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Sampling DC offset of a composite signal

Discussion in 'Electronic Design' started by Stephen Boulet, Dec 20, 2004.

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  1. I have a DC voltage with an AC signal superimposed on it. Is there a
    simple circuit to get the DC offset? If a digital volt meter can do it ...

    I'm trying to use the DC voltage to null out a DC offset in a
    differential signal. Our application uses only positive voltages.

    Stephen
     
  2. Yes. A resister and a capacitor LP filter!!!

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  3. I read in sci.electronics.design that Stephen Boulet
    How about a low-pass filter?
     
  4. 1) Chopper stabilize the signal by chopping the AC source.

    2) Filter the *&(@# out of it with an LP filter.

    3) If the AC signal is fixed frequency then integrate the signal
    for one (or more) AC cycles or sample the signal at 1/2 the
    AC frequency and average the two measurements.

    4) Use an integrating measurement of sufficient length that the
    AC signal is insignificant.

    5) Get a better signal source.

    If you sample fast enough all the filtering/integrating can
    be done in software.

    However, I am not quite sure what you mean by 'only positive
    voltages'. After you subtract the DC offset from the signal
    the remaining AC will be negative half the time. If the system
    is strictly analog you will need to have a virtual ground somewhere.
     
  5. The signals are differential, each centered at 1.75 V. The AC signal
    superimposed on them can go as high as 0.8 Vpp, so the overall voltage
    is always positive. The bandwidth of the signal is from about 3 kHz to
    75 kHz.

    I'm not so used to working down at audio frequencies. Taking your idea 4
    (since it seems an active filter is needed), I can make a 1st order
    filter with an op amp in the inverting configuration (say 1 Mohm in
    parallel with 0.1 uF in the feedback path and a 1 Mohm resistor going
    into the inverting terminal). That would give me about 0.8 mV of ripple,
    not even taking into account part tolerances. I am feeding a dc offset
    into the non-inverting terminal, so the fact that it's an inverting amp
    is ok.

    I'll probably need a second order filter -- my max suggested error in
    common mode voltage is 1 mV.

    Thanks for the ideas.

    Stephen
     
  6. Ah, that's different. I thought you had a DC offset voltage to be
    gotten rid of.

    If you want the AC w/o the DC why not just use a differential
    amplifier? A _real_ diff. amp works better that the usual 1-trick
    pony. You will need to reference to a virtual ground, though ...

    If you want the DC without the AC why not just add the two signals.

    A small 1:1 audio transformer will remove any common mode DC, and
    is possibly the most sensible solution.

    To eliminate common mode errors due to the DC I would suggest using
    capacitive coupling.
     
  7. Mac

    Mac Guest

    View the following ascii-art schematic with courier or a similar
    fixed-width font.

    diff in + ----------+
    |
    \
    / R Low offset
    \ voltage op-amp.
    / |\
    | | \
    +---------+------------|+ \_______ Vout
    | | +--|- / |
    \ --- C | | / |
    / R --- | |/ |
    \ | +----------+
    / |
    | --- GND
    diff in - ----------+

    Vout will be your DC offset, if I understand you right.

    This is a simple low-pass filter with a time constant of RC/2. So you
    should choose R and C such that the frequency is well below the lowest
    frequency of interest. Also, you have to be careful that R is not too
    big because op-amps do have input bias current, and you want the
    maximum input bias current (see the datasheet) to be much less than the
    DC current flowing through R. So you want an op-amp with a low offset
    voltage and a low bias current.

    What the heck. We might as well choose R and C.

    We probably don't want to use a capacitor any bigger than 10 uF. So we'll
    set C to 10 uF and choose R based on that. We'll set f to 300 Hz, since
    that is one tenth of your cutoff frequency. This will assure minimal
    attenuation at frequencies of interest.

    f = 2 * pi * R/2 * C
    R = f / (pi * C)
    R = 300 / (pi * 10uF)
    R = 30 / (pi * 1uF)

    This is around 10 * 1,000,000 which is 10 Megohms. You will need to find
    an op-amp with a very low input bias current, or choose an even larger
    capacitor.

    So, tentatively, we have:
    R = 10 Megohm
    C = 10 uF

    You could also use less uF or megohms, and accept more attenuation
    near the low end of your band. Or you could cascade multiple op-amp
    low-pass stages to get good performance with more reasonable values for R
    and C. For example, you could put your cutoff frequency at 1000 Hz, and go
    through three stages. You might want to put together a spreadsheet to try
    different scenarios.

    And you should double-check all my calculations. I've been known to make
    mistakes!

    HTH

    --Mac
     
  8. Rich Grise

    Rich Grise Guest

    <AOL>
    Me, Too!
    </AOL>

    ;-)
    Rich
     
  9. Ban

    Ban Guest

    Use a DC-servo made from a precision opamp and a first order LP(to maintain
    stability) feeding back into the reference input of the existing amp. You
    will need 2circuits for both sides of the bridge. You do not need a second
    order filter. If there is some ripple because of the finite attenuation, you
    will not change the DC-level, but attenuate the lower frequencies, just like
    a high-pass. actually if you choose the input capacitor a little bigger, you
    can compensate for this.
     
  10. Ban

    Ban Guest

    That is very true.
    the time constant is R*C
    fg= 1/(2pi*R*C)
    with C= 0.1uF and 100Hz I get here 15k915, just take 15k.
     
  11. I read in sci.electronics.design that Stephen Boulet
    With one op-amp you can make a third-order filter. But be careful which
    op-amp you choose. And audio op-amp won't make a good filter at 75 kHz.
    You need a faster one.
     
  12. Fred Bloggs

    Fred Bloggs Guest

    You have two inputs V1= VDC + Vac/2 and V2= VDC-Vac/2 and Vac is the
    "signal"? -the only possible interpretation of "signals are
    differential." So what the hell do you want to pick off, VDC or Vac? If
    it's VDC then you've been given your answer, a summer. Or is it Vac, in
    which case you use an ac-coupled differential amplifier. What is your
    problem? Is this a single supply application you can't figure out?- or
    homework? Take it to sci.electronics.basics.
     
  13. And the problem with a low pass filter is...?

    --
    Many thanks,

    Don Lancaster
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    voice: (928)428-4073 email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  14. Al

    Al Guest

    Uhh, you analog guys is hard to please ;-)

    Al
     
  15. Mac

    Mac Guest

    Boy, I really screwed that one up. Thanks for correcting me. The only
    thing I got right was using R/2 instead of the usual R, because, if you
    look at my schematic, the resistors are in parallel with respect to the
    common mode input.

    So you should use:
    fq= 1/(pi*R*C)

    15k and 0.1uF will still work fine if the lowest frequency of interest is
    3000 Hz.

    --Mac
     
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