aluminum wire shunts have been around since the 1930's, used on a few C-6 series light sets put out by Noma.

 

Hope I've solved Your stumper! Merry Christmas, Rick Delair.

Thank you Rick! The resistance of the shunt does change when the bulb burns out, because the high line voltage is then applied to the aluminum oxide coating of the shunt wire which "burns through" and transforms it from an effective insulator to a conductor. I didn't understand how the shunt resistance could change without there being some kind of more-expensive active sensing device. At least I got the question right! An oxidized aluminum wire shunt is a cheap and easy solution. I think I do understand Christmas lights now!