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I have a rudimentary understanding of volts and amps. (Volts is like pressure and Amps are like the bandwidth of that pressure.) So just reasoning about it, I imagine charging to be an event where the car battery pack is being connected to the power source as an open system. Meaning that the source is pushing electrons into the car battery against the strain of the car battery which is pushing back on the power source. And, thus, achieving a charge implies having more pressure going into the battery than out. Is this correct so far?

So that leads me to wonder: Would I be expending more “effort” (energy) to charge with a weaker 110v push from the power source over a longer period of time than with a stronger 220v push over a shorter period of time?

Further, I understand that there is only about 90% efficiency when charging — some electricity is lost in the process as heat (?) and... (something else?). But I don’t know if that 90% number is relative to the voltage of the power source.

In short: Is it more cost effective to charge my Model 3 at home (in the 20% -- 80%) range on a higher voltage circuit, or is it basically even? And if it’s not even, how can we determine the difference other than by empirical measurement?

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    A naive take suggests it could be more efficient, as losses in resistive components scale with the square of the current, and your 110V circuit is specified as 15A. However I don't propose that as an answer because the manufacturer has the ability to optimise the charger design for certain conditions, and that can be a bigger effect. – Chris H Dec 6 '19 at 6:44
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Let's take a look at resistance!

With 220 volts, the resistance energy waste is R*(30 A)2. Charging power is 220 V * 30 A. So, resistance waste divided by charging power is R * (30 A) / (220 V).

With 110 volts, the fraction is R * (15 A) / (110 V).

Because 15 divided by 110 is the same as 30 divided by 220, if you are using the same cable, the power wasted to resistance is the same.

Now, if you could charge at 220 V and 15 A, you should be getting better efficiency with the high voltage.

However, this is just the cable resistance. The charger could be optimized for a particular voltage, as Chris H noted. It also could be optimized for a particular current.

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