# Lighting A Caboose



## Bwells (Mar 30, 2014)

I am trying to light a caboose with one interior light and two marker lights. I have the markers and they are 1.5 volt incandescent. The interior can be an LED or bulb. It will be run on DCC track power so I figure I need to change it to DC voltage using a Bridge rectifier and behind that will be a 7805 IC to cut the voltage to 5 VDC along with a couple of caps. Then it is off to the lights. I don't think I will have 5 VDC filtered but I'm not sure if it matters. I'm not happy with my bulb design so if anyone has some ideas on this schematic and a suggestion as to whether putting the markers in series and then parallel with with some kind of interior light bulb I would appreciate it.


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## gregc (Apr 25, 2015)

Bwells said:


> I am trying to light a caboose with one interior light and two marker lights.


there are two advantages of DCC, constant track voltage and constant alternating voltage (equal time both polarities).

the fact that the voltage is alternating means that it is not absolutely necessary to have a bridge rectifier. Of course, incandescent bulbs would only be half as bright (lazy man's light dimmer). You may need a single diode in series.

and since the alternating polarity is of constant voltage (not sinusoidal) the large caps aren't needed.

instead of using a 7805, why not try LM317 which is adjustable. Since you suggested using 1.5V bulbs, why not wire them in series and configure the lm317 to output 4.5V. By wiring the bulbs in series, only a 1/3 the current and power needed compared to three parallel circuits.

a different approach is to use LEDs. Not sure if three 3.4V LEDs can be wired in series, but they can be different colors and monochrome LEDs may only be 1.4V. LEDs can be made just as bright even if powered only half the time (double the current). And the lm317 can also be configured as a current limiter suitable for LED circuits.


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## Bwells (Mar 30, 2014)

The reason for the bridge rectifier is I thought 7805's were only for DC. I looked at the LM317 and I think the adjustment comes from the size of the resistors. This looks a little more complicated. Thanks for the tip on the LM317, I'll do more research on it. As for the bulbs, maybe one 2 volt and the two 1.5 volt bulbs, all in series.If I can use the 7805 on AC then that should work, I think!


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## gregc (Apr 25, 2015)

yes, the bridge provides a positive DC voltage to the circuit regardless of track polarity.

the September 1991 Railroad Model Craftsman has an article "a new constant lighting circuit" using LM317s. (I assume you have 7805s in hand and would prefer to use them).

The lm317 has an 1.25 internal reference used to maintain its output terminal 1.25V above its common terminal. external resistors can be used to adjust its common terminal voltage so that the output voltage is maintained at more than 1.25V above ground.

since there are bulbs that operate close to 1.25V, the article (written for DC) suggests using the lm317s without any adjustment. But in order to handle both polarities it suggest using two lm317s wired opposite one another (i assume the lm317 is cheaper than a bridge or drops less voltage when reverse biased).

the input of each is tied to opposite track inputs. the common terminal of each wired to the output terminal of the other. The bulb is wired across the output of both. Depending on the polarity of the track voltage, one lm317 regulates its output while the "lamp current [passes] through [the] internal reverse voltage protection circuit" of the other


i also believe the 7805 can be tricked into regulating a higher voltage similar to the way the lm317 can be adjusted. If you tie a 1k resistor between its output and common terminal and another 1k resistor between common and ground, there should be ~5v across each resistor and the 7805 output will be at 10V.

if you need 6V output to drive three 2V lamps, maybe a 1k and 200 ohm resistor will work. (learn to use what you got. don't be afraid to experiment).


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## Bwells (Mar 30, 2014)

Greg, thanks for the help here. I found this circuit and with a 680 ohm resistor for R1, V out would be about 4.97 volts according to a table I found. I didn't work through the equation, but I should be able to drop it to 4.5 by changing r1 values on a bread board. What kind of a connection is the arrow at R1?


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## gregc (Apr 25, 2015)

the schematics shows R1 as a potentiometer (pot) with the center wiper connected to one end. This allows the resistance between the center wiper and the bottom terminal to be adjusted between 0 - 5K.

you would wire up this circuit and adjust the pot to get the desired output voltage, then measure the resistance across the pot and replace it with a fixed resistor.

i'm not so sure the capacitors are need if the circuit is drawing a constant current and not not switching. A digital circuit needs both large and small capacitors to smooth out the high frequency (or short term) current transients. (don't sweat it)


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## Bwells (Mar 30, 2014)

Greg: This is my bread board try at the 317 diagram with R2 at 220 and R1 at 680 and no caps. I'll go to the train room and hook it to track power and check the output. I know I have a bunch of 1.5 volt "grain of wheat" somewhere, just have to locate them. Thanks again for your help, Brian.


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## Bwells (Mar 30, 2014)

Update: I hooked this thing up to track power and get an output that starts at 4.4 VAC and slowly drops, it got down to 3.8 VAC when I unhooked it. Not sure why except that there was no load, only my meter. I may add the caps but I'm not sure caps work with AC as they are polarity sensitive. Any thoughts?


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## gregc (Apr 25, 2015)

i'm not sure, but it may need a load.

try removing R1 and putting a resistor across the output in place of the bulb. try using the 220 ohm resistor as a load, hopefully you measure 1.25V. If you have 68 ohm 1/4 watt resistor might be better.

If that works put back R1 back and see what you measure across the load.


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## wvgca (Jan 21, 2013)

if you have exactly what's on the breadboard hooked to dcc track power, it won't work ... the 317 is a DC device, needs a bridge in front ... caps are just for ripple smoothening not needed for incandescent bulbs.. most of the 1.5v bulbs are 30ma, so a T0220 case LM317 won't need a heat sink ... if there are room issues a LM317M is a lot smaller, and still has a maximum 1/2 amp rating


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## Bwells (Mar 30, 2014)

I added a 220 ohm across the output without removing R1 and now have a reading that holds steady at .653 VAC. It's off to some friends for game day but when I get back, I will series 3 1.5's and see what happens. Thanks


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## Bwells (Mar 30, 2014)

WVGCA: I totally missed your post. I have a bridge in my hand and will try it in front of the LM317t circuit and see what happens. Question, are the 317, 7805, 7812 as well as caps only for DC circuits?


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## Bwells (Mar 30, 2014)

BINGO!!! I put a Bridge rectifier in front of the LM317 circuit and get a reading of 4.24 VDC at the output. Totally happy with that as 3 1.5 v incandescents in series will burn at less than 1.5, not by much, but shouldn't be as bright as full blast. once I find my bulbs, I can adjust brightness by changing a resistor, hopefully.
Thanks to both of you on the help.
I'll assume the 317, 7805,7812 are all DC components.


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## wvgca (Jan 21, 2013)

Bwells said:


> I'll assume the 317, 7805,7812 are all DC components.


Yup, DC positive output, and the caps if they have a black [negative] stripe, or a circular indent ring on one end .., by running series you will have more adjustment room to get the brightness that you prefer, in parallel the 317 has a lower threshold of 1.25V , which is barely below the 1.5v bulb at full brightness ..if you want to reduce size, you can get the 317 in SMD, about the size of those bulbs, but I think those are only rated for 100ma, a little shy for safety margin


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