22-Elec-B8 Power Electronics and Drives · December 2013

Question 4 of 6: Clamping Capacitors, Smoothing Reactors, and a Basic Chopper

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

Notes on this paper

Paper format. National Exams, December 2013 — 07-Elec-B8 Power Electronics and Drives. Open-book, three hours, any non-communicating calculator permitted. Six problems are printed and any five constitute a complete paper; all are of equal value (20 points each). Because this set is a study resource, all six problems are solved in full below, every lettered sub-part included.

Reference texts for this subject

M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. — inverters and PWM schemes (Ch. 6), dc–dc converters (Ch. 5), controlled rectifiers (Ch. 10), drives (Ch. 15).
N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed. — switch-mode inverters and harmonic analysis (Ch. 8).
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — synchronous machines (Ch. 5), induction machines (Ch. 7), dc machines (Ch. 8–9).
B. K. Bose, Modern Power Electronics and AC Drives — constant volts-per-hertz induction-motor drives and PWM side effects.
R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control — converter-fed dc drives.
C. W. Lander, Power Electronics, 3rd ed. — choppers and controlled rectifiers.
J. J. Grainger and W. D. Stevenson, Power System Analysis — salient-pole power-angle equations and capability curves.
IEEE Std 519 and CSA C22.1 (Canadian Electrical Code, Part I) — harmonic limits and harmonic-related conductor sizing in Canadian practice.

Check — two anomalies carried over from the printed paper. (i) Problem 6 is headed “December 2010” on the December 2013 question sheet and is solved here as printed. (ii) Problem 5 supplies a total leakage inductance of 1.25 mH but its accompanying formula describes $L_T$ as the total leakage reactance. The formula is dimensionally consistent only if $L_T$ is an inductance, so it is read as an inductance throughout, which is also what the numbers require.

Question 4: Clamping Capacitors, Smoothing Reactors, and a Basic Chopper (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) — clamping capacitors and smoothing reactors

Both components exist to protect switching devices from the consequences of stray energy storage, but they act on opposite quantities. A smoothing reactor is placed in series with the load or in the dc link and acts on current. Its first duty is to limit $di/dt$ through the semiconductors: a thyristor that is turned on faster than its gate plasma can spread across the junction concentrates the whole current in a small area and fails thermally, so manufacturers specify a maximum $di/dt$ and the series inductance is chosen to respect it. Its second duty is to smooth the load current ripple, keeping conduction continuous in a converter-fed drive; continuous conduction linearises the converter’s control characteristic, reduces the peak-to-average current ratio in the machine, and cuts the harmonic heating and torque pulsation the machine would otherwise suffer. In a chopper such as the one analysed below, the same inductance is what stores energy during the on-time and sustains load current through the free-wheeling diode during the off-time.

A clamping capacitor acts on voltage. Every practical circuit has stray inductance in its busbars, wiring and device packages, and when a switch interrupts current that inductance tries to sustain it by developing $L\,di/dt$ across the opening switch. Left unchecked the resulting spike exceeds the device’s safe operating area. The clamping capacitor — usually a low-inductance film capacitor mounted directly across the device or the dc link, often with a diode and a discharge resistor to form a snubber — provides a low-impedance path into which that trapped energy commutates, so the device voltage rises only to a defined clamp level and at a controlled $dv/dt$. Limiting $dv/dt$ matters independently, because a fast-rising anode voltage can retrigger a thyristor through its own junction capacitance. In multilevel topologies the same component takes on a structural role: the flying capacitors of a flying-capacitor inverter and the clamping capacitors of a neutral-point-clamped bridge hold the intermediate voltage levels that let the inverter synthesise a stepped output with reduced $dv/dt$ and lower harmonic content.

Parts (b), (c) and (d) — chopper current levels

Given.

Quantity	Symbol	Value
Chopper input voltage	$V_i$	30 V
Load resistance	$R$	0.275 Ω
Load time constant	$\tau = L/R$	1.475 ms
Maximum output current	$I_{\max}$	92 A
On-time	$t_{\text{on}}$	2.4 ms
Raised resistance (part d)	$R'$	0.375 Ω

Find. (b) the chopper period; (c) the minimum output current; and (d) both current levels after the resistance is raised to 0.375 Ω with the inductance, period and on-time unchanged.

Figure 4 — Chopper output current in periodic steady state, exponential rise while the chopper conducts and free-wheel decay while the diode carries. Raising R from 0.275 Ω to 0.375 Ω lowers both levels and worsens the relative ripple.

Approach. Use the steady-state periodic solution of the series $R$–$L$ circuit — exponential rise towards $V_i/R$ while the chopper conducts, exponential decay towards zero while the diode free-wheels — and impose the condition that the current returns to the same value each cycle.

Recover the inductance and the asymptotic current. The time constant fixes the inductance, $$L = R\tau = 0.275 \times 1.475\times10^{-3} = 0.4056 \ \text{mH},$$ and the current the load would eventually reach if the chopper stayed on indefinitely is $V_i/R = 30/0.275 = 109.09$ A. That the specified maximum of 92 A falls short of this confirms the chopper switches off well before the transient completes, so the exponential form must be kept — a linear-ripple approximation would be badly in error here, since $\tau$ is comparable to the period.
Part (b) — write the steady-state maximum. Over one cycle the current rises from $I_{\min}$ to $I_{\max}$ during $t_{\text{on}}$ and decays back from $I_{\max}$ to $I_{\min}$ during $t_{\text{off}}$. Eliminating $I_{\min}$ between the two expressions gives the standard steady-state result $$I_{\max} = \frac{V_i}{R}\left[\frac{1 - e^{-t_{\text{on}}/\tau}}{1 - e^{-T/\tau}}\right].$$
Solve for the period. With $t_{\text{on}}/\tau = 2.4/1.475 = 1.6271$, so $e^{-t_{\text{on}}/\tau} = 0.19650$, rearranging for the only unknown gives $$1 - e^{-T/\tau} = \frac{V_i}{R}\cdot\frac{1 - e^{-t_{\text{on}}/\tau}}{I_{\max}} = 109.09\times\frac{0.80350}{92} = 0.95277 .$$ Hence $e^{-T/\tau} = 0.04723$, $T/\tau = \ln(1/0.04723) = 3.0529$, and $$\boxed{\;T = 3.0529 \times 1.475 \ \text{ms} = 4.503 \ \text{ms}\;}$$ corresponding to a chopping frequency of 222 Hz and a duty ratio of $t_{\text{on}}/T = 0.533$.
Part (c) — use the free-wheeling interval. Once the chopper turns off the source is disconnected, so the current simply decays towards zero through the diode with no forcing term: $$I_{\min} = I_{\max}e^{-t_{\text{off}}/\tau}, \qquad t_{\text{off}} = T - t_{\text{on}} = 4.503 - 2.400 = 2.103 \ \text{ms}.$$ Substituting $t_{\text{off}}/\tau = 1.4258$ and $e^{-1.4258} = 0.24035$, $$\boxed{\;I_{\min} = 92 \times 0.24035 = 22.11 \ \text{A}\;}$$
Verify by an independent route. The companion closed form for the minimum, derived from the on-interval instead of the off-interval, must agree: $$I_{\min} = \frac{V_i}{R}\left[\frac{e^{t_{\text{on}}/\tau} - 1}{e^{T/\tau} - 1}\right] = 109.09\left[\frac{5.0892 - 1}{21.173 - 1}\right] = 109.09(0.20270) = 22.11 \ \text{A},$$ which confirms both the period and the minimum. The ripple ratio $I_{\min}/I_{\max} = 0.240$ is large, as expected when $\tau$ and $T$ are of the same order.
Part (d) — recompute the circuit constants for the raised resistance. The inductance is held at its original value, so raising the resistance shortens the time constant and lowers the asymptote together: $$\tau' = \frac{L}{R'} = \frac{0.4056\times10^{-3}}{0.375} = 1.0817 \ \text{ms}, \qquad \frac{V_i}{R'} = \frac{30}{0.375} = 80.0 \ \text{A}.$$ Both changes act to reduce the current, so the answer must come out below the original 92 A.
Apply the same steady-state formulae with $T$ and $t_{\text{on}}$ unchanged. Now $t_{\text{on}}/\tau' = 2.2186$ and $T/\tau' = 4.1631$, so $e^{-t_{\text{on}}/\tau'} = 0.10874$ and $e^{-T/\tau'} = 0.01557$: $$\boxed{\;I'_{\max} = 80.0\left[\frac{1 - 0.10874}{1 - 0.01557}\right] = 80.0(0.90546) = 72.43 \ \text{A}\;}$$ and, decaying over the same 2.103 ms off-time with $e^{-t_{\text{off}}/\tau'} = 0.14319$, $$\boxed{\;I'_{\min} = 72.43 \times 0.14319 = 10.37 \ \text{A}\;}$$
Sanity-check the trend. Both levels have fallen, as the lower asymptote demands, and the ripple has become relatively worse — the ratio $I'_{\min}/I'_{\max}$ drops from 0.240 to 0.143 — because the shorter time constant lets the current decay further in the same off-time. This is the practical reason a smoothing reactor is added when a chopper feeds a low-inductance load: the ripple is set by $t_{\text{off}}/\tau$, and only $\tau$ is available to the designer once the switching frequency is fixed.

Quantity	Symbol	Result
Load inductance	$L$	0.4056 mH
(b) Chopper period	$T$	4.503 ms (222 Hz)
(b) Off-time and duty ratio	$t_{\text{off}}$, $k$	2.103 ms, 0.533
(c) Minimum output current	$I_{\min}$	22.11 A
(c) Ripple ratio	$I_{\min}/I_{\max}$	0.240
(d) Time constant with $R' = 0.375$ Ω	$\tau'$	1.0817 ms
(d) Maximum output current	$I'_{\max}$	72.43 A
(d) Minimum output current	$I'_{\min}$	10.37 A

Topic. Steady-state current levels in a step-down (buck) chopper feeding a series $R$–$L$ load with a free-wheeling diode.

Key relations. $\tau = L/R$; on-interval $i(t) = (V_i/R)(1-e^{-t/\tau}) + I_{\min}e^{-t/\tau}$; off-interval $i(t) = I_{\max}e^{-t/\tau}$; steady state $I_{\max} = (V_i/R)(1-e^{-t_{\text{on}}/\tau})/(1-e^{-T/\tau})$ and $I_{\min} = (V_i/R)(e^{t_{\text{on}}/\tau}-1)/(e^{T/\tau}-1)$.

Why this works. Within each interval the circuit is linear and time-invariant, so the current is a single exponential heading for whatever asymptote the interval imposes — $V_i/R$ with the switch closed, zero with the diode conducting. Periodic steady state means the current at the end of the off-interval equals the current at the start of the next on-interval; imposing that one condition closes the two-equation system and yields the closed forms above. Because the off-interval has no source, its decay depends only on $t_{\text{off}}/\tau$, which is why a ratio of currents alone is enough to find a time in problems of this family — and why the minimum can be found here without ever computing $I_{\min}$ from the on-interval.

Common pitfall. Approximating the ripple linearly as $\Delta i = (V_i - I R)t_{\text{on}}/L$ is tempting and quite wrong when $\tau$ is comparable to $T$, as it is here; the linearised estimate misses by tens of amperes. The other frequent slip in part (d) is recomputing $\tau'$ from the old inductance-and-resistance product rather than holding $L$ fixed — the question changes $R$ only, so $L = R\tau$ must be evaluated once, from the original data, and then reused.

Source. Rashid, Power Electronics, 4th ed., Ch. 5 (dc–dc converters, §5.3 step-down converter with $RL$ load); Lander, Power Electronics, 3rd ed., Ch. 6.

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