Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

linear power supply with backup battery

Status
Not open for further replies.

logicnibble

New Member
Hello!

I want to build a linear power supply (AC/DC) to power a uC board and some external equipment. I also want to have a backup battery to keep powering the board when the mains power fails.

After some research, I decided to use a basic linear power supply: transformer + diode bridge + filter capacitor + voltage regulator IC. The battery would be a 12V / 7Ah SLA.

The voltage regulator would be the LM317 at 13.8V. At this point I would connect both the battery and the rest of the circuit: the battery would have a float charge and the rest of the circuit would have the 13.8V and another voltage regulator to 5V.

This way, if the mains power fails, I would have the battery powering the rest of the circuit.
So far so good. (I guess...)

But, if after a power fail the battery is really down, the charge current would be big and only limited by the LM317 internal limitation: min 1.5A, typical 2.2A. In this case, I would have to dimension the power supply (including a big heatsink) to this power.
Since this would be somehow overdimensioning I want to dimension it to 1A.

I can use the L200 with current limitation defined by a resistor value.

The question is: Can I still use the LM317 and define the filter capacitor such as the ripple would be so big at more than 1A that the minimum voltage before the LM317 would be much less than 13.8V + 2V LM317 dropout, hence decreasing the output voltage and work as a current limitation? The LM317 is much cheaper than L200, that's why I'm trying to use it.

Thanks!
 
Hello!

I want to build a linear power supply (AC/DC) to power a uC board and some external equipment. I also want to have a backup battery to keep powering the board when the mains power fails.

After some research, I decided to use a basic linear power supply: transformer + diode bridge + filter capacitor + voltage regulator IC. The battery would be a 12V / 7Ah SLA.

The voltage regulator would be the LM317 at 13.8V. At this point I would connect both the battery and the rest of the circuit: the battery would have a float charge and the rest of the circuit would have the 13.8V and another voltage regulator to 5V.

This way, if the mains power fails, I would have the battery powering the rest of the circuit.
So far so good. (I guess...)

But, if after a power fail the battery is really down, the charge current would be big and only limited by the LM317 internal limitation: min 1.5A, typical 2.2A. In this case, I would have to dimension the power supply (including a big heatsink) to this power.
Since this would be somehow overdimensioning I want to dimension it to 1A.

I can use the L200 with current limitation defined by a resistor value.

The question is: Can I still use the LM317 and define the filter capacitor such as the ripple would be so big at more than 1A that the minimum voltage before the LM317 would be much less than 13.8V + 2V LM317 dropout, hence decreasing the output voltage and work as a current limitation? The LM317 is much cheaper than L200, that's why I'm trying to use it.

Thanks!

You may put a diode from the battery (anode) to your circuit (cathode) in parallel with a resistor.

The resistor limits the battery's charging current, while the diode allows the battery to power the rest of the circuit.
 
Thanks!

That solution is good, but won't it damaged the battery? Manufacturers recommend charging with constant voltage (13.8V) but this way the voltage will decrease as the current decrease.

Another question: What would be the best protection for battery reverse polarity connection?
Check this circuit:
**broken link removed**
It has a diode in parallel which will cause the fuse to blow.
 
Nice web page, too bad about the circuit. :confused:

That circuit will charge the battery to 13v5, but the 7812 (depending on brand) may need 14v0 to 14v5 to operate properly. When the power is out, the battery quickly drops towards 12v, making things worse. :mad:
 
Wasting half of the power from a battery isn't good idea, have you considered using a switching regulator?
 
Status
Not open for further replies.

New Articles From Microcontroller Tips

Back
Top