I have a circuit that runs on 3.3v. I want it to be able to operate on two AA batteries down to about 2.2v (or until batteries are dead), and also with a 9V power supply. When the power supply is plugged in, the batteries should be disconnected.

The power jack being used has a switching ability, and would work fine if both power sources were at the same voltage. But when battery power is active, the voltage must be boosted to 3.3v, and when the power supply is plugged in the voltage must be regulated down to 3.3v.

I've not been able to find a single chip that will accept 2-9v and regulate at a 3.3v output.

What other solutions are there? Minimal parts count and cost is certainly preferred.

  • \$\begingroup\$ I think theres 2 connection points on the jack that can be served to connect grounds together when the jack is removed and then you can use that as battery ground. As for dropping voltages, a regulator that has a 3.3 extension to it (like LM1117-3.3) can drop the input voltage to 3.3V output. Just make sure you have a 10uF capacitor at input and 22uF capacitor at output of the regulator. It may help to add a series resistor at output to limit max current. \$\endgroup\$ Sep 14, 2019 at 5:41
  • \$\begingroup\$ What is the maximum current your circuit needs at 3.3V? \$\endgroup\$ Sep 14, 2019 at 6:45
  • \$\begingroup\$ ti.com/product/TPS63070 \$\endgroup\$ Sep 14, 2019 at 6:52
  • \$\begingroup\$ @Mike Makes perfect sense about the power jack, but I realized the reason that can't work is it is a center negative power supply. So the switched terminal is actually switching power and not ground. \$\endgroup\$ Sep 14, 2019 at 14:06
  • \$\begingroup\$ @BruceAbbott Unfortunately that is cost prohibitive. The circuit will need about 100mA. \$\endgroup\$ Sep 14, 2019 at 14:07

1 Answer 1


The TPS63070 buck-boost converter accepts 2-16V, however in the comments you say it is 'cost prohibitive', so a cheaper solution is required.

Here is a module which sells on eBay for US$2.75, with free shipping!

enter image description here

Your power jack can switch between the battery and power adapter with a circuit like this:-


simulate this circuit – Schematic created using CircuitLab

The diode is not strictly necessary, but is cheap protection against reversed input voltage (which might happen if someone plugs in the wrong power supply).

Some other modules are even cheaper, but have a lower maximum input voltage (eg. 5V). For these you would need another regulator to reduce the 9V. As you only need 100mA a Zener diode shunt regulator might work (the Zener diode would be wired between the buck-boost regulator's input and ground, and D1 replaced with a resistor to limit shunt current). Or you could use a negative linear regulator such as the LM7905. However these solutions would increase the parts count, and might not be any cheaper overall.


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