IR emitter /detector question

Discussion in 'The Projects Forum' started by zerocool5878, Jul 6, 2011.

  1. zerocool5878

    Thread Starter New Member

    Jun 12, 2009
    9
    0
    Maybe someone can be kind enough to help me. I would like to build a circut that runs a 1.5v-3v motor when the detector CAN see the emitter. When it can NOT it cuts the power to the motor.

    This is how I have it wired with just an LED for now
    [​IMG]

    There is 2 problems I have

    1) it is working backwards from what I need. ( when the beam is broken it delivers the power instead of kills it)

    2) it is not able to draw enough current to power the motor


    I think i need to use a transistor, a relay or an IC to get what I need but I am not sure which way I should be looking into.

    Any help or suggestions would be great.

    PS. I do not know a ton about electronics (still learning) but i can usually follow a schematic pretty well
     
  2. RFactor

    Active Member

    May 1, 2009
    33
    3
    When there is no light on the photo transistor, the current flows through the resistor and the motor. When the light is on the photo transistor, the current flows throught the resistor and the transistor is now conducting taking the current. If you put the motor where r2 is it would work... if the photo transistor is capable of carring the current. You could put a lower value resistor in series to keep from overloading the photo transistor.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Try it like this:

    [​IMG]

    Current through the phototransistor is limited by R2.
    Q2 is capable of sinking up to ~500mA current.
    The Schottky diode takes care of the motors' reverse-EMF.
    R3 makes certain that Q2 turns off when the IR light received is low.
    [eta]
    One big problem with a circuit this simple is that it has no way to differentiate between ambient light and the IR emitter. Television remote controls get around this problem by looking for a modulated signal in the 36kHz to 44kHz range.
     
    Last edited: Jul 6, 2011
    zerocool5878 likes this.
  4. zerocool5878

    Thread Starter New Member

    Jun 12, 2009
    9
    0
    This works perfect... Exactly what I needed to do. I did not have the Schottky diode so instead of using the motor to test i just used an LED. One question.. Do I have another option for that diode? I do not think radioshack carries it so I will have to order it from digikey if its my only option. I have a pack of diodes they are 1n914 but I think those work differently because I mostly see them used on the positive wire only not both as in your schematic

    Also as far as the ambient light it will not be a problem because this whole circuit will be housed in a PVC pipe and should not have any ambient light at all unless some penetrates thru the white pvc which i dont think will be a problem.

    Thanks again for your help
     
  5. zerocool5878

    Thread Starter New Member

    Jun 12, 2009
    9
    0
    Just going to add one more thing. I just noticed that the motor that I bought for this project is rated
    Current (load) 0.98A

    I don't know that the 2n2222 will be enough to power it. Should I look for a transistor rated 1a?

    Thanks again
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The 1N914/1N4148 is a very fast switching computer diode. Unfortunately, it's only rated for 100mA current, and your motor apparently draws 10x that much - so you'll need to use something like the Schottky diode I suggested, or other fast-recovery diode. The 1N400x series and 1N540x series diodes that Radio Shack sells are too slow to use.

    You've been pretty incomplete about the motor specs - at what voltage does it draw .98A? You haven't stated at what voltage you're going to be running the circuit with. I'd suggest replacing the 2N2222 with a TIP120 and increasing the value of R2, but the TIP120 is a Darlington and would have a relatively high Vce (0.7v to 1.6v) as compared to a typical transistor (0.1v to 0.4v), which might leave you with an underpowered motor.

    Motors usually draw much more current when starting from a stop, or when the rotor is locked.
     
  7. zerocool5878

    Thread Starter New Member

    Jun 12, 2009
    9
    0
    Here is the specs for the motor I have. The package didn't have too much info but I found this online.

    http://support.radioshack.com/support_supplies/doc19/19830.htm

    Im going to try to explain what I want to do with this in detail and maybe there is a better way to do it.


    The application for this is sort of a contraption for geocaching. The geocachers GPS brings him to a 2" PVC pipe mounted to a telephone pole about 30" long. the bottom is open and the top has a cap. When the cap is removed they have access to a small box they have to put 2 AA batteries in and hold down on a Dpdt momentary switch.

    This will spin the motor and lower one of these out the bottom.

    [​IMG]

    now my problems were that i need to have a stop for the up and down so when it goes to far down it doesn't start coming back up. and when its far up it doesn't bind or get caught. That's why i choose the ir emitter/detectors


    Here is a sketch of my plan for this


    http://oi51.tinypic.com/5fgj7m.jpg
    (the circle is a ball that triggers the IR sensors)

    I hope this is understandable

    Thanks again
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You're really complicating the heck out of something that should be simple as possible; a PVC pipe with a screw-on lid and some trinkets inside.

    Rubber bands won't last long at all outside in a tube; they will rapidly decay and become brittle or gummy. The steel parts on the motor will corrode. Someone will put batteries in backwards. Things will break. Someone will get frustrated and bust up your project; they won't have batteries so won't be able to get the trinket. People are supposed to leave a trinket, but it'll have to be a specific size and type for your geocache.

    The motor you've chosen draws too much current for a simple circuit that should be low-power.
     
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