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.

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I think the easiest thing to try, would be stacking three of those little flashlight modules in series, and in series with your 12V battery. And if that gives you as much brightness as you want, then you could probably just use them that way.

Then the next easiest thing to try would 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 one of 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
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.
RodPlunger (author)  Jack A Lopez5 years ago
This makes a lot of sense. Thanks for the very clear explanation and diagram. Now I just have to assemble my parts and give it a whirl...
RodPlunger (author)  RodPlunger5 years ago
One other question, before I install this setup, can I use any 12v battery for testing? Or do I have to use an automotive battery specifically? I know 12V is 12V, but the amps will be different across the batteries. Does that make a difference? It would just be a lot easier to bench test using a cheap 12V battery. Thanks.
There might be difference in brightness, depending on the particular battery you use, but I am guessing that the difference will be small. If it is more convenient to test your setup with a small 12V lead acid battery, that's fine.  Then after you've tried that you could attach it to a cigarette lighter plug, then just go outside and plug it into your truck, and see how that looks.  Here I am guessing that plugging into the cigarette lighter would be easier than wherever it was you were planning on permanently attaching it.

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.
iceng5 years ago
IF the battery pack measurement is 4V then Use three...........

BTW here is a pic how to measure current by using a strip of two sided PCB
inserted between the normal current path.

RodPlunger (author)  iceng5 years ago
So use 3 units in series with no resistor? I can probably do that.. Thanks!
Yes no resistor needed nor worth any concern.

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 :-)

iceng5 years ago
3AA cells is a 4½ VDC flash light.... The truck is 12VDC.....

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.

RodPlunger (author)  iceng5 years ago
I think I measured them the way you want. I measured the battery pack at 4.0V output. Then I measured the voltage with the LED in line and came up with a draw of 2.37V for the unit. (4.0 battery only/ 1.63 battery with LEDs= 2.37 difference) I may be mixing/confusing voltage and current, but since I am really just winging this, I don't know the right technical jargon.

"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 into trouble, but not enough to get out of it!