Making a Small Deadbolt for a Cabinet door

Discussion in 'The Projects Forum' started by torea, Jul 12, 2012.

  1. torea

    Thread Starter New Member

    Jul 3, 2012
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    Hey all, first post here! I've been digging through relevant threads but couldn't find anything that was what I needed.

    I have a small cabinet door that I'd like to lock/unlock with a hidden deadbolt. Basically, I want to put the bolt on the back of the door (hidden in the cabinet) and have it unlock with a solenoid triggered by a reed switch. It doesn't need to withstand anything huge, the cheap MDF door would probably break before then. I just want to have a place to put some small valuables, feel safe while they're there, and build something while I'm at it :D I understand there are better ways to secure valuables, but I feel like building something.


    So I understand how the reed switch would trigger the circuit, but I don't know how to hook up the solenoid nor where to get the right one.
    Also, is there a better way to do this? The solenoids I've seen on DigiKey and Mouser are a bit more than I want to spend for this project, and also don't look like I thought they should.

    Would something like this work? - http://www.parts-express.com/pe/showdetl.cfm?Partnumber=330-010&FTR=solenoid


    Thanks for the help! I'm more than willing to do the legwork if someone can point me in the right direction.
     
  2. strantor

    AAC Fanatic!

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    yes actually that solenoid would work. I've used that solenoid before for a pig trap. I thought it would be chinese crap, but it turned out to be a chinese treasure.

    when you say
    , which circuit are you referring to? You didn't post one. If you're thinking of passing the current of the solenoid through the redd switch, it would probably kill the reed switch. They are not meant for carrying any high current (usually).
     
  3. torea

    Thread Starter New Member

    Jul 3, 2012
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    Ok, let me back track. I'm very new to this and might need more help than I thought.

    I want the reed switch to activate the solenoid, allowing the door to be opened. The reed switch, switched with a magnet, would allow the lock to be hidden. Basically, stick the magnet to the door where the reed switch is which would activate the solenoid pulling the deadbolt out, open the door, then remove the magnet until I need to close the door again. So the solenoid wouldn't be active for very long, just enough to open the door. I'm hoping that would allow me to use a battery for the circuit.

    Can you point me in the direction of how to pull this off?
     
  4. KJ6EAD

    Senior Member

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  5. #12

    Expert

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    How to use a switch to fire a relay to fire a solenoid.
     
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  6. strantor

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    sure. The application you describe is totally doable, but it might take a few more components than you'd expect, depending on the robustness of your reed switch. Why don't you post the reed switch that you intend to use.

    If you don't have one selected, look for one that has a high amp capacity. I don't remember how many amps that solenoid pulls, but it's a lot. Maybe 15A or more? To get an idea how many amps a door lock solenoid pulls, sit in your car, with the engine off and the head lights on, then electrically lock & unlock the doors; see how the lights dim? Lots of amps. If you have an ammeter, you could actually test how many amps it pulls by measuring as you lock & unlock.
     
  7. Sensacell

    Well-Known Member

    Jun 19, 2012
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    The door lock actuator is much better solution than the solenoid.

    Solenoids are very inefficient and actually tricky to work with due to their highly non-linear force / displacement curve.

    The actuator has a motor, gearbox and limit switches inside, so it will be easy to operate a door bolt with this thing.

    I would use a small relay to drive the actuator, the contacts of a small reed switch cannot carry much current.

    I don't know how the actuators are controlled, two inputs? lock/unlock? or reversing the polarity? either way, the relay makes it simple.

    (I looked a the actuator again, it has only two wires, you need to reverse the polarity, a DPDT relay is a must.)
     
  8. strantor

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    Yes, a disconnect is also a must, or else it will stay constantly energized. I assume it will run on battery. also the actuator is intermittent duty I believe.

    maybe 2 reed switches, 2 SPDT relays, one set for lock, one for unlock?
     
  9. #12

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    I was assuming a spring loaded solenoid. Not lucky tonight. I keep making drawings and then finding out the circuit is going to be different from what I expected.
     
  10. strantor

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    DOH! well, could be as simple as your circuit X2.
     
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  11. torea

    Thread Starter New Member

    Jul 3, 2012
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    Thanks for the responses!

    KJ6EAD, thanks for the links! Very helpful, though I don't know if any of those will work for me. Just to make sure, if I were to use this one, turning it on would pull the 3" plunger back (i.e. unlocking)? Is the plunger just sitting in there (not held in). Then, if it were mounted facing downwards on the door, I could just a small magnet to help hold it in place once gravity pulls it down?
    - http://www.allelectronics.com/make-a-store/item/SOL-58/24-VDC-PULL-TYPE-SOLENOID/1.html


    #12, thanks for the circuit diagram. But wouldn't that be fairly taxing on the battery? Like that, the relay is always on when the reed switch isn't activated. But I could put the relay in series with the reed switch, so when I flipped the switch, it would activate the relay and turn on the solenoid, pulling the plunger back. Right?

    Strantor, looks like a good idea. I actually have a relay that would work for this, so I think I'll just use the reed switch to activate it.

    Sensacell, the description says it's you reverse the polarity to push/pull. Does that mean it always needs power? Or will it stay in one place until the power polarity is reversed?


    Thanks for all the help!
     
  12. Sensacell

    Well-Known Member

    Jun 19, 2012
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    Arghh! I don't own a car - I guess the motor just runs until it stalls at then end of travel, probably a thermal PTC fuse in series to prevent melt-down of the motor windings.

    2 reed switches sounds like the way to go then- unless, (gasp) - we use ELECTRONICS!
     
  13. strantor

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    the solenoid you linked to has a spring that pushes it out, so it will default to the "locked" position with no power.

    You got #12's circuit wrong. when the magnet is not present, the relay will not be energized and the circuit will not draw power.

    The actuator you originally linked to (push/pull) has weak "return to center" springs that will work if the actuator is not connected to anything, but with any type of load, it will just remain in the "push" or "pull" position after power is removed. it can also be pushed in or pulled out by hand with no power.
     
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  14. #12

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    The relay IS in series with the reed switch. I could only find a drawing of a closed switch, but it's really open until you apply the magnet.

    Edit: strantor already covered this but we were both typing at the same time.
     
  15. torea

    Thread Starter New Member

    Jul 3, 2012
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    Ahhhhh durp! My bad, I got confused by the way it was drawn. Can't believe I missed that. I get it now.

    So the solenoid would work better than the door lock actuator, because it wouldn't require anything more than #12's circuit? The door lock actuator would need the polarity to be flipped to 'lock'.
     
  16. #12

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    That's my position. I could swear I've seen some small, spring loaded solenoids, like way smaller than 10 or 15 amps. The goal right now is to find one or build one.
     
  17. strantor

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    Might be good to put a cap on the other side of the relay to provide actuating current for that solenoid.
     
  18. torea

    Thread Starter New Member

    Jul 3, 2012
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    Like a .1uf cap between the 12V battery and the solenoid?
     
  19. strantor

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    Like this:

    [​IMG]

    You haven't said what type of battery you'll use but I'm guessing it's probably going to be AA batteries or the like, which can't put out much current, especially when they are <100% charged. A capacitor's rapid discharge should provide the current needed initially (actuating current is a lot more than holding current). .1uF seems awful small. I was thinking 1000uF at least, but that's just a guess.
     
  20. torea

    Thread Starter New Member

    Jul 3, 2012
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    Ah, I was thinking .1uf because I thought you meant it was for something else.

    I was originally thinking an A23 battery. Might be too small though. I'm guessing a pair of 4-pack series AA holders in series would be better?
     
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