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enter image description here

The general question is how useful it is to use weak batteries with a boost converter - say, boosting 3AA of 1V each to 5V?

According to this image it seems that only 5% of the energy is left when the voltage has dropped to 1.1V but about 50% when it's 1.2V and dead for most uses. Still this usefullness of boosting depends on its efficiency.

Is there a rule of thumb when boosting is advisable based on battery type, the remaining voltage and the required current?

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    \$\begingroup\$ It looks like you've answered your own question. Was there anything else? \$\endgroup\$ – Dave Tweed Oct 11 '19 at 12:42
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    \$\begingroup\$ Dave is right. What are you really trying to ask? \$\endgroup\$ – Dr_Bunsen Oct 11 '19 at 13:05
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    \$\begingroup\$ How sensitive is the load to supply voltage? Does it still function as it should when the cells have dropped to 1.1V? If so there's no point in suffering the efficiency penalty of the converter. \$\endgroup\$ – Phil G Oct 11 '19 at 13:25
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    \$\begingroup\$ Where did you get that graph? Can you include a link? I'd like to see more on that. \$\endgroup\$ – Bort Oct 11 '19 at 13:56
  • \$\begingroup\$ Here's the link: upload.wikimedia.org/wikipedia/commons/thumb/2/2c/…discharge_current_100mA.svg/725px-AA_Alkaline_battery_energy_usage-_discharge_current_100mA.svg.png \$\endgroup\$ – OMGsh Oct 11 '19 at 14:08
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It is hard to answer your exact question, since you want information on a scenario that has no commercial relevance, so companies are not going to test almost drained batteries just to know how much energy can be harvested from them.

Those discharge curves they publish are used to design battery-powered products: the curves allow to estimate how long a completely new battery can make the product work before the "low-batt" signal kicks-in.

Although, your question has no commercial/industrial relevance, it could have some environmental relevance: if your intent is to reuse almost dead batteries in lower power devices, yours is a good question, but a difficult to answer one.

There is no easy answer. An alkaline drained battery cannot be characterized only by its voltage and internal resistance. Its past discharge history can have some impact on the residual energy content, and so does its exact chemistry and internal construction, which depend on the manufacturer.

For what is worth, if you have some environmentalist concerns, you could do like me. I try to drain the most out of old batteries: when the new ones are drained I measure their voltage and if it is above 1.1V (single cells) I keep them for low power devices, such as kitchen timers/clocks, wireless computer mice or the like.

When they are too weak to work for these low power devices, I reuse them in a Joule thief circuit powering an LED as a night light. Joule thieves are quite inefficient step-up converters, but they can be powered using cells with ~0.7V terminal voltage, and once powered they can drain the battery until it reaches something like 0.2V-0.3V (YMMV, it depends on the actual circuit).

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Generally using a nearly-flat AA/AAA battery with a boost converter is not going to give you much, especially if the thing you're powering requires substantial power. When the battery goes down, not only its voltage decreases, it's internal resistance increases as well, sometimes dramatically. So the battery may show, say, 1.1V when measured with a multimeter (and almost no current is drawn from the battery), but can quickly collapse to less than 1.0V when under 200 mA load:

battery discharge graph

(Source: Battery university)

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Based strictly on the graph you provided:

The curve intersects 1.2V at about 16 hours, by my reading, and hits zero at about 25.5 hours. That's about 63% of the lifetime, or 63% of the coulombs. Without trying to calculate area under the curve exactly, we can observe that it's monotone decreasing, and therefore the joules available before 1.2V are more than 63% of the total. My eyeball says it's probably about 75%. Call it nitpicking, but this is a significant difference from "about 50%".

But as others have said, the real problem with using a boost converter on a battery, if your load needs any significant power, isn't necessarily efficiency. Your graph is for a constant-current discharge, but the lower the battery voltage gets, the more current the booster will need to draw from it to provide a constant power. The more current the booster draws, the further the battery voltage drops. For any given power level, there is a point beyond which the battery won't deliver that much power at any voltage (consider the maximum power transfer theorem with the battery's internal resistance as the source resistance), and even a 100% efficient boost converter will be no help. And being imprecise, that point is really not very far past the 1.2V point, so your ability to work miracles is strictly limited.

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