Register to reply

Op-amp stability

by kpachar
Tags: opamp, stability
Share this thread:
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 ?
Phys.Org News Partner Engineering news on
Creative adaptation of a quadcopter
Filling up could cost less thanks to biomolecular engineering breakthrough
Tesla says decision on battery factory months away

Register to reply

Related Discussions
EU stability Current Events 17
What determines whether or not a matrix is stable? Linear & Abstract Algebra 2
Stability physics problem Introductory Physics Homework 4
Equilibrium stability Classical Physics 3
Question about stability Calculus & Beyond Homework 4