Why Charging current is getting limited and is it due to Battery BMS or Vehicle VCU in ev vehicles?

Limitations of Charging Current from Fast Charger

In electric vehicles (EVs), both the Battery Management System (BMS) and the Vehicle Control Unit (VCU) play critical roles in regulating charging current, but from different perspectives and with distinct responsibilities.

Battery Management System (BMS) Strategy:

The BMS is responsible for protecting and optimising the battery pack. It limits charging current based on:

Cell voltage and temperature: Prevents overcharging and thermal runaway.

State of Charge (SOC): Ensures cells are charged uniformly and safely.

State of Health (SOH): Adjusts charging parameters based on battery ageing.

Fault detection: If any cell anomaly is detected, BMS can restrict or halt charging.

Cell voltage imbalance: If nearing maximum voltage thresholds.

Vehicle Control Unit (VCU) Strategy

The VCU is the central brain of the EV, coordinating multiple subsystems including the BMS, motor controller, and onboard charger. It limits charging current based on:

Power availability from the grid or charger: Adjusts current to match source capacity.

Thermal management: Coordinates with cooling systems to allow safe and fast charging.

Driving mode and user settings: May prioritise slower charging as per battery state.

Communication with charging station: Utilises protocols such as ISO 15118 to negotiate the optimal current.

Other Common Power Limitations while Charging EV Vehicles:

Charger Limitations: The charger itself is limited to a maximum supply limited kW.

Load management: Shared Power in the charger itself in case of a Fast Charger.

Input Power Limitations from Grid: AC 1-phase vs. AC 3-phase supplied from the grid to the charger. Even unstable grid conditions like voltage drop can cause the charging current to reduce.

Why did residual voltages available on the load side after contactors open in a high-voltage (HV) battery pack?

Load side voltages available at high voltage battery packs after contactors/relays open are due to residual voltages because of link capacitors in the inverter circuitry of the EV vehicles.

What are Residual Voltages?

Due to the capacitance of the stored charge in the link capacitors, the voltage appeared on the load side even though the contactors remained open.

To discharge those voltages, some active and passive methods are employed by OEMs to drain the charges accumulated at capacitors.

Active Discharge: Using separate Contactors along with resistors to drain the charge.

Passive Discharge: Using only resistors, which are connected between DC High-voltage Buses, acts as a passive discharge to drain the voltages.

As per industry standards, which are followed worldwide, the maximum allowed residual voltage at the load side is between <60V DC, and for the safer side, some OEMs like TATA use <30VDC as a safe limit.

As per Ohm's Law, since the resistor we choose has very high resistance, only a small, minuscule current flows through it and doesn't have much impact while the contactors are closed.

But when the contactors are open, the resistors will act as a load to drain the charge buildup at the DC link capacitor.

In all modern electric vehicles, active residual voltage draining methods are employed to drain the charge accumulated in the DC link capacitors as soon as possible.

The minimum time for which the charges are drained will be between ( 5-10Seconds ).

Example Calculation of time required to drain the available Voltage at the DC link Capacitor:

The voltage across a discharging capacitor at any given time (t) can be calculated using the formula:

V(t)=V0​×e−t/τ

The formula for the RC time constant is:

$\tau =R\times C$

Let R = 1 MegaOhm (1,000,000 Ohms) and C = 1 mF (0.001 Farads).

These are often much smaller in real systems. 

For example: $\tau =1000000\Omega \times 0.001F=1000seconds$

•  Say the available voltage at the DC link capacitor is $100V$:

• We want to solve for $t$.

  Step 1: Isolate the exponential term

  $$\frac{60}{100}=e^{-\frac{t}{1000}}$$$$0.6=e^{-\frac{t}{1000}}$$

  Step 2: Take natural logarithm on both sides

  $$\ln (0.6)=\ln \left(e^{-\frac{t}{1000}}\right)$$$$\ln (0.6)=-\frac{t}{1000}$$

  Step 3: Calculate ln⁡(0.6)

  $$\ln (0.6)\approx -0.5108$$

  Step 4: Solve for tt

  $$-0.5108=-\frac{t}{1000}$$

  Multiply both sides by $-1000$:

  $$t=0.5108\times 1000=510.8$$

  Final answer: 

                                                                            t ≈ 511 ( 8.5 min)

• Say the available voltage at the DC link capacitor is 800V:

We want to solve for $t$.

Step 1: Isolate the exponential term

 Divide both sides by 800:

$$\frac{60}{800}=e^{-\frac{t}{1000}}$$$$0.075=e^{-\frac{t}{1000}}$$

Step 2: Take the natural logarithm on both sides.

$$\ln (0.075)=\ln \left(e^{-\frac{t}{1000}}\right)$$$$\ln (0.075)=-\frac{t}{1000}$$

Step 3: Calculate ln⁡(0.075);

$$\ln (0.075)\approx -2.59027$$

Step 4: Solve for t;

$$-2.59027=-\frac{t}{1000}$$

Multiply both sides by $-1000$:

$$t=2.59027\times 1000=2590.27$$

Final answer:

$$\boxed{{\displaystyle t\approx 2590.27}}$$

t ≈ 2590.27 ( 43 Min)

We can refer to ISO 6469-3 or ECE R100 for standards on residual voltage.