Is there a way to wire a two-light (two LED) signal to a turnout without using additional contacts? Maybe using diodes and the necessary dropping resistors? The common (white) to the frog and the red and other color to the feeder rails?

Any help appreciated.

Glen Duke

-- Glen Duke (Glen_Duke@141.net), January 11, 2005

Answers

Yes. The answer is similar to what you have already described if you have a polarity switched powered frog. For a 3 lead, 2-LED assembly, (common cathode or common anode) attach the lead common to both LEDs through a current limiting 1K resistor to the powered frog. Check LED polarity orientation and provide a 1N400x back voltage protective diode in series with each one of the other LED connections and wire those to the outer stock rails of the turnout. When the frog is at the polarity of the 'straight' stock rail (throwbar in the "R" position), the LED attached to the curved stock rail will light from constant DCC voltage on the rails. With throwbar and frog in the opposite position the other LED will light up. DonV

-- Don Vollrath (dvollrath@magnetek.com), January 13, 2005.