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Simple question about lighting structures

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Simple question about lighting structures
Posted by HarryHotspur on Thursday, May 15, 2008 10:40 PM

I have dozens (or so it seems) of strands of small, clear Christmas bulbs with 50 bulbs per strand. It seems like this would be a cheap and easy way to light the interior of about 12 structures on my layout.

But a couple of guys told me, "DO NOT use 110 volts of do it yourself wiring on your layout. You will burn your house down and probably electocute yourself." Okay, makes sense, I won't. But is it possible to run these bulbs (about 12 of them) off a spare MRC Tech 2400 I have? It says 17 VA maximum.

The bulbs are 3.0 volts and are wired in series. Thanks for any advice and/or explanations of what will/won't work. The only other indiciation on the strand is the bottom of the wall plug says 1 amp max.

- Harry

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Posted by Rotorranch on Thursday, May 15, 2008 10:54 PM

Yeah, you can do that.

If they are 3.0v, you can also get a 3v tranformer from somewhere like Radio Shack, and then wire some in series, and some parallel to vary the brightness.

I'd avoid wiring too many together in series. If one blows, the whole stirng would go down, leaving a trouble shooting nightmare.

Rotor

 Jake: How often does the train go by? Elwood: So often you won't even notice ...

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Posted by HarryHotspur on Thursday, May 15, 2008 11:02 PM
 Rotorranch wrote:

Yeah, you can do that.

If they are 3.0v, you can also get a 3v tranformer from somewhere like Radio Shack, and then wire some in series, and some parallel to vary the brightness.

Rotor

Great! Thanks. I think I prefer the powerpack because I guess it could act like a dimmer switch. Is there a formula that will allow me to determine how many bulbs I can power? Also, do I need to worry about sending too much voltage and blowing out the bulbs?

Sorry for all the questions, but as you can tell I'm an electrical dummy. 

- Harry

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Posted by tomikawaTT on Thursday, May 15, 2008 11:06 PM

I have thousands of 2.5v minibulbs, wired up in strings of 50 for use on Ye Olde Tannenbaum.  Outfit I was working for dumped them for about two cents a lamp after Christmas, plus I got a 10% employee discount.  What I love is the cute little sockets, which make changing dead lamps a simple pull and replace proposition (look, Ma, no solder!)

The key is, each bulb is designed to take 2.5v (but will give a nice, cheery, yellowish light on 1.6v.)  I wire them in series of four and power them from the 6.3v tap of a Radio Shack filament transformer.  Three in series are brighter, if you want that harsh, industrial light look in your model factory or warehouse.

I did wire them into one control panel as route indicators, with dropping resistors.  My more recent control panel will have the additional series bulbs placed to (dimly) illuminate the turnouts and close-clearance points of some hidden track (visible through windows in the fascia.)  The blasted resistors, at a dime each, cost more than the three additional bulbs and sockets needed for voltage regulation, and don't give off anything except heat.

As long as you can't see the lamp itself, and leave clearance and ventilation to prevent heat buildup, mini-lamps are great.

Chuck (modeling Central Japan in September, 1964)

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Posted by Rotorranch on Thursday, May 15, 2008 11:13 PM
 HarryHotspur wrote:
 Rotorranch wrote:

Yeah, you can do that.

If they are 3.0v, you can also get a 3v tranformer from somewhere like Radio Shack, and then wire some in series, and some parallel to vary the brightness.

Rotor

Great! Thanks. I think I prefer the powerpack because I guess it could act like a dimmer switch. Is there a formula that will allow me to determine how many bulbs I can power? Also, do I need to worry about sending too much voltage and blowing out the bulbs?

Sorry for all the questions, but as you can tell I'm an electrical dummy. 

Yes, there is a formula...but electronics class was a long time ago. Whistling [:-^]

Yes, if you crank the MRC up it WILL blow the bulbs, possibly leaving the layout in a blackout! Wink [;)]

I'm sure one of the electronics whizzes here can step in with the formula.

Rotor

 Jake: How often does the train go by? Elwood: So often you won't even notice ...

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Posted by HarryHotspur on Thursday, May 15, 2008 11:31 PM

That's good to know Chuck. And, yes, I'm looking for that yellowish glow to light small houses.

 Just as a starting point, I hooked up the 50 bulb strand to the MRC, turned off the room lights, and let 'er rip. Nothing. Zilch. Not even a faint glow. The MRC is "known good" because I used it to run a train before and after the bulb trial.

So I guess I need to reduce the number of bulbs. I certainly don't expect you or anyone here to teach me basic electronics, but is there a website you could point me to? Google produces mega hits of course, but the ones I looked at aren't on point with what I'm trying to learn.

I fact, since I only know the voltage of the bulbs and the max output of the MRC, is that enough info to do any calculations?

- Harry 

- Harry

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Posted by Rotorranch on Friday, May 16, 2008 12:28 AM

Just a WAG...If all 50 bulbs are in series on the same circuit, and it wont light on 12v, try 1/10 of the string, or 5 bulbs, since you are operating at 1/10 the voltage.

As I said, just a WAG.

Rotor

 Jake: How often does the train go by? Elwood: So often you won't even notice ...

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Posted by jeffrey-wimberly on Friday, May 16, 2008 3:43 AM
I have structures that are lit by old Christmas lights of the type you describe. Each structure contains two 3 volt bulbs wired in series on a 5 volt feed so each bulb gets 2.5 volts. They've been burning 24/7 for many years and haven't gone burned out yet.

Running Bear, Sundown, Louisiana
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Dr. Frankendiesel aka Scott Running Bear
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beatus homo qui invenit sapientiam


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Posted by HarryHotspur on Friday, May 16, 2008 12:54 PM

 jeffrey-wimberly wrote:
I have structures that are lit by old Christmas lights of the type you describe. Each structure contains two 3 volt bulbs wired in series on a 5 volt feed so each bulb gets 2.5 volts. They've been burning 24/7 for many years and haven't gone burned out yet.

That's great. Good to know. Thanks. 

- Harry

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Posted by HEdward on Friday, May 16, 2008 1:13 PM

Formula?  Babies are eating people food now, no more formula, thanks!

Here's a formula, equal parts charcoal, sulpher and potassium nitrate.  Mix carefully.  Ahhh....maybe not.

V=IR

I is the current for a reason that I've forgotten.

V equal volts.

R is the resistance.

Proud to be DD-2itized! 1:1 scale is too unrealistic. Twins are twice as nice!
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Posted by HarryHotspur on Friday, May 16, 2008 8:10 PM

Okay, somebody tell me if I've got this right. Assume the following: I have a strand of 50 bulbs rated at 2.5 volts each. My powerpack is rated at 12 volts, 17 VA max. So, if I wired 6 bulbs in series, each bulb would receive 2 volts (12volts/6bulbs). Right?

Then instead if I wired a bus, not sure if that's the right term, but I meanone wire from each terminal extending along the layout AND then wired 5 separate loops of 6 bulbs each in series, with the end of each loop connected to the bus, I would have 50 bulbs each receiving 2 volts. Correct?

If it is correct, how would I calculate the current draw to know if the powerpack could handle it? 

 - Harry

 

- Harry

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Posted by jeffrey-wimberly on Friday, May 16, 2008 8:18 PM
Can't help ya there. I use a computer power supply to run mine. It can handle a bunch of bulbs. I had it lighting almost a hundred of those little Christmas lights on my last layout.

Running Bear, Sundown, Louisiana
          Joined June, 2004

Dr. Frankendiesel aka Scott Running Bear
Space Mouse for president!
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Collector of Apple //e's
Running Bear Enterprises
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Posted by HarryHotspur on Friday, May 16, 2008 8:44 PM

 jeffrey-wimberly wrote:
Can't help ya there. I use a computer power supply to run mine. It can handle a bunch of bulbs. I had it lighting almost a hundred of those little Christmas lights on my last layout.

That's another option. I've Lord knows how many of those things around, and good ones too because I always bought the best quality parts (as far as I could tell) and assembled my own. But how could you work a dimmer in with that? 

- Harry

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Posted by jbinkley60 on Friday, May 16, 2008 8:45 PM
 HarryHotspur wrote:

Okay, somebody tell me if I've got this right. Assume the following: I have a strand of 50 bulbs rated at 2.5 volts each. My powerpack is rated at 12 volts, 17 VA max. So, if I wired 6 bulbs in series, each bulb would receive 2 volts (12volts/6bulbs). Right?

Right

 HarryHotspur wrote:

Then instead if I wired a bus, not sure if that's the right term, but I meanone wire from each terminal extending along the layout AND then wired 5 separate loops of 6 bulbs each in series, with the end of each loop connected to the bus, I would have 50 bulbs each receiving 2 volts. Correct?

By the term bus you refer to being in parallel electronically.  There would be 30 not 50 bulbs (5 loops of 6 each)

 HarryHotspur wrote:

If it is correct, how would I calculate the current draw to know if the powerpack could handle it? 

 - Harry

Not enough information has been provided to answer this question.  What you would need to know is the current draw for one of the lamps at 2 volts.  Then each "loop of 6 bulbs" would draw that amount of current, since they loops are wired in series.  So you would multiply the amount of current in one loop by 5 (each loop would draw the same amount of current) to get the total current draw.  A 17VA power pack can supply 17/12 = 1.42A of current or each loop can draw up to 1.42A / 5 = 283 ma.  283 ma is generally quite a lot for a 2V bulb.  Most bulbs draw much less than this but it depends upon the bulb.

 

Engineer Jeff NS Nut
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Posted by HarryHotspur on Friday, May 16, 2008 9:22 PM
 jbinkley60 wrote:
 HarryHotspur wrote:

Okay, somebody tell me if I've got this right. Assume the following: I have a strand of 50 bulbs rated at 2.5 volts each. My powerpack is rated at 12 volts, 17 VA max. So, if I wired 6 bulbs in series, each bulb would receive 2 volts (12volts/6bulbs). Right?

Right

Great. I'm finally making progress. 

 HarryHotspur wrote:

Then instead if I wired a bus, not sure if that's the right term, but I meanone wire from each terminal extending along the layout AND then wired 5 separate loops of 6 bulbs each in series, with the end of each loop connected to the bus, I would have 50 bulbs each receiving 2 volts. Correct?

By the term bus you refer to being in parallel electronically.  There would be 30 not 50 bulbs (5 loops of 6 each)

Yes, I meant parallel electronically, but my math was awful.

 HarryHotspur wrote:

If it is correct, how would I calculate the current draw to know if the powerpack could handle it? 

 - Harry

Not enough information has been provided to answer this question.  What you would need to know is the current draw for one of the lamps at 2 volts.  Then each "loop of 6 bulbs" would draw that amount of current, since they loops are wired in series.  So you would multiply the amount of current in one loop by 5 (each loop would draw the same amount of current) to get the total current draw.  A 17VA power pack can supply 17/12 = 1.42A of current or each loop can draw up to 1.42A / 5 = 283 ma.  283 ma is generally quite a lot for a 2V bulb.  Most bulbs draw much less than this but it depends upon the bulb.

 

Thanks a lot! That's a big help. 

- Harry

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Posted by HarryHotspur on Friday, May 16, 2008 11:47 PM

One last try.

On the tag it said use GE W2 35/50CL bulbs. I found on the net these bulbs are rated at 0.34 watts each. Is that enough info to calculate the number of bulbs my powerpack will handle?

BTW, I tried to do it myself, but V=IR doesn't seem to include watts.

- Harry 

- Harry

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Posted by jbinkley60 on Saturday, May 17, 2008 5:04 AM
 HarryHotspur wrote:

One last try.

On the tag it said use GE W2 35/50CL bulbs. I found on the net these bulbs are rated at 0.34 watts each. Is that enough info to calculate the number of bulbs my powerpack will handle?

BTW, I tried to do it myself, but V=IR doesn't seem to include watts.

- Harry 

P = IE   or .34W = I x 2V  or I = 170ma.   The only thing that isn't clear is if that is .34W of power at 2V.  I think you'll be fine.

 

Engineer Jeff NS Nut
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Posted by HarryHotspur on Saturday, May 17, 2008 2:27 PM
 jbinkley60 wrote:
 HarryHotspur wrote:

One last try.

On the tag it said use GE W2 35/50CL bulbs. I found on the net these bulbs are rated at 0.34 watts each. Is that enough info to calculate the number of bulbs my powerpack will handle?

BTW, I tried to do it myself, but V=IR doesn't seem to include watts.

- Harry 

P = IE   or .34W = I x 2V  or I = 170ma.   The only thing that isn't clear is if that is .34W of power at 2V.  I think you'll be fine.

 

Thanks again. Just so that I MIGHT learn how to do this myself, here is the info I found:

Appendix B

Test of Electrical Consumption of Christmas Lights

A digital power analyzer was used to examine the wattage of several strings of commercially available Christmas lights. Five different strings of various types were tested. Line voltage was 121 V during the test. Each was classified as a "Decorative Lighting String" by UL and were manufactured in China or Thailand:

Desctiption Type # of bulbs Watts W/bulb
1) Small clear bulbs
(GE W2-35/50CL)
Indoor/outdoor
Rated: 36 W
100 34.1 0.34
2) Small "midget globe
bulbs"
Indoor
Rated: 36 W
50 11.0 0.22
3) Small colored/clear bulbs
(GW-W2-35/50)
Indoor/outdoor
Rated: 36 W
100 33.4 0.33
4) Colored light set
C 7.5 5 W bulbs
Indoor/outdoor
Rated: 125 W
25 128.0 5.12
5) Large 10 W C- bulbs Outdoor
Rated: 500 W
50 504.4 10.17

Does this alter the answer my question? Sorry, I should have posted it the first time.

Link if needed is here.

 - Harry 

 

 

- Harry

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Posted by jbinkley60 on Saturday, May 17, 2008 6:55 PM
 HarryHotspur wrote:
 jbinkley60 wrote:
 HarryHotspur wrote:

One last try.

On the tag it said use GE W2 35/50CL bulbs. I found on the net these bulbs are rated at 0.34 watts each. Is that enough info to calculate the number of bulbs my powerpack will handle?

BTW, I tried to do it myself, but V=IR doesn't seem to include watts.

- Harry 

P = IE   or .34W = I x 2V  or I = 170ma.   The only thing that isn't clear is if that is .34W of power at 2V.  I think you'll be fine.

 

Thanks again. Just so that I MIGHT learn how to do this myself, here is the info I found:

Appendix B

Test of Electrical Consumption of Christmas Lights

A digital power analyzer was used to examine the wattage of several strings of commercially available Christmas lights. Five different strings of various types were tested. Line voltage was 121 V during the test. Each was classified as a "Decorative Lighting String" by UL and were manufactured in China or Thailand:

DesctiptionType# of bulbsWattsW/bulb
1) Small clear bulbs
(GE W2-35/50CL)
Indoor/outdoor
Rated: 36 W
10034.10.34
2) Small "midget globe
bulbs"
Indoor
Rated: 36 W
5011.00.22
3) Small colored/clear bulbs
(GW-W2-35/50)
Indoor/outdoor
Rated: 36 W
10033.40.33
4) Colored light set
C 7.5 5 W bulbs
Indoor/outdoor
Rated: 125 W
25128.05.12
5) Large 10 W C- bulbsOutdoor
Rated: 500 W
50504.410.17

Does this alter the answer my question? Sorry, I should have posted it the first time.

Link if needed is here.

 - Harry 

So 121V / 50 bulbs= 2.42V / bulb .  I'd say you are fine.

 

Engineer Jeff NS Nut
Visit my layout at: http://www.thebinks.com/trains/

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Posted by HarryHotspur on Saturday, May 17, 2008 9:36 PM
Thanks Jeff. I really appreciate the help.

- Harry

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