Need help with modeling an overloaded battery

Question

I wish to model a battery for a small computer game, a battery which will be subjected to heavy loads and deep discharge. The model doesn't have to be high-fidelity, just a rough approximation will do. I want to display both the current and voltage, and watch how they change as the battery runs out of juice.

I have a loose grasp of the basics, V=IR and so on, and can calculate these for a steady state circuit with an ideal power source but am lost when dealing with a more realistic battery model.

If I understand right, I can add add load to my battery, and figure out the current being drawn, and deduct that from the battery capacity in Ah, giving me a decreasing value for the "juice" left in the battery. But I get a bit lost thinking about what happens when the load is such that the current being supplied is very high. I assume the voltage should sag, but how can i model this? And is this current limit and voltage sag also connected to the available capacity of the battery?

I have read about representing a battery as a voltage source and a series resistor, but am a bit lost in how to use this. Should I be increasing the series resistance in relation to the decreasing capacity? What is the nature of that relationship, if so?

My apologies if this is not very clear, I have tried my best with the terminology.

Answer 1

You could model the power consuming "things" as constant resistance loads, all in parallel. A high resistance is a light load, and vice versa.

To find the total resistance of all parallel loads, use this equation:

R = 1/((1/R1) + (1/R2) + (1/R3) ...etc...)

If you have multiple batteries in parallel, you find the total resistance of all batteries using the same equation for the battery resistances.

The load resistance(s) is in series with the battery resistance(s), so next step is to simply add these two totals together. Rbatt + Rload.

Then use the equation Power = Voltage*Voltage/Resistance to determine the performance of each load. The resistance here is the resistance of the individual load. Power equals performance.

As the battery voltage drops, you will see that the performance of each load will drop drastically. Also, when you have many loads connected, the battery resistance will be significant (it will cause the voltage to sag). With few loads, the battery resistance will have less of an impact.

If you want to model a Lithium battery, start out at 4.2 V when full. When the battery reaches 3 V; it's "flat", and the power should be cut. If you want to model a NiMH battery, start at 1.5V and cut at 0.8V or so. Or you could set these values anywhere you want.

The tweakable values here are

• full battery voltage
• empty battery voltage
• load resistance (can be different for each individual load)
• battery resistance (can be different for each type of battery)
• battery capacity (mAh, different for each type of battery)

If you want to stay somewhat realistic, you do not want to combine batteries with different max and min voltages in parallel.

However, if you implement the option to put batteries in series as well as parallel, you open up a whole new realm of possibilities: The output voltage is doubled, the battery resistance is doubled, the capacity in mAh stays the same.

To model how the battery gets depleted, you indeed need a charge state vs voltage curve. Find a real one on the Internet and build yourself a dictionary, or make an equation that approximates it, or something to that effect.

If you want to simplify this a lot, you can just say that the battery is at max voltage (e.g. 4.2V) when it's full (has all the mAhs), and at min voltage (e.g. 3V) when it's empty (has zero mAh left). At any charge state between 0 mAh and full mAh, you can interpolate between these two voltages to find the battery voltage at that charge state. This will give you a linear discharge curve, not perfect, but it will serve as a placeholder.

Let's say that the battery capacity is 1000 mAh == 1 Ah == 60 amps-minute == 3600 amps-second. Let's also say that the load is 1 Ohm and the internal resistance of the battery is 0.1 Ohm. We start with a full battery at 4.2 V.

Then the algorithm could go like this, with one second ticks:

• Find the current, the first tick it is 4.2V / 1.1 Ohm = 3.82A.
• The load is producing 3.82A * 3.82A * 1 Ohm = 14.5924W, so you know the performance.
• While 3.82A * 3.82A * 0.1 Ohm = 1.45924W is lost as heat in the battery.
• The battery started out with 3600 amp-seconds, but now it only has 3600 - 3.82 = 3596.18 amp-seconds left.
• Now use your discharge curve to find the new battery voltage, let's say it's already dropped to 4.15V.
• Rinse and repeat.

You can now model the charge state, the battery voltage, and the appliance performance second by second.

Answer 2

I believe what you are looking to model is something like the following. It is an extremely crude representation, and the values I have should be ignored(for game balance purposes these can be tweaked), however it will reduce the voltage as current increases and/or the battery discharges. The initial voltage across the cap can be whatever value you need for the fully charged battery(again for game balance).

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

You haven't said what sort of battery. To a rough approximation, a standard dry cell battery (zinc carbon or alkaline) is a constant voltage source in series with a resistor. The resistance rises as the battery discharges.

In reality, the voltage drops, but not by a great deal until the battery is totally flat. Also, with a real battery, if the large load is removed, the resistance will drop a bit as the battery recovers over a few hours.

Other battery types could be modelled similarly, but tweaking the parameters. A lithium ion battery has a much lower internal resistance, but the voltage drops significantly (from about 3.7V down to 3.0V) as it runs down.