Customizing Incandescent Light Strings

[newby]

I'm trying to find out if I can customize my incandescent light strings.  Can I make them into light strings of less than multiplies of 50?  I know that I can modify them into multiples of 50 and that there shouldn't be less than 35 lights on a string, but I'm wondering if I can use resistors in the strings and then modifiy them to specific lengths?

[ItsMeBobO]

Careful ... The bulbs are different power draw on 50 vs 35.            in 50-bulb circuits, the bulbs run at 120V / 50 = 2.4V.  35-bulb circuits use 3.4V bulbs, and a 10-bulb circuit uses 12V bulbs

 

Maybe you could use the resistor that comes with it.  Called a bulb and just cover them up. 

[ericm]

i tried this and the bulbs get really bright and everyone of the blow after about a minute.  If you need 35, they have strings of 35 you can buy, otherwise buy the blackout covers.

[Max-Paul]

i tried this and the bulbs get really bright and everyone of the blow after about a minute.  If you need 35, they have strings of 35 you can buy, otherwise buy the blackout covers.

And you figured out why this happened? Remember Kirchhoff's  law. The voltage applied must drop across all of the loads. In short the voltage is 120V. Now if you have as Bob points out 50 lamps then each lamp must drop 2.4 volts and if you have 35 lamps. Then each lamp  must drop 3.4 each. If you reduce a 50 lamp string to 25 lamps. Now each lamp will have 4.8 volts across each lamp. up from 2.4 volts, so it has doubled. Think that this might be the reason your lamps burnt out?

[ericm]

I figured it out afterwards, but I'm still a newbie in this this was my first year... But learned on one small strand...

[newby]

So in essence I would have to replace each bulb I removed with a resistor of equal value, so just covering them would accomplish the same thing.  Would this cause the covered bulbs to overheat?  Also, could I replace the bulbs I removed with one resistor that would drop the voltage an amount equal to all the bulbs that I removed.  Example:  If I had a 50 bulb string, could I shorten the string to 40 bulbs with a resistor in the string that would drop 24V?

[Grinch]

The bulbs are in series which means Yes one large enough  resister would act exactly like 10 individual resisters.  Current * Resistance = Voltage and just rearrange the formula to get the unknown.  This will not be the same for lights wired in parallel. 

[newby]

If I used the resistors in the string, would I have to use a different voltage bulb?

[ItsMeBobO]

No need to change the bulbs.  In your first post you said you could remove bulbs from 50 down to 35.  I think that may have come from the knowledge that you can buy 35 bulb strings.   the 35 strings use a different voltage bulb so you cant do that.   

The goal is to keep the total resistance the same for the string.  If you need to remove 15 bulbs, you need to replace them with 15 bulbs of resistance.  2.4v for each bulb removed from a 50.  If you dont do this then the string will fail as ericm pointed out.

[Max-Paul]

So in essence I would have to replace each bulb I removed with a resistor of equal value, so just covering them would accomplish the same thing.  Would this cause the covered bulbs to overheat?  Also, could I replace the bulbs I removed with one resistor that would drop the voltage an amount equal to all the bulbs that I removed.  Example:  If I had a 50 bulb string, could I shorten the string to 40 bulbs with a resistor in the string that would drop 24V?

Yes, but there is one item still missing. How much current is being drawn in this circuit? Ohms law requires to items to calculate the 3rd item. WE have voltage of 24 volts. Now if we knew the current then we could figure out the resistance. Formula would be 24 divided by current = resistance. Current  is in the form of Amps, most lamps would be in mA. So use .xxx format.

 

Then there is one more issue. Once you know the current flowing in the circuit and the voltage you will be dropping across the single resistor. You need to figure out the wattage of the resistor. Example lets use the 24 volts you have called out and I am going to pull a current value out of thin air as an example. Lets use .033 Amps. So the wattage of the resistor is calculated as such, 24 X .003 = .792 watts. So, it looks like you would need a 1 Watt resistor in this example. Remember this. You can always use a resistor that has a higher than needed wattage value. But if you do not get one that is rated high enough. It will get very hot, might cause a fire or if you are lucky it will just burn its self up and open up.

Edited January 1, 2015 by Max-Paul

[newby]

Thank you all.  This has been very helpful.  I'll try to shorten some strings and see what happens.  One more thing, should I cover the resistor with heatshrink or something?

[Max-Paul]

What would you do if a critter chewed off the isolation from an extension cord? Would you leave the wire bare or would you tape it over? Just talking about a single wire, not both.

[newby]

I would tape it over. I guess I'm asking if heatshrink tubing would be an apropriate covering.  I was just wondering if it would keep the resistor from disapating its heat.

[Max-Paul]

Yes heat shrink is make just for the purpose of isolating the voltage from something that you do not want to energize. And yes your thinking is right, the heat shrink will act like a blanket and will keep some of the heat within the resistor. That is why I will always use the next size up watt rating of the resistor. Cost is small to insure that there will be no failure. When I rebuild an ican string into an LED string I am not building it to make money thus no short cuts in my strings.

[a31ford]

Because a resistor might get warm (or even hot) in this application, I would use a capacitor (or a pair of electrolitics back to back)

 

the charging of the NP (non polarized) Cap. can be figured out using Ohm's law...  (for the voltage drop required) and the frequency of the circuit (50 or 60hz).

 

This is an old trick, it will NOT get warm (or even hot), however, after about 15-20 years the capacitors will eventually dry out and the voltage to the lamps will climb.

 

I'd have to look up the old data on using Cap's so if you decide to go this route, let me know , and I will find the reference material. somewhere in my notes.

 

Greg

Edited January 11, 2015 by a31ford

[newby]

Would the capacitor hold the voltage and need to be discharged?

[a31ford]

Technically yes, however a high value resistor (one that would NOT even get warm) across the cap, would bleed the voltage down. something up in the 20-50 K ohm range would do it (That's 50,000 ohms, just so we are seeing the "K").

 

it would have no effect on the actual circuit, but would bleed off the residual voltage....

 

let me go find the article on using a cap as a voltage divider..... and I'll post it here.

 

EDIT >

 

a better one > http://www.electronics-tutorials.ws/capacitor/capacitive-voltage-divider.html

 

Wiki version (somewhat lame)  http://en.wikipedia.org/wiki/Voltage_divider

 

 

Greg

Edited January 13, 2015 by a31ford

[Max-Paul]

a31ford,

 

Have you done this with an LED string that has been cut down? Might work with an incan string, but have to wonder if it would work on a string of LED that only conduct in one direction. In order for a cap to work in this application the voltage has to be able to reverse on the two plates of the cap. Have to admit, during my school days Xc and XL kind of kicked my butt to grasp the concept behind it. The diagrams showing two caps dont make sense. I can see one cap in series with a string of lamps.

 

Could you shine a wee bit more light on this? Thanks

 

BTW wonder if this will work with the PWM voltage from a LOR controller? Something in the back of my head ask about fading and wasn't there an issue strings with caps catching on fire?

Edited January 13, 2015 by Max-Paul