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I have 12 of these LEDs and a 5v Arduino Pro Mini - like this one. This is going to be a very small project I need the smallest power source possible for this. Would I be able to run this with a few coin cell batteries?

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There are a few different calculations you can do that will help figure this out.

Energy consumption

The first is how much power you'll need to run the system. Each LED has a voltage drop of 2.0 V. Assuming you run the LEDs at their maximum current (20 mA), they'll each use 2.0 V * 0.020 A = 0.040 W. For 12 LEDs, that's 480 mW, assuming the rest of your system is perfectly efficient. In an hour, you would use 480 mWh, or 0.480 Wh.

As a first estimate, let's say your system will be only 50% efficient, so you need 2 * 0.480 Wh (roughly 1 Wh) to run for an hour. (You can probably do better, but we'll get to that.) A CR1225 coin cell is a 3 V battery rated for 47 mAh. That's the same as 3 V * 0.047 Ah = 0.141 Wh. That's about 1/7th of what you need, so your LEDs will last around 1/7th of an hour, i.e. 8 or 9 minutes, with one CR1225 coin cell.

Efficiency

Can you do better than just guessing 50% efficiency? Sure.

It depends on how you wire the LEDs and what current-limiting resistors you use. You could make 6 strings of 2 LEDs each and run them off of a 6 V pack made from two coin cells in series. Each string would have a 4 V drop, so you'd need to drop the remaining 6 - 4 = 2 V over a resistor. At 20 mA per string, that's a 2 V / 0.020 A = 100 ohm resistor.

In this case, 1/3 of your power would be dissipated in the resistors (since 2 V is 1/3 of 6 V), so your efficiency would be 66%.

You might try 4 strings of 3 LEDs with no ballast resistors, but I suspect that their brightness would drop off quickly as the battery voltage sagged. Your efficiency would be 100%, but the brightness would suffer.

You could also try running off of the 5 V supply you mentioned; with the right resistor, you could hit 80% efficiency.

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  • \$\begingroup\$ So if I used the 5v battery back all I would need to do is drop the current from the battery pack with a resistor and don't use any resistors on the LED's? \$\endgroup\$
    – davetheguy
    Commented Aug 18, 2010 at 3:51
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    \$\begingroup\$ @davetheguy Using higher voltage than you need and then dropping the voltage is wasteful if you're trying to get the most efficient. \$\endgroup\$
    – davr
    Commented Aug 18, 2010 at 4:07
  • \$\begingroup\$ Well I am really just trying to get this thing to run. Long life isn't needed honestly. \$\endgroup\$
    – davetheguy
    Commented Aug 18, 2010 at 4:32
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    \$\begingroup\$ You could also use brighter LEDs with less current. They tend to be more expensive, but that may be well worth it in this case. \$\endgroup\$
    – starblue
    Commented Aug 18, 2010 at 13:24
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    \$\begingroup\$ If you find that the LED drive losses are significant, there are some switching (hysteretic) LED drivers that can probably do better. You can make the MCU run more efficiently by smart coding and controller selection (keep it in sleep mode as much as possible). Either way, the LEDs themselves will probably be the power hog. Unless you can switch to HE LEDS as said or some other display technology (I presume this is display). Hard to say anything wihtout knowing the actual application. \$\endgroup\$
    – XTL
    Commented Aug 18, 2010 at 15:49
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With enough coin cells, certainly. But, if you want the batteries to last you may need to save some power.

You might try a small LiPo battery. But, you'll also need a charger.

You may want to buy some low power LEDs. That could cut the power budget.

Some variants of the ArduinoProMini can run down at 3.3V which would help.

You could put the Arduino's AVR CPU to sleep to save on batteries while it waits for a button press or it idles here's a tutorial. But, with all those LEDs on it'd be slim pickin's.

Perhaps you don't care how long the battery lasts, but you only want to use one cell?

If you could get away with under 100mA, you could use a single 1.5V cell and a DC to DC step up like this. But at 20mA per LED, that will be a problem.

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