Battery/Outlet power bypass

[tinkerpunked]

I am hooking some LEDs into a picture frame. I’m running them on 12 volts. I want the option of either using 8AA batteries or a 12 volt power adapter to plug into an outlet. If I hook the batteries and adapter in parallel, when plugged into the wall, I imagine it would still drain the batteries, but at half the rate as if it was running on only batteries…is this correct? And how would I hook it up so that when plugged into an outlet, it would bypass the batteries and use only the adapter, saving the batteries. Is there a simple circuit that could do this?

Thanks.

 

[Gryd3]

I am hooking some LEDs into a picture frame. I’m running them on 12 volts. I want the option of either using 8AA batteries or a 12 volt power adapter to plug into an outlet. If I hook the batteries and adapter in parallel, when plugged into the wall, I imagine it would still drain the batteries, but at half the rate as if it was running on only batteries…is this correct? And how would I hook it up so that when plugged into an outlet, it would bypass the batteries and use only the adapter, saving the batteries. Is there a simple circuit that could do this?

Thanks.

You can use a relay to do this for you

Find either a 'normally closed' relay, or a 'double throw' relay.
Wire the batteries in series with the switching side of the relay, then connect to the LEDs.
Wire the AC-DC Adaptor plug to the coil side of the relay, then connect in parallel with the LEDs *with a diode*.

When AC-DC powered, the relay will disconnect the batteries. When battery powered, the diode is required to prevent the batteries from disconnecting themselves with the relay.

 

[AnalogKid]

If I hook the batteries and adapter in parallel, when plugged into the wall, I imagine it would still drain the batteries, but at half the rate as if it was running on only batteries…is this correct?

Fortunately for you - no.

Using either a relay or some diodes, automatic all-wall-power / all-battery-power switching is pretty simple. Is the 12 V power adapter consistently higher output voltage that the battery pack? If so, you can do this with two small diodes. If the two power sources are within 1/2 volt of each other, you'll get more reliable operation with either a SPDT relay or a small circuit controlling an electronic switch. What is your skill set? Also, what is the max total LED current?

ak

 

[HellasTechn]

I am hooking some LEDs into a picture frame. I’m running them on 12 volts. I want the option of either using 8AA batteries or a 12 volt power adapter to plug into an outlet. If I hook the batteries and adapter in parallel, when plugged into the wall, I imagine it would still drain the batteries, but at half the rate as if it was running on only batteries…is this correct? And how would I hook it up so that when plugged into an outlet, it would bypass the batteries and use only the adapter, saving the batteries. Is there a simple circuit that could do this?

Thanks.

Do not connect the wall adapter in parallel with the batteries because A) if the batteries are non rechargeable they will run hot and eventually will explode and B) if they are rechargeable they will be damaged due to lack of charge controller. A relay would be best or a DC jack like the one the older hi-fi radio had (when plug is inserted power comes from the transformer when it is not then power is drained from the batteries.

 

[Gryd3]

Do not connect the wall adapter in parallel with the batteries because A) if the batteries are non rechargeable they will run hot and eventually will explode and B) if they are rechargeable they will be damaged due to lack of charge controller. A relay would be best or a DC jack like the one the older hi-fi radio had (when plug is inserted power comes from the transformer when it is not then power is drained from the batteries.

Good point with the potential damage, but perhaps you can share a link to a socket that would automatically disconnect the batteries with the plug was inserted?
So far, all recommendations are for building a simple circuit, or including safety components to prevent reverse current flow through either power supply.

 

[HellasTechn]

Good point with the potential damage, but perhaps you can share a link to a socket that would automatically disconnect the batteries with the plug was inserted?
So far, all recommendations are for building a simple circuit, or including safety components to prevent reverse current flow through either power supply.

Something like that :

http://www.ebay.com/itm/10pcs-5-5-x...448279?hash=item4d338e5757:g:dG4AAOSwpDdVdbPs

 

[Gryd3]

Something like that :

http://www.ebay.com/itm/10pcs-5-5-x...448279?hash=item4d338e5757:g:dG4AAOSwpDdVdbPs

That does not look like it has a built in switch to disconnect batteries when the AC-DC adaptor is plugged in.

or a DC jack like the one the older hi-fi radio had (when plug is inserted power comes from the transformer when it is not then power is drained from the batteries.

 

[HellasTechn]

That is what the 3rd terminal is for. so yes i think it can switch. If i am not wrong i think what it does is that it actually removes ground from the battery when pluged in.

 

[Gryd3]

That is what the 3rd terminal is for. so yes i think it can switch. If i am not wrong i think what it does is that it actually removes ground from the battery when pluged in.

Sounds good.
All of the DC Jacks I have seen like that usually have two pins tied together on ground, but I must admit to never actually testing to see if one of the two pins switched.
I recall a socket for a 3.5mm audio plug having an internal switch, but they have never been a component I required.

 

[HellasTechn]

Never needed one like that also but in cases like this they may come in handy.

 

[AnalogKid]

That does not look like it has a built in switch to disconnect batteries when the AC-DC adaptor is plugged in.

With that third terminal on the side it almost certainly does, but there's no way to know without a head-on shot that shows the inside around the center post.

ak

 

[HellasTechn]

He can just use a multimeter and a plug to test it !