Question 4 of 6: Inverter Auxiliaries and Chopper Ripple with a Changed Load Resistance

Question 4: Inverter Auxiliaries and Chopper Ripple with a Changed Load Resistance (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

A clamping capacitor limits the rate of rise and the peak of the voltage that appears across a switching device at turn-off. When a switch interrupts current in a circuit that unavoidably contains stray and load inductance, the collapsing current generates $L\,di/dt$ across the device; without a path for that energy the device is driven past its repetitive peak rating and fails. A capacitor connected across the device (usually with a series diode and a discharge resistor, forming a snubber) absorbs the trapped energy, clamps the excursion to a safe level and, by slowing the voltage rise, keeps the device inside its safe operating area during the switching transition. In a thyristor inverter it does a second job: it holds the $dv/dt$ below the value at which the device would self-trigger through its own internal junction capacitance. Clamping capacitors also stabilise a d.c. link against the pulsed current the bridge draws, which keeps the link voltage stiff between commutations.

A smoothing reactor is inductance deliberately added in series in the d.c. path. It reduces ripple in the d.c. link or armature current, which lowers the rms heating for a given mean current, reduces torque pulsation in a machine, and keeps a converter in continuous conduction down to light load — important because the control characteristic of a phase-controlled rectifier becomes non-linear and load-dependent the moment conduction goes discontinuous. It also limits the fault $di/dt$ through the semiconductors during a shoot-through or a commutation failure, giving the protection time to act, and it attenuates conducted harmonic current flowing back into the supply. In a current-source inverter the d.c. reactor is not an auxiliary at all but the element that defines the topology, since it is what makes the link behave as a current source.

Together they perform complementary duties: the capacitor holds the voltage stiff and bounded, and the reactor holds the current smooth and bounded. Both are sized from the energy stored in the parasitic elements and the ripple that the load can tolerate, not from the mean power flow.

Parts (b), (c) and (d) — chopper period and the effect of doubling R

Given.

Find. The chopper period, the minimum output current, and both current extremes after the resistance is doubled with the timing unchanged. There is no back e.m.f. in this problem.

Approach. Invert the closed form for $I_{\max}$ to get the period, evaluate $I_{\min}$ at that period, then recompute the time constant for the new resistance and re-evaluate both with the timing held fixed.

1. Fix the on-time and note that the inductance is what is really held constant. $t_{on}=0.9\tau=0.9\times 1.5=1.35$ ms, and $L=R\tau=0.2\times 1.5\times 10^{-3}=0.30$ mH. Part (d) keeps this inductance, not the time constant — that distinction is the whole content of the last sub-part.
2. Invert the peak-current expression for the period. With no back e.m.f. the steady-state peak is$$I_{\max}=\frac{V_{i}}{R}\,\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}}$$so the only unknown, $T$, is isolated by rearranging:$$1-e^{-T/\tau}=\frac{V_{i}}{R\,I_{\max}}\left(1-e^{-t_{on}/\tau}\right)=\frac{150}{90}\times 0.59343=0.98905$$
3. Solve for the period (part b). Taking logarithms, $T/\tau=-\ln(1-0.98905)=4.5145$, hence$$\boxed{T=1.5\ \text{ms}\times 4.5145=6.77\ \text{ms}}$$a switching frequency of 148 Hz and a duty ratio of only $\delta=1.35/6.77=0.199$. The off-time is more than three time constants long, which is why the ripple is so deep.
4. Evaluate the minimum current (part c).$$I_{\min}=\frac{V_{i}}{R}\,\frac{e^{t_{on}/\tau}-1}{e^{T/\tau}-1}=150\times\frac{1.4596}{90.324}$$$$\boxed{I_{\min}=2.42\ \text{A}}$$Equivalently, the free-wheel decay from 90 A over $t_{off}/\tau=3.615$ gives $90\,e^{-3.615}=2.42$ A — a useful independent route.
5. Recompute the time constant for the doubled resistance (part d). The inductance is unchanged, so$$\tau^{\prime}=\frac{L}{R^{\prime}}=\frac{0.30\times 10^{-3}}{0.4}=0.75\ \text{ms}$$and with $T$ and $t_{on}$ held fixed the normalised times both double: $t_{on}/\tau^{\prime}=1.80$ and $T/\tau^{\prime}=9.029$.
6. Re-evaluate the two extremes. With $V_{i}/R^{\prime}=75$ A:$$\boxed{I_{\max}=75\times\frac{0.83470}{0.99988}=62.61\ \text{A},\qquad I_{\min}=75\times\frac{5.0496}{8338.4}=0.045\ \text{A}}$$The peak falls by 30 per cent while the minimum collapses by a factor of more than fifty.
7. Check with the mean and interpret. The mean must be $\delta V_{i}/R^{\prime}=0.199\times 30/0.4=14.95$ A, which lies inside the band $0.045$ to $62.61$ A. The load is now within 0.05 A of losing conduction entirely: the off-time is 9.0 time constants and the current has essentially decayed away before the next pulse arrives.