# Electric Vehicle Charging [closed]

I have a few queries related to EV charging:

(1) When we refer to 'X kW' charger, is it that the charger is taking in X kW power and delivering power = X kW times the efficiency of the charger to the battery OR is it that X kW is what is being delivered and so the actual power being consumed from mains is X kW divided by efficiency.

(2) If the nominal voltage of car battery is 375V and one wants to charge the battery at a fast charging station which has chargers rated at 800V, how is the charging going to take place? Is the charger itself going to convert it to (say) 400V before delivering the power or is there already a DC/DC converter inside the car which converts 800V (or any other voltage) to 400V for charging?

(3) Does an EV have three different ports for 1-phase AC charging, 3-phase AC charging and DC fast charging or is that a single port can accommodate both single and three phase AC charging and then there is another one for DCFC?

## closed as off-topic by Nick Alexeev♦Jun 13 '18 at 18:12

• This question does not appear to be about electronics design within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• One question per question, please. Adding the 3rd question after the answers have been posted is an overstep. – Nick Alexeev Jun 13 '18 at 18:12
• Ok, should I rollback and ask the third one separately? – Pikachu Jun 13 '18 at 18:13

In general the power of a charger is the maximum delivered output power of a device. The power it takes from the mains will be higher and is often printed near the mains input.

So say we have a 10 kW charger with a 90 % efficiency then it will take up to 11.1 kW power from the mains.

90% * 11.1 kW = 10 kW

This is my engineer's view on power and efficiency, this is how I think it should be. It is possible that Marcus' answer is closer to the truth and that would mean that at a 10 kW charging station at 90 % you actually get only 9 kW. It is entirely possible that they make you pay for the 10% of power lost in the charger.

Different cars have different batteries at different voltages. For proper charging the actual voltage across the batteries must be monitored while controlling the charging current. The charging current depends on the type, state of charge, temperature and rated capacity of battery. The charging station does not have all this information so it cannot charge the battery properly.

It is much safer and simpler to let the charge controller in the car decide how the battery must be charged. The fast charging station only has to deliver the power.

There might be differences in what input voltages the car charger can support, it might communicate that to the charging station. If the car says: "I can handle 800 V, give me 800 V" only then will the charging station output 800 V. Without the confirmation it might stick to 400 V.

The charge controller in the car will be responsible for converting the 400V, 800 V or whatever voltage to the voltage required by the batteries.

The above is my best guess about how this should/could work. Not all (fast) charging might work the same everywhere in the world. Several charging standards are described on Wikipedia.

• Suppose there is a charger rated 72V DC. Is it possible to charge a 375V battery with such a charger? – Pikachu Jun 13 '18 at 17:55
• That depends. A charger can be designed and build to work with an input voltage of say 50 V to 1000 V AC or DC. But that would make the charger expensive and certainly inefficient. What would be the point of charging a high power, high voltage battery from 72 V DC? That would require upconverting the voltage and that always results in power being lost. There is no reason why a car should be able to charge from 72 V DC as almost no one needs it. Those that do should get a converter. Also charging from 72 V means power is limited so charging will take forever. – Bimpelrekkie Jun 13 '18 at 18:39

When we refer to 'X kW' charger, is it that the charger is taking in X kW power and delivering power = X kW times the efficiency of the charger to the battery OR is it that X kW is what is being delivered and so the actual power being consumed from mains is X kW divided by efficiency.

Since this is usually not formally defined (unless it is, but that would require a law), assume marketing.

Marketing means that the numbers need to look as shiny as possibly without getting into a liability lawsuit. That means, for example:

A 12 kW charger will deliver up to 90% of 12 kW, and deliver what stays after it gets mightily hot, into whatever load maximizes the power consumption. Probably a short circuit.

If the nominal voltage of car battery is 375V and one wants to charge the battery at a fast charging station which has chargers rated at 800V, how is the charging going to take place? Is the charger itself going to convert it to (say) 400V before delivering the power or is there already a DC/DC converter inside the car which converts 800V (or any other voltage) to 400V for charging?

We need to distinguish between "charging station" and "charger". Usually, the charging controller is part of the car or even the battery pack. The charging station typically is but a power meter and pay station with a power outlet, more or less. Exception are DC charging stations, which get instructions on how much current to deliver continously from the charge controller in the car.

That system is... easy to trick: https://gonium.net/schwarzladen.html