8. Control of Synchronous Machine

1. AC Brushless, DC Brushless and Switch Reluctance Machine

BLDC, BLAC and SR Machines are all synchronous machines,

BLDC and BLAC are the machine within permanent magnet, SR machine does not have permanent magnets
BLDC is the PMSM that optimized the design for apply the DC controller
BLAC is the PMSM that optimized the design for apply the AC controller

1.1 The difference between BLAC and BLDC

BLAC	BLDC
distributed winding	concentrated winding
sinusoidal wave magnetic field	triangular wave magnetic field
sinusoidal wave EMF	square wave EMF

2. Surface Mounted PMSM

For BLAC machine the magnetic field have sinusoidal waveform, with a surface mounted structure, to obtain this magnetic field, the field strength of the magnetic is different in different position,

3. Internal Mounted PMSM

\[ \begin{aligned} \bar v_s &= R_s \bar i_s + p \bar \psi_s \\ \bar \psi_s &= L_s \bar i_s + \bar \psi_{pm} \end{aligned} \]

at steady state,

\[ \bar v_s = R_s \bar i_s + p(L_s \bar i_s + \bar \psi_{pm}) = R_s \bar i_s + j\omega L_s \bar i_s + j\omega \bar \psi_{pm} \]

\[ \begin{aligned} \psi_{sd} &= L_d i_{sd} +\psi_{pm} \\ \psi_{sq} &= L_q i_{sq} \\ L_d &\gg L_q \end{aligned} \]

4. FOC Control

The dynamics of synchronous machine is:

\[ \begin{aligned} \bar v_s &= R_s \bar i_s + p \bar \psi_s \\ \bar v_s &= R_s \bar i_s + p \bar \psi_s + j \dot \theta_s \bar \psi_s \\ &= R_s \bar i_s + p \bar \psi_s + j \dot \theta_m \bar \psi_s \\ \end{aligned} \]

For dq axis, there have:

\[ \begin{aligned} v_d + jv_q &= R_s(i_{sd} + ji_{sq}) + p(\psi_{sd} + j\psi_{sq}) + j\dot \theta_m(\psi_{sd} + j\psi{sq}) \\ \psi_{sd} &= L_{sd}i_{sd} + \psi_{pm} \\ \psi_{sq} &= L_{sq}i_{sq} \\ v_d + jv_q &= R_si_{sd} + jR_si_{sq} + p\psi_{sd} + jp\psi_{sq} + j\dot \theta_m(L_{sd}i_{sd} + \psi_{pm} + jL_{sq}i_sq) \\ \hfill \\ \Rightarrow&\left\{\begin{aligned} v_{sd} &= R_si_{sd} + L_{sd}pi_{sd} - \dot \theta_m L_{sq}i_{sq} \\ v_{sq} &= R_si_{sq} + L_{sq}pi_{sq} + \dot \theta_m L_{sd}i_{sd} + \psi_{pm}\dot \theta_m \end{aligned}\right. \end{aligned} \]

\[ \Re(\bar v_s i_s) = \Re(R_s \bar i_s \bar i_s^*) + \Re(p\psi_s \bar i_s^*) + \Re(j\dot \theta_m \bar \psi_s \bar i_s^*) \]

\[ \begin{aligned} P_m &= -\dot \theta_m \Im(\bar i_s \bar i_s^*) \\ &= -\dot \theta_m \Im(L_{sd}i_{sd} + \psi_{pm} + jL_{sq}i_{sq})(i_{sd} - ji_{sq}) \\ &= -\dot \theta_m \Im(L_{sd}i_{sd}^{2} + \psi_{pm}i_{sd} + j(L_{sq}i_{sq}i_{sd} - \psi_{pm}i_{sq} + L_{sq}i_{sq}^{2} - L_{sd}i_{sd}i_{sq})) \\ &= -\dot \theta_m \Im(L_{sd}i_{sd}^2 + L_{sd}i_{sq}^2 + \psi_{pm}i_{sd} + j(-\psi_{pm}i_{sq} + (L_{sq} - L_{sd})i_{sd}i_{sq})) \end{aligned} \]

\[ \begin{aligned} v_{sd} + jv_{sq} &= j \dot \theta_m (L_{sd}i_{sd} + \psi_{pm} + j L_{sq}i_{sq}) \\ &= j \dot \theta_m L_{sd} i_{sd} + j \dot \theta_m \psi_{pm} - \dot \theta_m L_{sq} i_{sq} \\ v_{sd} &= - \dot \theta_m L_{sq} i_{sq} \\ v_{sq} &= \dot \theta_m L_{sd}i_{sd} + \dot \theta_m \psi_{pm} \end{aligned} \]

We consider \(L_{sd} = L_{sq}\), and we let \(i_{sd} = 0\)

Warning

Not all PMSM can apply the field weakening control, if the hysteresis is nonlinear, after many times of magnetizing and demagnetizing, the magnetics will loss its magnetic field.

5. BLDC Machine Control

5.1 2-Phase Working Condition

\[ \begin{aligned} \psi &= \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} B(\theta)Rld\theta = BRl\int_{\frac{-\pi}{2} + 2\theta_m}^{\frac{\pi}{2}}d\theta \\ &= BRl(\pi - 2\theta_m) \end{aligned} \]

The flux have a triangular shape, and because \(e = p\psi\), the EMF have a square wave shape. And because the flux cannot have a perfect triangular wave, we approximate the EMF with trapezoidal wave. When designing the BLDC machine, it is necessary to make sure there have \(120^\circ\) constant part.

The power of the BLDC have:

\[ \begin{aligned} P &= e_a i_a + e_b i_b + e_c i_c \\ &= 2K_e \Omega_m I_c \end{aligned} \]

And the generated torque have:

\[ T = \frac{P}{\Omega_m} = 2K_e I_c \]

So we only need 3 hall sensor to identify the position of the rotor

5.2 3-Phase Working Condition

\[ \begin{aligned} &\begin{aligned} &V_{DC} = R_s i_b + L_s pi_b + e_b - e_a -L_s pi_a - R_s i_a \\ &0 = R_s i_a + L_s pi_a + e_a - e_c - L_s pi_c -R_s i_c \\ &i_a + i_b + i_c = 0 \end{aligned} \\ \Rightarrow 0 &= R_s i_a + L_s pi_a + e_a - e_c - L_s p(-i_a -i_b) - R_s(-i_a -i_b) \\ L_s pi_b + R_s i_b &= -2R_s i_{sa} - 2L_s i_{sa} + e_c - e_a \\ V_{DC} &= -2R_s i_{sa} - 2L_s i_{sa} + e_c - e_a + e_b - e_a - L_s pi_a - R_s i_a\\ V_{DC} &= -3R_s i_{sa} - 3L_s i_{sa} - 2k\Omega_m \\ R_s i_{sa} + L_s i_{sa} &= \frac{V_{DC} + 2k\Omega_m}{3} \\ R_s i_{sb} + L_s i_{sb} &= \frac{2V_{DC} - 2k\Omega_m}{3} \\ \end{aligned} \]

When phase A and phase B have the same speed of transient,

\[ -\frac{V_{DC} + 2k\Omega_m}{3} = \frac{2V_{DC} - 2k\Omega_m}{3} \]