I got a dwarf signal which has a red and green LED. Has a common wire and a red and green wire. It says to use 1/2 w resistor in both red and green line, and shows white going to a common + and the red and white to separate -. Talks only about DC, not DCC. I want to use this with DCC, a Caboose ground throw with built in SPST, and hand laid turnout. No problem wiring the active frog and turnout to the ground throw. 1. Don’t know where the red, green and white (common) on the dwarf goes to. (Ground throw SPST terminals or the turnout some how)? Where common is on the throw or the turnout is confusing to me. 2. Will the dwarf be ok w/DCC and is the resitors of right size and to be in series to LEDS. Thanks much, Hal
So you are using track power for the signals?
Do you mean a SPDT. … So the ground throw points are providing power to the turnout frog?
Assuming the two things above are true, this won’t work because the power going to the frog, reguardless of which way the turnout is thrown, will be a DCC bi-polar (almost AC) current. Each LED in the signal will get 1/2 the power all the time. {EDIT} The first solution I posted here won’t work. {EDIT again} The only solution (using only the SPDT contacts) I can think of involves adding a second DC power supply for the signal diodes. Finally, another note for anyone else trying to figure this out, after further reading the description of the signal LEDs, it sounds like they are in parallel in the same polarity with the common.
It depends on the voltage that your DCC system is set to. The higher the voltage your unit is set to the higher the resistance value will need to be.
Ok I think I’ve figured it out.
Put the white common wire to the frog.
Put the resistor into each of the green and red wires. Then run the other side of the resistor in the red wire to the outside(opposite the frog) diverging rail. Run the other side of the resistor in the green wire to the outside of the straight rail.
This assumes the diverging rail is the side track. If not swap the red and green.
Theory babble - Since the diodes have the same polarity facing the common, they will not short the DCC signal. Each diode will only be lit by 1/2 the DCC power signal.
Texas Zepher… thank you so much for the time you took to figure this out for me. Armed with your help, I’m ready to wire it up. Hal