@cyuI’ll answer in several comments.
One general rule is that the cost to run the inverter, onboard charger (for AC—L1 and L2—charging), and other electronics is essentially constant. Lower current through those usually means longer run-time, which means that constant power draw by the electronics leads to lower efficiency.
Example:
Using an L1 EVSE vs a low-amperage L2 EVSE vs a mid-amperage L2 EVSE vs a max-amperage L2 EVSE. Assume the onboard charger (AC to DC converter) requires 200 watts. This will be different for different brands, etc.; this is just illustrative.
There are other losses, so these numbers are inaccurate as you go to higher currents, mostly because of joule heating. So you won’t see much better energy efficiency above about 48 A. And this isn’t the efficiency of the entire system, just that one component. You’re going to see around 95-97% efficiency maximum.
- L1 = 1.4 kW draw from the wall. 200 watts goes to the onboard charger, 50 watts goes to the EVSE. That means, discounting any heat generated, only 1.15 kW goes to the battery. So you get an 82% efficiency.
- L2 @ 15 A = 3.6 kW draw from the wall. 200 Watts to onboard charger, 50 Watts to EVSE, ignore heat generated. 93% efficiency
- L2 @ 25 A = 6 kW: 96% efficiency
- L2 @ 48 A = 11.5 kW: 98% efficiency
L2 @ 80A = 19.2 kW: 99% efficiency- L2 @ 80A = 19.2 kW: 97% efficiency
I was thinking of a family jewels joke but realized that would be inappropriate. Same boat but I always appreciate Space educating us.I won’t pretend to understand all of this but I like to read it anyway and look some stuff up…
I did date a Jules in college and she was hot, but I don’t think that counts here…
![]()
I’m thinking @cyure is probably just mumbling math, math, math…..