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Induction motor vector control stator current vector

  1. Aug 8, 2015 #1

    Cdz

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    I am new to flux oriented control and i read that this method is to decouple the stator current, one component for flux producing and other for torque,so that so that we can control these two current components to control an AC motor like a DC one(right?)

    However when i try to dig into detail there are many problems.
    1.The two components of stator current, the one producing flux i can understand but the other component for controlling torque, i don't, shouldn't it be a rotor current since it's the rotor that in interacts with the magnetic flux to produce torque, not the stator.

    2.To control the stator current is to control the current vector represented as:
    FOC-1-26-12-13.gif
    where ia,ib and ic are three phase stator currents with balanced load. Since this three currents are thee vectors of equal magnitude and 120°difference, the sum of them should be zero,but it's not.Did i misunderstand the three phase currents?
     
  2. jcsd
  3. Aug 8, 2015 #2

    Hesch

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    The motor is digitally controlled by a computer. When the computer supplies 3 phase voltage at some amplitude/frequency, it measures the phase currents which will have an amplitude and a phase. These currents are decomposed into Id (not yielding torque) and Iq (yielding torque) by some Clarck/Park-transformation. Id is of no use, but it is there due to selfinduction in the stator/rotor and due to counter emf from the magnitized rotor.

    So the computer simply calculates what's going on inside the motor, which it uses to maximize efficiency of the motor, increase dynamic behaviour, and so on. It calculates slip, torque and has some tuned look-up-table to decide: What to do now? How to increase efficiency? It has two (three) possibilities: Frequency/(phase) and amplitude of the supplied voltage.

    The Id must be there, and is used to magnitize iron. The Iq provides torque.
    The sum should be zero. ( Error in some SW or HW? ).
     
    Last edited: Aug 8, 2015
  4. Aug 10, 2015 #3

    Cdz

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    Thank you for your answer.
    You mean the vector proxy.php?image=http%3A%2F%2Fwww.electrical4u.com%2Fequations%2FFOC-1-26-12-13.gif should be zero? But it's the vector that we need to control,if it's zero, how can it be controlled? I read about the vector here http://www.electrical4u.com/field-oriented-control/
    By the way, what do you mean by SW or HW error?
     
  5. Aug 10, 2015 #4

    Hesch

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    SoftWare / HardWare error.

    The HW error could be an error in an AD-converter or in some PWM-driver.
    In your link you have missed something important:

    With the mathematical processing capabilities offered by the microcontrollers, digital signal processors and FGPA, advanced control strategies can be implemented to decouple the torque generation and the magnetization functions in an AC induction motor. This decoupled torque and magnetization flux is commonly called rotor Flux Oriented Control(FOC).

    This is what you have to control. Remember that the three phase current are not passing the same coil or passing three coils in parallel ( at the same location ). The coils are distributed with mutual mechanical angles = 120°. So though the sum of the currents = 0, they will provide a magnetic field, that is expressed by a vector . This vector ( rotor flux vector ) is the one to be controlled (amplitude + phase) by means of the phase currents (amplitude + phase), but the sum of the phase currents is "normally" = 0.

    Say that the controller has detected some error in one of the phases (fuse blown), it would be "smart" to induce a DC current through the other two phases: That would result in a safety breaking of the motor. But even then, the sum of the currents could be 0.
     
    Last edited: Aug 10, 2015
  6. Aug 10, 2015 #5
    They're not, though, and the complex products don't sum to zero. You need a stator current vector to produce a (rotating) stator field.

    This is a balanced set of currents:
    ##
    i_a = \cos(\omega_s t)\\
    i_b = \cos(\omega_s t + \frac{2\pi}{3})\\
    i_c = \cos(\omega_s t - \frac{2\pi}{3})\\
    ##
    Try plugging those into the space vector transformation and see what you get.

    See here for an illustration.
     
  7. Aug 10, 2015 #6

    Hesch

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    The sum of the currents is indeed zero.

    Say the motor is Δ-coupled: It will run perfect, though having no neutral to compensate for a sum ≠ 0.

    It's the mechanical arrangement as of the three phase-coils that creates a flux-vector ≠ 0. If you taped those three coils together on a stack, so to say, the summed flux from the three coils would be zero. But in a rotating motor, the coils are arranged in three "planes" with a mutual angle = 120 degrees. Therefore the magnetization flux vector becomes ≠ 0.
     
  8. Aug 10, 2015 #7

    Cdz

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    I made a mistake of claiming that three current vectors are of the same magnitude with 120o difference in space while actually they are not of the same magnitude.
    I think that wrong notion comes from the fact that the magnitudes of three effective currents for balanced load are the same which i mistook for instantaneous currents.
    Yes, sum of currents is zero,but milesyoung means sum of current vectors.
    Thank you guys:woot:
     
  9. Aug 11, 2015 #8
    I wrote about the space-vector transformation, not phase/line currents.

    The terms ##i_a, i_b e^{j2\pi/3}, i_c e^{-j2\pi/3}## do not sum to zero if ##\{i_a, i_b, i_c\}## is a balanced set.
     
  10. Jul 25, 2016 #9

    Cdz

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    Thank you, man. :smile:
     
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