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

Problem 3: Single-Pulse-Modulated Bridge 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.

Given. D.c. link voltage $V_d = 220\ \text{V}$; single-phase full-bridge inverter with single-pulse modulation of width $\delta$; motor load at fundamental frequency $R = 8\ \Omega$ and $X_1 = \omega L = 6\ \Omega$; the modulation angle is chosen so that $b_5/b_3 = 0.3$. Harmonic reactance scales with order, $X_n = n\omega L$.

Find. (a) the distinguishing features of current-fed and voltage-fed inverters; (b) a proof of the printed harmonic-ratio identity; (c) the third-to-fundamental voltage ratio for the selected δ; (d) the fundamental, third and fifth harmonic currents drawn by the motor.

Approach. Part (b) is pure substitution of the two supplied trigonometric identities into the given Fourier coefficient. Part (c) reduces the resulting expression to a quadratic in $\sin^2(\delta/2)$, which has two admissible roots; the wide-pulse root is the operating point. Part (d) then evaluates $b_n$ and divides by the harmonic impedance $Z_n = R + jn\omega L$.

Part (a) — voltage-fed versus current-fed inverters. A voltage-fed (voltage-source) inverter is supplied from a stiff d.c. voltage — a large capacitor across the d.c. link, or a battery — so the link behaves as a near-ideal voltage source with very low internal impedance. Its output voltage waveform is imposed by the switching pattern and is essentially independent of the load; the output current is whatever the load impedance draws. Each switch must therefore be paired with an antiparallel feedback diode to carry reactive load current, the link must never be short-circuited (shoot-through is destructive, so dead time is mandatory), and regeneration requires either a second converter or a dynamic braking resistor because the link capacitor cannot reverse polarity.

A current-fed (current-source) inverter is supplied through a large series d.c. link inductor from a controlled rectifier, so the link behaves as a near-ideal current source. The inverter now imposes the current waveform — typically a quasi-square block of constant amplitude — while the output voltage is determined by the load and by the commutating capacitors. Feedback diodes are not required; instead series blocking diodes and commutating capacitors are used, and the output must never be open-circuited because the link inductor would develop a destructive voltage. Its great practical advantages are inherent protection against short circuits (the link inductor limits di/dt), robust four-quadrant operation — regeneration needs only a reversal of the rectifier voltage, with no extra hardware — and the ability to use simple thyristors. Against that, the drive is inherently less stable in open loop, cannot easily run more than one machine from one inverter, and the commutating capacitors make it machine-specific. Voltage-fed inverters dominate general-purpose PWM drives; current-fed inverters remain attractive for large, single-machine, regenerative drives such as mine hoists and rolling-mill mains.

1. Part (b) — form the ratio and substitute the identities. From the given coefficient, the $V_d$ and $\pi$ factors cancel: $$\frac{b_5}{b_3} = \frac{\dfrac{4V_d}{5\pi}\sin\dfrac{5\delta}{2}}{\dfrac{4V_d}{3\pi}\sin\dfrac{3\delta}{2}} = \frac{3}{5}\,\frac{\sin(5\theta)}{\sin(3\theta)}, \qquad \theta \equiv \frac{\delta}{2}.$$ Introducing the two supplied 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 exactly the printed result: $$\boxed{\frac{b_5}{b_3} = \frac{3}{5}\left[\frac{5\sin\frac{\delta}{2} - 20\sin^3\frac{\delta}{2} + 16\sin^5\frac{\delta}{2}}{3\sin\frac{\delta}{2} - 4\sin^3\frac{\delta}{2}}\right]}$$ which completes the required demonstration.
2. Part (c) — reduce to a quadratic in sin²(δ/2). Cancel one factor of $\sin\theta$ from numerator and denominator and write $u \equiv \sin^{2}(\delta/2)$: $$\frac{b_5}{b_3} = \frac{3}{5}\,\frac{5 - 20u + 16u^{2}}{3 - 4u} = 0.3 .$$ Cross-multiplying, $3(5 - 20u + 16u^{2}) = 1.5(3 - 4u)$, i.e. $48u^{2} - 54u + 10.5 = 0$, or after dividing by 1.5, $$32u^{2} - 36u + 7 = 0 \;\Longrightarrow\; u = \frac{36 \pm 20}{64} = 0.875 \;\text{or}\; 0.250 .$$
3. Part (c) — select the physically meaningful root. Both roots satisfy the constraint exactly, so the choice must be made on operating grounds. Using $b_3/b_1 = \frac{1}{3}\sin 3\theta/\sin\theta = (3-4u)/3$: the narrow root $u = 0.250$ gives $\delta = 60^{\circ}$ and $b_3/b_1 = +0.667$ — a third harmonic two-thirds the size of the fundamental, which no practical inverter would be operated at; the wide root $u = 0.875$ gives $\delta = 2\sin^{-1}\sqrt{0.875} = 138.59^{\circ}$ and $$\boxed{\frac{b_3}{b_1} = \frac{3 - 4(0.875)}{3} = -\frac{1}{6} = -0.1667}$$ The negative sign means the third harmonic is in antiphase with the fundamental; its magnitude is 16.67 per cent of the fundamental. This wide-pulse root is the operating point, and the narrow root is recorded and rejected.
4. Part (d) — evaluate the voltage harmonics. With $4V_d/\pi = 4(220)/\pi = 280.11\ \text{V}$ and $\theta = 69.295^{\circ}$: $$b_1 = 280.11\sin 69.295^{\circ} = 262.02\ \text{V}, \quad b_3 = \frac{280.11}{3}\sin 207.885^{\circ} = -43.67\ \text{V}, \quad b_5 = \frac{280.11}{5}\sin 346.475^{\circ} = -13.10\ \text{V}$$ (peak values). The checks close: $b_3/b_1 = -0.1667$ and $b_5/b_3 = 0.300$ as required. In rms terms these are 185.28 V, 30.88 V and 9.26 V.
5. Part (d) — divide by the harmonic impedances. The motor reactance rises in proportion to harmonic order, $Z_n = R + jn\omega L = 8 + j6n$: $$Z_1 = 10.00\,\angle 36.87^{\circ}\ \Omega, \quad Z_3 = 19.70\,\angle 66.04^{\circ}\ \Omega, \quad Z_5 = 31.05\,\angle 75.07^{\circ}\ \Omega .$$ Hence the current harmonics (peak, then rms) are $$\boxed{I_1 = 26.20\ \text{A peak} = 18.53\ \text{A rms}}$$ $$\boxed{I_3 = 2.217\ \text{A peak} = 1.568\ \text{A rms}, \qquad I_5 = 0.422\ \text{A peak} = 0.298\ \text{A rms}}$$ The load inductance has attenuated the harmonic content dramatically: a voltage spectrum in which the third harmonic is 16.7 per cent of the fundamental produces a current spectrum in which it is only 8.5 per cent, and the fifth falls from 5.0 to 1.6 per cent.