- Joined Jan 20, 2013
Why would they rate it at 8000mah @ 3.7v on the box then? That seems illogical if internally there were several cells in series.They use 3 or 4 high C rate cells in series to get a bit over 12 volts at high amps out
Advertizing in rarely logical.Why would they rate it at 8000mah @ 3.7v on the box then? That seems illogical if internally there were several cells in series.
Right.Amp hour ratings are the same regardless of number of equal sized cells in series. Only the voltages change.
Probably like this, but 1 more in series..Right.
8000mah @ 3.7v = four 2000mah lithium ion cells in parallel
2000mah @ 14.8v = four 2000mah lithium ion cells in series
The box says 8000mah @ 3.7v, so to get the voltage to boost an engine starter it would need 4 cells in series, which would mean each cell is only 2000mah each. Which is very typical for 18650 cells.
But the C rating would still be based on the 2000mah @ 14.8v output. So those cells would need to be 100C to 175C?
That's very impressive!Probably like this, but 1 more in series..
Don't forget you will only get a couple of minutes out of the little lithium guy. Maybe 4 -5 times longer from the big old lead acid.That's very impressive!
8ah * 30c @ 11.1v = 2.6kW
A typical 50 pound car battery might be rated 500 cold cranking amps, which would mean 500a @ 7.2v voltage dropped = 3.6kW
When the discharge C is calculated for lithium (polymer) ion batteries, is there some sort of standard as far as voltage drop under load? When it's providing 30C, as spec'ed, what is the expected voltage drop?
|Thread starter||Similar threads||Forum||Replies||Date|
|Seeking 12V 200A DPDT relay||General Electronics Chat||14|
|A||Designning H Bridge for 200A||General Electronics Chat||14|
|R||200A - Battery Charger Protection||Power Electronics||9|
|R||10Kw motor needs 200A start, is there a capacitor for this?||Power Electronics||6|
|T||200A neutral return grounded twice||General Electronics Chat||13|
by Jake Hertz
by Jake Hertz