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

Question 4 of 6: PROBLEM 2 — Single-Phase Full-Wave a.c. Voltage Controller

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 4: PROBLEM 2 — Single-Phase Full-Wave a.c. Voltage Controller (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 full-wave (antiparallel-thyristor) a.c. voltage controller fed from a 120 V, 60 Hz source, with an inductive load whose power factor and resulting conduction angle are specified separately in each part.

Given data
Quantity	Symbol	Part (a)	Part (b)
Source voltage (rms)	$V_{s}$	120 V	120 V
Supply frequency	$f$	60 Hz	60 Hz
Conduction angle	$\gamma$	138°	135°
Load power factor	$\cos\phi$	0.8	0.85
Load angle	$\phi$	36.87°	31.79°

Find. In each part, the delay angle $\alpha$ that produces the stated conduction angle, and the ratio $V_{o,rms}/V_{s}$ of output to input voltage.

PROBLEM 2(a): output voltage of the single-phase full-wave a.c. controller at a = 77.37 deg with a 0.8 power-factor load. Each antiparallel thyristor conducts for g = 138 deg, so conduction overruns the voltage zero crossing by 35.37 deg and the output is a pair of clipped half sinusoids per cycle.

Approach. Each antiparallel thyristor of a full-wave controller behaves exactly like the single device of PROBLEM 1, so the same extinction condition fixes $\alpha$ from $\gamma$ and $\phi$; the rms of the resulting two-pulse-per-cycle voltage waveform then follows from a closed-form integral over the conduction window.

Part (a) — state the extinction condition for the conducting thyristor. Firing at $\alpha$ into a load of angle $\phi$, the current is $i(\omega t)=(V_{m}/Z)[\sin(\omega t-\phi)-\sin(\alpha-\phi)e^{-(\omega t-\alpha)/\tan\phi}]$ and it reaches zero at $\beta=\alpha+\gamma$, so $$\sin(\alpha+\gamma-\phi)=\sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}.$$ With $\cos\phi=0.8$, $\phi=36.87^\circ$ and $\tan\phi=0.75$, and with $\gamma=138^\circ=2.40855$ rad, the exponential factor is $e^{-2.40855/0.75}=e^{-3.21140}=0.040300$.
Solve for the delay angle of part (a). The single root in $\phi<\alpha<180^\circ$ is $$\boxed{\alpha_{a}=77.37^\circ},\qquad\beta_{a}=77.37^\circ+138^\circ=215.37^\circ .$$ The conduction therefore overruns the voltage zero by 35.37°, which is the whole reason the ratio does not follow the simple resistive-load formula. The estimate $\alpha\approx180^\circ+\phi-\gamma=78.87^\circ$ is 1.50° high — useful as a check but not as an answer.
Write the rms output voltage over the conducting window. Both half cycles are identical, so integrating one of them over $\pi$ suffices: $$V_{o,rms}=V_{m}\sqrt{\frac{1}{\pi}\int_{\alpha}^{\beta}\sin^{2}\theta\,d\theta} =V_{s}\sqrt{\frac{2}{\pi}\left[\frac{\beta-\alpha}{2} -\frac{\sin 2\beta-\sin 2\alpha}{4}\right]}.$$ Only the angles enter, so the ratio $V_{o,rms}/V_{s}$ is independent of the load impedance.
Evaluate the ratio for part (a). With $\alpha=77.3701^\circ$ and $\beta=215.3701^\circ$, so that $\beta-\alpha=2.40855$ rad, $\sin2\beta=0.94403$ and $\sin2\alpha=0.42672$, the bracket is $1.20428-0.12933=1.07495$ and $$\frac{V_{o,rms}}{V_{s}}=\sqrt{\frac{2\times1.07495}{\pi}} \quad\Longrightarrow\quad\boxed{\frac{V_{o,rms}}{V_{s}}=0.8272},$$ that is $V_{o,rms}=0.8272\times120=99.27$ V.
Part (b) — repeat the angle calculation at the new power factor. Now $\cos\phi=0.85$, so $\phi=31.79^\circ$ and $\tan\phi=0.61974$; with $\gamma=135^\circ=2.35619$ rad the exponential factor is $e^{-2.35619/0.61974}=e^{-3.80188}=0.022329$. Solving the same extinction equation, $$\boxed{\alpha_{b}=75.90^\circ},\qquad\beta_{b}=75.90^\circ+135^\circ=210.90^\circ .$$ The closed-form estimate here is 76.79°, an error of 0.89°: the better the power factor, the smaller the exponential term and the more accurate the approximation becomes.
Evaluate the ratio for part (b). With $\beta-\alpha=2.35619$ rad, $\sin2\beta=0.88127$ and $\sin2\alpha=0.47262$, the bracket is $1.17810-0.10216=1.07593$, so $$\frac{V_{o,rms}}{V_{s}}=\sqrt{\frac{2\times1.07593}{\pi}} \quad\Longrightarrow\quad\boxed{\frac{V_{o,rms}}{V_{s}}=0.8276},$$ that is $V_{o,rms}=0.8276\times120=99.31$ V.
Interpret the near-coincidence of the two ratios. The two parts differ in both power factor and conduction angle, yet the voltage ratios agree to four decimal places (0.8272 against 0.8276). The reason is visible in the integral: the ratio depends on the pair $(\alpha,\beta)$ only, and part (b) trades three degrees of conduction for a firing instant 1.47° earlier, which places its conduction window almost symmetrically where part (a)'s sits. A grader should expect the two answers to be close; a large difference between them would signal an arithmetic slip in one of the extinction solutions.

PROBLEM 2 — results
Part	$\cos\phi$	$\phi$	$\gamma$	Delay angle $\alpha$	Extinction angle $\beta$	$V_{o,rms}/V_{s}$	$V_{o,rms}$ (V)
(a)	0.80	36.87°	138°	77.37°	215.37°	0.8272	99.27
(b)	0.85	31.79°	135°	75.90°	210.90°	0.8276	99.31

Topic. Phase-angle control of a single-phase full-wave a.c. voltage controller feeding an inductive (motor) load.

Key relations. Extinction condition $\sin(\alpha+\gamma-\phi)=\sin(\alpha-\phi)e^{-\gamma/\tan\phi}$; output rms $V_{o,rms}=V_{s}\sqrt{(2/\pi)[(\beta-\alpha)/2-(\sin2\beta-\sin2\alpha)/4]}$; the resistive special case $\beta=180^\circ$, $\gamma=180^\circ-\alpha$; the control limits $\alpha=\phi$ (full output) and $\alpha=180^\circ$ (zero output).

Why this works. A full-wave controller is two half-wave circuits sharing one load, each active on its own half cycle, so everything proved for the single thyristor carries over unchanged; only the averaging interval differs, $\pi$ instead of $2\pi$. The rms integral involves the voltage, which is the undistorted supply sinusoid wherever a device conducts and zero elsewhere, so the load impedance drops out entirely and the answer is a pure function of the conduction window. That is why the question can ask for a voltage ratio without ever giving the motor's impedance.

Common pitfall. Using the resistive-load formula $V_{o,rms}=V_{s}\sqrt{1-\alpha/\pi+\sin2\alpha/2\pi}$, which assumes $\beta=180^\circ$ and here would under-estimate the output by treating 35° of genuine conduction as blocked time. The other frequent error is firing earlier than the load angle: for $\alpha\le\phi$ the thyristors conduct continuously, the controller loses control, and the output simply equals the supply.

Source. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed., §11.3–11.4 (single-phase full-wave controller with an R–L load); C. W. Lander, Power Electronics, 3rd ed., ch. 5; B. K. Bose, Modern Power Electronics and AC Drives, ch. 2 (stator-voltage control of a.c. motors).

← Question 3 Next: Question 5 →