How to calculate right current to charge 2 cell series 8.4V 4000mAh battery lipo?

Question

I have 4000mAh 2 cell series 8.4V lipo battery

I need make a custom charger I am thinking of using LT3650-8.4 2A charging Ic

if I set 2A how long will it take to charge this fully /if timer is disabled timer?

is it (1st cell 4000mAh + 2nd cell 4000mAh) / 2000mA = 4h ?

if I set 1A how long will it take to charge this fully /if timer is disabled timer?

is it (1st cell 4000mAh + 2nd cell 4000mAh) / 1000mA = 8h ?

What is the maximum capacity of 2 cell battery this IC can charge properly? without a timer?

as I know we have to manage the full charge cycle (5.6-8.4) within 4h at least, so to charge it within 4h with the set current to 2A as in the figure, how long will it take to fully charge a 4000mAH 2 cell battery?

I think everybody has misunderstood the question I am asking. now everybody should be able to understand the question

Answer 1

Summary:

• Charging 2 x 4Ah cells in series with 2A current will charge them to ABOUT
  time = Ah/Charge_rate x 90% = 4/2 x 0.90 = 1.8 hours

• At that stage they will be at 8.4V and ABOUT 90% of capacity.

• Beyond that, if you wish to continue charging you must hold voltage at 4.2V/cell maximum and allow current to decrease under control of battery chemistry.

• Reduction of current to about C/5 will take around an additional 2 hours. Stopping at V=4.2V/cell without CC taper charging greatly increases battery life in return for a small reduction in per cycle capacity.

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LOOK at the LT3650 data sheet that Andrew cited.
READ what it says.

The graph on the 1st page shows the theoretical and practical result of CCCV LiIon charging. You are dangerously short on theoretical knowledge to build your own charger.

If charging current flows in series through multiple identical cells it will charge all of them at the same rate and in the same time. The voltage needed will increase with the cell count. You can only charge a cell at CC (constant current) until it reaches Vmax (typically 4.2V/celll).
Then you MUST either cease charging or hold Vcell constant at Vmax and allow the current to "taper"(reduce) under control of the cell chemistry. If you hold I at 2A then after a while Vcell will rise above Vmax and the cell will be damaged or destroyed.

IF the cell is rated at 1C = 4A charge max then you can charge at C/2 = 2A until cell reaches Vmax. Cell capacity will be about 90% of max. If there are 2 cells in series the time taken will be the same but the required voltage will be double.