22-Elec-B8 Power Electronics and Drives · December 2014

Question 1 of 6: A.C. Voltage Controller Feeding an Inductive Motor Load

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: A.C. Voltage Controller Feeding an Inductive Motor Load (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Effect of load inductance on the controller output

With a purely resistive load the load current is in phase with the supply voltage, so each thyristor extinguishes naturally at the supply zero crossing. Conduction therefore runs from the firing instant $\alpha$ to $180^\circ$, the conduction angle is simply $\gamma = 180^\circ - \alpha$, and the rms output voltage falls monotonically from the full supply value at $\alpha = 0$ to zero at $\alpha = 180^\circ$. The whole $0$–$180^\circ$ range is usable and the output is a simple, closed-form function of the firing angle alone.

Adding series inductance changes this in four connected ways. First, the current lags the voltage, and the energy stored in the inductance keeps the device conducting after the supply voltage has reversed. Extinction now occurs at an angle $\beta > 180^\circ$ fixed not by the supply but by the load, through the transcendental condition $\sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-(\beta-\alpha)/\tan\phi}$, where $\phi = \tan^{-1}(\omega L/R)$. Because conduction extends past the zero crossing into the region where the supply voltage opposes the current, the load sees a segment of negative voltage each half cycle; nevertheless, for a given firing angle the rms output voltage is higher than it would be with a resistive load of the same rms current, because the conduction window is longer.

Second, and most important in practice, the usable control range shrinks. For any firing angle $\alpha \le \phi$ the current never returns to zero before the anti-parallel device is fired: conduction becomes continuous, $\beta = \alpha + 180^\circ$, and the controller degenerates into a closed switch delivering the full sinusoidal supply. Control is only obtained over $\phi < \alpha < 180^\circ$, so a highly inductive load such as an induction motor at light load leaves very little voltage range to work with. Third, the gating requirement changes: a narrow gate pulse fired at $\alpha$ may arrive while the opposite device is still conducting and therefore while the incoming device is still reverse-biased, so a pulse train or a wide gate pulse extending to the actual current zero is required for reliable turn-on. Fourth, the harmonic spectrum changes: the inductance smooths the current, the notch in the current waveform is less abrupt than the notch in the voltage waveform, and the total harmonic distortion of the current is markedly lower than that of the voltage at the same firing angle. Two consequences follow for the designer — the rms output voltage can no longer be read from a single $\alpha$-only curve but needs a family of curves indexed by $\phi$, and the device rms and average current ratings must be taken from the true conduction interval rather than from a half-sine assumption.

Parts (b), (c) and (d) — Verification and quantities

Given. A single-phase full-wave a.c. voltage controller supplies a 5-hp motor from a 120 V (rms), 60 Hz source; the firing angle is set to $66^\circ$ and the resulting conduction angle is measured as $150^\circ$, and the motor efficiency is 0.92.

Find. Confirm that a load angle of $36.87^\circ$ is consistent with the stated firing and conduction angles, then obtain the rms output voltage of the controller and the average current carried by each thyristor.

Approach. Use the conduction angle to locate the extinction angle, invert the RL extinction condition to recover the load angle, integrate $v_s^2$ over the true conduction window for the rms voltage, and obtain the device average current from the motor power balance together with the dimensionless shape factor of the conduction waveform.

1. Part (b) — locate the extinction angle. The conduction angle is measured from firing to extinction, so $$\beta = \alpha + \gamma = 66^\circ + 150^\circ = \boxed{216^\circ}$$ Conduction persists $36^\circ$ beyond the supply zero crossing, which already signals a substantially inductive load.
2. Write the extinction condition for an RL load. Solving $L\,di/dt + Ri = V_m\sin\omega t$ from $i(\alpha) = 0$ gives $$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{(\alpha - \omega t)/\tan\phi}\right]$$ and setting $i(\beta) = 0$ removes $V_m/Z$ entirely, leaving the load-angle condition $$\sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-\gamma/\tan\phi}$$ The impedance magnitude has cancelled, which is exactly why the question can ask about the power factor without ever quoting $R$ or $L$.
3. Read the condition approximately first. At a power factor near 0.8, $\tan\phi \approx 0.75$ and $\gamma = 2.618\ \text{rad}$, so the exponential is $e^{-2.618/0.75} = 0.0305$ — small. The right-hand side is therefore nearly zero, which forces $\beta \approx 180^\circ + \phi$, i.e. $\phi \approx \alpha + \gamma - 180^\circ = 36^\circ$ and $\cos\phi \approx 0.809$. The stated value of 0.8 is already confirmed to within one per cent by a single line of algebra.
4. Now solve the condition exactly. Substituting $\alpha = 66^\circ$ and $\beta = 216^\circ$ and solving $\sin(216^\circ - \phi) - \sin(66^\circ - \phi)e^{-2.618/\tan\phi} = 0$ numerically gives $$\phi = 36.85^\circ, \qquad \cos\phi = \boxed{0.800}$$ The load power factor is therefore 0.8 as claimed; the $0.02^\circ$ gap from the quoted $36.87^\circ$ is rounding in the paper's own data, not an inconsistency.
5. Part (c) — integrate over the real conduction window. The output voltage equals the supply during $\alpha \le \omega t \le \beta$ of each half cycle and is zero elsewhere, so $$V_o = \sqrt{\frac{1}{\pi}\int_{\alpha}^{\beta} \left(V_m\sin\omega t\right)^2 d(\omega t)} = V_s\sqrt{\frac{1}{\pi}\left[\gamma - \frac{\sin 2\beta - \sin 2\alpha}{2}\right]}$$ Note the integral runs to $\beta$, not to $180^\circ$; using the resistive-load limit here is the single most common error in this question.
6. Substitute. With $\sin 2\beta = \sin 432^\circ = 0.9511$, $\sin 2\alpha = \sin 132^\circ = 0.7431$ and $\gamma = 2.6180\ \text{rad}$, $$V_o = 120\sqrt{\frac{2.6180 - 0.1040}{\pi}} = 120 \times 0.8946 = \boxed{107.35\ \text{V}}$$ which is 89.5 per cent of the supply — a modest reduction, consistent with a firing angle only $29^\circ$ above the load angle.
7. Part (d) — get the load rms current from the power balance. The motor delivers $5 \times 745.7 = 3728.5$ W, so it absorbs $P_{in} = 3728.5/0.92 = 4052.7$ W and the apparent power at the controller output is $S = P_{in}/\cos\phi = 4052.7/0.8 = 5065.9$ VA. Hence $$I_{rms} = \frac{S}{V_o} = \frac{5065.9}{107.35} = 47.19\ \text{A}$$
8. Convert rms load current to average device current with the waveform shape factor. Each thyristor conducts once per supply cycle, over $\alpha \le \omega t \le \beta$. Writing the normalised conduction waveform $u(\omega t) = \sin(\omega t - \phi) - \sin(\alpha - \phi)e^{(\alpha-\omega t)/\tan\phi}$, the unknown $V_m/Z$ divides out of the ratio $$k = \frac{\dfrac{1}{2\pi}\displaystyle\int_{\alpha}^{\beta} u\,d(\omega t)}{\sqrt{\dfrac{1}{\pi}\displaystyle\int_{\alpha}^{\beta} u^2\, d(\omega t)}} = 0.4136$$ so the load impedance never has to be known. Therefore $$I_{T(avg)} = k\,I_{rms} = 0.4136 \times 47.19 = \boxed{19.52\ \text{A}}$$
9. Sanity-check the device rating. If the conduction pulse were treated as a clean half sine of the same rms value, the average would be $\sqrt{2}\,I_{rms}/\pi = 21.24$ A. That crude estimate runs 8.8 per cent high because the real pulse is skewed by the exponential term, so it is a safe — conservative — number for selecting a device, but the 19.52 A figure is the one to quote.