When you drop a battery-powered device, the impact can open internal contacts for as long as 10ms, producing a momentary loss of power that can cause a false low battery indication. You can insure continuity of power by adding a large capacitor across the battery, but the capacitor must provide a certain amount of voltage headroom as margin against the discharge of load current. The capacitor voltage varies with discharge current as dV = Idt/C, so insufficient headroom calls for capacitance.
Another problem is that large capacitors tend to be leaky. Capacitor leakage is usually not a problem during normal operation, but during "sleep mode" it can be a substantial fraction of the total quiescent current, and thereby represent a significant reduction in battery life. The circuit of Figure 1 solves all of these problems.
Figure 1: This circuit removes discontinuities in power by backing up the battery (two AA cells) with charge on a reserve capacitor.
Two AA batteries provide 3V power, which is boosted to 3.3V by a step up dc-dc converter (U3). The large reserve capacitor of 2mF or 4mF is charged from the 3.3V output via a SPST CMOS analog switch (U1). The output of this 175W switch charges the reserve capacitor and drives the input of a low-dropout linear regulator (U2). U2's output is set to provide 1.68V when the battery is removed. The output is also divided down by the 80k/120k divider to trip an internal comparator connected to the low-battery input (LBI). The comparator's open-drain output (LBO) feeds back to U1's digital input (DIN), which turns the switch on (high) and off (low).
Figures 2-5 show the circuit performance for different values of reserve capacitor and load current. Figures 5-6 remove switch-response time by wiring the switch in the ON position.
Figure 2: The circuit with 4mF reserve capacitor and 100uA load: after removing the battery, power remains for 8.7s.
Figure 3: The circuit with 4mF reserve capacitor and 100mA load: after removing the battery, power remains for 10.8ms.
Figure 4: The circuit with 2mF reserve capacitor and 100uA load: after removing the battery, power remains for 1.48s.
Figure 5: circuit with 2mF reserve capacitor and 100mA load: after removing the battery, power remains for 920uS.
Figure 6: circuit with 2mF reserve capacitor, 100uA load, and switch wired closed: after removing the battery, power remains for 544ms.
Figure 7: The circuit with 2mF reserve capacitor, 100mA load, and switch wired closed: after removing the battery, power remains for 1.08ms.