Question 3 of 6: Current-Fed vs Voltage-Fed Inverters, and Single-Pulse-Modulation Harmonics

Question 3: Current-Fed vs Voltage-Fed Inverters, and Single-Pulse-Modulation Harmonics (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 is made entirely on the dc side. A voltage-fed (voltage-source) inverter is supplied from a stiff voltage source: a large capacitor across the dc link holds the link voltage essentially constant against the pulsating current the switches draw. A current-fed (current-source) inverter is supplied from a stiff current source, realised as a large series reactor in the link fed from a phase-controlled rectifier whose firing angle regulates the link current. Everything else follows from that one choice.

In a voltage-fed inverter the switches impose the output voltage waveform — quasi-square, or pulse-width modulated — and the load impedance decides the current. Each switch needs an antiparallel freewheel diode so that lagging load current has a return path, and the two devices in a leg must never conduct together, so a dead time is inserted between their gate signals. A short circuit at the terminals is catastrophic because the link capacitor can deliver enormous fault current, so fast protection is mandatory. Regeneration back into the ac supply needs either a second antiparallel rectifier or a braking chopper, since the link current must reverse while the link voltage cannot. On the other hand one voltage-fed inverter can supply several motors in parallel, it operates well from no-load to full load, and its output impedance is low.

In a current-fed inverter the switches steer the constant link current into the load in a quasi-square current waveform, and the load impedance decides the voltage. The devices see no freewheel diodes but do need series blocking diodes and commutating capacitors, because the constant link current must be transferred from one device to the next. The topology is inherently robust against an output short circuit — the link reactor limits $\mathrm{d}i/\mathrm{d}t$ and the current is already regulated — and regeneration is natural: to return power the link voltage simply reverses while the current keeps its direction, which the source-side bridge accommodates by moving past 90° firing into the inverting region. The penalties are that the drive is essentially single-motor, that light-load operation is poor because the fixed link current over-fluxes an unloaded machine, and that the open-loop operating point sits on the unstable side of the torque-slip curve, so closed-loop slip or flux regulation is not optional. Current-fed thyristor inverters accordingly found their home in large, single-motor, regenerating drives — mine hoists, rolling mills, large pumps — while voltage-fed PWM inverters with IGBTs dominate everything else.

Part (b) — Deriving the harmonic-ratio expression

Given. The single-pulse-modulated bridge produces a quasi-square output whose Fourier coefficients are $b_n=(4V_d/n\pi)\sin(n\delta/2)$, together with the two supplied identities for $\sin3\theta$ and $\sin5\theta$.

Find. A closed-form expression for $b_5/b_3$ in terms of $\sin(\delta/2)$ alone.

1. Form the ratio directly from the coefficient. The amplitude $4V_d/\pi$ is common to both harmonics and cancels; only the $1/n$ factors and the sines survive: $$\frac{b_5}{b_3}=\frac{(4V_d/5\pi)\sin(5\delta/2)}{(4V_d/3\pi)\sin(3\delta/2)} =\frac{3}{5}\cdot\frac{\sin(5\delta/2)}{\sin(3\delta/2)} .$$
2. Substitute the supplied identities with $\theta=\delta/2$. Writing $\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta$ and putting $\theta=\delta/2$ gives the required 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 is exactly the expression stated in the question.
3. Reduce it to a form that can be solved. Divide numerator and denominator by $\sin(\delta/2)$ and write $u=\sin^{2}(\delta/2)$: $$\frac{b_5}{b_3}=\frac{3}{5}\cdot\frac{5-20u+16u^{2}}{3-4u}.$$ The ratio depends on the modulation angle only through $u$, and the relation is a quadratic in $u$. That fact governs part (c).

Parts (c) and (d) — Modulation angle, harmonic ratio and load currents

Given. A single-phase full-wave bridge with single-pulse modulation, dc supply $V_d=220$ V, feeding a motor modelled at fundamental frequency as $R=8\ \Omega$ in series with $\omega L=6\ \Omega$; the modulation angle is chosen so that $b_5/b_3=0.225$.

Find. The ratio $b_3/b_1$, and then the fundamental, third and fifth harmonic components of the current drawn by the motor.

Approach. Set the ratio derived in part (b) equal to 0.225, solve the resulting quadratic in $u=\sin^2(\delta/2)$, screen the two roots on physical grounds, then evaluate the coefficients at the surviving root and divide each by the impedance seen at its own harmonic order.

1. Turn the specification into a quadratic. Setting $(3/5)(5-20u+16u^{2})/(3-4u)=0.225$ and clearing the denominator, $$5-20u+16u^{2}=0.375\,(3-4u)\quad\Longrightarrow\quad 16u^{2}-18.5u+3.875=0 .$$
2. Solve, and note that both roots are genuine. $$u=\frac{18.5\pm\sqrt{18.5^{2}-4(16)(3.875)}}{32}=\frac{18.5\pm9.7082}{32} \;\Rightarrow\; u_1=0.88151,\quad u_2=0.27474 .$$ Substituting either value back reproduces $b_5/b_3=0.225$ exactly, so the algebra cannot choose between them.
3. Screen the roots on the third-harmonic content. From the same reduction, $b_3/b_1=(3-4u)/3$. The narrow root $u_2$ corresponds to $\delta=63.22^\circ$ and gives $b_3/b_1=+0.634$ — a third harmonic almost two-thirds the size of the fundamental, which no inverter would be operated at, and which also delivers very little fundamental output. The wide root $u_1$ gives $$\delta=2\sin^{-1}\sqrt{0.88151}=139.73^\circ,$$ and is the operating point intended. The narrow root is recorded and rejected rather than ignored.
4. Evaluate the required ratio. At $u_1=0.88151$, $$\frac{b_3}{b_1}=\frac{3-4(0.88151)}{3}=1-\frac{4}{3}(0.88151),$$ $$\boxed{\frac{b_3}{b_1}=-0.1753}$$ i.e. the third harmonic is 17.53 per cent of the fundamental and in antiphase with it. The negative sign is the answer, not a slip: at pulse widths beyond 120° the third harmonic reverses, and it passes through zero exactly at $\delta=120^\circ$. For completeness $b_5/b_1=(16u^{2}-20u+5)/5=-0.0394$, and $(-0.0394)/(-0.1753)=0.225$ as required.
5. Compute the voltage harmonics. With $V_d=220$ V and $4V_d/\pi=280.11$ V, $$b_1=\frac{4V_d}{\pi}\sin\frac{\delta}{2}=280.11\times0.93889=262.99\ \text{V (peak)},$$ $$b_3=\frac{4V_d}{3\pi}\sin\frac{3\delta}{2}=-46.11\ \text{V},\qquad b_5=\frac{4V_d}{5\pi}\sin\frac{5\delta}{2}=-10.38\ \text{V}.$$ As a check on the modulation angle, the total rms output is $V_o=V_d\sqrt{\delta/\pi}=193.84$ V.
6. Find the impedance at each harmonic order. The resistance is constant but the reactance scales with frequency, so $$Z_n=R+jn\omega L=8+j6n:\quad |Z_1|=10.00\ \Omega,\quad |Z_3|=19.70\ \Omega, \quad |Z_5|=31.05\ \Omega ,$$ with phase angles 36.87°, 66.04° and 75.07° respectively.
7. Divide to obtain the current harmonics. Each harmonic acts independently, so $$I_1=\frac{262.99}{10.00}=26.30\ \text{A},\qquad I_3=\frac{-46.11}{19.70}=-2.341\ \text{A},\qquad I_5=\frac{-10.38}{31.05}=-0.334\ \text{A}\ \ \text{(peak)} ,$$ $$\boxed{I_1=26.30\ \text{A},\quad |I_3|=2.341\ \text{A},\quad |I_5|=0.334\ \text{A}\ \ (\text{peak})}$$ or in rms terms 18.60 A, 1.655 A and 0.236 A.

The comparison in Figure 3.2 is the engineering point of the problem. In the voltage spectrum the third harmonic is 17.5 per cent of the fundamental; in the current spectrum it is only 8.9 per cent, and the fifth falls from 3.9 per cent to 1.3 per cent. A series-inductive load is therefore its own harmonic filter, which is why a wide-pulse quasi-square inverter driving a motor produces a current waveform far closer to sinusoidal than its voltage waveform suggests, and why torque pulsation in such a drive is much milder than the voltage distortion implies.