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

Question 5 of 6: PROBLEM 3 — Step-Down Chopper Table

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 5: PROBLEM 3 — Step-Down Chopper Table (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 basic step-down (buck) chopper fed from a 24 V d.c. source into a series R–L load whose time constant is quoted directly in each row; the switch is operated so that the peak load current just reaches the allowed maximum of 20 A in steady state.

Given data
Quantity	Symbol	Value
Input voltage	$V_{i}$	24 V
Maximum allowed current	$I_{max}$	20 A
Case 1	$T$, $T_{on}$, $\tau$	2.40 ms, 1.8 ms, 1.5 ms → find $R$
Case 2	$T_{on}$, $\tau$, $R$	2.40 ms, 1.45 ms, 1.25 $\Omega$ → find $T$
Case 3	$T$, $\tau$, $R$	1.80 ms, 1.5 ms, 0.9 $\Omega$ → find $T_{on}$

Find. The one missing entry in each of the three rows, namely the load resistance of case 1, the chopper period of case 2 and the on-time of case 3.

PROBLEM 3, case 1: steady-state load current of the step-down chopper with T = 2.40 ms, Ton = 1.80 ms and tau = 1.50 ms. The current rises towards Vi/R = 22.84 A during Ton and decays towards zero during Toff; the peak is pinned at the allowed maximum of 20 A, which is what fixes R = 1.0507 ohm.

Approach. Every row is the same steady-state relation between peak current, duty and time constant, inverted for a different symbol; before inverting it for the period, screen the row against the physical bound $R<V_{i}/I_{max}$, because one row of this table has no solution at all.

Derive the steady-state peak-current relation. During $T_{on}$ the current rises towards $V_{i}/R$ from $I_{min}$; during $T_{off}=T-T_{on}$ it decays from $I_{max}$ towards zero (the free-wheel path carries no source voltage): $$I_{max}=\frac{V_{i}}{R}\left(1-e^{-T_{on}/\tau}\right)+I_{min}e^{-T_{on}/\tau}, \qquad I_{min}=I_{max}\,e^{-T_{off}/\tau}.$$ Eliminating $I_{min}$ between the two gives the single working formula $$I_{max}=\frac{V_{i}}{R}\cdot\frac{1-e^{-T_{on}/\tau}}{1-e^{-T/\tau}} .$$
Read the feasibility bound straight off that formula. Because $T_{on}<T$, the fraction is always less than unity, so $$I_{max}<\frac{V_{i}}{R}\qquad\Longleftrightarrow\qquad R<\frac{V_{i}}{I_{max}}=\frac{24}{20}=1.20\ \Omega .$$ Any row whose resistance exceeds 1.20 Ω cannot reach 20 A at any duty ratio whatsoever. This one division is worth doing before touching a logarithm, and it decides case 2 below.
Case 1 — solve for the load resistance. Rearranging step 1 with $T=2.40$ ms, $T_{on}=1.8$ ms and $\tau=1.5$ ms: $$R=\frac{V_{i}}{I_{max}}\cdot\frac{1-e^{-T_{on}/\tau}}{1-e^{-T/\tau}} =1.20\times\frac{1-e^{-1.2}}{1-e^{-1.6}} =1.20\times\frac{0.698806}{0.798103},$$ so that $$\boxed{R_{1}=1.0507\ \Omega}.$$ The result satisfies the bound of step 2 with room to spare, as it must.
Complete case 1 with the quantities that check it. The duty ratio is $\delta=1.8/2.40=0.75$, so the mean output voltage is $V_{o}=\delta V_{i}=18.00$ V and the current would settle at $V_{i}/R_{1}=22.84$ A if the switch stayed closed. The valley current is $I_{min}=I_{max}e^{-T_{off}/\tau}=20\,e^{-0.6/1.5}=13.41$ A, giving a peak-to-peak ripple of 6.59 A — about a third of the peak, which is consistent with the modest $T/\tau=1.6$ of this row.
Case 2 — apply the physical screen first. The row prints $R=1.25\ \Omega$, and $1.25>1.20$, so by step 2 the load can never carry more than $V_{i}/R=19.2$ A even with the switch permanently closed. Case 2 has no solution. The algebra confirms it: the bracket $$\frac{V_{i}}{R\,I_{max}}\left(1-e^{-T_{on}/\tau}\right) =\frac{24}{1.25\times20}\left(1-e^{-2.4/1.45}\right)=0.96\times0.808941=0.77658$$ does lie inside $(0,1)$, so the logarithm exists and returns $T=-\tau\ln(1-0.77658)=2.173$ ms — but that period is shorter than the row's own 2.40 ms on-time, which is impossible. The bracket test alone would have passed this row; only the resistance bound exposes it.
Offer one worked repair for case 2 rather than a silent patch. If the examiner intended the load of case 1 to be carried across, then with $R=1.0507\ \Omega$, $T_{on}=2.40$ ms and $\tau=1.45$ ms the bracket becomes 0.92389 and $$T=-1.45\ln(1-0.92289)=3.735\ \mathrm{ms},$$ a period comfortably longer than the on-time and a duty of 0.643. That is the answer to report as the plausible intent, clearly labelled, alongside the finding that the row as printed is infeasible.
Case 3 — solve for the on-time. With $T=1.80$ ms, $\tau=1.5$ ms and $R=0.9\ \Omega$ (which satisfies the bound), invert step 1 for $T_{on}$: $$1-e^{-T_{on}/\tau}=\frac{I_{max}R}{V_{i}}\left(1-e^{-T/\tau}\right) =\frac{20\times0.9}{24}\times0.698806=0.524104,$$ $$T_{on}=-\tau\ln\left(1-0.524104\right)=-1.5\ln(0.475896) \quad\Longrightarrow\quad\boxed{T_{on}=1.1138\ \mathrm{ms}} .$$ The duty is then 0.6188, the mean output voltage $V_{o}=14.85$ V, the valley current $I_{min}=12.66$ A and the ripple 7.34 A.
Back-substitute every completed row. Feeding each finished row into the original relation returns $I_{max}=20.00$ A exactly for cases 1 and 3, and the free check $I_{min}<\delta V_{i}/R<I_{max}$ holds in both (case 1: 13.41 < 17.13 < 20; case 3: 12.66 < 16.50 < 20). Case 2 fails every one of these tests, which is the independent confirmation that its infeasibility is a property of the printed data and not an arithmetic slip.

PROBLEM 3 — completed table (bold entries are the requested answers)
Case	$T$ (ms)	$T_{on}$ (ms)	$\tau$ (ms)	$R$ ($\Omega$)	Duty $\delta$	$V_{o}$ (V)	$I_{min}$ (A)
1	2.40	1.80	1.5	1.0507	0.750	18.00	13.41
2	no solution (2.173 formally, < $T_{on}$)	2.40	1.45	1.25 (> 1.20 limit)	—	—	—
2 (repaired with $R=1.0507\ \Omega$)	3.735	2.40	1.45	1.0507	0.643	15.42	7.97
3	1.80	1.1138	1.5	0.9	0.619	14.85	12.66

Check: case 2 as printed is physically impossible — a 1.25 Ω load on a 24 V source cannot reach 20 A, since even a permanently closed switch gives only 19.2 A. The answer above reports the row as infeasible, quotes the bound $R<V_{i}/I_{max}=1.20\ \Omega$, and offers the case-1 resistance as the single most likely intended repair. A candidate in the hall should do the same and state the assumption, as page-1 Note 1 explicitly invites.

Topic. Steady-state current limits of a step-down chopper feeding an R–L load, and the feasibility screening of a “complete the table” question.

Key relations. $I_{max}=(V_{i}/R)\left(1-e^{-T_{on}/\tau}\right)/\left(1-e^{-T/\tau}\right)$; $I_{min}=I_{max}e^{-T_{off}/\tau}$; $\delta=T_{on}/T$ and $V_{o}=\delta V_{i}$; feasibility bound $R<V_{i}/I_{max}$; the small-ripple approximation $\Delta I\approx V_{i}\delta(1-\delta)T/L$, valid only when $\tau\gg T$.

Why this works. In steady state the current is periodic, so whatever the inductor gains during $T_{on}$ it must lose during $T_{off}$. Writing the exponential solution for each interval and imposing that periodicity eliminates $I_{min}$ and leaves one algebraic relation among five quantities — $V_{i}$, $R$, $\tau$, $T$, $T_{on}$ — at the current limit. A table that gives four of them and asks for the fifth is therefore a single equation each time, and the only thing that changes row to row is which symbol sits alone. The bound in step 2 falls out for free because a ratio of two positive quantities, the smaller over the larger, cannot exceed one.

Common pitfall. Trusting the logarithm. The bracket $(V_{i}/RI_{max})(1-e^{-T_{on}/\tau})$ can sit safely inside $(0,1)$ and still produce a period shorter than the on-time it was built from, which is exactly what happens in case 2; a candidate who only checks that the logarithm is defined ships 2.173 ms and loses the marks. The second pitfall is applying the linear-ripple shortcut here: with $T/\tau$ of order 1.6 the exponentials are strongly curved and the shortcut is in error by tens of per cent.

Source. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed., §5.3–5.5 (step-down converter with an R–L load); N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics, 3rd ed., ch. 7; C. W. Lander, Power Electronics, 3rd ed., ch. 4.

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