When I short an AA battery, I have observed that it's voltage drops a couple hundred millivolts and when I break the short, its voltage starts to "replenish" at a decreasing speed.
What is the reason for this behavior?
Electronic – Why voltage drops when I short a battery
batteriesvoltage
Related Solutions
That's how batteries work. As you draw current from them, you deplete the chemical reactants near the plates, and it takes time for new reactants to take their place. This creates a gradient in the reactant concentration within the volume of the battery.
The terminal voltage is a function of both the reactant concentration at the plates and the voltage drop across the internal resistance of the battery (which rises a bit with reactant depletion). If you let it run long enough, the terminal voltage should stabilize at some lower value, although the overall depletion of the battery will eventually become a factor.
When you switch off, the voltage drop across the internal resistance disappears immediately, but it takes some time for the reactant concentration gradient to disappear, which is why you see the sudden jump up followed by the slow recovery to the original terminal voltage.
A battery has some internal resistance that for Li-ion batteries is usually in the 0.15 \$\Omega\$ region when fresh.
As you should know from Ohm's Law, \$V = I \times R\$ so if you put a load on the battery this will cause a current to drain from the battery and this will cause, depending on the current, a small voltage drop over the battery's internal resistance as it would on any normal resistor in a circuit. This can be explained visually like so:
simulate this circuit – Schematic created using CircuitLab
As the battery gets used and 'ages' this internal resistance increases so the voltage drop over it will also increase for a constant current. The 'open-circuit' voltage of the battery will always be more or less the initial voltage as the 'load' will be near infinity so all the voltage drop will be over this air-gap between the batteries two terminals.
As you mentioned the drop in voltage is less with a larger load, this is because the current draw will be less and hence the drop over the internal resistance will also be less.
You should look into a constant current source using a low voltage op-amp (I did something similar to this a couple of months a go and ran into the same issue - hence that question) and you'll be able to work out the mAh easily enough as the current should be more or less constant and you'll know the time taken for the voltage to fall.
Hopefully this helps and answers your question - it is expected that a voltage drop will be seen when applying a load to a battery.
Related Topic
- Electronic – Multimeter battery test
- Electronic – arduino – Measuring current drops voltage
- Electrical – Battery voltage drops when connected to a motor
- Electronic – short circuit current battery
- Electronic – Voltage drop on batteries when connected to ups/inverter
- Electronic – Full charged 12V battery with AC converter “empty” within one second – voltage drops, converter beeps
Best Answer
Actually, if you are truly shorting the battery, then its voltage will be 0 by definition.
It seems you are talking about the battery voltage shortly after abuse, like shorting. Batteries work on chemical reactions, which don't happen instantaneously. During heavy load, like being shorted, the battery is pumping charge around as much as it possibly can. Internally, charge is moved around by ions physically moving thru some electrolyte. Under heavy load, various things can happen to get in the way of efficient battery operation. Some of these things are permanent, but others can recover after some time for things to diffuse around inside the battery. This is why the open circuit voltage will be lower immediately after abuse, but will usually recover somewhat over time after that.
If you are asking why the voltage of a battery goes down with load, then the answer is in part internal resistance. A very simplistic model of a battery is a fixed voltage source with a resistance in series, also known as a Thevenin source. Even if the voltage source remains at the same level, the terminal voltage will go down with current since the drop across the resistance goes up with current.
There are also again chemical affects, so the voltage source doesn't stay fixed either. As the battery ages, both the voltage source will go down and the internal resistance will go up.
Batteries are complicated.