Mobile charger to charge LiIon batteries, Usng Voltage Divider to drop voltage

USSR

Newly Enlightened
Joined
Jan 7, 2012
Messages
97
Location
Subcontinent
hii have a lot of mobile phone charges that say "output 5v .7A" i want to use it to charge a liion battery..both prismatic and cylindrical 18650.i have few question..What does the figures in output mean ? does it means its a CC/CV output ?Can i use a Voltage Divider Circuit to Drop the extra voltages and get 4.2volts and then connect those 4.2 volts to the output of protected 18650 cell or unprotected liion cell ( saftey against over charge is gauranteed by firstly the fact that when the battery voltage equals the charger output voltage no more current flows and secondly i will be constantly observing it...)while reading about Voltage divider network in the book "Electic Circuits" by Nilsson it mentioned that the as long as the Load Resistance is very very greater then R2 ( wheare as r1 and r2 are the two series resistors and r2 is the resistor accros which i will be geting output ) ..the output voltage remains fairly constant.. ? so in my case the load is cell what should be its resistance ?Thanks
 

BringerOfLight

Newly Enlightened
Joined
May 23, 2011
Messages
95
hii have a lot of mobile phone charges that say "output 5v .7A" i want to use it to charge a liion battery..both prismatic and cylindrical 18650.i have few question..What does the figures in output mean ? does it means its a CC/CV output ?
All sorts of variations exist. At best, that thing is current limited.
Can i use a Voltage Divider Circuit to Drop the extra voltages and get 4.2volts and then connect those 4.2 volts to the output of protected 18650 cell or unprotected liion cell ( saftey against over charge is gauranteed by firstly the fact that when the battery voltage equals the charger output voltage no more current flows and secondly i will be constantly observing it...)while reading about Voltage divider network in the book "Electic Circuits" by Nilsson it mentioned that the as long as the Load Resistance is very very greater then R2 ( wheare as r1 and r2 are the two series resistors and r2 is the resistor accros which i will be geting output ) ..the output voltage remains fairly constant.. ? so in my case the load is cell what should be its resistance ?Thanks
Voltage dividers don't work for a meaningful amount of current (and it would be a bad idea in the first place).

Those things that call themselves mobile phone chargers are usually just standard DC power supplies and may have current limiting. You should look at 5V / USB powered charger boards:
http://www.candlepowerforums.com/vb/showthread.php?335736-Really-cheap-true-cc-cv-usb-charging-board
http://dangerousprototypes.com/2011/11/02/partlist-wednesday-lithium-polymer-battery-charger-chips/

or this particular Samsung phone charger:
http://www.candlepowerforums.com/vb/showthread.php?308664-Cheap-LiIon-charger-with-Samsung-Travel-charger-for-mobile-phone-(TCH137)
 

USSR

Newly Enlightened
Joined
Jan 7, 2012
Messages
97
Location
Subcontinent
i have boufght 3 samsung chargers just to mod them..
but evertime i buy one i get something different..

i want o obtain a constant 4.15-4.2 volts from that 5 volts ..so its already regulated etc.. and it can be obtined using VDR but the nilsson bok gave me a bit confussion which says that load resistance >> r2..
 

shadowjk

Enlightened
Joined
Oct 21, 2007
Messages
451
The resistance of a Li-Ion cell is somehere at around 0.1 Ohms. So you need a voltage divider with very very smaller resistance than 0.1 Ohm. This would mean your 5V powersupply would have to be able to provide a few hundred Amps, and your resistors would be dissipating a few thousand watts of heat. Getting a powersupply that strong, and resistors that can handle that amount of power, would be very very expensive. It's easier and cheaper to get a charging board.
 
Top