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Comment by idiotsecant | Hacker News Reader

Internal heating is proportional to current. Higher voltage means lower current. Charging at the same wattage implies lower current.

So: higher voltage implies lower heating.

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The cells need to charge whether they're in series or parallel. The efficiency they can absorb charge at high speeds without heating up is not primarily determined by how they're wired. You could wire 5,000 cells in series, charge them with 20kv at 4 amps, or wire them all in parallel, and charge with 4 volts at 20 ka. Each cell will produce the same amount of heat either way, they only charge with about 95% efficiency. Higher voltage doesn't really reduce the need for active cooling if you want to keep the cells under 40-50C.

Energy loss though resistance in the pack's internal wiring is likely a lot less than the loss due to the chemistry not being 100% efficient at absorbing (or delivering) charge without heating up. But it does allow for thinner wires to get max power out of the battery.

But I'm pretty sure the voltage seen by each individual battery is always the same, regardless of the distribution system voltage.

There should be less heating for higher voltages but if most if the heating is in the battery vs the distribution system then the higher voltages will not help much.

Also, if they make all wires smaller to save money and weight then there might not be any change in heating.

The current per cell is still the same. 800V charging just means that you put cells in series banks to achieve an ~800V module-level voltage. Current is reduced in the main charging cables, charge port, and pack fuse/contactor, but not in the individual cells.

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