Send a quick 12v pulse to a relay that I want to keep on so the light will stay on.

Discussion in 'Automotive Electronics' started by A20, May 11, 2016.

  1. A20

    Thread Starter New Member

    May 11, 2016
    4
    0
    hi everyone. Let me start off by saying what I'm trying to do.

    I am trying to turn a light on. When I tap my high beams on my car. That will send a quick 12v pulse to a relay that I want to keep on so the light will stay on. The only way to turn it off should be the turn off the car. Now I had someone draw me a circuit it seems like it will work but I have two questions. Can I have a common ground? Like in the picture in red. And he put in a second diode. I really don't understand the purpose of it being there. I get the first diode so 12v doesn't flow the wrong way. It should work fine without the second diode and have nothing there at all?


    Hopefully I have explained what I am trying to do.
    Thanks
     
  2. SLK001

    Well-Known Member

    Nov 29, 2011
    811
    224
    Can you have a common ground? YES

    Don't remove the diode. It shunts the back EMF from the coil collapse.
     
  3. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    1. There is only one power supply, so the ground of two sides are the common ground.

    2. The second diode is protecting the lamp, it should be there, the purpose is to discharge for the relay when it is off (the coil no current flows through from positive to the negative), when the coil lost the power then it will reverse the voltage and the voltage could be several times of 12V.

    3. Why you want that function?
     
  4. A20

    Thread Starter New Member

    May 11, 2016
    4
    0
    Thanks for the answers. I want the function because I rather modify the headlight wires than touch the whole wiring harness in the car. It's only to turn on an amber bulb so people can be careful.

    I'm no expert by any means I do appreciate the info about the second diode. Although I'm still lost about it haha.
     
  5. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    Next time, If you have some other questions, when you create a new thread, please make sure the title should be match the question, thank you.
     
  6. A20

    Thread Starter New Member

    May 11, 2016
    4
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  7. cork_ie

    Member

    Oct 8, 2011
    348
    58
    Your circuit should work , 85 is relay coil ground, 86 activates the relay. 30 is ign switched power and 87 is output to your light.

    Once you initially trigger your relay with the light switch the contacts 30 and 87 will join and terminal 87 will be +12V.
    This will feed terminal 86 and your relay will stay closed until 12V+ is removed from terminal 30.
    To avoid overheating your relay I would suggest putting a resistor of between terminal 87 and 86 so that the voltage at terminal 86 sits at about 7-8V. This will be sufficient to hold the relay closed. The power required to pull in a relay is considerably more than what s required to hold it closed.
     
  8. DC_Kid

    Distinguished Member

    Feb 25, 2008
    638
    9
    i would use a SCR, and if the load is hefty then a nFet to handle the load.

    but automotive usually does sink (low side) switching. if the hi beam is sink switching then you can use a pFet or pnp bjt in front of a std SCR, etc.
     
  9. A20

    Thread Starter New Member

    May 11, 2016
    4
    0
    Would you be kind enough to draw me a circuit with the scr? That would replace the relay right?

    I built the relay one it's working fine. But if there is a smarter/smaller way to do the same thing I would love to learn about it.
     
  10. DC_Kid

    Distinguished Member

    Feb 25, 2008
    638
    9
    sure, 3 small components needed.
    PS1 is car batt
    S2 is whatever is the pulse from turning on high beam
    S1 is whatever relay the ignition provides for the voltage, which opens when ignition goes off
    R1 is your light bulb load
    R2 and R3 , something like 1k each (i like to use pull down R3 but technically not needed
    D1 is SCR sized for the load, a heatsink may be needed depending on item chosen

    so basically its a solid state latching relay that can be done with just two discrete components. no relay coil flyback and nothing mechanical to worry about. i have assumed here from your previous posts that you somehow provide a positive pulse to turn the thing on. if R1 load is big (more than a few amps, then you'll need a FET or BJT to handle the load.


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