Why are we moving to higher voltage packs? We know that the battery cell is not a perfect current source as it also has an **internal resistance**.

Symbolically we can show a cell with the internal resistance as a resistor in series. R_{int} is the DC internal resistance, sometimes abbreviated as DCIR.

The DCIR is not just a single number for any given cell as it varies with State of Charge, State of Health, temperature and discharge time.

This means that when we increase the number of cells in series the resistance of the battery pack increases. We know from Ohm’s Law, that the voltage is proportional to current times resistance (V=IR).

That also means that as we increase the number of cells in series the voltage swing will increase.

The difference between the maximum charge voltage and minimum discharge voltage will increase with the pack nominal voltage.

In simple terms that is just the number of cells in series multiplied by the cell maximum and minimum voltage.

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The problem is the cells have an internal resistance and voltage limits. If we have an OCV of 3.7V @ 50% SOC and an internal resistance of 0.025Ω and we draw 10A from the cell the voltage will drop 0.25V

This is simply Ohms Law.

V = 3.7V – 10A x 0.025Ω = 3.45V

Hence the voltage of the cell under a 10A load will be 3.45V.

We can also calculate the maximum current we can draw taking the cell down to the minimum voltage:

2.5V = 3.7V – I x 0.025Ω

I = (3.7V – 2.5V) / 0.025Ω = 48A

These numbers are quite typical of a 5Ah NMC cell. Peak discharge is around 10C.

If we want more power then we need more voltage or more current. We could:

- use a large battery cell
- put more cells together in series / parallel

The problem is Joule Heating = I^{2}R. This means that if we double the current the heat losses in every resistive element increase by a factor of 4. That includes:

- busbars
- fuses
- contactors
- joints
- connectors

As Power = IV this means to increase power we increase current or voltage. Increasing current increases losses due to heating, increasing the voltage means we can keep the heating losses fixed. It does though mean we need more cells in series and higher voltages brings other constraints once we go above the safe working voltage of 60V DC. As we increase the voltage further then creepage and clearance distances have to be increased. Hence it is not without problems, however, high power means high voltage.

At an individual cell level the maximum current, resultant voltage drop and heating don’t change. The cell heat output will be the same whether it is in a 12V, 48V or 800V pack as it is defined by the discharge / charge current.

However, all of those other elements will have a maximum continuous current rating or maximum temperature (eg busbar insulating layer).

This means that as we increase the power requirements we increase the battery and system voltage.

From the above plot we see a general trend of increasing power and increasing nominal battery pack voltage. However, we have to consider the battery and how it operates with the system voltage limits. Also plotted on this graph are lines of continuous current.

In order to design a battery pack it is essential early on to determine the continuous current requirement as this is a key design factor.

As the pack size increases the rate at which it will be charged and discharged will increase. In order to manage and limit the maximum current the battery pack voltage will increase.

When we plot the nominal battery voltage versus pack total energy content we can see the voltage increasing in steps.

Typical nominal voltages:

- 3.6V
- 12V
- 48V
- 400V
- 800V