Question 8.1: A 4/2 VRM is shown in Fig. 8.3. Its dimensions are R = 3.8 ......

a. Using the magnetic circuit techniques of Chapter 1, we see that the maximum inductance L_{max} for phase 1 occurs when the rotor axis is aligned with the phase-1 magnetic axis.

From Eq. 1.31, we see that L_{max} is equal to

L= \frac{N^2}{(\mathrm{g}/μ_0 A_{\mathrm{g}})}=\frac{N^2 μ_0A_{\mathrm{g}}}{\mathrm{g}}\quad \quad \quad (1.31)

L_{max}= \frac{N^2 μ_0 α R D}{2 \mathrm{g}}

where α R D is the cross-sectional area of the air gap and 2g is the total gap length in the magnetic circuit. For the values given,

\begin{aligned}L_{max} & =\frac{N^2 μ_0 α R D}{2 \mathrm{g}}\\ &= \frac{(100)^2(4π × 10^{-7})(π/3)(3.8 × 10^{-2})(0.13)}{2 × (2.54 × 10 ^{-4})}\\& = 0.128 ~H\end{aligned}

Neglecting fringing, the inductance L(θ_m) will vary linearly with the air-gap cross-sectional area as shown in Fig. 8.4a. Note that this idealization predicts that the inductance is zero when there is no overlap when in fact there will be some small value of inductance, as shown in Fig. 8.2.

b. From Eq. 8.7, the torque consists of two terms

T_{mech}=\frac{1}{2} i_1^2 \frac{dL_{11} (θ_m)}{dθ_m}+ \frac{1}{2}i_2^2 \frac{dL_{11} (θ_m  –  90°)}{dθ_m}

and dL_{11}/dθ_m can be seen to be the stepped waveform of Fig. 8.4b whose maximum values are given by ±L_{max}/α (with α expressed in radians!). Thus the torque is as shown in Fig. 8.4c.

c. The peak torque due to each of the windings is given by

T_{max}= \left(\frac{L_{max}}{2α}\right) i^2= \left( \frac{0.128}{2(π/3)}\right) 5^2=1.53 ~N·m

(i) From the plot in Fig. 8.4c, at θ_m = 0°, the torque contribution from phase 2 is clearly zero. Although the phase-1 contribution appears to be indeterminate, in an actual machine the torque change from T_{max_1}  \text{to}  -T_{max_1}  \text{at}  θ_m = 0° would have a finite slope and the torque would be zero at θ = 0°. Thus the net torque from phases 1 and 2 at this position is zero.

Notice that the torque at θ_m = 0 is zero independent of the current levels in phases 1 and 2. This is a problem with the 4/2 configuration of Fig. 8.3 since the rotor can get “stuck” at this position (as well as at θ_m = ±90 °, ±180°), and there is no way that electrical torque can be produced to move it.

(ii) At θ_m = 45° both phases are providing torque. That of phase 1 is negative while that of phase 2 is positive. Because the phase currents are equal, the torques are thus equal and opposite and the net torque is zero. However, unlike the case of θ_m = 0°, the torque at this point can be made either positive or negative simply by appropriate selection of the phase currents.

(iii) At θ_m = 75° phase 1 produces no torque while phase 2 produces a positive torque of magnitude T_{max_2}. Thus the net torque at this position is positive and of magnitude 1.53 N · m. Notice that there is no combination of phase currents that will produce a negative torque at this position since the phase-1 torque is always zero while that of phase 2 can be only positive (or zero).