I'd assume that 3-cells could lead to a simple trickle-charge circuit?
3*1.35 == 4.05V, which is pretty close to the 5V power that you take from the USB, seems like there's enough room to get a regulator to turn 5V into a simple charging circuit. There'd also be enough output to probably run 3.3V circuits with a simple regulator.
4*1.35 == 5.4V, which is beyond the capacity for 5V to charge (unless you got a boost-converter).
---
2-cells == 2.7V, which means you probably need a boost-converter to convert the battery-pack into something useful?
Not that I'm the designer or anything, but I like discussing these issues :-) I haven't done any electronics since my college days.
What voltage does the device operate at? The main cutoffs these days seem to be 1.8V, 3.3V, and 5V right? 3-cells and targeting 3.3V + charging from a USB (at 5V) seems like the most obvious cutoff to me personally.
Newer STM32 lines go as far down as 1.7V. Or they would if I was able to find other chips (notably Bluetooth) that would suit me and could pull the same trick. As it stands, the circuit can work down to ~1.85V which is still quite comfortable with two cells.
I admit there is no particular reason to use 2 cells where I could use 3 (they would still fit comfortably inside keyboard enclosure). The only reason is vanity -- I hope it will look sexier to other EEs:)
I dislike built in cells. I like ability to just change batteries rather than have to plug in to charge. Nothing worse than a peripheral like headphones that you have to stop using just because charge ran out. Why do I need to finish my couch session just because my keyboard can't take a simple AA battery or two? Seems like completely unnecessary nuisance.
Also, if you are talking about cell voltage, you need to remember they loose voltage very quickly upon discharging and most of the discharging happens at much lower than max voltage. For NiMh you really want to make sure that your circuit works with at least 1.2V per cell (as an absolute minimum, in perfect conditions of room temperature and very low discharge). Any higher than that and you aren't actually discharging the cells properly and are not using their capacity.
If your circuit works in short bursts of high current or can be used outdoors, be prepared that the voltage can drop very quickly especially when the cell is older (has been through a lot of cycles).
3*1.35 == 4.05V, which is pretty close to the 5V power that you take from the USB, seems like there's enough room to get a regulator to turn 5V into a simple charging circuit. There'd also be enough output to probably run 3.3V circuits with a simple regulator.
4*1.35 == 5.4V, which is beyond the capacity for 5V to charge (unless you got a boost-converter).
---
2-cells == 2.7V, which means you probably need a boost-converter to convert the battery-pack into something useful?
Not that I'm the designer or anything, but I like discussing these issues :-) I haven't done any electronics since my college days.
What voltage does the device operate at? The main cutoffs these days seem to be 1.8V, 3.3V, and 5V right? 3-cells and targeting 3.3V + charging from a USB (at 5V) seems like the most obvious cutoff to me personally.