Question 3 of 6: Current-fed versus voltage-fed inverters, and single-pulse modulation harmonics

Question 3: Current-fed versus voltage-fed inverters, and single-pulse modulation harmonics (20 points)

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) — Voltage-fed versus current-fed inverters

The two families are distinguished by what the dc link presents to the inverter bridge, and almost every other difference follows from that one choice.

A voltage-fed (voltage-source) inverter is supplied from a stiff dc voltage: a large electrolytic capacitor across the link holds $V_d$ essentially constant and gives the link a low internal impedance. The bridge therefore imposes a voltage waveform on the load — a square, quasi-square or pulse-width-modulated voltage — and the load impedance decides what current flows. The output current is consequently smooth and roughly sinusoidal into an inductive motor, while the voltage is the chopped quantity. Because the link cannot change polarity (the capacitor voltage cannot reverse), regeneration requires either a second antiparallel rectifier bridge or a braking chopper and resistor. A shoot-through of one leg short-circuits the capacitor and is catastrophic, so dead-time and fast overcurrent protection are essential. Each switch needs an antiparallel feedback diode to carry the reactive load current.

A current-fed (current-source) inverter is supplied through a large series dc-link reactor from a controlled rectifier, so the link presents a stiff, nearly ripple-free current with a high internal impedance. The bridge now imposes a current waveform — typically a six-step block of current — and the load decides the voltage that appears at the terminals. The output current is the chopped quantity and the terminal voltage is comparatively smooth, though it carries commutation spikes that must be absorbed by capacitors across the machine terminals. Because the link current cannot reverse but the link voltage can, regeneration is inherently available simply by driving the source rectifier into inversion, which is why current-fed drives are favoured for large single-motor loads that must brake regeneratively — mine hoists, rolling-mill drives and large fans. The dc reactor limits fault current, so a momentary short is survivable rather than destructive, and no feedback diodes are needed on the switches.

The practical consequences: a voltage-fed inverter can feed several motors from one inverter and gives good performance across the speed range, but it needs fast switching devices and shoot-through protection; a current-fed inverter is rugged, naturally four-quadrant and well suited to a single large machine, but it is load-dependent, tends to be unstable open-loop (see Problem 5), and produces torque pulsations at low speed because of the stepped current.

Parts (b), (c) and (d) — The single-pulse inverter

Given.

Find. A proof of the supplied $b_5/b_3$ identity; the modulation angle and $b_3/b_1$ that satisfy $b_5/b_3 = 0.225$; and the fundamental, third and fifth harmonic components of the current drawn by the motor.

Approach. Substitute the supplied identities to prove the ratio, reduce it to a quadratic in $u = \sin^{2}(\delta/2)$, screen the two roots on operating sense, then divide each harmonic voltage by the impedance at its own frequency.

1. Part (b) — Form the ratio from the given coefficient. Substituting $n = 5$ and $n = 3$ into $b_n = \frac{4V_d}{n\pi}\sin\frac{n\delta}{2}$ and dividing, the common factor $4V_d/\pi$ cancels and only the reciprocal orders survive: $$\frac{b_5}{b_3} = \frac{\frac{4V_d}{5\pi}\sin\frac{5\delta}{2}}{\frac{4V_d}{3\pi}\sin\frac{3\delta}{2}} = \frac{3}{5}\cdot\frac{\sin\frac{5\delta}{2}}{\sin\frac{3\delta}{2}}.$$
2. Expand both sines with the supplied identities. Writing $\theta = \delta/2$ so that $5\delta/2 = 5\theta$ and $3\delta/2 = 3\theta$, the identities give directly $\sin 5\theta = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. Substituting, $$\boxed{\frac{b_5}{b_3} = \frac{3}{5}\left[\frac{5\sin\frac{\delta}{2} - 20\sin^3\frac{\delta}{2} + 16\sin^5\frac{\delta}{2}}{3\sin\frac{\delta}{2} - 4\sin^3\frac{\delta}{2}}\right]}$$ which is the required result.
3. Part (c) — Reduce the ratio to a quadratic. Divide numerator and denominator of the bracket by $\sin\theta$ and put $u = \sin^{2}(\delta/2)$: $$\frac{b_5}{b_3} = \frac{3}{5}\cdot\frac{5 - 20u + 16u^{2}}{3 - 4u}.$$ Setting this equal to 0.225 and clearing fractions, $3(5 - 20u + 16u^{2}) = 1.125(3 - 4u)$, i.e. $$48u^{2} - 55.5\,u + 11.625 = 0.$$
4. Solve, and screen the two roots on physical grounds. The discriminant is $55.5^{2} - 4(48)(11.625) = 848.25$, so $u = (55.5 \pm 29.1247)/96$, giving $$u_1 = 0.881508 \;\Rightarrow\; \delta = 139.731^\circ, \qquad u_2 = 0.274742 \;\Rightarrow\; \delta = 63.223^\circ .$$ Both satisfy the stated ratio exactly, so the constraint alone cannot choose between them and the choice must be made on operating sense. Evaluate the third-harmonic content at each before deciding.
5. Obtain the third-to-fundamental ratio. With $b_3/b_1 = \frac{1}{3}\cdot\frac{\sin 3\theta}{\sin\theta} = \frac{3 - 4u}{3}$, the narrow root gives $b_3/b_1 = +0.6337$ and the wide root gives $b_3/b_1 = -0.1753$. A third harmonic worth 63% of the fundamental is not an inverter operating point — it would be an unusable output — so the narrow root is recorded and rejected, and the wide pulse is selected: $$\boxed{\delta = 139.731^\circ,\qquad \frac{b_3}{b_1} = -0.1753}$$ i.e. the third harmonic is 17.53% of the fundamental and is in antiphase with it. The negative sign is the answer, not a slip: for pulse widths above 120° the third harmonic reverses, and it vanishes identically at $\delta = 120^\circ$.
6. Confirm the selection against the fifth harmonic. The companion identity is $b_5/b_1 = (16u^{2} - 20u + 5)/5$, which at $u_1$ gives $-0.039452$. Then $b_5/b_3 = (-0.039452)/(-0.175343) = 0.22500$, reproducing the stated ratio to five figures and confirming both the algebra and the root.
7. Part (d) — Evaluate the voltage harmonics. With $V_d = 220$ V and $\delta = 139.731^\circ$, $$b_1 = \frac{4(220)}{\pi}\sin(69.865^\circ) = 262.99\ \text{V (peak)},$$ and the ratios just found give $b_3 = -46.11$ V and $b_5 = -10.38$ V, both peak values. It is worth stating explicitly that the supplied $b_n$ are peak amplitudes, not rms.
8. Divide each harmonic voltage by the impedance at its own frequency. The motor reactance scales with order, so $Z_n = R + jn\omega L$: $$Z_1 = 10 + j12.5 = 16.008\angle 51.34^\circ\ \Omega,\quad Z_3 = 10 + j37.5 = 38.810\angle 75.07^\circ\ \Omega,\quad Z_5 = 10 + j62.5 = 63.295\angle 80.91^\circ\ \Omega.$$ Dividing the peak voltages, $$\boxed{I_1 = 16.429\ \text{A peak} = 11.617\ \text{A rms}},\qquad \boxed{I_3 = 1.188\ \text{A peak} = 0.840\ \text{A rms}},\qquad \boxed{I_5 = 0.164\ \text{A peak} = 0.116\ \text{A rms}}.$$
9. Read the physical message. The third harmonic is 17.5% of the fundamental in voltage but only 7.2% in current, and the fifth falls from 3.9% to 1.0%. The series inductance acts as a low-pass filter whose impedance rises with order, so the machine itself does most of the harmonic filtering — which is exactly why a motor tolerates a square-wave supply far better than a resistive load would.

Final results.