Question 8.5: Using the techniques of Chapter 3 and neglecting saturation ......

In general, the zero-torque positions of the motor can be found by setting the torque expression to zero and solving for the resultant rotor position. Thus setting

T_{mech} = T_0 (i_1 \sin θ_m  –  i_2 \cos θ_m)=0

gives

i_1 \sin θ_m  –  i_2 \cos θ_m=0

or

θ_m=\tan ^{-1} \left( \frac{i_2}{i_1} \right)

Note that not all of these zero-torque positions correspond to stable equilibrium positions. For example, operation with i_1= I_0  and  i_2 = 0 gives two zero-torque positions: θ_m = 0°  and  θ_m = 180°. Yet only the position θ_m = 0° is stable. This is directly analogous to the case of a hanging pendulum which sees zero torque both when it is hanging downward (θ = 0°) and when it is sitting inverted (θ = 180°). Yet, it is clear that the slightest perturbation of the position of the inverted pendulum will cause it to rotate downwards and that it will eventually come to rest in the stable hanging position.

Stable rest positions of the rotor are determined by the requirement that a restoring torque is produced as the rotor moves from that position. Thus, a negative torque should result if the rotor moves in the +θ_m direction, and a positive torque should result for motion in the -θ_m direction. Mathematically, this can be expressed as an additional constraint on the torque at the rest position

\left.\frac{∂T_{mech}}{∂θ_m}\right|_{i_1,i_2}  < 0

where the partial derivative is evaluated at the zero-torque position and is taken with the phase currents held constant. Thus, in this case, the rest position must satisfy the additional constraint that

\left.\frac{∂T_{mech}}{∂θ_m}\right|_{i_1,i_2}= – T_0 (i_1 \cos θ_m + i_2 \sin θ_m) < 0

From this equation, we see for example that with i_1 = I_0  and  i_2 = 0, at  θ_m = 0°, ∂T_{mech}/∂θ_m < 0 and thus θ_m = 0° is a stable rest position. Similarly, at θ_m = 180°, ∂T_{mech}/∂θ_m > 0  \text{and thus}  θ_m = 180° is not a stable rest position.

Using these relationships, Table 8.1 lists the stable rest positions of the rotor for the various combinations of phase currents.

From this table we see that this drive results in a step size of 45°.