Solved Problems · Set 39

Energy Storage and EV Charging

Part 7 · Renewables and Grid Interface — the technology both renewables depend on. Six problems on batteries, bidirectional converters and charging.

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

Set 39 — Energy Storage and EV Charging

Set 38 rode through a grid fault by storing half a megajoule in a spinning rotor for a sixth of a second. Extend that idea from milliseconds to hours and from kinetic to electrochemical, and it becomes the technology both renewable sources ultimately depend on — because both produce power when the resource is available rather than when it is wanted.

A battery is a source unlike any so far. It is bidirectional, so the converter must be too. Its voltage swings 40% with state of charge. It degrades according to how it is used, so the control strategy is part of the product's lifetime. And at the currents modern charging demands — hundreds of amps — the losses that were negligible elsewhere become the design's dominant term.

Part 7 · Chapter 29 · 6 solved problems

i Method Recap
  • A pack is defined by its series and parallel counts:

    \[ V_{pack} = N_sV_{cell}, \qquad E = V_{pack}Q, \qquad \text{C-rate} = \frac{I}{Q} \]
  • A bidirectional half bridge is a boost one way and a buck the other:

    \[ D_{boost} = 1-\frac{V_{bat}}{V_{link}}, \qquad \Delta I = \frac{V_{bat}D}{f_sL} \]
  • Interleaving \(n\) phases divides the current and cancels the ripple:

    \[ I_{phase} = \frac{I}{n}, \qquad \Delta I_{total} \ll n\Delta I_{phase} \]
  • A dual active bridge transfers power by phase shift:

    \[ P = \frac{V_1V_2'\,\phi\left(\pi-\left|\phi\right|\right)}{2\pi^2f_sL} \]
  • Doubling the system voltage quarters the conduction loss:

    \[ I = \frac{P}{V}, \qquad P_{loss} \propto I^2 \;\Longrightarrow\; \text{loss} \propto \frac{1}{V^2} \]
  • Pre-charge limits the inrush into the link capacitance:

    \[ \tau = RC, \qquad E_R = \tfrac12CV^2, \qquad t \approx 3\tau \]
Problem 1CoreThe Pack and How to Fill It

An EV pack uses 96 lithium cells in series, nominal 3.7 V, 3.0–4.2 V range, storing 75 kWh. Find the pack voltages, the capacity in ampere-hours, the current and C-rate for a 150 kW DC fast charge, and describe the CC–CV profile and why it exists.

Solution

The pack voltages:

\[ V_{nom} = (96)(3.7) = 355\ \text{V} \]
\[ V_{max} = (96)(4.2) = 403\ \text{V}, \qquad V_{min} = (96)(3.0) = 288\ \text{V} \]
\[ \frac{403}{288} = 1.40 \]

A 40% voltage swing from empty to full — wider than the PV string's temperature range in Set 37, and every converter connected to the pack must work across all of it. That range is the single most demanding specification a battery converter faces.

The capacity:

\[ Q = \frac{75{,}000\ \text{Wh}}{355\ \text{V}} = 211\ \text{Ah} \]

Which for a 96-series pack means each cell is 211 Ah — achieved either with large-format cells or by paralleling many small ones. The choice affects the balancing problem, the thermal design and the failure modes, but not the electrical interface.

The fast-charge current:

\[ I = \frac{150{,}000}{355} = 423\ \text{A} \]
\[ \text{C-rate} = \frac{423}{211} = \mathbf{2C} \]

Two C — the whole pack's energy delivered in half an hour. Four hundred amps is what makes charging cables heavy and, above about 200 A, liquid-cooled. It also makes every milliohm in the path matter: 1 mΩ of contact resistance dissipates 179 W.

The CC–CV profile:

PhaseControlEnds whenSoC reached
Constant currenthold \(I\); voltage rises\(V_{cell} = 4.2\) V~80%
Constant voltagehold \(V\); current tapers\(I < Q/20\)100%

The split explains the familiar charging curve. The first 80% arrives at full power; the last 20% takes as long again as the current tapers away. That is why manufacturers quote "10 to 80%" times — not marketing, but the physics of the CV phase.

Why the voltage must be held rather than the current. The cell voltage limit is a chemical boundary:

\[ V_{cell} > 4.2\ \text{V} \;\Longrightarrow\; \text{electrolyte oxidation, lithium plating, gas evolution} \]

Exceeding it does not merely reduce life — it deposits metallic lithium on the anode, which is both a permanent capacity loss and a route to an internal short. The CV phase exists because the cell's terminal voltage includes its internal \(IR\) drop, so holding a safe terminal voltage while the current falls lets the true cell potential approach the limit asymptotically.

Cell balancing. Series cells are never identical:

\[ \text{the weakest cell reaches } 4.2\ \text{V first and stops the charge for all 96} \]
MethodMechanismTrade
Passivebleed the high cells through a resistorsimple; wastes energy, slow
Active (capacitive)shuttle charge between cellsefficient; more components
Active (inductive)a small converter per cell groupfastest; most expensive

A 2% spread across 96 cells means the pack delivers 2% less than the sum of its cells until balanced. Passive balancing at a few hundred milliamps takes hours — which is acceptable, because a parked vehicle has hours.

The pack's 40% voltage swing is the specification that shapes every converter around it. A PV string moved 50% across a year and a battery moves 40% within a single discharge — so a battery converter must hold full power at 288 V and at 403, which is why buck and boost capability in one topology is the norm rather than the exception.
Answera 288–403 V, nominal 355 V   b\(Q = 211\ \text{Ah}\)   c 423 A = 2C   d CC to 80%, then CV as the current tapers
Problem 2CorePower in Both Directions

A bidirectional converter links the pack to a 700 V DC link at 150 kW. Find the duty ratio range, size the inductor for 20% ripple at 20 kHz, and show what four-phase interleaving achieves.

288– 403 V L 700 V link discharge: boost → ← charge: buck
One leg, two directions — boost discharging, buck charging
Solution

The topology. A single half-bridge leg with an inductor on the battery side serves both directions:

DirectionOperates asActive device
Dischargingboost (battery → link)lower switch
Chargingbuck (link → battery)upper switch

The same two devices, the same inductor, the same duty ratio relationship — only which device is modulated changes. It is the two-quadrant chopper of Set 17 with the machine replaced by a battery, and the transition between directions is seamless because the inductor current simply crosses zero.

The duty ratio range:

\[ D = 1-\frac{V_{bat}}{V_{link}} \]
State of charge\(V_{bat}\)\(D\)
Empty288 V0.589
Nominal355 V0.493
Full403 V0.424

A comfortable range around 0.5 — nowhere near the extremes where a boost converter's efficiency and right-half-plane zero cause trouble. Choosing the link voltage well above the pack's maximum is what buys that comfort.

The inductor, for 20% ripple at the worst case:

\[ I = \frac{150{,}000}{355} = 423\ \text{A}, \qquad \Delta I = (0.2)(423) = 84.5\ \text{A} \]
\[ L = \frac{V_{bat}D}{f_s\Delta I} = \frac{(355)(0.493)}{\left(20\times10^3\right)(84.5)} = 104\ \mu\text{H} \]

Only a hundred microhenries — but carrying 423 A. That combination is what makes it hard: the core must not saturate at 465 A peak, and the winding must carry 423 A with the skin and proximity effects of Set 24 at 20 kHz. It is the exact mirror of the PV inductor in Set 37 — 4.6 mH at 8.9 A.

Interleaving four phases attacks both difficulties at once:

\[ I_{phase} = \frac{423}{4} = 106\ \text{A} \]
BenefitDetail
Current per phase quartered106 A — standard devices and inductors
Ripple cancellationphases shifted 90° — total ripple far below \(4\Delta I\)
Effective frequency \(4\times\)80 kHz at the link — smaller capacitors
Conduction loss\(4\times(I/4)^2R = I^2R/4\)quartered
Thermal spreadingfour heat sources, not one
Phase sheddingrun fewer phases at light load for better efficiency

Set 18 met ripple cancellation in a two-phase buck; here it is doing four jobs at once. The last row is particularly valuable in a vehicle, where the converter spends most of its life far below rated power — shutting down three phases removes their switching and gate-drive losses entirely.

Where the ripple cancellation is perfect:

\[ n\ \text{phases: total ripple} \to 0\ \text{at } D = \frac{k}{n},\ k = 1\dots n-1 \]
\[ n = 4:\ \text{nulls at } D = 0.25,\ 0.50,\ 0.75 \]

And the operating range of 0.42–0.59 straddles the null at \(D = 0.5\) — so the input ripple is near zero at nominal state of charge and modest at the extremes. That is not a coincidence; the link voltage was chosen to put the nominal duty ratio there.

The device choice:

RequirementValue
Blocking voltage700 V + margin → 1200 V
Current per device106 A phase, 120 A peak
TechnologySiC MOSFET
Why SiClower loss, higher \(f_s\), smaller magnetics, no tail current
Body diodeused in synchronous rectification — check \(Q_{rr}\)

Silicon carbide dominates this application, and Set 39's numbers show why: at 423 A the conduction loss is everything, and a SiC MOSFET's resistive conduction beats an IGBT's fixed \(V_{CE(sat)}\) handsomely at partial load — which is where a vehicle lives.

A battery converter is the mirror image of a PV converter. Solar gave 4.6 mH at 8.9 A; a battery gives 104 µH at 423 A. The same equations produce opposite engineering problems — and interleaving four phases is what makes the high-current case buildable at all.
Answera\(D = 0.42\text{–}0.59\)   b\(L = 104\ \mu\text{H}\) at 423 A   c four phases give 106 A each, quartered loss, ripple null at \(D = 0.5\)
Problem 3Exam levelThe Onboard Charger

An 11 kW onboard charger uses an active front end and an isolated dual active bridge from a 700 V link to the 400 V pack at 100 kHz. Find the turns ratio and the transformer leakage inductance for a 45° phase shift at rated power, and explain why a DAB rather than an LLC.

Solution

The charger's two stages:

\[ \text{AC grid} \to \text{PFC rectifier} \to 700\ \text{V} \to \text{isolated DC–DC} \to \text{pack} \]
StageFunctionWhy
Active front endsinusoidal input current, unity PFSet 10's diode bridge would draw 100% THD
Isolated DC–DCgalvanic isolation, wide voltage rangesafety — the pack must float

Isolation is a safety requirement, not a convenience: an unisolated charger would connect the pack's chassis-referenced structure directly to the mains. Set 22's isolated topologies exist for exactly this.

The turns ratio:

\[ n = \frac{V_1}{V_2} = \frac{700}{400} = 1.75 \]
\[ V_2' = nV_2 = 700\ \text{V (referred to the primary)} \]

Choosing \(n\) so that the referred voltages match at nominal is deliberate. A DAB is most efficient when \(V_1 = V_2'\) — the circulating current grows as they diverge, which is why the pack's 40% swing is a real efficiency problem for this topology.

The power transfer law. A DAB moves power by phase-shifting one bridge's square wave against the other's:

\[ P = \frac{V_1V_2'\,\phi\left(\pi-\left|\phi\right|\right)}{2\pi^2f_sL} \]

The inductor \(L\) — usually just the transformer's leakage — is the energy transfer element, exactly as it was in the LLC of Set 29. Power flows from the leading bridge to the lagging one, so reversing the sign of \(\phi\) reverses the power. That is the whole bidirectional mechanism: one variable, both directions.

Size the inductance for rated power at a 45° shift:

\[ \phi = \frac{\pi}{4}:\quad \phi(\pi-\phi) = (0.785)(2.356) = 1.851 \]
\[ L = \frac{V_1V_2'\left[\phi(\pi-\phi)\right]}{2\pi^2f_sP} = \frac{(700)(700)(1.851)}{2\pi^2\left(100\times10^3\right)(11{,}000)} \]
\[ L = 41.8\ \mu\text{H} \]

Forty-two microhenries, which a transformer of this rating can provide as leakage alone — so no external inductor is needed. As with the LLC, a parasitic has become a design element.

Why 45° and not more. The power law peaks at \(\phi = \pi/2\):

\(\phi\)Relative powerNote
30°0.74
45°0.94design point — headroom retained
60°1.00
90°1.13 (maximum)no control margin; high circulating current

Designing at 90° would leave no headroom — any dip in link voltage could not be compensated. Forty-five degrees keeps a comfortable margin while staying in the region where circulating current, and therefore conduction loss, is modest.

DAB against the LLC of Set 29:

PropertyDABLLC
Bidirectionalinherently — reverse \(\phi\)no (CLLC needed)
Control variablephase shiftfrequency
Switching frequencyfixedvariable
Wide voltage rangemoderate — ZVS is lost off-nominalgood
Peak efficiency~97%~98%
Circulating currenthigher off-nominallower

The first row decides it for a charger that must also support vehicle-to-grid. A DAB reverses power by changing the sign of one variable; an LLC's rectifier is a diode bridge and cannot. The CLLC — a symmetric resonant tank on both sides — is the LLC's bidirectional answer, and modern chargers increasingly use it.

The complete 11 kW specification:

ItemValue
Input3-phase 400 V, 16 A
Front end6-switch active rectifier, THD < 5%
Link700 V
IsolationDAB or CLLC at 100 kHz
Turns ratio1.75
Leakage inductance42 µH
Output288–403 V, up to 38 A
Efficiency~95% overall (two stages)
DevicesSiC throughout
One variable, both directions. A dual active bridge transfers power by phase-shifting two square waves against a leakage inductance, so reversing the sign of the shift reverses the power — which is why vehicle-to-grid capability costs nothing in a DAB and requires a different topology in an LLC.
Answera\(n = 1.75\)   b\(L = 41.8\ \mu\text{H}\) at \(\phi = 45^\circ\)   c DAB is inherently bidirectional; an LLC is not
Problem 4Exam levelWhy 800 Volts

Compare 400 V and 800 V vehicle architectures for a 350 kW charge: find the currents and the relative conduction loss, and identify what else changes and what it costs.

Solution

The currents:

\[ 400\ \text{V}:\ I = \frac{350{,}000}{400} = 875\ \text{A} \]
\[ 800\ \text{V}:\ I = \frac{350{,}000}{800} = 438\ \text{A} \]

Eight hundred and seventy-five amps through a cable a person must lift and plug in. That number alone is why 800 V architectures exist — it is at the edge of what liquid cooling can manage in a handheld connector.

The loss ratio:

\[ \frac{P_{loss,400}}{P_{loss,800}} = \left(\frac{875}{438}\right)^2 = \mathbf{4} \]

Four times the conduction loss for the same delivered power, in every resistive element: the cable, the connector, the contactors, the busbars, the device channels and the motor windings. It is the same argument that put transmission lines at hundreds of kilovolts, applied inside a car.

What that buys, item by item:

BenefitDetail
Charging cablelighter, possibly not liquid-cooled
Conduction losses\(4\times\) lower everywhere
Cable and busbar masshalved — significant in a vehicle
SiC devices become natural1200 V SiC is a mature, high-performance part
Motor efficiencylower current, thinner conductors, less copper
Charging time350 kW becomes practical rather than heroic

The fourth row is a quiet but decisive advantage. At 400 V the natural device is a 650 V part, where silicon IGBTs and MOSFETs remain competitive; at 800 V it is a 1200 V part, where SiC's advantage over silicon is large. The architecture change and the device change reinforce each other.

And what it costs:

CostDetail
More cells in series~192 instead of 96 — more balancing channels
Higher-voltage componentscontactors, fuses, insulation, sensors
Compatibility with 400 V chargersneeds a boost stage or a split-pack trick
Insulation coordinationcreepage and clearance throughout
Service safety800 V DC is substantially more hazardous
Auxiliary loads12 V and 400 V accessories need conversion

The third row is the practical obstacle. Most installed public chargers are 400 V class, so an 800 V vehicle must be able to accept them — solved either with an extra boost converter or, elegantly, by reconfiguring the traction inverter and motor windings as a boost stage during charging, using hardware the car already carries.

The charging standards landscape:

ModePowerConverter location
AC Level 11.4–2 kWonboard
AC Level 27–22 kWonboard — Problem 3
DC fast50–150 kWoff-board
DC ultra-fast150–350 kWoff-board, liquid-cooled cable
Megawatt (trucks)> 1 MWoff-board, 1500 V class

The division is about mass. An onboard charger must be carried everywhere and is therefore limited to a few tens of kilograms; a DC charger sits on the ground and can weigh whatever it needs to. That is why DC charging bypasses the vehicle's converter entirely and talks straight to the pack.

What limits charging speed in practice. Not the converter:

\[ \text{cell chemistry} \;\Longrightarrow\; \text{lithium plating above a chemistry- and temperature-dependent rate} \]
LimitConsequence
Lithium platingpermanent capacity loss; worse when cold
Heat generation\(I^2R_{internal}\) — needs active cooling
CV taperthe last 20% cannot be hurried — Problem 1
Cell temperaturethe pack is pre-conditioned before a fast charge

A well-designed 350 kW charger will still deliver far less than 350 kW to a cold pack, because the battery management system limits the current to protect the cells. Vehicles now pre-condition the pack en route to a charger for exactly this reason — heating the battery so it can accept the power on arrival.

Doubling the voltage quarters the loss in every conductor. It is the transmission argument brought inside a vehicle — 875 A down to 438 at 350 kW — and it also moves the natural device from 650 V silicon to 1200 V SiC, where the semiconductor advantage is largest.
Answera 875 A at 400 V against 438 A at 800   b conduction loss \(4\times\) lower   c costs: more cells in series, higher-voltage parts, 400 V charger compatibility
Problem 5ChallengeWhat It Does to the Grid

Assess the impact of a site with four 350 kW chargers on a distribution network, describe how stationary storage buffers it, and explain what vehicle-to-grid offers and what it costs the battery.

Solution

The connected load:

\[ 4\times350\ \text{kW} = 1.4\ \text{MW} \]

Comparable with a small factory, appearing at a motorway service area or a supermarket car park — locations whose existing supply was sized for lighting and refrigeration. The connection upgrade, not the chargers, is usually the dominant project cost.

What makes it harder than a factory load:

CharacteristicConsequence
Very low diversityall four can start within seconds of each other
Fast ramps0 to 350 kW in seconds — voltage step
Coincident with evening peakdrivers charge when they arrive home or stop
Low utilisation1.4 MW connection used a few hours a day
Harmonicsmanageable — active front ends

The fourth row is the economic problem. Paying for a 1.4 MW connection that is used at 15% capacity factor is expensive, which is precisely the gap that on-site storage fills.

Buffering with stationary storage. Size a battery to cover a charging session from a modest grid connection:

\[ \text{grid supply } 300\ \text{kW}; \qquad \text{peak demand } 1.4\ \text{MW}; \qquad \text{deficit } 1.1\ \text{MW} \]
\[ \text{for a 20-minute session: } E = (1.1)\left(\frac{20}{60}\right) = 367\ \text{kWh} \]

A 370 kWh buffer — about five vehicle packs — lets a 300 kW connection serve a 1.4 MW site, recharging between sessions. The economics compare the storage cost against the connection upgrade, and at many sites the battery wins outright.

Vehicle-to-grid inverts the relationship. A parked fleet is a large distributed battery:

\[ \text{a car is parked} \approx 95\%\ \text{of the time} \]
V2G serviceValueEnergy throughput
Frequency responsehighest paidvery low — seconds at a time
Peak shavingmoderatemoderate
Energy arbitragelowesthigh — full cycles
Backup power (V2H)situationaloccasional

Read the first and third rows together. Frequency response pays best and cycles the battery least — a few kilowatt-hours moved back and forth per day — while arbitrage pays least and consumes real cycle life. The economics therefore favour the service that barely touches the battery.

The cost to the battery, which V2G proposals often understate:

\[ \text{degradation} \approx \frac{\text{pack cost}}{\text{cycle life}\times\text{usable energy}} \]
\[ \approx \frac{\pounds8000}{(2000)(60\ \text{kWh})} = \pounds0.067\ \text{per kWh cycled} \]

Nearly seven pence per kilowatt-hour of throughput, before any efficiency loss — which exceeds the arbitrage spread in most markets. That single figure explains why V2G deployment has concentrated on frequency response and why energy arbitrage from vehicle batteries remains largely theoretical.

What makes it work anyway:

EnablerDetail
Bidirectional chargerthe DAB of Problem 3 — no extra cost
Aggregationthousands of vehicles as one dispatchable resource
Smart charging firstshifting when to charge captures most of the value with no degradation
Falling cell costreduces the degradation charge per kWh
Second-life packsretired vehicle batteries as stationary storage

The third row deserves emphasis. Simply choosing when to charge — overnight rather than at the evening peak — delivers most of the grid benefit of V2G with none of the extra cycling. Unidirectional smart charging is the low-hanging fruit, and V2G is the refinement.

Shifting when a vehicle charges is worth almost as much as discharging it, and costs nothing. V2G's degradation charge of around seven pence per kilowatt-hour exceeds most arbitrage spreads, so the services that pay are the ones that barely cycle the battery — and smart unidirectional charging captures most of the grid value for free.
Answera 1.4 MW with almost no diversity   b 370 kWh of storage lets a 300 kW connection serve it   c V2G degradation ~£0.067/kWh favours frequency response over arbitrage
Problem 6ChallengeKeeping It Safe

Specify the battery management and safety system: state-of-charge estimation, the pre-charge circuit for a 500 µF link at 400 V, isolation monitoring and the contactor sequence. Compute the pre-charge time and resistor energy.

Solution

State of charge estimation. No sensor measures it directly, so it must be inferred:

MethodPrincipleWeakness
Coulomb counting\(\text{SoC} = \text{SoC}_0-\frac{1}{Q}\int i\,dt\)drifts — an open integrator
Open-circuit voltagelookup against a curveneeds hours of rest; flat for LFP
Impedancemeasure internal resistancetemperature-dependent
Kalman filtera model corrected by voltage measurementthe standard approach

Coulomb counting is accurate in the short term and drifts without bound — the same integrator problem as the sensorless flux estimator of Set 35. Combining it with an OCV correction in a Kalman filter gives the short-term accuracy of the integral with the long-term anchoring of the measurement.

The pre-charge problem. Closing a contactor onto a discharged link capacitor is a short circuit:

\[ i = \frac{V_{bat}}{R_{wiring}} \quad\text{— thousands of amps, limited only by parasitics} \]

The contactor welds shut, which is a dangerous failure: a welded main contactor cannot disconnect the pack in a crash or a fault. Pre-charge is a safety system, not merely a component protection.

The pre-charge circuit:

\[ C = 500\ \mu\text{F},\quad V = 400\ \text{V},\quad R = 100\ \Omega \]
\[ i_{peak} = \frac{400}{100} = 4.0\ \text{A} \quad\text{— against thousands} \]
\[ \tau = RC = (100)\left(500\times10^{-6}\right) = 50\ \text{ms} \]
\[ t_{95\%} \approx 3\tau = 150\ \text{ms} \]
\[ E_R = \tfrac12CV^2 = \tfrac12\left(500\times10^{-6}\right)(400)^2 = 40\ \text{J} \]

Exactly half the energy delivered ends up in the resistor regardless of its value — a classic result. Forty joules in 150 ms is 267 W average, so the resistor is rated for a pulse rather than continuously, exactly like the soft starter of Set 30 and the wind chopper of Set 38.

The contactor sequence:

StepActionCheck
1Close the negative contactor
2Close the pre-charge contactor
3Wait 150 msverify \(V_{link} > 95\%\) of \(V_{bat}\)
4Close the main positive contactor
5Open the pre-charge contactor
6Enable the inverter

Step 3's check is essential. If the link voltage does not rise, something is drawing current — a short in the inverter, a failed capacitor — and closing the main contactor onto that fault would be destructive. The pre-charge doubles as a diagnostic.

Isolation monitoring. The pack floats relative to the vehicle chassis:

\[ R_{iso} \ge 100\ \Omega/\text{V} \;\Longrightarrow\; \ge 40\ \text{k}\Omega\ \text{for a 400 V pack} \]

Measured continuously, because a single insulation failure is not itself hazardous — a floating system tolerates one fault — but a second one completes a circuit through the chassis. The monitor's job is to catch the first fault before the second arrives, which is the same philosophy as an IT earthing system in a hospital.

The complete BMS:

FunctionPurpose
Cell voltage monitoringall 96, to a few millivolts
Temperature monitoringmultiple points — gradients matter
Current measurementfor coulomb counting and protection
SoC and SoH estimationrange prediction and degradation tracking
Cell balancingProblem 1
Thermal managementcooling under charge, heating before fast charge
Contactor and pre-charge controlthe sequence above
Isolation monitoringcontinuous
Charge/discharge current limitsas functions of SoC and temperature
Fault detection and loggingdiagnostics and warranty

The last functional row is what actually governs the vehicle's behaviour: the BMS publishes a maximum permitted current that varies with temperature and state of charge, and every converter in the vehicle obeys it. The 350 kW charger of Problem 4 delivers whatever the BMS allows.

The battery management system, not the converter, sets the vehicle's limits. A 350 kW charger and a 150 kW traction inverter both take their instructions from a controller whose real job is protecting cells — and half the energy in every pre-charge ends up in a 100 Ω resistor, whatever value is chosen.
Answera Kalman-filtered coulomb counting with OCV correction   b\(\tau = 50\ \text{ms}\), 150 ms to 95%, 40 J in the resistor   c\(R_{iso} \ge 40\ \text{k}\Omega\)
Formulas

Key Formulas

QuantityRelationNotes
Pack voltage\(V = N_sV_{cell}\)288–403 V for 96S
Capacity\(Q = E/V_{nom}\)211 Ah
C-rate\(I/Q\)2C at 150 kW
CC–CVCC to \(V_{max}\), then CV~80% at the transition
Bidirectional duty\(D = 1-V_{bat}/V_{link}\)0.42–0.59
Inductor\(L = \dfrac{V_{bat}D}{f_s\Delta I}\)104 µH at 423 A
Interleaving\(I_{phase} = I/n\); loss \(\propto 1/n\)Ripple nulls at \(D = k/n\)
DAB power\(P = \dfrac{V_1V_2'\phi(\pi-|\phi|)}{2\pi^2f_sL}\)Sign of \(\phi\) reverses it
DAB inductance\(L = \dfrac{V_1V_2'\phi(\pi-\phi)}{2\pi^2f_sP}\)41.8 µH
DAB maximumat \(\phi = \pi/2\)Design at 45° for margin
Voltage and loss\(P_{loss} \propto 1/V^2\)800 V quarters it
Fast-charge current\(I = P/V\)875 A at 400 V, 438 at 800
Buffer storage\(E = \left(P_{peak}-P_{grid}\right)t\)367 kWh
Degradation cost\(\dfrac{\text{pack cost}}{\text{cycles}\times\text{usable kWh}}\)~£0.067/kWh
Pre-charge\(\tau = RC,\ t \approx 3\tau,\ E_R = \tfrac12CV^2\)50 ms, 150 ms, 40 J
Isolation\(R_{iso} \ge 100\ \Omega/\text{V}\)40 kΩ at 400 V
Pitfalls

Common Mistakes

  1. Designing a battery converter for nominal voltage only. The pack swings 40% from empty to full — Problem 1.

  2. Assuming charge time scales inversely with power. The CV taper adds as long again past 80% — Problem 1.

  3. Ignoring cell balancing. The weakest cell terminates the charge for all 96 — Problem 1.

  4. Sizing a single-phase converter for 423 A. Interleaving makes the inductor and devices ordinary — Problem 2.

  5. Choosing a DAB turns ratio away from the nominal match. Circulating current and loss of ZVS follow — Problem 3.

  6. Designing a DAB at 90° phase shift. It is the maximum, so no control margin remains — Problem 3.

  7. Using an LLC where bidirectional flow is needed. Its output rectifier blocks — a CLLC or DAB is required — Problem 3.

  8. Assuming an 800 V vehicle can ignore 400 V chargers. Most public infrastructure is 400 V class — Problem 4.

  9. Costing V2G without battery degradation. Around 7p per kWh cycled exceeds most arbitrage spreads — Problem 5.

  10. Omitting pre-charge. The main contactor welds, and a welded contactor cannot disconnect the pack — Problem 6.

Looking Ahead

Six problems on the technology that both renewable sources depend on. A 40% voltage swing shaped every converter around the pack; a 423 A inductor became four ordinary ones through interleaving; a dual active bridge reversed 11 kW by changing the sign of one variable; doubling the system voltage quartered the loss in every conductor; and a battery management system turned out to set the limits that every converter in the vehicle obeys.

What connects Sets 37 to 39 is that all three sources are converter-connected. A photovoltaic array, a wind turbine and a battery each reach the grid through a controlled inverter rather than through a synchronous machine bolted to it — and that changes the network fundamentally. A converter contributes almost no inertia, limits its fault current to about its rating, and can be told to do things a generator cannot: inject reactive power at zero output, respond to frequency in milliseconds, or transmit power along a cable that no AC line could use.

Next: Set 40 closes the book — HVDC, FACTS and Power Quality, where a modular multilevel converter stacks 1920 submodules to reach 321 output levels with no filter at all, a STATCOM is shown to hold its current where an SVC's collapses with voltage, series compensation doubles a line's capability and risks resonating with a turbine shaft, and an active filter is sized against the very rectifier that Set 10 built.