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I was hoping to power a 1 watt LED (would be great if this is capable of powering 2 or 3 watt LEDs) with a 3 volt coin cell.

  1. Will the coin cell be able to output voltages and amperages that are within acceptable levels for the LED's operation? Essentially will the LED or battery fail faster than is considered normal?
  2. If I need to add resistors to the circuit, what kind of heat levels will they produce?

The battery would ideally be a CR 2302 lithium cell, chosen due to a size requirement.

The LED has a forward voltage of 3.2-3.4 volts and a forward current of 350 mA.

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    \$\begingroup\$ I'll suggest just grab a pair and try it. \$\endgroup\$ – Passerby Apr 8 at 15:09
  • \$\begingroup\$ To give you an idea of how far out your are in size, an 18650 Lithium Ion rechargeable, which has excellent energy density, is a good match for 1W-5W LEDs. 5W willl run a flashlight fairly hot for an hour and a half to two hours on a single 18650. Worth noting brightness and total power decrease as the flashlight heats up, which may or may not be comparable or better/worse than your case if you have size constraints. \$\endgroup\$ – K H Apr 9 at 3:48
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This won't work. You're asking a source of power (battery) that's designed to maybe supply 60 mW to supply 1 W.

End of story – you need a different power source, no circuit is able to create power (physics wouldn't allow that).

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  • \$\begingroup\$ There's no conservation of power law, only conservation of energy. It's definitely possible to run a multi-watt LED from a coin cell - just not at 100% duty cycle. \$\endgroup\$ – pericynthion Apr 9 at 18:39
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The answer to this question is just to look in the datasheet. You'll see that this cell is hopelessly wrong for what you want to do.

The Panasonic CR2032 states 225 mAh capacity and typical max. load current of 200 uA.

So it can't remotely supply the load current. Even if it could, it would be flat in 40 mins.

You must search for parts (battery) by their characteristics (capacity, max current) and then choose the right part. You can't start with the part and try and make it fit your application.

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    \$\begingroup\$ "Continuous drain: 0.2mA." The discharge curve ends at 2mA -it only shows discharge of 2mA and lower. \$\endgroup\$ – JRE Apr 8 at 11:21
  • \$\begingroup\$ @JRE, thanks - my fast and careless reading on a busy day - I've taken the next table column value, which is 20.0 for 'dia (mm)' :-) It felt wrong when I took it, daft. \$\endgroup\$ – TonyM Apr 8 at 11:58
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If you want a tiny battery with a flat shape that can deliver that kind of power (but not for long) you'll have to use a pouch LiPo.

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It will not light it up to maximum level (very low current due to high internal resistance of the battery), but using a Joule Thief circuit one may get some light out of this diode. It is simple circuit with 5 components, but needs a custom transformer.

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