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Vector Control of the Induction Motor
First draft report (November 2014)

EJ2201 Electrical Machines and Drives

Group 20
Albane Schwob
Tim Mertens

Vector Control of the Induction Motor
Summary
Unlike a DC motor, an AC induction motor is much more complicated to control in a
fast way. The stator current (field winding) and the rotor current (armature winding) can
be changed independently from each other in a DC motor. This is not possible in an
induction motor, where stator and rotor current are dependent on the same
(stator)voltage (𝒖𝑠𝑠 in figure 1). An increase of this voltage will only have a relatively
slow response on the magnetising current because of the large Lm.
Vector control makes it possible to split up the stator current in a part that is responsible
for the magnetising of the air gap and a part that handles the torque demand. The former
is held a constant (because of the slow response), while the latter is used to control the
torque or the speed. In this way torque and speed in an AC motor can be changed as
quickly and efficient as a DC motor.

𝒊𝑠𝑠
+
𝒖𝑠𝑠

Rs

Lrl

Lsl

𝒊𝑟𝑠

𝑠
𝒊𝑚

Lm

Figure 1 Dynamic equivalent scheme

Rr’

+ s
jwyr

1 Methods
1.1 Machine Parameters
To obtain the parameters shown in the equivalent circuit above, the induction machine
used for the simulation model was first submitted to a no-load and a short-circuit test in
𝐿
a laboratory setup. (𝐿𝑠𝑙 = 𝐿𝑟𝑙 = 2𝑙)
In order to use these parameters in the simulation model, they have to be normalised in
per unit values.
The base values for the impedance and the inductance are:
𝑹𝒃𝒂𝒔𝒆 = 𝒁𝒃𝒂𝒔𝒆 =

𝑼𝒃𝒂𝒔𝒆
𝑰𝒃𝒂𝒔𝒆
𝒁𝒃𝒂𝒔𝒆

𝑳𝒃𝒂𝒔𝒆 = 𝝎

𝒃𝒂𝒔𝒆

=

𝟐𝟑𝟎
√𝟑∙𝟕,𝟖

= 𝟏𝟕. 𝟎𝟐𝟐𝟒 W

= 𝟎. 𝟎𝟓𝟒𝟏𝟗 𝑯

The normalized parameters can then be calculated by:
𝑹𝒏𝒐𝒓𝒎 =

𝑹
𝑹𝒃𝒂𝒔𝒆

𝑳𝒏𝒐𝒓𝒎 =

𝑳
𝑳𝒃𝒂𝒔𝒆

1.2 Simulating No-load and Short-circuit Conditions
In order to check the accuracy of the used induction machine model, a simulation of the
tests done in the laboratory exercise can be performed. Then a comparison can be made
between the simulated values and the measured values.
The simulations are all done in Simulink using a model shown in figure 2. Which has
the 2-phase stator voltage components as inputs, and the 2- phase stator current
components as outputs. Both are expressed in the stator reference frame.

The stator voltage and -current are by definition equal to:
2
𝒊𝑠 = ∙ (𝑖𝑎 + 𝒂 ∙ 𝑖𝑏 + 𝒂2 ∙ 𝑖𝑐 )
3
2
𝒖𝑠 = ∙ (𝑢𝑎 + 𝒂 ∙ 𝑢𝑏 + 𝒂2 ∙ 𝑢𝑐 )
3
With 𝒂 = 𝑒 𝑗∙120° .
2
Because of the factor 3, the magnitude of is and us are equal to the peak values of the
phase current and voltage in steady state (SS), and can therefore be used to calculate the
total impedance of the equivalent circuit of figure 1 . This impedance can then be
calculated by:
𝒖𝑠,𝑆𝑆
𝒁=
𝒊𝑠,𝑆𝑆
When using the 2-phase components (directly provided by the simulation) instead of the
3-phase components, the stator current and voltage can be represented by the following
complex numbers.
𝒊𝑠 = 𝑖𝑠𝛼 + 𝑗 ∙ 𝑖𝑠𝛽
𝒖𝑠 = 𝑢𝑠𝛼 + 𝑗 ∙ 𝑢𝑠𝛽
The induction motor can be simulated for different modes of operation by adjusting the
viscous damping constant, b. For the no-load test b will be set to 0, and for the shortcircuit test b is equal to a very large number (let’s say 10 000).
Taking into account the specific equivalent schemes of the two modes of operation, it
can be seen that:
𝒁𝑁𝐿 = 𝑅𝑠 + 𝑗 ∙ (𝑋𝑚 + 𝑋𝑠𝑙 )
𝒁𝑆𝐶 = (𝑅𝑠 + 𝑅𝑟′ ) + 𝑗 ∙ 𝑋𝑙
The machine parameters can then easily be determined by solving these two equations.
!Remember that in the per unit system wSS = 1, thus X = L

1.3 Implementation of a Current Control Loop
The first thing that has to be done to implement a current control loop is to make the
transformation from the stator reference frame to the synchronous reference frame dq
(Appendix A).
The stator current components in the synchronous reference frame are then submitted to
a negative feedback loop, where they are compared to their respective reference values.
In a next stage a PI control will adjust the input stator voltage according to the error
observed in the feedback loop. Finally the stator voltage is transformed back to the
stator reference frame ab (Appendix B).
From the specifications we know that a rise time of 15 ms has to be obtained, this
implies in the per unit system:
𝑡𝑛 = 𝑡 ∙ 𝜔𝑏𝑎𝑠𝑒 = 15 ∙ 2𝜋 ∙ 50 ∙ 10−3 = 4.71 𝑝. 𝑢.

This rise time can be reached by changing the P and I constants by trial and error. The
simulation model is shown in Appendix C.

1.4 Cross-coupling cancellation
i) We replace the constant entries of the block by step entries and then we simulate. The
simulation model is shown in Appendix D.
The feedback loop of the d current component is compared to a 25 p.u step. The
feedback loop of the q current component is compared to a 75 p.u step. We chose to set
the first step at 25 p.u and the second step up time at 75 p.u so that each time, the
current has time to reach its steady state value before the step is applied. The feedback
loop of the d current component is compared to a 25 p.u step. The feedback loop of the
q current component is compared to a 75 p.u step.
ii) We implement a cross coupling cancellation. The simulation model is shown in
Appendix E. The step blocks are set with the same values as in i).

2 Results and Discussion
2.1 Machine Parameters
From the laboratory exercise, the following values were determined:
𝑅𝑠 = 0.767 Ω
𝑅𝑟 = 1.21 Ω
𝐿𝑚 = 0.0759 H
𝐿𝑙 = 0.0127 H
After normalising these parameters to per unit values
𝑅𝑠𝑛 = 0.045 𝑝. 𝑢.
𝑅𝑟′𝑛 = 0.0711 𝑝. 𝑢.
𝐿𝑛𝑚 = 1,4 𝑝. 𝑢.
𝐿𝑛𝑙 = 0.234 𝑝. 𝑢.
All these values lie within the typical range for induction machine parameters, so they
make up a realistic representation of the induction motor.

2.2 Simulating No-load and Short-circuit Conditions
By conducting the no-load and the short-circuit tests, following steady state values for
ZNL and ZSC were obtained:
𝒁𝑁𝐿 = 0.045 + 𝑗 ∙ 1.526
𝒁𝑆𝐶 = 0.1054 + 𝑗 ∙ 0.2368
From these values the induction machine parameters can be derived as:

𝑅𝑠 = 0.045 𝑝. 𝑢.
𝑅𝑟′ = 0.0604 𝑝. 𝑢.
𝐿𝑙 = 0.2368 𝑝. 𝑢.
𝐿𝑚 = 1.289 𝑝. 𝑢.
It can be seen that the values for Rs and Ll are very accurate, but the values for R’r and
Lm have a relative error of respectively 15% and 8%.

2.3 Implementation of a Current Control Loop
Following graphs show respectively id and iq obtained by the simulation of the current
control loop of Appendix C. It can be seen from these graphs that the rise time is indeed
4.71 p.u.

2.4 Current components after step and cross coupling
i) Current components after step
The following graph show id and iq obtained by the simulation of the current control loop of
Appendix D.

As expected, the d component of the current steps up after 25 p.u and the q component of the
current steps up after 75 p.u. However, it has to be noticed that when the id steps up at 25 p.u,
it has an influence on iq which steps down. Similarly, when iq steps up at 75 ms, it has an
influence on id, which also steps up. This phenomen is called a cross-coupling effect. As id
and iq are coupled in the control loop, they influence each other. Unfortunately, we would
want both steps to be independent, that is why we are going to implement a cross coupling
cancellation in the following part.
ii) Cross coupling cancellation
The following graph show id and iq obtained by the simulation of the current control loop of
Appendix E.

We observe that at 25 p.u the d component of the current steps up and at 75 p.u, the q
component of the current steps up. Moreover, the q component of the current has a small
overshoot at the beginning of the simulation but then quickly reaches its steady state value,
thanks to the control loop.

Appendices

Appendix A Transformation Model ab-to-dq

With:

In1 – the alpha component.
In2 – the beta component.
In3- the transformation angle q.
Out1 – the d component.
Out2 – the q component.

Appendix B Transformation Model dq-to-ab

With:
 In1 – the d component.
 In2 – the q component.
 In3- the transformation angle q.
 Out1 – the alpha component.
 Out2 – the beta component.

Appendix C Current Control Loop

Appendix D

Current control loop with steps

Appendix E Current control loop with cross-coupling
cancellation

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