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How to use a 3.7V lithium battery as a 5V power supply for qi wireless charger

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SujithZis

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I heard of people suggesting for boost converter and step up converter circuit , and i don't know what components to start with .
btw its actually a qi magicdisk wireless charger , it needs 5V 2A input ,
 
Also one guy recommended me this IC , but i don't have no clue on how to use it
**broken link removed**
 
It will probably be cheaper to buy an off-the-shelf DC-DC boost converter than to make one. Bear in mind that the battery must be able to supply nearly 3A to give a 2A output at 5V.
 
i tried everything from sparkfun , and they dont ship to india ,
i tried searching in ebay.in and aliexpress , but iam not able to select a good one , everything is with output current 1~1.5A
 
Alot of searching and now iam thinking of using 2*3.7V 2A batteries and step down using 7805(but i have no clue on rest of the parts that should be attachd to it )


which method would be better and easier and compact
 
now iam thinking of using 2*3.7V 2A batteries and step down using 7805
Liner regulators like that wast a lot of power.
You still should use a DC to DC converter to get your batteries to last longer.
 
SujithZis said:
ramuna , those are too big , i need something a like a mint booster something few centimetres long ,

Well, the qi magicdisk wireless charger is normally powered using the USB port of a computer, so I'm guessing that you want to supply power to the charger independently from a computer. Assume that you obtain your ideal "few centimetres long" booster, if you add in the size of the required batteries to power the boost converter, will this total size be less than or greater than the plug-in power supplies I mention in post #6 ? Also, will these be single use batteries or rechargeable ? If the former, then the solution would be expensive, if the latter then all you are doing is to introduce an unnecessary and expensive additional step in the process of transferring power from the mains to the device (cell phone ?) being charged, in comparison with the solution proposed in post #6.
 
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