driving two different color leds with 2 wires only

Discussion in 'General Electronics Chat' started by bug13, Dec 6, 2017.

  1. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    Hi guys

    I need to drive two different color leds with 2 wires only over ~40M away. I am planing to do this as below, is there a better way to do it? Or any feedback/suggestion/critics etc...

    Thanks guys!

    Capture2.PNG
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    What is the voltage drop budget for the 80M of cable?
    What is the Vf of the LEDs?
    How fast do you need to turn the Transistors ON and OFF?
     
  3. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    I see, forgot to say they are just low power white leds.

    ILed is ~10mA max, that’s a lot, only need 5mA.

    Vf is ~3V

    I haven’t measure the cable the cable resistant, but I expect no more than 10ohm total resistance, 10mA max current. So I expect 0.1V drop.

    Driving them 100Hz max, that’s plenty to fool our human eyes :)
     
  4. Tonyr1084

    Well-Known Member

    Sep 24, 2015
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    Not certain how to answer your question. But here's what I'm seeing: Two LED's and two controlling inputs. Assume inputs are labeled A & B. If A is high and B is low then one LED (lets call it LED1) will light up. If B is high and A is low then LED2 will light up. If both A & B are high then neither LED will light up. If both A & B are LOW then neither LED will light up.

    Is that what you want?

    And clarify please: You said you have two white LED's of differing colors ? ? ?
     
  5. ebeowulf17

    Well-Known Member

    Aug 12, 2014
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    What were you planning for resistor values? I haven't simmed or calculated anything for real, but off the top of my head it seems like you're going to be dumping a LOT of current through whichever top resistor (R1 or R2) is above the active transistor.

    Maybe there's a combination of resistance values that balances better than I'm imagining, or maybe the wasted current doesn't bother you, but it certainly looks strange to me.

    I would've expected one of two likely solutions:
    • H-bridge arrangement
    • One pin controlling on/off function (bjt, MOSFET, relay, whatever) and the other pin controlling polarity reversal through a DPDT relay.
     
  6. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    Yes, that's what I want.
    And sorry for the confusion, they are both white LEDs, but with different color cover/dome.
     
  7. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    Current consumption is not a main concern, but I do agree it's a bit high. I do want to keep the cost to minimum, so I don't really want to use a h-bridge, relay etc. I deally, I would want to control it with 3.3V or 5V logic.
     
  8. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Since it can only be reversed biased D2 will never turn on.
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    Reduce the values of R3 and R4, and in series two 1N4148 with two led and be careful their polarity.
     
  10. ebeowulf17

    Well-Known Member

    Aug 12, 2014
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    Q1 active, Q2 off lights D2.
     
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  11. Tonyr1084

    Well-Known Member

    Sep 24, 2015
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    Kind of low on the voltage for that long a run. May work. Might have to get a "Good" wire. Something like a good ethernet like wire, one that is rated for long runs of data transmission. You may even need a power source at the lights just so you can switch them on and off. 3.3 volts is already close to the forward voltage of a white LED; depending on the exact LED you have. And I don't know if you're going to be able to turn on a regular transistor with just 5 volts. Might have to go with a logic level MOSFET. Of course, then the solution becomes even more simple. I'd bang out a drawing but it's close to my bed time and I have a few things to do before I hit the hay.
     
  12. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    Why do you suggest adding 1n4148 in series with R3 and R4?
     
  13. bug13

    Thread Starter Senior Member

    Feb 13, 2012
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    Sorry I mean to say the output voltage of my MCU to the transistor inputs are 3.3V or 5V. The 40M wires are referring to the two wire between the two LEDs and my transistors.
     
  14. dendad

    Active Member

    Feb 20, 2016
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    You could get rid of R3 and R4 then select R1 and R2 to set the currents.
    Try R1 = R2 = 1Kohm.
    The cable should not amount to vert much and it is not critical anyway.
     
    bug13 likes this.
  15. ScottWang

    Moderator

    Aug 23, 2012
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    Because when anyone of the led is turn on and the other one is turn off, but the polarity was reversed, that will causes the led damaged anytime.
     
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  16. ebeowulf17

    Well-Known Member

    Aug 12, 2014
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    I thought that each LED allowed enough current to flow that it limited the voltage across it to its forward voltage rating (Vf.) The Vf of each LED should be much lower than the reverse voltage it would take to a damage an LED, so effectively each LED is protecting the other LED from dangerous reverse-voltage levels.
     
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  17. ScottWang

    Moderator

    Aug 23, 2012
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    If we considering two situations that the first is pin to pin and one of led reverse the polarity and the second is like the first posted, now we discuss the first posted, what do you think that how many volts will be cross on the led which is turn off and comparing to the led which turn on ?
     
  18. dendad

    Active Member

    Feb 20, 2016
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    To protect the LEDs, you connect the LEDS back to back without R3 and R4. Then each LED will protect the other.
     
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  19. ebeowulf17

    Well-Known Member

    Aug 12, 2014
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    Sorry, I get it now. @dendad set me straight with that last post. I apologize.
     
  20. DickCappels

    Moderator

    Aug 21, 2008
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    Some LEDs cannot tolerate any reverse bias without suffering long term degredation. Check the datasheets for the LEDs you intend to use before you settle on a design.
     
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