Simple electrical question

Discussion in 'Getting Started' started by alex_mrrkb, Feb 2, 2003.

  1. alex_mrrkb

    alex_mrrkb Member

    Nov 13, 2002
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    Hi !!

    Probably an easy one ... but I'm not into electronics, so I must ask before I do something stupid ...

    I'm preparing to do some control panel wiring. I have some bi-colored leds that are gonna show to switches positions (feedback comes from a micro-switch mechanically connected to the turnout).

    My power supply is a computer power supply. I use the +5V connections from this beast. Right now, when connecting 2 of these leds, I use a 150 ohms resistor and the brightness is perfect.

    Question: There will be about 50 of theses leds later on the layout ... could that be too much for this 20amps power supply ? What I fear is that with each new led I connect, will all the others get a little darker ... until it barely shows ?

    Also: My feedback system uses +5V *AND* -5V from that power supply ... +5V mean switch is thrown, and -5V means switch is not thrown. The -5V output is only 0.5amps ... will there be a problem there, like the one I just stated ?

    Thanks !!!

    P.S.: Explanation of why I use +5V and -5V: With each turnout, a microswitch is mechanically connected to the switch machine... so when the turnout is operated, the microswitch "knows" the position of the turnout. There are 3 pins on these microswitch ( DPST) .. I send +5V on the outer pin, -5V on the opposite pin, and the switch sends me back +5V or -5V via it's middle pin depending on the turnout position. All I have to do is connect this "unknown" voltage to one side of a led, and the other leg to the power supply GND ... and voila ! The led light's up red or green accoring to the turnout position ! I fiddled a lot with this idea on my prototype borad (breadboard) and it gave great results ... I can't wait to see it in action on the control panel !!
  2. RailRon

    RailRon Active Member

    Nov 23, 2002
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    Hi Alex,

    when I apply Ohm's law to your functioning LED setup (5 Volt, resistance 150 Ohm), the current of a LED is 33 mA (Milliampere).

    Now if the potential of your +5V source is really 20 Amps, you could connect about 600 LEDs :eek: in parallel, all getting enough power for a bright signal light. :cool:
    That's the good news.

    On the other side, when your -5V source only drizzles out 0.5 A (500 mA), then you can only drive 15 LEDs from it. So the bad news is that if you want to realise your proposed system, you could install only about 15 turnout indicators. :( :mad: :(

    Could you install double pole, double throw microswitches (DPDT, two input and four output wires) to your turnouts? With them you could simply reverse polarity for the bicolor-LED when throwing the switch.

    The alternative is to use your existing SPDT switch, but then install TWO LEDs per turnout on your panel, a red and a green one. Input +5V (plus the 150 Ohm resistor) on the middle pin, then the output wires from both outer pins to the two LEDs, and from them one common return to GND. (That's how I did it on my former layout.)

    In both cases you'll have more than enough power to light all your LEDs (and then some more...) only by using the +5V and GND form your power supply! :) :) :)

  3. cidchase

    cidchase Active Member

    May 15, 2002
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    Hi alex,
    Your LED's probably draw in the vicinity of 20mA (or 0.02 Amps) each.
    If 25 of your turnouts (50%) are switched to the "Red" side, then the load on the -5V section would be 500mA (or 0.5 Amps).
    If you're operational scenario never calls for more switches than this to be one way or the other, then you should be OK as is. Check the voltage drop across the LED. If it's less than 2V then you'll have to run fewer than 25 LED's at a a time on the -5V side or increase the resistor value a little bit to prevent pushing the capacity of the power supply.

    If you can get your hands on another 5V 2A supply, you can use it instead of the -5V section of the one you have. Hook the new +5V side to the common of the old one, then the common of the new one will be your -5VDC @ 2 Amps. Just make sure that if there's an true earth ground anywhere, it's only at one point or terminal. Your power supply "common" may or may not be grounded, is it actually labelled "GND"??
    (not trying to confuse; you're prolly OK just like you have it now).

    BTW, I agree with Ron that the DPDT switch will solve your problem, just takes another wire to the turnout.
    I also agree that 2 LED's is probably a more standard method of annunciation.
  4. kf4jqd

    kf4jqd Active Member

    Jan 14, 2001
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    I also use LED's with a laptop computer's powersupply. I use 1K Ohm reistor in series. As for multi color LED's, does it have 2 or 3 wires coming out of it? If it's only 2, connect the resistor to any of the 2 wires. If it's the 3 wire, connect it to the center wire. I hope this helps.

  5. Paul Davis

    Paul Davis Member

    Dec 30, 2002
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    Here's an idea for you. The other high current output on your power supply is +12V So if you use the +5V output as your GND, connect +12 to one side of the microswitch and the real gnd to the other. to the led it'll look like -5V and +7V.

    seeing as you're just using the 2 pin led then one side is going to get more voltage and will aslo be slightly brighter. You probably want the green to have +7V as it tends to be dimmer than the red.

    This may actaully work better as those computer power supplys are designed to have something on all the outputs. They don't always like having High current on the +5V with nothing on the -12V

    Another option would be to use led's with 3 leads. that way you could connect +5V to the common terminal of the microswitch and have the other two terminals connected to the two of the leads of the LED while the 3rd goes through a resistor to ground.

    You can get 3 lead, rectangular led's 10 for a dollar or 100 for $8 at