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A couple of years ago I got an iPod touch as a gift and I decided to buy a USB charger for it. So I bought a really cheap one but it never worked. The iPod, once connected, did not like it and did not want to charge. Since it was so cheap I just let it sit in a drawer forgetting about it.
A little while ago then I stumbled on this very good article: http://www.ladyada.net/make/mintyboost/icharge.html
In which they decribe how they produced a battery powered USB charger. After reading that article I took my cheap USB charger and decided to modify it.
This will be a really, really easy modification and I think that anybody with a soldering iron could do it.
A little while ago then I stumbled on this very good article: http://www.ladyada.net/make/mintyboost/icharge.html
In which they decribe how they produced a battery powered USB charger. After reading that article I took my cheap USB charger and decided to modify it.
This will be a really, really easy modification and I think that anybody with a soldering iron could do it.
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Signing UpStep 1A little Theory
A USB connector has 4 pins: +V, D-, D+, GND. The +V pin with the GND give the +5 V that aliments the phone; while the D- and D+ pins are used for communications. Old USB electronic devices did not care about D+/- pins as long as the other two did give nourishment.
Nowadays the iPhone expects a certain voltage on those two pins to decide how much current to absorb from the charger. Putting a 2.0 V voltage on both the pins the iPhone will absorb about 500 mA, while with 2.8 V on D- and 2.0 V on D+ it will absorb about 1000 mA.
The same behaviour I expected to be observed on my iPod.
On the images there are the schemes for the two configurations. As you can see, using an opportune couple of resistors it is possible to get the voltage required. Obviously the 1000 mA configuration is better if you want your phone charged quicker, but it is possible that your power supply can not support that much current.
Nowadays the iPhone expects a certain voltage on those two pins to decide how much current to absorb from the charger. Putting a 2.0 V voltage on both the pins the iPhone will absorb about 500 mA, while with 2.8 V on D- and 2.0 V on D+ it will absorb about 1000 mA.
The same behaviour I expected to be observed on my iPod.
On the images there are the schemes for the two configurations. As you can see, using an opportune couple of resistors it is possible to get the voltage required. Obviously the 1000 mA configuration is better if you want your phone charged quicker, but it is possible that your power supply can not support that much current.
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i have 100ohm and 150 ohm resistor
which one would you use?
(the voltage is 5 volt)
i think 6ohm should have been chosen as
r=v/i
(5-2)/500mA
and that is definetly not 220 or 330 ohms
The voltage divider is used only for setting the right voltage on the data lines, they should not drain much current. But unfortunately I do not know what current is required on that line so I used the approach of trial and error. I used those two values just because they were matching and those were the only ones that I had.
il try it a few more times, if it dont, then il try it your way
thanks alot btw
You could increase the current, changing the voltages on the data lines.
If you read the linked Web page on the introduction you can find the right voltages. But you must double check if the charger is able to provide that current.
I was wondering how to implement the schematics in real life. I have never read schematics before and i'm looking to repair an old ipod charger that i broke. I took out my multimeter and measured the two small resistors on the small chip board and it measure around 50k and 75k resistance as in your diagram.
However, what i don't understand about the schematic is where the end point goes to. For example Ground, do i solder the three ground points together or do i leave them alone? I just can't understand the ground part and i can't seem to find a definite answer. It would be a great help! Thanks and sorry for bothering you.
Zane
Do not worry, I am glad if I can help you.
Yes, as you guessed the three GND points should be connected together, as all the +5 V points.
That is the usual way to express where the power pins should be connected. Since power and GND pins are usually all around the schematics, you just put a tag that tells that they are all connected together, instead of having traces that pass all over the place making a mess.
Caffeinomane
So in that case, i will be left with only two cables.
A +5 V Cable and a GND Cable right?
Zane
Am I right? If so, yes that is right.
Zane
Zane
V = i R
If you put a couple of resistors, say R1 and R2, in series:
GND |----/\/\/\(R1)----(A)----/\/\/\(R2)----| +5 V
you have:
V = i ( R1 + R2 ).
While the tension at point (A) is
VA = i R1.
Solving for i we get
V / ( R1 + R2 ) = VA / R1
VA = V R1 / ( R1 + R2 ).
So that is why I used 220 Ohm and 330 Ohm. If you are asking me why I used those two instead of 50 Ohm and 75 Ohm, the reason is that I did not have them while I had the other two. The difference between the two choices is the current that passes through the resistors. In fact, I did not know how much current was needed so I tested it and it worked fine.
Sorry, have been out of touch with Physics for awhile. Been like 2 years.
Thanks for the explanation. :)
I could try to measure those currents but to do that I would have to destroy a USB extension cable, which I do not have, unfortunately.