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How To: DC +/- Power Supply?

Discussion in 'Electronic Basics' started by Jeremy, Nov 8, 2004.

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  1. Jeremy

    Jeremy Guest


    Please excuse my stupidity here; I've checked the web, but can't seem
    to grasp the concepts.

    I'm working on a project that says it requires +/- 15volts DC to
    operate. It has a (common), (+), and (-) in.

    The 15 volt power supply that I have is the standard wall-wart style,
    and only shows a (+) and (-) out.

    How can I create the necessary supply for this device? I read
    something about needing to use a pair of voltage regulators, and other
    misc. parts to build a "rail splitter." Is this applicable in this
    situation of am I overcomplicating things?

    Any assistance in helping me understand what this means would be
    greatly appreciated! Thanks in advance,

  2. BobGardner

    BobGardner Guest

    Opamps usualy run off +15 and -15. You need 2 power supplies... both +15.....
    to hook one up as -15, hook the + to the gnd of the other one. Or just buy a
    +-15V supply.... check jameco.....
  3. CFoley1064

    CFoley1064 Guest

    Subject: How To: DC +/- Power Supply?
    Hi, Jeremy. Good newbie question.

    You're stuck with 15 volts potential between the two wires coming out of your
    wall wart, and a +/-15V supply has 30V potential between the + and -.

    One possible solution is a "rail splitter", which sets an artificial common
    halfway between the two. As long as you didn't require very much current, you
    might try this with two low ohm resistors splitting the difference like this
    (view in fixed font or M$ Notepad):

    +o---o-------o +7.5V
    | |22 ohms 3 Watt
    | |
    15V |
    | |22 ohms 3 Watt
    | |
    -o---o-------o -7.5V
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    Of course this wastes a lot of power just to get a low output impedance for the
    artificial GND. A better solution would be something like this:

    | |
    .-. .----|----.
    | |10K | | |
    | | | |\| |
    '-' '--|-\ |
    15V | | >---o-----oCOM
    | |/|
    .-. |
    | |10K |
    | | |
    '-' |
    | |
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    The op amp creates a low impedance for the virtual GND at the expense of 1 IC
    and a couple of resistors.

    These are the two basic ways to do a "rail splitter". But neither will get you
    where you want to go, which is a +/- 15V supply. In order to do that, you can
    do a bit of a trick with an oscillator and a few caps and diodes like this:

    +15V +15V +15V
    | + +
    .-. | |
    | |24K .---o----o---. C = 100uF 25V
    | | | 8 4 | D = 1N4002
    '-' | | C
    | | |
    o-----o7 | +|| D -14V
    | | 3o---||--o----|<--o----o
    .-. | | || | |
    | |56K | 555 | V C ---
    | | .--o6 | D - ---
    '-' | | | | +|
    | | | | | |
    o--o--o2 | === ===
    | | | GND GND
    --- | 1 5 |
    --- '---o----o---'
    | 3.3nF |
    === ===
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    This is called a voltage inverter, and isn't as tricky as it looks. Note that
    a cap blocks DC, and lets AC through. The output of the 555 is AC coupled by
    the first 100uF cap. The AC is clamped by the first diode so it doesn't exceed
    0.6V or so. This 15V peak-to-peak signal, with to bound at +0.6V, is then
    rectified by the second diode, and filtered by the second cap. It will provide
    a -15V, less the voltage drop across the two diodes. This will give you a
    -14VDC, which should be good for your op amps. Actually, this supply is
    unregulated, and will have greater ripple as you increase the current draw. I
    would limit the current from the negative supply to less than 20 mA with this

    If you need more current, you'll have to go with a more complicated or
    expensive setup. One thing you can do is buy a DC to DC converter which will
    efficiently provide you more current. Or, you can just chalk it up to
    experience and get the right power supply from Jameco, as another post has

    If you need more help, feel free to post again.

    Good luck
  4. I'm working on a project that says it requires +/- 15volts DC to
    If you open the wall-wart you will probably find a transformer, a
    rectifier, and a capacitor, which produces the 15 Volt DC output.

    You can use the transformer to get plus minus 15 Volt with a suitable
    rectifier stage and two capacitors.

    If you add regulators you can get plus minus 12 Volt, which is good for
    most op-amps and other projects.

    uits/diode_appl.html> Scroll down to 70 percent of the web page. Look at
    "Doubler - Version 2". That is a way to get plus minus outputs from a
    single winding.

    Building your own power supply is a good first project in electronics,
    and you will have use for it for many years.
  5. rwiehler

    rwiehler Guest

    look in any commercial power supply catalogue and you will see
    they are classified as single, dual or even triple voltage output.
    what you want is a dual voltage supply.
  6. A really cheap and simple way to get a dual supply:
    Get two 15 Volt wall-warts, connect the minus output of one to the plus
    output of the other and call that connection Ground.

    Now you have a dual power supply, which has three connections, Ground,
    plus 15 and minus 15 Volt.

    This dual supply is now unregulated.

    Add regulators to get a regulated dual power supply.
    The regulators need a few Volt to work, so you can now have +-12 Volt, if
    you used 12 Volt regulators.
  7. If the two wallwarts are differently sized, if one is a 1 Amp and the
    other is a 300 mAmp, for example, put the heavier one as the positive
    supply, because you will need 0-15 Volt for some parts of your circuits,
    and +- 15 (12) Volt for other parts of your circuits.

    You can use the unregulated 15 Volt for some purposes, where power is
    more important than smothness, and the regulated +-12Volt for other parts
    of your circuits, like powering op-amps.
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