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Modifying a wall-wart

Discussion in 'Electronic Design' started by Norm Dresner, Oct 21, 2003.

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  1. Norm Dresner

    Norm Dresner Guest

    I have a Elpac wall-wart which is spec'd as follows:
    5V @ 380 ma regulated
    +-12V @ 180 ma regulated

    and I'd like to modify it to change the 5V output to an (unregulated)
    voltage somewhere around 8V.

    I'm not particularly afraid of opening the thing up and changing the
    circuitry but I was wondering if anyone could give me some idea of what I'm
    likely to find when I do so I can plan what I'm going to do to it.

    As an alternative, I could use a wall-wart or external power supply which
    provides 8V and +-12V instead but I haven't found anything in searching
    through various catalogs. I think that something as high as 9V would
    function too.

  2. Rein Wiehler

    Rein Wiehler Guest

    I would expect a to220 format regulator like 7805 for the 5 volt
    output. Remove and bridge the input to output, maybe you get 8VDC and up
    to 15VDC. hard to predict. could also effect the -12Vdc output depending
    how it is wired.
  3. Uns Lider

    Uns Lider Guest

    For a quick one-off hack how about regulating down the +12V to 9V with
    a 7809? If the +12 on your wallwart isn't beefy enough, you could use
    an off-the-shelf switching power supply with [email protected]>=500mA.

    What's your application, anyway?

    -- uns
  4. Tim Shoppa

    Tim Shoppa Guest

    A small boost converter on the 5V output, or regulating the +12V output down,
    may satisfy your needs without opening it up. How much current do
    you need at 8V?

    Many of the Elpac's are internally switching regulators. They aren't
    particularly cheap, either.
    I would think that starting with an unregulated wall-wart or
    transformer(s) and doing all the regulation yourself would be far
    easier than modifying an existing Wall-wart. Doing the modification pretty
    much nullifies whatever UL/CSA/ EMI advantages you had by using a wall-wart
    in the first place.

  5. Norm Dresner

    Norm Dresner Guest

    I have no specs or documentation on the unit as the manufacturer is no
    longer in business :-( but I know that this input feeds a TO-7805 with a
    minimal heat sink and a TO-220 LM350 (nominal 3A regulator) with only a
    slightly larger one. I can't believe that the heat sink for the LM350 would
    be adequate for even a few watts of dissipation since there's effectively no

    As a first step -- since it's the minimal work path -- I'm going to open the
    wall wart and -- if I can -- jumper around the 5V regulator. If this works,
    great! If not, I'll create a real power supply with the necessary voltages.

  6. Norm,

    This would work with older wall-warts that use a 50/60Hz transformer, a
    bridge and a 7805. But I'm pretty sure the type you mentioned has an SMPS
    inside which is very hard to modify. You would have to reverse-engineer the
    circuit to find out how the output voltage is set and then has to poke
    around in a very conpactly build PCB covered with SMD components. Eventually
    you may find out that it cannot be modified at all. Best solution I can
    imagine is finding a +-12V supply that can provide enough current from the
    +12V to power +8V or +9V via a 7808 respectively 7809.

  7. Uns Lider

    Uns Lider Guest

    It might be instructive to power the thing for a while with an adjustable
    power supply, measure the current it uses, and finger-test the heatsinks
    after it's been running for a while with various voltages.
    It could be the minimal work path, but maybe not. It's high-risk because you
    have to reverse-engineer the wallwart before you can know how much work it's
    actually going to be. I'd go the route of adding a 7809 because I can know
    ahead of time exactly how much work that involves.

    -- uns
  8. Norm Dresner

    Norm Dresner Guest

    To recap, I got a piece of equipment (about 2 years ago on eBay) with what
    has turned out to be the wrong wall wart power supply. The WW outputs +-12V
    @180 mA and +5V @ 380 mA. The two +-12V inputs are used directly but the
    third input feeds a pair of IC regulators, an LM7805 and an LM350. From the
    resistors on the LM350 I believe that the input should be about 8-9V,
    obviously unregulated. I did a quick test and substituted a 9V 350 mA
    unregulated wall wart for the third (+5) input and its voltage was pulled
    down to around 6.25 V by the combined load of the two regulators and the
    circuitry they feed, but the equipment seemed to be closer to working.
    From this, I conclude that it's absolutely impossible to use the +12 @180 mA
    input to feed the third input as well as its own load too.

    I see two -- and only two possibilities -- for using this piece of equipment

    1. Open the wall wart and modify it. If I'm lucky, it'll be a linear
    supply (Note 1) and I could try jumpering the input to the +5 regulator to
    its output and see if that's enough to power the whole equipment. But since
    the LM350 is a 3A variable regulator (and they didn't use an LM317), I
    suspect that this might not work.

    2. (Assuming #1 fails) Construct a completely new power supply. While the
    +-12V outputs of the existing wall wart might be salvaged to feed the
    corresponding inputs, I'm going to need a completely new lower voltage
    supply. I've identified a few candidates and while I'll probably
    test-with-a-variable-supply first, I'm also inclined to just get a 9V
    transformer, bridge, and capacitor instead.

    Approach #1 is the next thing to do and if I find anything interesting
    inside, I'll post a follow-up.

    Thanks to all who posted suggestions and comments.

    Note 1: The wall wart is rated at 13.5W input but the outputs total 6.22 W,
    and since the wall wart is 3.5"x2.5"x2" (17.5, I strongly suspect a
    transformer and linear regulators inside.
  9. Wade Hassler

    Wade Hassler Guest

    If the thing is a switching regulator, try this:
    Since it's mass-produced, most of the parts used will be standard
    values. The divider that samples the output for feedback, however,
    will have at least one resistor whose value is dictated by arithmetic.
    Find the oddball resistor and replace it with a pot to get a new
    value. This has worked for me a couple (all right: exactly two) of
    Wade Hassler
  10. Norm Dresner

    Norm Dresner Guest

    I finally opened up the wall wart and found visible a transformer, a few
    caps, and three TO-220 IC's or transistors. A first guess is that they are
    an LM7812, an LM7912, and an LM7805 and it looks like it will be easy to
    remove the '05 and jumper its input to the output.

    I think that the large physical size of this unit was the key to predicting
    that it was a linear device and not a switcher.

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