22-Elec-B8 Power Electronics and Drives · December 2015

Question 1 of 6: SCR Static Characteristic and A.C. Voltage Controller Delay Angles

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

Question 1: SCR Static Characteristic and A.C. Voltage Controller Delay Angles (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) — The four regions of the SCR anode characteristic

[Figure not reproduced: Figure 1.1 — Static anode characteristic of a thyristor, redrawn from figure (1) of the paper. Each gate-current contour has its own forward breakover voltage; larger gate current moves the knee to the left. See the official exam paper.]

The characteristic plots anode current $I_A$ against anode–cathode voltage $V_{AK}$, and the family of curves in the first quadrant is parameterised by gate current, $I_{G0} < I_{G1} < I_{G2}$. The four labelled regions are the four electrical states the device can occupy.

Region 1 — forward conduction (the on-state). Once the device has latched, all three junctions are forward biased and the thyristor behaves as a closed switch: the curve is almost vertical, so a large anode current is carried at an on-state drop of only one to two volts. The current is set entirely by the external circuit, not by the gate, and the gate has lost control — this is the region in which the device does its useful work and dissipates its conduction loss. The device leaves region 1 only when the anode current falls below the holding current $I_H$, which is why the SCR is described as a current-operated latch.

Region 2 — forward blocking (the off-state). With $V_{AK}$ positive but the device not yet triggered, the centre junction $J_2$ is reverse biased and only a small forward leakage current flows. This is the region a phase-controlled converter lives in between the supply zero crossing and the firing instant, and its width in voltage is what allows the delay angle to exist at all. The device leaves region 2 either by gate injection or, undesirably, by the anode voltage reaching the forward breakover value $V_{BO}$. Raising the gate current lowers $V_{BO}$; with a sufficiently large gate current the breakover knee collapses onto the origin and the characteristic degenerates into that of a simple diode. The negative-resistance fold-back that joins region 2 to region 1 is traversed too quickly to be a stable operating locus.

Region 3 — reverse blocking. With $V_{AK}$ negative, junctions $J_1$ and $J_3$ are reverse biased and only the small reverse leakage current $I_R$ flows. This region is what lets a thyristor withstand the negative half cycle of an a.c. supply without conducting, and it is the property that makes natural (line) commutation possible.

Region 4 — reverse avalanche breakdown. Beyond the reverse breakdown voltage the reverse current rises without limit. Unlike forward breakover, which merely turns the device on, reverse avalanche is normally destructive because the current crowds into a small area of the junction. Region 4 therefore fixes the repetitive peak reverse voltage rating $V_{RRM}$, and it is the reason converters carry voltage-grading networks, RC snubbers and surge arresters, and are specified with a safety factor of roughly two to two-and-a-half on the peak circuit voltage.

Parts (b) and (c) — Delay angle of the a.c. voltage controller

Given.

Find. The delay (firing) angle $\alpha$ that produces a $135^\circ$ conduction window, first when the load power factor is 0.707, and then when the controller delivers 80 % of the supply voltage.

Approach. Because a thyristor turns off at the zero of current and not at the zero of voltage, the conduction angle is fixed by the extinction condition of the R–L load; case (b) is that condition read backwards for $\alpha$, while case (c) is the rms-output integral over the real conduction window read backwards for $\alpha$.

1. Convert the stated power factor into a load angle. For a series R–L load the impedance angle is $$\phi = \cos^{-1}(0.707) = 45.00^\circ , \qquad \tan\phi = 1.000 .$$
2. Write the extinction condition. With the thyristor gated at $\omega t = \alpha$, the instantaneous current is the sum of a forced sinusoid and a natural exponential, $$i(\omega t) = \frac{V_m}{Z}\Big[\sin(\omega t - \phi) - \sin(\alpha-\phi)\,e^{(\alpha-\omega t)/\tan\phi}\Big].$$ Conduction ends at $\omega t = \beta = \alpha + \gamma$, where $i = 0$. Setting the bracket to zero and writing $\beta - \alpha = \gamma$ gives the relation that ties $\alpha$, $\gamma$ and $\phi$ together: $$\sin(\alpha+\gamma-\phi) = \sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}.$$
3. Substitute the case (b) data. With $\gamma = 2.35619$ rad and $\tan\phi = 1$ the decay factor is $e^{-2.35619} = 0.09476$. Since $\alpha + 135^\circ - 45^\circ = \alpha + 90^\circ$, the left side is $\cos\alpha$, and $\sin(\alpha-45^\circ) = (\sin\alpha-\cos\alpha)/\sqrt{2}$, so $$\cos\alpha = 0.09476\,\frac{\sin\alpha - \cos\alpha}{\sqrt{2}} = 0.06699\,(\sin\alpha - \cos\alpha).$$ Collecting the cosine terms, $\cos\alpha\,(1 + 0.06699) = 0.06699\sin\alpha$, hence $\tan\alpha = 1.06699/0.06699 = 15.927$ and $$\boxed{\alpha_{(b)} = 86.41^\circ}$$
4. Check it against the closed-form approximation. When $e^{-\gamma/\tan\phi}$ is negligible the extinction condition collapses to $\alpha \approx 180^\circ + \phi - \gamma = 180^\circ + 45^\circ - 135^\circ = 90.00^\circ$. Here the decay factor is 0.0948, not negligible, so the approximation is $3.59^\circ$ high. The exact root is the answer; the approximation is a useful sanity check, not a substitute. The extinction angle is $\beta = 86.41^\circ + 135^\circ = 221.41^\circ$, and the resulting output ratio at this firing angle is 0.7823.
5. Set up case (c) from the rms output integral. The controller passes the supply sinusoid only between $\alpha$ and $\beta$, twice per cycle, so $$V_o = V_s\sqrt{\frac{1}{\pi}\left[\gamma - \frac{\sin 2\beta - \sin 2\alpha}{2}\right]}, \qquad \gamma \text{ in radians}.$$ Setting $V_o/V_s = 0.80$ gives $\gamma - (\sin 2\beta - \sin 2\alpha)/2 = 0.64\pi = 2.010619$, so $$\sin 2\beta - \sin 2\alpha = 2\,(2.356194 - 2.010619) = 0.691150 .$$
6. Reduce it to a single trigonometric equation. With $\beta = \alpha + 135^\circ$, $2\beta = 2\alpha + 270^\circ$ and $\sin(2\alpha+270^\circ) = -\cos 2\alpha$, so the condition becomes $-\cos 2\alpha - \sin 2\alpha = 0.691150$, that is $$\sqrt{2}\,\sin(2\alpha + 45^\circ) = -0.691150 \;\Longrightarrow\; \sin(2\alpha+45^\circ) = -0.48872 .$$ Taking the branch $2\alpha + 45^\circ = 209.26^\circ$, $$\boxed{\alpha_{(c)} = 82.13^\circ}$$
7. Screen the second algebraic branch. The other branch, $2\alpha + 45^\circ = 330.74^\circ$, returns $\alpha = 142.87^\circ$ and satisfies the rms equation exactly as well. It is nevertheless discarded: sweeping the extinction condition of step 2 over the whole range $0^\circ < \phi < 90^\circ$ shows that no load angle whatever produces a $135^\circ$ conduction window at $\alpha = 142.87^\circ$, so that root corresponds to no realizable load. Only $\alpha = 82.13^\circ$ can actually be built.
8. Confirm the physical consistency of the retained root. Solving the extinction condition at $\alpha = 82.13^\circ$ and $\gamma = 135^\circ$ for the load angle gives $\phi = 39.32^\circ$, i.e. a load power factor of 0.774. This is a slightly less inductive load than case (b), which is exactly why it can deliver a higher output ratio (0.800 against 0.782) at a smaller delay angle. The two answers are therefore mutually consistent rather than contradictory.