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sforston

Shortening Mini-strings

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If you were to shorten them to 16, it would put double the voltage across each of them and they would burn out VERY quickly. :(

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Dont listen to the nae sayers. Anything is possible, well almost anything. But it is going to require you to some research.

First, is this a full wave or half wave? What color are these LED? What wire warts do you see if any.

Here is the thing. Different color LEDs drop (yes think drop) different amount of volts across them. As I just posted in a different thread, Kirhoff's law says that that total voltage drops will equal the total voltage applied to the circuit. So even though 1 LED drops or requires 3.2 volts. If you apply 120 volts to it, the total voltage will be 120 volts applied. Now you have 2 LEDs and apply 120v. You will have 60 volts applied to both LEDs. Still about 33 LEDs to few to get 3.6 volts across each LED. Now with another law called Ohm's Law we can do some figuring and add a resistor to drop some voltage in place of LEDs. And example of why you would want to use a resistor to help lower the voltage so you have the right voltage across your LED string. If I remember correctly the following colors are 2.2 volts, red, org, and yel. If you take a string of 35, 2.2 volt LEDs and add them all up, you will get 77 volts. Thats about 43 volts to low for our 120 volts applied. So we need a resistor that will drop 43 volts across it. With Ohm's law and one more factor we can figure out what the value of the resistor needs to be. I am going to pull a number that is close but not really the right number. I am going to use 20mA as the last factor. Ohm's law says that resistance equals voltage divided by current. So we will divide the 43 volts by the .02 Amps and get a resistance of 2150 Ohms. But you wont find that value anywhere, so lets try 2.2K which is a common value. Now we have one last issue. Resistors are built in different physical sizes, and this is in part due to the wattage the resistor can handle. Wattage in this case is HEAT. Heat generated from voltage passing through the resistor. So Watt's law (ya another law) says that Watts equal voltage dropped times current. So 43 times .02 equals .86 watts. Again we will need to round this up to 1 watt resistor.

Now I gave you some insight to the process. But I have to make this disclaimer. All of the above is actual for D.C. circuits not A.C. There are more factors to be taken into account when working with dirty D.C. that has still the peak voltages. Or as some call A.C. superimposed on D.C.

Anyway we can keep this in the open or your welcome to IM me. But I will need answers to my above questions answered.

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Dont listen to the nae sayers. Anything is possible, well almost anything. But it is going to require you to some research.

First, is this a full wave or half wave? What color are these LED? What wire warts do you see if any.

Anyway we can keep this in the open or your welcome to IM me. But I will need answers to my above questions answered.

When did he ever say these were LED's?

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Thanks! I thought it would double them and burn 'em out really quickly...just thought I'd try!

Speaking of doubling, would it be possible to use two for each position?

This would leave you with 32 lights and you could just black out 3 of them. If this is too bright, you add a diode in series which would set them near half brightness.

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This is the LED part of the forum.

No, actually it's not. There's nothing LED-specific (or even LED suggestive) about this forum.

That said, the process for calculating resitor values, etc, is not very different between LEDs and incandescents - mainly that you need a higher wattage registor for incandescents due to the higher currents.

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My understanding is that you can shorten a 100ct set of minis to 50ct ... of you cut in the middle, and seal off the end.

Somewhere else, I believe they said you could also extend a 100ct to a 150ct ... again talking mini 100's only here, which are really (1) sets of 50 each.

I believe folks would heat shrink the ends after the cut.

But this would not apply to the 35 ct minis.

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No, actually it's not. There's nothing LED-specific (or even LED suggestive) about this forum.

That said, the process for calculating resitor values, etc, is not very different between LEDs and incandescents - mainly that you need a higher wattage registor for incandescents due to the higher currents.

I have to back track and take back the part about this being the LED section of the forum. Got myself confused where I was. Otherwise I still stand by someone trying to stir it up (see BMs comments).

But CarlD I will also have to disagree with your comment about LED and incandescent bulbs. With incandescent bulbs the filament is slow to glow and takes the surges of the peak voltage and current. Where as an LED responds instantly to peak voltage and current. It is fragile physically compared to the filament. And you best take into account the difference between Peak and RMS values or you will surely pop your LED in short order. I suggest your read up on RMS (root mean square). And dont forget that I am applying this to RAW D.C. not filtered D.C. Makes a world of difference what values you are going to use to calculate your resistor value.

Dont believe me. Fine! But I will continue to advise as I see it and no one will blow out a string of LEDs following my advice. That sir is the main difference between LED and incandescent lamps.

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Speaking of doubling, would it be possible to use two for each position?

This would leave you with 32 lights and you could just black out 3 of them. If this is too bright, you add a diode in series which would set them near half brightness.

I thought about that, but I don't think it would work with what I want to do. I picked up some silly eyeball glasses from wal-mart post halloween, and was thinking about making some creepy bush eyes before I got everything put away. 8 pairs of glasses, 16 holes = 16 lights needed. I was just trying to avoid using so many black out caps or using a 50 count set and having so many I couldn't spread them out over the yard. No biggie! Black out caps it is!

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There's 2 things i'd do if i needed 16 bulbs:

1: Get a set and hide the extra (or find a way to use all 20 :) )

2: If it really had to be 16, again get a set of 20, shorten to 16 sockets, replace all the

remaining 16 bulbs with 12-volt ones...this would work just fine and be 100% safe (but would

be slightly dim)

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Bulbs have different voltage values hence the different length of strings. I have purchased light strings as small as a count of 20.

You just need to make sure you use the correct bulb when replacing them.

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