Sealed lead-acid battery charging question.

Discussion in 'General Electronics Chat' started by nevarez_ga, Jul 10, 2011.

  1. nevarez_ga

    Thread Starter New Member

    Dec 9, 2009
    2
    0
    Hello, I don't know for sure if this is my first post because I have been reading this site for years and I found here all the answers to my questions even before asking. But this is a doubt I have been having for a long time.

    We all know that to charge a battery, say a 12v lead acid battery, voltage should be around 13.5 to 15 v, depending on state of charge and charging current.

    I found a 12v lead battery laying around, voltage was 9.8 when checked with the multimeter. I don't have a specific charger for this battery, so I used a variable voltage battery eliminator, yeah those that have a selectable voltage 3, 4.5, 6, 7.5, 9, 12.

    The one I have is a very cheap one, it's rated 1 ampere. When I selected 7.5 volts, the multimeter read 14.00 volts, or about the right charging voltage. When I measured its current capabilities (measureing current with no load), it read 1.5 amps. So I thought, it gives 1.5 amps at 14 volts, that's exactly what I need.

    But then when I connected the charger to the battery, the meter would read 10 volts, just a bit more than the already discharged battery. I thought the power source wasn't really able to give the 1.5 amps. So I selected 12v in the eliminator, with no load the meter read 23.5v and 1.25 amps. Again I connected it to the battery and read the voltage, 13.8 volts, exactly what I needed, and current was .37amps.

    The different voltages and currents indicated were measured while charging the battery, to measure the current I disconnected one terminal and then put the meter in series.

    The problem here is that I don't know which is REALLY the charging voltage, is the battery receiving a 1.25amps @ 23.5v charge? Or a .37amps @ 13.8 v charge? Why are the current and voltage lower when charging if the power source is able to handle that?

    thank you in advance, sorry for being repetitive but I wanted to be as clear as possible.
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,440
    3,361
    When you say you are measuring the current capabilities of the charger with no load you are actually applying a load of almost zero ohms - you are shorting across the charger.

    When you connect the 23V charger to the battery, the battery is loading the charger down to 13.8V while drawing 0.37A. This is ok for charging. As the battery charges up the voltage will rise and the current will drop. Keep monitoring the voltage and stop charging when the voltage rises to about 14.5V.

    A word of caution. Keep checking the temperature of both the charger and the battery with your hand. If either is too hot to touch, stop charging.
     
  3. pistnbroke

    Member

    May 9, 2011
    32
    1
    You must not let the battery go over 14v ...as it charges the current will drop so the load on your charger will drop and its voltage goes up. If you select the output that gives you 14v with no load and just connect that you can safely leave it for 24 hrs to charge and it will not go over 14v
     
  4. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    A 12v battery @ 9.8v is very flat & if its been that way for any length of time is probably unlikely to charge up properly as the plates will have sulphated.
     
  5. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    I'll second this statement. It is good however that the battery may be taking a charge at all. You may get some limited use out of it, but don't put it in your car or even your riding lawn mower, you probably won't be able to start either!
     
  6. nevarez_ga

    Thread Starter New Member

    Dec 9, 2009
    2
    0
    Thank you all guys for the prompt answer. This is exactly the information I was looking for. Right now the voltage is around 13.9v and current dropped to .15amps. The battery has a capacity of 4.5Ah and is going to be used for a laser alarm circuit which consumes 15mA when it's activated, and about 650mA (including the siren) when it goes off, as long as the battery can drive the speaker for 1 minute, it would suffice. I'm actually planning on selling some of these, hope it works!! Again thank you all for the info.
     
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