View Full Version : Resistor values for light strings

Elmo2resc

05-29-2008, 12:53 AM

I am building some animated wire frames. My problem is that I am using 70 bulb strings. The string is divided in half with 120volts feeding each. I want to keep the same intesity for all the bulbs, but I need to shorten strings to make it work. There would be to many unused bulbs otherwise. I need 1 string with only 12 bulbs and 2 strings with only 22 bulbs each. My though was to use resistors to make up for the missing bulbs. I only wanted to use the least amount of resistors as possible. I figured with a meter that 35 bulbs at 120 volts uses: 20 watts, .19 amps. I know this doen't add up exactly, but this is what my meter read when I tested the string. ok, I need 12 bulbs to complete one of the strings. that leaves me with 23 bulbs that need removed.

.57 watts per bulb

3.42 volts per bulb

.0054 amps per bulb

78.66 volts / .1242 amps = 634.35 ohms @ 13.11 watts

Is this correct? Can it be done?

Can resistors be ran in series to increase the values?

Im trying to learn this stuff, but sometimes I think I just confuse myself more then I already am.

Thanks for any help. Jerome

ppohlman

05-29-2008, 10:42 AM

... Can resistors be ran in series to increase the values? ...

Resistors can be combined to change the overall total resistance.

Series:

http://www.facstaff.bucknell.edu/mastascu/eLessonsHtml/Resist/Resist2A4.gif

RSeries = Ra + Rb

Parallel:

http://www.facstaff.bucknell.edu/mastascu/eLessonsHtml/Resist/Resist2A6.gif

1/RParallel = 1/Ra + 1/Rb

or for only two resistors in parallel

RParallel = (Ra * Rb) / (Ra + Rb)

Hope this helps.

omzig

05-29-2008, 11:16 AM

Also worth noting is that if resistors are connected in parallel, the current (and therefore power dissipated) is split among them.

I am also faced with this situation. I have several animations that are going to require short sections of lights. I was looking at using resistors too, but I don't like the idea that you have to use rather large power resistors to handle that kind of power. They get very warm and you are just wasting electricity as heat.

What I think that I am going to do instead is use transformers to provide lower voltages for the shorter strings. A while ago the company that I used to work for was throwing away a bunch of stuff and I snagged a few things including a rather large multi-tap transformer that has 60, 48, 36, and 24 volt outputs. Figuring a 2.4V drop per bulb this will give me the ability to use 25, 20, 15, and 10 count strings respectively. I haven't actually tried this yet, but I don't see why it wouldn't work.

rlilly

05-29-2008, 11:45 AM

For 12 bulbs @ 3.42 v/bulb, your voltage drop is 41.04. You'll need a resistor that'll drop 120- 41= 78.96 v

12 bulbs @ .0054 amp will use .0648 amp.

Since V/I=R, your resistor will need to be 78.96/.0648 = 1218 ohm.

The wattage of the resistor will be 78.96*.0648= 5.12 watts.

A quick look at Mouser shows a part# 588-20J-1.25K.

This is a 10 watt 1.25k resistor which should do the trick. (There nothing between 5 & 10 watts, so pick the higher value.)

You can use the above method to calculate the proper resistor for the 22 bulb strings.

Elmo2resc

05-29-2008, 08:06 PM

For 12 bulbs @ 3.42 v/bulb, your voltage drop is 41.04. You'll need a resistor that'll drop 120- 41= 78.96 v

12 bulbs @ .0054 amp will use .0648 amp.

Since V/I=R, your resistor will need to be 78.96/.0648 = 1218 ohm.

The wattage of the resistor will be 78.96*.0648= 5.12 watts.

A quick look at Mouser shows a part# 588-20J-1.25K.

This is a 10 watt 1.25k resistor which should do the trick. (There nothing between 5 & 10 watts, so pick the higher value.)

You can use the above method to calculate the proper resistor for the 22 bulb strings.

Bob, wouldn't I use the amps of the remaining number of lights (23 @ .0054 and not 12 @ .0054) to figure out the ohms of resistance needed? I'm a little confused here at how this works. Thanks for anymore help.

Jerome

also thanks to Omzig and Ppohlman for your help.

ErnieHorning

05-29-2008, 08:28 PM

This will work until a bulb burns out. Each time a bulb is lost, the voltage on each will go up a lot faster then it would with 70 in series. Loosing just one bulb may take out the whole string before you even notice it. Just something to think about.

rlilly

05-30-2008, 08:35 AM

For 12 bulbs @ 3.42 v/bulb, your voltage drop is 41.04. You'll need a resistor that'll drop 120- 41= 78.96 v

12 bulbs @ .0054 amp will use .0648 amp.

Since V/I=R, your resistor will need to be 78.96/.0648 = 1218 ohm.

The wattage of the resistor will be 78.96*.0648= 5.12 watts.

A quick look at Mouser shows a part# 588-20J-1.25K.

This is a 10 watt 1.25k resistor which should do the trick. (There nothing between 5 & 10 watts, so pick the higher value.)

You can use the above method to calculate the proper resistor for the 22 bulb strings.

Actually, I messed up in the above analysis.

You'll pass .19 amp regardless of how many bulbs there are.

The voltage drop remains 41.04 for the bulbs, and 78.96 for the resistor.

The value of the resistor is 78.96 (V)/ .19 (A) = 415 ohms.

Wattage for the resistor is 78.96*.19 = 15 watts.

Mouser 588-TN15P400RFE is a 400 ohm/15 watt.

You still need to think about Ernie's comment and Omzig's method.

Dan Ross

05-30-2008, 11:14 AM

Why not just order 6v lights for the 22 count strands and 12v lights for the 12 count strand. You won't get full brightness out of them, but you won't have to worry about resistors burning up or adding transformers. I found some at this site, I'm sure there are many more.

http://www.bettyschristmashouse.com/lights/mini_replace.htm

It looks like a pack of 20 lights is $1.49

omzig

05-30-2008, 12:05 PM

Why not just order 6v lights for the 22 count strands and 12v lights for the 12 count strand. You won't get full brightness out of them, but you won't have to worry about resistors burning up or adding transformers. I found some at this site, I'm sure there are many more.

http://www.bettyschristmashouse.com/lights/mini_replace.htm

It looks like a pack of 20 lights is $1.49

Hey Dan, have you actually bought any of these? A while back I bought some 6V replacement mini bulbs that I was going to use in a 50 count string that I had shortened. I figured that the bulb bases would be different, so I was going to put the new bulbs in the old bases, but the new bulbs were a smaller diameter so it didn't work too well.

Then I bought some 20ct strings but I wasn't happy with them because they weren't as bright as the other minis that I was using in the same animation. It's kind of weird that the 6V bulbs were dimmer. I would have figured that the higher voltage bulbs would be brighter if anything...go figure.

ErnieHorning

05-30-2008, 01:16 PM

I would have figured that the higher voltage bulbs would be brighter if anything...go figure.Itâ€™s the wattage not the voltage, just like the different lights in your house. The current will be less for the same wattage though; ohms law and all.

Elmo2resc

05-30-2008, 05:58 PM

Actually, I messed up in the above analysis.

You'll pass .19 amp regardless of how many bulbs there are.

The voltage drop remains 41.04 for the bulbs, and 78.96 for the resistor.

The value of the resistor is 78.96 (V)/ .19 (A) = 415 ohms.

Wattage for the resistor is 78.96*.19 = 15 watts.

Mouser 588-TN15P400RFE is a 400 ohm/15 watt.

You still need to think about Ernie's comment and Omzig's method.

Thanks, now I understand, that makes sense. I would use the transformer method but then I would have to buy three per figure. I didn't want to have that junk strapped to it or cost. Why wouldn't the bulbs drop at the same rate? They are still running at 3.42 volts each. Isn't the resistor making up for the bulbs that are removed? If the resistor burns wouldn't it shut the string down?

Jerome

Elmo2resc

06-22-2008, 07:28 PM

Just a quick update. The resistors work great. The intensity is the same on all strings. Now just to see how long they will last. Thanks for everyone's help.

Jerome

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