I'm planning a new small layout and like the look of those little 2 or 3 color dwarf signal lights. While most use them in conjunction with a turnout system, I plan to use them mainly for decorative purposes, and wish to power them using a 12VAC transformer. My plan is to hook up a two or three way switch to control the which color bulb is illuminated. Are these dwarf lights typically 3V (DC or AC?) LED's.
Powering "dwarf" track lights
- Thread starter Toolbelt
- Start date
-
- Tags
- dwarf signal voltage
ctclibby
Well-Known Member
Don't confuse yourself here. LED's are just diodes. There are LED + Resistor sets out there for [pick your fav] 3V, 6V, 12V ...etc. If you are rolling your own, this might help.
Note that you *usually* want to set the current @ 15 to 20mA and use a resistor to do just that. Assume Yellow LED, 5VDC power supply: (VPowerSupply - Vf ) / 0.020 ~ 145 Ohms. You can not get a 145Ohm resistor, 147 or 151 where the 151 will make the LED just a tad dimmer. These can be 1/8Watt parts too.
Later
Note that you *usually* want to set the current @ 15 to 20mA and use a resistor to do just that. Assume Yellow LED, 5VDC power supply: (VPowerSupply - Vf ) / 0.020 ~ 145 Ohms. You can not get a 145Ohm resistor, 147 or 151 where the 151 will make the LED just a tad dimmer. These can be 1/8Watt parts too.
Later
Last edited:
So I just need to throw a 150 ohm resistor on a 12vac supply wire then?
Iron Horseman
Well-Known Member
Are these current production signals, or are they old stock? Depending on how old, they could be anything from 16v to 1.5v bulbs or as ctclibby pointed out, LEDs. If you have specific ones in mind, it would help focus the discussion.
Iron Horseman
Well-Known Member
Ok, with a 12 vac power supply I would start with a 1K ohm resistor. If they are too bright and glare then increase to 1200 ohm, etc. until desired brightness is achieved.
Iron Horseman
Well-Known Member
nope, doesn't matter. With AC they will basically be turning on and off 60 times a second, but your eyes won't notice it.
Last edited:
Smudge617
Well-Known Member
I've so many of these dwarf signals I've lost count.
If you're going to use 2 aspect LED dwarf signals it's easy, there are 3 wires, one is your common or positive, this is the wire you need to fit a resistor to if not already fitted, and the other two are the ground/earth wires for the two LEDs (usually Red/Green).
The common positive connects directly to your power supply (if installing multiple signals a Distribution Board is useful).
Using an SPDT switch connect the two outer pins of the switch to the two ground wires on the signal LEDs, and the centre pin on the switch needs to go to the negative/earth on the power supply to complete the circuit.
When the switch is in the centre position, the lights will be off, moving the switch up or down will illuminate one or the other of the signal LEDs
If you're going to use 2 aspect LED dwarf signals it's easy, there are 3 wires, one is your common or positive, this is the wire you need to fit a resistor to if not already fitted, and the other two are the ground/earth wires for the two LEDs (usually Red/Green).
The common positive connects directly to your power supply (if installing multiple signals a Distribution Board is useful).
Using an SPDT switch connect the two outer pins of the switch to the two ground wires on the signal LEDs, and the centre pin on the switch needs to go to the negative/earth on the power supply to complete the circuit.
When the switch is in the centre position, the lights will be off, moving the switch up or down will illuminate one or the other of the signal LEDs
Thanks for that. On my small layout, my plan is to install maybe 6 signals around the tracks, and I'll likely have them each illuminated (either red or green) all the time, which I'll have to factor in when choosing a resistor value, and use a single switch to alternate the colours. After reading up more on LED's, it seems running them on DC is the better choice. So if I place the proper resistor on the positive feed to my switch, I should be able to simplify the system to avoid needing multiple resistors and switches. Does that make sense?
Smudge617
Well-Known Member
Ok, so your idea is to add the resistor between your power supply and the DB so you don't have to add multiple resistors to the signals, and using only one switch so they all would show red or green. Have I got this right ?
If you don't want to use a DB, I assume you intend to daisy chain the LEDs ?
Iron Horseman
Well-Known Member
Are the dwarfs 3 wires or 4. or are you using individual LEDs and putting them in the dwarf yourself?
If the cathode '+' is the common wire to a pre-wired dwarf as Smudge617 described, here is one way to do it. It could also be done as you suggest with the resistor on the common center part of the electrical switch. Doesn't matter. You could also flip the switches to the wire positive side instead of being on the negative side. Once again doesn't matter. There are many ways to do this, I don't know that one has any advantage over another.
yes, exactly that. I was thinking I would, for example, have 4 reds and 2 greens activated when I throw the switch one way, and 2 greens and 4 reds activated when throwing the switch the other way. It's more for appearance than function, just to add a little interest to the layout. If I daisy 6 leds, possibly I won't even need a resistor?
The ones I noted earlier are 2 led (red/green) units with three wires.
Smudge617
Well-Known Member
A daisy chain will work, but you will need a resistor, as you intend to shove 12V or about 1.25 amps down a set of diodes that each need about
70 milliamps to work.
Use Ohms Law to determine the minimum size of the resistor needed, which should give you full brightness on all the LEDs, you would then have to gradually step up the resistor if they're too bright, or use a dimmer switch.
The downside is that if you get anything wrong, you could blow some, if not all the LEDs, the signals, not having their own resistors, have no protection from an accidental overload.
I would still advise you to use an SPDT switch tho' (On-Off-On) at least then you have a choice as to whether you want them on all the time.
If you don't know Ohms Law, (which I fail miserably to even comprehend the basics of) here's a handy calculator.
Ohms Law Calculator
This free Ohm's Law calculator solves for any of the variables in the Ohm's Law equation using various units of measurement and gives out the solving steps.
www.calculator.net
Last edited:
Based on my limited research, it seems that LEDs themselves require DC voltage, so I will likely have to convert my AC voltage to DC voltage or just use a DC transformer instead. I will have to experiment with what voltage/resistors provide the best appearance. I want the dwarf lights visible, but just barely glowing.
Smudge617
Well-Known Member
A 12v wall wart is relatively inexpensive, but being LEDs your going to have to use a high resistor, or you could simply paint the LED faces with acrylic paint.
Iron Horseman
Well-Known Member
Yes but No, an AC power supply will work just fine. LEDs activate on DC but that doesn't mean they require a DC supply. That is to say, on an AC circuit they will light on the 1/2 of a cycle that is in the matching polarity, and they will block and go dark the other half. In effect making their own DC power. One doesn't see the flicker because our eyes can't see that fast of a change. This is exactly like the LED Christmas Tree lights work, they plug right into the AC wall outlet and just use 1/2 the power. Buy the way, LEDs work on DCC power also. All of the automatic dwarf lights on turnouts on my former layout drew power straight off the track.
As Smudge617 said though, to get them the desired brightness (or dimness in this case) use a higher resistor. The higher the resistor value the dimmer the LEDs will be.
Last edited:
GGNInNScale
Granite Gorge and Northern Again
OK, time to chime in- DO NOT use such high currents in the LEDs. They may be a bit brighter but they are getting hot which will degrade the output. You only need about 3-5milliamps for really bright lights. So, R = Voltage/Current. Use about 1kOhm for 3V, 1.5kOhms for 5V, and 3.3kOhms for 10-12Volts. Not really critical. The resistors can be 1/8W or SMD versions, really does not matter, depending on space available. I would attach a picture of an 0201 with an SMD resistor, but my camera cannot see it.... Resistor substitution boxes can be found on the internet for under 10 bucks if you want to play around easily. Remember that LEDs are diodes, so they are polarized devices- just check your connections with the diode tester in a multimeter.
