etriker
100 kW
Jeremy Harris said:etriker said:Discharge the laptop cell hard at 5 amps and watch the voltage drop then drop the discharge to 1 amp and watch the voltage go up while discharging at 1 amp.![]()
Of course it does.
Most cells behave like this when discharged beyond their rating for a time then have the discharge rate decreased. It's pretty fundamental to the way cells work, and is down to a couple of effects. The first is that higher than rated discharge current causes the cell to heat up from power loss (I²R loss) primarily, that is the resistive loss in the cell. Cell internal resistance (IR) is temperature dependent - make the cell hotter and its IR increases, which increases the I²R loss, which then makes the cell hotter still, etc, etc. Lower the discharge current and the cell starts to cool, as the I²R loss will have dropped (in the case of going from 5A to 1A the I²R loss will drop a great deal). As the cell cools, IR drops, reducing the resistive loss so terminal voltage increases.
I'm shocked that someone so obviously expert on the best cell charging methodology as yourself should find this surprising....................... :wink:
That may be the misunderstanding. I Have been in electronics repair for over 25 years and the more I do it the more I know that I am far from knowing it all.
I do know enough to build battery packs and get down the road using cheap discarded cells. Don't have to hire HK or Ping or cellman to build my packs. Works for me !
Some people think the IR goes down when the cell gets hotter.
http://www.buchmann.ca/chap9-page3.asp
https://docs.google.com/viewer?a=v&q=cache:Nec_uFeqtnEJ:www.mdpi.com/1424-8220/10/6/5604/pdf+li+ion+internal+resistance+increases+with+heat&hl=en&gl=us&pid=bl&srcid=ADGEESjcvVWuyY7OntEYcctC7m8EBA4Y7Wo9I3-YW-VTRlEByqRLVI69BG8-vedyy3Nt1HRKbjdJVZOP6xj2WJfkpxxIJrBFgQEg6PHw3djcduA3J96HyCxbQzp3eQyORCUg0baRYJUL&sig=AHIEtbSDP0OEljz9fPZG89u0ElWRDpUemQ
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090023862_2009023573.pdf
I am thinking the 5 amp discharge makes the cell get hot, lowers the IR then when the current discharge is lowered to 1 amp the cell is hot, has a lower IR and the voltage goes up because of the lower IR caused by heating up the cell ?