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Electrical Help Needed

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  • Member since
    July 2006
  • From: Mannheim, Germany
  • 26 posts
Electrical Help Needed
Posted by coblesan on Sunday, August 31, 2008 8:44 AM

Ok, so it's my first post.  I read these forums almost daily and I am amazed on the knowledge that is available here. 

A little background.  I currently live in Germany and am modeling a prototypical German railroad.  I want to incorporate all the things that everyone wants to put in their layout.  I would like to add Walthers Cornerstone Series Street Lamps (already purchased #933-1082 and #933-1095) to my layout.  The problem I am having is trying to figure out how to hook them up to the Power Pack.  My current power pack output is 14 - 16 Volts.  The instructions with the street lamps state that you must connect them to a power source of no more that 12 Volts.  I will be using a power bus throught the layout to connect the lamps. 

When reading the forums, I know the smaller the amount of voltage the longer the lights will last.  I also am a rookie at electrical stuff.  I have been reading about connecting resistors to each lamp to lower voltage (I think this is correct).  I do not know how many amps each street light will need or draws, so I do not know how much resistance is needed for each street lamp to work correctly.  Trying to find the information with Google is not really helping.

Any and all help is greatly appreciated.

Trying to make it work in Mannheim, Germany

Mike

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Posted by retsignalmtr on Sunday, August 31, 2008 9:00 AM
try connecting two lamps in series and see if they are bright enough for you. this will drop the voltage on each lamp to 7-8 volts. we use powerpacs with the same voltage here and i have found  that the voltage on the accessory terminals is rarely what is stated on the lable. i've used street lamps directly connected to a power buss with no problems of burnout without a resistor in line with them. try using an automobile turn signal bulb as a ballast resistor.
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Posted by Phoebe Vet on Sunday, August 31, 2008 9:04 AM

If you wire two lights in series (power lead one to light lead one, light lead two to second light, second light lead two to the other side of the power) it will solve the over voltage.  Current is not relevant.  Current is a limit.  the total amp draw of all devices must not exceed the amp limit of the supply device, but the lamp will only draw what it needs. (That is not a true description of current draw, but is a good illustration for people who do not understand electricity.)

It would be a better idea to use a separate device to power your lights, switches, etc.

Dave

Lackawanna Route of the Phoebe Snow

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Posted by cacole on Sunday, August 31, 2008 9:17 AM

You didn't say how many lamps you have.  The answers so far assume that you have only two lamps, but if you have more, you should wire them in series-parallel groups; i.e., three lamps in series as a group, wired in parallel with a second set of three wired in series.

It would be advisable to wire the lamps in sets of three or use an automotive bulb as the third lamp if you have only two, because when the power pack is turned on and off, there will be a very high voltage spike sent through the lamps that could burn them out or cause them to weaken over time.

A power pack typically puts out 18-19 Volts AC on the accessory terminals, but the voltage spike could approach 30 volts or more through cold lamp filaments.  Using the lamps in groups of three will make them a little dimmer but they will also last a lot longer.

Where are you in Germany?  I spent three years in Munich (1975-78) and three years in Frankfurt (1980-83) on Army assignments.

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Posted by Phoebe Vet on Sunday, August 31, 2008 9:28 AM

Three 12 volt lamps in series is too many for an 18 volt power supply.

I assumed that he would have the common sense to know that more bulbs would be treated separately in sets of two.

Dave

Lackawanna Route of the Phoebe Snow

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Posted by BlueHillsCPR on Sunday, August 31, 2008 9:35 AM

My first advice would be to use a separte power supply for your layout lighting and accessories.  Often modelers will use a "wall wart"...you know those black power transformers that plug into the wall and provide DC to all kinds of electronic gadgets.  Alternately some guys use an old computer power supply, slightly modified. 

Get one that is 12V and use resistors to reduce the power going to each lamp or buy a 9V supply and use it without resistors.  A transformer rated at 500 ma, (milliamps) will power a ton of lights.

EDIT: The maximum number of lights powered will be subject to the current draw of each light.  Also it is wise to only load the transformer to %75 of rated capacity for safety and durability reasons. 

Do a search in the layout building forum and in the electronics/dcc forum for "layout lighting".  You should find a bunch of threads that will be very helpful.

Here are a few links that may be usefull.

Layout Lighting 

Practical Electronics 

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Posted by Phoebe Vet on Sunday, August 31, 2008 9:40 AM
I agree with Blue Hills.  A 9 volt separate power supply would be perfect for 12 volt lights.

Dave

Lackawanna Route of the Phoebe Snow

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Posted by Mark R. on Sunday, August 31, 2008 9:52 AM
 BlueHillsCPR wrote:

.... A transformer rated at 500 ma, (milliamps) will power a ton of lights.

Actually, that's not necessarily true. A lot of the "cheaper" 12 volt mini bulbs will have a draw of about .050 amps (50 milliamps) .... a 500 milliamp transformer will only power 10 of those bulbs to reach the transformers capacity !

Mark.

¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ

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Posted by MisterBeasley on Sunday, August 31, 2008 10:11 AM

I run my Cornerstone street lights at 10 volts.  They don't provide a lot of light at that voltage, but the image is about what I'm looking for.  As I recall, these are rated at 16 volts.  If you run them well below their rated voltage, they'll last a long, long time.

I have an old train transformer with a lot of different voltage taps on it.  It's a Lionel, and it puts out AC.  This works fine for lights.  I just found the 10-volt tap and I use that for most of my lighting.

In looking at these lights, I don't think they are designed for bulb replacement.  You might be able to do it, but it's not going to be an easy job.  The socket mount also implies that the idea is to replace the whole unit if the bulb goes - another reason to be very careful with your voltage!

 

It takes an iron man to play with a toy iron horse. 

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Posted by Johnnny_reb on Sunday, August 31, 2008 10:25 AM

If you have a little skill in wiring you can use a "computer power supply" to meet all of your layout accessory/light needs. On my web site I detail how this can be done without opening the power supply case. Thus no shock hazard!

 http://johnnny.reb49a.googlepages.com/hobbypowersupply

With the cost of new power supplies so cheep. If need be you could build a few of them for less then what you would pay for "wallwarts" (wall mount transformers)

Johnnny_reb Once a word is spoken it can not be unspoken!

My Train Page   My Photobucket Page   My YouTube Channel

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Posted by BlueHillsCPR on Sunday, August 31, 2008 11:32 AM
 Mark R. wrote:
 BlueHillsCPR wrote:

.... A transformer rated at 500 ma, (milliamps) will power a ton of lights.

Actually, that's not necessarily true. A lot of the "cheaper" 12 volt mini bulbs will have a draw of about .050 amps (50 milliamps) .... a 500 milliamp transformer will only power 10 of those bulbs to reach the transformers capacity !

Mark.

You are right.  I should have qualified my statement by mentioning the draw of the lamps determining the maximum number of bulbs which could be powered.  Also it should be noted that it is wise to only load a tranformer to around %75 of maximum capacity for safety and longevity. Thumbs Up [tup]

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Posted by Seamonster on Sunday, August 31, 2008 12:11 PM

Another way to knock down voltage is to use diodes.  A rectifier diode will drop 0.7 volts.  The beauty of this is that it is completely independent of the current drawn by the lamp.  There are no calculations involved at all.  You just keep adding diodes in series until you've knocked down the required number of volts, i.e. 2 diodes will reduce the voltge by 1.4, three diodes by 2.1, etc.  The only thing to keep in mind is the forward current capacity of the diodes.  The smallest is usually 1 ampere and they go up from there.  So if you have 10 lamps drawing 50 mA each, a 1 amp diode is plenty.  This method, of course, works only with a D.C. (direct current) supply.  It will not work with A.C. power.

 

..... Bob

Beam me up, Scotty, there's no intelligent life down here. (Captain Kirk)

I reject your reality and substitute my own. (Adam Savage)

Resistance is not futile--it is voltage divided by current.

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Posted by Phoebe Vet on Sunday, August 31, 2008 12:31 PM
For what possible reason would you use a bunch of diodes in place of a 15 cent resistor?

Dave

Lackawanna Route of the Phoebe Snow

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Posted by Mark R. on Sunday, August 31, 2008 12:39 PM

 Phoebe Vet wrote:
For what possible reason would you use a bunch of diodes in place of a 15 cent resistor?

The diodes would be installed right at the power source creating the desired output voltage with no modification to each individual bulb. Resistors on the other hand would need to be added to each light source, and depending on the power source may require a larger wattage size to dissipate the heat.

.... Ultimately .... six of one, half a dozen of the other ! Wink [;)]

Regarding diode dropping and AC .... the method works equally well if you use a bridge rectifier to convert the AC to DC. By doing this, you already introduce two diodes in both waves creating an auto-matic voltage drop of 1.4 volts on the DC output.

Mark.

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Posted by Phoebe Vet on Sunday, August 31, 2008 12:48 PM
... or you could just use the correct power supply istead of jerry rigging everything with silly work arounds.

Dave

Lackawanna Route of the Phoebe Snow

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Posted by Mark R. on Sunday, August 31, 2008 12:54 PM

.... But we like to use the junk we have laying around as opposed to buying the "right" item !  Laugh [(-D]

Mark.

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Posted by Seamonster on Sunday, August 31, 2008 1:53 PM

 Phoebe Vet wrote:
For what possible reason would you use a bunch of diodes in place of a 15 cent resistor?

15 cents for a resistor?!  Can you say "ripoff" children?

Yeah, I agree, using the correct voltage power supply is the best way to go, but you can't always get one of the voltage you want.  Besides, like someone said, we like to use what we've got.  At least with diodes, the arithmetic is a lot more simple and you don't have to worry about heat buildup or a change in voltage if the number of connected lamps changes.

 

..... Bob

Beam me up, Scotty, there's no intelligent life down here. (Captain Kirk)

I reject your reality and substitute my own. (Adam Savage)

Resistance is not futile--it is voltage divided by current.

  • Member since
    April 2004
  • From: Ontario Canada
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Posted by Mark R. on Sunday, August 31, 2008 2:04 PM

If you rely on Radio Shack for your electrical components, 15 cents each is a bargain !!!  Whistling [:-^]

Mark.

¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ

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