I have been trying to theoretically analyse the operation of a 741 op-amp from a stability point of view by firstly deriving its open-loop transfer function.

Here is the magnitude plot of a 741 op-amp.

741 Frequency Response

From the above plot, I generated the open-loop transfer function.

Open Loop Transfer Function

From the open-loop transfer function I have derived the closed-loop transfer function and to keep things simple I have assumed that the op-amp is configured as a unity gain buffer ( β = H(s) = 1 ).

Closed Loop Transfer Function


Now, the Root Locus method plots the loci of the closed-loop poles on the s plane as some parameter is varied from 0 to infinity. I vary the dc open-loop gain k which is set to 100000 for the 741. The values of the poles at various values of dc open-loop gain can be found by equating the denominator of the closed-loop transfer function (the characteristic equation) to zero and solving for s. That is to say, the roots of the characteristic equation are equal to the poles of the closed-loop transfer function.

Here is the characteristic equation which is equated to zero.

Characteristic Equations

By entering a range of values for k into the above equation and solving for s using the following equation….

Quadratic Solving Formula

….I get the pole plot below on the s plane. The arrows represent increasing k values. I’m pretty confident about this plot because I have seen the same plot for a second-order system on various web sites.

Pole plot on s plane

My problem is this. I know from fig.1 above and from Nyquist stability criteria that increasing the dc open-loop gain will reduce gain and phase margins. As open-loop gain k increases, the loop gain will get down to unity at a higher frequency where the loop phase lag is nearer -180ᴼ thereby reducing stability margins and moving the op-amp closer to oscillation. (The phase lag should never quite reach -180ᴼ because this is a theoretical 2 pole amplifier). Why then doesn’t the root locus plot converge on the imaginary axis on the s plane as the dc open-loop gain k is increased. The plot is vertical. I think I am misunderstanding something.


1 Answer 1


The only thing you're misunderstanding is that those vertical root loci do, in fact, imply and ever-diminishing phase margin. They're describing a system that goes resonant, with a Q that increases (or \$\zeta\$ that decreases) as the gain goes up.

That increasing Q is indicative of ever-diminishing margins.

(And just to note, even though you seem to understand this already: the real circuit has more poles, off to the left of the one at 1MHz. In the root-locus view those poles will push the locus to the right, because root loci tend to go away from poles and toward zeros; that would make the system unstable. In the Bode-plot view, of course, those poles would add more phase shift; that's just a different view of the same phenomenon.)

  • \$\begingroup\$ Thanks Tim. I can appreciate the amp is getting nearer to the point of oscillation the further that the loci get from the real axis as zeta reduces, zeta reaching zero at infinity. What was confusing me was that, even though the amp is getting less stable, the loci are not moving any closer to the jw axis, the point which divides a stable system from an unstable one. \$\endgroup\$
    – user173271
    Commented Dec 13, 2019 at 12:52
  • \$\begingroup\$ What's important in this case is the angle from \$s = 0$ to the poles, which is getting ever less. Another way of looking at it is to think of the root locus as dimensionless, and just keep zooming it out so that you can see it -- then it starts to look bad. And that "bad" is rightly so -- because with a 2nd-order system like that, the higher in frequency you get, the more that unmodeled poles lurking far off in the left-hand plane will start having an effect, bending the locus out of the stability region. \$\endgroup\$
    – TimWescott
    Commented Dec 13, 2019 at 16:35

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