22-Elec-B8 Power Electronics and Drives · December 2015

Question 2 of 6: Harmonic Sources and a Full-Wave Controlled Rectifier Feeding a Back E.M.F.

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

Question 2: Harmonic Sources and a Full-Wave Controlled Rectifier Feeding a Back E.M.F. (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) — Three routes by which harmonics enter a distribution system

Power-electronic converters. Every line-commutated or switch-mode converter draws current in blocks rather than sinusoids. A six-pulse three-phase bridge presents characteristic orders $6k \pm 1$ — the fifth, seventh, eleventh and thirteenth — each with an amplitude roughly $1/n$ of the fundamental, while single-phase diode-capacitor front ends in electronic ballasts, computer supplies and small chargers draw a narrow current pulse rich in third harmonic. The third is a zero-sequence order, so the three phase contributions add arithmetically in the neutral instead of cancelling, and an apparently balanced office load can put more current in the neutral than in any phase conductor. This is the dominant harmonic source in a modern distribution network, and it is what IEEE 519 was written to bound.

Saturating magnetic circuits. Transformers, reactors and motors operated near or above the knee of their magnetisation curve draw a peaked, non-sinusoidal magnetising current even from a perfectly sinusoidal supply. Sustained over-voltage, energisation inrush and ferroresonance between a transformer magnetising branch and cable capacitance all drive the core further into saturation and enrich the odd orders, particularly the third, which circulates in any delta winding it can reach. The same mechanism explains why lightly loaded transformers on a high-voltage tap can be surprisingly poor harmonic citizens.

Arcing loads. Electric arc furnaces, arc and resistance welders, and discharge lighting with magnetic ballasts all have a strongly non-linear and time-varying arc voltage–current relation. Because the arc reignites at a slightly different instant every half cycle, these loads produce not only integer harmonics but a broad band of interharmonics and voltage flicker, and their spectrum is stochastic rather than fixed. Rotating machines contribute a smaller fourth mechanism through slotting and non-sinusoidal MMF distribution.

Parts (b) to (e) — The rectifier calculation

Given.

Find. The counter e.m.f. $E_c$, the delay angle $\alpha$, the average load current, and the average power absorbed by the counter e.m.f.

Approach. A counter e.m.f. load can only take current while the instantaneous rectified supply exceeds it, so $E_c$ follows directly from the earliest permissible firing instant, the extinction angle follows from the supply falling back through $E_c$, and the mean current is the integral of $(v_s - E_c)/R$ over the conduction window.

1. Interpret the minimum permissible delay angle. Firing before the rectified supply has risen to $E_c$ achieves nothing: the thyristor sees a reverse net voltage $E_c - v_s$ across the load loop and cannot latch. The earliest instant at which conduction can begin is therefore the instant at which $|v_s| = E_c$, and that instant is $\alpha_{\min}$. Hence $$E_c = V_m \sin\alpha_{\min} = 169.71 \times \sin 15^\circ = 169.71 \times 0.258819$$ $$\boxed{E_c = 43.92\ \text{V}}$$
2. Locate the extinction angle. Because the load is purely resistive apart from the constant e.m.f., there is no stored magnetic energy to prolong conduction: the current reaches zero at the moment the supply falls back through $E_c$ on the descending side of the half cycle. That is the mirror image of the first crossing, so $$\beta = 180^\circ - \alpha_{\min} = 180^\circ - 15^\circ = 165^\circ .$$ Note that $\beta$ is fixed by the load, not by the gate — delaying the firing later does not move it.
3. Extract the delay angle from the conduction window. The conduction angle is the width of that window, $\gamma = \beta - \alpha$, so $$\alpha = \beta - \gamma = 165^\circ - 145^\circ$$ $$\boxed{\alpha = 20.00^\circ}$$ which comfortably respects the $15^\circ$ minimum, confirming the operating point is admissible.
4. Average the load current over the conduction window. The instantaneous current is $i = (V_m\sin\omega t - E_c)/R$ for $\alpha \le \omega t \le \beta$ and zero elsewhere. In a full-wave bridge this window repeats twice per cycle, so the averaging interval is $\pi$, not $2\pi$: $$I_{avg} = \frac{1}{\pi R}\int_{\alpha}^{\beta}\big(V_m\sin\omega t - E_c\big)\,d(\omega t) = \frac{V_m(\cos\alpha - \cos\beta) - E_c(\beta-\alpha)}{\pi R}.$$
5. Substitute and evaluate. With $\cos 20^\circ = 0.939693$ and $\cos 165^\circ = -0.965926$, $$V_m(\cos\alpha-\cos\beta) = 169.71 \times 1.905619 = 323.40\ \text{V},$$ $$E_c(\beta-\alpha) = 43.92 \times 2.53073 = 111.16\ \text{V},$$ $$I_{avg} = \frac{323.40 - 111.16}{\pi \times 2} = \frac{212.25}{6.2832}$$ $$\boxed{I_{avg} = 33.78\ \text{A}}$$
6. Take the average power into the counter e.m.f. $E_c$ is a constant, so the average of the product is the product of the constant and the average current: $$P_{E} = E_c I_{avg} = 43.92 \times 33.78$$ $$\boxed{P_{E} = 1483.7\ \text{W}}$$
7. Close the power balance as a check. The rms load current over the same window is $I_{rms} = 41.02$ A, so the resistor dissipates $I_{rms}^2 R = 41.02^2 \times 2 = 3365.2$ W. Total power drawn from the supply is $1483.7 + 3365.2 = 4848.8$ W, of which only 30.6 % reaches the e.m.f. The form factor $I_{rms}/I_{avg} = 1.214$ is the reason: highly discontinuous current is an inefficient way to feed a back e.m.f., and it is why practical d.c. drives insert a smoothing reactor.