4

I am trying to understand why we can simply just calculate the usable energy that a battery has by multiplying the capacity and the nominal voltage.

Since the capacity (available electrical charge) in a battery depends on the drawn current (unlike in a capacitor), we can't actually know what energy is in the battery because we don't know the nominal current at which the capacity was set. So in order to really know how much energy we can get from a battery, we have to measure/calculate it for every possible load.

Here is my thinking and please correct me where i am wrong:

1

Most simple way of modeling a battery is with a simple internal resistance. Internal resistance does not affect the battery capacity because the charge is conserved. Constant current draw can be modeled by an ideal current source.

2

  • With a current-source as a load, the current is independent of the voltage of the battery (at least when the battery operates in its operating voltage)
  • Power drawn from the battery in this model does depend on the voltage.
  • Available electrical charge at the certain current draw linearly falls with time.

3

For a LiOn battery, we usually see a graph like this - how the voltage depends on left-over capacity. This capacity does kinda relate to how much electrical charge is left in the battery in the form of ions, but it has mainly to do where we define the usefulness of the battery. We defined the charge/capacity to be 0 when the voltage falls to 3V and for it to be at e.g. 2000mAh, when the voltage is at 4.2V. In reality however, there is more charge in the battery if we further drain the battery under 3V or charge it above 4.2, but it's either unusable (voltage too low), unsafe or it damages the battery in some way.

4 My doubt about the meaning of the nominal voltage

The energy received can be calculated with integrating Voltage and Current through time OR with integrating the Voltage through electrical charge.

The energy is represented by the surface under this graph. Nominal voltage is defined to be at the point, where the surface can simply be calculated with a simple multiplication instead of integration. Graphs

KrNeki
  • 41
  • 1

1 Answers1

1

The capacity of the battery should be specified at some given rate of discharge. If not, look for a battery supplier that will specify the battery correctly.

So, the amount of energy available can be estimated by multiplying the specified capacity by the specified final (terminal) voltage for that capacity rating. If the battery manufacturer says that the battery will provide 6Ah when discharged at 1A to a terminal voltage of 12V, then you can estimate that you will obtain at least 72Wh from the battery if you discharge it at 1A.

Of course this is only an estimate. Rarely is there an application where the discharge is truly a constant current and there will be some variation in the actual capacity of individual battery units. But such an estimate is usually suitable (good enough) for the typical application.

Michael Karas
  • 56,889
  • 3
  • 70
  • 138
Elliot Alderson
  • 31,192
  • 5
  • 29
  • 67
  • Can you add Energy to your answers. As in - Capacity seems to be different depending on the discharge. The charges are conserved, so they have to go somewhere. And wherever they go, the energy follows. So what happens with the energy? – KrNeki Dec 02 '20 at 10:11
  • No, the charges don't "have to go somewhere". What happens to the energy in a burning log if you pour water on it? The source of energy is a chemical reaction. Energy can be lost as heat in the battery itself, or because the chemical reactions become less efficient. – Elliot Alderson Dec 02 '20 at 13:32
  • I think the log comparison brings more questions in my mind. The energy that is already in heat goes to the water, which heats up and then warmed up leaks to the ground. The potential energy of the log is still there in the same amount if the log didn't absorb any water. If it did, then it's potential energy is now lower (because if you burn it now, it will use it's energy to evaporate water out of itself). So - different usage of the battery (fast, slow..) warms up the battery in the different ways because /reasons/ - ergo - the energy is lost as heat in the battery itself? – KrNeki Dec 03 '20 at 10:03
  • I guess that makes sence, since internal resistance is load dependent. Thanks. – KrNeki Dec 03 '20 at 10:05