Question 6 of 6: Bridge-Fed Separately Excited D.C. Drive

Question 6: Bridge-Fed Separately Excited D.C. Drive (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.

Given. A three-phase fully controlled bridge supplied from a 230 V (line-to-line) source feeds the armature of a separately excited d.c. motor whose armature current is held at 120 A at every operating point by the current loop.

Find. The types of d.c. drive classified by input supply and the variables controlled in a variable-speed d.c. drive; then the armature voltage at point A, the armature circuit resistance, output power and torque at point B, and the speed at point C.

Approach. Use the six-pulse bridge relation to convert each firing angle into an armature voltage. Because the armature current, and hence the resistive drop, is the same at every point, differencing two operating points cancels $I_aR_a$ and isolates the back-emf constant; one loop equation then yields $R_a$, and a third firing angle needs no new data.

Lead item — Types of d.c. drive and the controlled variables

Classified by the input supply, d.c. drives fall into three families. Single-phase drives take a single-phase a.c. supply through a half-wave, semi-converter, full converter or dual converter, and are used up to roughly 10–15 kW where only a single-phase supply is available; the half-wave and semi-converter forms are one-quadrant, the full converter is two-quadrant (it can invert and so regenerate but cannot reverse current), and the dual converter is four-quadrant. Three-phase drives take a three-phase supply through the same converter family, and are the standard choice above a few kilowatts because the six-pulse output has far lower ripple, a higher ripple frequency and much better supply utilisation and power factor. Chopper drives take a d.c. input — a battery, a traction third rail, or a rectified and filtered a.c. supply — and control the armature with a d.c. – d.c. converter; they dominate battery-electric traction and material handling, where regenerative braking through a two- or four-quadrant chopper is a defining advantage.

Three variables are controlled in a variable-speed d.c. drive, and they nest. Armature voltage is the primary speed control below base speed: with the field held at rated flux, speed is proportional to $V_a-I_aR_a$, and this region delivers constant available torque. Field current, and therefore flux, is the control above base speed: weakening the field raises speed beyond base at approximately constant power, with torque falling in inverse proportion. Armature current is the innermost controlled variable, held by a fast current loop inside the speed loop; because torque is proportional to $\Phi I_a$, controlling current is controlling torque, and it is also what enforces the commutator and thermal limits during acceleration. The three together give the classical constant-torque-then-constant-power speed envelope.

Parts (a) to (c) — Quantitative solution

1. Part (a) — convert the firing angle to an armature voltage. For a three-phase fully controlled bridge with continuous current, the mean output voltage is $$V_a=\frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.35047\,V_{LL}\cos\alpha .$$ With $V_{LL}=230$ V the bridge constant is $1.35047\times230=310.61$ V, so at $\alpha_1 = 42.5^\circ$ $$V_{a1}=310.61\cos 42.5^\circ=310.61\times0.73728=\boxed{229.0\ \text{V}} .$$ (The alternative form $3\sqrt{3}/\pi$ goes with the peak phase voltage and would give an answer high by $\sqrt{3}$; the line-to-line form is the one to use here.)
2. Part (b) — evaluate the second operating point and difference the two. At $\alpha_2 = 57^\circ$, $V_{a2}=310.61\cos 57^\circ = 169.17$ V. The armature loop is $V_a = E + I_aR_a$ with $E = k_e n$, and since $I_a$ is the same 120 A at both points the resistive drop cancels on subtraction: $$k_e=\frac{V_{a1}-V_{a2}}{n_1-n_2}=\frac{229.00-169.17}{1720-1000} =\frac{59.84}{720}=0.083104\ \text{V/rpm} .$$
3. Recover the armature circuit resistance. The back emfs are $E_1 = 0.083104\times1720 = 142.94$ V and $E_2 = 0.083104\times1000 = 83.10$ V, so $$R_a=\frac{V_{a1}-E_1}{I_a}=\frac{229.00-142.94}{120}=\boxed{0.717\ \Omega},$$ and the second point returns the same value, $(169.17-83.10)/120 = 0.717\ \Omega$, which confirms the differencing step.
4. Evaluate output power and torque at 1000 rpm. The developed mechanical power is the back emf times the armature current, never the terminal voltage times the current, because the copper loss must not be counted as output: $$P_{out}=E_2I_a=83.10\times120=\boxed{9.97\ \text{kW}} .$$ At $\omega = 2\pi(1000)/60 = 104.72$ rad/s the developed torque is $$T=\frac{P_{out}}{\omega}=\frac{9972.5}{104.72}=\boxed{95.2\ \text{N}\cdot\text{m}} .$$
5. Part (c) — find the speed at the third firing angle. No new data are needed. At $\alpha_3 = 65^\circ$, $V_{a3}=310.61\cos 65^\circ = 131.27$ V, and the same loop equation gives $$E_3=V_{a3}-I_aR_a=131.27-120(0.71721)=45.20\ \text{V},$$ $$n_3=\frac{E_3}{k_e}=\frac{45.20}{0.083104}=\boxed{544\ \text{rpm}} .$$
6. Structural check on the whole table. Since torque is $T=k_e(60/2\pi)I_a$ and both $k_e$ and $I_a$ are constant, every operating point in this problem develops the same 95.2 N$\cdot$m; only the speed, and therefore the power, changes. A torque that differed from point to point would be an arithmetic slip, and this one-line test catches it immediately.