Vector Control of Permanent Magnet Synchronous Motor (2)————Control Principle and Coordinate Transformation

2. Control principle of permanent magnet synchronous motor

2.1 Starting from the PMSM mathematical model

$\small dq$ Shaft voltage equation:

$\small \begin{bmatrix} u_{d} \\ u_{q} \end{bmatrix}=\begin{bmatrix} R_{s} &-\omega _{e} L_{q}\\ \omega _{e} L_{d}& R_{s} \end{bmatrix}\begin{bmatrix} i_{d}\\ i_{q} \end{bmatrix}+\frac{\mathrm{d} }{\mathrm{d} t}\begin{bmatrix} \Psi _{d}\\ \Psi _{q} \end{bmatrix}+\begin{bmatrix} 0\\ \omega _{e}\Psi _{f} \end{bmatrix}$

$\small dq$ Shaft flux equation:

$\small \begin{bmatrix} \Psi_{d} \\ \Psi_{q} \end{bmatrix}=\begin{bmatrix} L_{d} &0\\ 0& L_{q} \end{bmatrix}\begin{bmatrix} i_{d}\\ i_{q} \end{bmatrix}+\begin{bmatrix} \Psi _{f}\\ 0 \end{bmatrix}$

$\small dq$ Shaft torque equation:

$\small T_{e}=\frac{3}{2}p\left ( \Psi _{d}i_{q}-\Psi _{q}i_{d} \right )=\frac{3}{2}p\left [ \Psi _{f}i_{q}+\left ( L_{d}-L_{q}\right )i_{d}i_{q} \right ]$

$\small dq$ Axis motion equation:

$\small T_{e}=T_{L}+\frac{J}{n_{p}}\cdot \frac{\mathrm{d} \omega _{g} }{\mathrm{d} t}$

Analyze the above equation, if we can control $\small i_{d}=0$

Then the voltage equation can be simplified to:

$\small \left\{\begin{matrix} u_{q}=Ri_{q}+L\frac{\mathrm{d}i_{q} }{\mathrm{d} t}+\Psi _{f}\omega _{e}\\ u_{d}=-\omega _{e}Li_{q} \end{matrix}\right.$

The torque equation is:

$\small \frac{\mathrm{d} \omega _{m}}{\mathrm{d} t}=\frac{K_{t}}{J}i_{q}-\frac{B}{J}\omega_{m}-\frac{1}{J}T_{L}$

In the above formula: $\small \Psi _{f}$ is the permanent magnet flux linkage, $\small R$ and $\small L$ is the resistance inductance of the stator winding, $\small \omega _{e}$ is the electrical angular velocity $\small \omega _{m}$ of the motor , is the mechanical angular velocity of the motor, $\small P$ is the number of pole pairs, $\small K_{t}$ is the torque constant, $\small J$ is the moment of inertia, $\small B$ is the friction coefficient, and $\small T_{L}$ is the load factor .

From the above equation, we can see that $\small i_{q}$ we can control the magnitude of the torque only by controlling , and the $\small d$ shaft voltage is only $\small i_{q}$ related to it, which is extremely beneficial to our control.

And, when $\small i_{d}$ =0, it is equivalent to a typical separately-excited DC motor, the stator has only the quadrature axis component, and the space vector of the stator magnetomotive force is exactly orthogonal to the space vector of the permanent magnet magnetic field. Therefore, in order to reduce the loss, it is completely possible to set $\small i_{d}$ =0 to reduce the copper loss.

The vector control block diagram is shown below:

summary:

The principle of vector control is to try to simulate the torque control law of a DC motor on a permanent magnet synchronous motor. After coordinate transformation, the current vector is decomposed into a current component that generates magnetic flux and a current component that generates torque. The two components are perpendicular to each other. ,Independent. In this way, they can be adjusted individually, similar to the double closed-loop control system of a DC motor.

2.2 Coordinate transformation*

2.2.1 Reasons for coordinate conversion

In a permanent magnet synchronous motor, the angle between the stator magnetic potential $\small F_{s}$ , the rotor magnetic potential $\small F_{r}$ , and the air gap magnetic potential is not $\small 90^{\circ}$ , and the coupling is strong, and it is impossible to independently control the magnetic field and electromagnetic torque.
The excitation magnetic field of the DC motor is perpendicular to the armature magnetic potential, and the two are independent and do not affect each other.
There are various DC motor control strategies, which can make it cope with different occasions

Therefore, after analyzing the mathematical model of the permanent magnet synchronous motor, the coordinate transformation is performed to simulate it as a DC motor for control, which will greatly improve the controllability and operating efficiency of the motor.

2.2.2 Basic idea of coordinate transformation**

The principle of the equivalence of different motor models: the magnetomotive force generated in different coordinate systems is completely consistent.

As shown in a) in the above figure, when the motor is supplied with a three-phase balanced sinusoidal current, the resulting composite magnetomotive force is a rotating magnetomotive force, which is sinusoidally distributed in space and $\small \omega _{1}$ proceeds in the order of ABC with synchronous speed . Spin. The rotating magnetomotive force is not only generated by three-phase windings, but balanced multi-phase currents can generate the desired rotating electromagnetic field, and the two-phase is the simplest. The $\small 90^{\circ}$ rotating magnetic field can be generated only by passing in a balanced alternating current that is checked in time . If the magnitude and speed of the rotating magnetomotive force in control a) and b) are the same, then the two can be considered equivalent.

Look at the figure c) again. Two mutually perpendicular windings M and T, through which current flows, produce a composite magnetomotive force F. Obviously, this magnetomotive force is fixed relative to the M and T windings. At this time, if the two windings are artificially combined The entire iron core including the windings rotates at the above synchronous speed, then a rotating magnetic field equivalent to the three-phase winding can be generated. If someone is standing on this iron core and looking at it, the model of this motor is completely equivalent to a DC motor.

The equivalence of magnetomotive force also represents the equivalence of current. He is $\small i_{A }/i_{B }/i_{C },i_{a }/i_{b },i_{m }/i_{t }$ equivalent. The three of them can generate the same magnetomotive force. The most important task now is to find the accurate equivalent relationship between the above three sets of currents.

2.3 Three-phase static-two-phase static transformation-3/2 transformation

Physical basis: magnetomotive force of each phase = effective number of turns * current size

As shown in the figure above, for the sake of convenience, the $\small A$ phase and the $\small \alpha$ phase are superimposed to $\small ABC$ be the three-phase static magnetomotive force vector diagram and $\small \alpha \beta$ the two-phase static magnetomotive force vector diagram.

When the two sets of magnetomotive force are equal, $\small \alpha \beta$ the projections of the two sets of windings' instantaneous magnetomotive force on the axis are equal.

That is, the following relationship:

$\small N_{2}i_{\alpha }=N_{3}i_{A}-N_{3}i_{B}cos60^{\circ}-N_{3}i_{C}cos60^{\circ}=N_{3}\left ( i_{A}-\frac{1}{2} i_{B}-\frac{1}{2} i_{C}\right )$

$\small N_{2}i_{\beta }=N_{3}i_{B }sin60^{\circ}-N_{3}i_{C}sin60^{\circ}=\frac{\sqrt{3}}{2}N_{3}\left ( i_{B}-i_{C}\right )$

It is proved in Appendix 4 of Chen Boshi's book that when the power is unchanged before and after conversion, the turns ratio of three-phase and two-phase is:

$\small \frac{N_{3}}{N_{2}}=\sqrt{\frac{2}{3}}$

Combining the above two formulas, the transformation matrix can be obtained as:

$\small \begin{bmatrix} i_{\alpha }\\ i_{\beta } \end{bmatrix}=\sqrt{\frac{2}{3}}\begin{bmatrix} 1& -\frac{1}{2}& -\frac{1}{2} \\ 0& \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}\begin{bmatrix} i_{A }\\ i_{B }\\ i_{C } \end{bmatrix}$

If the three-phase winding is Y-shaped connection without a zero line, then $\small i_{a}+i_{b}+i_{c}=0$ the transformation matrix can be obtained by substituting the above formula:

$\small \begin{bmatrix} i_{\alpha }\\ i_{\beta } \end{bmatrix}=\begin{bmatrix} \sqrt{\frac{3}{2}}&0 \\ \frac{1}{\sqrt{2}}& \sqrt{2}\end{bmatrix}\begin{bmatrix} i_{A }\\ i_{B }\end{bmatrix}$

2.4 Two-phase static-two-phase rotating transformation-2s/2r transformation

As shown in the figure above, it $\small \alpha \beta$ is a two-phase stationary coordinate system and $\small MT$ a two-phase rotating coordinate system;

$\small MT$ Coordinate system to the synchronous speed $\small \omega_{1}$ of rotation, and $\small i_{t}$ , and $\small i_{m}$ of constant length (Since approximately equal to the number of turns).

The $\small \alpha \beta$ coordinate system is stationary, $\small \alpha$ and $\small M$ the angle between the axis and the axis $\small \varphi$ changes with time.

From this, it can be deduced that if the magnetomotive force of the two is equivalent, $\small i_{t}$ and the projection $\small i_{m}$ on the $\small \alpha$ axis and the $\small \beta$ axis should be equivalent to $\small i_{a}$ and $\small i_{b}$ , then:

$\small \left\{\begin{matrix} i_{\alpha}=i_{m}cos\varphi -i_{t}sin\varphi \\ i_{\beta }=i_{m}sin\varphi +i_{t}cos\varphi \end{matrix}\right.$

Thus, the transformation matrix of two-phase rotating and two-phase stationary is:

$\small C_{2r/2s}=\begin{bmatrix} cos\varphi &-sin\varphi \\ sin\varphi & cos\varphi \end{bmatrix}$

By transforming the matrix or changing the positions of the two sides of the formula, the two-phase static and two-phase rotating coordinate system can be obtained as:

$\small C_{2s/2r}=\begin{bmatrix} cos\varphi &sin\varphi \\ -sin\varphi & cos\varphi \end{bmatrix}$

Summary: The
permanent magnet synchronous motor system is a non-linear system. Using a mathematical model to simulate this system into a separately-excited DC motor model for control will greatly reduce the control difficulty, which is the core of the control strategy.

The core of coordinate transformation is that different coordinate systems produce the same magnetomotive force. Through the equivalent relationship between each coordinate system, the transformation matrix we need is calculated.

With the coordinate transformation and the simulated separately excited DC motor model, our next step is to design the current loop and speed loop.