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

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 3: Resistors

I needed a couple of resistors that could give me a 2.0 V out of 5.0 V. The ones that I chose are: 220 Ohm and 330 Ohm.
Both the pins should be at the same voltage thus only 2 pairs of resistors are needed.

Step 4: Drilling Holes

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!

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50 Discussions

Does anybody know about higher current defintion resistor values? Newer Iphones draw currents up to 2A.

Unfortunately, some chargers are inside (closed) connected to each middle pins D + from D-.
To use such a clear connection.

My note 3 phone stops drawing current at around 6.5 volts meaning it has overvoltage protection per the USB specs. With 4 volts on both data lines the Note 3 charges at its fastest rate. This I observed on my work bench. The current is controlled on demand by the phone once it auto recognizes what DC volts are present on the data lines. If I change the voltage I need to unplug the phone then restart the auto recognize sequence. What we could use is a way to tell the phone to send more current to the battery even when its almost charged, this would be a great app to have handy when I need to top off the battery before I hit the road. My next step is to log real time current on my oscilloscope. This is because current shifts or bounces all over the place as I can see on my 5A bench supply.

2 replies

You tested this? Charging at 6.5V? Ballsy. I hope it wasn't "your" phone...

RE: boost charging an almost full battery.

Most "experts" (and I say that because they don't all agree) claim that when a LiPo is approaching full, you should.

- dramatically taper off current

- pulse charge

Besides it not being recommended, I think you'd find that a 5V source would have trouble pushing a substantial current into an almost full LiPo. As it approaches full (4.2V) you need to be feeding it 4.3V to get any current to flow. I don't know what voltage you'd need to get, say 1.0A to flow. 0.6V of headroom doesn't leave much for supervisory/regulatory circuitry.

I suspect the other reason is that while you're punching current into an almost full LiPo, it's very hard to monitor its real "fullness" even if you pulse charge (you'd have to load it between pulses to get an accurate read).

The dangers of overcharging (bloating/fire) are just too high.

The battery management system will only ever supply the correct/safe/specified charging currents and voltages to the battery, it has its own DC-DC converters to power battery charging and phone operation at separate regulated voltages (independent from USB), and registers total-delivered and drawn charge to provide the battery gauge and further inform the charging algorithm's cutoff points.

The more total power available to the BMS from the USB port, the more total current it can put into the battery at 4.2V or whatever the specified voltage is, while also operating the phone.

DC-DC converters generally have a wide input, and with 6.5V you have 30% more power going into the BMS at any given available current.

Thanks for this tutorial some great knowledge in here!
Just a note...if the power source can handle it you can use a 270 ohm resistor in place of the 330 to get 2.75v across both data lines to charge at upto 2.1A without using loads of different resistors

No worries. At 2 volts, there is basically no chance of injury. This is not enough voltage to overcome the resistance of your skin. The exact voltage that may be dangerous depends on many factors, but if you are under about 20 volts, you are almost certainly safe.

4 replies

Voltage doesn't kill you! What kills you is the current ( amperes) which goes through your body and could BOIL your blod or maybe even stop your heart.

Correct. But you still have to have enough voltage to push the current through your skin resistance. Current will only cause damage if it can get past the resistance. This kind of voltage will cause no problems, even at 2000 amps!

I've done this with no resistor (you shouldn't do this but i was desperate) I have an emergency cable, Since I Assume usb is 5V (or 5v tolerant atleast don't hold me on it) from the voltage rail I just hooked D+ and D- to the 5v rail repectivley at first and worked on my ipod 2g, ipad 1st gen, iphone 4 and 4s, now I've added some diodes for proctection and adds a ~1v(per diode i use 1-2 per line since it was all I had other than resistors any diode [not zener] that can handle the voltages/currents used should probably work I assume) drop cause it doesn't feel comfortable to put 5v directly on the data lines

If I remember correctly the pros are using a 43K/49.9K resistor for both data lines. Remember the data line has a 27K pull down resistor in some devices.

Hi, My Samsung Galaxy TAB2 10" uses a charger which puts out:

5.1V on '+'(Red wire) and '-' (black wire),

1.27V between '-' (black) and both D+ (Green) and D- (White) so they must be bridged together (shorted).

3.87V between '+' (Red) and both D+ (Green) and D- (White)

My Solution was a USB charger (3.1A) but it needed a 38K resistor between + (red) and the shorted D+/D- wires. And a 10K resistor between - (black) and D+/D-

Voltage was not exactly the original values, but my tablet accepted it.

One note - the Samsung Tablets require 1.2v on both D+ and D- to do the 2.1A (I'm seeing 1.7 off the power supply, but close enough).

A second problem I had was with non-samsung cables, but it turned out to be the loss over a few feet of thin wires - samsung was closer to 5v on the far end, and a shorty cable was even better.

Props for a sweet cheap hack and clean package.

Question for you and forgive me- I'm a mechanical engineer toying with EE. Does applying 2.8V and 2.0V allow it to charge 1000mA or force it to 1000mA- bypassing a feedback control? I believe I read somewhere that the iphone was designed to charge 0-80% in one hour and 80-100% in another hour, or something like that. Sounds to me like they reduce current as it nears 100%.

I know electronics draw the current they need loosely speaking, so I guess what I'm asking is if this is in any way unsafe for the iphone, or if it will not know the difference and only see a higher current option, but not forced 1000mA?

My project is a solar panel. I got 5.6V and 3.6Asc and cut an iphone cable, soldered the red to positive with a diode and the black to the negative. Result was about 5.2V and I decided to try plugging in my phone. Nada. I could apply your information and with resistors create 2.8V and 2V, or 2V and 2V (only in full sun) but I want to know it wouldn't be overloading the phone. I expect them to have an auto-shutoff if something isn't right, but I'd rather not test that theory on my daily!