A1. The built-in protection for li-ion cells should be considered "emergency protection"--it should not be relied upon for normal cycle V & A protection. In a "safe" battery design, V&A protection should "always" be included in your design.
A2. A thermistor protected battery system is wise. Sometimes li-ion cells do not dis/charge at the same rate & cause an imbalance in the series/parallel configuration of larger Ah batteries. A thermistor can be used to detect an overheating condition, prevent a fire, slow the dis/charge rate/s, or indicate a dis/charging fault in a smart battery system. I consider it very important in a safe battery system, but, it is often overlooked or ignored when ICR type li-ion cells are used. If you want to make a safe battery pack, then I suggest you always include a thermistor & supporting circuitry--don't allow the incomplete/shoddy designs of others to affect your design decisions.
You are not a million miles away with your estimate.
BUT you need to think about what your device is going to be doing to make a real estimate. Use your calculations of:
500mAh / 59mA = 8.47h if your device is only going to be used as a receiver.
Or use the higher value of:
500mAh / 278mA = 1.8h if the device is going to be used as a transmitter.
But you may want to take some other possibilities into your calculations, if you dont need to device actually doing anything for some of the time, you can use the hibernate mode on the device. The current consumption in this mode is 4uA. So if your device lived in this hibernation mode for the whole of its life cycle:
500mAh / 0.004mA = 125,000h or 14 years...
but in general the low power deep sleep is 250uA so:
500mAh / 0.25mA = 2000h so a measly 8.3 days.
One other consideration is the battery life itself, when you look at datasheets, the battery capacity changes with the discharge current. For instance, in this datasheet, at 300mA discharge, the battery has just under 400mAh of capacity, a reduction of 20% from your calculations! but at 25mA, the battery could last over 600mAh, an increase of over 20% from your calculations.
One final thing to consider is that your device doesn't run at 9v, it runs at 3. This means that you need to do some trickery with the voltage. If you want to use regulator, you will instantly reduce your battery life by 60% due to current losses. Another way is to look into buck converters, where you could get a decent efficiency, you might only lose 20% of the battery life. The final consideration would be to change the batteries; if you were to use AA batteries, not only would you increase battery life (much higher mAh rating), you wouldn't get the same losses related to regulation because the voltage would already be lower.
Also, have you thought about using multiple chips instead of a single chip solution? may be easier to quantise current draw for each part. Just a thought
Best Answer
Get your circuit working and then turn off the power. Put your meter into current mode; the 200 mA scale is probably sufficient. Put your meter in series with your circuit by attaching the red lead of meter to the power supply, the black lead of meter to the input of your circuit, and the negative lead of your circuit to the power supply's negative terminal. Apply power; you will see the actual current usage of your circuit. Without circuit details I can't know the consumption, but I bet it will be 50 mA or less. Watch it as you operate all of the circuit's functions to see if the demand changes with different activities.
When you choose a battery or power regulator for your circuit, make sure it's rated higher than the minimum to account for surge at power-on and other power spikes. 500 mA would probably be good enough for a circuit like yours, but that's just a guess.
There are all kinds of complexities that can change things; if you are using motors, power consumption can go much higher than when the circuit is still, and there are other issues. But that is too complex for this conversation.