When USB was invented, I think the inventors had never thought that it would be used for so many purposes. The name Universal in USB has certainly come true. My guesstimate that much if not most of the USB cables in use today are being used just for charging devices. So I have to add my two cents’ worth about this use.
The USB specifications call for a five meter or sixteen foot limit on the length of the cable because longer cables have too much signal loss and interference and that could cause problems for the transfer of data. But using the cables just for charging doesn’t have any problem with data transfer because there is no data being transferred. So the 5 meter length of the cable does not really apply.
However the USB specification for the maximum current from the USB port is 0.5 amps or 500 milliamps. Using a longer cable can cause more losses in the charging current due to the resistance of the wires in the cable. So I think it’s a good idea to go over some simple math to find out how this could affect the 5 volts that is used for charging.
I have taken apart several USB cables and I’ve found that the charging wires are usually thicker than the signal wires. I have not examined the wires closely enough to find out exactly what size they are, so I will assume that they are 24 AWG or 0.5 mm diameter. The wire tables give the resistance of this wire as 25.7 ohms per k feet, or 25.7 milliohms per foot.
Let’s assume that the user decides to use a single 16 foot micro USB cable for charging his device at 0.5 amps. The power wires are 24 AWG and 16 feet long, but there are two wires so they have a total length of 32 feet. The resistance is 0.0257 ohms per foot times 32 feet, or .8224 ohms. At 1/2 amp flowing, this is 0.4112 volts drop. And this does not include any voltage drop in the connectors or other points. So the actual voltage at the device being charged is less than 4.6 volts.
That’s an excessive amount of loss in the cable. The amount of power loss is going to be 0.5 times 0.4112 or 0.2056 watts. The cable is carrying 5 volts at 0.5 amps, which is 2.5 watts total. The 0.2056 watts loss divided by the total power is 0.08224 or 8.224 percent of the total.
Suppose the user decides to add another 16 foot extension to the 16 existing foot cable, so the total is 32 feet or 10 meters. Now the voltage drop in the cable is over 0.8 volts and the voltage at the device being charged is down to 4.18 volts or less. The power loss is going to be double, or 16.24 percent of the total.
But there is more to this than what I’ve said. Most of the USB charging cables are plugged into the AC adapter that converts down to 5VDC at more than a half amp. Some adapters put out 1 amp, some for tablets put out 2 amps or even more. So the charging cable loss could greatly increase to double or quadruple what I calculated earlier.
Solutions
Well, I guess the most obvious solution is to keep the USB cable short and add an extension cord from the wall to the charger. Of course this doesn’t work if the charger is built into the wall socket (yes they’re available) or the USB jack is on the front of a desktop PC or similar. If the situation demands that the USB cable be longer, then keep it as close to the length it needs to be and don’t allow extra cable.
Another solution is to use a heavy cable to extend the USB cable. This could be 18 AWG speaker wire or similar. The resistance of 18 AWG wire is about 1/4 that of 24 AWG, so a cable could be four times longer, with the same resistance. A 1 meter USB cable could be 4 meters long without any noticeable increase, and even longer with a tolerable amount of increased resistance. The problem is that the user would have to make his own cable, since it doesn’t appear that any cables with heavy power wires are available for purchase. I would have to take a USB cable, cut off the ends and splice them onto a length of heavier wire.
There is another solution that may seem obvious to the reader. Why not just increase the output voltage to 5.4 volts, so that the other end would then be 5 volts under load? That is possible, but there is a risk of damaging the charging circuit in the device being charged by the excessive voltage. As soon as the battery in the device reaches full charge, the charging current drops and the voltage at the device starts to rise. If the voltage gets too high, the device may be damaged.
There is a solution to this rise in voltage. This is often used in high quality and high current power supplies. It is called remote sense. The cable has two wires for the current, and two more wires to connect the far end to the voltage regulator. When the far end voltage drops, the sense wires send back the drop and the regulator circuit adjusts the voltage to compensate for the cable loss. This requires a four wire cable, but the power wires do not have to be heavy gauge wire because any losses in them are corrected. But the cable has to be made especially for this application, and the power adapter has to be made to have this feature. It would be safest to connect the cable permanently to the adapter so that the cable cannot be disconnected and used for something else.
I have a power supply that has remote sensing, so I could make up a cable and connect it to this power supply. But I think I’ll just use a heavier cable instead. Carrying a tiny charging adapter around is a lot easier than carrying around a big bench power supply.
I once had an external 2.5 inch hard drive enclosure that plugged into the USB ports for power. Yes, it used two USB ports and had a ‘Y’ cable with two plugs on it. But this still doesn’t eliminate the voltage drop in a long cable, it just reduces the loss somewhat. But most adapters do not have two ports, and I don’t think it’s a good idea to take up two USB ports on the computer to charge something.
Update 2014 Nov 7 – I have been investigating this problem and I’ve blogged it lately. See more in my blog.