Question 6 of 6: Types of DC Drive, and a Bridge-Fed Separately Excited DC Motor

Question 6: Types of DC Drive, and a Bridge-Fed Separately Excited DC 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 of dc drives, and the controlled variables

Classification by input supply. DC drives divide into two families according to what supplies them. The first is the ac-fed, line-commutated converter drive, in which a phase-controlled rectifier converts the ac mains directly to a variable dc armature voltage. Within that family the distinctions that matter are the supply phase count (single-phase for fractional and small integral horsepower, three-phase above roughly 10 kW, because the ripple frequency is six times the line frequency rather than twice and the armature inductance needed is correspondingly smaller); the bridge configuration (a half-controlled bridge is cheaper and gives better input power factor but cannot invert, so it is one-quadrant; a fully-controlled bridge can invert and so gives two-quadrant operation with regenerative braking in one direction); and the number of converters (a dual converter of two fully-controlled bridges in antiparallel gives full four-quadrant operation with reversal of both torque and speed, either with a circulating current or in circulating-current-free mode). The second family is the dc-fed chopper drive, supplied from a battery, a traction third rail, a fuel cell, or a diode-rectified dc link, in which a dc–dc converter provides the variable armature voltage. Choppers are classified by quadrant capability as class A (one quadrant, motoring), class B (regenerative braking), class C (two quadrant), class D and class E (four quadrant). Traction and battery-electric applications are almost exclusively chopper drives; industrial mill and machine-tool drives fed from the mains are converter drives.

Variables to be controlled. A separately excited dc machine obeys $E=k_e\Phi\,n$ and $T=k_t\Phi\,I_a$, with $V_a=E+I_aR_a$, so it presents exactly two independent handles and a drive controls both. Armature voltage is the primary speed control below base speed: raising $V_a$ at constant field raises the back EMF and hence the speed, while the flux stays at rated value so that full torque remains available — this is the constant-torque region. Field current, and hence flux, is the control above base speed: with the armature voltage already at its ceiling, weakening the field raises the speed further, but torque falls in proportion to the flux while the product $T n$ stays roughly constant — the constant-power region. Armature current is controlled in its own right, because it is proportional to torque and because commutation and thermal limits impose a hard ceiling on it; every practical drive therefore uses a cascaded structure with an inner current (torque) loop that enforces the limit and an outer speed loop that commands it, with a position loop outside that in servo applications. In converter drives the manipulated variable that closes these loops is the firing angle of the rectifier; in chopper drives it is the duty ratio.

Parts (b) to (d) — Armature voltage, circuit resistance, output and speed

Given. A three-phase fully controlled bridge supplies the armature of a separately excited dc motor from a 230 V line-to-line source; the armature current is held at 150 A at every operating point, and three firing angles with two known speeds are specified.

Find. The armature voltage at the first operating point; the armature-circuit resistance, output power and torque at the second; and the speed at the third.

Approach. Because the armature current is identical at every operating point, the $I_aR_a$ drop is identical too. Differencing two operating points therefore cancels it exactly and isolates the machine constant; a single loop equation then yields the resistance, and the third firing angle needs no new data.

1. Armature voltage from the bridge. For a three-phase fully-controlled bridge in continuous conduction the mean output is $$V_a=\frac{3\sqrt2}{\pi}V_{LL}\cos\alpha=1.35047\,V_{LL}\cos\alpha =310.61\cos\alpha\ \text{V}.$$ At $\alpha_1=43^\circ$, $$\boxed{V_{a1}=310.61\cos43^\circ=227.17\ \text{V}} .$$ Note the coefficient: $3\sqrt2/\pi$ goes with the line-to-line rms voltage. The alternative form $3\sqrt3/\pi$ goes with the peak phase voltage, and mixing them inflates every answer by 22 per cent.
2. Armature voltage at the second operating point. $$V_{a2}=310.61\cos58^\circ=164.60\ \text{V}.$$
3. Eliminate the resistive drop by differencing. At both points $V_a=k_e n+I_aR_a$ with the same $I_aR_a$, so subtracting, $$k_e=\frac{V_{a1}-V_{a2}}{n_1-n_2}=\frac{227.17-164.60}{1720-1000}=\frac{62.57}{720},$$ $$k_e=0.086899\ \text{V per rev/min}\quad(=0.82965\ \text{V}\cdot\text{s/rad}).$$
4. Armature-circuit resistance from either loop equation. Using the first point, $E_1=k_en_1=149.47$ V, so $$R_a=\frac{V_{a1}-E_1}{I_a}=\frac{227.17-149.47}{150},\qquad \boxed{R_a=0.518\ \Omega} .$$ The second point gives $E_2=86.90$ V and $(164.60-86.90)/150=0.518\ \Omega$ — identical, as it must be.
5. Output power at 1000 rev/min. The mechanical power developed is the product of the back EMF and the armature current — not the terminal voltage and the current, which would include the copper loss: $$P_{\text{out}}=E_2I_a=86.90\times150=\boxed{13.03\ \text{kW}} .$$
6. Torque at 1000 rev/min. With $\omega=2\pi n/60=104.72$ rad/s, $$T=\frac{P_{\text{out}}}{\omega}=\frac{13\,035}{104.72}=\boxed{124.5\ \text{N}\cdot\text{m}} .$$ Equivalently $T=k_e'I_a=0.82965\times150=124.4$ N·m using the machine constant in V·s/rad, which is the same statement.
7. Speed at the third firing angle. The armature voltage falls to $$V_{a3}=310.61\cos65^\circ=131.27\ \text{V},$$ and since the current is unchanged the resistive drop is still $I_aR_a=77.70$ V, so the back EMF is $$E_3=131.27-77.70=53.57\ \text{V}\quad\Rightarrow\quad n_3=\frac{E_3}{k_e}=\frac{53.57}{0.086899},$$ $$\boxed{n_3=616\ \text{rev/min}} .$$
8. Sanity check on the trend. The three points lie on the same straight line in the $V_a$–$n$ plane, offset from the origin by the constant 77.70 V drop: 227.17 V at 1720 rev/min, 164.60 V at 1000 rev/min, 131.27 V at 616 rev/min. The slope between the first and third points, $(227.17-131.27)/(1720-616.5)=0.0869$, recovers $k_e$ and closes the problem.