Electronics > Beginners

Car battery confusion.

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IDEngineer:
FWIW, commerical LED bars for vehicles often have switchers in them. I've varied the supply voltage from well under 12VDC to well over 36VDC on some of them and you can see the input current vary inversely with voltage while the brightness remains steady. And some of these were Chinesium eBay units well under $50, so it appears to be the "norm".

james_s:

--- Quote from: Siwastaja on June 21, 2019, 07:35:39 pm ---
--- Quote from: Benta on June 21, 2019, 04:38:34 pm ---Actually, an automotive circuit should be able to withstand +/- 60 V, this can happen during a load dump.

--- End quote ---

Load dump probability is exaggerated. Safety-critical and high-reliability systems require load dump protection, but for a DIY led contraption, I wouldn't necessarily bother. Many cheap gadgets don't have it.


Just make sure it survives continuous 14.5V without overheating or blowing up, as you can normally see this voltage.

Putting fewer LEDs in series (reducing the total voltage over the LEDs) and increasing the resistor value to drop more voltage over the resistor makes the current regulation better against varying voltage, but with the cost of dropped efficiency.

Remember to calculate the power dissipation in the resistor, use a larger resistor size, and add proper fusing for fire safety.

--- End quote ---

I think the load dump thing was more an issue back when cars had primitive electrical systems with electromechanical voltage regulators and the only loads were DC motors and incandescent lamps. Modern cars (as in just about anything made in the last ~40 years) are quite a bit more refined, you can get a lot of spikes and noise but it's not going to be anything huge. Loads of electronics get operated in cars without issues, heck I've worked on ECUs and ignition controllers that had a bog standard 7805 in them.

bdunham7:

--- Quote ---I think the load dump thing was more an issue back when cars had primitive electrical systems with electromechanical voltage regulators and the only loads were DC motors and incandescent lamps. Modern cars (as in just about anything made in the last ~40 years) are quite a bit more refined, you can get a lot of spikes and noise but it's not going to be anything huge. Loads of electronics get operated in cars without issues, heck I've worked on ECUs and ignition controllers that had a bog standard 7805 in them.

--- End quote ---

Actually the load dump has to do with the time constant of the alternator (generator nowadays, of course) field winding and the impedance of the battery--it's ability to absorb large current spikes.  Alternators have time constants of several hundred milliseconds and very high open circuit voltage (hundreds of volts), so if they are operating a significant load that abruptly shuts off, they continue producing currents that can exceed 100 amps for a brief time.  Ordinarily, a properly functioning lead acid battery can absorb huge current spikes without much voltage gain, but if they are either defective, missing or both discharged and very cold, the voltage can go very high.  I've seen this happen many times with cars that were jump-started in cold weather. 

viperidae:
A load dump happens then the battery fails open circuit or gets disconnected while the alternator is charging it.
The standard test for withstanding a load dump in a 12v system is 120V for 400ms.

Some cars have load dump protection systems, some don't.

 As for the original question, your car is going to fluctuate when the engine is running between 13.8V and 14.4V when it switches between charging and float charging.

With the engine of it's going to be a little above 12V.

You're best option is going to be a constant current led driver, one designed to boost the voltage.

andrewlapham:
Could i just use an lm317 to cap the voltage to around 6v then use 2x Leds in series with a 1ohm resistor?

Solution 0: 2 x 10 array uses 20 LEDs exactly
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms
    +----|>|----|>|---/\/\/----+  R = 1 ohms

The wizard says: In solution 0:
  each 1 ohm resistor dissipates 0.4 mW
  the wizard thinks ¼W resistors are fine for your application
  together, all resistors dissipate 4 mW
  together, the diodes dissipate 1280 mW             
  total power dissipated by the array is 1284 mW     
  the array draws current of 200 mA from the source.


Would this work?

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