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12v 3amp battery charger lead acid

Discussion in 'Power Electronics' started by Suny, Sep 22, 2016.

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  1. Suny

    Suny

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    Sep 20, 2016
    I have a 12v 3amp transformer. I want to make a chsrger for my 12v 9ah lead acid battery. Pls give a simplest circuit. Also with charge indicator.
     
  2. davenn

    davenn Moderator

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    Sep 5, 2009

    [​IMG]
     
    CDRIVE and duke37 like this.
  3. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Dave, I like the circuit but why isn't D6 connected directly to the battery? Am I missing something?

    Chris
     
  4. Suny

    Suny

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    Sep 20, 2016
    I want a 3amp circuit
     
  5. davenn

    davenn Moderator

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    this was to give you something to work from
    1N5404 diodes instead of 1N4007
    LM388 instead of the LM317
     
  6. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    Just use the transformer, 4 diodes and a resistor.
    The 3A transformer is 2A DC
     
  7. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    How does 3AC (RMS) convert to 2ADC?

    Chris
     
  8. Colin Mitchell

    Colin Mitchell

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    The volts go up and the amp come down - the watts (VA) stay the same.
     
  9. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    I'm referring to the statement in bold. If a transformer has 3A rating it's specifying 3A RMS output not input. How does 3A RMS output translate to 2A DC?

    Chris
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Imagine you have a 10V RMS AC 100VA transformer. We will assume nice sine waves and that the current drawn is 10A RMS.

    We know that RMS measurements create an equivalency in the ability to heat something, and thus they allow us to measure power.

    If we assume that current and voltage are exactly in sync, we can deliver 100W of energy. If voltage and current are 90 degrees out of phase, the load is still 100VA, but the power is zero watts. We will initially assume everything is nicely in phase and track where that 100W is spent.

    Lets assume we have a perfect bridge rectifier and a perfect capacitor, and a perfect linear regulator.

    The capacitors charge up to the peak voltage of 14.14V. Then the regulator reduces the voltage to 10VDC which is passed to a load.

    Let's assume the load draws 7.07A.

    What is the total power being delivered by the transformer? Well it is equivalent to the voltage on the caps multiplied by the current through the load. 14.14 x 7.07 = 99.97W. And remember that the peak power our transformer can supply is 100W.

    In this case 71W goes to the load and 29W is dissipated by the regulator.

    Assuming a linear regulator, no matter what the voltage output, the output current should not exceed 0.707 times the rated transformer current.

    It's also interesting to note that the actual current drawn from the transformer may peak at 30 or 40 A as ALL the power is delivered in short high current pulses at the peak of each half cycle. The larger the filter capacitors, the shorter duration and higher current these peaks are.

    Different transformer, rectifier, and filtering topologies have different multipliers, but bridge rectification and capacitive filtering is so common that the figures get etched into your mind. (a different set would be etched into your mind if you were supplying HV using a vacuum tube double diode, a centre tapped transformer, a small capacitor, and the field coil of your loudspeaker as an inductor).
     
    HellasTechn likes this.
  11. Colin Mitchell

    Colin Mitchell

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    The 3A transformer is 2A DC.
    It's not the fault of the transformer.
    It's just that can only take 2 amps from the output of a bridge circuit that has a filter capacitor because the output voltage will be 40% higher than the voltage-rating of the transformer.
    This means we will be getting about 18v from a 12v transformer. We are happy with the voltage increase of 40% but we must take 30% less current so that the multiplication of the two values remains the same.
     
    Last edited: Sep 25, 2016
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    40% higher, therefore 30% lower. It's multiplication and division, not addition and subtraction.
     
  13. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Steve I get that but this statement says nothing about filter caps, voltage regulators or anything else. 3A RMS is equivalent to 3A DC.

    Chris
     
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  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yes, it requires filter caps for this statement to hold true.

    Its the filling up of the bits between the half waves which causes the issue.
     
  15. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    That makes a hell of a lot more sense. The way Colin wrote that is misleading,.. to say the least. That said a filter cap is a superfluous component for a basic lead acid battery charger that Colin's describing. On that note I wouldn't use a resistor as he described either. An incandescent lamp is a better choice. It also seconds as a charging indicator. ;)

    Chris
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I prefer some form of current limited voltage source. Sure, an incandescent bulb is a great current limiter, but without some form of voltage limit, it is very easy to overcharge lead acid batteries. For flooded cells this can actually be useful, but not for gel cells.
     
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