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Sony Ericsson CST-60 Charger

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

  1. Tesla

    Tesla

    165
    2
    May 10, 2010
    Truely amazing work Mitchejk !!!

    Could this be Q1? It is NPN and rated for high voltage.
    http://www.datasheetcatalog.org/datasheet/fairchild/KSE13001.pdf

    On the proprietary Ericsson connector the 2 bare pins stick out and are fully exposed. It would be quite easy to short them together. What's funny is that the marketing docs say the charger has "protection against short circuits".

    I'm still puzzled by the lack of a bridge rectifier as the main AC hits the board. The AC plug isn't keyed, so it can actually be plugged in either way.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Yes, whatever is shorting the terminals is protected by the power supply which fails promptly.
     
  3. TimeManx

    TimeManx

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    0
    May 11, 2010
    R1's supposed to be 0.5 Ω but the DMMs reading 1.5 Ω.
    Similar with R2. Should be 10 Ω but the DMM says it's open.
    R3's clearly burnt.
    R4 is reading fine (200 kΩ).

    Nope, no continuity in any way.

    How do I check the optocoupler?

    Well actually the wire got pulled so hard that it broke the solder from the connectors and consequently shorted.
    My phone was even attached to the charger. It immediately switched off at that moment.
     
  4. Mitchekj

    Mitchekj

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    0
    Jan 24, 2010
    Edit: Couple caveats, I was drinking last night. ;)

    The component designators may be wrong, I numbered the ones I couldn't see from top to bottom, just to help keep things together in my head.
    That schematic may be off as far as the transistors go, I couldn't find the exact ones, so I made guesses as to the pinouts based on what looked like it would work as a switcher. I do know that Q1 is HV rated, and Q2 is small signal. BCE/ECB etc, I'm not certain.
    Oh yeah, keep meaning to add: the output diode may be dead, too... if he saw too much current during the short. (The one closest to the xfmr.)

    You'd test an optocoupler just like any other semiconducter... forward/reverse checks. The secondary side should act like a diode, and the primary side should act like a transistor's collecter-emitter. Any shorts/opens are no good.

    How'd they get this thing to pass UL/CSA without a fuse or any kind of OCP? Rely on resistors to smoke to stop a fire? Wow.
     
    Last edited: May 12, 2010
  5. Tesla

    Tesla

    165
    2
    May 10, 2010
  6. TimeManx

    TimeManx

    31
    0
    May 11, 2010
    That diode is fine.

    Seems to me that the optocoupler is fine. One side reads like a normal diode & the other side reads slight different resistances on changing polarities. I presume that the side facing the input is the primary side.


    BTW I think I won't be able to find the 13001 6D331 transistor too easily in the stores. I found a 13001 HJ P1 ETA transistor in an old non-working SE charger I had lying around. I wonder if that can be used.
     
  7. Tesla

    Tesla

    165
    2
    May 10, 2010
    Transistors

    So on the C945 ... notice how the C945 is printed at the top (maybe also larger, bolder ... more predominately) ... that's how you know it's a C945 and you should ignore the other alphanumerics because they are just lot codes (and can confuse the real name when in comes to cross-referencing).

    Now, on the bad Q1 ... in the same fashion ... 13001 is the name?

    If so, and the other transistor you found in the old board is printed the same way (with 13001 as the real name) I would say it's probably the same (but who really knows). You might also notice if it has one leg connected to the hot AC. After it's removed, you might want to check it with an ohm meter (to see if it reads better than the old one).
     
  8. TimeManx

    TimeManx

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    May 11, 2010
    OK, & what I said about the optocoupler. Is that OK?
     
  9. Mitchekj

    Mitchekj

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    Jan 24, 2010
    Usually a failed optoisolator will read as a short... so you're probably good there. And yeah, side facing the AC input is the primary (photoreceiver) side in this case.
     
  10. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    Looking at the circuit more, it looks like R3 was supposed to act as part of an OCP scheme... R2 is the primary current sense, and if the voltage developed across R2 is above a certain threshold (too much current through the primary and switch Q1, like say a shorted output) it should trigger Q2 to shut the switcher down until the fault is cleared. Problem is, if you burn up R3, well... now there's no way for the OCP to work anymore, so you start smoking things like Q1.

    The opto is doing the same thing, but it's an OVP type setup... if you have any current flow through the zener (which means your voltage output is higher than the zener voltage) the resistor below the zener will start dropping voltage, which biases the opto's photodiode, which turns Q2 on by applying the voltage from the aux coil via D3 to Q2's base... turning the switcher off, etc.

    All theory at this point though. I don't have the PCB to scope around on. :)
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Just looking at the board again, it seems that there's a resistor directly across the output of the opto. Your diagram shows something different.

    It is possible I'm not lining things up correctly, so you may be right :)
     
  12. Mitchekj

    Mitchekj

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    Jan 24, 2010
    Nope, you're right... I've messed up that section. Told ya I was drinking. lol.

    Attached is my second try, hopefully my eyes are working a bit better now. :)
     

    Attached Files:

    Last edited: May 13, 2010
  13. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    Still looks like replacing Q1, Q2, R3, and R2 may get this thing working again... theoretically. :)

    I re-drew the output clamp, too... based on the PCB pics and fresh eyes. Now that section makes no sense to me, but I've gone over it a few times and the schematic I drew appears correct. (Aside from the +/- silkscreen seeming to be backwards?) The cap even supports what I've drawn, stripe toward the negative side, marked as "+" on the silk screen. Flumoxed.

    As to there not being a bridge on the input, it can be done. D1 is still "rectifying" the input, though you'd be missing half of it. I'd hate to see the power factor on this thing (I'm thinking something like 0.2 - 0.3) A large enough cap (C1) should fill in the gaps when the transformer demands current. Out of curiosity, what are the values for C1? It's gotta be huge (like 2200uF/400V,) but the size doesn't look all that large, which leads me to believe they're using something like 22-47uF. But if it works, it works, I guess... I'm no expert, after all.

    Edit: Also, R1 has me stumped... I can't think of any reason it'd be there other than to lower the efficiency. Maybe it's some kind of feeble attempt at EMI suppression? This thing must be one hell of an EMI radiator, by the bye.
     
    Last edited: May 13, 2010
  14. TimeManx

    TimeManx

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    May 11, 2010
    It's just 2.2uF/400V. This is crappy electronics, isn't it.

    I'll be replacing the components today. I'm hoping it'll start working again.
     
  15. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    R1 must be a fuse resistor.
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
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    Jan 21, 2010
    Either that or it limits the switch on current to 220A :)

    That may be the fabled "protection against short circuits", however it's clearly not the first thing to blow.

    Edit:

    It's built to a price.

    That 2.2uF capacitor seems about OK for up to around 300mA, assuming the efficiency is around 50%. (That assumes the capacitor provides the power for 1 1/2 half-cycles before it starts to get charged again).
     
    Last edited: May 13, 2010
  17. Tesla

    Tesla

    165
    2
    May 10, 2010
    So D1 must be a Schottky diode? By rectifying just one leg of the AC voltage you loose the isolation effect required to make AC and it becomes high voltage DC? Or, does D2 work with D1 to essentially make a complete bridge rectifier?

    So this isn’t a SMPS is it? It’s just a linear PS with a rudimentary feedback protection circuit. Still, it looks familiar … like the “hot ground” on the left side of the transformer. If there was a chassis and chassis ground, it would only apply the left side of the transformer (low voltage side), right?

    I had previously said that ZD1 on the output might be to keep the battery from discharging into the PS. But now that I’ve seen the board and schematic … it appears to be a protection diode. In case of a major malfunction in the PS, it should prevent excessive voltage from reaching the phone (even if it has the sacrifice itself).
     
  18. Tesla

    Tesla

    165
    2
    May 10, 2010
    I thought they were marked differently somehow...

    When we started this, I thought for sure the pico-fuse (maybe axial) was blown, but we could never find it.

    The case does have an UL mark on it ... strange a AC device would get that without a proper fuse.
     
  19. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    D1 is probably a standard 1N4007 or similar. It's just half-wave rectifying the AC in... with a large enough cap you will end up with a usable HV DC (HV bulk, Vrec, whatever you wish to call it.) Though your power factor is going to be horrendous since the input current waveform will have huge spikes each cycle, worse than using a full bridge. D2 is part of a snubber to protect the switch (Q1), nothing to do with the AC input.

    It's definitely looking like an SMPS to me. They've just managed to use discretes instead of an IC to control it. (All in the name of saving those few cents!) The RC on the base of the switcher is being fed by the aux coil (once it starts up, thanks to the 1M pull up) which is turning the transistor on/off at some unknown frequency (probably 60-100kHz I'd guess.) The aux coil outputs the same waveform as the primary, and secondary (which would be 180 out of phase), It's chopped up DC, just at a different voltage. The primary is being connected/disconnect from ground each cycle, so it can do it's inductorly duties and get the flyback in the secondary going. But hell, I may be comepletely wrong about how that whole section works... I've only had experience with supplies with an actual controller IC.

    The fabled "short circuit protection" is supposed to be Q2, sensing the current through R2 (and hence the primary) which is supposed to stop the switching if there's an over current problem. But like I said before, if an over current condition will smoke R3, well now you have no way of stopping the switch before it detonates.

    Q2 looks like it may even be some type of duty cycle control for Q1, which would make sense, so that you can control the output. Who knows. :)

    Still no idea how this passed UL. :)

    Edit: Oh yeah, I never did ask... what is the output voltage/current? If it's low enough, having a horrible power factor may not be an issue. (So they're using a 2.2uF bulk cap? the PF might not be all that bad to begin with.)
     
    Last edited: May 13, 2010
  20. Tesla

    Tesla

    165
    2
    May 10, 2010
    it's marked ...

    Input:100-240 VAC ~ 100mA
    Output: 4.9 VDC ~ 450mA

    BTW, thanks for the circuit explaination... it was very informative.
     
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