ajdelange
Well-Known Member
- First Name
- A. J.
- Joined
- Aug 1, 2019
- Threads
- 9
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- 2,883
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- 2,317
- Location
- Virginia/Quebec
- Vehicles
- Tesla XLR+2019, Lexus, Landcruiser, R1T
- Occupation
- EE Retired
Let's look at a pair of 3 volt batteries powering a 3 volt motor rated at 6 watts. It draws 2 A from the batteries which we would connect in parralell. Each would clearly supply 1 amp. If the battery impedance is 0.1 Ω then I^2R losses in the battery would be 0.1 W each for a total of 0.2 W. If the motor resistance is 0.15 Ω then the I^2R losses in the motor would be .15*2*2 = 0.6 W. Now let's rewind the motor for 6 volts. It would require twice the length of wire and if we use the same size wire the resistance will double to 0.3 Ω but the motor will only draw 1 ampere from a 6 V supply and the losses in this new motor will be 0.3*1*1 = 0.3 W or half what they were in the 3 V motor. But our wire cost has doubled and the motor is more expensive so we would very probably go to a smaller wire size. One with twice the resistance per unit length would result in wire resistance of 0.6 Ω which, with 1 amp current would dissipate 0.6W which is where we were with the 3 V motor. So with respect to motors there is a potential advantage with respect to loss depending on what the designer is willing to spend on copper.
In the batteries deliver 3 watts each whether they are connected in series or parallel so each is delivering 1 A in either configuration and the internal loss is therefore 0.1 W each for a total of 0.2. It's the same with charging.
In the interconnection between batteries and motors if you keep the same amount of copper the current drops in half and the power lost goes down to a quarter. So obviously you are going to look at thinner wire. unlike the case with the motor you don't have to increase the wire length so if you go to wire of half the cross section (with double the resistance you wind up loss reduced by a factor of (2)(1/4) = 1/2. Now I saw a blurb for one 900 volt advocate (can't remember which one) saying that this would save 60 lbs of copper per car. That is appreciable.
Finally we have the inverters. A transistor switching 1 amp will dissipate 1/4 of the heat dissipated by a transistor switching 2. That is a substantial reduction, of course, and could have significant implication on the cooling system. But it would require semiconductors with PIV tolerance twice what the 400 V designs need.
And I guess we need to mention the chargers. The manufacturers want to charge at rates that require so much copper in the cables that they become too unwieldy. The solution (which has been used) of liquid cooling the cables isn't too appealing. Doubling the voltage means, with wire half the size, half the power dissipation. There is clear advantage there.
In the batteries deliver 3 watts each whether they are connected in series or parallel so each is delivering 1 A in either configuration and the internal loss is therefore 0.1 W each for a total of 0.2. It's the same with charging.
In the interconnection between batteries and motors if you keep the same amount of copper the current drops in half and the power lost goes down to a quarter. So obviously you are going to look at thinner wire. unlike the case with the motor you don't have to increase the wire length so if you go to wire of half the cross section (with double the resistance you wind up loss reduced by a factor of (2)(1/4) = 1/2. Now I saw a blurb for one 900 volt advocate (can't remember which one) saying that this would save 60 lbs of copper per car. That is appreciable.
Finally we have the inverters. A transistor switching 1 amp will dissipate 1/4 of the heat dissipated by a transistor switching 2. That is a substantial reduction, of course, and could have significant implication on the cooling system. But it would require semiconductors with PIV tolerance twice what the 400 V designs need.
And I guess we need to mention the chargers. The manufacturers want to charge at rates that require so much copper in the cables that they become too unwieldy. The solution (which has been used) of liquid cooling the cables isn't too appealing. Doubling the voltage means, with wire half the size, half the power dissipation. There is clear advantage there.
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