stopcrazypp
Well-Known Member
You are talking about something different. He is talking about peak power available. For example, the new battery fuse can allow 1500A. Ignoring voltage sag and conversion losses for simplicity using your examples:lolachampcar
I do know that but what I was trying to say is; isn't there a transformer somewhere in between the pack and the motor that regulates DC voltage into AC? I am assuming here that AC motor always likes a specific range of voltage and it gets regulated. To give an example; (all assumptions)
SoC 100% --> Pack voltage 400V
You demand power from the motor with your right foot, 403V DC gets converted to 375V AC (why not 375 for the sake of the example) and your foot decides the amps you'll get out of it.
SoC 50% --> Pack voltage 370V (again, all assumptions)
You demand power with your right foot, 370V DC converted to 375 AC again (via the transformer) and your foot decides the amps for the kW needed.
But there's far more into this than I don't know. Heck even charging voltage we see whilst supercharging and pack voltage when discharging is different. Also, when you floor it, there's this battery phenomena called pulsing where voltage drops a lot and gets back up after you stop flooring it. All of this is just stuff I read about online but of course lack the expertise and detail. So I'd love anyone with more and thorough knowledge to chime in and any suggested reading is much appreciated.
SoC 100% --> Pack voltage 400V*1500A = 600kW max -> 600kW/375V AC = 1600A max AC
SoC 50% --> Pack voltage 370V*1500A = 555kW max -> 555kW/375V AC = 1480A max AC
Above examples will have to be adjusted for voltage sag depending on load and also conversion losses, but the general idea is there that the more power available from battery, the more power available to the motor.