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Formula for voltage output produced by an alternator?

  1. Sep 24, 2015 #1

    I have a science investigatory project issued by my school. For my (actually our) project, we decided we will make a some sort of concept alternator powered by water collected from rain. Our goal is pretty small, to power up a small 5-volt bulb. Assuming the rotor rotates 2-3 times a second, how many turns will the coil need to attain 5 volts or more?

    Edit: I meant magneto by alternator.

    an 8th grader
    Last edited by a moderator: Sep 24, 2015
  2. jcsd
  3. Sep 24, 2015 #2


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  4. Sep 24, 2015 #3
    Thanks for the link, but I have the concept of the project already laid out. I just need the specific values of things such as mentioned before, how much turns of coil do I need to reach 5 volts when the magnet turns 2-3 times per second?
  5. Sep 24, 2015 #4


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    A number of turns that will produce 5 volts unloaded will do you absolutely no good at all if the bulb loads it well beyond its capacity to generate current. That not too likely to happen but my point is that your having totally ignored the current (/power) requirements implies that you don't quite grasp the fundamentals so it would be a good idea for you to study up a bit on basic electricity.

    Taking on such a project at 8th grade is a terrific thing to do, so I don't mean to discourage you, I'm just suggesting that there is maybe an hour's worth of reading that you should do on some site that explores the basics of electricity (the concepts of voltage / current / power and so forth).
  6. Sep 24, 2015 #5
    I'm trying to power up a 3-volt bulb.
  7. Sep 28, 2015 #6
    In my opinion you could use the simple formula as:

    Induced emf Eph = 4.44* f *Flux*Tph*kw then:

    turns per phase Tph = Eph / (4.44* f*Flux* kw)

    Frequency of generated emf f = p*NS/120 = p*ns/2,

    Ns = Synchronous speed in rpm[rotations/minute]

    ns = synchronous speed in rps[ rotations/second]

    p = no of pole pairs.

    kw = winding factor

    Air gap flux per pole Flux= Bav*PI()*D*L/p where:

    Bav = Average [magnetic] flux density

    D=stator diameter; L=stator length

    Let’s take an example:

    Let's take a pole of Neodymium Magnet N30 Br=1.12 T[Wb/m^2]


    Let's average magnetic flux density Bav=Br

    Let's take a stator of 3cm inside diameter and 10 cm length

    Let's take 2 poles[p=number of pole pairs=1]

    Let's take winding factor kw=0.955

    Flux= Bav*PI()*D*L/p=1.12[Wb/m^2]*pi()*0.03[m]*0.1[m]/1=0.010556 Wb

    f=p*ns/2=1*3/2=1.5 Hz

    Tph = Eph / (4.44* f*Flux* kw)=5/(4.44*1.5*0.010556*0.955)=75 turns/phase
  8. Sep 28, 2015 #7


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    Then why are you trying to produce 5 volts instead of 3? If your source has decent power capability you'll either burn out the bulb or, more reasonably, have to burn off some of the excess as heat in a resistive voltage divider
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