Chapter 10: Source Inductance, Dual Converters, and Power Factor — Power Electronics

Where this sits

Part 2 · AC–DC Converters
Chapter 10 of 30 — the last of Part 2

You should already know

\(V_{dc} = 1.35V_L\cos\alpha\), the firing sequence, inverter mode and the margin angle from Chapter 9.

By the end you can

Compute the overlap angle and the voltage droop it causes, design a dual converter for four-quadrant duty, and evaluate power-factor remedies.

Time

≈ 60 min reading · ≈ 55 min problems

i What you'll learn

Why current cannot transfer instantly between devices, and how source inductance creates the overlap angle \( \mu \).
The resulting voltage droop \( \frac{3\omega L_s}{\pi}I_o \) — which looks like a resistance but dissipates nothing.
How overlap eats into the inverter margin, and why that turns a nuisance into a hazard.
Voltage notching on the supply, and why other equipment on the same busbar suffers.
The dual converter: two bridges back to back giving four-quadrant operation, and the condition \( \alpha_1 + \alpha_2 = 180^\circ \).
Circulating-current and non-circulating control, and the trade between reversal speed and hardware.
Every practical route to a better power factor, from a line reactor to an active front end.

Section 10-1

Commutation Takes Time

Every analysis so far has assumed that when one device stops conducting, another instantly takes over the full load current. Real supplies do not permit it.

Between the ideal voltage source and the converter there is always inductance — transformer leakage reactance, cable inductance, the source impedance of the network. Call it \( L_s \). And an inductor's defining property is that its current cannot change instantaneously.

◇Intuition · handing over a baton at speed

Two runners exchanging a baton cannot do it at a mathematical instant. There is a stretch of track where both are holding it — one accelerating, one slowing.

Commutation is the same. When the incoming thyristor fires:

Its current must rise from 0 to \( I_o \), and the outgoing device's must fall from \( I_o \) to 0.
Both are limited by \( di/dt = v/L_s \), so neither can jump.
During that interval both devices conduct. The two supply phases are momentarily connected together through the two devices — a deliberate, controlled short between phases.
The output voltage during the overlap is not either phase; it is the average of the two.

The interval, expressed as an angle, is the overlap angle \( \mu \) — also called the commutation angle. Typical values are 5° to 25°, growing with load current and with source inductance.

The consequence is a notch. For \( \mu \) degrees out of every 60°, the output sits at the mean of two line voltages instead of at the higher one. Area is lost from under the curve, so the average output falls.

Applying the loop equation during the overlap and integrating gives the two results that define this chapter. For a three-phase full converter:

🔑

Overlap angle and the droop it causes

\[ \cos\alpha - \cos(\alpha+\mu) = \frac{2\omega L_s I_o}{V_{mL}}, \qquad V_{dc} = \frac{3V_{mL}}{\pi}\cos\alpha - \frac{3\omega L_s}{\pi}I_o \]

The single-phase full converter obeys the same pattern with different constants: \( \cos\alpha - \cos(\alpha+\mu) = \dfrac{2\omega L_sI_o}{V_m} \) and \( V_{dc} = \dfrac{2V_m}{\pi}\cos\alpha - \dfrac{2\omega L_s}{\pi}I_o \).

Interactive · commutation overlap

A three-phase full converter on a 415 V, 50 Hz supply at \( \alpha = 30^\circ \). Increase the source inductance or the load current and watch the notches open up at every commutation — and the output voltage sag with them.

Source \( L_s \) 1.0 mH

Load current 100 A

Overlap μ10.6°

Ideal V_dc485 V

Actual V_dc455 V

Droop6.2%

—

Section recap. Source inductance prevents instantaneous current transfer, so two devices conduct together for the overlap angle \( \mu \). During it the output is the mean of two line voltages, cutting area from under the curve: \( V_{dc} = \frac{3V_{mL}}{\pi}\cos\alpha - \frac{3\omega L_s}{\pi}I_o \).

Section 10-2

A Resistance That Is Not One

Look again at the droop term. It is proportional to \( I_o \), so it behaves exactly like a series resistance:

Equivalent commutation resistance

\[ R_{\mu} = \frac{3\omega L_s}{\pi}\ \text{(three-phase)}, \qquad R_{\mu} = \frac{2\omega L_s}{\pi}\ \text{(single-phase)} \]

So a converter's terminal characteristic droops linearly with load, and the equivalent circuit is an ideal source \( 1.35V_L\cos\alpha \) behind a resistance \( R_\mu \). That model is genuinely useful — and it is also a trap.

!Common mistake

"The overlap causes a voltage drop, so it must be dissipating \( I_o^2R_\mu \) watts."

Wrong Voltage drop times current is power lost, so \( L_s \) is heating up somewhere.

Right An ideal inductor dissipates nothing. Energy stored during one half of the commutation is returned during the other. \( R_\mu \) is a modelling convenience, not a resistor.

So where did the volt-seconds go? They were not converted to heat — they were never delivered to the DC side. During overlap, part of the supply voltage appears across \( L_s \) rather than across the load. The energy is exchanged with the source, not consumed.

What it costs you instead is reactive power and displacement. The supply current is delayed by roughly \( \mu/2 \) beyond the firing delay, so the displacement factor becomes approximately \( \cos(\alpha + \mu/2) \) rather than \( \cos\alpha \). The converter draws more reactive power without doing more work.

Why the distinction matters practically: you cannot cool your way out of it, and you cannot find the "loss" with a thermal camera. The remedy is a stiffer supply — a larger transformer or lower impedance — not a bigger heatsink.

⚙Voltage notching, and why the neighbours complain

The notch does not only appear at the converter's DC output. During overlap two supply phases are momentarily short-circuited through the two conducting devices, so a notch appears in the supply voltage itself — at the converter terminals and, attenuated by the impedance ratio, all the way back to the point of common coupling.

Six notches per cycle, each of depth set by the ratio of converter-side to source-side inductance. The consequences are a standard part of industrial power-quality work:

Zero-crossing detectors misfire. A notch that crosses zero creates extra apparent zero crossings, confusing other equipment's synchronisation — including, embarrassingly, other converters' firing circuits.
Capacitors fail. The steep notch edges are rich in high-frequency content, and a power-factor-correction capacitor is a low impedance to it.
Electronics reset. Deep notching upsets sensitive control equipment on the same busbar.
Audible noise from transformers and reactors excited at the notch frequency.

The standard remedy is a line reactor in series with each converter. It seems paradoxical — adding inductance makes the overlap longer — but it works because it moves the notch. The reactor becomes the dominant impedance between the converter and the supply, so most of the notch depth appears across the reactor rather than at the point of common coupling. The neighbours see a shallower notch; the converter sees a slightly larger droop. IEEE 519 sets limits on notch depth and area for precisely this reason.

Section recap. The droop \( R_\mu = 3\omega L_s/\pi \) looks like a resistance but dissipates nothing — the volt-seconds are exchanged with the source, not lost as heat. The real costs are extra reactive power, a displacement factor closer to \( \cos(\alpha+\mu/2) \), and notching of the supply voltage that affects everything else on the busbar.

Section 10-3

Overlap and the Inverter Margin

Chapter 9 introduced the margin angle without quantifying \( \mu \). Now it can be closed properly, and the result explains why inverting converters fail in ways that rectifying ones do not.

🔑

The margin, with overlap accounted for

\[ \gamma = 180^\circ - \alpha - \mu \;\ge\; \omega t_q + \text{safety} \]

Overlap subtracts directly from the available extinction angle. And \( \mu \) is not a constant — it grows with load current and shrinks with supply voltage, which is exactly the wrong way round for safety.

→Why the failure mode is a conspiracy, not a single cause

Consider a regenerating drive at \( \alpha = 145^\circ \) with \( \mu = 10^\circ \), so \( \gamma = 25^\circ \) — apparently a huge margin against the 2° or so that \( \omega t_q \) requires. Now let three ordinary things happen together:

The supply dips 15%. Commutation is driven by the line voltage, so \( \mu \) grows — roughly as \( 1/V \) for a given current.
The load surges. \( \mu \) grows with \( I_o \) too, and the two effects multiply rather than add.
The junction is hot. \( t_q \) worsens significantly at high temperature, so the required margin grows at the same moment the available margin is shrinking.

Each effect alone is survivable. Together they can consume 25° of margin in a few cycles — and then the outgoing thyristor fails to recover, both devices in a leg conduct, and the DC-side EMF feeds a short circuit across the supply.

What competent controllers actually do, rather than relying on a fixed \( \alpha_{max} \):

Measure the supply voltage and retard \( \alpha \) automatically when it falls.
Use the current reference to predict \( \mu \) and limit \( \alpha \) accordingly — an \( \alpha \) limit that moves with load.
Detect the actual extinction angle from device voltage measurements and close a loop on \( \gamma \) itself. This is standard practice in HVDC, where it is called constant extinction angle control.

The general lesson generalises well beyond converters: when a margin depends on several variables that all move together under stress, a fixed limit computed at nominal conditions is not a safety measure.

Section recap. \( \gamma = 180^\circ - \alpha - \mu \), and \( \mu \) grows with load current and with falling supply voltage — precisely the conditions of a fault. A fixed \( \alpha_{max} \) is therefore not enough; real controllers adapt the limit or close a loop on \( \gamma \) directly.

Section 10-4

Four Quadrants: The Dual Converter

A single full converter reaches two quadrants: its voltage can reverse, its current cannot. That is enough to brake a motor running in one direction — but not to drive it the other way.

A reversing mill, a mine hoist, a machine-tool spindle and a lift all need four quadrants: torque and speed in both directions independently. The answer is two full converters connected in anti-parallel across the same load.

Two bridges facing opposite ways — and every combination of torque and speed becomes reachable

Converter 1 supplies current in the forward direction, covering quadrants I (forward motoring) and IV (forward braking, by inverting). Converter 2 is connected the other way round and covers quadrants III and II. Between them every combination of voltage and current polarity is available.

One condition governs the arrangement:

🔑

The dual-converter condition

\[ V_{dc1} = -V_{dc2} \;\Longrightarrow\; \cos\alpha_1 = -\cos\alpha_2 \;\Longrightarrow\; \boxed{\alpha_1 + \alpha_2 = 180^\circ} \]

Both converters are connected across the same load, so their average output voltages must match in magnitude and oppose in polarity. If they do not, the mismatch appears across whatever impedance lies between them and drives a large circulating current — limited by nothing but the source impedance.

!Common mistake

"\( \alpha_1 + \alpha_2 = 180^\circ \) means the instantaneous voltages cancel, so no current can circulate."

It equalises the averages, not the instantaneous waveforms — and that distinction is the entire design problem.

Both converters produce six-pulse waveforms with 4.2% ripple, but fired at different angles, so their ripples do not match. The instantaneous difference \( v_{o1} + v_{o2} \) is a non-zero ripple voltage at six times line frequency, and it drives a circulating current around the loop formed by the two bridges.

That current is limited only by the loop impedance, which is small. So a dual converter must either:

Insert a reactor to limit it deliberately — the circulating-current mode, or
Ensure only one converter is ever gated — the non-circulating mode.

Section 10-5 is the trade between those two, and it is one of the cleaner examples in the book of a design decision with no dominant answer.

Section recap. Two full converters in anti-parallel reach all four quadrants, with \( \alpha_1 + \alpha_2 = 180^\circ \) equalising their average voltages. Their instantaneous ripples still differ, so a circulating current flows unless it is limited by a reactor or prevented by blocking one converter.

Section 10-5

Circulating Current, or Not

The two ways to run a dual converter
Property	Circulating-current mode	Non-circulating (blocking) mode
Both converters gated?	Yes, always	No — one is blocked
Extra hardware	Current-limiting reactors, often two or four	None
Reversal time	Immediate — the other converter is already conducting	2–10 ms dead time while current is verified zero
Conduction at zero current	Always continuous — the circulating current keeps both alive	Discontinuous near zero, with the control problems of Ch. 8
Losses	Higher — circulating current is real current	Lower
Control complexity	Simple logic, more hardware	Complex logic, less hardware
Risk	Reactor saturation if control fails	Shoot-through if both are gated at once
Typical use	Fast reversing mill drives, servo drives	Hoists, general reversing drives, most modern designs

◇Intuition · keeping the engine running at the traffic light

The two modes are the same choice a driver makes at a red light.

Circulating current — engine idling Both converters stay in conduction, with a small current flowing round the loop through the reactors. When the command reverses, the other converter is already alive and takes over instantly. You burn a little fuel to be ready.

Non-circulating — engine off Only one converter is gated. To reverse, you must stop it, confirm the current has actually reached zero, wait a dead time, then start the other. Nothing is wasted while idle, but starting takes a few milliseconds.

Why the dead time cannot be shortened much. Before gating the second converter you must be certain the first has stopped conducting — not merely that its gate pulses were removed. A thyristor keeps conducting until its current falls below \( I_H \) and it recovers (Chapter 3). If the second converter is gated while the first is still on, the two bridges form a direct short across the supply.

So the dead time is set by measurement confidence, not by control speed: current-zero detection, plus a margin for \( t_q \), plus safety. Two to ten milliseconds is typical — imperceptible on a hoist, unacceptable on a fast reversing mill where torque must reverse in under a millisecond.

Which is why both modes survive. The choice is bought with reversal speed and paid for in reactors and losses, and different applications draw the line differently.

In circulating-current mode the reactor is sized from the instantaneous voltage difference. Over a commutation interval the peak circulating current is approximately

Peak circulating current, three-phase dual converter

\[ i_{r(max)} \approx \frac{V_{mL}}{\omega L_r}\Bigl(1 - \cos\frac{\pi}{3}\Bigr) = \frac{0.5\,V_{mL}}{\omega L_r} \]

so the reactor is chosen to hold that peak to a modest fraction — typically 10 to 30% — of rated load current. Larger reactors mean smaller circulating current and more copper, iron, cost and space; this is a straightforward engineering compromise, and it is why circulating-current designs are physically bulky.

Section recap. Circulating-current mode keeps both converters conducting through limiting reactors, giving instant reversal and continuous conduction at the price of extra hardware and losses. Non-circulating mode blocks one converter, costing a 2–10 ms dead time and discontinuous conduction near zero, but needs no reactors.

Section 10-6

Improving the Power Factor

Part 2 has arrived at an uncomfortable result. A phase-controlled converter is efficient, robust, cheap at high power and capable of four-quadrant operation — and it treats the supply badly, with \( \mathrm{PF} = 0.955\cos\alpha \) falling toward zero exactly where drives spend much of their working life.

Every remedy attacks one of the two factors. It is worth being explicit about which, because that determines whether a remedy can work at all.

Remedies, and which factor each one attacks
Remedy	Attacks	How it works	Cost
Line reactor / DC choke	Distortion	Widens the conduction angle, rounding the current blocks. THD falls from ~31% to ~25%	Cheap; adds droop
Capacitor bank	Displacement	Supplies the lagging reactive current locally	Cheap, but resonance risk — see below
Tuned passive filter	Distortion	Low-impedance path at the 5th and 7th, absorbing them	Bulky; must be designed for the specific network
Extinction-angle control	Displacement	Fires late but commutates early, so the current leads rather than lags	Needs force-commutated devices
Symmetrical-angle control	Displacement	Centres the conduction block on the voltage peak, so displacement stays near zero	Needs force-commutated devices
Sequence control	Displacement	Two converters in series; one runs fully on, only the other is phase-controlled, so the average \(\alpha\) stays small	Two transformers and two bridges
12-pulse connection	Distortion	Cancels the 5th and 7th (Ch. 7)	Two-secondary transformer
PWM active front end	Both	Self-commutating bridge shapes the current sinusoidally and in phase	Highest — IGBTs, control, filters

?Think about it

Chapter 6 was emphatic that a capacitor bank cannot fix a diode rectifier's power factor, because the deficit is pure distortion. Yet the table above lists capacitors as a legitimate remedy for a phase-controlled converter. Is that a contradiction?

A way in

No — and the difference is exactly the point that makes power factor worth splitting into two factors.

Diode bridge: \( \mathrm{PF} = 0.955 \times \cos 0° = 0.955 \). The displacement factor is already unity. There is no lagging fundamental current to compensate, so a capacitor has nothing to work on.
Phase-controlled converter at \( \alpha = 60^\circ \): \( \mathrm{PF} = 0.955 \times \cos 60° = 0.478 \). Now the displacement factor is 0.5 — a genuine lagging reactive current that capacitors can supply locally.

So the answer depends entirely on which factor is failing. Correcting the displacement here could lift the power factor from 0.478 to about 0.955 — the distortion floor, which no capacitor can pass.

But the resonance warning from Chapter 6 still applies, and more sharply. A capacitor bank forms a parallel resonant circuit with the supply inductance. If the resonant frequency lands near the 5th or 7th harmonic — which is common, because the required kVAr often puts it there — the harmonic current is amplified, possibly by an order of magnitude. The result is failed capacitors, overheated transformers, and worse distortion than before.

What practitioners actually do is fit a detuned bank: a small series reactor placed with each capacitor step, chosen to put the resonance safely below the 5th harmonic (typically around the 3.8th or 4.2nd order). The bank then corrects displacement as intended and presents an inductive impedance at every harmonic of interest, so it cannot resonate with the supply.

The transferable lesson: diagnose which factor is failing before choosing a remedy, and check what the remedy does to the frequencies you were not thinking about.

⚙Where Part 2 hands over to the rest of the book

Every remedy above except the last is a way of making a line-commutated converter behave better. The last one abandons line commutation entirely, and it is where the syllabus goes next.

The chopper-fed DC drive (Chapter 11). Rectify uncontrolled at \( \alpha = 0 \), so the input power factor stays at 0.955 regardless of speed, then control the motor voltage with a DC–DC converter's duty ratio. The reactive-power problem simply does not arise.
The PFC boost stage (Chapter 12). A single controlled switch after a diode bridge, shaping the input current to follow the voltage. Power factor above 0.99, and it is now mandatory in most equipment above 75 W.
The active front end (Chapter 17). A fully controlled IGBT bridge run with PWM, drawing sinusoidal current at any phase angle you choose — including leading, so it can supply reactive power to the network rather than consume it. This is also the basis of voltage-source HVDC.

Part 2 is not obsolete — thyristor converters still hold the records for voltage, current and on-state efficiency, and they will run mill drives and HVDC links for decades. But its central limitation, that commutation must be borrowed from the supply and paid for in reactive power, is precisely what Parts 3 and 4 were developed to escape.

Section recap. Diagnose the failing factor first. Reactors, filters and multipulse connections attack distortion; capacitors, extinction-angle, symmetrical-angle and sequence control attack displacement; only a PWM active front end fixes both. Capacitors work here — unlike on a diode bridge — but must be detuned against resonance.

Section 10-7

Worked Examples

1 Overlap angle and droop

Problem. A three-phase full converter on 415 V, 50 Hz runs at \( \alpha = 30^\circ \) with a source inductance of 1.0 mH per phase and a load current of 100 A. Find the overlap angle, the ideal and actual output voltages, and the droop.

Solution. \( V_{mL} = \sqrt2(415) = 587 \) V and \( \omega L_s = 2\pi(50)(0.001) = 0.314\ \Omega \).

Working

\[ \cos\alpha - \cos(\alpha+\mu) = \frac{2(0.314)(100)}{587} = 0.107 \]

\[ \cos(30°+\mu) = 0.866 - 0.107 = 0.759 \;\Rightarrow\; 30° + \mu = 40.6° \;\Rightarrow\; \mu = 10.6^\circ \]

\[ V_{dc(ideal)} = 1.35(415)\cos30° = 485\ \text{V} \]

\[ V_{dc} = 485 - \frac{3(0.314)}{\pi}(100) = 485 - 30 = 455\ \text{V} \quad (6.2\%\ \text{droop}) \]

The equivalent commutation resistance is \( R_\mu = 3(0.314)/\pi = 0.30\ \Omega \) — but no power is dissipated in it. Confirm with the explorer above: set 1.0 mH and 100 A and the readouts should match.

2 Regulation across the load range

Problem. For the converter of Example 1, find the output at no load, half load (50 A) and full load (100 A). Then find the firing angle needed to hold 455 V at half load.

Solution. The droop is linear in \( I_o \), so \( V_{dc} = 485 - 0.30\,I_o \):

Working

\[ I_o = 0:\ 485\ \text{V}; \qquad I_o = 50:\ 470\ \text{V}; \qquad I_o = 100:\ 455\ \text{V} \]

\[ \text{To hold 455 V at 50 A:}\quad 1.35(415)\cos\alpha = 455 + 0.30(50) = 470 \]

\[ \cos\alpha = \frac{470}{560} = 0.839 \;\Rightarrow\; \alpha = 33.0^\circ \]

So the controller must advance \( \alpha \) from 33.0° to 30.0° as the load rises from 50 A to 100 A, purely to compensate the overlap droop.

This is exactly what a voltage or current control loop does automatically — which is why the droop is rarely visible in a closed-loop drive. It shows up instead as reduced headroom: a converter designed with no margin for \( R_\mu \) will run out of firing-angle range at full load, and the drive will fail to reach rated speed under load while behaving perfectly on test.

3 Overlap eating the inverter margin

Problem. The converter of Example 1 now inverts at \( \alpha = 150^\circ \) with the same 100 A. Find the extinction angle. Then find the current at which the margin falls to \( 10^\circ \).

Solution. At \( \alpha = 150^\circ \) recompute the overlap, since it depends on \( \alpha \):

Working

\[ \cos150° - \cos(150°+\mu) = 0.107 \;\Rightarrow\; \cos(150°+\mu) = -0.866 - 0.107 = -0.973 \]

\[ 150° + \mu = 166.7° \;\Rightarrow\; \mu = 16.7^\circ, \qquad \gamma = 180° - 150° - 16.7° = 13.3^\circ \]

Note that the same inductance and current give a larger overlap when inverting (16.7° against 10.6°), because the commutating voltage available near \( \alpha = 150^\circ \) is smaller.

For a 10° margin: we need \( \mu = 20^\circ \), so \( \cos150° - \cos170° = 0.119 = 2\omega L_sI_o/V_{mL} \), giving \( I_o = 0.119(587)/(2 \times 0.314) = 111 \) A.

Only an 11% current overload consumes a third of the remaining margin. At 50 Hz, \( 13.3^\circ \) is 0.74 ms — still far more than a converter-grade \( t_q \) of 100 µs, so this design is safe. But the sensitivity is the point: margin disappears quickly under overload, which is why the adaptive limits of Section 10-3 exist.

4 Setting up a dual converter

Problem. A dual converter on 415 V drives a reversing mill. Converter 1 is at \( \alpha_1 = 40^\circ \). Find \( \alpha_2 \), both output voltages, and state which quadrant the drive is in if the load current flows through converter 1.

Solution.

Working

\[ \alpha_2 = 180° - 40° = 140° \]

\[ V_{dc1} = 560\cos40° = +429\ \text{V}, \qquad V_{dc2} = 560\cos140° = -429\ \text{V} \]

The magnitudes match and the polarities oppose, so no average voltage drives circulating current — the condition is satisfied.

With current flowing through converter 1 and \( V_{dc1} \) positive, the product \( VI \) is positive: power flows into the machine. The drive is in quadrant I — forward motoring.

What happens on a reversal command: in circulating-current mode, the controller sweeps \( \alpha_1 \) past 90° while \( \alpha_2 \) follows to keep the sum at 180°; converter 2 is already conducting and simply takes over as the current transfers. In non-circulating mode, converter 1's pulses are removed, the current is verified to have reached zero, a dead time elapses, and only then is converter 2 gated.

5 Sizing the circulating-current reactor

Problem. The dual converter above is to run in circulating-current mode with the peak circulating current limited to 15% of the 200 A rating. Size the reactor.

Solution. The target peak is \( 0.15(200) = 30 \) A. Using the estimate of Section 10-5:

Working

\[ i_{r(max)} \approx \frac{0.5\,V_{mL}}{\omega L_r} \;\Rightarrow\; L_r \approx \frac{0.5(587)}{2\pi(50)(30)} = \frac{293}{9425} = 31\ \text{mH} \]

A 31 mH reactor carrying 30 A of ripple plus a share of the 200 A load current is a substantial iron-cored component — physically large, and it must not saturate at peak load or the circulating current runs away.

This is the honest cost of circulating-current mode. You are buying instantaneous torque reversal with a large, expensive, lossy reactor. For a mill stand reversing many times a minute that is money well spent; for a hoist that reverses twice an hour it plainly is not, and the non-circulating mode's few milliseconds of dead time are free by comparison.

6 Evaluating a power-factor remedy

Problem. A 415 V drive runs at \( \alpha = 65^\circ \) drawing 150 A DC. Find the real, apparent and reactive power and the power factor. Evaluate (a) a detuned capacitor bank, (b) a chopper-fed drive.

Solution. Supply current, then the three powers:

Working

\[ I_s = 0.816(150) = 122\ \text{A}, \qquad S = \sqrt3(415)(122) = 87.7\ \text{kVA} \]

\[ \mathrm{PF} = 0.955\cos65° = 0.404, \qquad P = 0.404(87.7) = 35.4\ \text{kW} \]

\[ Q \approx S\sin65° = 87.7(0.906) = 79.5\ \text{kVAr} \]

(a) Detuned capacitor bank. Compensating the 79.5 kVAr lifts the displacement factor to about unity, leaving the 0.955 distortion floor — power factor rises from 0.404 to roughly 0.95, and apparent power falls from 87.7 to about 37 kVA. Effective and cheap, but it must be detuned (reactor in series, resonance placed below the 5th harmonic), it must switch in steps as the drive's angle changes, and it does nothing about the 31% current distortion.

(b) Chopper-fed drive. Rectify at \( \alpha = 0 \) and control on the DC side: the power factor becomes 0.955 at every speed, with no compensation equipment, no switching steps and no resonance risk. Apparent power falls to 37 kVA directly.

Verdict: for an existing installation the capacitor bank is the pragmatic retrofit. For a new design the chopper is better on every count — which is why Chapter 11 exists, and why phase-controlled DC drives have been steadily replaced.

Review

Summary & Formula Sheet

✓ Overlap

Source inductance stops current transferring instantly, so two devices conduct for \( \mu \). Output is then the mean of two line voltages.

✓ Droop, not loss

\( R_\mu = 3\omega L_s/\pi \) behaves like a resistance but dissipates nothing. The cost is reactive power and displacement.

✓ Margin

\( \gamma = 180° - \alpha - \mu \), and \( \mu \) grows with current and with falling voltage — so a fixed \( \alpha_{max} \) is not a safety measure.

✓ Notching

Overlap shorts two phases briefly, notching the supply voltage. A line reactor moves the notch away from the busbar.

✓ Four quadrants

Two converters in anti-parallel with \( \alpha_1 + \alpha_2 = 180° \). Averages match; instantaneous ripples do not.

✓ Fixing PF

Diagnose the failing factor first. Only a PWM active front end fixes both distortion and displacement.

Formula sheet · Chapter 10

Overlap angle, three-phasesolve for \(\mu\)

\( \cos\alpha - \cos(\alpha+\mu) = \dfrac{2\omega L_s I_o}{V_{mL}} \)

Overlap angle, single-phasesame form, phase peak

\( \cos\alpha - \cos(\alpha+\mu) = \dfrac{2\omega L_s I_o}{V_m} \)

Output with overlap, three-phasethe design equation

\( V_{dc} = \dfrac{3V_{mL}}{\pi}\cos\alpha - \dfrac{3\omega L_s}{\pi}I_o \)

Output with overlap, single-phase

\( V_{dc} = \dfrac{2V_m}{\pi}\cos\alpha - \dfrac{2\omega L_s}{\pi}I_o \)

Equivalent commutation resistancedissipates nothing

\( R_\mu = \dfrac{3\omega L_s}{\pi} \ \text{(3-φ)}, \qquad \dfrac{2\omega L_s}{\pi} \ \text{(1-φ)} \)

Displacement with overlapworse than \(\cos\alpha\)

\( \cos\phi_1 \approx \cos\!\left(\alpha + \dfrac{\mu}{2}\right) \)

Extinction anglemust exceed \(\omega t_q\)

\( \gamma = 180^\circ - \alpha - \mu \)

Dual-converter conditionequalises averages only

\( \alpha_1 + \alpha_2 = 180^\circ \)

Circulating-current reactorestimate for sizing

\( i_{r(max)} \approx \dfrac{0.5\,V_{mL}}{\omega L_r} \)

Power factor of the converterbefore any remedy

\( \mathrm{PF} = 0.955\cos\alpha \ \text{(3-φ)}, \qquad 0.900\cos\alpha \ \text{(1-φ)} \)

Reactive power drawnwhat the capacitor bank must supply

\( Q \approx S\sin\alpha, \qquad S = \sqrt3\,V_LI_s \)

Key terms

Source inductance, \(L_s\): Transformer leakage plus line inductance between the ideal source and the converter. Sets the overlap angle.
Overlap (commutation) angle, \(\mu\): The interval during which both the outgoing and incoming devices conduct. Grows with load current, shrinks with supply voltage.
Commutation resistance, \(R_\mu\): The equivalent series resistance modelling the overlap droop. A modelling device — it dissipates no power.
Voltage notching: Notches in the supply voltage caused by the momentary phase-to-phase short during overlap. Limited by IEEE 519.
Line reactor: Series inductance at the converter input. Deepens the notch locally so it is shallower at the point of common coupling, and reduces THD.
Dual converter: Two full converters in anti-parallel across one load, giving all four quadrants of torque and speed.
Circulating current: Current flowing round the loop of a dual converter because the two instantaneous ripple waveforms differ, even when their averages match.
Non-circulating mode: Only one converter gated at a time. No reactors needed, at the price of a 2–10 ms dead time at reversal.
Detuned capacitor bank: Capacitors with a series reactor placed to put the resonance below the 5th harmonic, so they correct displacement without amplifying harmonics.
Active front end: A self-commutating PWM rectifier drawing sinusoidal current at a chosen phase angle. Fixes distortion and displacement together.

Self-assessment

Test Yourself

Six questions closing Part 2. If these come easily, you can analyse any line-commutated converter you meet.

Chapter 10 · six questions answers hidden until you ask

A converter's output sags from 485 V at no load to 450 V at 120 A. Is the supply transformer losing 4.2 kW?

No. The droop is \( 35 \) V, so \( 35 \times 120 = 4.2 \) kW appears to be missing — but almost none of it is dissipated.

The droop is caused by commutation overlap, not by resistance. During overlap, part of the supply voltage appears across the source inductance instead of across the load. An ideal inductor stores and returns energy; it does not consume it. The volt-seconds were never delivered to the DC side, so no heat is generated.

Where the energy actually goes: it is exchanged with the supply as reactive power. The converter draws more apparent power without doing more work, and the displacement factor worsens from \( \cos\alpha \) to about \( \cos(\alpha + \mu/2) \).

How to tell the two apart in practice: resistive droop would heat something you could find with a thermometer, and would be present at any firing angle in proportion to \( I^2 \). Commutation droop is proportional to \( I \), varies with supply stiffness, and leaves everything cool. Of course, real transformer winding resistance also contributes a genuine loss — the two effects coexist, and only the \( R_\mu \) part is loss-free.

Why does adding a line reactor — which makes overlap worse — improve power quality for other equipment on the busbar?

Because it changes where the notch appears, not whether it happens.

During overlap, two phases are shorted together through the converter. The resulting voltage notch divides between the impedances on either side of the converter terminals. Without a line reactor the only impedance is the supply's own, so the full notch propagates back to the point of common coupling.

Insert a reactor and it becomes the dominant impedance in that divider. Most of the notch depth now appears across the reactor, and only a small fraction reaches the busbar.

What you pay: a longer overlap angle and a larger voltage droop, since \( R_\mu = 3\omega L_s/\pi \) has grown. There is also a small copper loss.

The bonus: the same reactor widens the current conduction blocks, so THD falls from about 31% to 25–30%. A 3–5% impedance line reactor is one of the highest-value components in an industrial drive installation — cheap, passive, and it improves two problems at once.

A dual converter has \( \alpha_1 = 30^\circ \) and \( \alpha_2 = 120^\circ \). What is wrong, and what happens?

\( \alpha_1 + \alpha_2 = 150^\circ \), not 180°. The condition is violated, and the two converters no longer produce equal and opposite averages:

\( V_{dc1} = 560\cos30° = +485 \) V, \( V_{dc2} = 560\cos120° = -280 \) V.

The sum around the loop is \( 485 - 280 = 205 \) V of net DC voltage driving current round the loop formed by the two bridges — limited only by the loop resistance and reactor.

The consequences depend on the mode:

Circulating-current mode: the reactor limits \( di/dt \), but a sustained 205 V across it ramps the current up until the reactor saturates — after which there is essentially nothing limiting it. Device failure follows within cycles.
Non-circulating mode: only one converter is gated, so no loop exists and nothing happens immediately. But the moment both are briefly gated during a changeover, the same fault appears.

Why this is a realistic failure rather than a contrived one: the two firing-angle references are generated by separate circuits, and a calibration drift, a failed op-amp or a software sign error in one of them produces exactly this. It is why dual-converter controllers derive \( \alpha_2 \) from \( \alpha_1 \) by subtraction rather than controlling both independently.

Why does the overlap angle come out larger when inverting at \( \alpha = 150^\circ \) than when rectifying at \( \alpha = 30^\circ \), with the same current and inductance?

Because commutation is driven by the instantaneous difference between the outgoing and incoming line voltages, and that difference is much smaller near \( \alpha = 150^\circ \).

The relation \( \cos\alpha - \cos(\alpha+\mu) = 2\omega L_sI_o/V_{mL} \) has a fixed right-hand side. The left-hand side is the change in \( \cos \) over the overlap interval — and \( \cos \) changes fastest near 90° and slowest near 0° and 180°.

So near \( \alpha = 150^\circ \) a larger \( \mu \) is needed to produce the same change in cosine. Working Example 3 gives 16.7° against 10.6° for identical conditions.

Why this is the worst possible arrangement: inverting is exactly where margin is scarce, and it is exactly where overlap is largest. The two effects compound — \( \gamma = 180° - \alpha - \mu \) loses ground at both ends. This is the single strongest reason the practical inverter limit sits near 150° rather than 170°, and why inverter-mode operation demands adaptive limits rather than a fixed end-stop.

A plant fits a capacitor bank to correct its drives' power factor, and six months later the capacitors start failing. What most likely happened?

Harmonic resonance between the capacitor bank and the supply inductance, almost certainly near the 5th harmonic.

The bank forms a parallel resonant circuit with the source inductance at \( f_r = f\sqrt{S_{sc}/Q_c} \), where \( S_{sc} \) is the short-circuit level and \( Q_c \) the bank rating. For typical industrial values that frequency often lands between the 4th and 7th harmonic — and the drives are injecting a large 5th harmonic current.

At resonance the harmonic current circulating between the bank and the supply is amplified, sometimes by an order of magnitude. The capacitors see currents far above rating, heat up, and fail — often one at a time, which makes the diagnosis harder because the resonant frequency shifts as each step drops out.

The correct design is a detuned bank: a small reactor in series with each capacitor step, sized to place the series resonance at about 3.8 to 4.2 times line frequency. The bank is then inductive at the 5th and above, so it cannot resonate with the supply, and it still supplies the fundamental reactive power it was bought for.

The lesson generalises: a remedy that works on the fundamental can be actively harmful at harmonic frequencies. Always ask what a new component does at the frequencies you were not designing for.

Part 2 is now complete. In one sentence each: what can a line-commutated converter do supremely well, and what can it not do at all?

What it does supremely well: convert between AC and DC at the highest voltages, currents and efficiencies available from any controllable device — with free commutation from the supply, no auxiliary circuits, and the ability to reverse power flow under software control. Nothing else runs a 1000 MW HVDC terminal or a 10 MW mill drive as cheaply or as reliably.

What it cannot do: control its reactive power. Commutation is borrowed from the supply, and the price of borrowing it is a lagging current whose angle is fixed by the firing delay. A line-commutated converter therefore always absorbs reactive power, always draws distorted current, and always gets worse as it is turned down.

Everything that follows in this book is an answer to that second sentence:

Part 3 puts the control on the DC side, so the rectifier can run at \( \alpha = 0 \).
Part 4 builds converters from self-commutating devices, which owe the supply nothing and can place their current wherever the controller chooses.

The thyristor converter is not obsolete — it is specialised, and it holds the top of the power range. But it is the last converter in this book that needs the grid's permission to switch off.

Practice

Problems

Three habits carry through this chapter:

Compute \( \mu \) at the actual operating point — it depends on \( \alpha \), \( I_o \) and the supply voltage, all of which move.
Never treat the droop as a loss. It is a reactive effect, and no heat accompanies it.
For power-factor questions, name the failing factor first — distortion or displacement decides which remedies can work.

Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design questions worth discussing in a tutorial.

A three-phase full converter on 415 V, 50 Hz has \( L_s = 0.8 \) mH and runs at \( \alpha = 25^\circ \) with 80 A. Find \( \mu \), the ideal and actual \( V_{dc} \), and the percentage droop.
For the converter of Problem 1, find the equivalent commutation resistance and sketch the terminal characteristic from no load to 150 A.
A single-phase full converter on 230 V with \( L_s = 1.5 \) mH runs at \( \alpha = 45^\circ \) with 25 A. Find \( \mu \) and \( V_{dc} \).
A converter inverts at \( \alpha = 145^\circ \) with \( L_s = 1.0 \) mH, \( I_o = 60 \) A on 415 V. Find \( \mu \) and \( \gamma \), and state whether a thyristor with \( t_q = 90\ \mu\text{s} \) is adequate.
A dual converter operates with \( \alpha_1 = 55^\circ \). Find \( \alpha_2 \) and both output voltages on a 415 V supply.
Explain why the overlap angle is larger at \( \alpha = 150^\circ \) than at \( \alpha = 30^\circ \) for identical current and inductance, and compute both for \( L_s = 1 \) mH, \( I_o = 100 \) A, 415 V.
A drive at \( \alpha = 70^\circ \) draws 200 A DC from a 415 V supply. Find \( P \), \( S \), \( Q \) and the power factor. Size a capacitor bank to correct the displacement, and state two precautions.
A dual converter must limit its peak circulating current to 20% of a 300 A rating on a 415 V supply. Size the reactor, and comment on its physical practicality.
Compare circulating-current and non-circulating dual converters for (a) a reversing mill stand reversing every 4 seconds, (b) a mine hoist reversing every 20 minutes. Justify a different answer for each.
A converter's supply voltage dips 20% while its load current rises 30%. Estimate the new overlap angle if it was 12° at nominal conditions, and discuss the consequence if the converter was inverting at \( \alpha = 150^\circ \).
An installation has four 415 V, 200 kW phase-controlled drives averaging \( \alpha = 55^\circ \). Estimate the total reactive demand, evaluate three remedies on cost and effectiveness, and recommend one with reasons.
Explain, with reference to commutation, why a line-commutated converter can never achieve a leading power factor, and what changes in a self-commutating converter to make it possible.

Part 2 is complete. You can now analyse any line-commutated rectifier: pick the pulse number for the ideal output, apply \( \cos\alpha \) for phase control, subtract \( R_\mu I_o \) for source inductance, and check the margin if it inverts. Part 3 changes the question entirely — instead of borrowing commutation from the supply, the converter will turn itself off, and control will move from a firing angle to a duty ratio.