Power from 2 Sources

[AtomSoft]

Hey guys I want to have the option of powering a device from a wall or from batteries. I wanted to know whats the best way to have both options on without damaging the batteries...

I can make a 7805 power circuit pretty good but i dont know whats the best way to disconnect batteries from circuit when powered by wall.

Whats the best way?

ADDED:

The image below is what i have.. I want to replace SJ-1 Which is circled in YELLOW with some automatic way to disconnect battery from circuit...

Or some way to make them work together without frying batteries

 

[bigal_scorpio]

Hi mate,

I use the DC barrel connectors that have 3 pins on the rear, one is live from the center pin, the other two are shorted when there is no plug in so you run these in line with your jump. Then when the plug is inserted the connection to the battery is broken and the external power takes over.

I can give you a cat number if you like.

Al

20 1096 on rapid electronics

 

[AtomSoft]

Yes Please! heh Thanks

Any other thoughts?

 

[Nigel Goodwin]

You just need either one or two diodes, depending on exactly where you switch - no need for mechanical switches at all.

 

[AtomSoft]

can you please provide more info? As diode would reduce my cost since i can buy in bulk also heh.... Thanks Nigel

 

[AtomSoft]

20 1096 on rapid electronics ....

Ill take a look at that now also...

 

[Hero999]

Does the voltage always need to be regulated?

I'd recommend connecting the battery before the regulator?

What voltage battery are you using?

You might need a low drop-out regulator.

 

[AtomSoft]

Im using 3 AAA batteries. The regulator will be a 5v regulator... Its to power a MCU

 

[Electronworks]

Regarding Nigel's suggestions with the diodes: Put the anode of one diode to the output of your 7805 and the anode of your other diode to the output of your battery. Schottkys work best as they have a lower voltage drop. Then short the 2 cathodes together. When you wall cube is connected (as long as the output voltage of your 7805 is higher than that of the batteries) its diode conducts (and reverse biasses the battery diode). When your wall cube is disconnected, the battery diode conducts. Regarding the battery diode, you might want to make this up of 2 series diodes to stop any reverse leakage current (that might be present in a single diode) charging your batteries.

Alternatively, use the wall cube to drive a relay that disconnects the battery voltage. If you did this, you might still want to add a series resistor coming from the battery just in case your 7805 voltage is much higher than your battery. When you disconnect the wall cube (and hence reconnect the battery), any capacitance across the load will dump itself on the battery. if the voltage is higher, it might try to charge the battery...