22-Elec-B8 Power Electronics and Drives · May 2013

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

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

Question 6: DC Drive Speed Control, 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.

(a) Speed control below and above base speed

The behaviour of a separately excited dc machine is captured by two equations, $E = k\phi\,n$ for the back-EMF and $T = k\phi\,I_a$ for the torque, together with the armature loop equation $V_a = E + I_a R_a$. Combining the first and the third gives the speed directly:

There are therefore only two ways to change speed — alter the armature voltage $V_a$ or alter the flux $\phi$ — and a dc drive uses them in sequence, with base speed as the dividing line. Base speed is the speed reached at rated armature voltage with rated flux.

Below base speed: armature-voltage control. The field is held at its rated value, so $\phi$ is constant and speed is very nearly proportional to $V_a$. The converter — here a phase-controlled three-phase bridge, or in a low-voltage drive a chopper — is used to set $V_a$ anywhere from zero to rated. The crucial consequence is that with $\phi$ fixed, the torque available at rated armature current is also fixed: this is the constant-torque region. Because output power is $T\omega$, the available power rises in proportion to speed and reaches its rated value exactly at base speed. This is the region used for hoists, mill drives, traction from standstill and anything else needing full torque at low speed.

Above base speed: field weakening. Once $V_a$ has reached the converter’s ceiling, further increase in speed cannot come from voltage. Instead the field current is reduced, weakening $\phi$; since $n \approx V_a/k\phi$, the speed rises in inverse proportion to the flux. But $V_a$ and $I_a$ are now both at their limits, so the input power — and hence the output power — cannot increase: this is the constant-power region, and the available torque falls as $1/n$. Physically the drive is trading torque for speed.

The two regions differ in what is held constant, in what is varied, and in what the machine can deliver, and the field-weakening range is additionally limited in a way that the voltage range is not. Weakening the flux increases the armature current needed for a given torque and worsens both armature reaction and commutation, so the practical range is usually two or three to one. It also introduces a hazard absent below base speed: complete loss of field drives $\phi$ toward zero and the speed toward destruction, so field-failure protection is mandatory. Finally, the field circuit is much more inductive than the armature, so field-weakening response is slow — another reason to prefer armature control wherever the required speed permits.

(b), (c), (d) Bridge-fed separately excited dc motor

Given. A three-phase full-wave (six-pulse) bridge rectifier feeds the armature of a separately excited dc motor, with the data below. The field is constant throughout, and the armature current is held at 170 A at every operating point.

Find. The armature voltage at 48°; the armature circuit resistance, output power and torque at 60° and 1000 rpm; and the speed at 65°.

Approach. Get the mean armature voltage at each firing angle from the six-pulse bridge relation. Because $I_a$ and the flux are both constant, the difference between two operating points eliminates the $I_aR_a$ drop and yields the back-EMF constant; the drop and hence $R_a$ then follow, after which power, torque and the third speed are direct.

1. Mean output voltage of the six-pulse bridge. For a three-phase full-wave controlled bridge with continuous current the mean output voltage is$$V_a = \frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.3503 \times 230\cos\alpha = 310.56\cos\alpha\ \text{V}$$so 310.56 V is the ceiling available at zero firing angle. Each firing angle in the question simply scales this by its cosine.
2. Armature voltage at 48° — part (b). Substituting the first firing angle,$$V_{a1} = 310.56\cos 48^\circ = 310.56(0.66913)$$which gives$$\boxed{V_{a1} = 207.81\ \text{V at }1750\text{ rpm}}$$
3. Armature voltage at 60°. Similarly, for the second operating point,$$V_{a2} = 310.56\cos 60^\circ = 310.56(0.5) = 155.28\ \text{V}\quad\text{at }1000\text{ rpm}$$These two operating points are what make the problem solvable, since two unknowns ($k\phi$ and $R_a$) need two equations.
4. Extract the back-EMF constant — part (c). The armature loop gives $V_a = E + I_aR_a$ at each point, with $E = k_e n$. Because $I_a$ is 170 A at both points, the $I_aR_a$ term is identical and subtracting the two equations cancels it entirely:$$V_{a1} - V_{a2} = k_e(n_1 - n_2) \quad\Longrightarrow\quad k_e = \frac{207.81 - 155.28}{1750 - 1000} = \frac{52.53}{750}$$so $k_e = 0.07004$ V/rpm. The two back-EMFs are then $E_1 = 0.07004(1750) = 122.57$ V and $E_2 = 0.07004(1000) = 70.04$ V.
5. Armature circuit resistance. Returning to either loop equation — the first, say — and solving for the resistance,$$R_a = \frac{V_{a1} - E_1}{I_a} = \frac{207.81 - 122.57}{170} = \frac{85.24}{170}$$so that$$\boxed{R_a = 0.5015\ \Omega}$$Substituting back into the second loop equation confirms it: $70.04 + 170(0.5015) = 155.28$ V, matching $V_{a2}$ exactly.
6. Output power and torque at 1000 rpm. The power converted from electrical to mechanical form is the product of back-EMF and armature current,$$P = E_2 I_a = 70.04 \times 170 = 11\,907\ \text{W} = 11.91\ \text{kW}$$and dividing by the mechanical angular speed $\omega_m = 2\pi(1000)/60 = 104.72$ rad/s gives$$\boxed{P = 11.91\ \text{kW},\qquad T = \frac{11\,907}{104.72} = 113.7\ \text{N}\cdot\text{m}}$$The torque can be checked independently from $T = k_t I_a$ with $k_t = 60k_e/2\pi = 0.6688$ N·m/A, giving $0.6688(170) = 113.7$ N·m as before. Note that because flux and armature current are both constant, this torque applies at every operating point in the question — only the power changes with speed.
7. Speed at 65° — part (d). The armature voltage falls again with the cosine,$$V_{a3} = 310.56\cos 65^\circ = 310.56(0.42262) = 131.27\ \text{V}$$The current is still 170 A, so the drop is unchanged at 85.24 V and the back-EMF is $E_3 = 131.27 - 85.24 = 46.01$ V. Dividing by the back-EMF constant,$$n_3 = \frac{E_3}{k_e} = \frac{46.01}{0.07004}$$which gives$$\boxed{n_3 = 657\ \text{rpm}}$$The trend is right and worth noting: advancing the firing angle from 48° to 60° to 65° drops the speed from 1750 to 1000 to 657 rpm. The last 5° costs 343 rpm because the cosine is steepest near 90° and, more importantly, because the fixed 85 V resistive drop is consuming an ever larger share of a shrinking armature voltage.