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Adding LEDs to Turnouts

I have an N-gauge train with some remote-controlled switch tracks, "snap-switch" type.

This is an older style train, not the DCC type.

I would like to add red and green LEDs to the switch tracks. The switch tracks are operated from the "accessories" terminals of a small power supply which are nominally 17 V AC, (I measured it open-circuit at 17.7 V AC).

The switch tracks are operated by pressing momentarily on a momentary SPDT switch.

I'd like a simple way to hook up the LEDs either at the remote location or at the SPDT switch. Since power to the switch track is applied only momentarily, I suppose that some sort of latching mechanism is needed.

The 17 V AC is part of the complication, it seems.

For reference, the switch tracks are Atlas #'s 2580 and 2581; the power supply is Model Rectifier Corp. Railine 370N.

The switch tracks are apparently solenoid powered.

The power for the tracks only is 15 V DC. I wouldn't want to power the LEDs from that terminal pair because it turns off when the train is stopped, and includes the speed control.

Anybody here ever do this stuff with this type of remote turnout?

My guesses are some sort of relay or solid state relay, but I have no idea what the specific part number or circuit would be.

Thanks.

--- Joe

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Reply to
Joseph Sroka-10.2.8
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The switches I have are fitted with an extra SPDT on the other end that selects one side depending on which end of the track is selected. This is purely mechanical. Maybe you could fit something like this on the moving part of the tracks, so that it pushes the inverter one way or the other depending on which track is selected. Then you can put whatever voltage in the inverter, it is not electricaly connected to your solenoid.

Reply to
OBones

I read in sci.electronics.design that Joseph Sroka-10.2.8 wrote (in ) about 'Adding LEDs to Turnouts', on Thu, 14 Apr 2005:

You mean 2-colour light signals?

Yes, you have the right idea, but it isn't a simple thing for you to do if, as seems to be the case, you don't have much experience of electronics.

You need to get an 'always-on' DC supply of around 5 V minimum, preferably 12 V for easy LED-driving. Can you get at the 17 V AC supply? It's probably used to produce the 15 V DC as well, so make sure that you never have ANY connection between the two supplies. 17 V AC will give you 24 V DC when rectified, so you will need to halve it with a potential divider (two 1 kohm 0.5 W resistors). You don't need much current, so an unregulated 12 V supply with significant internal impedance should be OK.

Alternatively, just get a cheap 12 V DC regulated wall-wart. Regulated, in this case, because an unregulated one will give you around 16 V DC on light load, and that is too much.

You can use 4000 series or 74HC series logic chips to do what you want. You need flip-flops to 'latch' the LED drive signals. The tricky thing may be how to get the signals you need to tell the flip-flops to change state. You could take a sniff off the momentary solenoid-drive signal, but that's 'fail-dangerous' - if the flip-flops get out of step with the solenoids, you'll get a green signal on the wrong track!

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Reply to
John Woodgate

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