Electronic – low voltage battery disconnect from regulator

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.

There are several low quiescent current adjustable output low drop-out linear regulators that meet the criteria given in the question. The parametric search on the web sites of several of the usual LDO manufacturers would yield some options.

For instance, Texas Instruments TPS76201-Q1:

• Adjustable Output Voltage: 0.7 V to 5.5 V
• 27 uA Quiescent Current at 100 mA load
• 1 uA Quiescent Current if the enable pin is driven high

The typical application circuit is simple as well:

At the stated operating conditions, power dissipation in such linear regulators is:

P = (4.1V - 2.05V) * 0.127 mA = 0.26035 mW in a worst-case calculation.

P = (4.1V - 2.05V) * 0.1 mA + (4.1V * 0.027 mA) = 0.3157 mW in a worst-case calculation. (Thanks, markrages, for the correction)

Still well under the 1 mW power budget specified in the question.