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    SOLVED PIR sensor to cut backlight power

    Troubleshooting
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    • MrH
      MrH last edited by

      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
        L_i_v_e last edited by L_i_v_e

        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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        • MrH
          MrH last edited by MrH

          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
            TomasZmuda last edited by

            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

            MrH 1 Reply Last reply Reply Quote 2
            • MrH
              MrH @TomasZmuda last edited by

              @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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