fritter63

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I will probably some flax for this, but on my EV Conversion one of my Tesla S modules had (4) 18650 cells fail in the same cell group. Rather than wasting a $1600 module I connected 4 matched 18650 cells in a parrallel group and connect that to to the bus plate that included the damaged cells. I approximated an appropriate fuse size. I has worked for 6 months now.
PLEASE tell us that you used duct tape and not bailing wire for this??? :cool:
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TechOps

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lol, BABC .. also known as a step-up transformer.

I can think of two ways to retain backward compatibility with V1-V3 superchargers, and also allow 900V V4 superchargers:

Option 1. A 2:1 step-up transformer would boost incoming DC voltage from 450V to 900V on V1-V3. (As mentioned by Luke42, that would imply more weight, metal, and cost on-board the CT)

Option 2. They could have two sets of connection points for the battery pack toward the incoming DC charger, by simply changing how many battery strings are in series. Connections 1 would be at 900V and connections 2 would be at 450V. (I see this as much more likely due to the downsides of option 1 .. "the best part is no part.")

Option 3. Some other clever idea Tesla's electrical engineers come up with.

Everything beyond the battery pack could be at 900V, which would give the benefits mentioned by Elon, mainly a reduction in wire gauge and material weight.
 

LDRHAWKE

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When describing voltage, current, and resistance, a common analogy is a water tank. In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, and current is represented by the water flow. So for this analogy, remember:

  • Water = Charge
  • Pressure = Voltage
  • Flow = Current
Consider a water tank at a certain height above the ground. At the bottom of this tank there is a hose.



Tesla Cybertruck Cybertruck 900V architecture confirmed? 5113d1c3ce395fc87d000000



The pressure at the end of the hose can represent voltage. The water in the tank represents charge. The more water in the tank, the higher the charge, the more pressure is measured at the end of the hose.

We can think of this tank as a battery, a place where we store a certain amount of energy and then release it. If we drain our tank a certain amount, the pressure created at the end of the hose goes down. We can think of this as decreasing voltage, like when a flashlight gets dimmer as the batteries run down. There is also a decrease in the amount of water that will flow through the hose. Less pressure means less water is flowing, which brings us to current.
 

fritter63

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When describing voltage, current, and resistance, a common analogy is a water tank.
Exactly how I always think (and explain) it.

Remember, a pressure washer can take the paint off your car.....:ROFLMAO:
 


MikeF

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I checked through the thread to see if this Webcast was already referenced, and it doesn't appear to be. It has a discussion from the Tesla Q1 Financial Results and Q&A Webcast, and includes Elon, during a discussion of 800V Architecture. Maybe since this was discussed during the Q1 Earnings Call, more study revealed that even more advantage could be gained from 900V Architecture. Like many CT related issues, they are still somewhat fluid.

 


SwampNut

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Eh... if sized and applied correctly transformers typically have an efficiency rate in the high 90's.
You seem to be suggesting that's trivial. I don't care if it's just 3% loss, that's a ton. Plus more heat, something battery systems already struggle to control.
 

Bill906

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You seem to be suggesting that's trivial. I don't care if it's just 3% loss, that's a ton. Plus more heat, something battery systems already struggle to control.
You said there is a lot of heat and losses. "a lot" is subjective. I personally would not use "a lot" or "a ton" to describe the heat loss in a well applied transformer. Any voltage change is going to have some losses. If the correct method is applied and sized correctly, it would not, relatively speaking, be "a lot".
 

Jescocom

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Can someone please explain what this means? Does it mean better charging times? More utility, better safety?
Power, i.e. Wttage, equals volts times amps. (P=E*I, ohms law). So, the more voltage you have ,the less amperage you need to produce the same amount of power. More Voltage with the same amount of amps would mean more power! I have no Idea if it would effect charging, but it would definitely effect torque!
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