pulse motor circuit

Discussion in 'The Projects Forum' started by floomdoggle, Sep 9, 2008.

  1. floomdoggle

    Thread Starter Senior Member

    Sep 1, 2008
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    Hello to all,
    This is my first post. To start with, my electronics knowledge is High School in 1972. We built a super heterodyne radio. With tubes.
    I am trying to build the electronic section of a monopole pulse motor. Try not to laugh too hard. I have been searching this forum and other places, but just don't seem to either see what I need, or get the drift.
    Thanks in advance.
    Dan
     
  2. blocco a spirale

    AAC Fanatic!

    Jun 18, 2008
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    This won't work because the base of the 2n3904 is tied to -V.

    When should the coil be energised - when the beam is broken?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    You have the base of the 2N3904 connected to the emitter of the ITR8102. It should be connected between the collector and the 1k resistor instead. As things are right now, the base of the 2N3904 won't ever get any current, as it's connected to ground.

    With that change, the transistors will get turned ON when there is an object in the path of the IR LED.

    If you want the transistors ON when there is no object in the path of the IR LED, then leave the connection on the emitter of the ITR8102, and remove the connection of the emitter to ground.
     
    Last edited: Sep 9, 2008
  4. floomdoggle

    Thread Starter Senior Member

    Sep 1, 2008
    217
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    @blocco-
    The timing wheel is slotted, so to pulse @ light

    @Sarge
    I re-did the circuit. Is this better?
    Thanks again
    Dan
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    The connection you have is not what sgtwookie was suggesting.

    The emitter of the phototransistor should have stayed connected to ground and the base of the first transistor needs to be connected to the collector of the phototransistor.


    hgmjr
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    That's better, Dan. :)

    You could drop the 680 Ohm current limit resistor as low as 540 Ohms, which would give you 20mA current through the IR LED; that's figuring on the typical Vf (forward voltage) of 1.2v @20mA.

    Rlimit = (VoltageSupply - Vf) / Current Desired = (12-1.2)/20mA = 10.8/0.02 = 540 Ohms

    hgmjr,
    Dan chose the 2nd option that I suggested.
     
  7. hgmjr

    Moderator

    Jan 28, 2005
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    Keep in mind that if you elect to go with your latest connection, the sense of the photo-detector will reversed from the previous arrangement that sgtwookie recommended.

    The coil will be energized when the sensor is unblocked and de-energized when the sensor is blocked.

    hgmjr
     
  8. floomdoggle

    Thread Starter Senior Member

    Sep 1, 2008
    217
    2
    @hgmjr
    I though sgt wookie wrote if i used the light position, to use the emitter. I'm all ears. Thanks for the reply.
    Dan
     
  9. floomdoggle

    Thread Starter Senior Member

    Sep 1, 2008
    217
    2
    @Sgt Wookie,
    Will do. Now, I think I'll go fry some transistors, and let you know how it works.

    @hgmjr
    Yes, I want to use the unblocked mode, seems to make more sense to my limited experience. If you have any other design. I'm all ears.

    Thanks again, all.
    Dan
     
  10. blocco a spirale

    AAC Fanatic!

    Jun 18, 2008
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    Have you considered what will happen to the coil if the motor is stalled?
     
  11. floomdoggle

    Thread Starter Senior Member

    Sep 1, 2008
    217
    2
    Yes, the transistors will burn. A kill switch will be on the final circuit. Also, a neon light will go from collector to emitter to warn of overly high voltage. This motor is not a leave-it-alone type. It must be monitored like an automobile. Eventually it will power a bicycle.
    Oh yeah, I hooked up the circuit, and immediately burned the 2n3904. I had a 2n2502 lying around, and it worked fine. Thanks again to all who helped!
    Dan
     
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