3 ms·

I think you're overestimating the difficulty of moving 500A. As GP said, liquid-cooled cables are the solution and come in reasonably-sized cable thicknesses (C

I think you're overestimating the difficulty of moving 500A. As GP said, liquid-cooled cables are the solution and come in reasonably-sized cable thicknesses (CCS form factor). This is ITT Cannon's product as an example: https://ittcannon.com/core/medialibrary/ittcannon/website/literature/catalogs-brochures/itt-cannon-evc-dc-liquid-cooled-brochure.pdf https://ittcannon.com/core/medialibrary/ittcannon/website/li... Also note that 1000V isn't an upper bound; many vehicles under design now will be ~1000V, and I expect that to push up to at least 1500V within the next few years.

imtringued 6y ago
I'm not an electrical engineer but I ran into the I^2R heat loss problem with a simple mosfet circuit. If I drive the mosfets at 48V (voltage is actually irrelevant) and 70A and the RDS of the mosfet is 0.002 Ohm then the mosfet will lose 9.8W of power to heat. For a small compact SMD mosfet that's a lot of heat and requires a heatsink. Once you scale up to 500A you are basically losing 500A x 500A x 0.002 Ohm = 250W of power just on a single mosfet. I'm not very experienced in this subject but I noticed that mosfets with lower voltage ratings tend to also have lower drain source resistance (RDS) which means doing this at 1000V can only get harder.
- aetherspawn 6y ago
  As the parent comment said, moving high currents in cables isn’t that difficult. Mosfets are small devices, but cables have a lot of surface area and mass, so even dissipating kW into a cable is not an issue and can be solved with liquid cooling. Imagine we used 35mm2 at 0.55 Ohm/km and we have 2 poles. If the charging cable is 3m including inside the device, that’s 2x3 = 6m. That’s 6.6 mOhms total, so I2R on that at 500A is only 1.65kW, which is nothing.