As I sit here watching it rain cats and dogs I started pondering again on a question Grey One brought up in a previous thread: http://www.trainboard.com/grapevine/showthread.php?129820-How-Do-You-Track-Your-Points-er-Which-Way-They-Are-Thrown I had all but given up on my plan to use track voltage to power Bi-colored LEDs due to the amount of current needed for all the turnouts I need to wire. (old plan below) But then I thought of another angle, using some type of latching integrated circuit and a separate power source for the LEDs. Here’s a schematic of what I think this IC should look like inside. (One IC per turnout) And now the questions: 1. Do any of you with electronic backgrounds know if any such ICs exist? 2. I drew solenoids that activate and direct the LED power. (Obviously there is no solenoid in an IC). Would this type of circuit draw less power from the track than a LED? I was hoping that it would draw virtually 0 current from the track once the LED voltage was directed. 3. Cost for such a circuit? Any thoughts on my idea? Brian
How about this method? See post #4 http://www.nscale.net/forums/showthread.php?23148-Unitrack-Turnout-Control if you cannot see the picture on the 4th post. here it is:
All my trunouts are manually thrown so switch machines aren't an option. I was hoping for a simple IC that would sense the current flow of the turnout and then direct a seperate current source in a desired direction to the LED. Brian
Are you going to be using DC or DCC? If you use DCC, then creating the circuit will be easier. If you use DC, it will be more complicated due to the track polarity switching
DCC. Buying an IC that will do what I want is preferrable but I could build a circuit if I had a schematic and parts list. Brian
Could you use a 3 lead bi-polar led with a covential DC power pack and a DCC system? Following the circuit on Gartner's website (http://www.wiringfordcc.com/signaling.htm), it appears that two leads of the led are connected to the two stock rails while the third lead is connected to the switch on a manual ground throw. I did awful in my circuits class, but from what I have gathered on led's, the change in color is due to a change in voltage, or lack thereof (either +12V to -12V or +12V to +12V, no drop). I have noticed a couple of posts in this thread that are concerned with the current and voltage draw of the led's. If they could be continuously powered with a DC powerpack, you could control the voltage out to the led without sacrificing anything from a DCC booster. I realize that you would have to keep track of which rail is which on the layout, but to me, that sounds simpler than building a circuit, even if it is just a couple of diodes and resistors. Doug
If you look at my original schematic it is pretty much what the 3rd option shows in your link. (sans the frog power switch since I am using peco power routing turnouts). Problem is after you power over 60 bi-colored LEDs with this method, and 5+ locomotives using track power I am pushing the current limits of my DCC system. Thats why I was wondering if the IC I am searching for would draw any track power (at least less than a LED) after it directs the seperate LED power source to the LED(s). Thanks though for your thoughts. Brian
Well Rats!!! I was trying to improve on my IC schematic pic in my original post and seem to have lost the picture. To make matters worse I cannot edit my original post to insert a new one. So heres the new pic of the IC Schematic I am looking for.
If you're using caboose industry ground throws. Then they sell a ground throw that has contacts in it and 3 leads to power frogs,signals, switch indicators, you name it. Just provide one cheap DC power pack to power everything. Or for that matter if you don't want to use a power pack with the LEDS. You can buy a cheap charger for a cell phone. Some are rated at just the right voltage for LEDS. make sure its an old one though because the new ones also carry digital info. Cut off the jack pack and wire the LEDS to it, worked wonders for me.
Yes Caboose ground throws are the mechanical answer to the problem but for reasons of appearance this isn't an option for me. No offense to all the guys that use these, but they just don't do it for me visually. Thats why I'm looking for an electronic option if one exists. Good idea on the phone charger though. If I come up with the correct IC I will look at all the phone warts I've collected over the years. Brian
no offense taken. I choose them because I've seen great reliability out of the caboose ground throws. For true operation flexibility I'd say the Atlas Switch Machines (the giant ones with the slides) are the best because they are both electric and mechanical, but I won't use them for the same reason as you, they look to awkward on a layout.
If you are doing manual turnouts, why not combine the two. How about this: You use a spdt(Single poll double throw) slider switch to control the turnout LEDs and throw the switch. You drill a small hole in the switch head(black area) to secure a stiff metal wire(i.e. paperclip) You bend the wire like so, to give the wire tension(like a spring). You then attach the wire to the switch. You can use this design also for a reversing loop. Instead of throw a light the switch would change the power. Hope this helps
Yes this another mechanical way to do what I am trying to accomplish but I am looking for a solid state (ie. IC, electronic component only) method that would be essentially invisible when viewed from the surface of the layout. The circuit I am trying to assemble must be available I just don't know how to describe it in a google search. Brian
Correct me if I'm wrong, but aren't you using a 5 amp Lenz system and running N-Scale? If so, I don't think you will be as close to the limits as you think. Most LED's are plenty bright enough at 10 ma, and most N-scale locomotives draw less than .2 amps with the wheels slipping. If you were to allow .25 amps per locomotive and 10 ma per LED, you could have 100 LED's and 10 locomotives with power to spare[(100 x .01) + (10 x .25) = 3.5 amps. Regardless, to do what you want you need a photocoupler, also sometimes called an optoisolator or optocoupler. Here is one example. It is basically an LED and phototransistor packaged together. You power the LED(but with less current than needed for the indicator LED) and the light it emits triggers the phototransistor to conduct, turning on the indicator LED. You would need a resistor on the photocoupler input to protect the photocoupler's LED and make sure the photocoupler can handle AC current. I can come up with a circuit later if you need me to.
I have been researching using the optocoupler, but I am still a little confused. Can you give an example for a problem such as this?
Here is an example of a circuit using an optoisolator(the section within the blue square is a single integrated circuit that is the optoisolator) : Note that the optoisolator shows two LED's in reverse parallel(often called an AC input optoisolator). Most optoisolators only have one, but when dealing with DCC you have to either have the two LED's or circuitry to protect the optoisolator from reverse voltages. Depending on the selection of optoisolator, LED, DCC voltage and accessory voltage, this circuit would power the LED at around 10 ma while only drawing around 1 ma from the DCC power.
Correct. You can get IC's that have multiple optoisolators in a single package. For example, you can get a 16 pin DIP IC with 4 individual optoisolators in it. One thing I forgot to mention is one if the specs that is important when using optoisolators in this way is the current transfer ratio, which is the ratio of the output current to the input current. In the example I posted above, you would need a current transfer ratio of 1000% (10ma / 1ma x 100). You can get the same result with a low current transfer ratio optoisolator by driving a transistor which in turn drives the LED, but, of course, that adds parts and complexity to the circuit.
CSX Robert. I knew there was some kind of IC or home made circuit that would do what I am looking for. THANK YOU!!!! So, after looking at the link you included to a retailer that sells these things, I think I need the 8 pin version. Heres how I think it should be wired. Your opinion (or anyones for that matter) would be greatly appreciated. I didn't show any resistors on the DC side since I still don't know whether I'm going to buy bi-color LEDs with a common cathode or anode. Brian ps. sorry I haven't responded sooner. I've been up to my "you know where" in alligators.
There are some problems with the circuit you have shown. The optoisolator diagram that you show does not have the two LED's in reverse parallel on the inputs, meaning it would be damaged by the reverse voltages encountered with DCC. The output side of the optoisolator is also different from the one I showed. It appears to have internal amplifiers(that's what the two hollow triangles in the diagram are) driving the output transistors. There are several different output designs for optoisolators, and depending on the specs for the particular optoisolator that one might work in this situation, but the wiring would be different. The amplifiers need to be powered, which would come in on pins 8(+) and 5(GND). The LED would have to be common anode and the cathodes would be connected to pins 6 and 7.
