22-Elec-B8 Power Electronics and Drives · Undated paper

Question 3 of 6: PROBLEM 1 — Controlled Half-Wave Rectifier with an R–L Load

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

Notes on this paper

Paper format. National Examinations, May 2019 — 16-Elec-B8 Power Electronics and Drives. Three hours; open book; any non-communicating calculator, whose make and model must be written on the first inside sheet of the work book. The paper is in two parts and the candidate must attempt all parts: Part 1-A is ten five-point short-answer items, Part 1-B is ten five-point multiple-choice items each requiring a written explanation, and Part 2 is four thirty-point problems. Page-1 Note 4 states that the maximum total score is 220 points and that 150 points is a full mark (100 per cent), so there are 70 points of built-in bonus. Note 5 warns that an answer without its working scores nothing. All a.c. voltages and currents below are rms unless stated otherwise, and three-phase voltages are line-to-line. Every part of every question is solved below, in the exam's own order, with the six sections mapping to Part 1-A, Part 1-B and PROBLEMs 1 to 4.

Reference texts.

M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. — the primary reference for this exam code (devices ch. 4, controlled rectifiers ch. 3 and 10, choppers ch. 5, a.c. voltage controllers ch. 11, d.c. drives ch. 15).
N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed. — PWM, switching-converter waveforms, snubbers and utility applications.
C. W. Lander, Power Electronics, 3rd ed. — compact treatments of phase control and of the chopper current equations used in PROBLEM 3.
B. K. Bose, Modern Power Electronics and AC Drives, and R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control — drive context for PROBLEMs 2 and 4.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — transformer core loss (Part 1-A item b) and the separately excited d.c. machine (PROBLEM 4).
J. J. Grainger and W. D. Stevenson, Power System Analysis, and IEEE Std 519 — harmonics, HVDC economics and static VAR compensation.

Two printing anomalies in the paper itself, both worth a line in the answer book. First, Part 1-B Question 9 repeats Question 5 word for word (“AC voltage controllers convert…”) — ten of the fifty Part 1-B points are the same item asked twice. Both are answered below, the second with the quantitative detail the first does not need. Second, the arithmetic of the cover page does not close in the candidate's favour by accident: 50 + 50 + 120 = 220 points are on offer against a full mark of 150, so a candidate should attempt every part and let the surplus absorb the inevitable slips rather than budgeting time as if 100 points were the target.

Question 3: PROBLEM 1 — Controlled Half-Wave Rectifier with an R–L Load (30 points)

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 controlled half-wave rectifier supplies a series R–L load from a 220 V rms sinusoidal source; the load power factor is 0.8 lagging, and the table fixes the conduction angle in each case together with one further quantity.

Given data
Quantity	Symbol	Value
Supply voltage (rms)	$V_{s}$	220 V
Supply peak	$V_{m}=\sqrt{2}V_{s}$	311.13 V
Load power factor	$\cos\phi$	0.8 lagging
Load angle	$\phi=\arccos 0.8$	36.87°
Case A conduction angle / mean current	$\gamma_{A}$, $I_{A}$	150°, 27.5 A
Case B conduction angle / resistance	$\gamma_{B}$, $R_{B}$	147.5°, 1.3 Ω

Find. For case A the delay angle $\alpha$ and the load resistance $R$; for case B the delay angle $\alpha$ and the average value of the d.c. output current.

Case A of PROBLEM 1: the thyristor is gated at a = 66.02 deg and the inductive load current free-wheels through the device until b = 216.02 deg, giving the printed conduction angle g = 150 deg. The current pulse is the shaded area; its mean over the FULL 360 deg period is the 27.5 A quoted in the table.

Approach. The extinction angle of an R–L half-wave circuit is fixed by the load angle alone, so solving the current-zero condition for the printed conduction angle gives $\alpha$ without any knowledge of $R$; the mean of the current pulse over the full $2\pi$ period then links $V_{m}/Z$ to the tabulated current and yields whichever of $R$ and $I$ is missing.

Part A and Part B — write the load current of the conducting interval. With the thyristor gated at $\alpha$ and the source $v=V_{m}\sin\omega t$, the solution of $L\,di/dt+Ri=V_{m}\sin\omega t$ with $i(\alpha)=0$ is $$i(\omega t)=\frac{V_{m}}{Z}\left[\sin(\omega t-\phi)-\sin(\alpha-\phi)\, e^{-(\omega t-\alpha)/\tan\phi}\right],\qquad Z=\sqrt{R^{2}+(\omega L)^{2}},\;\;\phi=\arctan\frac{\omega L}{R}.$$ Here $\cos\phi=0.8$ gives $\phi=36.87^\circ$ and $\tan\phi=0.75$, so the exponential decay constant is fixed for both cases.
Impose the current zero to relate $\alpha$ and $\gamma$. The device turns off naturally at the extinction angle $\beta=\alpha+\gamma$, where $i=0$: $$\sin(\beta-\phi)=\sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}.$$ This is a transcendental equation in $\alpha$ alone — $Z$, $V_{m}$ and $R$ have all cancelled — which is why the two unknown resistances never enter the angle calculation. Note in passing that the two cases are independent: nothing in the question says they share a load.
Solve case A ($\gamma=150^\circ=2.61799$ rad). Substituting $e^{-\gamma/\tan\phi}=e^{-3.49066}=0.030481$ and solving numerically, $$\boxed{\alpha_{A}=66.02^\circ}\qquad\Rightarrow\qquad \beta_{A}=66.02^\circ+150^\circ=216.02^\circ .$$ The familiar closed-form estimate $\alpha\approx180^\circ+\phi-\gamma=66.87^\circ$ is within 0.85°, which is the expected accuracy at this power factor and conduction angle.
Average the current pulse over the full period. A half-wave rectifier conducts once per supply cycle, so the mean is taken over $2\pi$ and not over the conduction window: $$I_{avg}=\frac{1}{2\pi}\int_{\alpha}^{\beta}\frac{V_{m}}{Z} \left[\sin(\omega t-\phi)-\sin(\alpha-\phi)e^{-(\omega t-\alpha)/\tan\phi}\right]d(\omega t) =\frac{V_{m}}{Z}\,K,$$ where $K$ depends only on $\alpha$, $\gamma$ and $\phi$. Evaluating the integral for case A gives $K_{A}=0.241766$.
Extract the case-A impedance and resistance. Rearranging the previous step with the printed $I_{A}=27.5$ A, $$Z_{A}=\frac{V_{m}K_{A}}{I_{A}}=\frac{311.127\times0.241766}{27.5}=2.7353\ \Omega, \qquad R_{A}=Z_{A}\cos\phi .$$ Hence $$\boxed{R_{A}=2.7353\times0.8=2.188\ \Omega}$$ with reactance $X_{A}=Z_{A}\sin\phi=1.641\ \Omega$, i.e. $L_{A}=4.35$ mH at 60 Hz.
Solve case B ($\gamma=147.5^\circ$) for the delay angle. The same transcendental equation with $e^{-\gamma/\tan\phi}=e^{-3.43248}=0.032307$ gives $$\boxed{\alpha_{B}=68.40^\circ},\qquad \beta_{B}=68.40^\circ+147.5^\circ=215.90^\circ,$$ against the closed-form estimate 69.37°. As expected, a shorter conduction angle requires a later firing instant.
Convert the given resistance into the mean current. For case B the impedance follows from the printed resistance, $Z_{B}=R_{B}/\cos\phi=1.3/0.8=1.625$ Ω, and the integral evaluated at $\alpha_{B}$, $\gamma_{B}$ gives $K_{B}=0.234379$. Therefore $$I_{B}=\frac{V_{m}K_{B}}{Z_{B}}=\frac{311.127\times0.234379}{1.625} \quad\Longrightarrow\quad \boxed{I_{B}=44.87\ \mathrm{A}} .$$ Since case B has the smaller impedance, the larger current is exactly what the physics demands even though its conduction angle is slightly shorter.
Cross-check with the rms current and the load power. Repeating the integration with the square of the current gives $I_{rms,A}=47.02$ A and $I_{rms,B}=77.35$ A, so the real power absorbed is $I_{rms}^{2}R = 4.84$ kW and 7.78 kW respectively. The form factors $I_{rms}/I_{avg}$ are 1.71 and 1.72, which is the value expected of a single 150° pulse per cycle and confirms that the two integrations are consistent with one another.

PROBLEM 1 — completed table (the two requested entries per row are boxed above; rms current and power are added as a check)
Case	Delay angle $\alpha$	Conduction angle $\gamma$	Mean current $I$ (A)	Load resistance $R$ ($\Omega$)	$I_{rms}$ (A)	Load power (W)
A	66.02°	150° (given)	27.5 (given)	2.188	47.02	4837
B	68.40°	147.5° (given)	44.87	1.3 (given)	77.35	7779

Check: the question gives a load power factor of 0.8 but no frequency, so $\phi=36.87^\circ$ is treated as the load impedance angle and $R=Z\cos\phi$ throughout; the inductance quoted in step 5 assumes 60 Hz and is offered only as a physical check, not as an answer. The two rows are also treated as independent operating cases, which is the only reading consistent with case A yielding 2.188 Ω while case B is told 1.3 Ω.

Topic. Phase-controlled half-wave rectification of an inductive load, and the transcendental link between delay angle, extinction angle and conduction angle.

Key relations. $i(\omega t)=(V_{m}/Z)[\sin(\omega t-\phi)-\sin(\alpha-\phi)e^{-(\omega t-\alpha)/\tan\phi}]$; extinction condition $\sin(\beta-\phi)=\sin(\alpha-\phi)e^{-\gamma/\tan\phi}$ with $\gamma=\beta-\alpha$; mean current $I_{avg}=(1/2\pi)\int_{\alpha}^{\beta}i\,d(\omega t)$; $Z=R/\cos\phi$; the approximation $\alpha\approx180^\circ+\phi-\gamma$.

Why this works. The R–L circuit responds to a sinusoid with a steady-state component lagging by $\phi$ plus a decaying transient whose amplitude is set by the initial condition $i(\alpha)=0$. Because both terms carry the same factor $V_{m}/Z$, the shape of the current pulse — and therefore the extinction angle — depends only on $\alpha$ and $\phi$, while the size of the pulse is set by $V_{m}/Z$. That separation is what makes the problem solvable in two clean stages: angles first from the zero-crossing condition, magnitudes second from the mean. When the exponential factor $e^{-\gamma/\tan\phi}$ is small — here 0.030 — the transient has almost died before extinction and the closed-form estimate $\alpha\approx180^\circ+\phi-\gamma$ becomes a reliable sanity check.

Common pitfall. Dividing the current integral by the conduction angle, or by $\pi$, instead of by $2\pi$: a half-wave converter delivers one pulse per cycle, so the average must be spread over the whole $360^\circ$, and using $\pi$ inflates $I_{avg}$ by a factor of two and halves the deduced resistance. The second frequent error is assuming $\beta=180^\circ$, which holds only for a resistive load; with $\gamma=150^\circ$ and $\alpha=66^\circ$ the current in fact persists 36° past the voltage zero.

Source. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed., §3.6 (single-phase controlled rectifier with an R–L load) and §7.4; C. W. Lander, Power Electronics, 3rd ed., ch. 2; N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics, 3rd ed., ch. 5.

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