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AC/DC problem

Discussion in 'Electronic Design' started by Alex, Mar 25, 2007.

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  1. Alex

    Alex Guest

    I have to make a AC to DC circuit. I have 10 uV to 1 mV (+ or -) on
    the input. the output have to be 50mV to 5 V,(one polarity only) to
    drive a VCO.
    The idea is to measure a DC current to or from a battery over
    time .The result will then be read out in Ah ,and the display count up
    when charging, and down when using current.
    I have all working very well except the AC/DC ,where I get too big
    differences between in and out voltage , that is, at low currents
    equal 50 uV or less the difference is 1:2 but ok when the voltage is
    the solution I seek can have nonliearity ,and 20% difference between
    positiv and negative voltage is also acceptable.
    After trying several solutions with op-ams, like application notes on
    a "absolute value amp " from NS ,- a typical AC/DC with rectifier for
    one half wave and adding the other halfwave,- and my own idea of
    halfwave rectifier for each polarity ,and inverting one and then
    adding the two signals,I have come to a dead period with no better
    solutions,, can you give me som ideas, maybe some aplication notes or
    some different solutions, ... THANK IN ADVANCE
  2. Fred Bloggs

    Fred Bloggs Guest

    This is not really called an "AC to DC circuit" and you cannot achieve
    DC amplification of those minute signal levels with ordinary op amps,
    you must go to chopper amps or zero offset sampling of some kind.
  3. MassiveProng

    MassiveProng Guest

    Why can't so many of you retarded fucks learn how to set your PC
    clocks correctly?
  4. Jamie

    Jamie Guest

    you need a precision rectifier.
    look that up on the net.
    you'll find lots of examples.
    as far as you forcing both - and + to + only, how would you decide on
    which direction it's going?
    i guess you could use 2 half wave precision rectifiers..
  5. If I'm reading it correctly it's more like an offset circuit. Zero
    current at mid range, full negative below that and full positive above.
    Otherwise he'll have to pass a separte sign signal to accumulate amps.
    Looks like a capacity meter. Of course it leaves out little things like
    charge losses and self discharge.

  6. John  Larkin

    John Larkin Guest

    That's not exactly AC/DC, it's more like an absolute value circuit...
    the semantic difference being that the polarity reversals are probably
    infrequent and not periodic.

    You can make a very fairly simple circuit that multiplies a signal by
    +1 or -1, and drive that from a comparator that detects the input
    polarity. But at 50 uV, all the parts - amp, polarity switch,
    comparator - have to be very good.

    What does the final data - absolute value of battery charge/discharge
    - actually mean?

  7. Alex

    Alex Guest

    Thank you for your time and very good help allready.
    a few comments to your replies.
    Yes, agree that the correct name must be "absolute value "
    Have now looked for precision recifiers,, have simply forgotten to use
    the web.. happens now and then..
    The polarity ; using an opamp with high or open loop as polarity
    detector. the polarity gives output for count up/down circuits.
    The final data means, hmm if I understand the question correct then ;
    In small solar systems, the charge and discharge of the batteries will
    give much information to the owner on how much there can be used ,
    furthermore, you can decide if more batteries is usefull or you need
    more solar panels.
    I'm aware that the voltage level I used, could be a problem but then I
    got some good ideas from you. :) think I will be able to accept 50mV
    drop across a shunt , for 20 Ampere This would give a loss of 1
    watt ,and the voltage drop will still be without any influence on the
    The presicion is not that important as it will be a problem to know
    exactly the efficiency in the battery, witch I guess, also varies with
    temperature, and how big current that is drawn ,and maybe other
    factors too. but the better the counter works the better result after
  8. How many are you making? If it's only a handfull look real hard at hall
    effect current sensors. Allegro has IC style ones in the current range
    you are talking about.

    Easier output voltages.

    Also remember your effirncy error is going to accumulate as you run,
    probably quite quickly.

  9. John Barrett

    John Barrett Guest

    Want to avoid all these problems ?? I posted a link to a hall effect current
    sensor that works for AC or DC a few weeks ago, with models rated from 20 to
    100 amps .. all of which give a 0-5v output -- stick one on the charging
    circuit feed to the bank.. stick another on the output side and monitor both
    currents independently... have your microcontroller ADC the 2 values and
    compute the difference for net charge/discharge rate

    I could dig up the link again if you are interested
  10. GPG

    GPG Guest

  11. Alex

    Alex Guest

    Thanks again for new inputs.. thank to John Barret and Robert Adsett
    for giving me the idea of using hall element sensors.. have asked for
    samples at Allegro allready.. they seem a good choise for the project,
    as I can omit the DC/DC supply to the "absolute value amplifier "too.
    And the currents can be quite high without problems .
    (the "high end current sensors" from Maxim was considered earlier in
    the design, but the one that gave out polarity was not to get anymore,
    and this circuit was not used because of this.)
    The texas link from GPG is very intersting.. the device checks all
    about the battery ,too bad its only for LIon batteries..
    The efficiency faktor is a problem that I had given some thoughts..
    think the Ah monitor have to be reset now and then to get a usefull
    picture of the charge/discharge. beside an adjustment of discharge /
    charge is added, so the owner can find a point where the meter will
    show a usefull result over longer timeinterval.
    I will return when I have found a solution
    Thank you all for help.
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