I have a portable Bluetooth speaker that broke. The unit functions fine off of AC but will not run off of the battery (12V SLA).
When plugged in it IS charging the battery.

I would like some assistance in designing a circuit that would :
A. Have two DC Voltage inputs. One from the OEM power supply (it's fairly steady at about 14.4 volts).
The other is the 12v SLA battery.
B. Give the AC power supply priority. If it's plugged in I want it to power from the main.
C. Charge the battery while plugged in.
D. I'm looking for something more on the simpler side. minimal surface mount components anyway...

Any assistance would be greatly appreciated. Thanks in advance.

 

It might be as simple as this:

 

The unit functions fine off of AC but will not run off of the battery (12V SLA).

I don't understand. If it won't run off the battery, what are we doing?

 

My first impression is, "You already have a machine that does what you want, if you just repair it."
After all the free Uninterruptible Power Supplies I have received with nothing wrong but a dead battery, that's probably what's wrong here.

 

The isolation between the 2 sources is faulty. However, the 2 pin header where the AC supply connects to the board functions fine. So instead of trying to component level troubleshoot a board with smd all over it, I figured I can build my own switch/isolation/charging circuit with output connected to original 2 pin header...and learn.

 

Battery tests good and I have a brand new one
I would love to be able to repair it. It's a little daunting looking at all that smd, however.
In the first circuit posted, what are the concerns as far as overcharging battery?

 

Looking at diagram it appears the diode in series with battery will reduce battery voltage to 13.7 which is right in between the 13.6-13.8 float range specified by manufacturer. I have some 40+ SS34 Schottky diodes, rated to 40V @ 3A...think those will do the trick(they are surface mount but they are bulky).

 

The isolation between the 2 sources is faulty. However, the 2 pin header where the AC supply connects to the board functions fine. So instead of trying to component level troubleshoot a board with smd all over it, I figured I can build my own switch/isolation/charging circuit with output connected to original 2 pin header...and learn.

Ah, I think I see. So you want to power it from a battery connected via the AC adapter pins, as opposed to the original battery connections? Have you tried that to see if it works that way? If so, you "just" need to solve the overcharging issue.

 

Maybe not. Shottky diodes are famous for having a lower voltage drop than old silicon rectifier diodes.

 

Looking at diagram it appears the diode in series with battery will reduce battery voltage to 13.7 which is right in between the 13.6-13.8 float range specified by manufacturer. I have some 40+ SS34 Schottky diodes, rated to 40V @ 3A...think those will do the trick(they are surface mount but they are bulky).

They have a lower ∆V than a silicone diode, so you may need twice as many in series to get the drop you need.

 

Maybe not. Shottky diodes are famous for having a lower voltage drop than old silicon rectifier diodes.

You are correct. I believe these drop about .235 V . What would happen if I put two in series with battery?

 

0.235 + 0.235 = 0.47

 

Or I may be able to find better diodes off of a salvage board I have lying around.

 

0.235 + 0.235 = 0.47

Or 3 in series...lol

 

I have plenty of room inside enclosure. I could probably even get away with adding a cooling fan. I have heat sinks that should work fine and less trouble

 

Going to give this a try tomorrow. Thanks for the help and suggestions. Very much appreciated.

 

So...last question. If I place 2 diodes in parallel where the diodes are will I in effect double the current rating approx.? Mostly the one in series with battery since current for battery charge and running the unit both go through that one. Thank you again.

 

I would not expect it to fully double the rating, because it's inevitable that they won't be identical and one will end up carrying more current than the other.

But in your favor is the fact that a diode's voltage drop increases with heat and current. So as one gets hot, it will resist current a bit more and the other diode will come into play.