Solved Problems · Set 40

HVDC, FACTS and Power Quality

Part 7 · Renewables and Grid Interface — the largest converters ever built, and the last set. Six problems closing the book where it began: a switch, a waveform, and a choice.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 40 — HVDC, FACTS and Power Quality

Sets 37 to 39 shared something beyond their subject matter. A photovoltaic array, a wind turbine and a battery all reach the grid through a controlled converter rather than through a synchronous machine bolted directly to it — and that changes the network itself.

A converter contributes almost no inertia. It limits its fault current to roughly its own rating rather than delivering several times it. And it can be instructed to do things no generator can: supply reactive power at zero real output, respond to frequency within a cycle, or move power along a cable that no AC line could use at all. This last set is about the converters built specifically to exploit that — the largest power electronics ever constructed — and it closes the book where it opened, on a switch, a waveform and a choice.

Part 7 · Chapter 30 · 6 solved problems

i Method Recap
  • HVDC removes the reactive burden of long AC lines:

    \[ I_{charging} = \omega C V\ell \quad\text{— zero for DC} \]
  • An MMC's submodule count follows the link voltage:

    \[ N = \frac{V_{dc}}{V_{SM}}\ \text{per arm}, \qquad \text{levels} = N+1, \qquad \text{total} = 6N \]
  • A shunt compensator's reactive power follows the voltage difference:

    \[ Q = \frac{3V_{ph}\left(E_{ph}-V_{ph}\right)}{X} \]

    A STATCOM's current is independent of voltage; an SVC's falls as \(V^2\).

  • Series compensation shortens the line electrically:

    \[ P = \frac{V_1V_2\sin\delta}{X_L-X_C} \]
  • And introduces an electrical resonance that can meet a shaft mode:

    \[ f_{er} = f_0\sqrt{\frac{X_C}{X_L}}, \qquad f_{complement} = f_0-f_{er} \]
  • An active filter is rated for the harmonic current alone:

    \[ I_h = I_1\cdot THD, \qquad S_{APF} = \sqrt3\,V\,I_h \]
Problem 1CoreWhen DC Beats AC

A ±500 kV bipolar HVDC link transmits 3000 MW. Find the pole current, explain the 12-pulse converter arrangement, and identify the distances at which DC becomes cheaper than AC for overhead and submarine routes.

Solution

The pole current:

\[ I = \frac{P}{2V_{pole}} = \frac{3000\times10^6}{2\left(500\times10^3\right)} = 3000\ \text{A} \]

Three kiloamps per pole, with the two poles at +500 and −500 kV so that normal operation needs no earth return. Losing one pole halves the capacity rather than losing it entirely — the reason bipolar arrangements dominate.

The converter. Line-commutated HVDC uses the six-pulse bridge of Set 14, doubled:

\[ \text{two 6-pulse bridges in series, fed by }\Delta\text{ and }Y\ \text{transformer secondaries} \]

The 30° phase shift between the secondaries cancels the 5th and 7th harmonics on the AC side, exactly as Set 11 showed for a 12-pulse rectifier. Several such units in series reach the full pole voltage, and each is a valve group that can be bypassed for maintenance.

Why DC wins over distance. Three separate effects:

EffectACDC
Cable charging current\(\omega CV\ell\) — grows with lengthzero
Conductors32
Skin effectreduces usable conductor areanone
Insulationrated for peak, used at rmsrated and used at the same value
Stability limit\(P = V_1V_2\sin\delta/X\)none — no angle
Terminal costlow — transformershigh — converter stations

The last row is why distance matters. DC has a large fixed cost at each end and a lower cost per kilometre, so the two curves cross — and where they cross depends entirely on whether the route is overhead or submarine.

The breakeven distances:

RouteBreakevenReason
Overhead line600–800 kmline cost savings must repay two converter stations
Submarine or underground cable50–80 kmcharging current is far higher in cable
Asynchronous interconnectionany distance — even zeroAC cannot do it at all

The third row is the case where distance is irrelevant. Two grids at different frequencies, or at the same nominal frequency but not synchronised, cannot be joined by AC at any length — so back-to-back HVDC stations exist with no transmission line between them, purely to make the connection.

What line commutation costs. Set 31 established the mechanism and its price:

LimitationConsequence
Consumes reactive power50–60% of the rating in filters and capacitors
Needs a strong AC systemcannot start into a dead network
Power reversal needs voltage reversalawkward for multi-terminal schemes
Large filterscharacteristic harmonics on both sides
Commutation failurean AC dip can trip the link — Set 14

The second row is decisive for offshore wind. An LCC needs an existing AC voltage to commutate against, and an offshore platform surrounded by converter-connected turbines has no such source — which is exactly why VSC-HVDC exists, and Problem 2.

What LCC retains:

\[ \text{losses} \approx 0.7\%\ \text{per station}, \qquad \text{ratings to } 12\ \text{GW at }\pm1100\ \text{kV} \]

Nothing else moves that much power. Line-commutated thyristors switch at line frequency with no switching loss and carry thousands of amps per device, so the largest links in the world — bulk hydro transmission across China and Brazil — remain LCC and will for the foreseeable future.

HVDC's economics are a fixed cost against a distance cost, and the crossover moves by an order of magnitude between overhead line and cable. Six hundred kilometres overhead, sixty in cable — and for two grids that are not synchronised, zero, because AC cannot make the connection at any length.
Answera\(I = 3000\ \text{A per pole}\)   b 12-pulse groups in series, \(\Delta\) and \(Y\) secondaries   c breakeven ~700 km overhead, ~60 km cable, zero for asynchronous links
Problem 2CoreNineteen Hundred Submodules

A VSC-HVDC link uses a modular multilevel converter at ±320 kV with 2 kV submodules. Find the submodules per arm, the total count and the number of output levels, and explain what the MMC does that a two-level converter cannot.

Solution

The submodule count:

\[ V_{dc} = 640\ \text{kV} \;\Longrightarrow\; N = \frac{640{,}000}{2000} = 320\ \text{per arm} \]
\[ \text{six arms} \;\Longrightarrow\; 6N = \mathbf{1920\ submodules} \]

Nearly two thousand independent half-bridge cells, each with its own capacitor, gate drive, local controller and communication link — and each is essentially the cascaded cell of Set 29 scaled up. The MMC is the cascaded H-bridge idea taken to its conclusion.

The output levels:

\[ \text{levels} = N+1 = 321 \]

Three hundred and twenty-one voltage levels per phase. Each step is 2 kV out of 640 — 0.3% — so the output is sinusoidal to well within any harmonic limit.

What that buys:

PropertyTwo-level VSCMMC
Output THDhigh — needs large filters< 1% — no AC filters at all
\(dv/dt\)\(V_{dc}\) per transition2 kV per step
Device switching frequency~1 kHz~100 Hz — each cell switches rarely
Losses~1.5% per station~1% per station
Series device matchingcritical — hundreds in seriesnot required
Scalabilitylimited by series connectionadd cells
Redundancynonespare cells bypass on failure

The fifth row is what made VSC-HVDC practical. A two-level converter at 640 kV needs hundreds of IGBTs in series switching within nanoseconds of each other — the string problem of Set 5, at a scale where a single mismatch destroys the valve. The MMC sidesteps it: each cell manages its own 2 kV independently, and they need not switch together at all.

What VSC does that LCC cannot:

CapabilityWhy it matters
Black start into a dead networkself-commutating — needs no existing AC voltage
Independent P and Q controlacts as a STATCOM as well as a link
Power reversal without voltage reversalenables multi-terminal DC grids
Connects to weak AC systemsno commutation failure mode
Compactno large filters or capacitor banks

The first row is exactly the offshore wind requirement from Set 38. An offshore platform surrounded by converter-connected turbines has no synchronous source to commutate against, so the converter must create the AC voltage itself — then the turbines synchronise to it. Only a VSC can do this.

The MMC's own problems:

ChallengeSolution
Capacitor voltage balancingsort cells by voltage; insert those that charge in the needed direction
Circulating current between armsa dedicated suppression controller
Communication to 1920 cellsoptical fibre, hierarchical
Capacitor energy storagetens of kJ per MVA — large cells
DC fault currenthalf-bridge cells cannot block it — see below

The first row is the MMC's defining control task and it has an elegant solution: at each switching decision, sort the arm's cells by capacitor voltage and insert whichever ones the current direction will move towards balance. The redundancy that Set 28's zero states provided appears here as a choice of which cells to use.

The DC fault problem, which is genuinely hard:

\[ \text{a half-bridge cell's antiparallel diodes conduct regardless of the gates} \]

On a DC-side short circuit, the AC system feeds the fault through those diodes as an uncontrolled rectifier — and the converter cannot stop it. Full-bridge cells can block DC fault current but cost twice the devices and more loss; the alternative is a DC circuit breaker, which must interrupt thousands of amps with no natural current zero. Both are active areas of development, and it is the main obstacle to true multi-terminal DC grids.

The MMC solved the series-device problem by refusing to have one. Instead of hundreds of IGBTs switching in unison at 640 kV, 1920 independent 2 kV cells each switch about a hundred times a second — giving 321 output levels, no AC filters, and a converter that can create a grid from nothing.
Answera 320 submodules per arm, 1920 total   b 321 levels, THD < 1% with no filter   c black start, independent P and Q, connection to weak grids
Problem 3Exam levelHolding the Voltage Up

A 50 MVAr STATCOM connects to a 33 kV bus through a 0.15 pu reactance. Find the converter voltage needed for full capacitive output and the current, then compare its behaviour with an SVC as the bus voltage collapses.

Solution

The base quantities:

\[ Z_{base} = \frac{V^2}{S} = \frac{33{,}000^2}{50\times10^6} = 21.78\ \Omega \]
\[ X = (0.15)(21.78) = 3.267\ \Omega, \qquad V_{ph} = \frac{33{,}000}{\sqrt3} = 19.05\ \text{kV} \]

The mechanism. A STATCOM is a voltage source behind a reactance, exactly like a synchronous condenser:

\[ Q = \frac{3V_{ph}\left(E_{ph}-V_{ph}\right)}{X} \]
\[ E_{ph}-V_{ph} = \frac{QX}{3V_{ph}} = \frac{\left(50\times10^6\right)(3.267)}{3\left(19{,}053\right)} = 2858\ \text{V} \]
\[ E_{ph} = 21.91\ \text{kV} \;\Longrightarrow\; E_{LL} = 37.95\ \text{kV} = \mathbf{1.15\ pu} \]

Fifteen per cent above the bus voltage delivers 50 MVAr into it. Making \(E\) lower than \(V\) by the same amount absorbs 50 MVAr — symmetric operation with no change of hardware, which is one of the STATCOM's advantages over a switched capacitor bank.

The current:

\[ I = \frac{Q}{\sqrt3\,V} = \frac{50\times10^6}{\sqrt3\left(33{,}000\right)} = 875\ \text{A} \]

Which is also the converter's rating, and here is the key point: that current is set by the converter's own capability, not by the bus voltage.

The comparison that matters. Consider what each device does as the bus voltage falls — the situation in which reactive support is most needed:

\[ \text{SVC (a controlled susceptance): } Q = BV^2 \;\Longrightarrow\; Q \propto V^2 \]
\[ \text{STATCOM (a controlled current source): } Q = \sqrt3VI \;\Longrightarrow\; Q \propto V \]
Bus voltageSVC outputSTATCOM output
1.0 pu50 MVAr50 MVAr
0.8 pu32 MVAr40 MVAr
0.5 pu12.5 MVAr25 MVAr
0.3 pu4.5 MVAr15 MVAr

At half voltage the STATCOM delivers twice what an SVC can. An SVC is a capacitor, and a capacitor's current falls with the voltage across it — so precisely when the network is collapsing, an SVC withdraws its support. A STATCOM holds its rated current down to very low voltage and keeps pushing.

The full comparison:

PropertySVCSTATCOM
Technologythyristor-switched L and CVSC with a DC capacitor
Output at low voltage\(\propto V^2\)\(\propto V\)
Response time~20 ms~5 ms
Footprintlarge — reactors and capacitor bankscompact
Harmonicsneeds filtersmultilevel — minimal
Active powernonepossible with storage
Costlowerhigher

The sixth row hints at the modern direction: add a battery to a STATCOM's DC link and it becomes a device that supports voltage and frequency — combining this problem with Set 39.

The FACTS family, for orientation:

DeviceConnectionControls
SVCshuntvoltage — thyristor-based
STATCOMshuntvoltage — VSC-based
TCSCseriesline reactance — Problem 4
SSSCseriesinjected series voltage
UPFCbothvoltage, angle and impedance together

All five are converters placed on a transmission system to control quantities that were historically fixed by its geometry. The unified power flow controller of the last row is the most general — a shunt and a series converter sharing a DC link, able to control all three parameters independently.

A capacitor abandons the network exactly when it is needed. An SVC's output falls as \(V^2\) during the voltage collapse it exists to prevent; a STATCOM, being a current source, holds twice as much at half voltage. That single difference is why voltage support has moved from switched capacitors to converters.
Answera\(E = 1.15\ \text{pu} = 37.95\ \text{kV}\)   b\(I = 875\ \text{A}\)   c at 0.5 pu the STATCOM gives 25 MVAr against the SVC's 12.5
Problem 4Exam levelShortening the Line

A transmission line is series-compensated to 50%. Find the effect on power transfer capability, compute the electrical resonant frequency and its complement, and explain the sub-synchronous resonance hazard and how a TCSC addresses it.

Solution

The power transfer relation. For a line between two buses:

\[ P = \frac{V_1V_2\sin\delta}{X} \]

Inserting a series capacitor reduces the effective reactance:

\[ X_{eff} = X_L-X_C = X_L(1-k), \qquad k = \frac{X_C}{X_L} = 0.5 \]
\[ P_{new} = \frac{P_{old}}{1-0.5} = \mathbf{2P_{old}} \]

The line's capability doubles — without a single new tower, conductor or right of way. Series compensation is the cheapest capacity increase available on a transmission system, which is exactly why it is used despite what follows.

The additional benefits:

BenefitMechanism
Higher transfer limitlower effective \(X\)
Improved transient stabilitya larger synchronising torque coefficient
Better voltage regulationless reactive drop along the line
Load sharing between parallel pathscompensate the one to be favoured
Self-regulating\(X_C\)'s compensation grows with the current

The last row is elegant: the capacitor's voltage rises with the line current, so its compensating effect automatically increases exactly when the line is most heavily loaded and most needs it.

Now the hazard. A series capacitor and the line's inductance form a series resonant circuit:

\[ f_{er} = f_0\sqrt{\frac{X_C}{X_L}} = (50)\sqrt{0.5} = 35.4\ \text{Hz} \]

Below the power frequency — hence sub-synchronous. A disturbance excites a current at 35.4 Hz in the electrical network, and that current interacts with the generator rotor.

The complement is what reaches the shaft:

\[ f_{complement} = f_0-f_{er} = 50-35.4 = 14.6\ \text{Hz} \]

A 35.4 Hz stator current produces a torque component at \(50-35.4 = 14.6\) Hz on the rotor. A turbine-generator shaft is not a rigid body — it is a chain of masses on a torsional spring, with natural modes typically between 10 and 50 Hz. If one of them lies near 14.6 Hz, the electrical and mechanical systems exchange energy at that frequency.

And the exchange can grow. Three distinct phenomena:

PhenomenonMechanismSeverity
Induction generator effectnegative resistance at \(f_{er}\)growing electrical oscillation
Torsional interactiona shaft mode is negatively dampedgrowing shaft oscillation
Torque amplificationa fault transient hits a shaft modeimmediate shaft damage

The consequences are mechanical and permanent. The Mohave incidents of the 1970s cracked two generator shafts before the mechanism was understood, and SSR studies have been mandatory for series-compensated systems ever since.

The TCSC solution. Placing a thyristor-controlled reactor in parallel with the series capacitor makes the reactance adjustable:

\[ X_{TCSC}(\alpha)\ \text{— continuously variable, and \emph{apparently inductive} at sub-synchronous frequencies} \]
TCSC capabilityValue
Variable compensationmatch the level to the loading
Power flow controlsteer flow between parallel paths
SSR mitigationappears resistive-inductive below \(f_0\) — damps rather than excites
Damping of power swingsmodulate to oppose the oscillation
Detuningshift \(f_{er}\) away from a known shaft mode

The third row is the important one and is not obvious: the thyristor-controlled branch makes the assembly behave as a capacitor at 50 Hz — where compensation is wanted — and not as a capacitor at 35 Hz, where it would resonate. One device, two different characters at two frequencies.

Halving a line's reactance doubles its capability and creates an electrical resonance below the power frequency. That resonance appears on the shaft at its complement — 14.6 Hz here — where turbine torsional modes live, and the resulting exchange has cracked generator shafts. A TCSC is a capacitor at 50 Hz that deliberately is not one at 35.
Answera power transfer doubles   b\(f_{er} = 35.4\ \text{Hz}\), complement 14.6 Hz   c SSR if a shaft mode is near 14.6 Hz; a TCSC detunes it
Problem 5ChallengeCancelling the Harmonics

A 415 V rectifier load draws 100 A of fundamental current at 30% THD. Size a shunt active filter to correct it, determine the switching frequency needed to cancel up to the 25th harmonic, and compare with the passive filter of Set 32.

Solution

The load's harmonic content:

\[ I_h = I_1\cdot THD = (100)(0.30) = 30\ \text{A rms of harmonics} \]
\[ I_{rms} = I_1\sqrt{1+THD^2} = (100)\sqrt{1.09} = 104.4\ \text{A} \]

The load is a capacitor-input rectifier of the kind Set 10 analysed — the harmonics are not incidental, they are what such a load draws by construction.

The active filter's principle. Measure the load current, extract its harmonic content, and inject the negative of it:

\[ i_{source} = i_{load}+i_{APF} = i_{load}-i_{h,load} = i_1 \]

The supply then sees a purely sinusoidal current in phase with the voltage, while the harmonics circulate between the load and the filter. It is an inverter — the same six switches as Set 26 — controlled to produce a specific distorted current rather than a sinusoid.

The rating:

\[ S_{APF} = \sqrt3\,V\,I_h = \sqrt3(415)(30) = 21.6\ \text{kVA} \]
\[ S_{load} = \sqrt3(415)(104.4) = 75.0\ \text{kVA} \]
\[ \frac{21.6}{75.0} = \mathbf{29\%\ of\ the\ load\ rating} \]

Under a third — because the filter handles only the harmonic current, never the fundamental. That is the same partial-rating argument that made the DFIG attractive in Set 38, and it is what makes active filtering economically viable.

The bandwidth requirement. To cancel a harmonic, the filter must be able to produce it:

\[ 25\text{th harmonic} = (25)(50) = 1250\ \text{Hz} \]
\[ f_s \ge 10\times1250 = 12.5\ \text{kHz} \quad\to\quad \text{use } 20\ \text{kHz} \]

A factor of ten between the switching frequency and the highest harmonic to be cancelled is the practical rule — the current loop needs that much margin to track. Cancelling to the 50th would require 25 kHz of bandwidth and 50 kHz of switching, which at this power is demanding.

Against the passive filter of Set 32:

PropertyPassive (tuned)Active
Harmonics addressedone per branchall, simultaneously
Tuning driftages; must be detuned deliberatelynone
Resonance with the supplya real hazard — Sets 21 and 32none
Adapts to load changesnoyes
Reactive compensationyes — a by-productyes — on command, either sign
Lossesvery low~2% of its rating
Costlowhigh
Overload behaviourabsorbs whatever arriveslimits at its rating

Rows two and three are the passive filter's fundamental weaknesses, and they were the subject of Set 32, Problem 4: a tuned branch drifts upward in frequency as it ages and can become an amplifying parallel resonance. An active filter has no resonance to drift and no tuning to maintain.

The hybrid answer, which is what large installations actually use:

\[ \text{passive branch for the 5th and 7th} + \text{small active filter for the rest} \]

The passive branches carry the bulk of the harmonic current cheaply, and a much smaller active filter handles the residual, the higher orders, and — importantly — damps the passive branch's resonance. The active unit can then be rated at perhaps 10% of the load rather than 29%.

What the standards require:

StandardApplies toTypical limit
IEEE 519the point of common coupling5% current TDD for a typical \(I_{sc}/I_L\)
IEC 61000-3-2equipment below 16 Aabsolute limits per harmonic
IEC 61000-3-12equipment 16–75 Arelative to short-circuit ratio
G5/5 (UK)connection agreementsvoltage distortion at the PCC

Note that IEEE 519 applies at the point of common coupling — the shared connection — not at each item of equipment. It is a limit on what a customer injects into the network, which is why the responsibility falls on the installation as a whole and why site-level filtering is the usual answer.

An active filter handles only the harmonic current, so it costs a third of the load's rating. That partial rating is the same argument that justified the DFIG's converter — and unlike a tuned passive branch, it cannot drift into a resonance, adapt to nothing, or amplify the very harmonic it was installed to remove.
Answera\(I_h = 30\ \text{A},\ S_{APF} = 21.6\ \text{kVA}\) — 29% of the load   b\(f_s \ge 12.5\ \text{kHz}\) for the 25th   c no tuning drift and no resonance, unlike a passive filter
Problem 6ChallengeA Grid Made of Converters

As synchronous generation is displaced by converter-connected sources, three properties of the power system change fundamentally. Identify them, explain the consequences, and describe what grid-forming control does about it — then trace how every part of this book contributed.

Solution

What a synchronous machine provides for free:

PropertyMechanism
Inertiaa spinning mass resists frequency change — instantly, physically
Fault current5–7× rated, sustained — operates protection
Voltage source behaviouran EMF behind a reactance — sets the grid's voltage and angle
Reactive capabilityfrom field excitation
Harmonic sinklow impedance at harmonic frequencies

None of the first three is designed in — they are consequences of the machine's physics, and the entire power system was built assuming they would always be present.

What a conventional converter provides instead:

PropertyConverterConsequence
Inertiaessentially nonefaster frequency excursions
Fault current~1.1× ratedprotection may not detect a fault
Behaviourcurrent source, following the gridneeds a grid to follow
Reactive capabilityexcellent, instantaneousa genuine gain
Harmonicsa source, not a sinkneeds filtering — Set 32

The third row is the deepest problem. A grid-following converter uses a phase-locked loop to synchronise to an existing voltage — and if every source is grid-following, there is nothing for any of them to follow. The system cannot start, and at high converter penetration it becomes unstable well before that point.

The consequences, quantified:

\[ \text{RoCoF} = \frac{\Delta P}{2H S_{base}}f_0 \]

The rate of change of frequency after a generation loss is inversely proportional to system inertia \(H\). As synchronous plant retires, \(H\) falls, and the same loss produces a steeper frequency fall — giving protection and response services less time to act, and risking cascading disconnection of generation on RoCoF protection.

Grid-forming control reverses the converter's role:

Grid-followingGrid-forming
Behaves ascurrent sourcevoltage source behind a reactance
SynchronisationPLL tracks the gridits own internal oscillator
Needs an existing gridyesno
Black startnoyes
Provides inertiasynthetic onlyinherent in the control law
Supports other convertersnoyes — forms the reference

A grid-forming converter emulates a synchronous machine's behaviour in software — an internal angle that responds to power imbalance with a chosen inertia constant. It still needs energy behind it to deliver that inertia, which is where Set 39's storage comes in.

The converter-dominated toolkit that replaces what was lost:

Lost propertyReplacementWhere in this book
Inertiasynthetic inertia, grid-forming controlSets 38, 39
Fault currentnew protection philosophies; synchronous condensersthis set
Voltage sourcegrid-forming convertersthis problem
Reactive supportSTATCOM, and every inverterProblem 3
Frequency responsestorage, demand response, V2GSet 39
Long-distance transferHVDCProblems 1 and 2
Power qualityactive filtersProblem 5

Every entry in the middle column is a power electronic converter. The transition is not from one generation technology to another — it is from a grid whose properties emerged from spinning iron to one whose properties are chosen, in software, by the engineers who write the control laws.

Which is where this book has been going. Trace it back:

PartContribution
1 — Devicesthe switch, its losses and its thermal limit
2 — RectifiersAC to DC, and the harmonics it costs
3 — DC–DCvolt-second balance, and control loops around it
4 — InvertersDC to AC, modulation, and the hexagon
5 — AC–AC and EMCdirect conversion, and the parasitics that follow every edge
6 — Drivesa reference frame, and a machine that obeys it
7 — Gridthe same converters, at the scale of a network
\[ \text{Set 1: } V_{out} = DV_{in} \;\longrightarrow\; \text{Set 40: } 1920\ \text{submodules at }640\ \text{kV} \]

The MMC of Problem 2 is 1920 switching cells, each obeying the volt-second balance of Set 1, arranged by the multilevel argument of Set 29, modulated by the vector reasoning of Set 28, protected by the thermal analysis of Set 8 and filtered by the principles of Set 32. Nothing in this last set required an idea that was not already built.

The grid's properties used to emerge from spinning iron; now they are chosen. Inertia, fault current and voltage-source behaviour were free consequences of a synchronous machine's physics, and every one of them must now be written into a control law — which makes the converter designer responsible for the stability of the network itself.
Answera inertia, fault current and voltage-source behaviour are lost   b RoCoF steepens as \(H\) falls   c grid-forming control makes the converter a voltage source with a chosen inertia
Formulas

Key Formulas

QuantityRelationNotes
HVDC pole current\(I = P/(2V_{pole})\)3000 A
Cable charging\(I = \omega CV\ell\)Zero for DC
Breakeven distance600–800 km overhead; 50–80 km cableZero if asynchronous
MMC submodules\(N = V_{dc}/V_{SM}\) per arm320 per arm
Total submodules\(6N\)1920
Output levels\(N+1\)321 — no AC filter
Shunt compensator\(Q = \dfrac{3V_{ph}(E_{ph}-V_{ph})}{X}\)\(E = 1.15\) pu
SVC output\(Q \propto V^2\)Collapses with voltage
STATCOM output\(Q \propto V\)Holds rated current
Compensated transfer\(P = \dfrac{V_1V_2\sin\delta}{X_L(1-k)}\)Doubles at \(k = 0.5\)
Electrical resonance\(f_{er} = f_0\sqrt{X_C/X_L}\)35.4 Hz
Complement\(f_0-f_{er}\)14.6 Hz — the shaft sees this
Active filter current\(I_h = I_1\cdot THD\)30 A
Active filter rating\(\sqrt3VI_h\)29% of the load
Bandwidth rule\(f_s \ge 10f_{h,max}\)12.5 kHz for the 25th
RoCoF\(\dfrac{\Delta P}{2HS_{base}}f_0\)Steepens as \(H\) falls
Pitfalls

Common Mistakes

  1. Quoting one breakeven distance for HVDC. Overhead and cable differ by an order of magnitude — Problem 1.

  2. Proposing LCC-HVDC for offshore wind. It cannot commutate without an existing AC source — Problems 1 and 2.

  3. Counting MMC submodules per leg rather than per arm. Six arms, not three — Problem 2.

  4. Assuming an MMC can block DC faults. Half-bridge cells conduct through their diodes regardless of the gates — Problem 2.

  5. Treating a STATCOM and an SVC as equivalent. One holds current, the other collapses as \(V^2\) — Problem 3.

  6. Adding series compensation without an SSR study. The complement frequency lands where shaft modes live — Problem 4.

  7. Confusing \(f_{er}\) with the frequency seen by the shaft. The shaft sees \(f_0-f_{er}\) — Problem 4.

  8. Rating an active filter for the full load current. It carries only the harmonic component — Problem 5.

  9. Choosing a switching frequency too close to the highest harmonic. A factor of ten is needed — Problem 5.

  10. Assuming a grid of grid-following converters can operate. They all need something to follow — Problem 6.

The End of the Series

Forty sets and two hundred and forty-one problems. It began with a single switch, a duty ratio and the observation that an inductor's voltage must average to zero. It ends with 1920 switching cells at 640 kilovolts, obeying that same rule.

The through-line was never the topologies. It was that a small number of ideas keep reappearing wearing different clothes: volt-second balance became dwell-time calculation on a hexagon; the string problem of series thyristors became the reason for cascaded cells; the 31.08% distortion of a 120° quasi-square wave turned up as a rectifier's current, an inverter's voltage and a current-source inverter's output because they are the same waveform seen from different terminals; and the cascade of a fast current loop inside a slow outer loop served a DC machine, an induction machine, a magnet machine and a grid-tied inverter without alteration.

The recurring engineering lesson was that every gain is bought. Faster switching improved every efficiency figure in this book and worsened every problem in Set 32. Permanent magnets removed the magnetising current and removed the ability to switch the field off. A DC link bought voltage boost, ride-through and simple commutation, and cost a capacitor that is the shortest-lived component in any converter. Nothing here was free, and the skill was always in choosing which price to pay.

And the last problem closed a circle. The properties that made the power system work — inertia, fault current, a voltage to synchronise to — were free consequences of spinning iron, and are now written into control laws. The converter designer has become responsible for the behaviour of the network itself. Every one of the forty sets is a piece of how that is done.

Back to the full index for all forty problem sets, or return to Set 1 to start again with a switch and a duty ratio.