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2021 Refresh: Model S (and X?) Changes to 12v System

oktane

Active Member
Oct 25, 2016
2,067
2,981
USA
Calculate the load and throw in a resistor in series to get the needed voltage drop 😂😂😂😂😂😂🤣🤣

For example if the cig lighter voltage is 15.5V, and needed voltage is 13V, and the radar detector draws 300mA, a 8 ohm resistor (rated at 0.75W) wired in series should do the trick.

V=IR so 0.3 amps x 8 ohms = 2.5 volts dropped.

P=IV = 2.5 volts x 0.3 amps = 0.72 wattage resistor required

The caution here is I'm assuming a constant 300mA current draw by the radar detector, but it could be more for startup and less for continuous operation, which means the voltage drop across the resistor and the voltage to the radar detector will fluctuate, as will the power across the resistor, so a higher rated resistor is better.

Please use any of this information at your own risk it might be completely wrong. Also remember, a resistor dissipating energy will heat up a little.
 
Calculate the load and throw in a resistor in series to get the needed voltage drop 😂😂😂😂😂😂🤣🤣

For example if the cig lighter voltage is 15.5V, and needed voltage is 13V, and the radar detector draws 300mA, a 8 ohm resistor (rated at 0.75W) wired in series should do the trick.

V=IR so 0.3 amps x 8 ohms = 2.5 volts dropped.

P=IV = 2.5 volts x 0.3 amps = 0.72 wattage resistor required

The caution here is I'm assuming a constant 300mA current draw by the radar detector, but it could be more for startup and less for continuous operation, which means the voltage drop across the resistor and the voltage to the radar detector will fluctuate, as will the power across the resistor, so a higher rated resistor is better.

Please use any of this information at your own risk it might be completely wrong. Also remember, a resistor dissipating energy will heat up a little.
This is a tricky thing to do as V drop across the resister will vary based on current. An alternative is to use a diode of sufficient power rating and get a 0.7V drop per diode. These solutions to burn power and may work for low power accessories, but is not suitable for mid to high power applications such as tire compressor or inverter.
 

4SUPER9

2022 Model S
Jun 6, 2013
3,372
3,308
So Cal
This is a tricky thing to do as V drop across the resister will vary based on current. An alternative is to use a diode of sufficient power rating and get a 0.7V drop per diode. These solutions to burn power and may work for low power accessories, but is not suitable for mid to high power applications such as tire compressor or inverter.
Perhaps my diodes are a little on the wimpy side. I put one in line and got a drop of at most 0.2V. It may be a little silly to line up 7 or 8 of these. Might some one have a recommendation for something a little more robust?
 
Perhaps my diodes are a little on the wimpy side. I put one in line and got a drop of at most 0.2V. It may be a little silly to line up 7 or 8 of these. Might some one have a recommendation for something a little more robust?
Something is off, if you use a normal silicon based diode (almost all of them are that), your voltage drop would be around 0.7V when current is flowing. Now, if no current if flowing at all, then you won’t get any drop. But as soon as you start to flow current it will drop 0.7v. Make sure you measure it when some current is flowing, even just a little (say 100mA)
 

4SUPER9

2022 Model S
Jun 6, 2013
3,372
3,308
So Cal
Something is off, if you use a normal silicon based diode (almost all of them are that), your voltage drop would be around 0.7V when current is flowing. Now, if no current if flowing at all, then you won’t get any drop. But as soon as you start to flow current it will drop 0.7v. Make sure you measure it when some current is flowing, even just a little (say 100mA)
So a voltage meter does not qualify as flow? Please forgive my ignorance. That is what I am using and got a .25 drop with 1 diode, and about .5 drop running 2 diodes in series.
 

oktane

Active Member
Oct 25, 2016
2,067
2,981
USA
Something is off, if you use a normal silicon based diode (almost all of them are that), your voltage drop would be around 0.7V when current is flowing. Now, if no current if flowing at all, then you won’t get any drop. But as soon as you start to flow current it will drop 0.7v. Make sure you measure it when some current is flowing, even just a little (say 100mA)
Not a bad idea, or maybe put a resistor in series with a zener diode in reverse bias and use the fixed voltage across the diode to power the load.
 
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Battpower

Active Member
Supporting Member
Oct 10, 2019
2,395
2,439
Uk
I don't get why with all the tech in the car, Tesla couldn't just stick in a small buck converter to run any ancillary equipment. 12v is the automotive standard everywhere for cars. Voltage range is well established and while 16v might appear momentarily in some legacy cases, it would be verging on a fault condition.

Unlike resistive / diode based solutions that just dissipate energy as heat, a little buck converter will run cool, likely stabilise the voltage to within a few mV and also limit the current to a level you may well be able to set depending on the convertor. All for the price of a couple of beers.

But again, why on a car so tech laden not just provide 12v ancillary power that meets well established norms?
 
I don't get why with all the tech in the car, Tesla couldn't just stick in a small buck converter to run any ancillary equipment. 12v is the automotive standard everywhere for cars. Voltage range is well established and while 16v might appear momentarily in some legacy cases, it would be verging on a fault condition.

Unlike resistive / diode based solutions that just dissipate energy as heat, a little buck converter will run cool, likely stabilise the voltage to within a few mV and also limit the current to a level you may well be able to set depending on the convertor. All for the price of a couple of beers.

But again, why on a car so tech laden not just provide 12v ancillary power that meets well established norms?
They wouldn't have needed to do that even. They could have used a 4s 12V lifepo battery, which would have kept the 12V voltage right at 12.8V. I have no idea why they went with Li-ion, especially since that battery is cycling at least once a day, if not more. Lifepo would be better suited for thousands of charge / discharge cycles.
 
But again, why on a car so tech laden not just provide 12v ancillary power that meets well established norms?
They also get a performance improvement by raising the 12V voltage, the current goes down. All the accessories perform better. They only need to guarantee that all their stuff works up to 15V. I don't think they care about aftermarket accessories. The peak voltage is 16.8V, but I hope they never charge the Li-ion battery that high. They shouldn't go above 16V (4V/cell or ~80%) to maximize the battery life.
 

Battpower

Active Member
Supporting Member
Oct 10, 2019
2,395
2,439
Uk
They wouldn't have needed to do that even. They could have used a 4s 12V lifepo battery, which would have kept the 12V voltage right at 12.8V. I have no idea why they went with Li-ion, especially since that battery is cycling at least once a day, if not more. Lifepo would be better suited for thousands of charge / discharge cycles.
I agree. I was working from the assumption that there must have been some other reason for the choice but not one I could think of!
They also get a performance improvement by raising the 12V voltage, the current goes down. All the accessories perform better. They only need to guarantee that all their stuff works up to 15V. I don't think they care about aftermarket accessories. The peak voltage is 16.8V, but I hope they never charge the Li-ion battery that high. They shouldn't go above 16V (4V/cell or ~80%) to maximize the battery life.
Do you really think power loss from I² matters for low power electronics?

Agree there is a theoretical benefit, but in the efficiency vs compatibility balance in this case I would have thought sticking to convention would win hands down.

Are there actually any devices that have components that genuinely don't function (as effectively) bellow say 15v? I saw LED headlights mentioned. My Kona has projector LEDs that equal or outperform the headlights on my other cars and they run at 12v just fine apparently.

We are only talking a difference of what, 2 to 3 volts. Even at 200 watts load power the additional power lost in the cables isn't going to be much.
 
I agree. I was working from the assumption that there must have been some other reason for the choice but not one I could think of!

Do you really think power loss from I² matters for low power electronics?

Agree there is a theoretical benefit, but in the efficiency vs compatibility balance in this case I would have thought sticking to convention would win hands down.

Are there actually any devices that have components that genuinely don't function (as effectively) bellow say 15v? I saw LED headlights mentioned. My Kona has projector LEDs that equal or outperform the headlights on my other cars and they run at 12v just fine apparently.

We are only talking a difference of what, 2 to 3 volts. Even at 200 watts load power the additional power lost in the cables isn't going to be much.
I imagine everything on the 12V bus is designed to work down to 10V and up to 16V, at least. So, everything should work fine at any voltage in that range. It probably wasn't much work to verify that everything works at 16V.

i2r loss for stuff like the seat heaters or other high current 12V stuff might be something.

Going from 12V to 16V reduces current 25%, so i2r loss in wiring goes down ~44%. Not bad for very little cost. For a 30A circuit, that might add up to something. And it probably improves performance on stuff like the stereo amplifier, etc.
 

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