Question 3 of 6: Harmonics in a Single-Pulse-Modulation Inverter

Question 3: Harmonics in a Single-Pulse-Modulation 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.

Part (a) — three harmful effects of harmonics

Additional heating and derating of cables, transformers and machines. Copper loss scales with the square of the total rms current, so harmonic current heats a conductor without doing useful work, and the skin and proximity effects make the effective resistance rise with frequency. Transformer core loss rises faster still, roughly as $n^{2}$ for eddy-current loss, which is why transformers feeding converter loads are specified by K-factor and are commonly derated. In a motor, negative-sequence harmonic fields produce loss and pulsating torque without contributing to mean torque.

Resonance with power-factor-correction capacitors. A capacitor bank and the upstream source inductance form a parallel resonant circuit, typically somewhere between the fifth and thirteenth harmonic. If a converter injects current at that order, the resulting circulating current and voltage magnification can be several times the injected value, blowing capacitor fuses and distorting the bus voltage for every other customer on it. Detuning reactors exist precisely to move that resonance below the lowest injected order.

Malfunction of protection, metering and sensitive equipment. Distorted waveforms have multiple zero crossings, which upsets zero-crossing-synchronised firing circuits, electronic relays and clocks. Induction-disc energy meters mis-register, current transformers saturate, and the neutral conductor of a four-wire system carries the zero-sequence triplen harmonics, which add arithmetically rather than cancelling — a neutral can run hotter than the phases it serves. In Canada, IEEE Std 519 is the customary planning limit invoked by utility interconnection agreements, with CSA C22.1 governing the neutral sizing that follows from it.

Part (b) — deriving the printed harmonic ratio

Approach. Form the ratio directly from the given coefficient, then expand the two multiple-angle sines with the identities supplied on the paper.

1. Take the ratio of the two coefficients. With $b_{n}=(4V_{d}/n\pi)\sin(n\delta/2)$ the supply voltage and the $4/\pi$ cancel and only the order and the sine survive:$$\frac{b_{5}}{b_{3}}=\frac{(4V_{d}/5\pi)\sin(5\delta/2)}{(4V_{d}/3\pi)\sin(3\delta/2)}=\frac{3}{5}\,\frac{\sin(5\delta/2)}{\sin(3\delta/2)}$$
2. Expand both sines about the half-angle. Writing $\theta=\delta/2$ and using the two printed identities, $\sin 5\theta=5\sin\theta-20\sin^{3}\theta+16\sin^{5}\theta$ and $\sin 3\theta=3\sin\theta-4\sin^{3}\theta$, gives the required form:$$\boxed{\frac{b_{5}}{b_{3}}=\frac{3}{5}\left[\frac{5\sin\tfrac{\delta}{2}-20\sin^{3}\tfrac{\delta}{2}+16\sin^{5}\tfrac{\delta}{2}}{3\sin\tfrac{\delta}{2}-4\sin^{3}\tfrac{\delta}{2}}\right]}$$
3. Reduce it to a form that can be solved. Dividing numerator and denominator by $\sin\theta$ and writing $u=\sin^{2}(\delta/2)$ turns the ratio into a rational function of a single variable, which is what makes part (c) tractable:$$\frac{b_{5}}{b_{3}}=\frac{3}{5}\,\frac{5-20u+16u^{2}}{3-4u}$$

Parts (c) and (d) — selecting the pulse width and finding the currents

Given.

Find. The ratio $b_{3}/b_{1}$ that follows from $b_{5}/b_{3}=0.2$, and the fundamental, third and fifth harmonic components of the current drawn by the motor.

Approach. Impose the harmonic ratio on the reduced form from part (b), solve the resulting quadratic in $u$, select the physically admissible root, then divide each harmonic voltage by the impedance the motor presents at that order.

1. Impose the ratio and clear the fraction. Setting the reduced form equal to 0.2:$$\begin{aligned}3\left(5-20u+16u^{2}\right)&=3-4u\\48u^{2}-56u+12&=0\\12u^{2}-14u+3&=0\end{aligned}$$a genuine quadratic, so two roots must be expected and screened.
2. Solve and record both roots. The discriminant is $196-144=52$, so$$u=\frac{14\pm\sqrt{52}}{24}\;\Longrightarrow\;u_{1}=0.8838,\qquad u_{2}=0.2829$$Taking $\delta=2\arcsin\sqrt{u}$ on the principal branch gives $\delta_{1}=140.12^{\circ}$ (wide pulse) and $\delta_{2}=64.24^{\circ}$ (narrow pulse). Both satisfy $b_{5}/b_{3}=0.2$ exactly — the constraint alone does not choose between them.
3. Select the root on inverter physics. The third-harmonic ratio follows from the same substitution, $b_{3}/b_{1}=(3-4u)/3=1-\tfrac{4}{3}u$. The narrow root gives $b_{3}/b_{1}=+0.623$ — a third harmonic almost two-thirds of the fundamental, which no inverter would be operated at and which the motor could not tolerate. The wide root gives a small third harmonic and is the operating point intended:$$\boxed{\delta=140.12^{\circ},\qquad\frac{b_{3}}{b_{1}}=-0.1784}$$The negative sign is part of the answer, not a slip: past $\delta=120^{\circ}$ the third harmonic reverses phase relative to the fundamental, and it passes through zero exactly at $\delta=120^{\circ}$.
4. Evaluate the harmonic voltages (peak). With $4V_{d}/\pi=280.11$ V and $\sin(\delta/2)=0.9401$:$$\begin{aligned}b_{1}&=280.11\times 0.9401=263.34\ \text{V}\\b_{3}&=b_{1}\times(-0.1784)=-46.98\ \text{V}\\b_{5}&=0.2\,b_{3}=-9.40\ \text{V}\end{aligned}$$These are peak values, because the printed coefficient is the peak of a sine term; divide by $\sqrt{2}$ for rms.
5. Form the impedance the motor presents at each order. The resistance is frequency-independent and the reactance scales with the order, so $Z_{n}=R+jn\omega L=7.5+j10n$:$$Z_{1}=12.50\angle 53.13^{\circ}\ \Omega,\quad Z_{3}=30.92\angle 75.96^{\circ}\ \Omega,\quad Z_{5}=50.56\angle 81.47^{\circ}\ \Omega$$
6. Divide to obtain the harmonic currents.$$\boxed{\begin{aligned}I_{1}&=21.07\ \text{A peak }(14.90\ \text{A rms})\\I_{3}&=1.52\ \text{A peak }(1.07\ \text{A rms})\\I_{5}&=0.186\ \text{A peak }(0.131\ \text{A rms})\end{aligned}}$$Each harmonic current lags its own voltage by the angle of $Z_{n}$, so the third lags by $75.96^{\circ}$ and the fifth by $81.47^{\circ}$.
7. Interpret the result. The inductive load is a first-order low-pass path, so the harmonic current spectrum is far cleaner than the harmonic voltage spectrum: the third harmonic voltage is 17.8 per cent of fundamental but the third harmonic current only 7.2 per cent, and the fifth falls from 3.6 per cent of voltage to 0.9 per cent of current. This is the standard argument for tolerating a coarse inverter voltage waveform when the load is a motor.