Wednesday, June 26, 2013

Sealed Lead Acid Battery And How to Charge It Using LM317

Good evening and welcome back to my blog! In an earlier post I've written about the integrated circuit LM317 and how you can make a variable voltage regulator .Now I'm going to write about how this integrated circuit can be used to charge Sealed Lead Acid Battery or which is often called by SLA battery and also put some information about these type of battery.

What is a SLA battery? 




A generic SLA Battery


In the picture above what you are seeing is a sealed lead acid battery. Lead acid battery is the probably the most common and well known type of rechargeable battery and it is here for like 150 years, maybe a little less but I'm pretty sure they will survive way more than 150 years. These type of cells are used in computer UPS, Small DC Fan, powering traction motor of small vehicle and many other applications. The main reason for this is they are cheap and works reliably. It also can give you a good amount of power for example the picture that you see here is rated at 12V 8.2Ah. Which means if you draw 8.2A current from this battery at 12V will provide power for an hour. In other words you can draw about 98.4W for an hour. That's pretty good amount of power.

Sealed Lead Acid Battery does not need maintenance that is why they are often called maintenance free lead acid battery. The standard lead acid battery, I mean the flooded one needs water every couple months as water evaporates. But in sealed lead acid you just charge properly and regularly and that's it. Don't overload, short circuit it will run for about 2 years, I mean they do offer like 200-300 cycles but for normal usage they can easily last 2 years. I will put some very useful links at the end of the post, make sure to check that to learn more.

  Now lets look at the circuit diagram.



From My previous post or from the internet you may know that the output voltage is defined by the following relation.

Vout = 1.25*(1+R2/R1)  + Iadj*R2

Here, R1 is 120ohm & 220ohm, R2 is 470ohm+1KohmVR or 5Kohm VR. (VR=Variable Resistor)

For lead acid battery you have to charge it with 1/10th of its Ah rating. That means a 8.2Ah battery needs to be charged at about 0.82A and of-course with proper voltage which is 13.8V in this circuit.

Now how does the circuit work?

The LM317 will provide proper voltage for charging as it is very important.In the first diagram the charging current for the battery is controlled by Q1 ,R1,R4& R5. Variable resistor  VR1 can be used to set the charging current.As the battery gets charged the the current through R1 increases .This changes the conduction of Q1.Since collector of Q1 is connected to adjust pin of IC LM 317 the voltage at the output of of LM 317 increases.When battery is fully charged charger circuit reduces the charging current and this mode is called trickle charging mode.

In the second diagram same thing is happening too. The diode IN4007 is a bleeder  diode.

Important notes here is , diodes drops about 0.7V each so make sure to use proper transformer for this circuit. This regulator may also drop about 3V. So, its good to provide 18V-20V as input.

 LM317 needs a moderately sized heat-sink. Apply small amount of thermal paste & fit it with a heat-sink. If you need more power, use a LM338.

Calibration :

Connect the circuit, start the power & connect a voltmeter to its output lead before connecting a battery. Then use the variable Resistor to find out proper 13.8V. Then attach the battery. About the Resistor R use value defined by 0.6/Max Current. So, if you need 1A max current you will be needing a 0.6ohm 5W Resistor for that.

Hope You have enjoyed the writing & good luck with your works.

Resources:

1. Learn more about lead acid battery.
2. Wikipedia article about lead acid battery.
3. Voltage regulator using LM317.
4. Link to my other posts.
 

3 comments:

  1. Thank you for this diagram, i have some question. My requirement is atleast 3amps, what capacity do i need on resistor side? Any requirement for powersupply amps? Thank you sir!

    ReplyDelete
    Replies
    1. You can simply use a 0.2ohm resistor, make sure to use one with high capacity as charging at 3A will make it really hot. Moreover you can't use the LM317 for 3A, go for LM338 and last but not the least make sure that your main transformer that is feeding this circuit can provide that much power.

      Delete
  2. Hi You,

    There’s a new way to bring nearly any type of old battery back to life ...so it’s just like new again.

    This method works with nearly every type of battery out there ...and it’s simple and quick.

    >> Click here to learn this secret battery reconditioning method <<

    In case you’re wondering, you’ll be able to bring car, phone, and laptop batteries back to life with this.

    It even works with solar/off-grid, marine, golf cart, and forklift batteries. Plus, many more!

    >> Click here to learn how to bring your dead batteries back to life again <<

    With this recondition battery secret, you won’t have to buy new expensive batteries anymore. You can just recondition your old, used batteries and save a lot of money!

    And >> This new video presentation << shows you how:


    Best regards.






    .

    ReplyDelete