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Modify a cheap USB charger to feed an iPod, iPhone or Samsung Galaxy

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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.
 
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Step 1: A 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.

Step 2: Opening and checking

Picture of Opening and checking
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To see if the data pins of my USB charges were at the wrong voltage, I used a hobby knife to open the plastic case.
The cases were glued together so I had to break open them. The Car USB chargher, though, had the top that was screwed on.

I was happy to find fuses inside the two: my iPod should be safe!

As you can see on the pictures the data pins are floating and that is why my iPod did not want to charge.
The specifications of the stickers on the items say that the wall plug can provide no more that 500 mA while the car charger no more than 1000 mA. I decided to stick to 500 mA for both of them.

Step 4: Drilling holes

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I drilled 4 holes on the wall plug board with my Dremel for my resistors. The holes are near the USB pins to make the soldering easier.
The car charger has pleny of space so I did not need to drill holes.

Step 5: Soldering

I then added the new resistors to the boards. I short-circuited the data pins with a little bit of solder, since I wanted both of them at 2.0 V. Unfortunately I am not a very skilled solderer, as you can see on the car charger pictures.

Step 6: Checking

When I finished the soldering I checked all the connections with a multimeter. Then I plugged the chargers and checked if the tension was good. I managed to get about 2.1 V on both of them. This was a really dangerous step because I had them powered and uncovered, touching them could lead to injuries.

Step 7: Testing

I then tested the chargers on my iPod, Samsung Galaxy and on some friends' iPods. They seem to charge just fine and do not complain.

While charging the Samsung Galaxy, though, the charger got really hot. I am not sure if it is good or not, since I have never used it without my "fix". I should measure the current that the Galaxy drains, to see if it is in the parameters of the charger.

My conclusion is that: It Works!
steve.rand1 month ago

I used a slightly different guide to do this to my 3amp charger for my Note 2. Before, with a genuine cable it was drawing 460ma and 1amp off a 1amp iphone charger.

I cut an old usb cable and measured voltages out of the Apple charger and my genuine Samsung charger. The Samsung charger put out about 1.5v on both data wires, and the Apple put out 2.3 and 1.8v.

The way I did it was to use a 10k (-) and 33k (+) resistor to drop the voltage down to 1.5v, and tapped that onto the D+ and D- wires that I soldered together on the end facing the phone. This tells the phone that the charger is capable of higher amperage draws, and it simply draws more current out of the 5v+ wiring. This does not increase voltage, it does not draw more power using the data wiring... it simply is a trigger the phone uses to verify that it won't melt your charger because it is one that is rated for 2amp draws.

To sort out some comments below:

1) 30v to fry a phone is possible, probably using a transistor to double/triple voltage. Don't be stupid though.

2) If your charger can do 800mah but your phone can only draw 500mah... you will only ever draw a maximum of 500mah. You can have 1 million amp potential but you will never draw it if your hardware is not set up to draw it.

3) Shorting the data wires facing the phone did increase my phones charging rate to just under 1 amp. Interesting, and a cheap mod for anyone who wants a quick boost out of a generic charger and who has a phone that can draw 1amp (iphone etc)

Lastly, I would mod a decent cable so as to not ruin a charger, and to prevent someone possibly killing their device by using your modded charger.

How can I make one send 30 volts through an iPhone in order to fry it?

anirwan3 months ago
my car charger support upto 800 ma....so I want my charger to give 700 or 800 ma so please suggest me what kind of resistor I use and how to connect them???it is same as just yours..............
wrexy19946 months ago
I have a IPod classic 160gb, and found that by shorting the data lines together, it charges at about 350mA. So I suspect that this is the maximum it will draw from any 5V supply. Not recommended to be connected to a USB port (max. 100mA).
wasteinc7 months ago
Thanks for the instructable, I also have the same problem with the resistors (75KOhms and 50Kohms are not widely used). But a voltage divider providing 5Vin 2Vout can be obtained with the more commonly used 68KOhm and 47Kohm resistors (R1 and R2 respectively), and those values are closer -current wise- to the "original" spec, while still using widely available resistors.

I wonder if the only reason they used not so commonly used resistors , is to make things more difficult for tinkers (thats why they put the voltage through D+ D- on the first place)
Comwiz9910 months ago
Sorry, I meant the voltage over R1 in a voltage divider is V*R1(R1 + R2). Also English isn't my first language, so ignore my spelling and grammer errors :P
Comwiz9910 months ago
I would advise against using 220 and 330 ohm resistors, as I believe they are a bit small. Essentially what this circuit is, is 2 simple voltage dividers. The formula for the voltage over a resistor R1 in a voltage devider is R1/(R1 + R2), so only the ratio of the resistors are important. By using 2 x (330 + 220) resistors in parallel you are essentially creating an equivalent resistance of 275 ohms which conducts a current of about 18mA between 5v and ground, and this current cannot be "used" for charging. However, I am making the assumption that the device charging does not require to sink a lot of current through the D lines. The original circuit uses resistors in the XX K range, which is more suitable.
dewsky611 year ago
I was thinking.....If you take your schematic, and to the wiring in the middle of two male and female USB connectors, wouldn't that just make a completely portable plug that you could insert into almost ANY USB port to charge the devices? I believe it would work best in Low Power mode, but would work I believe.
Simple breadboard and a small case......Just a thought.....I might make one......
Hi! Yeah this idea does work. i made one a few months ago but I kept it to 500 mA so I can use the m/f dongle more safely on more chargers. Works great.
rmasip1 year ago
Good day,
I did the modification on a car charger rated at 2A with the following resistances (33k,33k,22k,27k)
When I plug the iPhone (4S) it works but the charger powers off intermittently (like 1sec on, 1/4sec off, and so on)
What have I done wrong? Any issue with the charger being of 2A?
Nice schematics, from MintyBoost? :)



Having chopped a USB cable, I have checked the voltages of the iPod 2.1A 5.1V charger. D+ is 2.8V, and D- is 2.0V.
kwang62 years ago
how many ohm i need for d+ and d-?
i have 100ohm and 150 ohm resistor
which one would you use?
(the voltage is 5 volt)
ojasvi12 years ago
please can you tell me that why did u choose 220 and 330 ohm resistors
i think 6ohm should have been chosen as
r=v/i
(5-2)/500mA
and that is definetly not 220 or 330 ohms
Caffeinomane (author)  ojasvi12 years ago
Your calculation is right, but you are missing the point that the current that charges the battery does not come from the data lines. That current is taken from the +5 V line.

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.
i tried with 100k ohm, and it seems to work, but it charges slowly,
il try it a few more times, if it dont, then il try it your way
thanks alot btw
Caffeinomane (author)  ojasvi12 years ago
Well, if you put 2 V on the data lines you will limit the charging current to 500 mA, that is less than what official chargers provide. I think that the official charger provides about 1 A. So it will charge slower.

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.
ZaneChua2 years ago
Hey Caffeinomane,

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
Caffeinomane (author)  ZaneChua2 years ago
Hello!

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
Oh.

So in that case, i will be left with only two cables.
A +5 V Cable and a GND Cable right?

Zane
Caffeinomane (author)  ZaneChua2 years ago
Let me sum up what I understood from your post: you are trying to build the charger on a board by yourserlf. So you soldered a USB connector to the board, soldered the Data pins to the voltage divider, and the +5 V and GND to a couple of cables. Now you want to connect the two cables to a power supply that gives you 5 V.

Am I right? If so, yes that is right.
Ah. Alright. Thanks so much! :)

Zane
Aha, i have yet another question. On Page 3 you said that you used the 220 and 330 ohm resistors to get 2.0v out of 5.0v. However, isn't Resistance Calculated by R=V/I ? If i followed that formula, i would get something vastly wrong. How did you calculate the amount of ohms needed?

Zane
Caffeinomane (author)  ZaneChua2 years ago
I am not sure what were your calculations but the choice of the resistors was based on the principle of the voltage divider. Given the Ohm formula:

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.
Ah Okay.

Sorry, have been out of touch with Physics for awhile. Been like 2 years.

Thanks for the explanation. :)
Vermin2 years ago
Great instructable. The only addition you could possibly make would be to explain why you used 220/330R instead of the 49.9/75k shown in the schematic and the practical lower ohms limit (power dissipation) and upper ohms limit (sense current required) for resistor divider choice.
Caffeinomane (author)  Vermin2 years ago
That is a good question. I used those resistors just because they were the only two kinds that I had that matched the tension that I needed. I do not know the limits of the current that the data pins will absorb, because I did not find the specifications of the iPod.

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.
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