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Comment by octoberfranklin | Hacker News Reader

You basically dodged his question there.

Increasing I^2R heat loss by 5% doesn't let you shrink the cable much. You go from "humongously phat cable" to "really phat cable". The problem remains.

Also remember that the cost of the petroleum distribution infrastructure is shared between in-town drivers and road-trip drivers. If these "fast chargers" are only supposed to be used by people who need more than one batteryful of driving in a given day (i.e. people on a road trip), the budget available to pay for the refueling infrastructure is going to be tiny compared to the petroleum infrastructure. EVs will simply never "catch on" for non-commuter use if we go down that road.

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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/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.

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.

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.

Let's just go superconducting cable and we're good to go!

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