# Help needed, Powering LED 12V 10W CHIP using a car battery...



## eklogite (Aug 18, 2014)

Hello every one.

I have planned to use some 12V LED chips to for lighting in my garden.
I will be using a 55Ah Car battery and lots of LED chips.
My question is that will the 12.6V and high current blow my leds?
Do I need to place some resistors or LM317 to regulate current ?

Thank you.
eklogite


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## jason 77 (Aug 18, 2014)

Yes you are going to have to limit the current each LED chips gets. What are the specs on the LED chips you have?


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## eklogite (Aug 18, 2014)

Hi there,
I have checked manufacturer's info.
The specs are as follows:
Power: 10W, 
Lumens: 900-1000Lm 
Current: 800-900 mA 
Voltage: 9-12V (Seller recommended a 1.5 Ohm resistor for 12V)
They are array of 3x3 led inside the chip.

Is a 1W 1.5 Ohm resistor okay for every led chip I have ?
On 12.6V they get quite hot to touch... (over 100 Degrees Celsius)

Also other product:
3W high power LED, 180-200 Lumens...
700mA at 3.2 - 3.4 Volts.
What is your recommendation for using the 3.2 Volt led chips with a 4.2 18650 Li-Ion battery?
I am planning to use a 18650 battery bank which has a capacity around 10000 mAh
Can I use a voltage rectifier LM317 circuit for all chips on this line ?
They are 50 chips...

Sincerely.
Eklogite


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## poolman966554 (Aug 18, 2014)

eklogite said:


> Hi there,
> I have checked manufacturer's info.
> The specs are as follows:
> Power: 10W,
> ...




The seller saying use a 1.5ohm resistor seems kinda vague to me.. That will change with slight voltage variance, heat and other factors. That said, a resistor is a poor way of driving an led. But, lets do it.
First off, this is all calculated by using Ohms Law.
lets do the math. We need to know the power supply's voltage. I will give an Example using 12.6V power supply.

Formula to calc resistor is "R= V / I"

Ex: say you have 12.6v supply voltage, and the led wants "12V" at given amperage of .800 amps..
(12.6-12)/.800 = .75 ohm resistor


Ex 2: to calculate wattage rating of resistor, use formula "Watts= V x A"
So, here we will take supply voltage, and subtract led forward voltage(12.6-12) x .800 = .75 watts

So, if you had a 12.6v supply, you would need a .75ohm/ 1 watt resistor. 
Its always better to round up on the wattage of a resistor if you have the room too.

personally, id look on ebay for a "Constant Current" driver within the specs your after. 
i say that because with the heat thats generated, it will lower the voltage required, and increase the amperage, = more heat made. rinse and repeat until it frys itself.
A constant current driver will keep amperage in check.

Also, ensure you heatsink is adequate enough to support the led


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## DIWdiver (Aug 18, 2014)

eklogite said:


> My question is that will the 12.6V and high current blow my leds?



As others have suggested, yes.



eklogite said:


> Do I need to place some resistors or LM317 to regulate current ?



Resistors, yes. Other options include LED drivers and other current limited regulators.

The LM317 is a poor choice in this situation because it needs several volts across it before it works properly, and you may not have quite that much.


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## DIWdiver (Aug 18, 2014)

poolman966554 said:


> That will change with slight voltage variance, heat and other factors. That said, a resistor is a poor way of driving an led.



Agreed. But it's cheap and easy, and works okay if you have enough voltage differential and aren't concerned with high efficiency or constant output.



poolman966554 said:


> Ex: say you have 12.6v supply voltage, and the led wants "12V" at given amperage of .800 amps..
> (12.6-12)/.800 = .75 ohm resistor



But the chip wants 1.5 ohms at 12V, 800-900 mA. That means the chip wants less than 12V directly. At 800 mA, it would be 12V - (1.5ohms * 0.8A) = 10.8V. At 900 mA it would be 10.65V. That would be about 3.55-3.6V per die, which is pretty believable. In fact, the range is probably somewhat greater than that, but I'd go with 3.6V * 3 = 10.8V for calculations, and make sure I leave some room for error.



poolman966554 said:


> Ex 2: to calculate wattage rating of resistor, use formula "Watts= V x A"
> So, here we will take supply voltage, and subtract led forward voltage(12.6-12) x .800 = .75 watts


 
Or (12.6-10.8) * 0.8 = 1.44W. Use a 2W resistor.



poolman966554 said:


> Its always better to round up on the wattage of a resistor if you have the room too.



Absolutely!

As you can see here, a relatively small change in one of the voltages can result in a large change in some of the calculations. This is extremely important because the battery voltage isn't constant. When trickle charging it's well over 13V, and if hard charging it can be over 14V. When nearly discharged, it's well below 12V, probably closer to 10V under load. Run some numbers and you'll see it's a designer's nightmare. Plenty of complexity vs. performance tradeoff to be made here!


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## eklogite (Aug 19, 2014)

Hello all,
Thanks for your great info.
I appreciate that.

I have searched a lot on the net and found some voltage regulators based on LM2596 on Chinese sellers site Aliexpress...
Unit price is around ~ $1.3 each and have the following tech specs what I see in manufaturer's site...

LM2596S DC-DC step-down module
The input voltage
3.2 V ~ 40 V
The output voltage
35 V ~ 1.25 V
The output current
3 a (Max)
Conversion efficiency
92% (the highest)
The output ripple
< 30 mv
Switching frequency
65 KHZ
Working temperature
- 45 ~ + 85
The size
43 mm * 21 * 14 mm mm (length * width * height) 

 Connection port:
IN + enter the positive - negative input IN!
The positive OUT - output negative OUT + output
Range of input voltage: 3.2 V dc to 40 V, the input voltage must be higher than to the output voltage of 1.5 V or more. Can't booster)
Range of output voltage: 1.25 V dc to 35 V voltage continuously adjustable, high efficiency (92%) of the maximum output current is 3 a.
Measured input 12 v and 5 v 1 a output current at the time of the load regulation is less than 1%
Note: this is step-down module, the input voltage must be higher than the output voltage.
2 A output can long time don't need to add heat sink under the output current is greater than 2.5 A (or output power greater than 10 w) work please add heat sink for A long time.


I can use this and set the voltage to 9.8 Volts for 10W led chips and 3.3 volts for the tiny 3.2-3.4 Volt 200 lumen led chips.
Since it can regulate about 2A or more, I can use one for two 10 V chips and 3 smaller chips at 3.2V 700mA...

What do you think about it ?

Thanks.
Eklogite


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## DIWdiver (Aug 19, 2014)

Better to get one with adjustable current, then you don't have to worry about getting the voltage right.


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