22-Elec-B8 Power Electronics and Drives · December 2014

Question 3 of 6: Single-Pulse-Modulation Inverter — Harmonic Content and Load Currents

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

Question 3: Single-Pulse-Modulation Inverter — Harmonic Content and Load Currents (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-control techniques for a single-phase bridge inverter are single-pulse modulation, in which one pulse of adjustable width $\delta$ is produced per half cycle; multiple-pulse (uniform) modulation, in which $p$ equal pulses of common width are distributed symmetrically across each half cycle; and sinusoidal pulse-width modulation, in which the pulse widths follow a sinusoidal envelope. To these should be added selective harmonic elimination, where a small number of switching angles are computed to null chosen harmonics, and, in three-phase bridges, space-vector modulation.

Sinusoidal PWM is the technique worth explaining in detail because it is the basis of nearly every modern drive. A sinusoidal reference (modulating) wave of the desired output frequency $f_1$ and amplitude $A_r$ is compared continuously against a triangular carrier of frequency $f_c \gg f_1$ and amplitude $A_c$. Whenever the reference exceeds the carrier the upper device of a bridge leg is gated on and the lower device off; the comparison reverses on the other leg. The resulting train of pulses has widths proportional to the instantaneous value of the reference, so the average voltage taken over one carrier period tracks the sine. The modulation index $M = A_r/A_c$ sets the fundamental amplitude linearly, $V_1 = M V_d$ for $M \le 1$, so amplitude and frequency are controlled independently — exactly what a constant volts-per-hertz drive needs.

The advantage is spectral. Because the switching instants are determined by a high-frequency carrier, the dominant unwanted components move up to sidebands centred on $f_c$, $2f_c$, $3f_c$ and so on, rather than sitting at the third and fifth harmonic of the output. Low-order harmonics — which are the ones that produce torque pulsation and require bulky filters — are almost absent, and the small high-order residue is filtered very effectively by the motor's own leakage inductance. The price is switching loss proportional to $f_c$, and the parasitic effects of a fast $dv/dt$ discussed in Question 5. Single-pulse modulation, which this question goes on to analyse, sits at the opposite extreme: one switching event per half cycle gives the lowest possible loss but leaves a third harmonic that can approach the fundamental in magnitude.

Part (b) — Derivation of the third-harmonic ratio

Take the ratio of the given coefficients directly. With $b_n = (4V_d/n\pi)\sin(n\delta/2)$,

$$\frac{b_3}{b_1} = \frac{\dfrac{4V_d}{3\pi}\sin\dfrac{3\delta}{2}}{\dfrac{4V_d}{\pi}\sin\dfrac{\delta}{2}} = \frac{1}{3}\cdot\frac{\sin(3\delta/2)}{\sin(\delta/2)}$$

The supply voltage and the common factor $4/\pi$ cancel, leaving a pure function of the pulse width. Applying the triple-angle identity $\sin 3x = 3\sin x - 4\sin^3 x$ with $x = \delta/2$,

$$\frac{\sin(3\delta/2)}{\sin(\delta/2)} = \frac{3\sin(\delta/2) - 4\sin^3(\delta/2)}{\sin(\delta/2)} = 3 - 4\sin^2\frac{\delta}{2}$$

and therefore

$$\boxed{\frac{b_3}{b_1} = \frac{1}{3}\left[3 - 4\sin^2\frac{\delta}{2}\right]}$$

as required. Two features of this result are worth noting before using it. The ratio equals 1 at $\delta \to 0$ (a vanishingly narrow pulse is an impulse, whose harmonics are all equal), falls through zero at $\sin^2(\delta/2) = 3/4$, i.e. $\delta = 120^\circ$, and becomes negative for wider pulses — the third harmonic then being in antiphase with the fundamental. A single-pulse inverter is therefore usually operated near $\delta = 120^\circ$ when third-harmonic content matters, and the same algebra explains why the quasi-square wave of a three-phase bridge, which has $120^\circ$ conduction, is free of triplen harmonics.

Parts (c) and (d) — Pulse width, spectrum and load currents

Given. A single-phase full-bridge single-pulse-modulation inverter runs from a 220 V dc supply into an ac motor paralleled with a power-factor-correcting capacitor, and the pulse width is trimmed until the third harmonic of the output voltage is one quarter of the fundamental.

Find. The pulse width that produces the required third-harmonic ratio and the corresponding fifth-harmonic ratio, then the fundamental, third and fifth harmonic components of the current the inverter must deliver into the parallel combination.

Approach. Invert the ratio derived in part (b) to get $\delta$, use the quintuple-angle identity for the fifth harmonic, then evaluate each harmonic voltage from the given Fourier coefficient and divide by the harmonic impedance of the parallel network.

1. Part (c) — solve for the pulse width. Setting the derived ratio equal to 0.25, $$\frac{1}{3}\left[3 - 4\sin^2\frac{\delta}{2}\right] = 0.25 \;\Longrightarrow\; \sin^2\frac{\delta}{2} = \frac{3 - 0.75}{4} = 0.5625$$ so $\sin(\delta/2) = 0.75$ and $$\delta = 2\sin^{-1}(0.75) = \boxed{97.18^\circ}$$ The inverse sine has a second branch, $\delta/2 = 131.41^\circ$, but it gives $\delta = 262.8^\circ$, which exceeds the $180^\circ$ half cycle available and is rejected as unphysical.
2. Part (c) — express the fifth harmonic in the same variable. By the same route as part (b), and using $\sin 5x = 16\sin^5 x - 20\sin^3 x + 5\sin x$, $$\frac{b_5}{b_1} = \frac{1}{5}\cdot\frac{\sin(5\delta/2)}{\sin(\delta/2)} = \frac{1}{5}\left[16\sin^4\frac{\delta}{2} - 20\sin^2\frac{\delta}{2} + 5\right]$$ The problem has been reduced to a polynomial in the single quantity already known.
3. Part (c) — substitute. With $\sin(\delta/2) = 0.75$, so $\sin^2 = 0.5625$ and $\sin^4 = 0.31641$, $$\frac{b_5}{b_1} = \frac{16(0.31641) - 20(0.5625) + 5}{5} = \frac{5.0625 - 11.25 + 5}{5} = \boxed{-0.2375}$$ The negative sign is physically meaningful, not a slip: at this pulse width the fifth harmonic is in antiphase with the fundamental. Its magnitude, 23.75 per cent, is close to that of the third — the characteristic weakness of single-pulse modulation.
4. Part (d) — evaluate the harmonic voltages. Substituting $V_d = 220$ V and $\sin(n\delta/2)$ into the given coefficient, the peak amplitudes are $$b_1 = \frac{4(220)}{\pi}(0.75) = 210.08\ \text{V}, \quad b_3 = 0.25\,b_1 = 52.52\ \text{V}, \quad b_5 = -0.2375\,b_1 = -49.90\ \text{V}$$ These are peak values; the corresponding rms voltages are smaller by $\sqrt{2}$.
5. Part (d) — form the harmonic admittance of the parallel network. At order $n$ the motor branch is $Z_{m,n} = R + jn\omega L$ and the capacitor branch has admittance $Y_{c,n} = jn\omega C$, so the inverter sees $$Y_n = \frac{1}{R + jn\omega L} + jn\omega C$$ Because the resistance is small, the motor admittance falls roughly as $1/n$ while the capacitor admittance rises as $n$ — the capacitor comes to dominate.
6. Part (d) — evaluate order by order. At the fundamental $Z_{m,1} = 0.12 + j0.09\ \Omega$, $|Z_{m,1}| = 0.15\ \Omega$, giving $Y_{m,1} = 5.333 - j4.000$ S; adding $j3$ S for the capacitor yields $Y_1 = 5.333 - j1.000$ S, $|Y_1| = 5.4263$ S at $-10.62^\circ$. The capacitor has cancelled three quarters of the motor's lagging reactive current, which is precisely its purpose. Repeating at $n = 3$ and $n = 5$ gives $|Y_3| = 6.0650$ S at $+76.90^\circ$ and $|Y_5| = 12.9372$ S at $+87.55^\circ$.
7. Part (d) — multiply out the currents. $I_n = b_n Y_n$, so $$I_1 = 210.08 \times 5.4263 = 1139.98\ \text{A (peak)} = \boxed{806.1\ \text{A rms}}$$ $$I_3 = 52.52 \times 6.0650 = 318.54\ \text{A (peak)} = \boxed{225.2\ \text{A rms}}$$ $$I_5 = 49.90 \times 12.9372 = 645.50\ \text{A (peak)} = \boxed{456.4\ \text{A rms}}$$
8. Interpret the result. The fifth-harmonic voltage is only 23.75 per cent of the fundamental, yet the fifth-harmonic current reaches 57 per cent of the fundamental. The reason is entirely in the capacitor: at $n = 5$ its susceptance is 15 S against a motor-branch admittance of only 2.15 S, so the capacitor alone sets the fifth-harmonic current. A power-factor-correcting capacitor placed across a non-sinusoidal source is a harmonic sink, and this is the design lesson the question is built around.