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A New Chapter for MagicMirror: The Community Takes the Lead
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PIR sensor to cut backlight power

Scheduled Pinned Locked Moved Solved Troubleshooting
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  • M Offline
    MrH
    last edited by Aug 15, 2016, 5:00 PM

    Hi, need some help from you guys :)

    I’m planing to use a PIR sensor to cut power to the backlight board of my LCD. I dont want to shut down my monitor or the signal to it due to the anoying “no signal” and “Input” screens.

    Dont think I need any code for this as I should be able to use the pir without any action from the Pi, I just draw power from it.

    As my LCD driver board has an add-on board just for the backlight it should be easy. As I can gather I have two options.

    1. Use a relay to cut the 12v connection between lcd driverboard and inverter.
    2. Disregard the relay and connect the PIR directly to inverter as it has a ON/OFF wire connection set as HIGH when ON

    As my skills in schematics and power distrubution is rather limited I could use your inputs on my two options and please have a look on my fancy pictures, the only connection that is in place is the one connecting LCD driver board and inverter. Have I missed anyting? should that wiring work?

    0_1471280355179_Untitled presentation.jpg

    0_1471280387057_Untitled presentation (1).jpg

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    • L Offline
      L_i_v_e
      last edited by L_i_v_e Aug 16, 2016, 11:02 AM Aug 16, 2016, 11:02 AM

      Or, you could simply remove power from the board, so 0w consumation whien you don’t need it. Add the relay to the power cable of the board (my board needs 1 sec to turn all on this way)

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      • M Offline
        MrH
        last edited by MrH Aug 16, 2016, 11:16 AM Aug 16, 2016, 11:15 AM

        Indeed, but then I get that annoying OSD window telling me what input is selected :) And by keeping power to the main board but cut power to the light inverter gives me a drop from 38w to 6-7w. For me, thats a win if it means I get rid of that OSD input window ;)

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        • T Offline
          TomasZmuda
          last edited by Aug 16, 2016, 11:25 AM

          I’m using the second option, because LCD’s boot up time is about 8 sec.

          But you have to hook up ground, as i updated your schema. Otherwise the PIR will have zero ground.

          0_1471346555046_untitled_ldvs.jpg

          M 1 Reply Last reply Aug 16, 2016, 11:34 AM Reply Quote 2
          • M Offline
            MrH @TomasZmuda
            last edited by Aug 16, 2016, 11:34 AM

            @TomasZmuda

            Great! Thank you so much!

            option 1 only cuts 12v to the inverter, not the main board so should have the same effect as option 2 but if I can drop the relay it’s one less thing to power so with your help I will go for option 2 for sure :)

            thanks again!

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