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True Zero to 10 vdc PS

Discussion in 'Electronic Design' started by V8TR4, Oct 28, 2003.

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  1. V8TR4

    V8TR4 Guest

    Hello everyone,

    I need to make a Power supply that mimicks a 0-10vdc out that is common on
    industrial equipment outputs like chart recorders and stuff. I need very
    little current, not more then 100ma ever, and 10ma is probably what is the

    I have built one using an LM317 that then outputs to a voltage devider. The
    problem was it doesn't get down to low volatge, much less zero.

    Is there a slicker and not very complicated PS design out there to do what I
    want? I was considering using a 555 as a PWM and then using a 2n2222 to vary
    a voltage source. Any comments or ideas would be much appreicated. I have
    just started playing around with op-amps and it seems at first glance that
    these could be used to do what I want as well. Not sure of the details yet
    but think it can be done.

    This unit is for testing some instruments I have and is not for commercial

  2. cpemma

    cpemma Guest

    One simple way is to regulate to 10.7v or so, put that across a 1k pot, then
    take the wiper to the base of a small high-gain NPN transistor connected as
    an emitter-follower. The final output is input less 0.7 volts, so you'll get
    your zero and a fairly linear scale compared to a potential divider alone.

    But I don't see why your potential divider didn't go to zero, are you sure
    you didn't connect the pot as a variable resistance?

  3. There is a way to make your LM317 go down to 0 Volts. Check the typical
    applications on the data sheet:

    Alternatively, you could use an op-amp configured as a buffer, -5V/+15V
    rails, and a multi-turn pot from GND to 10V regulated voltage, with moving
    pin to the op-amp's [+] input. Like this:

    .. ___
    .. .----|___|----.
    .. +10V | 3k9 |
    .. | | |
    .. | | +15V |
    .. | | |\| |
    .. | '----|-\ |
    .. .-. | >-----+----o 0 - 10V
    .. | |<----+----|+/
    .. 10k| | | |/|
    .. '-' | -5V
    .. | ---
    .. | ---100nF
    .. | |
    .. | |
    .. +------+
    .. |
    .. ===
    .. GND

    You can then choose an op-amp that can deliver the output current you want,
    or use a pass transistor to amplify it.

    Costas Vlachos Email:
    SPAM-TRAPPED: Please remove "-X-" before replying
  4. I suspect the problem was that the 'target' instrument, has a small amount
    of 'pull-up' behaviour. I have seen this with many instrument inputs of this
    type. Assuming that the potential divider has a significant resistance, then
    the output will never go to zero.
    Really, I'd look at having a symmetrical power supply (perhaps +/-15v), and
    using a power op-amp. Something like an LM675, can be driven from a pot
    across a voltage reference, or even a D-A converter, and will give more
    current than needed, and will maintain the output very close to the required
    voltage, into a wide range of different loads.

    Best Wishes
  5. If you use an op-amp in voltage follower configuration, the negative
    input connected to the output, you can steer the output by changing
    the voltage on the positive input, for example with a linear pot.

    If you have +-10Volt power to the op-amp you can put the input pot
    between these power rails and get a +-10Volt output.
    If you want 0- +10Volt you put the pot between +10 and ground.

    Most op-amps can deliver 10mA at the output, but if you need more you
    can use two transistors to buffer the output so it can deliver more

    Something like this:The two transistors emitters connected together
    becomes the output. The negative feedback can be moved from the opamp
    output to the new output.
    But this buffering with transistors will reduce the maximum output
    voltage, it will not reach exactly the full power rail voltage.

    Not all opamps do that either, so you must choose opamp carefully if
    you need to get close to the rails. It is easier to choose power rails
    which are more than the maximum voltage you will need.
    Power the circuit from +-15 if you want +-10 output, for example.

    o---------. | |>
    | | | V |
    | | |\| | - |
    | '--|-\ | | |
    .-. | >|--o--o o-------
    | |<---|+/ | |
    | | |/| V |
    '-' - |
    | | |<
    created by Andy´s ASCII-Circuit v1.21 Beta
  6. budgie

    budgie Guest

    Using the KISS principle, why not just a wirewound pot across a
    regulated 10V source? That'll certainly go all the way from 10V down
    to 0V even if the load has a pull-up or pull-down characteristic.
  7. Finally, a good answer. ;-)

    Maybe put a small resistor in series with the regulated source (12V)
    to limit the current through the wiper if it gets shorted. Measure the
    actual voltage into the instrument under calibration/test with an
    appropriate DVM.

    Best regards,
    Spehro Pefhany
  8. Jeff

    Jeff Guest

    Refernce the pot to negative 1.2V with respect to ground. This works great
    with these cheap regulators. If you have a negative voltage source avalible,
    it's also really easy to do with a 1.2 V voltage refernce!
  9. cpemma

    cpemma Guest

    At the worst case 100mA draw that's a high wattage pot and very poor
    linearity. Much more expensive than a transistor or two and a standard pot.
  10. Yeah, but 100mA is just plain silly if you're at all familiar with
    these things. 1mA *maximum* is more like it.

    Best regards,
    Spehro Pefhany
  11. Baphomet

    Baphomet Guest

    I'm sorry...that's too simple. Let H.P. design it and they'll find a way to
    do the same thing with 57,000 transistors and a partridge in a pear tree ;-)
  12. budgie

    budgie Guest

    I doubt if it will get anywhere near that current.
    Source linearity is academic in any solution, as testing and/or
    calibration would mandate that a suitable meter be connected at the
  13. gagir

    gagir Guest

    There are lots of opamps that can source 100mA.

    A voltage follower maybe?

    Eric Girard
  14. GPG

    GPG Guest

    Eliminate diodes, put resistor between bases and emitters so that v
    across will turn transistors on
  15. You don't need the second transistor if you don't need negative voltage
    But that's not what your little diagram shows here!
    You've got the feedback connection at the output of the op amp, which is
    going to result in less voltage available at the output of the circuit...
    If 15 volt supplies are used then the output should reach 10 volts, no
  16. mike

    mike Guest

    If thems is diodes in the base, you ain't gonna get ANY output. I'd
    call that a problem.

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  17. Yes, but I gave him a more general solution, just in case he wanted a
    more generally useful variable voltage supply.
    That is why I wrote it in the text, I was too lazy to change the
    schematic after adding the two transistors.
    It wasn't a complete schematic anyhow, more an illustration of the
    building blocks used.
  18. Yes, and note that the second transistor is needed for any
    negative currents, even when the voltage is positive. Both
    transistors allow for full four-quadrant operation.

    However, your diodes are going to be a problem. How are you
    going to turn on the PNP transistor? You need at least biasing
    resistors to get base current. I'd leave the diodes out
    completely. You'll get some cross-over distortion as the op-amp
    slews from +.7V to -.7V (it will be relatively quick since it's
    open gain), but that shouldn't be a problem for a power supply.

    Like: ;-)

    | |
    | |
    | | |
    | |/ |
    | |\| +----| |
    | +---+|-\ | |> |
    .-. | >---++ | |
    | |<----------|+/ | o--------+------------o
    | | |/| | |
    '-' | |<
    | o----|
    created by Andy´s ASCII-Circuit v1.22.310103 Beta
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