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Hello everybody,

Charging devices off USB ports on computers has always been a quite a bit of a hassle to me. This I say because I own 2 mp3 players and soon I’m going to own and Amazon Kindle. So I began on building up a power adapter to recharge my devices rather than buying it from somewhere. The process at first was quite frustrating, partly because when I first built the circuit which gave an output of 4.97V on the positive rail of the USB port, my mp3 player wasn’t accepting the charge although the voltage was absolutely within range. The second part that frustrated me was I had build a Printed Circuit Board from scratch for the first circuit, btw I used a LM317L, ah, what a waste of time...

Since the first try was a bust, I began on researching about USB (wikipedia’s great), and, well since the Kindle manual says “use a charger that complies with USB Battery Charging Rev.1.2” , I downloaded USB Battery Charging 1.2.pdf (from USB Charging Rev. 1.2). That was a hell of a technical document, phew!

Anyways, after all that I found out that certain devices(Apple iPhone, iPod) require different voltages on the USB rails to induce them into charging, for my mp3 players(Transcend MP330K) the voltages on the rails is a must. But this isn't true for all devices, some cheap phones might accept only the 5.0V the USB has to provide. But now this is the ultimate fail-safe in USB charging.

Typically a USB cable/port has 4 wires/pinouts.

1. VBUS - Red (or orange) +5V

2. D - - White (of gold) Data-

3. D + - Green Data+

4. GND - Black (or blue) Ground

In the case of USB charging of certain devices that require additional voltages, +5V on the bus rail isn’t enough. The D- and D+ needs to be in a certain voltage level to accept a charge off the port, and the voltage levels the D+ and D- are in determines the amout of current the devices would get off from the charging port.

If you want to charge your device using a maximum of 500mA, which is typical for small devices, providing a voltage of +2V for both the D+ and D- will make the device draw out a maximum of 0.5A.

If you want a faster charge you could make the device draw out a maximum of 1A by setting the voltage of the D+ to 2.00V and the D- to 2.80V. You could go even upto 2A by setting D+ to 2.80V and D- to 2.00V, but keep in mind 2A can’t be acheived using the circuit that I’m about to demonstrate. It is important to maintain these voltages even when the charger is under heavy loads.

Step 1: The Circuit (how It Works)

So with all that in mind, I designed a circuit that fulfills all above declarations.

It uses two LM7806 linear voltage regulators. RG1 delivers power to the VBUS rail, a max of +5.17V. RG2 delivers +2V on the D+ and D- rails. So the drop in the voltage under a heavy load does not affect the D+ and D- rails’ voltage. VR1 and VR2 are 1kΩ variable resistors set to 670Ω and 940Ω respectively. You could manipulate the voltages on the D+ and D- rails using VR1 and VR2. D5 is a 1W Zenner diode (1N4733, 5.1V), it controls the voltage on the VBUS rail, and it emits 436.76mW of heat (,way below its rated 1W). D3 is a simple rectifier diode to drop the voltage down a notch which reduces the heat output of D5 and it disables the back-flow of current in certain circumstances.

A fuse, F1, rated 1A is used for additional protection on the LM7806 as it is rated a max of 1A. RG1 heats up under a heavy load, so using a simple heat-sink is advisable. RG2 doesn't heat up pretty much. Use a Female Type A USB port on CN1.

Step 2: The PCB

Print the PCB using the Artwork displayed below/above. Be sure to get the pin-outs correct. Refer USB wiki if you are uncertain about the pin-outs. The dimensions of the PCB is 7cm x 7cm.

Step 3: Check Before You Leap!

*Standard voltage on the USB VBUS rail is +5v with a deviance of ±0.25V, so 5.17V max will not harm your USB device in any way. Please check the voltages on the USB rails before plugging in ANY USB device using a DIGITAL MULTIMETER.

Step 4: Try It! and Thanks.

So this concludes the “Advanced USB Charger” instructable. Thank you.

Amod Sachintha.

You could have put both voltage regulators in parallel. Or just get ride of RG2 altogether and use higher ohm resistors connected directly to your source as your voltage divide.
<p>Sometimes the voltage at the source drops considerably when under load, so if resistors were to be used instead of RG2, those voltages would also drop consequently. I use a 240V-16V step-down transformer plus I have attached a small fan to the input, the voltmeter shows the input voltage under operation (without a load at the USB) as 10.62V. Once under a load this drops even more (8.82v), so if resistors were to be used, either at one poin,t you could tweak the voltage to 2.00v but then, when under a load it would drop, or you could tweak the voltage higher than 2.00v (say 2.30v) and once you connect a load to the USB it might drop to 2.00V. But in both cases the D+ and D- lines are over or under voltage for a split second, and that should not happen. Hence the necessity for a second Regulator.</p>
<p>You did a good job overall making this a simple and to the point circuit. Thought, just so you know. If you had connected the voltage regulators in parallel they would have shared the loads. Meaning they would have been sharing the current that is supplied to both the charging circuit and the D+/D- voltage divide. The only change that you would need to make it to connect the Vout pins on the regulators. This will help prevent RG1 from getting so hot.</p>
<p>Point taken, thanks in advance. :)</p>
Good, but lots of smartphones &amp; most tablets require the 2.1 amp output.
<p>Yeah, it's a matter of changing the RG1 Regulator. But I'd personally charge my devices with a low current (1A max) but with a precise voltage cause it would lengthen the battery life, it takes time but it's worth when it comes to certain devices.</p>
awesome! I had known idea about the voltages on the data lines telling it how much to pull.

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