• State TI Thinks Resolved
  • Locked Locked
  • Replies 6 replies
  • Answers 2 answers
  • Subscribers 49 subscribers
  • Views 2540 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

OPAMP Stability - Observing phase margin

Abderrezak HADJI
Prodigy 50 points

Hello TI,

I am following some of greatest lab you have online. Now I am interested in OPAMP stability and I am watching "Precision Labs - Op amps: Stability".

We know that for stability of a system, the denominator of the transfer function must be ≠ from 0 (math is useful :D), which leads to loop gain different from -1 (AOL*ß≠-1 in case of an amplifier). I understand that the Phase margin tells how much the AOL*ß is close to -1.

Until here I am good, but when I have seen video 10.2 TI Precision Labs - Op Amps: Stability - Phase margin (8:07 picture bellow) I can not see why the phase margin is 5°. From the Gain graph |AOL*ß|=0dB and Phase(AOL*ß)≈0°. Putting this back to complex annotation gives AOL*ß≈1 and not -1. What I see is this example is 175° far from the instability point. Am I misunderstanding something?

  • 0
    Guru 48395 points

    The op amp dominant Aol pole is giving a 90deg phase shift. The Noise Gain shown here has a zero leading to an eventual 90deg phase lead, you subtract the that phase from the Aol phase to get the loop phase shift which is trending towards -180deg around the loop here. 

  • 0 in reply to Michael Steffes
    Prodigy 50 points

    From the Gain graph |AOL*ß|db=|AOL|db-|1/ß|db=0dB=1 and from phase graph Arg(AOL*ß)=0° => AOL*ß=1. Is this correct? In other words, if the Phase(AOL*ß) graph is correct (and it seems), I can not see how the Phase Margin is 5° and in same time Phase(AOL*ß)=5° @fc

  • 0 in reply to Abderrezak HADJI
    TI__Mastermind 27460 points

    Hi Abderrezak,

    Phase Margin is the phase(AOL*ß) @fc, hence why they are the same value. I'm not seeing where you see the Phase graph Arg(AOL*ß) is giving you 0° at AOL*ß=1 (fc).

    Best Regards,

    Robert Clifton 

  • 0 in reply to Robert Clifton56
    Prodigy 50 points

    It is clear from the three graphs of the phases! Phase of Aol @fc is +90°, phase of 1/ß is +90°. Subtracting phase(1/ß) from phase(Aol) gives phase(Aol*ß) and it is 0° (or roughly, indeed +5°), which is well done on phase(Aol*ß) graph. At fc phase is +5°, putting back to complex annotation fives Aol*ß(jw)=|Aol*ß|*e^(j*phase(Aol*ß))=1*e^(j*5°) and this gives roughly 1 not -1.

    Also why you say that phase margin is the same as phase of Aol*ß. Phase margin is the amount of phase shift that is left until you hit 180° measured at the unity gain point. isn't it?

  • 0 in reply to Abderrezak HADJI
    TI__Mastermind 27460 points

    Hi Abderrezak,

    Sorry to keep you waiting. The phase of Aol @fc is at 90°, but the phase of 1/ß hasn't completed it's phase shift at fc and is only at 85°. 

    As for the second part, let me further take a look into this. I'm getting a little confused as well! 

    Best Regards, 

    Robert Clifton 

  • 0 in reply to Robert Clifton56
    TI__Guru**** 165126 points

    Hello Abderrezak,

    The phase axis in the chart is margin view which makes finding margin very easy.

    In actuality, the 180 degree label is -180 degrees and the 0 degree label is -360 degrees. -180 degrees from inversion and another -180 degrees from two poles. One pole is inside the op amp. See AOL curve. The other pole is the capacitor on inverting input.

    So the 5 degree of phase margin result is really a -355 degree loop delay. It's not a dependable oscillator, but it could oscillate. It would certainly ring and have gain peaking.