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6v SLA charger

Discussion in 'General Electronics Discussion' started by bigone5500, Aug 9, 2014.

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  1. bigone5500

    bigone5500

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    Apr 9, 2014
    I am going to build a simple circuit to charge my 6v lead acid batteries using a 12v solar charger. The circuit I have came across is this:

    [​IMG]
    I would like to know how to calculate the charge rate and limit it to either 500mA or 1A. 1A preferably. I would also like to know what transistor to use in this circuit.

    R3 has a note which on the web site is this: This circuit sets up the crest current through R3, the charging current can reach 0.6A when R3 is 1 ohm However, I don't know how that value was achieved.

    Thanks.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Essentially that circuit creates an output voltage source that is 1.25 * (1100+240)/240 Volts. That calculates as close to 7V.

    R3 limits the current by reducing the voltage so as to maintain the voltage across R3 to less than about 0.7V (the note you give uses 0.6V).

    Thus, what you have here is a current limited voltage source.

    This is an acceptable, but not perfect way to charge lead acid batteries.

    A couple of things are not dealt with in this circuit:
    1. The voltage on a lead acid battery varies with temperature. This circuit will likely overcharge or undercharge depending on the temperature.
    2. Flooded cells can be charged to a higher voltage than sealed cells, and maintaining a sealed lead acid cell at 7V lay be too high for float charging (although for intermittent charging it may be acceptable).
    3. Connecting the battery backwards would be very bad juju.
    4. The tolerance of the LM317 is such that you may want to be able to trim the output voltage (typically by changing R2 into (say) a 1k resistor in series with a 200 ohm pot. This would give you a voltage adjustment range of 6.5V to 7.5V if the 317 has a reference voltage of 1.25, or alternately allow the reference voltage to be between 1.17V and 1.35V for an output of 7V. The datasheet indicates the actual variation should be a little less than this, but there are tolerances in other components as well.
    The transistor could probably be any run of the mill small signal NPN transistor. 2N2222, BC548, etc. It's not critical. Note that R3 cannot easily be made variable because it must carry the full charge current.

    The LM317 may need to be attached to a large heatsink f the current is high and/or if the input voltage (which would need to be 9V or higher) is large enough that the power dissipation becomes significant. This weill be even more of an issue if the battery is very flat or has a dead cell in it.

    Oh, and there are different LM317's with different current limits. Depending on the package they are typically rated fro 1A or 1.5A, and will shut down if they get too hot. This will protect the device but is not a recommended way to design a product -- i.e. use a too-small heatsink and let the device overheat and go into thermal shutdown.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Oh, the max current is approximately 0.7/R3 - so 1 ohm gives you 0.7A (the exact value depends on a number of issues)
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    In addition to what Steve said, I recommend adding a diode (e.g. 1N400x or 1N540x) across the LM317 to protect it from damage if the input supply is removed - connect the anode to the output pin and the cathode to the input pin - and a low-value resistor e.g. 100Ω in series with the base of the transistor to protect it from possible damage as well.
     
  5. bigone5500

    bigone5500

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    Apr 9, 2014
    Thank you for all the input. It seems that this is probably just to supply a voltage to a device. Not to use to charge a battery.
     
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