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I'm looking at an LED light and it says it takes 2.2 watts with a 3.7V / 900 mAh battery. How do I calculate how long the LED light will last if it's on continuously?

The item in question is this

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    \$\begingroup\$ Given that information, you can't. You also need to know the voltage. Then it's easy. \$\endgroup\$
    – user16324
    Sep 3, 2017 at 10:40
  • \$\begingroup\$ I might be wrong but considering your LED is 3.3v: 0,9Ah * 3.3V = 2,97W of power you've available, meaning your it will last 1,32h because the LED will take 2.2W every hour. \$\endgroup\$
    – TCB13
    Sep 3, 2017 at 10:40
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    \$\begingroup\$ @TCB13 I'm being pedantic here but a watt is power and power is instantaneous so you can't "take 2.2W every hour". The battery stores energy not power so you can take 2.2W continuously for an hour \$ (2.2\text{ watt} \cdot \text{hour}) \$ or you can express this energy in Joules. \$\endgroup\$ Sep 3, 2017 at 10:51
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    \$\begingroup\$ @TCB13 Yes, you're wrong. You're multiplying the forward voltage of the LED by the battery capacity (in Ah. It's a charge, not a current!), this does not mean anything. You need also the battery voltage, to know how much energy the battery has (battery capacity * battery voltage). When you have that energy, you divide it by the LED's consumption. That's the best time you can hope for, assuming a 100% efficiency of the DC/DC convertery you'll probably must insert between the battery and the light. \$\endgroup\$
    – next-hack
    Sep 3, 2017 at 10:54
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    \$\begingroup\$ @TCB13 it's not "not extremely correct". It's plain wrong. \$\endgroup\$ Sep 3, 2017 at 11:14

2 Answers 2

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If your battery voltage is V, its capacity (in Ah) is Q, and your lamp's power is P, and if you're using a DC/DC converter with efficiency: $$\eta$$

Then, the time is simply: $$\eta\frac{QV}{P}$$

Of course, to get the numerical result in hours, Q must be in Ah and P in W, and V in Volts.

This is a simplification, as it assumes a constant battery voltage V, until the battery is fully depleted.

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  • \$\begingroup\$ This is the item rads.stackoverflow.com/amzn/click/B072QL568M 3.7V \$\endgroup\$
    – Jack
    Sep 3, 2017 at 11:30
  • \$\begingroup\$ then looking at the specs put $$\eta=1$$ and V = 3.7. \$\endgroup\$
    – next-hack
    Sep 3, 2017 at 11:39
  • \$\begingroup\$ I get 1.513... Is that supposed to be in hours? \$\endgroup\$
    – Jack
    Sep 3, 2017 at 12:29
  • \$\begingroup\$ @Jack Yes. Still I guess it could be an overestimation. \$\endgroup\$
    – next-hack
    Sep 3, 2017 at 13:02
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Considerations: Your LED runs at 3.3v.

  • mAh how much energy you have on your battery;
  • Know how much you'll be using every hour;
  • Consider that an amperage is a coulomb per second;

The Math: Calculate how long the battery will last by dividing its total charge in mAh by your LED load in mA.

900 mAh / 666mA = 1,36h (I've considered your LED is 3.3V meaning 2,2/3,3 = 666mA).

If you're unsure about my math, you can also use this online calculator that will give your the same result: http://ncalculators.com/electrical/battery-life-calculator.htm

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    \$\begingroup\$ This assumes, that there is a linear regulator and not a switching converter, otherwise the answer is unknown without the battery voltage. \$\endgroup\$
    – Arsenal
    Sep 3, 2017 at 11:08
  • \$\begingroup\$ This is the item rads.stackoverflow.com/amzn/click/B072QL568M 3.7V \$\endgroup\$
    – Jack
    Sep 3, 2017 at 11:32

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