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How do I convert 0-10V to -5V +5V differential?

Discussion in 'Electronic Basics' started by [email protected], Aug 31, 2005.

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  1. Guest


    I have a signal generator that is putting out a staircase wave 0-10V. I
    want this wave to drive a laser galvanometer, but the input is
    differential (-5V to +5V) how would I go about converting the signal
    from one format to another.

    Frequency is low (30-40Hz)

  2. BobG

    BobG Guest

    Need to arithmetically add -5V. Try a couple of Rs and a pot to -10V
  3. Guest

    I'm a complete amatuer

    - so I think what you are saying is that I need to provide a -10V
    supply and combine the signals somehow?

    If someone could do a quick ascii sketch of the layout, that would

  4. John Fields

    John Fields Guest

    View with a non-proportional font like Courier:

    | | >--+-->Vout
    [R2] +-----|-/ |
    | | | |
    GND +--[R3]-----+
    | |
    | -V

    R1 = R2
    R3 = R4
  5. Chris

    Chris Guest

    Hi, Steve. First off, many signal generators have a dial that can
    control the DC level of an AC signal. It's usually labelled "DC" or
    "DC Offset". See if you have that adjustment on your signal generator.
    If you do, your problem is solved. Just tweak the dial to get the
    appropriate DC offset. If you don't have that adjustment, you might
    want to scrounge around and find a signal generator that does -- most
    do, and this is the easiest and best solution.

    If your signal generator is homebrew, or you don't have a DC Offset
    adjustment available, and the input to your galvo amp is high
    impedance, you might be able to get away with just a DC blocking
    capacitor like this (view in fixed font or M$ Notepad):

    ` C1 .-----------.
    ` +|| | |
    ` 0 - 10V .-----------||-------o-------o---oIn |
    ` 30 Hz | || | | | |
    ` / \ 100uF | .-. | Galvo |
    ` (A C) /-/ 10K| | | Amp |
    ` \_/ 1N4733 ^ | | | |
    ` | Vz=5.1V | '-' | |
    ` | | | | |
    ` | V | | |
    ` | /-/ | | |
    ` | | | | |
    ` | | | | |
    ` | | | | |
    ` '--------------------o-------o---oGND |
    ` | |
    ` '-----------'
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    This may not work for you if the galvo input isn't high impedance, or
    if you need the laser to remain in one position. There will also be
    some minor distortion of the input voltage, even once the DC voltage
    stabilizes. But it's easy, the components are available at Radio Shack
    or any other hobbyist source, and it's all passive components -- no
    extra power supply. Just one cap, two zener diodes and a resistor.
    Note that some galvo amp inputs require a DC path to GND, so don't
    forget the 10K resistor, or the galvo might drift away.

    If you want instantaneous, precise translation of your 0 to 10VDC
    staircase into a -5V to 5V staircase, which will work from DC to audio
    frequency, you'll need an op amp and split supplies (+/-15V, for
    instance). Since the galvo uses these, they're frequently available on
    the terminal block or external connector. Just RTFM (Read The Fine
    Manual). You would use the op amp (it really doeesn't matter which one
    -- an LM741 available at Radio Shack would work just fine here) and a
    handful of resistors to make what's called a difference amplifier.
    This just subtracts 5V from your input voltage. It would look
    something like this (view in fixed font or M$ Notepad):
    ` ___
    ` .---|___|---.
    ` | 22K |
    ` | |
    ` 22K | +15V |
    ` ___ | |\| |
    ` 5VDC o---|___|-o---|-\ | Output
    ` ___ | >----o---o
    ` 0 to 10Vo---|___|-o---|+/ -5V to 5V
    ` 22K | |/|
    ` | -15V
    ` .-.
    ` 22K| |
    ` | |
    ` '-'
    ` |
    ` ===
    ` GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    The circuit works by subtracting the 5V (you can make the 5V by putting
    a pot between the + supply and GND) from your input signal. That's why
    it's called a difference amplifier.

    But judging from your own evaluation of your skill level (complete
    newbie) this might be a bit of a stretch.

    Let us know if any of these solutions look good, or if you need more

    Good luck
  6. Guest

    Thanks for you all your help - I have all the parts to build the op-amp
    solution at home - the only thing that puzzles me a bit is the
    +15V/-15V power to the 741 - how do I create +15 and -15V from a
    standard transformer.

    Is there a book on this somewhere...?


  7. John Fields

    John Fields Guest

    Since the 741 can take an input supply of plus and minus 22 volts,
    max, you can easily get less than that that from a simple,
    unregulated, half wave supply using a 12V transformer, like this:

    | |
    | |
    MAINS>--+ || +--[<1N4001]--+-------|---->-17V
    | || | | |
    P || S | |+
    R || E [C1] [C2]
    I || C |+ |
    | || | | |
    MAINS>--+ || +-------------+-------+---->GND

    Since your output is only going to swing between +5 and -5V, and the
    741 needs about 3V above and below that to let that happen, you've
    got about 9V of headroom to either rail so C1 and C2 can be fairly
    modest, depending on what else you're going to use the supply for.

    What can you tell us about the load?
  8. Chris

    Chris Guest

    Mr. Fields has given the best solution. The OP might want to add a
    couple of protection diodes to keep the inputs in the range of the
    supplies and the caps from being reverse-biased at turn-on, like this
    (view in fixed font or M$ Notepad):

    ` D
    ` o------o-------->|----o-----o------o+14VDC
    ` | +| |
    ` | C --- - D
    ` o------|---. --- ^
    ` | | | |
    ` 10.2VAC | '----------o-----o------o
    ` from | +| |
    ` Wall Wart | C --- - D
    ` | --- ^
    ` | D | |
    ` '--------|<----o-----o------o-14VDC
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    Radio Shack P/N:
    Wall wart:
    9-13VAC/800mA AC-to-AC Adapter $16.99 Model: 273-1631 (set for 10.2VAC)

    470µF 35V 20% Axial-Lead Electrolytic Capacitor $1.39 Catalog #:

    25 Rectifier Diodes (1N400X) $2.59 Catalog #: 276-1653

    Any 9VAC to 12VAC wall wart rated for over 50mA and up to 1 amp will do
    fine, as well as any 1N400X diode and any cap over 470uF, rated at
    25VDC or above. The OP should be able to scrounge the components he
    needs from his junkbox.

    However, the OP really should look at the docs on his galvo amp again.
    Nearly all of the ones I'm familiar with have the dual supplies
    accessible from a terminal block or connector somewhere on the chassis.

    Good luck
  9. John Fields

    John Fields Guest

    Since we have current being steered into either one cap or the other
    by the main rectifying diodes, I don't see how either of them could
    become reverse-biased, since there's no available path.

    Likewise, I don't understand what you mean by "to keep the inputs in
    the range of the supplies."
  10. Chris

    Chris Guest

    Hi, Mr. Fields. It's possible the two extra diodes are "gilding the
    lilly", but possibly not.

    First, the whole circuit without the extra diodes:
    ` ___
    ` .--|___|--.
    ` ph A D | |
    ` o-----o----->|---o-----------------------. |
    ` | +| ___ | |\| |
    ` | C1--- in(-)o-|___|-o--|-\ |
    ` o-----|---. --- ___ | >---o---o
    ` ph B | | | in(+)o-|___|-o--|+/
    ` | '------o---. | |/|
    ` | +| | .-. |
    ` | C2--- === | | |
    ` | --- GND | | |
    ` | D | '-' |
    ` '-----|<---o---------. | |
    ` | === |
    ` | GND |
    ` '-------------'
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    Let's assume that, at turn-on, ph A is more positive than ph B. That
    means C1 will start charging up in a positive direction. Several mA of
    current will start to flow through the op amp Vcc pin to the Vee pin.
    For that half cycle, the - of C2 will be more positive than GND. As ph
    B goes negative with respect to ph A, C2 will start to charge
    negatively with respect to GND, relieving the problem.

    Since there are 22K resistors between the inputs and GND, the
    possibility of latchup on turn-on is not there for an LM741. It
    becomes more of a hazard if one of the op amp inputs is grounded. And
    it becomes more of a hazard as the DC current draw of the circuit
    increases. Obviously the several mA of one op amp won't charge up the
    470uF cap to a big voltage in 1/2 line cycle. But for some op amps or
    4000-series logic using unregulated dual supplies, or for heavier DC
    loads, this can be a problem.

    The problem is exacerbated when you have two different circuits with
    two different power supplies. If the signal generator turns on when
    this one is still off, this phantom reverse charging of the power
    supply can be a more significant failure mode, because it can be
    continuous. The 0-10VDC ramp voltage would cause this phantom reverse
    charging on C2 through the non-inverting input to the Vee pin if the
    signal generator is on before the wall wart is turned on -- that is,
    unless you had the extra diode to keep the cap at less than -0.7V.
    Again, with the series resistors, I don't believe you'd get latchup at
    turn-on with most op amps. _This_ was my main concern, and the reason
    I threw in the extra diodes. Any reverse charging is current limited,
    so the cap would also probably be OK, too.

    The thing is, once old dogs like me learn good tricks (usually the hard
    way), we tend to stick with them. But it's good to get some feedback
    from somebody who's thinking things through. It's probable the two
    extra diodes aren't necessary, if the OP uses an LM741 (and he could be
    using a CMOS op-amp, which would mean the diodes just might be
    necessary -- they sometimes latch up at lower currents). I guess I
    just like to use them on unregulated dual supplies like the above,
    especially if there's more than one power source lurking around. And
    after all, the OP's buying a pack of 25 diodes. That means 23 spares
    -- might as well.

    Thanks for the heads up, and the obviously more efficient solution.

  11. John Fields

    John Fields Guest

    That's interesting. I didn't even see that path!

    Looking at it a little more closely, though, the voltage which will
    be developed across the cap, with an average current of 3mA for half
    a cycle of 60 Hz will be:

    I dt 3E-3A * 8.3E-3s
    E = ------ = ----------------- ~ 0.053V
    C 4.7E-4F

    So that's only 53mV away from ground.
    Possibly, but I believe the Vee terminal would have to go to
    something more positive than a diode drop above ground for the opamp
    to be in danger, even if one of the inputs was tied directly to
    ground. For a 470µF cap, that implies a current of:

    C dv 4.7E-4F * 7E-1V
    I = ------ = ----------------- ~ 0.04A
    dt 8.3E-3s

    That's not an awful lot of current, so the diodes may well be
    warranted. Either that or larger caps... ;)
  12. Chris

    Chris Guest

    John Fields wrote:
    Back in the day, it used to be somewhat common practice to use split
    supplies with a 4016 or 4066 for low level +/- analog input voltages.
    Combined with a CA3130 op amp and some logic ICs, you could do some Q&D
    signal conditioning.

    Thanks for the spot, Mr. Fields. It's good to know we have a few pros
    like yourself, Mr. Holme and Mr. Popelish here among the hoi-polloi,
    keeping us all honest. ;-)

  13. Pi

    Pi Guest

    I don't know how precise it must be? I have drawn an standard opamp
    schematic here, important: use courier as font to read it (or copy it
    to notepad and use Courier there):
    ---| 10k |--
    +5V | ----- |
    | ----- | |\ |
    --| 10k |---| \ |
    ----- |- \ |
    0-10V | \---------- -5 to +5V out
    | ----- | /
    --| 10k |-- |+ /
    ----- | | /
    | |/
    | | 10k
    | |
    | |
    --- gnd, connect gnds of 0-10V, +5V, opamp power supply gnd
    You need a fixed +5V, generate it with a 78L05 or something else.
    Depends on how precise you want to to be. Supply the opamp with +/-
    15V. Any opamp will do, use one that does not need a cap in the
    feedback, or use a small cap across the feedback resistor.

    Feel free to ask any questions, by email: as I do
    not check this newsgroup often and may miss answers.

    Pieter Hoeben
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