Question 4 of 6: Series Smoothing Reactors, and a Basic Chopper Feeding an R–L Load

Question 4: Series Smoothing Reactors, and a Basic Chopper Feeding 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.

Part (a) — Why series smoothing reactors are used in inverter circuits

A series smoothing reactor is an inductor placed in the dc link between the source (or the source-side rectifier) and the inverter bridge. It is there to make the link behave like the idealised source the inverter topology assumes, and it earns its place for several separate reasons.

The first and most fundamental is that a current-fed inverter requires a stiff current source. The bridge steers the link current from one output leg to another every 60 or 120 electrical degrees; without a large series inductance the link current would collapse and rebuild at every commutation, and the quasi-square current waveform the topology depends on would not exist. The reactor stores enough magnetic energy that the link current is essentially constant over a switching period, so the rectifier sets its magnitude and the inverter sets only its routing.

Second, the reactor limits $\mathrm{d}i/\mathrm{d}t$ through the switching devices. Thyristors have a finite rate at which the conducting area spreads from the gate, and exceeding the rated current slope destroys them by local heating; the series inductance guarantees the slope stays inside the rating during both normal commutation and a fault. It also buys time during a fault: with a large link inductance the current rises slowly enough that protection can detect and clear the fault before the devices are damaged, which is one reason current-fed drives tolerate an output short circuit that would destroy a voltage-fed one.

Third, the reactor decouples the source from the inverter's switching-frequency current pulses. The bridge draws a rectangular, harmonic-rich current; without smoothing that ripple flows back into the rectifier and the ac supply, distorting the supply voltage at the point of common coupling, causing commutation notches, and heating the source transformer. The reactor presents a rising impedance to those harmonics and confines most of the ripple to the local loop. It also keeps the rectifier in continuous conduction at light load, which is what makes the familiar $V_a=1.35\,V_{LL}\cos\alpha$ relationship valid down to low output.

Fourth, in the natural commutation of a thyristor inverter the stored energy in the link reactor supplies the current that charges the commutating capacitors and forces the outgoing device into reverse bias for long enough to regain its blocking capability. In that sense the reactor is not merely a filter but an active participant in commutation. The trade-offs are cost, volume, copper loss and, most importantly, a slower current-loop response: a large link inductance makes the drive's torque response sluggish, so the value is chosen as the smallest that satisfies the ripple, $\mathrm{d}i/\mathrm{d}t$ and commutation requirements.

Parts (b) to (d) — Chopper time constant, current limits and time-domain waveforms

Given. A basic (step-down) chopper switches a 24 V source into a series R–L load at a 2 ms period, and the current ripple is specified as a ratio rather than as a pair of magnitudes.

Find. The load time constant and the on-time; the maximum and minimum currents; the time-domain expressions for both sub-intervals and the current at $t=1$ ms and $t=1.5$ ms.

Approach. The free-wheel interval has no driving source, so the current decays purely exponentially and the ratio alone fixes the off-time with no knowledge of the current magnitudes whatever. Only after the timing is known does the steady-state boundary condition set the actual levels.

1. Load time constant. $$\tau=\frac{L}{R}=\frac{0.45\times10^{-3}}{1.8}=0.25\times10^{-3}\ \text{s} =\boxed{0.250\ \text{ms}} .$$ The period is eight time constants long, which is worth noticing: the load will very nearly reach its steady value during the on-time.
2. Off-time from the current ratio alone. During the free-wheel interval the source is disconnected and the current decays from $I_{\max}$ toward zero: $i=I_{\max}e^{-t'/\tau}$. Setting $i=I_{\min}$ at $t'=t_{\text{off}}$, $$\frac{I_{\min}}{I_{\max}}=e^{-t_{\text{off}}/\tau} \;\Rightarrow\; t_{\text{off}}=-\tau\ln(0.75)=0.25\times0.28768 =0.0719\ \text{ms} .$$ No current magnitude was needed, only the ratio.
3. On-time and duty ratio. $$t_{\text{on}}=T-t_{\text{off}}=2.000-0.0719=\boxed{1.9281\ \text{ms}},\qquad \delta=\frac{t_{\text{on}}}{T}=0.9640 .$$ A duty ratio this high is the direct consequence of a load time constant eight times shorter than the period: the current falls very fast once the source is removed, so almost the whole period must be spent charging.
4. Maximum current from the steady-state boundary condition. Requiring the waveform to repeat exactly period after period gives the standard result $$I_{\max}=\frac{V_i}{R}\cdot\frac{1-e^{-t_{\text{on}}/\tau}}{1-e^{-T/\tau}} =13.3333\times\frac{1-e^{-7.7123}}{1-e^{-8}} ,$$ $$\boxed{I_{\max}=13.332\ \text{A}} .$$
5. Minimum current. Directly from the specified ratio, $$I_{\min}=0.75\,I_{\max}=\boxed{9.999\ \text{A}} ,$$ and the free-wheel expression reproduces it, $13.332\,e^{-0.0719/0.25}=9.999$ A. The mean current $\delta V_i/R=12.854$ A lies between the two limits, as it must.
6. Time-domain expression during the on-interval. With the switch closed the load sees the full 24 V and the current rises from $I_{\min}$ towards $V_i/R$: $$i(t)=\frac{V_i}{R}+\left(I_{\min}-\frac{V_i}{R}\right)e^{-t/\tau} =13.333-3.334\,e^{-t/0.25\ \text{ms}},\qquad 0\le t\le 1.9281\ \text{ms}.$$
7. Time-domain expression during the free-wheel interval. With the switch open the current circulates through the freewheel diode and decays to zero with the same time constant, measured from the instant of turn-off: $$i(t')=I_{\max}e^{-t'/\tau}=13.332\,e^{-t'/0.25\ \text{ms}},\qquad 0\le t'\le 0.0719\ \text{ms},\quad t'=t-t_{\text{on}} .$$
8. Evaluate at the two requested instants. Both 1 ms and 1.5 ms are less than $t_{\text{on}}=1.9281$ ms, so both fall in the on-interval and the rising expression applies to each: $$i(1\ \text{ms})=13.333-3.334\,e^{-4}=13.333-0.061=\boxed{13.272\ \text{A}},$$ $$i(1.5\ \text{ms})=13.333-3.334\,e^{-6}=13.333-0.008=\boxed{13.325\ \text{A}} .$$ Screening the instants against $t_{\text{on}}$ before choosing an expression is the whole trick of part (d); a lower duty ratio would have put one of them in the free-wheel interval, and using the wrong branch returns a value larger than $I_{\max}$, which is itself the tell that a mistake has been made.