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

Question 4 of 6: Single-Pulse-Modulated Inverter Feeding a Motor with Power-Factor Correction

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

Question 4: Single-Pulse-Modulated Inverter Feeding a Motor with Power-Factor Correction (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) — Techniques for inverter operation

Three widely used voltage-source inverter control techniques are single-pulse (quasi-square-wave) modulation, in which each half cycle carries one pulse of adjustable width $\delta$; multiple-pulse or uniform PWM, in which the half cycle is divided into $p$ equal slots each carrying one pulse of the same width; and sinusoidal PWM, in which a sinusoidal reference is compared continuously against a triangular carrier. Selective harmonic elimination and space-vector modulation are two further members of the same family.

Taking sinusoidal PWM as the technique to develop: a reference sinusoid at the wanted output frequency $f_1$ is compared with a triangular carrier at $f_c = m_f f_1$, and each device is switched whenever the two waveforms cross. The width of every resulting pulse is therefore proportional to the instantaneous value of the reference, so the pulse pattern carries a sinusoidal area distribution and the fundamental of the output follows the reference. Writing the modulation index as $m_a = \hat{V}_{ref}/\hat{V}_{tri}$, the fundamental output amplitude is $m_aV_d$ for $m_a \le 1$, which gives linear, continuously variable voltage control at a fixed d.c.-link voltage — exactly what a constant-flux variable-frequency drive needs, since voltage and frequency can then be varied together from one modulator. The dominant residual harmonics are pushed up into sidebands clustered around $m_f$ and its multiples, far above the fundamental, where the load inductance attenuates them strongly; choosing $m_f$ odd and a multiple of three suppresses even orders and cancels triplen orders in a three-phase load. Driving $m_a$ above unity enters over-modulation, in which the output amplitude grows more slowly, low-order harmonics reappear, and the waveform degenerates continuously into the square wave that gives the maximum obtainable fundamental.

Part (b) — Deriving the third-to-fundamental ratio

Approach. Form the ratio directly from the given coefficient and reduce the resulting $\sin 3x/\sin x$ with the triple-angle identity.

1. Form the ratio from the stated coefficient. Writing $x = \delta/2$ and applying $b_n = 4V_d\sin(n\delta/2)/(n\pi)$ at $n = 3$ and $n = 1$, the factor $4V_d/\pi$ cancels: $$\frac{b_3}{b_1} = \frac{\dfrac{4V_d}{3\pi}\sin 3x}{\dfrac{4V_d}{\pi}\sin x} = \frac{1}{3}\cdot\frac{\sin 3x}{\sin x}.$$
2. Expand the triple angle. The standard identity is $\sin 3x = 3\sin x - 4\sin^3 x$, so dividing through by $\sin x$ (non-zero for any usable pulse width) gives $\sin 3x/\sin x = 3 - 4\sin^2 x$. Substituting $x = \delta/2$, $$\boxed{\dfrac{b_3}{b_1} = \dfrac{1}{3}\left[3 - 4\sin^2\dfrac{\delta}{2}\right]}$$ which is the required result.

Parts (c) and (d) — Harmonic voltages and currents

Given.

Find. The fifth-to-fundamental voltage ratio at the modulation angle that gives $b_3/b_1 = 0.25$, and the fundamental, third and fifth harmonic components of the current the inverter must deliver into the parallel motor-and-capacitor load.

Approach. Invert the part (b) relation for the modulation angle, use the companion fifth-harmonic identity to get $b_5/b_1$, then drive each harmonic voltage through the parallel admittance of motor and capacitor evaluated at its own order.

1. Solve for the modulation angle. Writing $u = \sin^2(\delta/2)$, the part (b) result is linear in $u$: $$\frac{b_3}{b_1} = \frac{3-4u}{3} = 0.25 \;\Longrightarrow\; 3 - 4u = 0.75 \;\Longrightarrow\; u = 0.5625,$$ so $\sin(\delta/2) = \sqrt{0.5625} = 0.7500$ and $\delta/2 = 48.59^\circ$, giving $$\boxed{\delta = 97.18^\circ}$$ Because the relation is linear in $u$ there is only one value of $u$ to consider; the supplementary arcsine branch, $\delta/2 = 131.41^\circ$, would require $\delta = 262.82^\circ$, wider than the half cycle it must fit inside, and is discarded on that ground alone.
2. Write the fifth-harmonic ratio in the same variable. Applying the coefficient at $n = 5$ and using $\sin 5x = 16\sin^5x - 20\sin^3x + 5\sin x$, $$\frac{b_5}{b_1} = \frac{1}{5}\cdot\frac{\sin 5x}{\sin x} = \frac{16u^2 - 20u + 5}{5}.$$
3. Evaluate at $u = 0.5625$. $$\frac{b_5}{b_1} = \frac{16(0.31641) - 20(0.5625) + 5}{5} = \frac{5.0625 - 11.2500 + 5}{5} = \frac{-1.1875}{5}$$ $$\boxed{\frac{b_5}{b_1} = -0.2375}$$ The negative sign is a genuine result and not a slip: at this pulse width the fifth harmonic is in antiphase with the fundamental. Its magnitude, 23.75 %, is very nearly as large as the third.
4. Compute the harmonic voltage amplitudes. The fundamental peak follows from the given coefficient: $$b_1 = \frac{4V_d}{\pi}\sin\frac{\delta}{2} = \frac{4\times220}{\pi}\times0.7500 = 280.11\times0.7500 = 210.08\ \text{V (peak)},$$ and therefore $b_3 = 0.25\,b_1 = 52.52$ V and $b_5 = -0.2375\,b_1 = -49.90$ V, all peak values.
5. Assemble the load admittance at each order. The inverter feeds both branches, so the current it must supply is set by the parallel admittance. At harmonic order $n$ the motor reactance scales with $n$ and the capacitor susceptance likewise scales with $n$: $$Y_n = \frac{1}{R + jn\omega L} + jn\omega C .$$
6. Evaluate the three admittances. For $n = 1$, $1/(0.12+j0.09) = 5.333 - j4.000$ S, and adding $j3$ gives $5.333 - j1.000$, so $|Y_1| = 5.4263$ S. For $n = 3$, $1/(0.12+j0.27) = 1.375 - j3.093$ S, and adding $j9$ gives $1.375 + j5.907$, so $|Y_3| = 6.0650$ S. For $n = 5$, $1/(0.12+j0.45) = 0.553 - j2.075$ S, and adding $j15$ gives $0.553 + j12.925$, so $|Y_5| = 12.9372$ S.
7. Multiply through to obtain the harmonic currents. Since the $b_n$ are peak values, so are the products; the rms values follow on dividing by $\sqrt{2}$: $$I_1 = 210.08\times5.4263 = 1139.98\ \text{A peak} = 806.09\ \text{A rms},$$ $$I_3 = 52.52\times6.0650 = 318.54\ \text{A peak} = 225.24\ \text{A rms},$$ $$\boxed{I_5 = 49.90\times12.9372 = 645.50\ \text{A peak} = 456.44\ \text{A rms}}$$
8. Read the ranking. The harmonic current ranking does not follow the harmonic voltage ranking. Although $|b_5/b_1|$ is only 0.2375, the fifth-harmonic current is 56.6 % of the fundamental, and it exceeds the third-harmonic current by a factor of two. The reason is visible in step 6: at $n = 5$ the capacitor contributes 15 S while the motor branch contributes only 2.15 S, so the capacitor, installed to correct the fundamental power factor, has become the dominant sink for the high-order harmonics.