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Need a negative voltage

Discussion in 'Electronic Basics' started by [email protected], Oct 4, 2005.

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  1. Guest

    In an attempt to repair a presumably faulty preamp I realised the
    construction deals with two voltages; +14 and -14 (as well as ground).
    I don't have access to the transformer unit, so thought I could
    construct my own.

    Am I correct in my assumption that what I need is a 28 volt power
    source, and to connect two resistors in series across the two leads,
    and I'll end up with +14 volts on the positive lead, -14 on the
    negative, and a 0 volt "ground" where the two resistors join?
    Kinda like this (hoping the ASCII-schematic looks like it's meant to)..

    --------- +14
    |
    __|__
    | |
    | R1 |
    |____|
    |
    |---- 0/"ground"
    __|__
    | |
    | R2 |
    |____|
    |
    |
    --------- -14

    So how do I calculate the resistors? I need less than 1 A (probably
    much less as 1 A is what the fuses in the amplifier are rated at).

    Cheers!
     
  2. Probably not a good idea.

    To get the zero volts, you'd need R1=R2. The thevenin equivalent
    would then be a 0V source through an R of (R1)/2. To keep the
    variation of your "ground" to, let's say, 1/2V about ground at up to
    one amp you'd need to have (R1)/2*(1A) = .5V. This means R1 of 1 ohm.

    So, R1 and R2 would each need to be at least able to handle 200 watts.
    Just in case, make them 1 kW resistors. Your 28V supply should need
    to support about 15 amps. About 500 watts worth. This would then
    supply your up-to-1A pre-amplifier, assuming it could tolerate the .5V
    variations in the ground.

    You will probably want to avoid that solution, though.

    Jon
     
  3. PeteS

    PeteS Guest

    Using a resistor pair is not a particularly good idea for any
    significant current (significant being more than a few milliamps)

    A better solution might be to use about 36V in total across a centre
    tapped transformer. i.e. each side relative to the centre tap has 18V
    above the ripple after rectification and *under load*. Use the centre
    tap as your local common (ground).

    Then use Linear regulators (if you are less than 1A and you stay close
    to the drop out - about 2.5V each - then the heat sinking won't be too
    bad).

    I know the drop out is a little highly spec'd here, but that's standard
    for the old style 78xx beasts (and various others). A low dropout
    regulator has it's own very special problems that will bite you in the
    behind.

    Cheers

    PeteS
     
  4. Bob Monsen

    Bob Monsen Guest

    Using your idea, you'll end up with something that looks like this:

    ----o-----------o------ +14
    | |
    __|__ __|__
    | | | |
    | R1 | | L1 |
    |____| |____|
    | |
    o-----------o------ 'gnd'
    __|__ __|__
    | | | |
    | R2 | | L2 |
    |____| |____|
    | |
    | |
    ----o-----------o---- -14


    Where L1 and L2 are indeterminate impedances. GND will only be at 0V
    in your diagram when L1=L2, which you probably can't guarantee.

    If you use very low value resistors for R1 and R2, then you can be
    sure that those values will cause L1 and L2 to be large in comparison.

    However, a better way to go would be to use a couple of transistors,
    like this:

    ----o---------o-----o------ +14
    | | |
    __|__ | __|__
    | | |/ | |
    | R1 | .-| NPN | L1 |
    |____| | |> |____|
    | | | |
    o-----o o-----o------ 'gnd'
    __|__ | | __|__
    | | | | | |
    | R2 | | |> | L2 |
    |____| '-| PNP |____|
    | |\ |
    | | |
    ----o---------o-----o---- -14


    If you do this, it'll keep GND pretty much at 0V, even if the
    impedances of L1 and L2 are not equal.

    Note that the current through R1 and R2 should be about 1/10 of what
    the maximum current through either transistor is. This will depend on
    how different L1 and L2 are. If L1 is not there, for example (an
    infinite impedance) and L2 is 14 ohms, then it'll draw 1A through
    the NPN transistor, and no current through the PNP transistor. So,
    you'll need about 100mA through R1/R2, meaning R1 and R2 should be
    140 ohm 2W resistors. The Transistors should be beefy, since they may be
    dissipating 14W. Big TO-3 cases and heatsinks are probably a good idea.

    If you use an opamp to drive the transistors, with feedback from the
    virtual ground, it'll be far more precise. Something like this:

    ----o------------------------------o------o------ +14
    | | |
    __|__ | __|__
    | | |/ | |
    | R1 | .-----| NPN | L1 |
    |____| | | |> |____|
    | |\ | | |
    o--------+| \ | o------o-----------.
    __|__ | \ | | __|__ |
    | | | >--------o | | | |
    | R2 | .-o| / | |> | L2 | |
    |____| | | / '-----| PNP |____| |
    | | |/ |\ | |
    | | | | |
    ----o------)-----------------------o------o---- -14 |
    | Rf |
    '---------/\/\/\/--------------------------'

    With this circuit, your virtual ground would be far more accurate, and
    the resistors R1 and R2 can be large, like 100k. Make R1 = R2, and Rf
    = R1/2. That way, your opamp inputs will see similar impedance.

    An even better way would be to just generate the right voltages to
    begin with. You can get 'center tapped' transformers, which provide AC
    voltages which are mirror images of each other, and a ground. If you
    use one of these, you can generate +-14V fairly easily.

    For example, assuming a 1A current draw on either or both, you'll need
    a 28VAC transformer that is rated for something like 35 VA. Using
    this, use the following circuit:

    .---------.
    | o-----|>|----o--------- 19 max
    | | |
    | | --- + C
    | | ---
    | | |
    | | |
    ---o + | |
    | | |
    | o------------o--------- gnd
    ---o - | |
    | | |
    | | --- + C
    | | ---
    | | |
    | | |
    | | |
    | o----|<|-----o--------- -19 max
    '---------'

    The unregulated output goes into linear regulators, maybe LM317 and
    LM337 linear regulators, suitably configured. (see the datasheet at
    www.national.com)

    These regulators require the supply to be about 2V above or below,
    respectively. Thus, the supply must stay above 16V. That means that
    the droop across the smoothing capacitor must be less than 3V in 1/60
    of a second. So, using the famous formula:

    1A = C * 3*60

    we compute that each C needs to be at least 5600uF, and rated for at
    least 25V. Using a transformer that puts out more volts will allow a
    smaller C, at the cost of more dissipation in the regulator. You are
    already looking at an average power of 3.5W, which is pushing an LM317
    unless it is heatsinked.

    This supply will waste far less energy than the one above, because it
    will not have to dissipate energy in the case where the positive load
    is more than the negative load.

    You can also buy cheap +- 15V supplies. I've seen them for $10
    mailorder from www.mpja.com. Here is one for $15 that might work for
    you:

    http://www.mpja.com/productview.asp?product=15695+PS
     
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