Among mainstream battery technologies, LiPo has the best energy density, so you will probably want to go this way. In your case, since you will use a regulator anyway, every configuration possible (including single cell combined with a step-up boost regulator). Once you have chosen the total energy capacity you need for your application and the cell configuration, there are two parameters to consider:
- Quality: for the best specific energy, you will need to get high quality cells, which are going to cost more.
- Discharge rating: cells are usually rated for a given maximum discharge current, usually expressed relatively to its capacity (for example, a 10C rating for a 1000mAh cell means that the max current is 10*1000mA = 10A). While higher C rating may mean a lower ESR, there is a trade-off and increasing C rating significantly lowers the specific energy. As an example, a 40C 2000mAh cell from a high quality chinese manufacturer I was looking at recently had a specific energy of about 140 Wh/kg while the 15C version had a specific energy of 170 Wh/kg. This difference can largely offset any loss due to slightly increased ESR.
Whatever configuration you will end up using, your require that the battery lasts at least 5h. Assuming little loss in DC-DC regulation, this means that the average current would be at most 0.2C and the peak current (from your numbers) about 0.6-1C. I would thus look for 5C-rated good-quality cells, which should get you 180Wh/kg or more.
It does not appear in your equation because this equation assumes you're using the battery at its output voltage during the whole usage without conversion.
This is not the case here, because you're using a step-down converter. So, to build the correct equation you:
- get Vavgbat: the average voltage of the battery during the whole discharge cycle: the discharge graph of the battery datasheet shows it's around 3.6V for the low currents, such as the one you use.
- get Iavgbat: the current you'll draw from the battery, in average, during the whole cycle. It is not the current you use at the output of the DC-DC converter (that is where you missed something, I think). If we say the converter output current is Iout, then Iavgbat = (Iout * Vout / Vavgbat) / efficiency (efficiency being the efficiency of the DC-DC converter, usually around 80-90%, check the datasheet).
- then you apply the forumla you mentioned: time = capacity / Iavgbat.
So you have: $$time = \frac{capacity}{Iout \frac{Vout}{Vavgbat}} efficiency$$
Now, you see the output voltage in the formula.
So, if capacity = 3.4Ah, Iout = 400µA and efficiency = 85%, we have:
- time = 8670 hours (about one year) for a 3V output
- time = 14450 hours (more than a year and a half) for a 1.8V output
One more thing: given the large times resulting, I think you have to account for the batteries self discharge (or leakage current), which may be significant. Unfortunately, I didn't see it mentioned in the batteries datasheet.
Additional detail: Where does the Iavgbat = (Iout * Vout / Vavgbat) / efficiency formula comes from ?
It comes from the fact a DC-DC converter, unlike a linear regulator, is able to output (almost) as much power as it draws from its input. So Pin = Pout / efficiency. If we say Pin = Vavgbat * Iavgbat and Pout = Vout * Iout, we can obtain the above formula.
On the opposite, with a linear regulator, the voltage is dropped without any consequence on the input/output current. So Iavgbat would be equal to Iout (not accounting for the quiescent current), which was your initial (inaccurate) assumption.
Best Answer
If there's a significant difference in measured capacity between 500mA and 1A discharge currents, you should stick to lower value, i.e. 500mA (or even less). Unless you're interested in capacity under specific load, you should aim to measure the maximum value, and cells typically have higher capacity at smaller discharge current.
The cut-off voltage cannot be reasonably estimated and should be taken from the datasheet (idem for the full-charge voltage). If the additional capacity between 3.0V and 2.8V is insignificant to you, you can stop at 3.0V for all cells (or even 3.2V, which already correspond to 95-98% discharge; IMO 5% precision is plenty for the purpose of sorting used cells). What you should avoid is discharging cells with 3.0V cut-off down to 2.8V.