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positve and negative power supply

Discussion in 'Electronic Design' started by Sean Bartholomew, Nov 25, 2004.

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  1. im building a circuit using the 4052 cmos chip.
    i need to connect the 4052 Vee to at least as negative as -4V for
    audio reasons. it cant be at 0 for my purposes.
    i would like the unit to fit into the 9V supply format (widely used in
    the guitar fx world) but ive pursuaded myself to go to a 12V supply as
    i need a regulated +5V supply (not applicable with a +9V trans if i
    intend to have -4V unless i get a REGULATED 9V

    how do i get a negative voltage without floating the ground (virtual

    what i had in mind was to take a 12V in. regulate it to 9V and 4V with
    a 78L09 and a 317T respectively.
    assign the +4 to ground (virtual). this would bring the +9V down to
    +5V (connected to all the components) and the 0V would then become -4V
    which i would assign to the Vee pin of the 4052. i may even give the
    4052 power pin a +8V feed (from the +12V on the transformer directly)
    for more audio headroom but the fact that its unregulated may not be a
    satisfactory solution and +5V is fine.

    this would mean that i cannot connect the sleeves of the outside audio
    connections to internal ground as it really is +4V with respect to the
    outside ground world but i COULD make sure that they at least all
    connect together albeit isolated from the internal circuitry.

    my trusted friend however does not recommend in the slightest that i
    have this virtual ground setup.

    any ideas?
  2. Tam/WB2TT

    Tam/WB2TT Guest

    This is easy if you have access to the transformer. Connect one side of the
    transformer secondary through a series capacitor to a clamp diode. Then half
    wave rectify the negative clamped voltage. Obviously, the voltage will
    depend on the transformer.

  3. john jardine

    john jardine Guest

    The idea can work perfectly OK in practice. What you ideally want though is
    that the incoming 12V supply is 'floating' (like from a wall-wart). Then
    your (+4V) assigned ground can actually be tied to a true ground point such
    as the metalwork and incoming signal connectors.
    Other than that it is sometimes as easy just to take in a 9V supply.
    Regulate it to give +5V and 0V and then use '7660' chip and 2 capacitors to
    electronically invert some of the +5V rail to give a -5V rail. Doesn't
    matter then if the incoming 9V is floating or not.
  4. ok im liking this. both possibilities intrigue me,
    1st, ur saying that BECAUSE its a wall wart transformer, it IS
    floating. i didnt think of that. there is no direct connection to true
    ground. nice. so i assign anything i want to ground and connecting
    that to outside ground from other devices is fine. ok. ill definitely
    try that.

    2nd, this 7660 chip INVERTS power? interesting. i was looking at the
    MAXIM chip that delivers +-9V from a single 9V supply but all audio
    experts inform me that its too noisy for audio usage.
    is this 7660 chip noisy? or would it introduce noise in the 5V reg
  5. john jardine

    john jardine Guest


    Yes, like the Maxim chip, the 7660 can be a bit noisy. Though a lot depends
    on how much care is taken with cleaning the supply rails.
    For simplicity it's easier to go the linear route.
  6. also a thought just occured to me.
    theoretically, if i build TWO regulated 5V supplies from a single 9V
    transformer supply, i should be able to connect the positive of one to
    the negative (0) of the other treating THAT as ground ref. then id
    have +5V, 0 and -5V.

    but wouldnt that:
    A- simply just short out the ground and the +5V of one of the supplies
    since both their grounds are connected?
    B- theoretically give me 10V diff total WITHOUT any reclocking from a
    9V supply?
  7. john jardine

    john jardine Guest

    Yes. They can be stacked like that but you'll then need at least a 12V-14V
    input as the voltages are now sitting on top of each other.

    Unfortunately an awkward problem pops up when stacking positive voltage
    They can only source current and not sink it. That opamp powered from across
    the outer +/-5V rails is fine. The "0V" centre rail is fine.
    Trouble comes when the opamp drives to drive a load connected to the "0"V

    When the ouput swing goes below 0V then it's no problem, as current is taken
    OUT of the mid rail regulator and passes via the opamp back down the real 0v
    But ... when the opamp tries to swing a positive output voltage to its load,
    the top regulator is quite happy to feed current via the opamp into the load
    but this current then can't go back via the pretend 0V rail into the mid
    voltage regulator output pin. Essentially the circuit can only run as a half
    wave rectifier.
    The inefficient work-around is to stick a resistor across the middle rail
    and the bottom rail which constantly passes a current that's a tad more than
    the opamp will be liable to ask for when driving its load at maximum. (say
    20ma's worth).
  8. hmmm sounds like i should stick to my original plan then what with the
    half wave of stacking 2 5V and the noise of the 7660.
    so 9V and 4V here i come. id totally use 10V and 5V but then ill have
    to have a supply greater than 12V.
  9. Tam/WB2TT

    Tam/WB2TT Guest

    You guys are still doing it the hard way. If you want +5 and about -5 V from
    a 12 V floating supply, connect a 4.7 V zener diode in series with the
    negative supply lead, and connect the other end to "ground ". Now connect a
    series regulator between ground and the positive supply lead. The original
    negative supply lead will be -4.7V. This will work so long as the positive
    supply current is more than the negative supply current.

  10. Don Pearce

    Don Pearce Guest

    The way you describe this, it is impossible for the positive and
    negative currents to be different, because they are in series with
    each other. Also, with a voltage regulator from + to ground, and a
    zener from - to ground, there is nothing to limit the zener current,
    and it will go pop.

    Pearce Consulting
  11. <silly thinking>
    Perhaps you could use the audio signal, amplify it, ac coupled,
    schottkey rectifier, and charge a supercap to get your negative
    supply. It may sound a bit distorted when all is new, but it
    will get better after a minute or two and stay that way ;)
    After all, Vee does not draw much current. If it is just for hobby
    and personal use, I'd might even drop in a lithium button cell
    or two.
    </silly thinking>
  12. thanks for all the help guys. i love this forum.
  13. Ken Smith

    Ken Smith Guest

    The Intersil 7660 blows up too easily. I suggest the Linear equivelent.
    You can get it from digikey.

    With both these parts, there cab be a problem if your circuit pulls the
    output above ground. Adding a small MOSFET as a common gate stage can
    protect it from this.
    These chips draw their current in spikes. If you RC decouple it from the
    +5V, there shouldn't be much trouble with that. You can also add a filter
    on the -5V produced.
  14. Ken Smith

    Ken Smith Guest

    I think he means this:

    -----! LM7805 !-------------- +5V
    +----------------- GND
    /-/ 4.7V
    --------+----------------- -5V (almost)

    The circuit works if the +5V's load is greater than (the -5's load minus
  15. Tam/WB2TT

    Tam/WB2TT Guest

    The zener current (plus the negative load current) will be exactly equal to
    the positive load current. So, if the positive load current is 250 ma, and
    the negative load is 5 ma, the zener current will be 245 ma.

  16. Fred Bloggs

    Fred Bloggs Guest

    View in a fixed-width font such as Courier.
  17. Rich Grise

    Rich Grise Guest

    No it won't. The Zener diode will pass the quiescent current of the
    regulator when there's no load. You just stick it in the '-' lead and pick
    up 0V at their junction.

    But it is true, that for the circuit to work, the current through the
    "positive half" of the load has to be greater than the "negative half" by
    a value which is inside the operating range of the Zener current.

    A possibly more effective/efficient way to make a virtual ground is with
    an opamp follower driven from that voltage divider someone has mentioned.

  18. Fred Bloggs

    Fred Bloggs Guest

    Well- if you have 9V regulated then you only need one standard
    regulator. If Vee of the 4052 is your only load, then this current is
    all leakage- you keep the 79L05 in regulation with a small 500uA current
    through the 10K which guarantees the 5V loading always exceeds the -4V
    current draw.
    View in a fixed-width font such as Courier.
  19. YD

    YD Guest

    If your wallwart supplies AC rather that DC you can do it like this:

    -------+---|<---+------| ----- +5V
    | | '-----'
    From | | |
    Xfmr | --- ===
    | --- GND
    --+ | | 100uF
    | | |
    === | ===
    GND | GND
    | 7905
    | .-----.
    ---->|---+------| |---- -5V
    | '-----'
    | |
    --- ===
    --- GND
    | 100uF

    created by Andy´s ASCII-Circuit v1.22.310103 Beta

    Of course, you can always open the wallwart and remove any rectifiers
    and filters inside it. If you don't expect to draw more than 100 mA
    use 7xL05's.

    - YD.
  20. Ken Smith

    Ken Smith Guest

    Never use 7XLXX regulators. If you are going to the bother of wiring up 3
    pins use a 7XXX regulator. They don't cost enough more to worry about,
    they are harder to break and the heatsink hole makes them easier to mount.
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