I just ordered this Motorola charger to use as a spare charger for a host of usb devices in my home like iphone, mp3 players etc. I did not look too closely at the specs before ordering, but now I see it supplies 5.1 V instead of the regular 5 V for usb chargers. Will that damage devices connected to it? And why would Motorola design a 5.1 V charger for usb charged devices, as they work at 5 V?
No damage, but you may notice your phone heats up slightly during the charging due to the 0.1V voltage drop. It should not cause long term damage.
Thank you. So, it should be fine. That is what I wanted to hear. I have done some further research and apparently devices that are usb powered can be safely charged in the range of 4.9-5.1 V. Maybe even more, depending on the device.
All chargers supply a higher voltage than the batteries they charge. That is pretty much how they work. There has to be a voltage differential to produce the necessary current flow in the correct direction to charge the battery. If you look at your car, it has a 12V battery, but typical alternators provide 13.8 to 14.4V charging voltage to the battery. Batteries are very robust devices. You have nothing to worry about.
Thanks, that helps. I was just thrown by the fact that a vast majority of usb chargers are rated for exact 5 V. The folks at Motorola must be slightly weird to have theirs supply 5.1 V instead. You are correct as far as a battery is concerned, but normally it is the small circuitry in between the charging current and the battery that is blown by the higher voltage. That circuitry's main function, among other functions, is to sense when the battery is fully charged and cut off the charging current to prolong the battery's life. That circuitry is rated for a fixed voltage and if you apply a higher voltage to it, a capacitor (I can not remember what it is called exactly, I think coupling capacitor maybe) usually gets blown. When this happens the device is practically rendered dead as it is very difficult to fix. I remember a rule that "current is pulled from the charger by the device" whereas "voltage is pushed from the charger to the device" Supplying a higher voltage to a small electronic gadget like a mp3 player or a phone will most likely kill it. Or is my reasoning flawed?
Your reasoning, in terms of theory (vs real-world) is right. But you still cannot fully charge a 5V battery with a 5V charger. There's has to be a "difference in potential" in order for current to flow. Now will will blow something if you attempt to charge a 5V battery with a 10V charger? Most likely. But here, in practice, we are talking about a 5.1V charger for a 5 volt battery.
If we consider the batter as a "capacitor" then the charging automatically stops when the capacitor voltage is also 5V. Thus, it should be possible to charge a battery to 5V using a 5V charger. As for higher voltages, it has more to do with heat than voltage. If the component is rated for 5V then the excess becomes power loss/voltage loss (=heat) since the circuit is designed to use only a certain current and a certain voltage. The heat will damage the components.
Nope. You can charge it, but it will never get fully charged because the closer the battery gets to fully charged, the resistance to current flow in that direction increases. A charger's voltage HAS to be higher than the battery it is charging for the battery to become fully charged. Remember, a fully charged 5V battery will read a bit higher than 5V. And a battery will become pretty much useless if the voltage drops just a little bit below 5V - perhaps as little as 4.5V and the battery is useless. A little research on google will bare this all out.
