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

Question 2 of 6: Harmonic Sources, and a Controlled Rectifier Feeding a Back-EMF Load

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

Question 2: Harmonic Sources, and a Controlled Rectifier Feeding a Back-EMF Load (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) Three causes of harmonics in a distribution system

Power-electronic converters. Line-commutated rectifiers, variable-frequency drives, uninterruptible supplies, electric-vehicle chargers and the switch-mode power supplies inside virtually every modern electronic load draw current in switched bursts rather than sinusoidally. A six-pulse bridge, the workhorse of industrial rectification, draws characteristic harmonics of order $h = 6k \pm 1$ — the 5th, 7th, 11th and 13th — with the 5th typically the largest at roughly 20 % of fundamental. Single-phase rectifiers with capacitive filters are worse still and are rich in the triplen (3rd, 9th, 15th) harmonics that add arithmetically in the neutral of a four-wire feeder.

Magnetic saturation of iron-cored apparatus. Transformers, reactors and rotating machines are deliberately worked near the knee of the $B$–$H$ curve for economy. Because that characteristic is non-linear, a sinusoidal applied voltage demands a peaked, non-sinusoidal magnetising current dominated by the third harmonic. The effect grows sharply with overexcitation, so a lightly loaded feeder whose voltage has drifted high becomes a significant harmonic source in its own right, as do transformer inrush transients.

Arcing and discharge loads. Arc furnaces, arc welders, and discharge lighting on magnetic ballasts have a voltage–current characteristic that is both non-linear and time-varying, since the arc voltage depends on the arc length as the scrap moves. These produce a broad, partly non-integer (interharmonic) spectrum together with flicker. In Canadian practice the resulting distortion is assessed against IEEE Std 519 limits at the point of common coupling, and CSA C22.1 (the Canadian Electrical Code) governs the neutral-conductor sizing that triplen harmonics make necessary.

(b) to (e) Controlled rectifier with a counter-EMF load

Given. A single-phase full-wave controlled rectifier supplies a resistance in series with a constant counter-EMF, with the data below.

Find. The minimum firing angle that will let the thyristors turn on at all; the firing angle that produces the stated 135° conduction; the resulting mean load current; and the mean current when the firing angle is reduced to 12°.

Approach. With no inductance the load current is algebraic: it exists only while the instantaneous supply exceeds the counter-EMF. That single fact fixes both the earliest firing angle and the extinction angle, and the mean current then follows from integrating $(v_s - E_c)/R$ across the conduction window.

1. Write the peak supply voltage. The 120 V is an rms value, so$$V_m = \sqrt{2}\,V_s = \sqrt{2}\times 120 = 169.71\ \text{V}$$and the instantaneous supply seen by the load is $v_s = 169.71\sin\omega t$ over each half cycle.
2. Minimum permissible delay angle — part (b). A thyristor will not turn on unless it is forward-biased, and here that requires the supply to have already climbed above the counter-EMF. The threshold is therefore $V_m\sin\alpha_{min} = E_c$, so$$\alpha_{min} = \arcsin\!\left(\frac{E_c}{V_m}\right) = \arcsin\!\left(\frac{30}{169.71}\right) = \arcsin(0.17678)$$which evaluates to$$\boxed{\alpha_{min} = 10.19^\circ}$$Firing any earlier than this simply wastes the gate pulse: the device stays blocked because the load opposes conduction.
3. Locate the extinction angle. Since the load has no inductance there is no stored energy to maintain current, so conduction ceases the instant the supply falls back to the counter-EMF. By the symmetry of the sine that happens at$$\beta = 180^\circ - \alpha_{min} = 180^\circ - 10.19^\circ = 169.81^\circ$$This is the key structural point of the problem: $\beta$ is set by the load, not by the firing circuit.
4. Delay angle for 135° conduction — part (c). The conduction angle is the span between firing and extinction, $\gamma = \beta - \alpha$. Rearranging and substituting,$$\alpha = \beta - \gamma = 169.81^\circ - 135^\circ$$so that$$\boxed{\alpha = 34.81^\circ}$$A check is worthwhile: at 34.81° the supply stands at $169.71\sin 34.81^\circ = 96.9$ V, comfortably above the 30 V counter-EMF, so the current jumps abruptly to $(96.9-30)/2.25 = 29.7$ A at turn-on, as a resistive load must.
5. Mean load current — part (d). During conduction the current is $i = (V_m\sin\theta - E_c)/R$ and it is zero elsewhere. Averaging over the half-cycle period of $\pi$ radians that a full-wave rectifier repeats on,$$I_{avg} = \frac{1}{\pi R}\Big[V_m(\cos\alpha - \cos\beta) - E_c(\beta - \alpha)\Big]$$Substituting $\alpha = 0.6076$ rad, $\beta = 2.9642$ rad, $\cos\alpha = 0.82116$ and $\cos\beta = -0.98424$ gives $V_m(1.80540) = 306.37$ and $E_c(2.35655) = 70.70$, so$$I_{avg} = \frac{306.37 - 70.70}{\pi \times 2.25} = \frac{235.67}{7.0686}$$and hence$$\boxed{I_{avg} = 33.34\ \text{A}}$$Sense check on the loop: the mean load voltage must equal $E_c + I_{avg}R = 30 + 33.34\times 2.25 = 105.0$ V, and integrating the terminal voltage directly (supply while conducting, $E_c$ while blocked) returns the same 105.0 V.
6. Mean current with the firing angle advanced to 12° — part (e). Reducing the enforced minimum to 12° and operating there widens the conduction window, because extinction is still pinned by the load at $\beta = 169.81^\circ$:$$\gamma_{new} = 169.81^\circ - 12^\circ = 157.81^\circ$$Re-evaluating the same integral with $\alpha = 12^\circ = 0.20944$ rad and $\cos 12^\circ = 0.97815$,$$I_{avg} = \frac{169.71(0.97815 + 0.98424) - 30(2.75473)}{7.0686} = \frac{333.03 - 82.64}{7.0686}$$giving$$\boxed{I_{avg} = 35.42\ \text{A}}$$Advancing the firing by 22.8° buys only 6 % more current, because the extra conduction is taken near the zero crossings where $v_s - E_c$ is small.