Cordless Charger Circuit Help

Discussion in 'The Projects Forum' started by smokeater, Mar 14, 2010.

  1. smokeater

    Thread Starter New Member

    Mar 14, 2010
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    I have three chargers for a cordless grease Gun. I have been working on them trying to figure out how this circuit is working. I have checked the transistors and the diodes. All seem fine. I am not sure what is going on. The diode used to light up and now it won't. The batteries show 13v. and the charger is outputing 14v without load. The used to come on till they were charged.. I have attached photos for help. I figured with three units I could get one to work. I can't understand how the LED lights. It appears the battery actuates the relay when connected.. Could someone help me understand the circuit. Not sure what the transistors are doing? It appears I am getting good DC voltage from the regulator section
     
  2. Bychon

    Member

    Mar 12, 2010
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    Maybe the light is only supposed to be on when it is charging a battery, and the batteries are charged so the charger refuses to try to charge a full battery. The board is simple enough that I could draw the schematic from looking at the board and turning it other side up about a hundred times, but you have the board. Try tracing out the circuit and scanning in a drawing for us. Try to read the part numbers on the transistors. Knowing the polarity of the transistors makes this kind of job easier.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Tried to find a datasheet for the relay, but this is as close as I could get:
    http://www.relay-rayex.com/downloadfiles/LEG-F_SERIES.pdf
    I don't know if your particular relay has a reverse-EMF protection diode in it or not, but if it doesn't, that's probably why the LEDs don't work anymore; they likely got "zapped" when the power to the charger was cut and the relay coil still had current flowing through it.

    It's kind of a funky design anyway; looks like the charge voltage limit is set by the relay pull-in voltage. That could vary a good bit from relay to relay. For the 24v version (which that one is), the "maximum pickup" is 16.8v.
     
  4. smokeater

    Thread Starter New Member

    Mar 14, 2010
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    I think I have the circuit drawn out. It doesn't appear the relay has any diode according to the specs I found. Q1 is A1015, and Q2 is C945. The fuse is 5A. The relay is LEG-24T. I am confused as to what the "T" terminal on the battery is doing. There is also a thermal circuit breaker in the battery pack from the "T" terminal to the negative.. I am still not able to figure out what is wrong. With the battery connected, I am getting around 13v across the battery + and - I get a -6v across the battery - and the "T" terminal. I am getting a combined 19v across the "T" terminal and the + battery.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    I've re-drawn your circuit to be a bit more conventional in appearance. Not knowing what the transistors are, I just threw in a couple of standard American transistors. Yours probably are Asian.

    I think that the "T" connection goes to a thermally controlled charging terminal (basically, a thermal switch) inside the battery itself. When the battery heats up, the connection is opened.

    Anyway, see if the schematic makes more sense to you when drawn this way.

    [​IMG]

    If there is no diode across the relay, Q2 is probably dead.
     
  6. smokeater

    Thread Starter New Member

    Mar 14, 2010
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    Many Thanks for taking the time to answer my questions. I appreciate the help more than you can know...

    To better understand how this circuit works. Does the circuit compare the voltage of the battery and once the voltabe is high enough, it turns off Q1. When you connect the battery the relay activates. Would this still do it if the voltage on the battery is too low? Would it improve the circuit safety to add a diode across the relay coil? I have three of these chargers, and it seems they don't hold up very well.

    I have checked the transistors with my meter and it seems that they are working, but I will try swapping them. I have access to a bunch of old SK parts. The current Q1 is a A1015 PNP . I found ns SK3466, and a SK3114A that looks like they may cross. They are more power, but would they work.

    I also have Q2 is a C945 NPN. I found an SK9229 or SK3845 that may work based on the power. I have several other SK parts, but can't find any cross reference online for these old parts.
     
    Last edited: Mar 16, 2010
  7. SgtWookie

    Expert

    Jul 17, 2007
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    I would definitely use a diode across the relay coil. A 1N4004 diode would work OK.
    I still think that Q1 or D1 (the LED) could get blasted by the coil.
    If your meter has a diode check function, see if you can test the LEDs with it.

    I am not certain that the schematic is correct. Please double-check against the actual boards.
     
  8. Bychon

    Member

    Mar 12, 2010
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    I am attempting to attach a PDF. If it arrives I will type some more.
     
  9. Bychon

    Member

    Mar 12, 2010
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    No luck with the image, but it shows that Sgt. Wookie made at least 2 mistakes in re-drawing. From my re-drawing the drawing from Smokeater, post 4 here, I see:

    The .22 resistor and the fuse are in series with the + terminal of the charger. If more than 3.18 amps peak is flowing, the .7 volts developed across the .22 will send current through the 100 ohm resistor and the pnp will allow current to the LED. The LED will only light if the battery is charging by allowing peak input pulses of 3 amps or more. Actually, a bit less is required because the fuse has some resistance.

    If at least .7 volts arrive at the - terminal of the charger, it will flow through the 3 k resistor and the npn transistor will close the relay. The capacitor holds that voltage steady so the transistor doesn't release the relay between charging pulses.

    Closing the relay flops the relay contact at the T terminal, which apparently grounds the T terminal and allows major current to flow through the battery from its T terminal to ground.

    So, plugging in the battery puts voltage on the - terminal, which flops the relay and closes the current ground at the T terminal.

    The relay definitely needs a diode across its coil, Positive terminal of the diode to the + output and negative terminal of the diode to the collector of the npn transistor. The manufacturer might have included that diode in the relay...or not.

    The relay pull in voltage is not related to how much voltage the battery has, except the battery must have at least .7 volts to activate the npn transistor. The relay coil is energized by the positive charging pulse voltage, which happens every 8.33 milliseconds. Apparently that is fast enough to keep the relay from fluttering. If the battery contains most of its rated voltage, that will also help sustain the relay coil current between pulses.

    Please redraw the drawing in post 4, declaring the negative terminal of the 4 diode bridge as ground. Things get a lot easier to figure if the drawing makes sense by conforming to the shapes and layout you are accustomed to.
     
  10. smokeater

    Thread Starter New Member

    Mar 14, 2010
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    I have updated the drawing based on everyones comments. Again I really appreciate this..
    I have also include voltage readings with the battery connected. The relay is kicking in and out when I disconnect the battery.
    The led never seems to come on, or if it does it is a very slight flicker once on a while..
    I have replace both transistors. Q2, C945 has been replaced with a SK3854, and Q1 A1015 has been replaced with a SK3466.
    It seems like it may be charging as the voltage on the battery did rise a little over time.. but the diode doesn't seem to do anything..
     
  11. Bychon

    Member

    Mar 12, 2010
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    It looks like my first answer was right. The batteries are full. If they aren't accepting 3 amp pulses, the LED won't light. It's not an "ON" light. It's a "charging" light.
     
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