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Offset voltage on voltage follower

Discussion in 'Electronic Design' started by Tuurbo46, Aug 20, 2004.

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  1. Tuurbo46

    Tuurbo46 Guest

    Hi
    I need a bit of help on a circuit im trying to build. I will try to explain
    my problem below.

    Circuit requirements:

    *input to op-amp = 0 to 5v
    *output from op-amp = 0.5v to 3.5v


    Im am currently using a CA3140E opamp with a 0.5v offset in it, which is
    quoted in the data sheet.

    Pin wirring on chip:

    Pin 1: to left leg of 10k pot
    pin 2: connected to pin 6 with no series resistor
    pin 3: varible input voltage, 0 to 5v
    pin 4: to middle leg of above 10k pot and connected to GND
    pin 5: to right leg of above 10k pot
    pin 6: connected to pin 2 with no series resistor, and then this voltage is
    dropped across 2 resistors in series to produce the output voltage ratio


    Im at the stage in the circuit, that when you increase the input voltage
    from o to 5v, the output is alway 0.5v higher than the input which is
    correct. At this point i drop the output voltage over a potential divider
    and i get the correct output voltage ratio over this. The problem starts
    when a 1k load resistor is taken from the potential divider. At this point
    the voltage drops. When the load resistor is removed the circuit produces
    the correct output voltage again.

    I think the problem is the circuit cant provide enough corrent, but i cant
    see how to change this.

    Any ideas?

    Anyone have any better ideas?
     
  2. Tim Wescott

    Tim Wescott Guest

    I would not be very trusting of a 1/2-volt offset produced by severely
    yanking the "offset _null_" line of an Op-Amp.

    Your circuit indeed cannot provide current with it's potential divider
    without it's output voltage changing. Assuming that your input
    impedance must be high, I would suggest that you use the first stage as
    a voltage buffer, with a potential divider in the middle, followed by
    another voltage buffer. Further, I would put the 1/2 volt offset in the
    potential divider, like so:


    Vref
    +
    |
    |
    .-.
    | |
    | |
    input |\ '-'
    -----------|+\ ___ |
    | >--o---|___|---o |\
    .--|-/ | o----------|+\ output
    | |/ | | | >---o----------
    | | .-. .---|-/ |
    '--------' | | | |/ |
    | | | |
    '-' '----------'
    |
    |
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Use a dual op-amp and you won't increase your package size. You'll need
    to use a reference voltage that's right for the level of precision that
    you need; this may range anywhere from whatever you're using for your
    raw supply voltage to a high-precision 2.5V reference. Setting the
    values of the three resistors in the voltage drop/shift network are left
    as an exercise to the reader, of course.
     
  3. I made a similar comment in .basics, where the OP also asked this question
    separately.
    I said as much there, too.

    Without really considering the load on the input source, assuming a 5V only
    supply, and trying to keep to just one rail-to-rail opamp, I got something like
    this:

    (0-3V)
    30k 27k 90k
    0V to 5V input >-----/\/\/----+----/\/\/----, ,-----/\/\/-----,
    | | | |
    | | | +5 |
    \ | | |\| |
    / 45k | +-----|-\ |
    +5V \ | | | >------+--> OUT
    --- | | | ,--|+/
    | | | \ | |/|
    | --- | 45k/ | |
    \ gnd | \ | ---
    / 180k | | | gnd
    \ | | |
    | | --- |
    | (0.5V) 27k | gnd |
    +------------/\/\/----------+ |
    | +-------'
    | |
    \ \
    / 20k / 45k
    \ \
    | |
    | |
    --- ---
    gnd gnd

    :)

    Jon
     
  4. Tim Wescott

    Tim Wescott Guest

    I was going to send him to 'basics -- I should have looked, I suppose.

    I think I win the "fewest resistor" contest -- you can remove my first
    opamp and still use my little network, assuming low source impedance.
     
  5. Or leave out both op-amps if the load is high-Z.


    +5.0V
    o
    |
    .-.
    | |
    | |24.0K 1%
    '-'
    0-5V ___ | 0.5-3.5V
    Vin o ----|___|----+-------> Vout (eg. to ADC)
    |
    4.02K 1% .-.
    | |
    | | 8.06K 1%
    '-'
    |
    |
    ===
    GND

    Best regards,
    Spehro Pefhany
     
  6. Tim Wescott

    Tim Wescott Guest

    The OP complained about his Z=1k load pulling the voltage down, so you
    at least need the follower.

    Nice of you to calculate the values, though.
     
  7. John Fields

    John Fields Guest


    ---
    LOL!!!

    I knew it was there somewhere, but I couldn't wrap my head around
    it...

    Very damned clever!


    +5 +5 +5
    | | |
    [4.02K] [24.0K] [3.443K]
    | | |
    +---------+ = +----3.497V
    | |
    [8.06K] [8.06K]
    | |
    GND GND

    and

    +5 +5
    | |
    [24.0K] [24.0K]
    | |
    +---------+ = +----0.503V
    | | |
    [4.02K] [8.06K] [2.682K]
    | | |
    GND GND GND
     
  8. I read in sci.electronics.design that Tim Wescott
    Just make the resistors ohms instead of kohms.
     
  9. How did you find the values? Brute force with three nested for 0->96 loops?
     
  10. Just so the OP knows what's going on here:

    Vin * R1 * R2 + 5.0 * R2 * R3 + 0.0 * R1 * R3
    Vout = ---------------------------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    Vin * R1 * R2 + 5.0 * R2 * R3
    = -----------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    Vin * R1 * R2 5.0 * R2 * R3
    = --------------------------- + ---------------------------
    R1 * R2 + R1 * R3 + R2 * R3 R1 * R2 + R1 * R3 + R2 * R3

    Since this also must be:

    Vout = 0.5 + Vin * (3/5)

    it's then clear that:

    3 R1 * R2
    - = ---------------------------
    5 R1 * R2 + R1 * R3 + R2 * R3


    R2
    0.6 = R1 * ---------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    and also,

    5.0 * R2 * R3
    0.5 = ---------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    R2 * R3
    0.1 = ---------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    R2
    0.1 = R3 * ---------------------------
    R1 * R2 + R1 * R3 + R2 * R3


    which means that,

    R1 = 6 * R3
    R2 = 2 * R3

    Call R3 just R, for now, and recheck:

    Vin * 6 * R * 2 * R + 5.0 * 2 * R * R
    Vout = -------------------------------------
    6 * R * 2 * R + 6 * R * R + 2 * R * R


    Vin * 12 * R^2 + 10 * R^2
    = ----------------------------
    12 * R^2 + 6 * R^2 + 2 * R^2


    Vin * 12 * R^2 + 10 * R^2
    = -------------------------
    20 * R^2


    Vin * 12 + 10
    = -------------
    20


    = (3/5) * Vin + 0.5

    Just the desired answer!

    So,

    +5V
    |
    |
    \
    / 6*R
    \
    |
    0V to 5V R | 0.5V to 3.5V
    Vin O---/\/\/----+-------> Vout
    |
    |
    \
    / 2*R
    \
    |
    |
    ===
    GND


    Jon
     

  11. Beautiful ;-)
     
  12. No, I did it numerically with a steepest descent algorithm, and added
    the constraint that the output impedance should be just lower than
    2.5K (arbitrary, but the maximum recommended for many PIC ADCs, as you
    know) and the resistor values standard 1% series. But Jonathan's
    solution is FAR nicer. ;-)

    Best regards,
    Spehro Pefhany
     
  13. Output impedance is:

    R*(2R)*(6R) 12*R^3 3
    ----------------------- = ------ = - * R
    R*(2R)+R*(6R)+(2R)*(6R) 20*R^2 5

    So, R = (5/3) * desired impedance, in this case.

    Not much left for the OP to do, I suppose, except get rid of that opamp-twisting
    pot on the offset null pins.

    Jon
     
  14. Fred Bloggs

    Fred Bloggs Guest

    Well if Vout=3/5*Vin + 0.5 then the sum of currents at the Vout node :

    View in a fixed-width font such as Courier.



    3 3 3
    - Vin + 0.5 - Vin - Vin + 0.5 - 5 - Vin + 0.5
    5 5 5
    ------------------- + ----------------- + ------------- =0

    R R R
    3 1 2



    Since this must be true for 0<=Vin<=5, the Vin and constant

    coefficients must both be zero:



    3 1 1 2 1 1 1 4.5
    - ( -- + --) - - -- =0 and 0.5 (-- + --) - --- =0
    5 R1 R2 5 R3 R3 R2 R1



    You call R3=R then call x=1/R1 and y= 1/R2 so that the above

    equations become:

    2 1 1
    x + y = - - and 9x - y = -
    3 R R

    5 1 1 1
    so that by inspection 10x= - - or x= - -
    3 R 6 R

    2 1 1 1 1 1
    and then y= - - - - - = - -
    3 R 6 R 2 R




    which makes R1=6R and R2=2R confirming your solution.
     
  15. -----
    I like that observation. Neatly tidies things up. And... of course!
    -----

    -----
    Also, you can now go back and re-verify your observation -- the part you
    excluded because it 'must be zero' given that Vin can vary over that range:

    0.5 4.5 0.5
    --- - --- + --- = 0
    R3 R1 R2

    Substituting:

    0.5 4.5 0.5
    --- - --- + --- = 0
    R3 6R3 2R3

    6 9 3
    --- - --- + --- = 0
    R3 R3 R3

    Which, of course, it is for all values of R3.

    Nice when you can flexibly change points of view with equal success.

    Jon
     
  16. I read in sci.electronics.design that Frank Bemelman <[email protected]
    Three unknowns, so three equations are needed. For example, zero and
    maximum input voltage, and any other value.
     
  17. Rich Grise

    Rich Grise Guest

    It's a straight line, so you can cancel out at least one unknown.

    Y = m * x + b

    Cheers!
    Rich
     
  18. I read in sci.electronics.design that Rich Grise <>
    The straight line criterion is equivalent to a third equation. There is
    NO way of solving two equations with three unknowns.
     
  19. It sure is. I have to admit I'm surprised that it results in such
    a nice 1-2-6 ratio. Too good to be true, sort of ;)
     
  20. Fred Bloggs

    Fred Bloggs Guest

    The most common pencil paper approach would probably be:

    View in a fixed-width font such as Courier.



    Thevenin Equivalent


    R RTH
    Vin >---/\/\--+--/\/\----< Vth
    |
    |
    |
    +---------> Vout



    Vin x RTH VTH x R
    Vout= --------- + ---------
    R + RTH R + RTH



    Equate coefficients to required Vout=3/5*Vin + 0.5 :


    (1) (2)
    RTH VTH x R
    --------- =3/5 and --------- = 0.5
    R + RTH R + RTH



    By inspection:


    (1)=> RTH=R/( 5/3-1)=3/2*R and


    (2)=> VTH*(1-3/5)=0.5 or VTH=0.5*5/2



    Then equate coefficients:



    R1 R2
    5V >---/\/\--+--/\/\----0V
    |
    |
    |
    +---------> VTH



    R1*R2 3 5 *R2 5
    (1)=> ------- = - R (2)=> ------- = 0.5 * -
    R1 + R2 2 R1 + R2 2



    R1 3 2 1
    (1)/(2) => -- = - R - --- or R1= 6R
    5 2 5 0.5


    1 2 1 1
    Then -- = --- - --- = --- or R2=2R
    R2 3*R 6*R 2*R
     
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