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DC Generator Output

  1. Oct 1, 2013 #1
    1) My understanding is that the output of a DC generator has the form Vmax*|sin(2πft)| for 2 poles. Having more poles makes the output smoother but it would still have ripples. Is this correct?

    2) In many places, I have seen the EMF of a DC generator written as: [itex]\phi[/itex] * (NZ/60) * (P/A).
    What value is given by this expression - is it the maximum, rms or mean value of the voltage?
     
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  3. Oct 1, 2013 #2

    Baluncore

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    1) Yes, it will always have some theoretical ripple.
     
  4. Oct 2, 2013 #3
    What about the expression for EMF generated..does it represent the maximum or mean value?
     
  5. Oct 2, 2013 #4

    Baluncore

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    I can't answer that part because I do not know what your symbols represent.
     
  6. Oct 2, 2013 #5
  7. Oct 2, 2013 #6

    jim hardy

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    That would be a sinusoid, wouldn't it? Perhaps absolute value of one if I understand your symbols...

    The commutator does its rectification (absolute value) by picking voltage at opportune time in the cycle
    and generally the more commutator segments you have the more closely you will be to smooth DC output.
    Observe on page 5 of your second link the two segment commutator provides only half wave rectification, which I would say is that first formula you posted. (fig 1.3A)

    1.3B shows smoothing effect of more commutator segments.

    When I took DC machines we went straight to the practical machines of the day with many commutator segments. We used the simpler equation for generated voltage Eg

    Eg = K X [itex]\Phi[/itex] X RPM

    where K lumps several parameters and constants and can be calculated or measured . We had a lab where we did both.
    It was easier to spin the machine and measure with a voltmeter than to disassemble it and measure physical dimensions and try to calculate flux.
    Nowadays I see model airplane motors with three segment commutators, and automobile fuel pumps with five . Looking forward to your explanation of those;

    old jim
     
  8. Oct 2, 2013 #7
    Thanks for the reply, Jim. (Btw looks like "my" symbols seem to be alien over here!)

    What I gather from ur reply is that when you worked with the DC machines, their output was nearly constant ("practical machines of the day with many commutator segments"). Therefore it was easy to use a voltmeter to see what the reading was.

    But even the formula that you used assumes a constant value for magnetic flux (which it would be if the machine has lots of poles). But what about the 2-poles machine? It seems to me that EMF = ϕ * (NZ/60) * (P/A) or K X Φ X RPM would give the maximum value if we use the maximum value of ϕ (that the armature winding cuts during a rotation) in these expressions. Does it make sense?
     
  9. Oct 2, 2013 #8

    jim hardy

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    The brushes pick off the voltage near the peak

    see fig 6 here
    http://www.reliance.com/mtr/mtrthrmn.htm

    I think you are writing the formula for voltage in an individual winding
    rather than after commutation

    more poles just increases frequency for given RPM
    and lets you add more brushes

    The more commutator segments the nearer the peak you pick off the voltage
    so the less ripple and the higher the mean

    also they can shape flux under poles so as to produce a flatter waveshape than sine, but that's verging on the exotic details..
     
  10. Oct 2, 2013 #9

    jim hardy

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    Your math is sharp

    try rewriting your formula from absolute value of sine (two commutator segments)

    to 360/n degrees either side of peak , where N = number commutator segments
    I think that'll approximate real world commutation

    and maybe that'll tie our formulas together..

    i'm really not academic enough for this site, but do try to contribute on practical points

    and I just love it when math and practicality converge

    old jim
     
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