The sequence a harmonic belongs to

The half the neutral does not carry decomposed an unbalanced star load into symmetrical components and found two things where one was expected: a zero-sequence current that the neutral carries three times over, and a negative-sequence current that no neutral carries and only a meter on the supply could see. It closed by naming its sharpest limit. Every number on that page was a single-frequency statement, because symmetrical components are defined on phasors, and the commonest load in a modern installation is not a single frequency: the neutral that carries more than a line found three balanced rectifier loads putting $\sqrt3$ times a line current into the neutral, with no imbalance anywhere.

Sequence components can be computed harmonic by harmonic, which that essay called a different and larger calculation and did not do. This one does it, and the calculation turns out to be smaller than expected: for a balanced load every harmonic is exactly one sequence, which one is a rule of arithmetic, and the sequence then decides where in a three-phase system the harmonic can travel. A delta stops one sequence; a delta–wye transformer turns the other two in opposite directions; and the three connections a designer chooses between are three ways of filtering a spectrum by sequence.

The loads and the decomposition #

Each load draws its current in a pulse either side of each voltage peak, as a capacitor-input rectifier does — the direct voltage that is a sawtooth measured the reservoir that makes the pulse — modelled here as a rectangle of ten amperes, sixty degrees wide, built on whole samples of a cycle and mirrored for the negative half-cycle so the waveform has exact half-wave symmetry. The three phases are the same waveform a third of a cycle apart. Each phase is transformed into its harmonics, and at each harmonic the three phases’ phasors are decomposed into zero, positive and negative sequence by the same three lines of arithmetic the single-frequency essays used.

At every harmonic the decomposition puts all of the current into one sequence and none into the other two, to the arithmetic’s floor. The reason is one line. A waveform delayed by a third of a cycle has its $h$-th harmonic delayed by $h$ thirds of that harmonic’s own cycle, $h \times 120°$. For $h = 3k$ that is a whole number of turns — no shift at all — so the three phasors are equal, which is zero sequence. For $h = 3k + 1$ the shift is 120 degrees beyond whole turns, the fundamental’s own pattern: positive sequence. For $h = 3k - 1$ it is 240, which is −120: the phasors rotate the other way, negative sequence. A balanced set of identical waveforms cannot produce anything else.

The amounts follow from the pulse’s spectrum. At sixty degrees the zero-sequence harmonics — third, ninth, fifteenth and so on — together are 72.5 per cent of the fundamental, the negative-sequence ones 24.1 and the positive ones beyond the fundamental 16.7. That ordering is the important number: most of a narrow-pulse load’s distortion is zero sequence, which is exactly the part a three-wire system has no path for.

A delta has no path for a zero sequence #

Connected from each line to neutral, each line carries its load’s whole current and the neutral carries the sum, which is three times the zero-sequence part: the triplens add, as the earlier essay found. Connected from line to line, each line carries the difference of the two branch currents meeting at its terminal. A zero-sequence harmonic is the same in every branch, so the difference of two of them is nothing, and the line carries no triplen at all — to the arithmetic’s floor, not approximately. Every other harmonic is a balanced set whose branches differ in phase by 120 degrees, so their difference is $\sqrt3$ times either, exactly as the fundamental’s is.

The line current’s distortion falls from 78.2 per cent to 29.4: the triplens were most of it. That is why a three-phase rectifier — six diodes on three wires, no neutral — draws a line current with no third harmonic whatever its pulse shape, and why distribution engineers worry about triplens in four-wire systems and almost never in three-wire ones.

Where the triplens go #

The triplens have not disappeared. A delta is a loop, and a current that is the same in all three branches can flow round it without passing through any terminal.

What the three branches share — their sum, divided by three — flows round the delta, and the figure shows it as a waveform of its own: 3.324 amperes rms, a train of pulses at three times the mains frequency. Its rms is exactly the root-sum-square of the branch current’s triplen harmonics, computed from five hundred and thirty-nine of them, which is the second route to the same current: one from the time-domain sum, one from the spectrum. The branch current is 5.76 amperes rms and the line current, the difference of two branches, 8.14 — $\sqrt2$ times the branch, not the $\sqrt3$ a sinusoid would give, because the line has lost the triplen part the branch keeps.

So a delta does not filter the triplens; it short-circuits them inside itself. Whatever forms the delta carries them and dissipates in them — the windings of a delta-connected transformer, the loads of a delta-connected bank — and nothing outside it can see them. A designer who measures the line currents of a delta-connected transformer feeding rectifiers will find a clean, low-distortion current and a transformer running hotter than that current explains; the current that sizes the transformer found a rectifier’s pulse heating a winding by its rms, and inside a delta the rms includes current the line ammeter never shows.

Turning the other two sequences opposite ways #

The negative- and positive-sequence harmonics do reach the lines, and a delta does not stop them. A transformer with a delta on one side and a star on the other does something to them instead: it shifts a positive-sequence set by thirty degrees one way and a negative-sequence set by thirty degrees the other.

Two identical groups of loads, one fed straight and one through a delta–wye transformer whose secondary lags by thirty degrees, put their harmonics into the same supply line. The second group’s currents are the first’s delayed by a twelfth of a cycle, and the transformer carries them to its primary as the difference of two line currents over $\sqrt3$ — the delta winding’s arithmetic, done on the waveforms rather than assumed. In the supply the fifth, seventh, seventeenth and nineteenth cancel to the arithmetic’s floor, and the eleventh and thirteenth remain at their single-group proportions, 9.6 and 7.2 per cent.

The sequence rule says why without the waveforms. The thirty-degree delay turns harmonic $h$ by $h \times 30°$; the transformer then turns it back by thirty degrees if it is positive sequence and forward by thirty if negative. The fifth, negative, ends at $-150° - 30° = -180°$, opposite to the first group’s fifth, and cancels; the seventh, positive, at $-210° + 30° = -180°$, and cancels; the eleventh, negative, at $-330° - 30° = -360°$, and adds. That is a twelve-pulse arrangement, and it is a statement about sequences: it cancels the harmonics with $h = 12k \pm 6 \pm 1$ and keeps $h = 12k \pm 1$.

The three connections, against the pulse’s width #

At every conduction angle the order is the same: phase to neutral worst, line to line better, the twelve-pulse pair best. Each connection removes one more sequence’s worth of harmonics, and the amounts removed depend on how the pulse distributes its distortion among them. For narrow pulses the triplens are most of the distortion and the delta’s removal of them is worth the most — 1.90 to 1.44 at twenty degrees.

The curves are not monotone in the angle, and that is a fact about rectangular pulses rather than a defect of the measurement. Each harmonic of a rectangle passes through nulls as the rectangle widens, so the harmonics a connection leaves rise and fall with the width: the twelve-pulse pair leaves 0.18 at thirty degrees and 0.33 at forty-five, as its eleventh and thirteenth climb out of a null. The distortion is computed from the whole waveform’s rms against its fundamental rather than from a truncated list of harmonics, so every harmonic the samples hold is counted.

The heat a delta hides #

The circulating current is not small beside the current it rides on. A branch of 5.76 amperes rms whose triplen part is 3.324 amperes has a non-triplen part of $\sqrt{5.76^2 - 3.324^2} = 4.70$ amperes, which is what the branch would carry if the triplens could leave. The branch’s heating goes as the square of its rms, so it runs $(5.76/4.70)^2 = 1.50$ times hotter than the part that reaches the lines would make it. A delta-connected transformer winding feeding narrow-pulse rectifier loads dissipates half as much again as a calculation from its line currents predicts, and nothing measured outside it shows why.

That is the heat the earlier essay’s star connection put into the neutral conductor, moved into the transformer. The neutral of a four-wire system carries the triplens of all three phases added together; the delta of a delta–wye transformer carries them circulating. The total triplen current has a path either way, and the choice of connection decides only which conductor is rated for it — the neutral, which can be oversized by the electrician, or the winding, which is fixed by the manufacturer.

The power factor the supply sees #

A load’s true power factor from the supply’s side is its fundamental’s in-phase share of the whole current, and for loads whose pulses sit on the voltage peaks the displacement is nearly nothing, so the power factor is the distortion factor, $1/\sqrt{1 + \mathrm{THD}^2}$. Connected phase to neutral at sixty degrees, that is $1/\sqrt{1 + 0.782^2} = 0.788$. Connected line to line, with the triplens gone from the line, $1/\sqrt{1 + 0.294^2} = 0.959$. As a twelve-pulse pair, about 0.99.

The same loads, drawing the same power, look to the supply like three different power factors, and the capacitor that was right once is the warning about what a capacitor can do about the first: nothing. A power factor below one because of distortion is not a phase angle a reactance can correct; it is current at frequencies the voltage does not have — the current that does no work computed reactive power three ways and none of them has a term for it — and the connections in the figures are the remedy that works, because they change which harmonics reach the line rather than trying to cancel them with the fundamental.

What a meter reads #

What a meter multiplies by found an averaging meter’s reading wrong by a factor set by the waveform’s shape, and a line current with its triplens removed has a different shape from the branch current that produced it. A clamp meter on a delta-fed line reads a current with a crest factor and form factor of its own, nearer a sine’s than the load’s pulse; a meter on a branch inside the delta reads the pulse. Both readings are honest, and they are readings of different currents, which is the practical form of this essay’s point: in a three-phase system the same load has different waveforms in different conductors, and the conductor decides which sequences are present.

What a motor on the same supply feels #

The negative-sequence harmonics matter for a reason the neutral never shows. A three-phase machine on the same supply as the rectifier loads sees the voltage distortion their currents cause through the supply’s impedance, and each harmonic of that distortion has the sequence of the current that caused it. A positive- sequence harmonic drives a field that rotates the same way as the fundamental’s, at seven or thirteen times its speed; a negative-sequence harmonic drives one that rotates backwards, at five or eleven times. A machine’s rotor is a short-circuited winding to any field that is not turning with it, so both kinds induce rotor currents and heat, and the backwards ones produce a braking torque besides.

The earlier essay found a motor’s impedance to a negative-sequence field lower than to a positive one, so a small negative-sequence voltage drives a disproportionate current. The fifth harmonic, the largest of the negative-sequence set at 20.6 per cent of a single group’s fundamental here, is therefore the harmonic a motor feels most, and it is exactly the one a twelve-pulse arrangement removes from the supply. A designer choosing between connections for a rectifier load that shares a supply with motors is choosing, among other things, whether the motors see the fifth.

Why one sequence per harmonic is not the whole story #

The single-sequence result needs the three loads to be identical. Three phases, and the wire that carries nothing found the neutral carrying 5 × 10⁻¹⁵ amperes for three equal linear loads; the same balance is what makes each harmonic pure. With one load different, each harmonic’s three phasors are no longer a rotated copy of each other, and each harmonic has all three sequences in some proportion — so an unbalanced set of rectifier loads puts a little fifth harmonic into the neutral and a little third harmonic into a delta’s lines. The earlier essay’s imbalance arithmetic then applies harmonic by harmonic, and the larger calculation it anticipated is the sum of one small calculation per harmonic.

What the decomposition leaves out #

The pulses are rectangles of fixed height, the supply is stiff and each load is independent. A real rectifier’s pulse is shaped by the supply’s own impedance, and a soft supply rounds the pulses, which lowers their high harmonics; it also couples the loads, since each pulse sags the voltage the others see. The transformer is ideal: no leakage inductance, which in a real twelve-pulse arrangement both rounds the currents and limits how well the fifth and seventh cancel, since the two groups then see different source impedances. And the twelve-pulse cancellation needs the two groups to draw equal currents; a mismatch between them leaves a residue of the fifth and seventh proportional to it.

The number to keep #

Three balanced loads give each harmonic a single sequence, set by $h \bmod 3$: zero for triplens, positive for $3k + 1$, negative for $3k - 1$. Connected line to line, the loads put no triplen into a line — they circulate round the delta, 3.32 A of them inside branches of 5.76 A — and the line’s distortion falls from 78 to 29 per cent. A delta–wye transformer turns positive and negative sequences opposite ways by thirty degrees, so a second group fed through it cancels the fifth and seventh in the supply and leaves the eleventh and thirteenth.

Still open: the loads that differ, the supply that is soft, and the transformer that leaks #

Loads that are not identical. With one load drawing more than the others each harmonic acquires all three sequences, and a little of the fifth reaches the neutral. How large an imbalance puts a stated fraction of the triplens into a delta’s lines, or the fifth into the neutral, would join this essay’s arithmetic to the single-frequency imbalance the earlier one measured.

A soft supply. A supply’s impedance rounds each pulse and couples the three loads through the voltage they share. Whether the sequence rule survives — and how much it changes the triplen current circulating in a delta — is the measurement with the supply’s impedance in the netlist.

A transformer that leaks. Leakage inductance in the delta–wye transformer gives the second group a different source impedance from the first, so the two groups’ pulses differ and the fifth and seventh no longer cancel exactly. The residue against the leakage would say how good a twelve-pulse arrangement is with the transformer a real one.