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How to calculate motor parts and energy requirements?

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by AverageJoe, Dec 27, 2017.

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  1. AverageJoe

    AverageJoe New Member

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    Dec 27, 2017
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    Hello Everybody, I’ve been wanting to ask this question for a very long time.


    My main thing is knowing if a low power, low current motor is possible using a BPW34 photodiode. I hope you guys can help. Thank you!



    I purchased a 3 in 1 soil tester meter https://www.ebay.com/p/3-in-1-Soil-Tester-Meter-for-Garden-Lawn-Plant-Pot-Moisture-Light-Ph-Sensor-Tool/1712433238?iid=311911276636


    that comes with an analog panel meter


    https://www.jameco.com/z/AVM7030-Velleman-Analog-Voltage-Panel-Meter-30-Volt-DC2-8-X-2-4_316654.html


    In the center it has a photodiode BPW34


    https://www.digikey.com/catalog/en/partgroup/bpw34/12351



    The needle of this meter turns at very low voltages and current.

    I measured everything and will post the exact info below.


    My question is


    How do engineers calculate how much voltage, current, coil resistance, and magnet strength a motor will need to be able to make a full turn?


    Let’s suppose this magnet was smaller, and the engineer wanted to make a smaller meter. How would he know how many coils turns (resistance), voltage, and current he would need to make this thing turn?


    Like in this science project


    https://www.education.com/science-fair/article/no-frills-motor/


    How did this person know how many turns (resistance) he needed, voltage, current, and magnet strength to make that motor turn?


    In my meter, when you apply .5 mV and .5uA it will turn until stopped by a piece of metal at the bottom that holds the magnet.


    The coil measures 794 ohms. The armature (coil) and needle weighs .415 grams.

    The photodiode causes the coil to turn to full 180 degrees when it reaches .5 volts and 500uA (micro amps).

    The bipolar circular magnet weighs 1.81 grams and has about 12 grams of lift force.


    https://www.flickr.com/photos/153503439@N02/39315572332/in/dateposted-public/

    https://www.flickr.com/photos/153503439@N02/39315594942/in/dateposted-public/

    https://www.flickr.com/photos/153503439@N02/39315601022/in/dateposted-public/

    https://www.flickr.com/photos/153503439@N02/38637464474/in/dateposted-public/




    Another question that I have is


    Since the magnet will stop at 180 degrees when the poles of the coil attract the opposite poles of the magnet, is there a small ic I could use to invert the current so that it would continue to spin? Don’t worry about the stopper.
     
  2. kubeek

    kubeek Well-Known Member

    Joined:
    Mar 11, 2006
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    Location:
    Prague, Czechia (not Chechnya)
    The force that the coil makes is counteracted by a small spring, so that is where you have to look for the balance. Force of spring versus the force on the coil produced by coil current ad the field produced by the permanent magnet.
     

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