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 points)

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 variables they control

Types of dc drive by input supply. Dc drives divide first according to whether the incoming supply is alternating or already direct.

AC-fed (phase-controlled, rectifier-type) drives take an ac supply and use a line-commutated thyristor converter to produce a variable dc armature voltage. Within this family the subdivisions are by phase number and converter configuration: single-phase drives — half-wave for very small motors (fractional kW), the half-controlled or semi-converter bridge for two-quadrant motoring with a lower device count and better input power factor, and the fully-controlled bridge where inversion is required; and three-phase drives — the half-controlled bridge, the fully-controlled six-pulse bridge (the configuration of this question, used from a few kW to several MW), and the twelve-pulse arrangement for very large drives where supply harmonics must be reduced. Where four-quadrant operation is needed, two fully-controlled bridges are connected back-to-back as a dual converter, either with circulating current through reactors or in circulating-current-free mode. Line-commutated converters are simple and rugged, but they draw a lagging displacement current that worsens as the firing angle increases, and they inject characteristic harmonics into the supply.

DC-fed (chopper) drives take an existing dc supply — a battery, a traction third rail or overhead line, a fuel cell, or a diode rectifier with a dc link — and use a dc-to-dc converter to vary the armature voltage. Classified by quadrant capability these are the class-A (first-quadrant) chopper for motoring only, the class-B (second-quadrant) chopper for regenerative braking, the class-C two-quadrant chopper, and the class-E four-quadrant chopper for full reversing service. Chopper drives switch at several hundred hertz to several kilohertz rather than at line frequency, so they give a much faster current loop, smaller armature current ripple for the same inductance, and a near-unity input displacement factor; they are standard in traction and battery vehicles.

Variables controlled in a dc variable-speed drive. The separately excited dc machine obeys $E_b = k\Phi\omega$ and $T = k\Phi I_a$, with $V_a = E_b + I_aR_a$. These three equations identify what is worth manipulating.

• Armature voltage $V_a$ — the principal speed control below base speed. With the field held at rated flux, speed is very nearly proportional to $V_a$, and the full rated torque remains available at every speed, so this is the constant-torque region. In a rectifier drive $V_a$ is set by the firing angle $\alpha$, which is therefore the actual manipulated variable.
• Field current, i.e. flux $\Phi$ — used above base speed, once $V_a$ has reached its ceiling. Weakening the field raises speed in inverse proportion but reduces available torque for the same armature current, giving the constant-power region. The field is fed from its own small controlled rectifier.
• Armature current $I_a$ — controlled in an inner, faster loop inside the speed loop. Because torque is proportional to $I_a$ at fixed flux, this loop is the torque loop; it also enforces the current limit that protects the commutator and the converter during acceleration and stalls.
• Speed $\omega$ (and, in servo applications, shaft position) — the outermost loop, closed on a tachogenerator or encoder, whose output is the current reference.

The cascade structure — speed loop commanding a current loop commanding a firing angle — is what allows a dc drive to hold speed accurately while never exceeding a safe armature current, and it is the arrangement used in the drive analysed below.

Part (b) — Completing the drive table

Given.

Find. The rectifier output voltage, armature-circuit resistance, back emf, shaft speed, firing angle and output torque needed to complete all four rows.

Approach. Write the bridge transfer characteristic and the armature loop equation; difference the two constant-current cases to eliminate $I_aR_a$ and isolate $k_e$; back-substitute for $R_a$; then complete each remaining row by forward substitution.

1. Part (b) — Write the converter transfer characteristic. For a three-phase fully-controlled bridge the mean output voltage in continuous conduction is $$V_{dc} = \frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.35047\times230\times\cos\alpha = 310.61\cos\alpha\ \text{V}.$$ Note the constant: $3\sqrt{2}/\pi$ goes with the line-to-line rms voltage supplied here. Evaluating for the three given angles, $$V_A = 219.63\ \text{V},\qquad V_B = 155.30\ \text{V},\qquad V_C = 131.27\ \text{V}.$$
2. Exploit the fact that the armature current is the same in cases A and B. The armature loop equation is $V_{dc} = E_b + I_aR_a$ with $E_b = k_e n$. Because $I_a = 125$ A in both cases, the term $I_aR_a$ is identical in both, and differencing the two loop equations cancels it entirely: $$V_A - V_B = k_e(n_A - n_B) \;\Longrightarrow\; k_e = \frac{219.63 - 155.30}{1720 - 1000} = \frac{64.33}{720} = \boxed{0.089346\ \text{V/rpm}}.$$ This is the step that makes the problem solvable with no motor nameplate data at all.
3. Recover the armature-circuit resistance. Substituting $k_e$ back into either loop equation, $$R_a = \frac{V_A - k_en_A}{I_a} = \frac{219.634 - 153.675}{125} = \frac{65.959}{125} = \boxed{0.5277\ \Omega},$$ and case B independently returns $(155.305 - 89.346)/125 = 0.5277\ \Omega$, confirming the pair.
4. Fill in the back emfs for the two given cases. $$E_A = k_en_A = 0.089346\times1720 = \boxed{153.68\ \text{V}},\qquad E_B = k_en_B = 0.089346\times1000 = \boxed{89.35\ \text{V}}.$$
5. Case C — back emf and shaft speed. The firing angle is given and the current is unchanged, so the loop equation runs forwards: $$E_C = V_C - I_aR_a = 131.269 - 125(0.52767) = 131.269 - 65.958 = \boxed{65.31\ \text{V}},$$ $$n_C = \frac{E_C}{k_e} = \frac{65.31}{0.089346} = \boxed{731.0\ \text{rpm}}.$$
6. Establish the torque constant and fill the torque column. Shaft torque is developed power divided by mechanical speed, and the developed power is $E_bI_a$ — not $V_{dc}I_a$, which includes the armature copper loss. Since $E_b = k_en$ with $n$ in rpm, $$T_a = \frac{E_bI_a}{\omega} = \frac{k_e n I_a}{2\pi n/60} = k_e\frac{60}{2\pi}I_a = 0.85319\,I_a.$$ The torque therefore depends only on the current, so cases A, B and C all develop the same torque: $$T_A = T_B = T_C = 0.85319\times125 = \boxed{106.65\ \text{N}\cdot\text{m}}.$$ As a check, computing case A the long way gives $153.68\times125/(2\pi\times1720/60) = 106.65$ N·m, and case C gives the same, which validates $k_e$, $R_a$ and $n_C$ together.
7. Case D — identify the missing constraint. Case D supplies only $I_a = 100$ A and asks for both $\alpha$ and $n$; one loop equation cannot determine two unknowns, so a further condition is required. The physically natural one, and the one adopted here, is that the drive's speed loop holds the shaft at the case-C speed of 731.0 rpm while the load torque falls, which is exactly what a speed-regulated drive does when the load lightens. The general locus without that assumption is $n = \left[310.61\cos\alpha - 100(0.5277)\right]/0.089346$.
8. Case D — complete the row under that assumption. Holding the speed holds the back emf, so $E_D = E_C = 65.31$ V and $$V_D = E_D + I_aR_a = 65.31 + 100(0.5277) = \boxed{118.08\ \text{V}},$$ $$\cos\alpha_D = \frac{118.077}{310.609} = 0.38015 \;\Longrightarrow\; \boxed{\alpha_D = 67.66^\circ},$$ $$T_D = 0.85319\times100 = \boxed{85.32\ \text{N}\cdot\text{m}}.$$ For contrast, had the firing angle instead been left at 65° while the load fell to 100 A, the machine would have accelerated to 878.6 rpm — the figure is quoted so a marker can see the alternative reading was considered.

Final results — completed table.

Speed constant $k_e = 0.089346$ V/rpm; torque constant $k_t = k_e(60/2\pi) = 0.85319$ N·m/A.