Finding 12v source

Thread Starter

Chrisbux

Joined Apr 30, 2022
3
Hi.. I have this magnetic door control unit. The supply is 240v but i want to add a 12v battery backup. So need to tap into the pcb the first point the voltage has been converted from 240v to 12v dc. This will hopefully allow all functions to operate from a battery. Can anyone help please.16513234876985287900241679727200.jpg20220430_134130.jpg20220430_134229.jpg
 

MrChips

Joined Oct 2, 2009
26,099
Welcome to AAC!

Pins 4 and 5 are +12V and GND connections.
Measure the voltage across these two points.

What type of battery are you planning on using?
Depending on the battery voltage you will either be constantly charging or discharging the battery. A common solution is to use two diodes from each power source so that only the higher of the two sources will be supplying current to the load.
 

Thread Starter

Chrisbux

Joined Apr 30, 2022
3
Welcome to AAC!

Pins 4 and 5 are +12V and GND connections.
Measure the voltage across these two points.

What type of battery are you planning on using?
Depending on the battery voltage you will either be constantly charging or discharging the battery. A common solution is to use two diodes from each power source so that only the higher of the two sources will be supplying current to the load.
Thanks for your reply.
I have a rechargable battery and a dc control unit that may be of some use. I have no idea on how to create a circuit for this to work though.20220430_143902.jpgScreenshot_20220430-145049_Amazon Shopping.jpg
 

Irving

Joined Jan 30, 2016
2,668
Looking at the PCB, +12v (pin4) is connected internally to one side of the relay coil and possibly COM (on the relay ) and then the 12v rail continues back into the innards of the board. In theory 12v externally connected to pin 5 will power the internal circuit that provides the relay timing, but if that source was a lead-acid battery it wouldn't be charged properly. An 11.1v LiPo battery would be a better choice, as long as it has an integral BMS, or you use one of the many 3S protection boards available on eBay.

Before doing that you need to verify the controller works with 12v fed back into it with the mains power off. Ideally you should do this with a 12v bench supply with the current limiter set low and gradually increase the current limit to verify correct operation.
 
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Irving

Joined Jan 30, 2016
2,668
That auto-backup board is of no use to you unless you can unequivocally find a single point on the internal PCB where the power supply function connects to the relay control function and looking at the PCB I don't think there is one, or its not obvious.
 

MrChips

Joined Oct 2, 2009
26,099
A lead acid battery has a standing voltage of about 12.8V.
This is going to force current into the PCB 12V power supply.
What you need are two diodes to isolate the two power sources that looks something like this.
Where it shows USB POWER, this is your on-board 12V supply.
You may need to break the 12V tracks at some point to separate the 12V power supply from the load.

You might be able to get this to work with just the 12V battery and one diode connected to the 12V and GND, pins 4 and 5.
For starters, I would experiment with two diodes in series with the 12V battery. The reasoning here is that the battery will be in standby mode and will only kick in if you lose AC power.


1651328874822.png
 

Irving

Joined Jan 30, 2016
2,668
Pin-3 (COM) is connected to pin-5 (GND).
Yes - I'd just realized that myself... I meant that the COM of the relay is connected to 12v, so NC (2) or NO (1) are powered outputs, so COM on the relay appears to be connected to one side of the relay coil and then to the 12v out.
 

Irving

Joined Jan 30, 2016
2,668
You might be able to get this to work with just the 12V battery and one diode connected to the 12V and GND, pins 4 and 5.
For starters, I would experiment with two diodes in series with the 12V battery. The reasoning here is that the battery will be in standby mode and will only kick in if you lose AC power.


View attachment 266174
But that won't charge the battery...

And it can't charge unless you give it 13+ volts... a SLA isn't going to be much use here without a secondary scharger.
 

Irving

Joined Jan 30, 2016
2,668
The power supply can be turned up to 13.4 - 13.8v (12v +15% from the given spec) so it can charge & maintain the SLA battery on a float charge, when connected across pins 4 and 5.

Looking at the components and layout this is a fairly simple switch-mode power supply. The output of the switchmode inverter, the secondary side of the transformer, is rectified by diode U11 and smoothed by C12 and the series inductor (and C20, though its hard to see from the angle of the photo). The 3-terminal voltage reference to the lower left of the transformer and the trimmer potentiometer KR1 set the output voltage, feeding control back to the hot side via optocoupler U3.

When the mains power fails the battery will continue to power the circuitry on the secondary of the transformer. The diode U11 will block the battery from feeding current into the transformer secondary winding.

Check this by measuring continuity between +12 out on pin 4 and the centre lead of U11 - it should be <200milliohm (0.2 ohm) on a good test meter.
 
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