Power Factor, Harmonics and Dual Converters

The current waveform and its fundamental:

A square wave contains every odd harmonic, falling as \(1/n\):

Contrast with the three-phase bridge. The difference is not a matter of degree:

The entire difference is the triplens. A three-wire three-phase supply offers them no return path, so they simply cannot flow; a single-phase supply has a neutral and they flow freely. That one structural fact accounts for the whole 48.3% against 31.1%.

Why the third harmonic is the dangerous one. In a four-wire installation the third harmonics of all three phases are in phase with each other:

Thirty-six amps in a neutral conductor, from three single-phase converters each drawing 36 A of fundamental. The neutral is traditionally sized smaller than the phases and carries no overcurrent protection — so it heats without any indication.

Power and reactive power:

Note that \(P\) and \(Q\) do not account for \(S\): \(\sqrt{P^2+Q^2} = 8.28\) kVA against 9.20 actual. The missing 4.0 kVA is the distortion power, carried entirely by harmonics that transfer no energy at all.

Size the capacitor from the reactive power it must supply:

Now find the new supply current. The harmonic content is completely unaffected — the capacitor cannot supply harmonic current at the right phase to cancel anything:

The fundamental falls, because less of it is reactive:

Recombine:

Improved from 0.450 to 0.700 — useful, but nowhere near the 0.95 the capacitor was sized for. The displacement factor is now 0.95; the power factor is not, because distortion is still there.

The ceiling on this approach. Even perfect displacement correction leaves the distortion factor:

So no capacitor bank can take this converter past 0.900, and reaching even that needs full correction to unity displacement. The harmonic current of 17.4 A flows regardless — through the supply, through the transformer, and now also through the capacitor.

And here is the danger. The capacitor and the source inductance form a parallel resonant circuit:

The resonance sits at the 5.4th harmonic — between the 5th at 250 Hz and the 7th at 350 Hz, and uncomfortably close to the 5th. The converter injects 7.2 A at exactly that frequency into a circuit that is near resonance, and the resulting current can be several times larger than what was injected.

What that produces in practice:

The last row is the diagnostic signature. A plant that adds power-factor correction and immediately starts losing capacitors has almost certainly created a resonance near a harmonic the existing converters were already injecting.

The fix is a detuning reactor in series with each capacitor, placing the resonance safely below the lowest injected harmonic:

Above its tuning frequency the branch is inductive, so it cannot resonate with the source at any higher harmonic. The bank still corrects displacement at 50 Hz and now also absorbs a little 5th-harmonic current instead of amplifying it.

The arrangement. Two bridges are connected back to back across the same load, pointing in opposite directions. Converter 1 supplies positive current, converter 2 supplies negative current, and between them the load can be driven in all four quadrants:

Converter 2 must produce the same output so that the load sees a consistent voltage whichever bridge is active. Because it faces the load the other way, its own output must be the negative of converter 1's:

This is the fundamental relation of a dual converter. Note that when converter 1 is rectifying at 60°, converter 2 is set to invert at 120° — it is standing by, ready to accept current in the opposite direction the instant it is needed.

Why only one may conduct. The two bridges are connected directly across each other. Their average voltages cancel, but their instantaneous waveforms do not:

Both are 6-pulse waveforms and the ripple components do not cancel — they are 30° apart in phase. With no impedance between them, that difference voltage would drive an unlimited current straight from one bridge to the other, bypassing the load entirely. In circulating-current-free mode the answer is simply to keep one bridge fully blocked.

The changeover sequence, which is where the mode's cost appears:

Step 4 is the irreducible one. A thyristor that has just stopped conducting still holds stored charge; firing the opposing bridge before it has recovered short-circuits the supply through both converters — the same failure as a commutation failure, and with the same consequences.

What the dead time costs. During steps 1 to 6 the load current is zero, so:

Torque disappears entirely for several milliseconds. For a rolling mill or a hoist that is acceptable; for a position servo it is a dead zone in the control loop that produces limit cycling around zero torque. That is the whole argument for the circulating-current mode of Problem 4.

The driving voltage. With both bridges gated, the voltage acting around the loop they form is the sum of their two outputs:

Zero on average — which is what makes the mode possible at all. If the averages did not cancel, a DC circulating current would build without limit.

But the ripple does not cancel. Both outputs are 6-pulse waveforms at 300 Hz, and at \(\alpha_1 = 60^\circ\) and \(\alpha_2 = 120^\circ\) their ripples are displaced. The sum is a 300 Hz alternating voltage of substantial amplitude:

There is no simple closed form, because \(v_r\) is a piecewise cosine. Integrating numerically over one 300 Hz ripple period gives the peak-to-peak excursion, and since the circulating current can only be positive — thyristors again — that excursion is the peak.

Evaluate. Integrating the ripple sum for a trial \(L_r = 10\) mH:

Since \(i_r \propto 1/L_r\), scaling to the 40 A target:

What the reactor has to be. This is not a small component:

A 12 mH reactor that must stay linear at 240 A is an iron-cored component weighing tens of kilograms, with an air gap sized for the DC bias. In practice two reactors are used, one in each converter's leg, so that each carries only its own current.

The running cost. The circulating current flows through the thyristors of both converters and does no useful work:

That current adds to the conduction loss in every device and to the reactor's own copper loss, all for the sake of never letting the current reach zero. Note also that both converters are now drawing from the supply simultaneously, so the reactive loading is worse than in the free mode.

What it buys:

The last two rows are the real prize. Because the circulating current keeps both converters conducting at all times, neither ever enters the discontinuous mode of Set 13 — so the transfer function stays linear from full forward torque to full reverse, with no gain change and no dead zone. For a servo that is worth a heavy reactor.

The machine quantities. At 100 rad/s the back EMF magnitude is fixed:

Apply the armature equation in each quadrant. The rule is mechanical: the resistance always opposes the current, so \(I_aR_a\) always adds in the direction of current flow:

Quadrants I and IV use converter 1 (positive current); II and III use converter 2. The sign of \(E\) follows the speed and the sign of \(I_a\) follows the torque.

Quadrant I — forward motoring:

Quadrant II — forward braking. The machine still turns forward, so \(E\) stays positive, but the current must reverse — which means converter 2 takes over, and from its own terminals the machine looks like a source:

The machine generates 90 kW, the armature resistance takes 6 kW, and 84 kW goes back to the supply. Note the resistance term now subtracts from the magnitude, because the current and the EMF have opposite signs.

The complete map:

Read the \(\alpha\) column: the two motoring quadrants use firing angles below 90° and the two braking quadrants use angles above it. Rectification and inversion map exactly onto motoring and braking.

Where the transitions are hard. Two of the four boundaries require a converter changeover:

The first is the demanding one, because it happens at full speed with full torque reversing — a hoist catching a descending load, a mill reversing its pass. In circulating-current-free mode there are several milliseconds of zero torque there; in circulating-current mode there are none. That single transition is what decides which mode a drive is built with.

The baseline, from Problem 1:

Every remedy attacks either the 0.900 (distortion) or the \(\cos60^\circ\) (displacement), and it is worth being clear which before comparing them.

Semiconverter — attacks displacement, by cutting the current block short:

Nearly doubles the power factor and reduces the current by 18%, for free — but gives up quadrant IV entirely, so it cannot be used on any drive that must brake.

Capacitor correction — attacks displacement only, from Problem 2:

Works on any converter without changing it, but leaves 17.4 A of harmonic current untouched, and introduces a resonance at the 5.4th harmonic that needs a detuning reactor.

Twelve-pulse — attacks distortion only:

Barely helps — because at \(\alpha = 60^\circ\) the problem is overwhelmingly displacement, and pulse number does nothing about that. Twelve-pulse is a harmonic remedy, and the harmonics were already the smaller part of the loss.

Active front end — attacks both, by controlling the shape and the phase of the input current directly:

Less than half the supply current for the same delivered power, with THD under 5% and bidirectional power flow. It is a PWM converter running at kilohertz, using the self-commutating devices of Set 7 — and it is the only method here that is not a compromise.

Everything together:

Only the last row addresses both terms, and it does so by abandoning line commutation altogether. Everything above it is a way of making the best of a converter whose current shape and phase are dictated by the supply.

What Part 2 has actually established. The line-commutated converter is efficient, robust, cheap and scalable to hundreds of megawatts — and it draws a current it cannot control:

That coupling is the root of every power-factor result in Part 2. Breaking it requires a switch that can be turned off at will, at a frequency far above the line — which is exactly the device Part 1 ended with, and exactly what Part 3 begins to use.