Question 6 of 6: Three-Phase Bridge Speed Control of a Separately Excited D.C. Motor

Question 6: Three-Phase Bridge Speed Control of 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) — the bridge and how it controls speed

A three-phase full-wave bridge uses six thyristors in two groups of three. The upper group connects the most positive line to the positive d.c. terminal and the lower group connects the most negative line to the negative terminal, so at any instant one device from each group conducts and the load sees a line-to-line voltage. Each device conducts for $120^{\circ}$ and the conduction pattern advances every $60^{\circ}$, giving six pulses of output ripple per supply cycle at 360 Hz on a 60 Hz system. Delaying each gate pulse by an angle $\alpha$ measured from the natural commutation instant reduces the mean output to$$V_{a}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha=1.3505\,V_{LL}\cos\alpha$$which is continuously adjustable from $1.3505V_{LL}$ down through zero at $\alpha=90^{\circ}$ and negative beyond it. The six-pulse ripple is small and at high frequency, so the armature inductance alone usually holds the current continuous, which is the condition under which the cosine law above is valid.

For a separately excited motor the steady-state relations are $V_{a}=E+I_{a}R_{a}$ with $E=k\phi\,n$ and $T=k\phi\,I_{a}$. Below rated speed the field is held at its rated value and the armature voltage is varied by $\alpha$: since the flux is constant, torque per ampere is constant, and the drive delivers constant torque capability all the way down to standstill. This is the region the numerical parts of this question sit in. Above rated speed the armature voltage has already reached its ceiling at $\alpha\to 0$, so further speed is obtained by weakening the field with a second, smaller converter. Speed then rises as $1/\phi$ while the available torque falls as $\phi$, giving a constant-power region. The two regions together form the familiar constant-torque then constant-power envelope of a thyristor d.c. drive.

One further property matters in practice: with $\alpha>90^{\circ}$ the mean armature voltage reverses and the bridge inverts, returning energy to the a.c. line. A single bridge can therefore regenerate while driving in one direction; four-quadrant operation needs a second anti-parallel bridge or a field reversal.

Parts (b), (c) and (d) — machine constants, power and speed

Given.

Find. The armature voltage at $45^{\circ}$; the armature-circuit resistance, output power and torque at $55^{\circ}$; and the speed at $65^{\circ}$.

Approach. Convert each firing angle to an armature voltage with the bridge cosine law, difference two operating points to eliminate the unknown resistive drop and isolate the back-e.m.f. constant, then use one loop equation to recover the resistance and the third angle to find the new speed.

1. Evaluate the bridge constant once. $3\sqrt{2}/\pi=1.35047$, so $V_{a}=1.35047\times 220\cos\alpha=297.10\cos\alpha$ volts. Note that this is the form using the line-to-line rms voltage; the alternative $3\sqrt{3}/\pi$ belongs with the peak phase voltage and would inflate every answer by $\sqrt{3}$.
2. Answer part (b) directly.$$\boxed{V_{a1}=297.10\cos 45^{\circ}=210.1\ \text{V}}$$at 1750 rev/min. For part (c) the same law gives $V_{a2}=297.10\cos 55^{\circ}=170.4$ V at 1200 rev/min.
3. Difference the two loop equations. Writing $V_{a}=k_{e}n+I_{a}R_{a}$ at each point and subtracting, the resistive drop cancels identically because $I_{a}$ is the same at both:$$k_{e}=\frac{V_{a1}-V_{a2}}{n_{1}-n_{2}}=\frac{210.08-170.41}{1750-1200}=0.07213\ \text{V per rev/min}$$This is the step the question is built around; without the constant-current statement there would be two unknowns and only two equations containing three.
4. Recover the armature-circuit resistance. The back e.m.f. at the first point is $E_{1}=0.072135\times 1750=126.2$ V, so$$\boxed{R_{a}=\frac{V_{a1}-E_{1}}{I_{a}}=\frac{210.08-126.24}{165}=0.508\ \Omega}$$The second loop equation returns the same figure, as it must.
5. Compute the output power and torque at 1200 rev/min. The developed (mechanical) power is the back e.m.f. times the armature current, never the terminal voltage times the current — the difference is the copper loss. With $E_{2}=0.072135\times 1200=86.56$ V:$$\begin{aligned}P_{\text{out}}&=E_{2}I_{a}=86.56\times 165=14.28\ \text{kW}\\\omega_{m}&=\frac{2\pi\times 1200}{60}=125.66\ \text{rad/s}\end{aligned}$$$$\boxed{P_{\text{out}}=14.28\ \text{kW},\qquad T=\frac{14282}{125.66}=113.7\ \text{N}\cdot\text{m}}$$
6. Find the speed at the third firing angle. $V_{a3}=297.10\cos 65^{\circ}=125.54$ V, and the resistive drop is unchanged at $I_{a}R_{a}=165\times 0.50818=83.85$ V, so$$E_{3}=125.54-83.85=41.69\ \text{V}\;\Longrightarrow\;\boxed{n_{3}=\frac{41.69}{0.072135}=578\ \text{rev/min}}$$No new data were needed — the two constants found above carry the whole characteristic.