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Op-amp stability

  1. Mar 6, 2004 #1
    I am confused about the stability of an op-amp voltage follower.

    In the text that follows, I will arrive at a conclusion that
    the op-amp voltage follower is unstable.

    Please find out the flaw in the argument.

    Consider this :
    Four systems, A, B, C and D.
    C and D are charectarized by an input-output
    relationship of the form:
    C: y(t)=x(t-p), p is a positive real number.
    (y = output , x = input)
    D: y(t)=x(t-q), q is a positive real number.
    (y = output , x = input)

    The systems A, B, C and D are connected as shown below:

    X and R are the inputs to the system A.
    X is an external input whereas R is the output of D.
    Output of A is M
    M is input to C
    N is output of C
    N is input to B
    O is output of B
    O is input to D
    R is output of D

    (Copy this text into notepad and set off wrod wrap to view it correctly)
    ______ ______ ______ ______

    Input=X-->| | M | | N | | O | | R
    |-->| A |--->| C |--->| B |--->| D |--->|
    | | | | | | | | | |
    ^ ------- ------- ------- ------- |

    As is evident from the diagram above, A is a system whose output
    is a function of two signals.
    M, N, and O, R are intermediate signals as shown.

    Next, I define the input-output relationships for
    the sytems A and B as follows:

    A: y(t) = a*(v(t)-u(t)) , a is a real number.
    (v and u are its two inputs)

    B: y(t)= b*x(t), b is a real number.

    Finally, the equations describing all the signals under the
    connections as shown above are:

    M(t) = a*(X(t)-R(t))
    N(t) = M(t-p)
    O(t) = b*(N(t))
    R(t) = O(t-q)

    Back substituting,
    M(t) = a*(X(t) - O(t-q))
    M(t) = a*(X(t) - b*(N(t-q))
    M(t) = a*(X(t) - b*(M(t-p-q))
    let p+q = r
    then, M(t) = a*X(t) - b*(M(t-r)) ---(1)

    ie, the present value of M is equal to the difference of a
    scaled version of the present input and a scaled version of
    one of its past outputs.

    Let all systems be causal and initially relaxed so that
    M(t)=0 for t<0.
    Thus, until the time t=r, M(t) = a*X(t) (for 0 < t < r).

    Now if a is large, then M(t) rises to formidable values even
    for values of X(t) close to zero, during the time 0 < t < r.

    When t just grows above r, M(t) drops down as it obeys (1).

    Now take the case of X(t)=u(t), the unit step.
    From (1), M(t) = a - b*M(t-r)

    For 0 < t < r, M(t) = a.

    For r < t < 2r , M(t) = a - b*a since M(t-r) = a for r < t < 2r.
    = a*(1-b)
    = a - ab

    For 2r< t < 3r,M(t)= a - b*(a - ab))
    (since M(t-r)=(a - ab) for this interval)
    M(t)= a - ab +ab*b
    M(t)= a - ab(1-b)

    Thus (M(t) for 2r < t < 3r) > (M(t) for r < t < 2r)
    provided b<1

    For the next interval,
    M(t) = a - b*(a-ab(1-b)
    = a -ab +ab -ab*b
    = a -ab*b
    > a - (ab-ab*b),M(t) in 2r < t < 3r
    provided b<1

    This pattern continues for every time interval of width r.
    It is easy to see that M(t) keeps on increasing for b>1.

    Substituting b = 1, we get M(t) alternating between 0 and a.

    It is obvious that the systems described above collectively
    represent an op-amp with negative feedbak, time delays between the
    input and output having been taken into account.
    With resistive feedback networks, the b<1 case holds.
    For the voltage follower, the b=1 case holds.

    In both the cases the results observed in the laboratory
    and the ideal, zero time delay models deviate enormously from this
    more realistic model of the op-amp.

    How is this explained ? Is there a flaw in the above derivation ?
    Last edited: Mar 6, 2004
  2. jcsd
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