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

Question 1: A.C. Voltage Controller Feeding an Induction Motor (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.

Given. A single-phase back-to-back (full-wave) thyristor controller fed from a 220 V, 60 Hz supply drives a 20 hp induction motor whose fundamental power-factor angle is $\phi = 20^\circ$ lagging, and the measured conduction angle of each device is $\gamma = 160^\circ$.

Find. The qualitative effect of load inductance on the controller output; then the delay angle $\alpha$ consistent with $\gamma = 160^\circ$, the rms output voltage, and the average current carried by each thyristor.

Approach. Write the load current of the R–L branch during conduction, impose $i(\beta)=0$ to link $\alpha$ and $\gamma$, integrate the chopped sinusoid for the rms output, and obtain the device average current from the motor power balance combined with the normalised current waveform — a route in which the (unstated) load impedance cancels.

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

With a purely resistive load the current is in phase with the voltage, so it extinguishes exactly at the supply zero. The extinction angle is then fixed at $\beta = 180^\circ$, the conduction angle is simply $\gamma = 180^\circ - \alpha$, and the rms output falls smoothly from the full supply value at $\alpha = 0$ to zero at $\alpha = 180^\circ$.

Inductance changes this completely, because inductance stores energy. The current cannot be extinguished while the winding still holds flux, so conduction continues after the supply voltage has reversed, until the volt–seconds returned to the source equal those absorbed. Four consequences follow, and all of them matter when a controller is sized for a motor rather than for a heater.

First, the conduction window extends past the supply zero, so the load terminals see a negative-going tail of voltage. For a given delay angle the rms output is therefore higher than the resistive-load value, and the negative tail subtracts from the useful volt–seconds while adding to the heating. Second, the low end of the control range disappears: for any $\alpha \le \phi$ the current from the previous half cycle has not yet reached zero when the opposite device is fired, the two thyristors conduct continuously, and the output equals the supply. Control exists only over $\phi \le \alpha \le 180^\circ$, so an inductive load with $\phi = 20^\circ$ throws away the first twenty degrees of the range.

Third, the control law becomes nonlinear. The conduction angle is no longer $180^\circ - \alpha$ but the root of a transcendental extinction equation, so equal increments of firing angle no longer give equal increments of output voltage — a closed-loop controller must either linearise or work from measured feedback. Fourth, gating becomes harder: at the firing instant the device current is zero, so a single narrow gate pulse may not raise the anode current to the latching value before the pulse ends. A pulse train, a wide pulse, or a pulse transformer with a long volt–second capability is required.

The same physics is what makes the exam's next three sub-parts non-trivial: because $\beta > 180^\circ$, neither $\alpha$ nor the rms output can be read off a resistive-load formula.

Parts (b) to (d) — Quantitative solution

1. Part (b) — write the conduction-interval current and impose extinction. For $\alpha \le \omega t \le \beta$ the series R–L branch obeys $L\,di/dt + Ri = V_m \sin \omega t$, whose solution with $i(\alpha)=0$ is $$i(\omega t)=\frac{V_m}{Z}\left[\sin(\omega t-\phi)-\sin(\alpha-\phi)\, e^{(\alpha-\omega t)/\tan\phi}\right],\qquad Z=\sqrt{R^2+(\omega L)^2}.$$ Setting $i(\beta)=0$ with $\beta = \alpha+\gamma$ gives the extinction condition $$\sin(\alpha+\gamma-\phi)=\sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}.$$ The load impedance $Z$ divides out, which is why the problem is solvable without it.
2. Evaluate the damping factor. With $\phi = 20^\circ$, $\tan\phi = 0.36397$, and $\gamma = 160^\circ = 2.79253$ rad, so $$e^{-\gamma/\tan\phi}=e^{-2.79253/0.36397}=e^{-7.6724}=4.664\times10^{-4}.$$ The right-hand side is three orders of magnitude smaller than the left, so the extinction condition is very nearly $\sin(\alpha+\gamma-\phi)=0$.
3. Solve for the delay angle. Solving the full transcendental equation numerically over $21^\circ \le \alpha \le 89^\circ$ returns $$\boxed{\alpha = 39.99^\circ \approx 40^\circ},\qquad \beta=\alpha+\gamma=199.99^\circ .$$ The exam's stated value is therefore verified. The closed-form estimate obtained by dropping the exponential, $\alpha \approx 180^\circ+\phi-\gamma = 40.00^\circ$, is accurate to $0.01^\circ$ here precisely because the damping factor is so small; at poorer power factors it is not, and it must be treated as a sanity check rather than an answer.
4. Part (c) — integrate the chopped sinusoid for the rms output. Each half cycle contributes an identical conduction window, so $$V_o^2=\frac{1}{\pi}\int_{\alpha}^{\beta}2V_s^2\sin^2\theta\,d\theta =\frac{V_s^2}{\pi}\left[(\beta-\alpha)-\frac{\sin 2\beta-\sin 2\alpha}{2}\right].$$ Substituting $\beta-\alpha = 2.79253$ rad, $\sin 2\alpha = \sin 79.98^\circ = 0.98477$ and $\sin 2\beta = \sin 399.98^\circ = 0.64247$: $$V_o=220\sqrt{\frac{2.79253+0.17115}{\pi}}=220\sqrt{0.94338} =\boxed{213.7\ \text{V (rms)}}.$$ That is 97.1 per cent of the supply — a large conduction angle removes very little voltage, which is the practical reason a controller of this kind is a starter rather than a speed control.
5. Part (d) — obtain the load rms current from the motor power balance. The mechanical output is $P_{out}=20\times 745.7 = 14\,914$ W, so the electrical input is $$P_{in}=\frac{P_{out}}{\eta}=\frac{14\,914}{0.87}=17\,142.5\ \text{W}.$$ Working at the fundamental power factor, the apparent power is $S = P_{in}/\cos\phi = 17\,142.5/0.93969 = 18\,242\ \text{VA}$ and hence $$I_{rms}=\frac{S}{V_o}=\frac{18\,242}{213.68}=85.37\ \text{A}.$$
6. Convert rms current to device average current through the normalised waveform. Write the conduction-interval shape as $u(\theta)=\sin(\theta-\phi)-\sin(\alpha-\phi)e^{(\alpha-\theta)/\tan\phi}$, so that $i = (V_m/Z)\,u(\theta)$. Each thyristor of the back-to-back pair conducts once per cycle, so its average is taken over $2\pi$ while the rms of the load current is taken over $\pi$: $$\frac{I_{T,avg}}{I_{rms}} =\frac{\dfrac{1}{2\pi}\displaystyle\int_{\alpha}^{\beta}u\,d\theta} {\sqrt{\dfrac{1}{\pi}\displaystyle\int_{\alpha}^{\beta}u^2\,d\theta}}=0.42856 .$$ The factor $V_m/Z$ cancels identically, so the unknown motor impedance never enters.
7. Evaluate the thyristor average current. $$I_{T,avg}=0.42856\times 85.37=\boxed{36.6\ \text{A}}.$$ As a device-selection cross-check, idealising the conduction interval as a half sine of the same rms value gives $\sqrt{2}\,I_{rms}/\pi = 38.4$ A, five per cent high. The shortcut is conservative, which is the direction one wants when choosing an SCR, but the integrated value is the one to quote.

Note that the ratio of total rms quantities is not the fundamental impedance: here $V_o/I_{rms} = 2.503\ \Omega$ against an implied fundamental $|Z| = 2.457\ \Omega$, a 1.9 per cent gap. The chopped waveform's harmonics see $nX$ rather than $X$, so the load draws proportionally less harmonic current than harmonic voltage.