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

Question 6 of 6: Three-Phase Bridge Converter Driving a Separately Excited D.C. Motor

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

Question 6: Three-Phase Bridge Converter Driving a Separately Excited D.C. Motor (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) — Classification and controlled variables

Types by input supply. D.C. drives are classified first by what feeds the converter. Single-phase converter drives take a single-phase a.c. supply and use a half-wave, semi-converter, full converter or dual-converter bridge; they are limited to small ratings, typically up to about 15 kW, and their high ripple frequency of two or four times the supply frequency demands a substantial smoothing reactor. Three-phase converter drives take a three-phase supply and are the workhorse for medium and large ratings; the six-pulse bridge gives a ripple frequency of six times the supply frequency, so the armature current is far smoother, continuous conduction is easier to maintain, and the converter can be built for hundreds of kilowatts. Chopper (d.c.–d.c.) drives take a fixed d.c. supply — a battery, a traction third rail, or a diode-rectified link — and vary the armature voltage by duty ratio; these dominate battery-electric traction and mobile equipment. Within each class, a single converter gives one- or two-quadrant operation, while dual converters and four-quadrant choppers add reversal and regeneration.

Variables to be controlled. The primary controlled variable below base speed is the armature voltage, adjusted through the firing angle or duty ratio; because the flux is held at its rated value, this gives constant-torque capability from standstill to base speed. Above base speed the field current, and hence the flux, is reduced — field weakening — which extends the speed range at constant power at the cost of proportionally reduced torque. The armature current is controlled in an inner loop, both because it is the direct analogue of torque and because it must be limited during acceleration and fault conditions to protect the commutator and the devices. In a full drive these appear as a cascaded structure: an outer speed loop generating a current reference, an inner current loop generating the firing angle, and a separate field regulator; the direction of rotation and the ability to regenerate are handled by converter topology rather than by an additional loop.

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

Given.

Find. The armature voltage at $45^\circ$; the armature-circuit resistance, developed power and torque at $55^\circ$ and 1000 rev/min; and the speed reached at $65^\circ$.

Approach. The bridge fixes the armature voltage from the firing angle alone; two operating points at the same armature current then give two equations in the two unknowns $k_e$ and $R_a$, after which the third firing angle is a direct substitution.

1. Write the converter transfer relation. For a fully controlled six-pulse bridge in continuous conduction, the mean output voltage is $$V_a = \frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.35\,V_{LL}\cos\alpha = 310.61\cos\alpha\ \ \text{V}.$$
2. Evaluate at the first firing angle. $$V_{a1} = 310.61\times\cos 45^\circ = 310.61\times0.70711$$ $$\boxed{V_{a1} = 219.63\ \text{V at }1700\ \text{rev/min}}$$
3. Evaluate at the second firing angle. $V_{a2} = 310.61\times\cos 55^\circ = 310.61\times0.57358 = 178.16$ V, this time at 1000 rev/min.
4. Set up the two armature equations. For a separately excited machine at constant flux, $V_a = E + I_aR_a$ with $E = k_eN$. Both operating points carry the same 120 A, so $$219.63 = 1700k_e + 120R_a, \qquad 178.16 = 1000k_e + 120R_a .$$
5. Eliminate the resistive drop. Subtracting removes $I_aR_a$ entirely — this is exactly why the constant-current statement in the question matters: $$41.48 = 700k_e \;\Longrightarrow\; k_e = 0.059251\ \text{V per rev/min}.$$ Equivalently $k_t = k_e\times60/(2\pi) = 0.5658$ N·m/A.
6. Back out the armature-circuit resistance. At 1700 rev/min the back e.m.f. is $E_1 = 0.059251\times1700 = 100.73$ V, so $$\boxed{R_a = \frac{V_{a1}-E_1}{I_a} = \frac{219.63-100.73}{120} = 0.991\ \Omega}$$ Substituting back into the 1000 rev/min equation reproduces 178.16 V exactly, confirming the pair is consistent.
7. Find the developed power and torque at 1000 rev/min. The back e.m.f. is $E_2 = 0.059251\times1000 = 59.25$ V, and the power converted to mechanical form is the product of back e.m.f. and armature current, the copper loss having already been accounted for in $R_a$: $$\boxed{P_{out} = E_2I_a = 59.25\times120 = 7110\ \text{W}}$$ With $\omega = 2\pi\times1000/60 = 104.72$ rad/s, $$\boxed{T = \frac{P_{out}}{\omega} = \frac{7110}{104.72} = 67.90\ \text{N}\cdot\text{m}}$$ The same figure follows from $T = k_tI_a = 0.5658\times120 = 67.90$ N·m, which is a useful cross-check because it uses $k_e$ rather than the power.
8. Advance the firing angle to $65^\circ$ and find the new speed. The armature voltage becomes $V_{a3} = 310.61\times\cos 65^\circ = 131.27$ V. The armature current is stated to remain 120 A, so the resistive drop is unchanged at $120\times0.991 = 118.91$ V and $$E_3 = 131.27 - 118.91 = 12.36\ \text{V},$$ $$\boxed{N_3 = \frac{E_3}{k_e} = \frac{12.36}{0.059251} = 208.6\ \text{rev/min}}$$
9. Sanity-check the trend. Ten degrees of extra delay between $45^\circ$ and $55^\circ$ cost 700 rev/min; the next ten degrees, between $55^\circ$ and $65^\circ$, cost 791 rev/min. The acceleration of the loss is the cosine steepening as $\alpha$ approaches $90^\circ$, and it is why phase-controlled drives lose controllability — and power factor — at low speed.