Question 4 of 5: Single-phase a.c. voltage controller feeding a motor

Question 4: Single-phase a.c. voltage controller feeding a 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.

Find. (a) confirmation that the firing delay is $\alpha = 40^\circ$; (b) the rms output voltage; (c) the mean current in each thyristor and in the controller as a whole.

Approach. An inductive load holds current past the supply zero, so $\alpha$ and $\gamma$ are tied by the transcendental extinction condition; solve that for $\alpha$, integrate the chopped sine over the real conduction window for the rms output, then get the load current from the motor's power balance and the device current from the shape of the conduction pulse.

1. Part (a) — write the extinction condition. During conduction the load current obeys $$i(\theta) = \frac{V_m}{|Z|}\left[\sin(\theta - \phi) - \sin(\alpha - \phi)\,e^{(\alpha-\theta)/\tan\phi}\right]$$ Current extinguishes at $\theta = \beta = \alpha + \gamma$, so setting the bracket to zero gives the condition that links the two angles: $$\sin(\alpha + \gamma - \phi) = \sin(\alpha - \phi)\,e^{-\gamma/\tan\phi}$$ The load impedance $|Z|$ has cancelled: the conduction window depends on the load angle alone.
2. Evaluate the damping factor and solve. With $\gamma = 160^\circ = 2.7925\ \text{rad}$ and $\tan 20^\circ = 0.36397$, $$e^{-\gamma/\tan\phi} = e^{-7.672} = 4.655 \times 10^{-4}$$ which is small enough that the right-hand side nearly vanishes and the condition becomes $\sin(\alpha + 140^\circ) \approx 0$, i.e. $\alpha + \gamma - \phi \approx 180^\circ$. Solving the full equation numerically over $21^\circ < \alpha < 89^\circ$ gives $$\boxed{\,\alpha = 39.99^\circ \approx 40^\circ \quad (\beta = \alpha + \gamma = 199.99^\circ)\,}$$ The stated delay angle is therefore verified.
3. Confirm with the closed form, and say when it may be trusted. Because the exponential term is negligible, the condition collapses to $$\alpha \approx 180^\circ + \phi - \gamma = 180^\circ + 20^\circ - 160^\circ = 40^\circ$$ which agrees with the numerical root to $0.01^\circ$. That agreement is a consequence of the damping factor being $4.7 \times 10^{-4}$: the approximation degrades rapidly at poorer power factor or smaller conduction angle (at $\cos\phi = 0.5$ and $\gamma = 135^\circ$ the same estimate is in error by almost nine degrees). Here it is exact enough to be the intended route, and the numerical solution merely confirms it.
4. Part (b) — integrate the chopped sine over the real window. The load voltage is the supply between $\alpha$ and $\beta$ and zero elsewhere, so over a half period $$V_o = V_s\sqrt{\frac{1}{\pi}\left[(\beta - \alpha) - \frac{\sin 2\beta - \sin 2\alpha}{2}\right]}$$ Substituting $\beta - \alpha = \gamma = 2.7925\ \text{rad}$, $\sin 2\beta = \sin 399.98^\circ = 0.6428$ and $\sin 2\alpha = \sin 79.98^\circ = 0.9848$, $$V_o = 220\sqrt{\frac{2.7925 + 0.1710}{\pi}} = 220\sqrt{0.9433} = 220 \times 0.9713$$ $$\boxed{\,V_o = 213.7\ \text{V (rms)}\,}$$ The controller is delivering 97.1 per cent of the supply voltage, which is what a $160^\circ$ conduction angle out of a possible $180^\circ$ should give.
5. Part (c) — get the load current from the motor's power balance. The motor delivers 20 hp at 87 per cent efficiency and $\cos\phi = 0.940$, so $$P_{\text{in}} = \frac{20 \times 745.7}{0.87} = \frac{14\,914}{0.87} = 17\,142.5\ \text{W}, \qquad S = \frac{P_{\text{in}}}{\cos\phi} = \frac{17\,142.5}{0.9397} = 18\,242.7\ \text{VA}$$ $$I_{\text{rms}} = \frac{S}{V_o} = \frac{18\,242.7}{213.7} = 85.37\ \text{A}$$
6. Take the mean-to-rms ratio from the shape of the conduction pulse. The load impedance is never given and is not needed: define the normalised pulse $u(\theta) = \sin(\theta-\phi) - \sin(\alpha-\phi)e^{(\alpha-\theta)/\tan\phi}$ on $[\alpha, \beta]$. Each thyristor conducts once per cycle, so its mean is taken over $2\pi$ while the load rms is taken over $\pi$, and the unknown $V_m/|Z|$ cancels in the ratio: $$\frac{I_{T,\text{avg}}}{I_{\text{rms}}} = \frac{\dfrac{1}{2\pi}\displaystyle\int_\alpha^\beta u\,d\theta} {\sqrt{\dfrac{1}{\pi}\displaystyle\int_\alpha^\beta u^2\,d\theta}} = \frac{1.8154/2\pi}{\sqrt{1.4279/\pi}} = 0.42856$$ $$\boxed{\,I_{T,\text{avg}} = 0.42856 \times 85.37 = 36.59\ \text{A per thyristor}\,}$$
7. Scale to the controller as a whole. The two devices are in inverse parallel and each carries one half cycle, so the mean of the rectified (magnitude) current through the controller is twice the per-device figure: $$\boxed{\,\overline{|i_o|} = 2 \times 36.59 = 73.18\ \text{A}\,}$$ It is worth stating explicitly that the signed average of the controller's output current is zero, as it must be for an a.c. waveform; 73.18 A is the quantity a moving-coil meter behind a bridge, or a thermal rating calculation for the pair, would use.
8. Sanity-check the device rating. Approximating the pulse as a half sine gives $\sqrt{2}I_{\text{rms}}/\pi = \sqrt{2}(85.37)/\pi = 38.43\ \text{A}$, which is 5 per cent above the exact 36.59 A. The shortcut errs on the conservative side, so it is a legitimate first pass for selecting a device but not for reporting an answer.

Final results.