Question 4 of 6: Harmonics of a Single-Pulse-Modulated Inverter

Question 4: Harmonics of a Single-Pulse-Modulated 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.

Given. A single-phase full-wave bridge inverter uses single-pulse modulation of width $\delta$, so its output harmonics are $b_n=(4V_d/n\pi)\sin(n\delta/2)$. The d.c. link is 220 V and the motor load is represented at fundamental frequency by a series $R = 9\ \Omega$ with $\omega L = 5\ \Omega$.

Find. Three harmful effects of harmonics; the algebraic proof of the $b_5/b_3$ expression; the ratio $b_3/b_1$ at which $b_5/b_3 = 0.2$; and the fundamental, third and fifth harmonic currents drawn by the motor.

Approach. The proof is a direct substitution of the supplied identities. Imposing $b_5/b_3 = 0.2$ then reduces to a quadratic in $u=\sin^2(\delta/2)$; the two roots must both be examined and one rejected on physical grounds before the harmonic voltages, and hence the currents through $R+jn\omega L$, can be evaluated.

Lead item — Three harmful effects of harmonics in distribution systems

1. Additional losses, heating and equipment derating. Conductor resistance rises with frequency through skin and proximity effects, so harmonic current produces more $I^2R$ loss per ampere than fundamental current does, while transformer eddy-current loss scales roughly as the square of the harmonic order. A transformer supplying a heavily distorted load must be derated or specified with a K-factor. The worst case is the neutral of a four-wire system: triplen harmonics are zero-sequence and add arithmetically in the neutral, which can therefore carry more current than any phase conductor even though it is often sized smaller.

2. Resonance with power-factor-correction capacitors. A capacitor bank and the supply inductance form a parallel resonant circuit. If its resonant frequency lands near a harmonic the system produces — the fifth and seventh are the usual offenders — the harmonic voltage and current are magnified many times over, and the result is capacitor dielectric failure, nuisance fuse operation and severe voltage distortion across the whole bus. The remedy, a detuning reactor, is standard practice precisely because this failure mode is so common.

3. Malfunction of connected equipment and interference. Distorted voltage produces multiple zero crossings, which upsets equipment that synchronises to the supply, including thyristor gate-control circuits and some protective relays. Induction-disc and average-responding meters misread, motors suffer torque pulsation and extra rotor heating from negative-sequence harmonic fields, and the higher-order components couple into adjacent communication circuits. In Canada, harmonic limits are applied at the point of common coupling under IEEE 519 as referenced by utility connection standards, alongside the installation rules of the Canadian Electrical Code (CSA C22.1).

Parts (a), (c) and (d) — Quantitative solution

1. Part (a) — form the harmonic ratio from the given coefficient. Writing $\theta = \delta/2$, the two coefficients are $b_5=(4V_d/5\pi)\sin 5\theta$ and $b_3=(4V_d/3\pi)\sin 3\theta$, so the constant $4V_d/\pi$ cancels and $$\frac{b_5}{b_3}=\frac{(1/5)\sin 5\theta}{(1/3)\sin 3\theta} =\frac{3}{5}\cdot\frac{\sin 5\theta}{\sin 3\theta}.$$
2. Substitute the supplied identities. With $\sin 3\theta = 3\sin\theta-4\sin^3\theta$ and $\sin 5\theta = 5\sin\theta-20\sin^3\theta+16\sin^5\theta$, $$\boxed{\frac{b_5}{b_3}=\frac{3}{5}\left[\frac{5\sin\frac{\delta}{2} -20\sin^3\frac{\delta}{2}+16\sin^5\frac{\delta}{2}} {3\sin\frac{\delta}{2}-4\sin^3\frac{\delta}{2}}\right]},$$ which is the printed result. The proof is complete.
3. Part (c) — reduce the condition to a quadratic. Dividing numerator and denominator by $\sin(\delta/2)$ and writing $u=\sin^2(\delta/2)$ gives $$\frac{b_5}{b_3}=\frac{3}{5}\cdot\frac{16u^2-20u+5}{3-4u}=k .$$ Clearing the denominator produces $$48u^2+(20k-60)u+(15-15k)=0 ,$$ and with $k=0.2$ this is $48u^2-56u+12=0$, i.e. $12u^2-14u+3=0$.
4. Solve and screen the two roots. The quadratic gives $$u=\frac{14\pm\sqrt{52}}{24}=0.88380 \quad\text{or}\quad 0.28287 ,$$ corresponding to $\delta = 140.14^\circ$ and $\delta = 64.26^\circ$ respectively. Both satisfy $b_5/b_3=0.2$ exactly, so the ratio alone cannot choose between them. The third-harmonic content decides it: from $b_3/b_1=(3-4u)/3$, the narrow pulse gives $b_3/b_1=+0.623$ — a third harmonic almost two-thirds of the fundamental, which is not an inverter operating point — whereas the wide pulse gives a modest and usable value. The wide root is selected and the narrow one is recorded and rejected.
5. Report the third-to-fundamental ratio. With $u=0.88380$, $$\frac{b_3}{b_1}=\frac{3-4(0.88380)}{3}=\boxed{-0.1784},$$ that is, a third harmonic of 17.8 per cent of the fundamental, in antiphase with it. The negative sign is part of the answer, not a slip: for pulse widths beyond $120^\circ$ the third harmonic reverses, and it passes through zero exactly at $\delta = 120^\circ$. For completeness, $b_5/b_1=(16u^2-20u+5)/5=-0.03568$, whose ratio to $b_3/b_1$ returns the required 0.2000.
6. Part (d) — evaluate the harmonic voltages. With $\delta = 140.138^\circ$, $\sin(\delta/2)=0.94010$ and $4V_d/\pi = 280.113$ V, so $$b_1=280.113\sin 70.07^\circ=263.34\ \text{V},\quad b_3=-46.98\ \text{V},\quad b_5=-9.396\ \text{V}\ \ \text{(peak)}.$$
7. Divide each by the impedance at its own frequency. The motor branch presents $Z_n=R+jn\omega L = 9+j5n$: $$Z_1=10.296\angle 29.05^\circ\ \Omega,\quad Z_3=17.493\angle 59.04^\circ\ \Omega,\quad Z_5=26.571\angle 70.20^\circ\ \Omega .$$ Hence the peak harmonic currents are $$\boxed{I_1=25.58\ \text{A},\quad I_3=2.686\ \text{A},\quad I_5=0.3536\ \text{A}}\ \ \text{(peak)},$$ equivalently 18.09 A, 1.899 A and 0.2500 A rms.
8. Interpret the result. The inductive reactance rises in proportion to the harmonic order while the harmonic voltage falls as $1/n$ multiplied by a shrinking sine factor, so the current spectrum is far cleaner than the voltage spectrum: the third harmonic is 17.8 per cent of the fundamental in voltage but only 10.5 per cent in current, and the fifth falls from 3.6 per cent to 1.4 per cent. The machine's own leakage inductance is doing the filtering, which is the standard argument for why a motor tolerates square-wave inverter drive far better than a capacitive load would.