Operational
Amplifier Stability
Do You Need To
Worry About Operational Amplifier Stability?
The Circuit
Analyzing The
Circuit
Applying
The Nyquist Stability Criterion
The
OpAmp Transfer Function
Stability
Implications
Discussion
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 The Nyquist Stability Criterion  Applied to Operational Amplifier Circuits
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Do
You Need To Worry About Operational Amplifier Stability?
Some time ago the author was doing some consulting work for an organization
that will remain nameless. He needed a simple operational amplifier
circuit, the one shown below. There was only one kind of operational
amplifier in the stockroom  a fifteen minute walk  and despite many attempts
 often with new opamps  this circuit would not work. It oscillated
unmercilesly.
To make a long story short, I did some Nyquist analysis, went to the local
"Cheapo" electronic parts store, overpaid for a 741, and got the circuit
to work. The Nyquist analysis was the key. I could have been
there forever otherwise.
Goals
For This Lesson
Your goal for this lesson is simple
Given an operational amplifier circuit of the type shown below,
Be able to apply the Nyquist stability criterion (NSC) to the circuit to
predict stability. Your prediction will be done using the NSC working
with Bode' plot data.
You will need to have a good working knowledge
of the Nyquist stability criterion and how to apply it, particularly how
to apply it using Bode' plot data.
The
Circuit
We are going to examine a particular circuit, the one shown below.
This
is a fairly general circuit if we permit the two impedances to have capacitors
and inductors as well as resistors. To determine stability our plan
of attack is as follows.

Analyze the circuit accounting
for the finite gain of the amplifier and the frequency dependent gain.

Apply the Nyquist stability
criterion to the result.
Analyzing
The Circuit
We are going to examine a particular
circuit, the one shown below.
Apply
KCL at the node at the inverting input to the amplifier  the red dot.
We will use the symbol V_{n}(s) to represent that voltage
at the "negative" (inverting) node.
Do
not assume that the voltage at the inverting node, V_{n},
is zero.
Applying
KCL, we find:
[V_{out}(s)
 V_{n}(s)]/Z_{f}(s) + [V_{1}(s)
 V_{n}(s)]/Z_{1}(s) = 0
V_{n}(s)
is the voltage at the inverting terminal.
V_{out}(s)
= A(s)[V_{p}(s)  V_{n}(s)]
A(s) is the transfer
function of the amplifier.
Note: We assume here that all of the variables and impedances are
functions of s. We will not show that functional dependence explicitly
until we need to.
The KCL equation relates three voltages including the voltage,V_{n}
, at the inverting input node. In the usual analysis that voltage
is assumed to be zero (ground) in this circuit and the inverting input
node is called a virtual ground. We will not make that assumption
here because we are not going to assume that the gain of the operational
amplifier is infinite. Instead, we will assume that the gain of the
operational amplifier is finite and that it is frequency dependent.

We will assume that the
voltage at the noninverting node, Vp, is zero.
Many
real operational amplifiers have a transfer function that can be approximated
with a second order model. Later, we will modify that claim since
behavior at higher frequencies is not always best modelled with a time
constant. In any event, we would expect the form of the transfer
function for the operational amplifier to be like this.
A(s) = G_{dc}/[(st_{1}
+ 1)(st_{2}
+ 1)]
To account for A(s), rewrite the KCL equation

Note that V_{out}
= A(s)V_{n}

V_{n} =
V_{out}/A(s).
First write out KCL eliminating
the terms with V_{n}, by replacing V_{n}
with V_{out}/A(s).

[V_{out}(s)
 V_{n}(s)]/Z_{f}(s) + [V_{1}(s)
 V_{n}(s)]/Z_{1}(s) = 0

[V_{out}(s)
+ V_{out}/A(s)]/Z_{f}(s) + [V_{1}(s)
+ V_{out}/A(s)]/Z_{1}(s) = 0
Next,
solve for the output voltage in terms of everything else. First collect
all the output voltage terms on the left side of the equation and put the
remaining input term on the right side of the equation.

V_{out}(s)[1/Z_{f}(s)
+ A(s)/Z_{f}(s) + A(s)/Z_{1}(s)] =  V_{1}(s)/Z_{1}(s)

So:

The result is an expression that relates the
output voltage and the input voltage in terms of the gain and the impedances,
and we only need to solve for the output voltage once we know the input
voltage

As a check on our solution, let the gain, A,
become very large. If you do that in the expression at the right,
term multiplying Z_{f}/Z_{1} becomes one
and we have the familiar expression we know from the infinite gain assumption.
Applying
The Nyquist Stability Criterion
The important thing to recognize in the transfer function above is that
the transfer function is of the general form:
For the amplifier circuit, the expression A(s)Z_{1}(s)/[Z_{1}(s)
+ Z_{f}(s)] plays the role
of KG(s), so we need to apply the Nyquist Stability Criterion to that expression.
First, before we do anything, we will need to know the transfer function
for a typical operational amplifier. We'll look at the 741 which
is the most common operational amplifier used. Usually, on a complete
spec sheet for an opamp there will be an openloop frequency response.
That's really A(jw)
that is being represented. Here is a typical plot. This is
typically what you find on a spec sheet.Here
are a few links to spec sheets that give frequency response characteristics.
What you see on the plot is that there seem to be two corner frequencies.
That would lead to a transfer function model for A(s) of the form:

A(s) = G_{dc}/[(st_{1}
+ 1)(st_{2}
+ 1)]

G_{dc}
looks to be larger that 10^{5}, since the DC gain is over
100 db.

The low frequency corner
seems to be a little over 3 Hz, which would give a time constant of about
.1 sec.

The high frequency corner
seems to be a little over 3 MHz, which would give a time constant of about
.1 msec.

For these parameter values,
the transfer function is:

A(s) = 2x10^{5}/[(.1s
+ 1)(10^{7}s + 1)]

These are rough values,
but they indicate the general form of the transfer function.
The
real question is whether this transfer function realistically models the
opamp. There are times when the high frequency phase characteristic
really drops off more than this model would indicate.
Implications
For Stability
If we look at the frequency response of the amplifier, the phase response
is particularly interesting.

The phase of A(jw)
quickly reaches 90^{o} and stays there until the frequency
is near a megahertz.

If the other factor, Z1/(Z1
+ Zf) adds more phase we have the possibility of a low phase margin!
Let's
take a look at one possibility that could cause problems. Here's
an interesting circuit. You might be tempted to use this circuit
as a differentiator. Let's examine what happens when the gain of
the amplifier is taken into account.

First substitute values
for the impedances into the equivalent openloop gain expression.

Next, simplify and plot
the Bode' plot for the equivalent openloop gain expression.
If
we simplify the equivalent openloop gain, we obtain the expression below.
We can examine a Bode' plot for this combination
and look at the phase margin for the system.

Note that the opamp gain, A(s), is multiplied
by a factor that adds another pole.

The added pole will give more phase lag.

There can be a problem if the phase lag gets
too large. Click
here for more info.
We will start by selecting some component values.
We will choose some values arbitrarily and then determine the effect of
those choices.

Choose R = 100kW.
That's fairly large and still practical.

Choose C = 0.1mf.
Again, large and still reasonable.

That gives an infinite
gain transfer function of jwRC
= .01jw.

The factor we will use
to multiply A(s) is:
Now,
we need to examine the implications of the finite gain on stability.
We will plot a Bode' plot for the transfer function with the added factor
of 1/(.01s + 1). Here is that plot.

Note the steep dropoff,
at 40 db/decade. That occurs from around 10 Hz until you can't see
the plot any more.

The phase plot shows phase
right at 180^{o} for frequencies above 100 Hz. The
phase eventually goes below 180^{o} at higher frequencies.

The system is just marginally
stable at best, and it might be unstable.

If the system is stable,
it has a very low phase margin and it will exhibit a great deal of ringing.
Discussion
What does this all mean? Well, what you have learned in this lesson
is that the Nyquist stability criterion can be applied to operational amplifier
circuits. In the process you should see that you may not want to
believe the conclusions you would draw from the infinite gain model because
that model doesn't tell the whole store. There is a process you can
use to determine how well the system You've learned the process in this
introduction and you should be able to do the following:

Apply the Nyquist criterion
to other operational amplifier circuits with the Z_{1}(s)
& Z_{f}(s) configuration for which we derived
the output expression.

Extend the concept to
other operational amplifier configurations where you suspect finite, frequencydependent
gain in the opamp may have repercussions.