lunes, 14 de junio de 2010

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 op-amps - 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 Vn(s) to represent that voltage at the "negative" (inverting) node. Do not assume that the voltage at the inverting node, Vn, is zero.
        Applying KCL, we find:

[Vout(s) - Vn(s)]/Zf(s) + [V1(s) - Vn(s)]/Z1(s) = 0 Vn(s) is the voltage at the inverting terminal. Vout(s) = A(s)[Vp(s) - Vn(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,Vn , 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) = Gdc/[(st1 + 1)(st2 + 1)]

        To account for A(s), rewrite the KCL equation
  • Note that Vout = -A(s)Vn
  • Vn =  -Vout/A(s).
First write out KCL eliminating the terms with Vn, by replacing Vn with -Vout/A(s).
  • [Vout(s) - Vn(s)]/Zf(s) + [V1(s) - Vn(s)]/Z1(s) = 0
  • [Vout(s) + Vout/A(s)]/Zf(s) + [V1(s) + Vout/A(s)]/Z1(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.
  • Vout(s)[1/Zf(s) + A(s)/Zf(s) + A(s)/Z1(s)] = - V1(s)/Z1(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 Zf/Z1 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:
  • KG(s)/[1 + KG(s)]
For the amplifier circuit, the expression A(s)Z1(s)/[Z1(s) + Zf(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 op-amp there will be an open-loop 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) = Gdc/[(st1 + 1)(st2 + 1)]
    • Gdc looks to be larger that 105, 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) = 2x105/[(.1s + 1)(10-7s + 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 op-amp.  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 -90o 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 open-loop gain expression.
  • Next, simplify and plot the Bode' plot for the equivalent open-loop gain expression.
        If we simplify the equivalent open-loop 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 op-amp 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:
    • 1/(.01s + 1)
        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 drop-off, 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 -180o for frequencies above 100 Hz.  The phase eventually goes below  -180o 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.

        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 Z1(s) & Zf(s)  configuration for which we derived the output expression.
  • Extend the concept to other operational amplifier configurations where you suspect finite, frequency-dependent gain in the op-amp may have repercussions

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