1
\$\begingroup\$

I've been looking for the answer, but everything I've found is assuming the LED is of a lesser voltage, or a fixed voltage. More than anything, I'm trying to determine if the concept below is feasible before I begin blowing the poor little guys up.

I'm building a control panel for a railroad layout. Below is a brief example of what I'm putting together. My apologies for the poor schematic. For reference, the screw terminals represent the power input and relay output (sans the center, unused terminal) on a Vellman Mk111. Please note that the label "To ?V Lighing" is 12VDC (haven't changed that yet).

enter image description here

Starting from the left, the first and second are for two separate 12VDC lighting connections. The extra to their right is representative of four Vellman MK111 Timers (to save space, only one is shown). The SPDT switch shown there represents a NC momentary switch that will be added to each of these timers. There will be one LED there to show power to those circuits. The power supply is 12VDC, 1A. Each Timer is rated at 100ma.

The separate diagram on the far right begins at the output of a MRC Railpower 1300, which has a variable output between 0-15VDC, up to 7A. The datasheet I found does not specify if the 7A is supplied to the DC output, the A/C output (which I will not be using) or a combination of both. Having tested each locomotive running with a voltmeter (to log the voltage needed to run at scale speeds), they will be running between 9 to 11VDC maximum. The LED indicator has been added to indicate when the Timer is active to a small section of electrically isolated track, allowing the train to move forward onto the rest of the layout, which is powered constantly, only to stop once it reaches this track until the timer activates several minutes later. The SPDT switch allows continuous operation in the event I need it.

The LEDs that will be used are supplied by All Electronics, part number LED-12B, 12V Blue Diffused LED, 5MM (T1 3/4). Data from their website indicates "built-in resistors for 12Vdc, 12mA operation. No external resistor required. Works well on 4-12Vdc. Dimmer at lower voltages." Not worried about the dimmer operation.

Again, I'm looking to verify the feasibility of using a 12V LED in such an application, and as to whether or not it is being applied properly. My apologies for the long explanation, but I felt some may find it useful to help determine an answer and try to give as much info as possible to eliminate confusion.

Knowing my luck, I just created a lot more confusion. I can be good at that!

\$\endgroup\$
3
  • 2
    \$\begingroup\$ That LED should work fine. It differs from many other LED's in that it has a resistor integrated into the LED to limit current. make sure you observe correct polarity. Reverse polarity at 12V could permanently damage the LED. I did not read your whole post because my attention span started to fail me, but someone else may come along and write a proper answer. \$\endgroup\$
    – user57037
    Commented Mar 26, 2017 at 3:15
  • \$\begingroup\$ Thanks! I think I was born with the knack to bore people (even myself). Also forgot to add to the post that I do remember the importance of the anode and cathode. See, I bored myself so bad I forgot to mention that! \$\endgroup\$
    – Tubabuhda
    Commented Mar 26, 2017 at 3:20
  • 1
    \$\begingroup\$ Haha! Don't take it that way. Better to provide more info than less, and if you were enough of an expert to know the exact amount of info needed, you probably wouldn't need to ask the question in the first place. \$\endgroup\$
    – user57037
    Commented Mar 26, 2017 at 3:25

2 Answers 2

1
\$\begingroup\$

The LEDS are fine. But I do not think you have the ones on the timer output circuits hooked up quite right.

If I understand your schematic, BIG IF, the two terminals at the bottom act as a switch on the 12V.

enter image description here

If so the LED is connected across the switch. As such the LED is actually applying a little current to the track and will glow when a loco is on the track, but the loco will not move, or may just crawl with a dimmer LED.

When the timer closes the LED will be bypassed and will go out.

Assuming you want the LED to light when the timer turns the power on, you need to arrange it this way.

schematic

simulate this circuit – Schematic created using CircuitLab

BTW: I am not sure how you are speed controlling, and I am assuming you never reverse the voltage to the track.

If the track power is reversed you need to add a bridge rectifier for each LED as shown below.

schematic

simulate this circuit

\$\endgroup\$
10
  • \$\begingroup\$ Thanks for the catch Trevor. I missed the polarity issue entirely. Speed controlling is done through one of these, as is reversing: modeltrainstuff.com/ProductDetails.asp?ProductCode=MRC-AA300 \$\endgroup\$
    – Tubabuhda
    Commented Mar 26, 2017 at 14:18
  • \$\begingroup\$ Let me redo that answer then... \$\endgroup\$
    – Trevor_G
    Commented Mar 26, 2017 at 14:25
  • \$\begingroup\$ Funny thing is, the way you show the LED in series is how I had it to begin with! That LED will need to be removed since reversing voltage is something that is on the itinerary, unless a suitable bi-directional LED is available. Or is there another alternative without having to resort to another form of indicator lamp. \$\endgroup\$
    – Tubabuhda
    Commented Mar 26, 2017 at 14:27
  • \$\begingroup\$ I do realize that adding a second LED with reversed polarity would just blow the one with the cathode connected to the positive terminal to useless trash. That idea is right out. \$\endgroup\$
    – Tubabuhda
    Commented Mar 26, 2017 at 14:29
  • 1
    \$\begingroup\$ Most likely fly-wiring, with the aide of shrink wrap and/or liquid wraps to prevent shorts. I threw the idea of using resistors out the window a while ago after testing them to determine start/stall voltages. Those values are so low (generally in the range of 2.5-3V), the availability of cheap, local parts is non-existent, and my horrible math skills.... Let's just say I believe in the KISS principle. Dropping the voltage, even by 1V is currently shelved till well AFTER the layout is built. Been 20yrs since I last tried building circuits. Much to relearn first. \$\endgroup\$
    – Tubabuhda
    Commented Mar 26, 2017 at 15:12
0
\$\begingroup\$

If you have reversing voltage, you can put two leds in opposite direction, each protected against reverse voltage by a schottky diode. You can use led of different colour to see which direction the current flows to. Only one led will shine at a time.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.