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!

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Jim Thompson 14 years ago

The data sheet on such switches tells nothing. What do they do, close a contact between the extra pin and the barrel? ...Jim Thompson

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Rich Grise 14 years ago

By throwing away two-thirds of its capacity in a regulator? Unless you've got some kind of switcher going there.

Thanks, Rich

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Jim Thompson 14 years ago

While you're at it... pick a three-legged regulator, test it's output with input floating (but ground connected)... apply a current-limited voltage to the output, gradually up from zero to say 3.5V and see what current, if any, is drawn from your test supply. ...Jim Thompson

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panfilero 14 years ago

Hmmm.... I didn't realize they were so bad, I'll have to think about that.

To my original question, I've come up with this kind of configuration, I'm not sure if it would work.... basically when no 9V battery is plugged into the jack, it's switch is closed and I'm using a comparator to trigger some transistors to route my 3V, then when the

9V is plugged in, the switch opens and the 9V feeds the regulator.... this seem reasonable?

sketch of circuit is here:

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thanks

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John Fields 14 years ago

No, they open an internal fixed contact and connect the moving contact to the barrel, just like in a switching phone jack. Here's a typical case: http://products.cui.com/CUI_PJ-030C-SMT_Datasheet.pdf?fileID=5957 What usually happens is that the internal battery + is connected to pin 1, battery - is connected to pin 3, and circuit ground is connected to pin 2. Then, when the external supply is plugged in, its + is connected to pin 1, the circuit between battery minus and circuit ground is broken, and the external supply minus will be connected to pin 2. What I don't understand is Panfilero's scheme.

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Fred Bloggs 14 years ago

If it's only 15 ma then you can use a small analog switch like shown below, it is a SPDT implemented in transmission gates from Fairchild, or anything similar. Depending on the load you may want some hold-up capacity on the output a schmitt intervening the regulator output and the S(elect) input to the switch.

Please view in a fixed-width font such as Courier.

. . . . . . 3V Pin >-----------------+------------. . | | . | | FSA3157 . | --------- . ------- | | VCC | . | | | | | . 9V BATT>---|3V REG |---------+---|S | . | | | | | | . | | | +---+B1 A|--> 3V . | | | | | | . ------- '---|---|B0 | . | | | | . | | | GND | . | [100K] --------- . | | | . | | | . -----------+-------------+--------+------- . . .

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Jim Thompson 14 years ago

Does the switch OPEN when barrel is plugged in? The only description I've found for a barrel connector with switch say switch is NORMALLY OPEN and closes when barrel is inserted...

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...Jim Thompson

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Jim Thompson 14 years ago

By your reference, is pin 2 connected to pin 3 until the plug is inserted? Or vice versa? ...Jim Thompson

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Jim Thompson 14 years ago

This link implies 2-3 is open until plug insertion, then 2-3 is closed...

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If I can get the switch polarity correct (as in my reference) I think I have a simple solution to panfilero's problem. ...Jim Thompson

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Fred Bloggs 14 years ago

This would be the usual battery backup type of circuit, you don't say where the 3V PIN comes from, but I assume it's from the line. So remove the line power and the 9V kicks in...

Please view in a fixed-width font such as Courier.

. . . . . . 3V Pin >----------------------+---|>|--+------. . | | | . .----|---|>|--' | FSA3157 . | | --------- . | | | VCC | . | | | | . | +----------|S | . | | | | . ------- | +----------+B1 A|--> 3V . | | | | | | . 9V BATT>---|3V REG |-----+----|----------|B0 | . | | | | | . | | | | GND | . | | [100K] --------- . ------- | | . | | | . -----------+--------------+---------------+------- . . .

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Jamie 14 years ago

How about this idea? Maybe the reg can handle out side voltage on its input?

3.3V | | | + +-----------------------+||-+ | ||-> depletion mode PMOS | ||-+ | + | | | | | ----------- | | | | | 9V+--+------+ +----------+----------+ | 3.3V | | REG | | | | | | | | |- ----+-----| | | | + GND (created by AACircuit v1.28.6 beta 04/19/05

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Jamie

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John Fields 14 years ago

In the top circuit you've turned on the top FET and turned off the bottom one, but you're back-feeding 3V into the regulator since you haven't broken its return to ground. Is that bad? Probably, but it needs a little more study on your part. In the bottom circuit you've turned off the top FET and turned on the bottom one, but you've opened up the mechanical switch, so the regulator is referenced to ground, but floating.

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John Fields 14 years ago

Yes, of course, at which time the jack floats the battery and lets the external supply take over.

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ehsjr 14 years ago

Use a relay?

+9-+-----------------Coil---Gnd | +-------[Reg+3]---v ---o-- to ckt ^ | Batt +3 -------------+

Ed

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Fred Bloggs 14 years ago

e

C

=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 |

------+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 GND =A0 |

=A0 =A0 =A0 =A0 ---------

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

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

-

ut?

=A0 3.3V

=A0 |

=A0 +

=A0depletion mode PMOS

+

|

That's good, and if the reg can't handle it then a little help like so would work: Please view in a fixed-width font such as Courier.

. . . . 3.3V . | . | . | . + . +-------------------------+---------+||-+ . | | ||-> depletion mode PMOS . | | ||-+ . | | + . | | | . | | | . | ----------- G | . | | | | . 9V+--+------+ +----- S D ----------+----------+ . | | 3.3V | . | | REG | N-CH . [100K] | | ENHANCEMENT . | | | 2N7000 e.g. . | | | . | | |- . | ----+-----| . | | . --+----------+ . | . + . GND . . . But I wouldn't use this for a 15mA circuit, a little micro sized switch comes in cheaper, this would be good for bigger loads.

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Spehro Pefhany 14 years ago

Careful. If the regulator backfeeds the 3V to the input, it's possible the relay won't drop out when the 9V is removed (typical guaranteed dropout is something like 5%-10% of nominal voltage (0.45~0.9V on a 9V relay)- less at low temperatures). A diode at the regulator input would isolate it (and protect against the common occurence of a momentarily reversed 9V battery)

One could replace the electromechanical relay with an appropriate form CMOS SSR, but it will have a bit of voltage drop.

Best regards, Spehro Pefhany

"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

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Jamie 14 years ago

It was a thought to be able to switch power with no interruptions.

I don't think in the case here a back flow in the reg is going to cause any problems, which is why I didn't bother with the CT pole switch.

Jamie

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ehsjr 14 years ago

Good point. The 3 volts from the battery can't get to the relay coil, but if there's a big cap on the input to the circuit, intended to keep the circuit powered for the brief time that the relay takes to drop out, it could possibly keep the relay energized if the reg backfeeds. The isolation diode is good cheap protection against that. Thanks!

Ed

Modified diagram:

+9-+-----------------Coil---Gnd | +-->|---[Reg+3]---v ---o-- to ckt ^ | Batt +3 -------------+

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Jim Thompson 14 years ago

Here you go...

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I didn't have anything better than a BSS84 in my library, so it struggles at 3V gate drive.

Likewise I had no 3-legged regulator, so I used LM317. With a

3-legged regulator I doubt the need for the 2N7000, its gate resistor and the diode.

But it's something to start with.

Good exercise, I learned how to make a symbol act like a real jack ;-) ...Jim Thompson

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Jim Thompson 14 years ago

At your current levels, replacing FET's with 2N3904 and 2N3906 works better. ...Jim Thompson

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Auto-detecting battery power for circuit

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