Stupid question I ought to be able to solve myself

Discussion in 'General Electronics Chat' started by marshallf3, Apr 24, 2011.

  1. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    But I need a circuit that when given 12V it will drive a 200 mA load for about two seconds then turn back off until the 12V is removed for at least 10 seconds.

    Obviously a 555 timer based idea, I just can't figure out how to come up with the proper circuit/values as I'm still fairly foggy what with the medications they've got me on after I had that 36 stitch operation.
     
  2. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    If you google for "555 monostable calculator" you can get an online calc where you enter "2 seconds" and it shows you the parts values.

    Then you juts need to hook up a power transistor or relay to the output of the 555.
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    I see two monostables, 1st one triggers off the 12V, the second one detects power down for 10 seconds. A 3rd part, a latch and hold, keeps the 1st 555 inactive until the second 555 is fully triggered, which resets the latch and hold. A SCR maybe.

    Does the 12V go completely to ground (or open) when it isn't there?
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Is the 12V your sole source of power? You might then need a different approach to the second timer, unless you can store enough charge (in a very large capacitor?) to power the second 555.
     
  5. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    12V is either present or open circuit. A pulldown resistor/bleeded on the B+ input to the 555 would be possible, if needed.

    Just started my day with a 20 mg Oxycontin and 2 Loritabs so I hope you understand why I'm not thinking clearly during the day.
     
  6. THE_RB

    AAC Fanatic!

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    Sorry I missed that point "the 12v must remain off for 10 seconds" and it's a little difficult as your circuit will have no power for those 10 seconds.

    You might be able to add a large cap and a FET, the cap would hold the FET on for 10 seconds after power is removed and inhibit the next cycle.
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    I'm making the assumption the circuit will have power for those 10 seconds, from some other source. Not a valid assumption?

    Would you like me to draw up what I described?
     
  8. eblc1388

    Senior Member

    Nov 28, 2008
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    I think the problem is well described as follows, but only marshallf3 will know for sure.

    [​IMG]
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
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    A timing diagram is a jolly good idea, eblc1388.

    However I am not sure of this yet.

    The 12 volts is either a signal (pulse) or it is a power supply in which case it is always available and therefore needs to be definitely disconnected.

    So what is needed is a controlled 2 way or double throw switch. The number of poles depends upon whether the 12 volts is signal or supply or both.

    The 2 ways of the switch are wired to be disconnected one way and connecting the load to the supply the other and also controlling the 12 volts if necessary.

    Such a switch can easily be had either as a relay or a solid state (matrix) switch.

    The control circuitry can follow the suggestions already made about monostables.

    go well
     
    Last edited: Apr 30, 2011
  10. eblc1388

    Senior Member

    Nov 28, 2008
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    What happens then?

    Is the problem solved by the OP? If so, why didn't he say so and share the solution to other members?
     
  11. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    Thanks, I think I've got plenty of info now I hadn't thought of.
     
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