Chapter 29 of 30 — the second of Part 7
- Why storage is an energy and a power specification, and why confusing the two is the most common sizing error.
- The battery model you actually need: open-circuit voltage against state of charge, plus an internal resistance — and what each term costs you.
- Why charging ends with a constant-voltage tail that takes as long as the constant-current phase, and why fast charging cannot avoid it.
- The realisation that makes bidirectional conversion almost free: a synchronous buck is a boost running the other way.
- The dual active bridge — isolated, bidirectional, soft-switched — controlled by a single phase-shift variable.
- Why a DAB loses zero-voltage switching at light load, and why the loss is asymmetric when the voltages are mismatched.
- What the EV charging levels and connector standards actually specify, and where the AC/DC boundary sits.
- V2G: what makes it technically straightforward and commercially difficult.
Energy and Power Are Different Specifications
Chapter 28 ended with two sources that produce power when conditions allow rather than when it is wanted. That mismatch is what storage exists to bridge, and it is why storage moved from a niche to a central concern of power electronics within about fifteen years.
Before any converter is drawn, one distinction has to be settled.
- A 2 MW / 8 MWh system is 0.25C — four hours of discharge. It is an energy system: shifting solar into the evening.
- A 2 MW / 0.5 MWh system is 4C — fifteen minutes. It is a power system: frequency response and smoothing.
Both are "2 MW battery storage". They use different cell chemistries, different thermal designs and different converters, and one cannot do the other's job. Ask for the duration, always.
| Service | Timescale | Typical C-rate | What decides the design |
|---|---|---|---|
| Frequency response | seconds | 2–4C | Power and response time; energy is almost irrelevant |
| Smoothing renewables | minutes | 1–2C | Cycle life — thousands of shallow cycles per year |
| Energy shifting | hours | 0.25–0.5C | Energy cost per kWh, and round-trip efficiency |
| Backup / UPS | minutes to hours | 0.5–2C | Reliability and transfer time; the system may never cycle |
| EV traction | hours to discharge, minutes to charge | 0.3C out, 1–3C in | Energy density on discharge, thermal limits on charge |
The Battery as a Circuit Element
A power electronics engineer does not need an electrochemical model. Two elements carry almost all the behaviour that matters for converter design.
Positive when charging, negative when discharging. Everything a charger must cope with follows from the shapes of those two terms.
- \( V_{oc}(SoC) \) is a curve, not a constant. For lithium iron phosphate it is famously flat — about 3.2 V across most of the range — which makes the cell easy on the converter and very hard to estimate state of charge from. For NMC it slopes usefully from about 3.6 V to 4.2 V.
- \( R_{int} \) is small but decisive. A few milliohms per cell, rising sharply at low temperature and as the cell ages. It sets the heating, the voltage sag under load, and the round-trip efficiency.
- Write the loss
Ohmic loss in the cell\[ P_{loss} = I^2R_{int} \]
- Express it as a fraction of the power being handled
Loss fraction is proportional to current, so doubling the C-rate doubles the percentage lost — and quadruples the heat to be removed.Working\[ \frac{P_{loss}}{P} = \frac{I^2R_{int}}{VI} = \frac{IR_{int}}{V} \]
- Get the round-trip efficiency
Energy is lost on the way in and again on the way out:
This is why the same battery is quoted at 95% round trip at 0.25C and 88% at 2C. The number is meaningless without the rate.Round trip\[ \eta_{rt} \approx \left(1 - \frac{I_{ch}R_{int}}{V}\right)\left(1 - \frac{I_{dis}R_{int}}{V}\right) \]
Making Power Flow Both Ways
Bidirectional conversion sounds like a new topology. Usually it is not — it is a recognition about a circuit that was already there.
| Topology | Isolated? | Control variable | Typical use |
|---|---|---|---|
| Synchronous buck / boost | no | duty ratio | EV traction inverter DC–DC, battery to DC link |
| Four-switch buck-boost | no | two duty ratios | Where the two voltages overlap in range |
| Interleaved synchronous | no | duty ratio, phase-shifted | High current — ripple cancellation and shared thermals |
| Dual active bridge | yes | phase shift | On-board chargers, solid-state transformers, storage interfaces |
| CLLC resonant | yes | frequency (and phase) | On-board chargers where efficiency dominates |
| Bidirectional PFC / active front end | no | current reference | The grid interface — the stage that makes V2G possible |
The Dual Active Bridge
When isolation is required as well — and for anything touching a vehicle or the grid it is — the dual active bridge is the standard answer. Two full bridges face each other across a transformer, and the transformer's leakage inductance is not a parasitic to be minimised but the energy transfer element itself.
Both bridges produce square waves. Shift one relative to the other and power flows.
- Write what the inductor sees The inductance \( L \) has a square wave of \( \pm V_1 \) on one side and \( \pm V_2' = \pm nV_2 \) on the other, phase-shifted by \( \varphi \). The voltage across it is the difference — a staircase that changes value at each edge.
- Integrate to get the current at the two switching instants
with \( \omega = 2\pi f_s \) and \( \varphi \) in radians.Working, using half-wave symmetry \( i(\pi) = -i(0) \)\[ i(0) = \frac{\pi(V_2'-V_1) - 2V_2'\varphi}{2\omega L}, \qquad i(\varphi) = \frac{2V_1\varphi + \pi(V_2'-V_1)}{2\omega L} \]
- Average the transferred power over a cycle
The single-phase-shift power equation\[ P = \frac{nV_1V_2\,\varphi\left(\pi - |\varphi|\right)}{2\pi^2f_sL} \]
- Find the maximum
Note what is absent: the duty ratio. Power depends only on the phase shift, the two voltages, the frequency and the inductance. Reverse the sign of \( \varphi \) and the power reverses — the same equation, the same hardware, no mode change.Result\[ \varphi = \frac{\pi}{2} \;\Longrightarrow\; P_{max} = \frac{nV_1V_2}{8f_sL} \]
An 8 kW on-board charger: 400 V DC link, 25 μH of leakage inductance, 100 kHz. Move the phase shift through zero to reverse the power. Then mismatch the voltage ratio and watch one bridge lose zero-voltage switching before the other.
—
EV Charging, and Where the Rectifier Lives
Charging standards look like a thicket of names. One question cuts through nearly all of it: which side of the cable does the AC–DC conversion happen on?
- AC charging. The cable carries AC. The rectifier is the on-board charger in the vehicle, so its power is limited by what a car can carry and cool — in practice 7 to 22 kW.
- DC charging. The cable carries DC. The rectifier is in the roadside cabinet, where mass and cooling are cheap, so 50 to 350 kW is available. The vehicle contributes only contactors and the battery management system.
Everything else — connector shapes, communication protocols, national variants — is detail layered on that division.
| Level | Supply | Typical power | Rectifier location | What limits it |
|---|---|---|---|---|
| AC Level 1 | single-phase 120–230 V | 1.4–3.7 kW | on board | The domestic socket. Overnight charging only. |
| AC Level 2 | single or three-phase, 230/400 V | 7–22 kW | on board | Mass, volume and cooling of the on-board charger. |
| DC fast | three-phase, off-board conversion | 50–150 kW | off board | Battery thermal limits, and the site's grid connection. |
| DC ultra-fast | three-phase, off-board | 150–350 kW | off board | Cable current — above about 500 A the cable must be liquid-cooled. |
- See why the industry moved to 800 V packs
Charging at 350 kW into a 400 V pack requires
which is an impractical cable — heavy, stiff, and needing active cooling.Working\[ I = \frac{350{,}000}{400} = 875\ \text{A} \]
- Double the voltage
Still large, but achievable with a cable a person can lift.Result\[ I = \frac{350{,}000}{800} = 437\ \text{A} \]
- Count the losses, which is the stronger argument Conduction loss goes as \( I^2R \), so halving the current cuts the loss in the cable, the connector, the contactors and the pack's own resistance by a factor of four — and that loss appears as heat inside the battery, which is precisely what limits the charging rate.
Worked Examples
Problem. A 20 MW solar farm needs storage for (a) smoothing cloud-induced ramps, holding output steady for up to 90 seconds, and (b) shifting 4 hours of evening generation. Size both, and comment on the converter in each case.
Same power rating, 160 times the energy. The two systems have almost nothing in common:
- The 40C system is impractical for lithium — cells rated for 40C exist but are expensive and short-lived. This job goes to supercapacitors or to a lithium titanate pack sized by power rather than energy, and in practice the battery would be oversized on energy simply to keep the C-rate reasonable.
- The 0.25C system is a straightforward LFP installation. Its cost is dominated by the cells, and its converter is a modest 20 MW at a comfortable current.
The converter differs too, and not in the way one might guess. Both are 20 MW, so the semiconductors are similar. But the smoothing system's converter must reverse from full charge to full discharge in well under a second and spends its life at partial load, so its control bandwidth and its light-load efficiency matter enormously. The shifting system's converter runs near full load for hours in one direction, so peak efficiency and thermal design dominate.
And the pricing consequence that follows from the C-rate: a 20 MW / 0.5 MWh system may cost less than a tenth of a 20 MW / 80 MWh one, because storage cost is dominated by energy, not power. Quoting either as "a 20 MW battery" conceals a factor of ten in price.
Problem. An 11 kW on-board charger works from a 400 V DC link to a battery that ranges 250–450 V. Choose the transformer ratio, find the leakage inductance for full power at 60° phase shift with \( f_s = 100 \) kHz, and check the ZVS range at both battery extremes.
Now the two extremes, which is the point of the exercise. With \( n = 1.143 \), the reflected battery voltage is \( nV_2 \):
So the charger hard-switches below roughly a third of full phase shift at either extreme. Putting a number on the worse of the two: at \( d = 0.714 \) with \( L = 16.2\ \mu \)H, a phase shift of 25.7° corresponds to
so bridge 2 hard-switches below about 39% of full power whenever the battery is near empty.
Why that is worse than it sounds. The CV tail is exactly the low-power, high-battery-voltage condition, and it is where the charger spends the last third of every session. The efficiency quoted on the datasheet — measured at full power, mid-range battery — is not the efficiency that determines the energy actually consumed.
The options, and what each costs:
- Extended-phase-shift modulation. Add the bridge duty ratios as control variables to restore ZVS at low power. No hardware change; a substantially harder controller.
- Reduce \( L \) and use a larger phase shift for the same power. This raises the circulating current and the conduction loss — a real trade, not a free improvement.
- Accept it. At 5 kW the absolute switching loss may be tolerable even without ZVS, because the currents are small. This is a legitimate answer, and it must be a calculation rather than an assumption.
Problem. A 77 kWh pack accepts 175 kW to 50% state of charge, then tapers linearly to 50 kW at 80%, then follows a CV tail to 100% taking 35 minutes. Starting at 10%, find the time to 80% and to 100%, and the average power in each phase.
Read the two halves of that result against each other. The first 70% of the pack took 23 minutes; the last 20% took 35 minutes. The final fifth of the charge consumed 60% of the total time.
This single asymmetry shapes the entire charging industry. Route planners target 80%. Networks price by time as well as by energy, or impose idle fees, because a vehicle occupying a 350 kW bay while drawing 26 kW is wasting an expensive asset. And a manufacturer's "350 kW charging" claim describes a peak that may be held for only a few minutes.
The honest way to compare vehicles is not peak power but average power from 10% to 80% — 141 kW here. A car that peaks at 350 kW but tapers immediately can easily lose to one that holds 150 kW steadily, and the second is the better machine for a long journey.
And note where the taper comes from. It is not the charger being cautious. It is \( V_{oc}(SoC) \) rising towards the cell voltage limit while \( R_{int} \) also rises — the physics of Section 29-2, appearing directly in a commercial specification.
Summary and Formula Sheet
Chapter 29 in five sentences:
- Energy and power are independent specifications, and their ratio — the C-rate — determines the chemistry, the thermal design and the converter.
- An open-circuit voltage curve plus an internal resistance is enough to size a charger, and loss fraction rises in proportion to current.
- Charging is constant-current then constant-voltage, and the tail is slow because the driving voltage difference collapses as the cell fills.
- Replacing every diode with a transistor makes a converter bidirectional; the hardware becomes symmetric and the difficulty moves into the control.
- A dual active bridge transfers power through leakage inductance under phase-shift control, and its zero-voltage switching is an operating-point property that is lost at light load and mismatched voltages.
Key terms
- C-rate
- Power divided by energy capacity. 1C means the whole capacity in one hour.
- State of charge
- Remaining capacity as a fraction. Hard to measure on a flat \( V_{oc} \) curve such as LFP.
- CC–CV charging
- Constant current to a voltage limit, then constant voltage until the current decays. The tail is unavoidable.
- Lithium plating
- Metallic lithium deposited on the anode when charged too hard or too cold. Irreversible, and the reason for the voltage limit.
- Round-trip efficiency
- Energy out divided by energy in, over a full cycle. Only meaningful when the C-rate is stated.
- Dual active bridge
- Two bridges across a transformer, transferring power through its leakage inductance under phase-shift control.
- Voltage ratio d
- \( nV_2/V_1 \) in a DAB. Unity is ideal; departure from it costs one bridge its zero-voltage switching.
- Zero-voltage switching
- Turning a device on after its output capacitance has already been discharged by the circuit current. A property of the operating point.
- On-board charger
- The AC–DC converter carried in the vehicle. Limited to about 22 kW by mass and cooling.
- Active front end
- A bidirectional PFC rectifier presenting sinusoidal unity-power-factor current to the grid. The stage that makes export possible.
- V2G
- Vehicle-to-grid: exporting from a vehicle battery. Technically solved; limited by warranties, standards and settlement.
Test Yourself
Two suppliers both quote "a 5 MW battery". Why is that not enough information to compare them, and what single question resolves it?
Ask for the duration. Power alone says nothing about energy, and energy is what dominates the cost.
The two extremes of what "5 MW" might mean:
- 5 MW / 1.25 MWh (4C) — fifteen minutes of discharge. A frequency-response asset. Sized by power; the cells are chosen for rate capability rather than energy density.
- 5 MW / 20 MWh (0.25C) — four hours. An energy-shifting asset. Sized by energy; the cells are chosen for cost per kWh and cycle life.
Sixteen times the energy for the same power, and since installed cost is dominated by cells, roughly an order of magnitude in price.
What else changes with the C-rate, and is often overlooked:
- Chemistry. High-rate duty pushes towards LTO or high-power NMC; energy duty towards LFP or high-nickel NMC.
- Thermal design. Loss goes as \( I^2R \), so a 4C system dissipates sixteen times the heat of a 1C one for the same pack. Cooling may move from air to liquid.
- Round-trip efficiency. Loss fraction is \( IR_{int}/V \), so it is proportional to rate. The same cells might give 95% at 0.25C and 88% at 2C.
- Degradation. High rates and deep cycles both age cells faster, and a frequency-response asset may perform thousands of shallow cycles a year against a shifting asset's few hundred deep ones.
- The converter's duty. Fast reversal and good light-load efficiency for one; peak efficiency at sustained full load for the other.
And the same confusion in the other direction is just as common: a "100 kWh home battery" that can only deliver 5 kW cannot start a heat pump, however much energy it holds. Both numbers, always.
Why does the last 20% of an EV charge take longer than the first 70%, and why can a better charger not fix it?
Because the limit is inside the cell, not in the charger. The charger is a voltage source that has run out of voltage difference.
Trace it through the model. The terminal voltage is \( V_{oc}(SoC) + IR_{int} \), and there is a hard ceiling — about 4.2 V per cell for NMC — above which lithium plates onto the anode irreversibly.
- Early in the charge, \( V_{oc} \) is low, so a large \( IR_{int} \) can be added before reaching the ceiling. Full current is available.
- As the cell fills, \( V_{oc} \) rises towards the ceiling and the permitted \( IR_{int} \) shrinks in proportion. The current must fall.
- In the CV phase, \( I = \left(V_{limit}-V_{oc}(SoC)\right)/R_{int} \), and the numerator collapses as the cell fills. The decay is roughly exponential, so the last few percent take disproportionately long.
Two effects make it worse than the simple picture: \( R_{int} \) itself rises towards the top of the range, and lithium diffusion into the electrode becomes the limiting process rather than the external circuit. Pushing harder does not fill the cell faster; it plates lithium.
So a better charger cannot help, because the charger is not the constraint. What can help:
- Warm the pack. \( R_{int} \) can be three times higher at 0 °C than at 25 °C, so a cold pack reaches the voltage ceiling at a much lower current. Pre-conditioning is worth several minutes and costs only some energy.
- Higher pack voltage. An 800 V architecture halves the current for the same power, cutting \( I^2R \) heating fourfold — and heat is what forces the taper in the first place.
- Better cell design — thinner electrodes, better electrolytes — which raises the rate at which lithium can intercalate. This is where the real progress is being made.
- Stop at 80%. Usually correct, and it also reduces calendar ageing.
And the practical consequence for comparing vehicles: peak charging power is close to meaningless. The useful figure is average power from 10% to 80%, which captures the taper.
Why is a synchronous buck already a bidirectional converter, and what actually has to change to use it that way?
Because each switch position already conducts in both directions — the transistor one way, its body diode the other — and the direction of power flow is decided by the sign of the average inductor current, which is a control decision.
The two views of the same circuit:
- From the high-voltage side: a buck. \( V_{low} = DV_{high} \), power flows down.
- From the low-voltage side: a boost. \( V_{high} = V_{low}/(1-D') \), power flows up.
The switches, the inductor and the capacitors are identical. Nothing is added.
The general rule this illustrates: a converter is bidirectional when every switch position can conduct in both directions and block in one. A transistor with an antiparallel diode does exactly that, so replacing every diode with a transistor makes the power stage bidirectional. Since synchronous rectification is now standard for efficiency reasons, most modern converters are bidirectional whether or not it was intended.
What genuinely has to change, and where the engineering effort goes:
- Continuous conduction through zero. In one direction the converter would normally enter discontinuous conduction at light load. A bidirectional converter must stay continuous, allowing the inductor current to go negative — which changes the small-signal model and therefore the compensator.
- Deciding what is regulated. Charging may regulate current; discharging may regulate the DC-link voltage. The controller must switch objective without a transient at the handover.
- Dead time hurts twice. With current flowing either way, the dead-time-induced voltage error changes sign with the current, distorting the transfer characteristic near zero — the same effect that causes zero-crossing distortion in inverters.
- Protection is symmetric now. Overcurrent, precharge and fault response must all work in both directions, and a fault during reverse flow is not the mirror image of one during forward flow.
The summary worth remembering: the power stage is nearly free; the control is the product.
A DAB is described as soft-switched. Under what conditions is that false, and why are those the conditions a charger actually lives in?
ZVS is a property of the operating point, not of the topology. It is lost near zero phase shift, and asymmetrically whenever the voltage ratio departs from unity.
The requirement. Before a device turns on, the inductor current must already be flowing in the direction that discharges its output capacitance. From the edge currents:
with \( d = nV_2/V_1 \). At \( d = 1 \) both reduce to \( |\varphi| > 0 \) and ZVS holds everywhere except exactly zero power. Any mismatch opens a dead band around zero, and it hurts bridge 1 when \( d>1 \) and bridge 2 when \( d<1 \).
Now why a charger lives there. Both bad conditions arrive together:
- Low power means small \( \varphi \) — and the CV tail is a long, low-power phase.
- High battery voltage at the end of charge means \( d \) well above unity — and that is exactly when the CV tail happens.
- Low battery voltage at the start means \( d \) well below unity, costing bridge 2 its ZVS during the initial low-power ramp.
So the converter is soft-switched precisely where it was measured for the datasheet — full power, mid-range battery — and hard-switched over a substantial part of a real session.
The remedies, and their prices:
- Extended (dual or triple) phase-shift modulation — use the bridge duty ratios as extra control freedoms. Restores ZVS over a much wider range; the controller becomes considerably harder and the modulation boundaries must be handled carefully.
- Choose \( n \) for the middle of the battery range, so \( d \) is near unity for most of the session and mismatched only at the ends.
- Add a regulating stage in front so the DAB always sees \( d \approx 1 \). Costs an extra conversion.
- Use a CLLC resonant converter, which retains soft switching more naturally across load — at the cost of frequency control and a harder magnetic design.
And the habit worth forming: when a converter is quoted at 98% efficiency, ask at what power and what voltage ratio, and ask to see the curve. Peak efficiency is a single point on a surface.
Why did the industry move to 800 V battery packs, when 400 V systems were well established and every component had to be requalified?
Because charging current, not charging voltage, is what everything else is limited by — and doubling the voltage halves it.
What 875 A would require. A cable of that rating is too heavy and too stiff for a person to handle, and needs active liquid cooling along its whole length. Connectors, contactors and fuses at that current are large and expensive. The practical ceiling for an air-cooled cable is around 200–250 A; liquid cooling extends it to about 500 A.
But the stronger argument is the losses, because they go as the square:
- Halving the current cuts \( I^2R \) by four — in the cable, the connector, the contactors, the busbars and the pack's own internal resistance.
- The loss inside the pack appears as heat in the cells, and cell heating is precisely what forces the charging taper. Less heat means the high-current phase can be held longer.
- The same benefit applies to the traction inverter and the motor for the whole life of the vehicle, not only while charging.
What it cost, honestly:
- New semiconductors. 650 V silicon IGBTs no longer suffice; 1200 V devices are needed, and silicon carbide became the practical choice — initially at a significant price premium.
- Insulation and creepage requirements rise throughout, affecting motor windings, connectors and the entire high-voltage distribution.
- Backward compatibility. Most installed fast chargers were 400 V, so an 800 V vehicle needed a boost stage — often implemented cleverly by reusing the traction inverter and motor windings as the boost converter rather than adding hardware.
- Service and safety training across the whole aftermarket.
And the direction of travel: heavy vehicles are already looking beyond 800 V, because the same argument applies with more force at megawatt charging. The reasoning is not specific to cars — it is the same reason transmission is at 400 kV rather than 4 kV.
The hardware for V2G is already fitted to millions of vehicles. Why has it not happened, and what does that tell you about engineering more generally?
Because none of the remaining obstacles are power electronics problems. Both stages of an on-board charger — the active front end and the isolated DC–DC — are inherently bidirectional. Exporting requires a control change, not a hardware change.
What is actually blocking it:
- Degradation, and who bears it. Grid cycling consumes cycle life the owner paid for. Whether the service payment exceeds the degradation cost depends on tariff, chemistry and cycle depth — and many warranties simply exclude it, which ends the discussion before the economics start.
- Standards. ISO 15118 defines the communication; CHAdeMO supported bidirectional operation years before CCS did; and every national grid code adds its own protection and certification requirements. A capable vehicle may be prohibited in a given jurisdiction.
- Metering and settlement. Export needs an export meter, a licence or exemption, and a market that will pay for a resource that can drive away. This is slow institutional work with no technical content.
- Protection. Everything in Chapter 28's anti-islanding discussion applies, plus the awkward case of deliberately islanding to power a house during an outage — a different mode with different requirements.
- Availability. A grid operator contracts for a resource that will be present. A car is not guaranteed to be.
Where it already works, because these objections weaken:
- Vehicle-to-home and vehicle-to-load. No export, so no licence and no settlement. Commercially available for several years.
- Fleets. Predictable schedules, one owner bearing both the degradation and the revenue — every objection weakens at once, which is why almost every serious V2G deployment is a bus or van fleet.
- Smart charging. Simply controlling when to charge captures much of the grid value with none of the difficulty, and is being deployed at scale now.
And the general lesson, which is a good one to leave this chapter with: for a mature technology the binding constraint is frequently not technical. Recognising when your converter is no longer the problem — and that the remaining work is standards, contracts and warranties — is part of engineering judgement, not a departure from it.
Problems
Three habits for storage and charging work:
- Always ask for both numbers — kW and kWh. A single figure describes nothing.
- Check the operating point, not the topology, when soft switching is claimed.
- Follow the current, not the power, when sizing cables, connectors and thermal paths.
Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.
- A storage system is specified as 3 MW / 12 MWh. Find its C-rate and duration, and state one service it suits and one it does not.
- A pack of 96 series cells has \( V_{oc} = 3.7 \) V and \( R_{int} = 2.2 \) mΩ per cell. Find the pack voltage and resistance, the terminal voltage at 150 A discharge, and the loss as a percentage of delivered power.
- For the pack of Problem 2, find the round-trip efficiency at 0.5C and at 2C, given a 100 Ah capacity, and comment on the difference.
- A DAB has \( V_1 = 400 \) V, \( n V_2 = 400 \) V, \( f_s = 120 \) kHz and \( L = 20\ \mu \)H. Find the maximum power, the phase shift for 4 kW, and the peak inductor current there.
- For the DAB of Problem 4, the battery falls to \( nV_2 = 320 \) V. Find the new maximum power, the ZVS boundary for each bridge, and the power below which one bridge hard-switches.
- A 22 kW on-board charger is proposed for a vehicle with a 400 V pack. Estimate the AC input current on a three-phase 400 V supply, and explain what limits on-board chargers to roughly this power.
- A 64 kWh pack charges at 120 kW to 55% state of charge, then tapers linearly to 40 kW at 80%. Find the time from 15% to 80% and the average power, and compare with a vehicle that holds 90 kW flat to 80%.
- Explain why a lithium iron phosphate pack is easy on the converter but hard on the battery management system, and identify two consequences for a storage installation.
- A bidirectional converter transitions from charging at 50 A to discharging at 50 A. Identify three control problems at the crossing, and propose a mitigation for each.
- Design the power electronics for a 250 kW DC fast charger serving both 400 V and 800 V vehicles: choose the topology of each stage, decide where isolation sits, specify the cable current and cooling, and state how the two pack voltages are accommodated.
- A commercial building has a 500 kW peak demand, a 200 kW solar array and a fleet of ten electric vans. Design a storage and charging scheme to minimise the peak demand charge, size the battery and the converters, and state what happens on a grid outage.
- Compare a dual active bridge with a CLLC resonant converter for an 11 kW on-board charger with a 250–450 V battery range. Address soft switching across the range, control complexity, magnetic design and behaviour at light load, and recommend one with reasons.