Connect with us

Unusual 12V SLA Charger Circuit

Discussion in 'General Electronics Discussion' started by Old Steve, Sep 15, 2015.

Scroll to continue with content
  1. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    Like a dummy, I connected my SLA charger to the battery backwards last night, heard a 'crack' and smelled the magic smoke. I connected it to the battery without the mains connected yet, by the way.
    I pulled it apart and fixed it, (only a track was damaged), but in the process was reminded of it's unusual switched-capacitor configuration when I looked at the schematic. (I 'reverse-engineered' it years ago, and drew up a diagram.)

    I just thought someone might be interested in the circuit. Works really well, using only a 9VAC transformer.
    Also, it originally it charged to 14.9V, a bit high for my liking, so I modified it for 14.2V.
    The circuit, a bit rough but readable:-

    12V 800mA SLA Charger.JPG

    Luckily, due to it's design, the current path was back through the diode bridge, only allowing 1.2V of reverse voltage across other components for a moment until the track was no more, which did no damage.
    Working fine again now.
     
  2. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    That is a bit much for my brain cell.
    C1 and C2 provide smoothing at about 12V. Where does the extra voltage come from?
    Presumably the fet boosts the voltage in some way but it has a load of only 1k.
    Please describe the circuit if you have worked it out.
     
  3. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    I sort of follow it.
    First note that the MOSFET drain is attached to the unrectified transformer secondary, not to DC.
    So, when the AC waveform drops, the MOSFET switches off and C2 charges via the MOSFET's inherent source-drain diode, then, when the AC voltage rises, the MOSFET is switched on, pulling the bottom of C2 up and therefore 'pumping up' the voltage available to charge the battery.
     
  4. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    I'm not real sure about C1. I haven't looked at that circuit for many years until now, and this type of analogue circuit isn't my strong point. It works well though.

    Maybe C1 stops the bottom of C2 from going too low, so that the battery isn't pumped up too high when the MOSFET pulls it up.
    If anyone can explain the operation more clearly, please speak up.....
     
    Last edited: Sep 15, 2015
  5. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
    It look like a sort of charge pump circuit the A.C voltage rides on the D.C level from the rectifier. Not sure about C2, only thing I can think of is it is applying a load to the transformer when then battery is removed, crude regulation.
    Thanks
    Adam
     
    Martaine2005 likes this.
  6. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    I guess you mean C1? (The bottom one.)

    I wouldn't have thought smoothing was needed - the battery would already act like a big capacitor. Many old car battery chargers connect the rectified DC from the diode bridge directly to the battery without smoothing at all, and only bi-metal contacts for current regulation. No other components.

    The two capacitors could act like a voltage divider, ensuring that the bottom of C2 doesn't get higher than half the rectified voltage peak, so that when the MOSFET pulls the C1 / C2 junction high, only half the voltage is added to the existing DC level.

    I'm glad that someone besides me found it interesting.[/QUOTE]
     
  7. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
    Yes sorry Steve C1. It's not when the battery is on load I think I said when the battery is removed, it was just an idea. I don't think your idea about potential divider is correct, I might be wrong however, I bet if I remove C1 the voltage is still the same :). Let me have another look tonight and see if I shed a bit more light on the circuit. Unless someone else comes up with a better explanation. I like this sort of thing, thank you.
    Thanks
    Adam
     
  8. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    While it was apart I should have pulled out the oscilloscope and taken some measurements maybe, but the 'scope is deeply buried and I've been flat-out on other things, so I just quickly repaired it and put it back together.
    When I first pulled one apart, about 20 years ago, I was impressed by it, so took the time to trace the schematic. It's a great little charger, and has worked faultlessly for all these years. (Until I messed up the other night, at least.)

    Edit: It looks like a fairly easy circuit to simulate in LTSpice or another simulator and could be done with generic components, but I personally don't have time right now. Up to my neck in another project.
     
  9. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
    Ok so I have had another look at the circuit and I still think C1 has a smoothing effect. The top capacitor will be take energy from the battery and or the supply. Then as the voltage at the join point of the two capacitors rises this pushes the energy that's in the top capacitor out into the battery.

    The voltage at the join point decreases during the A.C cycle but this takes time and would not go negative, otherwise the bottom capacitor would have a reverse voltage applied across it which could damage it. So there must be a residual positive voltage left on the positive plate of the bottom capacitor. The cycle then repeats. It basically pumps energy from the bottom capacitor into the battery by adding it's voltage to the D.C level applied to the top capacitor. The connection of the rectifier to the battery is needed to sense the voltage of the battery during charging.

    I think :)
    Adam
     
  10. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    That's exactly what I thought. The top cap is charged by the supply, then the MOSFET pulls the C2/C1 junction high, pulling the top of C2 above the battery voltage and discharging the cap into the battery. Effectively a switched-capacitor system.

    I agree that the lower cap would keep some voltage and not go negative. I think that overall we're saying a similar thing, but in different ways. The way I see it is that the two caps are a voltage divider in which the lower cap has about half the battery voltage across it, or a little less, with the other half across the top cap. When the top cap is pulled up by the MOSFET, it dumps that voltage into the battery. The sensing is done the whole time, and the MOSFET is switched off when the battery voltage reaches 14.9V in the circuit as it's presented. In my charger, I shorted across R8 to alter the circuit so that the charge now stops at 14.2V.

    The parts for my current project have been held up and won't be here until at least Monday now, so over the next day or so I'll make up an LTSpice simulation of the charger circuit. I plan to stay up a little longer tonight, so I'll start on the simulation circuit now.
     
  11. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    Well, once I started, I couldn't stop. All done with the simulation, Adam.
    I'm not sure if you have LTSpice or not, so I'm attaching screen shots of the schematic and the trace output, plus the actual LTSpice *.asc schematic for you to run if you do have LTSpice.

    I needed to connect a 12.6V voltage source as a simulated 12.6V battery to get an approximation of what happens, but unfortunately that source doesn't 'charge up', but the results are great apart from that. I put traces at all of the relevant points in the circuit.
    N.B. I didn't know how much series resistance to give the 'battery' so settled on 1 ohm. Probably far less in reality. And we never get to see the charger cut off, because as mentioned, the 'battery' doesn't actually charge.

    If you don't have LTSpice, and want me to measure elsewhere in the circuit, let me know.
    The schematic:-
    Schematic.JPG

    The trace outputs:-
    Trace.JPG
    (The blue trace label is hard to read, but it's the AC input voltage, 9VAC RMS, measured between "Vac_top" and "Vac_bot".)

    And the *.asc file is attached below VV.
     

    Attached Files:

    Last edited: Sep 18, 2015
  12. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    I wasn't thinking clearly - of course I could simulate the battery charging. I started it's voltage at 11V and made it rise to 15.2V in this one. It shows the MOSFET switching off:-

    Battery Charged.JPG
     
  13. Old Steve

    Old Steve

    734
    169
    Jul 23, 2015
    I should be in bed asleep, but couldn't leave this alone. One last trace added - the current through the source of the MOSFET. 5A positive peaks, then 6A reverse current peaks when the MOSFET's internal diode conducts, discharging C1. (The grey trace):-

    MOSFET Source Current.JPG
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-