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

oktane

Active Member
Oct 25, 2016
1,872
2,842
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.
 

Snowstorm

Active Member
Dec 8, 2016
1,568
1,505
Ontario Canada
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

MSLR #RN11510 6/7/21
Jun 6, 2013
3,084
2,919
California
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?
 

Snowstorm

Active Member
Dec 8, 2016
1,568
1,505
Ontario Canada
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

MSLR #RN11510 6/7/21
Jun 6, 2013
3,084
2,919
California
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
1,872
2,842
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.
 
Last edited:

n2mb_racing

Active Member
Jun 14, 2014
1,309
1,136
durham, NC

n2mb_racing

Active Member
Jun 14, 2014
1,309
1,136
durham, NC
Any clue where you would power it from? My understanding is there are no posts on the battery. Could you pull power in the rear controller set up in the back?
If it is like the Model 3, you can get high current 12V directly from the DC/DC output terminal. Just be sure to add a fuse for your amp.
 

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