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Charging a lead acid battery

  1. Jul 27, 2011 #1
    I want to charge a led-acid battery
    it needs 12V DC
    1.i have transformer to convert 220V AC to 12V AC
    after that can i give 12V AC to bridge rectifier and from bridge rectifier
    to battery terminals directly or should i use a filter to remove the fluctuations in DC ?
    2. what should be the DC current while charging ?
    3. should current be pure DC or is it okay if there is fluctuation
    because I'm not using filter instead I'm giving directly from rectifier to battery terminals.
     
  2. jcsd
  3. Jul 27, 2011 #2

    russ_watters

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    Re: charghing a ledacid battery

    Have you checked the voltage on the battery? 12V is probably too low...
     
  4. Jul 27, 2011 #3
    Re: charghing a ledacid battery

    yes, its written on it
    "charging method : 12V constant voltage charging "
     
  5. Jul 27, 2011 #4

    vk6kro

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    Re: charghing a ledacid battery

    12 volts AC means 12 V RMS so this will have peaks of 1.414 times 12 volts or about 17 volts.

    A bridge rectifier will drop about 2 volts of this to give about 15 volts peak on load.

    So, you can do it and the battery will charge on voltage peaks.

    Depending on the transformer's resistance, you may need to add some resistance in series with the battery to limit the current on peaks. 12 volt lamps are a convenient way of doing this.

    You don't need to filter the output of the battery charger.
     
  6. Jul 27, 2011 #5
    Re: charghing a ledacid battery

    how to know what is the limiting current for charging ?
     
  7. Jul 27, 2011 #6

    vk6kro

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    Re: charghing a ledacid battery

    Depending on the size of the battery, the charging current from a battery charger can be up to, say, 10 amps or as directed by the battery maker. In a car, the charging current can be up to 50 amps or so for a short time.

    In your case, the limiting factor would be the current rating of the diodes. You need to limit this current so that you do not destroy the diodes.

    If you use a 12 Volt 3 amp (36 Watt) lamp in series with the battery, then the maximum current that will flow into a completely flat battery will be about 3 amps. This is safe for the battery and may be safe for your diode bridge if it is rated for 3 amps or more.
    The brightness of the lamp also gives you an indication of the charging current.
     
  8. Jul 27, 2011 #7
    Re: charghing a ledacid battery

    What is the capacity of your battery in AH ? Generally, the upper limit of charging current is said as Battery capacity in AH/8.

    I am not clear here. I think using a large enough capacitor at the output of battery charger will greatly decrease the charging time. This is because, say the battery is already charged to 11V DC. Now, if we apply an unfiltered Full-Wave rectified AC, then the battery will get charged only for the fraction of time when the Rectified AC is at 11V to 17V to 11V. Rest of the time when the Rectified AC is from 11V to 0V to 11V there is no charging (Diode reverse biased).
    If we instead use a capacitor large enough to always maintain the out-put voltage at around the peak (17V) then we can get continuous charging.
    Please correct me If I am wrong.
     
  9. Jul 27, 2011 #8

    vk6kro

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    Re: charghing a ledacid battery

    In the example above, the voltage across the battery does not rise to 17 volts. The voltage may rise by only a fraction of a volt.
    If you did put a capacitor across the battery, it would charge to the same voltage as the battery and it would be unable to charge the battery. So, you would just waste charge by doing this.

    If the limiting factor on current was the maximum rating of the bridge rectifier, then storing charge in a capacitor would only reduce the current available to the battery.
     
  10. Jul 27, 2011 #9

    mheslep

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    Re: charghing a ledacid battery

    A battery is, in part, already a capacitor.
     
    Last edited: Jul 27, 2011
  11. Jul 27, 2011 #10
    Re: charghing a ledacid battery

    From my practical knowledge, the 12V Lead-Acid battery should get charged to as high as 14Volts before it is called full-charged. Isn't it?
    But thats another matter.
    Oh yeah, the capacitor would only get charged to near the battery voltage. But, because, the battery has some internal resistance, and because of the rippling nature of input Voltage and hence the rippling nature of the Battery Current, the voltage drop across the battery's internal resistance will also be rippling nature
    V(int_res) = I(max)*Sin(wt)*R(int)
    Hence the voltage across the battery would be
    V(bat) = V(bat_nominal) + V(int_res)
    = 11 V + I(max)*Sin(wt)*R(int)
    So, the capacitor gets charged to the max of this Voltage.
    So, during when the source isn't providing charging current, the capacitor will be doing.
    I can understand that the effect am trying to describe is negligibly small for practical values of charging current, battery resistance, source resistance etc.So, I rest my case that including Capacitor will be beneficial. Thank you.

    However, including it shouldn't be dis-beneficial except for economic reason of its cost.
    Your above Bolded quote, I can't grasp. I am in thought, that whatever goes into the capacitor eventually, goes into the battery (after the first-time charging of the capacitor). Capacitor don't consume anything (Or are you talking about its losses?). Where would I waste charge?
     
  12. Jul 27, 2011 #11

    vk6kro

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    the effect am trying to describe is negligibly small for practical values of charging current
    So, I rest my case that including Capacitor will be beneficial. Thank you.


    !!!!!!!

    I am in thought, that whatever goes into the capacitor eventually, goes into the battery (after the first-time charging of the capacitor). Capacitor don't consume anything (Or are you talking about its losses?). Where would I waste charge?

    The charge in the capacitor would not be returned to the battery. If you used the battery until its voltage dropped to zero, then the charge in the capacitor would be dissipated in the load. This is no different to adding another battery in parallel with the first one.

    However this was a basic question about charging batteries, so we need to stay practical and deal with reality. Capacitors with capacitance anywhere near big enough to affect the charging of a lead acid battery would be very expensive.
     
  13. Jul 27, 2011 #12
    Forget everything I wrote. I couldn't write it clearly. Sorry.
    Just one question. If I placed a capacitor, How do I waste charge? After charging for the first time, it practically, just rests there doing nothing, right?
     
  14. Jul 27, 2011 #13

    vk6kro

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    Although we can make some general statements and approximations, you really need to say what size capacitor, what kind of bridge rectifier, and transformer and battery.

    After that, you can do a simulation and calculations and make some decisions.

    If there was limited current available from the power supply, then storing it in a capacitor and delivering it to the battery a few milliseconds later would not affect the final charge in the battery.
    As I pointed out, though, even this does not happen and the charge in the capacitor stays there and is not available to the battery. If this was a capacitor of several Farads, then this is considerable wasted energy as far as charging the battery is concerned.

    To get any real improvement, you might be able to find a switch mode device that can make better use of the output of the rectifier. I have seen such devices advertised and they give an output to a battery even when the input voltage is less than 3 volts.
     
  15. Jul 27, 2011 #14
    That is one-time only charge.You don't constantly need to be providing that charge, once charged for the first time, it rests there. Thats not much of a wasted charge, not much as to worth mentioning. I am not really in disagreement to anything you said, just wanted to make sure I wasn't missing anything.
    Thanks.

    And, one extension question. (I hope I am not sued for hi-jacking.)
    1. Suppose there is a switch mode charger, that produces a clean DC output and hence charges the battery at say a constant current of 8A.

    2. Suppose there is another regular Charger (like we discussed here) that charges only for fraction of time, has rippling DC output but have an average charging current of 8A (although the peak might be as high as 15A)

    Which is better as far as charging time, heating and life of the battery is concerned?
    I think Heating and charging time would be same. Life is improved by the rippling current.
     
  16. Jul 28, 2011 #15

    davenn

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    Lead Acid batteries are designed to be constant voltage charged. As they charge up the current level drops.

    If you try to constant charge a lead-acid battery it will have a very short life ... it will most likely overheat, cook and possibly explode.

    DONT DO IT !! :)

    Dave
     
  17. Jul 28, 2011 #16

    vk6kro

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    Again, you can't really say any of that without analysing it properly.

    A battery charger that uses a single resistor to limit charging current will produce a tapering current. So, the current drops off as the battery voltage increases.
    Also the current is supplied from a smaller part of the input cycle, so the dropoff in current is even more severe than just the resistor would produce.

    A car battery in a car doesn't get treated like this. It gets a brutal treatment where it may get 30 amps for 5 minutes and then nothing once it is fully charged. So, there is no chance of it exploding on a simple battery charger.

    If it is charged with a dedicated charging chip, it will get treated more like it does in a car. It would get a constant 8 amps, say, until it is charged and then nothing or maybe a trickle charge.
     
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