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I have an application with two AA/AAA batteries (so total of 3V or less) powering a unit with a micro controller. I want to design a low cost switch over circuit that can disconnect the battery supply when an external supply is connected with 3.3V. I also want to be able to detect with a micro controller when the external cable is connected supplying 3.3V. I am aware that this can be done with some source selector switch OR controller ICs like LTC4413. But these chips are relatively very expensive and the minimum input volt is only about 2.5V. I would need to go down as much as about 1.8V, as battery can go down to that level. Looking for some ideas how this can be done, thank you.

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  • \$\begingroup\$ Many of the common coaxial power connectors include suitable switches. \$\endgroup\$ – Peter Bennett Feb 25 '16 at 16:47
  • \$\begingroup\$ Thank you, I am planning to use just a two/four pin snap in connector to connect this cable between two units - one will supply 3.3V to a second battery powered unit. So, not planning to use any power connector jack with switch at the moment. \$\endgroup\$ – user101095 Feb 25 '16 at 17:06
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The most simple solution is to simply Diode-OR both supplies like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The downside is that you'll drop 0.7V across a regular silicon diode and about 0.15-0.2V across a Schottky, which may or may not be acceptable to you.

If you want something a little more complex, but will actually turn off any supply from the battery (and probably have less voltage drop when the battery is connected):

schematic

simulate this circuit

The FET should be some sort of low-RDSon FET. Either way you will still need some sort of power conditioning circuit, like a buck-boost so that you can accept voltages higher and lower than your required output voltage. These circuits take care of the power source arbitration.

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  • \$\begingroup\$ In the second example, keep in mind that the resistors will dissipate nearly 0.25 watts as heat when the power supply is connected, which is not the most efficient solution. Also, why are you using 1N4148s as power diodes? \$\endgroup\$ – 3871968 Feb 25 '16 at 17:16
  • \$\begingroup\$ The 10 ohm was supposed to be a 10k, the diodes were picked at random to demonstrate the concept, not an actual part selection. \$\endgroup\$ – Brendan Simpson Feb 25 '16 at 17:19
  • \$\begingroup\$ I think I wouldn't even need that diode as when cable is not connected, the 3.3V line would be floating and when connected, it would always be higher that the battery volt. Isn't it? \$\endgroup\$ – user101095 Feb 25 '16 at 17:26
  • \$\begingroup\$ I would make the pull down higher, like 100K to reduce current through the external line. So, can I make R1 as 10K and R2 as 100K or so? And have V1 as 3.3V instead of 5V? As gate would be higher than source volt, I think it should be ok. \$\endgroup\$ – user101095 Feb 25 '16 at 17:33
  • \$\begingroup\$ The schematic symbol for that MOSFET shows as an enhancement-mode P-channel device. I'm not seeing a 0.3V signal being enough to "close" most such FETs. However, if you changed to a 1M shorted across V1 & a depletion-mode MOSFET, you could use the VPP from V1 to actively "turn off" the FET & effectively remove your batt. from the circuit. \$\endgroup\$ – Robherc KV5ROB Feb 25 '16 at 17:34

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