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Single supply opamp calculations

Discussion in 'Electronic Basics' started by MRW, Jun 18, 2007.

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  1. MRW

    MRW Guest

    Hi, I'm doing some more opamp calculations just to get used to opamp
    golden rules some more. I decided to do an ideal analysis on an
    inverting, single-supply opamp: http://i14.tinypic.com/5xebi4w.jpg

    After applying some golden rules, I got the following equations:

    (V1 / R1) + (Vo / R2) - [ (Va * (R2 - R1) ] / (R1 * R2) = 0

    Re-arranging the terms, I get:

    Vo / V1 = -(R2 / R1) + (Va * V1) * [ (R2 - R1) * R1] / (R2 * R2)

    Did I make a mistake in my calculation? I was expecting something like
    this:

    Vo / V1 = -(R2 / R1) + V2

    Thanks!
     
  2. yes... your dimensions are not even correct.
    Va = Vb = V2 for an ideal op amp, so...


    I1 = (V2 - V1)/R1
    I2 = (Vo - V2)/R2

    since I1 = I2 for an ideal op amp we have

    (V2 - V1)/R1 = (Vo - V2)/R2

    or

    R2/R1*(V2 - V1) + V2 = Vo

    If V2 was 0 then you would just have -R2/R1*V1 = Vo which is the basic
    inverting op amp. In this case you have shifted by voltages by V2...

    i.e.

    -R2/R1*(V1 - V2) = (Vo - V2)

    Jon
     
  3. Bob Myers

    Bob Myers Guest

    Looks like you're trying to sum the currents at the node labelled
    "Va" - but don't forget, the current through R2 is NOT Vo/R2
    unless Va = 0...which cannot be the case here, given the offset
    voltage (V2) at the non-inverting (Vb) terminal. Nor is the current
    through R1 simply V1/R1, for the same reason.

    There are two ways to do this - assume (Va-Vb) is zero (which
    basically assumes infinite gain for the op amp), in which case the
    currents are (V1-V2)/R1 and (Vo-V2)/R2 - OR you can assume
    that there is an error voltage here (Ve = Va - Vb) which is amplified
    by the op-amp such that Vo = AVe (where A is the gain). The
    latter path gives you the more accurate solution, which can then
    be reduced to what you expect by assuming that A gets very, very
    large and noting which terms will them drop out.

    Bob M.
     
  4. MRW

    MRW Guest

    Thanks, Jon!

    I also had the assumption that I1 = I2 and that Va=Vb=V2, but I think
    I made a mistake when I split my initial equation into the following:

    I1 = - (V1 - V2) / R1
    I2 = - (V2 - Vo) / R2

    to this....

    (V1 / R1) - (V2 / R1) = (V2 / R2) - (Vo / R2)

    (V1 / R1) + (Vo / R2) - (V2 / R1) - (V2 / R2) = 0


    then I used the following relationship: (A / C) +/- (A / D) = [ A *
    (D +/- C) ] / CD on

    -(V2 / R1) - (V2 / R2) and turn it into

    -[ V2 * (R2 - R1) ] / (R1 * R2)

    so I had

    (V1 / R1) + (Vo / R2) - (-[ V2 * (R2 - R1) ] / (R1 * R2))

    .... I still can't quite pick out my exact mistake, yet.. but I'm sure
    after much scrutiny it'll hit me like a light bulb.
     
  5. MRW

    MRW Guest

    I understand this portion.

    I'm having a hard time grasping the last sentence in this paragraph.
    Sorry.
     
  6. He's simply taking into account that a real op amp does not have infinite
    gain... a needless assumption in this case.... don't worry about it. If you
    can't do it for the ideal case you can't do it for the non-ideal case.
     
  7. Bob Myers

    Bob Myers Guest

    Well, yes, I was trying to get into the non-ideal case, because I
    think it's important that people eventually see where all of these
    magic formulas actually come from. If we take the usual
    inverting-amp configuration with the non-inverting input grounded
    (in other words, I'm going to ignore what was originally called
    "V2" here as a needless complication at this point - all it will really
    do is offset the output, anyway), the analysis really isn't all
    that difficult.

    We'll call the input (at the "free" end of R1) Vin, the output Vout,
    and the voltage across the op-amp inputs Va. The "ideal"
    analysis would have you assume that Va is zero, or call the
    inverting input a "virtual ground," or some such, without really
    explaining why you do that. But again, it's really not all that hard
    to go through the full analysis, and what you'll wind up with is
    the general formula for any amplier used in this configuration, no
    matter how big the gain is.

    We will, however, make the assumption that no current enters the
    inverting input (i.e., the input impedance is way bigger than anything
    else involved, such that any current into the input is negligible), and
    that makes the summing of currents at this node very simple:

    (Vin - Va)/R1 = (Va - Vout)/R2

    But by definition, the output voltage must be the amp gain times
    the voltage at the amplifier's input, which is Va, so:

    If Vout = A(Va), then Va = Vout/A, and

    (Vin - Vout/A)/R1 = (Vout/A - Vout)/R2

    Solving for the overall gain (Vout/Vin), we get

    Vout/Vin = R2 / [(1/A)*(R1/R2 + 1) - R1]

    Note where the open-loop gain term "A" (i.e., the gain of the amplifier
    itself, without the R2 and R1 external components) winds up. If A
    gets very large, the 1/A term will approach zero, and take the whole
    first part of the denominator with it. If that's the case (and it's a
    reasonable
    assumption for an op-amp, where the open-loop gain is very commonly
    in the tens if not hundreds of thousands), then that first term drops
    out completely, and the whole thing winds up with the familiar

    Vout/Vin = - (R2/R1)

    Simple, no?

    Bob M.
     
  8. MRW

    MRW Guest

    I'm following so far.
    I'm understanding this portion.
    This is clearer to me.
    That's much better! Thanks, Bob!
     
  9. redbelly

    redbelly Guest

    Found your mistake, read on:

    Here is your error:

    - (V2 / R1) - (V2 / R2)
    = - [(V2 / R1) + (V2 / R2)]
    = -[ V2 * (R2 + R1) ] / (R1 * R2)

    Note the "+" sign in the (R2+R1) term, and the over-riding "-" sign in
    front of the expression.
    Regards,

    Mark
     
  10. MRW

    MRW Guest


    Thanks, Mark!
     
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