Project 1
1. Design a battery pack for a car roughly 150 Kw with 120 V. Use 3500 mAh 3.6V nominal NMC chemistry cell.
a. Design the battery pack configuration.
Ans.
Given,
Battery Pack
Power = 150 kW
Pack Voltage = 120 V
Cell
Capacity = 3500 mAh
Voltage = 3.6 V
So now as we know,
P = V*I
I = P/V
I = 150kW/120 V = 1250 A
If the battery is used for an hour at a 1C rate
Pack Capacity = 1250A *1h = 1250Ah
Now when we have the Capacity and Voltage values of both battery pack and cell as well
For Series connection,
Number of cells in series (S) = Pack Voltage / Nominal Cell voltage
mS = 120/3.6 = 33.3 ≈ 34 cells in series
Number of cells in parallel = Pack Ah/ Cell Ah
nP = 1250 / 3.5 = 357.14 ≈ 357 cells in parallel
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b. Draw the BMS topology for this battery pack.
Ans. Since in part (a) we calculated the number of cells required for the required battery pack, which is the of the configuration of 357P34S.
For such battery configuration, Modular Topology would be preferred because
- The 34 cells in series
- The application of the battery is higher i.e. 150kW.
2 Sets of 17 cells would be used.

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2. Build a 3S2P battery pack configuration with generic battery block using Matlab/Simulink.
Ans.
The 3S2P configuration in this 3 cells are connected in series and 2 sets of these 3 series cells are connected in parallel.
The parameter given inside the battery is given below:-
- Nominal voltage of a cell 7.2V
- Rated capacity of a cell 5.4Ah
Therefore ,
Nominal voltage of the battery pack = no. of cell in series * nominal voltage of a cell
= 3 * 7.2v
= 21.6 V
Rated capacity of the battery pack = no of cell in parallel* rated capacity of a cell
= 2 * 5.4Ah
= 10.8Ah
Li-ion Battery -

Resistance :



Observation-
The battery is showing practical symptoms of deterioration in SoC with the passing of time under load as seen in the Scope above.
Also, the Display shows the current as 4.349 A and the Voltage of the battery as 23.49V at the end of 300 sec (5 min) simulation.
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