2

Use
I am assuming that this is possible for any phone model's battery. I am making this vague on purpose (this isn't for a specific project). The goal is to have the most compact and long-lasting battery possible for my arduino projects.

Note
I don't care about the cost. This would be for personal use (not selling this or using it for school), so I don't care about patenting or copyright issues.

Question
I want to know if this even makes sense for a portable arduino power source option. I want to benefit from newer phones' long battery life, as well as their compact size.

Also:
Are there better alternatives to this using different batteries? (I have looked at Nick Gammon's posts over setting sleep cycles to extend battery life (http://www.gammon.com.au/power); If utilizing his methods on different batteries are better than using a phone battery please let me know...)

If Possible...
Does it make sense for something like a:

  • LED suit (lasting 12 hours)
  • Portable gaming system (lasting 8 hours) - similar to: (https://www.adafruit.com/products/2510)
    I Know this is for Raspberry Pi
  • BlueTooth Speaker (lasting 6 hours)

etc...

Improve this question
7
  • There's nothing inherently special about phone batteries; they're usually just Li-ion/Lipo batteries with an integrated thermistor.
    – Ignacio Vazquez-Abrams
    Commented Dec 18, 2015 at 16:32
  • @IgnacioVazquez-Abrams - I understand that, but I'm primarily interested in them for their low profile. Like the new iPhone and Samsung smartphones. Would it make sense to add these to an Arduino as a power source? (i.e. are there any implications to using them)
    – ITSUUUUUH
    Commented Dec 18, 2015 at 16:38
  • It makes sense to use a Li-ion battery as a power source for a microcontroller. It makes slightly less sense to use one with a bog-standard Arduino Uno or Mega due to all the other stuff on the board.
    – Ignacio Vazquez-Abrams
    Commented Dec 18, 2015 at 16:41
  • @IgnacioVazquez-Abrams - Are you saying that the "other stuff" draws too much power to make sense for a Li-ion battery?
    – ITSUUUUUH
    Commented Dec 18, 2015 at 16:51
  • The boards are fine for prototyping, but once you're done with that you should move to a breakout board or bare MCU.
    – Ignacio Vazquez-Abrams
    Commented Dec 18, 2015 at 16:56

2 Answers 2

Reset to default
1

Sounds like a good solution.

Just note that most cellphones use Lithium batteries. Those are typically around 3.7 volt. They require some special way of charging, or they may explode or catch on fire. They also get damaged if discharged too much.

These batteries may have some protection circuitry inside them. This would be ideal in your case, as it protects against over-discharge, and accidental shorts.

An alternative would be the cells used in notebooks called 18650s. These are available with protection and without. The advantage is that you have a standard size. You can buy battery holders for them. That way you can easily exchange batteries. You can also combine cells to create higher voltages.

As for the runtime; you can make a rough calculation of it.

Improve this answer
1
  • Thanks for the advice! I hadn't thought of using the batteries found in notebooks. I also appreciated the calculator.
    – ITSUUUUUH
    Commented Dec 18, 2015 at 16:57
1

I've done this before, but not very practical as a bare "flat-pack" outside its case presents other "issues", the battery "may" catch fire if punctured or shorted too long. (read fire hazard when punctured) Lithium fires are like thermite, you can't put them out, tossing water on it will only make it worse.

Then there's the voltage mismatch (unless you're using a DUE). A POL boost converter is needed for best efficiency to get 5-5.2V from a 3.7V cell. And a specialized charging IC is needed so that you don't overcharge/over-rate the current charging the pack. (constant-voltage/constant-current regulator type is needed). I think Adafruit has a lithium breakout board for the 3.7V lithium cell charger, it's for their lilypad stuff i think.

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.