Question 1 of 6: SCR Characteristic and a Full-Wave a.c. Voltage Controller

Problem 1: SCR Characteristic and a Full-Wave a.c. Voltage Controller (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) which gate-current ordering the characteristic supports and what X₁ and X₂ represent; (b) the firing delay angle α consistent with a 160° conduction angle into a 0.8-power-factor load; (c) the rms voltage the controller delivers; (d) the mean current carried by each thyristor.

[Figure not reproduced: Figure 1.1 — SCR forward and reverse V–I characteristic redrawn from Figure (1) of the paper. The breakover voltage collapses as gate current rises, so the innermost curve carries the largest gate current. See the official exam paper.]

Approach. Part (a) is read directly off the breakover-voltage ordering; parts (b)–(d) follow the standard inductive-load a.c.-controller chain — extinction condition → α, then the rms of the truncated sine over the conduction window, then the motor power balance to fix the load current and its mean value per thyristor.

1. Part (a) — decide between Statements A and B from the breakover ordering. Gate current supplies the extra carriers that trigger regenerative turn-on, so the more gate current is injected, the lower the forward voltage at which the device breaks over: $V_{BO}(I_{G0}) > V_{BO}(I_{G1}) > V_{BO}(I_{G2})$ requires $I_{G2} > I_{G1} > I_{G0}$. In Figure (1) the curve labelled $I_{G2}$ is the one nearest the origin and $I_{G0}$ is the outermost, so the ordering is exactly that of Statement A: $$\boxed{\text{Statement A is correct: } I_{G2} > I_{G1} > I_{G0}}$$ Statement B would require the breakover voltage to rise with gate drive, which contradicts the plotted family.
2. Part (a) — identify the two marked points. $X_1$ sits on the negative voltage axis at the knee where the reverse characteristic turns sharply downward, so it marks the reverse breakdown (avalanche) voltage $V_{BR}$ — the largest reverse voltage the SCR can block before destructive reverse conduction; the device must be applied well below it. $X_2$ sits on the positive voltage axis at the outermost knee, on the $I_{G0}$ (zero or smallest gate current) curve, so it marks the forward breakover voltage $V_{BO}$ — the anode voltage at which the SCR latches on without any gate signal. Together they bound the blocking capability of the device in the two directions.
3. Part (b) — fix the load angle. The motor presents a lagging load whose impedance angle is set by its power factor, $$\varphi = \cos^{-1}(0.8) = 36.87^{\circ}, \qquad \tan\varphi = 0.75 .$$ With an inductive load the thyristor current does not stop at the supply zero crossing; it decays to zero at an extinction angle $\beta = \alpha + \gamma$.
4. Part (b) — apply the extinction condition and solve for α. For $\alpha \le \omega t \le \beta$ the thyristor current is $$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \varphi) - \sin(\alpha - \varphi)\,e^{(\alpha - \omega t)/\tan\varphi}\right].$$ Setting $i(\beta) = 0$ with $\beta = \alpha + \gamma$ gives the standard conduction-angle relation $$\sin(\alpha + \gamma - \varphi) = \sin(\alpha - \varphi)\,e^{-\gamma/\tan\varphi}.$$ Substituting $\gamma = 160^{\circ} = 2.7925\ \text{rad}$ and $\tan\varphi = 0.75$ makes the right-hand exponential $e^{-3.7233} = 0.02415$, so $$\sin(\alpha + 123.13^{\circ}) = 0.02415\,\sin(\alpha - 36.87^{\circ}),$$ whose root in the physical range is $$\boxed{\alpha = 56.41^{\circ}}, \qquad \beta = \alpha + \gamma = 216.41^{\circ}.$$
5. Part (c) — rms output voltage over the conduction window. Both half-cycles are identical, so averaging the squared output over $\pi$ radians is enough: $$V_{o,rms}^{2} = \frac{V_m^{2}}{\pi}\int_{\alpha}^{\beta}\sin^{2}\omega t\;d(\omega t) = \frac{V_s^{2}}{\pi}\left[\gamma - \frac{\sin 2\beta - \sin 2\alpha}{2}\right].$$ With $\sin 2\alpha = \sin 112.81^{\circ} = 0.92183$ and $\sin 2\beta = \sin 432.81^{\circ} = 0.95564$, the bracket is $2.7925 - 0.01690 = 2.7756$, so $$V_{o,rms} = 230\sqrt{\frac{2.7756}{\pi}} = \boxed{216.2\ \text{V (rms)}}$$ i.e. the controller retains about 94 per cent of the supply voltage at this firing angle.
6. Part (d) — convert the motor rating into an rms line current. The mechanical output is $75 \times 746 = 55\,950\ \text{W}$, so the electrical input is $$P_{in} = \frac{55\,950}{0.92} = 60\,815\ \text{W}, \qquad S = \frac{P_{in}}{\cos\varphi} = \frac{60\,815}{0.8} = 76\,019\ \text{VA}.$$ Dividing by the rms voltage actually applied to the motor, $$I_{rms} = \frac{76\,019}{216.2} = 351.6\ \text{A}.$$
7. Part (d) — average current in each thyristor. Each SCR carries one conduction burst per supply cycle, so its mean current is the burst integral divided by $2\pi$, while the rms above is the same waveform measured over $\pi$. Taking the ratio of the two integrals of the normalised waveform $u(\omega t)$ of Step 4 removes the unknown impedance: $$\frac{I_{T,avg}}{I_{rms}} = \frac{\frac{1}{2\pi}\int_{\alpha}^{\beta}u\;d(\omega t)}{\sqrt{\frac{1}{\pi}\int_{\alpha}^{\beta}u^{2}\,d(\omega t)}} = \frac{0.27019}{0.63324} = 0.4267 .$$ Hence $$I_{T,avg} = 0.4267 \times 351.6 = \boxed{150.0\ \text{A}}$$ per thyristor, which is the figure the device data sheet must be checked against.