I am looking at the following figure (Maxim APPLICATION NOTE 5129
Stabilize Your Transimpedance Amplifier):

maxim figure
(source: maximintegrated.com)

I just can't understand the following: in the region half-way between fF and fi, the slope difference indicates a 180 degree phase shift between open-loop and the external network, albeit the gain is larger than unity (it is 1 only at the intersection). So, what prevents this compensated op amp from saturating to one rail in this case due to the positive feedback and over-unity gain in some frequency range ? Thanks.

  • \$\begingroup\$ You're right. I was mis-remembering something. \$\endgroup\$
    – The Photon
    Commented Aug 4, 2015 at 4:43

1 Answer 1


Cipocip - the phase of the loop gain does not assume a value of 180deg between those two frequencies. You must not look at the asymptotic lines only. The real magnitude of the loop gain does not have a roll-off of -40dB/dec. This is true because (a) the open-loop gain of the opamp and (b) the rise of 1/beta both never reach 20 dB/dec (corresponding to a 90deg shift). As you know, each first-order system has a phase shift of 90 deg for infinite frequencies only. In the present case, the situation is even less critical because the 1/beta function has a pole (following the zero). Hence, the maximum phase shift will be (perhaps) only 75...80 deg. - dpending on the distance between both frequencies.

More than that, a clear picture of the stability margin can be gained looking at the NYQUIST plot for the loop gain. The critical point never is encircled.

  • \$\begingroup\$ Hm, I cannot refute your argument. I wonder if other experts would agree with you. \$\endgroup\$
    – cipocip
    Commented Aug 4, 2015 at 8:04
  • \$\begingroup\$ Why not? It is clear that 1/beta NEVER reaches 90 deg. And the same applies for the open-loop gain (if it is app. a 1st order function in the active region). I was teaching feedback techniques and stability problems for 25 years. \$\endgroup\$
    – LvW
    Commented Aug 4, 2015 at 8:32
  • \$\begingroup\$ refute=to prove to be false or erroneous; I said cannot refute. \$\endgroup\$
    – cipocip
    Commented Aug 4, 2015 at 8:41
  • \$\begingroup\$ OK - you see, I have to improve my english knowledge. However, as I have mentioned - you can prove my statement using the NYQUIST plot and the corresponding criterion (if you trust it). \$\endgroup\$
    – LvW
    Commented Aug 4, 2015 at 9:57

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