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Thread: Shutter Signal Help...

  1. #1
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    Default Shutter Signal Help...

    Hi Everyone,
    I'm trying to get this....
    http://www.dacoinstruments.com/shutters/spring.shtml
    shutter to interface with the 5+ VDC shutter signal from my QM-2000. I tried a mechanical 5VDC relay and that obviously does not work. Not enough current. What would be the easiest circuit for running this shutter off of the 5V signal?
    Thanks!
    Adam

  2. #2
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    A small solid state relay would be perfect. You can get relays that have outputs either DC or AC depending on which solenoid you're using. You can get them from Digikey, Mouser, Jameco, etc.

    Tim

  3. #3
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    I would use a darlington transistor to drive this. You come out of the QM2000, through a resistor, to the base of a Darlington. Then you take the collector (or emitter, depending on how you want to wire it) and connect it to the shutter.

    I talked MechEng3 through this, and he wrote back to me that it worked perfect. If you could get his attention, I am sure he would share the details...

    Bill

  4. #4
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    This was sent to me from Bill Benner some time ago. It works great and all the parts on hand at your local rip-off shack.
    Attached Thumbnails Attached Thumbnails Shutter controller.JPG  

    You are the only one that can make your dreams come true....and the only one that can stop them...A.M. Dietrich

  5. #5
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    Quote Originally Posted by MechEng3 View Post
    This was sent to me from Bill Benner some time ago. It works great and all the parts on hand at your local rip-off shack.
    Thanks Bill and MechEng3,
    However, I cannot seem to get this circuit to work.... I built two just in case I had a bad TIP120. Any suggestions?
    Thanks again!
    Adam

  6. #6
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    Okay, I got it to work with a slightly different schematic though. I'll post it in a couple hours. My shop is a total mess right now.

  7. #7
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    I needed a break. Check it out...


  8. #8
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    You definitely want to put the diode across the coil (not between the emitter and collector of the darlington). When the transistor leaves conduction, there will be a very high voltage spike from the collector to the emitter, caused by the flyback action of the coil. Empirically we can say that the flyback voltage can be 10 times the initial coil voltage. This means that there could be 120V across the transistor, which may very well actually destroy the transistor.

    Also, the diode does another, not so obvious thing. When the actuator moves from its energized to non-energized position, it will generate back-EMF. The diode helps to provide damping, thus helping the actuator not overshoot the rest position.

    Best regards,

    William Benner

  9. #9
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    Quote Originally Posted by Pangolin View Post
    You definitely want to put the diode across the coil (not between the emitter and collector of the darlington). When the transistor leaves conduction, there will be a very high voltage spike from the collector to the emitter, caused by the flyback action of the coil. Empirically we can say that the flyback voltage can be 10 times the initial coil voltage. This means that there could be 120V across the transistor, which may very well actually destroy the transistor.

    Also, the diode does another, not so obvious thing. When the actuator moves from its energized to non-energized position, it will generate back-EMF. The diode helps to provide damping, thus helping the actuator not overshoot the rest position.

    Best regards,

    William Benner

    Hi Bill,
    So I could just add another diode across the coil and be okay, right? I probably actuated the shutter over 100 times last night during testing and didn't have any problems though. I think my problem with the original circuit was that the 10V power supply does not share the same ground as the 5V signal.
    Thanks Again!
    Adam

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