I have a low power IC (an RTC module) which draws about 2.5uA average draw at 2.5V. I want to be able to power this from a coin cell (2V to 3.1V) directly, or from a larger battery pack (3.5 - 20V).
It is also possible for both batteries to be present at the same time.

Here's a way to do that:


simulate this circuit – Schematic created using CircuitLab

Problem is, the load is very low power so if we use Schottky diodes and power the device from a coin cell, leaving the battery pack disconnected then the high leakage current of the schottky diode will likely dominate the overall power draw, flowing backwards through D2, into the regulator and to ground.

If a regular diode were used instead of a Schottky diode for D2, then a larger voltage drop would be introduced. If we were powering the devices from the Bat Pack then the 2.5V regulator will no longer work because the minimum voltage that the load will run at is 2.4V. If a 3.3V regulator was used instead, a low-dropout regulator (< 200mV) would be required. These are quite hard to find with quiescent currents of < 2.5uA. If the regulator has a high quiescent current then obviously the efficiency is reduced significantly. Not to mention the significant energy is dissipated in D2 due to the voltage drop across it.

Is there a better way to efficiently OR low current, low voltage power supplies together?

  • \$\begingroup\$ Some/many/most RTC chips have dedicated pins for main and battery supply and handle switchover internally. Doesn't yours? \$\endgroup\$
    – JimmyB
    Commented Oct 20, 2016 at 9:20

1 Answer 1


There might be some possible clever scheme using MOSFETs, but most likely the additional control circuitry will also have some leakage that will make it as inefficient as the simple scheme you planned.

I am sure the solution here is simply to find the appropriate components that, in the end, will lead to an acceptable compromise. For example:

Regarding the diode:

BAT54 has 0.5µA typ leakage at 25°C and 5V reverse voltage. The drop is higher than with more beefy schottkies though. About 0.2V at a 0.1mA. So if you keep using a 2.5V regulator, you may end up below the 2.4V min voltage required, so:

Regarding the voltage regulator:

You could use a 3.0V LDO with ultra low quiescent current to compensate for this higher diode drop. You can actually find some rather easily: TPS78330, for example, has 0.5µA Iq. NCP4681/NCP4684 claims 1µA typ. MCP1710 claims 20nA (!).

Just spend some more time on google/mouser/digikey and reading datasheets. What you want is achievable the way you planned it, with the right components.


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