Three-Phase Voltage Source Inverters

The line voltage is a quasi-square wave. Each leg sits high for 180°, and the difference between two legs 120° apart is \(+V_{dc}\) for 120°, zero for 60°, \(-V_{dc}\) for 120°, zero for 60°:

Which is exactly the quasi-square wave of Set 25, Problem 2 with \(\delta = 120^\circ\) — the three-phase inverter produces that waveform on every line pair without anyone choosing an angle. The geometry does it.

The phase voltage is the six-step shape, which no individual leg produces. With a floating star point, the load itself divides the applied voltages:

Two levels in each half cycle, six steps per cycle — hence "six-step inverter". The star point floats to whatever makes the three phase voltages sum to zero, and that floating is what creates the intermediate level.

The distortion:

Only \(6k\pm1\). The triplens are in phase in all three lines — zero sequence — so they cancel in every line-to-line difference and cannot flow at all in a three-wire load. The single-phase inverter's biggest harmonic simply does not exist here.

The output power:

Against 81.1% for the single-phase square wave of Set 25. Removing the triplens moved ten points of power from the harmonics into the fundamental, and it cost nothing.

The recurring 31.08%. This exact figure has now appeared three times:

All three are the same waveform — a 120° quasi-square wave — appearing as a current in one case and a voltage in the others. The rectifier and the inverter are duals, and their distortion figures are identical because their waveforms are.

What changes. Each device conducts for 120° instead of 180, so only two devices conduct at any instant instead of three — and each phase is disconnected for 60° in every half cycle:

The phase voltage is now a clean quasi-square wave with only two levels, rather than the six-step shape. Simpler — but that simplicity comes from disconnecting the load, which is exactly the problem.

The phase voltage:

The comparison:

Thirteen per cent less voltage, a quarter less power, and the same distortion — because both waveforms are 120° quasi-square shapes and both lack triplens. The only genuine advantage is in the fifth row.

The one real advantage. In 120° mode there is a 60° interval in every leg during which neither device is commanded on:

A shoot-through is structurally impossible, with no dead-time circuit required. In the era of slow thyristors with millisecond turn-off times — the forced-commutation inverters of Set 6 — that was decisive: the alternative was an elaborate commutation circuit on every leg.

Why it is now rare. Three reasons, all consequences of fast self-commutating devices:

The third row is the practical killer. With a motor load the current cannot stop when the device does, so it freewheels through the anti-parallel diodes — and the phase voltage during the supposed 60° gap is then whatever the diodes impose, not zero. The clean analysis above holds only for a resistive load.

Each leg is modulated against the same carrier with references 120° apart. Relative to the DC midpoint, a leg's fundamental is:

Note the \(V_{dc}/2\) — each leg is effectively a half bridge referenced to the midpoint, not a full bridge. That factor is where the three-phase inverter's utilisation problem begins.

The line voltage is \(\sqrt3\) larger:

The linear limit is \(m_a = 1\), where the reference peak equals the carrier peak:

Compare with six-step:

Sinusoidal PWM throws away more than a fifth of the DC link's capability. For a 415 V drive that is the difference between needing a 587 V link and a 678 V one — and the higher link means higher-voltage devices with worse conduction, plus more insulation stress on the motor.

Where the loss comes from. The three references are sinusoids, and their peaks constrain the modulation:

But the useful output is the line voltage, which is a difference of two phases — and the difference of two sinusoids 120° apart never peaks when either of them does. So the constraint is applied to a quantity the load never sees. Something can be added to all three references at once, changing each phase's peak without changing any difference at all — which is Problem 4.

Where the harmonics land. As in Set 25, they cluster around the carrier — but with a three-phase simplification:

So choosing \(m_f\) odd and a multiple of three — 21, 33, 39 — removes both the even harmonics and the dominant carrier component from the line voltages, leaving only the sidebands. It costs nothing and is standard.

The modified reference:

Added identically to all three phases — and since the third harmonic of a 120°-shifted sinusoid is in phase with the original (\(3\times120^\circ = 360^\circ\)), the same waveform serves all three. It is a zero-sequence component by construction.

Find the new peak. Differentiate:

The peak has fallen from 1.000 to 0.866 while the fundamental component is unchanged. The waveform is flattened — a saddle shape with equal maxima at 60° and 120° instead of a single peak at 90°.

Which allows a larger fundamental:

A 15.5% increase, and the result is exactly \(V_{dc}/\sqrt2\) — a satisfying figure that reappears in Set 28 as the radius of the largest circle that fits inside the space-vector hexagon.

Why the load never sees it. The added component is identical in all three phases, so it cancels in every difference:

The third harmonic terms subtract exactly. And in a three-wire load there is no path for a zero-sequence current in any case — it appears only between the load's star point and the DC midpoint, where nothing is connected.

The reference waveform, tabulated:

The injection pushes the middle of the waveform up and pulls its crest down, filling in the saddle. The area — and therefore the fundamental — is unchanged; only the peak moves, and the peak is what the carrier constrains.

Where this leaves the three modes:

Injection recovers about half the gap to six-step and keeps the clean spectrum. The remaining 9.3% is only reachable by overmodulating into the six-step waveform, which brings the low-order harmonics back — Set 27.

Define it. Relative to the DC link midpoint, each leg is at \(\pm V_{dc}/2\). The common-mode voltage is their average — equivalently, the voltage of the load's star point relative to the midpoint:

For a balanced three-phase set this would be zero. But the inverter produces eight discrete states, not a balanced set, and only two of them give zero.

Evaluate all eight, with 1 denoting the upper device on:

Four hundred volts peak-to-peak of common-mode square wave, appearing between the motor's star point and earth at every switching transition — and the two zero states, which the modulator uses constantly, are the two extremes.

The current it drives. The motor windings have capacitance to the stator frame, which is earthed:

Four ampere spikes into the earth conductor, thousands of times per second, from a drive whose useful current is a few tens of amps. They are brief — nanoseconds — so they carry little energy, but they are fast, and fast is what causes trouble.

What they damage:

The first three are one failure mode. The shaft is capacitively coupled to the stator, so it follows a fraction of \(v_{cm}\); when that exceeds the breakdown strength of the lubricant film — a few volts across a few microns — it arcs, and each arc removes a little metal. Bearings that should last a decade fail in months, and the characteristic fluting pattern identifies the cause immediately.

The remedies, and what each attacks:

The last row is the only one that attacks the cause. Using only the six active states restricts \(v_{cm}\) to \(\pm V_{dc}/6\) — a third of the swing — but the zero states are what allow the output voltage to be reduced below maximum, so the modulator loses most of its range. In practice the path-based remedies win.

Why this got worse, not better. Every modern improvement aggravates it:

Faster devices have shorter transitions, so a silicon-carbide drive switching in 20 ns produces five times the common-mode current of a 100 ns IGBT at the same voltage. The move to wide-bandgap devices, which Set 25 recommended for dead-time reasons, makes this problem five times worse — which is why SiC drives ship with common-mode chokes as standard.

The DC link, from the capacitor-filtered bridge of Set 10:

Falling to perhaps 550 V at full load as the capacitor's conduction angle widens. That sag is the constraint the whole design turns on.

Check each modulation scheme against the 415 V requirement:

Plain SPWM cannot produce rated motor voltage from a diode-rectified supply of the same nominal voltage. That single line is why every commercial drive uses space vector modulation or third-harmonic injection — it is not an optimisation, it is a requirement.

And it is only just adequate. With the link sagging to 550 V under load:

Standard practice, and the reason drives are specified to deliver rated torque only up to a base speed slightly below nominal, entering field weakening thereafter. A drive that must hold full voltage under a sagging supply needs either a boost stage or overmodulation into the six-step region.

The currents:

A 1200 V rating for a 587 V link looks generous and is standard for 415 V drives — the margin covers supply surges, regenerative link rise, and the inductive overshoot at turn-off. A 50 A module handles the 19.3 A peak with room for the 150% overload that variable-speed drives are expected to sustain.

The loss budget at 4 kHz:

Why 4 kHz. The frequency choice is squeezed from both sides:

At 4 kHz the drive is audibly noisy — the characteristic whine — but switching loss is already 63% of the total. Doubling to 8 kHz would add 92 W and take the efficiency to 97.0%, which is why larger drives switch slower, not faster, and accept the acoustic noise.