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

Question 3 of 6: Harmonic Generation, Mitigation, and Single-Pulse Modulation of a Bridge Inverter

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

Question 3: Harmonic Generation, Mitigation, and Single-Pulse Modulation of a Bridge Inverter (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) How harmonics arise, and three ways of mitigating them

Harmonics arise because a distribution system carries loads whose current is not proportional to the voltage across them. If a load is linear, a sinusoidal voltage produces a sinusoidal current at the same frequency and nothing else exists. If the load is non-linear — because it saturates, because it arcs, or because semiconductor switches connect and disconnect it during the cycle — then the current is still periodic at the supply frequency but is no longer a sinusoid. Any such periodic waveform can be decomposed by Fourier analysis into a fundamental plus components at integer multiples of it, and those components are physically real currents flowing in the network.

The mechanism that turns a local nuisance into a system-wide problem is the source impedance. Each harmonic current $I_h$ must flow back through the supply impedance, which is largely inductive and therefore rises with frequency as $Z_h \approx h\,\omega L_s$. It produces a harmonic voltage drop $V_h = I_h Z_h$, so the voltage at the point of common coupling itself becomes distorted. Every other customer on that feeder is then supplied with a distorted voltage even if their own loads are perfectly linear. The consequences are extra $I^2R$ and eddy-current heating in cables and transformers, overloaded neutrals where triplen harmonics add instead of cancelling, nuisance tripping, capacitor-bank failures through resonance, and torque pulsation and derating in motors.

Three mitigation measures. First, passive filtering. A series-tuned LC shunt trap placed at the offending load presents a near short circuit at its tuned frequency, so the 5th- and 7th-harmonic currents circulate locally instead of flowing into the supply; a series line reactor or dc-link choke complements this by raising the source impedance seen by the converter and flattening its current pulses. Second, phase multiplication. Increasing the converter pulse number cancels low-order harmonics by superposition: two six-pulse bridges fed through a 30° phase-shifting transformer form a twelve-pulse converter whose lowest characteristic harmonics are the 11th and 13th, and a delta winding traps the triplens so they never reach the supply. Third, active compensation. An active harmonic filter measures the load current, extracts its distorted part, and injects the equal and opposite current, so the supply sees a sinusoid; equivalently, replacing a diode front end with a PWM active rectifier makes the converter draw near-sinusoidal current at unity power factor in the first place. Where none of these is economic, the remaining option is accommodation — K-rated transformers, oversized neutrals and derating, sized in Canadian installations to CSA C22.1, with the distortion itself assessed against IEEE Std 519.

(b) Derivation of the fifth-to-third harmonic ratio

Given. The single-pulse-modulated bridge output has Fourier coefficients $b_n = (4V_d/n\pi)\sin(n\delta/2)$, where $\delta$ is the pulse width, together with the two trigonometric identities printed on the paper.

Find. A closed-form expression for $b_5/b_3$ in terms of $\sin(\delta/2)$ alone, matching the form quoted in the question.

Approach. Form the ratio directly from the general coefficient, then expand the two multiple-angle sines using the given identities with $\theta = \delta/2$.

1. Take the ratio of the two coefficients. Writing $b_5$ and $b_3$ from the general formula, the factor $4V_d$ cancels and the reciprocal harmonic orders invert:$$\frac{b_5}{b_3} = \frac{\dfrac{4V_d}{5\pi}\sin\dfrac{5\delta}{2}}{\dfrac{4V_d}{3\pi}\sin\dfrac{3\delta}{2}} = \frac{3}{5}\cdot\frac{\sin\dfrac{5\delta}{2}}{\sin\dfrac{3\delta}{2}}$$The 3/5 in front is therefore nothing more than the $1/n$ weighting of the Fourier series, and every trace of the dc supply voltage has vanished — the harmonic ratio depends only on pulse width.
2. Substitute $\theta = \delta/2$ and expand. With this substitution $5\delta/2 = 5\theta$ and $3\delta/2 = 3\theta$, so the printed identities apply verbatim:$$\sin 5\theta = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta,\qquad \sin 3\theta = 3\sin\theta - 4\sin^3\theta$$Both sides are now polynomials in the single variable $\sin\theta = \sin(\delta/2)$.
3. Assemble the required result. Putting the expansions back into the ratio,$$\boxed{\frac{b_5}{b_3} = \frac{3}{5}\left[\frac{5\sin\frac{\delta}{2} - 20\sin^3\frac{\delta}{2} + 16\sin^5\frac{\delta}{2}}{3\sin\frac{\delta}{2} - 4\sin^3\frac{\delta}{2}}\right]}$$which is the expression quoted in the question, as required. The result has been confirmed at several pulse widths against the original coefficient formula.

(c) Modulation angle for $b_5/b_3 = 0.4$, and the resulting $b_3/b_1$

Given. $V_d = 240$ V, and the pulse width $\delta$ is chosen so that $b_5/b_3 = 0.4$.

Find. The modulation angle $\delta$, and then the ratio of third-harmonic to fundamental output voltage, $b_3/b_1$.

Approach. Set the part-(b) expression equal to 0.4, clear the denominator to get a quadratic in $\sin^2(\delta/2)$, then select the physically usable root before evaluating $b_3/b_1$.

1. Reduce the constraint to a quadratic. Let $u = \sin^2(\delta/2)$ and cancel one factor of $\sin(\delta/2)$ from numerator and denominator. The condition $b_5/b_3 = 0.4$ becomes$$\frac{3}{5}\cdot\frac{5 - 20u + 16u^2}{3 - 4u} = 0.4 \quad\Longrightarrow\quad 3(5 - 20u + 16u^2) = 2(3 - 4u)$$Collecting terms gives the quadratic$$48u^2 - 52u + 9 = 0$$
2. Solve for the pulse width. The discriminant is $52^2 - 4(48)(9) = 976$, so$$u = \frac{52 \pm \sqrt{976}}{96} = 0.86709 \quad\text{or}\quad 0.21624$$Taking square roots, $\sin(\delta/2) = 0.93118$ or $0.46500$, which correspond to$$\delta = 137.23^\circ \quad\text{or}\quad \delta = 55.40^\circ$$Both genuinely satisfy $b_5/b_3 = 0.4$, so a choice must be made on engineering grounds.
3. Select the usable root. Evaluate the third-harmonic content each root would give, using $b_3/b_1 = \tfrac{1}{3}\sin(3\delta/2)/\sin(\delta/2)$. The narrow pulse $\delta = 55.40^\circ$ returns $b_3/b_1 = +0.712$ — a third harmonic 71 % of the fundamental, with a fundamental of only 142 V from a 240 V supply. That is not a waveform anyone would operate a motor on. The wide pulse $\delta = 137.23^\circ$ delivers a 284.5 V fundamental with far less distortion, so$$\boxed{\delta = 137.23^\circ}$$is the modulation angle intended, and the alternative root is recorded and rejected rather than ignored.
4. Evaluate the third-harmonic ratio. At $\delta = 137.23^\circ$ we have $\delta/2 = 68.62^\circ$ and $3\delta/2 = 205.85^\circ$, so $\sin(\delta/2) = 0.93118$ and $\sin(3\delta/2) = -0.43549$. Then$$\frac{b_3}{b_1} = \frac{1}{3}\cdot\frac{-0.43549}{0.93118} = -0.15589$$so the magnitude is$$\boxed{\left|\frac{b_3}{b_1}\right| = 0.156 \;\;(15.6\ \%)}$$The negative sign is physically meaningful and not a slip: at this pulse width the third harmonic is in antiphase with the fundamental, which is precisely why widening the pulse towards 180° suppresses it (at $\delta = 120^\circ$ it would vanish altogether).

(d) Fundamental, third and fifth harmonic output currents

Given. $V_d = 240$ V with $\delta = 137.23^\circ$ from part (c); the motor is represented at fundamental frequency by $R = 56\ \Omega$ in series with $\omega L = 4\ \Omega$.

Find. The 1st, 3rd and 5th harmonic components of the current drawn by the motor.

Approach. Evaluate each harmonic voltage from the coefficient formula, form the impedance at that harmonic order (remembering that only the reactance scales with $n$), and divide.

1. Evaluate the harmonic voltages. With $4V_d/\pi = 305.58$ V, the coefficient formula gives the peak amplitudes$$b_1 = 305.58\sin 68.62^\circ = 284.55\ \text{V},\qquad b_3 = \frac{305.58}{3}\sin 205.85^\circ = -44.36\ \text{V},\qquad b_5 = \frac{305.58}{5}\sin 343.09^\circ = -17.73\ \text{V}$$As a consistency check, $b_5/b_3 = 17.73/44.36 = 0.400$ and $b_3/b_1 = 0.156$, reproducing parts (b) and (c).
2. Form the impedance at each harmonic. The resistance is frequency-independent while the reactance is proportional to frequency, so $Z_n = R + jn\omega L = 56 + j4n$:$$Z_1 = 56 + j4 = 56.14\angle 4.09^\circ\ \Omega,\qquad Z_3 = 56 + j12 = 57.27\angle 12.09^\circ\ \Omega,\qquad Z_5 = 56 + j20 = 59.46\angle 19.65^\circ\ \Omega$$The rising impedance gives the winding a mild low-pass character, which attenuates the harmonic currents beyond the attenuation already present in the voltage spectrum.
3. Divide to obtain the harmonic currents. Taking the magnitude of each voltage over the magnitude of its impedance,$$I_1 = \frac{284.55}{56.14} = 5.068\ \text{A peak},\qquad I_3 = \frac{44.36}{57.27} = 0.775\ \text{A peak},\qquad I_5 = \frac{17.73}{59.46} = 0.298\ \text{A peak}$$Expressed as rms values, which is how motor current would be quoted,$$\boxed{I_1 = 3.584\ \text{A},\quad I_3 = 0.548\ \text{A},\quad I_5 = 0.211\ \text{A}\ \ (\text{rms})}$$Each harmonic current lags its own voltage by the corresponding impedance angle (4.09°, 12.09° and 19.65°). Relative to the fundamental the harmonic currents are 15.3 % and 5.9 %, so the distortion the motor actually experiences in current is a little smaller than the distortion in the applied voltage.