Power sharing with EV chargers

[Eclectic engineer]

A 3-phase 100A 120/208V panel contains 6 x 40A EV charging circuits connected to it. These EV chargers are 1-phase loads that operate at 208V. While the circuit breakers are rated for 40A, the max. amp draw is restricted to 32A because of the continuous load rule.
The 6 chargers are controlled by a software that restricts the total amperage draw from the panel to 80A. Each charging circuit uses a double-pole breaker that uses 2 out of the 3 phases of the panel. Following is the configuration -

Charger 1 - L1/L2
Charger 2 - L1/L2
Charger 3 - L2/L3
Charger 4 - L2/L3
Charger 5 - L1/L3
Charger 6 - L1/L3

My question -
When all 6 chargers are actively charging EV's, how will the amps be distributed across the 3 phases? Since the load limit is set to 80A, I am thinking that each charger will get 80/6 = 13.33A each. Is this right?
If it is 13.33A, then each phase is only loaded to 13.33 x 4 = 53.32A based on the connection combo. listed above.
I am a bit confused here. While each phase can go up to 80A, the calculation only adds up to 53.32A.
How can I ensure that each phase also totals up to 80A?

 

[wwhitney]

When all 6 chargers are actively charging EV's, how will the amps be distributed across the 3 phases? Since the load limit is set to 80A, I am thinking that each charger will get 80/6 = 13.33A each. Is this right?

No, assuming the software is programmed correctly.

If I1 is the current to charger 1, I2 to charger 2, etc, then on L1, currents I1 and I2 will add. Currents I3 and I4 won't contribute (not connected to L1). Currents I5 and I6 will add. And if I1 + I2 = I5 + I6, then the current on L1 will be sqrt(3)*(I1 + I2). When I1 + I2 differs from I5 + I6, the formula is more complicated.

So for the case that all the chargers are drawing the same current I, then the limitation should be 80A = sqrt(3) * 2 * I, or I = 23A.

Cheers, Wayne

 

[Eclectic engineer]

No, assuming the software is programmed correctly.

If I1 is the current to charger 1, I2 to charger 2, etc, then on L1, currents I1 and I2 will add. Currents I3 and I4 won't contribute (not connected to L1). Currents I5 and I6 will add. And if I1 + I2 = I5 + I6, then the current on L1 will be sqrt(3)*(I1 + I2). When I1 + I2 differs from I5 + I6, the formula is more complicated.

So for the case that all the chargers are drawing the same current I, then the limitation should be 80A = sqrt(3) * 2 * I, or I = 23A.

Cheers, Wayne

Hey Wayne. I am a bit lost with the equations here. Can you recommend a book or article where I can read more about this logic you described?

 

[wwhitney]

I don't have a good recommendation, maybe someone else here can suggest something.

The tricky point is that the current on L1 from an L1-L2 2-wire load is not in phase with the current on L1 from an L1-L3 2-wire load. So the current magnitudes do not add, you have to account for the phase difference. That means the magnitude of the sum of the currents will be less than the sum of the magnitudes.

That's a partial version of how current on the neutral cancels.

Cheers, Wayne

 

[Charged]

Dig around on chargepoint website. I have a print out I always look at but they have the standard power settings table that shows the limited kw each level 2 will get to accomplish different dwell times, short term , long term and overnight. I was doing some fleet charging sites where they would sit all night and I believe it got down to like 1.33kw per station and would still give them a full charge. If you compare that to a car acceptance rate , it works out. Clipper creek always has a good chart on there website that stays up to date with the acceptance rates for the different vehicles.
Another way to do it is set the limit on the panel and they will just give them as much kw as they can. There’s a lot more that goes into it , first car in gets more and how it’s distributed has cars get in system. Of course all systems are different. Chargepoint has a lot of literature though that you can base design practices on.