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555 timer question & Bounceless Garage door opener.

  1. Jan 29, 2014 #1
    555 timer question & "Bounceless" Garage door opener.

    When opening a garage door from a distance, it is sometimes necessary to push the button multiple times when beyond the range of the transmitter. One can inadvertnently push the button multiple times and cancel the door going up. To open the door, one must push the button three more times to go down, cancel, and go up.

    I resolved to solve this problem. I built a "bounceless" switch with a 555 timer and a relay.

    There are 3 leads going between the motor and radio receiver. 24 VAC, Common, and "relay". The radio receiver grounds the "relay" lead to signal the motor to close the door. All I needed to do was interrupt the "relay" leg. For my bounceless switch, the "relay out" of the radio receiver grounds the input of a 555 timer. The output of the 555 timer closes a contractor connected to "relay in" on the opener motor. I chose a resistor and capacitor that make the 555 hold the output down for about 4 seconds.

    Now, if I inadvertently hit the button multiple times, the motor does not receive multiple open signals because the 555 is still "holding down" the button.

    Here are the voltage requirements of my components.

    555 Chip - 4.5 to 15 VDC.
    Relay - 9 VDC

    Here are my available voltage sources
    24 VDC in radio receiver
    8 VDC in radio receiver.
    17 VDC repurposed transformer.

    Initially, I was hoping the 8VDC would be enough to close the relay, but it was not sufficient. Currently, I have the 17 VDC source powering both the chip and relay.

    Ok. Finally my question.

    Am I OK to have both the chip and relay powered by the 17 VDC source? The relay is on for only 4 seconds, so I'm not that worried about that burning up. The chip has a large input impedance, so it should be comfortable with 17 VDC.

    Is the 555 forgiving enough that I could even use the 24 VDC from the radio receiver? This would be a much more elegant solution.

    Attached Files:

  2. jcsd
  3. Jan 29, 2014 #2


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    There are a lot of variants of the 555. WHich are you using? If it is CMOS, the 17 VDC source is a no-no. If it is the LM555 from TI (probably the most popular) the max VCC rating is 18 VDC so you are pushing your luck at 17 VDC and 24 VDC will kill it. You could try powering a regulator to drive the 555's VCC or just use a voltage divider if you want a quick and dirty solution (assuming you're not using a battery)
  4. Jan 30, 2014 #3


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    The 555 should run on a lower voltage and drive a "open drain" FET that provides a path to ground for the higher relay voltage (through the relay coil). Don't drive the relay like that from the 555.

    Also, the relay needs a protection diode across it. When the switch opens, the relay coil produces a voltage spike (from the collapsing inductor field) that typically turns circuits into "switch once only" devices.

    Do a google IMAGE search for "555 drives a relay" and look at the IMAGES for one you like.
    Last edited: Jan 30, 2014
  5. Jan 30, 2014 #4
    Suppose the door is going down instead of up and you want to stop the door immediately. What will happen?
  6. Feb 4, 2014 #5
    Bro, I think it is perfect with 17 VDC but only if you want to upgrade then try 24 VDC. By the way my Automatic garage door opener is a very old unit. So, I want to ask does your switch would work with my opener?
  7. Feb 11, 2014 #6
    Ok. Regulating voltage to 12 VDC. Installed diode in parallel with relay coil side to handle collapsing inductor field.

    The 555 output is rated at 200 mA and I measured the relay drawing 20 mA at 12 VDC. Do I really need to make the 555 drive a transistor instead of driving the relay directly? Is there some other reason not to drive the relay with the 555?
  8. Feb 11, 2014 #7
    Door will bounce back when hitting an obstacle. Otherwise, you need to wait the 3 seconds for the 555 to reset. This is enough time for door to go through about 1/3 of its total travel.
  9. Feb 12, 2014 #8


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    Not really. I just don't like driving things from 555's directly. But the main reason was because of the supply voltage.

    At what output voltage is the 555 rated at 200ma, and at what VDD voltage?
  10. Feb 12, 2014 #9
    Which brand of garage door opener are you referring to? A garage door opener uses the breaking of the light beam as the first fault to cause the door to reverse. The door will also reverse if the door hits an object on the way down but the force it exerts on the object before stopping can be as high as 450 lbs. Garage doors have killed children.

    For this reason you may want to reverse the door before it strikes an object and your circuit may prevent you from doing that. By the way, I used to be Sr. Electronic Product Engineer for the world's largest garage door opener manufacturer.
  11. Feb 12, 2014 #10
    It's an older model, so it doesn't have the light beam. Manufacture is Allister. Radio receiver is delta 3.

    Not sure what the force required for bounce back is. I think I'll put some aluminum cans and see if it crushed them.

    The delay I built is only on the radio receiver. Hard wired switches go directly to the opener and allow for a quick stop of the door.

    I figure you'll only be closing the door with the remote when leaving. The door would typically opened just before you leave and closed as soon as you exit.

    I'm glad to have a garage door engineer reading my post!

    What about doors in commercial applications that close on their own after a period of time? Do they have the bounce back threshold set lower?
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