Model Railroad LED Locomotive direction indicator.

Sierra; CL2 are current limiters for LEDs. It is a two terminal current regulator, 20ma output with 5 to 90 volt input. Google CL2N3. 44 cents each from Mouser 689-CL2N3-G With these you can use any DC power supply regulated are not. You also can use current limiter resistors with a regulated supply, just tie the negitive side to the negitive side of the track power so the two rails will be the returns and witch one is negitive will give you direction. Run the positve of the supply out to the lights and hook each one to a rail using either a current limiting resistor or the CL2. You can still use your AC output for the 2nd supply. Its old school and simple.

True it might seem like overkill, but I anticipate that the layout will grow and I could use the extra power for other needs like uncouplers and machinery that might need a DC power source

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You should not use your varible DC output for accessories like that but a few LEDs would be OK. Most of your accessories will be AC anyway. Andy

Andy,

This sounds very intriguing. But hooking it up to the grounds is a little confusing. Are their any circuits on the web that might show this type circuit? I looked at the specs that Mouser has and can somewhat get to it. it seems that I would be using one rail, or electrically speaking, one side of the rail power for one side of the power supply for the LED's. Besides a whole lot of other stuff going on in terms of blocks and switches with the rails, I am thinking that there might be an effect with the rail being used as a conductor for yet another power supply, other than the one that power the locomotives.

Attached is a rough of the type power supply you need for the op amp circuit. You could use a 1 Amp version (the transformer). I did not look at what the chatter is about with the LEDs but you need a dual +/- power supply for the op amp. I believe what you posted was a single output power supply.

Ron

Sierra9093; You will not find any circuits on the web that might show this type circuit. Most rail signals are for block detection not indicating direction. This setup should work just fine. You can test easy. Andy

Ron and Andy,

Sorry about the delay, but my Comcast went on the fritz, and I am using an old standby dial up connection. Anyways, so I have to redesign the power supply portion, which is OK, but why won't the power supply that I designed work? It has a + and a -. I know that there must be a good reason. What would happen if I were to use it, as I designed it?

Thanks again fellas,
Ray

It's not worth designing a PS: See JE215: JAMECO VALUEPRO: Education & Hobby

Oh, I already have a PS (actually 2) that I built and it has an AC tap that I want to take advantage of. Oh, and I haven't gotten around to block detection yet, but I plan to. Thanks for the suggestion.

Sierra9093 said:

Ron and Andy,

Sorry about the delay, but my Comcast went on the fritz, and I am using an old standby dial up connection. Anyways, so I have to redesign the power supply portion, which is OK, but why won't the power supply that I designed work? It has a + and a -. I know that there must be a good reason. What would happen if I were to use it, as I designed it?

Thanks again fellas,
Ray

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No problem as I think the forums wend down last night.

You need a dual supply for the op amp and so both LEDs will work. When we say dual supply we mean two voltages referenced to common as in -12 0 +12 so it isn't just a matter of negative and positive single supply. That is keeping it simple. Also at $24 the suggested kit was a nice deal and adjustable.

Ron

They mentioned on CNN that Comcast also went down last night.

I thought that the UM741 solved that problem. Problem is that I do not have a common tap comming off the transformer, just 20 volts. Then the directional indicator has a driver power input for only two, a + and a -. Do I just run the two grounds off the 7812 regulator grounds? Also, if I get the $24 PS, won't I still have the problem of having just a - and +? Thanks

OP amps don't have a power ground UNLESS you make the (-) supply pin ground, but you need a dual supply.

The ground of the regulators will be the ground of the circuit, . The (+) and (-) supply from Jameco will have a ground point.

There are other ways to get a virtual ground such as using a rail splitter IC: **broken link removed** These are only good for 20 mA though. These IC's will take a DC voltage and create a ground exactly 1/2 the supply voltage. This might be cutting it close though. You could use multiple rail splitters if you wanted for each circuit.

I'll try to work up another design witht 2 op amps then and see what you guys think. Thanks again!

Sierra9093; If you are going to do all that extra work you mite as well do block detection. As I pointed out before you dont need all that circuitry just wire up the LEDs with a regular power supply. Why do all that work if you are going to do block detection someday anyway. Keep it simple. Andy

OK Ron, Andy and the rest of the gang, this is my latest design. The 20 VAC from the transformer does not have a center tap, so this design does not have one, so will it work this way? I used the grounds from the regulators the way Ron set them up without the center tap from the transformer.

Sierra: The Bipoler power supply is just to power op amps not so much the LEDs. Andy

Hi Ya

If you go back to my suggested power supply I laid out a dual +/- 12 Volt supply and I used the transformer center tap. I also used a 24 Volt center tapped (12-0-12) transformer. That is important. OK here is a little overview. The regulators need at least 2 volts above their regulating voltage to work. The transformer outputs are 12 volts AC RMS. Each half of the transformer is full wave rectified off the diode bridge. Those first capacitors will charge to a voltage of E PK or the RMS value times 1.414 = 16.9 volts. That allows the regulators to work. Each of the 12 volt outputs (+12V and -12V) based on the transformer I suggested will be capable of about 1 Amp more than enough for many op amps and LEDs.

As to the Op Amps based on input polarity the outputs will swing close to the rail voltages. Close to +12 or -12 based on the input which is the track polarity. Attached is an image showing a rough idea of whet will happen.

Hope that helps.

Ron