I thought I'd add a log on some of the considerations of adding a super-capacitor based backup power supply to a project.

In general, super capacitors aren't really different from electrolytic caps you're used to, except that they come in relatively high values.

My goal was to be able to preserve the almanac of the GPS module for at least a half an hour. The first step is to do some math to see what's necessary to achieve this.

The VBatt pin of the GPS module must be powered both in normal operation and when power is absent. For situations where battery backup is not required, this means you must tie the VBatt pin to Vcc. When power is not applied, however, the two must be separate. So we start with a schottky diode between the two. Vcc is allowed to power VBatt, but VBatt must never be allowed to flow into Vcc (for both the GPS module and the rest of the board). If we were using a primary cell, we'd also need to insure that Vcc was never applied to the primary cell, so a second schottky diode would go from the battery to VBatt. But when using supercaps or secondary cells, we need Vcc to charge the cell or cap.

Let's consider the case when the supercap is completely empty and power is first applied. Within the limits of ESR, the cap will resist the sudden change in voltage by attempting to draw every bit of power from the power supply it can get. Classically this is referred to as inrush, and it is a thing with heavily capacitative loads (this is characteristic of large logic systems with lots of bypass caps - they all add up). In the case of our supercap, this is definitely an undesirable outcome, so we must use a current limiting resistor. The result of this is that it will take some time for the cap to charge, but we expect the circuit to be mostly powered and the supercap to only supply backup power for brief outages. If we consider the turn-on case, our supply voltage will be 3.3 volts minus the forward voltage drop of our schottky diode. This drop will actually be quite low, because as the cap's voltage rises, the current will drop eventually to zero. In practice, it winds up being only around 100 mV or so. But at inrush time, it will be higher. We can call it 500 mV. At 2.8 volts, a 330Ω resistor will pass around 8 mA of current. That'll be fine. On the discharge side of the equation, the backup current draw is on the order of 33 µA, so we can expect only a few mV of drop during discharge - the resistor basically won't be a factor at all.

If we pick as a starting point a 100 mF supercap, we can figure that the RC time constant for charging it is 33 seconds. As a rule of thumb, we can call 2.5 time constants the time to "fully" charge the cap, so it will charge up in just under a minute and a half. There are various discharge calculators out there. If we pick one and plug in our values - 100 mF, acceptable voltage ranging from 3.3 to 2.5 volts, 33 µA current draw and 0.5 Ω ESR, you get just over 2400 seconds, or 40 minutes. Just fine for our purposes.