22-Elec-B8 Power Electronics and Drives · December 2015

Question 3 of 6: Smoothing Reactors and a Basic Step-Down Chopper

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

Question 3: Smoothing Reactors and a Basic Step-Down 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) — Why inverter circuits carry series smoothing reactors

Limiting $di/dt$ at turn-on. A thyristor begins conducting in a small area around the gate and the conducting region spreads outwards at a finite velocity. If the anode current rises faster than the device's rated $di/dt$, the initial current crowds into that small area and produces a local hot spot that destroys the junction long before the average current rating is approached. A series reactor holds $di/dt = v/L$ within the rating and is the standard protective measure in every line-commutated and force-commutated thyristor inverter.

Making the d.c. link behave as a stiff current source. In a current-source inverter the whole control philosophy depends on the link current being essentially constant over a switching period. The link reactor supplies that stiffness: it absorbs the difference between the rectifier's pulsating output voltage and the inverter's demand, so the link current changes only on the slow timescale of the speed loop and not with every commutation. Without it the inverter would see a current that collapses during each commutation interval and the machine-side waveform would lose its rectangular shape.

Ripple, commutation and decoupling. A larger link inductance reduces the peak-to-peak ripple in the link current, which lowers the rms heating in the devices and the machine, reduces the pulsating torque that ripple produces in the shaft, and eases the duty on the d.c.-link capacitor where one is present. The stored energy also assists forced commutation by maintaining current through the commutating capacitor for the required turn-off time. Finally, the reactor decouples the inverter from the supply: it slows the rate of rise of any fault or shoot-through current, buying time for the fuses and the electronic protection to act, and it attenuates the switching-frequency harmonic currents that would otherwise be injected back into the a.c. system.

Parts (b) to (d) — The chopper calculation

Given.

Find. The load time constant and the on time, the two limits of the steady-state ripple, the time-domain current in both sub-intervals, and the instantaneous current one millisecond into the period.

Approach. In the periodic steady state the exponential rise during the on time and the exponential decay during the off time must return the current to where it started, and that closure turns the stated ripple ratio into a direct statement about the off time alone.

1. Compute the load time constant. Both sub-intervals are governed by the same R–L pair, so $$\tau = \frac{L}{R} = \frac{0.3\times10^{-3}}{0.1}$$ $$\boxed{\tau = 3.000\ \text{ms}}$$ Note that $\tau$ is 1.25 times the chopping period, so the current cannot come close to either its final value or zero within one cycle — the ripple will be a small fraction of the mean.
2. Write the two steady-state limits. During the on time the switch connects the source and the current climbs from $I_{\min}$ towards $V_1/R$; during the off time the diode free-wheels the current and it decays from $I_{\max}$ towards zero (there is no back e.m.f. to arrest it). Requiring periodicity gives the standard pair $$I_{\max} = \frac{V_1}{R}\cdot\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}}, \qquad I_{\min} = I_{\max}\,e^{-t_{off}/\tau}.$$
3. Turn the ripple ratio into the off time. The second expression already contains the answer — the ratio depends on nothing but the off time: $$\frac{I_{\min}}{I_{\max}} = e^{-t_{off}/\tau} = 0.75 \;\Longrightarrow\; t_{off} = -\tau\ln 0.75 = 3.000 \times 0.287682 = 0.8630\ \text{ms}.$$ The on time is the remainder of the period, $$\boxed{t_{on} = T - t_{off} = 2.4 - 0.8630 = 1.5370\ \text{ms}}$$ corresponding to a duty ratio $\delta = t_{on}/T = 0.6404$.
4. Evaluate the maximum current. With $t_{on}/\tau = 0.512318$ and $T/\tau = 0.8$, so $e^{-0.512318} = 0.599090$ and $e^{-0.8} = 0.449329$, and with $V_1/R = 240$ A, $$I_{\max} = 240\times\frac{1-0.599090}{1-0.449329} = 240\times\frac{0.400910}{0.550671} = 240 \times 0.728034$$ $$\boxed{I_{\max} = 174.72\ \text{A}}$$
5. Recover the minimum from the stated ratio. $$I_{\min} = 0.75\,I_{\max} = 0.75\times174.72$$ $$\boxed{I_{\min} = 131.04\ \text{A}}$$ As an independent check, the approximate mean current $\delta V_1/R = 0.6404\times240 = 153.70$ A lies between the two limits, as it must, and the peak-to-peak ripple of 43.68 A is 28 % of the mean.
6. Write the on-interval current. Measuring $t$ from the start of the on time, the general R–L step response starting from $I_{\min}$ is $$i_{on}(t) = \frac{V_1}{R}\Big(1-e^{-t/\tau}\Big) + I_{\min}e^{-t/\tau} = 240 - 108.96\,e^{-t/3\,\text{ms}}\ \ \text{A}, \qquad 0\le t\le 1.5370\ \text{ms}.$$ At $t = t_{on}$ this returns $240 - 108.96\times0.599090 = 174.72$ A, matching $I_{\max}$.
7. Write the free-wheel current. Measuring $t'$ from the instant the switch opens, the source is disconnected and the current simply decays from $I_{\max}$: $$i_{off}(t') = I_{\max}e^{-t'/\tau} = 174.72\,e^{-t'/3\,\text{ms}}\ \ \text{A}, \qquad 0\le t'\le 0.8630\ \text{ms},$$ which returns 131.04 A at $t' = t_{off}$ and so closes the cycle exactly.
8. Evaluate the current one millisecond into the period. Since $1$ ms is less than $t_{on} = 1.5370$ ms, the instant lies in the on interval and the first expression applies: $$i(1\ \text{ms}) = 240 - 108.96\,e^{-1/3} = 240 - 108.96\times0.716531 = 240 - 78.07$$ $$\boxed{i(1\ \text{ms}) = 161.93\ \text{A}}$$