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Marker light hookup with soundtraxx long caboose decoder

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  • Member since
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Marker light hookup with soundtraxx long caboose decoder
Posted by JWK Railway on Thursday, January 26, 2017 4:17 PM

Has anyone had success hooking up Tomar bi-polar marker lights with a soundtraxx long caboose decoder? I am not electonically inclined and don't understand how to read prints. I say bi-polar since the white light on each marker light comes on when tested + or - on a AA battery. This will be used on a NCE DCC system.

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Posted by rrinker on Thursday, January 26, 2017 5:04 PM

 They are incandescent lamps, not LEDs, that's why they work no matter which way you hook up the battery. What they don't seem to mention on their web site is the current for the bulbs, which is what you need to know to calculate what resistors you will need. They do have markers with LEDs, the same part number with an L on the end is a standard LED, where a 1K resistor should be good.

                             --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

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Posted by JWK Railway on Thursday, January 26, 2017 5:24 PM

Thank you. I will call Tomar and then Soundtraxx for proper current set ups.

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Posted by DigitalGriffin on Thursday, January 26, 2017 9:45 PM

This is where a $8 voltmeter from Harbor freight comes in handy.  If you switch it to the Ohm (Omega symbol) setting, and put a probe across each wire, it will tell you what the resistence is. (Rtomar light)

VDCC = I(R1+ Rtomar light)
The tomar light is 1.5Volts so....
1.5V/Rtomar light = I
14 = VDCC

SO...

14 = (1.5/Rtomar light)(R1+ Rtomar light)
14 / (1.5/Rtomar light) = (R1+ Rtomar light)
(14 / (1.5/Rtomar light)) -  Rtomar light  =  R1

Ris the resistor you need to hook up.


So pratical example: Lets say you measure the leads across the bulb and it reads 15 Ohms

1.5 V/15 Ohms = .1 amps

14 = .1(R1+ 15)
14/.1 =R1+ 15
140 =   R1+ 15
140 - 15 = R1
125 = R1

So you need a 125 Ohm resistor in the above example.

Test
VDCC - (Vresistor) = Vtomar bulb
14 - (125 * .1) = 1.5
14 - 12.5 = 1.5
Confirmed.
 

Don - Specializing in layout DC->DCC conversions

Modeling C&O transition era and steel industries There's Nothing Like Big Steam!

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Posted by CSX Robert on Friday, January 27, 2017 7:27 AM

The problem with measuring the resistance of the light bulb is the cold resistance is going to be much lower than it's operating resistance.  Yo would be better off measuring the curent draw.

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Posted by JWK Railway on Friday, January 27, 2017 11:09 AM

Thanks for the technical advice, but not being a technical person, I called Tomar and found out the #818 incandecent bulbs that I had were 50 miliamp each. Then called soundtraxx and told to use a 470 Ohm 1/4 Watt resistor for each lamp to be hooked up to their decoder.

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Posted by JWK Railway on Friday, January 27, 2017 11:12 AM

I called Tomar and was told that the resistance would be the same either way.

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Posted by DigitalGriffin on Friday, January 27, 2017 1:29 PM

JWK Railway
I called Tomar and was told that the resistance would be the same either way



I thought so with the 1.5V.  But CSX robert is right.  A lot of bulbs (especially festoon bulbs) drastically alter in resistence when you put too much current in them)

That said, that means the bulb is 30 Ohms resistence.

14 = .05(R + 30)
14/.05 = R + 30
280 = R + 30
250 = R

Curious soundtraxx would suggest a resistor that big (around 470).  That's almost double what you need with typical DCC voltage.

You could start with 470...but I think that will be dim.  You might want to work your way down slowly to lower resistor values if it's too dim.

Don - Specializing in layout DC->DCC conversions

Modeling C&O transition era and steel industries There's Nothing Like Big Steam!

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Posted by hon30critter on Friday, January 27, 2017 2:44 PM

DigitalGriffin
You could start with 470...but I think that will be dim.

If the lights are still be bright enough with the 470 ohm resistor, using a larger resistor will extend the life of the bulbs.

Dave

I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!

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Posted by rrinker on Friday, January 27, 2017 4:58 PM

50mA through a 250 ohm resistor is .625 watts, you'd need a 1 watt resistor AND a place to put it so as not to melt the shell.

That's one reason to use the bigger resistor.

 Those tiny marker lights don't need full brightness to look nice, either - plus they are brass castings which, with a 50ma bulb at full brightness in each one, will get quite warm. Also potentially melting something, or softening the attachment point. And the bulbs will last longer. That's a second reason.

 Finally, a bulb has typically a lower resistence when cold, even these little ones, and the inrush current has been known to blow function outputs, especially when the running current is 50ma, the inrush can exceed a 100ma function limit easily. The bigger resistor takes that into account.

 I know when I go to buy some for a couple of cabooses (and a few of my locos - I should put them on ALL of my RS-3's but that's going to be a lot of work, it's a big fleet) I will definitely be getting the LED versions. Instead of the typical 1K resistor, they will probably need 4.7K or so.

                         --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
    December 2004
  • From: Pa.
  • 3,354 posts
Posted by DigitalGriffin on Saturday, January 28, 2017 3:10 PM

Good point on the 1 watt resistor.  That said 470 Ohms will lead to a .028 amp current.  30 * .028 = .84 Volts.  Those bulbs are dim at 1 V.  And I'm not worried about the heat.  They use 1.5Volts all the time for headlights in locos.

That said even if the bulb were 0 ohms, you aren't going to go past your 100ma rating on the function inputs.

So 330 would be a good trade off for bottom I think?  That would yield a 360 total resistance and a .038 total current.  That would be .47 watts or a 1/2 watt resistor.  That would leave 1.14 volts across the bulb.

Don - Specializing in layout DC->DCC conversions

Modeling C&O transition era and steel industries There's Nothing Like Big Steam!

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