Question 4 of 6: Series smoothing reactors, and a basic chopper with an R-L load

Question 4: Series smoothing reactors, and a basic chopper with an R-L load (20 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.

Part (a) — Series smoothing reactors in inverter circuits

A series smoothing reactor is a large inductance placed in the dc link between the source (usually a controlled rectifier or a battery) and the inverter bridge, and it is fitted for several distinct reasons that happen to be served by the same component.

The first and most important is to make the link behave as a current source. The rectifier's output is a rippling voltage, not a constant one; a reactor whose impedance $n\omega L$ at the ripple frequencies greatly exceeds the load's effective resistance holds the link current nearly constant over each cycle. In a current-fed inverter this is not an optional refinement — it is what makes the topology work, because the bridge must be handed a stiff current to commutate in blocks.

The second reason is protection. The reactor limits $\mathrm{d}i/\mathrm{d}t$ during commutation and during faults. If an inverter leg mis-fires or a load short develops, the rate of rise of fault current is held to $V_d/L$, which gives the protection time to act and keeps the fault current within the thyristors' $I^{2}t$ withstand. Without it, a shoot-through in a stiff-voltage link is limited only by stray inductance and is usually destructive.

The third reason is to decouple the source from the inverter's switching transients. The reactor blocks the high-frequency components generated by the bridge from propagating back into the rectifier and onto the supply, reducing the harmonic current injected into the ac system and the interference conducted along the dc cabling. Conversely, it prevents supply notches and rectifier commutation dips from disturbing the inverter's own commutation circuits.

Fourth, it provides energy storage. During each commutation interval the load must be handed current while the outgoing device recovers; the reactor supplies that current at essentially constant magnitude, which keeps the machine torque smooth and avoids the current collapse that would otherwise occur. Finally, in a rectifier-fed drive the reactor keeps conduction continuous down to light load: without it the current becomes discontinuous at low torque, the transfer characteristic of the converter becomes non-linear, the mean voltage rises unexpectedly and any speed loop tuned for continuous conduction becomes badly damped.

Parts (b), (c) and (d) — The chopper

Given.

Find. The load time constant and on-time; the maximum and minimum output currents; and the time-domain current expressions together with their values at $t = 1$ ms and $t = 1.5$ ms.

Approach. Use the free-wheel decay to convert the ripple ratio into the off-time, take the on-time from the period, then apply the periodic steady-state matching condition to fix the current levels and write the two exponential segments.

1. Part (b) — Load time constant. The load is a simple series R-L, so $$\tau = \frac{L}{R} = \frac{0.45\times10^{-3}}{1.8} = \boxed{0.25\ \text{ms}}.$$ This is one eighth of the 2 ms chopper period, which will turn out to dominate the character of the answer.
2. Get the off-time from the ripple ratio alone. When the switch opens, the load current free-wheels through the diode with no source in the loop, so it decays purely exponentially from $I_{max}$ towards zero: $i(t) = I_{max}e^{-t/\tau}$. At the end of the off interval it has reached $I_{min}$, hence $$\frac{I_{min}}{I_{max}} = e^{-t_{off}/\tau} \;\Longrightarrow\; t_{off} = -\tau\ln(0.75) = 0.25\times 0.287682 = 0.07192\ \text{ms}.$$ Notice that the current magnitudes never entered — the ratio by itself fixes the off-time.
3. The on-time follows from the period. $$t_{on} = T - t_{off} = 2.0 - 0.07192 = \boxed{1.92808\ \text{ms}},$$ a duty ratio of $\delta = t_{on}/T = 0.9640$. Such a high duty ratio is the direct consequence of $\tau \ll T$: with so fast a decay, only a very short free-wheel interval is needed to drop the current by 25%.
4. Part (c) — Maximum output current. In the steady state the current at the end of the ON interval must equal $I_{max}$, and matching the two exponentials around the cycle gives the standard result $$I_{max} = \frac{V_i}{R}\cdot\frac{1 - e^{-t_{on}/\tau}}{1 - e^{-T/\tau}} = 15.5556\times\frac{1 - 4.4728\times10^{-4}}{1 - 3.3546\times10^{-4}} = \boxed{15.554\ \text{A}}.$$
5. Minimum output current, and the mean as a check. The ripple ratio gives directly $$I_{min} = 0.75\,I_{max} = \boxed{11.665\ \text{A}}.$$ As an independent check, the mean voltage applied to the load is $\delta V_i$ and the mean voltage across $L$ over a period is zero, so $I_{mean} = \delta V_i/R = 0.9640\times 28/1.8 = 14.996$ A. That lies between $I_{min}$ and $I_{max}$ as it must, and sits close to $I_{max}$ because the current spends almost the whole period near its ceiling.
6. Part (d) — Time-domain expression during the ON interval. Measuring $t$ from the instant the switch closes, the load sees $V_i$ across $R + sL$ starting from $I_{min}$, so $$i_{on}(t) = \frac{V_i}{R}\left(1 - e^{-t/\tau}\right) + I_{min}e^{-t/\tau} = 15.556 - 3.891\,e^{-t/0.25\ \text{ms}},\qquad 0 \le t \le 1.92808\ \text{ms}.$$
7. Time-domain expression during the free-wheel interval. Measuring $t'$ from the instant the switch opens, $$i_{off}(t') = I_{max}e^{-t'/\tau} = 15.554\,e^{-t'/0.25\ \text{ms}},\qquad 0 \le t' \le 0.07192\ \text{ms}.$$ The two expressions are boxed together as the answer to the first half of part (d); note that the off-interval exponential is measured from turn-off, not from the start of the period.
8. Evaluate at the two requested instants. Both 1 ms and 1.5 ms are less than $t_{on} = 1.92808$ ms, so both fall inside the ON interval and $i_{on}$ is the expression to use in each case: $$i(1\ \text{ms}) = 15.556 - 3.891\,e^{-4} = 15.556 - 0.0713 = \boxed{15.484\ \text{A}},$$ $$i(1.5\ \text{ms}) = 15.556 - 3.891\,e^{-6} = 15.556 - 0.0096 = \boxed{15.546\ \text{A}}.$$ Checking the instant against $t_{on}$ before choosing the expression is essential: had 1.5 ms fallen in the free-wheel interval, the ON expression would have returned a value above $I_{max}$, which is itself the tell that the wrong branch was used.

Final results.