Auto-detecting battery power for circuit

panfilero · 2011-11-12T20:38:47+00:00

Hello,I have a circuit that works off of 3V. I get the 3V in two differentways, one is through a 9V battery that plugs into a DC power barrelreceptacle that goes to a 3V regulator, and the other is from a 3V pinthat is available on the circuit. I'm wondering if there is a way thecircuit could automatically use the 3V pin if there's no batteryplugged into the DC barrel receptacle, and if there is a batteryplugged in it would switch and use the 3V coming out of the 3Vregulator.At first I was thinking about OR'ing the power sources through somediodes... but since I'm working with 3V, I really don't want to giveup a diode voltage drop on my power. Next I started looking at the DCbarrel receptacles that have built in switches (I've never used thesebefore) and seems like this would be a good option if I was workingwith the same voltages, but since one of my voltage is 9V and theother is 3V, I need the 3V pin to bypass the voltage regulator, andthe 9V to go to the voltage regulator... so they are going todifferent places on my circuit (different sides of the regulatoressentially).any suggestions would be appreciated, much thanks!

E

ehsjr 14 years ago

Nice. I love PDF format you use, and the way you labeled V and I on your schematics. I would do the circuit differently (see below), now that you & JF figured out the jack switching and posted it in the schematics.

Ed

+3 +-----------------------+ | | --- +---[317]---+-->|---+ D1 isolates the 3V battery from R1-R2 - | | | D1 | to gnd, and the 317 from feedback from --- | | [R1] | the load, assuming a cap across the load - | | | [Load] to keep it powered during switching. | | +-----+ | Set Vout to 3.7, assuming a .7 drop in | | | | D1 at 15 mA. Use R1 = 240 and R2 = 470. | | [R2] | Or, replace R2 with a 500 ohm pot and | | 9V Gnd | | adjust until voltage to the load is | +---o o-----+-------+ 3 volts. | | | |>|

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J

Jim Thompson 14 years ago

Not clear, from the OP, if you can break +3V ground. I also thought of putting the +9 on the outside of the plug, rather than the middle pin.

BTW, The V/I you like, it's a feature of PSpice ;-) ...Jim Thompson

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B

Bitrex 14 years ago

I haven't been following the thread, but I didn't want to let an opportunity pass to say I _hate_ those type of shorting jacks! They're all over the place in guitar effects pedals and such, and they always seem to get worn and end up not making contact when the plug is pulled out. Your circuit does seem to fail safe if this happens, which makes things somewhat better.

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J

josephkk 14 years ago

different

pin..

Maybe switch at the 9V level?

?-)

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J

josephkk 14 years ago

different

pin..

Probsbly not that much of an issue with a bit of carefully thought out electronics.

?-)

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J

josephkk 14 years ago

different

the

the DC

these

say

=A0 | =A0 =A0 =A0 =A0| =A0 =A0 =A0|

=A0.----|---|>|--' =A0 =A0 =A0| FSA3157

=A0| =A0 =A0 =A0 =A0 =A0 ---------

=A0| =A0 =A0 =A0 =A0 =A0| =A0 VCC =A0 |

=A0| =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 |

=A0+----------|S =A0 =A0 =A0 =A0|

=A0| =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 |

=A0+----------+B1 =A0 =A0 =A0A|--> 3V

=A0| =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 |

|

=A0| =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 |

=A0| =A0 =A0 =A0 =A0 =A0| =A0 GND =A0 |

[100K] =A0 =A0 =A0 =A0 ---------

=A0 =A0 =A0 =A0 =A0 =A0 =A0 |

=A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 |

-----------+--------------+---------------+-------

input?

=A0 3.3V

=A0 |

=A0 +

=A0||-> =A0depletion mode PMOS

=A0||-+

=A0 +

=A0 |

=A0|

Just the same it is clear, clean, and supportable. I like it.

?-)

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Auto-detecting battery power for circuit

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