Go ahead and use the variable resistor suggested by Reid if your project is a one-off sand box project. If you are designing a product where you will be building many of these then I would select a resistor voltage divider made up of precision resistors (0.5 or 1% types are good for this application as the detector has a 1% tolerance). The two resistors in series will connect from battery supply to GND. The junction between the two resistors would connect to the V+ pin of your voltage detector. Pick the two resistors so that when the battery supply is just at 3.5V that the divided voltage (using the standard voltage divider equations) to the detector is just at 2.0 volts. Deploy the 2.0V version of the detector in your circuit. Also select the size range of the voltage divider resistors so that the current is approximately 15 to 20 times greater than the current draw of the detector so that the current drawn by the detector can be ignored when computing the divider ratio.
You want to use resistors like this so that when the battery discharges slowly during use that the divided voltage tracks linearly down along with the battery. At the computed threshold point the battery at 3.5V will trigger the under voltage detector at 2.0V.
Using an LDO is not appropriate because the LDO would give a constant output irregardless of the battery voltage. A diode to offset the battery voltage to the detector would work if you could find an almost perfect diode with the correct forward voltage drop but sadly perfect diodes are hard to find. The resistor divider is a much lower cost approach and will work nicely in this application.
The datasheet states that 2.2 uF is a minimum required for for stability. The typical output capacitance (Table 3) is stated as 10 uF. Adding more capacitance will smooth the output more and ensure better stability, but there are some side effects (below).
Just to make all of this clear as to why you need this capacitance, see the figure below. An LDO is simply a transistor (A) that is controlled by a feedback loop (B and C). The error amplifier is trying to make B equal to the internally generated Reference Voltage.
Now when a large load current is pulsed (like in a digital circuit), Vout will droop. This makes B too low, then C has to drive A on harder so it can compensate. This takes time. So if the pulses of load current are too fast, the loop will not be able to compensate quickly enough, and you can have oscillation.
The solution is to put a charge bank, i.e. a capacitor out there to source the quick pulses of current, and the slower job of controlling the average current is left to the regulator.
So why not go insanely big on this capacitance? There are a couple of reasons:
- Cost. Big caps cost more.
- Space. They take up more space.
- Charge/discharge time. When you turn the circuit on, the capacitor has to charge up. This will cause the voltage on Vout to more slowly rise. Some circuits don't like to turn on slowly. Also, the charge has to be discharged when powering down. This is usually less important.
Finally, the input capacitance question. This is usually less clearly defined. The main point that is important is that you have a good enough source so that the LDO internal reference works well. A good idea is to look at what the datasheet test circuits use. This one shows a 0.1 uF capacitor placed on the input.
Best Answer
The short answer is yes, but your voltage is going to track with the voltage of the battery. A slightly more elegant approach is the basic linear regulator shown below.
edit: just to be clear, this will still track with the battery voltage when the battery voltage is less than the breakdown voltage of the zener. If you can get away with a Vout of about 3V, you could select a zener with a breakdown voltage of 3.6V and get a steady output voltage over a relatively large section of your battery SOC.