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I have a trip in near future, I have a hub dynamo fit to my bicycle which has 6.0 V DC output of 3 Watts power. I have read few posts and realized that it is risky to charge my iPhone directly through the Dynamo hub.

I thought of an alternative, charging cheap AAA batteries (I can afford to lose some AAA batteries due to voltage fluctuation while charging than my iPhone itself!) and somehow charge my iPhone with it.

I need to know how to connect the AAA batteries so that I can charge my iPhone.

An iPhone needs one of the 2 configuration to be charged with:

Slow mode: 5 Volts, 0.5 Amps, 2.5 Watts
Fast mode: 5 Volts, 1.0 Amps, 5 Watts
(Reference: https://discussions.apple.com/docs/DOC-3511)

A AAA battery output is something like this:
1.25 Volt, 300-500 mAh.

Now how do I safely connect the batteries so that I get desired output to safely charge my iPhone?

What best do you suggest?

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closed as not a real question by Leon Heller, Nick Alexeev, Dave Tweed, Brian Carlton, Olin Lathrop Mar 27 '13 at 17:57

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Why is my question down voted? I have a genuine doubt and I need help here, if someone is down-voting it please let me know (comment) what is wrong in the question asked. \$\endgroup\$ – Raj Pawan Gumdal Mar 27 '13 at 8:42
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    \$\begingroup\$ Lack of research. Plug "charge iPhone with AAA battery" into Google and see what happens. \$\endgroup\$ – Nick Alexeev Mar 27 '13 at 8:54
  • \$\begingroup\$ Come on Nick, everyone would and should google before seeking out for help and I am one among everyone! All of the google results talk about the "readymade" product which would be capable of charging iPhone from AA / AAA batteries. But I am talking about how to build one cheaply as I do not have time to purchase one and also as I believe that the requirement is simple enough and should be possible to do it rather than buying it. I hope I have made my point clear here, still searching for answers. \$\endgroup\$ – Raj Pawan Gumdal Mar 27 '13 at 9:26
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    \$\begingroup\$ ladyada.net/make/mintyboost/icharge.html \$\endgroup\$ – apalopohapa Mar 27 '13 at 10:10
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    \$\begingroup\$ @Raj some constructive feedback: How would you solve your problem? You say "I believe that the requirement is simple enough" but on what basis? Putting up your own solution gets more up votes. \$\endgroup\$ – geometrikal Mar 27 '13 at 13:37
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The question title is misleading. You want to recharge your phone from the bicycle dynamo, and the AAA batteries are just a waystation for the energy to smooth out delivery to the phone. So you have multiple requirements for a circuit:

1) Accept input from the dynamo. Its output may be 6.0V while the bike is moving, but how does that voltage vary as the bike speed changes, or if its voltage output is regulated, how does the current output capability change with bike speed. What happens to your circuit when the bike is slow or stopped.

2) Charge a battery of some kind, which means the circuit must understand how to stop charging when the battery is full and stop discharging when it's empty (or the battery will be ruined).

3) Regulate the battery voltage to provide the proper output to the phone, which for iPhone means 5.0V at 500mA capability, along with some pullup/pulldown resistors on the data lines.

It's not clear how understanding all that, let alone designing, building, and testing it, is going to "save time" compared to buying something ready-made. Nonetheless let's take a look at what a circuit might look like.

The Minty Boost design mentioned above only does requirement 3). It's designed to take 2 x AA batteries and regulate their output to 5.0V for USB power. However it does not appear to turn itself off when the battery voltage gets below the "dead" threshold, so it's really only suitable to use with disposable batteries, as NiMH AA could be ruined if discharged below 1.0V per cell. For the same reason it wouldn't be suitable to use with Li-ion even if the 5.0V regulator still worked. The circuit does not accept power input from elsewhere nor charge batteries, so requirements 1) & 2) are unaddressed by Minty Boost.

To add battery charging, you'd need an additional regulator between the dynamo and the battery which limits voltage and current, and possibly controls charge termination depending on battery chemistry. For Li-ion you may use constant-current to constant-voltage charging which does not necessarily need a timeout, and also solves the problem that current into the battery may vary widely with bike speed. NiMH is harder to manage charging, and how to deal with fluctuating input to the charger may be difficult without making tradeoffs that dramatically shorten the cycle life.

With Li-ion you can pass the battery voltage through to the 5V regulator even while the battery is being charged, as fluctuating pass-through current only subtracts from input to the battery, which already fluctuates due to coming from a dynamo. For NiMH we already don't like fluctuating charge current so this issue only makes it worse.

You need a battery monitor circuit that at the least can turn off output from the battery to the phone when its voltage is too low. This is to say nothing of other safety features such as over/under-temperature, timeout, over-current protection with fuse or polyswitch, cell balancing if using more than one cell, and so on.

Overall you're looking at a decent sized project here, possibly months of work for a less-experienced engineer. Your question does not suggest a lot of experience and you may save a lot of time, money, and possibly be safer, buying something off the shelf.

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  • \$\begingroup\$ Thanks for all the information, I now realize that it is a decent sized project indeed! I would need and ask for more help after my trip to build such a circuit. \$\endgroup\$ – Raj Pawan Gumdal Mar 28 '13 at 5:54

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