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How could I build a 10 minute time delay relay?

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  • How could I build a 10 minute time delay relay?

    Hi.
    I am planning a carcomputer install using a program I found on this forum which hibernates the computer X minutes after pins 4 - 8 on the serial port are shorted. I want the program to hibernate my computer after 5 minutes. I think this is a great idea, however, I am still paranoid as to my battery draining incase my computer hangs for some reason. As many of you may know, software is not exactly the most reliable. So I was thinking maybe somehow build a cheap 10 minute time delay relay as a failsafe mechanism. When ACC power is lost from the engine shutting off, this would trigger a relay to short pins 4 - 8 causing the program to do a hibernate in 5 minutes. However, when ACC power is lost, this could also cause the 10 minute time delay relay to countdown. Incase the computer does not hibernate in 5 minutes, then at 10 minutes, the time delay relay could activate a main relay switch which would cut off power completely to my inverter, thereby doing a cold shutdown on the computer and not draining my battery. Ive done some searching but have not found anything useful. I was wondering if someone could suggest a way to accomplish this? Thanks.

  • #2
    Originally posted by nobb
    Hi.
    I am planning a carcomputer install using a program I found on this forum which hibernates the computer X minutes after pins 4 - 8 on the serial port are shorted. I want the program to hibernate my computer after 5 minutes. I think this is a great idea, however, I am still paranoid as to my battery draining incase my computer hangs for some reason. As many of you may know, software is not exactly the most reliable. So I was thinking maybe somehow build a cheap 10 minute time delay relay as a failsafe mechanism. When ACC power is lost from the engine shutting off, this would trigger a relay to short pins 4 - 8 causing the program to do a hibernate in 5 minutes. However, when ACC power is lost, this could also cause the 10 minute time delay relay to countdown. Incase the computer does not hibernate in 5 minutes, then at 10 minutes, the time delay relay could activate a main relay switch which would cut off power completely to my inverter, thereby doing a cold shutdown on the computer and not draining my battery. Ive done some searching but have not found anything useful. I was wondering if someone could suggest a way to accomplish this? Thanks.

    Sounds similar to a uSDC20D.
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    • #3
      Or, maybe just build it around a 555 IC. If you don't want a particularly complex circuit, just make two: one for the soft shutdown, and a second for the hard shutdown. Probably cost you maybe $5 in parts from the rat shack.
      Gen 1: Pentium 3 1GHz - ATX - 2005
      Gen 2: Pentium M 1.6GHz - ITX - 2006
      Gen 3: Pentium M 2.0GHz - 5.25" SBC - 2007
      Gen 4: (coming soon: Core2 Duo - 3.5" SBC - 2009)
      ...it never ends

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      • #4
        That shutdown controller cost too much. I think I'll try reading up on the 555. At this point, the schematics on the 555 seem really difficult to understand.

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        • #5
          Just approach it by section and function. Like this section of the circuit is the trigger, the other is the delay, etc. Just take a sample circuit and build on top of it. If I have some free time tonight after work, I can help.
          Gen 1: Pentium 3 1GHz - ATX - 2005
          Gen 2: Pentium M 1.6GHz - ITX - 2006
          Gen 3: Pentium M 2.0GHz - 5.25" SBC - 2007
          Gen 4: (coming soon: Core2 Duo - 3.5" SBC - 2009)
          ...it never ends

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          • #6
            Actually, why make it hard? RC time circuit... just a resistor and a capacitor. The voltage is constant, and so is the time needed, so it's just simple. Lessee... if using a 4700 uF cap, it's about 128k ohm resistor for 600 seconds.
            Gen 1: Pentium 3 1GHz - ATX - 2005
            Gen 2: Pentium M 1.6GHz - ITX - 2006
            Gen 3: Pentium M 2.0GHz - 5.25" SBC - 2007
            Gen 4: (coming soon: Core2 Duo - 3.5" SBC - 2009)
            ...it never ends

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            • #7
              Here you go.
              http://www.mp3car.com/vbulletin/showthread.php?t=54143
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              • #8
                Thanks for your tips. I actually considered just using a simple capacitor to hold the switch. But then, I remember I once made a bank of capacitors with some old capacitors lying around and this bank could only hold a led lit for a few minutes before it started to significantly dim. I figure that I would probably need quite a large bank of capacitors to keep the relay switch on for 10 minutes because I'm pretty sure the resistance of an average relay is not that high. Also, when you mentioned the RC time circuit what exactly is the resistor for? To slowly charge the capacitors right?

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                • #9
                  RC just simply stands for Resistor-Capacitor. A quick google search will give you all the nitty gritty of how it works. For your implementation, I only envision the RC as a replacement for the timer, not the whole circuit. It counts the 10 minutes which then triggers the rest of the circuit to shut down or pulse the power button, whatever.
                  Gen 1: Pentium 3 1GHz - ATX - 2005
                  Gen 2: Pentium M 1.6GHz - ITX - 2006
                  Gen 3: Pentium M 2.0GHz - 5.25" SBC - 2007
                  Gen 4: (coming soon: Core2 Duo - 3.5" SBC - 2009)
                  ...it never ends

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                  • #10
                    Originally posted by psyrex View Post
                    Actually, why make it hard? RC time circuit... just a resistor and a capacitor. The voltage is constant, and so is the time needed, so it's just simple. Lessee... if using a 4700 uF cap, it's about 128k ohm resistor for 600 seconds.
                    i'm bringing this thread back from the dead.
                    what combination of resistor cap would i need to make a 1-2 mintue timer?
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