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Op-amp stability

by kpachar
Tags: opamp, stability
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Mar6-04, 06:16 AM
P: 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 ?
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