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 ?