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Why am I so bad with op amp calculations?

Discussion in 'Electronic Basics' started by MRW, Jun 25, 2007.

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  1. MRW

    MRW Guest

    Hello all.. I have this circuit:
    http://bayimg.com/aAclCaabB

    It's just a basic summer circuit in single supply mode. I don't
    understand why I cannot resolve the equation similar to the one they
    have in Texas Instrument's Single Supply Op Amp Collection document.

    Please help me understand.

    I derived my calculations using superposition. First I derived V_out1
    by shorting V2 and V3. Then, I found V_out2 by shoring V2 and V1.
    Last, I shorted V1 and V3 to find V_out3. I got the following:

    V_out1 = - (R2 / R1) * V1
    V_out2 = - (R2 / R3) * V3

    V_out3 = V2 * [ (R3 || R1) + R2] / R2 = V2 * [ ( (R3 || R1) / R2) + 1]
    (I treated V_out3 like a basic non-inverting configuration)

    Now I have V_out = V_out1 + V_out2 + V_out3, which does not match with
    TI's document and with my simulation results.

    Any insights? Thank you very much!
     
  2. Eeyore

    Eeyore Guest

    It seems to me that you're making your equations too complicated.

    Graham
     
  3. MRW

    MRW Guest

    Hi Graham, how would your approach this?

    I threw away the original paper with my doodle and redid my work. I
    actually figured out my mistake this time. I made a mistake in the
    voltage divider portion in deriving V_out3. I switched the order of
    resistors. Tsk. Now it matches up to things.
     
  4. Guest

    My insight would be to ignore the printed sums, come in cold and
    just understand that when the opamp is working OK, both pins 2 and 3 -
    must- be the same voltage, the pins are are just voltage sensors and
    take no current.
    This means you can straight away work out the currents in R2 and R3
    and the direction the currents are travelling.
    The gives a resulting current that may be heading towards, or moving
    from pin 2. This current is the exact current that R2 must supply.to
    balance the input at 2.5V. Hence the output voltage can be worked out
    from R2 and that current.
     
  5. Eeyore

    Eeyore Guest

    Agreed.

    Graham
     
  6. MRW

    MRW Guest

    I am not quite sure why R1 is not represented in your text. To the
    best of my understanding, I have something like this.

    Something like this? http://bayimg.com/EACLNAAbD

    Using KCL, I got the following:

    V_o = - (R2 / R1)*(V1-V2) - (R2 / R3)*(V3-V2) + V2

    It took less than three steps and coincides with the simulation
    results. Very nice.

    Thanks!
     
  7. JeffM

    JeffM Guest

  8. Guest

    Yes. My "R2" should be R1.
    (for pictorial clarity I'd have the I2 I3 arrows going the other way)
     
  9. MRW

    MRW Guest


    Hmm.. now I have a problem. Here are my two equations so far:

    Using KCL:
    V_o = - (R2 / R1) * V1 - (R2 / R3) * V3 + (R2 / (R3 || R1)) * V2 + V2

    Using Superposition:
    V_o = - (R2 / R1) * V1 - (R2 / R3) * V3 + ((R2 / R1) + (R2 / R3) + 1)
    * V2

    But somehow when I compare the V2 coefficients:

    (R2 / (R3 || R1)) * V2 = ((R2 / R1) + (R2 / R3) * V2
    (R2 * (R2 + R3)) / (R3 * R1) = ((R2 / R1) + (R2 / R3)

    I don't get the same relationship:

    (R2 * (R3 +R1)) / (R1 * R3) =! (R2 * (R2 + R3)) / (R3 * R1) , this
    only works when R1 = R2

    :/
     
  10. MRW

    MRW Guest

    Never mind. The reason why I am bad with op amp calculations is
    because I always make a mistake somewhere. In this case, (R2 / (R3 ||
    R1)) was turned into R2*(R2 + R3) / (R3 * R1) on my paper instead of
    R2 * (R1 + R3) / (R3 *R1). I am stupid.
     
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