Three-Phase Diode Rectifiers

Establish the voltages. A three-pulse rectifier connects each phase to the load in turn, so the relevant peak is the phase peak:

Each phase conducts for 120° — from 30° before its peak to 30° after — because that is the interval during which it is the highest of the three. Averaging over that window:

Which is the general \(p\)-pulse formula with \(p = 3\): \(\left(3/\pi\right)\sin60^\circ = 0.827\). The output is 83% of the phase peak, against the single-phase half-wave's 32%.

The RMS, from the same window:

Form and ripple factor:

Against 121% for single-phase half-wave and 48.3% for the single-phase bridge. Three phases alone, with no filter and no extra devices beyond two more diodes, have cut the ripple by a factor of 2.6 relative to a full bridge.

Ripple frequency and blocking voltage:

The PIV is the line peak, not the phase peak, because a blocked diode has its own phase on the anode and the conducting phase on the cathode — and the difference between two phases is a line voltage. This catches people out reliably.

One defect remains. Each phase carries current in one direction only, so the supply sees a DC component in every phase:

That DC magnetises the supply transformer's core and drives it toward saturation, exactly as in the single-phase half-wave circuit of Set 9. It is the reason the three-pulse connection is rarely used on its own, and why the bridge of Problem 2 — which draws no DC at all — is the standard.

The bridge works on line voltages. At any instant one diode of the upper group and one of the lower group conduct, connecting the load between the most positive and the most negative phase — which is a line-to-line voltage:

Apply the \(p\)-pulse formula with \(p = 6\):

Equivalently, in terms of the phase peak:

Exactly twice the three-pulse figure, and 95.5% of the line peak — the output is almost the peak of the supply, before any filtering at all.

The RMS and ripple:

Note how close \(V_{rms}\) and \(V_{dc}\) have become — half a volt apart in 560. That closeness is why the ripple factor must be computed carefully: rounding either figure to four significant digits destroys the answer.

Ripple frequency and PIV:

Same PIV as the three-pulse circuit, but twice the output voltage from it — so the bridge uses its devices twice as effectively. And 300 Hz ripple needs a filter nine times smaller than the same attenuation at 100 Hz.

The full comparison:

Ripple falls roughly as \(1/p^2\) and the output rises toward the peak. Doubling the pulse number is worth about a factor of four in ripple — which is why Problem 5 finds twelve-pulse worthwhile despite the transformer it requires.

Output power:

Each diode conducts for 120° of every cycle — one third — carrying the full DC current while it does:

The \(\sqrt3\) is the familiar \(1/\sqrt{D}\) with \(D = 1/3\). As always the RMS is what rates the device, and it is 1.73 times the average.

The line current is a quasi-square wave. Each phase carries \(+I_{dc}\) for 120°, then zero for 60°, then \(-I_{dc}\) for 120°, then zero for 60°:

Less than the DC current, because each phase rests for 120° of every cycle. Note that unlike the three-pulse circuit, the waveform is symmetric — equal positive and negative blocks — so there is no DC in the transformer.

The fundamental, from the Fourier series of a 120° block:

Distortion and power factor:

Against 0.900 for the single-phase bridge and 0.29 for the capacitor-input circuit of Set 10. The gap in the waveform — the 60° rest between blocks — makes the current shape closer to a sine than a full square wave is.

Check against the apparent power:

The two routes agree, confirming both the line current and the fundamental. As with every diode rectifier, the displacement factor is unity and every bit of the shortfall is harmonic content — which Problem 4 identifies precisely.

Which harmonics exist. A \(p\)-pulse converter draws line-current harmonics only of order:

No even harmonics, because the waveform has half-wave symmetry; and no triplens, for a reason worth working out separately.

Amplitudes fall as \(1/n\):

The 5th alone is 20% of the fundamental — and it is a negative-sequence component, so in a motor it produces a torque pulsation at six times line frequency and heats the rotor without doing useful work.

Verify the THD from the series. Summing all the \(6k\pm1\) terms:

Which agrees with the closed form obtained from the RMS values in Problem 3:

Two completely independent routes — a harmonic sum and a ratio of RMS values — landing on the same figure. This is the standard check that a harmonic analysis is complete.

Why no triplens. Third-harmonic components in a three-phase system are in phase with each other — zero sequence — so they can only flow if there is a return path:

A three-wire supply has no neutral, so zero-sequence current has nowhere to go and the triplens are simply absent from the line current. This is the structural advantage over Set 10's single-phase circuit, whose third harmonic was its worst offender and whose triplens added in the neutral.

What this means for the installation:

The third of these is the dangerous one. A capacitor bank installed to correct displacement power factor can resonate with the supply inductance near 250 or 350 Hz, amplifying the very harmonic the rectifier injects — a common cause of capacitor failures in plants that added correction without a harmonic study.

Where the 30° comes from. A star secondary and a delta secondary on the same core produce line voltages 30° apart. Feeding two identical bridges from them gives two six-pulse outputs whose ripples are 30° out of step — half of the 60° ripple period.

So when one bridge's output is at a ripple peak, the other is at a trough, and the series sum has twelve smaller ripples per cycle instead of six large ones.

Output voltage, the two bridges in series:

The ripple factor, from the general formula with \(p = 12\):

Four times better than six-pulse, consistent with the \(1/p^2\) scaling. At 1% ripple and 600 Hz, the output filter for most loads becomes trivial or unnecessary.

The harmonic cancellation is the real prize. Applying the pulse rule with \(p = 12\):

The 5th and 7th have vanished. They are still drawn by each bridge individually, but the 30° phase shift makes them 150° and 210° apart on the primary side — effectively in antiphase — so they cancel in the transformer rather than reaching the supply.

The resulting distortion:

Halving the THD and quartering the ripple, for twice the devices and a purpose-wound transformer. At the multi-megawatt scale where the harmonic limits are strict and the transformer exists anyway, that trade is decisive — which is why HVDC converters and large electrolysis rectifiers are twelve-pulse or higher.

Why not go further? The same trick extends to 18 and 24 pulses with additional phase-shifted windings:

But the returns diminish and the transformer complexity does not. Beyond twelve pulses it is usually cheaper to fit a tuned filter or an active front end — and once an active front end is present it delivers under 5% THD and a controllable power factor, which no pulse number can.

Two diodes are always in the path. Both carry the full 200 A while conducting, so the drop is:

A 0.54% reduction — negligible here, unlike the motor drive of Set 9 where a back EMF amplified a similar drop into an 8.8% current error. With a constant-current load there is no amplification.

Device currents, from the 120° conduction of Problem 3:

Loss per diode, using the average for the offset and the RMS for the slope:

Note the split: 60 W from the junction offset and 40 W from the bulk resistance. Using the average current for both terms would give 73 W and understate the loss by 27% — the Set 2 error, still costing marks.

Cross-check against the terminal drop, which must agree:

Two independent routes agreeing exactly. A line-frequency rectifier is a very efficient machine, because there is no switching loss at all — the diodes commutate naturally at the crossings of the line voltages.

Device selection. Apply the constraints from Part 1 simultaneously:

The RMS again exceeds the average by \(\sqrt3\), and it is the one that sets the die size.

Close the loop with Set 8. One hundred watts per diode is a real thermal problem:

With a typical module \(R_{th(j-c)} = 0.35\) and interface 0.05, the heatsink must reach 0.40 K/W per device — or, for a six-pack module dissipating all 600 W through one baseplate, roughly 0.1 K/W overall. That is forced-air territory, exactly as Set 8's inverter turned out to be.