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Offset + gain on single opamp?

Discussion in 'Beginner Electronics' started by Z, Aug 1, 2006.

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  1. Z

    Z Guest

    I've been playing around with getting both gain and offset on a single non-
    inverting op amp and I'm not quite getting how to calculate the offset
    input voltage needed to get 1.27 volts out. I'm hoping someone might know
    what I'm missing.

    What I attempted was to get 2.5x gain with a +1.25 volt offset - so a zero
    volt input on the non-inverting input would give 1.25 volts out while a 1
    volt input would give 3.75 volts out.

    At first, I thought that since there was to be a 2.5x gain, I needed to add
    -1.25/2.5= -0.5V to the inverting input. I used a voltage divider to create
    the -0.5V. from the -5V supply.

    Next I went to tackle the gain. I calculated the thevenin equivilent
    resistance of the voltage divider then calculated the size of the feedback
    resistor and put it all together on a breadboard.

    I got the gain right - but with 0V in I get 0.750V out - which is only 1.5
    times the offset voltage I supplied. Ok, so my assumption about the affect
    of gain on the supplied offset voltage was wrong - I went looking on google
    to see if I could find out how to calculate the voltage required: two days
    ago!

    To sum up: I'm using an LMC6482 CMOS amp with 10 teraohm input impedance.
    Input signal is a 10k resistor to either ground or a voltage divider
    putting out 1V. I can calculate the gain with Thevenin's but I cannot
    figure out how to calculate the offset voltage required.

    Anyone able to steer me in the right direction?

    P.S. I have this working with two opamps (one chip). Its just bugging me
    that I cannot figure out how to do it with a single stage.
     
  2. Ban

    Ban Guest

    You have to use a non-inverting configuration. Vref is a negative reference,
    which can be derived from the negative rail as well. You can use trimmers in
    series with R2 and R3 to adjust gain and offset. The equations are the sum
    of currents in the inv.input node. This reminds me of homework though. :-(

    |\ 1.25/R1-Vref/R3=0
    o-------|+\ 3.75/R1-Vref/R3+1/R2=1
    Vin | >--+-----o
    +--|-/ | Vout
    | |/ .-.
    | | |R1
    | | |
    | '-'
    | |
    +--------+
    | |
    .-. .-.
    | |R2 | |R3
    | | | |
    '-' '-'
    | |
    === o
    GND -Vref
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
     
  3. Z

    Z Guest

    I don't doubt it does! But, I'm an old fart at 46 trying to use my hobby
    to make a 4-20mA transmitter for work. I looked into buying one, but its
    freakin' expensive!

    I have a design that uses three opamps and I recently improved that to
    two opamps. The first amp taking the input on the non-inverting input
    and the gain set at the inverting input (standard non-inverting with
    gain). The second amp is a unity gain voltage follower which I feed the
    offset voltage dialed up on a pot. I then take that buffered voltage and
    feed it to the gain-divider on the first amp in place of ground. Works
    like a charm.

    I saw Thevenin's theory a couple of years ago and, although I couldn't
    make heads nor tails of it at the time, I went back and tried again and
    finally managed to figure it out. I figured that I could use Thevenin's
    to get the offset and gain both on a single opamp and I was right,
    except that (of course) the offset voltage that worked in the other
    confuration doesn't work in this one.

    So anyway, I used the formula you supplied and came up with (what seems
    to be) a non-sensical value for R2. I used R3=100K and R1=15K and came
    up with one ohm for R2 while I was expecting a value somewhere around
    10-20k.

    Can you point me toward a site that maybe describes the principle?
    Formula are great but like most math, if you don't understand where the
    numbers came from then the answers don't make sense either 8)


    P.S. You mentioned the sum of currents - I keep thinking of voltages
    only, this is probably where I'm deficient.
     
  4. Z

    Z Guest

    Lol, what the hell am I talking about. I shouldn't post immediately
    after i wake up!

    4-20mA is what I'm working on presently and it has nothing to do with
    the circuit I posted here - though I am working on gain/offset etc on it
    too. That one I'm getting done by myself though.

    No, the circuit I was asking about is a buffer/gain/offset for a pH
    sensor. The pH sensor is a battery that actually puts out a positive
    voltage when its below neutral (acidic) and a negative voltage when it
    is above neutral (base). My ADC is single supply so it cannot handle the
    negative voltage.

    I'm using a 2.5V reference for the ADC and the pH sensor will never put
    out more than +/-450MV so I went with 2.5 times gain (+/- 1.125) and an
    offset that would put neutral (zero from sensor) smack dab in the middle
    of the ADC range. with the offset that would give an ADC input between
    ..125V and 2.375V.

    The range I'm actually interested in is pH7 to pH8 and I need two
    decimal places, so I'm using a 12 bit ADC. I would use a higher gain to
    have 6-9 PH span the entire ADC range, but I cannot figure out how to
    clamp the ADC input at zero volts. The ADC can withstand -0.5V on an
    input and I considered using a small signal diode from the ADC input to
    ground, but the voltage drop is larger than the -0.5V limit on the
    input.

    The three amp and two amp versions work perfect but I want to learn how
    to do it with one amp!
     
  5. Z

    Z Guest

    Oops, the ADC input tolerance is -0.3V, not -0.5. I'll look for germanium
    diodes though! -0.2V would work.

    The output impedence is somewhere around 500K-1Meg, which is why I'm using
    a 10 teraohm input impedenace op-amp. The sensor ground is simple connected
    to circuit ground.
     
  6. Z

    Z Guest

    I've considered that, but I don't know enough about batteries to know if
    that would damage the sensor or not.
     
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