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900 Volt Battery System

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Lucid Motors is touting a 900 Volt battery system for the upcoming Air.

Stretching my knowledge of electricity, I can wrap my head around Ohm's law - barely.

What are the upsides and downsides of the Lucid approach?
 
i think the standard answer is that you can send more power over a smaller wire, or equivalently have less loss in the same wire. But then you have to worry about sparks and other ways that the power can move from one place to the other overcoming resistance (eg wrapping the wire in an insulator). That has an impact on the wires in the magnets you put in the motor of course. That's a theoretical answer I'm not sure there's a clear cut practical answer.
 
One downside of higher voltages I see is that charging at home will be less efficient. Converting 120v(or 240v) AC to 900v DC seems like it has to be less efficient than 400v.

And then there is the marketing guys pushing this. Could be like the frequency rating on computer CPU's. Biggest number = bragging rights. Even when models with lower frequency may provide better performance.
 
I think the big driver is power transistors.

All modern EVs use one or another form of AC drive motors, with dedicated inverters feeding synthetic multiphase waveforms to them.

That means you have to have a lot of transistor power controlling high currents - usually parallel banks of high power mosfets or the like.

Each of those transistors is connected to the full battery voltage - and transistors that can safely handle higher voltages are harder to make and more expensive.

I think a major reason the industry settled on ~400V packs for the last decade was the best mass market power transistors had a 600V max.

Given Taycan, obviously there must be higher voltage options out there now; the cost and availability trade off may be worth it at some point.
 
It allows the pack to overcome more motor back-EMF.
This enables a combination of the following:
  • more windings which allows more torque for the same input current
  • more peak power any speed.
  • a higher top speed (due to more power at top end)
 
As far as charging is concerned the battery voltage should have no effect on charge speed. The car will charge in the time it takes to charge an individual cell in the battery. Efficiency gains due to reduced current should apply outside of battery modules, but individual cells have to see the same voltage and current they need regardless of the battery pack voltage. So losses within a cell should remain the same as always.

I suspect the push to higher voltages is mostly for lighter weight (smaller wires).
 
I think the big driver is power transistors.

All modern EVs use one or another form of AC drive motors, with dedicated inverters feeding synthetic multiphase waveforms to them.

That means you have to have a lot of transistor power controlling high currents - usually parallel banks of high power mosfets or the like.

Each of those transistors is connected to the full battery voltage - and transistors that can safely handle higher voltages are harder to make and more expensive.

I think a major reason the industry settled on ~400V packs for the last decade was the best mass market power transistors had a 600V max.

Given Taycan, obviously there must be higher voltage options out there now; the cost and availability trade off may be worth it at some point.

I was thinking it would be for faster charging - but not so.

You appear to have the answer, article just posted:

Full Page Reload
 
“The main loss in the battery is from impedance in the battery pack,” Rawlinson says. “Now in an [U.S. Environmental Protection Agency] test, the actual currents drawn are so small that those losses are tiny, so efficiency of the pack doesn’t really come to play. But in high-performance driving it matters: Double the current, you get 4 times the losses.”

For the same motor power output and pack capacity (cell count), putting the cells in series (double the output voltage) doesn't impact resistive pack loss.
Output power = I*V
With the 400 V connection of cells having a resistance of R:
loss in pack = I^2*R
Cut pack in half: each section has resistance of 2R, stack them in series: 2*2R = 4R
Voltage is doubled so current is halved: I*V= 2V*(1/2)*I
loss in pack = (1/2*I)^2*(4R) = (1/4)I^2*4R = I^2*R

Tesla's new tabless cell patent would reduce the resistance of each cell and thus boosts efficiency.
 
Charging stations that are only for 400V 500V systems are not directly compatible with 800V+ packs.
Porsche Taycan has standard 50kW DC-DC converter for those normal rapid charging stations.
Optional 150kW DC_DC converter is offered that allows charging Taycan's 800V pack at good speeds.
Though Ionity network supports Taycan correctly at 270kW.
Most "better than 50kW" stations support 1000V AFAIK.