I'm trying to do a simple, low power, low cost power "switch" for a custom board (I know, it will be complicated).

The device will be powered by default by a coin cell battery (3V3 and lower) and can also be powered by USB (5V). I will use a boost converter (TPS61291) for the battery, it will allow a constant 3V3, with a bypass for sleep mode controlled by a MCU.

This is a part of the power rails I designed:


simulate this circuit – Schematic created using CircuitLab

The battery is protected by the schottky diode (some nA can flow into it, but it is accepted by the manufacturer)

If the VUSB is not present, the diode D2 will avoid reverse current in USB connector. D4 is protecting the LDO itself from the reverse current by bypassing it.

The boost bypass will be activated by the MCU when it is in active or when USB is plugged.

Do you think it will work correctly?

  • 1
    \$\begingroup\$ I think D2 should probably be on the output of the regulator, and should be a schottky diode or other low-Vf diode. Putting D2 there should also make D4 unnecessary, so you save a few cents. \$\endgroup\$
    – Hearth
    Commented Feb 14, 2019 at 22:30
  • \$\begingroup\$ What is this "boost bypass" you are talking about? Please provide links to the datasheets for the LDO, battery, and boost converter. Please specify your power requirements out of the boost converter. \$\endgroup\$ Commented Feb 14, 2019 at 22:31
  • \$\begingroup\$ ...your boost converter idea seems unlikely to work, if you want to boost 3.0V to 3.3V (assuming 0.3V drop on the diode). Boost converters aren't usually designed to boost by such small amounts. \$\endgroup\$
    – Hearth
    Commented Feb 14, 2019 at 22:32
  • \$\begingroup\$ I have updated the question with the boost reference (TPS61291) \$\endgroup\$ Commented Feb 15, 2019 at 12:01
  • \$\begingroup\$ I would put the boost circuit between the battery and D3, and move D2 to the output of the LDO as Hearch suggested. For D2, D3 use UPS115UE3/TR7 1A, 0.22V Vf, digikey.com/product-detail/en/microsemi-corporation/… \$\endgroup\$
    – CrossRoads
    Commented Feb 15, 2019 at 13:53

1 Answer 1


Electronic industry already has a solid solution for this case. It is solved by a special class of devices called "power distribution switch", within PMIC group of IC. Search Digi-key for this type, "PMIC - Power Distribution Switches, Load Drivers", then select, say, "2:1" in the column "Ratio - Input:Output".

Essentially these are switches based on FETs, as "ideal diodes". Here is an example, FPF1320UCX.

enter image description here

A similar line of "PowerPath, Ideal Diodes & Load Switches" is offered by Analog Devices, see something like LTC4413

enter image description here

I am certain that many other manufacturers offer similar line of products. When comparing cost of solutions, you need to take into account not only the cost of individual components, but also the cost PCB area, and placing/soldering costs.

  • \$\begingroup\$ FPF1320 is a good component, I already found this one during my search. The problem is its quiescent current of 12µA. I plan to use 2µA in sleep mode (MCU is around 1uA, I added an extra 1uA). Concerning the LTC4413, its price is a little bit high. I will keep digging on digikey/mouser/farnell for these kind of PMIC. \$\endgroup\$ Commented Feb 15, 2019 at 12:37
  • \$\begingroup\$ @PierreOlivier, "I plan to use 2µA in sleep mode" - then you should state this explicitly that you target a deep sub-uA operation. The industry leaders were able to get correct control under 12 uA and 17 uA (LTC4413). It is probably a trade-off between losses in standby mode and losses in plain Schottky diode during active mode. \$\endgroup\$ Commented Feb 15, 2019 at 20:24

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