Three-Phase Controlled Converters and Inverter Mode

The output. Each conduction interval is delayed by \(\alpha\) from its natural commutation point, and averaging over the 60° window gives the cosine law:

Device currents are unchanged from Set 11. Each thyristor still conducts for 120°, just later:

Independent of \(\alpha\) — so the devices are rated once for the maximum load current and the firing angle never enters. That simplicity is one of the reasons the full converter, rather than a semiconverter arrangement, dominates at high power.

Line currents, the same 120° quasi-square wave:

The power factor now carries a displacement term:

Compare the single-phase full converter at the same 30°: 0.780. The three-phase bridge is better because its distortion factor is 0.955 rather than 0.900 — the 60° gaps in the line current make it closer to a sinusoid.

How the control range looks:

A single gate signal spans \(+560\) to \(-560\) V. Note that the line current is 40.8 A at every one of these points — the converter loads the supply identically whether it is delivering 28 kW or none.

The three-pulse cosine law, working on phase voltages:

Exactly half the bridge's 485.4 V, and for a structural reason: the bridge stacks an upper group and a lower group in series, each doing the job of one three-pulse converter. Their outputs add.

Device currents are the same as the bridge's, since each thyristor still conducts 120°:

Three devices instead of six — but each carries the same current while delivering half the power. The devices are used half as effectively, which is the first strike against the circuit.

The DC in the supply is the second. Each phase carries current in one direction only:

That DC magnetises the supply transformer's core toward saturation, so the transformer must be oversized or specially connected (zig-zag). The bridge draws symmetric positive and negative blocks and has no such problem.

A third limitation: the control range. With an inductive load the cosine law holds to 90°, but the ripple becomes severe long before that, and the ripple frequency is only 150 Hz:

Same PIV, same power per device, twice the ripple frequency and no core DC — the bridge wins on everything except device count, and device count is the cheapest axis. That is why the three-pulse circuit survives mainly as a teaching step and as half of a bridge.

Where it does appear. Two places, both for specific reasons:

This doubles the current capability rather than the voltage, which suits very high current, low voltage rectifiers such as electrolysis cells. And (ii) as the upper or lower half of a bridge in fault analysis, where the two groups are treated separately.

Power flows out, so the voltage must reverse. The current is fixed positive by the thyristors:

The firing angle:

The extinction angle \(\gamma\) is the interval between the last firing and the point at which the commutating voltage reverses. It is the time the outgoing thyristor has to regain its blocking ability.

Convert the device requirement into an angle:

A margin of nearly ten to one. But \(\gamma\) must also cover the commutation overlap angle \(\mu\), which source inductance produces and which Set 15 quantifies — typically 5 to 10° at full load. Even so, 26.9° leaves room.

The maximum inverting voltage, if the control is limited to \(\gamma_{min} = 15^\circ\):

So 96.6% of the full DC voltage is available in inversion — the extinction-angle limit costs only 3.4%. That is the usual outcome: the constraint bites hard on the angle but gently on the voltage, because the cosine is flat near 180°.

What commutation failure looks like. If \(\gamma\) falls below the requirement, the outgoing thyristor is still conducting when the supply drives it forward again:

And the failure is self-worsening. A momentary supply dip increases the overlap angle \(\mu\), reducing \(\gamma\) at exactly the moment the margin was already thin. This is why inverting converters are always given a generous \(\gamma_{min}\), and why HVDC terminals monitor it continuously.

The asymmetry worth noticing. Rectification has no equivalent limit:

Firing early is always safe; firing late risks the converter. A line-commutated converter is therefore fundamentally safer as a rectifier than as an inverter — and that asymmetry disappears entirely once the switches can turn themselves off, which is the argument for the active front ends of Set 24.

The RMS output as a function of firing angle. Integrating \(v^2\) over one 60° conduction window, delayed by \(\alpha\):

Note the \(\cos2\alpha\): the RMS falls much more slowly than the average, which falls as \(\cos\alpha\). That divergence is the whole result.

Evaluate at the three angles:

From 4.2% to 53.5% over a firing range of 60°. At \(\alpha = 90^\circ\) the average is zero while the RMS is still 172.6 V — the output is pure ripple.

Why the ripple grows so fast. Two effects compound. The conducting segments are taken from a steeper part of the sine wave, so each is more curved; and the average they are ripple about is falling. The first raises the numerator, the second lowers the denominator:

The absolute ripple voltage rises six-fold while the average falls by half — hence the twelve-fold worsening of the ratio.

What this does to the smoothing reactor. The current ripple is the ripple voltage divided by the reactor impedance at 300 Hz:

So a reactor sized for acceptable ripple at \(\alpha = 0\) gives six times the ripple at 60°. And since the critical current for continuous conduction rises with ripple — Set 13 — the drive is most likely to go discontinuous at exactly the large firing angles where the ripple is worst.

The design implication:

The middle row is right and the third is what is usually done, because the reactor for \(\alpha = 60^\circ\) is six times the inductance of the one for \(\alpha = 0\). Note that the 300 Hz ripple frequency helps enormously here: the same job in a single-phase drive, at 100 Hz, would need three times the reactance again.

Motoring at 500 V output:

Torque and power:

Now brake regeneratively at the same speed. Here is the obstacle: the thyristors force \(I_a > 0\), and the machine is still turning the same way, so \(E\) is still \(+480\) V. With both positive, power must flow into the machine, whatever the firing angle.

A single converter cannot reverse the current and cannot reverse the machine's EMF. It can reverse only its own output voltage — which by itself is not enough.

Reverse the field. Reversing \(K\phi\) reverses \(E\) without changing the speed or the direction of rotation:

The machine generates 48 kW, the armature resistance takes 2 kW, and 46 kW reaches the supply. The braking torque is \(-450\) N·m — full rated torque, in the opposite sense.

The catch with field reversal. The field winding is highly inductive, so its time constant is long:

Compared with an armature time constant of tens of milliseconds. A field-reversal drive therefore takes a second or more to change from motoring to braking, which is fine for a rolling mill and useless for a servo. Worse, the flux passes through zero during the reversal, so torque is momentarily lost entirely.

The four quadrants, and what reaches them:

A single converter with field reversal covers three quadrants slowly. Genuine four-quadrant operation at armature speed needs two converters in anti-parallel — one for each current direction — which is the dual converter of Set 16.

Output at full conduction. Two thyristors are always in the path:

Device currents and losses:

The \(di/dt\) constraint, and why the firing range matters. At turn-on the instantaneous commutating voltage appears across the loop inductance, and it is largest at the largest firing angle:

So a small series reactor is needed — about 1 µH added to the existing 2. Note that this constraint is set by the largest firing angle the converter will ever use, not by the operating point: at \(\alpha = 0\) the commutating voltage at turn-on is zero and there is no \(di/dt\) problem at all.

The thermal budget, closing back to Set 8:

With a typical \(R_{th(j-c)} = 0.10\) and interface 0.05, the heatsink must reach 0.32 K/W per device — or, for all six on one baseplate carrying 1080 W, roughly 0.06 K/W overall. That is forced air or water at this rating.

The complete specification:

Six of the eight rows come straight from Part 1. The converter analysis of Part 2 supplies the currents and the voltages; everything that decides whether the devices survive was established before any waveform was drawn.