Step 2: Case Preparation

Cut a hole in the side of the project case that approximately matches the size of your USB plug.

• Outline the usb plug with pencil on the outside of the case.

• Cut away the plastic using a Dremel or similar rotary tool.

• If your USB plug is rather large, like the one we used, you may need to cut away some of the lid, do this in the same fashion as above.

Note: If you have the means to do so, it may be easier if you clamp the project case to a work surface to hold it steady while you cut.
i want help in 2 points :-<br>1). If i dont want to remove battery when not in use.......<br>2). If i want to use it as powerbank and directly charge it and reuse......,<br><br>plzzz help<br>
I tried connecting directly without soldering. it didn't work, is it compulsory to solder for it to work.?? pls reply
<p>You shouldn't need to solder for any project to work. As long as it is making contact, it should work. However, this project is just wrong, as the 9v battery is directly connected to the USB female port, so you would be feeding 9v to your device which could easily damage it.</p>
<p>Simple and neat. Version 2.0 of this would be to include a battery charger circuit and use LiPO battery pack to source something like 6000 mAh. These are readily available for around $30, so would be good to calculate the cost of building one yourself. This then can be used to supply multiple boards (like RPi, Arduino etc.) simultaneously. Not sure if these 9V batteries cross 1000 mAh ratings. Another option would be to use AA NiCd or NiMH that are rechargeable and can provide around 1000 to 3000 mAh. There is always a tradeoff between cost and efficiency. :)</p>
can we use any other device like resistors instead of voltage regulator<br>
<p>how can i charge the power bank... can i directly charge with the electric charger..</p>
Hello everyone... i love what happens here on Instructables. I love building up stuffs like this cos i use an Android device and the battery seems to drain very easily. Pls i have a charging project i want to do. Can i talk to anyone to help me out with how?
Why hasn't this project been corrected? It is a travesty and irresponsible to leave instructions that could lead to frying someone's electronic equipment! <br> <br>DO NOT USE THIS CIRCUIT AS SHOWN! Having wrong information for the public is worse than nothing at all!
dwoods-1 is correct you WILL damage anything which requires 5 volts with these connections as you WILL be supplying 9 volts directly.<br>I cannot find any reference to a .7285 regulator but a 7805 would be fine.<br><br>Looking at the diagram you provide you need to connect both black wires to the centre pin. Connect the red wire from the battery clip to the top pin and the red wire of the usb cable to the bottom one.<br><br>See this diagram: http://www.talkingelectronics.com/projects/LogicDesigner/images/7805.gif
Scrap the earlier comment. Follow this schematic:<br><br>
This is configuration for sources, which can give you 1A. If you're not sure if it can acheve this, it's safer to use this schematic:&nbsp;<a href="http://www.ladyada.net/images/mintyboost/usb4res500mA.gif">http://www.ladyada.net/images/mintyboost/usb4res500mA.gif</a><br> because it limits current draw to 500mA, which is safer to regulator and lowers its temperature. I hope you understand what I mean :)
I forgot to mention that the above schematic is what is inside the Apple USB chargers. I figured since Apple uses it in their charger, then it would probably work for a lot of the newer apple devices. Although, this schematic is a few years old, so Apple might have changed the schematic around a little bit. Also, since it is line powered, it can draw the single amp that it needs. And for the heating problem, the common output current is up to 1 amp (I'm sure there are higher current regulators). Because 1 amp is the highest current output, and that is what the above schematic uses, there would be heat. A simple heat sink would suffice. I will agree that my schematic is obviously not suitable for a portable charger. Although, if it were powered by some &quot;D&quot; type batteries it is possible. Anyway, to anyone still reading this, use the circuit in domints comment.
http://www.raphnet.net/electronique/gc_n64_usb/lm317rev1g.pdf<br><br>The chip in your picture and sorry but as you point out there you basically hooking the usb power straight to USB @9v !!<br><br>would it not be better to use a 7805 5v roltage regulator in this configuration ?<br><br>http://forum.allaboutcircuits.com/image_cache/httpwww.sminntech.comimages7805datasheet.gif
you could add a switch between the battery and the regulator, it would save the trouble of actually removing the battery, but would accomplish the same thing.
Hi there, these kind of voltage regulators are not efficient at all. If you regulate from 9V down to 5V then you are burning 4V through a resistor, which is almost have of what the battery gives. It is better to use a switching voltage regulator such as the TL497. You can even make a design to keep charging even after you have less than 5V of potential differential in your battery.
Okay Ive seen this so many times before and have made this 2 different ways i understand the use of a resistor but have you tested it with and without one?? i'm curious about charge time as well as battery life both with and without the resistor.
I agree with domints. The way it's soldered would just supply 9 volts to the USB port. Regarding the pullup resistor, you will only need these if you want to charge newer apple products. A 10k-50k resistor is needed between ground and pin4(data line) of the USB port and a 10k-50k resistor between a 3 volt power source and pin3(data line) of the USB port. I am not 100% on this but I am pretty sure that it will work. Good ible!
Hey. Nice 'ible, but there's one think that you should change in this project. You should use some pullup resistors on data lines, 'cause this won't load Apple devices. You can read about this here: <br>http://www.ladyada.net/make/mintyboost/icharge.html<br><br>And what is full name for your voltage regulator? Even Google doesn't know much about 7285/.7285 . I think 7805 / LM7805 is more popular. With 7805 you can use even 2.8 pullups, because you can get 1A from 7805, but I'd use heatsink for regulator. With pullups for 0.5A it shouldn't be neccesary to use haetsink.
And connection to voltage regulator is a bit weird. Are you sure, you get out of this devide 5V? I think it isn't possible to achieve this, because you are just soldering battery clips to USB plug, with something connected in parallell to USB. I'd like to see how fast you'll bake your phone or iPod.
Hello Group2. Nice project. Is this for a school assignment?

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