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Does the following simple test deduced from Nyquist criterion also apply to positive feedback case or it needs to be modified?

If magnitude of loop gain i.e. |T (jω)| > 1 at the frequency where phase(T (jω))= −180◦, then the amplifier is unstable.

Reference: Analysis and Design of Analog Integrated Circuits, 5th Edition http://fa.ee.sut.ac.ir/Downloads/Ac...sign_of_Analog_Integrated_Circuits_5th_ed.pdf, pg 628 2nd last paragraph

https://www.allaboutcircuits.com/te...ve-feedback-part-4-introduction-to-stability/

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    \$\begingroup\$ isn't it just a sign convention? to call something positive feedback, the loop is the same, but you're now adding (or taking away) a -1 gain into its form, so after changing the "form" to positve, you'd then have to add another -1 into the loop to make it stable again... so thus it would flip the inequality? \$\endgroup\$
    – Pete W
    Feb 7 at 23:07
  • \$\begingroup\$ Like @PeteW said the condition is agnostic as to positive or negative feedback. However, the condition isn't correct. You can have positive loop gain with phase equal to -180 degrees, and as long as the phase isn't near -180 when the gain crosses zero dB, the loop will be stable. That's called "conditional stability". Usually not seen in amplifiers, but often in other circuits like voltage mode DC-DC converters. \$\endgroup\$
    – John D
    Feb 7 at 23:20

1 Answer 1

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If magnitude of loop gain i.e. |T (jω)| > 1 at the frequency where phase(T (jω))= −180◦, then the amplifier is unstable.

This is incorrect, but it is a very common mistake.

One can easily construct counter-examples to this proposed test. For example, assume the loop transfer function is:

$$F(s) = 100\frac{(s+10)^2}{(s+1)^3}$$

enter image description here

There are two frequencies where the phase is \$-180^\circ\$, and at both of those frequencies, the absolute value of the gain is greater than 1. Yet the closed loop transfer function is stable.

$$H(s) = \frac{F(s)}{1+F(s)} = \frac{100(s+10)^2}{s^3+103s^2+2003s + 10000}$$

\$H(s)\$ has three poles, all real, and all in the left half plane at

  • -79.344
  • -15.552
  • -8.1039

The linked All About Circuits page actually states the test (almost) correctly.

if the loop gain’s magnitude is less than unity at the frequency where the loop gain’s phase shift is 180°, the circuit is stable.

Note carefully the difference between that claim, and the claim in the question to which this is an answer:

If magnitude of loop gain i.e. |T (jω)| > 1 at the frequency where phase(T (jω))= −180◦, then the amplifier is unstable.


Does the following simple test deduced from Nyquist criterion also apply to positive feedback case or it needs to be modified?

Positive feedback is the same as negative feedback shifted by \$180^\circ\$. The unity gain negative feedback closed loop transfer H(s) corresponding to the loop transfer function F(s) is given by.

$$H(s)=\frac{F(s)}{1+F(s)}$$

On the other hand, the unity gain positive feedback closed loop transfer H(s) corresponding to the loop transfer function F(s) is given by.

$$H(s)=\frac{F(s)}{1-F(s)}$$

So, a stability test for negative feedback to the frequency response for -F(s), we will get the result corresponding to a stability test for positive feedback.

But what is the frequency response for -F(s)? The magnitude is identical, but the phase angle is shifted by \$180^\circ\$.

For example, if this is the frequency response for F(s)

enter image description here

Then this is the frequency response for -F(s)

enter image description here

So, if you want test the stability of a positive feedback system, you can shift the phase of the loop frequency response by \$180^\circ\$, and perform a stability test on the modified frequency response with the same test you would use for a negative feedback system.

If the open loop frequency response is stable (no RHP poles), and if the magnitude of the loop frequency response is less than 1 (or 0 dB), at every frequency where the phase is \$\mathbf{180^\circ}\$, then the closed loop system is stable.

(We cannot conclude that if the magnitude is >0 that the closed loop system is unstable, as we have shown a counter-example above).


At the risk of creating confusion, you could alternatively perform the following modified test on the original loop frequency response, instead of the original test on the modified loop frequency response. The MODIFIED version of the test goes like this:

If the open loop frequency response is stable (no RHP poles), and if the magnitude of the loop frequency response is less than 1 (or 0 dB), at every frequency where the phase is \$\mathbf{0^\circ}\$, then the closed loop system (closed with positive feedback) is stable.

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  • \$\begingroup\$ Thanks for your answer! From this thread: engineering.stackexchange.com/questions/32863/…, someone stated that one could proceed using Nyquist stability criterion for positive feedback system just by multiplying the gain factor H(s) by -1. \$\endgroup\$
    – Matt
    Feb 9 at 3:31
  • \$\begingroup\$ If this is the case and assume the loop gain for a positive feedback system is T(s) = G(s)H, does the test for positive feedback becomes "If magnitude of loop gain i.e. |-T(s)| > 1 at the frequency where phase(-T(s))= −180◦, then the amplifier is unstable." which is equivalent to "If magnitude of loop gain i.e. |T(s)| > 1 at the frequency where phase(T(s))= 0◦, then the amplifier is unstable."? \$\endgroup\$
    – Matt
    Feb 9 at 3:31
  • \$\begingroup\$ I have edited the answer to clarify how to perform the test when feedback is positive. Please be aware that your comment again has the test inversed, and there are counter examples where loop gain is > 1 at frequencies where phase is 180 degrees that are stable. That test is "if gain < 1, then stable", it is NOT "if gain > 1, then unstable". \$\endgroup\$ Feb 9 at 20:44

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