AVR Freaks

At a minimum, you should give up the linear regulator and go for a buck or buck-boost regulator, a type of dc-dc switch-mode-power-supply.  Pololu has a great number of regulators, many of which are likely drop-in replacements for your linear regulator, and are in the neighbourhood of 75% - 95% efficient (versus the 25% efficiency of your 3.3V reg running from an 11.1V source):

Check out their product selector.

Plenty of other source for these things, too.

Can you provide any information into why a 3.3 volt regulator can have a maximum input voltage of 15 volts if 11.1 volts is too much? Or is it more related to the maximum current discharge of the battery and the voltage?

Assuming you're talking a linear regulator, these work by dumping the extra energy as heat.

For example, if you've got 11.1 volts going in, and 3.3 volts going out, that's a drop of 7.8 volts.  If you draw 250 mA, the regulator will dissipate 7.8V * 0.25A = 1.95W of heat.  That's a lot.  probably too much, depending upon the regulator.

What's the part number?  How much current are you pulling?  Do you have a heat sink?

(11.1 - 3.3)*the current

"the current" is the current consumed/drawn by your AVR or ARM device and its externally connected peripherals. Let's say you have a board that draws 50 mA and a connected display that draws another 50mA... that makes a total of 100mA. So your linear regulator would have to tolerate/handle/dissipate (11.1-3.3)*0.1 = 780 milliwatts while your "load would only be handling/consuming 3.3*0.1 or 330 milliwatts. Of course as Joey has pointed out, you could replace the linear regulator with a switcher design and reduce that 780mW to around 80mW; a big saving indeed. But if you only need 3.3 volts, a 3.2v LiFePO4 battery can run your design without the need and expense of any regulator.

I agree though that if you need 5 volts you need another approach. In my case I use a boost switcher and go from 3.2 volts to 5 volts.

The "lesson" to be taken from this discussion is that you need to consider your actual needs and try to minimise wastage of power and money... at least that has always been my approach. YMMV.

Cheers,

Ross

Note: use a fuse on the LiPo. It can deliver enough current to melt pcb tracks and wires if there is a short circuit.

You're not kidding:

The battery will give way more than 42A into a short circuit!

It sure can.  I was only showing what the battery is rated for.  It could turn a PCB into a cinder without breaking a sweat.

As well, these batteries get a tad upset when mistreated. Youtube has plenty of vids.

Most recently, an e-cigarette exploded in the someone's pocket while paying for something (e-cigarette refills?) at a gas station:

https://www.youtube.com/watch?v=k1LjSuq0rk8

So get 1S for when you want to supply 3.3. Use 2S when you need 5.0V. Peak voltage at 1S is 4.2V with 3.7V nominal.

To be honest you may have a job to find a regulator that can take an input as low as 3.7 to provide 3.3 on the output.

As I was interested to know what the solution to this was myself I just googled "how to drop lipo voltage to 3.3v" and got some really terrific results. Seems you are not alone in this quest.

Remember the that 3.7V rating of a LiPo is 'nominal'.  What 'nominal' means is different for every battery and manufacturer.  It is based on a specific rated discharge current.  Discharge currents are rated in units called 'C', which represent the current required to (nominally) completely discharge the LiPo.  So for an 850 mAh LiPo a discharge current of 850 mA is 1 C, and a discharge current of 1.7 A is 2 C, etc.  However, both the mAh rating and the voltage specified by a manufacturer are generally based on a much lower discharge current, often 0.1 C or 0.05 C.  In the case of the latter, an 850 mAh LiPo will be able to provide only 42.5 mA at 3.7V over the stated capacity of the cell.  If you draw more than 0.05 C from such a battery, it's voltage will drop below 3.7V long before the cell is considered 'empty'.  In effect, there is charge  still left in the battery but you can't extract it at a higher current without the battery voltage dropping.  Few manufacturers even specify under which conditions their claims of capacity are measured.

Ultimately, you want to look at the datasheet for a given LiPo in concert with your requirements before making your selection.  Often datasheets lack the kind of detail you want, so you can look to third-party testing.  As an example (although the nominal voltage is different at 3.2 V) take a look at the link I posted in #10.  You'll want to select a LiPo which will remain at a high enough voltage under the expected current load, so that the voltage delivered to your selected regulator is above the sum of the regulator output plus the regulator's voltage drop (at the required current load), and to do so for as long as you require before needing a recharge.  Note also that the voltage drop on a regulator, even an LDO, is dependent upon the output current, so look at the datasheet for that, too.  Higher currents spell higher regulator drop out.  The reg you mention in post #9 has a drop out of 1.1 V @ 450 mA @ 25°C, and 1.0 V @ 100°C.  There are better LDO regs, but you aren't likely to find any with a 0.4 V drop out at 450 mA.

I'd say that if you want to run a 3.3V device from a nominal 3.7V LiPo, you'll want an appropriately sized buck-boost regulator, such as the ones you can find at the Pololu page I linked in post #3.  This will allow you to use all of the energy in the LiPo while maintaining a solid 3.3V output, and with minimal wasted energy.

Using a 3.2 V battery directly may be suitable, but again look at the link in #10.  With a discharge current of 500 mA, the voltage drops below 3.0 V after about 52 minutes, leaving 25% of the battery's energy unusable.  If your device can operate below 3V, then you get at a bit more of the energy.  However if it needs even 3.1V, the run time drops to about 46 minutes, leaving 30% of the battery's energy unreachable.  A buck-boost regulator will solve this.

Note that most but not all LiPo and other Li-based batteries include protection circuitry to (among other things) prevent a cell from over-discharging.  Over-discharge (and over-charge, and over-current) spells death for a LiPo cell, and often fireworks, so it's critically important that you have it.  If the cell itself doesn't have it (check datasheet), make certain that the power supply (whether linear, buck, boost, or buck-boost) has it.  Indeed, there are all-in-one solutions like this:

https://www.sparkfun.com/products/11231

... but note that that particular device is only capable of 200 mA at 3.3V.