Part 8 · Chapter 38

HVDC Transmission and FACTS

Alternating current transmits power at whatever angle the network happens to settle at; power electronics replaces that passive settlement with a commanded number — on a DC link the power is dialled in directly, and in a FACTS controller the line's own impedance, voltage and angle become adjustable quantities.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • The four reasons DC beats AC — no charging current, no stability angle, asynchronous coupling and cheaper line — and where each one bites.
  • How the break-even distance falls out of two straight lines on a cost-versus-distance plot, and why it is \(\sim\!800\;\mathrm{km}\) overhead but \(\sim\!100\;\mathrm{km}\) for cable.
  • Monopolar, bipolar, homopolar and back-to-back configurations, and what each is used for.
  • The six-pulse bridge from first principles: \(V_{d0} = \frac{3\sqrt2}{\pi}V_{LL}\), the \(\cos\alpha\) law, and the commutation overlap that turns \(X_c\) into an equivalent resistance \(3X_c/\pi\).
  • Inverter operation, the extinction angle \(\gamma\), and why commutation failure is the characteristic HVDC fault.
  • The control hierarchy — constant current at the rectifier, constant extinction angle at the inverter, the current margin and VDCOL.
  • Why twelve-pulse conversion removes the 5th and 7th harmonics, and how a single-tuned filter is sized.
  • The FACTS family: series compensation and the TCSC change \(X\); the SVC and STATCOM change \(V\); the UPFC changes all three terms of \(P = V_1V_2\sin\delta / X\).
Section 38-1

Why Direct Current

The war of the currents was settled in the 1890s on a single point: alternating voltage can be transformed, and direct voltage — at the time — could not. Everything else in this book has followed from that decision. Power electronics has now returned the argument, because a thyristor valve does for DC what a transformer does for AC, and once the conversion is possible the comparison must be made again on its merits.

Four advantages are decisive, and each traces back to something established earlier in this book.

There is no charging current. Chapter 7 gave a line a shunt capacitance and Chapter 12 showed what it does over distance: a long AC line draws a charging current \(V\omega C\ell\) whether or not it is loaded, and the Ferranti rise of Chapter 14 is the visible symptom. For a cable, where \(C\) is one to two orders of magnitude larger (Chapter 9), the charging current of a \(400\;\mathrm{kV}\) cable equals its full thermal rating at a length of only \(50\)–\(70\;\mathrm{km}\) — beyond which the cable is transmitting nothing but its own reactive power. At direct voltage \(\omega = 0\), the capacitance charges once at energisation and never again. Every long submarine link in the world is DC for this reason alone.

There is no stability angle. Chapter 26 wrote \(P = V_1V_2\sin\delta/X\) and Chapters 28 and 29 showed that this ties transmissible power to a rotor angle which must stay well short of \(90^\circ\). On a DC link \(P = V_d I_d\); there is no angle, no synchronising torque and no steady-state stability limit. The power a DC line can carry is set by conductor heating and by insulation, and by nothing else.

Two systems need not be synchronous. A DC link cares nothing about the frequency or the phase at either end. It can join a \(50\;\mathrm{Hz}\) network to a \(60\;\mathrm{Hz}\) one, or two \(50\;\mathrm{Hz}\) networks whose angles are not controlled relative to each other. India's five regional grids were joined by back-to-back HVDC stations before they were synchronised; Japan's \(50\;\mathrm{Hz}\) and \(60\;\mathrm{Hz}\) halves are still joined only by DC.

The line itself is cheaper. A bipolar DC line uses two conductors where a three-phase circuit uses three; there is no skin effect, so the whole cross-section carries current; and because insulation must withstand the peak, a DC line at \(\pm V_d\) needs the same insulation as an AC line whose peak line-to-earth voltage is \(V_d\). The tower is narrower, the right-of-way is narrower, and the corona and radio interference of Chapter 15 are lower for the same transmitted power.

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Two further operational gains
A DC link's power is a commanded variable, adjustable in milliseconds and reversible, and it contributes no fault current to the AC system it feeds.

Fast power modulation lets a link damp the inter-area oscillations of Chapter 28 or provide emergency support after a trip. Contributing no short-circuit current means a DC infeed strengthens a network's power balance without raising the breaker duty computed in Chapter 25 — an advantage no AC line can offer.

Against all this stands a short but heavy list of costs. The converter stations are expensive and complex. A line-commutated converter absorbs reactive power at both ends, typically \(50\)–\(60\%\) of the transmitted active power, so filters and capacitor banks must be installed to supply it. The converters generate characteristic harmonics on both the AC and DC sides. Interrupting DC is genuinely hard, because there is no natural current zero — the tapping of a DC line into a multi-terminal network was for decades limited by the absence of a practical DC circuit breaker. And a line-commutated inverter depends on the AC voltage at its terminals to commutate at all, so a weak receiving system causes commutation failures.

Section 38-2

The Break-Even Distance

The economic comparison is unusually clean, because the two technologies differ in where their money sits. AC has cheap terminals and an expensive line; DC has expensive terminals and a cheap line. Write each total cost as a straight line in the route length \(d\):

Total cost of each alternative
\[ C_{AC} = T_{AC} + c_{AC}\,d, \qquad C_{DC} = T_{DC} + c_{DC}\,d \]

with \(T\) the terminal cost (substations, or converter stations with their transformers, valves, filters and reactors) and \(c\) the cost per kilometre of line or cable. Since \(T_{DC} > T_{AC}\) and \(c_{DC} < c_{AC}\), the two lines cross exactly once. Setting them equal,

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Break-even distance
\[ d^{*} = \frac{T_{DC} - T_{AC}}{c_{AC} - c_{DC}} \]

Beyond \(d^{*}\), DC is cheaper; below it, AC is. Typical figures put \(d^{*}\) at \(600\)–\(800\;\mathrm{km}\) for overhead lines and at only \(50\)–\(100\;\mathrm{km}\) for cables — because \(c_{AC}\) for cable is several times \(c_{DC}\), so the denominator is large and the crossover comes early.

route length d, km total cost 0200400600 0 400 800 1200 1600 AC overhead DC overhead d* = 800 km AC cable DC cable 133 km DC pays its terminal premium back at the rate (c_AC − c_DC) per kilometre
Two straight lines and one crossing — the whole economics of the AC/DC choice

Three refinements matter in practice. Losses should enter the comparison, and they favour DC: for the same power and the same conductor cross-section a bipolar DC line has lower \(I^2R\) loss than a three-phase AC line, but converter losses of about \(0.7\%\) per station are added at each end, so the loss advantage begins only after a few hundred kilometres. Reactive compensation must be counted on the AC side — a long AC line needs intermediate switching stations with shunt reactors, and each of those adds a step to \(T_{AC}\), effectively bending the AC line upward. And in a great many modern projects the break-even calculation is never performed at all, because the requirement is asynchronous interconnection or a controlled flow, for which AC is not an alternative at any distance. That is precisely the case for a back-to-back link, where \(d = 0\).

The distance argument is the weakest of the four. Cost crossover explains the long overhead links; the cable, the asynchronous tie and the controllability arguments explain almost everything else that has been built since 2000. When a question asks "why HVDC?", the distance answer alone is incomplete.
Section 38-3

Converter Configurations

A link is described by how many conductors it uses and what serves as the return path.

Monopolar. One conductor, normally at negative polarity — negative because it suffers less corona loss and less radio interference (Chapter 15) — with the return through earth or sea via electrodes sited some kilometres from the converter stations. It is the cheapest arrangement, but a continuous earth current of a thousand amperes or more corrodes buried metalwork and biases transformer neutrals into saturation, so a metallic return conductor is often provided instead, at reduced insulation.

Bipolar. Two conductors, one at \(+V_d\) and one at \(-V_d\), each with its own converter, and the two converter groups joined at the earthed neutral point. In balanced operation the earth carries only the small difference current. Its great merit is redundancy: if one pole fails, the other continues to carry half the rated power using the earth return, so the link degrades rather than fails. This is the standard arrangement for all major transmission links.

Homopolar. Two or more conductors of the same polarity, always negative, with earth return. Loss of one converter leaves the others able to use both conductors, so the current-carrying capability is preserved; but the earth carries the full current continuously, which is unacceptable in most locations. It is rare.

Back-to-back. Rectifier and inverter in the same building, with a DC bus a few metres long. There is no line at all, so \(d = 0\) and the DC voltage can be chosen low with a correspondingly high current, minimising valve cost. The purpose is entirely the asynchronous coupling and the controlled flow between two networks.

Multi-terminal. Three or more converter stations on a common DC network, in series or in parallel. Parallel connection — all stations at the same DC voltage, sharing current — is the practical form, but it requires that a fault on any one branch be isolated without collapsing the whole DC network, which is why multi-terminal schemes waited on the development of the DC circuit breaker and of voltage-source converters.

ConfigurationConductorsReturnOn one-pole failureTypical application
Monopolar1 (usually \(-V_d\))Earth / sea, or metallicLink lostFirst stage of a staged project; short cable
Bipolar2 (\(+V_d, -V_d\))Earth, small unbalance only\(50\%\) power retainedAll large transmission links
Homopolar2 or more, all \(-V_d\)Earth, full currentFull current on remaining valvesRare
Back-to-backNone (\(d=0\))Link lostAsynchronous tie, controlled interchange
Multi-terminal2 (parallel taps)EarthDepends on DC breakerOffshore wind clusters, DC grids
Section 38-4

The Six-Pulse Bridge and Commutation Overlap

The converter itself is the Graetz bridge: six valves, arranged as an upper group whose cathodes are joined to the positive DC terminal and a lower group whose anodes are joined to the negative one. Each valve conducts for \(120^\circ\), two valves conduct at any instant — one from each group — and the six conduction intervals repeat every \(60^\circ\), which is why the DC-side ripple is at six times the supply frequency.

Start with diodes, that is, with zero firing delay. At any instant the upper group conducts through whichever phase is most positive and the lower group through whichever is most negative, so the DC output follows the upper envelope of the six line-to-line voltages. Its average is found by integrating a single \(60^\circ\) segment centred on the peak of one line voltage:

Ideal no-load direct voltage
\[ V_{d0} = \frac{1}{\pi/3}\int_{-\pi/6}^{\pi/6}\sqrt2\,V_{LL}\cos\omega t\;\mathrm{d}(\omega t) = \frac{3\sqrt2\,V_{LL}}{\pi}\Big[\sin\omega t\Big]_{-\pi/6}^{\pi/6} \]
\[ = \frac{3\sqrt2\,V_{LL}}{\pi}\left(2\times\frac12\right) = \frac{3\sqrt2}{\pi}V_{LL} = 1.35\,V_{LL} \]

Now replace the diodes by thyristors and delay each firing by an angle \(\alpha\) past the instant at which its diode would have conducted — the natural commutation point. The same \(60^\circ\) integration window simply slides by \(\alpha\):

The cosine law
\[ V_d = \frac{3}{\pi}\int_{-\pi/6+\alpha}^{\pi/6+\alpha}\sqrt2\,V_{LL}\cos\omega t\;\mathrm{d}(\omega t) = \frac{3\sqrt2 V_{LL}}{\pi}\cos\alpha = V_{d0}\cos\alpha \]
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One control angle does everything
\(V_d = V_{d0}\cos\alpha\) is positive for \(\alpha < 90^\circ\) (rectification) and negative for \(\alpha > 90^\circ\) (inversion), while \(I_d\) can only ever flow one way through the valves.

Since \(P = V_d I_d\), reversing \(V_d\) reverses the power flow with the current direction unchanged. This is why an HVDC link reverses power by changing firing angles, not by reversing current — and why the line's insulation must tolerate both polarities.

The idealisation that remains is instantaneous commutation. In reality the transfer of current from the outgoing valve to the incoming one is opposed by the leakage reactance \(X_c\) of the converter transformer, and both valves conduct together for an overlap angle \(\mu\), during which the DC terminal is momentarily connected to two phases at once and its voltage is the mean of the two. The lost area under the voltage waveform is what the analysis must account for.

During overlap the two commutating phases are short-circuited through \(2X_c\), and the commutating current rises from \(0\) to \(I_d\) driven by the line-to-line voltage between them. Integrating that circuit over the overlap interval gives the commutation equation:

The commutation equation
\[ I_d = \frac{\sqrt2\,V_{LL}}{2X_c}\Big[\cos\alpha - \cos(\alpha+\mu)\Big] \]

and the mean DC voltage becomes the average of the pre- and post-commutation values:

Direct voltage with overlap, and the equivalent circuit
\[ V_d = \frac{V_{d0}}{2}\Big[\cos\alpha + \cos(\alpha+\mu)\Big] \]
\[ \text{Eliminating } \mu:\qquad V_d = V_{d0}\cos\alpha - \frac{3X_c}{\pi}I_d = V_{d0}\cos\alpha - R_c I_d \]
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The equivalent commutating resistance
\[ R_c = \frac{3X_c}{\pi} \]

\(R_c\) behaves like a resistance in the DC circuit — it produces a voltage drop proportional to \(I_d\) — but it dissipates no power. The energy is not lost; it is exchanged with the AC system. A converter is therefore a voltage source \(V_{d0}\cos\alpha\) behind a lossless "resistance" \(R_c\), and that two-element model is all that is needed to analyse a complete link.

Overlap also fixes the AC-side power factor. The fundamental AC current lags the voltage by a displacement angle \(\phi\), and a short derivation from the same waveform gives the very useful approximation

Displacement power factor
\[ \cos\phi \;\approx\; \frac{\cos\alpha + \cos(\alpha+\mu)}{2} \;=\; \frac{V_d}{V_{d0}} \]

Since \(\alpha\) is never zero — a minimum of about \(5^\circ\) is held so that the valve has a positive forward voltage to fire into — the power factor is always lagging, and the converter always absorbs reactive power. At a typical operating point \(\cos\phi \approx 0.9\), so \(Q \approx 0.5P\): the reactive burden mentioned in Section 38-1 is a direct consequence of the cosine law and not an incidental defect.

SIX-PULSE GRAETZ BRIDGE a b c X_c gate (α) L_d +V_d −V_d I_d → DC OUTPUT, α AND μ v ωt α = 0 (diode) envelope delayed by α, notched by μ α μ V_d = V_d0 cos α − (3X_c/π) I_d
The bridge, the delay angle that scales the output, and the overlap that loads it down

Push \(\alpha\) beyond \(90^\circ\) and \(V_d\) goes negative: the bridge now opposes the current the other converter drives through it, absorbs power from the DC side and returns it to the AC system. It has become an inverter. Because a thyristor needs a definite reverse-bias interval to regain its blocking ability, an inverter is not described by \(\alpha\) but by the angle remaining after commutation ends and before the valve voltage reverses. Defining the advance angle \(\beta = 180^\circ - \alpha\) and the extinction angle \(\gamma = \beta - \mu = 180^\circ - \alpha - \mu\), the same algebra gives the inverter's DC voltage magnitude in two equivalent forms:

Inverter equations
\[ V_{di} = \frac{V_{d0i}}{2}\big[\cos\beta + \cos\gamma\big] = V_{d0i}\cos\gamma - R_{ci}I_d = V_{d0i}\cos\beta + R_{ci}I_d \]

The extinction angle must exceed the valve's turn-off requirement — expressed as a time, typically \(400\;\mu\mathrm{s}\), which at \(50\;\mathrm{Hz}\) is \(7^\circ\) — with margin for AC voltage dips and distortion. A minimum \(\gamma\) of \(15^\circ\)–\(18^\circ\) is standard. If \(\gamma\) falls below the turn-off requirement, the outgoing valve fails to regain blocking and conducts again when its voltage returns positive: a commutation failure, which short-circuits the DC side of the inverter and collapses \(V_{di}\) for a cycle or two. A single voltage dip in the receiving AC system is enough to cause one, which is why line-commutated inverters need a reasonably strong AC system — conventionally a short-circuit ratio above about \(2\) to \(3\).

Connecting the two converters through the line resistance \(R_L\) completes the link. With the rectifier driving and the inverter receiving,

The link equation
\[ I_d = \frac{V_{d0r}\cos\alpha - V_{d0i}\cos\gamma}{R_{cr} + R_L - R_{ci}} \]

The minus sign on \(R_{ci}\) is not a slip. It arises because the inverter's commutation drop, written with \(\cos\gamma\), acts to increase the current rather than oppose it. The denominator is therefore small — often just \(R_L\) when the two converters are identical — which means \(I_d\) is extremely sensitive to the numerator. A one-percent change in either AC voltage can swing the direct current by tens of percent. The link cannot be left to find its own operating point; it must be controlled, closely and quickly.

Section 38-5

Control of the Link

The sensitivity just identified sets the whole control philosophy. Each converter has one control variable — \(\alpha\) at the rectifier, \(\gamma\) (or \(\beta\)) at the inverter — and the link has two quantities to fix, the direct current and the direct voltage. The natural division of labour is: the rectifier controls current, the inverter controls voltage.

The rectifier runs constant current (CC) control. A regulator compares \(I_d\) with the current order \(I_{ord}\) and adjusts \(\alpha\) to close the error, subject to a lower limit \(\alpha_{\min}\approx5^\circ\). The characteristic in the \(V_d\)–\(I_d\) plane is therefore a nearly vertical line at \(I_d = I_{ord}\), with a steeply falling segment at the left where \(\alpha\) hits its minimum and the rectifier can do no more.

The inverter runs constant extinction angle (CEA) control, holding \(\gamma\) at its minimum safe value. Since \(V_{di} = V_{d0i}\cos\gamma - R_{ci}I_d\), constant \(\gamma\) traces a line of slope \(-R_{ci}\) in the same plane: nearly horizontal, and fixing the voltage. Holding \(\gamma\) at the minimum is not just a safety constraint — it also minimises the reactive power the inverter absorbs, since \(\cos\phi_i \approx V_{di}/V_{d0i}\).

The intersection of the two characteristics is the operating point. But an inverter also carries a current regulator, whose order is set below the rectifier's by the current margin \(\Delta I\), typically \(10\%\)–\(15\%\) of rated current:

Current margin
\[ I_{ord,\,inv} = I_{ord,\,rec} - \Delta I \]

In normal operation the inverter's current regulator is saturated and inactive, because the actual current \(I_{ord,rec}\) is above its own order. Now let the rectifier's AC voltage fall far enough that even at \(\alpha_{\min}\) it cannot maintain \(I_{ord,rec}\). Its characteristic drops, the intersection slides onto the inverter's current characteristic, and control transfers automatically: the inverter now sets the current at \(I_{ord,rec}-\Delta I\) and the rectifier, stuck at \(\alpha_{\min}\), sets the voltage. The link keeps running at \(85\%\)–\(90\%\) of ordered current instead of collapsing, and no communication between the two stations was required for the changeover.

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Why the margin exists
The current margin guarantees that exactly one station is in current control at any instant, and makes the handover between them automatic and stable.

Without it, both current regulators would fight over the same variable and the operating point would be indeterminate — the two nearly vertical characteristics would either coincide or fail to meet at all.

Two further limits complete the picture. A minimum current of around \(0.1\;\mathrm{pu}\) is enforced, because below it the current becomes discontinuous and the smoothing reactor can no longer keep the valves conducting. And the voltage-dependent current order limit, VDCOL, reduces the current order automatically when \(V_d\) falls below about \(0.6\)–\(0.8\;\mathrm{pu}\):

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VDCOL
When the direct voltage collapses, reduce the current order in proportion, so that the converters draw less reactive power from the very AC system that is already weak.

A depressed DC voltage means either an AC fault or a commutation failure. Continuing to demand rated current would force \(\alpha\) low and \(\mu\) large, increasing the reactive demand at exactly the wrong moment and risking repeated commutation failures. VDCOL backs the link off, lets the AC voltage recover, and then ramps the current order back up — it is the single most important feature for post-fault recovery.

V_d I_d rectifier at α_min rectifier CC I_ord inverter CEA (γ_min) inverter CC ΔI margin normal operating point VDCOL cuts the order as V_d collapses rectifier AC voltage depressed → control passes to the inverter whichever station cannot meet its order gives up current control to the other
The steady-state V–I characteristics: constant current, constant extinction angle and the margin that switches between them
Section 38-6

Twelve-Pulse Conversion, Harmonics and Filters

A six-pulse bridge draws a quasi-square current from the AC system, and the Fourier series of that waveform is exactly what the filter designer must deal with. With ideal smoothing on the DC side the current in each AC phase is a \(120^\circ\) rectangular block, whose harmonics are

AC-side harmonics of a six-pulse bridge
\[ I_1 = \frac{\sqrt6}{\pi}I_d = 0.7797\,I_d, \qquad I_h = \frac{I_1}{h}, \qquad h = 6k \pm 1 = 5,7,11,13,17,19,\dots \]

The triplen harmonics are absent because they are zero-sequence and the delta winding of the converter transformer traps them; the even harmonics are absent by half-wave symmetry. On the DC side the corresponding ripple orders are \(h = 6k = 6,12,18,\dots\). The total harmonic distortion of the AC current is \(31.1\%\) — utterly unacceptable on a transmission bus.

The elegant fix uses the phase shift a transformer connection provides for free. Feed two six-pulse bridges in series on the DC side from two transformer secondaries, one star-connected and one delta-connected. The delta secondary's line voltages are shifted \(30^\circ\) from the star's, so the two bridges fire \(30^\circ\) apart and their DC outputs interleave to give twelve pulses per cycle. On the AC side, the \(30^\circ\) shift reverses the phase of the \(5\)th and \(7\)th harmonic contributions of one bridge relative to the other, and they cancel in the common primary winding.

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Twelve-pulse harmonics
\[ \text{AC side: } h = 12k\pm1 = 11,13,23,25,\dots \qquad \text{DC side: } h = 12k = 12,24,\dots \]

The \(5\)th, \(7\)th, \(17\)th and \(19\)th vanish. The current THD falls from \(31.1\%\) to \(15.2\%\), the DC ripple falls sharply and moves to a higher frequency where the smoothing reactor is more effective, and the two filters that would have been needed for the 5th and 7th are saved. Every modern HVDC converter is twelve-pulse for this reason; the two bridges together are called a pole and their transformers deliberately have different vector groups.

The remaining harmonics are removed by shunt filters at the converter bus. A single-tuned filter is a series \(RLC\) branch resonant at the harmonic to be trapped: at that frequency its impedance falls to the resistance \(R\), shunting the harmonic current to earth. Sizing it is a two-step calculation. The capacitor is chosen first, because at fundamental frequency the branch is essentially capacitive and its reactive output is genuinely useful — it supplies part of the converter's reactive demand:

Single-tuned filter design
\[ Q_f = \frac{V^2}{X_C} = V^2\,\omega C \;\Longrightarrow\; C = \frac{Q_f}{\omega V^2}, \qquad \omega_h L = \frac{1}{\omega_h C} \;\Longrightarrow\; L = \frac{1}{h^2\omega^2 C} \]
\[ \text{equivalently}\quad X_L = \frac{X_C}{h^2}, \qquad Q_{\text{factor}} = \frac{h\,\omega L}{R} \;\;(\text{typically } 30\text{–}100) \]

Filters are provided for the two lowest characteristic orders — the \(11\)th and \(13\)th in a twelve-pulse scheme — plus a high-pass damped branch to cover \(23\)rd and above, and often a smaller branch tuned to the \(3\)rd for the non-characteristic harmonics produced by unbalanced AC voltages or unequal firing. Together with plain shunt capacitor banks they supply the \(50\)–\(60\%\) reactive demand. On the DC side a smoothing reactor of \(0.3\)–\(0.8\;\mathrm{H}\) plus DC filters limit the ripple current that would otherwise induce interference in nearby telephone circuits.

Section 38-7

VSC-HVDC

Everything so far assumed a line-commutated converter (LCC): thyristors, which can be turned on by a gate pulse but can only be turned off by the external circuit driving their current to zero. Every characteristic weakness of classical HVDC follows from that one limitation — the dependence on a strong AC voltage for commutation, the unavoidable lagging power factor, the inability to start into a dead network, and the fact that reversing power requires reversing the DC voltage.

A voltage-source converter uses IGBTs, which turn off on command. The converter can then synthesise, by pulse-width modulation or by the staircase of a modular multilevel converter, an AC voltage \(V_c\) of any magnitude and any phase relative to the network voltage \(V_s\). Across the interface reactance \(X\) the exchange is then given by the same two expressions we derived in Chapter 34 for any voltage source behind a reactance:

Independent control of P and Q
\[ P = \frac{V_s V_c \sin\delta}{X}, \qquad Q = \frac{V_s\big(V_s - V_c\cos\delta\big)}{X} \]

The angle \(\delta\) controls active power and the magnitude \(V_c\) controls reactive power, almost independently. The consequences are large. A VSC can supply or absorb reactive power at each terminal, so it behaves as a STATCOM in addition to transmitting power. It can energise a dead network — black start — because it makes its own voltage rather than borrowing one. It suffers no commutation failure. And it reverses power by reversing the current while keeping the DC voltage polarity fixed, which is exactly what a multi-terminal DC grid needs: every station can then be tapped off a common DC bus of fixed polarity, and extruded XLPE cable, which cannot tolerate polarity reversal, becomes usable.

PropertyLCC (thyristor)VSC (IGBT)
Rating availableUp to \(\sim\!12\;\mathrm{GW}\), \(\pm1100\;\mathrm{kV}\)Up to \(\sim\!3\;\mathrm{GW}\), \(\pm640\;\mathrm{kV}\)
Converter losses per station\(\approx 0.7\%\)\(\approx 1\%\) (MMC), higher for 2-level
Reactive powerAbsorbs \(50\)–\(60\%\) of \(P\)Independently controllable, either sign
FiltersLarge AC and DC filter yardsSmall or none with MMC
Weak AC system / black startNeeds SCR \(> 2\)–\(3\); cannot black startWorks into a passive network; can black start
Power reversalReverse \(V_d\), current unchangedReverse \(I_d\), \(V_d\) unchanged
DC fault behaviourRides through by firing controlDiodes feed the fault; needs DC breaker or full-bridge cells
FootprintLarge\(\approx\) half
Where each belongs. LCC still wins for bulk point-to-point transfer over great distance, where its lower losses and higher ratings dominate. VSC wins for offshore wind connections (a passive, dead network at one end), for city-centre infeeds (small footprint, cable), and for anything that must become a multi-terminal DC grid. Chapter 39 returns to the first of these when it takes up renewable integration.
Section 38-8

FACTS: Series Compensation and the TCSC

An HVDC link replaces the AC connection entirely. The alternative philosophy — Flexible AC Transmission Systems — keeps the AC line and uses power electronics to make its parameters adjustable. The target is the equation that has run through Chapters 26 to 34:

The three terms available for control
\[ P = \frac{V_1 V_2 \sin\delta}{X} \]

There are exactly three ways to raise \(P\): raise the voltages, reduce the reactance, or change the angle. Each defines a family of FACTS controllers, and the series family attacks \(X\).

Insert a capacitor of reactance \(X_C\) in series with a line of reactance \(X_L\). The effective reactance falls to \(X_L - X_C\), and defining the degree of compensation \(k = X_C/X_L\),

Series compensation
\[ X_{\text{eff}} = X_L(1-k), \qquad P = \frac{V_1V_2\sin\delta}{X_L(1-k)}, \qquad P_{\max} = \frac{V_1V_2}{X_L(1-k)} \]

Fifty percent compensation doubles the transmissible power at every angle, which is why a series capacitor is by a wide margin the cheapest way to increase the capability of an existing long line. It also improves stability directly: the equal-area criterion of Chapter 29 shows that raising \(P_{\max}\) enlarges the decelerating area and lengthens the critical clearing time. And because the compensating voltage \(I X_C\) is proportional to the current, the correction is automatically largest when the loading is heaviest — a self-regulating property no shunt device has.

Two dangers come with it. The first is that \(k\) is limited in practice to about \(70\%\); beyond that the line becomes so stiff that fault currents rise steeply and the protection of Chapter 36 — distance relays in particular, which measure impedance to the fault — is confused by a capacitor of negative reactance inside the measured loop. The second is subsynchronous resonance. The compensated line is a series \(LC\) circuit with a natural frequency

Electrical resonance and the complementary rotor frequency
\[ f_{er} = f\sqrt{\frac{X_C}{X_L}} = f\sqrt{k}, \qquad f_{\text{rotor}} = f - f_{er} \]

Since \(k < 1\), \(f_{er}\) is always below the power frequency. A current at \(f_{er}\) in the stator produces a torque on the rotor at \(f - f_{er}\); if that lands on one of the turbine-generator shaft's torsional natural frequencies, the shaft oscillation and the electrical oscillation feed each other. The Mohave incident of 1970, in which a shaft was twisted apart, made this a design constraint rather than a curiosity.

The thyristor-controlled series capacitor answers both problems. A thyristor-controlled reactor is placed in parallel with the series capacitor, and its firing angle varies the reactor's effective fundamental reactance \(X_L(\alpha)\). The parallel combination, taken as capacitive-positive, is

TCSC characteristic
\[ X_{TCSC}(\alpha) = \frac{X_C\,X_L(\alpha)}{X_L(\alpha) - X_C} \]

With the thyristors fully off, \(X_L(\alpha)\to\infty\) and \(X_{TCSC}\to X_C\): the plain capacitor. As the reactor is brought in, \(X_L(\alpha)\) falls, the denominator shrinks and \(X_{TCSC}\) grows — the capacitive vernier region, giving a boost of \(1\) to \(3\) times \(X_C\) with continuous control. Push further and \(X_L(\alpha) = X_C\) is the internal resonance, which is avoided by a blocked band of firing angles. Beyond it \(X_L(\alpha) < X_C\), the expression turns negative and the device is a controllable inductor, useful for limiting fault current or for damping.

🔑
What the TCSC adds over a fixed capacitor
A continuously variable series reactance, and — crucially — an apparent reactance at subsynchronous frequencies that is inductive rather than capacitive.

The vernier control is fast enough to modulate the series reactance in step with a power swing, damping inter-area oscillations directly. And because the closed-loop control presents an inductive impedance below the power frequency, the series resonance that drives SSR simply does not form — a TCSC is described as "SSR-neutral", and it is the reason a TCSC is preferred over a fixed capacitor on lines near large thermal units.

The fully electronic member of the series family is the static synchronous series compensator, a voltage-source converter that injects a voltage in quadrature with the line current. It emulates a capacitive or inductive reactance without any passive capacitor at all, and because the injected voltage is independent of the current magnitude it remains effective at light load, where a series capacitor does nothing.

Section 38-9

FACTS: Shunt Compensation and the UPFC

The shunt family attacks the voltage terms. Chapter 34 established the governing relation: injecting reactive power \(Q\) at a bus of short-circuit level \(S_{sc}\) raises its voltage by approximately

Voltage sensitivity to shunt reactive injection
\[ \frac{\Delta V}{V} \approx \frac{Q}{S_{sc}} \]

Mechanically switched capacitors do this in coarse steps and in hundreds of milliseconds. A static VAr compensator does it continuously and within a cycle or two. Its core is a thyristor-controlled reactor: an inductor in series with a pair of back-to-back thyristors, fired at an angle \(\alpha\) measured from the voltage zero so that conduction occupies \(\sigma = 2(\pi-\alpha)\) each half cycle. Fourier analysis of the resulting current gives the fundamental susceptance

Thyristor-controlled reactor
\[ B_{TCR}(\sigma) = \frac{\sigma - \sin\sigma}{\pi X_L}, \qquad \sigma = 2(\pi - \alpha) \]
\[ \alpha = 90^\circ \Rightarrow \sigma = 180^\circ,\; B_{TCR} = 1/X_L \text{ (full conduction)}; \qquad \alpha = 180^\circ \Rightarrow B_{TCR} = 0 \]

The reactor alone can only absorb. Pair it with a fixed capacitor bank or with thyristor-switched capacitors and the net susceptance becomes continuously variable through zero:

SVC output
\[ B_{SVC} = B_C - B_{TCR}(\sigma), \qquad Q_{SVC} = V^2 B_{SVC} \]

That last expression contains the SVC's fundamental limitation. It is a controlled susceptance, so its reactive output falls with the square of the voltage — and it is needed most precisely when the voltage has collapsed. At \(0.8\;\mathrm{pu}\) an SVC delivers only \(64\%\) of its rating.

The STATCOM removes that defect. It is a voltage-source converter shunt-connected through a reactance, synthesising a voltage \(V_c\) in phase with the bus voltage \(V_s\); the reactive exchange is then

STATCOM output
\[ Q = \frac{V_s\big(V_s - V_c\big)}{X}, \qquad \text{limited by } I_{\max}: \quad Q_{\max} = V_s I_{\max} \]

\(V_c > V_s\) makes it a capacitor, \(V_c < V_s\) an inductor, and the transition is continuous through zero. Because the machine is current-limited rather than susceptance-limited, its output falls only linearly with voltage: at \(0.8\;\mathrm{pu}\) a STATCOM still delivers \(80\%\) of rating, and it can be designed for a short-term overload beyond that. It is also faster — a quarter cycle rather than one to two cycles — and much smaller, since there is no large capacitor bank or reactor.

V (pu) capacitive I → ← inductive I 1.0 0.5 slope (droop) region SVC: Q ∝ V² STATCOM: Q ∝ V the shaded wedge is the support an SVC loses, and a STATCOM keeps, at depressed voltage
V–I capability: susceptance-limited SVC against current-limited STATCOM

One controller remains, and it is the general case. A unified power flow controller is two voltage-source converters sharing a common DC capacitor: one shunt-connected, one series-connected. The series converter injects a voltage \(V_{se}\) of arbitrary magnitude and arbitrary angle in the line — not merely in quadrature with the current, as an SSSC does. That single freedom subsumes all three control actions: a component in phase with the bus voltage is voltage regulation, a component in quadrature with the line current is series compensation, and a component in quadrature with the bus voltage is a phase shift.

Injecting an arbitrary voltage means the series converter exchanges real as well as reactive power with the line. That real power flows through the DC capacitor and is supplied by the shunt converter, drawn from the bus. The shunt converter meanwhile has spare capacity for independent reactive support, so it also acts as a STATCOM. The UPFC can therefore set the line's active and reactive flows to commanded values while holding its terminal voltage — the closest thing in AC transmission to the direct dispatch of a DC link.

ControllerConnectionTechnologyDirectly controlsPrincipal use
Series capacitor (fixed)SeriesPassive\(X\)Raise \(P_{\max}\) on long lines
TCSCSeriesThyristor\(X\), continuouslyPower flow, swing damping, SSR-neutral
SSSCSeriesVSCInjected \(V\) in quadratureCompensation independent of load
SVCShuntThyristor\(V\) via \(B\)Voltage support, \(Q \propto V^2\)
STATCOMShuntVSC\(V\) via injected \(V_c\)Fast voltage support, \(Q \propto V\)
TCPAR / phase shifterSeriesThyristor tap\(\delta\)Steer flow between parallel paths
UPFCBothTwo VSCs, common DC\(V\), \(X\) and \(\delta\)Full \(P\) and \(Q\) scheduling
The thread that closes Part 8. Chapter 34 controlled voltage with transformers and capacitor banks, on a timescale of seconds. Chapter 33 controlled frequency with governors, on a timescale of seconds to minutes. FACTS controllers act in a quarter of a cycle, which puts them inside the transient period of Chapter 29 rather than after it — and that is why they change stability limits, not merely operating points. An HVDC link goes one step further and removes the angle from the problem altogether.
Section 38-10

Worked Examples

1 Break-even distance, overhead and cable

Problem. For a \(2000\;\mathrm{MW}\) scheme, two AC terminal substations cost \(60\) M€ in total and a double-circuit \(400\;\mathrm{kV}\) overhead line costs \(0.60\) M€/km. Two HVDC converter stations cost \(300\) M€ and the DC overhead line costs \(0.30\) M€/km. (a) Find the break-even distance. (b) Repeat for a submarine route where the AC cable costs \(3.0\) M€/km and the DC cable \(1.2\) M€/km, terminal costs unchanged. (c) Comment on the difference.

Solution — (a).

Overhead route
\[ 60 + 0.60\,d = 300 + 0.30\,d \;\Longrightarrow\; 0.30\,d = 240 \;\Longrightarrow\; d^{*} = 800\;\mathrm{km} \]
\[ \text{Cost at break-even} = 60 + 0.60(800) = 540\;\text{M€} \]

(b) Submarine route.

Cable route
\[ 60 + 3.0\,d = 300 + 1.2\,d \;\Longrightarrow\; 1.8\,d = 240 \;\Longrightarrow\; d^{*} = 133\;\mathrm{km} \]

(c). The numerator — the converter-station premium of \(240\) M€ — is identical in both cases; only the rate at which DC pays it back has changed, from \(0.30\) to \(1.80\) M€ per kilometre. The cable break-even is six times shorter for that reason alone, and the true figure is shorter still, because at anything beyond \(60\)–\(70\;\mathrm{km}\) the AC cable's charging current leaves no capacity for load (Chapter 12), so the AC option must add compensation platforms or ceases to exist. The straight-line model here understates the DC advantage for cables and overstates it for short overhead routes, where AC needs no intermediate stations at all.

2 A six-pulse rectifier, end to end

Problem. A six-pulse bridge is fed from a converter transformer whose secondary line voltage is \(220\;\mathrm{kV}\) with a commutating reactance of \(X_c = 15\;\Omega\) per phase. It operates at \(\alpha = 15^\circ\) with \(I_d = 1200\;\mathrm{A}\). Find \(V_{d0}\), the direct voltage, the overlap angle, the displacement power factor, and the active and reactive power on the AC side.

Solution. The ideal no-load voltage and the commutating resistance are

Converter constants
\[ V_{d0} = \frac{3\sqrt2}{\pi}(220) = 1.3505 \times 220 = 297.1\;\mathrm{kV}, \qquad R_c = \frac{3X_c}{\pi} = \frac{45}{\pi} = 14.32\;\Omega \]
Direct voltage
\[ V_d = V_{d0}\cos\alpha - R_cI_d = 297.1(0.9659) - 14.32(1.2) = 287.0 - 17.2 = 269.8\;\mathrm{kV} \]

For the overlap angle use the commutation equation, rearranged:

Overlap
\[ \cos\alpha - \cos(\alpha+\mu) = \frac{2X_cI_d}{\sqrt2\,V_{LL}} = \frac{2(15)(1200)}{1.4142(220\,000)} = \frac{36\,000}{311\,127} = 0.1157 \]
\[ \cos(\alpha+\mu) = 0.9659 - 0.1157 = 0.8502 \;\Longrightarrow\; \alpha+\mu = 31.77^\circ \;\Longrightarrow\; \mu = 16.8^\circ \]
Power factor and power
\[ \cos\phi \approx \frac{\cos\alpha + \cos(\alpha+\mu)}{2} = \frac{0.9659+0.8502}{2} = 0.9081 \;\Longrightarrow\; \phi = 24.8^\circ \]
\[ P_d = V_dI_d = 269.8 \times 1.2 = 323.8\;\mathrm{MW} \]
\[ Q = P\tan\phi = 323.8 \times 0.4613 = 149.4\;\mathrm{MVAr} \]

Check on the AC side: \(I_1 = 0.7797(1200) = 935.6\;\mathrm{A}\), so \(S = \sqrt3(220)(0.9356) = 356.5\;\mathrm{MVA}\) and \(P = S\cos\phi = 356.5(0.9081) = 323.7\;\mathrm{MW}\), which reproduces \(P_d\) to rounding. The reactive demand is \(46\%\) of the active power at this modest firing angle — and it grows rapidly as \(\alpha\) is increased.

3 The complete link: rectifier, line and inverter

Problem. The rectifier of Example 2 feeds an inverter through a line of resistance \(R_L = 8\;\Omega\). The inverter has \(V_{d0i} = 290\;\mathrm{kV}\), \(X_{ci}=15\;\Omega\) and operates at \(\gamma = 18^\circ\). The link is to carry \(I_d = 1200\;\mathrm{A}\). Find the inverter direct voltage, the firing angle required at the rectifier, the power at each end, the line loss, and the reactive power the inverter absorbs.

Solution. Work from the inverter backwards, because \(\gamma\) is fixed by the CEA controller.

Inverter voltage
\[ V_{di} = V_{d0i}\cos\gamma - R_{ci}I_d = 290(0.9511) - 14.32(1.2) = 275.8 - 17.2 = 258.6\;\mathrm{kV} \]
Rectifier voltage and firing angle
\[ V_{dr} = V_{di} + I_dR_L = 258.6 + 1.2(8) = 268.2\;\mathrm{kV} \]
\[ V_{d0r}\cos\alpha = V_{dr} + R_{cr}I_d = 268.2 + 17.2 = 285.4\;\mathrm{kV} \]
\[ \cos\alpha = \frac{285.4}{297.1} = 0.9607 \;\Longrightarrow\; \alpha = 16.1^\circ \]
Powers
\[ P_r = 268.2(1.2) = 321.9\;\mathrm{MW}, \qquad P_i = 258.6(1.2) = 310.3\;\mathrm{MW} \]
\[ P_{\text{loss}} = I_d^2R_L = (1.2)^2(8) = 11.5\;\mathrm{MW} \quad (3.6\%) \;\text{✓} \]

The inverter's displacement factor follows from the same rule used at the rectifier:

Inverter reactive demand
\[ \cos\phi_i \approx \frac{V_{di}}{V_{d0i}} = \frac{258.6}{290} = 0.892 \;\Longrightarrow\; \phi_i = 26.9^\circ \]
\[ Q_i = P_i\tan\phi_i = 310.3(0.506) = 157.1\;\mathrm{MVAr} \]

Both ends absorb reactive power — the inverter as well as the rectifier, which is the point students most often get wrong. The inverter is delivering \(310\;\mathrm{MW}\) into the AC system while simultaneously drawing \(157\;\mathrm{MVAr}\) from it, and the filters and capacitor banks at the inverter station must supply that.

Finally, note the sensitivity warned of in Section 38-4. Using the link equation with \(R_{cr}=R_{ci}\), the denominator is just \(R_L = 8\;\Omega\), so a \(1\%\) fall in the rectifier's AC voltage — reducing \(V_{d0r}\cos\alpha\) by \(2.85\;\mathrm{kV}\) — would by itself change \(I_d\) by \(2850/8 = 356\;\mathrm{A}\), nearly \(30\%\). Only the current regulator prevents that.

4 Harmonics and a single-tuned filter

Problem. For the converter of Example 2, (a) find the fundamental and the \(5\)th, \(7\)th, \(11\)th and \(13\)th harmonic currents; (b) state what changes when two such bridges are combined into a twelve-pulse group; (c) design a single-tuned \(11\)th-harmonic filter at the \(220\;\mathrm{kV}\) bus rated \(40\;\mathrm{MVAr}\) at \(50\;\mathrm{Hz}\).

Solution — (a). With \(I_d = 1200\;\mathrm{A}\),

Characteristic harmonics
\[ I_1 = \frac{\sqrt6}{\pi}(1200) = 935.6\;\mathrm{A} \]
\[ I_5 = \frac{935.6}{5} = 187.1, \quad I_7 = \frac{935.6}{7} = 133.7, \quad I_{11} = 85.1, \quad I_{13} = 72.0\;\mathrm{A} \]
\[ \mathrm{THD} = \sqrt{\sum_{h=6k\pm1}\left(\frac1h\right)^2} = 31.1\% \]

(b). The \(30^\circ\) shift between the star- and delta-fed bridges reverses the \(5\)th and \(7\)th (and \(17\)th and \(19\)th) contributions of one bridge relative to the other, so they cancel in the primary. Only \(h = 12k\pm1\) survives, and the THD falls to \(15.2\%\). The DC ripple moves from the \(6\)th to the \(12\)th order, where the smoothing reactor's impedance is twice as large.

(c). Size the capacitor from the reactive output required at fundamental frequency:

Filter capacitor
\[ C = \frac{Q_f}{\omega V^2} = \frac{40\times10^6}{(314.16)(220\times10^3)^2} = \frac{40\times10^6}{1.5205\times10^{13}} = 2.63\;\mu\mathrm{F} \]
\[ X_C = \frac{1}{\omega C} = \frac{1}{(314.16)(2.63\times10^{-6})} = 1210\;\Omega \]
Filter reactor, tuned to \(h = 11\)
\[ X_L = \frac{X_C}{h^2} = \frac{1210}{121} = 10.0\;\Omega \;\Longrightarrow\; L = \frac{10.0}{314.16} = 31.8\;\mathrm{mH} \]

Check: at the \(11\)th harmonic, \(11X_L = 110\;\Omega\) and \(X_C/11 = 110\;\Omega\), which cancel exactly, leaving only the branch resistance. With a quality factor of \(50\), \(R = 11\omega L/50 = 110/50 = 2.2\;\Omega\), so the \(85.1\;\mathrm{A}\) of \(11\)th-harmonic current dissipates \((85.1)^2(2.2) = 15.9\;\mathrm{kW}\) — negligible — while being diverted from the network. At \(50\;\mathrm{Hz}\) the branch impedance is \(1210 - 10 = 1200\;\Omega\) capacitive, so it also contributes \((220\;\mathrm{kV})^2/1200 = 40.3\;\mathrm{MVAr}\) toward the converter's \(149\;\mathrm{MVAr}\) demand.

5 Series compensation raising transfer capability

Problem. A \(400\;\mathrm{kV}\), \(300\;\mathrm{km}\) line has a reactance of \(0.32\;\Omega/\mathrm{km}\) and negligible resistance. Both ends are held at \(400\;\mathrm{kV}\) and the maximum permitted angle is \(30^\circ\). (a) Find the power transfer and the theoretical maximum. (b) Find the degree of series compensation needed to carry \(1200\;\mathrm{MW}\) at \(30^\circ\), and the MVAr rating of the capacitor bank. (c) Find the subsynchronous resonance frequency and comment.

Solution — (a). \(X_L = 0.32 \times 300 = 96\;\Omega\).

Uncompensated
\[ P = \frac{V^2\sin\delta}{X_L} = \frac{(400)^2(0.5)}{96} = \frac{80\,000}{96} = 833\;\mathrm{MW} \]
\[ P_{\max} = \frac{(400)^2}{96} = 1667\;\mathrm{MW} \]

(b). To carry \(1200\;\mathrm{MW}\) at the same \(30^\circ\), the effective reactance must fall to

Required compensation
\[ X_{\text{eff}} = \frac{V^2\sin\delta}{P} = \frac{(400)^2(0.5)}{1200} = 66.7\;\Omega \]
\[ X_C = 96 - 66.7 = 29.3\;\Omega, \qquad k = \frac{29.3}{96} = 30.6\% \]
\[ P_{\max}' = \frac{(400)^2}{66.7} = 2400\;\mathrm{MW} \quad (\text{up from } 1667) \]

The line current at \(1200\;\mathrm{MW}\) is \(I = 1200\times10^6/(\sqrt3 \times 400\times10^3) = 1732\;\mathrm{A}\), so the three-phase rating of the bank is

Capacitor rating
\[ Q_C = 3I^2X_C = 3(1732)^2(29.3) = 264\;\mathrm{MVAr} \]

(c).

Resonance check
\[ f_{er} = 50\sqrt{k} = 50\sqrt{0.306} = 27.7\;\mathrm{Hz}, \qquad f_{\text{rotor}} = 50 - 27.7 = 22.3\;\mathrm{Hz} \]

A \(22\;\mathrm{Hz}\) torque component lands squarely in the range of turbine-generator torsional modes, which for a large thermal set typically lie between \(15\) and \(45\;\mathrm{Hz}\). If a large steam unit is radially connected to this line, the fixed capacitor must be replaced by a TCSC, or split so that no single section produces a resonance near a known torsional frequency. Note how cheaply the capability was bought: \(264\;\mathrm{MVAr}\) of series capacitor has raised the transfer at fixed angle by \(44\%\) and the stability limit by the same factor — far less than the cost of a second circuit.

6 Sizing an SVC, and why a STATCOM is smaller

Problem. A \(400\;\mathrm{kV}\) bus has a short-circuit level of \(8000\;\mathrm{MVA}\). A load increase depresses its voltage to \(0.94\;\mathrm{pu}\). (a) Find the reactive injection needed to restore \(1.0\;\mathrm{pu}\). (b) Find the rating an SVC needs to deliver that at \(0.94\;\mathrm{pu}\). (c) Repeat for a STATCOM, and compare.

Solution — (a). The Thévenin reactance is \(X_s = V^2/S_{sc} = (400)^2/8000 = 20\;\Omega\), and the sensitivity of Chapter 34 gives

Reactive injection required
\[ \frac{\Delta V}{V} = \frac{Q}{S_{sc}} \;\Longrightarrow\; Q = S_{sc}\frac{\Delta V}{V} = 8000(0.06) = 480\;\mathrm{MVAr} \]

(b) SVC. An SVC is a susceptance, so its output at voltage \(V\) is \(Q = B V^2\), i.e. \(Q_{\text{rated}}\) scaled by \(V^2\) in per unit. To deliver \(480\;\mathrm{MVAr}\) at \(0.94\;\mathrm{pu}\),

SVC rating
\[ Q_{\text{rated}} = \frac{480}{(0.94)^2} = \frac{480}{0.8836} = 543\;\mathrm{MVAr} \]

(c) STATCOM. A STATCOM is limited by its converter current, so \(Q = V I_{\max}\) falls only linearly:

STATCOM rating
\[ Q_{\text{rated}} = \frac{480}{0.94} = 511\;\mathrm{MVAr} \]

The STATCOM needs \(6\%\) less rating at \(0.94\;\mathrm{pu}\) — a modest saving. But the comparison widens fast: at \(0.8\;\mathrm{pu}\) the required ratings would be \(480/0.64 = 750\) and \(480/0.8 = 600\;\mathrm{MVAr}\), a \(20\%\) difference, and at \(0.6\;\mathrm{pu}\) it is \(40\%\). Since the whole purpose of the device is to act during a voltage collapse, the STATCOM's linear derating is worth much more than these round-figure comparisons at \(0.94\;\mathrm{pu}\) suggest. A plain \(480\;\mathrm{MVAr}\) capacitor bank, for comparison, would deliver only \(480(0.94)^2 = 424\;\mathrm{MVAr}\) and could not be varied at all.

Review

Chapter Summary

Why DC

No charging current, no stability angle, asynchronous coupling, cheaper line — and no fault-current contribution.

Break-even

\(d^{*}=(T_{DC}-T_{AC})/(c_{AC}-c_{DC})\): about \(800\;\mathrm{km}\) overhead, \(100\;\mathrm{km}\) for cable.

Configurations

Bipolar is standard — half the power survives a pole failure; back-to-back has \(d=0\) and exists for the tie alone.

The bridge

\(V_{d0}=\frac{3\sqrt2}{\pi}V_{LL}=1.35V_{LL}\); \(V_d = V_{d0}\cos\alpha - R_cI_d\) with \(R_c=3X_c/\pi\).

Inverter

\(V_{di}=V_{d0i}\cos\gamma - R_{ci}I_d\); \(\gamma \ge 15^\circ\)–\(18^\circ\) or commutation failure follows.

Control

Rectifier CC, inverter CEA, handover by the current margin, VDCOL for recovery.

Twelve-pulse

The \(30^\circ\) transformer shift kills the 5th and 7th; THD falls \(31.1\% \to 15.2\%\).

VSC

Independent \(P\) and \(Q\), black start, no commutation failure, reversal by current not polarity.

Series FACTS

\(X_{\text{eff}}=X_L(1-k)\) doubles \(P_{\max}\) at \(k=0.5\); the TCSC adds vernier control and SSR neutrality.

Shunt FACTS

SVC gives \(Q\propto V^2\), STATCOM \(Q\propto V\); the UPFC controls \(V\), \(X\) and \(\delta\) together.

Practice

Practice Problems

Take \(f = 50\;\mathrm{Hz}\) unless stated otherwise, and neglect converter losses. Problems 6 to 8 combine results from more than one section.

  1. An AC scheme costs \(45\) M€ in terminals plus \(0.55\) M€/km; the DC alternative costs \(280\) M€ plus \(0.28\) M€/km. Find the break-even distance and the total cost there. If DC converter losses of \(0.7\%\) per station and a loss valuation of \(0.05\) M€ per MW per year over \(25\) years are included for a \(1500\;\mathrm{MW}\) transfer, does the break-even distance rise or fall? Justify without full arithmetic.
  2. A six-pulse bridge has \(V_{LL}=132\;\mathrm{kV}\), \(X_c = 12\;\Omega\) and operates at \(\alpha = 20^\circ\) with \(I_d = 900\;\mathrm{A}\). Find \(V_{d0}\), \(V_d\), the overlap angle, the displacement power factor and the reactive power drawn.
  3. A twelve-pulse rectifier (two bridges in series) with \(V_{d0}\) of \(250\;\mathrm{kV}\) per bridge and \(X_c = 18\;\Omega\) per bridge delivers \(1500\;\mathrm{A}\) at \(\alpha = 18^\circ\). Find the total direct voltage and the transmitted power. Repeat with one bridge bypassed and comment on what happens to the harmonics.
  4. A link has \(V_{d0r} = 320\;\mathrm{kV}\), \(V_{d0i} = 310\;\mathrm{kV}\), \(R_{cr} = R_{ci} = 12\;\Omega\), \(R_L = 10\;\Omega\), \(\alpha = 15^\circ\) and \(\gamma = 18^\circ\). Find \(I_d\) and the power at each end. Then find the new \(\alpha\) that would restore the original current if the rectifier's AC voltage fell by \(5\%\).
  5. Design single-tuned filters for the \(11\)th and \(13\)th harmonics at a \(400\;\mathrm{kV}\) converter bus, each rated \(60\;\mathrm{MVAr}\) at \(50\;\mathrm{Hz}\). Give \(C\), \(L\) and \(R\) for a quality factor of \(60\), and state the total reactive support the two branches provide.
  6. A \(220\;\mathrm{kV}\), \(200\;\mathrm{km}\) line has \(X = 0.35\;\Omega/\mathrm{km}\). Find the series compensation needed to raise \(P_{\max}\) by \(60\%\), the resulting \(f_{er}\), and the MVAr rating of the bank if the line carries \(400\;\mathrm{MW}\) at unity power factor. State whether SSR is a concern and why.
  7. A bus with a short-circuit level of \(5000\;\mathrm{MVA}\) needs its voltage held within \(\pm 2\%\) against a swing of \(\pm 250\;\mathrm{MVAr}\) in load reactive demand. Size an SVC and a STATCOM for the duty, assuming the worst case occurs at \(0.9\;\mathrm{pu}\), and state the ratio of the two ratings.
  8. Two parallel paths of reactance \(20\;\Omega\) and \(50\;\Omega\) connect the same pair of buses, and \(1000\;\mathrm{MW}\) flows between them. Find how the flow divides. Then determine the series reactance a TCSC must insert in the stronger path so that the flow divides equally, and explain why a UPFC could achieve the same result in more than one way.
Tip: almost every HVDC calculation is one of three moves. Convert AC to DC with \(V_{d0}=1.35V_{LL}\); apply the control angle with \(\cos\alpha\) or \(\cos\gamma\); subtract the commutation drop \(R_cI_d = (3X_c/\pi)I_d\). Assemble the link by walking the DC circuit from one converter to the other, adding \(I_dR_L\) along the line. And when the question is about reactive power, remember that a line-commutated converter has no choice in the matter: \(\cos\phi \approx V_d/V_{d0}\), so the reactive demand is fixed the instant the direct voltage is fixed.