# 10x 3w led lights to run off 12v battery

i have ten 3w leds i need to build an array to run from a 12v car battery.

Led (forward voltage 3.6) (forward current 700 ma ) (Source voltage 12-13.9)

Will this work and if it will what wattage does the registers need to be? if not can you help with a design that will. Thank You

• LEDs have a range of behavior. Not simply exact values. Specify the exact LED you are getting and provide a link to the datasheet, if possible. Also add a note if this is going into a car (with an alternator and load-dump possibilities.)
– jonk
Commented Mar 27, 2017 at 23:54
• i got the led from ebay with no datasheet. it is going in a tractor with a alternator. i dont know what a load-dump is. Commented Mar 28, 2017 at 0:28
• Just curious, but why not just buy 12 V LED bulbs? They aren't expensive. Why do your own circuit for this? You do need to control the current and that takes circuitry, I think.
– jonk
Commented Mar 30, 2017 at 10:57

Simple Ohms Law. V = IR = .7 Amps * 10 Ohms = 7 Volts. Then P = V * I = 7 * .7 = 4.9 Watts. So you need more than that to make sure they don't fry themselves. 6 to 10 Watts each, and a heatsink for both the resistors and the leds. A fan may be a good idea.

Of course, that's ideal numbers. Your resistor may not be exactly 10 ohms, your battery not exactly 12 Volts, the leds not exactly 3.6V at 700 mA, so the resistor voltage and current change, etc. 3.6v + 3.6v + 7v is 14.4 volts. That's only on a car battery if being charged.

If you could do 9 leds instead, it's typically done with 3 leds in series with the resistor. Then you only need 1.2V across the resistor and then it's much less power wasted. 1.2 * 0.7 = 0.84 Watts, a 1 W resistor would be fine but use a 2 W just in case.

• Suppose the OP gets unlucky enough and they are really $3.2\:\textrm{V}$ each. Then there's $2.4\:\textrm{V}$ across the resistor instead of $1.2\:\textrm{V}$. Not a good situation. I think the OP is better off being safe and just ganging up two in series, not three. I don't trust the variations on LEDs enough to trim the margins down to the bare bones like that.
– jonk
Commented Mar 28, 2017 at 1:30
• how do i test to see if they are 3.2v or 3.6v? Commented Mar 28, 2017 at 1:51
• A constant current supply would be best. If not simply connect them to a suitable resistor and measure the voltage across the led and the resistor. Use ohms law to find out the current, and then you have the current at the voltage across the led, ie 3.x volts at n milliamps. Commented Mar 28, 2017 at 1:53
• Just try three, then two, see what works best for you.
– user76844
Commented Mar 30, 2017 at 21:31

You are proposing a $30\:\textrm{W}$ LED light operating on a tractor. I'll assume that this gets powered when you turn on the key or else activate a switch somewhere. [My compact tractor is a diesel JD 4320.]

Here is a short description of a "load dump" event. It can present a pretty high voltage (over $100\:\textrm{V}$ for a short time. LEDs can be sensitive when they are directly exposed to these events. But how much you want to worry about it is up to you. A lot of people don't worry about it and never complain. I'd worry about it more, especially when it came to expensive or critical circuitry.

I'm confused about why you don't just go buy some $12\:\textrm{V}$ LED bulbs. I think your LEDs are something like this:

Those are specified at $3.2-3.8\:\textrm{V}$ when operating at $750\:\textrm{mA}$ (which seems pretty darned close to your own specifications.) Here's the datasheet. (The usual $20\:\textrm{mm}$ diameter type.)

So if looking for something easy and commercial, this means you should look for either 400 lumen ($6\:\textrm{W}$) or 600 lumen ($9\:\textrm{W}$) LED bulbs designed to operate at $12\:\textrm{V}$. You can look them up in the common MR16 form factor with a GU5.3 base. Should be from USD 5 to USD 10, each. And you'd want five of them.

Just getting commercial units saves you a LOT of trouble and gives you the needed dissipation and form factor and standard sockets. Plus, they are designed to work well off of your source voltage.

If you serious, for some odd reason, in avoiding commercial units and you really want to fabricate your own (why???), then I'd try and emulate the above in some fashion. I'd set this up as five (5) separate circuits with separate current control and design it so that it is arranged with two-LED "bulbs", each "bulb" with an aluminum bar stock underneath used as dissipation, and with copper "knife" edges at each end used snap into a ceramic knife blade fuse mounting. This would make it easier for me to fabricate and easier to remove LED modules (each a pair of LEDs) that failed or otherwise were damaged and needed replacement. I would not design a single circuit to operate all 10 of the LEDs, though. Each one needs separate current control.

The above datasheet says that the voltage can vary from $3.2-3.8\:\textrm{V}$ each. That's quite a span.

Suppose you use a simple resistor to regulate your current. If you were to plan on $3.8\:\textrm{V}$ each, then your resistor value would be about $6.3\:\Omega$ to set the current at $700\:\textrm{mA}$. The resistor would dissipate about $3\:\textrm{W}$, too.

A circuit like the following could be used for each pair:

simulate this circuit – Schematic created using CircuitLab

(It's too little headroom [or worse than none] to pack in three such LEDs.)

$R_4$ sets the current through the LEDs, with the help of the TL431. In this case, it is $\frac{2.5\:\textrm{V}}{3.3\:\Omega}\approx 750\:\textrm{mA}$. I used a D45H11 for $Q_2$ because they are dirt cheap and widely available around the world. They are probably over-kill, as they can handle many times the specified current. But they will have plenty of remaining $\beta$ at these lower currents and I figured on keeping dissipation down to a few watts so that heat-sinking needs would be minimal. I added $R_6$ to siphon off some of the dissipation required by $Q_2$ and to provide additional current limits in case of part failures elsewhere. $R_4$ shows as a $5\:\textrm{W}$ resistor, but will only dissipate about $\frac{\left(2.5\:\textrm{V}\right)^2}{3.3\:\Omega}\approx 1.9\:\textrm{W}$.

There are other approaches. The above, though, has one huge advantage gained from using the TL431: nearly rock solid temperature compensation over a very wide range of temperatures.

Frankly, I think you are better off just buying commercial LED lighting, though.