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Sony Ericsson CST-60 Charger

Discussion in 'Troubleshooting and Repair' started by TimeManx, May 11, 2010.

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  1. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    How bout I attach an image of the circuit or maybe draw the diagram. Would that be in any way against the rules?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,839
    Jan 21, 2010
    No, that would be extremely helpful.
     
  3. Mitchekj

    Mitchekj

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    Jan 24, 2010
    If it died due to a short on the output, I have bad news...

    There was no OCP (over current protection) in the circuit, I would wager that your controller IC, at a minumum, has died. Your switching FET/BJT, maybe even the transformer, if it's not somewhat robust, would be suspect.

    The zener across the output is there to keep the the open voltage (the voltage when no load is connected) clamped to a reasonable level. An SMPS (in a constant current/power config.) no-load output voltage is much higher than when loaded, which could reflect back and destroy things on the primary side. A short would not be protected by this zener, obviously, since there would be very little voltage across a short, but MUCH current... enough to smoke things pretty quick. Most decent SMPS will incorporate OCP for just this reason, and it's usually integrated into the controller IC. Wonder what they're using for the controller...

    What is written on top of the controller IC? It's probably the only IC on the board, aside from the optocoupler.
     
    Last edited: May 12, 2010
  4. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    Nope, no other IC. Let me post the circuit pic.
     
  5. TimeManx

    TimeManx

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    May 11, 2010
    The back side image has been flipped horizontally.
    [​IMG]

    Input is at the top & output is at the bottom.
     
  6. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    Oh lol, definitely not any smps I've ever seen. I'd expect this is some kind of linear supply then. Disregard my post above. :)
     
  7. Tesla

    Tesla

    165
    2
    May 10, 2010
    Not even a bridge rectifier on the AC input (just a single diode)? Basically just a transformer doing the AC/DC conversion from the looks of it. Are you sure the "photo-coupler" doesn't have + ~ ~ - markings on it?

    The "diode" standing up next to the orange cap. and transformer ... is there a ring on one side or is it solid black?

    Is the small resistor right above and closest to the upper-most transistor burnt in the middle?

    Edit: What symbols are on the board for those 2 parts?
     
  8. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    No +/- markings.

    The diode has a ring. No symbol on the board for this part.

    Yea, you're right. It is burnt. The symbol is R3.
    So is this it???
     
  9. Tesla

    Tesla

    165
    2
    May 10, 2010
    With at least one lead of R3 unsoldered, what ohms do you read?

    ... what colors can you make out?
     
  10. Mitchekj

    Mitchekj

    288
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    Jan 24, 2010
    Now I'm intrigued. :) Can you do me a favor and measure continuity on the transformer? With the power off, mind you. Just trying to figure out which pins are making what winding. I'll try to whip up a quick schematic...

    For instance: the two pins on the output side should read continuity (very low resistance) since that's the secondary winding. How is the primary side wound? 1-2 and 3-4 or 1-4 and 2-3, etc.?

    Also, a part number or marking for those transistors would help. Since R3 is smoked, I'd be assuming Q2 is smoked also. Never know tho. :)
     
  11. TimeManx

    TimeManx

    31
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    May 11, 2010
    R3 reads 8.1K. and i'm guessing the colour might have been green, 2.5K?

    OK, I'll do that.
     
  12. Tesla

    Tesla

    165
    2
    May 10, 2010
    Yes ... Red, Green, Red = 2.5K ... So we know it's between 2.0K and 2.9K, right? Quite a ways from 8.1K. If those are 1/4 watt ... I would up-it to a 1/2 watter.

    I thought you had already done it, but if not (like he said) check all your diodes and transistors for shorts and opens.

    You marked the polarity on the DC out wires so you know where they go when it comes time to solder them back on, yes?
     
  13. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    Primary & secondary are both continuous.

    Primary side is 1-2 & 3-4. The pins on bottom left of the upper half are for the primary windings & the pins on the top left of the lower half are for the secondary winding.

    Q2 says C945 P331 & Q2 says 13001 6D331.
     
  14. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    I have already checked the diodes and all of them are fine.
    So I'll be checking the transistors now.
     
  15. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    A little confused... this is how I'd expect the xfmr to read, red lines meaning continuity:

    Is that what you're reading?
     

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  16. TimeManx

    TimeManx

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    May 11, 2010
    Yea that's what I said.
     
  17. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    Hmm, Q2 is totally shorted out & Q1 is totally open.
    What do you guys say?
     
  18. Tesla

    Tesla

    165
    2
    May 10, 2010
    Kinda makes sence ... it look a lot to burn that resistor.

    Q2 - C945 is a very popular NPN.

    Q1 - 13001 ... No continuity between any of the legs, even after reversing polarity?

    Replace those 3 parts and it might work.
     
    Last edited: May 12, 2010
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,839
    Jan 21, 2010
    Also check the optocoupler, the side closest the top when looking at the photos.

    By my estimation, the current required to toast that resistor would have had to pass though the optocoupler, (edit see below, probably not)

    It should measure as open circuit or very high resistance. If it has failed "open" it will be hard to establish from a simple resistance measurement. In circuit it should measure about 270K due to the resistor across it. If it's a low resistance, the optocoupler is shot too.

    That would be very interesting because it's certain that the optocoupler would have been OFF if the output was shorted. For it to turn on, the output has to rise significantly above the zener voltage, and it's the optocoupler that turns off the regulator to allow the voltage to fall. It is possible that it failed after the short was removed due to damage which caused the output to go excessively high.

    I would recommend drawing out the circuit and working from there. I have done part of it, but I had some stuff upside down so I've thrown it away in disgust :(

    The output looks unusual too. It looks to me that the diode on the output must be marked back to front. Check out the capacitor orientation, the zener orientation, and the +/- marking of the output.

    Then again, maybe I need more sleep.

    edit: or both transistors. (That was the part I had upside-down) The opto may not have been involved.
     
    Last edited: May 12, 2010
  20. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    Mein gott, it is an SMPS... a really cheap, crappy flyback at that.

    See attached, that would explain why there was no OCP, becuase there was no controller. This is an extremely low cost way to make a charger. Please do regard my post a couple of posts back about shorting the output.

    I would replace Q1, Q2, R2, R3 and R4. The optocoupler may have bit it, too. Though I can see why they would think the outputs would never be shorted toghether, if it were wired into a connecter... how did you manage to short the output, anyhow?

    Edit: And yeah, Steve, it seems like they got L/N and +/- backwards, eh? I just drew the schematic as though it would make sense... it could be the 180 flip he did w/ the photo that's messing me up.
     

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    Last edited: May 12, 2010
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