SLA Battery charger?

Discussion in 'General Electronics Chat' started by spinnaker, Dec 4, 2010.

  1. spinnaker

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    I'd like to drag an old post of Sarge's out of the file cabinet. Seems like a fairly easy to build charger.

    I'll be charging the battery from a wall wart. So I don't really care all that much about efficiency.

    I have a 12V 5AH battery. I have read the charge current should be around 10% of the AH rating so I'll need about .5 amps.


    In the post, he says R4 determines the maximum current. How do I determine the value for .5 amps?


    As far as R3 and R2, Is just a typical application for the LM317 where the formula can be found in the datasheet? Would R4 affect that formula?
     
  2. SgtWookie

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    In the schematic you linked to, R4 is the current sense resistor.

    When Vbe of Q1 reaches ~0.62v, it begins to conduct enough current to lower the LM317's output voltage to decrease the output current.

    To calculate the value needed for R4, use 0.62/Iout, where 0 < Iout <= 1.5A

    0.62v/500mA = 1.24 Ohms, which is the value you need for R4. You could use five 6.2 Ohm resistors in parallel. 1/2 Watt total power dissipation rating would be OK.
     
  3. spinnaker

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    Thanks Sarge. I finally did find the formula in the datasheet and it pretty much agrees with what you said but they use .6 v.

    What is confusing is that in the sample circuit they show a 4.7 resistor. And they show 120 ma as the result. It really should be 127 ma.

    As far as the voltage out, do I use R1, R2 and R3 in my formula?


    Also I would like to shut the battery down with a PIC, can I just place another 2n3904 in series with Q1 and use that to shut it down?

    And maybe a digital pot for R2? :)
     
  4. SgtWookie

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    It's an approximate formula, but I'm using what works in the simulation. It's the relationship between current flow through R4 and Q1's Vbe. The Vbe will be ~6.2v at the small amount of current required to pull the ADJ terminal down a bit.

    Like I mentioned above, it's approximate.

    Yes. Note that Vref can change somewhat over temperature. See the LM317 datasheet. If you use the values shown in my schematic, you should be able to adjust the output voltage via R2 to get the desired float voltage.

    What do you mean by "shut the battery down"? If you mean to disconnect the battery from the charger, you could put a logic-level N-ch MOSFET in the wire to the right of R3. It'll have from 0v (not charging) to 0.62v on it's source terminal (charging) due to R4, but with a 5v logic supply it should still be enough to turn on a decent logic-level MOSFET.

    Not the way it's configured. Digital pots generally require 5v across the "ends" of the pot, and output anywhere from 0v to 5v on the "wiper".
     
  5. spinnaker

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    Sorry I meant shut the charger down (LM317). Would I be better to have a MOSFET to shut the battery down? I have a couple of SPP18P06Ps on hand. The SPP18P06P is probably way more than I need. I need to make another order to Digikey anyway. So and suggestions are welcome.

    Or place a blocking diode so the battery does not you the circuit and just shutdown the LM317 as suggested in the LM317 datasheet.




    Maybe a selection of resistors then? That I can switch out with the PIC. It would probably be easier to implement any way, I have done this before.
     
  6. SgtWookie

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    The SPP18P06P is a P-ch MOSFET. You'd have to use a transistor or N-ch logic-level MOSFET to pull the gate low; otherwise you wouldn't be able to turn it off with a PIC. This implies more current consumption, as you'd need to use a pull-up resistor to turn the MOSFET off.

    A low-Vf Schottky diode would work to prevent discharging the battery when the LM317 was not providing current.

    Possibly. R3 would need to be two resistors in series, and the resistors tied in at the junction of R3a and R3b. The PIC could pull the other side of the resistors high, low, or set them to input (floating between 0v & Vdd)

    I'm not functioning very well today; I've been sick since Thanksgiving.
     
  7. spinnaker

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    I've done the voltage selection in my first PIC project so I am sure I can figure it out again.

    I have 1n5817, 1n5818 and 1n5820 on hand. I guess I can just look at the data sheets and see what the best choice is. But I am sure you know off the top of your head :) I am certain even in your non functioning state you still function better than me on a good day when it comes to this stuff. :)

    Hope you feel better. Next time don't eat so much turkey. :)
     
  8. SgtWookie

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    The 1N5820 will have the lowest Vf of the three you mentioned for a 500mA current.

    If you looked up the datasheet for your battery, you might discover that the max charging rate is closer to C/5. Some AGM SLAs are rated for C/3. Power dissipation in the regulator will be a problem at high currents. The plug supply output voltage needs to be 16v to 18v.
     
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