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

Question 4 of 6: Smoothing Reactors and Shunt Capacitors, and a Basic Chopper with an R-L Load

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

Question 4: Smoothing Reactors and Shunt Capacitors, and a Basic Chopper with an R-L Load (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.

(a) Functions of series smoothing reactors and shunt capacitors

Series smoothing reactors. A reactor placed in series with the load, or in the dc link, works because its impedance rises with frequency while it passes dc unimpeded. It therefore opposes the rate of change of current, and its first duty is ripple reduction: the switched voltage applied by a converter is smoothed into a nearly constant current, since the ripple is governed by $\Delta i \approx V\,\Delta t/L$. Four further duties follow from the same property. It keeps a phase-controlled converter in continuous conduction down to light load, which preserves the linear relationship between firing angle and mean output voltage and prevents the speed of a dc drive from running away as the load falls. It limits $di/dt$ through the semiconductors during turn-on and during a fault, protecting the devices and easing their snubber requirements. Placed on the ac side as a line reactor it raises the source impedance seen by the converter, flattening the current pulses and so reducing the harmonic current injected into the supply and the voltage notching imposed on other users. Finally, it limits inrush and short-circuit current, giving protection a chance to act.

Shunt capacitors. A capacitor across the dc link or the load does the dual job: it presents a low impedance to ripple and a high one to dc, so it holds voltage nearly constant. In a dc link it absorbs the rectifier ripple to produce a stiff bus, and just as importantly it supplies the sharp high-frequency current pulses that the inverter switches demand, which the supply and its wiring inductance could not deliver. That local low-inductance current loop is what keeps switching overvoltages within the devices’ ratings. It also provides somewhere for energy returned by an inductive or regenerating load to go during the free-wheel or braking interval, and it decouples the input rectifier from the output inverter so that each can be designed independently. On the ac side, shunt capacitors serve a different purpose — reactive-power compensation and power-factor correction — but there they must be checked against resonance with the supply inductance, because a capacitor bank tuned near a characteristic harmonic will amplify rather than absorb it.

Used together in the classic LC arrangement, the series reactor smooths current and the shunt capacitor smooths voltage, and the pair forms the low-pass filter that separates the switching behaviour of the converter from the smooth quantities the load and the supply require.

(b), (c), (d) Basic chopper with an R-L load

Given. A step-down (buck) chopper with a free-wheel diode supplies a series R-L load, with the data below.

Find. The load time constant and the on-time; the maximum and minimum output current; and the time-domain current expressions with their values at 1 ms and 3.1 ms.

Approach. In the steady state the current is periodic, so the exponential decay during the off-time alone determines the ripple ratio and therefore $t_{off}$. With $t_{on}$ known, matching the rise and decay over one period gives $I_{max}$ and $I_{min}$, and the standard first-order forms then supply the time-domain expressions.

1. Load time constant. The load is a simple series R-L combination, so$$\tau = \frac{L}{R} = \frac{0.5\times 10^{-3}}{0.27} = 1.852\ \text{ms}$$Note that $\tau$ is comparable to the 3.2 ms period, which is why the current visibly curves rather than ramping linearly.
2. On-time from the ripple ratio — part (b). When the switch opens, the source is disconnected and the current free-wheels through the diode into the R-L load, decaying purely exponentially from $I_{max}$ toward zero: $i = I_{max}e^{-t/\tau}$. At the end of the off-time it has reached $I_{min}$, so$$\frac{I_{min}}{I_{max}} = e^{-t_{off}/\tau} = 0.9 \quad\Longrightarrow\quad t_{off} = -\tau\ln 0.9 = 1.852 \times 0.10536 = 0.1951\ \text{ms}$$and since $t_{on} + t_{off} = T$,$$\boxed{\tau = 1.852\ \text{ms},\qquad t_{on} = 3.2 - 0.1951 = 3.005\ \text{ms}}$$The duty cycle is therefore $t_{on}/T = 0.939$ — very high, which is exactly what holding the ripple down to 10 % demands from a load whose time constant is shorter than the chopping period.
3. Maximum and minimum currents — part (c). During the on-time the current rises from $I_{min}$ toward the steady asymptote $V_i/R = 32/0.27 = 118.52$ A, reaching $I_{max}$ at $t = t_{on}$. Imposing periodicity on the rise and the decay together gives the standard closed form$$I_{max} = \frac{V_i}{R}\cdot\frac{1 - e^{-t_{on}/\tau}}{1 - e^{-T/\tau}} = 118.52 \times \frac{1 - 0.19735}{1 - 0.17769} = 118.52 \times 0.97608$$so that$$\boxed{I_{max} = 115.68\ \text{A},\qquad I_{min} = 0.9\,I_{max} = 104.12\ \text{A}}$$Both results were cross-checked the other way round: starting from $I_{min} = 104.12$ A and letting the current rise for 3.005 ms returns 115.68 A, and decaying from 115.68 A for 0.1951 ms returns 104.12 A. The mean load current is $(I_{max}+I_{min})/2 = 109.9$ A, consistent with a mean output voltage of $V_i t_{on}/T = 30.05$ V across 0.27 Ω.
4. Time-domain expressions — part (d). During the on-interval, measuring $t$ from the instant of turn-on, the current is the usual first-order rise from $I_{min}$ toward $V_i/R$:$$i_{on}(t) = \frac{V_i}{R} + \left(I_{min} - \frac{V_i}{R}\right)e^{-t/\tau} = 118.52 - 14.40\,e^{-t/1.852\,\text{ms}},\qquad 0 \le t \le 3.005\ \text{ms}$$During the off-interval, measuring $t'$ from turn-off, the source is out of circuit and the decay is toward zero:$$i_{off}(t') = I_{max}\,e^{-t'/\tau} = 115.68\,e^{-t'/1.852\,\text{ms}},\qquad 0 \le t' \le 0.1951\ \text{ms}$$These two branches are the answer to the first half of part (d).
5. Evaluate at the two requested instants. First identify which interval each instant falls in. Since $t_{on} = 3.005$ ms, the instant $t = 1$ ms lies within the on-time, so$$i(1\ \text{ms}) = 118.52 - 14.40\,e^{-1/1.852} = 118.52 - 14.40(0.58275) = 118.52 - 8.39$$whereas $t = 3.1$ ms lies after turn-off, at $t' = 3.1 - 3.005 = 0.0951$ ms into the free-wheel interval, so$$i(3.1\ \text{ms}) = 115.68\,e^{-0.0951/1.852} = 115.68(0.94994)$$Evaluating both,$$\boxed{i(1\ \text{ms}) = 110.13\ \text{A},\qquad i(3.1\ \text{ms}) = 109.89\ \text{A}}$$Both lie inside the 104.1–115.7 A ripple band, as they must. The near-coincidence of the two values is not a coincidence of arithmetic: 1 ms is partway up the rise and 3.1 ms is just past the peak on the way down, and the two happen to fall at almost the same height.