* NOTE SAFETY CAUTIONS AT END *
Caveat Emptor - No responsibility taken for advice given.
Playing with LiIons is an almost safe activity for careful competent adults.
Do it wrong and they die.
You may too, but probably not - but your equipment may.
The configuration should be 2p4s (or === 4s2p) as cells are nominal 3.7V each x 4 = 14.8V.
3.7V x 2.5 Ah c 8 cells = 74 Wh which matches your figure.
It is common for functionally dead LiIon battery packs to contain usefully alive LiIon cells.
It is a VERY GOOD IDEA [tm] to use cells which have inbuilt protection circuits. Many (probably most) do have, and if it's an HP battery pack as the part number suggests then they very probably will have.
If they are cylindrical 18650 cells rated at 2500 mAh then they will have at least 1C discharge capability (2.5A) and probably 2C (5A). YMMV!. With 2 strings in parallel that amounts to 5A or 10A.
Loaded typical voltage for a 4 cell pack will be in the 12V - 14V range and V after a full charge will be 4 x 4.2V = 16.8V.
As 12V = about as low as you should go x 5A (all cells) that's 60 W. That should be enough for your screwdriver - this can be measured easily enough.
You may be able to use the notebook charging circuitry if notebook death is unrelated to this. This may extract from the laptop functionally well BUT it's possible that the processor goy=t it's fingers in the pie due to gas gauging or whatever so you *may * find the circuitry is too embedded to be extractable. If you don't mind having a charger that looks suspiciously like a dead laptop then you can probably use the whole unit.
The battery will probably have monitoring points for each cell = at least 5 contacts (-ve, +1, +2, +3, +4 and may also have a thermistor for temperature monitoring - probably v+ and maybe ground referenced = 6 contacts. You could keep the battery pack whole and put it back in the laptop to charge (so you'd need a cord or other way of connecting to the s.driver. ) or you can run N leads into the laptop from the s.driver or ... .
You can buy N cell LiIon charger IC's relatively cheaply and you may find it easier to build a new charger from scratch.
If you can manage the lxury of accessing each cell separately (floating mini chargers) a LiIon cell is VERY easy to charge well when it is working OK.
- Charge at 1C (2.5A) or less until Vcell = 4.1V.
then
- Charge at 4.1V until I charge = C/x
where x depends on what you want to achieve. X=2 gives long cell life and notably lower capacity. C/4 or C/10 may be used. C/20 is for the energy density freaks.
- If you stop at the top of the constant current mode (when Vcharge = 4.1V then you will get ~= 70% capacity and far better cell life. This is also the optimum level to leave cells at when in storage or no use for a while.
DANGER WILL ROBINSON
I said 4.1V above. Official figure is 4.2V for full cell BUT if you do that yu MUST temperature compensate. Magic smoke, VERY thick and hot, may or may not happen not much above 4.2V.
You are probably aware that 'vent with flame" is a LiIon cells favorite parlor trick. They can self immolate just because they can on a bad day - but many many many last for years with no problems.
DO NOT SHORT.
DO NOT CHARGE PAST 4.2v
DO NOT CHARGE AT HIGH CURRENT (> 1C) WITHOUT DATASHEET SUPPORT.
Do not bend fold spindle mutilate staple - but especially in this context do not spindle - poking them with sharp objects may get exciting action.
People have had good results doing what you are trying to do.
Do it properly and you should get a good result.
Best Answer
Alrighty, so there will be the answer that we can extract from the information on the datasheet/discharge curve chart. That will give you a fairly good idea of what to expect. Or if you want more that than you can run tests on your specific cells to find out. But I think you are just needing a more of a ballpark answer. Ballpark is the only thing that is really of any value, unless you know exactly what temperature the cells will be at all the time as well as some other variables.
So, li-ion cells have a non-linear discharge curve. If we look up the discharge curve for your particular cells it will look something like this one.
Now you can understand why you would need to know the current draw if you wanted more than a ballpark guess; the capacity in W-H changes with load because different loads cause different voltage sags.
So anyways, lets assume we are talking about 2A draw only. Look at the 2A curve (the black line with a negative slope) You are saying you wanted to set the cutoff at 3v. Well since we are ballparking, we can notice that the discharge curve for the region you are interested in is close enough to linear that you could just easily calculate the area under the curve without an integral.
If you aren't sure if you need the capacity in WH or just AH well ah is easily found by just finding the capacity that relates to the cutoff voltage you want in the discharge curve.
If you do want WH though: If we say the voltage range is 4.1 to 3.1, and the capacity is 3.2ah well then you get a rectangle with a triangle on top of it that you can find the area of and know your Watt Hours. WH = (3.2*3.1) + 1/2(3.2* (4.1-3.1))