lionelsoniThe pins are probably meant for circuit-board mounting. You might check out what Radio Shack has that might work with them,
I found this breadboard at Radio Shack. I think it will work for connecting to the relays and will save on a lot of messy wires and soldering.
Terry Thomann Fredericksburg, Virginia That is me on the left. My brother got the train TCA 09-64381
Some afterthoughts:
It may not have been obvious that the same DC supplies will be used for all the turnouts on the layout--you don't need separate supplies for each turnout.
If you design your own LED indicators, you can arrange to power them from the same supply that you use to charge the capacitors.
You can power related turnouts from a single capacitor-lamp circuit. For example, the two turnouts of a crossover are always thrown together; so they might as well have a single control switch. The circuit we used here will usually have enough charge to do that; but, if not, you can increase the capacitance, perhaps by paralleling a second capacitor with the first one.
You can use a power resistor in place of the lamp; but it doesn't fully charge the capacitor as fast as the lamp does. As the capacitor charges, the charging rate drops using either a resistor or a lamp. But, as the lamp dims, its resistance drops, partially compensating for the reduced voltage difference between the capacitor and the supply.
I like to use the lamp in some lighted accessory for this purpose. For example, I have several lighted bumpers wired into the capacitive-discharge circuits of nearby turnouts. To isolate them from the track, I just mount them on a very short piece of track at the end of the siding, isolated by gaps from the rest of the siding track.
Don't forget the stud-and-probe scheme I touted earlier. It's a good way to make a control-panel map more compact.
You don't have to use green for the straight path and red for the curve. On a prototype railroad, the red really indicates the diverging path. Although they almost always make the diverging path curved, on a toy-train layout, for lack of space, we often need to do it the other way; so on a control panel, you might want to use red for a diverging path, even if it is straight.
Bob Nelson
Sounds good to me!
balidasGOOD JOB! I'd check this thread a couple times a day for updates. Once I'm done with a current project, I'll start on my turnouts.
GOOD JOB! I'd check this thread a couple times a day for updates. Once I'm done with a current project, I'll start on my turnouts.
Bob gets all the credit. I only bought some inexpensive parts and stumbled through his instructions to put one turnout together as a sample project. I would like to write this up to eliminate all my mistakes, add some pictures and a wiring diagram so it will be a clear and concise procedure, and either post it here or submit it to CTT as an article (after Bob has checked it for accuracy and with his permission). I think this is a fantastic way to operate turnouts and make a very efficient control panel.
lionelsoniCongratulations! It was a pleasure helping you with this. I'm curious: Did the red wire on the LED turn out to be the positive one?
Congratulations! It was a pleasure helping you with this.
I'm curious: Did the red wire on the LED turn out to be the positive one?
Bob,
Yes, the red lead was positive and I had no trouble finding out which pin on the relay to attach each LED so that I got a green light when the turnout was in the straight position and a red light for the curved track. Watch for an email from me about a write up I am working on for this project.
Thanks Bob for all your technical advise and patience with me to get this right. Also thanks to all others who added their valuable advice and comments.
Everything is working. Your suggestion of using the 3 batteries to check LED polarity worked like a charm. I have finished one turnout and it works perfectley. I love how the latching relay holds its position even when the power is turned off and when it is turned on again the correct indicator light is on. Now all I have to do is finish my layout and build a control panel. Thanks so much. This has been a great project.
He could use an ohmmeter, if its battery is no more than 5 volts. (I wouldn't risk more than that, even with a ballast resistor in the circuit.) However, the trick is knowing which ohmmeter lead is which. I have seen them with both polarities. Unless he has another voltmeter to check the ohmmeter or a spare diode to compare with the LED, he still won't know which end is up.
I thought of an experiment that might sort out the LEDs for you: The problem with just trying them out on 12 volts is that they can stand 5 volts of reverse voltage, but not much more. But red and green LEDs, unlike blue and white ones, need only a couple of volts of forward voltage to start to light up. So, get yourself three 1.5-volt dry cells (any size) and connect them in series (as they would be in a three-cell flashlight) to make a 4.5-volt DC supply. Connect an LED to that to see which way it lights up. The end of each cell with the nub on it is the positive terminal. The LED won't be very bright; but I think you'll see some light.
By the way, since there are no lumps in the wires in the vendor's picture, I think the ballast resistor must be inside the plastic case, which means that the lump in your wire may after all be a diode to protect the LED from reverse voltage. Maybe.
lionelsoniI would probably just solder directly to them.
I'll solder them. Give me some good practice. As soon as I figure out the LEDs I'll be done.
The turnout works much better with this DC circuit. With track power they were very sluggish.
The pins are probably meant for circuit-board mounting. You might check out what Radio Shack has that might work with them, like this, for example: http://www.radioshack.com/product/index.jsp?productId=2102845&filterName=Type&filterValue=Breadboards
I would probably just solder directly to them.
The LEDs I got do look a lot like the pictures but as you say the pictures don't show any lumps. I did send them a question about the polarity of the leads. They should get back to me in a day or two. I can't believe they didn't identify them. Not sure if I can get access to the LEDs. Everything seems sealed in a plastic case.
I did begin attaching leads to the latching relay. Boy are the pins on the relay small. Is it OK to solder to them? They look like they should go into some kind of circuit board that the leads would be attached to. Is there a better way to attach leads to the pins? This first turnout is just a practice one to learn the set up. If there is something I can plug all my latching relays into for the panel board that would make it a lot easier. So far though everything is working as you indicated it should. I could get addicted to this kind of project. Thanks for all your help. I could NEVER have figured any of this out for myself.
I will check previous posts to see if you said where you got them. There might be something on the internet. Or, if you see this first, post a link to the vendor.
I found your link. The web-site picture shows red and black leads, which is almost certainly positive and negative respectively. However, red and yellow is worrisome. Red might still be positive--or not. You might want to try to contact them. The thing in series with the lead is probably a ballast resistor, not a diode, unless there are two things in series.
I looked at the web site again. I don't see any lumps in their pictures. Do your LED assemblies resemble their picture at all? If you can get access to the actual LED, you may be able to identify the cathode, which often has a flat place on the LED's flange next to it. That will be the negative side.
lionelsoniConnect the negative terminal of both LEDs to the negative terminal of the 12-volt supply.
I am not sure which wire on the LED is negative. There is one red and one yellow. The yellow has something (possibly a diode?) in its line. Can't fine anything in the description of it on line to indicate which is which. I have several so if I ruin one by getting it wrong, at least then I'll know. I would have a 50/50 chance of getting it right. Any ideas?
I wired up the toggle switch to the turnout and it works perfectly. I did notice that in one direction, when the turnout coils retracted the operating rod it would get stuck like it is overtraveling. I was able to glue in a small block of wood to prevent overtravel, and it works fine. The turnout switches fine and should work perfectly this way.
Most of the early HO power packs use a rehostat(variable resistor) to control the output voltage. If you do not have a load(engine) connected to the power pack, the output will read 12V. Also, the DC is just rectified AC-- no filtering. For what you intend to use them for, they will work just fine.
A Day Without Trains is a Day Wasted
lionelsoniYou saved me the trouble of looking up the relay diagram--great!
That is the least I could do. you have given me a lot of help on this project, and I am learning a lot about train wiring from it. Thanks. I checked out my new 12 volt transformer today and it seems to be ok. It has a set of constant 12 volt DC terminals on the side as well as a set for cab 1 and cab 2. It also has a set of AC terminals. I will probably use the constant DC terminals. The cab 1 & 2 rheostats don't seem to work well. As soon as I start to turn the knobs they both go right to 12 volts and there isn't much variation in voltage output. Is this normal with these old transformers? Every one I get seems to work this way. No matter, It will work well for what I need. When I get home tonight I will complete the wireing of my first turnout and see how it works. I'm building a little sample control panel so I can see how it all works. Can't wait to see the turnout snap back and forth and watch the LEDs change from red to green. This is going to be so cool. Thanks again. I'll let you know how it turns out.
You saved me the trouble of looking up the relay diagram--great! Pins 1, 3, and 6 are the operating coils, which will be connected in parallel with the turnout coils. Pin 3 is common to both coils, so it gets connected to the new wire that you installed, or to the negative terminal of the capacitor, which is the same thing. Notice that it requires a negative polarity, which is why we installed the capacitor and the 18-volt power supply the way we did. Pins 1 and 6 get connected to the same insulated terminals of the turnout to which you connected the SPDT switch. (You did try that out, right?)
We will use pins 7, 10, and 12 to connect the LEDs, one or the other, to the 12-volt supply. There are two equally good ways to do this; but here's one: Connect the negative terminal of both LEDs to the negative terminal of the 12-volt supply. Connect relay pin 7 to the positive terminal of the 12-volt supply. Connect pin 10 to the positive terminal of one LED and pin 12 to the positive terminal of the other LED. If the colors don't match the turnout position the way you want, swap the LEDs.
Below is a schematic of the latching relay. It is Greek to me can you tell me what gets connected to each of the pins? This is exactly how my relay is numbered. This came from the seller's web site. I got my new 12 volt transformer today but I forgot it at work so will have to test it tomorrow.
Congratulations! You're right, the two outer terminals of the switch go to the two insulated turnout terminals. The center switch terminal goes to the outside rails generally. The switch does the same thing as the truck axles; it connects one or the other of the control rails to the outside rails generally.
Your relay has 3 terminals for the coils and 3 terminals for the contacts.
Got it right this time. Works perfectly. It is amazing how the turnout snaps into place when the truck hits the control rails. The Capacitor charges up just like it should. This is so much better than using track AC power. Can't thank you enough.
I should have my new 12 volt transformer in a couple days, maybe tomorrow. I have a SPDT toggle switch. I assume of the three terminals on the switch two go to the turnout and the center one I'm not sure what goes to it. I have a latching relay. It is 12 volt DC. It is a SPDT, but I am confused (as usual) because it has six terminals on the bottom. There is a black band at one end. Does this indicate polarity? The pins on the bottom are numbered 1 3 6 on one side and 12 10 7 on the other. 1 and 12 are at the end with the black band. Does this tell you anything?
I also have one red and one green LED lamp.
You have it wired wrong. Neither lamp terminal should connect to the outside rails. Only the + terminals of the capacitor and the 18-volt supply connect to the outside rails.
Capacitor: + to outside rails, - to new wire and one lamp terminal
18-volt supply: + to outside rails, - to other lamp terminal
I don't know your meter; but you're probably right that 1 is positive.
The lamp should fully charge the capacitor in 1 2/3 seconds.
lionelsoniThen you can go ahead and connect up the 18-volt supply that you already have, since that's the one to use to charge the capacitor. Connect the positive side of the supply to the outside rails, or etc. Connect the negative side of the supply to the free terminal on the lamp. The lamp will light only when the capacitor is charging. Then put a truck on the track over one or the other of the control rails, to test the capacitive discharge operation.
Something is amiss. I conntected the Negative from the Capacitor to the new wire to the turnout coils. The Positive side of the Capacitor is connected to one side of the lamp and to the outside rail. (I used a track connector CTC)
The other side of the lamp is connected to the Negative side of the Transformer. The Positive on the Transformer is connected to the outside rail via the same CTC track connector. The transformer is not marked positive or negative, but when I put my meter to the terminals with the negative lead on #2 and the positive lead on #1 it get an 18 Volt reading. I am assuming that #1 is positive and #2 is Negative.
When I apply power, the lamp goes on. How long does it take to charge the capacitor? I thought it would be pretty quick. When I run an engine's front trucks onto the control rails nothing happens. I have power on the track via a 1033 Lionel Transformer. The lamp for the turnout indicator lantern lights up. I do hear a very faint buzzing from the turnout coils.
Have I wired this correctly? Could it be that the turnout I selected is no good?
Thanks for your patience with me with this. I'd love to get this right so I can start on the toggle switch and latching relay part of this.
Any of the above; that is, connect the positive side of the capacitor to any outside rail other than the two control rails or to any wire or transformer post that is connected to the outside rails.
Then you can go ahead and connect up the 18-volt supply that you already have, since that's the one to use to charge the capacitor. Connect the positive side of the supply to the outside rails, or etc. Connect the negative side of the supply to the free terminal on the lamp. The lamp will light only when the capacitor is charging. Then put a truck on the track over one or the other of the control rails, to test the capacitive discharge operation.
Success!
The frog is drilled and tapped. The Negative end of the capacitor is soldered to a lamp socket and to the new wire from inside and the new wire is routed out the bottom of the turnout base. There was a nice slot so I didn't have to drill a hole. Will the lamp always be on, or will it only be on when the capacitor is charging? I used a lamp socket so I can change the bulb easily when it burns out.
When you say the Positive side of the capicator is attached to an outside rail does that mean I can connect it to any outside rail or to a buss wire under the table that connects to the outside rails?
I should have my new 12V transformer in a couple of days. Then I will be ready to connect the toggle switch to the turnout and to the indicator lights.
Thanks
Oil is a good idea. I've never had a bit break in one; but the tap gets stopped easily. Do a lot of back-and-forth, back the tap out frequently and clean out the chips, and don't try to advance more that about a quarter-turn without backing up a few turns.
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