# What Resistor do I need for an LED light bar in a 12V Auto System?

Hi All,

I am wanting to build an LED light bar from dollar store flashlights to use as driving lights on my truck. There is currently a 12V wiring harness with 15A relay that I plan to tap into. As for the lights I want to use, I have no idea how many mili amps they use for voltage. There are 7 LEDs inside of each flashlight head. I know that each flash unit head draws 2.37V when I test for voltage drop. My thinking is to run 3 head units in series to reach a total of 7.11V of resistance. My question is, how big of a resistor do I need to put on the power supply in order to not fry the lights? There will be a total of 6 lights altogether, but I just planned on running them in series of 3 units. Would it be better to run all 6 in series? If this doesn't make any sense, let me know and I will try and explain it better. Thanks.

I am wanting to build an LED light bar from dollar store flashlights to use as driving lights on my truck. There is currently a 12V wiring harness with 15A relay that I plan to tap into. As for the lights I want to use, I have no idea how many mili amps they use for voltage. There are 7 LEDs inside of each flashlight head. I know that each flash unit head draws 2.37V when I test for voltage drop. My thinking is to run 3 head units in series to reach a total of 7.11V of resistance. My question is, how big of a resistor do I need to put on the power supply in order to not fry the lights? There will be a total of 6 lights altogether, but I just planned on running them in series of 3 units. Would it be better to run all 6 in series? If this doesn't make any sense, let me know and I will try and explain it better. Thanks.

oldestThen

the next easiest thing to trywould be a stack of two modules in series, also in series with about 30 to 50 ohms of external resistors, and also in series with the 12V battery. Also I think the easiest way to find the amount of resistance you want, would be to start with a series stack of five 10 ohm resistors (5*10 = 50 ). Then if this is not bright enough, try a stack of four 10 ohm resistors (4*10 =40). If more brightness than that is desired, try three (3*10 =30)Another advantage of using a series stack of 10 ohm resistors for your test, is that you can measure the voltage across

oneof the 10 ohm resistors, and from that measurement infer the current flowing through your modules. E.g. if a 10 ohm resistor has a voltage 1.0V across it, then it must have a current of 0.1A = 100mA flowing through it. If it has voltage of 0.5V across it , then it has a current of 50mA flowing through it.I am hoping the attached picture-drawing

https://www.instructables.com/file/FEVIXORH1AH6VN9/

will help to clarify what I am describing.

Guessing that each one of these modules will be happy with something like 100mA to 200mA flowing through it, which works out to something like 15 to 30 mA per LED, assuming each of those 7 parallel LEDs in the module is drawing equal amounts of current.

Using 10 ohm resistors with a power rating of 0.5 watt, implies that the maximum current through that resistor is about 220 mA. Using 10 ohm resistors with a power rating of 0.25 watt, implies a maximum current of about 150 mA. So you should probably use the half-watt size 10 ohm resistors, if you can find those, rather than the quarter-watt ones.

There might also be a small difference in brightness that varies with the trucks battery voltage, which will vary with other influences, e.g. I would expect the battery voltage to be a few volts higher when the battery is charging.

I am guessing any such variation in brightness, due to varying battery voltage, will be small, and not troublesome.

BTW here is a pic how to measure current by using a strip of two sided PCB

inserted between the normal current path.

Something just slightly strange here

Each new cell is 1½ Volts, three should ( 1½+ 1½ + 1½ ) = 4½ V...

They drop voltage with use which is probable in your keeping..

The light is satisfactory for you at 4V then it will be better at 13½V to 14V

which is your truck's running voltage.

You made a good decision at first, post a pic when you have it made :-)

A

You can only put two in series ( 4½ + 4½ ) = 9 VDC.....

And three in series ( 4½ + 4½ +4½ ) = 13½ VDC would

not light at all when the engine wasn't running.

However when two are in series you need to provide a ( 12 - 9 ) = 3 volt

drop in series with the two lights.

Now you use ie mix current ( mili amps ) with voltage in the same

sentence.. You draw current Never Volts...

Just measure the current of one flashlight and We can calc the resistance or just place a 3 volt zener diode in series.

A

"You can only put two in series ( 4½ + 4½ ) = 9 VDC..... And three in series ( 4½ + 4½ +4½ ) = 13½ VDC would not light at all when the engine wasn't running."I assumed the same before I measured what I thought was the actual draw from the unit. Is it not 2.37V? If it is, then 3 in series would only be 7.11V and I need the difference in a resistor from the 12VDC off the battery.

Thanks for your help. I usually know just enough about stuff to get myself

intotrouble, but not enough toget outof it!