I am planning on buying this LED


This is for my my project of doing a video light that I can carry to the field. Since these are 10W 12V-14v, my math (P/V = I) (10/14) equals to 710mA (assuming 10W is achieved at 14v).

My plan is to buy 3 of them and wire them up in parallel. That means, (.71mA * 3) I will have a load of 2.13A for the project. If I compute this correct (7A/3.2A), I will have around 3.2Hr of use before I run out of juice. The 7A in the eq because I'm planning on using a 12v7ah lead acid battery, its a deep cycle battery. The reason I am going to use this is because they are cheap and available, though heavy, it's tolerable.

Although these computations is accurate if I'm going to use it at full capacity. But since I can't drive the LEDs at full capacity because my supply is only 12v. Also, I'm going also to use a step-down motor with a potentiometer to dim it when needed.


All-in all, I'm hoping if anyone can advise me of anything I had missed? Are my computations correct?


You stated your battery is 12V 7Ah battery.

LED (actually the bar light uses 4) is 12-14V (that's the drop), and they say can use up to 10W. Let's do the math: $$ I = \frac{P}{E} = \frac{10W}{12V} = 833 mA \\ $$

No sense in calculating at 14V, you're limited to 12V.

$$ 3 \cdot 0.833A = 2.499A $$

So, BEST case, 2.8 hours. But you really need to see the curve for your specific battery, and it's derating for thermal, etc, to see if it can realistically output 7Ah-- manufacturers (for marketing purposes) usually put the 'best case' information on the battery, but only the datasheet for it will provide the truth.

I'd expect 2 hours and change. If you put an ammeter on it when you get it, and see how much current it's actually drawing, that will let you get a better idea.

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