Inverter question

Discussion in 'Homework Help' started by PAB, Jul 11, 2007.

  1. PAB

    Thread Starter New Member

    Jun 21, 2007
    6
    0
    Can anyone help me with these questions
    Problem: I have a transformer with a (pri) -star || (sec) -star/delta (producing 2 x 3 phase supplies 30 electrical degrees apart) this in turn suppling a rectifier which runs @ 100% duty cycle and converts these 6 phases to DC Voltage.

    Question: i); What is the DC output voltage of the rectifier, (assuming there is no voltage drop accross the SCR's or anywhere in the system), and
    ii); what is the ripple voltage and frequency of the DC output if the supply frequency is 50Hz.
    iii); What capacitance (in Farads) would be required to remove this Ripple voltage.



    Answer:
    The best way, I found out is to draw a phasor diagram, for star an delta and calculate phase angle and voltage amplitude

    @ 380Vac phase to phase
    (6 phases)

    (sec) PhaseS STAR winding
    A: 380 sin 0 = 0V
    B: 380 sin 120 = 329V
    C: 380 sin -120 = -329V (remmember this phase will be inverted)

    (sec) Phases DELTA Winding
    D: 380 sin 30 = 190V
    E: 380 sin 150 = 190V
    F: 380 sin -90 = -380V (remmember this phase will be inverted)

    From this work out the complex number for each phase (with trig)

    A: 0/_ 0 = 0

    B: 329/_ 120 = (cos120 x 329) = -164.5 , (sin120 x 329) = j(284.9)

    c: 329/_ -120 (remmember this phase will be inverted)
    C: 329 /_ 120 = (cos120 x 329) = -164.5 , (sin120 x 329) = j(284.9)


    D: 190/_ 30 = (cos30 x 190) = 164.5 , (sin30 x 190) = j(95)

    E: 190/_ 150 = (cos150 x 190) = -164.5 , (sin150 x 190) = j(95)

    F: 380/_ -90 (remmember this phase will be inverted)
    F: 380/_ 90 = (cos90 x 380) = 0 , (sin90 x 380) = j(380)

    Now it it simple addition....-----------------------------------------
    = -329 , = j(1139.8)

    from -329+j1139.8 we get (pythagoras theorem) = 1186Vdc

    1186V will be the peak output voltage of the inverter across the DC bus.

    remmember this is unfiltered so there will be a small ripple voltage.


    To calculate this ripple

    There will be an intersecting of the phases every 15degrees therefore:

    (remmembering that @ 50Hz there will be 12 pulses every 0.02seconds)

    per phase
    (380 sin90) = 380V
    (380 sin75) = 367.05V

    Therefore Vripple = 12.95V or 3.4%ripple per phase

    3.4% of 1186V = 40.32Vripple @ (50Hz x12)

    Vripple= 40.32V @ 600Hz


    Is this correct? How do I work out the Capacitance value I need to add?
     
  2. Newton1Law

    Member

    Aug 22, 2007
    10
    0
    This would be a twelve pulse rectifier and would have a minimum voltage of 0.966 per unit (based on the peak value of your voltage) and thus a 0.034 per unit ripple at 720 Hertzs. The capacitor you would need is approximately given by C>= 0.7 * (I load) / (Vmax ripple voltage allowed)*(Frequency). There is no amount of capacitance that would give you zero ripple.
     
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