Question 8.3: Consider the idealized 4/2 VRM of Example 8.1. Assume that i......

a. From Eq. 8.15, the differential equation governing the current buildup in phase 1 is given by

v_{j}=\left[R_j+\frac{d L_{jj}\left(\theta_{\mathrm{m}}\right)}{d \theta_{\mathrm{m}}} \frac{d \theta_{\mathrm{m}}}{d t}\right] i_{j}+L_{jj}\left(\theta_{\mathrm{m}}\right) \frac{d i_{j}}{d t} \quad \quad \quad (8.15)

v_{1}=\left[R+\frac{d L_{11}\left(\theta_{\mathrm{m}}\right)}{d \theta_{\mathrm{m}}} \frac{d \theta_{\mathrm{m}}}{d t}\right] i_{1}+L_{11}\left(\theta_{\mathrm{m}}\right) \frac{d i_{1}}{d t}

At 4000 r/min,

\omega_{\mathrm{m}}=\frac{d \theta_{\mathrm{m}}}{d t}=4000 ~ \mathrm{r} / \mathrm{min} \times \frac{\pi}{30}\left[\frac{\mathrm{rad} / \mathrm{sec}}{\mathrm{r} / \mathrm{min}}\right]=\frac{400 \pi}{3} ~\mathrm{rad} / \mathrm{sec}

From Fig. 8.4 (for -60^{\circ} \leq \theta_{\mathrm{m}} \leq 0^{\circ})

\begin{aligned}L_{11}\left(\theta_{\mathrm{m}}\right) & =L_{l}+\frac{L_{\max }}{\pi / 3}\left(\theta_{\mathrm{m}}+\frac{\pi}{3}\right) \\&=0.005+0.122\left(\theta_{\mathrm{m}}+\pi / 3\right)\end{aligned}

Thus

\frac{dL_{11}\left(\theta_{\mathrm{m}}\right)}{d \theta_{\mathrm{m}}}=0.122 ~\mathrm{H} / \mathrm{rad}

and

\frac{dL_{11}\left(\theta_{\mathrm{m}}\right)}{d \theta_{\mathrm{m}}} \frac{d \theta_{\mathrm{m}}}{d t}=51.1 ~\Omega

which is much greater than the resistance R = 1.5 Ω

This will enable us to obtain an approximate solution for the current by neglecting the Ri term in Eq. 8.13. We must then solve

v_j=R_j i_j + \frac{d}{dt}[L_{jj}(θ_m) i_j ] \quad \quad \quad (8.13)

\frac{d\left(L_{11} i_{1}\right)}{d t}=v_{1}

for which the solution is

i_{1}(t)=\frac{\int_{0}^{t} v_{1} d t}{L_{11}(t)}=\frac{V_{0} t}{L_{11}(t)}

Substituting

\theta_{\mathrm{m}}=-\frac{\pi}{3}+\omega_{\mathrm{m}} t

into the expression for L_{11}\left(\theta_{\mathrm{m}}\right) then gives

i_{1}(t)=\frac{100 t}{0.005+51.1 t} \mathrm{~A}

which is valid until \theta_{\mathrm{m}}=0^{\circ} at t = 2.5 msec, at which point i_{1}(t)=1.88 \mathrm{~A}.

b. During the period of current decay the solution proceeds as in part (a). From Fig. 8.4, for 0^{\circ} \leq \theta_{\mathrm{m}} \leq 60^{\circ}, d L_{11}\left(\theta_{\mathrm{m}}\right) / d t=-51.1~ \Omega and the Ri term can again be ignored in Eq. 8.13.

Thus, since the applied voltage is -200 V for this time period (t ≥ 2.5 msec until \left.i_{1}(t)=0\right) in an effort to bring the current rapidly to zero, since the current must be continuous at timet_0=2.5 ~msec, and since, from Fig. 8.4 (for 0^{\circ} \leq \theta_{\mathrm{m}} \leq 60^{\circ})

\begin{aligned}L_{11}\left(\theta_{\mathrm{m}}\right) & =L_{l}+\frac{L_{\max }}{\pi / 3}\left(\frac{\pi}{3}-\theta_{\mathrm{m}}\right) \\& =0.005+0.122\left(\pi / 3-\theta_{\mathrm{m}}\right)\end{aligned}

we see that the solution becomes

\begin{aligned}i_{1}(t) & =\frac{L_{11}\left(t_{0}\right) i_{1}\left(t_{0}\right)+\int_{t_{0}}^{t} v_{1} d t}{L_{11}(t)} \\& =\frac{0.25-200\left(t-2.5 \times 10^{-3}\right)}{0.005+51.1\left(5 \times 10^{-3}-t\right)}\end{aligned}

From this equation, we see that the current reaches zero at t = 3.75 msec.

c. The torque can be found from Eq. 8.9 by setting i_{2}=0. Thus

\begin{aligned}T_{\mathrm{mech}} &=\frac{1}{2} i_{1}^{2} \frac{d L_{11}(θ_m)}{d \theta_{\mathrm{m}}}+\frac{1}{2} i_{2}^{2} \frac{d L_{22}(θ_m)}{d \theta_{\mathrm{m}}}\\&=\frac{1}{2} i_{1}^{2} \frac{d L_{11}(θ_m)}{d \theta_{\mathrm{m}}}+\frac{1}{2} i_{2}^{2} \frac{d L_{11}(θ_m  –  90°)}{d \theta_{\mathrm{m}}} \quad (8.9)\end{aligned}

T_{\mathrm{mech}}=\frac{1}{2} i_{1}^{2} \frac{d L_{11}}{d \theta_{\mathrm{m}}}

Using MATLAB and the results of parts (a) and (b), the current waveform is plotted in Fig. 8.11a and the torque in Fig. 8.11b. The integral under the torque curve is 3.35 × 10^{-4} ~N · m · sec while that under the positive portion of the torque curve corresponding to positive torque is 4.56 × 10^{-4} ~N· m · sec. Thus we see that the negative torque produces a 27 percent reduction in average torque from that which would otherwise be available if the current could be reduced instantaneously to zero.

Notice first from the results of part (b) and from Fig. 8.11a that, in spite of applying a negative voltage of twice the magnitude of the voltage used to build up the current, current continues to flow in the winding for 1.25 ms after reversal of the applied voltage. From Fig. 8.11b, we see that the result is a significant period of negative torque production. In practice, this may, for example, dictate a control scheme which reverses the phase current in advance of the time that the sign of dL(θ_m)/dθ_m reverses, achieving a larger average torque by trading off some reduction in average positive torque against a larger decrease in average negative torque.

This example also illustrates another important aspect of VRM operation. For a system of resistance of 1.5 Ω and constant inductance, one would expect a steady-state current of 100/1.5 = 66.7 A. Yet in this system the steady-state current is less than 2 A. The reason for this is evident from Eqs. 8.14

v_{j}=\{R_j+\frac{d}{dt} [L_{jj}\left(\theta_{\mathrm{m}}\right)]\}i_{j}+L_{jj}\left(\theta_{\mathrm{m}}\right) \frac{d i_{j}}{d t} \quad \quad \quad (8.14)

and 8.15 where we see that dL_{11}(θ_m)/dt = 51.1  Ω appears as an apparent resistance in series with the winding resistance which is much larger than the winding resistance itself. The corresponding voltage drop (the speed voltage) is of sufficient magnitude to limit the steady-state current to a value of 100/51.1 = 1.96 A.

Here is the MATLAB script: