22-Elec-B8 Power Electronics and Drives · December 2013

Question 3 of 6: Current-Fed versus Voltage-Fed Inverters, and Harmonic Analysis of a Single-Pulse-Modulated Bridge

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

Question 3: Current-Fed versus Voltage-Fed Inverters, and Harmonic Analysis of a Single-Pulse-Modulated Bridge (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) — current-fed versus voltage-fed inverters

The distinction lies in what the dc link presents to the bridge. A voltage-fed (voltage-source) inverter is supplied from a stiff dc voltage, enforced by a large electrolytic capacitor across the link. The bridge therefore imposes the output voltage waveform — a square wave, quasi-square wave or PWM pattern of amplitude $\pm V_d$ — and the load impedance decides what current flows. Because the load is normally inductive, the current lags and must be allowed to flow while the opposite voltage is applied, so every switch needs an antiparallel feedback diode. The cardinal operating rule is that the two devices in a leg must never conduct together: a shoot-through short-circuits the capacitor and destroys the bridge, which is why dead time is inserted between gate signals.

A current-fed (current-source) inverter is supplied through a large series inductor in the dc link, so the link current is held nearly constant and the bridge imposes the output current waveform — typically a quasi-square current of amplitude $I_d$ — while the load and the commutation capacitors decide the voltage. The consequences invert. An output short circuit is harmless because the link inductor limits the rate of rise of current, which makes the topology inherently rugged; but an open circuit is dangerous, since the inductor will develop whatever voltage is needed to keep current flowing. Feedback diodes are not required; instead series blocking diodes and commutating capacitors are used, and the devices see substantial voltage spikes at commutation.

Practically, the voltage-fed inverter dominates general-purpose ac drives because PWM is straightforward, the dynamic response is fast and multiple motors can share one link. The current-fed inverter is found in large drives (multi-megawatt synchronous motor drives, load-commutated inverters) where regeneration into the supply is wanted — it is achieved simply by reversing the dc link voltage, with no second converter — and where robustness against faults matters more than dynamic bandwidth. Its drawbacks are a bulky dc inductor, slower response, and a torque pulsation caused by the stepped current waveform.

Parts (b), (c) and (d) — harmonics of the single-pulse-modulated bridge

Given.

Find. (b) a proof of the stated $b_5/b_3$ identity; (c) the modulation angle satisfying $b_5/b_3 = 0.275$ and the resulting $b_3/b_1$; and (d) the fundamental, third and fifth harmonic components of the current drawn by the motor.

Approach. Form the ratio directly from $b_n$, expand $\sin(3\delta/2)$ and $\sin(5\delta/2)$ with the supplied identities, then solve the resulting quadratic in $\sin^{2}(\delta/2)$ and divide each harmonic voltage by the impedance the motor presents at that harmonic order.

1. Part (b) — form the ratio and eliminate the constants. Writing the coefficient twice and dividing, the supply voltage and the factor $4/\pi$ cancel and only the order and the sine survive: $$\frac{b_5}{b_3} = \frac{\dfrac{4V_d}{5\pi}\sin\dfrac{5\delta}{2}}{\dfrac{4V_d}{3\pi}\sin\dfrac{3\delta}{2}} = \frac{3}{5}\,\frac{\sin(5\delta/2)}{\sin(3\delta/2)} .$$
2. Substitute the multiple-angle identities. Put $\theta = \delta/2$, so that $5\delta/2 = 5\theta$ and $3\delta/2 = 3\theta$. The supplied identities then give $\sin 5\theta = 5\sin\theta - 20\sin^{3}\theta + 16\sin^{5}\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^{3}\theta$, whence $$\boxed{\;\frac{b_5}{b_3} = \frac{3}{5}\left[\frac{5\sin\frac{\delta}{2} - 20\sin^{3}\frac{\delta}{2} + 16\sin^{5}\frac{\delta}{2}}{3\sin\frac{\delta}{2} - 4\sin^{3}\frac{\delta}{2}}\right]\;}$$ which is the required result.
3. Part (c) — reduce the identity to a quadratic. Cancelling one factor of $\sin(\delta/2)$ from numerator and denominator and writing $u = \sin^{2}(\delta/2)$ turns the expression into a ratio of quadratics: $$k = \frac{b_5}{b_3} = \frac{3}{5}\left[\frac{5 - 20u + 16u^{2}}{3 - 4u}\right].$$ Cross-multiplying and collecting terms gives the compact form $$48u^{2} + (20k - 60)u + 15(1 - k) = 0 .$$
4. Solve for the modulation angle. With $k = 0.275$ the coefficients become $48u^{2} - 54.5u + 10.875 = 0$, whose discriminant is $54.5^{2} - 4(48)(10.875) = 2970.25 - 2088 = 882.25$, so $\sqrt{\Delta} = 29.7027$ and $$u = \frac{54.5 \pm 29.7027}{96} \quad \Rightarrow \quad u_1 = 0.87711, \qquad u_2 = 0.25831 .$$ Both roots satisfy the constraint exactly, giving $\delta_1 = 2\arcsin\sqrt{0.87711} = 138.96^\circ$ and $\delta_2 = 61.09^\circ$. The physically sensible one is the wide pulse: $$\boxed{\;\delta = 138.96^\circ\;}$$ because the narrow root produces $b_3/b_1 = 0.656$, a third harmonic two-thirds the size of the fundamental, which no inverter would be operated at.
5. Compute the required third-to-fundamental ratio. Using $b_n \propto (1/n)\sin(n\delta/2)$ with $\delta/2 = 69.48^\circ$, so that $3\delta/2 = 208.44^\circ$ and $\sin(3\delta/2) = -0.47619$: $$\frac{b_3}{b_1} = \frac{1}{3}\,\frac{\sin(3\delta/2)}{\sin(\delta/2)} = \frac{1}{3}\left(\frac{-0.47601}{0.93654}\right) = \boxed{\;-0.1695\;}$$ i.e. the third harmonic is 16.95 percent of the fundamental and is in antiphase with it. The negative sign is physical, not a slip: beyond $\delta = 120^\circ$ the factor $\sin(3\delta/2)$ has changed sign, and at exactly $\delta = 120^\circ$ the third harmonic vanishes altogether — the classic single-pulse harmonic-elimination point.
6. Part (d) — evaluate the harmonic voltages. With $4V_d/\pi = 4(220)/\pi = 280.11$ V, the peak coefficients follow directly: $$b_1 = 280.11\sin 69.48^\circ = 262.34 \ \text{V}, \quad b_3 = \frac{280.11}{3}\sin 208.44^\circ = -44.46 \ \text{V}, \quad b_5 = \frac{280.11}{5}\sin 347.39^\circ = -12.23 \ \text{V},$$ and the ratio check $b_5/b_3 = (-12.23)/(-44.46) = 0.275$ recovers the specification.
7. Find the impedance at each harmonic order. The resistance is unchanged with frequency but the reactance scales with $n$, so $Z_n = R + jn\omega L = 8.2 + j6.1n$: $$|Z_1| = \sqrt{8.2^{2}+6.1^{2}} = 10.220\ \Omega\ \angle 36.65^\circ, \quad |Z_3| = \sqrt{8.2^{2}+18.3^{2}} = 20.053\ \Omega\ \angle 65.86^\circ,$$ $$|Z_5| = \sqrt{8.2^{2}+30.5^{2}} = 31.583\ \Omega\ \angle 74.95^\circ .$$
8. Divide to obtain the harmonic currents. Each harmonic acts independently because the circuit is linear, so $$\boxed{\;I_1 = \frac{262.34}{10.220} = 25.67\ \text{A peak} \ (18.15\ \text{A rms})\;}$$ $$\boxed{\;I_3 = \frac{44.46}{20.053} = 2.22\ \text{A peak} \ (1.57\ \text{A rms}), \qquad I_5 = \frac{12.23}{31.583} = 0.39\ \text{A peak} \ (0.27\ \text{A rms})\;}$$ The load inductance has done most of the filtering for us: a third-harmonic voltage that is 17 percent of the fundamental produces a third-harmonic current of only 8.6 percent, and the fifth harmonic is down to 1.5 percent. That $1/n$ attenuation from the impedance, on top of the $1/n$ already present in $b_n$, is why an inductive machine tolerates a crude inverter waveform far better than a resistive load would.