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2 LED's circuit (not so simple as it sounds)

Discussion in 'LEDs and Optoelectronics' started by Indulgence, Jul 10, 2012.

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  1. Indulgence

    Indulgence

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    Jul 10, 2012
    Hi everyone,

    Currently I am working on this tricky project where the task is to make it work WITHOUT a switch. Anything can be used except a simple switch. So, here is the problem.

    There is one +14V node and another node with three possible cases 0V, +14V or open circuit. When these nodes are connected the green LED has to indicate if +14V is in the second node. The red LED has to indicate if there is 0V in that node. If it is open circuit, both LED's has to be off. Again, no switches are allowed. Any other components or extra energy sources can be used.

    I have attached a simple picture to visualize the problem.

    Please ask if anything is unclear. Any help will be appreciated.

    Looking forward to your help!!!
     

    Attached Files:

  2. GreenGiant

    GreenGiant

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    6
    Feb 9, 2012
    well to make things easier you are going to have to reduce to voltage down from 14V.

    for the green LED all you need to do is get the voltage to 5 or 12 (common LED ratings) and have a current limiting resistor in line with it to ground, this way the LED will light whenever the 14V is present

    For the red LED you can use a PNP transistor as a switch with power to the emitter (probably 5 volts) and have the 14/0V connected to the base, when there is 14V there the LED will be off, and when its not then there wont be any

    The problem is you will need a separate power supply for the 5V (since there will be 0V applied) therefor you will not be able to detect an open circuit as it will read 0V almost exactly the same as for the red LED

    The only other option I see is to use a microcontroller but that is a whole other animal
     
  3. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    I don't see any green LED in your diagram and what is "a clip" ?
     
  4. Indulgence

    Indulgence

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    Jul 10, 2012
    Thanks GreenGiant!

    Would you mind sketching me a circuit for that?
     
  5. Indulgence

    Indulgence

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    Jul 10, 2012
    CDRIVE, the green LED is not there yet since I do not know how and where to put it. You can ignore "a clip" it's just a piece of wire.
     
  6. BobK

    BobK

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    Jan 5, 2010
    Green Giant's solution requires a ground. If you have a ground, it is much easier that that. You simply connect the red LED between the node 1 and node 2 and the green one between node 2 and ground.

    But I am guessing that you have no ground available, only the two nodes, is that correct?

    Bob
     
  7. BobK

    BobK

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    Jan 5, 2010
    Here is the solution, assuming a red LED with 2.0V and 20ma and a green LED at 2.5V and 20ma. +!4 is the node that is at 14V and In is the node that is at one of 3 states.
     

    Attached Files:

  8. Indulgence

    Indulgence

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    Jul 10, 2012
    I have ground available. However, this solution is not suitable because when the second node is open circuit both green and red LED's will be in series and on at all times. In this case both LED's have to be off.
     
  9. CocaCola

    CocaCola

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    Apr 7, 2012
    5 and 12 volts are not common LED ratings, I have never come across LEDs in that range unless they where already paired with a resistor for that voltage... LEDs are current driven, and usually rated in the 2-4 volt @ about 10-20mA range... Running them off 5 volts or 14 volts only requires a different resistor value to limit the current to an acceptable level...
     
  10. BobK

    BobK

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    Jan 5, 2010
    If you have a ground available, there is no problem at all! The solution is totally obvious:
     

    Attached Files:

  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    A simple answer is to generate a 1/2 VCC rail using a voltage divider from +14V to 0V and connect the LEDs in reverse parallel, one side to the voltage divider and the other side to the probe.

    . . . . . . . . . 1K . . . . . . . 1K
    +14V --------/\/\/\/\-----------/\/\/\/\/----------0V
    . . . . . . . . . . . . . . . |
    . . . . . . . . . . . . . . . |------|>|-----
    . . . . . . . . . . . . . . . | . . . R . . |
    . . . . . . . . . . . . . . . | . . . G . . |
    . . . . . . . . . . . . . . . |------|<|----------- probe

    You can also use a bi-colour LED.
     
  12. BobK

    BobK

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    Jan 5, 2010
    That is the same as my circuit except the resistors are on the other side of the LEDs.

    Bob
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    BobK: No, the important difference is in the behaviour when the probe is not connected. The OP specified that in this case, neither of the LEDs should light. With your circuit, both of them will light when the probe is not connected to anything.
     
  14. BobK

    BobK

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    Jan 5, 2010
    No, when the probe (which is called In in my circuit) is high impedance, neither LED lights since both LEDs are connected ot the probe on one side.

    Edited: Never mind, you are correct!.

    Bob
     
    Last edited: Jul 10, 2012
  15. Indulgence

    Indulgence

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    Jul 10, 2012
    In this case if In is open circuit both diodes will be ON. If In is o/c both LED's must be off. Therefore, this is not a solution.
     
  16. BobK

    BobK

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    Jan 5, 2010
    Yep, I see that now. KrisBlueNZ solution does look right though.

    Bob
     
  17. Indulgence

    Indulgence

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    Jul 10, 2012
    Yes, I will try this solution tomorrow. Instead 1k resistors I will be using 10k and 5.5k to get +9V because I can't go lower otherwise the other part of my project won't work.

    Thank you people!
     
  18. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    You better do your testing in the dark because with those value resistors you may not see the LEDs light in the day time.
     
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