22-Elec-B8 Power Electronics and Drives · December 2014

Question 6 of 6: Three-Phase Bridge-Fed Separately Excited DC Drive

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

Question 6: Three-Phase Bridge-Fed Separately Excited DC 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.

Part (a) — Classes of dc drive and the controlled variables

Classified by the supply from which they work, dc drives fall into three families. Single-phase converter-fed drives take a single-phase ac supply through a half-wave, semi-converter, full-converter or dual-converter bridge; they are simple and inexpensive and are used up to roughly 15 kW, but their output ripple is large and they need substantial armature-circuit inductance to maintain continuous conduction. Three-phase converter-fed drives, the class in this question, take a three-phase supply through a six-pulse bridge; the six-times-supply-frequency ripple is much smaller, continuous conduction is far easier to maintain, and ratings extend to several hundred kilowatts, with a dual bridge added where four-quadrant operation is needed. Chopper-fed drives work from an existing fixed dc source — a battery, a traction line or a rectified and filtered supply — and vary the armature voltage by pulse-width modulation at kilohertz rates; they dominate in battery vehicles and rail traction, where regenerative braking back into the source is straightforward.

The variables to be controlled in a dc variable-speed drive follow from the two machine equations $E = k_e\Phi\,n$ and $T = k_t\Phi\,I_a$. Below base speed the armature terminal voltage is varied at full field, which gives constant available torque and a speed proportional to voltage; above base speed the armature voltage is held at its ceiling and the field current, and hence the flux, is weakened, giving constant available power and a speed inversely proportional to flux. Inside those outer laws, the armature current is the controlled variable of the inner loop, because it is directly proportional to torque and must be limited to protect the commutator and the converter; the speed itself is the outer-loop controlled variable, measured by tachometer or encoder. In practice the manipulated variable is the converter firing angle $\alpha$, which the current controller adjusts; the cascade of speed loop over current loop over firing angle is the standard architecture.

Parts (b), (c) and (d) — Drive calculations

Given. A three-phase full-wave bridge fed from a 220 V (line-to-line) supply feeds the armature of a separately excited dc motor which draws a constant 195 A, and three firing angles with two known speeds are quoted.

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

Approach. Write the bridge output voltage as a cosine law in the firing angle, then exploit the constancy of the armature current — which makes the $I_aR_a$ drop identical at every operating point — by subtracting the two loop equations to isolate the speed constant, after which everything else follows from a single loop equation.

1. Part (b) — the bridge output voltage. For a six-pulse fully controlled bridge fed with line-to-line voltage $V_{LL}$, the mean output at firing angle $\alpha$ is $$V_a = \frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.3505\,V_{LL}\cos\alpha = 297.10\cos\alpha\ \text{V}$$ Note the coefficient goes with the line-to-line voltage; the alternative form $3\sqrt{3}/\pi$ belongs with the peak phase voltage, and confusing the two is a 73 per cent error.
2. Part (b) — evaluate at $40^\circ$. $$V_{a1} = 297.10\cos 40^\circ = 297.10 \times 0.76604 = \boxed{227.60\ \text{V}}$$ Continuous conduction is assured because the question states the armature current is maintained at 195 A at all times, so this mean-voltage expression is valid.
3. Part (c) — the second operating point. At $\alpha = 60^\circ$, $$V_{a2} = 297.10\cos 60^\circ = 148.55\ \text{V}$$ The armature loop equation at each point is $V_a = E + I_aR_a$ with $E = k_e n$, the field being constant.
4. Part (c) — difference the two loop equations. Because $I_a$ is the same 195 A at both points, the term $I_aR_a$ is identical and cancels on subtraction: $$V_{a1} - V_{a2} = k_e(n_1 - n_2)$$ $$k_e = \frac{227.60 - 148.55}{1750 - 1000} = \boxed{0.10539\ \text{V/rpm}}$$ This is the pivot of the whole question: one subtraction removes the unknown resistance and delivers the machine constant directly.
5. Part (c) — recover the armature-circuit resistance. The back EMF at 1000 rpm is $E_2 = 0.10539 \times 1000 = 105.39$ V, so from the loop equation at that point $$R_a = \frac{V_{a2} - E_2}{I_a} = \frac{148.55 - 105.39}{195} = \boxed{0.2213\ \Omega}$$ Substituting back into the first point gives $0.10539(1750) + 195(0.2213) = 227.60$ V, reproducing $V_{a1}$ exactly, so both loop equations are satisfied.
6. Part (c) — output power and torque. Mechanical power is developed at the back EMF, not at the terminals: $$P_{out} = E_2 I_a = 105.39 \times 195 = \boxed{20.55\ \text{kW}}$$ and with $\omega = 2\pi(1000)/60 = 104.72$ rad/s, $$T = \frac{P_{out}}{\omega} = \frac{20\,551}{104.72} = \boxed{196.25\ \text{N}\cdot\text{m}}$$ Using $V_{a2}I_a$ instead of $E_2I_a$ would give 28.97 kW, 41 per cent too high, because it counts the resistive loss as useful output.
7. Part (d) — the third firing angle. With the same constant current the $I_aR_a$ drop is unchanged at $195 \times 0.2213 = 43.16$ V, so $$V_{a3} = 297.10\cos 65^\circ = 125.56\ \text{V}, \qquad E_3 = 125.56 - 43.16 = 82.40\ \text{V}$$ and therefore $$n_3 = \frac{E_3}{k_e} = \frac{82.40}{0.10539} = \boxed{782\ \text{rpm}}$$ Retarding the firing by a further $5^\circ$ has cost 218 rpm, more than the 750 rpm lost over the previous $20^\circ$ per degree, because the cosine law steepens as $\alpha$ approaches $90^\circ$.