I've read the numerous other questions addressing this topic, but I'm not sure if any of them apply to my situation.

I am designing a device that will run off a 3.0-3.3V supply voltage and draw no more than 40mA when active. When the device is connected to a USB port, it should draw its power from the bus. When no bus voltage is present, I want the device to be powered by a single 1.5V AAA battery stepped up to ~3.3V using a boost converter like a TPS61097 or a TLV61225.

I don't want to just diode-or the two supplies together: when USB power is connected, the boost converter should be inoperative and there should be (ideally) no current drawn from the battery. Ideally, the device's current in the inactive state should be < 10µA, so I don't want to add another "expensive" chip like the LTC4412.

I think the right answer is to run the battery voltage through a depletion-mode MOSFET that shuts off when the 5V USB supply is present on the gate, but I'm having trouble wrapping my head around some of these concepts, and I'm not sure how I'd go about choosing a part.

So, my questions are?

  • What do I put in place of the question mark in the diagram below? If it's a MOSFET, how can I choose the right part?
  • If I use an LDO to get 3.3V from the USB bus, I'd need to put a protection diode on it, right?
  • Even if there's no input current to the boost converter, do I still need a protection diode on its output?


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Simple: use a boost converter with a shutdown pin - when connected to USB the USB shuts down the boost converter. Oh look - both those ones you listed have one. \$\endgroup\$
    – Majenko
    Commented Oct 30, 2014 at 22:39
  • \$\begingroup\$ @Majenko-notGoogle caveat he may need to invert the USB 5V to make it the correct logic sense for enable \$\endgroup\$
    – vicatcu
    Commented Oct 30, 2014 at 22:47
  • \$\begingroup\$ I was envisioning a resistor pulling EN up to VIN (1.5V), then an NPN switching it LOW from the USB 5V input. I'll craft an answer with a schematic. \$\endgroup\$
    – Majenko
    Commented Oct 30, 2014 at 23:23
  • \$\begingroup\$ I know TI should have two input buck/boosts that already have input selection logic internally. \$\endgroup\$
    – Passerby
    Commented Oct 30, 2014 at 23:54

2 Answers 2


Both the boost regulators you have listed have an EN input. This is used to enable or disable the regulator, and can be very simply controlled by the USB's input power:


simulate this circuit – Schematic created using CircuitLab

Under normal battery operation R3 pulls the EN pin high, so the boost regulator is enabled. When you connect the USB current flows through R1 and turns Q1 on. this pulls EN low. R2 is used to ensure Q1 switches off when there is no USB connection allowing R3 to pull EN up again.

You may want to still keep the two diodes to prevent any back-flow of current through the regulator that isn't currently in use.

Note: I chose the resistors with ball-park figures. You may need to tweak to both ensure the EN pin gets the right voltages / currents to activate, and keep your quiescent current through R3 as low as you can while the USB is connected.

  • \$\begingroup\$ I don't think the EN pin can be used on the TPS61097--the datasheet says: "When the IC is disabled (EN = VIL) the Bypass Switch is turned on to provide a direct, low impedance connection from the input voltage (at the L pin) to the load (VOUT)." I'm pretty sure that's not desired behavior. Looking at the block diagram of the TLV61225, it might be able to handle reverse current fine. I guess I'll just have to get some parts and try it out. \$\endgroup\$ Commented Oct 31, 2014 at 0:13
  • \$\begingroup\$ With the diodes there, bypassing the boost regulator would just present 1.5V on the anode of D2, which would be blocked as the cathode will be higher, so nothing will flow. \$\endgroup\$
    – Majenko
    Commented Oct 31, 2014 at 10:47
  • \$\begingroup\$ Ah, I see. Could I use a MOSFET like a 2N7002 to pull the EN pin low so I wouldn't need R1 and R2? Will the existence of D2 cause any significant decrease in standby battery life? \$\endgroup\$ Commented Oct 31, 2014 at 20:16
  • \$\begingroup\$ You will still want R1 and R2 even with a MOSFET - R2 to pull the gate low when there's no USB, and R1 to limit the USB's inrush current. R1 is less critical, but still good to have. D2 will drop the voltage a little, it shouldn't affect the current. \$\endgroup\$
    – Majenko
    Commented Oct 31, 2014 at 20:36

After a decent amount of experimentation, I decided on the MCP16251 boost converter, as it has a low quiescent current, true output disconnect, and a low cost.

However, I found that even with the shutdown pin pulled low, and R3=1MΩ, current draw from the battery was around 2µA. Since ideally no current should be drawn from the battery when the device is powered by USB, I used a p-channel MOSFET (I used an Si2305DS, but the FDN340P appears to be a good choice as well) as a load switch, as described in this article.

I'm using an LDK220M36R LDO to regulate the USB bus voltage to 3.6V, and a 1SS367 Schottky diode connected to its output. The circuit appears to operate properly without a diode on the output of the MCP16251.


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