Modifying Power Source: Hook Up to Wall & Recharge

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A user seeks to modify a small electronic toy powered by AAA batteries to connect it to a wall outlet for recharging, considering a lithium battery option. While rechargeable AAA batteries and chargers are readily available, the user prefers a more complex solution for future projects. Concerns are raised about the dangers of building a custom lithium-ion battery charger due to the risks of improper charging. Alternatives like Ni-Cad batteries and their chargers are suggested as safer options. The discussion emphasizes the importance of proper battery management to avoid hazards.
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I have a small electronic toy that takes 4 regular triple a batteries. I would like to open it up and see how much current the batteries are supplying, and then I would like to hook up a battery that I don't have to throw away. I would like to hook up the device to the wall to recharge it. I think that a lithium battery would work, but what device would i buy so that the battery would recharge from the wall? thanks
 
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Why don't you just buy some rechargeable AAA batteries, and a recharger that fits them? You can buy such things at almost any store, including Wal-Mart and Sears and so on.

- Warren
 
While this is certainly true (and certainly easier), I want to do it for the tinker factor and the knowledge. I would like to use this in a more complicated project later. Thanks a lot.
 
You cannot (and should not) build your own lithium-ion battery charger. Lithium-ion batteries are very sensitive to both under- and over-charge situations, and will readily explode if not charged (or discharged) properly. It is a testament to modern electrical engineering that something so potentially dangerous can be so well controlled.

As far as Ni-Cad and other rechargeable chemistries, you can buy battery packs and chargers from any of a million places on the web. I suggest http://www.batteryspace.com

- Warren
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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