Homemade UPS

Go To Last Post
13 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hello guys..

I've had this posted at another forum as well, but i'd wanna run this by some experts on the topic here.
As the title sugguests, i tried making a simple UPS for a 12 Volts application. The specs are:

nominal current: 750mA
output voltage: 11~13.7V

The UPS should switch off when the battery voltage drops below 11V to prevent to battery from dying, and automaticly switch back on once the AC input is restored.
I came up with this schematic:


(click to enlarge)

EXT_12V is an external 12 Volts line (actual voltage is more like 16V and should be followed by a 7812 for example). It is only present when AC power is plugged.

I would like to know if i made any mistakes.

I've created a schmitt trigger using an op-amp to prevent the UPS to start oscillating once the battery is almost empty, and the MOSFET is switched off.
I've calculated that the power supply switches off at 10.95 Volts, and stays off until the voltage over the battery returns to 12.95V or higher. This should be enough to compensate when the .75 Amp application is disconnected and battery voltage will rise.
The output of the LM317 is set at 14.4V to have the battery being charged at roughly 13.7 Volts (0.65V drop for the 1N5400)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

For high side load switching, you need a P channel MOSFET. To drive that you need an N channel MOSFET or an NPN transistor. The LM358 is not suitable for driving the P channel MOSFET directly as the output doesn't swing close enough to the supply rails.

If you think education is expensive, try ignorance.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

How about driving an N channel MOSFET (BUZ11) to switch the ground line instead?
(resisters/zener around the opamp need to be moved to invert the output)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
How about driving an N channel MOSFET (BUZ11) to switch the ground line instead?

That will do the trick, in theory. But the LM358 is not a very good choice. Like emuler said.

But there are more things to consider:
1. The circuit takes care of battery-undervoltage, but not of battery overcurrent.
2. The LM317 will try to re-charge the battery with max. ~1.5A, if battery voltage is less than LM317 output. What type of battery are you using ?

I'd recommend to write functional specifications first. And THEN design.

Nard

A GIF is worth a thousend words   They are called Rosa, Sylvia, Tessa and Tina, You can find them https://www.linuxmint.com/

Dragon broken ? http://aplomb.nl/TechStuff/Dragon/Dragon.html for how-to-fix tips

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Plons wrote:

That will do the trick, in theory. But the LM358 is not a very good choice. Like emuler said.

But there are more things to consider:
1. The circuit takes care of battery-undervoltage, but not of battery overcurrent.
2. The LM317 will try to re-charge the battery with max. ~1.5A, if battery voltage is less than LM317 output. What type of battery are you using ?


Battery type: BFS 12-1,2 12V lead-acid battery, 1.2Ah.

Did i mention the circuitry draws some 700mA continueously with dips to 500 from time to time, so the current cannot be kept below 700.

Is there a good (and simple) way to limit the charging current, while still allowing 700mA to be drawn from the battery when power fails?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Take a look at the TL499 from Texas insruments. It is basically everything you need unless you want to lear how to create your own UPS.

Neat chip.

Jim

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"Step N is required before you can do step N+1!" - ka7ehk

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

Speak sweetly. It makes your words easier to digest when at a later date you have to eat them ;-)  - Source Unknown

Please Read: Code-of-Conduct

Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You should consider putting your comparator circuit on the low side of your switch. This way current won't be drained after the battery has discharged.

You could also consider just using a voltage divider / npn transistor to set the low voltage disconnect rather than the comparator. Its reasonably accurate for an application like this.

This may help
https://www.avrfreaks.net/modules/PNphpBB2/files/discharger_281.png

oddbudman

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

jgmdesign wrote:
Take a look at the TL499 from Texas insruments. It is basically everything you need unless you want to lear how to create your own UPS.
Looks like this is a DC/DC converter, that can deliver 100mA max... how will this help me to power .75 Amps at 12 Volts ?

I'll keep this little chip in mind if i need nice, small dc-dc converter, but for now i just need to find a way to limit charging current, while NOT limiting the battery's output current: it still needs to power the .75A system.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Go back and look at that LM317. You have some potential problems right there. Consider your max load current of 0.75A. That thing takes about 1.5V drop (input to output) to work. At 0.75A, there will be very close to 1W dissipated. That is pretty major "heat". If its more like 16V in, 12V out, the regulator will dissipate 3W. The regulator could (effectively) melt without a good heat sink.

That is one of the big problems with linear regulators.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Keep in mind that the battery charging voltage is considerably higher than the normal battery voltage (during the discharge cycle). Approx. 13.7V for a 12V lead acid battery. This usually means that a regulator is needed after the battery, as the voltage when the battery is almost dead is about 10V (or lower, depending on how close you want to drive the battery to a deep discharge condition).

You also need to do something to prevent overcharging the darned thing. :o

If you think education is expensive, try ignorance.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I've been thinking .. why not use 2 regulators? one for powering the circuitry, one for charging the battery:

The battery must be charged with 13.6~13.8 Volts, and i am using the design schematic from the LM317 datasheet, R1 limiting the charging current.
I need 0.65V more on the output of the regulator, because of D1.

To prevent the main regulator from charging the battery i need a second diode D2, and the output of the top regulator must be higher to make sure IC1 is the one powering the circuitry (13.9V or higher).

V_OUT will be roughly 13.2 volts when AC is plugged in, and drop to 0.7 volts below the battery voltage when unplugged (~12V).

The threshold for the "off switch" comparator can be lowered 0.7 volts to 10.3 (battery will still be at 11V at that time).

[[ This is all assuming my circuitry works correctly from 10.3 ... 13.2 Volts ]]

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Look at a TL431 to do the low voltage turn off - you'll find one of these in a dead PC power supply.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0