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Offset voltage on voltage follower

Discussion in 'Electronic Basics' started by tuurbo46, Aug 19, 2004.

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  1. tuurbo46

    tuurbo46 Guest


    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

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

    Any ideas?
  2. CFoley1064

    CFoley1064 Guest

    Subject: Offset voltage on voltage follower
    This type of thing is called level shifting. You might want to try something
    like this (view in fixed font or M$ Notepad):

    Summing Op Amp Level Shifter
    2K | |
    3K.-. | .----------.
    +5V | | | | VCC |
    + | | | | + |
    | '-' | | |\| |
    .-. | | '---|-\ |
    18K| | === | ___ | >---o
    | | GND '-|___|--o---|+/ Output (0.5V to 3.5V)
    '-' 33K | |/|
    | ___ | VEE
    | 33K
    2K| |
    | |

    created by Andy´s ASCII-Circuit v1.24.140803 Beta
    (By the way, the shareware program AAC is a good way to do this rather than
    listing nodes)

    Looking at the circuit above, you can see that your 0-5V signal is attenuated
    to a 0-3V signal (the span of your final output is 3V) by the 2K/3K divider.
    Then you add 0.5V with the 18K/2k divider. The two identical 33K resistors
    cause a summing action to get your 0.5 to 3.5V. This is buffered by your
    CA3140 set as a voltage follower to give you a low impedance output. Use 1%
    resistors for better accuracy.

    Good question. The non-inverting summing amplifier is your friend. One of the
    best basic tutorials on how to use op amps is National Semiconductor's App Note

    Read it -- a lot of what you need to use op amps is right there.

    Good luck
  3. This looks wrong. It boils down to:

    0-3V V1 O-----/\/\/----,
    0.5V V2 O-----/\/\/----+-----> V3, to opamp + input

    That should yield:

    V3 = (V1*34.8k + V2*34.2k) / (34.2k+34.8k)
    = 247.83mV + 0.50435*V1

    at the + input. I'd tend to guess, then, that the output would vary from about
    245mV to about 1.76V.

    Not quite the desired 0.5 to 3.5V.

  4. CFoley1064

    CFoley1064 Guest

    Subject: Re: Offset voltage on voltage follower
    It is wrong. I guess I was thinking (?) about an inverting summer with the
    virtual ground input, which isn't applicable. The OP should try something like
    this (view in fixed font or M$ Notepad):

    Vin (0-5V) | .---------.
    | | |
    | | |\| |
    | '--|-\ |
    | | >---o------o
    Vo o-----|+/
    +5V | |/|
    + |
    | |
    .-. |
    | | |
    | |3K |
    '-' |
    | ___ |
    | 150K
    | |
    | |1K

    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    The 1K and 3K resistors give a reference voltage of 1.25V. The 100K and 150K
    resistors give a summing voltage of 0.5V when the voltage input is 0V, and 3.5V
    when the voltage input is 5V. This assumes a small (about 1%) error for the
    source impedance of the 1.25V reference from the power supply. If that's
    unacceptable, a separate 1.25V reference can be devised which is not dependent
    on a resistive voltage divider or power supply variations. The math for a
    non-inverting summer is a little more complicated, but it's in the app note,

    Cheez. A senior moment, I suppose. Thanks for the spot, Jon.

  5. First, scale the input to 1V using a voltage divider. Then, use a
    non-inverting amplifier with a gain of 3 and a range of 0.5 to 3.5.

    | |
    .-. |
    Vin | | 2k |
    0-5V | | | |
    | I | '-' |
    .-. V | |
    4k | | | |
    | | | |\ |
    '-' o----|-\ |
    | | | >---------o--- Vout
    o---------|----|+/ 0.5 to 3.5V
    | | |/
    | |
    .-. .-.
    1k | | | | 1k
    | | | |
    '-' '-'
    | |
    GND Vx = -0.25V reference... :?

    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    Since Vin goes from 0 to 5, V+ goes from 0 to 1. Then, due to the
    negative feedback, V- goes from 0 to 1.

    At Vin = 0, we have V- = 0, so I = .25/1k = 250uA. Thus, Vout = 2k *
    250u + 0V = 0.5.

    At Vin = 5, V+ = 1, so I = 1.25/1k = 1.25mA. Thus, Vout = 1.25m * 2k +
    1 = 3.5V

    The problem, of course, is manufacturing the -1/4 V low impedance

    Bob Monsen
  6. I had been thinking more along the following lines, earlier, after reading
    Chris's post which had the right 'idea' but the wrong implementation details. I
    think I knew where Chris was coming from, though.

    I'll post the schematic I was thinking under another response here that is
    similar to Chris's, but I think arranged with the topology I believe he was
    reaching for.

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