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How to take "bounce" out of switch ??

Discussion in 'Electronic Basics' started by Thomas, Apr 26, 2005.

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  1. Thomas

    Thomas Guest

    I have a small magnetic switch that is triggering when I hit a bump or rough
    spot in the road. Moving the switch or changing it is nearing impossible so
    I need to take out the "false" triggers. Need something simple and reliable.
    I am thinking 1 second or less should do it... Any and all help is
    appreciated.
     
  2. John Fields

    John Fields Guest

    ---
    Can you tell us more about the circuit?

    What is the voltage across the switch?

    How much current does it switch?

    How is it used in the circuit? (can you post a schematic?)

    Is it a magnetic switch or a relay of some sort?
     
  3. Thomas

    Thomas Guest

    Its just a basic magnetic switch detecting when I have the lid closed.
    Operates on 5V DC. Due to the nature of the beast, it is not feasable to
    change the existing switch. I need to take the bounce out. I am currently
    using the switch to detect when the lid is open and when it is, a small DC
    relay is used to turn some lights on. The lid and/or switch can not be
    changed. Right now, potholes, speedbumps and the occasional varmit ;) cause
    the magnetic switch to detect the opening. I was thinking of a small
    circuit to put between the switch and realy to provide a time constant
    before the relay is pulled. Maybe something with a pot to vary the time from
    1 sec down. Would a one shot 555 work?
     
  4. The simplest circuit requires adding just one resistor and capacitor,
    as shown here:
    http://www.terrypin.dial.pipex.com/Images/MagSwitchDebounce.gif
     
  5. Thomas

    Thomas Guest

    Thanks for the input. Would the circuit vary if 12V instead of 5V would be
    used? Also, is that a 22 Ohm resistor in series with the pot? And again, I
    really appreciate the help!
     
  6. Thomas

    Thomas Guest

    Also Terry, do you have the formula for me to calculate on the resistor /
    cap compos to get the time variance?
     
  7. Lord Garth

    Lord Garth Guest

    It's there in case the pot is turned to one extreme thus putting the power
    supply directly
    across the capacitor. The time constant is T=RC and if 12 volts is applied,
    be sure you
    use a 16 volt capacitor.
     
  8. If everything was in proportion, then no, the circuit could be the
    same. IOW, if the pull-in voltage of the 12V relay was (12/5)*4 =
    9.6V, and the its DC resistance was (12/5)*400 = 960 ohms, then the
    timing would follow a similar curve.

    Yes, it is a 22 ohm resistor. With the pot at its maximum, the total
    resistance in series with the big cap is therefore 90 ohms.

    Dealing with your other post's query, a formula (which would be
    complex, as we're dealing with exponential voltage over time) would be
    of little practical value to you. As I said, your relay will not have
    the specs I've assumed.

    There are also other factors I didn't raise earlier, which will also
    have a bearing on the practical results, and reinforce my suggestion
    to use trial and error (assuming you want to use this ultra-simple
    circuit). In practice I suspect the lid won't be as co-operative as in
    my simplistic assumptions. IOW, it won't falsely open for 0.1 - 1
    second and then stay closed again for a second or so before opening
    again (giving the 4700uF capacitor plenty of time to discharge through
    the relay's coil). It will probably be opening and closing to some
    extent constantly - almost vibrating - with some jumps higher and
    therefore lasting longer than others. That sort of behaviour presents
    quite a complex signal at the input to the series resistor.
     
  9. John Homppi emailed me pointing out a careless mistake in my drawing:
    I had the cap shorted to ground! Thanks, John, well-spotted. Now
    corrected.
     
  10. Rich Grise

    Rich Grise Guest

    First, please don't use the term "bounce" for what is merely false
    triggering. Switch bounce is an entirely different thing - it's very
    fast spikes caused by the switch contacts themselves bouncing when
    they crash together. All you have is a switch that opens when you
    whack it.

    So, all you need is a time delay on "open" - a 555 and a little logic
    should be fine. And with a decent capacitor, you shouldn't even need
    a pot - just get values from the chart in the data sheet.

    Good Luck!
    Rich
     
  11. JeffM

    JeffM Guest

    I have a small magnetic switch that is triggering
    Can you use a stronger magnet?
     
  12. Thomas

    Thomas Guest

    I am using a 4700uF cap with a 20 Ohm resistor and it works GREAT. I get
    about a 1.5 second delay before the coil discharges. I really appreciate
    it!

    Many thanks!!!!
     
  13. Pleased to hear it.

    Coming from me, always keen to get electronics into any gadget, this
    is an odd observation - but I'm curious why you didn't go for some
    simpler solution? How about a simple catch that can be slipped off
    when you open the lid?
     
  14. ====================

    It's best to keep your posts in the original thread, not via email.

    I don't understand your follow-up question. The circuit I posted for
    you already delays the action of the relay, for a time determined by
    the capacitor and the resistance, due to the capacitor charging up
    from zero to the pull-in voltage. And then, when the lid closes, there
    will be a shorter delay before the relay contacts open again, while
    the capacitor discharges via the relay coil until its voltage is below
    the (lower) drop-out level.

    I was also surprised by your earlier post in which you said the
    discharge delay was as long as 1.5 s. Please draw your circuit and
    state the voltage supply, measured pull-in and drop-out voltages, and
    the resistance of the disconnected relay.
     
  15. Thomas

    Thomas Guest

    You are correct. I don't know what I was thinking. I really appreciated the
    help.
     
  16. Guest

    Thomas schreef:
    If you draw your problem on paper and scan it to my e-mail ( 100 dpi
    in JPG)

    maybe I can help you by sending you a other scan with the solution to
    your
    problem.


    to your service,

     
  17. Guest

    my email is williamdobAThotmail.com
     
  18. Did you not notice that this was sorted fairly conclusively a couple
    of days after Thomas posted his original enquiry (12 days ago)?
     
  19. Paul Jones

    Paul Jones Guest

    There is nothing ! How does it know what a "real" bump is ?!
    Your circuit needs to be intellegent enuf to reject a false bump

    You need a "mass" and spring and a pot to measure the travel an
    something to integrate all this .

    what you are doin is not simple
     
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