MINI LIGHTS AT LESS THAN 20

[shredocaaster]

Is there a way to make less than 20 minilights work so they dont blow.... say like adding a resistor or such somewhere in the line?



thanks

[Tim Fischer]

shredocaaster wrote:

Is there a way to make less than 20 minilights work so they dont blow.... say like adding a resistor or such somewhere in the line?



thanks

It's theoretically possible, but you need quite a resistor at that point in terms of power handling.

It's quite a bit easier to just black out the 'extra' lights. 20-light strings are already problematic in that when a bulb blows, the others are quite overvoltaged...

-Tim

[Max-Paul]

I agree with Tim, it is much easier to just get some duct tape and tape up 15 of 35 lamps. All though, this does lead to the hassle of finding a burnt out lamp. In the case of the resistor you need to know two things.

1) how many lamps are being cut out of a circuit, and the voltage of rating of the lamps. Example 15 lamps at 3.2 volts = 48 volts X 1.414 = 67.8

2) how much current is flowing through the lamps? I X 1.414 = Irms

then you can do this math example. 67.8 Irms = R. And to find the wattage of the resistor is 67.8 X Irms = Watts

Not knowing the current and actual number or lamps and their voltage rating. I can not suggest actual values.

[ErnieHorning]

The resistor method works until the first bulb burns out. Then they all go. I don't know of any simple 120 volt AC constant circuits at the 100-200 mA that is needed.

It's possible with LED's but you need to know what you're doing.

[Tim Fischer]

ErnieHorning wrote:

The resistor method works until the first bulb burns out. Then they all go.

Heck, that's not much different that a regular 20-light string. I've had strings where I've noticed a single bulb was out as I was leaving for somewhere, and come home to the string out and upon testing, virtually every bulb is dead (some completely silver inside instead of clear)

[ErnieHorning]

And the fuse didn't blow did it? ... Something I mentioned in another thread somewhere on this same topic.

[shredocaaster]

Thanks, guys.



It turns out the easiest way to do this is tell your Crazed wife right before she goes out shopping that 20 light LED strands don't exist... and guess what she came home with.

[beeiilll]

I was going to say "Why don't you just buy 20 light strands".

I use them for things myself.

[Max-Paul]

ErnieHorning wrote:

The resistor method works until the first bulb burns out. Then they all go. I don't know of any simple 120 volt AC constant circuits at the 100-200 mA that is needed.

It's possible with LED's but you need to know what you're doing.

Ernie, at this minute I can not see why using a resistor to replace X number of bulbs would make the string susceptible to major melt down if one bulb fails. Could you please explain what I am missing? I am thinking that the resistor is going to limit the current. And because we kept the lamps at a lower dropping voltage that the remaining lamps will only have to take a smaller portion each of the missing lamps voltage (see Kirchhoff law). So, why would the resistor increase the chance of mass failure. I could see that if we went from a 35 ct string to a 20 count string and changed the bulbs voltage how a single bulb could cause a mass meltdown.

So, Please enlighten me to what I am missing here.

Oh and what is this rambling about 100-200mA about? Never gave any hard current numbers in my post which I presume what the above quote is about, my post above.

[ErnieHorning]

Max-Paul wrote:

So, why would the resistor increase the chance of mass failure.

I had to look up Kirchhoff Law to find out it was again. It looks like it doesn't really apply though. It's for a closed loop, ie. a length of wire were you attach both ends together. There could be a voltage potential somewhere on the wire but there must also be an opposite voltage somewhere else that cancels it out.

The algebraic sum of the voltage (potential) differences in any loop must equal zero.

The OP wanted 20 lights. Typical mini light count is 100. It's pretty well known that a 100 count string is just two 50 count strings connected together so you're able to cut them in half. Let's say that a string requires .2 amps (200 mA) so when cut in half, each 50 count string uses .1 amps (100 mA).

A little history:

Mini strings originally came with only 48 lights. This is why if you 120 volts by 2.5, you get 48. 48 was confusing to add strings together so two more bulbs were added. This also benifits in that strings with 2.5 volt bulbs will last slightly longer.

So to get to 20 bulbs we need to substitute 30 bulbs with a resistor (120 volts /.1 amps = 720Ω) to waste 72 volts (remember each bulb only get 2.4 volts) leaving 48 volts accross 20 bulbs. Each bulb when lit has an equivilent resistance of 2.4 volts / .1 amps = 24Ω. 20 bulbs * 24Ω = 480Ω. Add the 720Ω resistor and you have 1200Ω total. 120 volts / 1200Ω = .1 amps, so the values check out.

Now one bulb burns out and the internal shunt shorts accross the filiment and the contunues to light. But now we only have 19 bulbs and the same resistor. So now 48 volts is apply to 19 bulbs. 48 volts / 19 bulbs = 2.52 volts. That's slightly over the rating of the bulb.

I don't have any hard fast numbers as to how long the bulbs will last at different voltages. Normal is 2.4 volts and I know from personal testing that they last for about 1 second at 3 volts. I'm told, by the president that of the company the makes the Light Keeper Pro, that a mini string is good for 3000 hours and that one bulb out will decrease the life by 1/3 and that would be 2.45 volts per bulb.

[ErnieHorning]

After further review of Kirchhoff Law, it basically states, what goes in must come out.

I guess I've been doing this for so long that I've figured that was a given.:?