Question 5.14.1: Derive an expression for the drive torque of the stepper mot...

The torque is derived in two steps. First, we obtain a field solution in the interior of an energized stator with no rotor present. Second, we reduce the rotor to an equivalent current distribution and then compute the torque on it using a Lorentz force approach with the stator field as an external field.

Stator field: Let L and R_{s} denote the length and inner radius of the stator, respectively. Each stator has N_{pole} teeth,  which equals the number of poles on the rotor. The stator is enclosed by an n-turn coil and a flux return ring. A cross section of the interior of a four-pole stator is shown in Fig. 5.49a. Let Φ_{t} denote the angular span of each tooth and Φ_{g} denote the angular expanse of the gap between each tooth. These angles are related by the following equation,

Φ_{g} = \frac{2π}{N_{pole}} -Φ_{t} .          (5.281)

We seek a field solution for the interior of the stator. To obtain an analytical solution, we assume that the stator is infinitely long. This assumption is not as radical as it might first seem because the stator tends to confine the flux and minimize leakage etc. In the interior of the stator,

∇ × H = 0.

Thus, we can represent H in terms of a scalar potential, H = – ∇ φ_{m } (Section 3.4). The field problem reduces to a boundary value problem (BVP) in cylindrical coordinates (Section 3.6.2). Specifically, we solve

∇^{2} φ_{m } = 0.            (5.282)

The general solution to Eq. (5.282) that is well behaved at r = 0 is of the form

φ_{m }(r, Φ) = \sum\limits_{p=1}^{∞} a_{p} r^{p} sin(pΦ) + b_{p} r^{p} cos(pΦ).                (5.283)

We choose a coordinate system in which Φ = 0 coincides with the center of a stator tooth (Fig. 5.51a). In this coordinate system φ_{m }(r, Φ) must be an even function of Φ. Thus, we discard the sin(nΦ) terms in Eq. (5.283) and obtain

φ_{m }(r, Φ) = \sum\limits_{p=1}^{∞}b_{p} r^{p} cos(pΦ).                (5.284)

We determine the coefficients b_{p} by matching the boundary conditions at the interior surface of the stator. We assume that the potential φ_{m}(R_{s}, Φ) along this surface is constant across each tooth and alternates from φ_{max} to –φ_{max} on successive teeth. As the gap between neighboring teeth is small (Φ_{g}<<Φ_{t}), we assume that the potential is linear between the teeth and this implies a profile for φ_{m }(R_{s}, Φ) as shown in Fig. 5.51b. To solve the BVP we need a Fourier series representation for φ_{m }(R_{s}, Φ). First, we derive an expression for φ_{max} in terms of the motor’s physical parameters. Let g denote the gap between adjacent stator teeth, and let H_{g} denote the field in the gap region (Figs. 5.49b and 5.50). We assume that the potential is linear between adjacent teeth. Therefore,

H_{g} = – \frac{1}{R_{s}} \frac{Δφ_{m }}{ΔΦ}               (5.285)

= – \frac{2φ_{max}}{g},              (5.286)

where ΔΦ = Φ_{g} and g = R_{s}Φ_{g}. We apply Eq. (3.138) from Section 3.5 to the dotted path in Fig. 5.50 and obtain

Eq. (3.138): \oint_{c}H ⋅dl = I_{tot},

H_{g} = \frac{ni}{g}.                (5.287)

Therefore,

φ_{max} =- \frac{ni}{2}.                (5.288)

Now that we know φ_{max}, we can represent the potential along the stator φ_{m }(R_{s}, Φ) in terms of the Fourier series

φ_{m }(R_{s}, Φ) = \sum\limits_{k=1,3,5,…}^{∞}A_{k}cos(α_{k}Φ),                (5.289)

where

A_{k} = -\frac{8}{πk^{2} N_{pole}} \frac{φ_{max}}{Φ_{g}} [cos(kπ) – 1 ] cos(α_{k} \frac{Φ_{t}}{2})

= \frac{4}{πk^{2} N_{pole}} \frac{ni}{Φ_{g}} [cos(kπ) – 1 ] cos(α_{k} \frac{Φ_{t}}{2}),             (5.290)

and α_{k}  = kN_{pole}/2. We evaluate Eq. (5.284) at r = R_{s}, compare this with Eq. (5.289), and obtain

b_{p} = A_{k} R_{s}^{-p} \qquad (p  =  \frac{kN_{pole}}{2}).

Thus, the potential in the interior of the stator is

φ_{m }(r, Φ) = \sum\limits_{k=1,3,5,…}^{∞}A_{k} (\frac{r}{R_{s}})^{α_{k}}cos(α_{k}Φ).                (5.291)

The field components are obtained from B = – μ_{0}∇φ_{m}. For example,

B_{r}(r, Φ) = -μ_{0}\sum\limits_{k=1,3,5,…}^{∞}\frac{α_{k}}{R_{s}^{α_{k}}} A_{k} r^{(α_{k} -1)}  cos(α_{k}Φ).          (5.292)

Drive torque: To compute the torque we represent the rotor as a distribution of equivalent currents. Recall that the rotor is radially polarized with

M = ± M_{s}\hat{r} .

Let R_{1} and R_{2} denote its inner and outer radius, respectively. Assume that it is initially positioned with one of its north poles centered about 0° (i.e., each pole is initially centered with respect to a stator tooth). The equivalent current densities are given by Eq. (3.95). We find that the volume current density is zero, J_{m} = ∇ × M = 0, and that the surface current density is

Eq. (3.95):Equivalent Currents
J_{m} = ∇ × M (A/m^{2}) (volume current density)
j_{m} = M × \hat{n} (A/m) (surface current density).

j_{m} = M × \hat{n}

= ± M_{s}\hat{z} .

These surface currents are located at the edges of each sector at angular positions

Φ_{n} = (n + \frac{1}{2}) \frac{2π}{N_{pole}}          (n = 0, 1, 2, . . . , N_{pole} – 1).

It is important to note that there are two such currents at these positions, one along the surface of each of the neighboring sectors. If the rotor is rotated by an angle θ, the positions of the surface currents change to

Φ_{n}(θ) = θ +(n+\frac{1}{2}) \frac{2π}{N_{pole}}                    (n = 0, 1, 2, . . . , N_{pole} – 1).

The surface current densities at these locations are

j_{m}(n) = \left\{\begin{matrix} M_{s}\hat{z} \qquad (n = 0, 2, 4, . . .) \\ – M_{s}\hat{z} \qquad (n = 1, 3, 5, . . .).\end{matrix} \right.            (5.293)

The torque is given by Eq. (3.100) from Section 3.3, that is,

Eq. (3.100): T = \int_{v} r ×(J_{m} ×B_{ext}) dv + \oint_{s} r × (j_{m} × B_{ext}) ds ,

T(θ) = 2L  \sum\limits_{n=0}^{N_{pole} -1} \int_{R_{1}}^{R_{2}} r× [j_{m}(n) × B_{stator} (r , Φ_{n}(Φ))] dr.              (5.294)

Factor 2 takes into account the contribution of each surface of the neighboring sectors. From Eq. (5.294) we find that the component of torque about the z-axis is

T_{z}(θ) = 2LM_{s} \sum\limits_{n=0}^{N_{pole} -1} (-1)^{n} \int_{R_{1}}^{R_{2}} B_{r}(r , Φ_{n}(Φ)) r dr.        (5.295)

Finally, substituting Eq. (5.292) into Eq. (5.295) gives

T_{z}(θ) = 2 μ_{0}LM_{s} \sum\limits_{n=0}^{N_{pole} -1} \sum\limits_{k =1,3,5,…}^{∞} (-1)^{n} \\ × A_{k} ( \frac{α_{k}}{α_{k} +1} )\frac{(R_{2}^{(α_{k}+1)} – R_{1}^{(α_{k}+1)}}{R_{s}^{α_{k}}} cos[α_{k}Φ_{n}(θ)],             (5.296)

where L and R_{s} are the length and interior radius of the stator and

A_{k} = \frac{4}{πk^{2} N_{pole}} \frac{ni}{Φ_{g}} [cos(kπ) – 1 ] cos(α_{k} \frac{Φ_{t}}{2}) .

Here,

α_{k} = \frac{kN_{pole}}{2},

and

Φ_{n}(Φ)= θ + (n + \frac{1}{2})  \frac{2π}{N_{pole}}.

Calculations: We demonstrate Eq. (5.296) with some sample calculations. The following parameters are used in the analysis:

M_{s} = 4.3 × 10^{5} A/m
N_{pole} = 10
n = 100 turns
i = 0.25 A
R_{1} = 6.0 mm
R_{2} = 8.0 mm
R_{s} = 8.1 mm
L = 2.0 mm.

The angular span of each stator tooth is taken to be Φ_{t} = 27° (75% of the span of a rotor pole). The angular span of the gap between neighboring teeth is Φ_{g} = 9° ( Φ_{g} = 2π/ N_{pole} – Φ_{t}). There are two phases to consider, which are labeled 1 and 2, respectively. Recall that the stator teeth of the two phases are offset from one another by an angular span equal to one half of the pole pitch, which in this case is 18°. The drive torque T_{z} (θ) for each phase is shown in Fig. 5.52. This plot shows the torque on the rotor as it rotates the angular span of one pole (0 ≤ θ ≤ 36°). When  θ = 0, the rotor is positioned as in Fig. 5.49c and d for phase 1 and phase 2, respectively. In this initial position, phase 1 exerts almost maximum torque, and phase two exerts zero torque. The peak torque for phase 1 is 1500 μN ·  m, which occurs at θ = 12°. When the rotor is rotated to θ = 18° (half a pole pitch) as shown in Fig. 5.49d, the torque of phase 1 is zero and the torque of phase 2 is near a maximum. For 18 ≤ θ ≤ 36°, the torque of phase 1 reverses direction as shown.