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.
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.
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.<br><br>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.
Excellent idea on how to burn down your house ?.
<p>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 </p>
<p>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.</p>
<p>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.</p>
<p>Voltage(V) = Current(I) X Resistance(R).........Remember Ohm's Law......</p>
Please watch this video.<br><br>https://www.youtube.com/watch?v=8xONZcBJh5A
<p>FYI The BC1.2 certification specification calls for D+ D- = 14.25K to 24.8K pull down resistor</p><p>Check out the TPS2511</p><p><a href="http://www.active-semi.com/sheets/ActiveSemi_Analog_Power_PSG_1Q2014.pdf" rel="nofollow">http://www.active-semi.com/sheets/ActiveSemi_Analog_Power_PSG_1Q2014.pdf</a></p><p><a href="http://komposter.com.ua/documents/BC1.2_FINAL.pdf" rel="nofollow">http://komposter.com.ua/documents/BC1.2_FINAL.pdf</a></p>
<p>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.</p>
<p>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.</p>
<p>FYI Check out the TPS2511</p><p><a href="http://www.active-semi.com/sheets/ActiveSemi_Analog_Power_PSG_1Q2014.pdf" rel="nofollow">http://www.active-semi.com/sheets/ActiveSemi_Analog_Power_PSG_1Q2014.pdf</a></p><p><a href="http://komposter.com.ua/documents/BC1.2_FINAL.pdf" rel="nofollow">http://komposter.com.ua/documents/BC1.2_FINAL.pdf</a></p>
<p>Hi, My Samsung Galaxy TAB2 10&quot; uses a charger which puts out:</p><p> 5.1V on '+'(Red wire) and '-' (black wire), </p><p>1.27V between '-' (black) and both D+ (Green) and D- (White) so they must be bridged together (shorted).</p><p>3.87V between '+' (Red) and both D+ (Green) and D- (White) </p><p>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- </p><p>Voltage was not exactly the original values, but my tablet accepted it.</p>
<p>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).</p><p>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.</p>
<p>Props for a sweet cheap hack and clean package.<br><br>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%.</p><p>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?</p><p>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!</p>
<p>To answer your question &quot;</p><p>Does applying 2.8V and 2.0V allow it to charge 1000mA or force it to 1000mA- bypassing a feedback control?&quot; - it will allow it to charge at 1000mA.</p><p>If your phone thinks it can draw 1A (1000mA), then it will do so. To <br>achieve this, you must provided the correct resistor values for the 2.7V <br> and 2.0V on the data lines. (I've done this before, with a 9V battery, <br>and the battery became very hot and drained in a matter of minutes).</p><p>However, if you only want to draw 0.5A (500mA), then you need to provided 2V on D+ and D-. Your phone will then realize that it can only draw 500mA.</p><p>In retrospect, if you phone only needs to draw an N amount of current, then it will only draw that much. You just have to ensure that your circuit can provide the needed current.</p>
<p>It really depends on the charger and the device. I've had my iPhone connected to a charger advertised as 2A, and neither got hot. You'd really have to measure the current draw of the device, and consider the rating of the charger you use. As was previously mentioned, if the device normally draws 500 mA and you have it connected to a 2A charger, it will still only draw 500 mA. It's kind of like putting racing tires on a Volkswagen Beetle; the tires may withstand much higher speeds but unless you change something else (like a bigger engine/different transmission) your Volkswagen is still going to top out at factory speeds. </p><p> Adding resistors in the case of an iPhone will allow it to draw up to 2A, and if the charger is only rated for 500mA or 1A then yes it can get hot, or blow the internal fuse, activate a thermal cutoff, etcetera. </p><p>Resistors: A 4.7 ohm and 5.1-ohm will produce the same voltage as a 47K and a 51K, a 4.7M and 5.1M, etcetera...The main purpose of a resistor is to change the flow of current, and the change in voltage is just a side effect. I agree that the usage of higher-value resistors is to restrict current flow over the data pins. It's hard to tell with reverse-engineering the phone, but usually the device checks the voltage present on the data pins and then allows a certain charge rate based on that voltage...It doesnt need much current over the data pins and Apple likely used higher-valued resistors to protect the phone/iPod/iPad.. Hope that clears the water somewhat.</p>
<p>I have a second comment. I found this website with a more accurate search: <a href="http://voltaicsystems.com/blog/choosing-usb-pin-voltages-for-iphones-and-ipads/" rel="nofollow">http://voltaicsystems.com/blog/choosing-usb-pin-vo...</a> and they support what you are saying.<br><br>This second question is simple- does your modified device get hot or overheat? My concern with them setting 500mA is that they see it as safe and no risk of catching a car on fire when neglected in the hot sun (lawsuit engineering? lol).</p><p>Again, thanks for the good hack!</p>
<p>The reason it is heating up might have something to do with the low resistance you took for the voltage divider. Take something in the range of 1k+ instead, so you can be sure that if some current is flowing it is only a small amount. If there is a voltage on the pins that is enough, so better to be safe and make it so that only a few milliamps can flow through.</p>
<p>Many many thanks for this charger upgradation...<br>Actually, not only cheap chargers, other original mobile chargers did not required the data pins modified.</p><p>I'm was trying to make a multi-USB ported charger... So, I can travel with only one charger for many device...</p><p>But, my two 'Sony' cameras didn't supporting any other charger, except the original &amp; PC...!!</p><p>Just modified a cheap charger... Cameras are charging perfectly...</p><p>3.1 V on D-</p><p>2.3 V on D+</p><p>Is this OK?...</p>
<p>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.</p><p>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.</p><p>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.</p><p>To sort out some comments below:</p><p>1) 30v to fry a phone is possible, probably using a transistor to double/triple voltage. Don't be stupid though.</p><p>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.</p><p>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)</p><p>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.</p>
<p>How can I make one send 30 volts through an iPhone in order to fry it?</p>
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..............
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).
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 &quot;original&quot; spec, while still using widely available resistors. <br> <br>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)
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
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 &quot;used&quot; 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.
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. <br>Simple breadboard and a small case......Just a thought.....I might make one...... <br>
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.
Good day, <br>I did the modification on a car charger rated at 2A with the following resistances (33k,33k,22k,27k) <br>When I plug the iPhone (4S) it works but the charger powers off intermittently (like 1sec on, 1/4sec off, and so on) <br>What have I done wrong? Any issue with the charger being of 2A?
Nice schematics, from MintyBoost? :) <br><br><br> <br>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.
how many ohm i need for d+ and d-?<br>i have 100ohm and 150 ohm resistor<br>which one would you use?<br>(the voltage is 5 volt)
please can you tell me that why did u choose 220 and 330 ohm resistors<br>i think 6ohm should have been chosen as<br>r=v/i<br>(5-2)/500mA<br>and that is definetly not 220 or 330 ohms
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.<br><br>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,<br>il try it a few more times, if it dont, then il try it your way<br>thanks alot btw
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.<br><br>You could increase the current, changing the voltages on the data lines.<br>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.
Hey Caffeinomane,<br><br>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.<br><br>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.<br><br>Zane
Hello!<br><br>Do not worry, I am glad if I can help you.<br>Yes, as you guessed the three GND points should be connected together, as all the +5 V points.<br><br>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.<br><br>Caffeinomane
Oh.<br><br>So in that case, i will be left with only two cables.<br>A +5 V Cable and a GND Cable right?<br><br>Zane
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.<br><br>Am I right? If so, yes that is right.
Ah. Alright. Thanks so much! :)<br><br>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?<br><br>Zane
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:<br><br> V = i R<br><br>If you put a couple of resistors, say R1 and R2, in series:<br><br> GND |----/\/\/\(R1)----(A)----/\/\/\(R2)----| +5 V<br><br> you have:<br><br> V = i ( R1 + R2 ).<br><br>While the tension at point (A) is<br><br> VA = i R1.<br><br>Solving for i we get<br><br> V / ( R1 + R2 ) = VA / R1<br><br> VA = V R1 / ( R1 + R2 ).<br><br>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.<br><br>Sorry, have been out of touch with Physics for awhile. Been like 2 years.<br><br>Thanks for the explanation. :)

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