Here is some great data on used battery capacity:

http://www2.ife.ee.ethz.ch/~rolfz/batak/ICBR2003_Zinniker.pdf

About 33% wasted is their figure.

Perhaps Batteroo would be kind enough to publish their data?

From the paper also, 10% were thrown away perfectly new.

Both figures can easily be attributed to just just human factors and multiple cell devices. The weak cell dies before the others get to zero and the user throws them all out.

I'm curious how much can be attributed to that, how much is due to an abnormally high device voltage cutoff, or simply a chemical battery recovery issue.

How much usable energy 'comes back' if a battery is discharged hard (like modern cameras) then allowed to 'recover'?

Hi there,

No energy really 'comes back', it is all there to begin with, just hiding a little.

In the more advanced electrical battery models there is a secondary storage component, a capacitor, that is smaller than the primary storage component which is a larger capacitor, and the resistance between the smaller cap and the larger cap is larger than the resistance from the battery terminal to the larger capacitor.

What this means is that when the battery is discharged, the larger capacitor voltage is what we measure when we measure the terminal voltage, while the smaller capacitor still has a slightly higher voltage across it because it does not discharge as much as the main capacitor due to the higher resistance.

When the load is finally disconnected, the smaller capacitor slowly recharges the larger main capacitor, and so we see a voltage rise across the battery terminals.

So the battery really looks sort of like two batteries, but they are not directly in parallel they are separated by a resistance that is larger than the so called series resistance of the battery, and the second smaller battery can charge the bigger main battery a little when the load is disconnected.

You might note however that the smaller battery gets it's energy from the regular charge process too just like the larger battery when the whole battery is charged normally, so there is no extra energy coming from anywhere.

This is still using only a simplified model with only two storage elements but that illustrates the main idea from a purely electrical perspective.

There are various battery models available on the web.

To get back to the main subject of the energy left in a battery when it is disconnected before all it's energy is depleted, here is a quick table of cutoff voltages and the total percent that could be attained if the rest of the energy was used too, compared to what is actually used up to that point...

1.4v 2.3hr 12.2x

1.3v 7.6hr 3.7x

1.2v 17.3hr 1.6x

1.1v 24.7hr 1.1x

1.0v 27.8hr 1.0x

For example in the above, if the cutoff voltage is 1.1v then we would only gain around 10 percent more if we depleted it all the way. If the cutoff voltage was 1.2v however then we would gain about 60 percent more run time.

These numbers come from a typical battery curve for a 100mw load but are not intended to be super accurate, just for comparison to get a feel for how important the voltage cutoff point really is.