0
\$\begingroup\$

enter image description here

I am planning to charge my 12V sealed Lead Acid battery from AC to DC Converter. The output of the AC-DC Converter is adjustable so that I can set my voltage to be 14.7V(considering diode voltage drop) in order to charge the battery. The resistor 10 Ω is used to limit the charge current of 210mA.12V is mainly used to drive fan and other small circuits

My questions :

1.Once the battery is fully charged, the output across the battery would be 12.7V.I do not want to disconnect the battery or switch it over the float charging (13.6V) as I want my design to keep it simple. My only concern is overcharging the battery if left connected. Does it have any impact on its life. Do I need specific charging circuitry to charge the lead acid battery.

2.In case of power failure ,my battery delivers the power required to the system. Since my voltage regulator requires minimum input voltage greater than 2V of expected output, my Linear regulator does not work. So I have to modify my design to use either LDO /Switching regulator. Another option is give the unregulated output of the battery to my fan(it can work anywhere from 10V) during normal conditions. So I can power small circuits using 5V. Does it work perfectly?Please give ur suugestions.

\$\endgroup\$
5
  • \$\begingroup\$ Your post is hard to follow but "The resistor 10 Ω is used to limit the charge current of 210mA" probably won't work like that. I don't know what internal resistance your battery has, but I bet is a lot less than 60 Ω (which added to those 10 would limit current to 210mA on a 14.7V source). (I've ignored the diode drop.) \$\endgroup\$
    – got trolled too much this week
    Commented Oct 22, 2015 at 9:00
  • \$\begingroup\$ With the diode drop considered 0.7V, you'd need 56.6 Ω internal resistance of the battery in order to limit the current to 210mA. \$\endgroup\$
    – got trolled too much this week
    Commented Oct 22, 2015 at 9:07
  • \$\begingroup\$ Datasheet says the internal resistance as "N/A". Sorry I miscalculated it. \$\endgroup\$
    – ANONYMOUS
    Commented Oct 22, 2015 at 9:21
  • 1
    \$\begingroup\$ The battery output is 11.89V if it is completely discharged.In that case,the voltage difference(14.4-11.89) would be around 2.51V.Then the current would be approximately 250mA \$\endgroup\$
    – ANONYMOUS
    Commented Oct 22, 2015 at 13:14
  • \$\begingroup\$ Under normal operating conditions, yes. But if your battery develops a shorted cell, your design will facilitate thermal runaway [of the remaining cells] and a big shazzam will ensue. \$\endgroup\$
    – got trolled too much this week
    Commented Oct 28, 2015 at 10:30

1 Answer 1

Reset to default
1
\$\begingroup\$

I think batteries usually follow a CC/CV or constant current/constant voltage system to charge up batteries. The LiPoly batteries need such a mechanism. So, this can be achieved by a system that sends out PWMs that control the current and voltage in a more controlled manner. Ofcourse lead acid charging will be different. But please read up on that as I believe you are bypassing that.

And yes over-charging is definetily an issue in your case. You can ofcourse have a dedicated IC or even a combination of op-amp/comparators to do the job for you. When the voltage in the battery reaches a specific value then have the comparator/opamp circuitry turn off or isolate it via a MOSFET(again check the current rating and heat dissipation here).

What you are doing is blindly charging the battery which can very well end up in a pile of ash.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.