December 18, 1997

Circuit automates switch-over
between battery and source

Len Sherman, Maxim Integrated Products, Sunnyvale, CA

The use of a logic supply or an ac wall adapter can extend battery life in a small portable system by offloading the battery. However, managing the switch-over between battery and external power can be messy. For high-cell-count batteries, you can do the job with simple and inexpensive diode switches, but for one- or two-cell batteries, the forward drop (even for Schottky diodes) adds an unacceptable voltage loss. Replacing the diodes with MOSFETs can reduce this loss, but the circuitry necessary to control the MOSFETs adds cost and complexity.

You can avoid these problems and eliminate the MOSFETs by switching at the regulator output rather than at the battery input (Figure 1). With no external power source, a low-power boost converter, IC₁, delivers as much as 100 mA at 3.3V from two AAA alkaline cells.

Then, connecting the external 5V source causes a small linear regulator, IC₂, to take over. R₁, R₂, and R₃ set the linear regulator's output (3.45V) higher than the boost converter's output (3.2V), so feedback to the boost converter at Pin 2 causes this converter to stop switching. The linear regulator delivers a V_OUT of 3.45V, and catch diode D₁ prevents current flow to the battery. IC₁ receives power at Pin 6 but does no switching because V_OUT is above its setpoint.

Because IC₂'s output must always be higher than that of IC₁, this circuit imposes a tolerance slightly wider than 5% on its nominal 3.3V output. The nominal output for IC₁ is 3.2V with the resistor values shown, and the resistors' 1% tolerance allows a minimum of 3.1V (6% below 3.3V). The nominal output for IC₂ is set at 3.45V, and its maximum is 3.55V (7.5% above 3.3V). When boosting from two AAA cells, the conversion efficiency is 80%, and the no-load battery drain is 160 µA. When operating from the applied 5V, the battery drain is essentially zero, and the no-load current drain from 5V is 90 µA. (DI #2130)

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