One thing is for sure, the principles of basic electricity can get very confusing very fast. I am attempting to shorten a string of christmas lights from 150 to 30. They are wired in series of 50. I have found out that by simply splicing the set of 30 and plugging them in, they are extra bright and in danger of blowing from excess voltage. What is the optimum way to go about reducing the voltage to normalize the brightness? What kind of calculations are required to solve this? By the way, I'm using the typical 2.5V 3A christmas lights. I also can't just cover the other lamps in the series, I need these 30 to be spliced by themselves. I know this might get ugly, but thanks for helping a confused technician out.
Shortening Christmas lights
- Thread starterNikgwolf
- Start date
Lotos
Active Member
Ditto.
Every time I've had to do this, I've found some way to 'hide' the excess lights...
Sometimes taping them, sometimes tucking them inside a flat, sometimes taping two lights together to form almost a single source...
If you *really* need to cut some off the end... It's a simple calculation.
X*2.5=Desired Voltage... Where X is the number you wish to have... So
30*2.5=75 Volts
Since the original number of 50, produces 125v from that... They're technically being run under voltage in their natural habitat... So these 30 will be a bit brighter than the 50 beside them...
As for the correct way to get 75 Volts? A Variac and an inline voltmeter (to measure the output... The numbers printed on the face of most old Variacs are just a 'guide') will do it... But you lose all ability to dim them like that. Well, aside from with the Variac itself, obviously...
There's a reason most of us just try and hide the excess
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Yea. Don't just chop the end off. My physics teacher in high school didn't even understand this concept so he taught it wrong, then told me I was wrong when I tried to correct him.
I'm leaving room for error because standard voltage in the US is roughly 110v-125v, but each outlet in a different area of a building and in a different part of the grid will be different, so exact calculations are a best-cast scenario.
Each light bulb on a standard string of 50 drops ~2.5v of the ~120v. Often you'll find multiples of 50 for the number in string based on the following diagram.
Each section of 50 bridges between a hot wire and a neutral wire. No matter what happens, 120v is going through that section when the lights are turned on. In a simple world, one light goes out in a section, all go out because the gap between the hot wire and the neutral wire is no longer bridged. Then, some excellent designers came up with a way for the filaments to fail such that no matter what, that entire section of the string wouldn't turn off. So if one 2.5v bulb on a string fails, those 2.5v then get diffused across the remaining 49 bulbs because 120v still needs to get from the hot wire to the neutral.
For the purposes of this next paragraph, we'll assume the voltage present is 125v which is divided equally among all light bulbs that work.
When one bulb goes out, that's not a problem because that means the rest of the bulbs are still getting (125v/49bulbs) = 2.55v each. Usually this is fine. But let's say you leave that string out for a while and another seven bulbs fail. Now it's (125v/42bulbs) = 2.97v. That's almost 3v for filaments that are designed for only 2.5v. Eventually each filament will drop so many volts that they will become brighter and hotter. As that happens, more filaments fail from the heat.
This leads us to a really cool scenario. Eventually so many bulbs will fail that as each one fails, the rest of the string becomes noticeably brighter. Again, brighter = hotter, therefore as each one fails, the chance of another failing grows. So over the course of a few minutes you could watch almost an entire string of lights become very bright as more fail until eventually all of the filaments have burned out. (a lot of people don't believe this until they see it)
In your case, cutting the string down takes two steps. First, cut out two of the three sections of 50 lamps. Refer to the diagram to figure out what I mean.
Now just chopping the other 20 bulbs off would result in more voltage being dropped across each of the filaments than they can handle. To prevent the bulbs from burning out, you would need to do one of two things -- you can drop the source voltage down in 2.5v decrements to be proportionate to the number of light bulbs that remain, or you can find a way to drop the rest of the volts in the system. In this case, the number of light bulbs removed is 20. At 2.5v each, you need to drop 50v across a resistance.
However, if I'm not mistaken, we need more information to determine the resistances. Your 3amps is either a mistake or your idea of standard Christmas lights is different than mine. That would mean one strand of 150 lights would produce (120v*3a) 360watts, and that is very, very high. A standard string of mini-bulbs would produce about 75w -- about 25w per 50 lamps. Assuming this information to be correct (source: HowStuffWorks "Christmas Lights Power Consumption"), let's continue.
25w/120v = .2083a
125v/.2038a = 576.09ohms
576.09ohms/50bulbs = 11.52ohms/bulb
11.52ohms*20bulbs = 230.4ohms
You would need a 230ohm resistance in line with your 30 light bulbs, and it would need to be rated for at least 10watts. There's a number of ways you could do that, including just putting (20) 11.5ohm resistors (with a rating of a minimum of 1/2w each, but you'd probably want them to be rated for 1w in this application) in series.
[Please don't actually use those numbers unless someone takes the time to check that they are accurate. It's entirely possible that somewhere along the way I made an error. I cross-referenced some data with that HWF article, so they look to be about accurate, but I make no guarantees.]
For all practical purposes, you're best off using a 50-bulb strand and finding a way to hide the other 20 lamps. Otherwise cut it down to 30 bulbs total and vary the source voltage appropriately as Lotos described. Using resistors to do what you want to would be a PITA. It would work, but it would be tedious to find the right components and solder them in, plus then you have to pack the entire rig so that no one can come in contact with conductive materials.
Also remember if you go the route of varying the source voltage, don't forget to mark your string of lights for future reference NOT to be plugged into a 120v wall receptacle. They'll be very bright for a short moment before every single filament goes supernova and the entire string is dead.
I'm leaving room for error because standard voltage in the US is roughly 110v-125v, but each outlet in a different area of a building and in a different part of the grid will be different, so exact calculations are a best-cast scenario.
Each light bulb on a standard string of 50 drops ~2.5v of the ~120v. Often you'll find multiples of 50 for the number in string based on the following diagram.
Each section of 50 bridges between a hot wire and a neutral wire. No matter what happens, 120v is going through that section when the lights are turned on. In a simple world, one light goes out in a section, all go out because the gap between the hot wire and the neutral wire is no longer bridged. Then, some excellent designers came up with a way for the filaments to fail such that no matter what, that entire section of the string wouldn't turn off. So if one 2.5v bulb on a string fails, those 2.5v then get diffused across the remaining 49 bulbs because 120v still needs to get from the hot wire to the neutral.
For the purposes of this next paragraph, we'll assume the voltage present is 125v which is divided equally among all light bulbs that work.
When one bulb goes out, that's not a problem because that means the rest of the bulbs are still getting (125v/49bulbs) = 2.55v each. Usually this is fine. But let's say you leave that string out for a while and another seven bulbs fail. Now it's (125v/42bulbs) = 2.97v. That's almost 3v for filaments that are designed for only 2.5v. Eventually each filament will drop so many volts that they will become brighter and hotter. As that happens, more filaments fail from the heat.
This leads us to a really cool scenario. Eventually so many bulbs will fail that as each one fails, the rest of the string becomes noticeably brighter. Again, brighter = hotter, therefore as each one fails, the chance of another failing grows. So over the course of a few minutes you could watch almost an entire string of lights become very bright as more fail until eventually all of the filaments have burned out. (a lot of people don't believe this until they see it)
In your case, cutting the string down takes two steps. First, cut out two of the three sections of 50 lamps. Refer to the diagram to figure out what I mean.
Now just chopping the other 20 bulbs off would result in more voltage being dropped across each of the filaments than they can handle. To prevent the bulbs from burning out, you would need to do one of two things -- you can drop the source voltage down in 2.5v decrements to be proportionate to the number of light bulbs that remain, or you can find a way to drop the rest of the volts in the system. In this case, the number of light bulbs removed is 20. At 2.5v each, you need to drop 50v across a resistance.
However, if I'm not mistaken, we need more information to determine the resistances. Your 3amps is either a mistake or your idea of standard Christmas lights is different than mine. That would mean one strand of 150 lights would produce (120v*3a) 360watts, and that is very, very high. A standard string of mini-bulbs would produce about 75w -- about 25w per 50 lamps. Assuming this information to be correct (source: HowStuffWorks "Christmas Lights Power Consumption"), let's continue.
25w/120v = .2083a
125v/.2038a = 576.09ohms
576.09ohms/50bulbs = 11.52ohms/bulb
11.52ohms*20bulbs = 230.4ohms
You would need a 230ohm resistance in line with your 30 light bulbs, and it would need to be rated for at least 10watts. There's a number of ways you could do that, including just putting (20) 11.5ohm resistors (with a rating of a minimum of 1/2w each, but you'd probably want them to be rated for 1w in this application) in series.
[Please don't actually use those numbers unless someone takes the time to check that they are accurate. It's entirely possible that somewhere along the way I made an error. I cross-referenced some data with that HWF article, so they look to be about accurate, but I make no guarantees.]
For all practical purposes, you're best off using a 50-bulb strand and finding a way to hide the other 20 lamps. Otherwise cut it down to 30 bulbs total and vary the source voltage appropriately as Lotos described. Using resistors to do what you want to would be a PITA. It would work, but it would be tedious to find the right components and solder them in, plus then you have to pack the entire rig so that no one can come in contact with conductive materials.
Also remember if you go the route of varying the source voltage, don't forget to mark your string of lights for future reference NOT to be plugged into a 120v wall receptacle. They'll be very bright for a short moment before every single filament goes supernova and the entire string is dead.
TimmyP1955
Well-Known Member
Use the mains transformer from a blown up power amplifier as a step-down, if you get lucky and find a blown amp with the right transformer
Use a bridged power amplifier with a 50Hz/60Hz sine wave as it's input. Something that's capable of 700W or more into 8 Ohms would have the necessary voltage swing.
Use a bridged power amplifier with a 50Hz/60Hz sine wave as it's input. Something that's capable of 700W or more into 8 Ohms would have the necessary voltage swing.
OK, now that we have had some very good theory and possible solutions, let's drop back to the KISS method.
10 each 2.5volt lamps will draw 25 volts. make three strings of 10 lamps and parallel those strings together, and connect to the secondary of a 24 or 25 volt transformer. It has been a while, but I believe that Radio Schack sells a 25 volt transformer at about 3 amps. 24 volt transformers are used in Air conditioning control and available in many different sizes of current. You will find them on the shelf at most all A/C supply houses. mpja.com has a large variety of transformers at very reasonable prices.
A little cutting soldering, some zip cord and a tranformer and you are in buisness. I do this stuff all of the time to make strings for star fields behind windows and window lights in apartment building skylines for scenery on stage.
10 each 2.5volt lamps will draw 25 volts. make three strings of 10 lamps and parallel those strings together, and connect to the secondary of a 24 or 25 volt transformer. It has been a while, but I believe that Radio Schack sells a 25 volt transformer at about 3 amps. 24 volt transformers are used in Air conditioning control and available in many different sizes of current. You will find them on the shelf at most all A/C supply houses. mpja.com has a large variety of transformers at very reasonable prices.
A little cutting soldering, some zip cord and a tranformer and you are in buisness. I do this stuff all of the time to make strings for star fields behind windows and window lights in apartment building skylines for scenery on stage.
FatherMurphy
Active Member
I'm assuming that you don't want 20 covered lights onstage because there's no place to hide them that isn't in view. However, since you still have to run power to them somehow, for an even simpler solution, merely splice some wire between bulbs 20 and 21 sufficiently long to put bulbs 21 through 50 onstage, leaving bulbs 1 through 20 offstage in a convenient hiding place. No math, no transformers, just a little care taken to make solid, safe connections.
Or, if you don't need remote control, there are six and twelve lamp strings out there that are powered off of AA battery packs, for use in craft items.
Or, if you don't need remote control, there are six and twelve lamp strings out there that are powered off of AA battery packs, for use in craft items.
If you go with the series resistance approach you could use a capacitor instead of the resistor if you can find one. As I remember, due to reactance the capacitor returns the "excess" power to the supply instead of dissipating it. For this situation put a 10uF capacitor in series with the lights.
In any case, I get the same value of the necessary resistance that MNicolai calculated.
In any case, I get the same value of the necessary resistance that MNicolai calculated.
Nikgwolf, what exactly are you using the Christmas lights as a source forhttp://www.controlbooth.com/forums/members/nikgwolf.html?
Nice job, BTW!
Nice job, BTW!
kicknargel
Well-Known Member
Wow, you all got the skin off your cats. Some with lasers, some with samurai swords, and some just looked around for a pre-skinned cat.
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